Mechanics of Materials Solutions Chapter06 Probs67 78

6.67 A hollow circular cold-rolled bronze [G1 = 6,500 ksi] tube (1) with an outside diameter of 2.00 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. 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. P6.67

Solution Section Properties: For tube (1) and core (2), the polar moments of inertia are:

I p1 = I p2 =

π

⎡⎣ (2.00 in.) 4 − (1.25 in.) 4 ⎤⎦ = 1.331112 in.4 32

π 32

(1.25 in.) 4 = 0.239684 in.4

Equilibrium: ΣM x = T1 + T2 − 20 kip-in. = 0

(a)

Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(b)

Geometry of Deformation Relationship: Since the tube and core are securely bonded together, the angles of twist in both members must be equal; therefore, φ1 = φ2 (c) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL = 2 2 G1 I p1 G2 I p 2

(d)

Solve the Equations: Solve Eq. (d) for T1: L2 G1 I p1 ⎛ 6,500 ksi ⎞ ⎛ 1.331112 in.4 ⎞ = T2 ⎜ = 2.887878 T2 T1 = T2 ⎟⎜ 4 ⎟ L1 G2 I p 2 ⎝ 12,500 ksi ⎠ ⎝ 0.239684 in. ⎠ and substitute this result into Eq. (a) to compute the torque T2 in core (2): T1 + T2 = 2.887878 T2 + T2 = 3.887878 T2 = 20 kip-in. ∴T2 = 5.144 kip-in. = 5.14 kip-in.

The torque in tube (1) is therefore: T1 = 20 kip-in. − T2 = 20 kip-in. − 5.144 kip-in. = 14.856 kip-in. = 14.86 kip-in.

Ans. Ans.

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(b) Maximum Shear Stress: The maximum shear stress in tube (1) is: T R (14.856 kip-in.)(2.00 in./ 2) τ1 = 1 1 = = 11.16 ksi I p1 1.331112 in.4 The maximum shear stress in core (2) is: TR (5.144 kip-in.)(1.25 in./ 2) τ2 = 2 2 = = 13.41 ksi I p2 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. (c)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. TL (14.856 kip-in.)(10 in.) Ans. φ1 = 1 1 = = 0.017170 rad = 0.01717 rad G1 I p1 (6,500 ksi)(1.331112 in.4 )

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6.68 A hollow circular cold-rolled bronze [G1 = 6,500 ksi] tube (1) with an outside diameter of 2.00 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. 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. P6.68

Solution Section Properties: For tube (1) and core (2), the polar moments of inertia are:

I p1 = I p2 =

π

⎡⎣ (2.00 in.) 4 − (1.25 in.) 4 ⎤⎦ = 1.331112 in.4 32

π 32

(1.25 in.) 4 = 0.239684 in.4

Equilibrium: ΣM x = T1 + T2 − T = 0

(a)

Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(b)

Geometry of Deformation Relationship: Since the tube and core are securely bonded together, the angles of twist in both members must be equal; therefore, φ1 = φ2 (c) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL = 2 2 G1 I p1 G2 I p 2

(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: TR T τ ∴ = τ= Ip Ip R which allows Eq. (d) to be rewritten as: τ 1 L1 τ 2 L2 = (e) G1 R1 G2 R2 Solve Eq. (e) for τ 1: L G R ⎛ 6,500 ksi ⎞ ⎛ (2.00 in./2) ⎞ τ1 = τ 2 2 1 1 = τ 2 ⎜ (f) ⎟ = 0.832τ 2 ⎟⎜ L1 G2 R2 ⎝ 12,500 ksi ⎠ ⎝ (1.25 in./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.

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: τ 1 = 0.832(60 ksi) = 49.92 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 τ2 = = 32.4519 ksi ≤ 60 ksi O.K. 0.832 Now that the maximum shear stresses in the tube and the core are known, the torques in each component can be computed: τ 1 I p1 (27 ksi)(1.331112 in.4 ) T1 = = = 35.940 kip-in. R1 ( 2.00 in./2 ) T2 =

τ 2 I p2 R2

=

(32.4519 ksi)(0.239684 in.4 ) = 12.445 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 = 35.940 kip-in. + 12.445 kip-in. = 48.385 kip-in.= 48.4 kip-in.

Ans.

