# Mechanics of Materials Solutions CH05

January 8, 2018 | Author: ArishChoy | Category: Deformation (Mechanics), Stress (Mechanics), Equations, Copyright, Building Engineering

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Mechanics of Materials Probs23-43...

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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: ΣFy = −2 F1 − F2 − 360 kN = 0 Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL e1 = 1 1 e2 = 2 2 (b) A1 E1 A2 E2 Geometry of Deformations Relationship For this configuration, the elongations of both members will be equal; therefore, e1 = e2

(c)

Compatibility Equation Substitute the force-deformation relationships (b) into the geometry of deformation relationship (c) 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 A E ⎤ ΣFy = −2 F1 − F2 = −2 F1 − F1 2 2 = − F1 ⎢ 2 + 2 2 ⎥ = 360 kN A1 E1 A1 E1 ⎦ ⎣ For this structure, the areas and elastic moduli are given below: A1 = (6 mm)(200 mm) = 1, 200 mm 2 A2 = (200 mm) 2 = 40, 000 mm 2

E1 = 200 GPa

(e)

(f)

E2 = 12 GPa

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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: F (−90 kN)(1,000 N/kN) = −75 MPa = 75 MPa (C) σ1 = 1 = A1 1,200 mm 2

σ2 =

F2 (−180 kN)(1,000 N/kN) = = −4.50 MPa = 4.50 MPa (C) A2 40,000 mm 2

Ans.

Ans.

(b) The shortening of the block is equal to the elongation (i.e., contraction in this instance) of member (2): FL (−180, 000 N)(1,200 mm) u B = e2 = 2 2 = = −0.450 mm = 0.450 mm ↓ Ans. 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 2,300 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 280 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: ΣFy = F1 − F2 − 280 kN = 0 (a) Force-Deformation Relationships: FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2 Geometry of Deformations: e1 + e2 = 0

(b)

(c)

Compatibility Equation: Substitute Eqs. (b) into Eq. (c) 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 ⎤ ΣFy = F1 − F2 = F1 − ⎢ − F1 1 2 2 ⎥ = F1 ⎢1 + 1 2 2 ⎥ = 280 kN L2 A1 E1 ⎦ ⎣ ⎣ L2 A1 E1 ⎦

(d)

(e)

(f)

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Notice that A1 = A2 and E1 = E2 for this structure. Therefore, F1 can be computed as: 280 kN 280 kN 280 kN F1 = = = = 178.182 kN L1 A2 E2 1.6 m 1.571429 1+ 1+ 2.8 m L2 A1 E1 and from Eq. (e), F2 has a value of L A E 1.6 m F2 = − F1 1 2 2 = −(178.182 kN) = −101.818 kN L2 A1 E1 2.8 m Normal Stresses: The normal stresses in each axial member can now be calculated: F 178,182 N σ1 = 1 = = 77.5 MPa = 77.5 MPa (T) A1 2,300 mm 2

σ2 =

F2 −101,818 N = = −44.3 MPa = 44.3 MPa (C) A2 2,300 mm 2

Ans.

Ans.

(b) The deflection of flange B is equal to the elongation (i.e., contraction in this instance) of member (2): FL (−101,818 N)(2,800 mm) = −0.620 mm = 0.620 mm ↓ u B = e2 = 2 2 = Ans. A2 E2 (2,300 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 0.50 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 2,300 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 280 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: ΣFy = F1 − F2 − 280 kN = 0 (a) Force-Deformation Relationships: FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2 Geometry of Deformations: e1 + e2 = 0.50 mm

(b)

(c)

Compatibility Equation: Substitute Eqs. (b) into Eq. (c) to derive the compatibility equation: F1 L1 F2 L2 + = 0.50 mm A1 E1 A2 E2 Solve the Equations: Solve Eq. (d) for F2: ⎡ FL ⎤ A E (0.5 mm)A2 E2 L A E F2 = ⎢ 0.5 mm − 1 1 ⎥ 2 2 = − F1 1 2 2 A1 E1 ⎦ L2 L2 L2 A1 E1 ⎣ and substitute this expression into Eq. (a) to determine F1: ⎡ (0.5 mm)A2 E2 L A E ⎤ ΣFy = F1 − F2 = F1 − ⎢ − F1 1 2 2 ⎥ = 280 kN L2 L2 A1 E1 ⎦ ⎣ Simplify this expression, gathering terms with F1 on the left-hand side of the equation:

