Philpot MoM 2nd Ch01-06 ISM

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Chapter 1  Stress    1.1 Introduction    1.2 Normal Stress Under Axial Loading    1.3 Direct Shear Stress    1.4 Bearing Stress    1.5 Stresses on Inclined Sections    1.6 Equality of Shear Stresses on Perpendicular Planes    Chapter 2  Strain    2.1 Displacement, Deformation, and the Concept of Strain    2.2 Normal Strain    2.3 Shear Strain    2.4 Thermal Strain    Chapter 3  Mechanical Properties of Materials    3.1 The Tension Test    3.2 The Stress‐Strain Diagram    3.3 Hooke’s Law    3.4 Poisson’s Ratio    Chapter 4  Design Concepts    4.1 Introduction    4.2 Types of Loads    4.3 Safety    4.4 Allowable Stress Design    4.5 Load and Resistance Factor Design    Chapter 5  Axial Deformation    5.1 Introduction    5.2 Saint‐Venant’s Principle    5.3 Deformations in Axially Loaded Bars    5.4 Deformations in a System of Axially Loaded Bars    5.5 Statically Indeterminate Axially Loaded Members    5.6 Thermal Effects on Axial Deformation    5.7 Stress Concentrations    Chapter 6  Torsion    6.1 Introduction    6.2 Torsional Shear Strain    6.3 Torsional Shear Stress    6.4 Stresses on Oblique Planes    6.5 Torsional Deformations    6.6 Torsion Sign Conventions    6.7 Gears in Torsion Assemblies    6.8 Power Transmission    6.9 Statically Indeterminate Torsion Members    6.10 Stress Concentrations in Circular Shafts Under Torsional Loadings    6.11 Torsion of Noncircular Sections    6.12 Torsion of Thin‐Walled Tubes: Shear Flow 

  Chapter 7  Equilibrium of Beams    7.1 Introduction    7.2 Shear and Moment in Beams    7.3 Graphical Method for Constructing Shear and Moment Diagrams    7.4 Discontinuity Functions to Represent Load, Shear, and Moment    Chapter 8  Bending    8.1 Introduction    8.2 Flexural Strains    8.3 Normal Strains in Beams    8.4 Analysis of Bending Stresses in Beams    8.5 Introductory Beam Design for Strength    8.6 Flexural Stresses in Beams of Two Materials    8.7 Bending Due to Eccentric Axial Load    8.8 Unsymmetric Bending    8.9 Stress Concentrations Under Flexural Loadings    Chapter 9  Shear Stress in Beams    9.1 Introduction    9.2 Resultant Forces Produced by Bending Stresses    9.3 The Shear Stress Formula    9 .4 The First Moment of Area Q    9.5 Shear Stresses in Beams of Rectangular Cross Section    9.6 Shear Stresses in Beams of Circular Cross Section    9.7 Shear Stresses in Webs of Flanged Beams    9.8 Shear Flow in Built‐Up Members    Chapter 10  Beam Deflections    10.1 Introduction    10.2 Moment‐Curvature Relationship    10.3 The Differential Equation of the Elastic Curve    10.4 Deflections by Integration of a Moment Equation    10.5 Deflections by Integration of Shear‐Force or Load Equations    10.6 Deflections Using Discontinuity Functions    10.7 Method of Superposition    Chapter 11  Statically Indeterminate Beams    11.1 Introduction    11.2 Types of Statically Indeterminate Beams    11.3 The Integration Method    11.4 Use of Discontinuity Functions for Statically Indeterminate Beams    11.5 The Superposition Method             

Chapter 12  Stress Transformations    12.1 Introduction    12.2 Stress at a General Point in an Arbitrarily Loaded Body    12.3 Equilibrium of the Stress Element    12.4 Two‐Dimensional or Plane Stress    12.5 Generating the Stress Element    12.6 Equilibrium Method for Plane Stress Transformation    12.7General Equations of Plane Stress Transformation    12.8 Principal Stresses and Maximum Shear Stress    12.9 Presentation of Stress Transformation Results    12.10 Mohr’s Circle for Plane Stress    12.11 General State of Stress at a Point    Chapter 13  Strain Transformations    13.1 Introduction    13.2 Two‐Dimensional or Plane Strain    13.3 Transformation Equations for Plane Strain    13.4 Principal Strains and Maximum Shearing Strain    13.5 Presentation of Strain Transformation Results    13.6 Mohr’s Circle for Plane Strain    13.7 Strain Measurement and Strain Rosettes    13.8 Generalized Hooke’s Law for Isotropic Materials    Chapter 14  Thin‐Walled Pressure Vessels    14.1 Introduction    14.2 Spherical Pressure Vessels    14.3 Cylindrical Pressure Vessels    14.4 Strains in Pressure Vessels    Chapter 15  Combined Loads    15.1 Introduction    15.2 Combined Axial and Torsional Loads    15.3 Principal Stresses in a Flexural Member    15.4 General Combined Loadings    15.5 Theories of Failure    Chapter 16  Columns    16.1 Introduction    16.2 Buckling of Pin‐Ended Columns    16.3 The Effect of End Conditions on Column Buckling    16.4 The Secant Formula    16.5 Empirical Column Formulas & Centric Loading    16.6 Eccentrically Loaded Columns   

1.1 A stainless steel tube with an outside diameter of 60 mm and a wall thickness of 5 mm is used as a compression member. If the normal stress in the member must be limited to 200 MPa, determine the maximum load P that the member can support.

Solution The cross-sectional area of the stainless steel tube is A



( D2  d 2 ) 



[(60 mm)2  (50 mm)2 ]  863.938 mm2 4 4 The normal stress in the tube can be expressed as P  A The maximum normal stress in the tube must be limited to 200 MPa. Using 200 MPa as the allowable normal stress, rearrange this expression to solve for the maximum load P Ans. Pmax   allow A  (200 N/mm2 )(863.938 mm2 )  172,788 N  172.8 kN

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1.2 A 2024-T4 aluminum tube with an outside diameter of 2.50 in. will be used to support a 27-kip load. If the normal stress in the member must be limited to 18 ksi, determine the wall thickness required for the tube.

Solution From the definition of normal stress, solve for the minimum area required to support a 27-kip load without exceeding a stress of 18 ksi P P 27 kips   Amin    1.500 in.2 A  18 ksi The cross-sectional area of the aluminum tube is given by A



( D2  d 2 )

4 Set this expression equal to the minimum area and solve for the maximum inside diameter d



4

[(2.50 in.)2  d 2 ]  1.500 in.2

(2.50 in.)2  d 2  (2.50 in.)2 

4



4



(1.500 in.2 )

(1.500 in.2 )  d 2

 d max  2.08330 in. The outside diameter D, the inside diameter d, and the wall thickness t are related by D  d  2t Therefore, the minimum wall thickness required for the aluminum tube is D  d 2.50 in.  2.08330 in. tmin    0.20835 in.  0.208 in. 2 2

Ans.

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1.3 Two solid cylindrical rods (1) and (2) are joined together at flange B and loaded, as shown in Fig. P1.3. The diameter of rod (1) is d1 = 24 mm and the diameter of rod (2) is d2 = 42 mm. Determine the normal stresses in rods (1) and (2). Fig. P1.3

Solution Cut a FBD through rod (1) that includes the free end of the rod at A. Assume that the internal force in rod (1) is tension. From equilibrium, Fx  F1  80 kN  0  F1  80 kN (T) Next, cut a FBD through rod (2) that includes the free end of the rod A. Assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): Fx  F2  140 kN  140 kN  80 kN  0

 F2  200 kN  200 kN (C)

From the given diameter of rod (1), the cross-sectional area of rod (1) is

A1 



(24 mm) 2  452.3893 mm 2

4 and thus, the normal stress in rod (1) is F (80 kN)(1,000 N/kN) 1  1   176.8388 MPa  176.8 MPa (T) A1 452.3893 mm 2

Ans.

From the given diameter of rod (2), the cross-sectional area of rod (2) is A2 



(42 mm) 2  1,385.4424 mm2

4 Accordingly, the normal stress in rod (2) is F (200 kN)(1,000 N/kN) 2  2   144.3582 MPa  144.4 MPa (C) A2 1,385.4424 mm 2

Ans.

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1.4 Two solid cylindrical rods (1) and (2) are joined together at flange B and loaded, as shown in Fig. P1.4. If the normal stress in each rod must be limited to 120 MPa, determine the minimum diameter required for each rod. Fig. P1.4

Solution Cut a FBD through rod (1) that includes the free end of the rod at A. Assume that the internal force in rod (1) is tension. From equilibrium, Fx  F1  80 kN  0  F1  80 kN (T) Next, cut a FBD through rod (2) that includes the free end of the rod A. Assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): Fx  F2  140 kN  140 kN  80 kN  0

 F2  200 kN  200 kN (C)

If the normal stress in rod (1) must be limited to 120 MPa, then the minimum cross-sectional area that can be used for rod (1) is F (80 kN)(1,000 N/kN) A1,min  1   666.6667 mm2 2  120 N/mm The minimum rod diameter is therefore A1,min 



4

d12  666.6667 mm 2

 d1  29.1346 mm  29.1 mm

Ans.

Similarly, the normal stress in rod (2) must be limited to 120 MPa. Notice that rod (2) is in compression. In this situation, we are concerned only with the magnitude of the stress; therefore, we will use the magnitude of F2 in the calculations for the minimum required cross-sectional area. F (200 kN)(1,000 N/kN) A2,min  2   1, 666.6667 mm 2 2  120 N/mm The minimum diameter for rod (2) is therefore A2,min 



4

d 22  1,666.6667 mm2

 d 2  46.0659 mm  46.1 mm

Ans.

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1.5 Two solid cylindrical rods (1) and (2) are joined together at flange B and loaded, as shown in Fig. P1.5. If the normal stress in each rod must be limited to 40 ksi, determine the minimum diameter required for each rod.

Fig. P1.5

Solution Cut a FBD through rod (1). The FBD should include the free end of the rod at A. As a matter of course, we will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, Fy   F1  15 kips  0

 F1  15 kips  15 kips (C) Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): Fy   F2  30 kips  30 kips  15 kips  0

 F2  75 kips  75 kips (C) Notice that rods (1) and (2) are in compression. In this situation, we are concerned only with the stress magnitude; therefore, we will use the force magnitudes to determine the minimum required cross-sectional areas. If the normal stress in rod (1) must be limited to 40 ksi, then the minimum crosssectional area that can be used for rod (1) is F 15 kips A1,min  1   0.375 in.2  40 ksi The minimum rod diameter is therefore A1,min 



d12  0.375 in.2  d1  0.69099 in.  0.691 in. Ans. 4 Similarly, the normal stress in rod (2) must be limited to 40 ksi, which requires a minimum area of F 75 kips A2,min  2   1.875 in.2  40 ksi The minimum diameter for rod (2) is therefore A2,min 



4

d 22  1.875 in.2

 d 2  1.545097 in.  1.545 in.

Ans.

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1.6 Two solid cylindrical rods (1) and (2) are joined together at flange B and loaded, as shown in Fig. P1.6. The diameter of rod (1) is 1.75 in. and the diameter of rod (2) is 2.50 in. Determine the normal stresses in rods (1) and (2).

Fig. P1.6

Solution Cut a FBD through rod (1). The FBD should include the free end of the rod at A. We will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, Fy   F1  15 kips  0

 F1  15 kips  15 kips (C) Next, cut a FBD through rod (2) that includes the free end of the rod at A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2):

Fy   F2  30 kips  30 kips  15 kips  0  F2  75 kips  75 kips (C) From the given diameter of rod (1), the cross-sectional area of rod (1) is A1 



(1.75 in.)2  2.4053 in.2

4 and thus, the normal stress in rod (1) is F 15 kips 1  1   6.23627 ksi  6.24 ksi (C) A1 2.4053 in.2

Ans.

From the given diameter of rod (2), the cross-sectional area of rod (2) is A2 



(2.50 in.) 2  4.9087 in.2

4 Accordingly, the normal stress in rod (2) is F 75 kips 2  2   15.2789 ksi  15.28 ksi (C) A2 2.4053 in.2

Ans.

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1.7 Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in Fig. P1.7. The diameter of aluminum rod (1) is 2.00 in., the diameter of brass rod (2) is 1.50 in., and the diameter of steel rod (3) is 3.00 in. Determine the normal stress in each of the three rods.

Fig. P1.7

Solution Cut a FBD through rod (1). The FBD should include the free end A. We will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, Fy   F1  8 kips  4 kips  4 kips  0  F1  16 kips  16 kips (C)

FBD through rod (1) FBD through rod (2)

FBD through rod (3) Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): Fy   F2  8 kips  4 kips  4 kips  15 kips  15 kips  0  F2  14 kips  14 kips (T) Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in rod (3) is: Fy   F3  8 kips  4 kips  4 kips  15 kips  15 kips  20 kips  20 kips  0

 F3  26 kips  26 kips (C) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

From the given diameter of rod (1), the cross-sectional area of rod (1) is A1 



(2.00 in.) 2  3.1416 in.2

4 and thus, the normal stress in aluminum rod (1) is F 16 kips 1  1   5.0930 ksi  5.09 ksi (C) A1 3.1416 in.2

Ans.

From the given diameter of rod (2), the cross-sectional area of rod (2) is A2 



(1.50 in.)2  1.7671 in.2

4 Accordingly, the normal stress in brass rod (2) is F 14 kips 2  2   7.9224 ksi  7.92 ksi (T) A2 1.7671 in.2

Ans.

Finally, the cross-sectional area of rod (3) is A3 



(3.00 in.) 2  7.0686 in.2

4 and the normal stress in the steel rod is F 26 kips 3  3   3.6782 ksi  3.68 ksi (C) A3 7.0686 in.2

Ans.

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1.8 Axial loads are applied with rigid bearing plates to the solid cylindrical rods shown in Fig. P1.8. The normal stress in aluminum rod (1) must be limited to 18 ksi, the normal stress in brass rod (2) must be limited to 25 ksi, and the normal stress in steel rod (3) must be limited to 15 ksi. Determine the minimum diameter required for each of the three rods.

Fig. P1.8

Solution The internal forces in the three rods must be determined. Begin with a FBD cut through rod (1) that includes the free end A. We will assume that the internal force in rod (1) is tension (even though it obviously will be in compression). From equilibrium, Fy   F1  8 kips  4 kips  4 kips  0  F1  16 kips  16 kips (C)

FBD through rod (1) FBD through rod (2)

FBD through rod (3) Next, cut a FBD through rod (2) that includes the free end A. Again, we will assume that the internal force in rod (2) is tension. Equilibrium of this FBD reveals the internal force in rod (2): Fy   F2  8 kips  4 kips  4 kips  15 kips  15 kips  0  F2  14 kips  14 kips (T) Similarly, cut a FBD through rod (3) that includes the free end A. From this FBD, the internal force in rod (3) is: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Fy   F3  8 kips  4 kips  4 kips  15 kips  15 kips  20 kips  20 kips  0  F3  26 kips  26 kips (C) Notice that two of the three rods are in compression. In these situations, we are concerned only with the stress magnitude; therefore, we will use the force magnitudes to determine the minimum required crosssectional areas, and in turn, the minimum rod diameters. The normal stress in aluminum rod (1) must be limited to 18 ksi; therefore, the minimum cross-sectional area required for rod (1) is F 16 kips A1,min  1   0.8889 in.2  1 18 ksi The minimum rod diameter is therefore A1,min 



4

d12  0.8889 in.2

 d1  1.0638 in.  1.064 in.

Ans.

The normal stress in brass rod (2) must be limited to 25 ksi, which requires a minimum area of F 14 kips A2,min  2   0.5600 in.2  2 25 ksi which requires a minimum diameter for rod (2) of A2,min 



4

d 22  0.5600 in.2

 d 2  0.8444 in.  0.844 in.

Ans.

The normal stress in steel rod (3) must be limited to 15 ksi. The minimum cross-sectional area required for this rod is: F 26 kips A3,min  3   1.7333 in.2  3 15 ksi which requires a minimum diameter for rod (3) of A3,min 



4

d32  1.7333 in.2

 d3  1.4856 in.  1.486 in.

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

1.9 Two solid cylindrical rods support a load of P = 50 kN, as shown in Fig. P1.9. If the normal stress in each rod must be limited to 130 MPa, determine the minimum diameter required for each rod.

Fig. P1.10

Solution Consider a FBD of joint B. Determine the angle  between rod (1) and the horizontal axis: 4.0 m tan    1.600   57.9946 2.5 m and the angle  between rod (2) and the horizontal axis: 2.3 m tan    0.7188    35.7067 3.2 m Write equilibrium equations for the sum of forces in the horizontal and vertical directions. Note: Rods (1) and (2) are two-force members. Fx  F2 cos(35.7067)  F1 cos(57.9946)  0 Fy  F2 sin(35.7067)  F1 sin(57.9946)  P  0

(a) (b)

Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the substitution method, Eq. (b) can be solved for F2 in terms of F1: cos(57.9946) F2  F1 (c) cos(35.7067) Substituting Eq. (c) into Eq. (b) gives cos(57.9946) F1 sin(35.7067)  F1 sin(57.9946)  P cos(35.6553)

F1  cos(57.9946) tan(35.7067)  sin(57.9946)   P  F1 

P P  cos(57.9946) tan(35.7067)  sin(57.9946) 1.2289

For the given load of P = 50 kN, the internal force in rod (1) is therefore: 50 kN F1   40.6856 kN 1.2289 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Backsubstituting this result into Eq. (c) gives force F2: cos(57.9946) cos(57.9946) F2  F1  (40.6856 kN)  26.5553 kN cos(35.7067) cos(35.7067) The normal stress in rod (1) must be limited to 130 MPa; therefore, the minimum cross-sectional area required for rod (1) is F (40.6856 kN)(1,000 N/kN) A1,min  1   312.9664 mm 2 2 1 130 N/mm The minimum rod diameter is therefore A1,min 



4

d12  312.9664 mm 2

 d1  19.9620 mm  19.96 mm

Ans.

The minimum area required for rod (2) is F (26.5553 kN)(1,000 N/kN) A2,min  2   204.2718 mm 2 2 2 130 N/mm which requires a minimum diameter for rod (2) of A2,min 



4

d 22  204.2718 mm2

 d 2  16.1272 mm  16.13 mm

Ans.

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1.10 Two solid cylindrical rods support a load of P = 27 kN, as shown in Fig. P1.10. Rod (1) has a diameter of 16 mm and the diameter of rod (2) is 12 mm. Determine the normal stress in each rod.

Fig. P1.10

Solution Consider a FBD of joint B. Determine the angle  between rod (1) and the horizontal axis: 4.0 m tan    1.600   57.9946 2.5 m and the angle  between rod (2) and the horizontal axis: 2.3 m tan    0.7188    35.7067 3.2 m Write equilibrium equations for the sum of forces in the horizontal and vertical directions. Note: Rods (1) and (2) are two-force members. Fx  F2 cos(35.7067)  F1 cos(57.9946)  0 Fy  F2 sin(35.7067)  F1 sin(57.9946)  P  0

(a) (b)

Unknown forces F1 and F2 can be found from the simultaneous solution of Eqs. (a) and (b). Using the substitution method, Eq. (b) can be solved for F2 in terms of F1: cos(57.9946) F2  F1 (c) cos(35.7067) Substituting Eq. (c) into Eq. (b) gives cos(57.9946) F1 sin(35.7067)  F1 sin(57.9946)  P cos(35.6553)

F1  cos(57.9946) tan(35.7067)  sin(57.9946)   P  F1 

P P  cos(57.9946) tan(35.7067)  sin(57.9946) 1.2289

For the given load of P = 27 kN, the internal force in rod (1) is therefore: 27 kN F1   21.9702 kN 1.2289 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Backsubstituting this result into Eq. (c) gives force F2: cos(57.9946) cos(57.9946) F2  F1  (21.9702 kN)  14.3399 kN cos(35.7067) cos(35.7067) The diameter of rod (1) is 16 mm; therefore, its cross-sectional area is: A1 



(16 mm)2  201.0619 mm 2 4 and the normal stress in rod (1) is: F (21.9702 kN)(1,000 N/kN) 1  1   109.2710 N/mm 2  109.3 MPa (T) 2 A1 201.0619 mm

Ans.

The diameter of rod (2) is 12 mm; therefore, its cross-sectional area is: A2 



(12 mm)2  113.0973 mm 2

4 and the normal stress in rod (2) is: F (14.3399 kN)(1,000 N/kN) 2  2   126.7924 N/mm 2  126.8 MPa (T) A2 113.0973 mm 2

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

1.11 A simple pin-connected truss is loaded and supported as shown in Fig. P1.11. All members of the truss are aluminum pipes that have an outside diameter of 4.00 in. and a wall thickness of 0.226 in. Determine the normal stress in each truss member.

Fig. P1.11

Solution Overall equilibrium: Begin the solution by determining the external reaction forces acting on the truss at supports A and B. Write equilibrium equations that include all external forces. Note that only the external forces (i.e., loads and reaction forces) are considered at this time. The internal forces acting in the truss members will be considered after the external reactions have been computed. The free-body diagram (FBD) of the entire truss is shown. The following equilibrium equations can be written for this structure: Fx  Ax  2 kips  0



 Ax  2 kips M A  By (6 ft)  (5 kips)(14 ft)  (2 kips)(7 ft)  0  By  14 kips Fy  Ay  By  5 kips  0  Ay  9 kips Method of joints: Before beginning the process of determining the internal forces in the axial members, the geometry of the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use the definition of the tangent function to determine AC and BC: 7 ft tan  AC   0.50  AC  26.565 14 ft 7 ft tan  BC   0.875  BC  41.186 8 ft Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Joint A: Begin the solution process by considering a FBD of joint A. Consider only those forces acting directly on joint A. In this instance, two axial members, AB and AC, are connected at joint A. Additionally, two reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed in each truss member. Fx  FAC cos(26.565)  FAB  Ax  0 (a) Fy  FAC sin(26.565)  Ay  0 (b) Solve Eq. (b) for FAC: Ay 9 kips FAC     20.125 kips sin(26.565) sin(26.565) and then compute FAB using Eq. (a): FAB   FAC cos(26.565)  Ax   (20.125 kips) cos(26.565)  ( 2 kips)  16.000 kips

Joint B: Next, consider a FBD of joint B. In this instance, the equilibrium equations associated with joint B seem easier to solve than those that would pertain to joint C. As before, tension forces will be assumed in each truss member. Fx   FAB  FBC cos(41.186)  0 (c) Fy  FBC sin(41.186)  By  0 (d) Solve Eq. (d) for FBC: By 14 kips FBC     21.260 kips sin(41.186) sin(41.186) Eq. (c) can be used as a check on our calculations: Fx   FAB  FBC cos(41.186)   ( 16.000 kips)  ( 21.260 kips) cos(41.186)  0

Section properties: For each of the three truss members: d  4.00 in.  2(0.226 in.)  3.548 in.

A

Checks!



(4.00 in.) 2  (3.548 in.) 2   2.67954 in.2 4

Normal stress in each truss member: F 16.000 kips  AB  AB   5.971 ksi  5.97 ksi (C) AAB 2.67954 in.2

Ans.

 AC 

FAC 20.125 kips   7.510 ksi  7.51 ksi (T) AAC 2.67954 in.2

Ans.

 BC 

FBC 21.260 kips   7.934 ksi  7.93 ksi (C) ABC 2.67954 in.2

Ans.

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1.12 A simple pin-connected truss is loaded and supported as shown in Fig. P1.12. All members of the truss are aluminum pipes that have an outside diameter of 60 mm and a wall thickness of 4 mm. Determine the normal stress in each truss member.

Fig. P1.12

Solution Overall equilibrium: Begin the solution by determining the external reaction forces acting on the truss at supports A and B. Write equilibrium equations that include all external forces. Note that only the external forces (i.e., loads and reaction forces) are considered at this time. The internal forces acting in the truss members will be considered after the external reactions have been computed. The freebody diagram (FBD) of the entire truss is shown. The following equilibrium equations can be written for this structure: Fx  Ax  12 kN  0

 Ax  12 kN M A  By (1 m)  (15 kN)(4.3 m)  0  By  64.5 kN Fy  Ay  By  15 kN  0  Ay  49.5 kN Method of joints: Before beginning the process of determining the internal forces in the axial members, the geometry of the truss will be used to determine the magnitude of the inclination angles of members AB and BC. Use the definition of the tangent function to determine AB and BC: 1.5 m tan  AB   1.50   AB  56.310 1.0 m 1.5 m tan  BC   0.454545   BC  24.444 3.3 m

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

Joint A: Begin the solution process by considering a FBD of joint A. Consider only those forces acting directly on joint A. In this instance, two axial members, AB and AC, are connected at joint A. Additionally, two reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed in each truss member. Fx  FAC  FAB cos(56.310)  Ax  0 (a) Fy  Ay  FAB sin(56.310)  0 (b) Solve Eq. (b) for FAB: Ay 49.5 kN FAB    59.492 kN sin(56.310) sin(56.310) and then compute FAC using Eq. (a): FAC   FAB cos(56.310)  Ax   ( 59.492 kN)cos(56.310)  ( 12 kN)  45.000 kN

Joint C: Next, consider a FBD of joint C. In this instance, the equilibrium equations associated with joint C seem easier to solve than those that would pertain to joint B. As before, tension forces will be assumed in each truss member. Fx   FAC  FBC cos(24.444)  12 kN  0 (c) Fy   FBC sin(24.444)  15 kN  0 (d) Solve Eq. (d) for FBC: 15 kN FBC   36.249 kN sin(24.444) Eq. (c) can be used as a check on our calculations: Fx   FAC  FBC cos(24.444)  12 kN  0   (45.000 kN)  ( 36.249 kN) cos(24.444)  12 kN  0

Section properties: For each of the three truss members: d  60 mm  2(4 mm)  52 mm

A

Checks!



(60 mm) 2  (52 mm) 2   703.7168 mm2 4

Normal stress in each truss member: F ( 59.492 kN)(1,000 N/kN)  AB  AB   84.539 MPa  84.5 MPa (C) AAB 703.7168 mm 2 F (45.000 kN)(1,000 N/kN)  AC  AC   63.946 MPa  63.9 MPa (T) AAC 703.7168 mm 2

 BC 

FBC ( 36.249 kN)(1,000 N/kN)   51.511 MPa  51.5 MPa (C) ABC 703.7168 mm 2

Ans. Ans. Ans.

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1.13 A simple pin-connected truss is loaded and supported as shown in Fig. P1.13. All members of the truss are aluminum pipes that have an outside diameter of 42 mm and a wall thickness of 3.5 mm. Determine the normal stress in each truss member.

Fig. P1.13

Solution

Overall equilibrium: Begin the solution by determining the external reaction forces acting on the truss at supports A and B. Write equilibrium equations that include all external forces. Note that only the external forces (i.e., loads and reaction forces) are considered at this time. The internal forces acting in the truss members will be considered after the external reactions have been computed. The free-body diagram (FBD) of the entire truss is shown. The following equilibrium equations can be written for this structure: Fy  Ay  30 kN  0

 Ay  30 kN

M A  (30 kN)(4.5 m)  (15 kN)(1.6 m)  Bx (5.6 m)  0



 Bx  19.821 kN Fx  Ax  Bx  15 kN  0 Ax  15 kN  Bx  15 kN  (19.821 kN)

 Ax  34.821 kN

Method of joints: Before beginning the process of determining the internal forces in the axial members, the geometry of the truss will be used to determine the magnitude of the inclination angles of members AC and BC. Use the definition of the tangent function to determine AC and BC: 1.6 m tan  AC   0.355556   AC  19.573 4.5 m 4m tan  BC   0.888889   BC  41.634 4.5 m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Joint A: Begin the solution process by considering a FBD of joint A. Consider only those forces acting directly on joint A. In this instance, two axial members, AB and AC, are connected at joint A. Additionally, two reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed in each truss member. Fx  Ax  FAC cos(19.573)  0 (a) Fy  Ay  FAC sin(19.573)  FAB  0 (b) Solve Eq. (a) for FAC: Ax 34.821 kN FAC    36.957 kN cos(19.573) cos(19.573) and then compute FAB using Eq. (b): FAB  Ay  FAC sin(19.573)

 (30.000 kN)  (36.957 kN)sin(19.573)  17.619 kN Joint B: Next, consider a FBD of joint B. In this instance, the equilibrium equations associated with joint B seem easier to solve than those that would pertain to joint C. As before, tension forces will be assumed in each truss member. Fx  Bx  FBC cos(41.634)  0 (c) Fy  FBC sin(41.634)  FAB  0 (d) Solve Eq. (c) for FBC: Bx ( 19.821 kN) FBC    26.520 kN cos(41.634) cos(41.634) Eq. (d) can be used as a check on our calculations: Fy  FBC sin(41.634)  FAB  (26.520 kN)sin(41.634)  (17.619 kN)  0 Section properties: For each of the three truss members: d  42 mm  2(3.5 mm)  35 mm

A

Checks!



(42 mm) 2  (35 mm) 2   423.3296 mm2 4

Normal stress in each truss member: F (17.619 kN)(1,000 N/kN)  AB  AB   41.620 MPa  41.6 MPa (T) AAB 423.3296 mm 2

Ans.

 AC 

FAC (36.957 kN)(1,000 N/kN)   87.301 MPa  87.3 MPa (T) AAC 423.3296 mm 2

Ans.

 BC 

FBC ( 26.520 kN)(1,000 N/kN)   62.647 MPa  62.6 MPa (C) ABC 423.3296 mm 2

Ans.

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1.14 The members of the truss shown in Fig. P1.14 are aluminum pipes that have an outside diameter of 4.50 in. and a wall thickness of 0.237 in. Determine the normal stress in each truss member.

Fig. P1.14

Solution Overall equilibrium: Begin the solution by determining the external reaction forces acting on the truss at supports A and B. Write equilibrium equations that include all external forces. Note that only the external forces (i.e., loads and reaction forces) are considered at this time. The internal forces acting in the truss members will be considered after the external reactions have been computed. The free-body diagram (FBD) of the entire truss is shown. The following equilibrium equations can be written for this structure: Fx  Ax  (15 kips)cos50  0

 Ax  9.642 kips M A  By (4 ft)  (15 kips)(4 ft)cos50  (15 kips)(18 ft)sin50  0  By  61.350 kips Fy  Ay  By  (15 kips)sin 50  0  Ay  49.859 kips Method of joints: Before beginning the process of determining the internal forces in the axial members, the geometry of the truss will be used to determine the magnitude of the inclination angles of members AB, AC, and BC. Use the definition of the tangent function to determine AB, AC, and BC: 6 ft tan  AB   1.5   AB  56.3099 4 ft 4 ft tan  AC   0.222222   AC  12.5288 18 ft 10 ft tan  BC   0.714286   BC  35.5377 14 ft

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Joint A: Begin the solution process by considering a FBD of joint A. Consider only those forces acting directly on joint A. In this instance, two axial members, AB and AC, are connected at joint A. Additionally, two reaction forces, Ax and Ay, act at joint A. Tension forces will be assumed in each truss member. Fx  FAC cos(12.5288)  FAB cos(56.3099)  Ax  0 (a) Fy  FAC sin(12.5288)  FAB sin(56.3099)  Ay  0 (b) Solve Eqs. (a) and (b) simultaneously to obtain: FAB  49.948 kips

FAC  38.259 kips Joint B: Next, consider a FBD of joint B. In this instance, the equilibrium equations associated with joint B seem easier to solve than those that would pertain to joint C. As before, tension forces will be assumed in each truss member. Fx  FBC cos(35.5377)  FAB cos(56.3099)  0 (c) Fy  FBC sin(35.5377)  FAB sin(56.3099)  By  0 (d) Solve Eq. (c) for FBC: cos(56.3099) cos(56.3099) FBC  FAB  ( 49.9484)  34.048 kips cos(35.5377) cos(35.5377) Eq. (d) can be used as a check on our calculations: Fy  FBC sin(35.5377)  FAB sin(56.3099)  By

 (34.0485 kips)sin(35.5377)  ( 49.9484 kips)sin(56.3099)  61.350 kips  0 Section properties: For each of the three truss members: d  4.50 in.  2(0.237 in.)  4.026 in.

A



(4.50 in.) 2  (4.026 in.) 2   3.17405 in.2 4

Normal stress in each truss member: F 49.948 kips  AB  AB   15.736 ksi  15.74 ksi (C) AAB 3.17405 in.2 F 38.259 kips  AC  AC   12.054 ksi  12.05 ksi (T) AAC 3.17405 in.2

 BC 

Checks!

FBC 34.048 kips   10.727 ksi  10.73 ksi (C) ABC 3.17405 in.2

Ans. Ans. Ans.

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1.15 Bar (1) in Fig. P1.15 has a crosssectional area of 0.75 in.2. If the stress in bar (1) must be limited to 30 ksi, determine the maximum load P that may be supported by the structure.

Fig. P1.15

Solution Given that the cross-sectional area of bar (1) is 0.75 in.2 and its normal stress must be limited to 30 ksi, the maximum force that may be carried by bar (1) is F1,max  1 A1  (30 ksi)(0.75 in.2 )  22.5 kips Consider a FBD of ABC. From the moment equilibrium equation about joint A, the relationship between the force in bar (1) and the load P is: M A  (6 ft)F1  (10 ft)P  0 P 

6 ft F1 10 ft

Substitute the maximum force F1,max = 22.5 kips into this relationship to obtain the maximum load that may be applied to the structure: 6 ft 6 ft P F1  (22.5 kips)  13.50 kips Ans. 10 ft 10 ft

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1.16 Two 6 in. wide wooden boards are to be joined by splice plates that will be fully glued on the contact surfaces. The glue to be used can safely provide a shear strength of 120 psi. Determine the smallest allowable length L that can be used for the splice plates for an applied load of P = 10,000 lb. Note that a gap of 0.5 in. is required between boards (1) and (2).

Fig. P1.16

Solution Consider a FBD of board (2). The glue on the splice plates provides resistance to the 10,000 lb applied load on both the top and bottom surfaces of board (2). Denoting the shear resistance on a glue surface as V, equilibrium in the horizontal direction requires Fx  P  V  V  0 V 

10, 000 lb  5, 000 lb 2

In other words, each glue surface must be large enough so that 5,000 lb of shear resistance can be provided to board (2). Since the glue has a shear strength of 120 psi, the area of each glue surface on board (2) must be at least 5, 000 lb Amin   41.6667 in.2 120 psi The boards are 6-in. wide; therefore, glue must be spread along board (2) for a length of at least 41.6667 in.2 Lglue joint   6.9444 in. 6 in. Although we’ve discussed only board (2), the same rationale applies to board (1). For both boards (1) and (2), the glue must be applied along a length of at least 6.9444 in. on both the top and bottom of the boards in order to resist the 10,000 lb applied load. The glue applied to boards (1) and (2) must be matched by glue applied to the splice plates. Therefore, the splice plates must be at least 6.9444 in. + 6.9444 in. = 13.8889 in. long. However, we are told that a 0.5-in. gap is required between boards (1) and (2); therefore, the splice plates must be 0.5-in. longer. Altogether, the length of the splice plates must be at least Ans. Lmin  6.9444 in.  6.9444 in.  0.5 in.  14.39 in.

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1.17 For the clevis connection shown in Fig. P1.17, determine the shear stress in the 22-mm diameter bolt for an applied load of P = 90 kN.

Fig. P1.17

Solution Consider a FBD of the bar that is connected by the clevis, including a portion of the bolt. If the shear force acting on each exposed surface of the bolt is denoted by V, then the shear force on each bolt surface is 90 kN Fx  P  V  V  0 V   45 kN 2 The area of the bolt surface exposed by the FBD is simply the cross-sectional area of the bolt: Abolt 



4

2 d bolt 



4

(22 mm)2  380.1327 mm 2

Therefore, the shear stress in the bolt is V (45 kN)(1,000 N/kN)    118.3797 N/mm 2  118.4 MPa Abolt 380.1327 mm 2

Ans.

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1.18 For the clevis connection shown in Fig. P1.18, the shear stress in the 3/8 in. diameter bolt must be limited to 36 ksi. Determine the maximum load P that may be applied to the connection. Fig. P1.18

Solution Consider a FBD of the bar that is connected by the clevis, including a portion of the bolt. If the shear force acting on each exposed surface of the bolt is denoted by V, then the shear force on each bolt surface is related to the load P by: Fx  P  V  V  0  P  2V The area of the bolt surface exposed by the FBD is simply the cross-sectional area of the bolt: Abolt 



4

2 d bolt 



4

(3 / 8 in.) 2 



4

(0.3750 in.) 2  0.1104466 in.2

If the shear stress in the bolt must be limited to 36 ksi, the maximum shear force V on a single crosssectional surface must be limited to V   Abolt  (36 ksi)(0.1104466 in.2 )  3.976078 kips Therefore, the maximum load P that may be applied to the connection is P  2V  2(3.976078 kips)  7.952156 kips  7.95 kips

Ans.

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1.19 For the connection shown in Fig. P1.19, determine the average shear stress in the 7/8-in. diameter bolts if the load is P = 45 kips.

Fig. P1.19

Solution The bolts in this connection act in single shear. The cross-sectional area of a single bolt is Abolt 



2 d bolt 



(7 / 8 in.)2 



(0.875 in.) 2  0.6013205 in.2 4 4 4 Since there are five bolts, the total area that carries shear stress is AV  5 Abolt  5(0.6013205 in.2 )  3.006602 in.2 Therefore, the shear stress in each bolt is P 45 kips    14.96706 ksi  14.97 ksi AV 3.006602 in.2

Ans.

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1.20 The five-bolt connection shown in Fig. P1.20 must support an applied load of P = 300 kN. If the average shear stress in the bolts must be limited to 225 MPa, determine the minimum bolt diameter that may be used in the connection.

Fig P1.20

Solution To support a load of 300 kN while not exceeding an average shear stress of 225 MPa, the total shear area provided by the bolts must be at least P (300 kN)(1,000 N/kN) AV    1,333.3333 mm 2 2  225 N/mm Since there are five single-shear bolts in this connection, five cross-sectional surfaces carry shear stress. Consequently, each bolt must provide a minimum area of AV 1,333.3333 mm 2 Abolt    266.6667 mm 2 5 5 The minimum bolt diameter is therefore Abolt  266.6667 mm2 



4

2 d bolt

 d bolt  18.4264 mm  18.43 mm

Ans.

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1.21 The three-bolt connection shown in Fig. P1.21 must support an applied load of P = 40 kips. If the average shear stress in the bolts must be limited to 24 ksi, determine the minimum bolt diameter that may be used in the connection.

Fig. P1.21

Solution The shear force V that must be provided by the bolts equals the applied load of P = 40 kips. The total shear area required is thus V 40 kips AV    1.66667 in.2  24 ksi The three bolts in this connection act in double shear; therefore, six cross-sectional bolt surfaces are available to transmit shear stress. AV 1.66667 in.2 Abolt    0.27778 in.2 per surface (2 surfaces per bolt)(3 bolts) 6 surfaces The minimum bolt diameter must be



4

2 d bolt  0.27778 in.2

 d bolt  0.59471 in.  0.595 in.

Ans.

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1.22 For the connection shown in Fig. P1.22, the average shear stress in the 12-mm-diameter bolts must be limited to 160 MPa. Determine the maximum load P that may be applied to the connection.

Fig. P1.22

Solution The cross-sectional area of a 12-mm-diameter bolt is Abolt 



2 d bolt 



(12 mm)2  113.097355 mm2

4 4 This is a double-shear connection. Therefore, the three bolts provide a total shear area of AV  2(3 bolts)Abolt  2(3 bolts)(113.097355 mm 2 )  678.58401 mm 2 Since the shear stress must be limited to 160 MPa, the total shear force that can be resisted by the three bolts is Vmax   AV  (160 N/mm 2 )(678.58401 mm 2 )  108,573.442 N In this connection, the shear force in the bolts is equal to the applied load P; therefore, Ans. Pmax  108.6 kN

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1.23 A hydraulic punch press is used to punch a slot in a 0.50-in. thick plate, as illustrated in Fig. P1.23. If the plate shears at a stress of 30 ksi, determine the minimum force P required to punch the slot.

Fig. P1.23

Solution The shear stress associated with removal of the slug exists on its perimeter. The perimeter of the slug is given by perimeter  2(3.00 in.) +  (0.75 in.)  8.35619 in. Thus, the area subjected to shear stress is AV  perimeter  plate thickness  (8.35619 in.)(0.50 in.)  4.17810 in.2 Given that the plate shears at  = 30 ksi, the force required to remove the slug is therefore Pmin   AV  (30 ksi)(4.17810 in.2 )  125.343 kips  125.3 kips

Ans.

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1.24 A coupling is used to connect a 2 in. diameter plastic pipe (1) to a 1.5 in. diameter pipe (2), as shown in Fig. P1.24. If the average shear stress in the adhesive must be limited to 400 psi, determine the minimum lengths L1 and L2 required for the joint if the applied load P is 5,000 lb.

Fig. P1.24

Solution To resist a shear force of 5,000 lb, the area of adhesive required on each pipe is V 5, 000 lb AV    12.5 in.2  adhesive 400 psi Consider the coupling on pipe (1). The adhesive is applied to the circumference of the pipe, and the circumference C1 of pipe (1) is C1   D1   (2.0 in.)  6.2832 in. The minimum length L1 is therefore A 12.5 in.2 Ans. L1  V   1.9894 in.  1.989 in. C1 6.2832 in. Consider the coupling on pipe (2). The circumference C2 of pipe (2) is C2   D2   (1.5 in.)  4.7124 in. The minimum length L2 is therefore A 12.5 in.2 L2  V   2.6526 in.  2.65 in. C2 4.7124 in.

Ans.

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1.25 A lever is attached to a shaft with a square shear key, as shown in Fig. P1.25. The force applied to the lever is P = 400 N. If the shear stress in the key must not exceed 90 MPa, determine the minimum dimension “a” that must be used if the key is 15 mm long.

Fig. P1.25

Solution To determine the shear force V that must be resisted by the shear key, sum moments about the center of the shaft (which will be denoted O):  50 mm  M O  (400 N)(750 mm)   V 0 V  12,000 N  2  Since the shear stress in the key must not exceed 90 MPa, the shear area required is V 12,000 N AV    133.3333 mm2 2  90 N/mm The shear area in the key is given by the product of its length L (i.e., 15 mm) and its width a. Therefore, the minimum key width a is A 133.3333 mm 2 a V   8.8889 mm  8.89 mm Ans. L 15 mm

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1.26 A common trailer hitch connection is shown in Fig. P1.26. The shear stress in the pin must be limited to 30,000 psi. If the applied load is P = 4,000 lb, determine the minimum diameter that must be used for the pin.

Fig. P1.26

Solution The shear force V acting in the hitch pin is equal to the applied load; therefore, V = P = 4,000 lb. The shear area required to support a 4,000 lb shear force is V 4, 000 lb AV    0.1333 in.2  30, 000 psi The hitch pin is used in a double-shear connection; therefore, two cross-sectional areas of the pin are subjected to shear stress. Thus, the cross-sectional area of the pin is given by AV 0.1333 in.2 AV  2 Apin  Apin    0.0667 in.2 2 2 and the minimum pin diameter is



4

2 d pin  0.0667 in.2

 d pin  0.2913 in.  0.291 in.

Ans.

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1.27 An axial load P is supported by a short steel column, which has a cross-sectional area of 11,400 mm2. If the average normal stress in the steel column must not exceed 110 MPa, determine the minimum required dimension “a” so that the bearing stress between the base plate and the concrete slab does not exceed 8 MPa.

Fig. P1.27

Solution Since the normal stress in the steel column must not exceed 110 MPa, the maximum column load is Pmax   A  (110 N/mm 2 )(11, 400 mm 2 )  1, 254, 000 N The maximum column load must be distributed over a large enough area so that the bearing stress between the base plate and the concrete slab does not exceed 8 MPa; therefore, the minimum plate area is P 1, 254, 000 N Amin    156, 750 mm 2 2 b 8 N/mm Since the plate is square, the minimum plate dimension a must be Amin  156, 750 mm2  a  a  a  395.9167 mm  396 mm

Ans.

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1.28 The steel pipe column shown in Fig. P1.28 has an outside diameter of 8.625 in. and a wall thickness of 0.25 in. The timber beam is 10.75 in wide, and the upper plate has the same width. The load imposed on the column by the timber beam is 80 kips. Determine (a) The average bearing stress at the surfaces between the pipe column and the upper and lower steel bearing plates. (b) The length L of the rectangular upper bearing plate if its width is 10.75 in. and the average bearing stress between the steel plate and the wood beam is not to exceed 500 psi. (c) The dimension “a” of the square lower bearing plate if the average bearing stress between the lower bearing plate and the concrete slab is not to exceed 900 psi.

Fig. P1.28

Solution (a) The area of contact between the pipe column and one of the bearing plates is simply the crosssectional area of the pipe. To calculate the pipe area, we must first calculate the pipe inside diameter d: D  d  2t  d  D  2t  8.625 in.  2(0.25 in.)  8.125 in. The pipe cross-sectional area is





 D 2  d 2   (8.625 in.)2  (8.125 in.) 2   6.5777 in.2 4 4 Therefore, the bearing stress between the pipe and one of the bearing plates is P 80 kips b    12.1623 ksi  12.16 ksi Ab 6.5777 in.2 Apipe 

Ans.

(b) The bearing stress between the timber beam and the upper bearing plate must not exceed 500 psi (i.e., 0.5 ksi). To support a load of 80 kips, the contact area must be at least P 80 kips Ab    160 in.2  b 0.5 ksi If the width of the timber beam is 10.75 in., then the length L of the upper bearing plate must be Ab 160 in.2 L   14.8837 in.  14.88 in. Ans. beam width 10.75 in. (c) The bearing stress between the concrete slab and the lower bearing plate must not exceed 900 psi (i.e., 0.9 ksi). To support the 80-kip pipe load, the contact area must be at least P 80 kips Ab    88.8889 in.2  b 0.9 ksi Since the lower bearing plate is square, its dimension a must be Ans. Ab  a  a  88.8889 in.2  a  9.4281 in.  9.43 in.

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1.29 A vertical shaft is supported by a thrust collar and bearing plate, as shown in Fig. P1.29. The average shear stress in the collar must be limited to 18 ksi. The average bearing stress between the collar and the plate must be limited to 24 ksi. Based on these limits, determine the maximum axial load P that can be applied to the shaft.

Fig. P1.29

Solution Consider collar shear stress: The area subjected to shear stress in the collar is equal to the product of the shaft circumference and the collar thickness; therefore, AV  shaft circumference  collar thickness   (1.0 in.)(0.5 in.)  1.5708 in.2 If the shear stress must not exceed 18 ksi, the maximum load that can be supported by the vertical shaft is: P   AV  (18 ksi)(1.5708 in.2 )  28.2743 kips Consider collar bearing stress: We must determine the area of contact between the collar and the plate. The overall cross-sectional area of the collar is

Acollar 



(1.5 in.) 2  1.7671 in.2

4 is reduced by the area taken up by the shaft Ashaft 



(1.0 in.) 2  0.7854 in.2

4 Therefore, the area of the collar that actually contacts the plate is Ab  Acollar  Ashaft  1.7671 in.2  0.7854 in.2  0.9817 in.2 If the bearing stress must not exceed 24 ksi, the maximum load that can be supported by the vertical shaft is: P   b Ab  (24 ksi)(0.9817 in.2 )  23.5619 kips

Controlling P: Considering both shear stress in the collar and bearing stress between the collar and the plate, the maximum load that can be supported by the shaft is Ans. Pmax  23.6 kips

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1.30 A structural steel bar with a 25 mm × 75 mm rectangular cross section is subjected to an axial load of 150 kN. Determine the maximum normal and shear stresses in the bar.

Solution The maximum normal stress in the steel bar is F (150 kN)(1,000 N/kN)  max    80 MPa A (25 mm)(75 mm) The maximum shear stress is one-half of the maximum normal stress

 max 

 max 2

 40 MPa

Ans.

Ans.

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1.31 A steel rod of circular cross section will be used to carry an axial load of 92 kips. The maximum stresses in the rod must be limited to 30 ksi in tension and 12 ksi in shear. Determine the required diameter for the rod.

Solution Based on the allowable 30 ksi tension stress limit, the minimum cross-sectional area of the rod is F 92 kips Amin    3.0667 in.2  max 30 ksi For the 12-ksi shear stress limit, the minimum cross-sectional area of the rod must be F 92 kips Amin    3.8333 in.2 2 max 2(12 ksi) Therefore, the rod must have a cross-sectional area of at least 3.8333 in.2 in order to satisfy both the normal and shear stress limits. The minimum rod diameter D is therefore



4

2 d min  3.8333 in.2

 d min  2.2092 in.  2.21 in.

Ans.

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1.32 An axial load P is applied to the rectangular bar shown in Fig. P1.32. The cross-sectional area of the bar is 400 mm2. Determine the normal stress perpendicular to plane AB and the shear stress parallel to plane AB if the bar is subjected to an axial load of P = 70 kN. Fig. P1.32

Solution The angle  for the inclined plane is 35°. The normal force N perpendicular to plane AB is found from N  P cos  (40 kN)cos35  57.3406 kN and the shear force V parallel to plane AB is V  P sin   (70 kN)sin35  40.1504 kN

The cross-sectional area of the bar is 400 mm2, but the area along inclined plane AB is A 400 mm 2 An    488.3098 mm 2 cos  cos35 The normal stress n perpendicular to plane AB is N (57.3406 kN)(1,000 N/kN) n    117.4268 MPa  117.4 MPa An 488.3098 mm 2 The shear stress nt parallel to plane AB is V (40.1504 kN)(1,000 N/kN)  nt    82.2231 MPa  82.2 MPa An 488.3098 mm 2

Ans.

Ans.



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1.33 An axial load P is applied to the 1.75 in. by 0.75 in. rectangular bar shown in Fig. P1.33. Determine the normal stress perpendicular to plane AB and the shear stress parallel to plane AB if the bar is subjected to an axial load of P = 18 kips. Fig. P1.33

Solution The angle  for the inclined plane is 60°. The normal force N perpendicular to plane AB is found from N  P cos  (18 kips)cos60  9.0 kips and the shear force V parallel to plane AB is V  P sin   (18 kips)sin 60  15.5885 kips

The cross-sectional area of the bar is (1.75 in.)(0.75 in.) = 1.3125 in.2, but the area along inclined plane AB is 1.3125 in.2 An  A / cos    2.6250 in.2 cos 60 The normal stress n perpendicular to plane AB is N 9.0 kips n    3.4286 ksi  3.43 ksi An 2.6250 in.2 The shear stress nt parallel to plane AB is V 15.5885 kips  nt    5.9385 ksi  5.94 ksi An 2.6250 in.2

Ans.

Ans.



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1.34 A compression load of P = 80 kips is applied to a 4 in. by 4 in. square post, as shown in Fig. P1.34. Determine the normal stress perpendicular to plane AB and the shear stress parallel to plane AB.

Fig. P1.34

Solution The angle  for the inclined plane is 55°. The normal force N perpendicular to plane AB is found from N  P cos  (80 kips) cos55  45.8861 kips and the shear force V parallel to plane AB is V  P sin   (80 kips)sin 55  65.5322 kips The cross-sectional area of the post is (4 in.)(4 in.) = 16 in.2, but the area along inclined plane AB is 16 in.2 An  A / cos    27.8951 in.2 cos 55

The normal stress n perpendicular to plane AB is N 45.8861 kips n    1.6449 ksi  1.645 ksi An 27.8951 in.2 The shear stress nt parallel to plane AB is V 65.5322 kips  nt    2.3492 ksi  2.35 ksi An 27.8951 in.2

Ans.

Ans.

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1.35 Specifications for the 50 mm × 50 mm square bar shown in Fig. P1.35 require that the normal and shear stresses on plane AB not exceed 120 MPa and 90 MPa, respectively. Determine the maximum load P that can be applied without exceeding the specifications.

Fig. P1.35

Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The cross-sectional area of the square bar is A = (50 mm)2 = 2,500 mm2, and the angle  for plane AB is 55°. The normal stress on plane AB is limited to 120 MPa; therefore, the maximum load P that can be supported by the square bar is found from Eq. (a): 2 A n 2(2,500 mm 2 )(120 N/mm 2 ) P   911,882 N 1  cos 2 1  cos 2(55) The shear stress on plane AB is limited to 90 MPa. From Eq. (b), the maximum load P based the shear stress limit is 2 A nt 2(2,500 mm 2 )(90 N/mm 2 ) P   478,880 N sin 2 sin 2(55) Thus, the maximum load that can be supported by the bar is Pmax  479 kN

Ans.

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1.36 Specifications for the 6 in. × 6 in. square post shown in Fig. P1.36 require that the normal and shear stresses on plane AB not exceed 800 psi and 400 psi, respectively. Determine the maximum load P that can be applied without exceeding the specifications.

Fig. P1.36

Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The cross-sectional area of the square post is A = (6 in.)2 = 36 in.2, and the angle  for plane AB is 40°. The normal stress on plane AB is limited to 800 psi; therefore, the maximum load P that can be supported by the square post is found from Eq. (a): 2 A n 2(36 in.2 )(800 psi) P   49,078 lb 1  cos 2 1  cos 2(40) The shear stress on plane AB is limited to 400 psi. From Eq. (b), the maximum load P based the shear stress limit is 2 A nt 2(36 in.2 )(400 psi) P   29, 244 lb sin 2 sin 2(40) Thus, the maximum load that can be supported by the post is Pmax  29,200 lb  29.2 kips

Ans.

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1.37 A 90 mm wide bar will be used to carry an axial tension load of 280 kN. The normal and shear stresses on plane AB must be limited to 150 MPa and 100 MPa, respectively. Determine the minimum thickness t required for the bar.

Fig. P1.37

Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The angle  for plane AB is 50°. The normal stress on plane AB is limited to 150 MPa; therefore, the minimum cross-sectional area A required to support P = 280 kN can be found from Eq. (a): P (280 kN)(1,000 N/kN) A (1  cos 2 )  (1  cos 2(50))  771.2617 mm 2 2 2 n 2(150 N/mm ) The shear stress on plane AB is limited to 100 MPa; therefore, the minimum cross-sectional area A required to support P = 280 kN can be found from Eq. (b): P (280 kN)(1,000 N/kN) A sin 2  sin 2(50)  1,378.7309 mm 2 2 nt 2(100 N/mm 2 ) To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin = 1,379.7309 mm2. Since the bar width is 90 mm, the minimum bar thickness t must be 1,378.7309 mm 2 tmin   15.3192 mm  15.32 mm Ans. 90 mm

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1.38 A rectangular bar having width w = 6.00 in. and thickness t = 1.50 in. is subjected to a tension load P. The normal and shear stresses on plane AB must not exceed 16 ksi and 8 ksi, respectively. Determine the maximum load P that can be applied without exceeding either stress limit. Fig. P1.38

Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The angle  for inclined plane AB is calculated from 3 tan    3   71.5651 1 The cross-sectional area of the bar is A = w×t = (6.00 in.)(1.50 in.) = 9.0 in.2. The normal stress on plane AB is limited to 16 ksi; therefore, the maximum load P can be found from Eq. (a): 2 A n 2(9.0 in.2 )(16 ksi) P   1, 440 ksi 1  cos 2 1  cos 2(71.5651) The shear stress on plane AB is limited to 8 ksi. From Eq. (b), the maximum load P based the shear stress limit is 2 A nt 2(9.0 in.2 )(8 ksi) P   240 kips sin 2 sin 2(71.5651) Thus, the maximum load that can be supported by the bar is Pmax  240 kips

Ans.

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1.39 In Fig. P1.39, a rectangular bar having width w = 1.25 in. and thickness t is subjected to a tension load of P = 30 kips. The normal and shear stresses on plane AB must not exceed 12 ksi and 8 ksi, respectively. Determine the minimum bar thickness t required for the bar. Fig. P1.39

Solution The general equations for normal and shear stresses on an inclined plane in terms of the angle  are P (a) n  (1  cos 2 ) 2A and P (b)  nt  sin 2 2A The angle  for inclined plane AB is calculated from 3 tan    3   71.5651 1 The normal stress on plane AB is limited to 12 ksi; therefore, the minimum cross-sectional area A required to support P = 30 kips can be found from Eq. (a): P 30 kips A (1  cos 2 )  (1  cos 2(71.5651))  0.2500 in.2 2 n 2(12 ksi) The shear stress on plane AB is limited to 8 ksi; therefore, the minimum cross-sectional area A required to support P = 30 kips can be found from Eq. (b): P 30 kips A sin 2  sin 2(71.5651)  1.1250 in.2 2 nt 2(8 ksi) To satisfy both the normal and shear stress requirements, the cross-sectional area must be at least Amin = 1.1250 in.2. Since the bar width is 1.25 in., the minimum bar thickness t must be 1.1250 in.2 tmin   0.900 in.  0.900 in. Ans. 1.25 in.

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1.40 The rectangular bar has a width of w = 3.00 in. and a thickness of t = 2.00 in. The normal stress on plane AB of the rectangular block shown in Fig. P1.40 is 6 ksi (C) when the load P is applied. Determine: (a) the magnitude of load P. (b) the shear stress on plane AB. (c) the maximum normal and shear stresses in the block at any possible orientation. Fig. P1.40

Solution The general equation for normal stress on an inclined plane in terms of the angle  is P n  (1  cos 2 ) 2A and the angle  for inclined plane AB is 3 tan    0.75   36.8699 4 The cross-sectional area of the rectangular bar is A = (3.00 in.)(2.00 in.) = 6.00 in.2.

(a)

(a) Since the normal stress on plane AB is given as 6 ksi, the magnitude of load P can be calculated from Eq. (a): 2 A n 2(6.0 in.2 )(6 ksi) P   56.25 kips  56.3 kips Ans. 1  cos 2 1  cos 2(36.8699) (b) The general equation for shear stress on an inclined plane in terms of the angle  is P  nt  sin 2 2A therefore, the shear stress on plane AB is 56.25 kips  nt  sin 2(36.8699)  4.50 ksi 2(6.00 in.2 ) (c) The maximum normal stress at any possible orientation is P 56.25 kips  max    9.3750 ksi  9.38 ksi A 6.00 in.2 and the maximum shear stress at any possible orientation in the block is P 56.25 kips  max    4.6875 ksi  4.69 ksi 2 A 2(6.00 in.2 )

Ans.

Ans.

Ans.

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1.41 The rectangular bar has a width of w = 100 mm and a thickness of t = 75 mm. The shear stress on plane AB of the rectangular block shown in Fig. P1.41 is 12 MPa when the load P is applied. Determine: (a) the magnitude of load P. (b) the normal stress on plane AB. (c) the maximum normal and shear stresses in the block at any possible orientation. Fig. P1.41

Solution The general equation for shear stress on an inclined plane in terms of the angle  is P  nt  sin 2 2A and the angle  for inclined plane AB is 3 tan    0.75   36.8699 4 The cross-sectional area of the rectangular bar is A = (100 mm)(75 mm) = 7,500 mm2.

(a)

(a) Since the shear stress on plane AB is given as 12 MPa, the magnitude of load P can be calculated from Eq. (a): 2 A nt 2(7,500 mm 2 )(12 N/mm 2 ) P   187,500 N  187.5 kN Ans. sin 2 sin 2(36.8699) (b) The general equation for normal stress on an inclined plane in terms of the angle  is P n  (1  cos 2 ) 2A therefore, the normal stress on plane AB is 187,500 N n  (1  cos 2(36.8699))  16.00 MPa 2(7,500 mm 2 ) (c) The maximum normal stress at any possible orientation is P 187,500 N  max    25.0 MPa A 7,500 mm 2 and the maximum shear stress at any possible orientation in the block is P 187,500 N  max    12.50 MPa 2 A 2(7,500 mm 2 )

Ans.

Ans.

Ans.

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2.1 When an axial load is applied to the ends of the bar shown in Fig. P2.1, the total elongation of the bar between joints A and C is 0.15 in. In segment (2), the normal strain is measured as 1,300 in./in. Determine: (a) the elongation of segment (2). (b) the normal strain in segment (1) of the bar. Fig. P2.1

Solution (a) From the definition of normal strain, the elongation in segment (2) can be computed as (1,300 10 6 )(90 in.) 0.1170 in. 2 2 L2

Ans.

(b) The combined elongations of segments (1) and (2) is given as 0.15 in. Therefore, the elongation that occurs in segment (1) must be 0.15 in. 0.1170 in. 0.0330 in. 1 total 2 The strain in segment (1) can now be computed: 0.0330 in. 1 0.000825 in./in. 825 μin./in. 1 L1 40 in.

Ans.

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2.2 A rigid steel bar is supported by three rods, as shown in Fig. P2.2. There is no strain in the rods before the load P is applied. After load P is applied, the normal strain in rod (2) is 1,080 in./in. Assume initial rod lengths of L1 = 130 in. and L2 = 75 in. Determine: (a) the normal strain in rods (1). (b) the normal strain in rods (1) if there is a 1/32-in. gap in the connections between the rigid bar and rods (1) at joints A and C before the load is applied. (c) the normal strain in rods (1) if there is a 1/32-in. gap in the connection between the rigid bar and rod (2) at joint B before the load is applied. Fig. P2.2

Solution (a) From the normal strain in rod (2) and its length, the deformation of rod (2) can be calculated: (1,080 10 6 )(75 in.) 0.0810 in. 2 2 L2 Since rod (2) is assumed to be connected to the rigid bar with a perfect connection, the rigid bar must move downward by an amount equal to the deformation of rod (2); therefore, vB 0.0810 in. (downward) 2 By symmetry, the rigid bar must remain horizontal as it moves downward, and thus, vB = vA = vC. Rods (1) are connected to the rigid bar at A and C, and again, perfect connections are assumed. The deformation of rod (1) must be equal to the deflection of joint A (or C); thus, 1 = 0.0810 in. The normal strain in rods (1) can now be calculated as: 0.0810 in. 1 0.0006231 in./in. 623 μin./in. Ans. 1 L1 130 in. (b) We can assume that the bolted connection at B is perfect; therefore, vB = 0.0810 in. (downward). Further, the rigid bar must remain horizontal as it deflects downward by virtue of symmetry. Therefore, the deflection downward of joints A and C is still equal to 0.0810 in. What effect is caused by the gap at A and C? When joint A (or C) moves downward by 0.0810 in., the first 1/32-in. of this downward movement does not stretch rod (1)—it just closes the gap. Therefore, rod (1) only gets elongated by the amount vA 0.03125 in. 1

0.0810 in. 0.03125 in. 0.04975 in.

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This deformation creates a strain in rod (1) of: 0.04975 in. 1 0.0003827 in./in. 383 μin./in. 1 L1 130 in.

Ans.

(c) The gap is now at joint B. We know the strain in rod (2); hence, we know its deformation must be 0.0810 in. However, the first 1/32-in. of downward movement by the rigid bar does not elongate the rod—it simply closes the gap. To elongate rod (2) by 0.0810 in., joint B must move down: vB 0.03125 in. 2

0.0810 in. 0.03125 in. 0.11225 in. Again, since the rigid bar remains horizontal, vB = vA = vC. The joints at A and C are assumed to be perfect; thus, vA 0.11225 in. 1 and the normal strain in rod (1) is: 0.11225 in. 1 0.00086346 in./in. 863 μin./in. 1 L1 130 in.

Ans.

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2.3 A rigid steel bar is supported by three rods, as shown in Fig. P2.3. There is no strain in the rods before the load P is applied. After load P is applied, the normal strain in rods (1) is 860 m/m. Assume initial rod lengths of L1 = 2,400 mm and L2 = 1,800 mm. Determine: (a) the normal strain in rod (2). (b) the normal strain in rod (2) if there is a 2mm gap in the connections between the rigid bar and rods (1) at joints A and C before the load is applied. (c) the normal strain in rod (2) if there is a 2mm gap in the connection between the rigid bar and rod (2) at joint B before the load is applied. Fig. P2.3

Solution (a) From the normal strain in rod (1) and its length, the deformation of rod (1) can be calculated: (860 10 6 )(2, 400 mm) 2.064 mm 1 1 L1 Since rod (1) is assumed to be connected to the rigid bar with a perfect connection, the rigid bar must move downward by an amount equal to the deformation of rod (1); therefore, vA vC 2.064 mm (downward) 1 By symmetry, the rigid bar must remain horizontal as it moves downward, and thus, vB = vA = vC. Rod (2) is connected to the rigid bar at B, and again, a perfect connection is assumed. The deformation of rod (2) must be equal to the deflection of joint B; thus, 2 = 2.064 mm. The normal strain in rod (2) can now be calculated as: 2.064 mm 2 0.0011467 mm/mm 1,147 μmm/mm Ans. 2 L2 1,800 mm (b) We know the strain in rod (1); hence, we know its deformation must be 2.064 mm. However, the joints at A and C are not perfect connections. The first 2 mm of downward movement by the rigid bar does not elongate rod (1)—it simply closes the gap. To elongate rod (1) by 2.064 mm, joint A must move down: vA 2 mm 2.064 mm 2 mm 4.064 mm 1 By symmetry, the rigid bar remains horizontal; therefore, vB = vA = vC. The joint at B is assumed to be perfect; thus, any downward movement of the rigid bar also elongates rod (2) by the same amount: vB 4.064 mm 2

and the normal strain in rod (2) is: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

2 2

L2

4.064 mm 1,800 mm

0.0022578 mm/mm

2, 260 μmm/mm

Ans.

(c) We now assume that the bolted connection at A (or C) is perfect; therefore, the rigid bar deflection as calculated in part (a) must be vA = 2.064 mm (downward). Further, the rigid bar must remain horizontal as it deflects downward by virtue of symmetry; thus, vB = vA = vC. Therefore, the deflection downward of joint B is vB = 2.064 mm What effect is caused by the gap at B? When joint B moves downward by 2.064 mm, the first 2 mm of this downward movement does not stretch rod (2)—it just closes the gap. Therefore, rod (2) only gets elongated by the amount vB 2 mm 2

2.064 mm 2 mm 0.064 mm This deformation creates a strain in rod (2) of: 0.064 mm 2 0.0000356 mm/mm 2 L2 1,800 mm

35.6 μmm/mm

Ans.

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2.4 A rigid bar ABCD is supported by two bars as shown in Fig. P2.4. There is no strain in the vertical bars before load P is applied. After load P is applied, the normal strain in rod (1) is −570 m/m. Determine: (a) the normal strain in rod (2). (b) the normal strain in rod (2) if there is a 1-mm gap in the connection at pin C before the load is applied. (c) the normal strain in rod (2) if there is a 1-mm gap in the connection at pin B before the load is applied. Fig. P2.4

Solution (a) From the strain given for rod (1) 1 1 L1

( 570 10 6 )(900 mm) 0.5130 mm Therefore, vB = 0.5130 mm (downward). From the deformation diagram of rigid bar ABCD vB vC 240 mm (240 mm 360 mm) 600 mm vC (0.5130 mm) 1.2825 mm 240 mm Therefore,

2 2

2

L2

= 1.2825 mm (elongation), and thus, from the definition of strain: 1.2825 mm 855 10 6 mm/mm 855 με 1,500 mm

Ans.

(b) The 1-mm gap at C doesn’t affect rod (1); therefore, 1 = −0.5130 mm. The rigid bar deformation diagram is unaffected; thus, vB = 0.5130 mm (downward) and vC = 1.2825 mm (downward). The rigid bar must move downward 1 mm at C before it begins to elongate member (2). Therefore, the elongation of member (2) is vC 1 mm 2

1.2825 mm 1 mm 0.2825 mm

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and so the strain in member (2) is 0.2825 mm 2 188.3 10 2 L2 1,500 mm

6

mm/mm

188.3 με

Ans.

(c) From the strain given for rod (1), 1 = −0.5130 mm. In order to contract rod (1) by this amount, the rigid bar must move downward at B by vB 0.5130 mm 1 mm 1.5130 mm From deformation diagram of rigid bar ABCD vB vC 240 mm (240 mm 360 mm) 600 mm vC (1.5130 mm) 3.7825 mm 240 mm and so 3.7825 mm 2 2,521.7 10 6 mm/mm 2 L2 1,500 mm

2,520 με

Ans.

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2.5 In Fig. P2.5, rigid bar ABC is supported by a pin connection at B and two axial members. A slot in member (1) allows the pin at A to slide 0.25-in. before it contacts the axial member. If the load P produces a compression normal strain in member (1) of −1,300 in./in., determine the normal strain in member (2).

Fig. P2.5

Solution From the strain given for member (1) 1 1 L1

( 1,300 10 6 )(32 in.) 0.0416 in. Pin A has to move 0.25 in. before it contacts member (1); therefore, vA = 1.080 mm (downward). vA 0.0416 in. 0.25 in.

0.2916 in. 0.2916 in. (to the left)

From the deformation diagram of rigid bar ABC vA vC 12 in. 20 in. 20 in. vC (0.2916 in.) 0.4860 in. (downward) 12 in. Therefore, 2 = 0.4860 in. (elongation). From the definition of strain, 0.4860 in. 2 3,037.5 10 6 in./in. 3,040 με 2 L2 160 in.

Ans.

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2.6 The sanding-drum mandrel shown in Fig. P2.6 is made for use with a hand drill. The mandrel is made from a rubber-like material that expands when the nut is tightened to secure the sanding sleeve placed over the outside surface. If the diameter D of the mandrel increases from 2.00 in. to 2.15 in. as the nut is tightened, determine (a) the average normal strain along a diameter of the mandrel. (b) the circumferential strain at the outside surface of the mandrel.

Fig. P2.6

Solution (a) The change in strain along a diameter is found from D 2.15 in. 2.00 in. 0.075 in./in. D D 2.00 in. (b) Note that the circumference of a circle is given by D. The change in strain around the circumference of the mandrel is found from C (2.15 in.) (2.00 in.) 0.075 in./in. C C (2.00 in.)

Ans.

Ans.

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2.7 The normal strain in a suspended bar of material of varying cross section due to its own weight is given by the expression y/3E where is the specific weight of the material, y is the distance from the free (i.e., bottom) end of the bar, and E is a material constant. Determine, in terms of , L, and E, (a) the change in length of the bar due to its own weight. (b) the average normal strain over the length L of the bar. (c) the maximum normal strain in the bar.

Solution (a) The strain of the suspended bar due to its own weight is given as y 3E Consider a slice of the bar having length dy. In general, = L. Applying this definition to the bar slice, the deformation of slice dy is given by y d dy dy 3E Since this strain expression varies with y, the total deformation of the bar must be found by integrating d over the bar length: L

y dy 3 E 0

y2 3E 2

L

0

L2 6E

(b) The average normal strain is found by dividing the expression above for L2 / 6 E L avg L 6E

Ans.

by the bar length L Ans.

(c) Since the given strain expression varies with y, the maximum normal strain occurs at the maximum value of y, that is, at y = L: y L Ans. max y L 3E y L 3E

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2.8 A steel cable is used to support an elevator cage at the bottom of a 2,000-ft deep mineshaft. A uniform normal strain of 250 in./in. is produced in the cable by the weight of the cage. At each point, the weight of the cable produces an additional normal strain that is proportional to the length of the cable below the point. If the total normal strain in the cable at the cable drum (upper end of the cable) is 700 in./in., determine (a) the strain in the cable at a depth of 500 ft. (b) the total elongation of the cable.

Solution Call the vertical coordinate y and establish the origin of the y axis at the lower end of the steel cable. The strain in the cable has a constant term (i.e., = 250 ) and a term (we will call it k) that varies with the vertical coordinate y. 250 10 6 k y The problem states that the normal strain at the cable drum (i.e., y = 2,000 ft) is 700 in./in. Knowing this value, the constant k can be determined 700 10 6 250 10 6 k (2, 000 ft)

700 10 6 250 10 6 0.225 10 6 /ft 2, 000 ft Substituting this value for k in the strain expression gives 250 10 6 (0.225 10 6 /ft) y k

(a) At a depth of 500 ft, the y coordinate is y = 1,500 ft. Therefore, the cable strain at a depth of 500 ft is Ans. 250 10 6 (0.225 10 6 /ft)(1,500 ft) 587.5 10 6 588 με (b) The total elongation is found by integrating the strain expression over the cable length; thus, 2000

dy

250 10

6

(0.225 10 6 /ft) y dy

0 6

(250 10 ) y

0.225 10 2

0.950 ft 11.40 in.

2000

6

y

2 0

Ans.

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2.9 The 16 × 22 × 25-mm rubber blocks shown in Fig. P2.9 are used in a double U shear mount to isolate the vibration of a machine from its supports. An applied load of P = 690 N causes the upper frame to be deflected downward by 7 mm. Determine the average shear strain and the shear stress in the rubber blocks.

Fig. P2.9

Solution Consider the deformation of one block. After a downward deflection of 7 mm: 7 mm tan 0.4375 0.412410 rad 16 mm and thus, the shear strain in the block is Ans. 0.412 rad 412,000 μrad Note that the small angle approximation most definitely does not apply here! The applied load of 690 N is carried by two blocks; therefore, the shear force applied to one block is V = 345 N. The area subjected to shear stress is the area that is parallel to the direction of the shear force; that is, the 22 mm by 25 mm surface of the block. The shear stress is 345 N (22 mm)(25 mm) 0.627 MPa 627 kPa

Ans.

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2.10 A thin polymer plate PQR is deformed such that corner Q is displaced downward 1/16-in. to new position Q’ as shown in Fig. P2.10. Determine the shear strain at Q’ associated with the two edges (PQ and QR).

Fig. P2.10

Solution Before deformation, the angle of the plate at Q was 90° or /2 radians. We must now determine the plate angle at Q′ after deformation. The difference between these angles is the shear strain. After point Q displaces downward by 1/16in., the angle ′ is 25 in. 25 in. tan (10 in. 1 / 16-in.) 10.0625 in. 68.075215 and the angle ′ is 4 in. 4 in. tan (10 in. 1 / 16-in.) 10.0625 in. 21.678589 After deformation, the angle of the plate at Q′ is 68.075215 21.678589 89.753804 1.566499 rad The difference in the plate angle at Q before and after deformation is the shear strain: 2

1.566499 rad

0.004297 rad

4,300 μrad

Ans.

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2.11 A thin polymer plate PQR is deformed so that corner Q is displaced downward 1.0 mm to new position Q’ as shown in Fig. P2.11. Determine the shear strain at Q’ associated with the two edges (PQ and QR).