(b) Torques in each component: As computed previously, the torque in tube (1) is: T1 = 35.9 kip-in.

Ans.

and the torque in core (2) is: T2 = 12.45 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. (c)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. TL (35.940 kip-in.)(10 in.) Ans. φ1 = 1 1 = = 0.041538 rad = 0.0415 rad G1 I p1 (6,500 ksi)(1.331112 in.4 )

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6.69 A composite assembly consisting of an inner steel [G = 80 GPa] shaft (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 = 1,500 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:

I p1 = I p2 =

π

⎡⎣ (50 mm) 4 − (40 mm) 4 ⎤⎦ = 362, 265 mm 4 32

π 32

(20 mm) 4 = 15, 708.0 mm 4

Equilibrium: ΣM x = T1 + T2 − 1,500 N-m = 0

(a)

Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(b)

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, φ1 = φ2 (c) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL = 2 2 G1 I p1 G2 I p 2

(d)

Solve the Equations: Solve Eq. (d) for T1: 4 L G I p1 ⎛ 28 GPa ⎞ ⎛ 362, 265 mm ⎞ T1 = T2 2 1 = T2 ⎜ = 8.071858 T2 ⎟⎜ 4 ⎟ L1 G2 I p 2 ⎝ 80 GPa ⎠ ⎝ 15, 708.0 mm ⎠ and substitute this result into Eq. (a) to compute the torque T2 in steel core (2): T1 + T2 = 8.071858 T2 + T2 = 9.071858 T2 = 1,500 N-m

∴T2 = 165.346 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 = 1,500 N-m − T2 = 1,500 N-m − 165.346 N-m = 1,334.654 N-m (a) Maximum Shear Stress: The maximum shear stress in aluminum tube (1) is: T R (1,334.654 N-m)(50 mm / 2)(1, 000 mm/m) τ1 = 1 1 = = 92.1 MPa I p1 362, 265 mm 4 The maximum shear stress in steel core (2) is: TR (165.346 N-m)(20 mm / 2)(1, 000 mm/m) τ2 = 2 2 = = 105.3 MPa I p2 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. (c)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. TL (1,334.654 N-m)(450 mm)(1, 000 mm/m) Ans. φ1 = 1 1 = = 0.059210 rad = 0.0592 rad G1 I p1 (28, 000 N/mm 2 )(362,265 mm 4 )

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6.70 A composite assembly consisting of an inner steel [G = 80 GPa] shaft (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 120 MPa, and the allowable shear stress of steel core (2) is 165 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:

I p1 = I p2 =

π

⎡ (50 mm) 4 − (40 mm) 4 ⎤⎦ = 362, 265 mm 4 32 ⎣

π 32

(20 mm) 4 = 15, 708.0 mm 4

Equilibrium: ΣM x = T1 + T2 − T = 0

(a)

Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(b)

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, φ1 = φ2 (c) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL = 2 2 G1 I p1 G2 I p 2

(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: TR T τ ∴ = τ= Ip Ip R 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.

τ 1 L1

=

τ 2 L2

(e) G1 R1 G2 R2 Solve Eq. (e) for τ 1: L G R ⎛ 28 GPa ⎞ ⎛ (50 mm/2) ⎞ τ1 = τ 2 2 1 1 = τ 2 ⎜ (f) ⎟ = 0.875τ 2 ⎟⎜ L1 G2 R2 ⎝ 80 GPa ⎠ ⎝ (20 mm/2) ⎠ Assume the core controls: If the shear stress in core (2) reaches its allowable value of 165 MPa (in other words, if the core controls), then the shear stress in tube (1) will be: τ 1 = 0.875(165 MPa) = 144.375 MPa ≥ 120 MPa N.G. which is larger than the 120 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 120 MPa allowable value: 120 MPa τ2 = = 137.1429 MPa ≤ 165 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: τ 1 I p1 (120 N/mm 2 )(362,265 mm 4 ) T1 = = = 1, 738,872 N-mm R1 ( 50 mm/2 ) T2 =

τ 2 I p2 R2

(137.1429 MPa)(15,708.0 mm 4 ) = = 215, 424 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, 738,872 N-mm + 215, 424 N-mm = 1,954, 286 N-mm= 1,954 N-m

Ans.