(d)

(e)

(f)

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⎡ (0.5 mm)A2 E2 L − F1 1 F1 − ⎢ L2 L2 ⎣ ⎡ L F1 ⎢1 + 1 ⎣ L2

A2 E2 ⎤ ⎥ = 280 kN A1 E1 ⎦ A2 E2 ⎤ (0.5 mm)A2 E2 ⎥ = 280 kN + A1 E1 ⎦ L2

Notice that A1 = A2 and E1 = E2 for this structure. Therefore, F1 can be computed as: (0.5 mm)(2,300 mm 2 )(200, 000 N/mm 2 ) 280, 000 N + 362,143 N 2,800 mm F1 = = = 230, 455 N 1,600 mm 1.571429 1+ 2,800 mm and from Eq. (a), F2 has a value of F2 = F1 − 280 kN = 230, 455 N − 280, 000 N = −49,545 N Normal Stresses: The normal stresses in each axial member can now be calculated: F 230, 455 N σ1 = 1 = = 100.2 MPa = 100.2 MPa (T) A1 2,300 mm 2

σ2 =

F2 −49,545 N = = −21.5 MPa = 21.5 MPa (C) A2 2,300 mm 2

(b) The downward deflection of flange B is equal to the elongation of member (1): FL (230, 455 N)(1,600 mm) u B = e1 = 1 1 = = 0.802 mm = 0.802 mm ↓ A1 E1 (2,300 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 0.625 in. and they are symmetrically positioned relative to the center rod (2) and the applied load P. Steel rod (2) has a diameter of 0.375 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 = 14 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: ΣFy = 2 F1 + F2 − P = 0 (a) Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL e1 = 1 1 e2 = 2 2 (b) A1 E1 A2 E2 Geometry of Deformations Relationship For this configuration, the elongations of all rods will be equal; therefore, e1 = e2

(c)

Compatibility Equation Substitute the force-deformation relationships (b) into the geometry of deformation relationship (c) to derive the compatibility equation: F1 L1 F2 L2 = (d) 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 Substitute this expression into equilibrium equation (a) and solve for F1: ⎡ L A E L A E ⎤ ΣFy = 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 = 14 kips, and the lengths, areas, and elastic moduli are given below: L1 = 24 in. L2 = 60 in.

π

(0.625 in.) 2 = 0.30680 in.2

π

(0.375 in.) 2 = 0.11045 in.2 4 4 E1 = 15, 000 ksi E2 = 30, 000 ksi Substitute these values into Eq. (f) and calculate F1 = 6.11888 kips. Backsubstitute into Eq. (e) to calculate F2 = 1.76224 kips. A1 =

A2 =

Normal Stresses The normal stresses in each axial member can now be calculated: F 6.11888 kips σ1 = 1 = = 19.94 ksi A1 0.30680 in.2

σ2 =

F2 1.76224 kips = = 15.96 ksi A2 0.11045 in.2

Ans.

Ans.

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5.27 A steel [E = 30,000 ksi] pipe column with a cross-sectional area of A1 = 5.60 in.2 is connected at flange B to an aluminum alloy [E = 10,000 ksi] pipe with a cross-sectional area of A2 = 4.40 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 steel pipe (1) and aluminum 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: ΣFx = − F1 + F2 − 60 kips = 0 (a) Force-Deformation Relationships: FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2

(b)