Fig. P2.11

Solution Before deformation, the angle of the plate at Q was 90° or /2 radians. We must now determine the plate angle at Q′ after deformation. The difference between these angles is the shear strain. After deformation, the angle ′ is 120 mm 120 mm tan (300 mm 1 mm) 301 mm 21.735741 and the angle ′ is 750 mm 750 mm tan (300 mm 1 mm) 301 mm 68.132764 Therefore, after deformation, the angle of the plate at Q′ is 21.735741 68.132764 89.868505 1.568501 rad The difference in the angle at Q before and after deformation is the shear strain: 2

1.568501 rad

0.002295 rad

2,300 μrad

Ans.

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2.12 A thin rectangular plate is uniformly deformed as shown in Fig. P2.12. Determine the shear strain xy at P.

Fig. P2.12

Solution Before deformation, the angle of the plate at P was 90° or /2 radians. We must now determine the plate angle at P after deformation. The difference between these angles is the shear strain. After deformation, the angle that side PQ makes with the horizontal axis is 1.4 mm tan 1,100 mm 0.072922 and the angle that side PR makes with the vertical axis is 0.7 mm tan 600 mm 0.066845 Therefore, the angle of the plate at P that was initially 90° has been reduced by the sum of and : 0.072922 0.066845 0.139767 0.002439 rad Since shear strain is defined as the change in angle between two initially perpendicular lines, the shear strain in the plate at P is Ans. 0.002439 rad 2,440 μrad P Note: Since these angles are small, we could have just as well used tan ≈ and tan the extra steps involved in using the inverse tangent function. Thus, 1.4 mm tan 0.001273 rad 1,100 mm 0.7 mm tan 0.001167 rad 600 mm 0.001273 rad 0.001167 rad 0.002439 rad 2,440 μrad P

≈

and saved

Ans.

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2.13 A thin rectangular plate is uniformly deformed as shown in Fig. P2.13. Determine the shear strain xy at P.

Fig. P2.13

Solution Before deformation, the angle of the plate at P was 90° or /2 radians. We must now determine the plate angle at P after deformation. The difference between these angles is the shear strain. After deformation, the angle that side PQ makes with the horizontal axis is 0.125 in. tan 0.397881 18 in. and the angle that side PR makes with the vertical axis is 0.125 in. tan 0.286477 25 in. By inspection, the angle of the plate at P has been increased by and decreased by ; thus, the angle at P after deformation is: 90 90 0.397881 0.286477 90.111404 The angle in the plate at P that was initially 90° has been increased to 90.111404° = 1.572741 rad. The shear strain in the plate at P is thus P

2

1.572741 rad

0.001944 rad

1,944 μrad

Ans.

Note: Since these angles are small, we could have just as well used tan ≈ and tan ≈ and saved the extra steps involved in using the inverse tangent functions. Thus, 0.125 in. tan 0.006944 rad 18 in. 0.125 in. tan 0.005000 rad 25 in. By inspection, the angle of the plate at P has been increased by and decreased by ; thus, the angle at P after deformation is: 0.006944 rad 0.005000 rad 0.001944 rad 2 2 2 Therefore, the shear strain in the plate at P is P = −0.001944 rad = −1,944 rad.

Ans.

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2.14 A thin polymer plate PQR is deformed so that corner Q is displaced downward 1.0 mm to new position Q’ as shown in Fig. P2.14. Determine the shear strain at Q’ associated with the two edges (PQ and QR).

Fig. P2.14

Solution Before deformation, the angle of the plate at Q was 90° or /2 radians. We must now determine the plate angle at Q′ after deformation. The difference between these angles is the shear strain. After deformation, the angle ′ is 180 mm 1.0 mm 181 mm tan 300 mm 300 mm 31.103981 and the angle ′ is 500 mm 1.0 mm 499 mm tan 300 mm 300 mm 58.985614

Therefore, after deformation, the angle of the plate at Q′ is 31.103981 58.985614 90.089595 1.572360 rad The difference in plate angle before and after deformation is the shear strain: 2

1.572360 rad

0.001564 rad

1,564 μrad

Ans.

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2.15 An airplane has a half-wingspan of 33 m. Determine the change in length of the aluminum alloy [ A = 22.5×10-6/°C] wing spar if the plane leaves the ground at a temperature of 15°C and climbs to an altitude where the temperature is –55°C.

Solution The change in temperature between the ground and the altitude in flight is T Tfinal Tinitial 55°C 15°C 70°C The thermal strain is given by T (22.5 10 6 /°C)( 70°C) T

0.001575 mm/mm

and thus the deformation in the 33-m wing is ( 0.001575 mm/mm)(33 m) 0.0520 m TL

52.0 mm

Ans.

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2.16 A large cement kiln has a length of 400 ft and a diameter of 20 ft. Determine the change in length and diameter of the structural steel [ S = 6.5×10-6/°F] shell caused by an increase in temperature of 350°F.

Solution The thermal strain is given by T (6.5 10 6 /°F)( 350°F) T

0.002275 in./in.

The change in length of the 225-ft kiln length is L T L ( 0.002275 in./in.)(400 ft) 0.9100 ft The change in diameter of the 20-ft diameter is D T L ( 0.002275 in./in.)(20 ft) 0.004550 ft

10.92 in.

0.546 in.

Ans.

Ans.

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2.17 A cast iron pipe has an inside diameter of d = 208 mm and an outside diameter of D = 236 mm. The length of the pipe is L = 3.0 m. The coefficient of thermal expansion for cast iron is I = 12.1×10-6/°C. Determine the dimension changes caused by an increase in temperature of 70°C.

Solution The thermal strain caused by a temperature increase of 70°C is given by T (12.1 10 6 /°C)(70°C) 0.000847 mm/mm T The dimension changes caused by a temperature increase of 70°C are D T D (0.000847 mm/mm)(236 mm) 0.1999 mm

d

T

d

(0.000847 mm/mm)(208 mm)

L

T

L (0.000847 mm/mm)(3.0 m) 0.002541 m

Ans. Ans.

0.1762 mm 2.54 mm

Ans.

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2.18 At a temperature of 40°F, a 0.08-in. gap exists between the ends of the two bars shown in Fig. P2.18. Bar (1) is an aluminum alloy [ = 12.5 × 10−6/°F] and bar (2) is stainless steel [ = 9.6 × 10−6/°F]. The supports at A and C are rigid. Determine the lowest temperature at which the two bars contact each other.

Fig. P2.18

Solution Write expressions for the temperature-induced deformations and set this equal to the 0.08-in. gap: 0.08 in. 1 T L1 2 T L2 T

L

1 1

2

L2

0.08 in.

0.08 in. 0.08 in. 77.821°F -6 (12.5 10 / °F)(40 in.) (9.6 10-6 / °F)(55 in.) 1 L1 2 L2 Since the initial temperature is 40°F, the temperature at which the gap is closed is 117.8°F. T

Ans.

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2.19 At a temperature of 5°C, a 3-mm gap exists between two polymer bars and a rigid support, as shown in Fig. P2.19. Bars (1) and (2) have coefficients of thermal expansion of = 140 × 10−6/°C and = 67 × 10−6/°C, respectively. The supports at A and C are rigid. Determine the lowest temperature at which the 3-mm gap is closed.

Fig. P2.19

Solution Write expressions for the temperature-induced deformations and set this equal to the 3-mm gap: 3 mm 1 T L1 2 T L2 T

L

1 1

2

L2

3 mm

3 mm 3 mm 30.084°C -6 (140 10 / °C)(540 mm) (67 10-6 / °C)(360 mm) 1 L1 2 L2 Since the initial temperature is 5°C, the temperature at which the gap is closed is 35.084°C = 35.1°C. Ans. T

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2.20 An aluminum pipe has a length of 60 m at a temperature of 10°C. An adjacent steel pipe at the same temperature is 5 mm longer. At what temperature will the aluminum pipe be 15 mm longer than the steel pipe? Assume that the coefficient of thermal expansion for the aluminum is 22.5×10-6/°C and that the coefficient of thermal expansion for the steel is 12.5×10-6/°C.

Solution The length of the aluminum pipe after a change in temperature can be expressed as Lfinal A Linitial A L A Linitial A A T Linitial A Similarly, the length of the steel pipe after the same change in temperature is given by Lfinal S Linitial S L S

(a)

Linitial S (b) S T Linitial S From the problem statement, we are trying to determine the temperature change that will cause the final length of the aluminum pipe to be 15 mm longer than the steel pipe. This requirement can be expressed as (c) Lfinal A Lfinal S 15 mm Substitute Eqs. (a) and (b) into Eq. (c) to obtain the following relationship Linitial A Linitial S 15 mm A T Linitial A S T Linitial S

Collect the terms with T on the left-hand side of the equation Linitial S Linitial A 15 mm A T Linitial A S T Linitial S Factor out T T A Linitial

A

S

Linitial

S

Linitial

S

Linitial

A

15 mm

and thus, T can be expressed as Linitial S Linitial A 15 mm T A Linitial A S Linitial S Convert all length dimensions to units of millimeters and solve 60, 005 mm 60, 000 mm 15 mm T 6 (22.5 10 /°C)(60, 000 mm) (12.5 10 6 /°C)(60, 005 mm) 20 mm 20 mm 1.35 mm/°C 0.75 mm/°C 0.60 mm/°C 33.3°C Initially, the pipes were at a temperature of 10°C. With the temperature change determined above, the temperature at which the aluminum pipe is 15 mm longer than the steel pipe is Ans. Tfinal Tinitial T 10°C 33.3°C 43.3°C

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2.21 Determine the movement of the pointer of Fig. P2.21 with respect to the scale zero in response to a temperature increase of 60°F. The coefficients of thermal expansion are 6.6×10-6/°F for the steel and 12.5×10-6/°F for the aluminum.

Fig. P2.21

Solution In response to the 60°F temperature increase, the steel pieces elongate by the amount (6.6 10 6 /°F)(60°F)(12 in.) 0.004752 in. Steel Steel T LSteel and the aluminum piece elongates by (12.5 10 6 /°F)(60°F)(12 in.) 0.009000 in. Alum Alum T LAlum From the deformation diagram of the pointer, the scale reading can be determined from similar triangles vpointer Alum Steel 1.5 in. vpointer

8.5 in. 8.5 in. Alum 1.5 in.

Steel

5.6667 0.009000 in. 0.004752 in.

0.0241 in. (upward)

Ans.

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2.22 Determine the horizontal movement of point A of Fig. P2.22 due to a temperature increase of 75°C. Assume that member AE has a negligible coefficient of thermal expansion. The coefficients of thermal expansion are 11.9×10-6/°C for the steel and 22.5×10-6/°C for the aluminum alloy.

Fig. P2.22

Solution In response to the 75°C temperature increase, the steel piece elongates by the amount (11.9 10 6 /°C)(75°C)(300 mm) 0.267750 mm Steel Steel T LSteel Thus, joint C moves to the right by 0.267750 mm. Next, calculate the deformation of the aluminum piece: (22.5 10 6 /°C)(75°C)(300 mm) Alum Alum T LAlum Joint E moves to the right by 0.50625 mm.

0.50625 mm

Take the initial position of E as the origin. The coordinates of E in the deflected position are (0.50625 mm, 0) and the coordinates of C are (0.267750 mm, 25 mm). Use the deflected-position coordinates of E and C to determine the slope of the pointer. yC yE 25 mm 0 25 mm slope 104.821803 xC xE 0.267750 mm 0.50625 mm 0.2385 mm A general equation for the deflected pointer can be expressed as y mx b 104.821803x b

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Use the known coordinates of point E to determine b: 0 104.821803(0.50625 mm) b b 53.066038 mm Thus, the deflected pointer can be described by the line y 104.821803x 53.066038 mm or rearranging to solve for x (53.066038 mm) y x 104.821803 At pointer tip A, y = 275 mm; therefore, the x coordinate of the pointer tip is (53.066038 mm 275 mm) 221.933962 mm Ans. x 2.11725 mm 2.12 mm 104.821803 104.821803 The x coordinate is the same as the horizontal movement since we took the initial position of joint E as the origin.

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2.23 At a temperature of 25°C, a cold-rolled red brass [ B = 17.6×10-6/°C] sleeve has an inside diameter of dB = 299.75 mm and an outside diameter of DB = 310 mm. The sleeve is to be placed on a steel [ S = 11.9×10-6/°C] shaft with an outside diameter of DS = 300 mm. If the temperatures of the sleeve and the shaft remain the same, determine the temperature at which the sleeve will slip over the shaft with a gap of 0.05 mm.

Solution The inside diameter of the brass sleeve after a change in temperature can be expressed as d final B dinitial B d B dinitial B B T d initial B Similarly, the outside diameter of the steel shaft after the same change in temperature is given by Dfinal S Dinitial S D S

(a)

Dinitial S (b) S T Dinitial S From the problem statement, we are trying to determine the temperature change that will cause the inside diameter of the brass sleeve to be 0.05 mm greater than the outside diameter of the steel shaft. This requirement can be expressed as (c) dfinal B Dfinal S 0.05 mm Substitute Eqs. (a) and (b) into Eq. (c) to obtain the following relationship dinitial B Dinitial S 0.05 mm B T dinitial B S T Dinitial S

Collect the terms with T on the left-hand side of the equation Dinitial S dinitial B 0.05 mm B T dinitial B S T Dinitial S Factor out T T B dinitial

B

S

Dinitial

S

Dinitial

S

dinitial

B

0.05 mm

and thus, T can be expressed as Dinitial S dinitial B 0.05 mm T B d initial B S Dinitial S Solve for the temperature change 300 mm 299.75 mm 0.05 mm T 6 (17.6 10 /°C)(299.75 mm) (11.9 10 6 /°C)(300 mm) 0.30 mm 0.30 mm 0.005276 mm/°C 0.003570 mm/°C 0.001706 mm/°C 175.9°C Initially, the shaft and sleeve were at a temperature of 25°C. With the temperature change determined above, the temperature at which the sleeve fits over the shaft with a gap of 0.05 mm is Ans. Tfinal Tinitial T 25°C 175.9°C 201°C

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3.1 At the proportional limit, a 2-inch gage length of a 0.375-in.-diameter alloy bar has elongated 0.0083 in. and the diameter has been reduced 0.0005 in. The total tension force on the bar was 4.75 kips. Determine the following properties of the material: (a) the modulus of elasticity. (b) Poisson’s ratio. (c) the proportional limit.

Solution (a) The bar cross-sectional area is A



D2 



(0.375 in.)2  0.110447 in.2 4 4 and thus, the normal stress corresponding to the 4.75-kip force is 4.75 kips   43.007204 ksi 0.110447 in.2 The longitudinal strain in the bar is  0.0083 in.  long    0.004150 in./in. L 2 in. The modulus of elasticity is therefore  43.007204 ksi E   10,363.2 ksi  10,360 ksi  long 0.004150 in./in.

(b) The longitudinal strain in the bar was calculated previously as  long  0.004150 in./in. The lateral strain can be determined from the reduction of the diameter: D 0.0005 in.  lat    0.001333 in./in. D 0.375 in. Poisson’s ratio for this specimen is therefore  0.001333 in./in.    lat    0.321285  0.321  long 0.004150 in./in.

Ans.

Ans.

(c) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore, Ans.  PL  43.0 ksi

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3.2 At the proportional limit, a 30 mm wide × 12 mm thick bar elongates 2.0 mm under an axial load of 41.5 kN. The bar is 1.5-m long. If Poisson’s ratio is 0.33 for the material, determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the change in each lateral dimension.

Solution (a) The bar cross-sectional area is A  (30 mm)(12 mm)  360 mm 2 and thus, the normal stress corresponding to the 41.5-kN axial load is (41.5 kN)(1,000 N/kN)   115.2778 MPa 360 mm2 The longitudinal strain in the bar is  2.0 mm  long    0.001333 mm/mm L (1.5 m)(1,000 mm/m) The modulus of elasticity is therefore  115.2778 MPa E   86, 458 MPa  86.5 GPa  long 0.001333 mm/mm

Ans.

(b) Based on the problem statement, the stress in the bar is equal to the proportional limit; therefore, Ans.  PL  115.3 MPa (c) Poisson’s ratio is given as  = 0.33. The longitudinal strain in the bar was calculated previously as  long  0.001333 mm/mm The corresponding lateral strain can be determined from Poisson’s ratio:  lat   long  (0.33)(0.001333 mm/mm)  0.000440 mm/mm Using this lateral strain, the change in bar width is width  lat (width)  (0.000440 mm/mm)(30 mm)  0.01320 mm and the change in bar thickness is thickness  lat (thickness)  (0.000440 mm/mm)(12 mm)  0.00528 mm

Ans. Ans.

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3.3 At an axial load of 22 kN, a 45-mm-wide × 15-mm-thick polyimide polymer bar elongates 3.0 mm while the bar width contracts 0.25 mm. The bar is 200 mm long. At the 22-kN load, the stress in the polymer bar is less than its proportional limit. Determine: (a) the modulus of elasticity. (b) Poisson’s ratio. (c) the change in the bar thickness.

Solution (a) The bar cross-sectional area is A  (15 mm)(45 mm)  675 mm 2 and thus, the normal stress corresponding to the 22-kN axial load is (22 kN)(1,000 N/kN)   32.592593 MPa 675 mm2 The longitudinal strain in the bar is  3.0 mm  long    0.0150 mm/mm L 200 mm The modulus of elasticity is therefore  32.592593 MPa E   2,173 MPa  2.17 GPa  long 0.0150 mm/mm

Ans.

(b) The longitudinal strain in the bar was calculated previously as  long  0.0150 mm/mm The lateral strain can be determined from the reduction of the bar width: width 0.25 mm  lat    0.005556 mm/mm width 45 mm Poisson’s ratio for this specimen is therefore  0.005556 mm/mm    lat    0.370370  0.370  long 0.0150 mm/mm (c) The change in bar thickness can be found from the lateral strain: thickness  lat (thickness)  (0.005556 mm/mm)(15 mm)  0.0833 mm

Ans.

Ans.

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3.4 A 0.75-in.-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B. The width of the bar is w = 3.0 in. Strain gages bonded to the specimen measure the following strains in the longitudinal (x) and transverse (y) directions: x = 840  and y = −250 . (a) Determine Poisson’s ratio for this specimen. (b) If the measured strains were produced by an axial load of P = 32 kips, what is the modulus of elasticity for this specimen? Fig. P3.4

Solution (a) Poisson’s ratio for this specimen is   250 με    lat   y    0.298  long x 840 με (b) The bar cross-sectional area is A  (3.0 in.)(0.75 in.)  2.25 in.2 and so the normal stress for an axial load of P = 32 kips is 32 kips   14.222222 ksi 2.25 in.2 The modulus of elasticity is thus  14.222222 ksi E   16,931.2 ksi  16,930 ksi  long  1 in./in.  (840 με)    1,000,000 με 

Ans.

Ans.

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3.5 A 6-mm-thick rectangular alloy bar is subjected to a tensile load P by pins at A and B. The width of the bar is w = 30 mm. Strain gages bonded to the specimen measure the following strains in the longitudinal (x) and transverse (y) directions: x = 900  and y = −275 . (a) Determine Poisson’s ratio for this specimen. (b) If the measured strains were produced by an axial load of P = 19 kN, what is the modulus of elasticity for this specimen?

Fig. P3.5

Solution (a) Poisson’s ratio for this specimen is   275 με    lat   y    0.306  long x 900 με (b) The bar cross-sectional area is A  (30 mm)(6 mm)  180 mm 2 and so the normal stress for an axial load of P = 19 kN is (19 kN)(1,000 N/kN)   105.556 MPa 180 mm2 The modulus of elasticity is thus  105.556 MPa E   117,284.0 MPa  117.3 GPa  long  1 mm/mm  (900 με)   1,000,000 με 

Ans.

Ans.

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3.6 A copper rod [E = 110 GPa] originally 350-mm long is pulled in tension with a normal stress of 180 MPa. If the deformation is entirely elastic, what is the resulting elongation?

Solution Since the deformation is elastic, the strain in the rod can be determined from Hooke’s Law,  180 MPa    0.0016364 mm/mm E 110,000 MPa The elongation in the rod is thus    L  (0.0016364 mm/mm)(350 mm)  0.573 mm

Ans.

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3.7 A 6061-T6 aluminum tube [E = 10,000 ksi;  = 0.33] has an outside diameter of 4.000 in. and a wall thickness of 0.065 in. (a) Determine the tension force that must be applied to the tube to cause its outside diameter to contract by 0.005 in. (b) If the tube is 84-in. long, what is the overall elongation?

Solution (a) The lateral strain associated with the given diameter contraction is 0.005 in.  lat   0.001250 in./in. 4.000 in. From the given Poisson’s ratio, the longitudinal strain in the tube must be  0.001250 in./in.  long   lat    0.003788 in./in.  0.33 and from Hooke’s Law, the normal stress can be calculated as   E long  (10, 000 ksi)(0.003788 in./in.)  37.878788 ksi The cross-sectional area of the tube is needed to determine the tension force. Given that the outside diameter of the tube is 4.000 in. and the wall thickness is 0.065 in., the inside diameter of the tube is 3.870 in. The tube cross-sectional area is thus



(4.000 in.)2  (3.870 in.)2   0.803541 in.2 4 and the force applied to the tube is F   A  (37.878788 ksi)(0.803541 in.2 )  30.437154 kips  30.4 kips A

Ans.

(b) The longitudinal strain was calculated previously. Use the longitudinal strain to determine the overall elongation: Ans.    long L  (0.003788 in./in.)(84 in.)  0.318 in.

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3.8 A metal specimen with an original diameter of 0.500 in. and a gage length of 2.000 in. is tested in tension until fracture occurs. At the point of fracture, the diameter of the specimen is 0.260 in. and the fractured gage length is 3.08 in. Calculate the ductility in terms of percent elongation and percent reduction in area.

Solution Percent elongation is simply the longitudinal strain at fracture:  (3.08 in.  2.000 in.) 1.08 in.     0.54 in./in. L 2.000 in. 2.000 in.

 percent elongation  54%

Ans.

The initial cross-sectional area of the specimen is A0 



(0.500 in.) 2  0.196350 in.2 4 The final cross-sectional area at the fracture location is Af 



(0.260 in.) 2  0.053093 in.2

4 The percent reduction in area is

percent reduction of area 

A0  Af A0

(100%)

(0.196350 in.2  0.053093 in.2 )  (100%)  73.0% 0.196350 in.2

Ans.

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3.9 A portion of the stress-strain curve for a stainless steel alloy is shown in Fig. P3.9. A 350mm-long bar is loaded in tension until it elongates 2.0 mm and then the load is removed. (a) What is the permanent set in the bar? (b) What is the length of the unloaded bar? (c) If the bar is reloaded, what will be the proportional limit?

Fig. P3.9

Solution

(a) The normal strain in the specimen is  2.0 mm    0.005714 mm/mm L 350 mm Construct a line parallel to the elastic modulus line that passes through the data curve at a strain of  = 0.005714 mm/mm. The strain value at which this modulus line intersects the strain axis is the permanent set: permanent set  0.0035 mm/mm Ans. (b) The length of the unloaded bar is therefore:    L  (0.0035 mm/mm)(350 mm)  1.225 mm L f  350 mm  1.225 mm  351.225 mm

(c) From the stress-strain curve, the reload proportional limit is 444 MPa .

Ans. Ans.

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3.10 The 16 by 22 by 25-mm rubber blocks shown in Fig. P3.10 are used in a double U shear mount to isolate the vibration of a machine from its supports. An applied load of P = 285 N causes the upper frame to be deflected downward by 5 mm. Determine the shear modulus G of the rubber blocks.

Fig. P3.10

Solution Consider a FBD of the upper U frame. The downward force P is resisted by two upward shear forces V; therefore, V = 285 N / 2 = 142.5 N. Next, consider a FBD of one of the rubber blocks. The shear force acting on one rubber block is V = 142.5 N. The area of the rubber block that is parallel to the direction of V is AV  (22 mm)(25 mm)  550 mm 2 Consequently, the shear stress in one rubber block is V 142.5 N    0.259091 MPa AV 550 mm 2 The shear strain associated with the 5-mm downward displacement of the rubber blocks is given by: 5 mm tan    0.312500    0.302885 rad 16 mm From Hooke’s Law for shear stress and shear strain, the shear modulus G can be computed:  0.259091 MPa   G G    0.855411 MPa  0.855 MPa  0.302885 rad

Ans.

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3.11 Two hard rubber blocks are used in an anti-vibration mount to support a small machine, as shown in Fig. P3.11. An applied load of P = 150 lb causes a downward deflection of 0.25 in. Determine the shear modulus of the rubber blocks. Assume a = 0.5 in., b = 1.0 in., and c = 2.5 in.

Fig. P3.11

Solution Determine the shear strain from the angle formed by the downward deflection and the block thickness a: 0.250 in. tan    0.500    0.463648 rad 0.50 in. Note: The small angle approximation tan  is not applicable in this instance. Determine the shear stress from the applied load P and the block area. Note that this is a double shear configuration; therefore, the shear force V acting on a single rubber block is half of the total load: V = P/2 = 75 lb. To determine the area needed here, consider the surface that is bonded to the plate. This area has dimensions of bc. The shear stress acting on a single block is therefore: V 75 lb    30 psi A (1.0 in.)(2.5 in.) The shear modulus G can be calculated from Hooke’s law for shear stress and strain:  30 psi   G G    64.704 psi  64.7 psi  0.463648 rad

Ans.

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3.12 Two hard rubber blocks [G = 350 kPa] are used in an antivibration mount to support a small machine, as shown in Fig. P3.12. Determine the downward deflection that will occur for an applied load of P = 900 N. Assume a = 20 mm, b = 50 mm, and c = 80 mm.

Fig. P3.12

Solution Note that this is a double shear configuration; therefore, the shear force V acting on a single rubber block is half of the total load: V = P/2 = 450 N. Determine the shear stress from the shear force V and the block area. To determine the area needed here, consider the surface that is bonded to the plate. This area has dimensions of bc. The shear stress acting on a single block is therefore: V 450 N    0.112500 MPa A (50 mm)(80 mm) Since the shear modulus G is given, the shear strain can be calculated from Hooke’s law for shear stress and shear strain:  0.112500 MPa   G     0.321429 rad G 0.350 MPa From the angle  and the block thickness a, the downward deflection  of the block can be determined from: tan  



a

   a tan   (20 mm) tan(0.321429 rad)  6.659512 mm  6.66 mm

Ans.

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3.13 A load test on a 6 mm diameter by 225 mm long aluminum alloy rod found that a tension load of 4,800 N caused an elastic elongation of 0.52 mm in the rod. Using this result, determine the elastic elongation that would be expected for a 24-mm-diameter rod of the same material if the rod were 1.2 m long and subjected to a tension force of 37 kN.

Solution The cross-sectional area of the 6-mm-diameter rod is A



(6 mm) 2  28.274334 mm 2

4 Thus, the normal stress in the rod due to a 4,800-N load is F 4,800 N    169.76527 MPa A 28.274334 mm2 The strain in the 225-mm long rod associated with a 0.52-mm elongation is  0.52 mm    0.0023111 mm/mm L 225 mm Therefore, the elastic modulus of the aluminum alloy is  169.76527 MPa E   73,456.128 MPa  0.0023111 mm/mm

The cross-sectional area of the 24-mm-diameter rod is  A  (24 mm) 2  252.389342 mm 2 4 Thus, the normal stress in the 24-mm-diameter rod due to a 37-kN load is F (37 kN)(1,000 N/kN)    81.787957 MPa A 452.389342 mm 2 From Hooke’s Law, the strain that would be expected is  81.787957 MPa    0.0011134 mm/mm E 73, 456.128 MPa Since the 42-mm-diameter rod is 1.2-m long, the expected elongation is    L  (0.0011134 mm/mm)(1.2 m)(1,000 mm/m)  1.336111 mm  1.336 mm

Ans.

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3.14 The stress-strain diagram for a particular stainless steel alloy is shown in Fig. P3.14. A rod made from this material is initially 800 mm long at a temperature of 20°C. After a tension force is applied to the rod and the temperature is increased by 200°C, the length of the rod is 804 mm. Determine the stress in the rod and state whether the elongation in the rod is elastic or inelastic. Assume the coefficient of thermal expansion for this material is 18 × 10−6/°C.

Fig. P3.14

Solution The 4-mm total elongation of the rod is due to a combination of load and temperature increase. The 200°C temperature increase causes a normal strain of:  T   T  (18 106 / C )(200C )  0.003600 mm/mm which means that the rod elongates  T  T L  (0.003600 mm/mm)(800 mm)  2.8800 mm The portion of the 4-mm total elongation due to load is therefore       T  4 mm  2.8800 mm  1.1200 mm The strain corresponding to this elongation is  1.1200 mm      0.001400 mm/mm L 800 mm By inspection of the stress-strain curve, this strain is clearly in the linear region. Therefore, the rod is elastic in this instance. For the linear region, the elastic modulus can be determined from the lower scale plot:  (400 MPa  0) E   200, 000 MPa  (0.002 mm/mm  0) Using Hooke’s Law (or directly from the - diagram), the stress corresponding to the 0.001400 mm/mm strain is   E  (200,000 MPa)(0.001400 mm/mm)  280 MPa

Ans.

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3.15 In Fig. P3.15, rigid bar ABC is supported by axial member (1), which has a cross-sectional area of 400 mm2, an elastic modulus of E = 70 GPa, and a coefficient of thermal expansion of  = 22.5 × 10−6 /°C. After load P is applied to the rigid bar and the temperature rises 40°C, a strain gage affixed to member (1) measures a strain increase of 2,150 . Determine: (a) the normal stress in member (1). (b) the magnitude of applied load P. (c) the deflection of the rigid bar at C. Fig. P3.15

Solution (a) The strain measured in member (1) is due to both the internal force in the member and the temperature change. The strain caused by the temperature change is  T   T  (22.5  106 / °C)(40°C)  0.000900 mm/mm Since the total strain is  = 2,150  = 0.002150 mm/mm, the strain caused by the internal force in member (1) must be     T  0.002150 mm/mm  0.000900 mm/mm  0.001250 mm/mm The elastic modulus of member (1) is E = 70 GPa; thus, from Hooke’s Law, the stress in the member is: Ans. 1  E  (70,000 MPa)(0.001250 mm/mm)  87.5 MPa (b) If the normal stress in member (1) is 87.5 MPa, the axial force in the member is F1   1 A1  (87.5 N/mm 2 )(400 mm 2 )  35,000 N Consider moment equilibrium of rigid bar ABC about joint A to determine the magnitude of P: M A  (1.75 m)(35,000 N)  (3.0 m)P  0

 P  20,417 N  20.4 kN

Ans.

(c) The strain in member (1) was measured as  = 2,150  = 0.002150 mm/mm; therefore, the total elongation of member (1) is 1  1L1  (0.002150 mm/mm)(3,750 mm)  8.0625 mm The deflection of the rigid bar at B is equal to this elongation; therefore, vB = 1 = 8.0625 mm (downward). By similar triangles, the deflection of the rigid bar at C is given by: vB v  C 1.75 m 3.0 m 3.0 m 3.0  vC  vB  (8.0625 mm)  13.821429 mm  13.82 mm  Ans. 1.75 m 1.75

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3.16 A tensile test specimen of 1045 hot-rolled steel having a diameter of 0.505 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown in Fig. P3.16. Determine (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.20% offset). (e) the fracture stress. (f) the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.392 in.

Fig. P3.16

Solution From the stress-strain curve, the proportional limit will be taken as  = 60 ksi at a strain of  = 0.0019. (Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.) (a) The modulus of elasticity is  60 ksi E   31,600 ksi  0.0019 in./in. (b) From the diagram, the proportional limit is taken as  PL  60 ksi (c) The ultimate strength is U  105 ksi (d) The yield strength is Y  68 ksi (e) The fracture stress is  fracture  98 ksi

Ans.

Ans. Ans. Ans. Ans.

(f) The original cross-sectional area of the specimen is A0 



(0.505 in.)2  0.200296 in.2

4 The cross-sectional area of the specimen at the fracture location is Af 



(0.392 in.) 2  0.120687 in.2 4 The true fracture stress is therefore 0.200296 in.2 true  fracture  (98 ksi)  162.6 ksi 0.120687 in.2

Ans.

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3.17 A tensile test specimen of stainless steel alloy having a diameter of 0.495 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown in Fig. P3.17. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.20% offset). (e) the fracture stress. (f) the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.350 in.

Fig. P3.17

Solution From the stress-strain curve, the proportional limit will be taken as  = 60 ksi at a strain of  = 0.002. (Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.) (a) The modulus of elasticity is  60 ksi E   30,000 ksi  0.002 in./in. (b) From the diagram, the proportional limit is taken as  PL  60 ksi (c) The ultimate strength is U  159 ksi (d) The yield strength is Y  80 ksi (e) The fracture stress is  fracture  135 ksi

Ans.

Ans. Ans. Ans. Ans.

(f) The original cross-sectional area of the specimen is A0 



(0.495 in.)2  0.192442 in.2

4 The cross-sectional area of the specimen at the fracture location is Af 



(0.350 in.) 2  0.096211 in.2

4 The true fracture stress is therefore 0.192442 in.2 true  fracture  (135 ksi)  270 ksi 0.096211 in.2

Ans.

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3.18 A bronze alloy specimen having a diameter of 12.8 mm and a gage length of 50 mm was tested to fracture. Stress and strain data obtained during the test are shown in Fig. P3.18. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.20% offset). (e) the fracture stress. (f) the true fracture stress if the final diameter of the specimen at the location of the fracture was 10.5 mm.

Fig. P3.18

Solution From the stress-strain curve, the proportional limit will be taken as  = 210 MPa at a strain of  = 0.002. (Obviously, there can be quite a bit of leeway in pulling numbers from such a limited plot.) (a) The modulus of elasticity is  210 MPa E   105,000 MPa  0.002 in./in. (b) From the diagram, the proportional limit is taken as  PL  210 MPa (c) The ultimate strength is U  380 MPa (d) The yield strength is Y  290 MPa (e) The fracture stress is  fracture  320 MPa

Ans.

Ans. Ans. Ans. Ans.

(f) The original cross-sectional area of the specimen is A0 



(12.8 mm) 2  128.679635 mm 2

4 The cross-sectional area of the specimen at the fracture location is  Af  (10.5 mm)2  86.590148 mm 2 4 The true fracture stress is therefore 128.679635 mm 2 true  fracture  (320 MPa)  476 MPa 86.590148 mm 2

Ans.

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3.19 An alloy specimen having a diameter of 12.8 mm and a gage length of 50 mm was tested to fracture. Load and deformation data obtained during the test are given. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.05% offset). (e) the yield strength (0.20% offset). (f) the fracture stress. (g) the true fracture stress if the final diameter of the specimen at the location of the fracture was 11.3 mm.

(kN)

Change in Length (mm)

0 7.6 14.9 22.2 28.5 29.9 30.6 32.0 33.0 33.3 36.8 41.0

0 0.02 0.04 0.06 0.08 0.10 0.12 0.16 0.20 0.24 0.50 1.00

Load

(kN)

Change in Length (mm)

43.8 45.8 48.3 49.7 50.4 50.7 50.4 50.0 49.7 47.9 45.1

1.50 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 fracture

Load

Solution The plot of the stress-strain data is shown below.

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(a) Using the data point for 28.5 kN load and 0.08 mm elongation, the modulus of elasticity can be calculated as  221.48 MPa Ans. E   138, 400 MPa  0.0016 mm/mm (b) From the diagram, the proportional limit is taken as  PL  234 MPa (c) The ultimate strength is (50.7 kN)(1,000 N/kN) U   394 MPa  (12.8 mm)2 4

Ans.

Ans.

(d) The yield strength by the 0.05% offset method is Y  239 MPa

Ans.

(e) The yield strength by the 0.2% offset method is Y  259 MPa

Ans.

(f) The fracture stress is (45.1 kN)(1,000 N/kN)  fracture   350.48 MPa  350 MPa  2 (12.8 mm) 4

Ans.

(f) The original cross-sectional area of the specimen is A0 



(12.8 mm) 2  128.679635 mm 2

4 The cross-sectional area of the specimen at the fracture location is

Af 



(11.3 mm)2  100.287492 mm 2 4 The true fracture stress is therefore 128.679635 mm 2 true  fracture  (356 MPa)  457 MPa 100.287492 mm 2

Ans.

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3.20 A 1035 hot-rolled steel specimen with a diameter of 0.500 in. and a 2.0-in. gage length was tested to fracture. Load and deformation data obtained during the test are given. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.05% offset). (e) the yield strength (0.20% offset). (f) the fracture stress. (g) the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.387 in.

(lb)

Change in Length (in.)

0 2,690 5,670 8,360 11,050 12,540 13,150 13,140 12,530 12,540 12,840 12,840

0 0.0009 0.0018 0.0028 0.0037 0.0042 0.0046 0.0060 0.0079 0.0098 0.0121 0.0139

Load

(lb)

Change in Length (in.)

12,540 12,540 14,930 17,020 18,220 18,820 19,110 19,110 18,520 17,620 16,730 16,130 15,900

0.0209 0.0255 0.0487 0.0835 0.1252 0.1809 0.2551 0.2968 0.3107 0.3246 0.3339 0.3385 fracture

Load

Solution The plot of the stress-strain data is shown below.

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(a) Using the data point for 12,540 lb load and 0.0042 in. elongation, the modulus of elasticity can be calculated as  63.866 ksi Ans. E   30, 400 ksi  0.0021 in./in. (b) From the diagram, the proportional limit is taken as  PL  63.8 ksi

Ans.

(c) The ultimate strength is U  97.3 ksi

Ans.

(d) The yield strength using the 0.05% offset method is Y  65.4 ksi

Ans.

(e) The yield strength using the 0.2% offset method is Y  63.8 ksi

Ans.

(f) The fracture stress is 15,900 lb  fracture   80,978 psi  81.0 ksi  2 (0.500 in.) 4

Ans.

(g) The original cross-sectional area of the specimen is A0 



(0.500 in.) 2  0.196350 in.2 4 The cross-sectional area of the specimen at the fracture location is Af 



(0.387 in.) 2  0.117628 in.2

4 The true fracture stress is therefore 0.196350 in.2 true  fracture  (80,978 psi)  135,172 psi  135.2 ksi 0.117628 in.2

Ans.

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3.21 A 2024-T4 aluminum test specimen with a diameter of 0.505 in. and a 2.0-in. gage length was tested to fracture. Load and deformation data obtained during the test are given. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.05% offset). (e) the yield strength (0.20% offset). (f) the fracture stress. (g) the true fracture stress if the final diameter of the specimen at the location of the fracture was 0.452 in.

Load (lb) 0 1,300 2,390 3,470 4,560 5,640 6,720 7,380 8,240 8,890 9,330 9,980 10,200 10,630

Change in Length (in.) 0.0000 0.0014 0.0023 0.0032 0.0042 0.0051 0.0060 0.0070 0.0079 0.0088 0.0097 0.0107 0.0116 0.0125

Load (lb) 11,060 11,500 12,360 12,580 12,800 13,020 13,230 13,450 13,670 13,880 14,100 14,100 14,100 14,100 14,100

Change in Length (in.) 0.0139 0.0162 0.0278 0.0394 0.0603 0.0788 0.0974 0.1159 0.1391 0.1623 0.1994 0.2551 0.3200 0.3246 fracture

Solution The plot of the stress-strain data is shown below.

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(a) Using the data point for 6,720 lb load and 0.0060 in. elongation, the modulus of elasticity can be calculated as  33.55 ksi Ans. E   11,180 ksi  0.003 in./in. (b) From the diagram, the proportional limit is taken as  PL  33.6 ksi

Ans.

(c) The ultimate strength is U  70.4 ksi

Ans.

(d) The yield strength using the 0.05% offset method is Y  44.4 ksi

Ans.

(e) The yield strength using the 0.2% offset method is Y  54.5 ksi

Ans.

(f) The fracture stress is  fracture  70.4 ksi

Ans.

(g) The original cross-sectional area of the specimen is A0 



(0.505 in.)2  0.200296 in.2

4 The cross-sectional area of the specimen at the fracture location is

Af 



(0.452 in.) 2  0.160460 in.2 4 The true fracture stress is therefore 0.200296 in.2 true  fracture  (70.4 ksi)  87.9 ksi 0.160460 in.2

Ans.

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3.22 A 1045 hot-rolled steel tension test specimen has a diameter of 6.00 mm and a gage length of 25 mm. In a test to fracture, the stress and strain data below were obtained. Determine: (a) the modulus of elasticity. (b) the proportional limit. (c) the ultimate strength. (d) the yield strength (0.05% offset). (e) the yield strength (0.20% offset). (f) the fracture stress. (g) the true fracture stress if the final diameter of the specimen at the location of the fracture was 4.65 mm.

(kN)

Change in Length (mm)

0.00 2.94 5.58 8.52 11.16 12.63 13.02 13.16 13.22 13.22 13.25 13.22

0.00 0.01 0.02 0.03 0.05 0.05 0.06 0.08 0.08 0.10 0.14 0.17

Load

(kN)

Change in Length (mm)

13.22 16.15 18.50 20.27 20.56 20.67 20.72 20.61 20.27 19.97 19.68 19.09 18.72

0.29 0.61 1.04 1.80 2.26 2.78 3.36 3.83 3.94 4.00 4.06 4.12 fracture

Load

Solution The plot of the stress-strain data is shown below.

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(a) Using the data point for 8.52 kN load and 0.03 mm elongation, the modulus of elasticity can be calculated as  301.3 MPa Ans. E   251, 000 MPa  0.0012 mm/mm (b) From the diagram, the proportional limit is taken as  PL  400 MPa (c) The ultimate strength is (20.72 kN)(1,000 N/kN) U   732.82 MPa  733 MPa  (6.00 mm)2 4

Ans.

Ans.

(d) The yield strength by the 0.05% offset method is Y  465 MPa

Ans.

(e) The yield strength by the 0.2% offset method is Y  465 MPa

Ans.

(f) The fracture stress is (18.72 kN)(1,000 N/kN)  fracture   662.08 MPa  662 MPa  2 (6.00 mm) 4

Ans.

(f) The original cross-sectional area of the specimen is  A0  (6 mm) 2  28.274334 mm 2 4 The cross-sectional area of the specimen at the fracture location is Af 



(4.65 mm) 2  16.982272 mm2 4 The true fracture stress is therefore 28.274334 mm 2 true  fracture  (662.08 MPa)  1,102 MPa 16.982272 mm 2

Ans.

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3.23 Rigid bar BCD in Fig. P3.23 is supported by a pin at C and by aluminum rod (1). A concentrated load P is applied to the lower end of aluminum rod (2), which is attached to the rigid bar at D. The crosssectional area of each rod is A = 0.20 in.2 and the elastic modulus of the aluminum material is E = 10,000 ksi. After the load P is applied at E, the strain in rod (1) is measured as 1,350  (tension). (a) Determine the magnitude of load P. (b) Determine the total deflection of point E relative to its initial position.

Fig. P3.23

Solution (a) From the measured strain, the stress in rod (1) is  1  E11  (10,000 ksi)(1,350  10 6 in./in.)  13.5 ksi and thus, the force in rod (1) is F1   1 A1  (13.5 ksi)(0.20 in.2 )  2.70 kips (T) Consider the equilibrium of the rigid bar, and write a moment equilibrium equation about C to determine the magnitude of load P: M C  (20 in.)(2.70 kips)  (30 in.)P  0  P  1.8 kips

Ans.

(b) From the measured strain, the elongation of rod (1) is 1  1L1  (1,350  106 in./in.)(50 in.)  0.0675 in. From similar triangles, the deflection of the rigid bar at D can be expressed in terms of the deflection at B: vB v  D 20 in. 30 in. 30 in. 30 in.  vD  vB  (0.0675 in.)  0.10125 in. 20 in. 20 in.

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The elongation of rod (2) due to the 1.8-kip load must be determined, also. The stress in rod (2) is F 1.8 kips 2  2   9 ksi A2 0.2 in.2 and consequently, the strain in rod (2) is  9 ksi 2  2   0.000900 in./in. E2 10,000 ksi

From the strain, the elongation in rod (2) can be computed:  2   2 L2  (0.000900 in./in.)(100 in.)  0.0900 in. The deflection of joint E is the sum of the rigid bar deflection at D and the elongation in rod (2):

vE  vD   2  0.10125 in.  0.09 in.  0.19125 in.  0.1913 in. 

Ans.

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3.24 Rigid bar ABC is supported by pin-connected axial member (1) and by a 0.75-in.-diameter double shear pin connection at C as shown in Fig. P3.24. Member (1) is a 2.75 in. wide by 1.25 in. thick rectangular bar made of aluminum with an elastic modulus of 10,000 ksi. When a concentrated load P is applied to the rigid bar at A, the normal strain in member (1) is measured as –880 . (a) Determine the magnitude of applied load P. (b) Determine the average shear stress in the pin at C.

Fig. P3.24

Solution Member (1) is a two-force member that is oriented at  with respect to the horizontal axis: 30 in. tan    0.75   36.870 40 in. From a FBD of rigid structure ABC, the following equilibrium equations can be written: Fx   F1 cos 36.870  Cx  0 (a) Fy   P  F1 sin 36.870  C y  0 (b) M C  P(80 in.)  ( F1 sin 36.870)(56 in.)  0

(c)

(a) The force in member (1) can be determined from the measured strain of –880 . From Hooke’s law, the stress in member (1) is: 1 in./in.  1  E1  (10,000 ksi)(  880 με)  8.800 ksi 1  106 με and thus, the force in member (1) is: F1  1 A1  ( 8.800 ksi)(2.75 in.)(1.25 in.)  30.250 kips Substitute this value into Eq. (c) to compute P: ( F sin 36.870)(56 in.) ( 30.250 kips)(sin 36.870)(56 in.) P 1  80 in. 80 in.  12.705 kips  12.71 kips

Ans.

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(b) Substitute the values for P and F1 into Eqs. (a) and (b) to obtain Cx and Cy: Cx  F1 cos36.870  ( 30.250 kips)cos36.870  24.200 kips C y  P  F1 sin 36.870  12.705 kips  ( 30.250 kips)sin 36.870  5.445 kips The resultant pin force at C is found from Cx and Cy: C  Cx2  C y2  (24.200 kips)2  ( 5.445 kips) 2  24.805 kips

The pin at C is supported in a double-shear connection. The cross-sectional area of the 0.75-in.diameter pin is: Apin 



4

(0.75 in.)2  0.4421786 in.2

The average shear stress in the pin is thus: C 24.805 kips    28.074 ksi  28.1 ksi AV (2 surfaces)(0.441786 in.2 /surface)

Ans.

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3.25 A concentrated load P is supported by two bars as shown in Fig. P3.25. Bar (1) is made of cold-rolled red brass [E = 16,700 ksi;  = 10.4 × 10−6 /°F] and has a cross-sectional area of 0.225 in.2. Bar (2) is made of 6061-T6 aluminum [E = 10,000 ksi;  = 13.1 × 10−6 /°F] and has a crosssectional area of 0.375 in.2. After load P has been applied and the temperature of the entire assembly has increased by 50°F, the total strain in bar (1) is measured as 1,400  (elongation). Determine: (a) the magnitude of load P. (b) the total strain in bar (2).

Fig. P3.25

Solution Consider a FBD of joint B. Members (1) and (2) are two-force members; therefore, the equilibrium equations can be written as: Fx  F2 cos 55  F1 cos 40  0 (a) Fy  F2 sin 55  F1 sin 40  P  0 (b) From Eq. (a): cos 40 F2  F1 cos55 Substitute this expression into Eq. (b) to obtain:  cos 40   F1 cos55  sin 55  F1 sin 40  P F1  cos 40 tan 55  sin 40  P

 P  F1 1.736812

(c)

(d)

(a) The total strain in bar (1) is measured as 1,400  (elongation). Part of this strain is due to the stress acting in the bar and part of this strain is due to the temperature increase. The strain caused by the temperature change is T   T  (10.4  106 / °F)(50°F)  520  106 in./in. Since the total strain is  = 1,400  = 0.001400 in./in., the strain caused by the stress in bar (1) must be:     T  0.001400 in./in.  0.000520 in./in.  0.000880 in./in. From Hooke’s law, the stress in bar (1) is therefore: 1  E1  (16,700 ksi)(0.000880 in./in.)  14.696 ksi and thus, the force in bar (1) is: F1   1 A1  (14.696 ksi)(0.225 in.2 )  3.3066 kips

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From Eq. (d), the applied load P is: P  F1 1.736812  (3.3066 kips)(1.736812)  5.7429 kips  5.74 kips

Ans.

(b) From Eq. (c), the force in bar (2) is: cos 40 F2  F1  (3.3066 kips)(1.335558)  4.4162 kips cos55 The normal stress in bar (2) is therefore: F 4.4162 kips 2  2   11.7764 ksi A2 0.375 in.2 The normal strain due to this stress is:  11.7764 ksi 2  2   1,177.64  10 6 in./in. E2 10,000 ksi The strain caused in bar (2) by the temperature change is: T   T  (13.1  106 / °F)(50°F)  655  10 6 in./in. and thus, the total strain in bar (2) is:     T  1,177.64  106 in./in.  655  10 6 in./in.  1,832.64  10 6 in./in.  1,833 με

Ans.

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3.26 The rigid bar AC in Fig. P3.26 is supported by two axial bars (1) and (2). Both axial bars are made of bronze [E = 100 GPa;  = 18 × 10−6 /°C]. The crosssectional area of bar (1) is A1 = 240 mm2 and the cross-sectional area of bar (2) is A2 = 360 mm2. After load P has been applied and the temperature of the entire assembly has increased by 30°C, the total strain in bar (2) is measured as 1,220  (elongation). Determine: (a) the magnitude of load P. (b) the vertical displacement of pin A.

Fig. P3.26

Solution (a) The total strain in bar (2) is caused partly by the axial force in the bar and partly by the increase in temperature. The strain caused by the 30°C temperature increase is: T   T  (18  10 6 / C)(30C)  0.000540 mm/mm The strain caused by the axial force in the bar is thus:  2,   2   2,T  0.001220 mm/mm  0.000540 mm/mm  0.000680 mm/mm The stress in bar (2) is  2  E2 2,  (100,000 MPa)(0.000680 mm/mm)  68 MPa and the force in bar (2) is F2   2 A2  (68 N/mm 2 )(360 mm 2 )  24, 480 N Next, consider a FBD of the rigid bar AC. Equilibrium equations for this FBD are: Fy  F1  F2  P  0

M A  (1, 400 mm)F2  (500 mm)P  0 which can be solve simultaneously to give: 500 mm F2  P  0.357143P 1,400 mm and F1  0.642857 P The applied load P can be expressed in terms of F2 as 1 P F2  2.8F2 0.357143

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and so the magnitude of load P is P  2.8F2  2.8(24,480 N)  68,544 N  68.5 kN

Ans.

(b) The force in bar (1) is F1  0.642857 P  (0.642857)(68,544 N)  44,064 N Thus, the stress in bar (1) is F 44,064 N 1  1   183.60 MPa A1 240 mm 2 The normal strain due to the axial force in bar (1) is  183.60 MPa 1,  1   0.001836 mm/mm E1 100,000 MPa The normal strain caused by the 30°C temperature increase is: 1,T   T  (18  106 /C)(30C)  0.000540 mm/mm Therefore, the total strain in bar (1) is 1  1,  1,T  0.001836 mm/mm  0.000540 mm/mm  0.002376 mm/mm and the elongation in bar (1) is 1  1, L1  (0.002376 mm/mm)(1,300 mm)  3.0888 mm Since rigid bar ABC is connected to bar (1) (with a perfect connection), joint A displaces downward by an amount equal to the elongation of bar (1); therefore,

vA  1  3.09 mm 

Ans.

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3.27 The rigid bar in Fig. P3.27 is supported by axial bar (1) and by a pin connection at C. Axial bar (1) has a cross-sectional area of A1 = 275 mm2, an elastic modulus of E = 200 GPa, and a coefficient of thermal expansion of  = 11.9 × 10−6 /°C. The pin at C has a diameter of 25 mm. After load P has been applied and the temperature of the entire assembly has been increased by 20°C, the total strain in bar (1) is measured as 925  (elongation). Determine: (a) the magnitude of load P. (b) the average shear stress in pin C. Fig. P3.27

Solution The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress:     T The normal strain due to the increase in temperature is: T   T  (11.9  106 /C)(20C)  0.000238 mm/mm Therefore, the normal stress in bar (1) causes a normal strain of:     T  0.000925 mm/mm  0.000238 mm/mm  0.000687 mm/mm From Hooke’s law, the normal stress in bar (1) can be calculated as: 1  E  (200,000 MPa)(0.000687 mm/mm)  137.4 MPa and thus the axial force in bar (1) must be: F1   1 A1  (137.4 N/mm 2 )(275 mm 2 )  37,785 N Next, consider a free-body diagram of the rigid bar. Write a moment equilibrium equation about pin C: M C  (260 mm)F1  (540 mm)P  (260 mm)(37,785 N)  (540 mm)P  0  P  18,192.8 N  18.19 kN

Ans.

Now that P is known, the horizontal and vertical reactions at C can be calculated: Fx  Cx  F1  0  Cx  F1  37,785 N

Fy  C y  P  0

 C y  P  19,192.8 N

The resultant force acting on pin C is: C  Cx2  C y2  (37,785 N)2  (18,192.8 N)2  41,936.659 N

Since the pin at C is a double shear connection, the shear force acting on one shear plane is half of the resultant force: V = 20,968.330 N. The area of one shear plane of the 25-mm-diameter pin at C (in other words, the cross-sectional area of the pin) is: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Apin 



(25 mm)2  490.874 mm2

4 and thus the average shear stress in pin C is: V 20,968.330 N C    42.716 N/mm 2  42.7 MPa 2 AV 490.874 mm

Ans.

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3.28 The rigid bar in Fig. P3.28 is supported by axial bar (1) and by a pin connection at C. Axial bar (1) has a crosssectional area of A1 = 275 mm2, an elastic modulus of E = 200 GPa, and a coefficient of thermal expansion of  = 11.9 × 10−6 /°C. The pin at C has a diameter of 25 mm. After load P has been applied and the temperature of the entire assembly has been decreased by 30°C, the total strain in bar (1) is measured as 925  (elongation). Determine: (a) the magnitude of load P. (b) the average shear stress in pin C. Fig. P3.28

Solution The total strain in bar (1) consists of thermal strain as well as normal strain caused by normal stress:     T The normal strain due to the decrease in temperature is: T   T  (11.9  106 /C)(  30C)  0.000357 mm/mm Therefore, the normal stress in bar (1) causes a normal strain of:     T  0.000925 mm/mm  ( 0.000357 mm/mm)  0.001282 mm/mm From Hooke’s law, the normal stress in bar (1) can be calculated as: 1  E  (200,000 MPa)(0.001282 mm/mm)  256.4 MPa and thus the axial force in bar (1) must be: F1   1 A1  (256.4 N/mm 2 )(275 mm 2 )  70,510 N Next, consider a free-body diagram of the rigid bar. Write a moment equilibrium equation about pin C: M C  (260 mm)F1  (540 mm)P  (260 mm)(70,510 N)  (540 mm)P  0  P  33,949.3 N  33.9 kN

Ans.

Now that P is known, the horizontal and vertical reactions at C can be calculated: Fx  Cx  F1  0  Cx  F1  70,510 N

Fy  C y  P  0

 C y  P  33,949.3 N

The resultant force acting on pin C is: C  Cx2  C y2  (70,510 N)2  (33,949.3 N) 2  78,257.3 N

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Since the pin at C is a double shear connection, the shear force acting on one shear plane is half of the resultant force: V = 39,128.7 N. The area of one shear plane of the 25-mm-diameter pin at C (in other words, the cross-sectional area of the pin) is: Apin 



(25 mm)2  490.874 mm2

4 and thus the average shear stress in pin C is: V 39,128.7 N C    79.712 N/mm 2  79.7 MPa AV 490.874 mm 2

Ans.

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4.1 A stainless steel alloy bar 25 mm wide by 16 mm thick is subjected to an axial load of P = 145 kN. Using the stress-strain diagram given in Fig. P4.1, determine: (a) the factor of safety with respect to the yield strength defined by the 0.20% offset method. (b) the factor of safety with respect to the ultimate strength.

Fig. P4.1

Solution (a) From the stress-strain curve, the yield strength defined by the 0.20% offset method is  Y  550 MPa The normal stress in the bar is F (145 kN)(1,000 N/kN)    362.5 MPa A (25 mm)(16 mm) Therefore, the factor of safety with respect to yield is  550 MPa FS  Y   1.517  actual 362.5 MPa (b) From the stress-strain curve, the ultimate strength is  U  1,100 MPa Therefore, the factor of safety with respect to the ultimate strength is  1,100 MPa FS  U   3.03  actual 362.5 MPa

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4.2 Six bolts are used in the connection between the axial member and the support, as shown in Fig. P4.2. The ultimate shear strength of the bolts is 300 MPa, and a factor of safety of 4.0 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 475 kN.

Fig. P4.2

Solution The allowable shear stress for the bolts is  300 MPa  allow  U   75 MPa FS 4 To support a load of P = 350 kN, the total area subjected to shear stress must equal or exceed: P 475,000 N AV    6,333.333 mm 2 2  allow 75 N/mm There are six bolts, and each bolt acts in double shear; therefore, the total area subjected to shear stress is



2 AV  (6 bolts)(2 surfaces per bolt) d bolt 4 The minimum bolt diameter can be found be equating the two expressions for AV:



2 (12 surfaces) d bolt  6,333.333 mm2 4  d bolt  25.923 mm  25.9 mm

Ans.

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4.3 A 14-kip load is supported by two bars as shown in Fig. P4.3. Bar (1) is made of cold-rolled red brass (Y = 60 ksi) and has a cross-sectional area of 0.225 in.2. Bar (2) is made of 6061-T6 aluminum (Y = 40 ksi) and has a cross-sectional area of 0.375 in.2. Determine the factor of safety with respect to yielding for each of the bars.

Fig. P4.3

Solution Consider a FBD of joint B. Members (1) and (2) are two-force members; therefore, the equilibrium equations can be written as: Fx  F2 cos50  F1 cos35  0 (a) Fy  F2 sin 50  F1 sin 35  14 kips  0 (b) From Eq. (a): cos35 F2  F1 cos50 Substitute this expression into Eq. (b) to obtain:  cos35   F1 cos50  sin 50  F1 sin 35  14 kips F1  cos35 tan 50  sin 35  14 kips

F1 1.549804  14 kips

F1  9.033401 kips Backsubstitute to obtain F2: cos35 F2  F1  (9.033401 kips)(1.274374)  11.511935 kips cos50 The normal stress in bar (1) is F 9.033401 kips 1  1   40.148452 ksi A1 0.225 in.2 Therefore, the factor of safety in bar (1) is  60 ksi FS1  Y   1.494454  1.494  1 40.148452 ksi The normal stress in bar (2) is F 11.511935 kips 2  2   30.698495 ksi A2 0.375 in.2 and thus, the factor of safety in bar (2) is  40 ksi FS2  Y   1.302995  1.303  2 30.698495 ksi

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4.4 A steel plate is to be attached to a support with three bolts as shown in Fig. P4.4. The cross-sectional area of the plate is 560 mm2 and the yield strength of the steel is 250 MPa. The ultimate shear strength of the bolts is 425 MPa. A factor of safety of 1.67 with respect to yield is required for the plate. A factor of safety of 4.0 with respect to the ultimate shear strength is required for the bolts. Determine the minimum bolt diameter required to develop the full strength of the plate. (Note: consider only the gross crosssectional area of the plate—not the net area.) Fig. P4.4

Solution The allowable normal stress is  250 MPa  allow  Y   149.7006 MPa FS 1.67 The full strength of the plate (based on the gross cross-sectional area) is therefore: Pmax   allow A  (149.7006 MPa)(560 mm 2 )  83,832.3353 N The allowable shear stress of the bolts is  425 MPa  allow  U   106.25 MPa FS 4.0 The minimum shear area required to support the maximum load P is P 83,832.3353 N AV  max   789.0102 mm 2 2  allow 106.25 N/mm There are three bolts, each acting in single shear, which provide a shear area of





2 2 AV  (3 bolts)(1 surface per bolt) d bolt  (3 surfaces) d bolt 4 4 Equating the two expressions for the shear area, the minimum bolt diameter can be computed:



2 (3 surfaces) d bolt  789.0102 mm2 4  d bolt  18.2994 mm  18.30 mm

Ans.

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4.5 In Fig. P4.5, member (1) is a steel bar with a crosssectional area of 1.35 in.2 and a yield strength of 50 ksi. Member (2) is a pair of 6061-T6 aluminum bars having a combined cross-sectional area of 3.50 in.2 and a yield strength of 40 ksi. A factor of safety of 1.6 with respect to yield is required for both members. Determine the maximum allowable load P that may be applied to the structure. Report the factors of safety for both members at the allowable load.

Fig. P4.5

Solution Consider a FBD of joint B and write the following equilibrium equations: Fx  F2 cos 65  F1  0 Fy  F2 sin 65  P  0 From Eq. (b): P F2  sin 65 Substituting Eq. (c) into Eq. (a) gives an expression for P in terms of F1: F2 cos 65  F1  0

(a) (b) (c)

 P    cos 65  F1  0 sin 65  P  F1 tan 65 and rearranging Eq. (c) gives an expression for P in terms of F2: P  F2 sin 65

(d) (e)

The allowable normal stress in member (1) is  50 ksi  allow,1  Y ,1   31.25 ksi FS1 1.6 and the allowable force in member (1) is Fallow,1   allow,1 A1  (31.25 ksi)(1.35 in.2 )  42.1875 kips

(f)

The allowable normal stress in member (2) is  40 ksi  allow,2  Y ,2   25.00 ksi FS2 1.6 and the allowable force in member (2) is Fallow,2   allow,2 A2  (25.00 ksi)(3.50 in.2 )  87.50 kips

(g)

Substitute the allowable force in member (1) from Eq. (f) into Eq. (d) to obtain the maximum load P based on the capacity of member (1): P  Fallow,1 tan 65  (42.1875 kips) tan 65  90.471386 kips (h)

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Repeat with the allowable force in member (2) from Eq. (g) substituted into Eq. (e) to obtain the maximum load P based on the capacity of member (2): P  Fallow,2 sin 65  (87.50 kips)sin 65  79.301931 kips (i) Compare the results in Eqs. (h) and (i) to find that the maximum load P that may be applied is controlled by the capacity of member (2): Ans. Pmax  79.3 kips The factor of safety in member (2) is thus: FS2  1.6 For an applied load of P = 79.3 kips, the force in member (1) can be computed from Eq. (d): P 79.301931 kips F1    36.979096 kips tan 65 tan 65 which results in a normal stress of F 36.979096 kips 1  1   27.391924 ksi A1 1.350 in.2 and thus, its factor of safety is:  50 ksi FS1  Y ,1   1.825 1 27.391924 ksi

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4.6 The rigid structure ABD in Fig. P4.6 is supported at B by a 35-mm-diameter tie rod (1) and at A by a 30-mm-diameter pin used in a single shear connection. The tie rod is connected at B and C by 24-mmdiameter pins used in double shear connections. Tie rod (1) has a yield strength of 250 MPa, and each of the pins has an ultimate shear strength of 330 MPa. A concentrated load of P = 50 kN acts as shown at D. Determine: (a) the normal stress in rod (1). (b) the shearing stress in the pins at A and B. (c) the factor of safety with respect to the yield strength for tie rod (1). (d) the factor of safety with respect to the ultimate strength for the pins at A and B.

Fig. P4.6

Solution Member (1) is a two-force member that is oriented at  with respect to the horizontal axis: 8m tan    1.509434 5.3 m   56.4755

From a FBD of rigid structure ABD, the following equilibrium equations can be written: Fx  P cos 60  F1 cos56.4755  Ax  0 Fy   P sin 60  F1 sin 56.4755  Ay  0

(a) (b)

M A  ( F1 cos56.4755)(8 m)  ( P cos 60)(3.4 m)  ( P sin 60)(7.5 m)  0 (c) From Eq. (c): (3.4 m)cos 60  (7.5 m)sin 60 (3.4 m)cos 60  (7.5 m)sin 60 F1  P  (50 kN)  92.7405 kN (8 m)cos56.4755 (8 m)cos56.4755

Substitute F1 into Eq. (a) to obtain Ax: Ax  F1 cos56.4755  P cos 60  (92.7405 kN)cos56.4755  (50 kN)cos 60  26.2199 kN and substitute F1 into Eq. (b) to obtain Ay: Ay  F1 sin 56.4755  P sin 60  (92.7405 kN)sin 56.4755  (50 kN)sin 60  120.6144 kN Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The resultant pin force at A is found from Ax and Ay: A

Ax2  Ay2  (26.2199 kN)2  (120.6144 kN)2  123.4314 kN

(a) The cross-sectional area of the 35-mm-diameter tie rod is: A1 



(35 mm)2  962.112750 mm2

4 and thus, the normal stress in rod (1) is: F 92,740.5 N 1  1   96.3925 MPa  96.4 MPa A1 962.112750 mm 2

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(b) The 30-mm-diameter single shear pin at A has a shear area of AV , A 



(30 mm)2  706.858347 mm2

4 Consequently, the shear stress in pin A is 123,431.4 N A   174.6197 MPa  174.6 MPa 706.858347 mm2

Ans.

The 24-mm-diameter double shear pins at B and C have a shear area of AV , B  (2 surfaces)



(24 mm) 2  904.778685 mm 2

4 The shear stress in pins B and C is 92,740.5 N B   102.5008 MPa  102.5 MPa 904.778685 mm 2

(c) The factor of safety for tie rod (1) is  250 MPa FS1  Y   2.59  1 96.3925 MPa (d) The factor of safety with respect to the ultimate strength for the pin at A is  330 MPa FS A  U   1.890  A 174.6197 MPa and the factor of safety for pin B is  330 MPa FSB  U   3.22  B 102.5008 MPa

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4.7 Rigid bar ABD in Fig. P4.7 is supported by a pin connection at A and a tension link BC. The 8-mmdiameter pin at A is supported in a double shear connection, and the 12-mm-diameter pins at B and C are both used in single shear connections. Link BC is 30-mm wide and 6-mm thick. The ultimate shear strength of the pins is 330 MPa and the yield strength of link BC is 250 MPa. (a) Determine the factor of safety in pins A and B with respect to the ultimate shear strength. (b) Determine the factor of safety in link BC with respect to the yield strength. Fig. P4.7

Solution Consider a free-body diagram of the rigid bar. Link BC is a two-force member that is oriented at an angle of  with respect to the horizontal axis: 0.6 m tan     53.1301 0.45 m

The equilibrium equations for the rigid bar can be written as: Fx   FBC cos53.1301  (8.2 kN)cos70  Ax  0 Fy  FBC sin 53.1301  (8.2 kN)sin70  Ay  0 M A  (0.75 m)FBC cos53.1301  (0.45 m)FBC sin 53.1301 (1.85 m)(8.2 kN)sin70  (0.75 m)(8.2 kN)cos70  0 Solving these three equations simultaneously gives: FBC  20.1958 kN Ax  9.3129 kN Ay  8.4511 kN

The resultant force at pin A is: A

Ax2  Ay2  (9.3129 kN)2  ( 8.4511 kN) 2  12.5758 kN

Before the factors of safety can be determined, we must compute the shear stresses in pins A and B as well as the normal stress in link BC. Pin shear stresses: The 8-mm-diameter pin at A is supported in a double shear connection; therefore, the shear force acting on one shear plane (which is simply equal to the cross-sectional area of the pin) is half of the resultant force at pin A: VA = 6.2879 kN. The cross-sectional area of the pin at A is: Apin A 



(8 mm)2  50.265 mm2 4 and therefore, the shear stress in pin A is: V (6.2879 kN)(1000 N/kN) A  A   125.0940 N/mm 2  125.0940 MPa 2 AV 50.265 mm Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The 12-mm-diameter pin at B and C is supported in a single shear connection and so the shear force acting on one shear plane is the entire force in link BC: VB = 20.1958 kN. The cross-sectional area of the pin at B is: Apin B 



(12 mm)2  113.097 mm2

4 and therefore, the shear stress in pin B is: V (20.1958 kN)(1000 N/kN) B  B   178.5697 N/mm 2  178.5697 MPa AV 113.097 mm 2

Normal stress in link: The normal stress in link BC is: F (20.1958 kN)(1000 N/kN)  BC  BC   112.1986 N/mm 2  112.1986 MPa ABC (30 mm)(6 mm) Factors of safety: For pin A, the factor of safety is:  330 MPa FS A  U   2.64  A 125.0940 MPa For pin B, the factor of safety is:  330 MPa FSB  U   1.848  B 178.5697 MPa The factor of safety in link BC is:  250 MPa FSBC  Y   2.23  BC 112.1986 MPa

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4.8 In Fig. P4.8, davit ABD is supported at A by a single shear pin connection and at B by a tie rod (1). The pin at A has a diameter of 1.25 in. and the pins at B and C are each 0.75-in.-diameter pins. Tie rod (1) has an area of 1.50 in.2. The ultimate shear strength in each pin is 80 ksi, and the yield strength of the tie rod is 36 ksi. A concentrated load of 25 kips is applied as shown to the davit structure at D. Determine: (a) the normal stress in rod (1). (b) the shearing stress in the pins at A and B. (c) the factor of safety with respect to the yield strength for tie rod (1). (d) the factor of safety with respect to the ultimate strength for the pins at A and B. Fig. P4.8

Solution Member (1) is a two-force member that is oriented at  with respect to the horizontal axis: 9 ft tan    0.75   36.870 12 ft From a FBD of rigid structure ABD, the following equilibrium equations can be written: Fx  (25 kips) cos 60  F1 cos 36.870  Ax  0 (a) Fy  (25 kips)sin 60  F1 sin 36.870  Ay  0 (b) M A  ( F1 cos 36.870)(9 ft)  (25 kips) cos 60(11 ft) (25 kips) sin 60(7 ft)  0

(c)

From Eq. (c):

(11 ft) cos 60  (7 ft)sin 60  40.1465 kips (9 ft) cos 36.870 Substitute F1 into Eq. (a) to obtain Ax: Ax  F1 cos 36.870  (25 kips) cos 60  (40.1465 kips) cos36.870  (25 kips) cos 60  19.6172 kips and substitute F1 into Eq. (b) to obtain Ay: Ay  F1 sin 36.870  (25 kips)sin 60  (40.1465 kips)sin 36.870  (25 kips)sin 60  45.7386 kips F1  (25 kips)

The resultant pin force at A is found from Ax and Ay: A  Ax2  Ay2  (19.6172 kips)2  (45.7386 kips)2  49.7680 kips

(a) The normal stress in tie rod (1) is: F 40.1465 kips 1  1   26.7643 ksi  26.8 ksi A1 1.50 in.2

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(b) The 1.25-in.-diameter single shear pin at A has a shear area of AV , A 



(1.25 in.)2  1.2272 in.2

4 Consequently, the shear stress in pin A is 49.7680 kips A   40.5541 ksi  40.6 ksi 1.2272 in.2

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The 0.75-in.-diameter double shear pins at B and C have a shear area of AV , B  (2 surfaces)



(0.75 in.) 2  0.8836 in.2

4 The shear stress in pins B and C is 40.1465 kips B   45.4352 ksi  45.4 ksi 0.8836 in.2

(c) The factor of safety for tie rod (1) is  36 ksi FS1  Y   1.345  1 26.7643 ksi (d) The factor of safety with respect to the ultimate strength for the pin at A is  80 ksi FS A  U   1.973  A 40.5541 ksi and the factor of safety for pin B is  80 ksi FSB  U   1.761  B 45.4352 ksi

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4.9 The pin-connected structure is subjected to a load P as shown in Fig. P4.9. Inclined member (1) has a crosssectional area of 250 mm2 and has a yield strength of 255 MPa. It is connected to rigid member ABC with a 16-mm-diameter pin in a double shear connection at B. The ultimate shear strength of the pin material is 300 MPa. For inclined member (1), the minimum factor of safety with respect to the yield strength is FSmin = 1.5. For the pin connections, the minimum factor of safety with respect to the ultimate strength is FSmin = 3.0. (a) Based on the capacity of member (1) and pin B, determine the maximum allowable load P that may be applied to the structure. (b) Rigid member ABC is supported by a double shear pin connection at A. Using FSmin = 3.0, determine the minimum pin diameter that may be used at support A.

Fig. P4.9

Solution The allowable normal stress for inclined member (1) is  255 MPa  allow  Y   170.00 MPa FS 1.50 and the allowable shear stress for the pins is  300 MPa  allow  U   100.0 MPa FS 3.00 (a) The allowable axial force in member (1) based on normal stress is Fallow   allow A1  (170.00 MPa)(250 mm 2 )  42,500 N Member (1) is connected with a 16-mm-diameter double shear pin. The shear area of this pin is:

(a)



AV  (2 surfaces) (16 mm) 2  402.1239 mm 2 4 Consequently, the shear force V (which is applied by the inclined member) that can be applied to the pin is limited to Vallow   allow AV  (100.0 MPa)(402.1239 mm 2 )  40, 212.4 N (b) Comparing the two values given in Eq. (a) and Eq. (b), the maximum allowable force in member (1) is F1  40, 212.4 N (c)

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Member (1) is a two-force member that is oriented at  with respect to the horizontal axis: 1.4 m tan    0.6087   31.3287 2.3 m From a FBD of rigid structure ABC, the following equilibrium equations can be written: Fx  Ax  F1 cos 31.3287  P  0 (d) Fy  Ay  F1 sin 31.3287  0 (e) M A  (3.2 m)P  ( F1 cos 31.3287)(1.4 m)  0

(f)

Substitute the value of F1 from Eq. (c) into Eq. (f) to obtain the maximum load P: (1.4 m)cos31.3287 P  F1 3.2 m (1.4 m)cos31.3287  (40,212.4 N) 3.2 m Ans.  15,027.8 N  15.03 kN

Substitute F1 and P into Eq. (d) to obtain Ax: Ax  P  F1 cos31.3287  (15,027.8 N)  (40, 212.4 N)cos31.3287  19,321.5 N and substitute F1 into Eq. (e) to obtain Ay: Ay  F1 sin 31.3287  (40, 212.4 N)sin 31.3287  20,908.3 N The resultant pin force at A is found from Ax and Ay: A

Ax2  Ay2  ( 19,321.5 N)2  (20,908.3 N) 2  28,468.9 N

The total shear area required to support the resultant pin force is A 28,468.9 N AV    284.689 mm2  allow 100 N/mm2 Since the pin at A is in a double shear connection, the shear area provided by the pin is  2 AV  (2 surfaces) d pin 4 Equate these two expressions and solve for the required minimum pin diameter:



2 (2 surfaces) d pin  284.689 mm 2 4  d pin  13.46 mm

Ans.