(b) Torques in each component: As computed previously, the torque in tube (1) is: T1 = 1, 739 N-m

Ans.

and the torque in core (2) is: T2 = 215 N-m

Ans.

(c) Angle of Twist: Since both the tube and the core twist exactly the same amount [i.e., Eq. (c)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. TL (1, 738,872 N-mm)(450 mm) Ans. φ1 = 1 1 = = 0.077143 rad = 0.0771 rad G1 I p1 (28, 000 N/mm 2 )(362,265 mm 4 )

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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 composite shaft is subjected to a torque of T = 750 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. P6.71

Solution Section Properties: For bronze sleeve (1) and steel core (2), the polar moments of inertia are: I p1 = I p2 =

π

⎡⎣ (35 mm) 4 − (25 mm) 4 ⎤⎦ = 108,974 mm 4 32

π 32

(25 mm) 4 = 38,349.5 mm 4

Equilibrium: ΣM x = T1 + T2 − 750 N-m = 0

(a)

Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(b)

Geometry of Deformation Relationship: Since the sleeve and core are securely bonded together, the angles of twist in both members must be equal; therefore, φ1 = φ2 (c) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL = 2 2 G1 I p1 G2 I p 2

(d)

Solve the Equations: Solve Eq. (d) for T1: 4 L2 G1 I p1 ⎛ 45 GPa ⎞ ⎛ 108,974 mm ⎞ T1 = T2 = T2 ⎜ = 1.598401T2 ⎟⎜ 4 ⎟ L1 G2 I p 2 ⎝ 80 GPa ⎠ ⎝ 38,349.5 mm ⎠ and substitute this result into Eq. (a) to compute the torque T2 in steel core (2): T1 + T2 = 1.598401T2 + T2 = 2.598401T2 = 750 N-m ∴T2 = 288.639 N-m The torque in bronze sleeve (1) is therefore: T1 = 750 N-m − T2 = 750 N-m − 288.639 N-m = 461.361 N-m

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(a) Maximum Shear Stress: The maximum shear stress in bronze sleeve (1) is: T R (451.351 N-m)(35 mm / 2)(1, 000 mm/m) τ1 = 1 1 = = 74.1 MPa I p1 108,974 mm 4 The maximum shear stress in steel core (2) is: TR (288.639 N-m)(25 mm / 2)(1, 000 mm/m) τ2 = 2 2 = = 94.1 MPa I p2 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. (c)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. TL (461.361 N-m)(360 mm)(1, 000 mm/m) Ans. φ1 = 1 1 = = 0.033869 rad = 0.0339 rad G1 I p1 (45, 000 N/mm 2 )(108,974 mm 4 )

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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 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. P6.72

Solution Section Properties: For bronze sleeve (1) and steel core (2), the polar moments of inertia are: I p1 = I p2 =

π

⎡⎣ (35 mm) 4 − (25 mm) 4 ⎤⎦ = 108,974 mm 4 32

π 32

(25 mm) 4 = 38,349.5 mm 4

Equilibrium: ΣM x = T1 + T2 − T = 0

(a)

Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(b)

Geometry of Deformation Relationship: Since the sleeve and core are securely bonded together, the angles of twist in both members must be equal; therefore, φ1 = φ2 (c) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL = 2 2 G1 I p1 G2 I p 2

(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: TR T τ ∴ = τ= Ip Ip R which allows Eq. (d) to be rewritten as:

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τ 1 L1

=

τ 2 L2

(e)

G1 R1 G2 R2 Solve Eq. (e) for τ 1: L G R ⎛ 45 GPa ⎞ ⎛ (35 mm/2) ⎞ τ1 = τ 2 2 1 1 = τ 2 ⎜ ⎟ = 0.7875τ 2 ⎟⎜ L1 G2 R2 ⎝ 80 GPa ⎠ ⎝ (25 mm/2) ⎠

(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: τ 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: τ 1 I p1 (118.125 N/mm 2 )(108,974 mm 4 ) = = 735,574 N-mm T1 = R1 ( 35 mm/2 ) T2 =

τ 2 I p2 R2

=

(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. (c)], either torque-twist relationship can be used to compute the angle of twist of the entire assembly. TL (735,574 N-mm)(360 mm) Ans. φ1 = 1 1 = = 0.054000 rad = 0.0540 rad G1 I p1 (45, 000 N/mm 2 )(108,974 mm 4 )