Geometry of Deformations: e1 + e2 = 0 Compatibility Equation: Substitute Eqs. (b) into Eq. (c) 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 ⎤ ΣFx = − F1 + F2 = − F1 + ⎢ − F1 1 2 2 ⎥ = − F1 ⎢1 + 1 2 2 ⎥ = 60 kips L2 A1 E1 ⎦ ⎣ ⎣ L2 A1 E1 ⎦ F1 can be computed as: 60 kips 60 kips 60 kips F1 = − = = = −49.2508 kips 2 L1 A2 E2 1.218254 ⎛ ⎞ 120 in. 4.40 in. 10, 000 ksi ⎛ ⎞ ⎛ ⎞ 1+ 1+ ⎜ ⎟⎜ ⎟ 2 ⎟⎜ L2 A1 E1 ⎝ 144 in. ⎠ ⎝ 5.60 in. ⎠ ⎝ 30, 000 ksi ⎠ and from Eq. (a), F2 has a value of F2 = F1 + 60 kips = −49.2508 kips + 60 kips = 10.7492 kips Normal Stresses: The normal stresses in each axial member can now be calculated: F −49.2508 kips σ1 = 1 = = −8.79 ksi = 8.79 ksi (C) A1 5.60 in.2

(c)

(d)

(e)

(f)

Ans.

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σ2 =

F2 10.7492 kips = = 2.44 ksi = 2.44 ksi (T) A2 4.40 in.2

Ans.

(b) The deflection of flange B is equal to the elongation (i.e., contraction in this instance) of member (1): F L (−49.2508 kips)(120 in.) u B = e1 = 1 1 = = −0.0352 in. = 0.0352 in. ← Ans. A1 E1 (5.60 in.2 )(30, 000 ksi)

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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: ΣFy = − F1 − F2 − 670 kN = 0 (a) Force-Deformation Relationships: FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2 Geometry of Deformations: e1 = e2

(b)

(c)

Compatibility Equation: Substitute Eqs. (b) into Eq. (c) 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 ⎤ ΣFy = − F1 − F2 = − F1 − ⎢ F1 1 2 2 ⎥ = − F1 ⎢1 + 1 2 2 ⎥ = 670 kN ⎣ L2 A1 E1 ⎦ ⎣ L2 A1 E1 ⎦

Since L1 = L2, the term L1/L2 = 1 can be eliminated and Eq. (f) simplified to ⎡ A E ⎤ F1 ⎢1 + 2 2 ⎥ = −670 kN ⎣ A1 E1 ⎦

(d)

(e)

(f)

(g)

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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

A2 = (4 bars)

π

(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 = 61,365.885 mm 2 F1 can now be computed as: ⎡ 1,134.115 mm 2 200 GPa ⎤ F1 ⎢1 + ⎥ = −670 kN 2 ⎣ 61,365.885 mm 29 GPa ⎦ ∴ F1 =

−670 kN = −594.258 kN 1.127457

and from Eq. (a), F2 has a value of F2 = − F1 − 670 kN = −(−594.248 kN) − 670 kN = −75.742 kN Normal Stresses: The normal stresses in each axial member can now be calculated. In the concrete, the normal stress is: F −594, 258 N σ1 = 1 = Ans. = −9.68 MPa = 9.68 MPa (C) A1 61,365.885 mm 2 and the normal stress in the reinforcing bars is F −75, 742 N σ2 = 2 = Ans. = −66.8 MPa = 66.8 MPa (C) A2 1,134.115 mm 2

(b) The shortening of the 1.5-m-long pier is equal to the elongation (i.e., contraction in this instance) of member (1) or member (2): FL (−594, 258 N)(1,500 mm) Ans. u = e1 = 1 1 = = −0.501 mm = 0.501 mm ↓ A1 E1 (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: ΣFy = − F1 − F2 − 670 kN = 0 (a) Force-Deformation Relationships: FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2 Geometry of Deformations: e1 = e2

(b)

(c)

Compatibility Equation: Substitute Eqs. (b) into Eq. (c) 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 ⎤ ΣFy = − F1 − F2 = − ⎢ F2 2 1 1 ⎥ − F2 = − F2 ⎢ 2 1 2 + 1⎥ = 670 kN ⎣ L1 A2 E2 ⎦ ⎣ L1 A2 E2 ⎦

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 ⎦

(d)

(e)

(f)

(g)

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From the problem statement, 20% of the total load must be carried by the reinforcing steel; therefore, F2 = 0.20(−670 kN) = −134 kN 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 = 27.586207 A2 ∴ A2 = 2,186.369 mm 2

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

π

4

2 Dbar ≥ 2,186.369 mm 2

∴ Dbar ≥ 26.4 mm

Ans.