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4.10 After load P is applied to the pin-connected structure shown in Fig. P4.10, a normal strain of  = +550  is measured in the longitudinal direction of member (1). The cross-sectional area of member (1) is A1 = 0.60 in.2, its elastic modulus is E1 = 29,000 ksi, and its yield strength is 36 ksi. (a) Determine the axial force in member (1), the applied load P, and the resultant force at pin B. (b) The ultimate shear strength of the steel pins is 54 ksi. Determine the minimum diameter for the pin at B if a factor of safety of 2.5 with respect to the ultimate shear strength is required. (c) Compute the factor of safety for member (1) with respect to its yield strength. Fig. P4.10

Solution (a) Given the strain in member (1), its stress can be computed from Hooke’s law: 1  E11  (29,000 ksi)(0.000550 in./in.)  15.9500 ksi The force in member (1) is the product of the normal stress and the cross-sectional area: F1  1 A1  (15.9500 ksi)(0.60 in.2 )  9.57 kips

Ans.

Next, consider a free-body diagram of horizontal member ABC. Member (1) makes an angle of  with respect to the horizontal axis: 3.50 ft tan     36.3844 4.75 ft Note that member (1) is a two-force member. Equilibrium equations for horizontal member ABC can be written as: Fx  Bx  F1 cos36.3844  0 Fy  By  F1 sin 36.3844  P  0 M B  (6.50 ft)F1 sin 36.3844  (9.50 ft)P  0 Substituting the value F1 = 14.79 kips into these equations gives: Bx  7.7044 kips By  9.5611 kips P  3.8842 kips  3.88 kips

Ans.

The resultant force at pin B is: B  Bx2  By2  (7.7044 kips)2  (9.5611 kips) 2  12.2789 kips  12.28 kips

Ans.

(b) The pin at B is supported in a double shear connection. Therefore, the shear force acting on one shear plane of the pin is half of the resultant force: VB = 6.1395 kips. The allowable shear stress for the pin is computed from the ultimate shear strength and the factor of safety:  54 ksi a  U   21.6 ksi FS 2.5 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Next, the cross-sectional area required for a single shear plane can be determined: V V 6.1395 kips   B  a  AB  B   0.2842 in.2 AB a 21.6 ksi To provide AB, the pin at B must have a diameter of at least:  2 d B  0.2842 in.2  d B  0.602 in. 4 (c) The factor of safety for member (1) with respect to its yield strength is:  36 ksi FS1  Y   2.26  1 15.9500 ksi

Ans.

Ans.

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4.11 The simple pin-connected structure carries a concentrated load P as shown in Fig. P4.11. The rigid bar is supported by strut AB and by a pin support at C. The steel strut AB has a crosssectional area of 0.25 in.2 and a yield strength of 60 ksi. The diameter of the steel pin at C is 0.375 in., and the ultimate shear strength is 54 ksi. If a factor of safety of 2.0 is required in both the strut and the pin at C, determine the maximum load P that can be supported by the structure.

Fig. P4.11

Solution From a FBD of the rigid bar, the following equilibrium equations can be written: Fx   F1  Cx  0 (a) Fy  C y  P  0 (b) M C  (10 in.)F1  (22 in.)P  0

(c)

From Eq. (c), express F1 in terms of the unknown load P: 22 in. F1  P  2.2 P (d) 10 in. Substitute this result into Eq. (a) to express Cx in terms of P: Cx  F1  2.2 P The resultant reaction force at pin C can now be expressed as a function of P: C  Cx2  C y2  (2.2P)2  ( P)2  2.41661P

(e)

Strut AB: The allowable normal stress for strut AB is  60 ksi  allow  Y   30 ksi FS 2.0 Therefore, the allowable axial force for strut AB is Fallow,1   allow A1  (30 ksi)(0.25 in.2 )  7.50 kips From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations on the strut normal stress is F 7.50 kips Pmax  allow,1   3.4091 kips (f) 2.2 2.2

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Pin C: The allowable shear stress for the pin at C is  54 ksi  allow  U   27 ksi FS 2.0 The 0.375-in.-diameter double shear pin at C has a shear area of



AV  (2 surfaces) (0.375 in.) 2  0.2209 in.2 4 thus, the allowable reaction force at C is C allow   allow AV  (27 ksi)(0.2209 in.2 )  5.9641 kips From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations on the pin shear stress is C allow 5.9641 kips Pmax    2.4680 kips (g) 2.41661 2.41661

Maximum load P: Comparing the results in Eqs. (f) and (g), the maximum load P that may be applied to the rigid bar is Ans. Pmax  2.4680 kips  2.47 kips

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4.12 In Fig. P4.12, rigid tee-beam ABC is supported at A by a single shear pin connection and at B by a strut, which consists of two 30 mm wide by 8 mm thick steel bars. The pins at A, B, and D are each 12 mm in diameter. The yield strength of the steel bars in strut (1) is 250 MPa, and the ultimate shear strength of each pin is 500 MPa. Determine the allowable load P that may be applied to the rigid bar at C if an overall factor of safety of 3.0 is required. Use L1 = 1.1 m and L2 = 1.3 m.

Fig. P4.12

Solution From a FBD of the rigid beam ABC, the following equilibrium equations can be written: Fx  Ax  0 (a) Fy  Ay  F1  P  0 (b) M A   L1 F1  ( L1  L2 )P  0

(c)

From Eq. (c), express F1 in terms of the unknown load P: L  L2 1.1 m  1.3 m F1   1 P P  2.181818P L1 1.1 m Substitute this result into Eq. (b) to express Ay in terms of P: Ay  P  F1  P  ( 2.181818P )  1.181818 P

(d)

The resultant reaction force at pin A can now be expressed as a function of P: A

Ax2  Ay2  (0)2  ( 1.181818P) 2  1.181818P

(e)

The allowable normal stress for strut (1) is  250 MPa  allow  Y   83.333333 MPa FS 3.0 Therefore, the allowable axial force for strut (1) is Fallow,1   allow A1  (83.333333 MPa)(2  30 mm  8 mm)  40,000 N From Eq. (d), the maximum load that may be applied to the rigid bar based on the strut normal stress limitation is Fallow,1 40,000 N Pmax    18,333 N (f) 2.181818 2.181818 The allowable shear stress for the pins at A, B, and D is  500 MPa  allow  U   166.666667 MPa FS 3.0 The cross-sectional area of a 12-mm-diameter pin is  Apin  (12 mm)2  113.097336 mm 2 4 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The pin at A is a single shear connection; therefore, AV,A = Apin. The allowable reaction force at A is A allow   allow AV , A  (166.666667 MPa)(113.097336 mm2 )  18,849.5560 N From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations of pin A is A allow 18,849.5560 N Pmax    15,949.627 N (g) 1.181818 1.181818 The pins at B and D are in double shear connections; therefore, AV,B = 2Apin. The allowable force in strut (1) based on the capacity of the pins at B and D is Fallow,1   allow AV , B  (166.666667 MPa)(2)(113.097336 mm2 )  37,699.1121 N From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations of the pins at B and D is Fallow,1 37,699.1121 N Pmax    17, 278.7611 N (h) 2.181818 2.181818

Comparing the results in Eqs. (f), (g), and (h), the maximum load P that may be applied to the rigid bar is Ans. Pmax  15,949.627 N  15.95 kN

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4.13 In Fig. P4.13, rigid bar ABC is supported at A by a single shear pin connection and at B by a strut, which consists of two 2 in. wide by 0.25 in. thick steel bars. The pins at A, B, and D each have a diameter of 0.5 in. The yield strength of the steel bars in strut (1) is 36 ksi, and the ultimate shear strength of each pin is 72 ksi. Determine the allowable load P that may be applied to the rigid bar at C if an overall factor of safety of 3.0 is required. Use L1 = 36 in. and L2 = 24 in.

Fig. P4.13

Solution From a FBD of the rigid beam ABC, the following equilibrium equations can be written: Fx  Ax  0 (a) Fy  Ay  F1  P  0 (b) M A   L1 F1  ( L1  L2 )P  0

(c)

From Eq. (c), express F1 in terms of the unknown load P: L  L2 36 in.  24 in. F1   1 P P  1.666667 P L1 36 in. Substitute this result into Eq. (b) to express Ay in terms of P: Ay  P  F1  P  (1.666667 P )  0.666667 P

(d)

The resultant reaction force at pin A can now be expressed as a function of P: A  Ax2  Ay2  (0)2  (0.666667 P) 2  0.666667 P

(e)

Strut (1): The allowable normal stress for strut (1) is  36 ksi  allow  Y   12.0 ksi FS 3.0 Therefore, the allowable axial force for strut (1) is Fallow,1   allow A1  (12.0 ksi)(2  2 in.  0.25 in.)  12.0 kips From Eq. (d), the maximum load that may be applied to the rigid bar based on the strut normal stress limitation is Fallow,1 12.0 kips Pmax    7.20 kips (f) 1.666667 1.666667

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Pins: The allowable shear stress for the pins at A, B, and D is  72 ksi  allow  U   24.0 ksi FS 3.0 The cross-sectional area of a 0.5-in.-diameter pin is Apin 



4

(0.5 in.)2  0.196350 in.2

Pin A: The pin at A is a single shear connection; therefore, AV,A = Apin. The allowable reaction force at A is A allow   allow AV , A  (24.0 ksi)(0.196350 in.2 )  4.7124 kips From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations of pin A is A allow 4.7124 kips Pmax    7.0686 kips (g) 0.666667 0.666667 Pin B and D: The pins at B and D are in double shear connections; therefore, AV,B = 2Apin. The allowable force in strut (1) based on the capacity of the pins at B and D is Fallow,1   allow AV , B  (24.0 ksi)(2)(0.196350 in.2 )  9.4248 kips From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations of the pins at B and D is Fallow,1 9.4248 kips Pmax    5.6549 kips (h) 1.666667 1.666667 Maximum load P: Comparing the results in Eqs. (f), (g), and (h), the maximum load P that may be applied to the rigid bar is Ans. Pmax  5.6549 kips  5.65 kips

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4.14 A concentrated load of P = 70 kips is applied to beam AB as shown in Fig. P4.14. Rod (1) has a diameter of 1.50 in., and its yield strength is 60 ksi. Pin A is supported in a double shear connection, and the ultimate shear strength of pin A is 80 ksi. (a) Determine the normal stress in rod (1). (b) Determine the factor of safety with respect to the yield strength for rod (1). (c) If a factor of safety of 3.0 with respect to the ultimate strength is specified for pin A, determine the minimum required pin diameter.

Fig. P4.14

Solution Tie rod (1) is a two-force member that is oriented at  with respect to the horizontal axis: 8.5 ft tan    0.7083   35.3112 12 ft From a FBD of beam AB, the following equilibrium equations can be written: Fx  Ax  F1 cos35.3112  0 (a) Fy  Ay  F1 sin 35.3112  P  0 (b)

 10 in.  M A  ( F1 cos35.3112)   12 in./ft  ( F1 sin 35.3112)(12 ft)  P(8 ft)  0

(c)

From Eq. (c):

(70 kips)(8 ft)  73.5272 kips (cos35.3112)  0.833333 ft   (sin35.3112)(12 ft) Substitute F1 into Eq. (a) to obtain Ax: Ax   F1 cos35.3112  (73.5272 kips)cos35.3112  60.0 kips and substitute F1 into Eq. (b) to obtain Ay: Ay  P  F1 sin 35.3112  70 kips  (73.5272 kips)sin 35.3112  27.50 kips The resultant pin force at A is found from Ax and Ay: F1 

A

Ax2  Ay2  (60.0 kips)2  (27.5 kips)2  66.0019 kips

(a) The cross-sectional area of the 1.5-in.-diameter tie rod (1) is A1 



4

(1.5 in.)2  1.7671 in.2

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The normal stress in tie rod (1) is: F 73.5272 kips 1  1   41.6079 ksi  41.6 ksi A1 1.7671 in.2 (b) The factor of safety for tie rod (1) is  60 ksi FS1  Y   1.442  1 41.6079 ksi

Ans.

Ans.

(c) The allowable shear stress is:  80 ksi  allow  U   26.6667 ksi FS 3.0 The shear area required for pin A is thus A 66.0019 kips AV    2.4751 in.2  ult 26.6667 ksi Since pin A is in a double shear connection, the minimum required pin area is AV 2.4751 in.2 Apin    1.2375 in.2 2 shear surfaces 2 Thus, the diameter of pin A must be



4

d A2  1.2375 in.2  d A  1.255 in.

Ans.

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4.15 Beam AB is supported as shown in Fig. P4.15. Tie rod (1) is attached at B and C with double shear pin connections while the pin at A is attached with a single shear connection. The pins at A, B, and C each have an ultimate shear strength of 54 ksi, and tie rod (1) has a yield strength of 36 ksi. A concentrated load of P = 16 kips is applied to the beam as shown. A factor of safety of 3.0 is required for all components. Determine: (a) the minimum required diameter for tie rod (1). (b) the minimum required diameter for the double shear pins at B and C. (c) the minimum required diameter for the single shear pin at A.

Fig. P4.15

Solution Tie rod (1) is a two-force member that is oriented at  with respect to the horizontal axis: 8.5 ft tan    0.7083   35.3112 12 ft From a FBD of beam AB, the following equilibrium equations can be written: Fx  Ax  F1 cos35.3112  0 (a) Fy  Ay  F1 sin 35.3112  P  0 (b)

 10 in.  M A  ( F1 cos35.3112)   12 in./ft  ( F1 sin 35.3112)(12 ft)  P(8 ft)  0

(c)

From Eq. (c):

(16 kips)(8 ft)  16.8062 kips (cos35.3112)  0.833333 ft   (sin35.3112)(12 ft) Substitute F1 into Eq. (a) to obtain Ax: Ax   F1 cos35.3112  (16.8062 kips)cos35.3112  13.7143 kips and substitute F1 into Eq. (b) to obtain Ay: Ay  P  F1 sin 35.3112  16 kips  (16.8062 kips)sin 35.3112  6.2857 kips The resultant pin force at A is found from Ax and Ay: F1 

A

Ax2  Ay2  (13.7143 kips)2  (6.2857 kips)2  15.0861 kips

(a) The allowable normal stress for tie rod (1) is  36 ksi  allow  Y   12.0 ksi FS 3.0 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The minimum cross-sectional area required for the tie rod is F 16.8062 kips Amin  1   1.400518 in.2  allow 12 ksi Therefore, the minimum tie rod diameter is



4

d12  1.400518 in.2

 d1  1.335363 in.  1.335 in.

Ans.

(b) The allowable shear stress for the pins at A, B, and C is  54 ksi  allow  U   18 ksi FS 3.0 The double-shear pin connection at B and C must support a load of F1 =16.8062 kips . The shear area AV required for these pins is F 16.8062 kips AV  1   0.933679 in.2  allow 18 ksi The pin diameter can be computed from



2 (2 surfaces) d pin  0.933679 in.2 4

 d pin  0.7710 in.  0.771 in.

Ans.

(c) The pin at A is a single shear connection; therefore, AV = Apin. The shear area AV required for this pin is A 15.0861 kips AV    0.838119 in.2  allow 18 ksi The pin diameter can be computed from



2 (1 surface) d pin  0.838119 in.2 4

 d pin  1.0330 in.  1.033 in.

Ans.

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4.16 In Fig. P4.16, the rigid member ABDE is supported at A by a single shear pin connection and at B by a tie rod (1). The tie rod is attached at B and C with double shear pin connections. The pins at A, B, and C each have an ultimate shear strength of 80 ksi, and tie rod (1) has a yield strength of 60 ksi. A concentrated load of P = 24 kips is applied perpendicular to DE, as shown. A factor of safety of 2.0 is required for all components. Determine: (a) the minimum required diameter for the tie rod. (b) the minimum required diameter for the pin at B. (c) the minimum required diameter for the pin at A.

Fig. P4.16

Solution Member (1) is a two-force member that is oriented at  with respect to the horizontal axis: 9 ft tan    1.5   56.310 6 ft From a FBD of rigid structure ABDE, the following equilibrium equations can be written: Fx  P cos 70  F1 cos 56.310  Ax  0 (a) Fy   P sin 70  F1 sin 56.310  Ay  0

(b)

M A  ( F1 cos 56.310)(9 ft)  ( P cos 70)[13 ft  (8 ft) sin 20] ( P sin 70)[(8 ft) cos 20]  0

(c)

From Eq. (c): (cos70)[13 ft  (8 ft)sin 20]  (sin 70)[(8 ft)cos 20] F1  P (cos56.310)(9 ft)

 (24 kips)

5.382084 ft  7.064178 ft  59.834286 kips 4.992293 ft

(d)

Substitute F1 into Eq. (a) to obtain Ax: Ax  F1 cos56.310  P cos 70

 (59.834286 kips) cos56.310  (24 kips) cos 70  24.981548 kips and substitute F1 into Eq. (b) to obtain Ay: Ay  F1 sin 56.310  P sin 70  (59.834286 kips)sin 56.310  (24 kips)sin 70  72.337797 kips Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The resultant pin force at A is found from Ax and Ay: A  Ax2  Ay2  (24.981548 kips)2  (72.337797 kips)2  76.529959 kips

(e)

(a) The allowable normal stress for tie rod (1) is  60 ksi  allow  Y   30.0 ksi FS 2.0 The minimum cross-sectional area required for the tie rod is F 59.834286 kips Amin  1   1.994476 in.2  allow 30.0 ksi Therefore, the minimum tie rod diameter is



4

d12  1.994476 in.2

 d1  1.593564 in.  1.594 in.

Ans.

(b) The allowable shear stress for the pins at A, B, and C is  80 ksi  allow  U   40.0 ksi FS 2.0 The double-shear pin connection at B and C must support a load of F1 = 59.834286 kips . The shear area AV required for these pins is F 59.834286 kips AV  1   1.495857 in.2  allow 40.0 ksi The pin diameter can be computed from



2 (2 surfaces) d pin  1.495857 in.2 4

 d pin  0.975855 in.  0.976 in.

Ans.

(c) The pin at A is a single shear connection; therefore, AV = Apin. The shear area AV required for this pin is A 76.529959 kips AV    1.913249 in.2  allow 40.0 ksi The pin diameter can be computed from



2 (1 surface) d pin  1.913249 in.2 4

 d pin  1.560777 in.  1.561 in.

Ans.

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4.17 Rigid bar ABC is subjected to a concentrated load P as shown in Fig. P4.17. Inclined member (1) has a cross sectional area of A1 = 2.250 in.2 and is connected at ends B and D by 1.00-in.-diameter pins in double shear connections. The rigid bar is supported at C by a 1.00-in.diameter pin in a single shear connection. The yield strength of inclined member (1) is 36 ksi, and the ultimate strength of each pin is 60 ksi. For inclined member (1), the minimum factor of safety with respect to the yield strength is FSmin = 1.5. For the pin connections, the minimum factor of safety with respect to the ultimate strength is FSmin = 2.0. Determine the maximum load P that can be supported by the structure. Fig. P4.17

Solution Member (1) is a two-force member that is oriented at  with respect to the horizontal axis: 40 in. tan    0.8   38.6598 50 in. From a FBD of rigid bar ABC, the following equilibrium equations can be written: Fx  F1 cos38.6598  Cx  0 (a) Fy   P  F1 sin 38.6598  C y  0 (b) M C  (80 in.)P  ( F1 sin 38.6598)(50 in.)  0

(c)

From Eq. (c), express F1 in terms of the unknown load P: (80 in.)P F1    2.561250 P (50 in.)sin 38.6598 Substitute this result into Eq. (a) to express Cx in terms of P: Cx  F1 cos38.6598  ( 2.561250 P)cos38.6598  2.0 P And similarly, express Cy in terms of P from Eq. (b): C y  P  F1 sin 38.6598  P  ( 2.561250 P)sin 38.6598  0.60 P

(d)

The resultant reaction force at pin C can now be expressed as a function of P: C  Cx2  C y2  (2.0 P)2  (0.60 P)2  2.088061P

(e)

Inclined member (1): The allowable normal stress for inclined member (1) is  36 ksi  allow  Y   24 ksi FS 1.5 Therefore, the allowable axial force for inclined member (1) is Fallow,1   allow A1  (24 ksi)(2.25 in.2 )  54 kips From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations on the inclined member normal stress is Fallow,1 54 kips Pmax    21.083455 kips (f) 2.561250 2.561250 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Pin B: The allowable shear stress for the pins is  60 ksi  allow  U   30 ksi FS 2.0 A 1.0-in.-diameter pin has a cross-sectional area of Apin 



(1.0 in.) 2  0.785398 in.2

4 For a double shear connection, the corresponding shear area is AV  2 Apin  2(0.785398 in.2 )  1.570796 in.2 Thus, the maximum allowable force in member (1) based on the pin at B is: Fallow,1   allow AV  (30 ksi)(1.570796 in.2 )  47.123880 kips From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations on the pin at B is Fallow,1 47.123880 kips Pmax    18.398782 kips (g) 2.561250 2.561250

Pin C: The pin at C is in a single shear connection; therefore, the shear area of pin C is equal to the crosssectional area of the 1.0-in.-diameter pin. The maximum allowable shear force in this pin is thus Fallow,C   allow AV  (30 ksi)(0.785398 in.2 )  23.561940 kips From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations on the pin shear stress is C allow 23.561940 kips Pmax    11.284125 kips (h) 2.088061 2.088061 Maximum load P: Comparing the results in Eqs. (f), (g), and (h), the maximum load P that may be applied to the rigid bar is Ans. Pmax  11.284125 kips  11.28 kips

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4.18 Rigid bar ABC is supported by pin-connected axial member (1) and by a pin connection at C as shown in Fig. P4.18. A 6,300-lb concentrated load is applied to the rigid bar at A. Member (1) is a 2.75 in. wide by 1.25 in. thick rectangular bar made of steel with a yield strength of Y = 36,000 psi. The pin at C has an ultimate shear strength of U = 60,000 psi. (a) Determine the axial force in member (1). (b) Determine the factor of safety in member (1) with respect to its yield strength. (c) Determine the magnitude of the resultant reaction force acting at pin C. (d) If a minimum factor of safety of FS = 3.0 with respect to the ultimate shear strength is required, determine the minimum diameter that may be used for the pin at C. Fig. P4.18

Solution Member (1) is a two-force member that is oriented at  with respect to the horizontal axis: 30 in. tan    0.75   36.870 40 in. From a FBD of rigid structure ABC, the following equilibrium equations can be written: Fx   F1 cos 36.870  Cx  0 (a) Fy  6,300 lb  F1 sin 36.870  C y  0 (b)

M C  (6,300 lb)(80 in.) (c) ( F1 sin 36.870)(56 in.)  0 (a) From Eq. (c), the axial force in member (1) is: (6,300 lb)(80 in.) F1    15,000 lb  15,000 lb (C) (56 in.)sin 36.870 (b) The normal stress magnitude in member (1) is F 15,000 lb 1  1   4,363.6364 psi A1 (2.75 in.)(1.25 in.) Therefore, its factor of safety with respect to the 36,000 psi yield stress is  36,000 psi FS1  Y   8.25  1 4,363.6364 psi

Ans.

Ans.

(c) Substitute F1 into Eq. (a) to obtain Cx: Cx  F1 cos36.870  (15, 000 lb) cos36.870  12, 000 lb and substitute F1 into Eq. (b) to obtain an expression for Ay in terms of the unknown load P: C y  6,300 lb  F1 sin 36.870  6,300 lb  (15, 000 lb)sin 36.870  2, 700 lb The resultant pin force at C is found from Cx and Cy: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

C  Cx2  C y2  ( 12,000 lb)2  ( 2,700 lb)2  12,300 lb

Ans.

(d) The allowable shear stress for the pins at C is  60,000 psi  allow  U   20,000 psi FS 3.0 The double-shear pin connection at C must support a load of 12,300 lb. The minimum shear area AV required to support this load is C 12,300 lb AV    0.6150 in.2  allow 20,000 psi The pin diameter can be computed from



2 (2 surfaces) d pin  0.6150 in.2 4

 d pin  0.625717 in.  0.626 in.

Ans.

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4.19 A rectangular steel plate is used as an axial member to support a dead load of 70 kips and a live load of 110 kips. The yield strength of the steel is 50 ksi. (a) Use the ASD method to determine the minimum cross-sectional area required for the axial member if a factor of safety of 1.67 with respect to yielding is required. (b) Use the LRFD method to determine the minimum cross-sectional area required for the axial member based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively.

Solution (a) The service load on the axial member is P  D  L  70 kips  110 kips  180 kips The allowable normal stress is  50 ksi  allow  Y   29.940 ksi FS 1.67 The minimum cross-sectional area required to support the service load is P 180 kips Amin    6.01 in.2  allow 29.940 ksi (b) The ultimate load for LRFD is U  1.2D  1.6L  1.2(70 kips)  1.6(110 kips)  260 kips The design equation for an axial member subjected to tension can be written in LRFD as t Y A  U Consequently, the minimum cross-sectional area required for the tension member is U 260 kips Amin    5.78 in.2 t Y 0.9(50 ksi)

Ans.

Ans.

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4.20 A 20-mm-thick steel plate will be used as an axial member to support a dead load of 150 kN and a live load of 220 kN. The yield strength of the steel is 250 MPa. (a) Use the ASD method to determine the minimum plate width b required for the axial member if a factor of safety of 1.67 with respect to yielding is required. (b) Use the LRFD method to determine the minimum plate width b required for the axial member based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively.

Solution (a) The service load on the axial member is P  D  L  150 kN  220 kN  370 kN The allowable normal stress is  250 MPa  allow  Y   149.70 MPa FS 1.67 The minimum cross-sectional area required to support the service load is P (370 kN)(1,000 N/kN) Amin    2, 471.61 mm 2  allow 149.70 N/mm 2 Since the plate is 20-mm-thick, the minimum plate width is therefore: A 2, 471.61 mm 2 bmin  min   123.6 mm t 20 mm (b) The ultimate load for LRFD is U  1.2D  1.6L  1.2(150 kN)  1.6(220 kN)  532 kN The design equation for an axial member subjected to tension can be written in LRFD as t Y A  U Consequently, the minimum cross-sectional area required for the tension member is U (532 kN)(1,000 N/kN) Amin    2,364.44 mm 2 2 t Y 0.9(250 N/mm ) Since the plate is 20-mm-thick, the minimum plate width is therefore: A 2,364.44 mm 2 bmin  min   118.2 mm t 20 mm

Ans.

Ans.

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4.21 A round steel tie rod is used as a tension member to support a dead load of 30 kips and a live load of 15 kips. The yield strength of the steel is 46 ksi. (a) Use the ASD method to determine the minimum diameter required for the tie rod if a factor of safety of 2.0 with respect to yielding is required. (b) Use the LRFD method to determine the minimum diameter required for the tie rod based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively.

Solution (a) The service load on the axial member is P  D  L  30 kips  15 kips  45 kips The allowable normal stress is  46 ksi  allow  Y   23.0 ksi FS 2.0 The minimum cross-sectional area required to support the service load is P 45 kips Amin    1.956522 in.2  allow 23.0 ksi and thus, the minimum tie rod diameter is



4

2 d min  1.956522 in.2

 d min  1.578 in.

Ans.

(b) The ultimate load for LRFD is U  1.2D  1.6L  1.2(30 kips)  1.6(15 kips)  60 kips The design equation for an axial member subjected to tension can be written in LRFD as t Y A  U Consequently, the minimum cross-sectional area required for the tension member is U 60 kips Amin    1.449275 in.2 t Y 0.9(46 ksi) and thus, the minimum tie rod diameter is



4

2 d min  1.449275 in.2

 d min  1.358 in.

Ans.

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4.22 A round steel tie rod is used as a tension member to support a dead load of 190 kN and a live load of 220 kN. The yield strength of the steel is 320 MPa. (a) Use the ASD method to determine the minimum diameter D required for the tie rod if a factor of safety of 2.0 with respect to yielding is required. (b) Use the LRFD method to determine the minimum diameter D required for the tie rod based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively.

Solution (a) The service load on the axial member is P  D  L  190 kN  220 kN  410 kN The allowable normal stress is  320 MPa  allow  Y   160 MPa FS 2.0 The minimum cross-sectional area required to support the service load is P (410 kN)(1,000 N/kN) Amin    2,562.500 mm 2  allow 160 N/mm 2 and thus, the minimum tie rod diameter is



4

2 d min  2,562.500 mm 2

 d min  57.1 mm

Ans.

(b) The ultimate load for LRFD is U  1.2D  1.6L  1.2(190 kN)  1.6(220 kN)  580 kN The design equation for an axial member subjected to tension can be written in LRFD as t Y A  U Consequently, the minimum cross-sectional area required for the tension member is U (580 kN)(1,000 N/kN) Amin    2, 013.889 mm 2 t Y 0.9(320 N/mm 2 ) and thus, the minimum tie rod diameter is



4

2 d min  2,013.889 mm2

 d min  50.6 mm

Ans.

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5.1 A steel [E = 200 GPa] rod with a circular cross section is 7.5-m long. Determine the minimum diameter required if the rod must transmit a tensile force of 50 kN without exceeding an allowable stress of 180 MPa or stretching more than 5 mm.

Solution If the normal stress in the rod cannot exceed 180 MPa, the cross-sectional area must equal or exceed P (50 kN)(1,000 N/kN) A   277.7778 mm 2 2  180 N/mm If the elongation must not exceed 5 mm, the cross-sectional area must equal or exceed PL (50 kN)(1,000 N/kN)(7,500 mm) A   375.0000 mm 2 2 E (200,000 N/mm )(5 mm) Therefore, the minimum cross-sectional area that may be used for the rod is Amin = 375 mm2. The corresponding rod diameter is



4

2 d rod  375 mm2

 d rod  21.8510 mm  21.9 mm

Ans.

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5.2 An aluminum [E = 10,000 ksi] control rod with a circular cross section must not stretch more than 0.25 in. when the tension in the rod is 2,200 lb. If the maximum allowable normal stress in the rod is 12 ksi, determine: (a) the smallest diameter that can be used for the rod. (b) the corresponding maximum length of the rod.

Solution (a) If the normal stress in the rod cannot exceed 20 ksi, the cross-sectional area must equal or exceed P 2.20 kips A   0.1833 in.2  12 ksi The corresponding rod diameter is



4

2 d rod  0.1833 in.2

 d rod  0.483 in.

Ans.

(b) If the elongation must not exceed 0.25 in., the aluminum control rod having a cross-sectional area of A = 0.1833 in.2 can have a length no greater than AE (0.1833 in.2 )(10,000 ksi)(0.25 in.) L   208.3333 in.  208 in. Ans. P 2.20 kips

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5.3 A 12-mm-diameter steel [E = 200 GPa] rod (2) is connected to a 30-mm-wide by 8-mm-thick rectangular aluminum [E = 70 GPa] bar (1), as shown in Fig. P5.3. Determine the force P required to stretch the assembly 10.0 mm. Fig. P5.3

Solution The elongations in the two axial members are expressed by FL FL 1  1 1 and 2  2 2 A1 E1 A2 E2 The total elongation of the assembly is thus FL F L uC  1   2  1 1  2 2 A1 E1 A2 E2 Since the internal forces F1 and F2 are equal to external load P, this expression can be simplified to  L L  uC  P  1  2   A1 E1 A2 E2  For rectangular aluminum bar (1), the cross-sectional area is A1  (30 mm)(8 mm)  240 mm 2 and the cross-sectional area of steel rod (2) is A2 



4

(12 mm)2  113.0973 mm2

The force P required to stretch the assembly 10.0 mm is thus uC P L1 L  2 A1E1 A2 E2 

10.0 mm

  450 mm 1,300 mm  (240 mm 2 )(70,000 N/mm 2 )  (113.09739 mm 2 )(200, 000 N/mm 2 )     118,682.6 N = 118.7 kN

Ans.

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5.4 Two polymer bars are connected to a rigid plate at B, as shown in Fig. P5.4. Bar (1) has a cross-sectional area of 1.65 in.2 and an elastic modulus of 2,400 ksi. Bar (2) has a cross-sectional area of 0.975 in.2 and an elastic modulus of 4,000 ksi. Determine the total deformation of the bar. Fig. P5.4

Solution Draw a FBD that cuts through member (1) to find that the internal axial force in member (1) is F1  8 kips (T) Similarly, draw a FBD that cuts through member (2) to find that the internal axial force in member (2) is F2  14 kips (T) The elongation in bar (1) can be computed as FL (8 kips)(20 in.) 1  1 1   0.040404 in. A1 E1 (1.65 in.2 )(2, 400 ksi) and the elongation in bar (2) can be computed as FL (14 kips)(30 in.) 2  2 2   0.107692 in. A2 E2 (0.975 in.2 )(4, 000 ksi) The total elongation of the bar is thus 1  2  0.040404 in.  0.107692 in.  0.1481 in.

Ans.

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5.5 An axial member consisting of two polymer bars is supported at C as shown in Fig. P5.5. Bar (1) has a cross-sectional area of 540 mm2 and an elastic modulus of 28 GPa. Bar (2) has a crosssectional area of 880 mm2 and an elastic modulus of 16.5 GPa. Determine the deflection of point A relative to support C. Fig. P5.5

Solution Draw a FBD that cuts through member (1) to find that the internal axial force in member (1) is F1  35 kN (T) Similarly, draw a FBD that cuts through member (2) and includes the free end of the axial member. From this FBD, the equilibrium equation is Fx  35 kN  50 kN  50 kN  F2  0 Therefore, the internal axial force in member (2) is F2  65 kN  65 kN (C) The deformation in bar (1) can be computed as F L (35 kN)(1,000 N/kN)(850 mm) 1  1 1   1.9676 mm A1 E1 (540 mm 2 )(28, 000 N/mm 2 ) and the deformation in bar (2) can be computed as FL (65 kN)(1,000 N/kN)(1,150 mm) 2  2 2   5.1481 mm A2 E2 (880 mm 2 )(16,500 N/mm 2 ) The deflection of point A relative to the support at C is the sum of these two deformations: uA  1  2  1.9676 mm  (5.1481 mm)  3.18 mm  3.18 mm 

Ans.

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5.6 The roof and second floor of a building are supported by the column shown in Fig. P5.6. The column is a structural steel W10 × 60 wide-flange section [E = 29,000 ksi; A = 17.6 in2]. The roof and floor subject the column to the axial forces shown. Determine: (a) the amount that the first floor will settle. (b) the amount that the roof will settle.

Fig. P5.6

Solution (a) Draw a FBD that cuts through member (1) and includes the free end of the column. From this FBD, the sum of forces in the vertical direction can be written as Fy  110 kips  155 kips  F1  0 Therefore, the internal axial force in member (1) is F1  270 kips  270 kips (C) The settlement of the first floor is determined by the deformation (i.e., contraction in this case) that occurs in member (1). FL u B  1  1 1 A1 E1



(270 kips)(16 ft)(12 in./ft) (17.6 in.2 )(29, 000 ksi)

 0.101567 in.  0.1016 in. 

Ans.

(b) Draw a FBD that cuts through member (2) and includes the free end of the column. From this FBD, the equilibrium equation is Fx  115 kips  F2  0 Therefore, the internal axial force in member (2) is F2  115 kips  115 kips (C) The deformation in member (2) (which will be contraction in this instance) can be computed as F2 L2 (115 kips)(14 ft)(12 in./ft)   0.037853 in. A2 E2 (17.6 in.2 )(29, 000 ksi) The settlement of the roof is found from the sum of the contractions in the two members:

2 

uC  1   2  0.101567 in.  (0.037853 in.)  0.139420 in.  0.1394 in. 

Ans.

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5.7 Aluminum [E = 70 GPa] member ABC supports a load of 28 kN, as shown in Fig. P5.7. Determine: (a) the value of load P such that the deflection of joint C is zero. (b) the corresponding deflection of joint B.

Fig. P5.7

Solution Cut a FBD that exposes the internal axial force in member (2): Fy  28 kN  F2  0  F2  28 kN Similarly, cut a FBD that exposes the internal axial force in member (1): Fy  28 kN  P  F1  0  F1  28 kN  P The deflection at joint C, which must ultimately equal zero, can be expressed in terms of the member deformations 1 and 2: FL F L uC  1   2  1 1  2 2  0 A1 E1 A2 E2 or FL FL uC  1 1  2 2 A1E1 A2 E2



(28,000 N  P)(1,000 mm)

 4

2

2

(50 mm) (70,000 N/mm )



(28,000 N)(1,300 mm)

 4

2

0 2

(32 mm) (70,000 N/mm )

The only unknown in this equation is P, which can be computed as P  80.5841 kN  80.6 kN The corresponding deflection at joint B can be found from the deformation in member (1): F L (28, 000 N  80,584 N)(1,300 mm) uB  1  1 1   0.497 mm  0.497 mm   A1 E1 2 2 (50 mm) (70, 000 N/mm ) 4

Ans.

Ans.

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5.8 A solid brass [E = 100 GPa] axial member is loaded and supported as shown in Fig. P5.8. Segments (1) and (2) each have a diameter of 25 mm and segment (3) has a diameter of 14 mm. Determine: (a) the deformation of segment (2). (b) the deflection of joint D with respect to the fixed support at A. (c) the maximum normal stress in the entire axial member.

Fig. P5.8

Solution (a) Draw a FBD that cuts through segment (2) and includes the free end of the axial member. From this FBD, the sum of forces in the vertical direction reveals the internal force in the segment: Fy  F2  14 kN  28 kN  0  F2  42 kN (T) The cross-sectional area of the 25-mm-diameter segment is  A2  (25 mm) 2  490.8739 mm 2 4 The deformation in segment (2) is thus FL (42,000 N)(1,200 mm) 2  2 2  A2 E2 (490.8739 mm 2 )(100, 000 N/mm 2 )  1.026740 mm  1.027 mm

Ans.

(b) The internal forces in segments (1) and (3) must be determined at the outset. From a FBD that cuts through segment (1) and includes the free end of the axial member: Fy  F1  40 kN  14 kN  28 kN  0  F1  82 kN (T) The deformation in segment (1) can be computed as FL (82,000 N)(1,800 mm) 1  1 1   3.006883 mm A1 E1 (490.8739 mm 2 )(100, 000 N/mm 2 )

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Similarly, consider a FBD that cuts through segment (3) and includes the free end of the axial member: Fy  F3  28 kN  0  F3  28 kN (T) The cross-sectional area of the 14-mm-diameter segment is A3 



(164 mm) 2  153.9380 mm 2

4 The deformation in segment (3) can be computed as FL (28,000 N)(1,600 mm) 3  3 3   2.810262 mm A3 E3 (153.9380 mm 2 )(100, 000 N/mm 2 )

The deflection of joint D with respect to the fixed support at A is found from the sum of the three segment deformations: u D  1   2   3  3.006883 mm  1.026740 mm  2.810262 mm  6.943885 mm  6.94 mm

Ans.

(c) Since segments (1) and (2) have the same cross-sectional area, the maximum normal stress in these two segments occurs where the axial force is greater; that is, in segment (1): F 82,000 N 1  1   167.0490 MPa (T) A1 490.8739 mm 2 The normal stress in segment (3) is F 28,000 N 3  3   181.8914 MPa (T) A3 153.9380 mm 2 Therefore, the maximum normal stress in the axial member occurs in segment (3): Ans.  max  181.9 MPa (T)

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5.9 A hollow steel [E = 30,000 ksi] tube (1) with an outside diameter of 2.75 in. and a wall thickness of 0.25 in. is fastened to a solid aluminum [E = 10,000 ksi] rod (2) that has a 2in.-diameter and a solid 1.375-in.-diameter aluminum rod (3). The bar is loaded as shown in Fig. P5.9. Determine: (a) the change in length of steel tube (1). (b) the deflection of joint D with respect to the fixed support at A. (c) the maximum normal stress in the entire axial assembly.

Fig. P5.9

Solution Before proceeding, it is convenient to determine the internal forces in each of the three axial segments. Segment (1): Draw a FBD that cuts through segment (1) and includes the free end of the axial member. From this FBD, the sum of forces in the horizontal direction gives the force in segment (1): Fx   F1  2(34 kips)  2(18 kips)  25 kips  0

 F1  57 kips Segment (2): Draw a FBD that cuts through segment (2) and includes the free end of the axial member. From this FBD, the sum of forces in the horizontal direction gives the force in segment (2): Fx   F2  2(18 kips)  25 kips  0  F2  11 kips Segment (3): Draw a FBD that cuts through segment (3) and includes the free end of the axial member. From this FBD, the sum of forces in the horizontal direction gives the force in segment (3): Fx   F3  25 kips  0  F3  25 kips (a) The cross-sectional area of the hollow steel tube (inside diameter = 2.25 in.) is



(2.75 in.)2  (2.25 in.) 2   1.963495 in.2 4 The deformation in segment (1) is thus FL (57 kips)(60 in.) 1  1 1   0.058060 in.  0.0581 in. A1 E1 (1.963495 in.2 )(30, 000 ksi) A1 

Ans.

(b) For segment (2), the cross-sectional area of the 2-in.-diameter aluminum rod is

A2 



(2 in.)2  3.141593 in.2

4 The deformation in segment (2) can be computed as FL (11 kips)(40 in.) 2  2 2   0.014006 in. A2 E2 (3.141593 in.2 )(10, 000 ksi) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

For segment (3), the cross-sectional area of the 1.375-in.-diameter aluminum rod is A3 



(1.375 in.) 2  1.484893 in.2

4 The deformation in segment (3) can be computed as FL (25 kips)(30 in.) 3  3 3   0.050509 in. A3 E3 (1.484893 in.2 )(10, 000 ksi)

The deflection of joint D with respect to the fixed support at A is found from the sum of the three segment deformations: uD  1   2   3  0.058060 in.  0.014006 in.  0.050509 in.

 0.094563 in.  0.0946 in.  0.0946 in. 

Ans.

(c) Compute the normal stress in each of the three segments: F 57 kips 1  1   29.0299 ksi (C) A1 1.963495 in.2

2 

F2 11 kips   3.5014 ksi (T) A2 3.141593 in.2

F3 25 kips   16.8362 ksi (C) A3 1.484893 in.2 Therefore, the maximum normal stress in the axial member occurs in segment (1):  max  29.0 ksi (C)

3 

Ans.

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5.10 A solid 5/8-in. steel [E = 29,000 ksi] rod (1) supports beam AB, as shown in Fig. P5.10. If the stress in the rod must not exceed 30 ksi and the maximum deformation in the rod must not exceed 0.25 in., determine the maximum load P that may be supported.

Fig. P5.10

Solution The area of the 5/8-in.-diameter rod is A1 



(0.625 in.)2  0.306796 in.2 4 If the stress in the rod must not exceed 30 ksi, then the maximum force that can be applied to rod (1) is F1   allow,1 A1  (30 ksi)(0.306796 in.2 )  9.203880 kips The length of rod (1) is

L1  (20 ft)2  (16 ft)2  25.612497 ft  307.35 in. If the maximum deformation in the rod must not exceed 0.25 in., the maximum internal force in the rod must be limited to 1 A1E1 (0.25 in.)(0.306796 in.2 )(29,000 ksi) F1    7.236932 kips L1 307.35 in. Therefore, the maximum internal force in rod (1) must not exceed 7.237 kips. Consider a FBD of the rigid beam. Rod (1) is a twoforce member at an orientation  defined by: 16 ft tan    0.8   38.660 20 ft Write the equilibrium equation for the sum of moments about A to find the relationship between F1 and P: M A  (20 ft)(F1 sin 38.660)  (8 ft)P  0 P 

(20 ft)(7.236932 kips)sin 38.660  11.302236 kips  11.30 kips 8 ft

Ans.

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5.11 A 1-in.-diameter by 16-ft-long cold-rolled bronze bar [E = 15,000 ksi and  = 0.320 lb/in3] hangs vertically while suspended from one end. Determine the change in length of the bar due to its own weight.

Solution An incremental length dy of the bar has an incremental deformation given by F ( y) d  dy AE The force in the bar can be expressed as the product of the unit density of the bronze (bronze) and the volume of the bar below the incremental slice dy: F  y    bronze A y Therefore, the incremental deformation can be expressed as  Ay  y d  bronze dy  bronze dy AE E Integrate this expression over the entire length L of the bar: L L L y    L2 1    bronze dy  bronze  y dy  bronze  y 2  0  bronze 0 E E 0 E 2 2E The change in length of the bar due to its own weight is therefore:  L2 (0.320 lb/in.3 )(16 ft 12 in./ft) 2   bronze   0.000393216 in.  0.000393 in. 2E 2(15, 000, 000 psi)

Ans.

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5.12 A homogenous rod of length L and elastic modulus E is a truncated cone with diameter that varies linearly from d0 at one end to 2d0 at the other end. A concentrated axial load P is applied to the ends of the rod, as shown in Fig. P5.12. Assume that the taper of the cone is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid. (a) Determine an expression for the stress distribution on an arbitrary cross section at x. (b) Determine an expression for the elongation of the rod.

P5.12

Solution (a) Determine an expression for the diameter of the truncated cone through the use of the similar triangles concept: 2d 0  d 0 d ( x )  d 0  L x x  (a)  d ( x)  d 0 1    L Next, consider equilibrium of the truncated cone:  F ( x)  P  Fx  F ( x)  P  0 The internal force F(x) can be expressed in terms of stress and area:

F ( x)    dA   ( x) A( x) A

Equating this expression to the equilibrium equation gives: F ( x)   ( x) A( x)  P

P A( x) From Eq. (a), the area of the cone at any location x is expressed as   ( x) 

2

x   d 1   2   d ( x) L A( x)   4 4 Therefore, the stress can be written as P 4P  ( x)   2 A( x) x   d 02 1    L 2 0

Ans.

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(b) The elongation in the axial member is found from Eq. (5.5): L L F ( x)    d   dx 0 0 A( x ) E For the truncated cone axial member: L 4P 4 P L dx  dx  2 2 0  Ed 02 0  x x   Ed 02 1   1   L L 4P   L  2 PL   2   x  Ed 0 1   Ed 02    L 0 L

Ans.

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5.13 Determine the extension, due to its own weight, of the conical bar shown in Fig. P5.13. The bar is made of aluminum alloy [E = 10,600 ksi and  = 0.100 lb/in.3]. The bar has a 2-in. radius at its upper end and a length of L = 20 ft. Assume the taper of the bar is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid.

Fig. P5.13

Solution An incremental length dy of the bar has an incremental deformation given by F ( y) d  dy AE The force in the bar can be expressed as the product of the unit density of the aluminum (alum) and the volume of the bar below the incremental slice dy. The volume below the slice dy is a cone. At y, the cross-sectional area of the base of the cone is: 2

y  Ay    r  L  and the volume of a cone is given by: 1 V  (area of base)(altitude) 3 The internal force at y can be expressed as 1 F ( y )   alum  Ay y 3 The incremental deformation can be expressed as  alum Ay y  y d  dy  alum dy 3 Ay E 3E Integrate this expression over the entire length L of the bar: L L L y    L2 1    alum dy  alum  y dx  alum  y 2  0  alum 0 3E 3E 0 3E 2 6E The change in length of the bar due to its own weight is therefore:  L2 (0.100 lb/in.3 )(20 ft 12 in./ft) 2   alum   9.0566  105 in.  90.6 106 in. 6E 6(10, 600, 000 psi)

Ans.

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5.14 Rigid bar ABCD is loaded and supported as shown in Fig. P5.14. Steel [E = 30,000 ksi] bars (1) and (2) are unstressed before the load P is applied. Bar (1) has a cross-sectional area of 0.75 in.2 and bar (2) has a cross-sectional area of 0.425 in.2. After load P is applied, the strain in bar (1) is found to be 780 . Determine: (a) the stresses in bars (1) and (2). (b) the vertical deflection of point D. (c) the load P.

Fig. P5.14

Solution From the strain in bar (1), the deformation in bar (1) is (780 10 6 in./in.)(60 in.) 0.0468 in. 1 1 L1 Since the joint at B is a perfect connection, the rigid bar deflection at B must equal the deformation of bar (1): vB 0.0468 in. 1 From a deformation diagram of the rigid bar, the vertical deflection of joint C is related to B by similar triangles: vB vC 40 in. 80 in. 80 in. vC vB 2vB 40 in. 2(0.0468 in.) 0.0936 in. The joint at C is also a perfect connection; therefore, the downward displacement of C also causes an equal deformation in bar (2): vC 0.0936 in. 2 (a) Now that the deformations in both bars are known, the normal stresses in each can be computed. The normal stress in bar (1) is F1L1 (0.0468 in.)(30,000 ksi) 1 L1 1 E1 23.4 ksi (T) Ans. 1 1 A1E1 E1 L1 60 in. and the normal stress in bar (2) is F2 L2 (0.0936 in.)(30,000 ksi) 2 L2 2 E2 31.2 ksi (T) Ans. 2 2 A2 E2 E2 L2 90 in. Alternatively, the stress in (1) could be calculated from Hooke’s Law: E1 1 (30,000 ksi)(780 10 6 in./in.) 23.4 ksi (T) 1

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(b) From a deformation diagram of the rigid bar, the vertical deflection of joint D is related to B and C by similar triangles: vB vC vD 40 in. 80 in. 110 in. 110 in. vD vB 2.75(0.0468 in.) 0.1287 in. Ans. 40 in. (c) The force in each bar can be determined from the stresses computed previously: F1 (23.4 ksi)(0.75 in.2 ) 17.55 kips 1 A1

F2

2

A2

(31.2 ksi)(0.425 in.2 ) 13.26 kips

Consider a FBD of the rigid bar. The equilibrium equation for the sum of moments about A can be used to determine the load P: M A (40 in.)F1 (80 in.)F2 (110 in.)P P

0

(40 in.)(17.55 kips) (80 in.)(13.26 kips) 110 in.

16.0255 kips

16.03 kips

Ans.

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5.15 Rigid bar ABCD is loaded and supported as shown in Fig. P5.15. Bars (1) and (2) are unstressed before the load P is applied. Bar (1) is made of bronze [E = 100 GPa] and has a crosssectional area of 520 mm2. Bar (2) is made of aluminum [E = 70 GPa] and has a cross-sectional area of 960 mm2. After the load P is applied, the force in bar (2) is found to be 25 kN (in tension). Determine: (a) the stresses in bars (1) and (2). (b) the vertical deflection of point A. (c) the load P. Fig. P5.15

Solution Given that the axial force in bar (2) is 25 kN (in tension), the deformation can be computed as: F2 L2 (25, 000 N)(800 mm) 0.297619 mm 2 A2 E2 (960 mm 2 )(70, 000 N/mm 2 ) Since the pin at C is a perfect connection, the deflection of the rigid bar at C is equal to the deformation of bar (2): vC 0.297619 mm 2 From a deformation diagram of the rigid bar, the vertical deflection of joint C is related to B by similar triangles: vB vC 1.6 m 0.5 m 1.6 m vB vC 3.2vC 0.5 m 3.2(0.297619 mm) 1.116071 mm The joint at B is also a perfect connection; therefore, the downward displacement of B also causes an equal contraction in bar (1): vB 1.116071 mm 1 Note that a downward displacement at B causes contraction (and hence, compression) in bar (1). (a) The normal stress in aluminum bar (2) can be computed from the known force in the bar: F2 25,000 N 26.041667 MPa 26.0 MPa (T) 2 A2 960 mm 2 The normal stress in bronze bar (1) can be computed from its contraction: F1L1 1 L1 1 A1E1 E1 E L1

1 1

1

( 1.116071 mm)(100,000 N/mm 2 ) 600 mm

186.011905 MPa

186.0 MPa (C)

Ans.

Ans.

(b) From a deformation diagram of the rigid bar, the vertical deflection of joint A is related to joint B by similar triangles: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

vA 2.0 m

vC 0.5 m vA

vC

2.0 m 0.5 m

4vC

4(0.297619 mm) 1.488095 mm

1.488 mm

Ans.

(c) The force in bar (2) is given as F2 = 25 kN. The force in bar (1) can be determined from the stress computed previously: F1 ( 186.011905 N/mm 2 )(520 mm 2 ) 96,726.1905 N 96.7262 kN 1 A1

Consider a FBD of the rigid bar. The equilibrium equation for the sum of moments about D can be used to determine the load P: M D (1.6 m)F1 (0.5 m)F2 (2.0 m)P 0 P

(0.5 m)(25 kN) (1.6 m)( 96.7262 kN) 2.0 m 77.5446 kN 77.5 kN

Ans.

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5.16 In Fig. P5.16, aluminum [E = 70 GPa] links (1) and (2) support rigid beam ABC. Link (1) has a cross-sectional area of 300 mm2 and link (2) has a crosssectional area of 450 mm2. For an applied load of P = 55 kN, determine the rigid beam deflection at point B.

Fig. P5.16

Solution From a FBD of the rigid beam, write two equilibrium equations: Fy F1 F2 55 kN 0 (a) M A (2, 200 mm)F2 (1, 400 mm)(55 kN) Solve Eq. (b) for F2: (1, 400 mm)(55 kN) F2 35 kN 2, 200 mm and backsubstitute into Eq. (a) to obtain F1: F1 55 kN F2 55 kN 35 kN 20 kN

0 (b)

Next, determine the deformations in links (1) and (2): F1L1 (20,000 N)(2,500 mm) 2.380952 mm 1 A1E1 (300 mm 2 )(70,000 N/mm 2 ) F2 L2 (35,000 N)(4,000 mm) 4.444444 mm 2 A2 E2 (450 mm 2 )(70,000 N/mm 2 ) Since the connections at A and C are perfect, the rigid beam deflections at these joints are equal to the deformations of links (1) and (2), respectively: vA 2.380952 mm and vC 4.444444 mm 1 2

The deflection at B can be determined from similar triangles: vC v A vB v A 2, 200 mm 1, 400 mm Solve this expression for vB: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

vB

1, 400 mm (vC v A ) v A 2,200 mm 1, 400 mm (4.444444 mm 2.380952 mm) 2.380952 mm 2,200 mm (0.636364)(2.053492 mm) 2.380952 mm = 3.69 mm

Ans.

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5.17 Rigid bar ABC is supported by bronze rod (1) and aluminum rod (2), as shown in Fig P5-17. A concentrated load P is applied to the free end of aluminum rod (3). Bronze rod (1) has an elastic modulus of E1 = 15,000 ksi and a diameter of d1 = 0.50 in. Aluminum rod (2) has an elastic modulus of E2 = 10,000 ksi and a diameter of d2 = 0.75 in. Aluminum rod (3) has a diameter of d3 = 1.0 in. The yield strength of the bronze is 48 ksi and the yield strength of the aluminum is 40 ksi. (a) Determine the magnitude of load P that can safely be applied to the structure if a minimum factor of safety of 1.67 is required. (b) Determine the deflection of point D for the load determined in part (a). (c) The pin used at B has an ultimate shear strength of 54 ksi. If a factor of safety of 3.0 is required for this double shear pin connection, determine the minimum pin diameter that can be used at B. Fig. P5.17

Solution Before beginning, the cross-sectional areas of the three rods can be calculated from the specified diameters: A1 0.196350 in.2 A2 0.441786 in.2 A3 0.785398 in.2 The allowable stress of the bronze is 48 ksi Y ,bronze 28.742515 ksi allow,1 FS 1.67 and the allowable stress of the aluminum is 40 ksi Y ,alum 23.952096 ksi allow,2 allow,3 FS 1.67 (a) From a FBD cut through rod (3), equilibrium requires that the internal force in rod (3) is F3 = P. From a FBD of the rigid bar, write two equilibrium equations: Fy F1 F2 F3 F1 F2 P 0

(a)

M A (4 ft)F2 (2.5 ft)F3 (4 ft)F2 (2.5 ft)P 0 (b) Solve Eq. (b) for F2 in terms of the unknown load P: 2.5 ft (c) F2 P 0.6250 P 4 ft Backsubstitute this expression into Eq. (a) to obtain F1 in terms of the unknown load P: F1 P F2 P 0.6250 P 0.3750 P (d)

Rearrange Eqs. (c) and (d) to express P in terms of F2 and F1, respectively: P 1.6000 F2 P 2.6667 F1

(e) (f)

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Based on the allowable stress for the bronze, the maximum allowable force that can be supported by rod (1) is: F1 (28.742515 ksi)(0.196350 in.2 ) 5.643579 kips allow,1 A1 Thus, from Eq. (f), the corresponding maximum load P is: P 2.6667 F1 2.6667(5.643579 kips) 15.049545 kips (g) Next, the maximum allowable force that can be supported by rod (2) based on the allowable stress for the aluminum is: F2 (23.952096 ksi)(0.441786 in.2 ) 10.581711 kips allow,2 A2 which leads to a corresponding maximum load P from Eq. (e): P 1.6000 F2 1.6000(10.581711 kips) 16.930738 kips (h) Finally, we should also check the capacity of rod (3): F3 (23.952096 ksi)(0.785398 in.2 ) 18.811931 kips allow,3 A3 which means that P 18.811931 kips

(i)

From a comparison of Eqs. (g), (h), and (i), the maximum load that may be applied to the structure is Ans. Pmax 15.049545 kips 15.05 kips Based on this maximum load P, the internal forces in the three rods are: F1 0.3750 P 0.3750(15.049545 kips) 5.643579 kips

F2

0.6250 P

0.6250(15.049545 kips)

F3

P 15.049545 kips

9.405966 kips

(b) Next, determine the deformations in rods (1), (2), and (3): F1L1 (5.643579 kips)(6 ft)(12 in./ft) 0.137964 in. 1 A1E1 (0.196350 in.2 )(15,000 ksi) 2

F2 L2 A2 E2

(9.405966 kips)(8 ft)(12 in./ft) (0.441786 in.2 )(10,000 ksi)

0.204391 in.

3

F3 L3 A3 E3

(15.049545 kips)(3 ft)(12 in./ft) (0.785398 in.2 )(10,000 ksi)

0.068982 in.

Since the connections at A and C are perfect, the rigid bar deflections at these joints are equal to the deformations of rods (1) and (2), respectively: vA 0.137964 in. and vC 0.204391 in. 1 2

The rigid bar deflection at B can be determined from similar triangles: vC vA vB vA 4 ft 2.5 ft

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Solve this expression for vB: 2.5 ft vB (vC vA ) vA 4 ft 2.5 ft (0.204391 in. 0.137964 in.) 0.137964 in. 4 ft (0.6250)(0.066427 in.) 0.137964 in.

0.179481 in. The deflection of joint D is equal to the deflection of the rigid bar at B plus the deformation in rod (3):

vD

vB

0.179481 in. 0.068982 in. 0.248463 in.

3

0.248 in.

Ans.

(c) The pin at B has an allowable shear stress of 54 ksi U 18 ksi allow FS 3.0 The force tending to shear this pin is equal to the load P = 15.049545 kips. The shear area AV required for this pin is thus V 15.049545 kips AV 0.836086 in.2 18 ksi allow Since the pin is used in a double shear connection, the shear area is equal to twice the cross-sectional area of the pin: AV

2 Apin

2

2 d pin

2

2 d pin

4 and so the minimum pin diameter is 4

0.836086 in.2

d pin

0.729568 in.

0.730 in.

Ans.

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5.18 Solve problem 5.14 when there is a clearance of 0.05 in. in the pin connection at C. 5.14 (Repeated here for convenience). Rigid bar ABCD is loaded and supported as shown in Fig. P5.14. Steel [E = 30,000 ksi] bars (1) and (2) are unstressed before the load P is applied. Bar (1) has a cross-sectional area of 0.75 in.2 and bar (2) has a cross-sectional area of 0.425 in.2. After load P is applied, the strain in bar (1) is found to be 780 . Determine: (a) the stresses in bars (1) and (2). (b) the vertical deflection of point D. (c) the load P. Fig. P5.14 (repeated)

Solution From the strain in bar (1), the deformation in bar (1) is (780 10 6 in./in.)(60 in.) 0.0468 in. 1 1 L1 Since the joint at B is a perfect connection, the rigid bar deflection at B must equal the deformation of bar (1): vB 0.0468 in. 1 From a deformation diagram of the rigid bar, the vertical deflection of joint C is related to B by similar triangles: vB vC 40 in. 80 in. 80 in. vC vB 2vB 40 in. 2(0.0468 in.) 0.0936 in. There is a 0.05-in. clearance in the connection at C; therefore, vC 0.05 in. 2 2

vC

0.05 in. 0.0936 in. 0.05 in. 0.0436 in.

(a) Now that the deformations in both bars are known, the normal stresses in each can be computed. The normal stress in bar (1) is F1L1 (0.0468 in.)(30,000 ksi) 1 L1 1 E1 23.4 ksi (T) Ans. 1 1 A1E1 E1 L1 60 in. and the normal stress in bar (2) is F2 L2 (0.0436 in.)(30,000 ksi) 2 L2 2 E2 14.5333 ksi 14.53 ksi (T) Ans. 2 2 A2 E2 E2 L2 90 in.

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(b) From a deformation diagram of the rigid bar, the vertical deflection of joint D is related to B and C by similar triangles: vB vC vD 40 in. 80 in. 110 in. 110 in. vD vB 2.75(0.0468 in.) 0.1287 in. Ans. 40 in. (c) The force in each bar can be determined from the stresses computed previously: F1 (23.4 ksi)(0.75 in.2 ) 17.55 kips 1 A1

F2

2

A2

(14.5333 ksi)(0.425 in.2 )

6.1767 kips

Consider a FBD of the rigid bar. The equilibrium equation for the sum of moments about A can be used to determine the load P: M A (40 in.)F1 (80 in.)F2 (110 in.)P P

0

(40 in.)(17.55 kips) (80 in.)(6.1767 kips) 110 in.

10.8739 kips

10.87 kips

Ans.

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5.19 The rigid beam in Fig. P5.19 is supported by links (1) and (2), which are made from a polymer material [E = 16 GPa]. Link (1) has a crosssectional area of 400 mm2 and link (2) has a crosssectional area of 800 mm2. Determine the maximum load P that may by applied if the deflection of the rigid beam is not to exceed 20 mm at point C.

Fig. P5.19

Solution Equilibrium: Consider a FBD of the rigid beam and assume tension in each link. Fy F1 F2 P 0 (a) M A (600 mm)F2 (900 mm)P From Eq. (b): 900 mm F2 P 1.5P 600 mm Backsubstituting into Eq. (a): F1 F2 P 1.5P P 0.5P

0

(b) (c) (d)

Force-deformation relationships: The relationship between internal force and member deformation for links (1) and (2) can be expressed as: F1L1 F2 L2 and 2 (e) 1 A1E1 A2 E2 Geometry of deformations: Consider a deformation diagram of the rigid beam. Using similar triangles, one way to express the relationships between vA, vB, and vC is: v A vC vC vB (f) 900 mm 300 mm Note: Here, vA, vB, and vC are treated as unsigned magnitudes in the directions shown on the deformation diagram. The rigid beam deflection at A will equal the deformation that occurs in link (1): vA 1 and similarly at B: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

vB 2 Equation (f) can now be rewritten in terms of e1 and e2 as: vC vC 1 2 900 mm 300 mm or 900 mm vC (vC 3vC 3 2 1 2) 300 mm Solving for vC gives: 2vC 3 2 vC 0.5 1 1.5 2 1 Substitute the force-deformation relationships from Eq. (e) to obtain: 0.5 F1L1 1.5 F2 L2 vC 0.5 1 1.5 2 A1E1 A2 E2 and then substitute Eqs. (c) and (d) to derive an expression for vC in terms of the unknown load P: 0.5(0.5P) L1 1.5(1.5P) L2 0.25L1 2.25L2 vC P A1 E1 A2 E2 A1 E1 A2 E2 The unknown load P is thus related to the rigid beam deflection at C by: vC P 0.25L1 2.25 L2 A1 E1 A2 E2 Substituting the appropriate values into this relationship gives the maximum load P that may be applied to the rigid beam at C without causing more than 20 mm deflection: 20 mm P (0.25)(1,000 mm) (2.25)(1, 250 mm) 2 2 (400 mm )(16,000 N/mm ) (800 mm 2 )(16,000 N/mm 2 )

77, 283 N

77.3 kN

Ans.

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5.20 Three aluminum [E = 10,000 ksi] bars are used to support the loads shown in Fig. P5.20. The elongation in each bar must be limited to 0.25 in. Determine the minimum cross-sectional area required for each bar.

Fig. P5.20

Solution Determine the inclination angles for bars (1), (2), and (3): 7 ft tan 1 34.992 1 10 ft 3 ft tan 2 18.435 2 9 ft 8 ft tan 3 57.995 1 5 ft The bar lengths are:

L1

(7 ft)2 (10 ft) 2

12.206556 ft 146.4787 in.

L2

(3 ft)2 (9 ft) 2

9.486833 ft 113.8420 in.

L3

(8 ft) 2 (5 ft) 2

9.433981 ft 113.2078 in.

Equilibrium: Consider a FBD of joint B and write two equilibrium equations: Fx F1 cos34.992 F2 cos18.435 0

Fy

F1 sin 34.992

F2 sin18.435

31 kips

0

Solve these two equations simultaneously to obtain F1 = 36.61967 kips and F2 = 31.62278 kips. Next, consider a FBD of joint C and write two additional equilibrium equations: Fx F2 cos18.435 F3 cos57.995 0

Fy

F2 sin18.435

F3 sin 57.995

38 kips

0

From this, the internal force in bar (3) is F3 = 56.60389 kips. The minimum cross-sectional area required for each bar is: F1L1 (36.61967 kips)(146.4787 in.) A1 2.15 in.2 e1E1 (0.25 in.)(10,000 ksi) A2

F2 L2 e2 E2

(31.62278 kips)(113.8420 in.) (0.25 in.)(10,000 ksi)

1.440 in.2

Ans. Ans.

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A3

F3 L3 e3 E3

(56.60389 kips)(113.2078 in.) (0.25 in.)(10,000 ksi)

2.56 in.2

Ans.

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5.21 A tie rod (1) and a pipe strut (2) are used to support an 80-kip load as shown in Fig. P5.21. Pipe strut (2) has an outside diameter of 6.625 in. and a wall thickness of 0.280 in. Both the tie rod and the pipe strut are made of structural steel with a modulus of elasticity of E = 29,000 ksi and a yield strength of Y = 36 ksi. For the tie rod, the minimum factor of safety with respect to yield is 1.5 and the maximum allowable axial elongation is 0.20 in. (a) Determine the minimum diameter required to satisfy both constraints for tie rod (1). (b) Draw a deformation diagram showing the final position of joint B.

Fig. P5.21

Solution The angles of inclination for members (1) and (2) are: 12 ft tan 1 26.565 1 24 ft 30 ft tan 2 51.340 2 24 ft The member lengths are:

L1

(12 ft)2 (24 ft) 2

26.8328 ft

321.9938 in.

L2

(30 ft)2 (24 ft) 2

38.4187 ft

461.0249 in.

(a) Equilibrium: Consider a FBD of joint B and write two equilibrium equations: Fx F1 cos 26.565 F2 cos51.340 0

Fy

F1 sin 26.565

F2 sin 51.340

80 kips

0

Solve these two equations simultaneously to obtain F1 = 51.1103 kips and F2 = −73.1786 kips. The allowable normal stress for tie rod (1) is 36 ksi Y 24 ksi allow FS 1.5 The minimum cross-sectional area required to satisfy the normal stress requirement is F1 51.1103 kips A 2.1296 in.2 24 ksi allow The maximum allowable axial deformation for the tie rod is 0.20 in. The minimum cross-sectional area required to satisfy the deformation requirement is F1L1 (51.1103 kips)(321.9938 in.) A 2.8374 in.2 (0.20 in.)(29,000 ksi) 1 E1 To satisfy both the stress and deformation requirements, the tie rod must have a minimum crosssectional area of A = 2.8374 in.2. The corresponding rod diameter is Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

4

2 d rod

2.8374 in.2

d rod

1.901 in.

Ans.

(b) The pipe has a cross-sectional area of A2 = 5.5814 in.2. The pipe deformation is F2 L2 ( 73.1786 kips)(461.0249 in.) 0.2084 in. 2 A2 E2 (5.5814 in.2 )(29,000 ksi) Rod (1) elongates 0.20 in. and pipe (2) contracts 0.2084 in. Therefore, the deformation diagram showing the final position of joint B is shown.

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5.22 Two axial members are used to support a load of P = 72 kips as shown in Fig. P5.22. Member (1) is 12-ft long, it has a cross-sectional area of A1 = 1.75 in.2, and it is made of structural steel [E = 29,000 ksi]. Member (2) is 16-ft long, it has a crosssectional area of A2 = 4.50 in.2, and it is made of an aluminum alloy [E = 10,000 ksi]. (a) Compute the normal stress in each axial member. (b) Compute the deformation of each axial member. (c) Draw a deformation diagram showing the final position of joint B. (d) Compute the horizontal and vertical displacements of joint B.

Fig. P5.22

Solution (a) Consider a FBD of joint B and write two equilibrium equations: Fx F1 F2 cos55 0

Fy

F2 sin 55

72 kips

0

Solve these two equations simultaneously to obtain F1 = 50.4149 kips and F2 = 87.8958 kips. The normal stress in member (1) is: F1 50.4149 kips 28.8 ksi 1 A1 1.75 in.2 and the normal stress in member (2) is: F2 87.8958 kips 19.53 ksi 2 A2 4.50 in.2 (b) The member deformations are F1L1 (50.4149 kips)(12 ft)(12 in./ft) 1 A1E1 (1.75 in.2 )(29,000 ksi) 2

F2 L2 A2 E2

(87.8958 kips)(16 ft)(12 in./ft) (4.50 in.2 )(10,000 ksi)

Ans.

Ans.

0.1430 in.

Ans.

0.375 in.

Ans.

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(c) Member (1) elongates 0.1430 in. and member (2) elongates 0.375 in. Therefore, the deformation diagram showing the final position of joint B is shown.

(d) The horizontal displacement of B is x 0.1430 in. The vertical displacement is found from 0.375 in. 0.1430cos55 sin 55 y 0.375 in. 0.1430cos55 y sin 55

Ans.

0.375 in. 0.0820 in. sin 55

0.558 in.

Ans.

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5.23 The 200 × 200 × 1,200-mm oak [E = 12 GPa] block (2) shown in Fig. P5.23 was reinforced by bolting two 6 × 200 × 1,200 mm steel [E = 200 GPa] plates (1) to opposite sides of the block. A concentrated load of 360 kN is applied to a rigid cap. Determine: (a) the normal stresses in the steel plates (1) and the oak block (2). (b) the shortening of the block when the load is applied.

Fig. P5.23

Solution Equilibrium Consider a FBD of the rigid cap. Sum forces in the vertical direction to obtain: Fy 2 F1 F2 360 kN 0 (a) Geometry of Deformations Relationship For this configuration, the deformations of both members will be equal; therefore, 1

2

(b)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): F1L1 F2 L2 (c) 1 2 A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (c) into the geometry of deformation relationship (b) to derive the compatibility equation: F1 L1 F2 L2 (d) A1 E1 A2 E2 Solve the Equations Solve Eq. (d) for F2, recognizing that steel plates (1) and oak block (2) have the same length: L AE L A E A E F2 F1 1 2 2 F1 1 2 2 F1 2 2 A1 E1 L2 L2 A1 E1 A1 E1 Substitute this expression into equilibrium equation (a) and solve for F1: A E A2 E2 Fy 2 F1 F2 2 F1 F1 2 2 F1 2 360 kN A1 E1 A1 E1

(e)

(f)

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For this structure, the areas and elastic moduli are given below: A1 (6 mm)(200 mm) 1, 200 mm2 A2 (200 mm)2

40,000 mm2

E1 200 GPa E2 12 GPa Substitute these values into Eq. (f) and calculate F1 = −90 kN. Backsubstitute into Eq. (e) to calculate F2 = −180 kN. Normal Stresses The normal stresses in each axial member can now be calculated: F1 ( 90 kN)(1,000 N/kN) 75 MPa 75 MPa (C) 1 A1 1,200 mm 2

2

F2 A2

( 180 kN)(1,000 N/kN) 40,000 mm 2

4.50 MPa

4.50 MPa (C)

Ans.

Ans.

(b) The shortening of the block is equal to the deformation (i.e., contraction in this instance) of member (2): F2 L2 ( 180,000 N)(1,200 mm) uB 0.450 mm 0.450 mm Ans. 2 A2 E2 (40,000 mm 2 )(12,000 N/mm 2 )

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5.24 Two identical steel [E = 200 GPa] pipes, each with a cross-sectional area of 1,475 mm2, are attached to unyielding supports at the top and bottom, as shown in Fig. P5.24. At flange B, a concentrated downward load of 120 kN is applied. Determine: (a) the normal stresses in the upper and lower pipes. (b) the deflection of flange B.

Fig. P5.24

Solution (a) Equilibrium: Consider a FBD of flange B. Sum forces in the vertical direction to obtain: Fy F1 F2 120 kN 0 (a) Geometry of Deformations: 0 1 2

(b)

Force-Deformation Relationships: F1L1 F2 L2 1 2 A1E1 A2 E2

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1 L1 F2 L2 0 A1 E1 A2 E2 Solve the Equations: Solve Eq. (d) for F2: L AE L A E F2 F1 1 2 2 F1 1 2 2 A1 E1 L2 L2 A1 E1 and substitute this expression into Eq. (a) to determine F1: L A E L A E Fy F1 F2 F1 F1 1 2 2 F1 1 1 2 2 L2 A1 E1 L2 A1 E1

(d)

(e)

120 kN

(f)

Notice that A1 = A2 and E1 = E2 for this structure. Therefore, F1 can be computed as: 120 kN 120 kN 120 kN F1 66.268657 kN L1 A2 E2 3.0 m 1.810811 1 1+ L2 A1 E1 3.7 m and from Eq. (e), F2 has a value of L A E 3.0 m F2 F1 1 2 2 (66.268657 kN) L2 A1 E1 3.7 m

53.731343 kN

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Normal Stresses: The normal stresses in each axial member can now be calculated: F1 66.268657 N 44.927903 MPa 44.9 MPa (T) 1 A1 1,475 mm 2

2

F2 A2

53,731.343 N 1,475 mm 2

36.428029 MPa

36.4 MPa (C)

Ans.

Ans.