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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: I p1 = I p2 =

π

⎡⎣(168 mm) 4 − (154 mm) 4 ⎤⎦ = 22,987,183 mm 4 32

π

⎡(114 mm) 4 − (102 mm) 4 ⎦⎤ = 5,954,575 mm 4 ⎣ 32

Equilibrium: ΣM x = −T1 + T2 + 20 kN-m = 0

(a)

Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2 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: φ1 + φ2 = 0 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL + 2 2 =0 G1 I p1 G2 I p 2

(b)

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1: L G I T1 = −T2 2 1 p1 L1 G2 I p 2 4 ⎛ 3 m ⎞ ⎛ 80 GPa ⎞ ⎛ 22,987,183 mm ⎞ = −T2 ⎜ ⎟⎜ ⎟⎜ 4 ⎟ ⎝ 3 m ⎠ ⎝ 80 GPa ⎠ ⎝ 5,954,575 mm ⎠ ⎛ 22,987,183 mm 4 ⎞ = −T2 ⎜ = −3.860424 T2 4 ⎟ ⎝ 5,954,575 mm ⎠

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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: T1 R1 (15.8851 kN-m)(168 mm / 2)(1, 000) 2 τ1 = = = 58.0 MPa I p1 22,987,183 mm 4 The maximum shear stress magnitude in member (2) is: TR (4.1149 kN-m)(114 mm / 2)(1, 000) 2 τ2 = 2 2 = = 39.4 MPa I p2 5,954,575 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, φ1 = φB − φ A Since joint A is restrained from rotating, φA = 0 and thus φ1 = φB The rotation angle at B can be determined by computing the angle of twist in member (1): TL φ1 = 1 1 G1 I p1 =

(15.8851 kN-m)(3,000 mm)(1, 000 mm/m)(1,000 N/kN) (80, 000 N/mm 2 )(22,987,183 mm 4 )

= 0.025914 rad = 0.0259 rad

Ans.

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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 20 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: I p1 = I p2 =

π

⎡⎣ (8.625 in.) 4 − (7.981 in.) 4 ⎤⎦ = 144.978 in.4 32

π

⎡(6.625 in.) 4 − (6.065 in.) 4 ⎦⎤ = 56.284 in.4 ⎣ 32

Equilibrium: ΣM x = −T1 + T2 + 20 kip-ft = 0

(a)

Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(b)

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: φ1 + φ2 = 0 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL + 2 2 =0 G1 I p1 G2 I p 2

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1: L G I T1 = −T2 2 1 p1 L1 G2 I p 2 ⎛ 25 ft ⎞ ⎛ 12, 000 ksi ⎞ ⎛ 144.978 in. ⎞ = −T2 ⎜ = −4.293050 T2 ⎟⎜ ⎟⎜ 4 ⎟ ⎝ 15 ft ⎠ ⎝ 12, 000 ksi ⎠ ⎝ 56.284 in. ⎠ and substitute this result into Eq. (a) to compute the torque T2: 4

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−T1 + T2 = − ( −4.293050 T2 ) + T2 = 5.293050 T2 = −20 kip-ft ∴ T2 = −3.778540 kip-ft The torque in member (1) is therefore: T1 = T2 + 20 kip-ft = −3.778540 kip-ft + 20 kip-ft = 16.221460 kip-ft

(a) Maximum Shear Stress: The maximum shear stress magnitude in member (1) is: T R (16.221460 kip-ft)(8.625 in./ 2)(12 in./ft) τ1 = 1 1 = = 5.79 ksi 144.978 in.4 I p1 The maximum shear stress magnitude in member (2) is: TR (3.778540 kip-ft)(6.625 in./ 2)(12 in./ft) τ2 = 2 2 = = 2.67 ksi 56.284 in.4 I p2

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, φ1 = φB − φ A Since joint A is restrained from rotating, φA = 0 and thus φ1 = φB The rotation angle at B can be determined by computing the angle of twist in member (1): TL φ1 = 1 1 G1 I p1 (16.221460 kip-ft)(15 ft)(12 in./ft) 2 = (12, 000 ksi)(144.978 in.4 ) = 0.020140 rad = 0.0201 rad

Ans.