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5.30 A load of P = 170 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: ΣFy = 2 F1 + F2 − P = 0 (a) Force-Deformation Relationships: The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL e1 = 1 1 e2 = 2 2 (b) A1 E1 A2 E2 Geometry of Deformations: For this configuration, the deflections of joints A, B, and C are equal: v A = vB = vC (c) The pin connections at A and C are ideal; therefore, the deflection of joints A and C will cause an identical elongation of rods (1): v A = e1 (d) The pin at B has a 3-mm clearance; thus, the elongation of rod (2) is related to rigid bar deflection vB by: vB = e2 + 3 mm (e) Substitute Eqs. (d) and (e) into Eq. (c) to obtain the geometry of deformation equation: e1 = e2 + 3 mm (f) Compatibility Equation: Substitute the force-deformation relationships (b) into the geometry of deformation relationship (f) to derive the compatibility equation: F1 L1 F2 L2 = + 3 mm A1 E1 A2 E2

(g)

Solve the Equations: Solve Eq. (g) for F1: ⎡F L ⎤ AE L A E (3 mm)A1 E1 F1 = ⎢ 2 2 + 3 mm ⎥ 1 1 = F2 2 1 1 + L1 A2 E2 L1 ⎣ A2 E2 ⎦ L1 Substitute this expression into Eq. (a) 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.

⎡ L A E (3 mm)A1 E1 ⎤ ΣFy = 2 F1 + F2 = 2 ⎢ F2 2 1 1 + ⎥ + F2 = P L A E L 1 2 2 1 ⎣ ⎦ and derive an expression for F2: ⎡ L A E ⎤ (3 mm)A1 E1 F2 ⎢ 2 2 1 1 + 1⎥ = P − 2 L1 ⎣ L1 A2 E2 ⎦

(3 mm)A1 E1 L1 ∴ F2 = (h) L A E 2 2 1 1 +1 L1 A2 E2 For this structure, P = 170 kN = 170,000 N, and the lengths, areas, and elastic moduli are given below: L1 = 2, 400 mm L2 = 1,800 mm P−2

π

(20 mm) 2 = 314.1593 mm 2

π

(24 mm) 2 = 452.3893 mm 2 4 4 E1 = 100, 000 MPa E2 = 200, 000 MPa Substitute these values into Eq. (h) and calculate F2 = 60,138 N. Backsubstitute into Eq. (a) to calculate F1 = 54,931 N. A1 =

A2 =

Normal Stresses: The normal stresses in each rod can now be calculated: F 54,931 N σ1 = 1 = = 174.9 MPa A1 314.1593 mm 2

σ2 =

F2 60,138 N = = 132.9 MPa A2 452.3893 mm 2

(b) The downward deflection of the rigid bar can be determined from the elongation of rods (1): FL (54,931 N)(2,400 mm) e1 = 1 1 = = 4.20 mm ↓ A1 E1 (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 =

π

⎡⎣(6.625 in.) 2 − (6.0650 in.) 2 ⎤⎦ = 5.5814 in.2 4

D2 = 4.00 in. A2 =

d 2 = 4.00 in. − 2(0.226 in.) = 3.5480 in.