(b) The deflection of flange B is equal to the deformation (i.e., contraction in this instance) of member (2): F2 L2 ( 53.731343 N)(3,700 mm) uB 0.673919 mm 0.674 mm Ans. 2 A2 E2 (1,475 mm 2 )(200,000 N/mm 2 )

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5.25 Solve Problem 5.24 if the lower support in Fig. P5.24 yields and displaces downward 1.0 mm as the load P is applied. 5.24 (Repeated here for convenience) Two identical steel [E = 200 GPa] pipes, each with a cross-sectional area of 1,475 mm2, are attached to unyielding supports at the top and bottom, as shown in Fig. P5.24. At flange B, a concentrated downward load of 120 kN is applied. Determine: (a) the normal stresses in the upper and lower pipes. (b) the deflection of flange B.

Fig. P5.24 (repeated)

Solution (a) Equilibrium: Consider a FBD of flange B. Sum forces in the vertical direction to obtain: Fy F1 F2 120 kN 0 (a) Geometry of Deformations: 1.0 mm 1 2

(b)

Force-Deformation Relationships: F1L1 F2 L2 1 2 A1E1 A2 E2

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1L1 F2 L2 1.0 mm A1E1 A2 E2 Solve the Equations: Solve Eq. (d) for F2: F1L1 A2 E2 (1.0 mm)A2 E2 F2 1.0 mm A1E1 L2 L2

F1

L1 A2 E2 L2 A1 E1

and substitute this expression into Eq. (a) to determine F1: (1.0 mm)A2 E2 L A E Fy F1 F2 F1 F1 1 2 2 120 kN L2 L2 A1 E1

(d)

(e)

(f)

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Simplify this expression, gathering terms with F1 on the left-hand side of the equation: (1.0 mm)A2 E2 L A E F1 F1 1 2 2 120 kN L2 L2 A1 E1

F1 1

L1 A2 E2 L2 A1 E1

120 kN

(1.0 mm)A2 E2 L2

Notice that A1 = A2 and E1 = E2 for this structure. Therefore, F1 can be computed as: (1.0 mm)(1,475 mm 2 )(200,000 N/mm 2 ) 120,000 N 199,729.7297 N 3,700 mm F1 110, 298.5075 N 3,000 mm 1.810811 1 3,700 mm and from Eq. (a), F2 has a value of F2 F1 120 kN 110, 298.507 N 120,000 N

9,701.493 N

Normal Stresses: The normal stresses in each axial member can now be calculated: F1 110, 298.507 N 74.778649 MPa 74.8 MPa (T) 1 A1 1,475 mm 2

2

F2 A2

9,701.493 N 1,475 mm 2

6.577283 MPa

6.58 MPa (C)

(b) The downward deflection of flange B is equal to the deformation of member (1): F1L1 (110, 298.507 N)(3,000 mm) uB 1.121680 mm 1.122 mm 1 A1E1 (1,475 mm 2 )(200,000 N/mm 2 )

Ans.

Ans.

Ans.

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5.26 A load P is supported by a structure consisting of rigid bar ABC, two identical solid bronze [E = 15,000 ksi] rods, and a solid steel [E = 30,000 ksi] rod, as shown in Fig. P5.26. The bronze rods (1) each have a diameter of 0.75 in. and they are symmetrically positioned relative to the center rod (2) and the applied load P. Steel rod (2) has a diameter of 0.50 in. If all bars are unstressed before the load P is applied, determine the normal stresses in the bronze and steel rods after a load of P = 20 kips is applied.

Fig. P5.26

Solution Equilibrium By virtue of symmetry, the forces in the two bronze rods (1) are identical. Consider a FBD of the rigid bar. Sum forces in the vertical direction to obtain: Fy 2 F1 F2 P 0 (a) Geometry of Deformations Relationship For this configuration, the deformations of all rods will be equal; therefore, (b) 1 2 Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): F1L1 F2 L2 (c) 1 2 A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (c) into the geometry of deformation relationship (b) to derive the compatibility equation: F1 L1 F2 L2 (d) A1 E1 A2 E2 Solve the Equations Solve Eq. (d) for F2: L AE F2 F1 1 2 2 A1 E1 L2

F1

L1 A2 E2 L2 A1 E1

Substitute this expression into equilibrium equation (a) and solve for F1: L A E L A E Fy 2 F1 F2 2 F1 F1 1 2 2 F1 2 1 2 2 P L2 A1 E1 L2 A1 E1

(e)

(f)

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For this structure, P = 20 kips, and the lengths, areas, and elastic moduli are given below: L1 50 in. L2 90 in.

(0.75 in.) 2 0.441786 in.2 A2 (0.50 in.) 2 0.196350 in.2 4 4 E1 15,000 ksi E2 30,000 ksi Substitute these values into Eq. (f) and calculate F1 = 8.019802 kips. Backsubstitute into Eq. (e) to calculate F2 = 3.960396 kips. A1

Normal Stresses The normal stresses in each axial member can now be calculated: F1 8.019802 kips 18.15 ksi 1 A1 0.441786 in.2

2

F2 A2

3.960396 kips 0.196350 in.2

20.2 ksi

Ans.

Ans.

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5.27 An aluminum alloy [E = 10,000 ksi] pipe with a cross-sectional area of A1 = 4.50 in.2 is connected at flange B to a steel [E = 30,000 ksi] pipe with a cross-sectional area of A2 = 3.20 in.2. The assembly (shown in Fig. P5.27) is connected to rigid supports at A and C. For the loading shown, determine: (a) the normal stresses in aluminum pipe (1) and steel pipe (2). (b) the deflection of flange B.

Fig. P5.27

Solution (a) Equilibrium: Consider a FBD of flange B. Sum forces in the horizontal direction to obtain: Fx F1 F2 90 kips 0 (a) Geometry of Deformations: 0 1 2

(b)

Force-Deformation Relationships: F1L1 F2 L2 1 2 A1E1 A2 E2

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1 L1 F2 L2 0 A1 E1 A2 E2

(d)

Solve the Equations: Solve Eq. (d) for F2: L AE L A E F2 F1 1 2 2 F1 1 2 2 A1 E1 L2 L2 A1 E1 and substitute this expression into Eq. (a) to determine F1: L A E L A E Fx F1 F2 F1 F1 1 2 2 F1 1 1 2 2 90 kips L2 A1 E1 L2 A1 E1 F1 can be computed as: 90 kips 90 kips 90 kips F1 35.273159 kips 2 L1 A2 E2 2.551515 160 in. 3.20 in. 30,000 ksi 1 1+ L2 A1 E1 220 in. 4.50 in.2 10,000 ksi and from Eq. (a), F2 has a value of F2 F1 90 kips 35.273159 kips 90 kips 54.726841 kips Normal Stresses: The normal stresses in each axial member can now be calculated: F1 35.273159 kips 7.838480 ksi 7.84 ksi (T) 1 A1 4.50 in.2

(e)

(f)

Ans.

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2

F2 A2

54.726841 kips 3.20 in.2

17.102138 ksi

17.10 ksi (C)

(b) The deflection of flange B is equal to the deformation of member (1): F1L1 (35.273159 kips)(160 in.) uB 0.125416 in. 0.1254 in. 1 A1E1 (4.50 in.2 )(10,000 ksi)

Ans.

Ans.

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5.28 The concrete [E = 29 GPa] pier shown in Fig. P5.28 is reinforced using four steel [E = 200 GPa] reinforcing rods, each having a diameter of 19 mm. If the pier is subjected to an axial load of 670 kN, determine: (a) the normal stress in the concrete and in the steel reinforcing rods. (b) the shortening of the pier.

Fig. P5.28

Solution (a) Equilibrium: Consider a FBD of the pier, cut around the upper end. The concrete will be designated member (1) and the reinforcing steel bars will be designated member (2). Sum forces in the vertical direction to obtain: Fy F1 F2 670 kN 0 (a) Geometry of Deformations: 1

2

Force-Deformation Relationships: F1L1 F2 L2 1 2 A1E1 A2 E2

(b)

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1 L1 F2 L2 A1 E1 A2 E2 Solve the Equations: Solve Eq. (d) for F2: L AE L A E F2 F1 1 2 2 F1 1 2 2 A1 E1 L2 L2 A1 E1 and substitute this expression into Eq. (a) to determine F1: L A E L A E Fy F1 F2 F1 F1 1 2 2 F1 1 1 2 2 L2 A1 E1 L2 A1 E1

(d)

(e)

670 kN

(f)

Since L1 = L2, the term L1/L2 = 1 can be eliminated and Eq. (f) simplified to A E F1 1 2 2 670 kN (g) A1 E1 The gross cross-sectional area of the pier is (250 mm)2 = 62,500 mm2; however, the reinforcing bars take up a portion of this area. The cross-sectional area of four 19-mm-diameter reinforcing bars is Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

A2

(4 bars)

(19 mm) 2

1,134.115 mm 2

4 and thus the area of the concrete is A1 62,500 mm 2 1,134.115 mm 2

F1 can now be computed as: 1,134.115 mm 2 200 GPa F1 1 61,365.885 mm 2 29 GPa

61,365.885 mm 2

670 kN F1

670 kN 1.127457

and from Eq. (a), F2 has a value of F2 F1 670 kN ( 594.248 kN) 670 kN

594.258 kN

75.742 kN

Normal Stresses: The normal stresses in each axial member can now be calculated. In the concrete, the normal stress is: F1 594, 258 N 9.68 MPa 9.68 MPa (C) Ans. 1 A1 61,365.885 mm 2 and the normal stress in the reinforcing bars is F2 75,742 N 66.8 MPa 66.8 MPa (C) Ans. 2 A2 1,134.115 mm 2 (b) The shortening of the 1.5-m-long pier is equal to the deformation (i.e., contraction in this instance) of member (1) or member (2): F1L1 ( 594, 258 N)(1,500 mm) u 0.501 mm 0.501 mm Ans. 1 A1E1 (61,365.885 mm 2 )(29,000 N/mm 2 )

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5.29 The concrete [E = 29 GPa] pier shown in Fig. P5.29 is reinforced using four steel [E = 200 GPa] reinforcing rods. If the pier is subjected to an axial force of 670 kN, determine the required diameter D of each rod so that 20% of the total load is carried by the steel.

Fig. P5.29

Solution (a) Equilibrium: Consider a FBD of the pier, cut around the upper end. The concrete will be designated member (1) and the reinforcing steel bars will be designated member (2). Sum forces in the vertical direction to obtain: Fy F1 F2 670 kN 0 (a) Geometry of Deformations: 1

(b)

2

Force-Deformation Relationships: F1L1 F2 L2 1 2 A1E1 A2 E2

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1 L1 F2 L2 A1 E1 A2 E2 Solve the Equations: Solve Eq. (d) for F1: L AE L A E F1 F2 2 1 1 F2 2 1 1 A2 E2 L1 L1 A2 E2 and substitute this expression into Eq. (a) to determine F2: L A E L A E Fy F1 F2 F2 2 1 1 F2 F2 2 1 2 1 L1 A2 E2 L1 A2 E2

(d)

(e)

670 kN

Since L1 = L2, the term L1/L2 = 1 can be eliminated and Eq. (f) simplified to A E F2 1 1 1 670 kN A2 E2

(f)

(g)

From the problem statement, 20% of the total load must be carried by the reinforcing steel; therefore, F2 0.20( 670 kN) 134 kN Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The gross cross-sectional area of the pier is Agross = (250 mm)2 = 62,500 mm2; however, the reinforcing bars take up a portion of this area. Therefore, the areas of the concrete (1) and steel (2) are related by: Agross A1 A2 and thus, the area of the concrete can be expressed as A1 = Agross − A2. Equation (g) can now be restated and solved for the area of the reinforcing steel bars: 62,500 mm 2 A2 29 GPa ( 134 kN) 1 670 kN A2 200 GPa 62,500 mm 2 A2

A2

27.586207 A2

2,186.369 mm 2

Knowing the area required from four bars, the diameter of each bar can be computed: 2 (4 bars) d bar 4

2,186.369 mm 2

d bar

26.4 mm

Ans.

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5.30 A load of P = 100 kN is supported by a structure consisting of rigid bar ABC, two identical solid bronze [E = 100 GPa] rods, and a solid steel [E = 200 GPa] rod, as shown in Fig. P5.30. The bronze rods (1) each have a diameter of 20 mm and they are symmetrically positioned relative to the center rod (2) and the applied load P. Steel rod (2) has a diameter of 24 mm. All bars are unstressed before the load P is applied; however, there is a 3-mm clearance in the bolted connection at B. Determine: (a) the normal stresses in the bronze and steel rods. (b) the downward deflection of rigid bar ABC. Fig. P5.30

Solution (a) Equilibrium: By virtue of symmetry, the forces in the two bronze rods (1) are identical. Consider a FBD of the rigid bar. Sum forces in the vertical direction to obtain: Fy 2 F1 F2 P 0 (a) Geometry of Deformations: For this configuration, the deflections of joints A, B, and C are equal: v A vB vC (b) The pin connections at A and C are ideal; therefore, the deflection of joints A and C will cause an identical deformation of rods (1): vA (c) 1 The pin at B has a 3-mm clearance; thus, the deformation of rod (2) is related to rigid bar deflection vB by: vB 3 mm (d) 2 Substitute Eqs. (c) and (d) into Eq. (b) to obtain the geometry of deformation equation: 3 mm (e) 1 2 Force-Deformation Relationships: The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): F1L1 F2 L2 (f) 1 2 A1E1 A2 E2 Compatibility Equation: Substitute the force-deformation relationships (f) into the geometry of deformation relationship (e) to derive the compatibility equation: F1 L1 F2 L2 3 mm A1 E1 A2 E2

(g)

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Solve the Equations: Solve Eq. (g) for F1: F2 L2 AE L A E (3 mm)A1 E1 F1 3 mm 1 1 F2 2 1 1 A2 E2 L1 L1 A2 E2 L1 Substitute this expression into Eq. (a) L A E (3 mm)A1 E1 Fy 2 F1 F2 2 F2 2 1 1 F2 P L1 A2 E2 L1 and derive an expression for F2: L A E (3 mm)A1 E1 F2 2 2 1 1 1 P 2 L1 A2 E2 L1

(3 mm)A1 E1 L1 L A E 2 2 1 1 1 L1 A2 E2

P 2 F2

(h)

For this structure, P = 100 kN = 100,000 N, and the lengths, areas, and elastic moduli are given below: L1 3,000 mm L2 1,500 mm

A1

(20 mm)2

314.159265 mm2

A2

(24 mm)2

452.389342 mm2

4 4 E1 100,000 MPa E2 200,000 MPa Substitute these values into Eq. (h) and calculate F2 = 27,588.728 N. Backsubstitute into Eq. (a) to calculate F1 = 36,205.636 N. Normal Stresses: The normal stresses in each rod can now be calculated: F1 36, 205.636 N 115.2 MPa 1 A1 314.1593 mm 2

2

F2 A2

27,588.728 N 452.3893 mm 2

61.0 MPa

(b) The downward deflection of the rigid bar can be determined from the deformation of rods (1): F1L1 (36, 205.636 N)(3,000 mm) v A vB vC 3.46 mm 1 A1E1 (314.1593 mm 2 )(100,000 MPa)

Ans.

Ans.

Ans.

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5.31 Two steel [E = 30,000 ksi] pipes (1) and (2) are connected at flange B, as shown in Fig. P5.31. Pipe (1) has an outside diameter of 6.625 in. and a wall thickness of 0.28 in. Pipe (2) has an outside diameter of 4.00 in. and a wall thickness of 0.226 in. If the normal stress in each steel pipe must be limited to 18 ksi, determine: (a) the maximum downward load P that may be applied at flange B. (b) the deflection of flange B at the load determined in part (a).

Fig. P5.31

Solution (a) Pipe section properties: The pipe cross-sectional areas are: D1 6.625 in. d1 6.625 in. 2(0.28 in.) 6.0650 in. A1 D2 A2

4

(6.625 in.) 2 (6.0650 in.) 2

4.00 in. 4

d2

5.5814 in.2

4.00 in. 2(0.226 in.)

(4.00 in.) 2 (3.5480 in.) 2

3.5480 in.

2.6795 in.2

Equilibrium: Consider a FBD of flange B. Sum forces in the vertical direction to obtain: Fy F2 F1 P 0 (a) Geometry of Deformations: 0 1 2 Force-Deformation Relationships: F1L1 F2 L2 1 2 A1E1 A2 E2

(b)

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1 L1 F2 L2 0 A1 E1 A2 E2

(d)

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For this problem, it is convenient to express the compatibility equation in terms of the normal stress 1 and 2: 1 L1 2 L2 0 (e) E1 E2 Solve the Equations: Solve Eq. (e) for 1: L2 E1 (f) 1 2 L1 E2 The elastic moduli of the two pipes are equal; therefore, E1/E2 = 1. The normal stresses in pipes (1) and (2) are limited to 18 ksi. Assume that the normal stress in pipe (2) controls, meaning that 2 = 18 ksi. Calculate the corresponding stress in pipe (1): L2 E1 16 ft (18 ksi) 28.80 ksi 1 2 L1 E2 10 ft Since this stress magnitude is greater than 18 ksi, our assumption is proven incorrect. We now know that the normal stress in pipe (1) actually controls; thus, 1 = −18 ksi (negative by inspection). The normal stress in pipe (1) can be found from Eq. (f) L1 E2 10 ft ( 18 ksi) 11.250 ksi (f) 2 1 L2 E1 16 ft Now that the stresses are known, the allowable forces F1 and F2 can be computed: F1 ( 18 ksi)(5.5814 in.2 ) 100.4652 kips 1 A1

F2

2

A2

(11.250 ksi)(2.6795 in.2 ) 30.1444 kips

Substitute these values into Eq. (a) to obtain the allowable load P: Pmax F2 F1 30.1444 kips ( 100.4652 kips) 130.6 kips

(b) The deflection of flange B is equal to the deformation of pipe (2): (11.250 ksi)(16 ft)(12 in./ft) 2 L2 uB 0.0720 in. 2 E2 30,000 ksi

Ans.

Ans.

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5.32 A solid aluminum [E = 70 GPa] rod (1) is connected to a solid bronze [E = 100 GPa] rod at flange B, as shown in Fig P5.32. Aluminum rod (1) has an outside diameter of 35 mm and bronze rod (2) has an outside diameter of 20 mm. The normal stress in the aluminum rod must be limited to 160 MPa, and the normal stress in the bronze rod must be limited to 110 MPa. Determine: (a) the maximum downward load P that may be applied at flange B. (b) the deflection of flange B at the load determined in part (a).

Fig. P5.32

Solution (a) Rod section properties: The rod cross-sectional areas are: A1

4

(35 mm)2

962.1128 mm2

A2

4

(20 mm)2

314.1593 mm2

Equilibrium: Consider a FBD of flange B. Sum forces in the vertical direction to obtain: Fy F2 F1 P 0 (a) Geometry of Deformations: 0 1 2 Force-Deformation Relationships: F1L1 F2 L2 1 2 A1E1 A2 E2

(b)

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1 L1 F2 L2 0 (d) A1 E1 A2 E2 For this problem, it is convenient to express the compatibility equation in terms of the normal stress 1 and 2: 1 L1 2 L2 0 (e) E1 E2 Solve the Equations: Solve Eq. (e) for 1: L2 E1 (f) 1 2 L1 E2 The normal stress in aluminum rod (1) is limited to 160 MPa. We will assume that the aluminum rod controls, and then, we will calculate the corresponding stress in bronze rod (2): Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

L1 E2 175 mm 100 GPa (160 MPa) 117.647 MPa L2 E1 340 mm 70 GPa This stress magnitude is greater than the 110 MPa allowable stress for the bronze rod; therefore, this calculation shows that the stress in the bronze rod actually controls. If the stress in the bronze rod is limited to 110 MPa, the normal stress in aluminum rod (1) can be found from Eq. (f) L2 E1 340 mm 70 GPa (110 MPa) 149.600 MPa (f) 1 2 L1 E2 175 mm 100 GPa 2

1

Now that the stresses are known, the allowable forces F1 and F2 can be computed: F1 ( 149.600 MPa)(962.1128 mm 2 ) 143,932.1 N 143.9321 kN 1 A1

F2

2

A2

(110 MPa)(314.1593 mm2 )

34,557.5 N

34.5575 kN

Substitute these values into Eq. (a) to obtain the allowable load P: Pmax F2 F1 34.5575 kN ( 143.9321 kN) 178.5 kN

(b) The deflection of flange B is equal to the deformation of rod (2): (110 MPa)(340 mm) 2 L2 uB 0.374 mm 2 E2 100,000 MPa

Ans.

Ans.

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5.33 A pin-connected structure is supported and loaded as shown in Fig. P5.33. Member ABCD is rigid and is horizontal before the load P is applied. Bars (1) and (2) are both made from steel [E = 30,000 ksi] and both have a crosssectional area of 1.25 in.2. A concentrated load of P = 25 kips acts on the structure at D. Determine: (a) the normal stresses in both bars (1) and (2). (b) the downward deflection of point D on the rigid bar.

Fig. P5.33

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin A gives the best information for this situation: M A (66 in.)F1 (168 in.)F2 (212 in.)P 0 (a)

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vB vC (b) 66 in. 168 in. Since there are no gaps, clearances, or other misfits at pins B and C, the deformation of member (1) will equal the deflection of the rigid bar at B and the deformation of member (2) will equal the deflection of the rigid bar at C. Therefore, Eq. (b) can be rewritten in terms of the member deformations as: 1

66 in.

2

168 in.

(c)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): F1L1 F2 L2 (d) 1 2 A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to derive the compatibility equation: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

1 F1L1 66 in. A1E1

1 F2 L2 168 in. A2 E2

(e)

Solve the Equations Solve Eq. (e) for F1: 66 in. L2 A1E1 66 in. L2 A1 E1 F1 F2 F2 168 in. A2 E2 L1 168 in. L1 A2 E2 Substitute this expression into equilibrium equation (a) and solve for F2: M A (66 in.)F1 (168 in.)F2 (212 in.)P 0

(66 in.)F2

66 in. L2 A1 E1 168 in. L1 A2 E2

F2 (66 in.)

66 in. L2 A1 E1 168 in. L1 A2 E2

(168 in.)F2

(212 in.)P

(168 in.)

(212 in.)P

(f)

(g)

For this structure, P = 20 kips, and the lengths, areas, and elastic moduli are given below: L1 80 in. L2 120 in. A1 1.25 in.2

A2 1.25 in.2

E1 30, 000 ksi E2 30, 000 ksi Substitute these values into Eq. (g) and calculate F2 = 25.617124 kips. Backsubstitute into Eq. (f) to calculate F1 = 15.095805 kips.

Normal Stresses The normal stresses in each axial member can now be calculated: F1 15.095805 kips 12.08 ksi 1 A1 1.25 in.2 F2 25.617124 kips 20.5 ksi 2 A2 1.25 in.2

Ans. Ans.

Deflections of the rigid bar Calculate the deformation of one of the axial members, say member (1): F1L1 (15.095805 kips)(80 in.) 0.032204 in. (h) 1 A1E1 (1.25 in.2 )(30,000 ksi) Since there are no gaps at pin B, the rigid bar deflection at B is equal to the deformation of member (1); therefore, vB = 1 = 0.032204 in. (downward). From similar triangles, the deflection of the rigid bar at D is related to vB by: vB vD (i) 66 in. 212 in. From Eq. (i), the deflection of the rigid bar at D is: 212 in. 212 in. vD vB (0.032204 in.) 0.1034 in. 66 in. 66 in.

Ans.

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5.34 A pin-connected structure is supported and loaded as shown in Fig. P5.34. Member ABCD is rigid and is horizontal before the load P is applied. Bars (1) and (2) are both made from steel [E = 30,000 ksi] and both have a crosssectional area of 1.25 in.2. If the normal stress in each steel bar must be limited to 18 ksi, determine the maximum load P that may be applied to the rigid bar.

Fig. P5.34

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin A gives the best information for this situation: M A (66 in.)F1 (168 in.)F2 (212 in.)P 0 (a)

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vB vC (b) 66 in. 168 in. Since there are no gaps, clearances, or other misfits at pins B and C, the deformation of member (1) will equal the deflection of the rigid bar at B and the deformation of member (2) will equal the deflection of the rigid bar at C. Therefore, Eq. (b) can be rewritten in terms of the member deformations as: 1

66 in.

2

168 in.

(c)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): F1L1 F2 L2 (d) 1 2 A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to derive the compatibility equation:

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1 F1L1 66 in. A1E1

1 F2 L2 168 in. A2 E2

(e)

Solve the Equations Since allowable stresses are specified, it is convenient to express Eq. (e) in terms of stress: 66 in. 2 L2 1 L1 E1 168 in. E2 and solve for 2: 168 in. L1 E2 2 1 66 in. L2 E1 Since both bars have the same elastic modulus, E1/E2 = 1. Substitute 1 = 18 ksi and solve for 168 in. L1 E2 168 in. 80 in. (18 ksi) 30.54546 ksi 18 ksi N.G. 2 1 66 in. L2 E1 66 in. 120 in. It is now evident that the stress in bar (2) controls. Substitute 2 = 18 ksi and solve for 66 in. L2 E1 66 in. 120 in. (18 ksi) 10.60714 ksi 18 ksi OK 1 2 168 in. L1 E2 168 in. 80 in.

(f)

2:

1:

Now that the stresses are known, the allowable forces F1 and F2 can be computed: F1 (10.60714 ksi)(1.25 in.2 ) 13.25893 kips 1 A1

F2

2

A2

(18 ksi)(1.25 in.2 )

22.5 kips

Substitute these values into Eq. (a) to obtain the allowable load P: M A (66 in.)(16.875 kips) (168 in.)(22.5 kips) (212 in.)P Pmax

21.95797 kips

22.0 kips

0

Ans.

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5.35 The pin-connected structure shown in Fig. P5.35 consists of a rigid beam ABCD and two supporting bars. Bar (1) is an aluminum alloy [E = 70 GPa] with a cross-sectional area of A1 = 800 mm2. Bar (2) is a bronze alloy [E = 100 GPa] with a cross-sectional area of A2 = 500 mm2. The normal stress in the aluminum bar must be limited to 70 MPa, and the normal stress in the bronze rod must be limited to 90 MPa. Determine: (a) the maximum downward load P that may be applied at B. (b) the deflection of the rigid beam at B. Fig. P5.35

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin D gives the best information for this situation: MD (3.5 m)F1 (1.2 m)F2 (2.7 m)P 0 (a)

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vC vA 3.5 m 1.2 m

(b)

Since there are no gaps, clearances, or other misfits at pins A and C, the deformation of member (1) will equal the deflection of the rigid bar at A and the deformation of member (2) will equal the deflection of the rigid bar at C. Therefore, Eq. (b) can be rewritten in terms of the member deformations as: 1

3.5 m

2

1.2 m

(c)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): F1L1 F2 L2 (d) 1 2 A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to derive the compatibility equation: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

1 F1 L1 3.5 m A1 E1

1 F2 L2 1.2 m A2 E2

(e)

Solve the Equations Since allowable stresses are specified, it is convenient to express Eq. (e) in terms of stress: L 3.5 m 2 L2 1 L1 2.9167 2 2 E1 1.2 m E2 E2 and solve for 1: L E 2.9167 2 2 1 1 L1 E2 Substitute 2 = 90 MPa and solve for 1: L E 3.3 m 70 GPa 2.9167 2 2 1 2.9167(90 MPa) 209.1 MPa 70 MPa N.G. 1 L1 E2 2.9 m 100 GPa It is now evident that the stress in bar (1) controls. Substitute 1 = 70 MPa and solve for 2: L1 E2 1 1 2.9 m 100 GPa (70 MPa) 30.13 MPa 90 MPa O.K. 2 1 2.9167 L2 E1 2.9167 3.3 m 70 GPa

(f)

Now that the stresses are known, the allowable forces F1 and F2 can be computed: F1 (70 MPa)(800 mm2 ) 56, 000 N 56.00 kN 1 A1

F2 (30.13 MPa)(500 mm2 ) 15, 065 N 15.065 kN 2 A2 Substitute these values into Eq. (a) to obtain the allowable load P: MD (3.5 m)(56.00 kN) (1.2 m)(15.065 kN) (2.7 m)P 0 Pmax

79.288 kN

79.3 kN

(b) Deflections of the rigid bar Calculate the deformation of one of the axial members, say member (1): F1L1 (56,000 N)(2,900 mm) 2.9000 mm 1 A1E1 (800 mm 2 )(70,000 MPa) Since there are no gaps at pin A, the rigid bar deflection at A is equal to the deformation of member (1); therefore, vA = 1 = 2.9000 mm (downward). From similar triangles, the deflection of the rigid bar at B is related to vA by: vA vB (h) 3.5 m 2.7 m From Eq. (h), the deflection of the rigid bar at B is: 2.7 m 2.7 m vB vA (2.9000 mm) 2.24 mm 3.5 m 3.5 m

Ans.

(g)

Ans.

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5.36 The pin-connected structure shown in Fig. P5.36 consists of a rigid beam ABCD and two supporting bars. Bar (1) is an aluminum alloy [E = 70 GPa] with a cross-sectional area of A1 = 800 mm2. Bar (2) is a bronze alloy [E = 100 GPa] with a cross-sectional area of A2 = 500 mm2. All bars are unstressed before the load P is applied; however, there is a 3-mm clearance in the pin connection at A. If a load of P = 54 kN is applied at B, determine: (a) the normal stresses in both bars (1) and (2). (b) the normal strains in bars (1) and (2). (c) determine the downward deflection of point A on the rigid bar.

Fig. P5.36

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin D gives the best information for this situation: MD (3.5 m)F1 (1.2 m)F2 (2.7 m)P 0 (a)

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vC vA (b) 3.5 m 1.2 m Since there is no clearance at pin C, the deformation of member (2) will equal the deflection of the rigid bar at C. However, there is a 3-mm clearance at A. Consequently, not all of the rigid beam deflection at A will go toward elongating bar (1). The relationship between rigid beam deflection and axial member deformation can be expressed: vA 3 mm 1 Equation (c) can be rewritten in terms of the member deformations as: 3 mm 1 2 3.5 m 1.2 m

(c) (d)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): F1L1 F2 L2 (e) 1 2 A1E1 A2 E2

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Compatibility Equation Substitute the force-deformation relationships (e) into the geometry of deformation relationship (d) to derive the compatibility equation: F1 L1 FL 3.5 m F2 L2 3 mm 2.9167 2 2 (f) A1 E1 1.2 m A2 E2 A2 E2 Solve the Equations Solve Eq. (f) for F2: A2 E2 F1 L1 F2 2.9167 L2 A1 E1

3 mm

F1

L1 A2 E2 2.9167 L2 A1 E1

(3 mm)A2 E2 2.9167 L2

and substitute into Eq. (a) to solve for F1: (3.5 m)F1 (1.2 m)F2 (3.5 m)F1 (1.2 m) F1

L1 A2 E2 2.9167 L2 A1 E1

(3.5 m)F1 (1.2 m)F1

(3 mm)A2 E2 2.9167 L2 L1 A2 E2 2.9167 L2 A1 E1

(2.7 m)(54 kN) (2.7 m)(54 kN) (2.7 m)(54 kN) (1.2 m)

(3 mm)A2 E2 2.9167 L2

(3 mm)A2 E2 2.9167 L2 L1 A2 E2 (3.5 m) (1.2 m) 2.9167 L2 A1 E1

(2.7 m)(54 kN) (1.2 m) F1

The value of F1 is thus calculated as:

(3 mm)(500 mm 2 )(100, 000 N/mm 2 ) 2.9167(3,300 mm) 2.9 m 500 mm 2 100 GPa 3.5 m (1.2 m) 2.9167(3.3 m) 800 mm 2 70 GPa 33, 247.4 N 33.2474 kN (2.7 m)(54, 000 N) (1.2 m)

F1

and backsubstituting into Eq. (a) gives F2: (2.7 m)(54 kN) (3.5 m)F1 F2 1.2 m (2.7 m)(54 kN) (3.5 m)(33.2474 kN) 1.2 m 24.5285 kN (a) Normal stresses: F1 33,247.4 N 1 A1 800 mm 2 2

F2 A2

24,528.5 N 500 mm 2

41.6 MPa (T)

Ans.

49.1 MPa (T)

Ans.

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(b) Normal strains: 41.5592 MPa 1 593.7 10 6 mm/mm 1 E1 70,000 MPa 49.0570 MPa 2 490.6 10 6 mm/mm 2 E2 100,000 MPa

594 με

Ans.

491 με

Ans.

(c) Deflections of the rigid bar Calculate the deformation of member (1): F1L1 (33, 247.4 N)(2,900 mm) 1.721739 mm 1 A1E1 (800 mm 2 )(70,000 MPa) The relationship between the rigid bar deflection at A and the deformation of member (1) was expressed in Eq. (c). Therefore, the rigid bar deflection at A is: vA 3 mm 1 1.721739 mm 3 mm 4.72 mm

Ans.

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5.37 The pin-connected structure shown in Fig. P5.37 consists of a rigid bar ABCD and two axial members. Bar (1) is bronze [E = 100 GPa] with a cross-sectional area of A1 = 620 mm2. Bar (2) is an aluminum alloy [E = 70 GPa] with a crosssectional area of A2 = 340 mm2. A concentrated load of P = 75 kN acts on the structure at D. Determine: (a) the normal stresses in both bars (1) and (2). (b) the downward deflection of point D on the rigid bar.

Fig. P5.37

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin C gives the best information for this situation: M C (825 mm)F1 (325 mm)F2 (1,100 mm)P

0

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vA vB vD 825 mm 325 mm 1,100 mm

(a)

(b)

Since there are no gaps or clearances at either pin A or pin B, the deformations of members (1) and (2) will equal the deflections of the rigid bar at A and B, respectively. 1

825 mm

2

325 mm

(c)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): F1L1 F2 L2 (d) 1 2 A1E1 A2 E2

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Compatibility Equation Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to derive the compatibility equation: F1L1 825 mm F2 L2 FL 2.538462 2 2 (e) A1E1 325 mm A2 E2 A2 E2 Solve the Equations Solve Eq. (e) for F1: L2 A1 E1 L1 A2 E2 and substitute into Eq. (a) to solve for F2: F1

2.538462 F2

(f) (825 mm)F1

(325 mm)F2

(1,100 mm)P

L2 A1 E1 L1 A2 E2

(325 mm)F2

(1,100 mm)P

(825 mm)(2.538462) F2

540 mm 620 mm 2 100 GPa F2 (825 mm)(2.538462) 900 mm 340 mm 2 70 GPa

325 mm

(1,100 mm)(75 kN) F2

22.9273 kN

Backsubstitute this result into Eq. (f) to compute F1: 540 mm 620 mm 2 100 GPa F1 (2.538462) (22.9273 kN) 900 mm 340 mm 2 70 GPa

90.9680 kN (a) Normal stresses: F1 90,968.0 N 1 A1 620 mm 2 2

F2 A2

22,927.3 N 340 mm 2

146.7 MPa (T)

Ans.

67.4 MPa (T)

Ans.

(b) Deflections of the rigid bar Calculate the deformation of member (1): F1L1 (90,968.0 N)(900 mm) 1 A1E1 (620 mm 2 )(100,000 MPa)

1.3205 mm

Since the pin at A is assumed to have a perfect connection, vA = 1,100 mm vD vA 1.333333(1.3205 mm) 1.761 mm 825 mm

1

= 1.3205 mm. From Eq. (b), Ans.

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5.38 The pin-connected structure shown in Fig. P5.38 consists of a rigid bar ABCD and two axial members. Bar (1) is bronze [E = 100 GPa] with a cross-sectional area of A1 = 620 mm2. Bar (2) is an aluminum alloy [E = 70 GPa] with a crosssectional area of A2 = 340 mm2. All bars are unstressed before the load P is applied; however, there is a 3-mm clearance in the pin connection at A. If a concentrated load of P = 90 kN acts on the structure at D, determine: (a) the normal stresses in both bars (1) and (2). (b) the normal strains in bars (1) and (2). (c) the downward deflection of point D on the rigid bar. Fig. P5.38

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin C gives the best information for this situation: M C (825 mm)F1 (325 mm)F2 (1,100 mm)P

0

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vA vB vD 825 mm 325 mm 1,100 mm

(a)

(b)

Since there is no clearance at pin B, the deformation of member (2) will equal the deflection of the rigid bar at B. However, there is a 3-mm clearance in the connection at A. Consequently, not all of the rigid bar deflection at A will go toward elongating bar (1). The relationship between rigid bar deflection and axial member deformation can be expressed: vA 3 mm 1 Equation (b) can be rewritten in terms of the member deformations as: 3 mm 1 2 825 mm 325 mm

(c) (d)

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Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): F1L1 F2 L2 (e) 1 2 A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (e) into the geometry of deformation relationship (d) to derive the compatibility equation: F1L1 825 mm F2 L2 FL 3 mm 2.538462 2 2 (f) A1E1 325 mm A2 E2 A2 E2 Solve the Equations Solve Eq. (f) for F2: A2 E2 F1L1 L1 A2 E2 (3 mm)A2 E2 F2 3 mm F1 2.538462 L2 A1E1 2.538462 L2 A1 E1 2.538462 L2 and substitute into Eq. (a) to solve for F1: (825 mm)F1 (325 mm)F2 (1,100 mm)P (825 mm)F1

(325 mm) F1

L1 A2 E2 2.538462 L2 A1 E1

(825 mm)F1

(325 mm)F1

(3 mm)A2 E2 2.538462 L2

L1 A2 E2 2.538462 L2 A1 E1

(1,100 mm)P (1,100 mm)P (325 mm)

(3 mm)A2 E2 2.538462 L2

(3 mm)A2 E2 2.538462 L2 L1 A2 E2 825 mm (325 mm) 2.538462 L2 A1 E1 (1,100 mm)P (325 mm)

F1

The value of F1 is thus calculated as: (3 mm)(340 mm 2 )(70,000 N/mm 2 ) 2.538462(540 mm) 900 mm 340 mm 2 70,000 N/mm 2 825 mm (325 mm) 2.538462(540 mm) 620 mm 2 100,000 N/mm 2 90, 495.6 N 90.4956 kN (1,100 mm)(90,000 N) (325 mm)

F1

and backsubstituting into Eq. (a) gives F2: (1,100 mm)(90 kN) (825 mm)F1 F2 325 mm (1,100 mm)(90 kN) (825 mm)(90.4956 kN) 325 mm 74.8957 kN

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(a) Normal stresses: F1 90, 495.6 N 1 A1 620 mm 2 F2 74,895.7 N 2 A2 340 mm 2

146.0 MPa (T)

Ans.

220 MPa (T)

Ans.

(b) Normal strains: F1 90, 495.6 N 1 A1E1 (620 mm 2 )(100,000 N/mm 2 ) 2

F2 A2 E2

74,895.7 N (340 mm 2 )(70,000 N/mm 2 )

1, 459.6 10

6

mm/mm

1, 460 με

Ans.

3,146.9 10

6

mm/mm

3,150 με

Ans.

(c) Deflections of the rigid bar Calculate the deformation of member (2): F2 L2 (74,895.7 N)(540 mm) 1.6993 mm 2 A2 E2 (340 mm 2 )(70,000 MPa) Since the pin at B is assumed to have a perfect connection, vB = 1,100 mm vD vB 3.384615(1.6993 mm) 5.75 mm 325 mm

2

= 1.6993 mm. From Eq. (b), Ans.

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5.39 The assembly shown in Fig. P5.39 consists of a solid aluminum alloy [E = 70 GPa] post (2) surrounded by a bronze [E = 100 GPa] tube (1). Before the load P is applied, there is a clearance of 2 mm between the post and the tube. The yield stress for the aluminum post is 260 MPa and the yield stress for the bronze tube is 340 MPa. Determine: (a) the maximum load P that may be applied to the assembly without causing yielding of either the post or the tube. (b) the downward displacement of rigid cap B. (c) the normal strain in the bronze tube.

Fig. P5.39

Solution Section properties: R1 35 mm r1

29 mm

R2 15 mm

A1 A2

(35 mm)2 (29 mm)2 (15 mm) 2

1, 206.372 mm2

706.858 mm 2

Equilibrium: Consider a FBD around rigid cap B after the gap has been closed. Sum forces in the vertical direction to obtain: Fy F1 F2 P 0 (a) Geometry of Deformations: 2 mm 2 1 Force-Deformation Relationships: F1L1 F2 L2 1 2 A1E1 A2 E2

(b)

(c)

Compatibility Equation Substitute the force-deformation relationships (c) into the geometry of deformation relationship (b) to derive the compatibility equation: F2 L2 F1 L1 2 mm (d) A2 E2 A1 E1 Solve the Equations Since allowable stresses are specified, it is convenient to express Eq. (d) in terms of stress: 2 L2 1 L1 2 mm E2 E1 and solve for 2:

(e)

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2

1

L1 E2 L2 E1

(2 mm)

E2 L2

600 mm 70, 000 MPa 70, 000 MPa (2 mm) 602 mm 100, 000 MPa 602 mm Substitute 1 = 340 MPa and solve for 2: 70, 000 MPa 0.697674(340 MPa) (2 mm) 469.8 MPa 260 MPa N.G. 2 602 mm It is now evident that the stress in post (2) controls. Substitute 2 = 260 MPa and solve for 1: 70, 000 MPa L2 E1 (2 mm) 1 2 602 mm L1 E2 1

70, 000 MPa 602 mm 100, 000 MPa 602 mm 600 mm 70, 000 MPa 340 MPa OK

260 MPa (2 mm) 39.333 MPa

(a) Maximum load: Now that the stresses are known, the allowable forces F1 and F2 can be computed: F1 (39.333 MPa)(1,206.372 mm2 ) 47, 450 N 47.450 kN 1 A1

F2 (260 MPa)(706.858 mm2 ) 183, 783 N 183.783 kN 2 A2 Substitute these values into Eq. (a) to obtain the allowable load P. By inspection, the forces in the post and the tube must be compression; therefore: Fy F1 F2 P 0 Pmax

( 47.450 kN) ( 183.783 kN)

231 kN

Ans.

(b) Displacement of cap B: The contraction of tube (1) is: F1L1 ( 47, 451 N)(600 mm) 0.2360 mm 1 A1E1 (1,206.372 mm 2 )(100,000 N/mm 2 ) The displacement of cap B is 2 mm greater:

uB

0.2360 mm 2 mm

2.2360 mm

2.24 mm

(c) Normal strain in tube (1): The strain in bronze tube (1) is: 39.333 MPa 1 393 με 1 E1 100,000 MPa

Ans.

Ans.

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5.40 A 4.5-m-long aluminum tube (1) is to be connected to a 2.4-mlong bronze pipe (2) at B. When put in place, however, a gap of 8 mm exists between the two members, as shown in Fig P5.40. Aluminum tube (1) has an elastic modulus of 70 GPa and a crosssectional area of 2,000 mm2. Bronze pipe (2) has an elastic modulus of 100 GPa and a cross-sectional area of 3,600 mm2. If bolts are inserted in the flanges and tightened so that the gap at B is closed, determine: (a) the normal stresses produced in each of the members. (b) the final position of flange B with respect to support A.

Fig. P5.40

Solution Equilibrium: Consider a FBD of flange B after the bolts have been tightened and the gap at B has been closed. Sum forces in the vertical direction to obtain: Fy F1 F2 0 F1 F2 (a) Geometry of Deformations: 8 mm 1 2

(b)

Force-Deformation Relationships: F1L1 F2 L2 1 2 A1E1 A2 E2

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1L1 F2 L2 8 mm A1E1 A2 E2

(d)

Solve the Equations: Substitute Eq. (a) into Eq. (d) and solve for F1: L1 L2 F1 8 mm A1E1 A2 E2

F1

8 mm 4,500 mm (2,000 mm 2 )(70,000 N/mm 2 ) 206,135.0 N

(a) Normal stresses: F1 206,135.0 N 1 A1 2,000 mm 2 2

F2 A2

206,135.0 N 3,600 mm 2

2,400 mm (3,600 mm 2 )(100,000 N/mm 2 ) (f)

103.0675 MPa

103.1 MPa (T)

Ans.

57.2597 MPa

57.3 MPa (T)

Ans.

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(b) The deflection of flange B is equal to the deformation of bronze pipe (2): F2 L2 (206,135.0 N)(2,400 mm) 1.374 mm 2 A2 E2 (3,600 mm 2 )(100,000 N/mm 2 )

Ans.

In its final position, flange B is located 2,401.374 mm away from support A.

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5.41 The assembly shown in Fig. P5.41 consists of a steel [E1 = 30,000 ksi; A1 = 1.25 in.2] rod (1), a rigid bearing plate B that is securely fastened to rod (1), and a bronze [E2 = 15,000 ksi; A2 = 3.75 in.2] post (2). The yield strengths of the steel and bronze are 62 ksi and 75 ksi, respectively. A clearance of 0.125 in. exists between the bearing plate B and bronze post (2) before the assembly is loaded. After a load of P = 65 kips is applied to the bearing plate, determine: (a) the normal stresses in bars (1) and (2). (b) the factors of safety with respect to yield for each of the members. (c) the vertical displacement of bearing plate B.

Fig. P5.41

Solution Deformation in rod (1) alone: Check to see if the bearing plate attached to rod (1) will contact the bronze post for the 65-kip load. F1L1 (65 kips)(14 ft)(12 in./ft) 0.2912 in. 0.125 in. gap contact will occur 1 A1E1 (1.25 in.2 )(30,000 ksi) This calculation proves that the bearing plate attached to rod (1) will contact the bronze post when the 65-kip load is applied; therefore, this structure must be analyzed as a statically indeterminate structure. Equilibrium: Consider a FBD of flange B after the gap at B has been closed. Sum forces in the vertical direction to obtain: Fy F1 F2 P 0 (a) Geometry of Deformations: 0.125 in. 1 2 Force-Deformation Relationships: F1L1 F2 L2 1 2 A1E1 A2 E2

(b)

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1L1 F2 L2 0.125 in. A1E1 A2 E2

(d)

Solve the Equations: Solve Eq. (d) for F1: F2 L2 A1E1 AE L A E F1 0.125 in. (0.125 in.) 1 1 F2 2 1 1 A2 E2 L1 L1 L1 A2 E2 and substitute the resulting expression into Eq. (a) to determine an expression for F2:

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(0.125 in.)

A1E1 L1

F2

L2 A1 E1 L1 A2 E2

F2

F2 1

65 kips L2 A1 E1 L1 A2 E2

0 (0.125 in.)

A1E1 L1

65 kips

A1E1 65 kips L1 L2 A1 E1 1 L1 A2 E2

(0.125 in.) F2

The axial force in post (2) is thus: (1.25 in.2 )(30,000 ksi) (0.125 in.) 65 kips (168 in.) F2 32.4609 kips 36 in. 1.25 in.2 30,000 ksi 1 168 in. 3.75 in.2 15,000 ksi and the axial force in rod (1) is: F1 F2 P 32.4609 kips 65 kips 32.5391 kips (a) Normal stresses: F1 32.5391 kips 1 A1 1.25 in.2 2

F2 A2

32.4609 kips 3.75 in.2

(b) Factors of safety: 62 ksi FS1 26.0313 ksi

2.38

26.0313 ksi

26.0 ksi (T)

8.6563 ksi

FS2

Ans.

8.66 ksi (C)

75 ksi 8.6563 ksi

Ans.

8.66

(c) Displacement of plate B: The displacement of plate B is equal to the deformation of rod (1). F1L1 (32.5391 kips)(168 in.) 0.145775 in. 1 A1E1 (1.25 in.2 )(30,000 ksi) uB

Ans.

Ans.

0.1458 in.

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5.42 A hollow steel [E = 30,000 ksi] tube (1) with an outside diameter of 3.50 in. and a wall thickness of 0.216 in. is fastened to a solid 2-in.diameter aluminum [E = 10,000 ksi] rod. The assembly is attached to unyielding supports at the left and right ends and is loaded as shown in Fig. P5.42. Determine: (a) the stresses in all parts of the axial structure. (b) the deflections of joints B and C.

Fig. P5.42

Solution Section properties: The steel tube cross-sectional area is: D1 3.50 in. d1 3.50 in. 2(0.216 in.) 3.068 in. (3.50 in.) 2 (3.068 in.) 2 2.2285 in.2 4 and the aluminum rod has a cross-sectional area of: D2 D3 2.00 in. A1

A2

A3

4

(2.00 in.) 2

3.1416 in.2

Equilibrium: Consider a FBD of flange B. Sum forces in the horizontal direction to obtain: Fx F1 F2 34 kips 0 (a)

Consider a FBD of flange C. Sum forces in the horizontal direction to obtain: Fx F2 F3 26 kips 0 (b) Geometry of Deformations: 0 1 2 3 Force-Deformation Relationships: F1L1 F2 L2 1 2 3 A1E1 A2 E2

(c)

F3 L3 A3 E3

(d)

Compatibility Equation: Substitute Eqs. (d) into Eq. (c) to derive the compatibility equation: F1 L1 F2 L2 F3 L3 0 A1 E1 A2 E2 A3 E3

(e)

Solve the Equations: Solve Eq. (a) for F1: F1 F2 34 kips

(f)

Solve Eq. (b) for F3: F3 F2 26 kips

(g)

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Substitute Eqs. (f) and (g) into compatibility equation (e): ( F2 34 kips)L1 F2 L2 ( F2 26 kips)L3 0 A1 E1 A2 E2 A3 E3 and expand terms: F2 L1 (34 kips)L1 F2 L2 F2 L3 (26 kips)L3 0 A1 E1 A1 E1 A2 E2 A3 E3 A3 E3 Regroup terms: L3 L1 L2 (34 kips)L1 F2 A1E1 A2 E2 A3 E3 A1E1 and solve for F2: (34 kips)L1 (26 kips)L3 A1 E1 A3 E3 F2 L3 L1 L2 A1 E1 A2 E2 A3 E3

(26 kips)L3 A3 E3

(34 kips)(48 in.) (26 kips)(60 in.) 2 (2.2285 in. )(30, 000 ksi) (3.1416 in.2 )(10, 000 ksi) 48 in. 60 in. 60 in. 2 2 (2.2285 in. )(30, 000 ksi) (3.1416 in. )(10, 000 ksi) (3.1416 in.2 )(10, 000 ksi) 16.3227 kips

Backsubstitute into Eqs. (f) and (g) to obtain F1 and F3: F1 F2 34 kips 16.3227 kips 34 kips 17.6773 kips

F3

F2 26 kips 16.3227 kips 26 kips

9.6773 kips

(a) Normal stresses: The normal stresses in each axial member can now be calculated: F1 17.6773 kips 7.9325 ksi 7.93 ksi (C) 1 A1 2.2285 in.2

2

F2 A2

16.3227 kips 3.1416 in.2

3

F3 A3

9.6773 kips 3.1416 in.2

5.1957 ksi

3.0834 ksi

5.20 ksi (T)

3.08 ksi (C)

Ans.

Ans.

Ans.

(b) Joint deflections: The deflection of flange B is equal to the deformation (i.e., contraction in this instance) of member (1): F1L1 ( 17.6773 kips)(48 in.) uB 0.012692 in. 0.01269 in. Ans. 1 A1E1 (2.2285 in.2 )(30,000 ksi) The deflection of flange C is equal to the sum of the deformations in members (1) and (2). The deformation of member (2) is:

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2

F2 L2 A2 E2

(16.3227 kips)(60 in.) (3.1416 in.2 )(10,000 ksi)

0.031174 in.

And thus, the deflection of flange C is: uC 0.012692 in. 0.031174 in. 0.018482 in. 1 2

0.01848 in.

Ans.

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5.43 Rigid bar ABCD in Fig. P5.43 is supported by a pin connection at A and by two axial bars (1) and (2). Bar (1) is a 30-in.-long bronze [E = 15,000 ksi] bar with a cross-sectional area of 1.25 in.2. Bar (2) is a 40-in.-long aluminum alloy [E = 10,000 ksi] bar with a cross-sectional area of 2.00 in.2. Both bars are unstressed before the load P is applied. If a concentrated load of P = 27 kips is applied to the rigid bar at D, determine: (a) the normal stresses in bars (1) and (2). (b) the deflection of the rigid bar at point D.

Fig. P5.43

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin A gives the best information for this situation: MA (36 in.)F1 (84 in.)F2 (98 in.)P 0 (a)

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vC vB vD (b) 36 in. 84 in. 98 in. There are no gaps, clearances, or other misfits at the pins in this structure. The vertical deflection of the rigid bar at C will produce deformation in member (2); however, and the vertical deflection of the rigid bar at B will create contraction in member (1) (see the discussion under the heading Structures with a rotating rigid bar in the text and examine Fig. 5.11). Therefore, Eq. (b) can be rewritten in terms of the member deformations as: 1

36 in.

2

84 in.

(c)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): F1L1 F2 L2 (d) 1 2 A1E1 A2 E2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Compatibility Equation Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to derive the compatibility equation: 1 F1 L1 1 F2 L2 (e) 36 in. A1 E1 84 in. A2 E2 Solve the Equations Solve Eq. (e) for F1: 36 in. L2 A1 E1 36 in. L2 A1 E1 F1 F2 F2 84 in. A2 E2 L1 84 in. L1 A2 E2 Substitute this expression into equilibrium equation (a) and solve for F2: MA (36 in.)F1 (84 in.)F2 (98 in.)P 0

(36 in.)F2

36 in. L2 A1 E1 84 in. L1 A2 E2

F2 (36 in.)

3 L2 A1 E1 7 L1 A2 E2

(84 in.)F2

(84 in.)

(f)

(98 in.)P

(98 in.)P

(g)

For this structure, P = 27 kips, and the lengths, areas, and elastic moduli are given below: L1 30 in. L2 40 in. A1 1.25 in.2

A2

2.00 in.2

E1 15, 000 ksi E2 10, 000 ksi Substitute these values into Eq. (g) and calculate F2 = 25.6183 kips. Backsubstitute into Eq. (f) to calculate F1 = −13.7241 kips.

(a) Normal Stresses The normal stresses in each axial member can now be calculated: F1 13.7241 kips 10.98 ksi (C) 1 A1 1.25 in.2 2

F2 A2

25.6183 kips 2.00 in.2

12.81 ksi (T)

Ans. Ans.

(b) Deflections of the rigid bar Calculate the deformation of one of the axial members, say member (2): F2 L2 (25.6183 kips)(40 in.) 0.051237 in. (h) 2 A2 E2 (2.00 in.2 )(10,000 ksi) Since there are no gaps at pin C, the rigid bar deflection at C is equal to the deformation of member (2); therefore, vC = 2 = 0.051237 in. (downward). From similar triangles [Eq. (b)], the deflection of the rigid bar at D is therefore: 98 in. 98 in. vD vC (0.051237 in.) 0.0598 in. Ans. 84 in. 84 in.

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5.44 A 25-mm-diameter by 3.5-m-long steel rod (1) is stress free after being attached to rigid supports, as shown in Fig. P5.44. At A, a 16-mm-diameter bolt is used to connect the rod to the support. Determine the normal stress in steel rod (1) and the shear stress in bolt A after the temperature drops 60°C. Use E = 200 GPa and  = 11.9 × 10−6/°C. Fig. P5.44

Solution Section properties: For the 25-mm-diameter rod, the cross-sectional area is: A1 



4

(25 mm) 2  490.873852 mm2

Force-Temperature-Deformation Relationship The relationship between internal force, temperature change, and deformation of an axial member is: FL 1  1 1  1T L1 A1E1 Since the rod is attached to rigid supports,  1 = 0. F1 L1  1T L1  0 A1 E1 Thus, the force produced in the rod by the temperature drop is: AE F1  1T L1 1 1  1T A1E1 L1  (11.9  106 / C)(  60C)(490.873852 mm 2 )(200,000 N/mm 2 )  70,096.786 N

The normal stress in the steel rod is: 70,096.786 N 1   142.8 MPa (T) 490.873852 mm 2

Ans.

The 16-mm-diameter bolt has a cross-sectional area of: Abolt 



(16 mm) 2  201.061930 mm 2

4 Since the bolt is loaded in double shear, the shear stress in the bolt is 70,096.786 N  bolt   174.3 MPa 2(201.061930 mm 2 )

Ans.

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5.45 A 0.875-in.-diameter by 15-ft-long steel rod (1) is stress free after being attached to rigid supports. A clevis-and-bolt connection, as shown in Fig. P5.45, connects the rod with the support at A. The normal stress in the steel rod must be limited to 18 ksi, and the shear stress in the bolt must be limited to 42 ksi. Assume E = 29,000 ksi and  = 6.6 × 10−6/°F and determine: (a) the temperature decrease that can be safely accommodated by rod (1) based on the allowable normal stress. (b) the minimum required diameter for the bolt at A using the temperature decrease found in part (a). Fig. P5.45

Solution Section properties: For the 0.875-in.-diameter rod, the cross-sectional area is:  A1  (0.875 in.) 2  0.601320 in.2 4 Force-Temperature-Deformation Relationship The relationship between internal force, temperature change, and deformation of an axial member is: FL 1  1 1  1T L1 A1E1 Since the rod is attached to rigid supports, 1 = 0. F1 L1  1T L1  0 A1 E1 which can also be expressed in terms of the rod normal stress: L  1 1  1T L1  0 E1 Solve for T corresponding to a 24 ksi normal stress in the steel rod: L 1  T   1 1  1 E1 1L1 1E1



18 ksi  94.0F (6.6  10 / F)(29,000 ksi) 6

Ans.

The normal force in the steel rod is: F1  (18 ksi)(0.601320 in.2 )  10.823768 kips If the allowable shear stress in the bolt is 42 ksi, the minimum diameter required for the double shear bolt is   2  10.823768 kips 2  d bolt   0.257709 in.2  42 ksi 4   d bolt  0.405 in.

Ans.

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5.46 A steel [E = 29,000 ksi and  = 6.6×10-6/°F] rod containing a turnbuckle has its ends attached to rigid walls. During the summer when the temperature is 82°F, the turnbuckle is tightened to produce a stress in the rod of 5 ksi. Determine the stress in the rod in the winter when the temperature is 10°F.

Solution The normal strain in the rod can be expressed as:





  T E Since the rod is attached to rigid walls, the rod strain after the temperature change is  = 0.





  T  0 E The change in temperature between the summer and winter is T  Twinter  Tsummer  10F  82F  72F Solve for the stress increase created by the 72°F drop in temperature.    T E



  (6.6  10 6 / F)(  72F)(29,000 ksi)  13.78 ksi (T)

In the summer, the rod had a tension normal stress of 5 ksi; therefore, the rod stress in the winter is: Ans.  winter  5 ksi  13.78 ksi  18.78 ksi (T)

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5.47 A high-density polyethylene [E = 120 ksi and  = 78 × 10−6/°F] block (1) is positioned in a fixture as shown in Fig. P5.47. The block is 2-in. by 2-in. by 32-in.-long. At room temperature, a gap of 0.10 in. exists between the block and the rigid support at B. Determine: (a) the normal stress in the block caused by a temperature increase of 100°F. (b) the normal strain in block (1) at the increased temperature.

Fig. P5.47

Solution If the polyethylene block were completely free to elongate, a temperature change of 100°F would cause an elongation of 1  1T L1  (78  10 6 / F)(100F)(32 in.)  0.2496 in. Since this elongation is greater than the 0.10-in. gap, the temperature change will cause the polyethylene block to contact the support at B, which will create normal stress in the block. Force-Temperature-Deformation Relationship The relationship between the internal force, temperature change, and the deformation of an axial member is: FL 1  1 1  1T1L1 A1E1 In this situation, the deformation of member (1) equals the 0.10-in. gap: F1 L1  1T1 L1  0.10 in. A1 E1 This relationship can be stated in terms of normal stress as  1 L1  1T1 L1  0.10 in. E1 (a) Normal stress: The normal stress in the block due to the 100°F temperature increase is: E 1   0.10 in.  1T1L1  1 L1

 0.10 in.  (78  106 / F)(100F)(32 in.) 

120 ksi  0.561 ksi  561 psi (C) 32 in.

(b) Normal strain: The normal strain in the polyethylene block is:  0.10 in. 1  1   0.003125 in./in.  3,125 με L1 32 in.

Ans.

Ans.

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5.48 The assembly shown in Fig. P5.48 consists of a brass shell (1) fully bonded to a solid ceramic core (2). The brass shell [E = 115 GPa,  = 18.7 × 10−6/°C] has an outside diameter of 50 mm and an inside diameter of 35 mm. The ceramic core [E = 290 GPa,  = 3.1 × 10−6/°C] has a diameter of 35 mm. At a temperature of 15°C, the assembly is unstressed. Determine the largest temperature increase that is acceptable for the assembly if the normal stress in the longitudinal direction of the brass shell must not exceed 80 MPa.

Fig. P5.48

Solution Section properties: The cross-sectional areas of brass shell (1) and ceramic core (2) are: A1 



(50 mm) 2  (35 mm) 2   1,001.3827 mm 2 4

A2 



4

(35 mm) 2  962.1128 mm2

Equilibrium Consider a FBD cut through the assembly. Sum forces in the horizontal direction to obtain: Fx   F1  F2  0  F2   F1 (a) Geometry of Deformations Relationship For this configuration, the deformations of both members will be equal; therefore, 1   2

(b)

Force-Temperature-Deformation Relationships The relationship between internal force, temperature change, and deformation of an axial member can be stated for members (1) and (2): FL FL 1  1 1  1T L1  2  2 2   2 T L2 (c) A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (c) into the geometry of deformation relationship (b) to derive the compatibility equation: F1 L1 FL  1T L1  2 2   2 T L2 (d) A1 E1 A2 E2 Solve the Equations Since a limiting stress is specified for brass shell (1), express Eq. (d) in terms of normal stress: L L  1 1  1T L1   2 2   2 T L2 E1 E2 Based on Eq. (a), the normal stress 2 can be expressed in terms of 1 as: F F F A A  2  2   1   1 1   1 1 A2 A2 A2 A1 A2 Substitute this expression into Eq. (e) to obtain L A L  1 1  1T L1   1 1 2   2 T L2 E1 A2 E2

(e)

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Rearrange terms L A L   1  1  1 2    2 T L2  1T L1  ( 2 L2  1L1 )T  E1 A2 E2  and solve for T, recognizing that both the shell and the core have the same length: L A L  1 A 1 1  1  1 2  1   1  E A2 E2  E A2 E2  T   1   1  2 L2  1 L1  2  1

(f)

Substitute the problem data into Eq. (f) to compute T that will produce a normal stress of 80 MPa in the brass shell:   1 1,001.3827 mm 2 1 (80 MPa)    2 115,000 MPa 962.1128 mm 290,000 MPa  T  3.1  106 / C  18.7  106 / C  62.9983C Since the problem asks for the largest temperature increase, Tmax  63.0C

Ans.

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5.49 At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown in Fig. P5.49. Bar (1) is an aluminum alloy [E = 10,000 ksi,  = 0.32,  = 12.5 × 10−6/°F] bar with a width of 3 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi,  = 0.12,  = 9.6 × 10−6/°F] bar with a width of 2 in. and a thickness of 0.75 in. The supports at A and C are rigid. Determine: (a) the lowest temperature at which the two bars contact each other. (b) the normal stress in the two bars at a temperature of 250°F. (c) the normal strain in the two bars at 250°F. (d) the change in width of the aluminum bar at a temperature of 250°F.

Fig. P5.49

Solution (a) Lowest Contact Temperature Before the gap is closed, only thermal strains and the associated axial elongations exist. Write expressions for the temperature-induced elongations and set this equal to the 0.04-in. gap: 1T L1   2 T L2  0.04 in. T 1 L1   2 L2   0.04 in.

0.04 in. 0.04 in.   48.6381°F -6 1 L1   2 L2 (12.5 10 / °F)(32 in.)  (9.6 10-6 / °F)(44 in.) Since the initial temperature is 60°F, the temperature at which the gap is closed is 108.6°F. T 

Ans.

(b) Equilibrium From the results obtained for part (a), we know that the gap will be closed at 250°F, making this a statically indeterminate axial configuration. Knowing this, consider a FBD at joint B. Assume that both internal axial forces will be tension, even though we know intuitively that both F1 and F2 will turn out to be compression. Fx   F1  F2  0 (a) Geometry of Deformations Relationship For this configuration, the sum of the deformations of members (1) and (2) must equal the initial gap: 1   2  0.04 in. (b) Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and the deformation of an axial member can be stated for members (1) and (2): FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 A1E1 A2 E2

(c)

Compatibility Equation Substitute the force-temperature-deformation relationships (c) into the geometry of deformation relationship (b) to derive the compatibility equation:

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F1 L1 FL  1T1 L1  2 2   2 T2 L2  0.04 in. A1 E1 A2 E2

(d)

Solve the Equations This part is no fun, but it must be done. From Eq. (a), F1 = F2. Substituting this into Eq. (d) gives: F1 L1 FL  1T1 L1  1 2   2 T2 L2  0.04 in. A1 E1 A2 E2

F1 L1 F1 L2   0.04 in.  1T1 L1   2 T2 L2 A1 E1 A2 E2  L L  F1  1  2   0.04 in.  1T1 L1   2 T2 L2  A1 E1 A2 E2  0.04 in.  1T1 L1   2 T2 L2 F1   L1 L2  AE  A E   1 1 2 2 

(e)

For this structure, the lengths, areas, and elastic moduli are given below. The temperature change is the same for both members; therefore, T1 = T2 = T = 190°F: L1  32 in. L2  44 in. A1  (3 in.)(0.75 in.)  2.25 in.2

A2  (2 in.)(0.75 in.)  1.50 in.2

E1  10,000 ksi

E2  28,000 ksi

1  12.5 10 / °F

 2  9.6 106 / °F

6

Substitute these values into Eq. (e) and calculate F1 = −47.0702 kips. Backsubstitute into Eq. (a) to calculate F2 = −47.0702 kips. Note that the internal forces are compression, as expected. Normal Stresses The normal stresses in each axial member can now be calculated: F 47.0702 kips 1  1   20.9201 ksi  20.9 ksi (C) A1 2.25 in.2

2 

F2 47.0702 kips   31.3801 ksi  31.4 ksi (C) A2 1.50 in.2

Ans.

(c) Normal Strains The force-temperature-deformation relationships were expressed in Eq. (b). By definition,  = /L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (c) by the respective member lengths: F F 1  1  1T1  2  2   2 T2 (f) A1 E1 A2 E2 Substitute the appropriate values to calculate the normal strains in each member:

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1  

F1  1T1 A1E1 47.0702 kips  (12.5  106 / °F)(190F) 2 (2.25 in. )(10,000 ksi)

 0.000283 in./in.  283 με

2  

Ans.

F2   2 T2 A2 E2 47.0702 kips  (9.6  106 / °F)(190F) (1.50 in.2 )(28,000 ksi)

 0.000703 in./in.  703 με

Ans.

(d) The change in width of the aluminum bar (1) is caused partly by the Poisson effect and partly by the temperature change. The longitudinal strain in the aluminum bar caused by the internal force F1 is: F 47.0702 kips  long,  1   2, 092.0110 6 in./in. 2 A1 E1 (2.25 in. )(10, 000 ksi) The accompanying lateral strain due to the Poisson effect is thus  lat,   long,  (0.32)(2,092.01106 in./in.)  669.44 106 in./in. The lateral strain caused by the temperature change is  lat,T  1T1  (12.5 106 / °F)(190F)  2,375 106 in./in. Therefore, the total lateral strain in aluminum bar (1) is  lat   lat,   lat,T

 669.44 106 in./in.  2,375 106 in./in.  3,044.44 106 in./in. The change is width of the aluminum bar is thus width  lat (width)  (3,044.44  106 in./in.)(3 in.)  0.00913 in.

Ans.

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5.50 At a temperature of 5°C, a 3-mm gap exists between two polymer bars and a rigid support, as shown in Fig. P5.50. Bar (1) is 50-mm wide and 20-mm thick [E = 800 MPa,  = 140 × 10−6/°C]. Bar (2) is 75-mm wide and 25-mm thick [E = 2.7 GPa,  = 67 × 10−6/°C] bar. The supports at A and C are rigid. Determine: (a) the lowest temperature at which the 3-mm gap is closed. (b) the normal stress in the two bars at a temperature of 60°C. (c) the normal strain in the two bars at 60°C.

Fig. P5.50

Solution (a) Before the gap is closed, only thermal strains and the associated axial elongations exist. Write expressions for the temperature-induced elongations and set this equal to the 3-mm gap: 1T L1   2 T L2  3 mm T 1 L1   2 L2   3 mm

3 mm 3 mm   24.038°C -6 1 L1   2 L2 (140 10 / °C)(700 mm)  (67 10-6 / °C)(400 mm) Since the initial temperature is 5°C, the temperature at which the gap is closed is 29.038°C = 29.0°C. Ans. T 

Equilibrium (b) From the results obtained for part (a), we know that the gap will be closed at 60°C, making this a statically indeterminate axial configuration. Knowing this, consider a FBD at joint B. Assume that both internal axial forces will be tension, even though we know intuitively that both F1 and F2 will turn out to be compression. Fx   F1  F2  0 (a) Geometry of Deformations Relationship For this configuration, the sum of the deformations of members (1) and (2) must equal the initial gap: 1   2  3 mm (b) Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and the deformation of an axial member can be stated for members (1) and (2): FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 A1E1 A2 E2 Compatibility Equation Substitute the force-temperature-deformation relationships (c) into the geometry of deformation relationship (b) to derive the compatibility equation: F1 L1 FL  1T1 L1  2 2   2 T2 L2  3 mm A1 E1 A2 E2

(c)

(d)

Solve the Equations This part is no fun, but it must be done. From Eq. (a), F1 = F2. Substituting this into Eq. (d) gives: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

F1 L1 FL  1T1 L1  1 2   2 T2 L2  3 mm A1 E1 A2 E2 F1 L1 F1 L2   3 mm  1T1 L1   2 T2 L2 A1 E1 A2 E2  L L  F1  1  2   3 mm  1T1 L1   2 T2 L2  A1 E1 A2 E2  3 mm  1T1 L1   2 T2 L2 F1   L1 L2  AE  A E   1 1 2 2 

(e)

For this structure, the lengths, areas, and elastic moduli are given below. The temperature change is the same for both members; therefore, T1 = T2 = T = 55°C: L1  700 mm L2  400 mm A1  (50 mm)(20 mm)  1,000 mm2

A2  (75 mm)(25 mm)  1,875 mm 2

E1  800 MPa

E2  2,700 MPa

1  140  10 / °C

 2  67  106 / °C

6

Substitute these values into Eq. (e) and calculate F1 = −4.050262 kN. Backsubstitute into Eq. (a) to calculate F2 = −4.050262 kN. Note that the internal forces are compression, as expected. Normal Stresses The normal stresses in each axial member can now be calculated: F ( 4.050262 kN)(1,000 N/kN) 1  1   4.050262 MPa  4.05 MPa (C) A1 1,000 mm 2

2 

F2 ( 4.050262 kN)(1,000 N/kN)   2.160140 MPa  2.16 MPa (C) A2 1,875 mm 2

Ans.

Normal Strains The force-temperature-deformation relationships were expressed in Eq. (c). By definition,  = /L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (c) by the respective member lengths: F F 1  1  1T1  2  2   2 T2 (f) A1 E1 A2 E2 Substitute the appropriate values to calculate the normal strains in each member: F 1  1  1T1 A1E1



( 4.050262 kN)(1,000 N/kN)  (140  106 / °C)(55C) 2 2 (1,000 mm )(800 N/mm )

 0.00263717 mm/mm  2,640 με

Ans.

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2  

F2   2 T2 A2 E2 ( 4.050262 kN)(1,000 N/kN)  (67  10 6 / °C)(55C) 2 2 (1,875 mm )(2,700 N/mm )

 0.00288495 mm/mm  2,880 με

Ans.

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5.51 The axial assembly shown in Fig. P5.51 consists of a solid 1-in.-diameter aluminum alloy [E = 10,000 ksi,  = 0.32,  = 12.5 × 10−6/°F] rod (1) and a solid 1.5-in.-diameter bronze [E = 15,000 ksi,  = 0.15,  = 9.4 × 10−-6/°F] rod (2). If the supports at A and C are rigid and the assembly is stress free at 0°F, determine: (a) the normal stress in both rods at 160°F. (b) the displacement of flange B. (c) the change in diameter of the aluminum rod.

Fig. P5.51

Solution Equilibrium Consider a FBD at joint B. Assume that both internal axial forces will be tension. Fx   F1  F2  0  F1  F2 (a)

Geometry of Deformations Relationship 1   2  0

(b)

Force-Temperature-Deformation Relationships FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 A1E1 A2 E2

(c)

Compatibility Equation F1 L1 FL  1T1 L1  2 2   2 T2 L2  0 A1 E1 A2 E2

(d)

Solve the Equations From Eq. (a), F1 = F2. The temperature change is the same for both members; therefore, T1 = T2 = T. Eq. (d) then can be written as: F1 L1 FL  1T L1  1 2   2 T L2  0 A1 E1 A2 E2 Solving for F1: F1 L1 F1 L2   1T L1   2 T L2 A1 E1 A2 E2  L L  F1  1  2   T 1 L1   2 L2   A1 E1 A2 E2  T 1 L1   2 L2  F1    L1 L2  AE  A E   1 1 2 2 

(e)

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For this structure, the lengths, areas, coefficients of thermal expansion, and elastic moduli are given below. L1  15 in. L2  22 in. A1 



(1 in.) 2  0.7854 in.2

4 E1  10, 000 ksi

A2 



(1.5 in.) 2  1.7671 in.2

4 E2  15, 000 ksi

1  12.5 106 / °F  2  9.4 106 / °F Substitute these values along with T = +160°F into Eq. (e) and calculate F1 = −23.0263 kips. From Eq. (a), F2 = F1 = −23.0263 kips. (a) Normal Stresses The normal stresses in each rod can now be calculated: F 23.0263 kips 1  1   29.318 ksi  29.3 ksi (C) A1 0.7854 in.2

2 

F2 23.0263 kips   13.030 ksi  13.03 ksi (C) A2 1.7671 in.2

Ans. Ans.

(b) Displacement of Flange B The displacement of flange B is equal to the deformation (i.e., contraction in this instance) of rod (1). The deformation of rod (1) is given by: FL 1  1 1  1T1L1 A1E1

( 23.0263 kips)(15 in.)  (12.5  106 / °F)(160°F)(15 in.)  0.013977 in. (0.7854 in.2 )(10,000 ksi) The displacement of flange B is thus: uB  1  0.01398 in.  

Ans.