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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 = 6 in., a shear modulus of 5,600 ksi, and an allowable shear stress of 15 ksi. Aluminum segment (2) has a diameter of 0.75 in., a length of L2 = 8 in., a shear modulus of 4,000 ksi, and an allowable shear stress of 12 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: I p1 = I p2 =

π

32

π 32

(1.00 in.) 4 = 0.098175 in.4 (0.75 in.) 4 = 0.031063 in.4

Equilibrium: ΣM x = −T1 + T2 + TB = 0

(a)

Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(b)

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: φ1 + φ2 = 0 (c) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL + 2 2 =0 G1 I p1 G2 I p 2

(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: TR T τ τ= ∴ = Ip Ip R 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.

τ 1 L1

=−

τ 2 L2

G1 R1 G2 R2 Solve Eq. (e) for τ 1: L G R ⎛ 8 in. ⎞ ⎛ 5, 600 ksi ⎞ ⎛ (1.00 in./2) ⎞ τ 1 = −τ 2 2 1 1 = −τ 2 ⎜ ⎟ = −2.488889τ 2 ⎟⎜ ⎟⎜ L1 G2 R2 ⎝ 6 in. ⎠ ⎝ 4,000 ksi ⎠ ⎝ (0.75 in./2) ⎠

(e)

(f)

Assume the shaft (2) controls: If the shear stress in shaft (2) reaches its allowable magnitude of 12 ksi, then the shear stress magnitude in shaft (1) will be: τ 1 = −2.488889(12 ksi) = 29.866667 ksi ≥ 15 ksi N.G. which is larger than the 15 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 15 ksi allowable magnitude: 15 ksi τ2 = = 6.026785 ksi ≤ 12 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: τ 1 I p1 (15 ksi)(0.098175 in.4 ) = = 2.94525 kip-in. T1 = R1 (1.00 in./2 )

τ 2 I p2

(6.026785 ksi)(0.031063 in.4 ) = 0.49923 kip-in. R2 ( 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 = −0.49923 kip-in. T2 =

=

(a) Total Torque: From Eq. (a), the total torque acting at flange B must not exceed: TB ,max = T1 − T2 = 2.94525 kip-in. − (−0.49923 kip-in.) = 3.44448 kip-in.= 3.44 kip-in.

Ans.

(b) Torques Magnitudes: As computed previously, the torque magnitude in shaft (1) is: T1 = 2.95 kip-in.

Ans.

and the torque magnitude in shaft (2) is: T2 = 0.499 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, φ1 = φB − φ A Since joint A is restrained from rotating, φA = 0 and thus φ1 = φB The rotation angle at B can be determined by computing the angle of twist in member (1): TL (2.94525 kip-in.)(6 in.) Ans. φ1 = 1 1 = = 0.032143 rad = 0.0321 rad G1 I p1 (5, 600 ksi)(0.098175 in.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.

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: I p1 = I p2 =

π

32

π 32

(18 mm) 4 = 10,306.0 mm 4 (24 mm) 4 = 32,572.0 mm 4

Equilibrium: ΣM x = −T1 + T2 + 270 N-m = 0

(a)

Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2 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: φ1 + φ2 = 0 Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL + 2 2 =0 G1 I p1 G2 I p 2

(b)

(c)

(d)

Solve the Equations: Solve Eq. (d) for T1: L G I T1 = −T2 2 1 p1 L1 G2 I p 2 4 ⎛ 165 mm ⎞ ⎛ 39 GPa ⎞ ⎛ 10,306.0 mm ⎞ = −T2 ⎜ = −0.309434 T2 ⎟⎜ ⎟⎜ 4 ⎟ ⎝ 235 mm ⎠ ⎝ 28 GPa ⎠ ⎝ 32,572.0 mm ⎠ and substitute this result into Eq. (a) to compute the torque T2:

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−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 R (63.8041 N-m)(18 mm / 2)(1, 000 mm/m) τ1 = 1 1 = = 55.7 MPa I p1 10,306.0 mm 4 The maximum shear stress magnitude in member (2) is: TR (206.1959 N-m)(24 mm / 2)(1, 000 mm/m) τ2 = 2 2 = = 76.0 MPa I p2 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, φ1 = φB − φ A Since joint A is restrained from rotating, φA = 0 and thus φ1 = φB The rotation angle at B can be determined by computing the angle of twist in member (1): TL φ1 = 1 1 G1 I p1 =