π

⎡⎣(4.00 in.) 2 − (3.5480 in.) 2 ⎤⎦ = 2.6795 in.2 4

Equilibrium: Consider a FBD of flange B. Sum forces in the vertical direction to obtain: ΣFy = F2 − F1 − P = 0 (a) Force-Deformation Relationships: FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2 Geometry of Deformations: e1 + e2 = 0

(b)

(c)

Compatibility Equation: Substitute Eqs. (b) into Eq. (c) 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 σ1 and σ2: σ 1 L1 σ 2 L2 + =0 (e) E1 E2 Solve the Equations: Solve Eq. (e) for σ1: L E σ 1 = −σ 2 2 1 (f) 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 σ2 = 18 ksi. Calculate the corresponding stress in pipe (1): L E 16 ft σ 1 = −σ 2 2 1 = −(18 ksi) = −28.80 ksi 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, σ1 = −18 ksi (negative by inspection). The normal stress in pipe (1) can be found from Eq. (f) L E 10 ft σ 2 = −σ 1 1 2 = −(−18 ksi) = 11.250 ksi (f) L2 E1 16 ft

Now that the stresses are known, the allowable forces F1 and F2 can be computed: F1 = σ 1 A1 = (−18 ksi)(5.5814 in.2 ) = −100.4652 kips F2 = σ 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 elongation of pipe (2): σ L (11.250 ksi)(16 ft)(12 in./ft) u B = e2 = 2 2 = = 0.0720 in. ↓ 30,000 ksi E2

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 100 mm and bronze rod (2) has an outside diameter of 50 mm. The normal stress in the aluminum rod must be limited to 180 MPa, and the normal stress in the bronze rod must be limited to 140 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 =

π

4

(100 mm) 2 = 7,853.982 mm 2

A2 =

π

4

(50 mm) 2 = 1,963.495 mm 2

Equilibrium: Consider a FBD of flange B. Sum forces in the vertical direction to obtain: ΣFy = F2 − F1 − P = 0 (a)

Force-Deformation Relationships: FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2 Geometry of Deformations: e1 + e2 = 0

(b)

(c)

Compatibility Equation: Substitute Eqs. (b) into Eq. (c) 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 σ1 and σ2: σ 1 L1 σ 2 L2 + =0 (e) E1 E2 Solve the Equations: Solve Eq. (e) for σ1: L E σ 1 = −σ 2 2 1 (f) L1 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.

The normal stress in aluminum rod (1) is limited to 180 MPa. We will assume that the aluminum rod controls, and then, we will calculate the corresponding stress in bronze rod (2): L E 500 mm 100 GPa σ 2 = −σ 1 1 2 = −(180 MPa) = −142.9 MPa L2 E1 900 mm 70 GPa This stress magnitude is greater than the 140 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 140 MPa, the normal stress in aluminum rod (1) can be found from Eq. (f) L E 900 mm 70 GPa σ 1 = −σ 2 2 1 = −(140 MPa) = −176.4 MPa (f) L1 E2 500 mm 100 GPa Now that the stresses are known, the allowable forces F1 and F2 can be computed: F1 = σ 1 A1 = (−176.4 MPa)(7,853.982 mm 2 ) = −1,385, 442 N = −1,385.4 kN F2 = σ 2 A2 = (140 MPa)(1,963.495 mm 2 ) = 274,889 N = 274.9 kN

Substitute these values into Eq. (a) to obtain the allowable load P: Pmax = F2 − F1 = 274.9 kN − (−1,385.4 kN) = 1, 660 kN

(b) The deflection of flange B is equal to the elongation of rod (2): σ L (140 MPa)(900 mm) u B = e2 = 2 2 = = 1.260 mm ↓ 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 cross-sectional area of 1.25 in.2. A concentrated load of P = 9 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: ΣM A = (54 in.)F1 + (108 in.)F2 − (132 in.)P = 0 (a) Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2 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 = (c) 54 in. 108 in. Since there are no gaps, clearances, or other misfits at pins B and C, the elongation of member (1) will equal the deflection of the rigid bar at B and the elongation of member (2) will equal the deflection of the rigid bar at C. Therefore, Eq. (c) can be rewritten in terms of the member elongations as: e1 e2 = 54 in. 108 in.