(c) Change in diameter of the aluminum rod The change in diameter of aluminum rod (1) is caused partly by the Poisson effect and partly by the temperature change. The longitudinal strain in the aluminum rod caused by the internal force F1 is: F 23.0263 kips  long,  1   2,931.78 106 in./in. 2 A1 E1 (0.7854 in. )(10, 000 ksi) The accompanying lateral strain due to the Poisson effect is thus  lat,   long,  (0.32)(2,931.78106 in./in.)  938.17 106 in./in. The lateral strain caused by the temperature change is  lat,T  1T1  (12.5 106 / °F)(160F)  2,000 106 in./in. Therefore, the total lateral strain in aluminum rod (1) is  lat   lat,   lat,T

 938.17 106 in./in.  2, 000 106 in./in.  2,938.17 106 in./in. The change in diameter of the aluminum rod is thus d1  lat d1  (2,938.17  106 in./in.)(1 in.)  0.00294 in.

Ans.

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5.52 The pin-connected structure shown in Fig. P5.52 consists of a rigid bar ABC, a solid bronze [E = 100 GPa,  = 16.9 × 10−6/°C] rod (1), and a solid aluminum alloy [E = 70 GPa,  = 22.5×10−6/°C] rod (2). Bronze rod (1) has a diameter of 24 and aluminum rod (2) has a diameter of 16 mm. The bars are unstressed when the structure is assembled at 25°C. After assembly, the temperature of rod (2) is decreased by 40°C while the temperature of rod (1) remains constant at 25°C. Determine the normal stresses in both rods for this condition. Fig. P5.52

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin B gives the best information for this situation: M B  (200 mm)F1  (350 mm)F2  0 (a)

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vA vC  (b) 200 mm 350 mm There are no gaps, clearances, or other misfits at pins A and C; therefore, Eq. (c) can be rewritten in terms of the member deformations as:

1

200 mm



2

350 mm

(c)

Force-Temperature-Deformation Relationships FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 A1E1 A2 E2 Compatibility Equation  F1L1   F2 L2  1 1  1T1L1     2 T2 L2    200 mm  A1E1  350 mm  A2 E2 

(d)

(e)

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Solve the Equations Solve Eq. (a) for F1: 350 mm F1   F2  1.75F2 200 mm

(f)

Substitute this result into Eq. (e):  ( 1.75F2 ) L1   F2 L2  1 1  1T1L1     2 T2 L2    200 mm  A1E1  350 mm  A2 E2  Note that T1 = 0°C; therefore, the equation simplifies to:  1.75F2 L1 200 mm  F2 L2     2 T2 L2   A1E1 350 mm  A2 E2  Solve this equation for F2:  1.75 F2 L1 200 mm  F2 L2     2 T2 L2   A1E1 350 mm  A2 E2  1.75 L1 200 L2  200 F2    2 T2 L2  A E 350 A E 350  1 1 2 2 200  2 T2 L2 350 (g) F2   1.75 L1 200 L2  A1E1 350 A2 E2 For this structure, the lengths, areas, coefficients of thermal expansion, and elastic moduli are given below. L1  250 mm L2  500 mm A1 



(24 mm) 2  452.3893 mm 2

4 E1  100,000 MPa

A2 



(16 mm) 2  201.0619 mm 2

4 E2  70,000 MPa

1  16.9  10 6 / °C  2  22.5  10 6 / °C Substitute these values along with T2 = −40°C into Eq. (g) and calculate F2 = −8,579.65 N. From Eq. (f), calculate F1 = −15,014.39 N. Normal Stresses The normal stresses in each axial member can now be calculated: F 15,014.39 N 1  1   33.2 MPa (C) A1 452.3893 mm 2 F 8,579.65 N 2  2   42.7 MPa (T) A2 201.0619 mm 2

Ans. Ans.

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5.53 Rigid bar ABC is supported by two identical solid bronze [E = 100 GPa,  = 16.9 × 10−6/°C] rods, and a solid steel [E = 200 GPa,  = 11.9×10−6/°C] rod, as shown in Fig. P5.53. The bronze rods (1) each have a diameter of 16 mm and they are symmetrically positioned relative to the center rod (2) and the applied load P. Steel rod (2) has a diameter of 20 mm. The bars are unstressed when the structure is assembled at 30°C. When the temperature decreases to –20°C, determine: (a) the normal stresses in the bronze and steel rods. (b) the normal strains in the bronze and steel rods. Fig. P5.53

Solution Equilibrium: By virtue of symmetry, the forces in the two bronze rods (1) are identical. Consider a FBD of the rigid bar. Sum forces in the vertical direction to obtain: Fy  2 F1  F2  0 (a) Geometry of Deformations: For this configuration, the deflections of joints A, B, and C are equal: v A  vB  vC All pin connections are ideal; therefore, the deflection of joints A and C will cause an identical deformation of rods (1): v A  1 and the rigid bar deflection vB will cause an identical deformation of rod (2): vB   2 Substitute Eqs. (c) and (d) into Eq. (b) to obtain the geometry of deformation equation: 1   2 Force-Temperature-Deformation Relationships: The relationship between internal force, temperature change, and deformation can be stated for members (1) and (2): FL FL 1  1 1  1T L1  2  2 2   2 T L2 A1E1 A2 E2 Compatibility Equation: Substitute the force-deformation relationships (f) into the geometry of deformation relationship (e) to derive the compatibility equation: F1 L1 FL  1T1 L1  2 2   2 T2 L2 A1 E1 A2 E2

(b)

(c) (d) (e)

(f)

(g)

Solve the Equations: For this situation, T1 = T2 = T. Solve Eq. (g) for F2: FL AE F2   1 1  1T1 L1   2 T2 L2  2 2  A1 E1  L2  F1

L1 A2 E2 AE  T 1 L1   2 L2  2 2 L2 A1 E1 L2

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Substitute this expression into Eq. (a) L A E AE 2 F1  F1 1 2 2  T 1L1   2 L2  2 2  0 L2 A1 E1 L2 and derive an expression for F1, noting that L1 = L2:  A E  F1  2  2 2   T 1   2  A2 E2 A1 E1   T 1   2  A2 E2 F1   A E 2 2 2 A1 E1 For this structure, the areas, coefficients of thermal expansion, and elastic moduli are given below:



A1 

(16 mm) 2  201.0619 mm 2

A2 



(h)

(20 mm) 2  314.1593 mm2

4 E1  100,000 MPa

4 E2  200,000 MPa

1  16.9  106 / °C

 2  11.9  106 / °C

Substitute these values along with T = −50°C into Eq. (h) and calculate F1: T 1   2  A2 E2 F1   A E 2 2 2 A1 E1 

( 50C) 16.9  10 6  11.9  10 6  (452.3893 mm 2 )(200,000 N/mm 2 )  314.1593 mm 2   200,000 MPa  2  201.0619 mm 2   100,000 MPa 

 3,064.97 N Backsubstitute this result into Eq. (a) to calculate F2 = –6,129.94 N.

(a) Normal Stresses: The normal stresses in each rod can now be calculated: F 3,064.97 N 1  1   15.24 MPa (T) A1 201.0619 mm 2

2 

F2 6,129.94 N   19.51 MPa (C) A2 314.1593 mm 2

Ans.

Ans.

(b) Normal Strains: The force-temperature-deformation relationships were expressed in Eq. (f). By definition,  = /L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (f) by the respective member lengths: F  F  1  1  1T1  1  1T1  2  2   2 T2  2   2 T2 (f) A1E1 E1 A2 E2 E2 Substitute the appropriate values to calculate the normal strains in each member:

1  

1 E1

 1T1

15.2439 MPa  (16.9  106 / °C)(  50C) 100,000 MPa

 0.00069256 mm/mm  693 με

Ans.

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2  

F2   2 T2 A2 E2 19.5122 MPa  (11.9  106 / °C)(  50C) 200,000 MPa

 0.00069256 mm/mm  693 με

Ans.

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5.54 A load of P = 85 kN is supported by a structure consisting of rigid bar ABC, two identical solid bronze [E = 100 GPa,  = 16.9 × 10−6/°C] rods, and a solid steel [E = 200 GPa,  = 11.9×10−6/°C] rod, as shown in Fig. P5.54. The bronze rods (1) each have a diameter of 20 mm and they are symmetrically positioned relative to the center rod (2) and the applied load P. Steel rod (2) has a diameter of 14 mm. The bars are unstressed when the structure is assembled. After the temperature of the structure has been increased by 45°C, determine: (a) the normal stresses in the bronze and steel rods. (b) the normal strains in the bronze and steel rods. (c) the downward deflection of rigid bar ABC. Fig. P5.54

Solution Equilibrium: By virtue of symmetry, the forces in the two bronze rods (1) are identical. Consider a FBD of the rigid bar. Sum forces in the vertical direction to obtain: Fy  2 F1  F2  P  0 (a) Geometry of Deformations: For this configuration, the deflections of joints A, B, and C are equal: v A  vB  vC All pin connections are ideal; therefore, the deflection of joints A and C will cause an identical deformation of rods (1): v A  1 and the rigid bar deflection vB will cause an identical deformation of rod (2): vB   2 Substitute Eqs. (c) and (d) into Eq. (b) to obtain the geometry of deformation equation: 1   2 Force-Temperature-Deformation Relationships: The relationship between internal force, temperature change, and deformation can be stated for members (1) and (2): FL FL 1  1 1  1T L1  2  2 2   2 T L2 A1E1 A2 E2 Compatibility Equation: Substitute the force-deformation relationships (f) into the geometry of deformation relationship (e) to derive the compatibility equation: F1 L1 FL  1T1 L1  2 2   2 T2 L2 A1 E1 A2 E2

(b)

(c) (d) (e)

(f)

(g)

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Solve the Equations: For this situation, T1 = T2 = T. Solve Eq. (g) for F2: FL AE F2   1 1  1T1 L1   2 T2 L2  2 2  A1 E1  L2 L1 A2 E2 AE  T 1 L1   2 L2  2 2 L2 A1 E1 L2 Substitute this expression into Eq. (a) L A E AE 2 F1  F1 1 2 2  T 1 L1   2 L2  2 2  P L2 A1 E1 L2 and derive an expression for F1:  L A E  AE F1  2  1 2 2   P  T 1L1   2 L2  2 2 L2 A1 E1  L2   F1

P  T 1L1   2 L2  F1 

A2 E2 L2

L1 A2 E2 L2 A1 E1 For this structure, P = 85 kN = 85,000 N, and the lengths, areas, and elastic moduli are given below: L1  1, 200 mm L2  2, 200 mm A1 



(h)

2

(20 mm) 2  314.1593 mm 2

4 E1  100,000 MPa

A2 



(14 mm) 2  153.9380 mm 2 4 E2  200,000 MPa

1  16.9  10 6 / °C  2  11.9  10 6 / °C Substitute these values along with T = +45°C into Eq. (h) and calculate F1: AE P  T 1L1   2 L2  2 2 L2 F1  L A E 2 1 2 2 L2 A1 E1 

(153.9380 mm 2 )(200,000 N/mm 2 ) 2, 200 mm 2  1, 200 mm   153.9380 mm   200,000 MPa  2  2, 200 mm   314.1593 mm 2   100,000 MPa 

(85,000 N)  (45C) (16.9  10 6 )(1, 200)  (11.9  10 6 )(2, 200) 

 35,002.53 N Backsubstitute this result into Eq. (a) to calculate F2 = 14,994.94 N.

(a) Normal Stresses: The normal stresses in each rod can now be calculated: F 35,002.53 N 1  1   111.4 MPa A1 314.1593 mm 2

2 

F2 14,994.94 N   97.4 MPa A2 153.9380 mm 2

Ans.

Ans.

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(b) Normal Strains: The force-temperature-deformation relationships were expressed in Eq. (f). By definition,  = /L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (f) by the respective member lengths: F F 1  1  1T1  2  2   2 T2 (f) A1 E1 A2 E2 Substitute the appropriate values to calculate the normal strains in each member: F 1  1  1T1 A1E1



35,002.53 N  (16.9  106 / °C)(45C) (314.1593 mm 2 )(100,000 N/mm2 )

 0.00187467 mm/mm  1,875 με

2  

Ans.

F2   2 T2 A2 E2 14,994.94 N  (11.9  10 6 / °C)(45C) 2 2 (153.9380 mm )(200,000 N/mm )

 0.0010225 mm/mm  1,023 με

Ans.

(c) Rigid bar deflection: The downward deflection of the rigid bar can be determined from the deformation of rods (1): FL vB  1  1 1  1T1L1 A1E1



(35,002.53N)(1,200 mm)  (16.9  106 / °C)(45°C)(1,200 mm) (314.1593 mm2 )(100,000 MPa )

 2.25 mm 

Ans.

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5.55 A solid aluminum [E = 70 GPa,  = 22.5 × 10−6/°C] rod (1) is connected to a solid bronze [E = 100 GPa,  = 16.9 × 10−6/°C] rod at flange B, as shown in Fig. P5.55. Aluminum rod (1) has a diameter of 40 mm and bronze rod (2) has a diameter of 120 mm. The bars are unstressed when the structure is assembled at 30°C. After the 300-kN load is applied to flange B, the temperature increases to 45°C. Determine: (a) the normal stresses in rods (1) and (2). (b) the deflection of flange B.

Fig. P5.55

Solution Equilibrium: Consider a FBD of flange B. Sum forces in the horizontal direction to obtain: Fx   F1  F2  300 kN  0 (a) Geometry of Deformations: 1   2  0

(b)

Force-Deformation Relationships: FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 A1E1 A2 E2

(c)

Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1 L1 FL  1T1 L1  2 2   2 T2 L2  0 A1 E1 A2 E2

(d)

Solve the Equations: For this situation, T1 = T2 = T. Solve Eq. (d) for F1: L A E AE F1   F2 2 1 1  T 1 L1   2 L2  1 1 L1 A2 E2 L1 Substitute this expression into Eq. (a): L A E AE F2 2 1 1  T 1 L1   2 L2  1 1  F2  300 kN L1 A2 E2 L1 and solve for F2: L A E  AE F2  2 1 1  1  300 kN  T 1 L1   2 L2  1 1 L1  L1 A2 E2  AE 300 kN  T 1 L1   2 L2  1 1 L1 F2   L2 A1 E1 1 L1 A2 E2

(f)

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Member lengths, areas, elastic moduli, and coefficients of thermal expansion are given below: L1  2, 600 mm L2  1, 000 mm A1 



(40 mm) 2  1, 256.637 mm 2

4 E1  70, 000 MPa

A2 



(120 mm) 2  11,309.734 mm 2

4 E2  100, 000 MPa

1  22.5 106 / °C  2  16.9 10 6 / °C Substitute these values along with T = +15°C into Eq. (f) and compute F2 = −328.4395 kN. Backsubstitute this result into Eq. (a) to find F1 = −28.4395 kN. (a) Normal Stresses: The normal stresses in each rod can now be calculated: F 28, 439.5 N 1  1   22.6 MPa (C) A1 1,256.637 mm 2

2 

F2 328,439.5 N   29.0 MPa (C) A2 11,309.734 mm 2

Ans.

Ans.

(b) Deflection of Flange B: The deflection of flange B can be determined from the deformation of rod (1): FL uB  1  1 1  1T1L1 A1E1



( 28,439.5 N)(2,600 mm)  (22.5  106 / °C)(15°C)(2,600 mm) (1,256.637 mm 2 )(70,000 MPa )

 0.0369 mm 

Ans.

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5.56 A steel [E = 30,000 ksi,  = 6.6 × 10−6/°F] pipe column (1) with a crosssectional area of A1 = 5.60 in.2 is connected at flange B to an aluminum alloy [E = 10,000 ksi,  = 12.5 × 10−6/°F] pipe (2) with a crosssectional area of A2 = 4.40 in.2. The assembly (shown in Fig. P5.56) is connected to rigid supports at A and C. It is initially unstressed at a temperature of 90°F. (a) At what temperature will the normal stress in steel pipe (1) be reduced to zero? (b) Determine the normal stresses in steel pipe (1) and aluminum pipe (2) when the temperature reaches –10°F.

Fig. P5.56

Solution Equilibrium: Consider a FBD of flange B. Sum forces in the horizontal direction to obtain: Fx   F1  F2  60 kips  0 (a) Geometry of Deformations: 1   2  0 Force-Temperature-Deformation Relationships: FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 A1E1 A2 E2 Compatibility Equation: Substitute Eqs. (c) into Eq. (b) to derive the compatibility equation: F1 L1 FL  1T1 L1  2 2   2 T2 L2  0 A1 E1 A2 E2

(b)

(c)

(d)

Solve the Equations: Set F1 = 0 and solve Eq. (a) to find F2 = 60 kips. Substitute these values for F1 and F2 into Eq. (d) along with the observation that the temperature change for both axial members is the same (i.e., T1 = T2 = T) and solve for T: FL F L (60 kips)(144 in.)  1 1 2 2 0 A1 E1 A2 E2 (4.40 in.2 )(10,000 ksi) T    75.758F 1 L1   2 L2 (6.6 106 / F)(120 in.)  (12.5 106 / F)(144 in.) Since the pipes are initially at a temperature 90°F, the temperature at which the normal stress in steel pipe (1) is reduced to zero is T  90F  75.748F  14.24F Ans. (b) Solve Eq. (a) for F2 to obtain F2  F1  60 kips (e) When the temperature reaches −10°F, the total change in temperature is T = −100°F. Substitute this value along with Eq. (e) into the compatibility equation [Eq. (d)] and derive an expression for F1:

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F1 L1 ( F1  60 kips)L2   1T L1   2 T L2 A1 E1 A2 E2  L L  (60 kips)L2 F1  1  2    T 1 L1   2 L2  A2 E2  A1 E1 A2 E2   L L  (60 kips)L2  F1  1  2   T 1 L1   2 L2   A2 E2  A1 E1 A2 E2  (60 kips)L2 T 1 L1   2 L2   A2 E2 F1  L1 L  2 A1 E1 A2 E2



and compute F1: F1 

(100F) (6.6 106 )(120 in.)  (12.5 10 6 )(144 in.)  

(60 kips)(144 in.) (4.40 in.2 )(10, 000 ksi)

120 in. 144 in.  2 (5.60 in. )(30, 000 ksi) (4.40 in.2 )(10, 000 ksi) 0.259200 in.  0.196364 in. 0.062836 in.   6 6 714.2857 10 in./kip  3, 272.7273 10 in./kip 3,987.0130 10 6 in./kip  15.7602 kips

From Eq. (a), F2 has a value of F2  F1  60 kips  15.7602 kips  60 kips  75.7602 kips (a) Normal Stresses: The normal stresses in each axial member can now be calculated: F 15.7602 kips 1  1   2.8143 ksi  2.81 ksi (T) A1 5.60 in.2

2 

F2 75.7602 kips   17.2182 ksi  17.22 ksi (T) A2 4.40 in.2

Ans.

Ans.

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5.57 A load P will be supported by a structure consisting of a rigid bar ABCD, a polymer [E = 2,300 ksi,  = 2.9 × 10−6/°F] bar (1) and an aluminum alloy [E = 10,000 ksi,  = 12.5 × 10−6/°F] bar (2), as shown in Fig. P5.57. Each bar has a cross-sectional area of 2.00 in.2. The bars are unstressed when the structure is assembled at 30°F. After a concentrated load of P = 26 kips is applied and the temperature is increased to 100°F, determine: (a) the normal stresses in bars (1) and (2). (b) the vertical deflection of joint D.

Fig. P5.57

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin A gives the best information for this situation: M A  (30 in.)F1  (84 in.)F2  (66 in.)P  0 (a)

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vB v (b)  D 30 in. 84 in. Since there are no gaps, clearances, or other misfits at pins B and D, the deformation of member (1) will equal the deflection of the rigid bar at B and the deformation of member (2) will equal the deflection of the rigid bar at D. Therefore, Eq. (b) can be rewritten in terms of the member deformations as:

1

30 in.



2

84 in.

(c)

Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and deformation of an axial member can be stated for members (1) and (2): FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 (d) A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to derive the compatibility equation:

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  1  F1L1 1  F2 L2  1T1L1     2 T2 L2    30 in.  A1E1  84 in.  A2 E2 

(e)

Solve the Equations Note that the temperature change for both axial members is the same (i.e., T1 = T2 = T). Solve Eq. (e) for F1:  F1 L1 30 in.  F2 L2    2 T L2   1T L1  A1 E1 84 in.  A2 E2  AE 30 in. L2 A1 E1 30 in.   2 T L2 1 1  1T A1 E1 84 in. L1 A2 E2 84 in. L1 Substitute this expression into equilibrium equation (a):  30 in. L2 A1 E1 30 in.  AE (30 in.)  F2   2 T L2 1 1  1T A1E1   (84 in.)F2  (66 in.)P  0 L1  84 in. L1 A2 E2 84 in.  and solve for F2: AE (30 in.)2 (66 in.)P   2 T L2 1 1  (30 in.)1T A1E1 84 in. L1 F2  2 (30 in.) L2 A1 E1  84 in. 84 in. L1 A2 E2 For this structure, P = 26 kips, and the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed below: L1  72 in. L2  96 in. F1  F2

A1  2.00 in.2

A2  2.00 in.2

E1  2,300 ksi

E2  10, 000 ksi

1  2.9 106 / °F

 2  12.5 106 / °F

(f)

(g)

Substitute these values along with T = 70°F into Eq. (g) and calculate F2 = 19.3218 kips. Backsubstitute into Eq. (f) to calculate F1 = 3.0991 kips. Normal Stresses The normal stresses in each axial member can now be calculated: F 3.0991 kips 1  1   1.550 ksi (T) A1 2.00 in.2

2 

F2 19.3218 kips   9.66 ksi (T) A2 2.00 in.2

Ans. Ans.

Deflections of the rigid bar Calculate the deformation of member (2): FL (19.3218 kips)(96 in.)  2  2 2   2 T2 L2   (12.5  10 6 / °F)(70°F)(96 in.)  0.1767 in. (h) 2 A2 E2 (2.00 in. )(10,000 ksi) Since there are no gaps at pin D, the rigid bar deflection at D is equal to the deformation of member (2); therefore:

vD   2  0.1767 in. 

Ans.

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5.58 A load P will be supported by a structure consisting of a rigid bar ABCD, an aluminum alloy [E = 10,000 ksi,  = 12.5 × 10−6/°F] bar (1) and a steel [E = 29,000 ksi,  = 6.5 × 10−6/°F] bar (2), as shown in Fig. P5.58. Each bar has a crosssectional area of 2.00 in.2. The bars are unstressed when the structure is assembled at 80°F. After a concentrated load of P = 42 kips is applied and the temperature is decreased to –10°F, determine: (a) the normal stresses in bars (1) and (2). (b) the vertical deflection of joint D. Fig. P5.58

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin A gives the best information for this situation: M A  (30 in.)F1  (84 in.)F2  (66 in.)P  0 (a) Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vB v (b)  D 30 in. 84 in. Since there are no gaps, clearances, or other misfits at pins B and D, the deformation of member (1) will equal the deflection of the rigid bar at B and the deformation of member (2) will equal the deflection of the rigid bar at D. Therefore, Eq. (b) can be rewritten in terms of the member deformations as:

1

30 in.



2

84 in.

(c)

Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and deformation of an axial member can be stated for members (1) and (2): FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 (d) A1E1 A2 E2 Compatibility Equation Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to derive the compatibility equation:   1  F1L1 1  F2 L2  1T1L1     2 T2 L2  (e)   30 in.  A1E1  84 in.  A2 E2  Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Solve the Equations Note that the temperature change for both axial members is the same (i.e., T1 = T2 = T). Solve Eq. (e) for F1:  F1 L1 30 in.  F2 L2    2 T L2   1T L1  A1 E1 84 in.  A2 E2  AE 30 in. L2 A1 E1 30 in.   2 T L2 1 1  1T A1 E1 84 in. L1 A2 E2 84 in. L1 Substitute this expression into equilibrium equation (a):  30 in. L2 A1 E1 30 in.  AE (30 in.)  F2   2 T L2 1 1  1T A1E1   (84 in.)F2  (66 in.)P  0 L1  84 in. L1 A2 E2 84 in.  and solve for F2: AE (30 in.)2 (66 in.)P   2 T L2 1 1  (30 in.)1T A1E1 84 in. L1 F2  2 (30 in.) L2 A1 E1  84 in. 84 in. L1 A2 E2 For this structure, P = 42 kips, and the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed below: L1  72 in. L2  96 in. F1  F2

A1  2.00 in.2

(f)

(g)

A2  2.00 in.2

E1  10,000 ksi

E2  29,000 ksi

1  12.5  10 / °F

 2  6.5  10 6 / °F

6

Substitute these values along with T = –90°F into Eq. (g) and calculate F2 = 25.4609 kips. Backsubstitute into Eq. (f) to calculate F1 = 21.1094 kips. Normal Stresses The normal stresses in each axial member can now be calculated: F 21.1094 kips 1  1   10.55 ksi (T) A1 2.00 in.2

2 

F2 25.4609 kips   12.73 ksi (T) A2 2.00 in.2

Ans. Ans.

Deflections of the rigid bar Calculate the deformation of member (2): FL  2  2 2   2 T2 L2 A2 E2

(25.4609 kips)(96 in.) (h)  (6.5  106 / °F)( 90°F)(96 in.)  0.01402 in. (2.00 in.2 )(29,000 ksi) Since there are no gaps at pin D, the rigid bar deflection at D is equal to the deformation of member (2); therefore: 

vD   2  0.01402 in. 

Ans.

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5.59 The pin-connected structure shown in Fig. P5.59 consists of a rigid bar ABCD and two axial members. Bar (1) is steel [E = 200 GPa,  = 11.7 × 10−6/°C ] with a cross-sectional area of A1 = 400 mm2. Bar (2) is an aluminum alloy [E = 70 GPa,  = 22.5 × 10−6/°C] with a crosssectional area of A2 = 400 mm2. The bars are unstressed when the structure is assembled. After a concentrated load of P = 36 kN is applied and the temperature is increased by 25°C, determine: (a) the normal stresses in bars (1) and (2). (b) the deflection of point D on the rigid bar. Fig. P5.59

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin C gives the best information for this situation: M C  (950 mm)F1  (600 mm)F2  (720 mm)(36 kN)  0

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vA vB vD   950 mm 600 mm 720 mm

(a)

(b)

Since there are no gaps or clearances at either pin A or pin B, the deformations of members (1) and (2) will equal the deflections of the rigid bar at A and B, respectively.

1

950 mm



2

600 mm

(c)

Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and deformation of an axial member can be stated for members (1) and (2) FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 (d) A1E1 A2 E2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Compatibility Equation Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to derive the compatibility equation:  F1L1   F2 L2  1 1 (e)  1T1L1     2 T2 L2    950 mm  A1E1 600 mm A E   2 2  Solve the Equations Solve Eq. (e) for F1:     950  F2 L2  AE F1     2 T2 L2   1T1L1  1 1      600  A2 E2  L1

(f)

and substitute this expression into Eq. (a):     950  F2 L2  A1E1 (950 mm)     T L    T L  (600 mm)F2  (720 mm)(36 kN)  2 2 2 1 1 1  600  A2 E2  L1   

(g)

For this structure, the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed below: L1  900 mm L2  900 mm A1  400 mm 2

A2  400 mm 2

E1  200,000 MPa

E2  70,000 MPa

1  11.7  10 / °C

 2  22.5  10 6 / °C

6

Substitute these values along with T1 = T2 = +25°C into Eq. (g) and calculate F2 = –3,989.18 N. Backsubstitute into Eq. (f) to calculate F1 = 29,803.69 N. (a) Normal stresses: F 29,803.69 N 1  1   74.5 MPa (T) A1 400 mm 2 F 3,989.18 N 2  2   9.97 MPa (C) A2 400 mm 2

Ans. Ans.

(b) Deflections of the rigid bar Calculate the deformation of member (1): FL 1  1 1  1T1L1 A1E1 (29,803.69 N)(900 mm)  (11.7  106 / C)(25C)(900 mm) 2 (400 mm )(200,000 MPa)  0.5985 mm 

Since the pin at A is assumed to have a perfect connection, vA = 1 = 0.5985 mm. From Eq. (b), 720 mm vD  vA  0.757895(0.5985 mm)  0.454 mm  950 mm

Ans.

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5.60 The pin-connected structure shown in Fig. P5.60 consists of a rigid bar ABCD and two axial members. Bar (1) is steel [E = 200 GPa,  = 11.7 × 10−6/°C ] with a cross-sectional area of A1 = 400 mm2. Bar (2) is an aluminum alloy [E = 70 GPa,  = 22.5 × 10−6/°C] with a crosssectional area of A2 = 400 mm2. The bars are unstressed when the structure is assembled. After a concentrated load of P = 36 kN is applied and the temperature is decreased by 50°C, determine: (a) the normal stresses in bars (1) and (2). (b) the deflection of point D on the rigid bar. Fig. P5.60

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin C gives the best information for this situation: M C  (950 mm)F1  (600 mm)F2  (720 mm)(36 kN)  0

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: vA vB vD   950 mm 600 mm 720 mm

(a)

(b)

Since there are no gaps or clearances at either pin A or pin B, the deformations of members (1) and (2) will equal the deflections of the rigid bar at A and B, respectively.

1

950 mm



2

600 mm

(c)

Force-Temperature-Deformation Relationships The relationship between the internal force, temperature change, and deformation of an axial member can be stated for members (1) and (2) FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 (d) A1E1 A2 E2

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Compatibility Equation Substitute the force-deformation relationships (d) into the geometry of deformation relationship (c) to derive the compatibility equation:  F1L1   F2 L2  1 1 (e)  1T1L1     2 T2 L2    950 mm  A1E1 600 mm A E   2 2  Solve the Equations Solve Eq. (e) for F1:     950  F2 L2  AE F1     2 T2 L2   1T1L1  1 1      600  A2 E2  L1

(f)

and substitute this expression into Eq. (a):     950  F2 L2  A1E1 (950 mm)     T L    T L  (600 mm)F2  (720 mm)(36 kN)  2 2 2 1 1 1  600  A2 E2  L1   

(g)

For this structure, the lengths, areas, elastic moduli, and coefficients of thermal expansion are listed below: L1  900 mm L2  900 mm A1  400 mm 2

A2  400 mm 2

E1  200,000 MPa

E2  70,000 MPa

1  11.7  10 / °C

 2  22.5  10 6 / °C

6

Substitute these values along with T1 = T2 = –50°C into Eq. (g) and calculate F2 = 23,855.47 N. Backsubstitute into Eq. (f) to calculate F1 = 12,217.60 N. (a) Normal stresses: F 12, 217.60 N 1  1   30.5 MPa (T) A1 400 mm 2 F 23,855.47 N 2  2   59.6 MPa (T) A2 400 mm 2

Ans. Ans.

(b) Deflections of the rigid bar Calculate the deformation of member (1): FL 1  1 1  1T1L1 A1E1 (12, 217.60 N)(900 mm)  (11.7  10 6 / C)( 50C)(900 mm) 2 (400 mm )(200,000 MPa)  0.3891 mm 

Since the pin at A is assumed to have a perfect connection, vA = 1 = –0.3891 mm (i.e., joint A moves to the left). From Eq. (b), 720 mm vD  vA  0.757895( 0.3891 mm)  0.295 mm  Ans. 950 mm

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5.61 Rigid bar ABCD is loaded and supported as shown in Fig. P5.61. Bar (1) is made of bronze [E = 100 GPa,  = 16.9 × 10−6/°C] and has a crosssectional area of 400 mm2. Bar (2) is made of aluminum [E = 70 GPa,  = 22.5 × 10−6/°C] and has a cross-sectional area of 600 mm2. Bars (1) and (2) are initially unstressed. After the temperature has increased by 40°C, determine: (a) the stresses in bars (1) and (2). (b) the vertical deflection of point A.

Fig. P5.61

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin D gives the best information for this situation: M D  (3 m)F1  (1 m)F2  0  F2  3F1 (a) Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: v vA v  B  C 4 m 3m 1m

(b)

There are no gaps, clearances, or other misfits at pins B and C; therefore, Eq. (b) can be rewritten in terms of the member deformations as: 1   2  1  3 2 (c) 3m 1m Note: To understand the negative sign associated with 1, see Section 5.5 for discussion of statically indeterminate rigid bar configurations with opposing members. Force-Temperature-Deformation Relationships FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 A1E1 A2 E2 Compatibility Equation FL  F1L1  1T1L1  3  2 2   2 T2 L2  A1E1  A2 E2 

(d)

(e)

Solve the Equations For this situation, T1 = T2 = T = 40°C. Substitute Eq. (a) into Eq. (e): Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

  3F  L  F1 L1  1T L1  3  1 2   2 T L2  A1E1  A2 E2  and solve for F1: T  3 2 L2  1 L1  F1   L1 9 L2  A1 E1 A2 E2 (40C) 3(22.5 106 / C)(920 mm)  (16.9 106 / C)(840 mm)  840 mm 9(920 mm)  2 2 (400 mm )(100, 000 N/mm ) (600 mm 2 )(70, 000 N/mm 2 )  13,990 N  13.990 kN Backsubstitute into Eq. (a) to find F2 = −41.970 kN. 

(a) Normal Stresses The normal stresses in each axial member can now be calculated: F 13,990 N 1  1   35.0 MPa (C) A1 400 mm 2

2 

F2 41,970 N   70.0 MPa (C) A2 600 mm 2

Ans. Ans.

(b) Deflection of the rigid bar at A Calculate the deformation of one of the axial members, say member (1): FL 1  1 1  1T1L1 A1E1 ( 13,990 N)(840 mm)  (16.9  106 / °C)(40C)(840 mm) 2 2 (400 mm )(100,000 N/mm )  0.27405 mm Since there are no gaps at pin B, the rigid bar deflection at B is equal to the deformation of member (1); therefore, vB = 1 = 0.27405 mm (upward). From similar triangles, the deflection of the rigid bar at A is related to vB by: vA v  B 4m 3m The deflection of the rigid bar at A is thus: 4m 4m vA  vB  (0.27405 mm)  0.365 mm  Ans. 3m 3m 

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5.62 Rigid bar ABCD in Fig. P5.62 is supported by a pin connection at A and by two axial bars (1) and (2). Bar (1) is a 30-in.-long bronze [E = 15,000 ksi,  = 9.4 × 10−6/°F] bar with a cross-sectional area of 1.25 in.2. Bar (2) is a 40-in.-long aluminum alloy [E = 10,000 ksi,  = 12.5 × 10−6/°F] bar with a crosssectional area of 2.00 in.2. Both bars are unstressed before the load P is applied. If a concentrated load of P = 27 kips is applied to the rigid bar at D and the temperature is decreased by 100°F, determine: (a) the normal stresses in bars (1) and (2). (b) the normal strains in bars (1) and (2). (c) the deflection of the rigid bar at point D.

Fig. P5.62

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin A gives the best information for this situation: M A  (36 in.)F1  (84 in.)F2  (98 in.)(27 kips)  0 (a) Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: v vB  C (b) 36 in. 84 in. There are no gaps, clearances, or other misfits at pins B and C; therefore, Eq. (b) can be rewritten in terms of the member deformations as: 1 2  (c) 36 in. 84 in. Note: To understand the negative sign associated with 1, see Section 5.5 for discussion of statically indeterminate rigid bar configurations with opposing members. Force-Deformation Relationships FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 A1E1 A2 E2 Compatibility Equation   1  F1 L1 1  F2 L2   1T1 L1     2 T2 L2    36 in.  A1 E1  84 in.  A2 E2 

(d)

(e)

Solve the Equations Solve Eq. (e) for F1: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

 36 in.  F2 L2  AE  F1      2 T2 L2   1T1 L1  1 1    84 in.  A2 E2  L1  3 F2 L2 3  AE     2 T2 L2  1T1 L1  1 1  7 A2 E2 7  L1

Substitute this expression into equilibrium equation (a) and solve for F2 using T1 = T2 = −100°F: (36 in.)F1  (84 in.)F2  (98 in.)(27 kips)   3 F2 L2 3  AE  (36 in.)      2 T2 L2  1T1 L1  1 1   (84 in.)F2  (98 in.)(27 kips)  L1    7 A2 E2 7

 3 L2 A1 E1  (36 in.)  84 in. F2  (98 in.)(27 kips)   7 L1 A2 E2  AE AE 3 (36 in.)  2 T2 L2 1 1  (36 in.)1T1 L1 1 1 7 L1 L1  3 40 in. 1.25 in.2 15, 000 ksi  (36 in.)  84 in. F2  (98 in.)(27 kips)  2  7 30 in. 2.00 in. 10, 000 ksi  3 (1.25 in.2 )(15, 000 ksi) (36 in.) (12.5 106 )(100F)(40 in.) 7 30 in. 2 (1.25 in. )(15, 000 ksi) (36 in.)(9.4 106 )(100F)(30 in.) 30 in. 2, 646 kip-in.  482.1429 kip-in.  634.5000 kip-in. 3,762.6429 kip-in.  19.2857 in.  84 in. 103.2857 in.  36.4295 kips Backsubstitute into Eq. (a) to find F1: (84 in.)(36.4295 kips)  (98 in.)(27 kips) F1   11.5022 kips 36 in. F2 

Normal Stresses The normal stresses in each axial member can now be calculated: F 11.5022 kips 1  1   9.20 ksi (T) A1 1.25 in.2

2 

F2 36.4295 kips   18.21 ksi (T) A2 2.00 in.2

Ans. Ans.

(b) Normal Strains The force-temperature-deformation relationships were expressed in Eq. (d). By definition,  = /L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (d) by their respective member lengths: F F 1  1  1T1  2  2   2 T2 A1 E1 A2 E2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Substitute the appropriate values to calculate the normal strains in each member: F 1  1  1T1 A1E1



11.5022 kips  (9.4  106 / °F)(  100F) 2 (1.25 in. )(15,000 ksi)

 326.55  106 in./in.  327 με

2  

Ans.

F2   2 T2 A2 E2 36.4295 kips  (12.5  106 / °F)(  100F) 2 (2.00 in. )(10,000 ksi)

 571.48  10 6 in./in.  571 με

Ans.

(c) Deflection of the rigid bar at D Calculate the deformation of one of the axial members, say member (2): FL (36.4295 kips)(40 in.)  2  2 2   2 T2 L2   (12.5  10 6 / °F)(  100F)(40 in.) A2 E2 (2.00 in.2 )(10,000 ksi)  0.022859 in. This deformation can also be determined from the strain in member (2):  2   2 L2  (571.48  106 in./in.)(40 in.)  0.022859 in.

Since there are no gaps at pin C, the rigid bar deflection at C is equal to the deformation of member (2); therefore, vC = 2 = 0.022859 in. (downward). From similar triangles, the deflection of the rigid bar at D is related to vC by: vC v  D 84 in. 98 in. The deflection of the rigid bar at D is thus: 98 in. 98 in. vD  vC  (0.022859 in.)  0.0267 in.  84 in. 84 in.

Ans.

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5.63 The pin-connected structure shown in Fig. P5.63 consists of a rigid bar ABC, a steel bar (1), and a steel rod (2). The cross-sectional area of bar (1) is 1.5 in.2 and the diameter of rod (2) is 0.75 in. Assume E = 30,000 ksi and  = 6.6 × 10−6/°F for both axial members. The bars are unstressed when the structure is assembled at 70°F. After application of a concentrated force of P = 20 kips, the temperature is decreased to 30°F. Determine: (a) the normal stresses in bar (1) and rod (2). (b) the normal strains in bar (1) and rod (2). (c) the deflection of pin C from its original position.

Fig. P5.63

Solution Equilibrium Consider a FBD of the rigid bar. Assume tension forces in members (1) and (2). A moment equation about pin B gives the best information for this situation: M B  (12 in.)F1  (20 in.)F2 (15 in.)(20 kips)  0

(a)

Geometry of Deformations Relationship Draw a deformation diagram of the rigid bar. The deflections of the rigid bar are related by similar triangles: v vA (b)  C 12 in. 20 in. There are no gaps, clearances, or other misfits at pins A and C; therefore, Eq. (b) can be rewritten in terms of the member deformations as: 1 2  (c) 12 in. 20 in. Note: To understand the negative sign associated with 1, see Section 5.5 for a discussion of statically indeterminate rigid bar configurations with opposing members. Force-Temperature-Deformation Relationships FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 A1E1 A2 E2

(d)

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Compatibility Equation   1  F1L1 1  F2 L2   1T1L1     2 T2 L2    12 in.  A1E1  20 in.  A2 E2 

(e)

Solve the Equations Solve Eq. (e) for F1:  12 in.  F2 L2  AE  F1      2 T2 L2   1T1 L1  1 1    20 in.  A2 E2  L1  12 F2 L2 12  AE     2 T2 L2  1T1 L1  1 1  20 A2 E2 20  L1 Substitute this expression into equilibrium equation (a) and solve for F2 using T1 = T2 = −40°F:

(12 in.)F1  (20 in.)F2  (15 in.)(20 kips)   12 F2 L2 12  AE  (12 in.)      2 T2 L2  1T1 L1  1 1   (20 in.)F2  (15 in.)(20 kips)  L1    20 A2 E2 20

Note that the area of rod (2) is A2 = /4(0.75 in.)2 = 0.4418 in.2.   12 L2 A1 E1  (12 in.)  20 in. F2  (15 in.)(20 kips)   20 L1 A2 E2  (12 in.)

AE AE 12  2 T2 L2 1 1  (12 in.)1T1 L1 1 1 20 L1 L1

 12 80 in. 1.5 in.2 30, 000 ksi   20 32 in. 0.4418 in.2 30, 000 ksi (12 in.)  20 in. F2  (15 in.)(20 kips)   12 (1.5 in.2 )(30, 000 ksi) (12 in.) (6.6 106 )( 40F)(80 in.) 20 32 in. 2 (1.5 in. )(30, 000 ksi) (12 in.)(6.6 106 )( 40F)(32 in.) 32 in. 300 kip-in.  213.84 kip-in.  142.56 kip-in. 656.40 kip-in.  61.113626 in.  20 in. 81.1136267 in.  8.092 kips

F2 

Backsubstitute into Eq. (a) to find F1: (20 in.)(8.092 kips)  (15 in.)(20 kips) F1   11.513 kips 12 in.

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(a) Normal Stresses The normal stresses in each axial member can now be calculated: F 11.513 kips 1  1   7.68 ksi (C) A1 1.5 in.2 F 8.092 kips 2  2   18.32 ksi (T) A2 0.4418 in.2

Ans. Ans.

(b) Normal Strains The force-temperature-deformation relationships were expressed in Eq. (d). By definition,  = /L. Therefore, the normal strain for each axial member can be determined by dividing the relationships in Eq. (d) by the respective member lengths: F F 1  1  1T1  2  2   2 T2 A1 E1 A2 E2 Substitute the appropriate values to calculate the normal strains in each member: F 1  1  1T1 A1E1



11.513 kips  (6.6  10 6 / °F)(  40F) (1.5 in.2 )(30,000 ksi)

 519.8  106 in./in.  520 με

2  

Ans.

F2   2 T2 A2 E2 8.092 kips  (6.6  106 / °F)(  40F) (0.4418 in.2 )(30,000 ksi)

 346.5  10 6 in./in.  347 με

Ans.

(c) Deflection of the rigid bar at C Calculate the deformation of one of the axial members, say member (2): FL  2  2 2   2 T2 L2 A2 E2 (8.092 kips)(80 in.)  (6.6  106 / °F)(  40F)(80 in.) 2 (0.4418 in. )(30,000 ksi)  0.027723 in. This deformation can also be determined from the strain in member (2):  2   2 L2  (346.5  10 6 in./in.)(80 in.)  0.027723 in. 

Since there are no gaps at pin C, the rigid bar deflection at C is equal to the deformation of member (2); therefore:

vC   2  0.0277 in. 

Ans.

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5.64 Three rods of different materials are connected and placed between rigid supports at A and D, as shown in Fig. P5.64. Properties for each of the three rods are given below. The bars are initially unstressed when the structure is assembled at 70°F. After the temperature has been increased to 250°F, determine: (a) the normal stresses in the three rods. (b) the force exerted on the rigid supports. (c) the deflections of joints B and C relative to rigid support A. Aluminum (1) L1 = 10 in. A1 = 0.8 in.2 E1 = 10,000 ksi  = 12.5×10-6/°F

Fig. P5.64

Cast Iron (2) L2 = 5 in. A2 = 1.8 in.2 E2 = 22,500 ksi  = 7.5×10-6/°F

Bronze (3) L3 = 7 in.

A3 = 0.6 in.2 E3 = 15,000 ksi  = 9.4×10-6/°F

Solution Equilibrium Consider a FBD at joint B. Assume that both internal axial forces will be tension. Fx   F1  F2  0  F1  F2 (a)

Similarly, consider a FBD at joint C. Assume that both internal axial forces will be tension. Fx   F2  F3  0  F3  F2 (b)

Geometry of Deformations Relationship 1   2   3  0 Force-Temperature-Deformation Relationships FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 A1E1 A2 E2 Compatibility Equation FL F1 L1 FL  1T1 L1  2 2   2 T2 L2  3 3   3 T3 L3  0 A1 E1 A2 E2 A3 E3

(c)

3 

F3 L3   3 T3 L3 A3 E3

(d)

(e)

Solve the Equations From Eq. (a), F1 = F2, and from Eq. (b), F3 = F2. The temperature change is the same for all members; therefore, T1 = T2 = T3 = T. Eq. (e) then can be written as:  F2  L1   T L  F2 L2   T L   F2  L3   T L  0 1 1 2 2 3 3 A1E1 A2 E2 A3 E3

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Solving for F2: F2 L1 F2 L2 F3 L3    1T L1   2 T L2   3T L3 A1 E1 A2 E2 A3 E3  L L  L F2  1  2  3   T 1 L1   2 L2   3 L3   A1 E1 A2 E2 A3 E3  T 1 L1   2 L2   3 L3  F2   L L1 L  2  3 A1 E1 A2 E2 A3 E3

(f)

Substitute the problem data along with T = +180°F into Eq. (f) and calculate F1 = −19.1025 kips. From Eq. (a), F1 = −19.1025 kips and from Eq. (b), F3 = −19.1025 kips. (a) Normal Stresses The normal stresses in each rod can now be calculated: F 19.1025 kips 1  1   23.878 ksi  23.9 ksi (C) A1 0.8 in.2

Ans.

2 

F2 19.1025 kips   10.613 ksi  10.61 ksi (C) A2 1.8 in.2

Ans.

3 

F3 19.1025 kips   31.838 ksi  31.8 ksi (C) A3 0.6 in.2

Ans.

(b) Force on Rigid Supports The force exerted on the rigid supports is equal to the internal axial force: RA  RD  19.10 kips

Ans.

(c) Deflection of Joints B and C The deflection of joint B is equal to the deformation (i.e., contraction in this instance) of rod (1). The deformation of rod (1) is given by: FL 1  1 1  1T1L1 A1E1

( 19.1025 kips)(10 in.)  (12.5  106 / °F)(180°F)(10 in.)  0.001378 in. (0.8 in.2 )(10,000 ksi) The deflection of joint B is thus: uB  1  0.001378 in.  The deformation of rod (2) is given by: FL  2  2 2   2 T2 L2 A2 E2 

( 19.1025 kips)(5 in.)  (7.5  106 / °F)(180°F)(5 in.)  0.004392 in. 2 (1.8 in. )(22,500 ksi) The deflection of joint C is: uC  uB   2  0.001378 in.  0.004392 in.  0.00301 in. 

Ans.



Ans.

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5.65 Three rods of different materials are connected and placed between rigid supports at A and D, as shown in Fig. P5.65. Properties for each of the three rods are given below. The bars are initially unstressed when the structure is assembled at 20°C. After the temperature has been increased to 100°C, determine: (a) the normal stresses in the three rods. (b) the force exerted on the rigid supports. (c) the deflections of joints B and C relative to rigid support A. Aluminum (1) L1 = 440 mm A1 = 1,200 mm2 E1 = 70 GPa  = 22.5×10-6/°C

Fig. P5.65

Cast Iron (2) L2 = 200 mm A2 = 2,800 mm2 E2 = 155 GPa  = 13.5×10-6/°C

Bronze (3)

L3 = 320 mm A3 = 800 mm2 E3 = 100 GPa  = 17.0×10-6/°C

Solution Equilibrium Consider a FBD at joint B. Assume that both internal axial forces will be tension. Fx   F1  F2  0  F1  F2 (a)

Similarly, consider a FBD at joint C. Assume that both internal axial forces will be tension. Fx   F2  F3  0  F3  F2 (b) Geometry of Deformations Relationship 1   2   3  0 Force-Temperature-Deformation Relationships FL FL 1  1 1  1T1L1  2  2 2   2 T2 L2 A1E1 A2 E2 Compatibility Equation FL F1 L1 FL  1T1 L1  2 2   2 T2 L2  3 3   3 T3 L3  0 A1 E1 A2 E2 A3 E3

(c)

3 

F3 L3   3 T3 L3 A3 E3

(d)

(e)

Solve the Equations From Eq. (a), F1 = F2, and from Eq. (b), F3 = F2. The temperature change is the same for all members; therefore, T1 = T2 = T3 = T. Eq. (e) then can be written as:  F2  L1   T L  F2 L2   T L   F2  L3   T L  0 1 1 2 2 3 3 A1E1 A2 E2 A3 E3

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Solving for F2: F2 L1 F2 L2 F3 L3    1T L1   2 T L2   3T L3 A1 E1 A2 E2 A3 E3  L L  L F2  1  2  3   T 1 L1   2 L2   3 L3   A1 E1 A2 E2 A3 E3  T 1 L1   2 L2   3 L3  F2   (f) L3 L1 L2   A1 E1 A2 E2 A3 E3 Substitute the problem data along with T = +80°C into Eq. (f) and calculate F1 = −148.80 kN. From Eq. (a), F1 = −148.80 kN and from Eq. (b), F3 = −148.80 kN.

(a) Normal Stresses The normal stresses in each rod can now be calculated: F 148,800 N 1  1   124.00 MPa  124.0 MPa (C) A1 1,200 mm 2

Ans.

2 

F2 148,800 N   53.143 MPa  53.1 MPa (C) A2 2,800 mm 2

Ans.

3 

F3 148,800 N   186.0 MPa  186.0 MPa (C) A3 800 mm 2

Ans.

(b) Force on Rigid Supports The force exerted on the rigid supports is equal to the internal axial force: RA  RD  148.8 kN

Ans.

(c) Deflection of Joints B and C The deflection of joint B is equal to the deformation (i.e., contraction in this instance) of rod (1). The deformation of rod (1) is given by: FL 1  1 1  1T1L1 A1E1

( 148,800 N)(440 mm)  (22.5  106 / °C)(80°C)(440 mm)  0.01257 mm (1,200 mm 2 )(70,000 N/mm 2 ) The deflection of joint B is thus: uB  1  0.01257 mm  

Ans.

The deformation of rod (2) is given by: FL  2  2 2   2 T2 L2 A2 E2



( 148,800 N)(200 mm)  (13.5  106 / °C)(80°C)(200 mm)  0.14743 mm 2 2 (2,800 mm )(155,000 N/mm )

The deflection of joint C is: uC  uB   2  0.01257 mm  0.14743 mm  0.1600 mm 

Ans.

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5.66 The machine part shown in Fig. P5.66 is 3/8-in. thick and is made of cold-rolled 18-8 stainless steel (see Appendix D for properties). Determine the maximum safe load P if a factor of safety of 2.5 with respect to failure by yield is specified. Fig. P5.66

Solution  allow 

Y FS



165 ksi  66 ksi 2.5

Fillet: D 4.0 in.   2.0 d 2.0 in. Pallow 

 allow Amin K



r 0.5 in.   0.25 d 2.0 in.

(66 ksi)(2.0 in.)(0.375 in.)  27.2 kips 1.82

Hole: d 1.25 in.   0.3125 D 4.0 in. Pallow 

 allow Anet K



 K  1.82

(a)

 K  2.34

(66 ksi)(4.0 in.  1.25 in.)(0.375 in.)  29.1 kips 2.34

Controlling Load: The fillet controls in this case; therefore, Pallow  27.2 kips

(b)

Ans.

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5.67 The machine part shown in Fig. P5.67 is 12 mm thick and is made of SAE 4340 heat-treated steel (see Appendix D for properties). The holes are centered in the bar. Determine the maximum safe load P if a factor of safety of 3.0 with respect to failure by yield is specified.  Fig. P5.67

Solution  allow 

Y FS



910 MPa  303.33 MPa 3.0

Small Hole: d 15 mm   0.1 D 100 mm Pallow 

 allow Anet K



(303.33 N/mm 2 )(100 mm  10 mm)(12 mm)  120,000 N  120.0 kN 2.73

Large Hole: d 35 mm   0.35 D 100 mm Pallow 

 allow Anet K



 K  2.73

(a)

 K  2.29

(303.33 N/mm 2 )(100 mm  35 mm)(12 mm)  103,319 N  103.3 kN 2.29

Controlling Load: The large hole controls in this case; therefore, Pallow  103.3 kN

(b)

Ans.

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5.68 A 100-mm-wide by 8-mm-thick steel bar is transmitting an axial tensile load of 3,000 N. After the load is applied, a 4mm-diameter hole is drilled through the bar, as shown in Fig. P5.68. The hole is centered in the bar. (a) Determine the stress at point A (on the edge of the hole) in the bar before and after the hole is drilled. (b) Does the axial stress at point B on the edge of the bar increase or decrease as the hole is drilled? Explain.

Fig. P5.68

Solution (a) Stress at point A: Before hole is drilled: P 3,000 N A    3.75 MPa A (100 mm)(8 mm) After hole is drilled: d 4 mm   0.04 D 100 mm

A 

Ans.

 K  2.89

PK (3,000 N)(2.89)   11.29 MPa Anet (100 mm  4 mm)(8 mm)

Ans.

(b) Stress at point B: The axial stress at point B decreases. Since the average stress changes very little with the introduction of the small hole and since the stress at A is larger than the average stress, the axial stress far from the hole must be less than the average stress.

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5.69 The machine part shown in Fig. P5.69 is 90-mm-wide by 12-mm-thick and is made of 2014-T4 aluminum (see Appendix D for properties). The hole is centered in the bar. Determine the maximum safe load P if a factor of safety of 1.50 with respect to failure by yield is specified.

Fig. P5.69

Solution  allow 

Y FS



290 MPa  193.33 MPa 1.50

Hole: d 30 mm   0.33 D 90 mm Pallow 

 allow Anet K

(193.33 N/mm 2 )(90 mm  30 mm)(12 mm)   60, 260 N  60.3 kN 2.31

Notches: r 10 mm   0.20 d 50 mm Pallow 

 allow Amin K

 K  2.31



D 90 mm   1.8 d 50 mm

 K  2.27

(193.33 N/mm 2 )(50 mm)(12 mm)  51,101 N  51.1 kN 2.27

Controlling Load: The notches control in this case; therefore, Pallow  51.1 kN

(a)

(b)

Ans.

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5.70 The machine part shown in Fig. P5.70 is 8-mm thick and is made of AISI 1020 cold-rolled steel (see Appendix D for properties). Determine the maximum safe load P if a factor of safety of 3 with respect to failure by yield is specified.

Fig. P5.70

Solution  allow 

Y FS



427 MPa  142.33 MPa 3

Fillets: D 120 mm   1.5 d 80 mm Pallow 

 allow Amin K



r 16 mm   0.2 d 80 mm

(142.33 MPa)(80 mm)(8 mm)  49.8 kN 1.83

Hole: d 15 mm   0.125 D 120 mm Pallow 

 allow Anet K



 allow Amin K



(a)

 K  2.67

(142.33 MPa)(120 mm  15 mm)(8 mm)  44.8 kN 2.67

Notches: r 10 mm   0.167 d 60 mm Pallow 

 K  1.83

D 800 mm   1.333 d 60 mm

 K  2.28

(142.33 MPa)(60 mm)(8 mm)  30.0 kN 2.28

Controlling Load: The notches control in this case; therefore, Pallow  30.0 kN

(b)

(c)

Ans.

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5.71 The machine part shown in Fig. P5.71 is 10 mm thick, is made of AISI 1020 coldrolled steel (see Appendix D for properties), and is subjected to a tensile load of P = 45 kN. Determine the minimum radius r that can be used between the two sections if a factor of safety of 2 with respect to failure by yield is specified. Round the minimum fillet radius up to the nearest 1-mm multiple. Fig. P5.71

Solution  allow  K

Y FS



 allow Amin Pallow

427 MPa  213.5 MPa 2

(213.5 N/mm2 )(40 mm)(10 mm)   1.90 45,000 N

D 80 mm  2 d 40 mm

Then, from Fig. 5.15 r  0.214 d

 rmin  (0.214)d  (0.214)(40 mm)  8.56 mm

 say rmin  9 mm

Ans.

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5.72 The 0.25-in.-thick bar shown in Fig. P5.72 is made of 2014-T4 aluminum (see Appendix D for properties) and will be subjected to an axial tensile load of P = 1,500 lbs. A 0.5625-in.-diameter hole is located on the centerline of the bar. Determine the minimum safe width D for the bar if a factor of safety of 2.5 with respect to failure by yield must be maintained. Fig. P5.72

Solution  allow 

Y FS



42 ksi  16.8 ksi  16,800 psi 2.5

By definition:  K  max  nom In this instance, the maximum stress equals the allowable stress:  16,800 psi K  max   K  nom  16,800 psi  nom  nom If we denote the net width of the bar at the hole location as w and the bar thickness at t, then w must satisfy: P P KP K  nom  K K  16,800 psi w  (a) Anet wt (16,800 psi) t And the bar width D is the sum of the net width w and the hole diameter d: (b) D  w d For a bar with a hole, the value of K is dependent on the ratio of d/D. Since K cannot be determined until D is known, and since the calculation of D from Eqs. (a) and (b) depends on K, a trial-and-error process can be established. One possible sequence is summarized in the table below. d D

d D

K [from Fig. 5.14]

w (in.) [from Eq. (a)]

D (in.) [from Eq. (b)]

0.25

2.43

0.868

1.430

0.393

0.39

2.25

0.804

1.366

0.412

0.41

2.23

0.796

1.359

0.414 (agrees with trial value)

Trial value of

Therefore, the minimum bar width is: Dmin  1.359 in.

Resulting value of

Ans.

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5.73 The stepped bar with a circular hole, shown in Fig. P5.73, is made of annealed 18-8 stainless steel. The bar is 12 mm thick and will be subjected to an axial tensile load of P = 70 kN. The normal stress in the bar is not to exceed 150 MPa. To the nearest millimeter, determine: (a) the maximum allowable hole diameter d. (b) the minimum allowable fillet radius r. Fig. P5.73

Solution (a) Maximum hole diameter d: By definition:  K  max  nom In this instance, the maximum stress equals the 150-MPa allowable stress:  150 MPa K  max   K  nom  150 MPa  nom  nom If we denote the net width of the bar at the hole location as w and the bar thickness at t, then w can be expressed as w = D – d. The nominal stress at the hole location depends on Anet, which in turn can be expressed in terms of w: P P P KP K nom  K K K  150 N/mm 2  (D  d )  Anet wt (D  d ) t (150 N/mm 2 ) t From this relationship, the hole diameter d is: KP d D (a) (150 N/mm 2 ) t For a bar with a hole, the value of K is dependent on the ratio of d/D. Since K cannot be determined until d is known, and since the calculation of d from Eq. (a) depends on K, a trial-and-error process can be established. One possible sequence is summarized in the table below. d D

d D

K [from Fig. 5.14]

d (mm) [from Eq. (a)]

0.10 0.18

2.73 2.56

23.83 30.44

0.183 0.234

0.23 0.26 0.28

2.47 2.42 2.39

33.94 35.89 37.06

0.29

2.37

37.83

0.261 0.276 0.285 0.291 (agrees with trial value)

Trial value of

Therefore, the maximum hole diameter is: dmax  37 mm

Resulting value of

Ans.

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(b) Minimum fillet radius r: D 130 mm   1.3 d 100 mm Therefore, the curve for D/d = 1.30 on Fig. 5.15 will be used to determine the stress concentration factor.  max  allow 150 N/mm2 K    2.571  nom P / Amin (70,000 N) / (100 mm)(12 mm) From Fig. 5.15, r  0.047 d

 rmin  0.047(100 mm)  4.7 mm

say rmin  5 mm

Ans.

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6.1 A solid circular steel shaft having an outside diameter of d = 0.75 in. is subjected to a pure torque of T = 650 lb-in. Determine the maximum shear stress in the shaft.

Solution The polar moment of inertia for the shaft is J



(0.75 in.) 4  0.031063 in.4

32 The maximum shear stress in the steel shaft is found from the elastic torsion formula: Tc (650 lb-in.)(0.75 in. / 2)  max    7,846.93 psi  7,850 psi J 0.031063 in.4

Ans.

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6.2 A hollow aluminum shaft with an outside diameter of 80 mm and a wall thickness of 5 mm has an allowable shear stress of 75 MPa. Determine the maximum torque T that may be applied to the shaft.

Solution The polar moment of inertia for the shaft is





 D 4  d 4   (80 mm)4  (70 mm) 4   1,664,062 mm 4 32 32 Rearrange the elastic torsion formula to determine the maximum torque T:  J (75 N/mm 2 )(1,664,062 mm 4 ) T  allow   3,120,117 N-mm  3,120 N-m c 80 mm / 2 J

Ans.

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6.3 A hollow steel shaft with an outside diameter of 100 mm and a wall thickness of 10 mm is subjected to a pure torque of T = 5,500 N-m. (a) Determine the maximum shear stress in the hollow shaft. (b) Determine the minimum diameter of a solid steel shaft for which the maximum shear stress is the same as in part (a) for the same torque T.

Solution The polar moment of inertia for the shaft is





 D 4  d 4   (100 mm)4  (80 mm) 4   5,796, 238 mm 4 32 32 (a) The maximum shear stress in the hollow steel shaft is found from the elastic torsion formula: Tc (5,500 N-m)(100 mm / 2)(1,000 mm/m)  max    47.445 MPa  47.4 MPa J 5,796, 238 mm 4 J

Ans.

(b) The polar moment of inertia for a solid shaft can be expressed as  J  d4 32 Rearrange the elastic torsion formula to group terms with d on the left-hand side:  d4 T  32 (d / 2)  and simplify to d 3 T  16  From this equation, the unknown diameter of the solid shaft can be expressed as 16T d3  To support a torque of T = 5,500 N-m without exceeding the maximum shear stress determined in part (a), a solid shaft must have a diameter of 16T 16(5,500 N-m)(1,000 mm/m) Ans. d3 3  83.891 mm  83.9 mm   (47.445 N/mm2 )

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6.4 A compound shaft consists of two pipe segments. Segment (1) has an outside diameter of 200 mm and a wall thickness of 10 mm. Segment (2) has an outside diameter of 150 mm and a wall thickness of 10 mm. The shaft is subjected to torques TB = 42 kN-m and TC = 18 kN-m, which act in the directions shown in Fig. P6.4. Determine the maximum shear stress magnitude in each shaft segment. Fig. P6.4

Solution

Equilibrium: M x  T1  42 kN-m  18 kN-m  0 M x  T2  18 kN-m  0 Section properties: J1 

J2 









 T1  24 kN-m  T2  18 kN-m

 D14  d14   (200 mm) 4  (180 mm) 4   54,019,686 mm 4 32  32 

 D24  d 24   (150 mm)4  (130 mm)4   21,551,281 mm 4 32  32 

Shear stress magnitudes: (24 kN-m)(200 mm / 2)(1,000 N/kN)(1,000 mm/m) 1   44.4 MPa 54,019,686 mm 4 (18 kN-m)(150 mm / 2)(1,000 N/kN)(1,000 mm/m) 2   62.3 MPa 21,661,281 mm 4

Ans. Ans.

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6.5 A compound shaft consists of two pipe segments. Segment (1) has an outside diameter of 10.75 in. and a wall thickness of 0.365 in. Segment (2) has an outside diameter of 6.625 in. and a wall thickness of 0.280 in. The shaft is subjected to torques TB = 60 kip-ft and TC = 24 kip-ft, which act in the directions shown in Fig. P6.5. Determine the maximum shear stress magnitude in each shaft segment. Fig. P6.5

Solution

Equilibrium: M x  T1  60 kip-ft  24 kip-ft  0 M x  T2  24 kip-ft  0 Section properties: J1 

J2 









 T1  36 kip-ft  T2  24 kip-ft

 D14  d14   (10.75 in.) 4  (10.02 in.) 4   321.4685 in.4 32  32 

 D24  d 24   (6.625 in.)4  (6.0650 in.) 4   56.2844 in.4 32  32 

Shear stress magnitudes: (36 kip-ft)(10.75 in. / 2)(12 in./ft) 1   7.22 ksi 321.4685 in.4 (24 kip-ft)(6.625 in. / 2)(12 in./ft) 2   16.95 ksi 56.2844 in.4

Ans. Ans.

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6.6 A compound shaft (Fig. P6.6) consists of brass segment (1) and aluminum segment (2). Segment (1) is a solid brass shaft with an outside diameter of 0.625 in. and an allowable shear stress of 6,000 psi. Segment (2) is a solid aluminum shaft with an outside diameter of 0.50 in. and an allowable shear stress of 9,000 psi. Determine the magnitude of the largest torque TC that may be applied at C.

Fig. P6.6

Solution Section properties: J1  J2 



32

(0.625 in.) 4  0.014980 in.4



32

(0.50 in.) 4  0.006136 in.4

Allowable internal torques:  J (6,000 psi)(0.014980 in.4 ) T1  1 1   287.621 lb-in. c1 0.625 in./2

T2 

 2 J2 c2



(9,000 psi)(0.006136 in.4 )  220.893 lb-in. 0.50 in./2

 controls

Equilibrium: The internal torque magnitude in each segment equals the external torque; therefore, T1 = T2 = TC. The controlling internal torque is T2 = 220.893 lb-in.; therefore, the maximum external torque TC that may be applied to the compound shaft is Ans. TC  220.893 lb-in.  221 lb-in.

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6.7 A compound shaft (Fig. P6.7) consists of brass segment (1) and aluminum segment (2). Segment (1) is a solid brass shaft with an allowable shear stress of 60 MPa. Segment (2) is a solid aluminum shaft with an allowable shear stress of 90 MPa. If a torque of TC = 23,000 N-m is applied at C, determine the minimum required diameter of (a) the brass shaft and (b) the aluminum shaft.

Fig. P6.7

Solution The polar moment of inertia for a solid shaft can be expressed as  J  d4 32 Rearrange the elastic torsion formula to group terms with d on the left-hand side:  d4 T  32 (d / 2)  and simplify to d 3 T  16  From this equation, the unknown diameter of the solid shaft can be expressed as 16T d3  Equilibrium: For this shaft, the internal torque magnitude in each segment equals the external torque; therefore, T1 = T2 = TC = 23,000 N-m. Minimum shaft diameters: (a) Brass shaft (1) 16T1 16(23,000 N-m)(1,000 mm/m) d1  3 3  125.0 mm  allow,1  (60 N/mm2 ) (b) Aluminum shaft (2) 16T2 16(23,000 N-m)(1,000 mm/m) d2  3 3  109.2 mm  allow,2  (90 N/mm 2 )

Ans.

Ans.

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6.8 A solid 0.75-in.-diameter shaft is subjected to the torques shown in Fig. P6.8. The bearings shown allow the shaft to turn freely. (a) Plot a torque diagram showing the internal torque in segments (1), (2), and (3) of the shaft. Use the sign convention presented in Section 6-6. (b) Determine the maximum shear stress magnitude in the shaft. Fig. P6.8

Solution Equilibrium:

M x  T1  10 lb-ft  0

 T1  10 lb-ft

M x  T2  10 lb-ft  50 lb-ft  0

T2  40 lb-ft

M x  T3  30 lb-ft  0

 T3  30 lb-ft

Section properties: J1  J 2  J 3 



32

(0.75 in.) 4  0.031063 in.4

Shear stress magnitudes: T c (10 lb-ft)(0.75 in. / 2)(12 in./ft) 1  1 1   1, 448.7 psi J1 0.031063 in.4

2 

T2c2 (40 lb-ft)(0.75 in. / 2)(12 in./ft)   5,794.7 psi J2 0.031063 in.4

3 

T3c3 (30 lb-ft)(0.75 in. / 2)(12 in./ft)   4,346.0 psi J3 0.031063 in.4

The maximum shear stress in the shaft occurs in segment (2):  max  5,790 psi

Ans.

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6.9 A solid constant-diameter shaft is subjected to the torques shown in Fig. P6.9. The bearings shown allow the shaft to turn freely. (a) Plot a torque diagram showing the internal torque in segments (1), (2), and (3) of the shaft. Use the sign convention presented in Section 6-6. (b) If the allowable shear stress in the shaft is 80 MPa, determine the minimum acceptable diameter for the shaft. Fig. P6.9

Solution

M x  160 N-m  T3  0

T3  160 N-m

M x  160 N-m  380 N-m  T2  0

T2  220 N-m

M x  160 N-m  380 N-m  330 N-m  T1  0

T1  110 N-m

The maximum torque magnitude in the shaft occurs in segment (2): Tmax = 220 N-m.

(b) The elastic torsion formula gives the relationship between shear stress and torque in a shaft. Tc  J In this instance, the torque and the allowable shear stress are known for the shaft. Rearrange the elastic torsion formula, putting the known terms on the right-hand side of the equation: J T  c  Express the left-hand side of this equation in terms of the shaft diameter D:



d4

32   d 3  T d / 2 16  and solve for the minimum acceptable diameter: 16 T 16(220 N-m)(1,000 mm/m) d3    14,005.635 mm3    (80 N/mm 2 )

 d  24.1 mm

Ans.

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6.10 A solid circular steel shaft having an outside diameter of 1.25 in. is subjected to a pure torque of T = 2,200 lb-in. The shear modulus of the steel is G = 12,000 ksi. Determine: (a) the maximum shear stress in the shaft. (b) the magnitude of the angle of twist in a 6-ft length of shaft.

Solution The polar moment of inertia for the shaft is J



32

(1.25 in.) 4  0.239684 in.4

(a) The maximum shear stress in the steel shaft is found from the elastic torsion formula: Tc (2,200 lb-in.)(1.25 in. / 2)  max    5, 736.7 psi  5.74 ksi J 0.239684 in.4 (b) The magnitude of the angle of twist in a 6-ft length of shaft is TL (2, 200 lb-in.)(6 ft)(12 in./ft)    0.055072 rad  0.0551 rad  3.16 JG (0.239684 in.4 )(12,000,000 psi)

Ans.

Ans.

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6.11 A solid circular steel shaft having an outside diameter of 35 mm is subjected to a pure torque of T = 640 N-m. The shear modulus of the steel is G = 80 GPa. Determine: (a) the maximum shear stress in the shaft. (b) the magnitude of the angle of twist in a 1.5-m length of shaft.

Solution The polar moment of inertia for the shaft is J



32

(35 mm)4  147,323.515 mm4

(a) The maximum shear stress in the steel shaft is found from the elastic torsion formula: Tc (640 N-m)(35 mm / 2)(1,000 mm/m)  max    76.023 MPa  76.0 MPa J 147,323.515 mm 4 (b) The magnitude of the angle of twist in a 1.5-m length of shaft is TL (640 N-m)(1.5 m)(1,000 mm/m) 2    0.081453 rad  0.0815 rad  4.67 JG (147,323.515 mm 4 )(80,000 N/mm 2 )

Ans.

Ans.

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6.12 A hollow steel shaft with an outside diameter of 85 mm and a wall thickness of 10 mm is subjected to a pure torque of T = 7,000 N-m. The shear modulus of the steel is G = 80 GPa. Determine: (a) the maximum shear stress in the shaft. (b) the magnitude of the angle of twist in a 2.5-m length of shaft.

Solution The polar moment of inertia for the shaft is J





 D 4  d 4   (85 mm)4  (65 mm) 4   3,372,303 mm 4 32 32

(a) The maximum shear stress in the steel shaft is found from the elastic torsion formula: Tc (7,000 N-m)(85 mm / 2)(1,000 mm/m)  max    88.219 MPa  88.2 MPa J 3,372,303 mm 4 (b) The magnitude of the angle of twist in a 2.5-m length of shaft is TL (7,000 N-m)(2.5 m)(1,000 mm/m) 2    0.064867 rad  0.0649 rad  3.72 JG (3,372,303 mm 4 )(80,000 N/mm 2 )

Ans.

Ans.

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6.13 A solid stainless steel [G = 12,500 ksi] shaft that is 72-in. long will be subjected to a pure torque of T = 900 lb-in. Determine the minimum diameter required if the shear stress must not exceed 8,000 psi and the angle of twist must not exceed 5°. Report both the maximum shear stress  and the angle of twist  at this minimum diameter.

Solution Consider shear stress: The polar moment of inertia for a solid shaft can be expressed as  J  d4 32 The elastic torsion formula can be rearranged to gather terms with d:  d4 d 3 T   32 ( d / 2) 16  From this equation, the unknown diameter of the solid shaft can be expressed as 16T d3  For the solid stainless steel shaft, the minimum diameter that will satisfy the allowable shear stress is: 16(900 lb-in.) d3  0.831 in.  (8,000 psi) Consider angle of twist: Rearrange the angle of twist equation: TL  4 TL  J  d  JG 32 G and solve for the minimum diameter that will satisfy the angle of twist limitation: 32TL 32(900 lb-in.)(72 in.) d4 4  0.882 in. G  (5)( rad/180)(12,500,000 psi) Therefore, the minimum diameter that could be used for the shaft is dmin  0.882 in.

Ans.

The angle of twist for this shaft is  = 5° = 0.087266 rad. To compute the shear stress in a 0.882-in.diameter shaft, first compute the polar moment of inertia: J



(0.882 in.)4  0.059404 in.4

32 The shear stress in the shaft is thus: (900 lb-in.)(0.882 in. / 2)  max   6,681.12 psi  6,680 psi 0.059404 in.4

Ans.

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6.14 A solid stainless steel [G = 86 GPa] shaft that is 2.0 m long will be subjected to a pure torque of T = 75 N-m. Determine the minimum diameter required if the shear stress must not exceed 50 MPa and the angle of twist must not exceed 4°. Report both the maximum shear stress  and the angle of twist  at this minimum diameter.