(63.8041 N-m)(235 mm)(1, 000 mm/m) (39, 000 N/mm 2 )(10,306.0 mm 4 )

= 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: I p1 = I p2 =

π

⎡⎣ (2.250 in.)4 − (1.750 in.) 4 ⎤⎦ = 1.595340 in.4 32

π

⎡(3.500 in.) 4 − (3.062 in.) 4 ⎦⎤ = 6.102156 in.4 ⎣ 32

Equilibrium: ΣM x = −T1 + T2 + 42 kip-in. = 0

(a)

Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(b)

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: φ1 + φ2 = 0 (c) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL + 2 2 =0 G1 I p1 G2 I p 2

(d)

Solve the Equations: Solve Eq. (d) for T1: L G I T1 = −T2 2 1 p1 L1 G2 I p 2 4 ⎛ 20 in. ⎞ ⎛ 12,500 ksi ⎞ ⎛ 1.595340 in. ⎞ = −T2 ⎜ = −0.291784 T2 ⎟⎜ ⎟⎜ 4 ⎟ ⎝ 40 in. ⎠ ⎝ 5, 600 ksi ⎠ ⎝ 6.102156 in. ⎠

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and substitute this result into Eq. (a) to compute the torque T2: −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 R (9.4868 kip-in.)(2.250 in./ 2) τ1 = 1 1 = = 6.69 ksi I p1 1.595340 in.4 The maximum shear stress magnitude in member (2) is: TR (32.5132 kip-in.)(3.500 in./ 2) τ2 = 2 2 = = 9.32 ksi I p2 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, φ1 = φB − φ A Since joint A is restrained from rotating, φA = 0 and thus φ1 = φB The rotation angle at B can be determined by computing the angle of twist in member (1): TL φ1 = 1 1 G1 I p1 =

(9.4868 kip-in.)(40 in.) (12,500 ksi)(1.595340 in.4 )

= 0.019029 rad = 0.01903 rad

Ans.

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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: I p1 = I p2 =

π

⎡ (2.250 in.)4 − (1.750 in.) 4 ⎤⎦ = 1.595340 in.4 32 ⎣

π

⎡(3.500 in.) 4 − (3.062 in.) 4 ⎦⎤ = 6.102156 in.4 32 ⎣

Equilibrium: ΣM x = −T1 + T2 + TB = 0

(a)

Torque-Twist Relationships: TL TL φ1 = 1 1 φ2 = 2 2 G1 I p1 G2 I p 2

(b)

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: φ1 + φ2 = 0 (c) Compatibility Equation: Substitute the torque-twist relationships [Eqs. (b)] into the geometry of deformation relationship [Eq. (c)] to obtain the compatibility equation: T1 L1 TL + 2 2 =0 G1 I p1 G2 I p 2

(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: TR T τ τ= ∴ = Ip Ip R 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.

τ 1 L1

=−

τ 2 L2

(e)

G1 R1 G2 R2 Solve Eq. (e) for τ 1: L G R ⎛ 20 in. ⎞ ⎛ 12,500 ksi ⎞ ⎛ (2.250 in./2) ⎞ τ 1 = −τ 2 2 1 1 = −τ 2 ⎜ ⎟ = −0.717474τ 2 ⎟⎜ ⎟⎜ L1 G2 R2 ⎝ 40 in. ⎠ ⎝ 5,600 ksi ⎠ ⎝ (3.500 in./2) ⎠

(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: τ 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 controls. Now that the maximum shear stress magnitudes in the two shafts are known, the torque magnitudes in each component can be computed: τ 1 I p1 (12.914541 ksi)(1.595340 in.4 ) = = 18.3139 kip-in. T1 = R1 ( 2.250 in./2 )

τ 2 I p2

(18 ksi)(6.102156 in.4 ) = 62.7650 kip-in. R2 ( 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. T2 =

=

(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, φ1 = φB − φ A Since joint A is restrained from rotating, φA = 0 and thus φ1 = φB The rotation angle at B can be determined by computing the angle of twist in member (1): TL (18.3139 kip-in.)(40 in.) Ans. φ1 = 1 1 = = 0.036735 rad = 0.0367 rad G1 I p1 (12,500 ksi)(1.595340 in.4 )

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