(b)

(d)

Compatibility Equation Substitute the force-deformation relationships (b) into the geometry of deformation relationship (d) 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 1 F2 L2 = 54 in. A1 E1 108 in. A2 E2

(e)

Solve the Equations Solve Eq. (e) for F1: 54 in. L2 A1 E1 54 in. L2 A1 E1 F1 = F2 = F2 108 in. A2 E2 L1 108 in. L1 A2 E2 Substitute this expression into equilibrium equation (a) and solve for F2: ΣM A = (54 in.)F1 + (108 in.)F2 − (132 in.)P = 0

(54 in.)F2

(f)

54 in. L2 A1 E1 + (108 in.)F2 = (132 in.)P 108 in. L1 A2 E2

⎡ ⎤ 1 L2 A1 E1 + (108 in.) ⎥ = (132 in.)P F2 ⎢ (54 in.) 2 L1 A2 E2 ⎣ ⎦

(g)

For this structure, P = 9 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 = 8 kips. Backsubstitute into Eq. (f) to calculate F1 = 6 kips.

Normal Stresses The normal stresses in each axial member can now be calculated: F 6 kips σ1 = 1 = = 4.80 ksi A1 1.25 in.2 F 8 kips σ2 = 2 = = 6.40 ksi A2 1.25 in.2

Ans. Ans.

Deflections of the rigid bar Calculate the elongation of one of the axial members, say member (1): FL (6 kips)(80 in.) e1 = 1 1 = = 0.01280 in. (h) A1 E1 (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 elongation of member (1); therefore, vB = e1 = 0.01280 in. (downward). From similar triangles, the deflection of the rigid bar at D is related to vB by: vB vD = (i) 54 in. 132 in. From Eq. (i), the deflection of the rigid bar at D is: 132 in. 132 in. vD = vB = Ans. (0.01280 in.) = 0.0313 in. ↓ 54 in. 54 in.

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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 cross-sectional 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: ΣM A = (54 in.)F1 + (108 in.)F2 − (132 in.)P = 0 (a) Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2 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 = (c) 54 in. 108 in. Since there are no gaps, clearances, or other misfits at pins B and C, the elongation of member (1) will equal the deflection of the rigid bar at B and the elongation of member (2) will equal the deflection of the rigid bar at C. Therefore, Eq. (c) can be rewritten in terms of the member elongations as: e1 e2 = 54 in. 108 in.

(b)

(d)

Compatibility Equation Substitute the force-deformation relationships (b) into the geometry of deformation relationship (d) 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 1 F2 L2 = 54 in. A1 E1 108 in. A2 E2 Solve the Equations Since allowable stresses are specifed, it is convenient to express Eq. (e) in terms of stress: σ 1 L1 54 in. σ 2 L2 1 σ 2 L2 = = E1 108 in. E2 2 E2 and solve for σ1: L E σ 2 = 2σ 1 1 2 L2 E1 Since both bars have the same elastic modulus, E1/E2 = 1. Substitute σ1 = 18 ksi and solve for σ2: L E 80 in. σ 2 = 2σ 1 1 2 = 2(18 ksi) = 24.0 ksi ≥ 18 ksi N.G. L2 E1 120 in.

(e)

(f)

It is now evident that the stress in bar (2) controls. Substitute σ2 = 18 ksi and solve for σ1: σ L E 18 ksi 120 in. σ1 = 2 2 1 = = 13.50 ksi ≤ 18 ksi OK 2 L1 E2 2 80 in. Now that the stresses are known, the allowable forces F1 and F2 can be computed: F1 = σ 1 A1 = (13.50 ksi)(1.25 in.2 ) = 16.875 kips F2 = σ 2 A2 = (18 ksi)(1.25 in.2 ) = 22.5 kips

Substitute these values into Eq. (a) to obtain the allowable load P: ΣM A = (54 in.)(16.875 kips) + (108 in.)(22.5 kips) − (132 in.)P = 0 ∴ Pmax = 25.3125 kips = 25.3 kips

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 = 2,400 mm2. Bar (2) is a bronze alloy [E = 100 GPa] with a cross-sectional area of A2 = 6,000 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: ΣM D = −(3.5 m)F1 − (1.2 m)F2 + (2.7 m)P = 0 (a) Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL e1 = 1 1 e2 = 2 2 (b) A1 E1 A2 E2 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 = C (c) 3.5 m 1.2 m