Solution Consider shear stress: The polar moment of inertia for a solid shaft can be expressed as  J  d4 32 The elastic torsion formula can be rearranged to gather terms with d:  d4 d 3 T   32 ( d / 2) 16  From this equation, the unknown diameter of the solid shaft can be expressed as 16T d3  For the solid stainless steel shaft, the minimum diameter that will satisfy the allowable shear stress is: 16(75 N-m)(1,000 mm/m) d3  19.70 mm  (50 N/mm2 ) Consider angle of twist: Rearrange the angle of twist equation: TL  4 TL  J  d  JG 32 G and solve for the minimum diameter that will satisfy the angle of twist limitation: 32TL 32(75 N-m)(2,000 mm)(1,000 mm/m) d4 4  22.46 mm G  (0.069813 rad)(86,000 N/mm 2 ) Therefore, the minimum diameter that could be used for the shaft is dmin  22.5 mm

Ans.

The angle of twist for this shaft is  = 0.069813 rad. To compute the shear stress in a 22.46-mmdiameter shaft, first compute the polar moment of inertia: J



(22.46 mm) 4  24,983.625 mm 4

32 The shear stress in the shaft is thus: (75 N-m)(22.46 mm / 2)(1,000 mm/m)  max   33.7 MPa 24,983.625 mm 4

Ans.

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6.15 A hollow steel [G = 12,000 ksi] shaft with an outside diameter of 3.50 in. will be subjected to a pure torque of T = 3,750 lb-ft. Determine the maximum inside diameter d that can be used if the shear stress must not exceed 8,000 psi and the angle of twist must not exceed 3° in an 8-ft length of shaft. Report both the maximum shear stress  and the angle of twist  for this maximum inside diameter.

Solution Consider shear stress: The polar moment of inertia for a hollow shaft can be expressed as   D 4  d 4  J 32 The elastic torsion formula can be rearranged to gather terms with D: 4 4   D  d  T  32 D / 2  Rearrange this equation to isolate the inside diameter d term: 32 T D 32 T D D4  d 4   d 4  D4  2 2 From this equation, the unknown inside diameter of the hollow shaft can be expressed as 32 T D d  4 D4  2 For the hollow steel shaft, the maximum diameter that will satisfy the allowable shear stress is: 32(3,750 lb-ft)(3.50 in.)(12 in./ft) d  4 (3.50 in.)4   2.6564 in. 2 (8,000 psi) Consider angle of twist: Rearrange the angle of twist equation: TL  TL  D 4  d 4    J  JG 32 G Rearrange this equation to isolate the inside diameter d term: 32TL d 4  D4   G and solve for the maximum inside diameter that will satisfy the angle of twist limitation: 32 TL 32(3,750 lb-ft)(8 ft)(12 in./ft) 2  4 (3.50 in.) 4   2.991 in.  G  (3)( /180)(12,000,000 psi) Therefore, the maximum inside diameter that could be used for the shaft is dmax  2.66 in. d  4 D4 

Ans.

The shear stress for this shaft is  = 8,000 psi. To compute the angle of twist in an 8-ft length of the hollow shaft, first compute the polar moment of inertia:





 D 4  d 4   (3.50 in.)4  (2.66 in.) 4   9.817318 in.4 32 32 The angle of twist in the shaft is thus: TL (3,750 lb-ft)(8 ft)(12 in./ft) 2    0.0367 rad  2.10 JG (9.817318 in.4 )(12,000,000 psi) J

Ans.

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6.16 A compound steel [G = 80 GPa] shaft (Fig. P6.16) consists of a solid 55-mmdiameter segment (1) and a solid 40-mmdiameter segment (2). The allowable shear stress of the steel is 70 MPa, and the maximum rotation angle at the free end of the compound shaft must be limited to C ≤ 3°. Determine the magnitude of the largest torque TC that may be applied at C.

Fig. P6.16

Solution Section properties: J1  J2 



32



32

(55 mm)4  898,360.5 mm4 (40 mm) 4  251,327.4 mm 4

Consider shear stress: The larger shear stress will occur in the smaller diameter segment; that is, segment (2).  J (70 N/mm 2 )(251,327.4 mm 4 ) Tmax    879,646 N-mm c 40 mm/2 Consider angle of twist: The rotation angle at C is the sum of the angles of twist in segments (1) and (2): TL TL C  1  2  1 1  2 2 J1G1 J 2G2 Since T1 = T2 = TC and G1 = G2, this equation can be simplified to Tmax  L1 L2      3 G  J1 J 2  and solved for the maximum torque: (3)( / 180)(80,000 N/mm 2 ) Tmax   739,395 N-mm 1,200 mm   800 mm  898,360.5 mm4  251,327.4 mm4    Therefore, the magnitude of the largest torque TC that may be applied at C is TC ,max  739,395 N-mm  739 N-m

Ans.

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6.17 A compound shaft (Fig. P6.17) consists of brass segment (1) and aluminum segment (2). Segment (1) is a solid brass [G = 5,600 ksi] shaft with an outside diameter of 1.75 in. and an allowable shear stress of 9,000 psi. Segment (2) is a solid aluminum [G = 4,000 ksi] shaft with an outside diameter of 1.25 in. and an allowable shear stress of 12,000 psi. The maximum rotation angle at the upper end of the compound shaft must be limited to C ≤ 4°. Determine the magnitude of the largest torque TC that may be applied at C.

Fig. P6.17

Solution From equilibrium, the internal torques in segments (1) and (2) are equal to the external torque TC. The elastic torsion formula gives the relationship between shear stress and torque in a shaft. Tc  J Consider shear stress: In this compound shaft, the diameters and allowable shear stresses in segments (1) and (2) are known. The elastic torsion formula can be rearranged to solve for the unknown torque. An expression can be written for each shaft segment: J  J T1  1 1 T2  2 2 c1 c2 For shaft segment (1), the polar moment of inertia is: J1 



(1.75 in.)4  0.920772 in.4

32 Use this value along with the 9,000 psi allowable shear stress to determine the allowable torque T1:  J (9,000 psi)(0.920772 in.4 ) (a) T1  1 1   9,470.8 lb-in. c1 (1.75 in./2)

For shaft segment (2), the polar moment of inertia is: J2 



(1.25 in.) 4  0.239684 in.4

32 Use this value along with the 12,000 psi allowable shear stress to determine the allowable torque T2:  2 J 2 (12,000 psi)(0.239684 in.4 ) (b) T2    4,601.9 lb-in. c2 (1.25 in./2)

Consider angle of twist: The angles of twists in segments (1) and (2) can be expressed as: TL TL 1  1 1 2  2 2 J1G1 J 2G2 The rotation angle at C is the sum of these two angles of twist: TL TL C  1  2  1 1  2 2 J1G1 J 2G2

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and since T1 = T2 = TC:  L L  C  TC  1  2   J1G1 J 2G2  Solving for TC gives:

TC 



C

L1 L  2 J1G1 J 2G2   rad  (4)   180  12 in. 18 in.  4 (0.920772 in. )(5,600,000 psi) (0.239684 in.4 )(4,000,000 psi)

 3,308.4 lb-in.

(c)

Compare the torque magnitudes in Eqs. (a), (b), and (c). The smallest torque controls; therefore, the maximum torque that can be applied to the compound shaft at C is TC = 3,308.4 lb-in. = 276 lb-ft. Ans.

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6.18 A compound steel [G = 80 GPa] shaft consists of solid 30-mm-diameter segments (1) and (3) and tube segment (2), which has an outside diameter of 60 mm and an inside diameter of 50 mm (Fig. P6.18). Determine: (a) the maximum shear stress in tube segment (2). (b) the angle of twist in tube segment (2). (c) the rotation angle of gear D relative to gear A. Fig. P6.18

Solution The torques in all shaft segments are equal; therefore, T1 = T2 = T3 = 240 N-m. Polar moments of inertia in the shaft segments will be needed for this calculation. For segments (1) and (3), which are solid 30-mm-diameter shafts, the polar moment of inertia is: J1 



(30 mm) 4  79,521.56 mm 4  J 3

32 For segment (2), which has a tube cross section, the polar moment of inertia is: J2 



(60 mm) 4  (50 mm) 4   658,752.71 mm 4 32 

Shear stress in tube segment (2): Use the elastic torsion formula to calculate the shear stress caused by an internal torque of T2 = 240 N-m. Note that the radius term used in the elastic torsion formula for the tube is the outside radius; that is, c2 = 60 mm/2 = 30 mm. Tc (240 N-m)(30 mm)(1,000 mm/m) 2  2 2   10.93 MPa Ans. J2 658,752.71 mm 4 Angle of twist in tube segment (2): Apply the angle of twist equation to segment (2). TL (240 N-m)(2,000 mm)(1,000 mm/m) 2  2 2   0.009108 rad  0.00911 rad J 2G2 (658,752.71 mm 4 )(80,000 N/mm 2 )

Ans.

Rotation angle of gear D relative to gear A: The angles of twist in segments (1) and (3) must be calculated. Since both segments have the same shear modulus, polar moment of inertia, and length, they will both have the same angle of twist: TL (240 N-m)(300 mm)(1,000 mm/m) 1  1 1   0.011318 rad  3 J1G1 (79,521.56 mm 4 )(80,000 N/mm 2 ) Since gear A is the origin of the coordinate system for this problem, we will arbitrarily define the rotation angle at gear A to be zero; that is, A = 0. The rotation angle of gear D relative to gear A is found by adding the angles of twist for the three segments to A:  D   A  1  2  3  0  0.011318 rad  0.009108 rad  0.011318 rad  0.031743 rad  0.0317 rad

Ans.

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6.19 A compound steel [G = 80 GPa] shaft consists of solid 40-mm-diameter segments (1) and (3) and tube segment (2), which has an outside diameter of 75 mm (Fig. P6.19). If the rotation of gear D relative to gear A must not exceed 0.01 rad, determine the maximum inside diameter that may be used for tube segment (2).

Fig. P6.19

Solution The polar moment of inertia for shaft segments (1) and (3) is: J1  J 3 



(40 mm) 4  251,327 mm 4

32 The rotation of gear D relative to gear A is equal to the sum of the angles of twist in segments (1), (2), and (3): D  1  2  3 The angles of twist in segments (1) and (3) is (240 N-m)(300 mm)(1,000 mm/m) 1  3   0.003581 rad (251,327 mm 4 )(80,000 N/mm 2 ) Therefore, the angle of twist that can allowed for segment (2) is D  1  2  3  0.01 rad 2  0.01 rad  2(0.003581 rad)  0.002838 rad

The minimum polar moment of inertia required for segment (2) is thus: T2 L2 (240 N-m)(2 m)(1,000 mm/m)2  0.002838 rad  J2   2,114,165 mm4 2 J 2G2 (80,000 N/mm )(0.002838 rad) Since the outside diameter of segment (2) is D2 = 75 mm, the maximum inside diameter d2 is:





 D24  d 24   (75 mm)4  d 24   2,114,165 mm4 32  32   d 2  56.4 mm

Ans.

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6.20 The compound shaft shown in Fig. P6.20 consists of aluminum segment (1) and steel segment (2). Aluminum segment (1) is a tube with an outside diameter of D1 = 4.00 in., a wall thickness of t1 = 0.25 in., and a shear modulus of G1 = 4,000 ksi. Steel segment (2) is a tube with an outside diameter of D2 = 2.50 in., a wall thickness of t2 = 0.125 in., and a shear modulus of G2 = 12,000 ksi. The compound shaft is subjected to torques applied at B and C, as shown in Fig. P6.20. (a) Prepare a diagram that shows the internal torque and the maximum shear stress in segments (1), and (2) of the shaft. Use the sign convention presented in Section 6-6. (b) Determine the rotation angle of B with respect to the support at A. (c) Determine the rotation angle of C with respect to the support at A.

Fig. P6.20

Solution Section properties: Polar moments of inertia in the shaft segments will be needed for this calculation. J1 

J2 









 D14  d14   (4.00 in.) 4  (3.50 in.) 4   10.4004 in.4 32  32 

 D24  d 24   (2.50 in.) 4  (2.25 in.) 4   1.3188 in.4 32  32 

(a) Equilibrium

M x  950 lb-ft  2,100 lb-ft  T1  0 T1  1,150 lb-ft

M x  950 lb-ft  T2  0 T2  950 lb-ft

Shear stress: T c (1,150 lb-ft)(4.00 in./2)(12 in./ft) 1  1 1   2,653.75 psi  2,650 psi J1 10.4004 in.4

2 

T2c2 ( 950 lb-ft)(2.50 in./2)(12 in./ft)   10,804.95 psi  10,800 psi J2 1.3188 in.4

Ans. Ans.

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(b) Rotation angle of B with respect to A: 1  B   A  B  0  B Therefore, TL (1,150 lb-ft)(9 ft)(12 in./ft)2 B  1  1 1   0.035826 rad  0.0358 rad J1G1 (10.4004 in.4 )(4,000,000 psi)

(c) Rotation angle of C with respect to A: 2  C  B C  B  2 The angle of twist in shaft (2) is T2 L2 (950 lb-ft)(6 ft)(12 in./ft)2 2    0.051864 rad J 2G2 (1.3188 in.4 )(12,000,000 psi) and thus, the rotation angle at C is C  B  2  0.035826 rad  (0.051864 rad)  0.016038 rad  0.01604 rad

Ans.

Ans.

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6.21 A solid 1.00-in.-diameter steel [G = 12,000 ksi] shaft is subjected to the torques shown in Fig. P6.21. (a) Prepare a diagram that shows the internal torque and the maximum shear stress in segments (1), (2), and (3) of the shaft. Use the sign convention presented in Section 6-6. (b) Determine the rotation angle of pulley C with respect to the support at A. (c) Determine the rotation angle of pulley D with respect to the support at A. Fig. P6.21

Solution Section properties: The polar moment of inertia for the solid 1.00-in.-diameter steel shaft segments will be needed for this calculation. J1 



32

d14 



32

(1.00 in.)4  0.098175 in.4  J 2  J 3

(a) Equilibrium

M x  240 lb-ft  280 lb-ft  130 lb-ft  T1  0 T1  90 lb-ft

M x  280 lb-ft  130 lb-ft  T2  0 T2  150 lb-ft

M x  130 lb-ft  T3  0 T3  130 lb-ft Shear stress: T c ( 90 lb-ft)(1.00 in./2)(12 in./ft) 1  1 1   5,500.4 psi  5,500 psi J1 0.098175 in.4

2 

T2c2 (150 lb-ft)(1.00 in./2)(12 in./ft)   9,167.3 psi  9,170 psi J2 0.098175 in.4

3 

T3c3 ( 130 lb-ft)(1.00 in./2)(12 in./ft)   7,945.0 psi  7,950 psi J3 0.098175 in.4

The maximum shear stress in the entire shaft is max = 9,170 psi.

Ans.

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(b) Rotation angle of C with respect to A: The angles of twist in the three shaft segments are: TL ( 90 lb-ft)(36 in.)(12 in./ft) 1  1 1   0.033002 rad J1G1 (0.098175 in.4 )(12,000,000 psi) TL (150 lb-ft)(36 in.)(12 in./ft) 2  2 2   0.055004 rad J 2G2 (0.098175 in.4 )(12,000,000 psi) T3 L3 ( 130 lb-ft)(36 in.)(12 in./ft)   0.047670 rad J 3G3 (0.098175 in.4 )(12,000,000 psi) The rotation angle of C with respect to A is found from the sum of the angles of twist in segments (1) and (2): Ans. C  1  2  0.033002 rad  0.055004 rad  0.022002 rad  0.0220 rad

3 

(c) Rotation angle of D with respect to A: The rotation angle of D with respect to A is found from the sum of the angles of twist in segments (1), (2), and (3):  D  1  2  3  0.033002 rad  0.055004 rad  ( 0.047670 rad)  0.025669 rad  0.0257 rad

Ans.

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6.22 A compound shaft supports several pulleys as shown in Fig. P6.22. Segments (1) and (4) are solid 25-mm-diameter steel [G = 80 GPa] shafts. Segments (2) and (3) are solid 50-mm-diameter steel shafts. The bearings shown allow the shaft to turn freely. (a) Determine the maximum shear stress in the compound shaft. (b) Determine the rotation angle of pulley D with respect to pulley B. (c) Determine the rotation angle of pulley E with respect to pulley A. Fig. P6.22

Solution Section properties: The polar moments of inertia for the shaft segments will be needed for this calculation. J1  J2 



32



32

d14  d 24 



32



32

(25 mm) 4  38,349.52 mm 4  J 4 (50 mm)4  613,592.32 mm 4  J 3

(a) Equilibrium

M x  T1  175 N-m  0 T1  175 N-m

M x  T2  2,100 N-m  175 N-m  0 T2  1,925 N-m

M x  650 N-m  225 N-m  T3  0 T3  425 N-m

M x  225 N-m  T4  0 T4  225 N-m

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Shear stress: T c ( 175 N-m)(25 mm/2)(1,000 mm/m) 1  1 1   57.041 MPa  57.0 MPa J1 38,349.52 mm 4 Tc (1,925 N-m)(50 mm/2)(1,000 mm/m) 2  2 2   78.432 MPa  78.4 MPa J2 613,592.32 mm 4

3 

T3c3 (425 N-m)(50 mm/2)(1,000 mm/m)   17.316 MPa  17.32 MPa J3 613,592.32 mm 4

4 

T4c4 ( 225 N-m)(25 mm/2)(1,000 mm/m)   73.339 MPa  73.3 MPa J4 38,349.52 mm 4

The maximum shear stress in the entire shaft is max = 78.4 MPa.

Ans.

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(b) Angles of twist: The angles of twist in the four shaft segments are: TL ( 175 N-m)(750 mm)(1,000 mm/m) 1  1 1   0.042781 rad J1G1 (38,349.52 mm 4 )(80,000 N/mm 2 )

2 

T2 L2 (1,925 N-m)(500 mm)(1,000 mm/m)   0.019608 rad J 2G2 (613,592.32 mm 4 )(80,000 N/mm 2 )

3 

T3 L3 (425 N-m)(625 mm)(1,000 mm/m)   0.005411 rad J 3G3 (613,592.32 mm 4 )(80,000 N/mm 2 )

4 

T4 L4 ( 225 N-m)(550 mm)(1,000 mm/m)   0.040336 rad J 4G4 (38,349.52 mm 4 )(80,000 N/mm 2 )

Rotation angle of pulley D with respect to pulley B: The rotation angle of D with respect to B is found from the sum of the angles of twist in segments (2) and (3): Ans. D / B  2  3  0.019608 rad  0.005411 rad  0.025019 rad  0.0250 rad (c) Rotation angle of pulley E with respect to pulley A: The rotation angle of E with respect to A is found from the sum of the angles of twist in all four segments:  E  1  2  3  4  0.042781 rad  0.019608 rad  0.005411 rad  ( 0.040336 rad)  0.058098 rad  0.0581 rad

Ans.

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6.23 A solid steel [G = 80 GPa] shaft of variable diameter is subjected to the torques shown in Fig. P6.23. The diameter of the shaft in segments (1) and (3) is 50 mm, and the diameter of the shaft in segment (2) is 80 mm. The bearings shown allow the shaft to turn freely. (a) Determine the maximum shear stress in the compound shaft. (b) Determine the rotation angle of pulley D with respect to pulley A. Fig. P6.23

Solution Section properties: The polar moments of inertia for the shaft segments will be needed for this calculation. J1  J2 



32



32

(50 mm) 4  613,592.32 mm 4  J 3 (80 mm) 4  4,021, 238.60 mm 4

(a) Equilibrium

M x  1, 200 N-m  T1  0 T1  1, 200 N-m

M x  1, 200 N-m  4,500 N-m  T2  0 T2  3,300 N-m

M x  T3  500 N-m  0 T3  500 N-m

Shear stress: T c (1, 200 N-m)(50 mm/2)(1,000 mm/m) 1  1 1   48.892 MPa  48.9 MPa J1 613,592.32 mm 4

2 

T2c2 ( 3,300 N-m)(80 mm/2)(1,000 mm/m)   32.826 MPa  32.8 MPa J2 4,021,238.60 mm 4

3 

T3c3 ( 500 N-m)(50 mm/2)(1,000 mm/m)   20.372 MPa  20.4 MPa J3 613,592.32 mm 4

The maximum shear stress in the compound shaft is  max  1  48.9 MPa

Ans.

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Angles of twist: The angles of twist in the three shaft segments are: T1L1 (1,200 N-m)(0.7 m)(1,000 mm/m)2 1    0.017112 rad J1G1 (613,592.32 mm4 )(80,000 N/mm2 )

T2 L2 (3,300 N-m)(1.8 m)(1,000 mm/m) 2 2    0.018464 rad J 2G2 (4,021,238.60 mm4 )(80,000 N/mm2 ) T3 L3 ( 500 N-m)(0.7 m)(1,000 mm/m)2 3    0.007130 rad J 3G3 (613,592.32 mm4 )(80,000 N/mm2 ) (b) Rotation angles:  A  0 rad

 B   A  1  0 rad  0.017112 rad  0.017112 rad C   B  2  0.017112 rad  (0.018464 rad)  0.001352 rad  D  C  3  0.001352 rad  ( 0.007130 rad)  0.008482 rad

Ans.

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6.24 A compound shaft drives three gears, as shown in Fig. P6.24. Segments (1) and (2) of the compound shaft are hollow aluminum [G = 4,000 ksi] tubes, which have an outside diameter of 3.00 in. and a wall thickness of 0.25 in. Segments (3) and (4) are solid 2.00-in.-diameter steel [G = 12,000 ksi] shafts. The bearings shown allow the shaft to turn freely. (a) Determine the maximum shear stress in the compound shaft. (b) Determine the rotation angle of flange C with respect to flange A. (c) Determine the rotation angle of gear E with respect to flange A. Fig. P6.24

Solution Section properties: The polar moment of inertia for the solid 1.00-in.-diameter steel shaft segments will be needed for this calculation. J1 

J3 





 D14  d14   (3.00 in.)4  (2.50 in.)4   4.117204 in.4  J 2 32  32 



32

d34 



32

(2.00 in.) 4  1.570796 in.4  J 4

(a) Equilibrium

M x  14 kip-in.  42 kip-in. 35 kip-in.  T1  0  T1  7 kip-in.

M x  14 kip-in.  42 kip-in.  T2  0 T2  28 kip-in.

M x  14 kip-in.  42 kip-in.  T3  0 T3  28 kip-in.

M x  14 kip-in.  T4  0 T4  14 kip-in.

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Shear stress: T c ( 7 kip-in.)(3.00 in./2) 1  1 1   2.55 ksi J1 4.117204 in.4 Tc ( 28 kip-in.)(3.00 in./2) 2  2 2   10.20 ksi J2 4.117204 in.4

3 

T3c3 ( 28 kip-in.)(2.00 in./2)   17.83 ksi J3 1.570796 in.4

4 

T4c4 (14 kip-in.)(2.00 in./2)   8.91 ksi J4 1.570796 in.4

The maximum shear stress in the compound shaft is  max   3  17.83 ksi

Ans.

(b) Angles of Twist: The angles of twist in the four shaft segments are: TL (7 kip-in.)(60 in.) 1  1 1   0.025503 rad J1G1 (4.117204 in.4 )(4,000 ksi)

2 

T2 L2 ( 28 kip-in.)(6 in.)   0.010201 rad J 2G2 (4.117204 in.4 )(4,000 ksi)

3 

T3 L3 ( 28 kip-in.)(18 in.)   0.026738 rad J 3G3 (1.570796 in.4 )(12,000 ksi)

4 

T4 L4 (14 kip-in.)(12 in.)   0.008913 rad J 4G4 (1.570796 in.4 )(12,000 ksi)

 A  0 rad  B   A  1  0 rad  0.025503 rad  0.0255 rad C   B  2  0.025503 rad  ( 0.010201 rad)  0.01530 rad  D  C  3  0.015302 rad  ( 0.026738 rad)  0.01144 rad  E   D  4  0.011436 rad  0.008913 rad  0.00252 rad The rotation angle of flange C with respect to A is C  0.01530 rad

Ans.

(c) Rotation angle of E with respect to A: E  0.00252 rad

Ans.

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6.25 A compound shaft drives several pulleys, as shown in Fig. P6.25. Segments (1) and (2) of the compound shaft are hollow aluminum [G = 4,000 ksi] tubes, which have an outside diameter of 3.00 in. and a wall thickness of 0.125 in. Segments (3) and (4) are solid 1.50-in.-diameter steel [G = 12,000 ksi] shafts. The bearings shown allow the shaft to turn freely. (a) Determine the maximum shear stress in the compound shaft. (b) Determine the rotation angle of flange C with respect to pulley A. (c) Determine the rotation angle of pulley E with respect to pulley A. Fig. P6.25

Solution (a) Equilibrium

M x  95 lb-ft  T4  0

 T4  95 lb-ft

M x  95 lb-ft  270 lb-ft  T3  0

 T3  175 lb-ft

Note: The internal torque in shaft segment (2) is the same as in segment (3); therefore, T2 = –175 lb-ft.

M x  T1  650 lb-ft  0

 T1  650 lb-ft

Polar moments of inertia in the shaft segments will be needed for this calculation. Segments (1) and (2) are hollow tubes with an outside diameter of 3.00 in. and an inside diameter of 3.00 in. – 2(0.125 in.) = 2.75 in. The polar moment of inertia of segments (1) and (2) is:





 D14  d14   (3.00 in.) 4  (2.75 in.) 4   2.33740 in.4  J 2 32 32 Segments (3) and (4) are solid 1.50-in.-diameter shafts, which have a polar moment of inertia of: J1 

J3 



32

d34 



32

(1.50 in.)4  0.49701 in.4  J 4

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Shear stresses: The shear stresses in each segment can be calculated by the elastic torsion formula: T c (650 lb-ft)(1.50 in.)(12 in./ft) 1  1 1   5,005.56 psi  5,010 psi J1 2.33740 in.4 Tc ( 175 lb-ft)(1.50 in.)(12 in./ft) 2  2 2   1,347.6 psi  1,348 psi J2 2.33740 in.4

3 

T3c3 ( 175 lb-ft)(0.75 in.)(12 in./ft)   3,169.0 psi  3,170 psi J3 0.49701 in.4

4 

T4c4 (95 lb-ft)(0.75 in.)(12 in./ft)   1,720.3 psi  1,720 psi J4 0.49701 in.4

The maximum shear stress in the compound shaft is  max  1  5,010 psi

Ans.

Angles of twist: TL (650 lb-ft)(72 in.)(12 in./ft) 1  1 1   0.060067 rad J1G1 (2.33740 in.4 )(4,000,000 lb/in.2 ) TL ( 175 lb-ft)(36 in.)(12 in./ft) 2  2 2   0.008086 rad J 2G2 (2.33740 in.4 )(4,000,000 lb/in.2 )

3 

T3 L3 ( 175 lb-ft)(54 in.)(12 in./ft)   0.019014 rad J 3G3 (0.49701 in.4 )(12,000,000 lb/in.2 )

4 

T4 L4 (95 lb-ft)(54 in.)(12 in./ft)   0.010322 rad J 4G4 (0.49701 in.4 )(12,000,000 lb/in.2 )

Rotation angles: The angles of twist can be defined in terms of the rotation angles at the ends of each segment: 1  B  A 2  C  B 3  D  C 4  E  D The origin of the coordinate system is located at pulley A. We will arbitrarily define the rotation angle at pulley A to be zero (A = 0). The rotation angle at B can be calculated from the angle of twist in segment (1): 1  B   A

 B   A  1  0  0.060067 rad  0.0601 rad Similarly, the rotation angle at C is determined from the angle of twist in segment (2) and the rotation angle of pulley B: 2  C  B  C  B  2  0.060067 rad  (0.008086 rad)  0.051981 rad  0.0520 rad The rotation angle at D is: 3  D  C  D  C  3  0.051981 rad  (0.019014 rad)  0.032967 rad  0.0330 rad and the rotation angle at E is: 4   E   D  E  D  4  0.032967 rad  0.010322 rad  0.043289 rad  0.0433 rad

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(b) Rotation angle of flange C with respect to pulley A: Using the rotation angles determined for the system, the rotation angle of flange C with respect to pulley A is simply: Ans. C  0.051981 rad  0.0520 rad (c) Rotation angle of pulley E with respect to pulley A: Using the rotation angles determined for the system, the rotation angle of flange C with respect to pulley A is simply: Ans. E  0.043289 rad  0.0433 rad

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6.26 A torque of TA = 460 lb-ft is applied to gear A of the gear train shown in Fig. P6.26. The bearings shown allow the shafts to rotate freely. (a) Determine the torque TD required for equilibrium of the system. (b) Assume shafts (1) and (2) are solid 1.5-in.diameter steel shafts. Determine the magnitude of the maximum shear stresses acting in each shaft. (c) Assume shafts (1) and (2) are solid steel shafts, which have an allowable shear stress of 6,000 psi. Determine the minimum diameter required for each shaft. Fig. P6.26

Solution (a) Torque TD required for equilibrium: 7 in. 7 in. TD TA (460 lb-ft) 805 lb-ft 4 in. 4 in.

Ans.

(b) Shear stress magnitudes if shafts are solid 1.5-in.-diameter: d4 (1.5 in.) 4 0.497010 in.4 32 32 T1c1 (460 lb-ft)(1.5 in./2)(12 in./ft) J1 0.497010 in.4

J 1

T2c2 J2

2

(805 lb-ft)(1.5 in./2)(12 in./ft) 0.497010 in.4

8,329.82 psi 14,577.18 psi

8,330 psi 14,580 psi

Ans. Ans.

(c) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter d:

Tc J

J c

d4

32 d /2

16

d3

T

Using this expression, solve for the minimum acceptable diameter of each shaft: 16 T1 16(460 lb-ft)(12 in./ft) d13 4.685522 in.3 (6,000 psi) allow d1 1.673347 in. d 23

16 T2 allow

d2

1.673 in.

16(805 lb-ft)(12 in./ft) (6,000 psi)

2.016502 in.

2.02 in.

Ans. 8.199663 in.3

Ans.

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6.27 A torque of TD = 450 N-m is applied to gear D of the gear train shown in Fig. P6.27. The bearings shown allow the shafts to rotate freely. (a) Determine the torque TA required for equilibrium of the system. (b) Assume shafts (1) and (2) are solid 30-mmdiameter steel shafts. Determine the magnitude of the maximum shear stresses acting in each shaft. (c) Assume shafts (1) and (2) are solid steel shafts, which have an allowable shear stress of 60 MPa. Determine the minimum diameter required for each shaft. Fig. P6.27

Solution (a) Torque TA required for equilibrium: 90 mm 90 mm TA TD (450 N-m) 150 mm 150 mm

Ans.

270 N-m

(b) Shear stress magnitudes if shafts are solid 30-mm-diameter: d4 (30 mm)4 79,521.564 mm4 32 32 T1c1 (270 N-m)(30 mm/2)(1,000 mm/m) J1 79,521.564 mm 4

J 1

T2c2 J2

2

(450 N-m)(30 mm/2)(1,000 mm/m) 79,521.564 mm 4

50.930 MPa 84.883 MPa

50.9 MPa

Ans.

84.9 MPa

Ans.

(c) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter d:

Tc J

J c

d4

32 d /2

16

d3

T

Using this expression, solve for the minimum acceptable diameter of each shaft: 16 T1 16(270 N-m)(1,000 mm/m) d13 22,918.312 mm3 2 (60 N/mm ) allow d1 d 23

28.405 mm 16 T2 allow

d2

28.4 mm

16(450 N-m)(1,000 mm/m) (60 N/mm 2 )

33.678 mm

33.7 mm

Ans. 38,197.186 mm3

Ans.

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6.28 The gear train system shown in Fig. P6.28 includes shafts (1) and (2), which are solid 1.375in.-diameter steel shafts. The allowable shear stress of each shaft is 8,000 psi. The bearings shown allow the shafts to rotate freely. Determine the maximum torque TD that can be applied to the system without exceeding the allowable shear stress in either shaft.

Fig. P6.28

Solution Section properties: J1

32

(1.375 in.) 4

0.350922 in.4

J2

Maximum torque in either shaft: (8,000 psi)(0.350922 in.4 ) allow J1 Tmax c1 1.375 in./2 Torque relationship: T1 T2 or T1 10 in. 6 in.

T2

10 in. 1.667 T2 6 in.

4,083.457 lb-in.

T1 controls

The torque in shaft (1) controls; therefore, T1 = 4,083.457 lb-in. Consequently, the maximum torque in shaft (2) must be limited to: 6 in. T2 T1 0.6(4,083.457 lb-in.) 2,450.074 lb-in. 204 lb-ft Ans. 10 in.

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6.29 The gear train system shown in Fig. P6.29 includes shafts (1) and (2), which are solid 20-mmdiameter steel shafts. The allowable shear stress of each shaft is 50 MPa. The bearings shown allow the shafts to rotate freely. Determine the maximum torque TD that can be applied to the system without exceeding the allowable shear stress in either shaft.

Fig. P6.29

Solution Section properties: J1

32

(20 mm)4

15,707.963 mm 4

J2

Maximum torque in either shaft: (50 N/mm2 )(15,707.963 mm4 ) allow J1 Tmax c1 20 mm/2 Torque relationship: T1 T2 200 mm 80 mm

or T1

T2

200 mm 80 mm

78,539.816 N-mm

2.50 T2

T1 controls

The torque in shaft (1) controls; therefore, T1 = 78,539.816 N-mm. Consequently, the maximum torque in shaft (2) must be limited to: 80 mm T2 T1 0.4(78,539.816 N-mm) 31,415.927 N-mm 31.4 N-m Ans. 200 mm

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6.30 In the gear system shown in Fig. P6.30, the motor applies a 210 lb-ft torque to the gear at A. A torque of TC = 300 lb-ft is removed from the shaft at gear C, and the remaining torque is removed at gear D. Segments (1) and (2) are solid 1.5-in.diameter steel [G = 12,000 ksi] shafts, and the bearings shown allow free rotation of the shaft. Determine: (a) the maximum shear stress in segments (1) and (2) of the shaft. (b) the rotation angle of gear D relative to gear B.

Fig. P6.30

Solution Section properties: J1

32

(1.5 in.) 4

0.497010 in.4

Torque relationship: TA TB hence TB 4 in. 10 in.

TA

J2

10 in. 4 in.

2.50 TA

2.50(210 lb-ft)

525 lb-ft

Shaft (1) has the same torque as gear B; therefore, T1 = 525 lb-ft. A torque of 300 lb-ft is removed from the shaft at gear C; therefore, TC = 300 lb-ft. The torque in shaft (2) must satisfy equilibrium: Mx T1 TC T2 0

T2 T1 TC To summarize: T1 525 lb-ft

525 lb-ft 300 lb-ft T2

225 lb-ft

225 lb-ft

(a) Maximum shear stress in shafts (1) and (2): T1c1 (525 lb-ft)(1.5 in./2)(12 in./ft) 9,506.856 psi 1 J1 0.497010 in.4

9,510 psi

T2c2 (225 lb-ft)(1.5 in./2)(12 in./ft) 4,074.367 psi 4,070 psi J2 0.497010 in.4 (b) Angles of twist in shafts (1) and (2): T1L1 (525 lb-ft)(60 in.)(12 in./ft) 0.063379 rad 1 J1G1 (0.497010 in.4 )(12,000,000 psi) 2

Ans. Ans.

T2 L2 (225 lb-ft)(40 in.)(12 in./ft) 0.018108 rad J 2G2 (0.497010 in.4 )(12,000,000 psi) Note: There is not enough information given to say whether these twist angles are positive or negative. However, both twist angles will have the same sign. say B 0 rad (since this is our reference) 2

D

B

1

2

0 rad 0.063379 rad 0.018108 rad

0.081487 rad

0.0815 rad

Ans.

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6.31 In the gear system shown in Fig. P6.31, the motor applies a 360 lb-ft torque to the gear at A. A torque of TC = 500 lb-ft is removed from the shaft at gear C, and the remaining torque is removed at gear D. Segments (1) and (2) are solid steel [G = 12,000 ksi] shafts, and the bearings shown allow free rotation of the shaft. (a) Determine the minimum permissible diameters for segments (1) and (2) of the shaft if the maximum shear stress must not exceed 6,000 psi. (b) If the same diameter is to be used for both segments (1) and (2), determine the minimum permissible diameter that can be used for the shaft if the maximum shear stress must not exceed 6,000 psi and the rotation angle of gear D relative to gear B must not exceed 0.10 rad. Fig. P6.31

Solution Torque relationship: The motor applies a torque of 360 lb-ft to gear A. Since gear B is bigger than gear A, the torque on gear B will be increased in proportion to the gear ratio: TA TB 10 in. hence TB TA 2.50 TA 2.50(360 lb-ft) 900 lb-ft 4 in. 10 in. 4 in. Let the positive x axis for shaft BCD extend from gear B toward gear D. In order for shaft BCD to be in equilibrium with the torques as shown on gears C and D, the torque at B must act in the negative x direction. Consequently, TB = −900 lb-ft. Draw a free-body diagram that cuts through shaft (1) and includes gear B. From this FBD, the torque in shaft (1) is: Mx 900 lb-ft T1 0 T1 900 lb-ft From the problem statement, we are told that a torque of 500 lb-ft is removed from the shaft at gear C. In other words, TC = 500 lb-ft. Draw a FBD that cuts through shaft (2) and includes both gears B and C. From this FBD, the torque in shaft (2) is: Mx 900 lb-ft 500 lb-ft T2 0

T2

400 lb-ft

To summarize: T1 900 lb-ft

T2

400 lb-ft

(a) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter d:

Tc J

J c

d4

32 d /2

16

d3

T

Using this expression, solve for the minimum acceptable diameter of each shaft: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

d13

16 T1 allow

d1 d 23

16(900 lb-ft)(12 in./ft) (6,000 psi)

2.092900 in. 16 T2 allow

d2

2.09 in.

16(400 lb-ft)(12 in./ft) (6,000 psi)

1.597178 in.

9.167325 in.3

1.597 in.

Ans. 4.074367 in.3

Ans.

(b) The rotation angle of gear D with respect to gear B is equal to the sum of the angles of twist in shaft segments (1) and (2): 0.10 rad D B 1 2 Since the same solid steel shaft is to be used for both segments (1) and (2): T1L1 T2 L2 1 0.01 rad T1L1 T2 L2 J1G1 J 2G2 JG or rearranging: (900 lb-ft)(60 in.)(12 in./ft) (400 lb-ft)(40 in.)(12 in./ft) J d4 1.883606 in.4 32 (12,000,000 psi)(0.1 rad) d

1.634 in.

Since the shaft has already been designed for stresses and since the diameter that is required to satisfy the rotation angle requirement is smaller than d1 determined in part (a), the minimum diameter required for a constant diameter shaft between gears B and D is: Ans. dmin 2.09 in.

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6.32 In the gear system shown in Fig. P6.32, the motor applies a torque of 220 N-m to the gear at A. A torque of TC = 400 N-m is removed from the shaft at gear C, and the remaining torque is removed at gear D. Segments (1) and (2) are solid 40-mm-diameter steel [G = 80 GPa] shafts, and the bearings shown allow free rotation of the shaft. (a) Determine the maximum shear stress in segments (1) and (2) of the shaft. (b) Determine the rotation angle of gear D relative to gear B.

Fig. P6.32

Solution Section properties: J1

32

(40 mm)4

Torque relationship: TA TB 100 mm 300 mm

251,327.41 mm 4

hence TB

TA

J2

300 mm 100 mm

3TA

3(220 N-m)

660 N-m

Shaft (1) has the same torque as gear B; therefore, T1 = 660 N-m. A torque of TC = 400 N-m is removed from the shaft at gear C. The torque in shaft (2) must satisfy equilibrium: Mx T1 TC T2 0

T2

T1 TC

660 N-m 400 N-m 260 N-m

To summarize: T1 660 N-m T2

260 N-m

(a) Maximum shear stress in shafts (1) and (2): T1c1 (660 N-m)(40 mm/2)(1,000 mm/m) 1 J1 251,327.41 mm 4 2

T2c2 J2

(260 N-m)(40 mm/2)(1,000 mm/m) 251,327.41 mm 4

52.521 MPa 20.690 MPa

52.5 MPa 20.7 MPa

Ans. Ans.

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(b) Angles of twist in shafts (1) and (2): T1L1 (660 N-m)(1.5 m)(1,000 mm/m) 2 1 J1G1 (251,327.41 mm4 )(80,000 N/mm2 )

0.049239 rad

T2 L2 (260 N-m)(1.0 m)(1,000 mm/m) 2 0.012931 rad 2 J 2G2 (251,327.41 mm4 )(80,000 N/mm 2 ) Note: There is not enough information given to say whether these twist angles are positive or negative. However, both twist angles will have the same sign. say

B D

0 rad B

(since this is our reference) 1

2

0 rad 0.049239 rad 0.012931 rad

0.062170 rad

0.0622 rad

Ans.

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6.33 In the gear system shown in Fig. P6.33, the motor applies a torque of 400 N-m to the gear at A. A torque of TC = 700 N-m is removed from the shaft at gear C, and the remaining torque is removed at gear D. Segments (1) and (2) are solid steel [G = 80 GPa] shafts, and the bearings shown allow free rotation of the shaft. (a) Determine the minimum permissible diameters for segments (1) and (2) of the shaft if the maximum shear stress must not exceed 40 MPa. (b) If the same diameter is to be used for segments (1) and (2), determine the minimum permissible diameter that can be used for the shaft if the maximum shear stress must not exceed 40 MPa and the rotation angle of gear D relative to gear B must not exceed 3.0°. Fig. P6.33

Solution Torque relationship: TA TB 100 mm 300 mm

hence TB

TA

300 mm 100 mm

3TA

3(400 N-m) 1, 200 N-m

Shaft (1) has the same torque as gear B; therefore, T1 = 1,200 N-m. A torque of TC = 700 N-m is removed from the shaft at gear C. The torque in shaft (2) must satisfy equilibrium: Mx T1 TC T2 0

T2 T1 TC To summarize: T1 1, 200 N-m

1, 200 N-m 700 N-m 500 N-m T2

500 N-m

(a) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter d:

Tc J

J c

d4

32 d /2

16

d3

T

Using this expression, solve for the minimum acceptable diameter of each shaft: 16 T1 16(1, 200 N-m)(1,000 mm/m) d13 152,788.745 mm3 2 (40 N/mm ) allow d1

d 23

53.460 mm

16 T2 allow

d2

53.5 mm

16(500 N-m)(1,000 mm/m) (40 N/mm 2 )

39.929 mm

39.9 mm

Ans. 63,661.977 mm3

Ans.

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(b) The rotation angle of gear D with respect to gear B is equal to the sum of the angles of twist in shaft segments (1) and (2): (3 )( /180 ) 0.052360 rad D B 1 2 Since the same solid steel shaft is to be used for both segments (1) and (2): T1L1 T2 L2 1 0.052360 rad T1L1 T2 L2 J1G1 J 2G2 JG or rearranging: J

32 d

d4

(1, 200 N-m)(1.5 m)(1,000 mm/m) 2 (500 N-m)(1.0 m)(1,000 mm/m) 2 (80,000 N/mm 2 )(0.052360 rad)

549,083.270 mm 4

48.6 mm

Since the shaft has already been designed for stresses and since the diameter that is required to satisfy the rotation angle requirement is smaller than d1 determined in part (a), the minimum diameter required for a constant diameter shaft between gears B and D is: Ans. dmin 53.5 mm

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6.34 In the gear system shown in Fig. P6.34, the motor applies a torque of 250 N-m to the gear at A. Shaft (1) is a solid 35-mm-diameter shaft, and shaft (2) is a solid 50-mm-diameter shaft. The bearings shown allow free rotation of the shafts. (a) Determine the torque TE provided by the gear system at gear E. (b) Determine the maximum shear stresses in shafts (1) and (2).

Fig. P6.34

Solution (a) Torque relationship: The torque on gear B is: 72 TB TA 3TA 3(250 N-m) 750 N-m 24 Shaft (1) has the same torque as gear B; therefore, T1 = 750 N-m and TC = 750 N-m. The torque on gear D is found from: 60 TD TC 2 TC 2(750 N-m) 1,500 N-m 30 Shaft (2) has the same torque as gear D; therefore, T2 = 1,500 N-m and Ans. TE 1,500 N-m To summarize: T1 750 N-m

T2

1,500 N-m

(b) Section properties: J1 J2

32 32

(35 mm) 4

147,323.51 mm 4

(50 mm) 4

613,592.32 mm 4

Maximum shear stress in shafts (1) and (2): T1c1 (750 N-m)(35 mm/2)(1,000 mm/m) 89.090 MPa 89.1 MPa 1 J1 147,323.51 mm 4 T2c2 (1,500 N-m)(50 mm/2)(1,000 mm/m) 61.115 MPa 61.1 MPa 2 J2 613,592.32 mm 4

Ans. Ans.

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6.35 In the gear system shown in Fig. P6.35, the motor applies a torque of 600 N-m to the gear at A. Shafts (1) and (2) are solid shafts, and the bearings shown allow free rotation of the shafts. (a) Determine the torque TE provided by the gear system at gear E. (b) If the allowable shear stress in each shaft must be limited to 70 MPa, determine the minimum permissible diameter for each shaft.

Fig. P6.35

Solution (a) Torque relationship: The torque on gear B is: 72 TB TA 3TA 3(600 N-m) 1,800 N-m 24 Shaft (1) has the same torque as gear B; therefore, T1 = 1,800 N-m and TC = 1,800 N-m. The torque on gear D is found from: 60 TD TC 2 TC 2(1,800 N-m) 3,600 N-m 30 Shaft (2) has the same torque as gear D; therefore, T2 = 3,600 N-m and Ans. TE 3,600 N-m To summarize: T1 1,800 N-m

T2

3,600 N-m

(b) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter d:

Tc J

J c

d4

32 d /2

16

d3

T

Using this expression, solve for the minimum acceptable diameter of each shaft: 16 T1 16(1,800 N-m)(1,000 mm/m) d13 130,961.782 mm3 2 (70 N/mm ) allow d1 d 23

50.783 mm

16 T2 allow

d2

50.8 mm

16(3,600 N-m)(1,000 mm/m) (70 N/mm 2 )

63.982 mm

64.0 mm

Ans. 261,923.563 mm3

Ans.

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6.36 In the gear system shown in Fig. P6.36, a torque of TE = 1,500 lb-ft is delivered at gear E. Shafts (1) and (2) are solid shafts, and the bearings shown allow free rotation of the shafts. (a) Determine the torque provided by the motor to gear A. (b) If the allowable shear stress in each shaft must be limited to 4,000 psi, determine the minimum permissible diameter for each shaft.

Fig. P6.36

Solution (a) Torque relationship: The torque on gear E creates a torque in shaft (2) of T2 = 1,500 lb-ft. Accordingly, the torque applied to gear D is also equal to 1,500 lb-ft. The torque on gear C is found with the gear ratio: 30 TC TD 0.5TD 0.5(1,500 lb-ft) 750 lb-ft 60 Shaft (1) has the same torque as gear C; therefore, T1 = 750 lb-ft and TB = 750 N-m. The torque on gear A is found from: 24 TC 750 lb-ft TA TB 250 lb-ft Ans. 72 3 3 To summarize: T1 750 lb-ft

T2

1,500 lb-ft

(b) Determine minimum diameters: The elastic torsion formula can be rearranged and rewritten in terms of the unknown diameter d:

Tc J

J c

d4

32 d /2

16

d3

T

Using this expression, solve for the minimum acceptable diameter of each shaft: 16 T1 16(750 lb-ft)(12 in./ft) d13 11.459156 in.3 (4,000 psi) allow d1 d 23

2.255 in.

16 T2 allow

d2

2.26 in.

16(1,500 lb-ft)(12 in./ft) (4, 000 psi)

2.840 in.

2.84 in.

Ans. 22.918313 in.3

Ans.

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6.37 In the gear system shown in Fig. P6.37, a torque of TE = 720 lb-ft is delivered at gear E. Shaft (1) is a solid 1.50-in.-diameter shaft, and shaft (2) is a solid 2.00-in.-diameter shaft. The bearings shown allow free rotation of the shafts. (a) Determine the torque provided by the motor to gear A. (b) Determine the maximum shear stresses in shafts (1) and (2).

Fig. P6.37

Solution (a) Torque relationship: The torque on gear E creates a torque in shaft (2) of T2 = 720 lb-ft. Accordingly, the torque applied to gear D is also equal to 720 lb-ft. The torque on gear C is found with the gear ratio: 30 TC TD 0.5TD 0.5(720 lb-ft) 360 lb-ft 60 Shaft (1) has the same torque as gear C; therefore, T1 = 360 lb-ft and TB = 360 N-m. The torque on gear A is found from: 24 TC 360 lb-ft TA TB 120 lb-ft Ans. 72 3 3 To summarize: T1 360 lb-ft

T2

720 lb-ft

(b) Section properties: J1 J2

32 32

(1.50 in.) 4

0.497010 in.4

(2.00 in.) 4

1.570796 in.4

Maximum shear stress in shafts (1) and (2): T1c1 (360 lb-ft)(1.50 in./2)(12 in./ft) 1 J1 0.497010 in.4 2

T2c2 J2

(720 lb-ft)(2.00 in./2)(12 in./ft) 1.570796 in.4

6,519.98 psi 5,500.396 psi

6,520 psi 5,500 psi

Ans. Ans.

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6.38 Two solid 30-mm-diameter steel shafts are connected by the gears shown in Fig. P6.38. The shaft lengths are L1 = 300 mm and L2 = 500 mm. Assume that the shear modulus of both shafts is G = 80 GPa and that the bearings shown allow free rotation of the shafts. If the torque applied at gear D is TD = 160 N-m, (a) determine the internal torques T1 and T2 in the two shafts. (b) determine the angles of twist 1 and 2. (c) determine the rotation angles B and C of gears B and C. (d) determine the rotation angle of gear D.

Fig. P6.38

Solution (a) Torque relationship: The torque on gear D creates a torque in shaft (2) of T2 = 160 N-m. M x TD T2 0 T2 TD 160 N-m Accordingly, the torque applied to gear C is also equal to 160 N-m. The torque on gear B is found with the gear ratio: 54 teeth TB TC 1.8 TC 1.8(160 N-m) 288 N-m 30 teeth Shaft (1) has the same torque as gear B; therefore, T1 = −288 N-m. To summarize: T1 288 N-m T2 160 N-m

Ans.

Section properties: (30 mm) 4 79,521.56 mm 4 J 2 32 (b) Angles of twist in shafts (1) and (2): T1L1 ( 288 N-m)(300 mm)(1,000 mm/m) 1 J1G1 (79,521.56 mm 4 )(80,000 N/mm 2 ) J1

2

T2 L2 J 2G2

(160 N-m)(500 mm)(1,000 mm/m) (79,521.56 mm 4 )(80,000 N/mm 2 )

0.013581 rad 0.012575 rad

0.01358 rad 0.01258 rad

Ans. Ans.

(c) Rotation angles of gears B and C: The rotation of gear B is found from the angle of twist in shaft (1): Ans. 0 rad ( 0.013581 rad) 0.01358 rad 1 B A B A 1 As gear B rotates, gear C also rotates but it rotates in the opposite direction. The magnitude of the rotation is dictated by the gear ratio: 54 teeth 54 teeth ( 0.013581 rad) 0.024446 rad 0.0244 rad Ans. C B 30 teeth 30 teeth The rotation of gear D is found from: Ans. 0.024446 rad 0.012575 rad 0.037021 rad 0.0370 rad 2 D C D C 2

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6.39 Two solid 1.75-in.-diameter steel shafts are connected by the gears shown in Fig. P6.39. The shaft lengths are L1 = 6 ft and L2 = 10 ft. Assume that the shear modulus of both shafts is G = 12,000 ksi and that the bearings shown allow free rotation of the shafts. If the torque applied at gear D is TD = 225 lb-ft, (a) determine the internal torques T1 and T2 in the two shafts. (b) determine the angles of twist 1 and 2. (c) determine the rotation angles B and C of gears B and C. (d) determine the rotation angle of gear D.

Fig. P6.39

Solution (a) Torque relationship: The torque on gear D creates a torque in shaft (2) of T2 = 225 lb-ft. M x TD T2 0 T2 TD 225 lb-ft Accordingly, the torque applied to gear C is also equal to 225 lb-ft. The torque on gear B is found with the gear ratio: 54 teeth TB TC 1.8 TC 1.8(225 lb-ft) 405 lb-ft 30 teeth Shaft (1) has the same torque as gear B; therefore, T1 = −405 lb-ft. To summarize: T1 405 lb-ft T2 225 lb-ft

Ans.

Section properties: (1.75 in.)4 0.920772 in.4 J 2 32 (b) Angles of twist in shafts (1) and (2): T1L1 ( 405 lb-ft)(6 ft)(12 in./ft)2 1 J1G1 (0.920772 in.4 )(12,000,000 psi) J1

2

T2 L2 J 2G2

(225 lb-ft)(10 ft)(12 in./ft)2 (0.920772 in.4 )(12,000,000 psi)

0.031669 rad 0.029323 rad

Ans.

0.0317 rad

Ans.

0.0293 rad

(c) Rotation angles of gears B and C: The rotation of gear B is found from the twist angle in shaft (1): Ans. 0 rad ( 0.031669 rad) 0.0317 rad 1 B A B A 1 As gear B rotates, gear C also rotates but it rotates in the opposite direction. The magnitude of the rotation is dictated by the gear ratio: 54 teeth 54 teeth ( 0.031669 rad) 0.057004 rad 0.0570 rad Ans. C B 30 teeth 30 teeth (d) Rotation angle of gear D:The rotation of gear D is found from: 0.057004 rad 0.029323 rad 0.086327 rad 2 D C D C 2

0.0863 rad

Ans.

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6.40 Two solid steel shafts are connected by the gears shown in Fig. P6.40. The design requirements for the system require (1) that both shafts must have the same diameter, (2) that the maximum shear stress in each shaft must be less than 6,000 psi, and (3) that the rotation angle of gear D must not exceed 3°. Determine the minimum required diameter of the shafts if the torque applied at gear D is TD = 345 lb-ft. The shaft lengths are L1 = 10 ft and L2 = 8 ft. Assume that the shear modulus of both shafts is G = 12,000 ksi and that the bearings shown allow free rotation of the shafts. Fig. P6.40

Solution Torque relationship: The torque on gear D creates a torque in shaft (2) of T2 = 1,000 lb-ft. M x TD T2 0 T2 TD 345 lb-ft 4,140 lb-in. Accordingly, the torque applied to gear C is also 4,140 lb-in. The torque on gear B is found with the gear ratio: 48 teeth TB TC 0.6667 TC 0.6667(4,140 lb-in.) 2,760 lb-in. 72 teeth Shaft (1) has the same torque as gear B; therefore, T1 = −2,760 lb-in. Diameters based on allowable shear stresses: The elastic torsion formula gives the relationship between shear stress and torque in a shaft. Tc J The torques in both shafts have been determined, and the allowable shear stress is specified. Rearrange the elastic torsion formula, putting the known terms on the right-hand side of the equation: J T c Express the left-hand side of this equation in terms of the shaft diameter D:

d4

32 d /2

16

d3

T

Solve for the minimum acceptable diameter in shaft (1): 16 T1 16(2,760 lb-in.) d13 2.342761 in.3 2 (6,000 lb/in. )

d1 1.328 in.

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Solve for the minimum acceptable diameter in shaft (2): 16 T2 16(4,140 lb-in.) d 23 3.514141 in.3 (6,000 lb/in.2 )

d 2 1.520 in. Of these two values, d2 controls. Therefore, both shafts could have a diameter of 1.520 in. or more and the shear stress constraint would be satisfied. The angles of twists in shafts (1) and (2) can be expressed as: T1L1 T2 L2 1 2 J1G1 J 2G2 The rotations of gears B and C are related since the arclengths turned by the two gears have the same magnitude. The gears turn in opposite directions; therefore, a negative sign is introduced. RC C RB B The rotation angle of gear B is equal to the angle of twist in shaft (1): B = 1. Therefore, the rotation angle of gear C can be expressed in terms of T1. RB RB RB T1L1 N B T1L1 C B 1 RC RC RC J1G1 N C J1G1 The rotation angle of gear D is equal to the rotation angle of gear C plus the twist that occurs in shaft (2): D

C

2

and so the rotation angle of gear D can be expressed in terms of the torques T1 and T2: N B T1L1 T2 L2 3 D N C J1G1 J 2G2 Shafts having the same diameters and the same shear moduli are required for this system; therefore, J1 = J2 = J and G1 = G2 = G. Factor these terms out to obtain: 1 NB 1 NB T1L1 T2 L2 3 J T1L1 T2 L2 JG NC G (3 ) NC Express the polar moment of inertia in terms of diameter to obtain the following relationship: 32 NB d4 T1L1 T2 L2 G (3 ) NC

32 (12,000,000 psi)(3 )

rad

180

48 teeth ( 2,760 lb-in.)(120 in.) (4,140 lb-in.)(96 in.) 72 teeth

4

10.022529 in. d 1.779 in.

Since this minimum diameter is greater than the diameter needed to satisfy the shear stress requirements, the rotation angle constraint controls. Therefore, the minimum shaft diameter that satisfies all requirements is d ≥ 1.779 in.

Ans.

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6.41 Two solid steel shafts are connected by the gears shown in Fig. P6.41. The design requirements for the system require (1) that both shafts must have the same diameter, (2) that the maximum shear stress in each shaft must be less than 50 MPa, and (3) that the rotation angle of gear D must not exceed 3°. Determine the minimum required diameter of the shafts if the torque applied at gear D is TD = 750 N-m. The shaft lengths are L1 = 2.5 m and L2 = 2.0 m. Assume that the shear modulus of both shafts is G = 80 GPa and that the bearings shown allow free rotation of the shafts. Fig. P6.41

Solution Torque relationship: The torque on gear D creates a torque in shaft (2) of T2 = 1,200 N-m. M x TD T2 0 T2 TD 750 N-m Accordingly, the torque applied to gear C is also 750 N-m. The torque on gear B is found with the gear ratio: 48 teeth TB TC 0.6667 TC 0.6667(750 N-m) 500 N-m 72 teeth Shaft (1) has the same torque as gear B; therefore, T1 = −500 N-m. Diameters based on allowable shear stresses: The elastic torsion formula gives the relationship between shear stress and torque in a shaft. This equation can be rearranged in terms of the outside diameter: Tc T d3 J 16 Solve for the minimum acceptable diameter in shaft (1): 16 T1 16(500 N-m)(1,000 mm/m) d13 50,929.582 mm3 2 (50 N/mm )

d1 37.067 mm Solve for the minimum acceptable diameter in shaft (2): 16 T2 16(750 N-m)(1,000 mm/m) d 23 76,394.373 mm3 2 (50 N/mm ) d 2 42.431 mm Of these two values, d2 controls. Therefore, both shafts could have a diameter of 42.431 mm or more and the shear stress constraint would be satisfied. The angles of twists in shafts (1) and (2) can be expressed as: T1L1 T2 L2 1 2 J1G1 J 2G2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The rotations of gears B and C are related since the arclengths turned by the two gears have the same magnitude. The gears turn in opposite directions; therefore, a negative sign is introduced. RC C RB B The rotation angle of gear B is equal to the angle of twist in shaft (1): B = 1. Therefore, the rotation angle of gear C can be expressed in terms of T1. RB RB RB T1L1 N B T1L1 C B 1 RC RC RC J1G1 N C J1G1 The rotation angle of gear D is equal to the rotation angle of gear C plus the twist that occurs in shaft (2): D

C

2

and so the rotation angle of gear D can be expressed in terms of the torques T1 and T2: N B T1L1 T2 L2 3 D N C J1G1 J 2G2 Shafts having the same diameters and the same shear moduli are required for this system; therefore, J1 = J2 = J and G1 = G2 = G. Factor these terms out to obtain: 1 NB 1 NB T1L1 T2 L2 3 J T1L1 T2 L2 JG NC G (3 ) NC Express the polar moment of inertia in terms of diameter to obtain the following relationship: 32 NB d4 T1L1 T2 L2 G (3 ) N C 32

48 teeth ( 500 N-m)(2.5 m) (750 N-m)(2.0 m) (1,000 mm/m) 2 72 teeth (80,000 N/mm 2 )(3 )( / 180 )

5,673,986.284 mm 4 d

48.806 mm

Since this minimum diameter is greater than the diameter needed to satisfy the shear stress requirements, the rotation angle constraint controls. Therefore, the minimum shaft diameter that satisfies all requirements is d ≥ 48.8 mm.

Ans.

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6.42 The driveshaft of an automobile is being designed to transmit 180 hp at 3,500 rpm. Determine the minimum diameter required for a solid steel shaft if the allowable shear stress in the shaft is not to exceed 6,000 psi.

Solution The torque in the driveshaft is: 550 lb-ft/s 180 hp 1 hp P T 3,500 rev 2 rad 1 min min 1 rev 60 s

270.109 lb-ft

The minimum diameter required for the shaft can be found from: T (270.109 lb-ft)(12 in./ft) d3 0.540217 in.3 16 6,000 psi allow d

1.401241 in.

1.401 in.

Ans.

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6.43 A tubular steel shaft transmits 225 hp at 4,000 rpm. Determine the maximum shear stress produced in the shaft if the outside diameter is D = 3.000 in. and the wall thickness is t = 0.125 in.

Solution The torque in the tubular steel shaft is: 550 lb-ft/s 225 hp 1 hp P T 4,000 rev 2 rad 1 min min 1 rev 60 s

295.431 lb-ft

The polar moment of inertia of the shaft is: J

32

D4

d4

32

(3.000 in.)4

(2.750 in.)4

and the maximum shear stress in the shaft is Tc (295.431 lb-ft)(3.000 in./2)(12 in./ft) J 2.337403 in.4

2.337403 in.4

2, 275.074 psi

2, 280 psi

Ans.

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6.44 A tubular steel shaft is being designed to transmit 225 kW at 1,700 rpm. The maximum shear stress in the shaft must not exceed 30 MPa. If the outside diameter of the shaft is D = 75 mm, determine the minimum wall thickness for the shaft.

Solution The torque in the tubular steel shaft is: 1,000 N-m/s 225 kW P 1 kW T 1,700 rev 2 rad 1 min min 1 rev 60 s

1, 263.877 N-m

The maximum shear stress in the shaft will be determined from the elastic torsion formula: Tc D4 d 4 where J J 32 Rearrange this equation, grouping the diameter terms on the left-hand side of the equation: D4 d 4 T

32 D / 2 Substitute the known values for T, , and D and solve for the inside diameter d: (75 mm) 4 d 4 (1, 263.877 N-m)(1,000 mm/m) 32

75 mm / 2 (75 mm) 4 d 4 d4

30 N/mm 2 16,092,181.76 mm 4 (75 mm)4 16,092,181.76 mm 4 d 62.7945 mm

15,548, 443.24 mm 4

The minimum wall thickness for the tubular steel shaft is thus D d 2t D d 75 mm 62.7945 mm t 6.10275 mm 6.10 mm 2 2

Ans.

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6.45 A solid 20-mm-diameter bronze shaft transmits 11 kW at 25 Hz to the propeller of a small sailboat. Determine the maximum shear stress produced in the shaft.

Solution The torque in the tubular steel shaft is: 1,000 N-m/s 11 kW P 1 kW T 70.028175 N-m 25 rev 2 rad s 1 rev The polar moment of inertia of the shaft is: J

32

(20 mm)4

15,707.963 mm4

and the maximum shear stress in the shaft is Tc (70.028175 N-m)(20 mm/2)(1,000 mm/m) J 15,707.963 mm 4

44.581 MPa

44.6 MPa

Ans.

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6.46 A steel propeller for a windmill transmits 5.5 kW at 65 rpm. If the allowable shear stress in the shaft must be limited to 60 MPa, determine the minimum diameter required for a solid shaft.

Solution The torque in the tubular steel shaft is: 1,000 N-m/s 5.5 kW P 1 kW T 65 rev 2 rad 1 min min 1 rev 60 s

808.017 N-m

The minimum diameter required for the shaft can be found from: T (808.017 N-m)(1,000 mm/m) d3 13, 466.957 mm3 2 16 60 N/mm allow d

40.934 mm

40.9 mm

Ans.

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6.47 A solid steel [G = 12,000 ksi] shaft with a 3-in. diameter must not twist more than 0.06 rad in a 16ft length. Determine the maximum horsepower that the shaft can transmit at 4 Hz.

Solution Section properties: The polar moment of inertia of the shaft is: J

32

(3 in.)4

7.952156 in.4

Angle of twist relationship: From the angle of twist relationship, compute the allowable torque: JG (0.06 rad)(7.952156 in.4 )(12,000 ksi) T 29.820585 kip-in. 2, 485.05 lb-ft L (16 ft)(12 in./ft) Power transmission: The maximum power that can be transmitted at 4 Hz is: 4 rev 2 rad P T (2, 485.05 lb-ft) 62, 456.088 lb-ft/s s 1 rev or in units of horsepower, 62, 456.088 lb-ft/s P 113.6 hp 550 lb-ft/s 1 hp

Ans.

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6.48 A tubular steel [G = 80 GPa] shaft with an outside diameter of D = 100 mm and a wall thickness of t = 6 mm must not twist more than 0.05 rad in a 7-m length. Determine the maximum power that the shaft can transmit at 375 rpm.

Solution Section properties: The polar moment of inertia of the shaft is: J

32

D4

d4

32

(100 mm) 4

(88 mm) 4

3,929,981.613 mm 4

Angle of twist relationship: From the angle of twist relationship, compute the allowable torque: JG (0.05 rad)(3,929,981.613 mm 4 )(80,000 N/mm 2 ) T L (7 m)(1,000 mm/m) 2,245,703.779 N-mm 2,245.704 N-m Power transmission: The maximum power that can be transmitted at 375 rpm is: 375 rev 2 rad 1 min P T (2, 245.704 N-m) 88,188.581 N-m/s 88.2 kW min 1 rev 60 s

Ans.

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6.49 A solid 3-in.-diameter bronze [G = 6,000 ksi] shaft is 7-ft long. The allowable shear stress in the shaft is 8 ksi and the angle of twist must not exceed 0.03 rad. Determine the maximum horsepower that this shaft can deliver (a) when rotating at 150 rpm. (b) when rotating at 540 rpm.

Solution Section properties: The polar moment of inertia of the shaft is: J

32

(3 in.)4

7.952156 in.4

Torque based on allowable shear stress: Compute the allowable torque if the shear stress must not exceed 8 ksi: J (8 ksi)(7.952156 in.4 ) T 42.4115 kip-in. (a) c 1.5 in. Torque based on angle of twist: Compute the allowable torque if the angle of twist must not exceed 0.03 rad: JG (0.03 rad)(7.952156 in.4 )(6,000 ksi) T 17.0403 kip-in. (b) L (7 ft)(12 in./ft) Controlling torque: From comparison of Eqs. (a) and (b), the maximum torque allowed for this shaft is: Tmax 17.0403 kip-in. 1, 420.028 lb-ft (a) Power transmission at 150 rpm: The maximum power that can be transmitted at 150 rpm is: 150 rev 2 rad 1 min P T (1, 420.028 lb-ft) 22,305.746 lb-ft/s min 1 rev 60 s or in units of horsepower, 22,305.746 lb-ft/s P 40.6 hp Ans. 550 lb-ft/s 1 hp (b) Power transmission at 540 rpm: The maximum power that can be transmitted at 540 rpm is: 540 rev 2 rad 1 min P T (1, 420.028 lb-ft) 80,300.684 lb-ft/s min 1 rev 60 s or in units of horsepower, 80,300.684 lb-ft/s P 146.0 hp Ans. 550 lb-ft/s 1 hp

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6.50 A hollow titanium [G = 43 GPa] shaft has an outside diameter of D = 50 mm and a wall thickness of t = 1.25 mm. The maximum shear stress in the shaft must be limited to 150 MPa. Determine: (a) the maximum power that can be transmitted by the shaft if the rotation speed must be limited to 20 Hz. (b) the magnitude of the angle of twist in a 700-mm length of the shaft when 30 kW is being transmitted at 8 Hz.

Solution Section properties: The polar moment of inertia of the shaft is: J

32

D4

d4

32

(50 mm)4

(47.5 mm)4

113,817.540 mm4

Torque based on allowable shear stress: Compute the allowable torque if the shear stress must not exceed 100 MPa: J (150 N/mm 2 )(113,817.540 mm 4 ) T 682,905.237 N-mm 682.905 N-m c 50 mm/2 (a) Power transmission: The maximum power that can be transmitted at 20 Hz is: 20 rev 2 rad P T (682.905 N-m) 85,816.403 N-m/s 85.8 kW s 1 rev

Ans.

(b) Angle of twist: The torque in the shaft when 30 kW is being transmitted at 8 Hz is: P 30,000 N-m/s T 596.831 N-m 8 rev 2 rad s 1 rev The corresponding angle of twist is: T L (596.831 N-m)(700 mm)(1,000 mm/m) JG (113,817.540 mm 4 )(43,000 N/mm 2 )

0.0854 rad

Ans.

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6.51 A tubular aluminum alloy [G = 4,000 ksi] shaft is being designed to transmit 400 hp at 1,500 rpm. The maximum shear stress in the shaft must not exceed 6 ksi and the angle of twist is not to exceed 5° in an 8-ft length. Determine the minimum permissible outside diameter if the inside diameter is to be threefourths of the outside diameter.

Solution Torque from power transmission equation: The torque in the shaft when 400 hp is being transmitted at 1,500 rpm is: 550 lb-ft/s (400 hp) 1 hp P T 1, 400.56 lb-ft 1,500 rev 2 rad 1 min min 1 rev 60 s Polar moment of inertia: The inside diameter of the shaft is to be three-fourths of the outside diameter; therefore, d = 0.75D. From this, the polar moment of inertia can be expressed as: D4 4 D4 4 4 4 4 4 J D d D (0.75D) 1 (0.75) 0.683594 0.067112 D 4 (a) 32 32 32 32 Diameter based on shear stress: The maximum shear stress must not exceed 6 ksi; thus, from the elastic torsion formula: Tc J T J c Use the results of Eq. (a) to simplify the left-hand side of this equation, giving an expression in terms of the outside diameter D: J 0.067112 D 4 0.134224 D3 c D/2 Solve for the outside diameter D: (1, 400.56 lb-ft)(12 in./ft) 0.134224 D3 2.801120 in.3 D 2.753175 in. (b) 6, 000 psi Diameter based on twist angle: The angle of twist is not to exceed 5° in an 8-ft length; thus, from the angle of twist formula: TL TL J JG G Use the results of Eq. (a) to simplify this equation and solve for the outside diameter D: (1, 400.56 lb-ft)(8 ft)(12 in./ft) 2 (c) 0.067112 D4 4.622180 in.4 D 2.881972 in. (5 )( /180 )(4,000,000 psi)

From the results of Eqs. (b) and (c), the minimum outside diameter D acceptable in this instance is: Ans. Dmin 2.88 in.

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6.52 A tubular steel [G = 80 GPa] shaft is being designed to transmit 150 kW at 30 Hz. The maximum shear stress in the shaft must not exceed 80 MPa and the angle of twist is not to exceed 6° in a 4-m length. Determine the minimum permissible outside diameter if the ratio of the inside diameter to the outside diameter is 0.80.

Solution Torque from power transmission equation: The torque in the shaft when 150 kW is being transmitted at 1,500 rpm is: P 150, 000 N-m/s T 795.7747 N-m 30 rev 2 rad s 1 rev Polar moment of inertia: The ratio of inside diameter to outside diameter for the shaft is to be d/D = 0.80 or in other words d = 0.80D. From this, the polar moment of inertia can be expressed as: D4 4 D4 4 4 4 4 4 J D d D (0.80 D) 1 (0.80) 0.590400 0.057962 D 4 (a) 32 32 32 32 Diameter based on shear stress: The maximum shear stress must not exceed 80 MPa; thus, from the elastic torsion formula: Tc J T J c Use the results of Eq. (a) to simplify the left-hand side of this equation, giving an expression in terms of the outside diameter D: J 0.057962 D 4 0.115925D3 c D/2 Solve for the outside diameter D: (795.7747 N-m)(1,000 mm/m) 0.115925D3 80 N/mm2

9,947.18375 mm3

D 44.1070 mm

(b)

Diameter based on twist angle: The angle of twist is not to exceed 6° in a 4-m length; thus, from the angle of twist formula: TL TL J JG G Use the results of Eq. (a) to simplify this equation and solve for the outside diameter D: (795.7747 N-m)(4 m)(1,000 mm/m) 2 0.057962 D4 379,954.43 mm 4 D 50.5996 mm (6 )( /180 )(80,000 N/mm2 )

(c)

From the results of Eqs. (b) and (c), the minimum outside diameter D acceptable in this instance is: Ans. Dmin 50.6 mm

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6.53 An automobile engine supplies 180 hp at 4,200 rpm to a driveshaft. If the allowable shear stress in the driveshaft must be limited to 5 ksi, determine: (a) the minimum diameter required for a solid driveshaft. (b) the maximum inside diameter permitted for a hollow driveshaft if the outside diameter is 2.00 in. (c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The weight of a shaft is proportional to its cross-sectional area.)

Solution Power transmission equation: The torque in the shaft is 550 lb-ft/s 180 hp 1 hp P T 225.0906 lb-ft 4,200 rev 2 rad 1 min min 1 rev 60 s (a) Solid driveshaft: The minimum diameter required for a solid shaft can be found from: T (225.0906 lb-ft)(12 in./ft) d3 0.540217 in.3 16 5,000 psi allow d

1.401241 in.

Ans.

1.401 in.

(b) Hollow driveshaft: From the elastic torsion formula: D4 d 4 Tc J T

J c 32 D / 2 Solve this equation for the maximum inside diameter d: (2.00 in.)4 d 4 (225.0906 lb-ft)(12 in./ft) 32

2.00 in. / 2 (2.00 in.)

4

5,000 psi d

4

d4

0.540217 in.3

5.502605 in.4 16 in.4 d

5.502605 in.4

10.497395 in.4 Ans.

1.800 in.

(c) Weight savings: The weights of the solid and hollow shafts are proportional to their respective cross-sectional areas. The cross-sectional area of the solid shaft is Asolid

(1.401241 in.)2

1.5421 in.2

4 and the cross-sectional area of the hollow shaft is

Ahollow

(2.00 in.)2 (1.80 in.) 2

0.5969 in.2

4 The weight savings can be determined from ( Asolid Ahollow ) weight savings (in percent) Asolid

(1.5421 in.2 0.5929 in.2 ) 1.5421 in.2

61.3%

Ans.

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6.54 The impeller shaft of a fluid agitator transmits 28 kW at 440 rpm. If the allowable shear stress in the impeller shaft must be limited to 80 MPa, determine: (a) the minimum diameter required for a solid impeller shaft. (b) the maximum inside diameter permitted for a hollow impeller shaft if the outside diameter is 40 mm. (c) the percent savings in weight realized if the hollow shaft is used instead of the solid shaft. (Hint: The weight of a shaft is proportional to its cross-sectional area.)

Solution Power transmission equation: The torque in the shaft is P 28, 000 N-m/s T 607.6825 N-m 440 rev 2 rad 1 min min 1 rev 60 s (a) Solid impeller shaft: The minimum diameter required for a solid shaft can be found from: T (607.6825 N-m)(1,000 mm/m) d3 7,596.0313 mm3 2 16 80 N/mm allow d

33.8209 mm

33.8 mm

Ans.

(b) Hollow driveshaft: From the elastic torsion formula: D4 d 4 Tc J T

J c 32 D / 2 Solve this equation for the maximum inside diameter: (40 mm) 4 d 4 (607.6825 N-m)(1,000 mm/m) 32

40 mm / 2 (40 mm) 4 d 4 d4

80 N/mm 1,547, 450.779 mm 4

2

7,596.0313 mm3

2,560,000 mm 4 1,547, 450.779 mm 4 d

31.7215 mm

1,012,549.221 mm 4

31.7 mm

Ans.

(c) Weight savings: The weights of the solid and hollow shafts are proportional to their respective cross-sectional areas. The cross-sectional area of the solid shaft is Asolid

(33.8209 mm) 2

898.3803 mm 2

4 and the cross-sectional area of the hollow shaft is (40 mm)2 (31.7215 mm) 2 466.3262 mm 2 4 The weight savings can be determined from ( Asolid Ahollow ) (898.3803 mm2 466.3262 mm2 ) weight savings (in percent) Asolid 898.3803 mm2 Ahollow

48.1%

Ans.

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6.55 A motor supplies 200 kW at 6 Hz to flange A of the shaft shown in Fig. P6.55. Gear B transfers 125 kW of power to operating machinery in the factory, and the remaining power in the shaft is transferred by gear D. Shafts (1) and (2) are solid aluminum [G = 28 GPa] shafts that have the same diameter and an allowable shear stress of = 40 MPa. Shaft (3) is a solid steel [G = 80 GPa] shaft with an allowable shear stress of = 55 MPa. Determine: (a) the minimum permissible diameter for aluminum shafts (1) and (2). (b) the minimum permissible diameter for steel shaft (3). (c) the rotation angle of gear D with respect to flange A if the shafts have the minimum permissible diameters as determined in (a) and (b). Fig. P6.55

Solution Power transmission: The torque on flange A is P 200, 000 N-m/s TA 5,305.165 N-m 6 rev 2 rad s 1 rev thus, the torque in shaft segment (1) is T1 = 5,305.165 N-m. The torque transferred by gear B is P 125, 000 N-m/s TB 3,315.728 N-m 6 rev 2 rad s 1 rev therefore, the remaining torque in segments (2) and (3) of the shaft is T2 TA TB 5,305.165 N-m 3,315.728 N-m 1,989.437 N-m (a) Minimum diameter for solid aluminum shaft: The minimum diameter required for the solid aluminum shaft can be found from: T1 (5,305.165 N-m)(1,000 mm/m) d13 132,629.125 mm3 2 16 40 N/mm 1,allow

d1

Ans.

87.7 mm

(b) Minimum diameter for solid steel shaft: The minimum diameter required for the solid steel shaft can be found from: T3 (1,989.437 N-m)(1,000 mm/m) d33 36,171.582 mm3 2 16 55 N/mm 3,allow

d3

Ans.

56.9 mm

(c) Angles of twist: The polar moments of inertia of the aluminum and steel shafts are: J1 J3

J2 32

32

(87.7 mm) 4

(56.9 mm) 4

5,807,621 mm 4

1,029,080 mm 4

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The angles of twist in the three shaft segments are: T1L1 (5,305.165 N-m)(0.8 m)(1,000 mm/m) 2 1 J1G1 (5,807,621 mm 4 )(28,000 N/mm 2 )

0.026100 rad

2

T2 L2 J 2G2

(1,989.437 N-m)(0.4 m)(1,000 mm/m) 2 (5,807,621 mm 4 )(28,000 N/mm 2 )

0.004894 rad

3

T3 L3 J 3G3

(1,989.437 N-m)(0.8 m)(1,000 mm/m)2 (1,029,080 mm 4 )(80,000 N/mm 2 )

0.019332 rad

Rotation angle at D: The rotation angle at D is equal to the sum of the angles of twist in the three shaft segments: D

1

2

3

0.026100 rad 0.004894 rad 0.019332 rad 0.050326 rad

0.0503 rad

Ans.

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6.56 A motor supplies 150 hp at 520 rpm to flange A of the shaft shown in Fig. P6.56. Gear B transfers 90 hp of power to operating machinery in the factory, and the remaining power in the shaft is transferred by gear D. Shafts (1) and (2) are solid aluminum [G = 4,000 ksi] shafts that have the same diameter and an allowable shear stress of = 6 ksi. Shaft (3) is a solid steel [G = 12,000 ksi] shaft with an allowable shear stress of = 8 ksi. Determine: (a) the minimum permissible diameter for aluminum shafts (1) and (2). (b) the minimum permissible diameter for steel shaft (3). (c) the rotation angle of gear D with respect to flange A if the shafts have the minimum permissible diameters as determined in (a) and (b). Fig. P6.56

Solution Power transmission: The torque on flange A is 550 lb-ft/s 150 hp 1 hp P TA 1,515.033 lb-ft 520 rev 2 rad 1 min min 1 rev 60 s thus, the torque in shaft segment (1) is T1 = 1,515.033 lb-ft. The torque transferred by gear B is 550 lb-ft/s 90 hp 1 hp P TB 909.020 lb-ft 520 rev 2 rad 1 min min 1 rev 60 s therefore, the remaining torque in segments (2) and (3) of the shaft is T2 TA TB 1,515.033 lb-ft 909.020 lb-ft 606.013 lb-ft (a) Minimum diameter for solid aluminum shaft: The minimum diameter required for the solid aluminum shaft can be found from: T1 (1,515.033 lb-ft)(12 in./ft) d13 3.030066 in.3 16 6,000 psi 1,allow

d1

2.49 in.

Ans.

(b) Minimum diameter for solid steel shaft: The minimum diameter required for the solid steel shaft can be found from: T3 (606.013 lb-ft)(12 in./ft) d33 0.909020 in.3 16 8,000 psi 3,allow

d3 1.667 in.

Ans.

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(c) Angles of twist: The polar moments of inertia of the aluminum and steel shafts are: J1 J3

J2 32

32

(2.49 in.) 4

(1.667 in.) 4

3.773960 in.4

0.758128 in.4

The angles of twist in the three shaft segments are: T1L1 (1,515.033 lb-ft)(2 ft)(12 in./ft)2 0.028904 rad 1 J1G1 (3.773960 in.4 )(4,000,000 psi) 2

T2 L2 J 2G2

(606.013 lb-ft)(1 ft)(12 in./ft)2 (3.773960 in.4 )(4,000,000 psi)

3

T3 L3 J 3G3

(606.013 lb-ft)(2 ft)(12 in./ft)2 (0.758128 in.4 )(12,000,000 psi)

0.005781 rad 0.019185 rad

Rotation angle at D: The rotation angle at D is equal to the sum of the angles of twist in the three shaft segments: D

1

2

3

0.028904 rad 0.005781 rad 0.019185 rad 0.053870 rad

0.0539 rad

Ans.

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6.57 A motor supplies sufficient power to the system shown in Fig. P6.57 so that gears C and D provide torques of TC = 800 N-m and TD = 550 N-m, respectively, to machinery in a factory. Power shaft segments (1) and (2) are hollow steel tubes with an outside diameter of D = 60 mm and an inside diameter of d = 50 mm. If the power shaft [i.e., segments (1) and (2)] rotates at 40 rpm, determine: (a) the maximum shear stress in power shaft segments (1) and (2). (b) the power (in kW) that must be provided by the motor as well as the rotation speed (in rpm). (c) the torque applied to gear A by the motor. Fig. P6.57

Solution Equilibrium: M x TD T2

T2

Mx

TC

0 TD

TD T1

550 N-m

T1 TC

0 TD

800 N-m 550 N-m 1,350 N-m

Section properties: J1

32

(60 mm) 4

(50 mm) 4

658,752.71 mm 4

(a) Shear stresses: T1c1 (1,350 N-m)(60 mm/2)(1,000 mm/m) 1 J1 658,752.71 mm 4 2

T2c2 J2

(550 N-m)(60 mm/2)(1,000 mm/m) 658,752.71 mm 4

J2

61.480 MPa

61.5 MPa

Ans.

25.047 MPa

25.0 MPa

Ans.

(b) Power transmission: Compute the power in shaft segment (1): 40 rev 2 rad 1 min P T1 (1,350 N-m) 5,654.867 N-m/s min 1 rev 60 s This power in the shaft must be provided by the motor. Therefore: Pmotor 5.65 kW

5.65 kW

Ans.

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The rotation speed of the motor is found with the use of the gear ratio: 72 teeth 3 B 3(40 rpm) 120 rpm A B motor 24 teeth (c) Motor torque: The motor must provide a torque of N 24 teeth TA TB A (1,350 N-m) 450 N-m NB 72 teeth

Ans.

Ans.

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6.58 A motor supplies sufficient power to the system shown in Fig. P6.58 so that gears C and D provide torques of TC = 100 lb-ft and TD = 80 lb-ft, respectively, to machinery in a factory. Power shaft segments (1) and (2) are hollow steel tubes with an outside diameter of D = 1.75 in. and an inside diameter of d = 1.50 in. If the power shaft [i.e., segments (1) and (2)] rotates at 540 rpm, determine: (a) the maximum shear stress in power shaft segments (1) and (2). (b) the power (in hp) that must be provided by the motor as well as the rotation speed (in rpm). (c) the torque applied to gear A by the motor. Fig. P6.58

Solution Equilibrium: M x TD T2

T2

Mx

TC

0 TD

TD T1

80 lb-ft

T1 TC

0 TD

100 lb-ft 80 lb-ft 180 lb-ft

Section properties: J1

32

(1.75 in.)4

(1.50 in.)4

0.423762 in.4

(a) Shear stresses: T1c1 (180 lb-ft)(1.75 in./2)(12 in./ft) 1 J1 0.423762 in.4 2

T2c2 J2

(80 lb-ft)(1.75 in./2)(12 in./ft) 0.423762 in.4

J2

4, 460.049 psi 1,982.244 psi

4, 460 psi 1,982 psi

Ans. Ans.

(b) Power transmission: Compute the power in shaft segment (1): 540 rev 2 rad 1 min P T1 (180 lb-ft) 10,178.760 lb-ft/s 18.507 hp min 1 rev 60 s This power in the shaft must be provided by the motor. Therefore: Pmotor 18.51 hp

Ans.

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The rotation speed of the motor is found with the use of the gear ratio: 72 teeth 3 B 3(540 rpm) 1,620 rpm A B motor 24 teeth (c) Motor torque: The motor must provide a torque of N 24 teeth TA TB A (180 lb-ft) 60 lb-ft 720 lb-in. NB 72 teeth

Ans.

Ans.

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6.59 A motor supplies 25 hp at 6 Hz to gear A of the drive system shown in Fig. P6.59. Shaft (1) is a solid 2.25-in.-diameter aluminum [G = 4,000 ksi] shaft with a length of L1 = 16 in. Shaft (2) is a solid 1.5-in.-diameter steel [G = 12,000 ksi] shaft with a length of L2 = 12 in. Shafts (1) and (2) are connected at flange C, and the bearings shown permit free rotation of the shaft. Determine: (a) the maximum shear stress in shafts (1) and (2). (b) the rotation angle of gear D with respect to gear B.

Fig. P6.59

Solution Power transmission: The torque acting on gear A is: 550 lb-ft/s 25 hp 1 hp P TA 364.730 lb-ft 6 rev 2 rad s 1 rev Equilibrium: The torque applied to gear B is: N 60 teeth TB TA B (364.730 lb-ft) 911.825 lb-ft NA 24 teeth The torque on shaft segments (1) and (2) as well as gear D is equal to the torque applied to gear B. T1 T2 TD TB 911.825 lb-ft Section properties: The polar moments of inertia for aluminum shaft (1) and steel shaft (2) are: J1

32

d14

32

(2.25 in.)4

2.516112 in.4

J2

(a) Maximum shear stresses: T1c1 (911.825 lb-ft)(2.25 in./2)(12 in./ft) 1 J1 2.516112 in.4 2

T2c2 J2

(911.825 lb-ft)(1.5 in./2)(12 in./ft) 0.497010 in.4

32

d 24

32

(1.5 in.)4

0.497010 in.4

4,892.326 psi

4,890 psi

Ans.

16,511.601 psi

16,510 psi

Ans.

(b) Angles of twist: The angles of twist in the two shaft segments are: T1L1 (911.825 lb-ft)(16 in.)(12 in./ft) 0.017395 rad 1 J1G1 (2.516112 in.4 )(4,000,000 psi) 2

T2 L2 J 2G2

(911.825 lb-ft)(12 in.)(12 in./ft) (0.497010 in.4 )(12,000,000 psi)

Rotation angle of D relative to B: 0.017395 rad 0.022015 rad D B 1 2

0.022015 rad

0.039410 rad

0.0394 rad

Ans.

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6.60 A motor supplies 20 kW at 400 rpm to gear A of the drive system shown in Fig. P6.60. Shaft (1) is a solid 50-mm-diameter aluminum [G = 28 GPa] shaft with a length of L1 = 1,200 mm. Shaft (2) is a solid 40-mm-diameter steel [G = 80 GPa] shaft with a length of L2 = 750 mm. Shafts (1) and (2) are connected at flange C, and the bearings shown permit free rotation of the shaft. Determine: (a) the maximum shear stress in shafts (1) and (2). (b) the rotation angle of gear D with respect to gear B.

Fig. P6.60

Solution Power transmission: The torque acting on gear A is: P 20,000 N-m/s TA 477.465 N-m 400 rev 2 rad 1 min min 1 rev 60 s Equilibrium: The torque applied to gear B is: N 60 teeth TB TA B (477.465 N-m) 1,193.662 N-m NA 24 teeth The torque on shaft segments (1) and (2) as well as gear D is equal to the torque applied to gear B. T1 T2 TD TB 1,193.662 N-m Section properties: The polar moments of inertia for aluminum shaft (1) and steel shaft (2) are: J1

32

d14

32

(50 mm)4

613,592.32 mm 4

J2

(a) Maximum shear stresses: T1c1 (1,193.662 N-m)(50 mm/2)(1,000 mm/m) 1 J1 613,592.32 mm 4 2

T2c2 J2

(1,193.662 N-m)(40 mm/2)(1,000 mm/m) 251,327.41 mm 4

32

d 24

32

(40 mm) 4

251,327.41 mm4

48.634 MPa

48.6 MPa

Ans.

94.989 MPa

95.0 MPa

Ans.

(b) Angles of twist: The angles of twist in the two shaft segments are: T1L1 (1,193.662 N-m)(1,200 mm)(1,000 mm/m) 0.083373 rad 1 J1G1 (613,592.32 mm 4 )(28,000 N/mm 2 ) 2

T2 L2 J 2G2

(1,193.662 N-m)(750 mm)(1,000 mm/m) (251,327.41 mm 4 )(80,000 N/mm 2 )

Rotation angle of D relative to B: 0.083373 rad 0.044526 rad D B 1 2

0.044526 rad

0.127899 rad

0.1279 rad

Ans.

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6.61 The motor shown in Fig. P6.61 supplies 10 hp at 1,500 rpm at A. The bearings shown permit free rotation of the shafts. (a) Shaft (1) is a solid 0.875-in.-diameter steel shaft. Determine the maximum shear stress produced in shaft (1). (b) If the shear stress in shaft (2) must be limited to 6,000 psi, determine the minimum acceptable diameter for shaft (2) if a solid shaft is used.

Fig. P6.61

Solution Power transmission: The torque acting in shaft (1) and on gear B is: 550 lb-ft/s 10 hp 1 hp P T1 TB 35.014 lb-ft 1,500 rev 2 rad 1 min min 1 rev 60 s Section properties: The polar moment of inertia for shaft (1) is: J1

32

d14

32

(0.875 in.)4

0.057548 in.4

(a) Maximum shear stress in shaft (1): T1c1 (35.014 lb-ft)(0.875 in./2)(12 in./ft) 1 J1 0.057548 in.4

3,194.258 psi

3,194 psi

Ans.

(b) Equilibrium: The torque applied to gear C is: N 30 teeth TC TB C (35.014 lb-ft) 21.884 lb-ft NB 48 teeth The torque on shaft segment (2) as well as gear D is equal to the torque applied to gear C. T2 TD TC 21.884 lb-ft The minimum diameter required for solid shaft (2) can be found from: T2 (21.884 lb-ft)(12 in./ft) d 23 0.043768 in.3 16 6,000 psi 2,allow d2

0.606328 in.

0.606 in.

Ans.

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6.62 The motor shown in Fig. P6.62 supplies 12 kW at 15 Hz at A. The bearings shown permit free rotation of the shafts. (a) Shaft (2) is a solid 35-mm-diameter steel shaft. Determine the maximum shear stress produced in shaft (2). (b) If the shear stress in shaft (1) must be limited to 40 MPa, determine the minimum acceptable diameter for shaft (1) if a solid shaft is used.

Fig. P6.62

Solution Power transmission: The torque acting in shaft (1) and on gear B is: P 12,000 N-m/s T1 TB 127.324 N-m 15 rev 2 rad s 1 rev The torque applied to gear C is: N 30 teeth TC TB C (127.324 N-m) NB 48 teeth

79.577 N-m

The torque on shaft segment (2) as well as gear D is equal to the torque applied to gear C. T2 TD TC 79.577 N-m Section properties: The polar moment of inertia for shaft (2) is: J2

32

d 24

32

(35 mm) 4

147,323.515 mm 4

(a) Maximum shear stress in shaft (2): T2c2 (79.577 N-m)(35 mm/2)(1,000 mm/m) 2 J2 147,323.515 mm 4 (b) Minimum diameter for solid shaft (1): T1 (127.324 N-m)(1,000 mm/m) d13 16 40 N/mm 2 1,allow d1

25.309 mm

25.3 mm

9.453 MPa

9.45 MPa

Ans.

3,183.099 mm3

Ans.

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6.63 The motor shown in Fig. P6.63 supplies 9 kW at 15 Hz at A. Shafts (1) and (2) are each solid 25-mm-diameter steel [G = 80 GPa] shafts with lengths of L1 = 900 mm and L2 = 1,200 mm, respectively. The bearings shown permit free rotation of the shafts. Determine: (a) the maximum shear stress produced in shafts (1) and (2). (b) the rotation angle of gear D with respect to flange A. Fig. P6.63

Solution Power transmission: The magnitude of the torque acting in shaft (1) and on gear B is: P 9,000 N-m/s T1 TB 95.493 N-m 15 rev 2 rad s 1 rev The magnitude of the torque applied to gear C is: N 54 teeth TC TB C (95.493 N-m) 143.239 N-m NB 36 teeth The torque magnitude in shaft segment (2) as well as the magnitude of the torque acting on gear D is equal to the torque applied to gear C. T2 TD TC 143.239 N-m To determine the proper signs for T1 and T2, consider equilibrium. First, consider a FBD that cuts through shaft (2) and includes gear D. M x TD T2 0 T2 TD 143.239 N-m Since T2 is positive, the torque in shaft (1) must be negative; therefore, T1 95.493 N-m Section properties: The polar moments of inertia for shafts (1) and (2) are equal: J1

32

d14

32

(25 mm) 4

38,349.520 mm 4

J2

(a) Maximum shear stress magnitudes: T1c1 (95.493 N-m)(25 mm/2)(1,000 mm/m) 1 J1 38,349.520 mm 4 2

T2c2 J2

(143.239 N-m)(25 mm/2)(1,000 mm/m) 38,349.520 mm 4

31.126 MPa 46.689 MPa

31.1 MPa 46.7 MPa

Ans. Ans.

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Angles of twist in shafts (1) and (2): T1L1 ( 95.493 N-m)(900 mm)(1,000 mm/m) 0.028013 rad 1 J1G1 (38,349.520 mm 4 )(80,000 N/mm 2 ) T2 L2 (143.239 N-m)(1,200 mm)(1,000 mm/m) 0.056027 rad 2 J 2G2 (38,349.520 mm 4 )(80,000 N/mm 2 ) Rotation angles of gears B and C: The rotation of gear B is found from the angle of twist in shaft (1): 0 rad ( 0.028013 rad) 0.028013 rad 1 B A B A 1 As gear B rotates, gear C also rotates but it rotates in the opposite direction. The magnitude of the rotation is dictated by the gear ratio: NB 36 teeth ( 0.028013 rad) 0.018676 rad C B NC 54 teeth (b) Rotation angle of gear D: The rotation of gear D with respect to flange A is found from: 0.018676 rad 0.056027 rad 0.074702 rad 0.0747 rad 2 D C D C 2

Ans.

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6.64 The motor shown in Fig. P6.64 supplies 150 hp at 1,800 rpm at A. Shafts (1) and (2) are each solid 2-in.-diameter steel [G = 12,000 ksi] shafts with lengths of L1 = 75 in. and L2 = 90 in., respectively. The bearings shown permit free rotation of the shafts. (a) Determine the maximum shear stress produced in shafts (1) and (2). (b) Determine the rotation angle of gear D with respect to flange A. Fig. P6.64

Solution Power transmission: The magnitude of the torque acting in shaft (1) and on gear B is: 550 lb-ft/s 150 hp 1 hp P T1 TB 437.676 lb-ft 1,800 rev 2 rad 1 min min 1 rev 60 s The magnitude of the torque applied to gear C is: N 54 teeth TC TB C (437.676 lb-ft) 656.514 lb-ft NB 36 teeth The torque magnitude in shaft segment (2) as well as the magnitude of the torque acting on gear D is equal to the torque applied to gear C. T2 TD TC 656.514 lb-ft To determine the proper signs for T1 and T2, consider equilibrium. First, consider a FBD that cuts through shaft (2) and includes gear D. M x TD T2 0 T2 TD 656.514 lb-ft Since T2 is positive, the torque in shaft (1) must be negative; therefore, T1 437.676 lb-ft Section properties: The polar moments of inertia for shafts (1) and (2) are equal: J1

32

d14

32

(2 in.) 4

1.570796 in.4

J2

(a) Maximum shear stress magnitudes: T1c1 (437.676 lb-ft)(2 in./2)(12 in./ft) 1 J1 1.570796 in.4 2

T2c2 J2

(656.514 lb-ft)(2 in./2)(12 in./ft) 1.570796 in.4

3,343.599 psi 5,015.399 psi

3,340 psi 5,020 psi

Ans. Ans.

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Angles of twist in shafts (1) and (2): T1L1 ( 437.676 lb-ft)(75 in.)(12 in./ft) 0.020897 rad 1 J1G1 (1.570796 in.4 )(12,000,000 psi) T2 L2 (656.514 lb-ft)(90 in.)(12 in./ft) 0.037615 rad 2 J 2G2 (1.570796 in.4 )(12,000,000 psi) Rotation angles of gears B and C: The rotation of gear B is found from the angle of twist in shaft (1): 0 rad ( 0.020897 rad) 0.020897 rad 1 B A B A 1 As gear B rotates, gear C also rotates but it rotates in the opposite direction. The magnitude of the rotation is dictated by the gear ratio: NB 36 teeth ( 0.020897 rad) 0.013932 rad C B NC 54 teeth (b) Rotation angle of gear D: The rotation of gear D with respect to flange A is found from: 0.013932 rad 0.037615 rad 0.051547 rad 0.0515 rad 2 D C D C 2

Ans.

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6.65 The gear train shown in Fig. P6.65 transmits power from a motor to a machine at E. The motor turns at a frequency of 50 Hz. The diameter of solid shaft (1) is 25 mm, the diameter of solid shaft (2) is 32 mm, and the allowable shear stress for each shaft is 60 MPa. Determine: (a) the maximum power that can be transmitted by the gear train. (b) the torque provided at gear E. (c) the rotation speed of gear E (in Hz). Fig. P6.65

Solution Section properties: J1 J2

32 32

d14

32

d 24

32

(25 mm) 4

38,349.52 mm 4

(32 mm) 4

102,943.71 mm 4

Allowable torques in the shafts: (60 N/mm2 )(38,349.52 mm4 ) allow J1 T1,allow c1 25 mm/2

T2,allow

allow

c2

J2

(60 N/mm2 )(102,943.71 mm4 ) 32 mm/2

184,078 N-mm 386,039 N-mm

Torque relationships at the gears: The magnitudes of the torques acting on the gears are related by: N N TB TA B TD TC D (a) NA NC Equilibrium: The magnitudes of the torques in the gears and shafts are related by: TB T1 TC TD T2 TE

(b)

Equations (a) and (b) express the relationship between the torque in shafts (1) and (2): N T2 T1 D NC

(c)

Controlling shaft torque: In Eq. (c), let T1 = T1,allow and compare the result with the allowable torque for shaft (2): N check T1,allow D T2,allow NC (184, 078 N-mm)

60 teeth 30 teeth

368,156 N-mm 386, 039 N-m

O.K.

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This calculation shows that the torque in shaft (1) controls the capacity of the gear train. Using this fact, the torques throughout the gear train can be determined: T1 184,078 N-mm T2 368,156 N-mm TB

TC

TA

TB

184,078 N-mm NA NB

(184,078 N-mm)

TD 24 teeth 72 teeth

TE

368,156 N-mm

61,359 N-mm

(a) Power transmission: The maximum power that can be transmitted by the gear train can be found from the torque and rotation speed at A: 50 rev 2 rad P TA A (61.359 N-m) 19, 276.571 N-m/s 19.28 kW Ans. s 1 rev (b) Torque at gear E: Equations (a) and (b) can be combined to express the relationship between the torque supplied by the motor at A and the torque available at gear E: N N N ND NB ND (d) TE TD TC D TB D TA B TA NC NC N A NC N A NC The torque available at gear E is thus: NB ND 72 teeth TE TA (61.359 N-m) N A NC 24 teeth

60 teeth 30 teeth

368.154 N-m

368 N-m

Ans.

This result could also be found from the torque in shaft (2), as calculated previously. (c) Rotation speed of gear E: There are a couple of ways to compute this speed. Using the gear ratios, the speed of gear E can be found from: N A NC 24 teeth 30 teeth (50 Hz) 8.33 Hz Ans. E A NB ND 72 teeth 60 teeth

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6.66 The gear train shown in Fig. P6.66 transmits power from a motor to a machine at E. The motor turns at 2,250 rpm. The diameter of solid shaft (1) is 1.50 in., the diameter of solid shaft (2) is 2.00 in., and the allowable shear stress for each shaft is 6 ksi. Determine: (a) the maximum power that can be transmitted by the gear train. (b) the torque provided at gear E. (c) the rotation speed of gear E (in Hz).

Fig. P6.66

Solution Section properties: J1 J2

32 32

d14

32

d 24

32

(1.50 in.) 4

0.497010 in.4

(2.00 in.) 4

1.570796 in.4

Allowable torques in the shafts: (6,000 psi)(0.497010 in.4 ) allow J1 T1,allow c1 1.50 in./2

T2,allow

allow

c2

J2

(6,000 psi)(1.570796 in.4 ) 2.00 in./2

3,976.1 lb-in. 9,424.8 lb-in.

Torque relationships at the gears: The magnitudes of the torques acting on the gears are related by: N N TB TA B TD TC D (a) NA NC Equilibrium: The magnitudes of the torques in the gears and shafts are related by: TB T1 TC TD T2 TE Equations (a) and (b) express the relationship between the torque in shafts (1) and (2): N T2 T1 D NC

(b)

(c)

Controlling shaft torque: In Eq. (c), let T1 = T1,allow and compare the result with the allowable torque for shaft (2): N check T1,allow D T2,allow NC (3,976.1 lb-in.)

60 teeth 30 teeth

7,952.2 lb-in. 9, 424.8 lb-in.

O.K.

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This calculation shows that the torque in shaft (1) controls the capacity of the gear train. Using this fact, the torques throughout the gear train can be determined: T1 3,976.1 lb-in. T2 7,952.2 lb-in. TB

TC

TA

TB

3,976.1 lb-in. NA NB

(3,976.1 lb-in.)

TD 24 teeth 72 teeth

TE

7,952.2 lb-in.

1,325.4 lb-in. 110.45 lb-ft

(a) Power transmission: The maximum power that can be transmitted by the gear train can be found from the torque and rotation speed at A: 2,250 rev 2 rad 1 min P TA A (110.45 lb-ft) 26,023.381 lb-ft/s 47.3 hp Ans. min 1 rev 60 s (b) Torque at gear E: Equations (a) and (b) can be combined to express the relationship between the torque supplied by the motor at A and the torque available at gear E: N N N ND NB ND (d) TE TD TC D TB D TA B TA NC NC N A NC N A NC The torque available at gear E is thus: NB ND 72 teeth TE TA (110.45 lb-ft) N A NC 24 teeth

60 teeth 30 teeth

662.7 lb-ft

663 lb-ft

Ans.

This result could also be found from the torque in shaft (2), as calculated previously. (c) Rotation speed of gear E: There are a couple of ways to compute this speed. Using the gear ratios, the speed of gear E can be found from: N A NC 24 teeth 30 teeth (2,250 rpm) 375 rpm Ans. E A NB ND 72 teeth 60 teeth

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6.67 A hollow circular cold-rolled bronze [G1 = 6,500 ksi] tube (1) with an outside diameter of 1.75 in. and an inside diameter of 1.25 in. is securely bonded to a solid 1.25-in.-diameter cold-rolled stainless steel [G2 = 12,500 ksi] core (2) as shown in Fig. P6.67. The allowable shear stress of tube (1) is 27 ksi, and the allowable shear stress of core (2) is 60 ksi. Determine: (a) the allowable torque T that can be applied to the tube-and-core assembly. (b) the corresponding torques produced in tube (1) and core (2). (c) the angle of twist produced in a 10-in. length of the assembly by the allowable torque T. Fig. P6.67

Solution Section Properties: For tube (1) and core (2), the polar moments of inertia are: J1  J2 



(1.75 in.) 4  (1.25 in.) 4   0.681087 in.4 32

 32

(1.25 in.) 4  0.239684 in.4

Equilibrium: M x  T1  T2  T  0

(a)

Geometry of Deformation Relationship: Since the tube and core are securely bonded together, the angles of twist in both members must be equal; therefore, 1  2 (b) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2  J1G1 J 2G2

(c)

(d)

Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as: Tc T     J J c which allows Eq. (d) to be rewritten as:  1L1  2 L2  (e) G1c1 G2c2 Solve Eq. (e) for  1: L G c  6,500 ksi   1.75 in./2  (f) 1   2 2 1 1   2   0.728 2  12,500 ksi   1.25 in./2  L1 G2 c2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Assume the core controls: If the shear stress in core (2) reaches its allowable value of 60 ksi (in other words, if the core controls), then the shear stress in tube (1) will be: 1  0.728(60 ksi)  43.68 ksi  27 ksi N.G. which is larger than the 27-ksi allowable shear stress for the tube. Therefore, the shear stress in the tube controls. Tube shear stress actually controls: Rearrange Eq. (f) to solve for the shear stress in core (2) given that the shear stress in tube (1) is at its 27-ksi allowable value: 27 ksi 2   37.088 ksi  60 ksi O.K. 0.728 Now that the maximum shear stresses in the tube and the core are known, the torques in each component can be computed:  J (27 ksi)(0.681087 in.4 ) T1  1 1   21.016 kip-in. c1 1.75 in./2 T2 

 2J2 c2

(37.088 ksi)(0.239684 in.4 )   14.223 kip-in. 1.25 in./2

(a) Total Torque: From Eq. (a), the total torque acting on the assembly must not exceed: Tmax  T1  T2  21.016 kip-in.  14.223 kip-in.  35.239 kip-in.  35.2 kip-in.

Ans.

(b) Torques in each component: As computed previously, the torque in tube (1) is: T1  21.0 kip-in.

Ans.

and the torque in core (2) is: T2  14.22 kip-in.

Ans.

(c) Angle of Twist produced in a 10-in. length: Since both the tube and the core twist exactly the same amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. TL (21.016 kip-in.)(10 in.) 1  1 1   0.047473 rad  0.0475 rad Ans. J1G1 (0.681087 in.4 )(6,500 ksi)

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6.68 A hollow circular cold-rolled bronze [G1 = 6,500 ksi] tube (1) with an outside diameter of 1.75 in. and an inside diameter of 1.25 in. is securely bonded to a solid 1.25-in.-diameter cold-rolled stainless steel [G2 = 12,500 ksi] core (2) as shown in Fig. P6.68. If a torque of T = 20 kip-in. is applied to the tube-and-core assembly, determine: (a) the torques produced in tube (1) and core (2). (b) the maximum shear stress in the bronze tube and in the stainless steel core. (c) the angle of twist produced in a 10-in. length of the assembly. Fig. P6.68

Solution Section Properties: For tube (1) and core (2), the polar moments of inertia are: J1  J2 



(1.75 in.) 4  (1.25 in.) 4   0.681087 in.4 32

 32

(1.25 in.) 4  0.239684 in.4

Equilibrium: M x  T1  T2  20 kip-in.  0

(a)

Geometry of Deformation Relationship: Since the tube and core are securely bonded together, the angles of twist in both members must be equal; therefore, 1  2 (b) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2  J1G1 J 2G2

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1:  0.681087 in.4   6,500 ksi  L2 J1 G1 T1  T2  T2   1.477632T2 L1 J 2 G2  0.239684 in.4   12,500 ksi  and substitute this result into Eq. (a) to compute the torque T2 in core (2): T1  T2  1.477632 T2  T2  2.477632 T2  20 kip-in.  T2  8.072 kip-in.  8.07 kip-in.

The torque in tube (1) is therefore: T1  20 kip-in.  T2  20 kip-in.  8.072 kip-in.  11.928 kip-in.  11.93 kip-in.

Ans. Ans.

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(b) Maximum Shear Stress: The maximum shear stress in tube (1) is: T c (11.928 kip-in.)(1.75 in. / 2) 1  1 1   15.32 ksi J1 0.681087 in.4 The maximum shear stress in core (2) is: Tc (8.072 kip-in.)(1.25 in. / 2) 2  2 2   21.0 ksi J2 0.239684 in.4

Ans.

Ans.

(c) Angle of Twist produced in a 10-in. length: Since both the tube and the core twist exactly the same amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. TL (11.928 kip-in.)(10 in.) 1  1 1   0.026943 rad  0.0269 rad Ans. J1G1 (0.681087 in.4 )(6,500 ksi)

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6.69 A composite assembly consisting of a steel [G = 80 GPa] core (2) connected by rigid plates at the ends of an aluminum [G = 28 GPa] tube (1) is shown in Fig. P6.69a. The cross-sectional dimensions of the assembly are shown in Fig. P6.69b. If a torque of T = 1,100 N-m is applied to the composite assembly, determine: (a) the maximum shear stress in the aluminum tube and in the steel core. (b) the rotation angle of end B relative to end A.

Fig. P6.69a Tube-and-Core Composite Shaft

Fig. P6.69b Cross-Sectional Dimensions

Solution Section Properties: For aluminum tube (1) and steel core (2), the polar moments of inertia are: J1  J2 



(50 mm) 4  (40 mm) 4   362, 265 mm 4 32 

 32

(20 mm) 4  15,708.0 mm 4

Equilibrium: M x  T1  T2  1,100 N-m  0

(a)

Geometry of Deformation Relationship: Since the tube and core are connected by rigid plates, the angles of twist in both members must be equal; therefore, 1  2 (b) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2  J1G1 J 2G2

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1:  362,265 mm4   28 GPa  L J G T1  T2 2 1 1  T2     8.071858T2 L1 J 2 G2  15,708.0 mm4   80 GPa  and substitute this result into Eq. (a) to compute the torque T2 in steel core (2): T1  T2  8.071858T2  T2  9.071858T2  1,100 N-m

T2  121.254 N-m Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

The torque in aluminum tube (1) is therefore: T1  1,100 N-m  T2  1,100 N-m  121.254 N-m  978.746 N-m (a) Maximum Shear Stress: The maximum shear stress in aluminum tube (1) is: T c (978.746 N-m)(50 mm / 2)(1,000 mm/m) 1  1 1   67.5 MPa J1 362, 265 mm 4 The maximum shear stress in steel core (2) is: Tc (121.254 N-m)(20 mm / 2)(1,000 mm/m) 2  2 2   77.2 MPa J2 15,708.0 mm 4

Ans.

Ans.

(b) Rotation Angle of B relative to A: Since both the tube and the core twist exactly the same amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. TL (978.746 N-m)(300 mm)(1,000 mm/m) 1  1 1   0.028947 rad  0.0289 rad Ans. J1G1 (362,265 mm 4 )( 28,000 N/mm 2 )

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6.70 A composite assembly consisting of a steel [G = 80 GPa] core (2) connected by rigid plates at the ends of an aluminum [G = 28 GPa] tube (1) is shown in Fig. P6.70a. The cross-sectional dimensions of the assembly are shown in Fig. P6.70b. The allowable shear stress of aluminum tube (1) is 90 MPa, and the allowable shear stress of steel core (2) is 130 MPa. Determine: (a) the allowable torque T that can be applied to the composite shaft. (b) the corresponding torques produced in tube (1) and core (2). (c) the angle of twist produced by the allowable torque T.

Fig. P6.70a Tube-and-Core Composite Shaft

Fig. P6.70b Cross-Sectional Dimensions

Solution Section Properties: For aluminum tube (1) and steel core (2), the polar moments of inertia are: J1  J2 



(50 mm) 4  (40 mm) 4   362, 265 mm 4 32

 32

(20 mm) 4  15,708.0 mm 4

Equilibrium: M x  T1  T2  T  0

(a)

Geometry of Deformation Relationship: Since the tube and core are connected by rigid plates, the angles of twist in both members must be equal; therefore, 1  2 (b) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2  J1G1 J 2G2

(c)

(d)

Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as: Tc T     J J c which allows Eq. (d) to be rewritten as:

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 1L1



 2 L2

(e) G1c1 G2c2 Solve Eq. (e) for  1: L G c  28 GPa   50 mm/2  1   2 2 1 1   2   0.875 2 (f)  80 GPa   20 mm/2  L1 G2 c2 Assume the core controls: If the shear stress in core (2) reaches its allowable value of 130 MPa (in other words, if the core controls), then the shear stress in tube (1) will be: 1  0.875(130 MPa)  113.75 MPa  90 MPa N.G. which is larger than the 90 MPa allowable shear stress for the tube. Therefore, the shear stress in the tube controls. Tube shear stress actually controls: Rearrange Eq. (f) to solve for the shear stress in core (2) given that the shear stress in tube (1) is at its 90 MPa allowable value: 90 MPa 2   102.857 MPa  130 MPa O.K. 0.875 Now that the maximum shear stresses in the tube and the core are known, the torques in each component can be computed: 1 J1 (90 N/mm 2 )(362,265 mm 4 ) T1    1,304,154 N-mm c1 50 mm/2 T2 

2J2 c2

(102.857 MPa)(15,708.0 mm 4 )   161,568 N-mm 20 mm/2

(a) Total Torque: From Eq. (a), the total torque acting on the assembly must not exceed: Tmax  T1  T2  1,304,154 N-mm  161,568 N-mm  1,465,721 N-mm  1,466 N-m (b) Torques in each component: As computed previously, the torque in tube (1) is: T1  1,304 N-m and the torque in core (2) is: T2  161.6 N-m

Ans.

Ans. Ans.

(c) Angle of Twist: Since both the tube and the core twist exactly the same amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. TL (1,304,154 N-mm)(300 mm) 1  1 1   0.038571 rad  0.0386 rad Ans. J1G1 (362,265 mm 4 )(28, 000 N/mm 2 )

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6.71 The composite shaft shown in Fig. P6.71 consists of a bronze sleeve (1) securely bonded to an inner steel core (2). The bronze sleeve has an outside diameter of 35 mm, an inside diameter of 25 mm, and a shear modulus of G1 = 45 GPa. The solid steel core has a diameter of 25 mm and a shear modulus of G2 = 80 GPa. The allowable shear stress of sleeve (1) is 180 MPa, and the allowable shear stress of core (2) is 150 MPa. Determine: (a) the allowable torque T that can be applied to the composite shaft. (b) the corresponding torques produced in sleeve (1) and core (2). (c) the rotation angle of end B relative to end A that is produced by the allowable torque T.

Fig. P6.71

Solution Section Properties: For bronze sleeve (1) and steel core (2), the polar moments of inertia are: J1  J2 



(35 mm) 4  (25 mm) 4   108,974 mm 4 32

 32

(25 mm) 4  38,349.5 mm 4

Equilibrium: M x  T1  T2  T  0

(a)

Geometry of Deformation Relationship: Since the sleeve and core are securely bonded together, the angles of twist in both members must be equal; therefore, 1  2 (b) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2  J1G1 J 2G2

(c)

(d)

Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as: Tc T     J J c which allows Eq. (d) to be rewritten as:  1L1  2 L2  (e) G1c1 G2c2 Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Solve Eq. (e) for  1: L G c  45 GPa   (35 mm/2)  1   2 2 1 1   2   0.7875 2  80 GPa   (25 mm/2)  L1 G2 c2

(f)

Assume the core controls: If the shear stress in core (2) reaches its allowable value of 150 MPa (in other words, if the core controls), then the shear stress in sleeve (1) will be:  1  0.7875(150 MPa)  118.125 MPa  180 MPa O.K. This calculation shows that the shear stress in the core does in fact controls. Now that the maximum shear stresses in the sleeve and the core are known, the torques in each component can be computed:  J (118.125 N/mm 2 )(108,974 mm 4 ) T1  1 1   735,574 N-mm c1 35 mm/2 T2 

 2J2 c2

(150 MPa)(38,349.5 mm 4 )   460,194 N-mm 25 mm/2

(a) Total Torque: From Eq. (a), the total torque acting on the assembly must not exceed: Tmax  T1  T2  735,574 N-mm  460,194 N-mm  1,195,769 N-mm  1,196 N-m (b) Torques in each component: As computed previously, the torque in sleeve (1) is: T1  736 N-m and the torque in core (2) is: T2  460 N-m

Ans.

Ans. Ans.

(c) Angle of Twist: Since both the sleeve and the core twist exactly the same amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. TL (735,574 N-mm)(360 mm) 1  1 1   0.054000 rad  0.0540 rad Ans. J1G1 (108,974 mm 4 )(45,000 N/mm 2 )

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6.72 The composite shaft shown in Fig. P6.72 consists of a bronze sleeve (1) securely bonded to an inner steel core (2). The bronze sleeve has an outside diameter of 35 mm, an inside diameter of 25 mm, and a shear modulus of G1 = 45 GPa. The solid steel core has a diameter of 25 mm and a shear modulus of G2 = 80 GPa. The composite shaft is subjected to a torque of T = 900 N-m. Determine: (a) the maximum shear stresses in the bronze sleeve and the steel core. (b) the rotation angle of end B relative to end A.

Fig. P6.72

Solution Section Properties: For bronze sleeve (1) and steel core (2), the polar moments of inertia are: J1  J2 



(35 mm) 4  (25 mm) 4   108,974 mm 4 32

 32

(25 mm) 4  38,349.5 mm 4

Equilibrium: M x  T1  T2  900 N-m  0

(a)

Geometry of Deformation Relationship: Since the sleeve and core are securely bonded together, the angles of twist in both members must be equal; therefore, 1  2 (b) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2  J1G1 J 2G2

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1:  108,974 mm4   45 GPa  L2 J1 G1 T1  T2  T2     1.598401T2 L1 J 2 G2  38,349.5 mm4   80 GPa  and substitute this result into Eq. (a) to compute the torque T2 in steel core (2): T1  T2  1.598401T2  T2  2.598401T2  900 N-m

T2  346.367 N-m The torque in bronze sleeve (1) is therefore: T1  900 N-m  T2  900 N-m  346.367 N-m  553.633 N-m

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(a) Maximum Shear Stress: The maximum shear stress in bronze sleeve (1) is: T c (553.633 N-m)(35 mm / 2)(1,000 mm/m) 1  1 1   88.9 MPa J1 108,974 mm 4 The maximum shear stress in steel core (2) is: Tc (346.367 N-m)(25 mm / 2)(1,000 mm/m) 2  2 2   112.9 MPa J2 38,349.5 mm 4

Ans.

Ans.

(b) Rotation Angle of B relative to A: Since both the sleeve and the core twist exactly the same amount [i.e., Eq. (b)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. TL (553.633 N-m)(360 mm)(1,000 mm/m) 1  1 1   0.040643 rad  0.0406 rad Ans. J1G1 (108,974 mm 4 )( 45,000 N/mm 2 )

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6.73 The composite shaft shown in Fig. P6.73 consists of two steel pipes that are connected at flange B and securely attached to rigid walls at A and C. Steel pipe (1) has an outside diameter of 168 mm and a wall thickness of 7 mm. Steel pipe (2) has an outside diameter of 114 mm and a wall thickness of 6 mm. Both pipes are 3-m long and have a shear modulus of 80 GPa. If a concentrated torque of 20 kN-m is applied to flange B, determine: (a) the maximum shear stress magnitudes in pipes (1) and (2). (b) the rotation angle of flange B relative to support A.

Fig. P6.73

Solution Section Properties: The polar moments of inertia for the two steel pipes are: J1  J2 



(168 mm) 4  (154 mm) 4   22,987,183 mm 4 32



(114 mm) 4  (102 mm) 4   5,954,575 mm 4 32

Equilibrium: M x  T1  T2  20 kN-m  0

(a)

Geometry of Deformation Relationship: Since the two steel pipes are securely attached to fixed supports at A and C, the sum of the angles of twist in the two pipes must equal zero: 1  2  0 Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1 L1 T2 L2  0 J1G1 J 2G2

(b)

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1: L J G T1  T2 2 1 1 L1 J 2 G2 4  3 m   22,987,183 mm   80 GPa   T2   3 m   5,954,575 mm 4   80 GPa 

 22,987,183 mm 4   T2   3.860424 T2  5,954,575 mm 4  Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

and substitute this result into Eq. (a) to compute the torque T2: T1  T2    3.860424 T2   T2  4.860424 T2  20 kN-m

T2  4.1149 kN-m The torque in member (1) is therefore: T1  T2  20 kN-m  4.1149 kN-m  20 kN-m  15.8851 kN-m (a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is: T1c1 (15.8851 kN-m)(168 mm / 2)(1,000)2 1    58.0 MPa J1 22,987,183 mm4 The maximum shear stress magnitude in member (2) is: Tc (4.1149 kN-m)(114 mm / 2)(1,000)2 2  2 2   39.4 MPa J2 5,954,575 mm4

Ans.

Ans.

(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends; hence, 1  B   A Since joint A is restrained from rotating, A = 0 and thus 1  B The rotation angle at B can be determined by computing the angle of twist in member (1): TL 1  1 1 J1G1



(15.8851 kN-m)(3,000 mm)(1,000 mm/m)(1,000 N/kN) (22,987,183 mm 4 )(80,000 N/mm 2 )

 0.025914 rad  0.0259 rad

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

6.74 The composite shaft shown in Fig. P6.74 consists of two steel pipes that are connected at flange B and securely attached to rigid walls at A and C. Steel pipe (1) has an outside diameter of 8.625 in., a wall thickness of 0.322 in., and a length of L1 = 15 ft. Steel pipe (2) has an outside diameter of 6.625 in., a wall thickness of 0.280 in., and a length of L2 = 25 ft. Both pipes have a shear modulus of 12,000 ksi. If a concentrated torque of 36 kip-ft is applied to flange B, determine: (a) the maximum shear stress magnitudes in pipes (1) and (2). (b) the rotation angle of flange B relative to support A.

Fig. P6.74

Solution Section Properties: The polar moments of inertia for the two steel pipes are: J1  J2 



(8.625 in.) 4  (7.981 in.) 4   144.978 in.4 32



(6.625 in.) 4  (6.065 in.) 4   56.284 in.4 32

Equilibrium: M x  T1  T2  36 kip-ft  0

(a)

Geometry of Deformation Relationship: Since the two steel pipes are securely attached to fixed supports at A and C, the sum of the angles of twist in the two pipes must equal zero: 1  2  0 Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1 L1 T2 L2  0 J1G1 J 2G2

(b)

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1: L J G T1  T2 2 1 1 L1 J 2 G2 4  25 ft   144.978 in.   12,000 ksi   T2   4.293050 T2  15 ft   56.284 in.4   12,000 ksi  and substitute this result into Eq. (a) to compute the torque T2:

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

T1  T2    4.293050T2   T2  5.293050T2  36 kip-ft  T2  6.801 kip-ft The torque in member (1) is therefore: T1  T2  36 kip-ft  6.801 kip-ft  36 kip-ft  29.199 kip-ft (a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is: T c (29.199 kip-ft)(8.625 in. / 2)(12 in./ft) 1  1 1   10.42 ksi J1 144.978 in.4 The maximum shear stress magnitude in member (2) is: Tc (6.801 kip-ft)(6.625 in. / 2)(12 in./ft) 2  2 2   4.80 ksi J2 56.284 in.4

Ans.

Ans.

(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends; hence, 1  B   A Since joint A is restrained from rotating, A = 0 and thus 1  B The rotation angle at B can be determined by computing the angle of twist in member (1): TL 1  1 1 J1G1



(29.199 kip-ft)(15 ft)(12 in./ft) 2 (144.978 in.4 )(12,000 ksi)

 0.036252 rad  0.0363 rad

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

6.75 The composite shaft shown in Fig. P6.75 consists of a solid brass segment (1) and a solid aluminum segment (2) that are connected at flange B and securely attached to rigid supports at A and C. Brass segment (1) has a diameter of 1.00 in., a length of L1 = 15 in., a shear modulus of 5,600 ksi, and an allowable shear stress of 8 ksi. Aluminum segment (2) has a diameter of 0.75 in., a length of L2 = 20 in., a shear modulus of 4,000 ksi, and an allowable shear stress of 6 ksi. Determine: (a) the allowable torque TB that can be applied to the composite shaft at flange B. (b) the magnitudes of the internal torques in segments (1) and (2). (c) the rotation angle of flange B that is produced by the allowable torque TB.

Fig. P6.75

Solution Section Properties: The polar moments of inertia for the two shafts are:  J1  (1.00 in.) 4  0.098175 in.4 32  J 2  (0.75 in.) 4  0.031063 in.4 32 Equilibrium: M x  T1  T2  TB  0

(a)

Geometry of Deformation Relationship: Since the two shafts are securely attached to fixed supports at A and C, the sum of the angles of twist in the two members must equal zero: 1  2  0 (b) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1 L1 T2 L2  0 J1G1 J 2G2

(c)

(d)

Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as: Tc T     J J c which allows Eq. (d) to be rewritten as:

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

 1L1



 2 L2

G1c1 G2c2 Solve Eq. (e) for  1: L G c  20 in.  5,600 ksi   1.00 in./2  1   2 2 1 1   2   2.488889 2  15 in.   4,000 ksi   0.75 in./2  L1 G2 c2

(e)

(f)

Assume the shaft (2) controls: If the shear stress in shaft (2) reaches its allowable magnitude of 6 ksi, then the shear stress magnitude in shaft (1) will be: 1  2.488889(6 ksi)  14.9333 ksi  8 ksi N.G. which is larger than the 8 ksi allowable shear stress magnitude for shaft (1). Therefore, the shear stress magnitude in shaft (1) must control. Shaft (1) actually controls: Rearrange Eq. (f) to solve for the shear stress magnitude in shaft (2) given that the shear stress magnitude in shaft (1) is at its 8 ksi allowable magnitude: 8 ksi 2   3.214 ksi  6 ksi O.K. 2.488889 Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes in each component can be computed:  J (8 ksi)(0.098175 in.4 ) T1  1 1   1.571 kip-in. c1 1.00 in./2

2J2

(3.214 ksi)(0.031063 in.4 )  0.266 kip-in. c2 0.75 in./2 Note that these are torque magnitudes. From inspection of the FBD of flange B and the associated equilibrium equation, it is apparent that T2 must act opposite to the direction assumed in the FBD, giving it a negative value. Therefore, by inspection T2  0.266 kip-in. T2 



(a) Total Torque: From Eq. (a), the total torque acting at flange B must not exceed: TB,max  T1  T2  1.571 kip-in.  (0.266 kip-in.)  1.837 kip-in.

Ans.

(b) Torques Magnitudes: As computed previously, the torque magnitude in shaft (1) is: T1  1.571 kip-in.

Ans.

and the torque magnitude in shaft (2) is: T2  0.266 kip-in.

Ans.

(c) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends; hence, 1  B   A Since joint A is restrained from rotating, A = 0 and thus 1  B The rotation angle at B can be determined by computing the angle of twist in member (1): TL (1.571 kip-in.)(15 in.) 1  1 1   0.042857 rad  0.0429 rad Ans. J1G1 (0.098175 in.4 )(5,600 ksi) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

6.76 The composite shaft shown in Fig. P6.76 consists of a solid brass segment (1) and a solid aluminum segment (2) that are connected at flange B and securely attached to rigid walls at A and C. Brass segment (1) has a diameter of 18 mm, a length of L1 = 235 mm, and a shear modulus of 39 GPa. Aluminum segment (2) has a diameter of 24 mm, a length of L2 = 165 mm, and a shear modulus of 28 GPa. If a concentrated torque of 270 N-m is applied to flange B, determine: (a) the maximum shear stress magnitudes in segments (1) and (2). (b) the rotation angle of flange B relative to support A. Fig. P6.76

Solution Section Properties: The polar moments of inertia for the two steel pipes are: J1  J2 



32

 32

(18 mm) 4  10,306.0 mm 4 (24 mm) 4  32,572.0 mm 4

Equilibrium: M x  T1  T2  270 N-m  0

(a)

Geometry of Deformation Relationship: Since the two steel pipes are securely attached to fixed supports at A and C, the sum of the angles of twist in the two pipes must equal zero: 1  2  0 Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1 L1 T2 L2  0 J1G1 J 2G2

(b)

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1: L J G T1  T2 2 1 1 L1 J 2 G2 4  165 mm   10,306.0 mm   39 GPa   T2   0.309434 T2  235 mm   32,572.0 mm 4   28 GPa 

and substitute this result into Eq. (a) to compute the torque T2: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

T1  T2    0.309434 T2   T2  1.309434 T2  270 N-m T2  206.1959 N-m The torque in member (1) is therefore: T1  T2  270 N-m  206.1959 N-m  270 N-m  63.8041 N-m (a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is: T c (63.8041 N-m)(18 mm / 2)(1,000 mm/m) 1  1 1   55.7 MPa J1 10,306.0 mm 4 The maximum shear stress magnitude in member (2) is: Tc (206.1959 N-m)(24 mm / 2)(1,000 mm/m) 2  2 2   76.0 MPa J2 32,572.0 mm 4

Ans.

Ans.

(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends; hence, 1  B   A Since joint A is restrained from rotating, A = 0 and thus 1  B The rotation angle at B can be determined by computing the angle of twist in member (1): TL 1  1 1 J1G1



(63.8041 N-m)(235 mm)(1,000 mm/m) (10,306.0 mm 4 )(39,000 N/mm 2 )

 0.037305 rad  0.0373 rad

Ans.

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6.77 The composite shaft shown in Fig. P6.77 consists of a stainless steel tube (1) and a brass tube (2) that are connected at flange B and securely attached to rigid supports at A and C. Stainless steel tube (1) has an outside diameter of 2.25 in., a wall thickness of 0.250 in., a length of L1 = 40 in., and a shear modulus of 12,500 ksi. Brass tube (2) has an outside diameter of 3.500 in., a wall thickness of 0.219 in., a length of L2 = 20 in., and a shear modulus of 5,600 ksi. If a concentrated torque of TB = 42 kip-in. is applied to flange B, determine: (a) the maximum shear stress magnitudes in tubes (1) and (2). (b) the rotation angle of flange B relative to support A.

Fig. P6.77

Solution Section Properties: The polar moments of inertia for the two tubes are: J1  J2 



(2.250 in.) 4  (1.750 in.) 4   1.595340 in.4 32



(3.500 in.) 4  (3.062 in.) 4   6.102156 in.4 32

Equilibrium: M x  T1  T2  42 kip-in.  0

(a)

Geometry of Deformation Relationship: Since the two tubes are securely attached to fixed supports at A and C, the sum of the angles of twist in the two tubes must equal zero: 1  2  0 (b) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1 L1 T2 L2  0 J1G1 J 2G2

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1: L J G T1  T2 2 1 1 L1 J 2 G2 4  20 in.  1.595340 in.   12,500 ksi   T2   0.291784 T2  40 in.   6.102156 in.4   5,600 ksi 

and substitute this result into Eq. (a) to compute the torque T2: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

T1  T2    0.291784 T2   T2  1.291784 T2  42 kip-in. T2  32.5132 kip-in. The torque in member (1) is therefore: T1  T2  42 kip-in.  32.5132 kip-in.  42 kip-in.  9.4868 kip-in. (a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is: T c (9.4868 kip-in.)(2.250 in. / 2) 1  1 1   6.69 ksi J1 1.595340 in.4 The maximum shear stress magnitude in member (2) is: Tc (32.5132 kip-in.)(3.500 in. / 2) 2  2 2   9.32 ksi J2 6.102156 in.4

Ans.

Ans.

(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends; hence, 1  B   A Since joint A is restrained from rotating, A = 0 and thus 1  B The rotation angle at B can be determined by computing the angle of twist in member (1): TL 1  1 1 J1G1



(9.4868 kip-in.)(40 in.) (1.595340 in.4 )(12,500 ksi)

 0.019029 rad  0.01903 rad

Ans.

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

6.78 The composite shaft shown in Fig. P6.78 consists of a stainless steel tube (1) and a brass tube (2) that are connected at flange B and securely attached to rigid supports at A and C. Stainless steel tube (1) has an outside diameter of 2.25 in., a wall thickness of 0.250 in., a length of L1 = 40 in., and a shear modulus of 12,500 ksi. Brass tube (2) has an outside diameter of 3.500 in., a wall thickness of 0.219 in., a length of L2 = 20 in., and a shear modulus of 5,600 ksi. The allowable shear stress in the stainless steel is 50 ksi, and the allowable shear stress in the brass is 18 ksi. Determine: (a) the allowable torque TB that can be applied to the composite shaft at flange B. (b) the rotation angle of flange B that is produced by the allowable torque TB.

Fig. P6.78

Solution Section Properties: The polar moments of inertia for the two tubes are: J1  J2 



(2.250 in.) 4  (1.750 in.) 4   1.595340 in.4 32 



(3.500 in.) 4  (3.062 in.) 4   6.102156 in.4 32 

Equilibrium: M x  T1  T2  TB  0

(a)

Geometry of Deformation Relationship: Since the two tubes are securely attached to fixed supports at A and C, the sum of the angles of twist in the two tubes must equal zero: 1  2  0 (b) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1 L1 T2 L2  0 J1G1 J 2G2

(c)

(d)

Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as: Tc T     J J c which allows Eq. (d) to be rewritten as:

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

 1L1



 2 L2

(e)

G1c1 G2c2 Solve Eq. (e) for  1: L G c  20 in.  12,500 ksi   2.250 in./2  1   2 2 1 1   2   0.717474 2  40 in.  5,600 ksi   3.500 in./2  L1 G2 c2

(f)

Assume the shaft (2) controls: If the shear stress in shaft (2) reaches its allowable magnitude of 18 ksi, then the shear stress magnitude in shaft (1) will be: 1  0.717474(18 ksi)  12.914541 ksi  50 ksi O.K. This calculation shows that the shear stress in shaft (2) does in fact control. Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes in each component can be computed:  J (12.914541 ksi)(1.595340 in.4 ) T1  1 1   18.3139 kip-in. c1 2.250 in./2

2J2

(18 ksi)(6.102156 in.4 ) T2    62.7650 kip-in. c2 3.500 in./2 Note that these are torque magnitudes. From inspection of the FBD of flange B and the associated equilibrium equation, it is apparent that T2 must act opposite to the direction assumed in the FBD, giving it a negative value. Therefore, by inspection T2  62.7650 kip-in.

(a) Allowable Torque TB: From Eq. (a), the total torque acting at flange B must not exceed: TB,max  T1  T2  18.3138 kip-in.  (62.7650 kip-in.)  81.0788 kip-in.  81.1 kip-in.

Ans.

(b) Rotation Angle of Flange B Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends; hence, 1  B   A Since joint A is restrained from rotating, A = 0 and thus 1  B The rotation angle at B can be determined by computing the angle of twist in member (1): TL (18.3139 kip-in.)(40 in.) 1  1 1   0.036735 rad  0.0367 rad Ans. J1G1 (1.595340 in.4 )(12,500 ksi)

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

6.79 The torsional assembly of Fig. P6.79 consists of a cold-rolled stainless steel tube connected to a solid cold-rolled brass segment at flange C. The assembly is securely fastened to rigid supports at A and D. Stainless steel tube (1) and (2) has an outside diameter of 3.50 in., a wall thickness of 0.120 in., and a shear modulus of G = 12,500 ksi. The solid brass segment (3) has a diameter of 2.00 in. and a shear modulus of G = 5,600 ksi. A concentrated torque of TB = 6 kip-ft is applied to the stainless steel pipe at B. Determine: (a) the maximum shear stress magnitude in the stainless steel tube. (b) the maximum shear stress magnitude in brass segment (3). (c) the rotation angle of flange C.

Fig. P6.79

Solution Section Properties: The polar moments of inertia for the stainless steel tube and the solid brass segment are: J1  J3 



(3.500 in.) 4  (3.260 in.) 4   3.643915 in.4  J 2 32

 32

(2.0 in.) 4  1.570796 in.4

Equilibrium: Consider a free-body diagram cut around joint B, where the external torque TB is applied: M x  T1  T2  TB  0 (a) and also consider a FBD cut around joint C. Although there is not an external torque applied at joint C, the section properties of the torsion structure change at C. M x  T2  T3  0 T2  T3 (b) Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero: 1  2  3  0 (c) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2

3 

T3 L3 J 3G3

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1L1 T2 L2 T3 L3   0 J1G1 J 2G2 J 3G3

(d)

(e)

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Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1 and T3 with equivalent expressions involving T2. From Eq. (a): T1  T2  TB (f) Substitute Eqs. (b) and (f) into Eq. (e) and simplify to derive an expression for T2: (T2  TB ) L1 T2 L2 T2 L3   0 J1G1 J 2G2 J 3G3  L L L  TL T2  1  2  3    B 1 J1G1  J1G1 J 2G2 J 3G3  TB L1 J1G1 T2   L1 L L  2  3 J1G1 J 2G2 J 3G3 Note that G1 = G2 and J1 = J2. Compute T2: (6 kip-ft)(42 in.)(12 in./ft) (3.643915 in.4 )(12,500 ksi) T2   42 in.  30 in. 24 in.  4 (3.643915 in. )(12,500 ksi) (1.570796 in.4 )(5,600 ksi)  15.4070 kip-in. Substitute this result into Eq. (f) to compute internal torque T1: T1  T2  TB  15.4070 kip-in.  (6 kip-ft)(12 in./ft)  56.5930 kip-in. and from Eq. (b): T3  T2  15.4070 kip-in.

(e)

Maximum Shear Stress: The maximum shear stress magnitudes in the three members are: T c (56.5930 kip-in.)(3.500 in. / 2) 1  1 1   27.2 ksi J1 3.643915 in.4

2 

T2c2 (15.4070 kip-in.)(3.500 in. / 2)   7.40 ksi J2 3.643915 in.4

3 

T3c3 (15.4070 kip-in.)(2.000 in. / 2)   9.81 ksi J3 1.570796 in.4

(a) Maximum Shear Stress Magnitude in Stainless Steel Tube: 1  27.2 ksi

Ans.

(b) Maximum Shear Stress Magnitude in Brass Shaft:  3  9.81 ksi

Ans.

(c) Rotation Angle of Flange C Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends: 1  B   A B   A  1 Similarly, the angle of twist in member (2) can be defined by: 2  C  B C  B  2

(f) (g)

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To derive an expression for C, substitute Eq. (f) into Eq. (g), and note that joint A is restrained from rotating; therefore, A = 0.  C  1  2 The angle of twist in member (1) is: TL (56.5930 kip-in.)(42 in.) 1  1 1   0.052184 rad J1G1 (3.643915 in.4 )(12,500 ksi) The angle of twist in member (2) is: TL ( 15.4070 kip-in.)(30 in.) 2  2 2   0.010148 rad J 2G2 (3.643915 in.4 )(12,500 ksi) The rotation angle of flange C is thus: C  1  2  0.052184 rad  (0.010148 rad)  0.042036 rad  0.0420 rad

Ans.

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6.80 The torsional assembly of Fig. P6.80 consists of a cold-rolled stainless steel tube connected to a solid cold-rolled brass segment at flange C. The assembly is securely fastened to rigid supports at A and D. Stainless steel tubes (1) and (2) have an outside diameter of 3.50 in., a wall thickness of 0.120 in., a shear modulus of G = 12,500 ksi, and an allowable shear stress of 30 ksi. The solid brass segment (3) has a diameter of 2.00 in., a shear modulus of G = 5,600 ksi, and an allowable shear stress of 18 ksi. Determine the maximum permissible magnitude for the concentrated torque TB. Fig. P6.80

Solution Section Properties: The polar moments of inertia for the stainless steel tube and the solid brass segment are: J1  J3 



(3.500 in.) 4  (3.260 in.) 4   3.643915 in.4  J 2 32

 32

(2.0 in.) 4  1.570796 in.4

Equilibrium: Consider a free-body diagram cut around joint B, where the external torque TB is applied: M x  T1  T2  TB  0 (a) and also consider a FBD cut around joint C. Although there is not an external torque applied at joint C, the section properties of the torsion structure change at C. M x  T2  T3  0 T2  T3 (b) Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero: 1  2  3  0 (c) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2

3 

T3 L3 J 3G3

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1L1 T2 L2 T3 L3   0 J1G1 J 2G2 J 3G3

(d)

(e)

Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to rewrite Eq. (e) in terms of stresses. In general, the elastic torsion formula can be rearranged as: Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Tc T    J J c which allows Eq. (e) to be rewritten as:  1L1  2 L2  3 L3   0 (f) G1c1 G2c2 G3c3 Also using the elastic torsion formula, Eq. (b) can be expressed in terms of stress:  2 J 2  3J3   J   3 2 2 (g) c2 c3 c3 c2 J 3 Substitute Eq. (g) into Eq. (f) to obtain 1L1  2 L2 L3  2 J 2   0 G1c1 G2c2 G3 c2 J 3 Simplify: L J   L2  3 2   L L L J  G c G3c2 J 3  1 1   2  2  3 2   1   2  2 2 L1 G1c1    G2c2 G3 c2 J 3    G1c1 Substitute values:   3.643915 in.4   30 in. 24 in.    (12,500 ksi)(3.50 in./2) (5,600 ksi)(3.50 in./2)  1.570796 in.4    1   2 42 in.     (12,500 ksi)(3.50 in./2)    3.6732  2 This calculation demonstrates that the shear stress in segment (1) of the stainless steel tube is much larger than the shear stress in segment (2). If the shear stress magnitude in segment (1) is 30 ksi, then the shear stress magnitude in segment (2) will be:



2  

1

 8.1673 ksi 3.6732 Next, we need to check the corresponding shear stress in brass shaft (3). From Eq. (g): c J 3  2 3 2 c2 J 3

(h)

Substitute the magnitude obtained for 2 in Eq. (h) into this expression and calculate the corresponding shear stress magnitude in shaft (3): 4  2.00 in./2   3.643915 in.   3  (8.1673 ksi)   10.8265 ksi  18 ksi O.K.   3.50 in./2   1.570796 in.4  Now that the maximum shear stresses in the three shaft segments are known, the torques in each component can be computed:  J (30 ksi)(3.643915 in.4 ) T1  1 1   62.4671 kip-in. c1 3.50 in./2 T2  T3 

2J2 c2

 3J3 c3



( 8.1673 ksi)(3.643915 in.4 )  17.0062 kip-in. 3.50 in./2

( 10.8265 ksi)(1.570796 in.4 )   17.0062 kip-in. 2.00 in./2

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Maximum Permissible Torque TB: From Eq. (a), the torque TB acting on the assembly must not exceed: Ans. TB,max  T1  T2  62.4671 kip-in.  (17.062 kip-in.)  79.4733 kip-in.  79.5 kip-in.

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6.81 The torsional assembly of Fig. P6.81a consists of a solid 75-mm-diameter bronze [G = 45 GPa] segment (1) securely connected at flange B to solid 75-mm-diameter stainless steel [G = 86 GPa] segments (2) and (3). The flange at B is secured by four 14-mm-diameter bolts, which are each located on a 120-mm-diameter bolt circle (Fig. P6.81b). The allowable shear stress of the bolts is 90 MPa, and friction effects in the flange can be neglected for this analysis. Determine: (a) the allowable torque TC that can be applied to the assembly at C without exceeding the capacity of the bolted flange connection. (b) the maximum shear stress magnitude in bronze segment (1). (c) the maximum shear stress magnitude in stainless steel segments (2) and (3).

Fig. P6.81a

Fig. P6.81b Flange B Bolts

Solution Section Properties: For bronze segment (1) and stainless steel segments (2) and (3), the polar moments of inertia are identical: J1  J 2  J 3 



32

(75 mm)4  3,106,311 mm4

Equilibrium: Consider a free-body diagram cut around flange B: M x  T1  T2  0 T1  T2 (a) and also consider a FBD cut around joint C, where the external torque TC is applied: M x  T2  T3  TC  0 (b) Capacity of Flange B: The flange at B is secured by four 14-mm-diameter bolts, which are each located on a 120-mm-diameter bolt circle. The allowable shear stress of the bolts is 90 MPa, and each bolt acts in single shear. The cross-sectional area of a single bolt is: Abolt 



(14 mm)2  153.9380 mm2

4 The shear force that can be provided by each bolt is: Vbolt  (90 N/mm 2 )(153.9380 mm 2 )  13,854.4236 N The total torque that can resisted by the bolts is thus: TB ,max  (4 bolts)(13,954.4236 N/bolt)(120 mm/2)  3,325, 062 N-mm

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The internal torques in segments (1) and (2) cannot exceed this magnitude; therefore, based on Eq. (a): T1  T2  3,325, 062 N-mm (c) From Eq. (b), we observe that the external torque applied to the shaft is related to internal torques T2 and T3. We must determine the relationship between these two internal torques by developing a compatibility equation, which is based on the geometry of deformations for this configuration. Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely attached to fixed supports at A and D, the sum of the angles of twist in the three segments must equal zero: 1  2  3  0 (d) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2

3 

T3 L3 J 3G3

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (e)] into the geometry of deformation relationship [Eq. (d)] to obtain the compatibility equation: T1L1 T2 L2 T3 L3   0 J1G1 J 2G2 J 3G3 Solve the Equations: Solve Eq. (f) for T3, noting that J1 = J2 = J3, G2 = G3, and T1 = T2: L G L  TL T L G J T3    1 1  2 2  3 3  T1  1 3  2   J1G1 J 2G2  L3  L3 G1 L3  Substitute the values determined for T1 and T2 in Eq. (c) into Eq. (g) to compute T3:  400 mm  86 GPa  750 mm  T3  (3,325, 062 N-mm)      400 mm  45 GPa  400 mm 

(e)

(f)

(g)

 (3,325, 062 N-mm)(3.786111)  12,589, 054 N-mm

(a) Allowable Torque TC: From equilibrium equation (b), TC  T2  T3  3,325,062 N-mm  ( 12,589,054 N-mm)  15,914,116 N-mm  15.91 kN-m

Ans.

(b) Shear Stress Magnitude in Bronze Segment (1): T c (3,325,062 N-mm)(75 mm / 2) 1  1 1   40.1 MPa J1 3,106,311 mm 4

Ans.

(c) Shear Stress Magnitude in Stainless Steel Segments (2) and (3): Tc (12,589,054 N-mm)(75 mm / 2) 3  3 3   152.0 MPa J3 3,106,311 mm 4

Ans.

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6.82 The torsional assembly shown in Fig. P6.82 consists of solid 2.50-in.-diameter aluminum [G = 4,000 ksi] segments (1) and (3) and a central solid 3.00-in.-diameter bronze [G = 6,500 ksi] segment (2). Concentrated torques of TB = T0 and TC = 2T0 are applied to the assembly at B and C, respectively. If T0 = 20 kip-in., determine: (a) the maximum shear stress magnitude in aluminum segments (1) and (3). (b) the maximum shear stress magnitude in bronze segment (2). (c) the rotation angle of joint C. Fig. P6.82

Solution Section Properties: The polar moments of inertia for the aluminum and bronze segments are: J1  J 3  J2 

 32



32

(2.50 in.) 4  3.834952 in.4

(3.00 in.) 4  7.952156 in.4

Equilibrium: Consider a free-body diagram cut around joint B, where the external torque TB is applied: M x  T1  T2  T0  0 (a) and also consider a FBD cut around joint C, where the external torque TC is applied: M x  T2  T3  2T0  0

(b)

Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero: 1  2  3  0 (c) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2

3 

T3 L3 J 3G3

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1L1 T2 L2 T3 L3   0 J1G1 J 2G2 J 3G3

(d)

(e)

Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1 and T3 with equivalent expressions involving T2. From Eq. (a): T1  T2  T0 (f) and from Eq. (b): T3  T2  2T0 (g) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Substitute Eqs. (f) and (g) into Eq. (e) and simplify to derive an expression for T2: (T2  T0 ) L1 T2 L2 (T2  2T0 ) L3   0 J1G1 J 2G2 J 3G3  L L L  2T L T L T2  1  2  3   0 3  0 1  J1G1 J 2G2 J 3G3  J 3G3 J1G1 2T0 L3 T0 L1  J 3G3 J1G1 T2  L1 L L  2  3 J1G1 J 2G2 J 3G3 Since G1 = G3 and J1 = J3, Eq. (h) can be further simplified and T2 can be computed as: T   2 L3  L1   T2  0   J1G1  L1  L3  L2  J 2G2   J1G1

(h)

20 kip-in. 2(12 in.)  12 in.   4   12 in.  12 in. 24 in. (3.834952 in. )(4,000 ksi)    4 4  (3.834952 in. )(4,000 ksi) (7.952156 in. )(6,500 ksi)   7.711461 kip-in. Backsubstitute this result into Eqs. (f) and (g) to obtain T1 and T3: T1  7.711461 kip-in.  20 kip-in.  27.711461 kip-in. T3  7.711461 kip-in.  2(20 kip-in.)  32.288539 kip-in. 

Maximum Shear Stress: The maximum shear stress magnitudes in the three segments are: T c (27.711461 kip-in.)(2.500 in. / 2) 1  1 1   9.03 ksi J1 3.834952 in.4 Tc (7.711461 kip-in.)(3.000 in. / 2) 2  2 2   1.455 ksi J2 7.952156 in.4

3 

T3c3 (32.288539 kip-in.)(2.500 in. / 2)   10.52 ksi J3 3.834952 in.4

(a) Maximum Shear Stress Magnitude in Aluminum Segments:  3  10.52 ksi

Ans.

(b) Maximum Shear Stress Magnitude in Bronze Segment:  2  1.455 ksi

Ans.

(c) Rotation Angle of Flange C Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends: 1  B   A B   A  1 Similarly, the angle of twist in member (2) can be defined by: 2  C  B C  B  2 To derive an expression for C, substitute Eq. (i) into Eq. (j), and note that joint A is restrained from rotating; therefore, A = 0. 

(i) (j)

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C  1  2 The angle of twist in member (1) is: TL (27.711461 kip-in.)(12 in.) 1  1 1   0.021678 rad J1G1 (3.834952 in.4 )(4,000 ksi) The angle of twist in member (2) is: TL (7.711461 kip-in.)(24 in.) 2  2 2   0.003581 rad J 2G2 (7.952156 in.4 )(6,500 ksi) The rotation angle of flange C is thus: C  1  2  0.021678 rad  0.003581 rad  0.025259 rad  0.0253 rad

Ans.

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6.83 The torsional assembly shown in Fig. P6.83 consists of solid 2.50-in.-diameter aluminum [G = 4,000 ksi] segments (1) and (3) and a central solid 3.00-in.-diameter bronze [G = 6,500 ksi] segment (2). Concentrated torques of TB = T0 and TC = 2T0 are applied to the assembly at B and C, respectively. If the rotation angle at joint C must not exceed 3°, determine: (a) the maximum magnitude of T0 that may be applied to the assembly. (b) the maximum shear stress magnitude in aluminum segments (1) and (3). (c) the maximum shear stress magnitude in bronze segment (2).

Fig. P6.83

Solution Section Properties: The polar moments of inertia for the aluminum and bronze segments are: J1  J 3  J2 

 32



32

(2.50 in.) 4  3.834952 in.4

(3.00 in.) 4  7.952156 in.4

Equilibrium: Consider a free-body diagram cut around joint B, where the external torque TB is applied: M x  T1  T2  T0  0 (a) and also consider a FBD cut around joint C, where the external torque TC is applied: M x  T2  T3  2T0  0 Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2

3 

(b)

T3 L3 J 3G3

Torque in Segment (3): From the problem statement, the rotation angle at C must not exceed 3°; consequently, the angle of twist in segment (3) must be the opposite of this value: 3  3  0.052360 rad From the torque-twist relationships, the internal torque in segment (3) can be computed: TL  J G (0.052360 rad)(3.834952 in.4 )(4,000 ksi) 3  3 3 T3  3 3 3   66.9325 kip-in. J 3G3 L3 12 in. Consider Rotation Angle of Flange C Relative to Support A: The rotation angle at C is given by the sum of the angles of twist in segments (1) and (2). This rotation angle must not exceed 3°.  TL TL C  1  2  1 1  2 2  3  0.052360 rad (c) J1G1 J 2G2 from Eq. (b): T2  T3  2T0 (d) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

and from Eq. (a): T1  T2  T0  (T3  2T0 )  T0  T3  3T0 (e) Substitute the expressions derived for T1 and T2 in Eqs. (d) and (e) into Eq. (c): (T3  3T0 ) L1 (T3  2T0 ) L2   0.052360 rad J1G1 J 2G2 Simplify  L  3L L  2L2  T3  1  2   T0  1    0.052360 rad  J1G1 J 2G2   J1G1 J 2G2  Solve for T0:  L L  0.052360 rad  T3  1  2   J1G1 J 2G2  T0  3L1 2 L2  J1G1 J 2G2 and compute:   12 in. 24 in. 0.052360 rad  ( 66.9325 kip-in.)   4 4 (3.834952 in. )(4,000 ksi) (7.952156 in. )(6,500 ksi)   T0  3(12 in.) 2(24 in.)  4 (3.834952 in. )(4,000 ksi) (7.952156 in.4 )(6,500 ksi)  41.4590 kip-in. Backsubstitute this value into Eq. (d) to compute T2: T2  T3  2T0  66.9325 kip-in.  2(41.4590 kip-in.)  15.9855 kip-in. and backsubstitute this value into Eq. (e) to compute T1: T1  T3  3T0  66.9325 kip-in.  3(41.4590 kip-in.)  57.4445 kip-in. (a) Maximum magnitude of T0: T0,max  41.5 kip-in.

Ans.

Maximum Shear Stress Magnitudes: The maximum shear stress magnitudes are: T c (57.4445 kip-in.)(2.500 in. / 2) 1  1 1   18.72 ksi J1 3.834952 in.4 Tc (15.9855 kip-in.)(3.000 in. / 2) 2  2 2   3.02 ksi J2 7.952156 in.4

3 

T3c3 (66.9325 kip-in.)(2.500 in. / 2)   21.8 ksi J3 3.834952 in.4

(b) Maximum Shear Stress Magnitude in Aluminum Segments:  3  21.8 ksi

Ans.

(c) Maximum Shear Stress Magnitude in Bronze Segment:  2  3.02 ksi

Ans.

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6.84 The torsional assembly shown in Fig. P6.84 consists of a solid 60-mm-diameter aluminum [G = 28 GPa] segment (2) and two bronze [G = 45 GPa] tube segments (1) and (3), which have an outside diameter of 75 mm and a wall thickness of 5 mm. If concentrated torques of TB = 9 kN-m and TC = 9 kN-m are applied in the directions shown, determine: (a) the maximum shear stress magnitude in bronze tube segments (1) and (3). (b) the maximum shear stress magnitude in aluminum segment (2). (c) the rotation angle of joint C. Fig. P6.84

Solution Section Properties: The polar moments of inertia for the aluminum and bronze segments are: J1  J 3  J2 

 32



(75 mm) 4  (65 mm) 4   1,353,830 mm 4 32

(60 mm) 4  1, 272,345 mm 4

Equilibrium: Consider a free-body diagram cut around joint B, where the external torque TB is applied: M x  T1  T2  TB  0 (a) and also consider a FBD cut around joint C, where the external torque TC is applied: M x  T2  T3  TC  0

(b)

Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero: 1  2  3  0 (c) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2

3 

T3 L3 J 3G3

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1L1 T2 L2 T3 L3   0 J1G1 J 2G2 J 3G3

(d)

(e)

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Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1 and T3 with equivalent expressions involving T2. From Eq. (a): T1  T2  TB (f) and from Eq. (b): T3  T2  TC (g) Substitute Eqs. (f) and (g) into Eq. (e) and simplify to derive an expression for T2: (T2  TB ) L1 T2 L2 (T2  TC ) L3   0 J1G1 J 2G2 J 3G3  L L L  TL T L T2  1  2  3    B 1  C 3 J1G1 J 3G3  J1G1 J 2G2 J 3G3  TB L1 TC L3  J1G1 J 3G3 T2   L1 L L  2  3 J1G1 J 2G2 J 3G3 Since G1 = G3 and J1 = J3, Eq. (h) can be simplified further and T2 can be computed as: 1  TB L1  TC L3   T2     J1G1  L1  L3  L2  J 2G2   J1G1

(h)

1  (9,000,000 N-mm)(325 mm)  (9,000,000 N-mm)(325 mm)  2   325 mm  325 mm 400 mm (1,353,830 mm )(45,000 N/mm )    4 4  (1,353,830 mm )(45,000) (1,272,345 mm )(28,000)   4,385, 216 N-mm Backsubstitute this result into Eqs. (f) and (g) to obtain T1 and T3: T1  4,385, 216 N-mm  9, 000, 000 N-mm  4, 614, 784 N-mm T3  4,385, 216 N-mm  9, 000, 000 N-mm  4, 614, 784 N-mm 

4

Maximum Shear Stress: The maximum shear stress magnitudes in the three segments are: T c (4,614,784 N-mm)(75 mm / 2) 1  1 1   127.8 MPa J1 1,353,830 mm 4 Tc (4,385, 216 N-mm)(60 mm / 2) 2  2 2   103.4 MPa J2 1, 272,345 mm 4

3 

T3c3 (4,614,784 N-mm)(75 mm / 2)   127.8 MPa J3 1,353,830 mm 4

(a) Maximum Shear Stress Magnitude in Bronze Segments: 1  127.8 MPa

Ans.

(b) Maximum Shear Stress Magnitude in Aluminum Segment:  2  103.4 MPa

Ans.

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(c) Rotation Angle of Joint C Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends: 1  B   A B   A  1 Similarly, the angle of twist in member (2) can be defined by: 2  C  B C  B  2 To derive an expression for C, substitute Eq. (i) into Eq. (j), and note that joint A is restrained from rotating; therefore, A = 0.  C  1  2 The angle of twist in member (1) is: TL (4,614,784 N-mm)(325 mm) 1  1 1   0.024618 rad J1G1 (1,353,830 mm 4 )(45,000 N/mm 2 ) The angle of twist in member (2) is: TL ( 4,385, 216 N-mm)(400 mm) 2  2 2   0.049237 rad J 2G2 (1, 272,345 mm 4 )(28,000 N/mm 2 ) The rotation angle of flange C is thus: C  1  2  0.024618 rad  (0.049237 rad)  0.024618 rad  0.0246 rad

(i) (j)

Ans.

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6.85 The torsional assembly shown in Fig. P6.85 consists of a solid 60-mm-diameter aluminum [G = 28 GPa] segment (2) and two bronze [G = 45 GPa] tube segments (1) and (3), which have an outside diameter of 75 mm and a wall thickness of 5 mm. If concentrated torques of TB = 6 kN-m and TC = 10 kN-m are applied in the directions shown, determine: (a) the maximum shear stress magnitude in bronze tube segments (1) and (3). (b) the maximum shear stress magnitude in aluminum segment (2). (c) the rotation angle of joint C. Fig. P6.85

Solution Section Properties: The polar moments of inertia for the aluminum and bronze segments are: J1  J 3  J2 

 32



(75 mm) 4  (65 mm) 4   1,353,830 mm 4 32

(60 mm) 4  1, 272,345 mm 4

Equilibrium: Consider a free-body diagram cut around joint B, where the external torque TB is applied: M x  T1  T2  TB  0 (a) and also consider a FBD cut around joint C, where the external torque TC is applied: M x  T2  T3  TC  0

(b)

Geometry of Deformation Relationship: Since the two ends of the torsion structure are securely attached to fixed supports at A and D, the sum of the angles of twist in the three shafts must equal zero: 1  2  3  0 (c) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2

3 

T3 L3 J 3G3

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1L1 T2 L2 T3 L3   0 J1G1 J 2G2 J 3G3

(d)

(e)

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Solve the Equations: The approach used here will be to reduce the variables in Eq. (e) by replacing T1 and T3 with equivalent expressions involving T2. From Eq. (a): T1  T2  TB (f) and from Eq. (b): T3  T2  TC (g) Substitute Eqs. (f) and (g) into Eq. (e) and simplify to derive an expression for T2: (T2  TB ) L1 T2 L2 (T2  TC ) L3   0 J1G1 J 2G2 J 3G3  L L L  TL T L T2  1  2  3    B 1  C 3 J1G1 J 3G3  J1G1 J 2G2 J 3G3  TB L1 TC L3  J1G1 J 3G3 T2   L1 L L  2  3 J1G1 J 2G2 J 3G3 Since G1 = G3 and J1 = J3, Eq. (h) can be simplified further and T2 can be computed as: 1  TB L1  TC L3   T2     J1G1  L1  L3  L2  J 2G2   J1G1

(h)

1  (6,000,000 N-mm)(325 mm)  (10,000,000 N-mm)(325 mm)  2   325 mm  325 mm 400 mm (1,353,830 mm )(45,000 N/mm )    (1,353,830 mm 4 )(45,000) (1,272,345 mm 4 )(28,000)    3,897,970 N-mm 

4

Backsubstitute this result into Eqs. (f) and (g) to obtain T1 and T3: T1  3,897,970 N-mm  6, 000, 000 N-mm  2,102, 030 N-mm T3  3,897,970 N-mm  10, 000, 000 N-mm  6,102, 030 N-mm Maximum Shear Stress: The maximum shear stress magnitudes in the three segments are: T c (2,102,030 N-mm)(75 mm / 2) 1  1 1   58.2 MPa J1 1,353,830 mm 4 Tc (3,897,970 N-mm)(60 mm / 2) 2  2 2   91.9 MPa J2 1, 272,345 mm 4

3 

T3c3 (6,102,030 N-mm)(75 mm / 2)   169.0 MPa J3 1,353,830 mm 4

(a) Maximum Shear Stress Magnitude in Bronze Segments:  3  169.0 MPa

Ans.

(b) Maximum Shear Stress Magnitude in Aluminum Segment:  2  91.9 MPa

Ans.

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(c) Rotation Angle of Joint C Relative to Support A: The angle of twist in member (1) can be defined by the difference in rotation angles at the two ends: 1  B   A B   A  1 Similarly, the angle of twist in member (2) can be defined by: 2  C  B C  B  2 To derive an expression for C, substitute Eq. (i) into Eq. (j), and note that joint A is restrained from rotating; therefore, A = 0.  C  1  2 The angle of twist in member (1) is: TL (2,102,030 N-mm)(325 mm) 1  1 1   0.011214 rad J1G1 (1,353,830 mm 4 )(45,000 N/mm 2 ) The angle of twist in member (2) is: TL ( 3,897,970 N-mm)(400 mm) 2  2 2   0.043766 rad J 2G2 (1, 272,345 mm 4 )(28,000 N/mm 2 ) The rotation angle of flange C is thus: C  1  2  0.011214 rad  (0.043766 rad)  0.032552 rad  0.0326 rad

(i) (j)

Ans.

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6.86 A solid 1.50-in.-diameter brass [G = 5,600 ksi] shaft [segments (1), (2), and (3)] has been stiffened between B and C by the addition of a cold-rolled stainless steel tube (4) (Fig. P6.86a). The tube (Fig. P6.86b) has an outside diameter of 3.50 in., a wall thickness of 0.12 in., and a shear modulus of G = 12,500 ksi. The tube is attached to the brass shaft by means of rigid flanges welded to the tube and to the shaft. (The thickness of the flanges can be neglected for this analysis.) If a torque of 400 lb-ft is applied to the shaft as shown in Fig. P6.86a, determine: (a) the maximum shear stress magnitude in segment (1) of the brass shaft. (b) the maximum shear stress magnitude in segment (2) of the brass shaft (i.e., between flanges B and C) (c) the maximum shear stress magnitude in the stainless steel tube (4). (d) the rotation angle of end D relative to end A.

Fig. P6.86a

Fig. P6.86b Cross Section Through Tube

Solution Section Properties: The polar moments of inertia for the brass shaft and the stainless steel tube are: J1  J 2  J 3  J4 



32

(1.50 in.) 4  0.497010 in.4



(3.50 in.) 4  (3.26 in.) 4   3.643915 in.4 32

(a) Shear Stress Magnitude in Brass Segment (1): T c (400 lb-ft)(1.500 in. / 2)(12 in./ft) 1  1 1   7, 243.3 psi  7, 240 psi J1 0.497010 in.4

Ans.

Statically Indeterminate Portion of the Shaft: The portion of the shaft between B and C is statically indeterminate. Equilibrium: Consider a free-body diagram that cuts through the shaft between B and C and includes the free end of the shaft at A: M x  T2  T4  400 lb-ft  0

(a)

Geometry of Deformation Relationship: Since the tube and shaft are securely connected by the rigid flanges, the angles of twist in both members must be equal; therefore, 2  4 (b) Torque-Twist Relationships: TL TL 2  2 2 4  4 4 J 2G2 J 4G4

(c)

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Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T2 L2 TL  4 4 J 2G2 J 4G4

(d)

Solve the Equations: Solve Eq. (d) for T2: 4 L J G  20 in.  0.497010 in.   5,600 ksi  T2  T4 4 2 2  T4   0.0611047 T4  20 in.  3.643915 in.4   12,500 ksi  L2 J 4 G4 and substitute this result into Eq. (a) to compute the torque T4 in the stainless steel tube (4): T2  T4  0.0611047 T4  T4  1.0611047 T4  400 lb-ft

T4  376.9656 lb-ft The torque in brass shaft segment (2) is therefore: T2  400 lb-ft  T4  400 lb-ft  ( 376.9656 lb-ft)  23.0344 lb-ft (b) Maximum Shear Stress in Brass Segment (2): The maximum shear stress in segment (2) is: Tc (23.0344 lb-ft)(1.50 in. / 2)(12 in./ft) 2  2 2   417.1 psi  417 psi Ans. J2 0.497010 in.4 (c) Maximum Shear Stress in Stainless Steel Tube (4): The maximum shear stress in tube (4) is: Tc (376.9656 lb-ft)(3.50 in. / 2)(12 in./ft) 4  4 4   2,172.5 psi  2,170 psi Ans. J4 3.643915 in.4 (d) Rotation Angle of End D Relative to End A: The angle of twist of segment (1) is: TL ( 400 lb-ft)(12 in.)(12 in./ft) 1  1 1   0.020695 rad J1G1 (0.497010 in.4 )(5,600,000 psi) The angle of twist in brass segment (3) has the same value: 3  0.020695 rad The angle of twist in brass segment (2) is: TL ( 23.0344 lb-ft)(20 in.)(12 in./ft) 2  2 2   0.0019863 rad J 2G2 (0.497010 in.4 )(5,600,000 psi) The rotation angle of end D relative to end A is the sum of these three angles of twist:  D  1  2  3  0.020695 rad  ( 0.0019863 rad)  ( 0.020695 rad)  0.0433763 rad  0.0434 rad

Ans.

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6.87 A solid 60-mm-diameter cold-rolled brass [G = 39 GPa] shaft that is 1.25-m long extends through and is completely bonded to a hollow aluminum [G = 28 GPa] tube, as shown in Fig. P6.87. Aluminum tube (1) has an outside diameter of 90 mm, an inside diameter of 60 mm, and a length of 0.75 m. Both the brass shaft and the aluminum tube are securely attached to the wall support at A. When the two torques shown are applied to the composite shaft, determine: (a) the maximum shear stress magnitude in aluminum tube (1). (b) the maximum shear stress magnitude in brass shaft segment (2). (c) the maximum shear stress magnitude in brass shaft segment (3). (d) the rotation angle of joint B. (e) the rotation angle of end C.

Fig. P6.87

Solution Section Properties: The polar moments of inertia for tube (1) and brass shafts (2) and (3) are: J1 



(90 mm) 4  (60 mm) 4   5,168,902 mm 4 32 

J 2  J3 

 32

(60 mm) 4  1, 272,345 mm 4

Equilibrium: Consider a free-body diagram cut around end C through segment (3): M x  T3  8 kN-m  0 T3  8 kN-m (a) and also consider a FBD cut around joint B: M x  T1  T2  T3  20 kN-m  0 T1  T2  T3  20 kN-m  12 kN-m

(b)

Eq. (b) reveals that the portion of the shaft between A and B is statically indeterminate. The five-step solution procedure will be used to determine T1 and T2 in this portion of the shaft. Geometry of Deformation Relationship: Since the aluminum tube and the brass shaft are securely bonded together, the angles of twist in both members must be equal; therefore, 1  2 (c) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2

(d)

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Compatibility Equation: Substitute the torque-twist relationships [Eqs. (d)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1L1 T2 L2  J1G1 J 2G2

(e)

Solve the Equations: Solve Eq. (e) for T1:  5,168,902 mm4   28 GPa  L J G T1  T2 2 1 1  T2     2.916667 T2 L1 J 2 G2  1,272,345 mm4   39 GPa  and substitute this result into Eq. (b) to compute the torque T2 in brass shaft (2): T1  T2  2.916667 T2  T2  3.916667 T2  12,000 N-m

T2  3,063.830 N-m The torque in aluminum tube (1) is therefore: T1  12, 000 N-m  T2  12, 000 N-m  3, 063.830 N-m  8,936.170 N-m (a) Maximum Shear Stress in Aluminum Tube: The maximum shear stress in aluminum tube (1) is: T c (8,936.170 N-m)(90 mm / 2)(1,000 mm/m) 1  1 1   77.8 MPa Ans. J1 5,168,902 mm 4 (b) Maximum Shear Stress Magnitude in Brass Shaft Segment (2): Tc (3,063.830 N-m)(60 mm / 2)(1,000 mm/m) 2  2 2   72.2 MPa J2 1, 272,345 mm 4

Ans.

(c) Maximum Shear Stress Magnitude in Brass Shaft Segment (3): Tc (8,000 N-m)(60 mm / 2)(1,000 mm/m) 3  3 3   188.6 MPa J3 1, 272,345 mm 4

Ans.

(d) Rotation Angle of Joint B: TL (8,936.170 N-m)(0.75 m)(1,000 mm/m) 2 1  1 1   0.046308 rad J1G1 (5,168,902 mm 4 )(28,000 N/mm 2 ) or

2 

T2 L2 (3,063.830 N-m)(0.75 m)(1,000 mm/m) 2   0.046308 rad J 2G2 (1,272,345 mm 4 )(39,000 N/mm 2 )

  B  0.0463 rad

Ans.

(e) Rotation Angle of End C: TL ( 8,000 N-m)(0.5 m)(1,000 mm/m) 2 3  3 3   0.080610 rad J 3G3 (1,272,345 mm 4 )(39,000 N/mm 2 )  C   B  3  0.046308 rad  ( 0.080610 rad)  0.0343 rad

Ans.

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6.88 The gear assembly shown in Fig. P6.88 is subjected to a torque of TC = 140 N-m. Shafts (1) and (2) are solid 20-mm-diameter steel shafts, and shaft (3) is a solid 25-mm-diameter steel shaft. Assume L = 400 mm and G = 80 GPa. Determine: (a) the maximum shear stress magnitude in shaft (1). (b) the maximum shear stress magnitude in shaft segment (3). (c) the rotation angle of gear E. (d) the rotation angle of gear C. Fig. P6.88

Solution Section Properties: The polar moments of inertia for the shafts are: J1  J 2  J3 

 32



32

(20 mm) 4  15,708.0 mm 4

(25 mm) 4  38,349.5 mm 4

Equilibrium: Consider a free-body diagram cut through shaft (2) and around gear C: M x  T2  TC  0  T2  TC  140 N-m (a) Next, consider a FBD cut around gear B through shafts (1) and (2). The teeth of gear C exert a force F on the teeth of gear B. This force F opposes the rotation of gear B. The radius of gear B will be denoted by RB for now (even though the gear radius is not given explicitly). M x  T2  T1  F  RB  0 (b) Finally, consider a FBD cut around gear E through shaft (3). The teeth of gear B exert an equal magnitude force F on the teeth of gear C, acting opposite to the direction assumed in the previous FBD. The radius of gear E will be denoted by RE for now. T M x  T3  F  RE  0 F   3 (c) RE The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give R T1  140 N-m  T3 B RE The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of gear teeth NB and NE: N  24 teeth  T1  140 N-m  T3 B  140 N-m  T3   140 N-m  0.4T3 (d)  60 teeth  NE

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Equation (d) summarizes the results of the equilibrium considerations, but there are still two unknowns in this equation: T1 and T3. Consequently, this problem is statically indeterminate. To solve the problem, an additional equation must be developed. This second equation will be derived from the relationship between the angles of twist in shafts (1) and (3). Geometry of Deformation Relationship: The rotation of gear B is equal to the angle of twist in shaft (1): B   1 and the rotation of gear E is equal to the angle of twist in shaft (3): E  3 However, since the gear teeth mesh, the rotation angles for gears B and E are not independent. The arclengths associated with the respective rotations must be equal, but the gears turn in opposite directions. The relationship between gear rotations can be stated as: RBB   REE where RB and RE are the radii of gears B and E, respectively. Since the gear rotation angles are related to the shaft angles of twist, this relationship can be expressed as: RB1   RE3 (e) Torque-Twist Relationships: TL TL 1  1 1 3  3 3 J1G1 J 3G3

(f)

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (f)] into the geometry of deformation relationship [Eq. (e)] to obtain: TL TL RB 1 1   RE 3 3 J1G1 J 3G3 which can be rearranged and expressed in terms of the gear ratio NB /NE: N B T1 L1 TL  3 3 (g) N E J1G1 J 3G3 Note that the compatibility equation has two unknowns: T1 and T3. This equation can be solved simultaneously with Eq. (d) to calculate the internal torques in shafts (1) and (3). Solve the Equations: Solve for internal torque T3 in Eq. (g):  N  L  J G  T3  T1  B   1   3   3   N E   L3   J1   G1  and substitute this result into Eq. (d): T1  140 N-m  0.4T3

  N   L   J  G   140 N-m  0.4  T1  B   1   3   3     N E   L3   J1   G1     24 teeth   38,349.5 mm 4    140 N-m  0.4 T1   4   60 teeth   15,708.0 mm    140 N-m  0.390624 T1

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Group the T1 terms to obtain: 140 N-m T1   100.6742 N-m 1.390624 Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3): T1  140 N-m  0.4T3 T1  140 N-m 100.6742 N-m  140 N-m   98.3146 N-m 0.4 0.4 (a) Maximum Shear Stress Magnitude in Shaft (1): T c (100.6742 N-m)(20 mm / 2)(1,000 mm/m) 1  1 1   64.1 MPa J1 15,708.0 mm 4  T3 

(b) Maximum Shear Stress Magnitude in Shaft (3): Tc (98.3146 N-m)(25 mm / 2)(1,000 mm/m) 3  3 3   32.0 MPa J3 38,349.5 mm 4

Ans.

Ans.

(c) Rotation Angle of Gear E: TL ( 98.3146 N-m)(1.25)(400 mm)(1,000 mm/m) 3  3 3   0.016023 rad J 3G3 (38,349.5 mm 4 )(80,000 N/mm 2 )   E  3  0.01602 rad

Ans.

(d) Rotation Angle of Gear C: TL (100.6742 N-m)(1.25)(400 mm)(1,000 mm/m) 1  1 1   0.040057 rad J1G1 (15,708.0 mm 4 )(80,000 N/mm 2 )

2 

T2 L2 (140 N-m)(400 mm)(1,000 mm/m)   0.044563 rad J 2G2 (15,708.0 mm 4 )(80,000 N/mm 2 )

 C   B  2  1  2  0.040057 rad  0.044563 rad  0.0846 rad

Ans.

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6.89 The gear assembly shown in Fig. P6.89 is subjected to a torque of TC = 1,100 lb-ft. Shafts (1) and (2) are solid 1.625-in.-diameter aluminum shafts and shaft (3) is a solid 2.00-in.-diameter aluminum shaft. Assume L = 20 in. and G = 4,000 ksi. Determine: (a) the maximum shear stress magnitude in shaft (1). (b) the maximum shear stress magnitude in shaft segment (3). (c) the rotation angle of gear E. (d) the rotation angle of gear C. Fig. P6.89

Solution Section Properties: The polar moments of inertia for the shafts are: J1  J 2  J3 

 32



32

(1.625 in.) 4  0.684563 in.4

(2.00 in.) 4  1.570796 in.4

Equilibrium: Consider a free-body diagram cut through shaft (2) and around gear C: M x  T2  TC  0  T2  TC  1,100 lb-ft (a) Next, consider a FBD cut around gear B through shafts (1) and (2). The teeth of gear C exert a force F on the teeth of gear B. This force F opposes the rotation of gear B. The radius of gear B will be denoted by RB for now (even though the gear radius is not given explicitly). M x  T2  T1  F  RB  0 (b) Finally, consider a FBD cut around gear E through shaft (3). The teeth of gear B exert an equal magnitude force F on the teeth of gear C, acting opposite to the direction assumed in the previous FBD. The radius of gear E will be denoted by RE for now. T M x  T3  F  RE  0 F   3 (c) RE The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give R T1  1,100 lb-ft  T3 B RE The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of gear teeth NB and NE: N  24 teeth  T1  1,100 lb-ft  T3 B  1,100 lb-ft  T3   1,100 lb-ft  0.4T3 (d)  60 teeth  NE

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Equation (d) summarizes the results of the equilibrium considerations, but there are still two unknowns in this equation: T1 and T3. Consequently, this problem is statically indeterminate. To solve the problem, an additional equation must be developed. This second equation will be derived from the relationship between the angles of twist in shafts (1) and (3). Geometry of Deformation Relationship: The rotation of gear B is equal to the angle of twist in shaft (1): B   1 and the rotation of gear E is equal to the angle of twist in shaft (3): E  3 However, since the gear teeth mesh, the rotation angles for gears B and E are not independent. The arclengths associated with the respective rotations must be equal, but the gears turn in opposite directions. The relationship between gear rotations can be stated as: RBB   REE where RB and RE are the radii of gears B and E, respectively. Since the gear rotation angles are related to the shaft angles of twist, this relationship can be expressed as: RB1   RE3 (e) Torque-Twist Relationships: TL TL 1  1 1 3  3 3 J1G1 J 3G3

(f)

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (f)] into the geometry of deformation relationship [Eq. (e)] to obtain: TL TL RB 1 1   RE 3 3 J1G1 J 3G3 which can be rearranged and expressed in terms of the gear ratio NB /NE: N B T1 L1 TL  3 3 (g) N E J1G1 J 3G3 Note that the compatibility equation has two unknowns: T1 and T3. This equation can be solved simultaneously with Eq. (d) to calculate the internal torques in shafts (1) and (3). Solve the Equations: Solve for internal torque T3 in Eq. (g):  N  L  J G  T3  T1  B   1   3   3   N E   L3   J1   G1  and substitute this result into Eq. (d): T1  1,100 lb-ft  0.4T3

  N   L   J  G   1,100 lb-ft  0.4  T1  B   1   3   3     N E   L3   J1   G1     24 teeth   1.570796 in.4    1,100 lb-ft  0.4 T1   4   60 teeth   0.684563 in.    1,100 lb-ft  0.367135T1

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Group the T1 terms to obtain: 1,100 lb-ft T1   804.602 lb-ft 1.367135 Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3): T1  1,100 lb-ft  0.4T3 T1  1,100 lb-ft 804.602 lb-ft  1,100 lb-ft   738.495 lb-ft 0.4 0.4 (a) Maximum Shear Stress Magnitude in Shaft (1): T c (804.602 lb-ft)(1.625 in. / 2)(12 in./ft) 1  1 1   11, 459.677 psi  11.46 ksi J1 0.684563 in.4  T3 

(b) Maximum Shear Stress Magnitude in Shaft (3): Tc (738.495 lb-ft)(2.00 in. / 2)(12 in./ft) 3  3 3   5,641.687 psi  5.64 ksi J3 1.570796 in.4

Ans.

Ans.

(c) Rotation Angle of Gear E: TL ( 738.495 lb-ft)(1.25)(20 in.)(12 in./ft) 3  3 3   0.035261 rad J 3G3 (1.570796 in.4 )(4,000,000 psi)   E  3  0.0353 rad

Ans.

(d) Rotation Angle of Gear C: TL (804.602 lb-ft)(1.25)(20 in.)(12 in./ft) 1  1 1   0.088151 rad J1G1 (0.684563 in.4 )(4,000,000 psi)

2 

T2 L2 (1,100 lb-ft)(20 in.)(12 in./ft)   0.096412 rad J 2G2 (0.684563 in.4 )(4,000,000 psi)

 C   B  2  1  2  0.088151 rad  0.096412 rad  0.1846 rad

Ans.

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6.90 A torque of TC = 460 N-m acts on gear C of the assembly shown in Fig. P6.90. Shafts (1) and (2) are solid 35-mm-diameter aluminum shafts and shaft (3) is a solid 25-mm-diameter aluminum shaft. Assume L = 200 mm and G = 28 GPa. Determine: (a) the maximum shear stress magnitude in shaft (1). (b) the maximum shear stress magnitude in shaft segment (3). (c) the rotation angle of gear E. (d) the rotation angle of gear C. Fig. P6.90

Solution Section Properties: The polar moments of inertia for the shafts are:  J1  J 2  (35 mm) 4  147,323.5 mm 4 32  J 3  (25 mm) 4  38,349.5 mm 4 32 Equilibrium: Consider a free-body diagram cut through shaft (2) and around gear C: M x  T2  TC  0  T2  TC  460 N-m (a) Next, consider a FBD cut around gear B through shafts (1) and (2). The teeth of gear E exert a force F on the teeth of gear B. This force F opposes the rotation of gear B. The radius of gear B will be denoted by RB for now (even though the gear radius is not given explicitly). M x  T2  T1  F  RB  0 (b) Finally, consider a FBD cut around gear E through shaft (3). The teeth of gear B exert an equal magnitude force F on the teeth of gear C, acting opposite to the direction assumed in the previous FBD. The radius of gear E will be denoted by RE for now. T M x  T3  F  RE  0 F   3 (c) RE The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give R T1  460 N-m  T3 B RE The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of gear teeth NB and NE: N  54 teeth  T1  460 N-m  T3 B  460 N-m  T3   460 N-m  1.285714 T3 (d)  42 teeth  NE Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Equation (d) summarizes the results of the equilibrium considerations, but there are still two unknowns in this equation: T1 and T3. Consequently, this problem is statically indeterminate. To solve the problem, an additional equation must be developed. This second equation will be derived from the relationship between the angles of twist in shafts (1) and (3). Geometry of Deformation Relationship: The rotation of gear B is equal to the angle of twist in shaft (1): B   1 and the rotation of gear E is equal to the angle of twist in shaft (3): E  3 However, since the gear teeth mesh, the rotation angles for gears B and E are not independent. The arclengths associated with the respective rotations must be equal, but the gears turn in opposite directions. The relationship between gear rotations can be stated as: RBB   REE where RB and RE are the radii of gears B and E, respectively. Since the gear rotation angles are related to the shaft angles of twist, this relationship can be expressed as: RB1   RE3 (e) Torque-Twist Relationships: TL TL 1  1 1 3  3 3 J1G1 J 3G3

(f)

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (f)] into the geometry of deformation relationship [Eq. (e)] to obtain: TL TL RB 1 1   RE 3 3 J1G1 J 3G3 which can be rearranged and expressed in terms of the gear ratio NB /NE: N B T1 L1 TL  3 3 (g) N E J1G1 J 3G3 Note that the compatibility equation has two unknowns: T1 and T3. This equation can be solved simultaneously with Eq. (d) to calculate the internal torques in shafts (1) and (3). Solve the Equations: Solve for internal torque T3 in Eq. (g):  N  L  J G  T3  T1  B   1   3   3   N E   L3   J1   G1  and substitute this result into Eq. (d): T1  460 N-m  1.285714 T3   N   L   J  G   460 N-m  1.285714  T1  B   1   3   3     N E   L3   J1   G1     54 teeth   38,349.5 mm 4    460 N-m  1.285714 T1   4   42 teeth   147,323.5 mm    460 N-m  0.430305 T1

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Group the T1 terms to obtain: 460 N-m T1   321.610 N-m 1.430305 Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3): T1  460 N-m  1.285714 T3  T3 

T1  460 N-m 321.610 N-m  460 N-m   107.637 N-m 1.285714 1.285714

(a) Maximum Shear Stress Magnitude in Shaft (1): T c (321.610 N-m)(35 mm / 2)(1,000 mm/m) 1  1 1   38.2 MPa J1 147,323.5 mm 4

Ans.

(b) Maximum Shear Stress Magnitude in Shaft (3): Tc (107.637 N-m)(25 mm / 2)(1,000 mm/m) 3  3 3   35.1 MPa J3 38,349.5 mm 4

Ans.

(c) Rotation Angle of Gear E: TL ( 107.637 N-m)(2)(200 mm)(1,000 mm/m) 3  3 3   0.040096 rad J 3G3 (38,349.5 mm 4 )(28, 000 N/mm 2 )   E  3  0.0401 rad

Ans.

(d) Rotation Angle of Gear C: TL (321.610 N-m)(2)(200 mm)(1,000 mm/m) 1  1 1   0.031186 rad J1G1 (147,323.5 mm 4 )(28,000 N/mm 2 )

2 

T2 L2 (460 N-m)(200 mm)(1,000 mm/m)   0.022303 rad J 2G2 (147,323.5 mm 4 )(28,000 N/mm 2 )

 C   B  2  1  2  0.031186 rad  0.022303 rad  0.0535 rad

Ans.

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6.91 A torque of TC = 40 kip-in. acts on gear C of the assembly shown in Fig. P6.91. Shafts (1) and (2) are solid 2.00-in.-diameter stainless steel shafts and shaft (3) is a solid 1.75-in.diameter stainless steel shaft. Assume L = 8 in. and G = 12,500 ksi. Determine: (a) the maximum shear stress magnitude in shaft (1). (b) the maximum shear stress magnitude in shaft segment (3). (c) the rotation angle of gear E. (d) the rotation angle of gear C. Fig. P6.91

Solution Section Properties: The polar moments of inertia for the shafts are: J1  J 2  J3 

 32



32

(2.00 in.) 4  1.570796 in.4

(1.75 in.) 4  0.920772 in.4

Equilibrium: Consider a free-body diagram cut through shaft (2) and around gear C: M x  T2  TC  0  T2  TC  40 kip-in. (a) Next, consider a FBD cut around gear B through shafts (1) and (2). The teeth of gear E exert a force F on the teeth of gear B. This force F opposes the rotation of gear B. The radius of gear B will be denoted by RB for now (even though the gear radius is not given explicitly). M x  T2  T1  F  RB  0 (b) Finally, consider a FBD cut around gear E through shaft (3). The teeth of gear B exert an equal magnitude force F on the teeth of gear C, acting opposite to the direction assumed in the previous FBD. The radius of gear E will be denoted by RE for now. T M x  T3  F  RE  0 F   3 (c) RE The results of Eqs. (a) and (c) can be substituted into Eq. (b) to give R T1  40 kip-in.  T3 B RE The ratio RB/RE is simply the gear ratio between gears B and E, which can also be expressed in terms of gear teeth NB and NE: N  54 teeth  T1  40 kip-in.  T3 B  40 kip-in.  T3   40 kip-in.  1.285714 T3 (d)  42 teeth  NE

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Equation (d) summarizes the results of the equilibrium considerations, but there are still two unknowns in this equation: T1 and T3. Consequently, this problem is statically indeterminate. To solve the problem, an additional equation must be developed. This second equation will be derived from the relationship between the angles of twist in shafts (1) and (3). Geometry of Deformation Relationship: The rotation of gear B is equal to the angle of twist in shaft (1): B   1 and the rotation of gear E is equal to the angle of twist in shaft (3): E  3 However, since the gear teeth mesh, the rotation angles for gears B and E are not independent. The arclengths associated with the respective rotations must be equal, but the gears turn in opposite directions. The relationship between gear rotations can be stated as: RBB   REE where RB and RE are the radii of gears B and E, respectively. Since the gear rotation angles are related to the shaft angles of twist, this relationship can be expressed as: RB1   RE3 (e) Torque-Twist Relationships: TL TL 1  1 1 3  3 3 J1G1 J 3G3

(f)

Compatibility Equation: Substitute the torque-twist relationships [Eqs. (f)] into the geometry of deformation relationship [Eq. (e)] to obtain: TL TL RB 1 1   RE 3 3 J1G1 J 3G3 which can be rearranged and expressed in terms of the gear ratio NB /NE: N B T1 L1 TL  3 3 (g) N E J1G1 J 3G3 Note that the compatibility equation has two unknowns: T1 and T3. This equation can be solved simultaneously with Eq. (d) to calculate the internal torques in shafts (1) and (3). Solve the Equations: Solve for internal torque T3 in Eq. (g):  N  L  J G  T3  T1  B   1   3   3   N E   L3   J1   G1  and substitute this result into Eq. (d): T1  40 kip-in.  1.285714 T3   N   L   J  G   40 kip-in.  1.285714  T1  B   1   3   3     N E   L3   J1   G1     54 teeth   0.920772 in.4    40 kip-in.  1.285714 T1   4    42 teeth   1.570796 in.    40 kip-in.  0.968994 T1

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Group the T1 terms to obtain: 40 kip-in. T1   20.3149 kip-in. 1.968994 Backsubstitute this result into Eq. (d) to find the internal torque in shaft (3): T1  40 kip-in.  1.285714 T3  T3 

T1  40 kip-in. 20.3149 kip-in.  40 kip-in.   15.3106 kip-in. 1.285714 1.285714

(a) Maximum Shear Stress Magnitude in Shaft (1): T c (20.3149 kip-in.)(2.00 in. / 2) 1  1 1   12.93 ksi J1 1.570796 in.4

Ans.

(b) Maximum Shear Stress Magnitude in Shaft (3): Tc (15.3106 kip-in.)(1.75 in. / 2) 3  3 3   14.55 ksi J3 0.920772 in.4

Ans.

(c) Rotation Angle of Gear E: TL ( 15.3106 kip-in.)(2)(8 in.) 3  3 3   0.021284 rad J 3G3 (0.920772 in.4 )(12, 500 ksi)   E  3  0.0213 rad

Ans.

(d) Rotation Angle of Gear C: TL (20.3149 kip-in.)(2)(8 in.) 1  1 1   0.016554 rad J1G1 (1.570796 in.4 )(12,500 ksi)

2 

T2 L2 (40 kip-in.)(8 in.)   0.016297 rad J 2G2 (1.570796 in.4 )(12,500 ksi)

 C   B  2  1  2  0.016554 rad  0.016297 rad  0.0329 rad

Ans.

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6.92 The steel [G = 12,000 ksi] pipe shown in Fig. P6.92 is fixed to the wall support at C. The bolt holes in the flange at A were supposed to align with mating holes in the wall support; however, an angular misalignment of 4° was found to exist. To connect the pipe to its supports, a temporary installation torque T′B must be applied at B to align flange A with the mating holes in the wall support. The outside diameter of the pipe is 3.50 in. and its wall thickness is 0.216 in. (a) Determine the temporary installation torque T′B that must be applied at B to align the bolt holes at A. (b) Determine the maximum shear stress initial in the pipe after the bolts are connected and the temporary installation torque at B is removed. (c) If the maximum shear stress in the pipe shaft must not exceed 12 ksi, determine the maximum external torque TB that can be applied at B after the bolts are connected. Fig. P6.92

Solution Section Properties: The polar moment of inertia for the pipe is: J1  J 2 



(3.50 in.)4  (3.068 in.)4   6.034313 in.4 32

(a) Temporary Installation Torque T′B: The shaft, which is fixed to the support at C, must be rotated 4° so that the connection at A can be completed. Consequently, segment (2) must be twisted 4° (or 0.069813 rad) by the temporary torque T′B. From the torque-twist relationship for shaft (2) TL 2  2 2 J 2 G2 the temporary torque T′B is: 2 J 2G2 (0.069813 rad)(6.034313 in.4 )(12,000 ksi) Ans. TB    28.1 kip-in. L2 (15 ft)(12 in./ft) (b) Maximum Shear Stress after Temporary Torque is Removed: After T′B is removed, the 4° angle of twist now applies to the total pipe length, i.e., 25 ft. The internal torque magnitude in the pipe due to the 4° misfit is thus:  JG (0.069813 rad)(6.034313 in.4 )(12,000 ksi) Tinitial  misfit   16.851 kip-in. ( L1  L2 ) (25 ft)(12 in./ft) and the maximum shear stress magnitude in the pipe due to this internal torque is: T c (16.851 kip-in.)(3.50 in. / 2)  initial  initial   4.89 ksi Ans. J 6.034313 in.4

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After Bolts Installed: After the bolts are installed, the pipe is analyzed as a statically indeterminate torsion structure. Equilibrium: M x  T1  T2  TB  0 (a) Note: By inspection, the internal torque T2 will end up having a negative value. Geometry of Deformation Relationship: The two shafts are securely attached to fixed supports at A and C; however, in this instance, the sum of the angles of twist in the two members must equal the misfit angle. Note: If the rotation angle at A is a positive 4° (as can be inferred from the problem sketch), then the pipe must twist in a negative sense to reach a zero rotation angle at C. 1  2  0.069813 rad (b) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2   0.069813 rad J1G1 J 2G2

(c)

(d)

Solve the Equations: Since the problem is expressed in terms of allowable stresses, it is convenient to rewrite Eq. (d) in terms of stresses. In general, the elastic torsion formula can be rearranged as: Tc T     J J c which allows Eq. (d) to be rewritten as: 1L1  L  0.069813 rad  2 2 (e) G1c1 G2c2 Solve Eq. (e) for  1: GR L G c  1  ( 0.069813 rad) 1 1   2 2 1 1 L1 L1 G2 c2 (12,000 ksi)(3.50 in./2)  15 ft   2   10 ft  (10 ft)(12 in./ft)  12.217305 ksi  1.5  2  ( 0.069813 rad)

(f)

Assume the shaft (2) controls: By inspection of the FBD, the internal torque T2 should have a negative value. Therefore, we can assume that the maximum shear stress in shaft (2) will be −12 ksi. If the shear stress in shaft (2) reaches this value, then the shear stress in shaft (1) will be: 1  12.217305 ksi  1.5( 12 ksi)  5.7827 ksi  12 ksi O.K. This calculation demonstrates that the shear stress magnitude in shaft (2) must control.

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Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes in each component can be computed: 1 J1 (5.7827 ksi)(6.034313 in.4 ) T1    19.9398 kip-in. c1 3.50 in./2 T2 

2J2 c2



( 12 ksi)(6.034313 in.4 )  41.3781 kip-in. 3.50 in./2

(c) Maximum Allowable Torque: From Eq. (a), the total torque acting at B must not exceed: TB,max  T1  T2  19.9398 kip-in.  (41.3781 kip-in.)  61.3179 kip-in.  61.3 kip-in.

Ans.

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6.93 The steel [G = 12,000 ksi] pipe shown in Fig. P6.93 is fixed to the wall support at C. The bolt holes in the flange at A were supposed to align with mating holes in the wall support; however, an angular misalignment of 4° was found to exist. To connect the pipe to its supports, a temporary installation torque T′B must be applied at B to align flange A with the mating holes in the wall support. The outside diameter of the pipe is 2.875 in. and its wall thickness is 0.203 in. (a) Determine the temporary installation torque T′B that must be applied at B to align the bolt holes at A. (b) Determine the maximum shear stress initial in the pipe after the bolts are connected and the temporary installation torque at B is removed. (c) Determine the magnitude of the maximum shear stress in segments (1) and (2) if an external torque of TB = 80 kip-in. is applied at B after the bolts are connected. Fig. P6.93

Solution Section Properties: The polar moment of inertia for the pipe is: J1  J 2 



(2.875 in.) 4  (2.469 in.) 4   3.059108 in.4 32

(a) Temporary Installation Torque T′B: The shaft, which is fixed to the support at C, must be rotated 4° so that the connection at A can be completed. Consequently, segment (2) must be twisted 4° (or 0.069813 rad) by the temporary torque T′B. From the torque-twist relationship for shaft (2) TL 2  2 2 J 2 G2 the temporary torque T′B is: 2 J 2G2 (0.069813 rad)(3.059108 in.4 )(12,000 ksi) Ans. TB    14.24 kip-in. L2 (15 ft)(12 in./ft) (b) Maximum Shear Stress after Temporary Torque is Removed: After T′B is removed, the 4° angle of twist now applies to the total pipe length, i.e., 25 ft. The internal torque magnitude in the pipe due to the 4° misfit is thus:  JG (0.069813 rad)(3.059108 in.4 )(12,000 ksi) Tinitial  misfit   8.5426 kip-in. ( L1  L2 ) (25 ft)(12 in./ft) and the maximum shear stress magnitude in the pipe due to this internal torque is: T c (8.5426 kip-in.)(2.875 in. / 2)  initial  initial   4.01 ksi Ans. J 3.059108 in.4

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After Bolts Installed: After the bolts are installed, the pipe is analyzed as a statically indeterminate torsion structure. Equilibrium: M x  T1  T2  TB  0 (a) Note: By inspection, the internal torque T2 will end up having a negative value. Geometry of Deformation Relationship: The two shafts are securely attached to fixed supports at A and C; however, in this instance, the sum of the angles of twist in the two members must equal the misfit angle. Note: If the rotation angle at A is a positive 4° (as can be inferred from the problem sketch), then the pipe must twist in a negative sense to reach a zero rotation angle at C. 1  2  0.069813 rad (b) Torque-Twist Relationships: TL TL 1  1 1 2  2 2 J1G1 J 2G2 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (c)] into the geometry of deformation relationship [Eq. (b)] to obtain the compatibility equation: T1L1 T2 L2   0.069813 rad J1G1 J 2G2

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1: JG L J G T1  ( 0.069813 rad) 1 1  T2 2 1 1 L1 L1 J 2 G2

(3.059108 in.4 )(12,000 ksi)  15 ft   T2   10 ft  (10 ft)(12 in./ft)  21.356600 kip-in.  1.5 T2 and substitute this result into Eq. (a) to compute the torque T2: T1  T2    21.356600 kip-in.  1.5T2   T2  21.356600 kip-in.  2.5T2  80 kip-in.  ( 0.069813 rad)

101.3566 kip-in.  40.5426 kip-in. 2.5 The torque in member (1) is therefore: T1  T2  80 kip-in.  40.5426 kip-in.  80 kip-in.  39.4574 kip-in.  T2 

(c) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is: T c (39.4574 kip-in.)(2.875 in. / 2) 1  1 1   18.54 ksi J1 3.059108 in.4 The maximum shear stress magnitude in member (2) is: Tc (40.5426 kip-in.)(2.875 in. / 2) 2  2 2   19.05 ksi J2 3.059108 in.4

Ans.

Ans.

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6.94 A stepped shaft with a major diameter of D = 1.375 in. and a minor diameter of d = 1.00 in. is subjected to a torque of 500 lb-in. A fillet with a radius of r = 3/16 in. is used to transition from the major diameter to the minor diameter. Determine the maximum shear stress in the shaft.

Solution Fillet: From Fig. 6.17b: D 1.375 in. 1.375 d 1.0 in.

r d

0.1875 in. 1.0 in.

0.1875

K

1.22

Section Properties of Minor Diameter Section: J

32

(1.00 in.) 4

0.098175 in.4

Maximum Shear Stress: Tc (500 lb-in.)(1.00 in./2) K (1.22) max J 0.098175 in.4

3,106.7 psi

3,110 psi

Ans.

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6.95 A stepped shaft with a major diameter of D = 20 mm and a minor diameter of d = 16 mm is subjected to a torque of 25 N-m. A full quarter-circular fillet having a radius of r = 2 mm is used to transition from the major diameter to the minor diameter. Determine the maximum shear stress in the shaft.

Solution Fillet: From Fig. 6.17b: D 20 mm 1.25 d 16 mm

r d

2 mm 16 mm

0.125

K

1.30

Section Properties of Minor Diameter Section: J

32

(16 mm) 4

6,434.0 mm 4

Maximum Shear Stress: Tc (25 N-m)(16 mm/2)(1,000 mm/m) K (1.30) J 6, 434.0 mm 4

40.4 MPa

Ans.

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6.96 A fillet with a radius of 1/2 in. is used at the junction in a stepped shaft where the diameter is reduced from 8.00 in. to 6.00 in. Determine the maximum torque that the shaft can transmit if the maximum shear stress in the fillet must be limited to 5 ksi.

Solution Fillet: From Fig. 6.17b: D 8.0 in. r 1.33 d 6.0 in. d

0.50 in. 6.0 in.

0.0833

K

1.41

Section Properties of Minor Diameter Section: J

32

(6.00 in.) 4

Maximum Torque: Tc K J allow J Tmax Kc

127.2345 in.4

(5 ksi)(127.2345 in.4 ) (1.41)(6.00 in./2)

150.4 kip-in.

Ans.

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6.97 A stepped shaft with a major diameter of D = 2.50 in. and a minor diameter of d = 1.25 in. is subjected to a torque of 1,200 lb-in. If the maximum shear stress must not exceed 4,000 psi, determine the minimum radius r that may be used for a fillet at the junction of the two shaft segments. The fillet radius must be chosen as a multiple of 0.05 in.

Solution Section Properties of Minor Diameter Section: J

32

(1.25 in.) 4

0.239684 in.4

Fillet Requirement: From Fig. 6.17b: Tc J (4,000 psi)(0.239684 in.4 ) K K J Tc (1, 200 lb-in.)(1.25 in. / 2) D 2.5 in. 2.00 d 1.25 in. r 0.17 d Minimum Fillet Radius: r rmin d (0.17)(1.25 in.) d

0.213 in.

say rmin

1.28

0.25 in.

Ans.

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6.98 A fillet with a radius of 16 mm is used at the junction in a stepped shaft where the diameter is reduced from 200 mm to 150 mm. Determine the maximum torque that the shaft can transmit if the maximum shear stress in the fillet must be limited to 55 MPa.

Solution Fillet: From Fig. 6.17b: D 200 mm 1.33 d 150 mm

r d

16 mm 150 mm

0.107

K

1.34

Section Properties of Minor Diameter Section: J

32

(150 mm) 4

Maximum Torque: Tc K J allow J Tmax Kc

49,700,978 mm 4

(55 N/mm 2 )(49,700,978 mm 4 ) (1.34)(150 mm/2)

27.1995 106 N-mm

27.2 kN-m

Ans.

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6.99 A stepped shaft with a major diameter of D = 50 mm and a minor diameter of d = 32 mm is subjected to a torque of 210 N-m. If the maximum shear stress must not exceed 40 MPa, determine the minimum radius r that may be used for a fillet at the junction of the two shaft segments. The fillet radius must be chosen as a multiple of 1 mm.

Solution Section Properties of Minor Diameter Section: J

32

(32 mm) 4

102,943.7 mm 4

Fillet Requirement: From Fig. 6.17b: Tc J (40 N/mm 2 )(102,943.7 mm 4 ) K K J Tc (210 N-m)(32 mm / 2)(1,000 mm/m) D 50 mm 1.56 d 32 mm r 0.19 d Minimum Fillet Radius: r rmin d (0.19)(32 mm) d

6.08 mm

say rmin

7 mm

1.23

Ans.

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6.100 A stepped shaft has a major diameter of D = 2.00 in. and a minor diameter of d = 1.50 in. A fillet with a 0.25-in. radius is used to transition between the two shaft segments. The maximum shear stress in the shaft must be limited to 9,000 psi. If the shaft rotates at a constant angular speed of 800 rpm, determine the maximum power that may be delivered by the shaft.

Solution Fillet: From Fig. 6.17b: D 2.00 in. 1.33 d 1.50 in.

r d

0.25 in. 1.50 in.

0.167

K

1.24

Section Properties of Minor Diameter Section: J

32

(1.50 in.) 4

0.497010 in.4

Maximum Torque: Tc K allow J

Tmax

allow

Kc

J

(9,000 lb/in.2 )(0.497010 in.4 ) (1.24)(1.50 in./2)

4,809.77 lb-in. 400.8143 lb-ft

Power transmission: The maximum power that can be transmitted at 800 rpm is: 800 rev 1 min 2 rad Pmax Tmax (400.8143 lb-ft) 33,578.54 lb-ft/s min 60 s 1 rev or in units of horsepower, 33,578.54 lb-ft/s Pmax 61.1 hp 550 lb-ft/s 1 hp

Ans.

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6.101 A stepped shaft has a major diameter of D = 100 mm and a minor diameter of d = 75 mm. A fillet with a 10-mm radius is used to transition between the two shaft segments. The maximum shear stress in the shaft must be limited to 60 MPa. If the shaft rotates at a constant angular speed of 500 rpm, determine the maximum power that may be delivered by the shaft.

Solution Fillet: From Fig. 6.17b: D 100 mm 1.33 d 75 mm

r d

10 mm 75 mm

0.133

K

1.29

Section Properties of Minor Diameter Section: J

32

(75 mm) 4

3,106,311.1 mm4

Maximum Torque: Tc K allow J

Tmax

allow

Kc

J

(60 N/mm 2 )(3,106,311.1 mm4 ) (1.29)(75 mm/2)

3,852,789 N-mm

3,852.789 N-m

Power transmission: The maximum power that can be transmitted at 500 rpm is: 500 rev 1 min 2 rad Pmax Tmax (3,852.789 N-m) min 60 s 1 rev 201,731.6 N-m/s 202 kW

Ans.

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6.102 A 2-in.-diameter shaft contains a ½-in.-deep U-shaped groove that has a ¼-in. radius at the bottom of the groove. The shaft must transmit a torque of T = 720 lb-in. Determine the maximum shear stress in the shaft.

Solution Groove: From Fig. 6.17a: d 2.0 in. 2(0.5 in.) 1.0 in. D 2.0 in. r 0.25 in. 2.00 d 1.0 in. d 1.0 in.

0.25

K

1.29

Section Properties at Minimum Diameter Section: J

32

(1.00 in.) 4

0.098175 in.4

Shaft Shear Stress: Tc (720 lb-in.)(1.00 in./2) K (1.29) max J 0.098175 in.4

4,730.34 psi

4,730 psi

Ans.

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6.103 A semicircular groove with a 6-mm radius is required in a 50-mm-diameter shaft. If the maximum allowable shear stress in the shaft must be limited to 40 MPa, determine the maximum torque that can be transmitted by the shaft.

Solution Groove: From Fig. 6.17a: d 50 mm 2(6 mm) 38 mm D 50 mm r 1.32 d 38 mm d

6 mm 38 mm

0.158

K

1.39

Section Properties at Minimum Diameter Section: J

32

(38 mm) 4

Maximum Torque: Tc K J allow J Tmax Kc

204,707.75 mm 4

(40 N/mm 2 )(204,707.75 mm 4 ) (1.39)(38 mm/2)

310,045.8 N-mm

310 N-m

Ans.

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6.104 A 40-mm-diameter shaft contains a 10-mm-deep U-shaped groove that has a 6-mm radius at the bottom of the groove. The maximum shear stress in the shaft must be limited to 60 MPa. If the shaft rotates at a constant angular speed of 22 Hz, determine the maximum power that may be delivered by the shaft.

Solution Groove: From Fig. 6.17a: d 40 mm 2(10 mm) D 40 mm 2.00 d 20 mm

20 mm r d

6 mm 20 mm

0.30

K

1.25

Section Properties at Minimum Diameter Section: J

32

(20 mm) 4

Maximum Torque: Tc K J allow J Tmax Kc

15,707.96 mm 4

(60 N/mm 2 )(15,707.96 mm 4 ) (1.25)(20 mm/2)

75.3982 N-m

Power transmission: The maximum power that can be transmitted at 22 Hz is: 22 rev 2 rad Pmax Tmax (75.3982 N-m) s 1 rev 10, 422.3 N-m/s 10.42 kW

Ans.

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6.105 A 1.25-in. -diameter shaft contains a 0.25-in.-deep U-shaped groove that has a 1/8-in. radius at the bottom of the groove. The maximum shear stress in the shaft must be limited to 12,000 psi. If the shaft rotates at a constant angular speed of 6 Hz, determine the maximum power that may be delivered by the shaft.

Solution Groove: From Fig. 6.17a: d 1.25 in. 2(0.25 in.) D 1.25 in. 1.67 d 0.75 in.

0.75 in. r d

0.125 in. 0.75 in.

0.167

K

1.39

Section Properties at Minimum Diameter Section: J

32

(0.75 in.) 4

Maximum Torque: Tc K J allow J Tmax Kc

0.031063 in.4

(12,000 psi)(0.031063 in.4 ) (1.39)(0.75 in./2)

715.1219 lb-in. 59.5935 lb-ft

Power transmission: The maximum power that can be transmitted at 6 Hz is: 6 rev 2 rad Pmax Tmax (59.5935 lb-ft) s 1 rev 2,246.6219 lb-ft/s or in units of horsepower, 2, 246.6219 lb-ft/s Pmax 4.08 hp 550 lb-ft/s 1 hp

Ans.

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6.106 A torque of magnitude T = 1.5 kip-in. is applied to each of the bars shown in Fig. P6.106. If the allowable shear stress is specified as allow = 8 ksi, determine the minimum required dimension b for each bar.

Fig. P6.106

Solution (a) Circular Section Rearrange the elastic torsion formula to group terms with d on the left-hand side:  d4 T d 3 T  which can be simplified to  32 ( d / 2)  16  From this equation, the unknown diameter of the solid shaft can be expressed as 16T d3  To support a torque of T = 1.5 kip-in. without exceeding the maximum shear stress of 8 ksi, the solid shaft must have a diameter (i.e., dimension b shown in the problem statement) of 16T 16(1.5 kip-in.) Ans. bmin  3 3  0.985 in.   (8 ksi) (b) Square Section From Table 6.1, b 1    0.208 a The maximum shear stress in a rectangular section is given by Eq. (6.22): T  max  2 a b For a square section where a = b, T 1.5 kip-in. b3    0.901442 in.3  bmin  0.966 in.  max (0.208)(8 ksi) ab

aspect ratio

(c) Rectangular Section From Table 6.1, 2b aspect ratio 2    0.246 b For a rectangular section where a = 2b, T 1.5 kip-in. b3    0.381098 in.3 2 max 2(0.246)(8 ksi)

 bmin  0.725 in.

Ans.

Ans.

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6.107 A torque of magnitude T = 270 Nm is applied to each of the bars shown in Fig. P6.107. If the allowable shear stress is specified as allow = 70 MPa, determine the minimum required dimension b for each bar.

Fig. P6.107

Solution (a) Circular Section Rearrange the elastic torsion formula to group terms with d on the left-hand side:  d4 T d 3 T  which can be simplified to  32 ( d / 2)  16  From this equation, the unknown diameter of the solid shaft can be expressed as 16T d3  To support a torque of T = 270 N-m without exceeding the maximum shear stress of 70 MPa, the solid shaft must have a diameter (i.e., dimension b shown in the problem statement) of 16T 16(270 N-m)(1,000 mm/m) Ans. bmin  3 3  27.0 mm   (70 N/mm2 ) (b) Square Section From Table 6.1, b 1    0.208 a The maximum shear stress in a rectangular section is given by Eq. (6.22): T  max  2 a b For a square section where a = b, T (270 N-m)(1,000 mm/m) b3    18,543.946 mm3  bmin  26.5 mm 2  max (0.208)(70 N/mm ) ab

aspect ratio

(c) Rectangular Section From Table 6.1, 2b aspect ratio 2    0.246 b For a rectangular section where a = 2b, T (270 N-m)(1,000 mm/m) b3    7,839.721 mm3 2 2 max 2(0.246)(70 N/mm )

 bmin  19.87 mm

Ans.

Ans.

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6.108 The bars shown in Fig. P6.108 have equal cross-sectional areas and they are each subjected to a torque of T = 160 N-m. Determine: (a) the maximum shear stress in each bar. (b) the rotation angle at the free end if each bar has a length of 300 mm. Assume G = 28 GPa.

Fig. P6.108

Solution Theoretical Basis: The maximum shear stress in a rectangular section is given by Eq. (6.22): T  max  2 a b and the angle of twist is given by Eq. (6.23) TL  3  a bG Rectangular Section 15 mm by 50 mm From linear interpolation of the values given in Table 6.1, 50 mm aspect ratio  3.333    0.272   0.269 15 mm Shear stress: T (160 N-m)(1,000 mm/m)  max  2   52.3 MPa  a b (0.272)(15 mm) 2 (50 mm) Angle of twist: TL (160 N-m)(300 mm)(1,000 mm/m)  3   0.0378 rad  a bG (0.269)(15 mm)3 (50 mm)(28,000 N/mm 2 ) Rectangular Section 25 mm by 30 mm From Table 6.1, 30 mm aspect ratio  1.2    0.219   0.166 25 mm Shear stress: T (160 N-m)(1,000 mm/m)  max  2   39.0 MPa  a b (0.219)(25 mm) 2 (30 mm) Angle of twist: TL (160 N-m)(300 mm)(1,000 mm/m)  3   0.0220 rad  a bG (0.166)(25 mm)3 (30 mm)(28,000 N/mm 2 )

Ans.

Ans.

Ans.

Ans.

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6.109 The allowable shear stress for each bar shown in Fig. P6.109 is 75 MPa. Determine: (a) the largest torque T that may be applied to each bar. (b) the corresponding rotation angle at the free end if each bar has a length of 300 mm. Assume G = 28 GPa.

Fig. P6.109

Solution Theoretical Basis: The maximum shear stress in a rectangular section is given by Eq. (6.22): T  max  2 a b and the angle of twist is given by Eq. (6.23) TL  3  a bG Rectangular Section 15 mm by 50 mm From linear interpolation of the values given in Table 6.1, 50 mm aspect ratio  3.333    0.272   0.269 15 mm Maximum torque: T   maxa2b  (75 N/mm2 )(0.272)(15 mm)2 (50 mm)  229,500 N-mm  230 N-m Angle of twist: TL (229,500 N-mm)(300 mm)  3   0.0542 rad  a bG (0.269)(15 mm)3 (50 mm)(28,000 N/mm 2 ) Rectangular Section 25 mm by 30 mm From Table 6.1, 30 mm aspect ratio  1.2    0.219   0.166 25 mm Maximum torque: T   maxa2b  (75 N/mm2 )(0.219)(25 mm)2 (30 mm)  307,979 N-mm  308 N-m Angle of twist: TL (307,979 N-mm)(300 mm)  3   0.0424 rad  a bG (0.166)(25 mm)3 (30 mm)(28,000 N/mm 2 )

Ans. Ans.

Ans. Ans.

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6.110 A solid circular rod having diameter D is to be replaced by a rectangular tube having cross-sectional dimensions D × 2D (which are measured to the wall centerlines of the cross section shown in Fig. P6.110). Determine the required minimum thickness tmin of the tube so that the maximum shear stress in the tube will not exceed the maximum shear stress in the solid bar. Fig. P6.110

Solution For the solid circular rod, the maximum shear stress is given by Tc T ( D / 2) 16T  max     4 J  D3 D 32 For the rectangular tube, the area enclosed by the median line is Am  ( D)(2 D)  2 D 2 The maximum shear stress for the thin-walled section is given by Eq. (6.25) T T T  max    2 2 Amt 2(2 D )t 4 D 2t Equate Eqs. (a) and (b) 16T T  3  D 4 D 2t and solve for t: D tmin  64

(a)

(b)

Ans.

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6.111 A 24 in. wide by 0.100 in. thick by 100 in. long steel sheet is to be formed into a hollow section by bending through 360° and welding (i.e., butt-welding) the long edges together. Assume a crosssectional medial length of 24 in. (no stretching of the sheet due to bending). If the maximum shear stress must be limited to 12 ksi, determine the maximum torque that can be carried by the hollow section if (a) the shape of the section is a circle. (b) the shape of the section is an equilateral triangle. (c) the shape of the section is a square. (d) the shape of the section is an 8 × 4 in. rectangle.

Solution The maximum shear stress for a thin-walled section is given by Eq. (6.25) T  max  2 Am t and thus, the maximum torque that can be carried by the hollow section is Tmax  2 max Amt (a) Circle:  d m  24 in. Am  Tmax



 d m  7.639437 in.

(7.639437 in.) 2  45.8366 in.2

4  2 max Amt  2(12 ksi)(45.8366 in.2 )(0.100 in.)  110.0 kip-in.

(b) Equilateral triangle: triangle sides are each 24 in./3  8 in. 1 1 Am  bh  (8 in.)(8 in.)sin 60  27.7128 in.2 2 2 Tmax  2 max Amt  2(12 ksi)(27.7128 in.2 )(0.100 in.)  66.5 kip-in.

Ans.

Ans.

(c) Square: sides of the square are each 24 in./4  6 in.

Am  bh  (6 in.)(6 in.)  36 in.2 Tmax  2 max Amt  2(12 ksi)(36 in.2 )(0.100 in.)  86.4 kip-in.

Ans.

(d) 8 × 4 in. rectangle: Am  bh  (8 in.)(4 in.)  32 in.2 Tmax  2 max Amt  2(12 ksi)(32 in.2 )(0.100 in.)  76.8 kip-in.

Ans.

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6.112 A 500 mm wide by 3 mm thick by 2 m long aluminum sheet is to be formed into a hollow section by bending through 360° and welding (i.e., butt-welding) the long edges together. Assume a crosssectional medial length of 500 mm (no stretching of the sheet due to bending). If the maximum shear stress must be limited to 75 MPa, determine the maximum torque that can be carried by the hollow section if (a) the shape of the section is a circle. (b) the shape of the section is an equilateral triangle. (c) the shape of the section is a square. (d) the shape of the section is a 150 × 100 mm rectangle.

Solution The maximum shear stress for a thin-walled section is given by Eq. (6.25) T  max  2 Am t and thus, the maximum torque that can be carried by the hollow section is Tmax  2 max Amt (a) Circle:  d m  500 mm Am 



 d m  159.155 mm

(159.155 mm) 2  19,894.382 mm 2

4  2 max Amt  2(75 N/mm 2 )(19,894.382 mm 2 )(3 mm)  8,952, 472 N-mm  8.95 kN-m

Ans.

(b) Equilateral triangle: triangle sides are each 500 mm/3  166.667 mm 1 1 Am  bh  (166.667 mm)(166.667 mm)sin 60  12,028.131 mm 2 2 2 Tmax  2 max Amt  2(75 N/mm 2 )(12,028.131 mm 2 )(3 mm)  5, 412,659 N-mm  5.41 kN-m

Ans.

Tmax

(c) Square: sides of the square are each 500 mm/4  125 mm Am  bh  (125 mm)(125 mm)  15,625 mm 2 Tmax  2 max Amt  2(75 N/mm 2 )(15,625 mm 2 )(3 mm)  7,031, 250 N-mm  7.03 kN-m

Ans.

(d) 150 × 100 mm rectangle: Am  bh  (150 mm)(100 mm)  15,000 mm 2 Tmax  2 max Amt  2(75 N/mm 2 )(15,000 mm 2 )(3 mm)  6,750,000 N-mm  6.75 kN-m

Ans.

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6.113 A torque of T = 150 kip-in. will be applied to the hollow, thin-walled aluminum alloy section shown in Fig. P6.113. If the maximum shear stress must be limited to 10 ksi, determine the minimum thickness required for the section. (Note: The dimensions shown are measured to the wall centerline.) Fig. P6.113

Solution The maximum shear stress for a thin-walled section is given by Eq. (6.25) T  max  2 Am t For the aluminum alloy section, Am  (6 in.)(8 in.) 



4

(6 in.)2  76.274 in.2

The minimum thickness required for the section if the maximum shear stress must be limited to 10 ksi is thus: T 150 kip-in. tmin    0.0983 in. Ans. 2 max Am 2(10 ksi)(76.274 in.2 )

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6.114 A torque of T = 2.5 kN-m will be applied to the hollow, thin-walled aluminum alloy section shown in Fig. P6.114. If the maximum shear stress must be limited to 50 MPa, determine the minimum thickness required for the section. (Note: The dimensions shown are measured to the wall centerline.) Fig. P6.114

Solution The maximum shear stress for a thin-walled section is given by Eq. (6.25) T  max  2 Am t For the aluminum alloy section, 1 Am   (100 mm)2  7,853.982 mm 2 4 The minimum thickness required for the section if the maximum shear stress must be limited to 50 MPa is thus: T (2.5 kN-m)(1,000 N/kN)(1,000 mm/m) tmin    3.18 mm Ans. 2 max Am 2(50 N/mm 2 )(7,853.982 mm 2 )

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6.115 A torque of T = 100 kip-in. will be applied to the hollow, thin-walled aluminum alloy section shown in Fig. P6.115. If the section has a uniform thickness of 0.100 in., determine the magnitude of the maximum shear stress developed in the section. (Note: The dimensions shown are measured to the wall centerline.) Fig. P6.115

Solution The maximum shear stress for a thin-walled section is given by Eq. (6.25) T  max  2 Am t For the aluminum alloy section, Am 



8

(10 in.)2  39.2699 in.2

The maximum shear stress developed in the section is: T 100 kip-in.  max    12.73 ksi 2 Amt 2(39.2699 in.2 )(0.100 in.)

Ans.

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6.116 A torque of T = 2.75 kN-m will be applied to the hollow, thin-walled aluminum alloy section shown in Fig. P6.116. If the section has a uniform thickness of 4 mm, determine the magnitude of the maximum shear stress developed in the section. (Note: The dimensions shown are measured to the wall centerline.) Fig. P6.116

Solution The maximum shear stress for a thin-walled section is given by Eq. (6.25) T  max  2 Am t For the aluminum alloy section, Am  (150 mm)(50 mm) 



2

(25 mm) 2  8, 481.748 mm 2

The maximum shear stress developed in the section is: T (2.75 kN-m)(1,000 N/kN)(1,000 mm/m)  max    40.5 MPa 2 Amt 2(8, 481.748 mm 2 )(4 mm)

Ans.

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6.117 A cross section of the leading edge of an airplane wing is shown in Fig. P6.117. The enclosed area is 82 in.2. Sheet thicknesses are shown on the diagram. For an applied torque of T = 100 kip-in., determine the magnitude of the maximum shear stress developed in the section. (Note: The dimensions shown are measured to the wall centerline.)

Fig. P6.117

Solution The maximum shear stress for a thin-walled section is given by Eq. (6.25) T  max  2 Am t The maximum shear stress developed in the section is: T 100 kip-in.  max    12.20 ksi 2 Amt 2(82 in.2 )(0.050 in.)

Ans.

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6.118 A cross section of an airplane fuselage made of aluminum alloy is shown in Fig. P6.118. For an applied torque of T = 1,250 kip-in. and an allowable shear stress of  = 7.5 ksi, determine the minimum thickness of the sheet (which must be constant for the entire periphery) required to resist the torque. (Note: The dimensions shown are measured to the wall centerline.)

Fig. P6.118

Solution The maximum shear stress for a thin-walled section is given by Eq. (6.25) T  max  2 Am t For the fuselage, Am  (30 in.)(20 in.)   (15 in.) 2  1,306.858 in.2 The minimum thickness required for the sheet if the maximum shear stress must be limited to 7.5 ksi is thus: T 1, 250 kip-in. tmin    0.0638 in. Ans. 2 max Am 2(7.5 ksi)(1,306.858 in.2 )

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