Since there are no gaps, clearances, or other misfits at pins A and C, the elongation of member (1) will equal the deflection of the rigid bar at A and the elongation of member (2) will equal the deflection of the rigid bar at C. Therefore, Eq. (c) can be rewritten in terms of the member elongations as: e1 e = 2 3.5 m 1.2 m

(d)

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Compatibility Equation Substitute the force-deformation relationships (b) into the geometry of deformation relationship (d) to derive the compatibility equation: 1 F1 L1 1 F2 L2 (e) = 3.5 m A1 E1 1.2 m A2 E2 Solve the Equations Since allowable stresses are specified, it is convenient to express Eq. (e) in terms of stress: σ 1 L1 3.5 m σ 2 L2 σ L = = 2.9167 2 2 E1 1.2 m E2 E2 and solve for σ1: L E σ 1 = 2.9167σ 2 2 1 L1 E2 Substitute σ2 = 90 MPa and solve for σ1: L E 3.3 m 70 GPa σ 1 = 2.9167σ 2 2 1 = 2.9167(90 MPa) = 209.1 MPa ≥ 70 MPa N.G. L1 E2 2.9 m 100 GPa It is now evident that the stress in bar (1) controls. Substitute σ1 = 70 MPa and solve for σ2: L E 1 1 2.9 m 100 GPa σ2 = σ1 1 2 = = 30.13 MPa ≤ 90 MPa (70 MPa) O.K. 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 = σ 1 A1 = (70 MPa)(2,400 mm 2 ) = 168, 000 N = 168.00 kN F2 = σ 2 A2 = (30.13 MPa)(6,000 mm 2 ) = 180, 780 N = 180.78 kN Substitute these values into Eq. (a) to obtain the allowable load P: ΣM D = −(3.5 m)(168.00 kN) − (1.2 m)(180.78 kN) + (2.7 m)P = 0

∴ Pmax = 298.12 kN = 298 kN (b) Deflections of the rigid bar Calculate the elongation of one of the axial members, say member (1): FL (168, 000 N)(2,900 mm) e1 = 1 1 = = 2.9000 mm A1 E1 (2, 400 mm 2 )(70, 000 MPa) Since there are no gaps at pin A, the rigid bar deflection at A is equal to the elongation of member (1); therefore, vA = e1 = 2.9000 mm (downward). From similar triangles, the deflection of the rigid bar at B is related to vA by: vA v (h) = B 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 = 2,400 mm2. Bar (2) is a bronze alloy [E = 100 GPa] with a cross-sectional area of A2 = 6,000 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 = 800 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: ΣM D = −(3.5 m)F1 − (1.2 m)F2 + (2.7 m)P = 0 (a) Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL e1 = 1 1 e2 = 2 2 (b) A1 E1 A2 E2 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 = C (c) 3.5 m 1.2 m

Since there is no clearance at pin C, the elongation 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 elongation can be expressed: v A = e1 + 3 mm Equation (c) can be rewritten in terms of the member elongations as: e1 + 3 mm e = 2 3.5 m 1.2 m

(d) (e)

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Compatibility Equation Substitute the force-deformation relationships (b) into the geometry of deformation relationship (e) to derive the compatibility equation: F1 L1 FL 3.5 m F2 L2 (f) + 3 mm = = 2.9167 2 2 A1 E1 1.2 m A2 E2 A2 E2 Solve the Equations Solve Eq. (f) for F2: ⎤ A2 E2 ⎡ F1 L1 L1 A2 E2 (3 mm)A2 E2 F2 = + 3 mm ⎥ = F1 + ⎢ 2.9167 L2 ⎣ A1 E1 2.9167 L2 A1 E1 2.9167 L2 ⎦ and substitute into Eq. (a) to solve for F1:

(3.5 m)F1 + (1.2 m)F2 = (2.7 m)(800 kN) ⎡ L1 A2 E2 (3 mm)A2 E2 ⎤ + (3.5 m)F1 + (1.2 m) ⎢ F1 ⎥ = (2.7 m)(800 kN) 2.9167 L2 ⎦ ⎣ 2.9167 L2 A1 E1 L1 A2 E2 (3 mm)A2 E2 = (2.7 m)(800 kN) − (1.2 m) (3.5 m)F1 + (1.2 m)F1 2.9167 L2 A1 E1 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)(800 kN) − (1.2 m) F1 = The value of F1 is thus calculated as:

(3 mm)(6,000 mm 2 )(100, 000 N/mm 2 ) 2.9167(3,300 mm) F1 = 2.9 m 6, 000 mm 2 100 GPa 3.5 m + (1.2 m) 2.9167(3.3 m) 2, 400 mm 2 70 GPa = 403,982 N = 404 kN (2.7 m)(800, 000 N) − (1.2 m)

and backsubstituting into Eq. (a) gives F2: (2.7 m)(800 kN) − (3.5 m)F1 F2 = 1.2 m (2.7 m)(800 kN) − (3.5 m)(403.982 kN) = 1.2 m = 621.719 kN (a) Normal stresses: F 403,982 N σ1 = 1 = = 168.3 MPa (T) A1 2,400 mm 2 F 621, 719 N σ2 = 2 = = 103.6 MPa (T) A2 6,000 mm 2

Ans. Ans.

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(b) Normal strains: σ 168.3 MPa ε1 = 1 = = 2.404 × 10−3 mm/mm = 2, 404 με E1 70,000 MPa σ 103.6 MPa ε2 = 2 = = 1.036 ×10−3 mm/mm = 1, 036 με E2 100,000 MPa

Ans. Ans.

(c) Deflections of the rigid bar Calculate the elongation of member (1): FL (403,982 N)(2,900 mm) e1 = 1 1 = = 6.9735 mm A1 E1 (2, 400 mm 2 )(70, 000 MPa)

The relationship between the rigid bar deflection at A and the elongation of member (1) was expressed in Eq. (d). Therefore, the rigid bar deflection at A is: v A = e1 + 3 mm = 6.9735 mm + 3 mm = 9.97 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 1,500mm-long bars. Bar (1) is steel [E = 200 GPa] with a cross-sectional area of A1 = 510 mm2. Bar (2) is an aluminum alloy [E = 70 GPa] with a crosssectional area of A2 = 1,300 mm2. A concentrated load of P = 22 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: ΣM C = (850 mm)F1 + (425 mm)F2 −(1, 000 mm)P = 0

(a)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL (b) e1 = 1 1 e2 = 2 2 A1 E1 A2 E2 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 = = 850 mm 425 mm 1, 000 mm

(c)

Since there are no gaps or clearances at either pin A or pin B, the elongations of members (1) and (2) will equal the deflections of the rigid bar at A and B, respectively. e1 e2 = (d) 850 mm 425 mm Compatibility Equation Substitute the force-deformation relationships (b) into the geometry of deformation relationship (d) 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.

F1 L1 850 mm F2 L2 FL = = 2.0 2 2 A1 E1 425 mm A2 E2 A2 E2

(e)

Solve the Equations Solve Eq. (e) for F1: L A E F1 = 2.0 F2 2 1 1 L1 A2 E2 and substitute into Eq. (a) to solve for F2: (850 mm)F1 + (425 mm)F2 = (1, 000 mm)P (850 mm)(2.0) F2

(f)

L2 A1 E1 + (425 mm)F2 = (1, 000 mm)P L1 A2 E2

⎡ ⎤ 1,500 mm 510 mm 2 200 GPa F2 ⎢(850 mm)(2.0) + 425 mm ⎥ = (1, 000 mm)(22 kN) 2 1,500 mm 1,300 mm 70 GPa ⎣ ⎦ ∴ F2 = 9.440 kN

Backsubstitute this result into Eq. (f) to compute F1: ⎡ 1,500 mm 510 mm 2 200 GPa ⎤ F1 = ⎢(2.0) (9.440 kN) 1,500 mm 1,300 mm 2 70 GPa ⎥⎦ ⎣ = 21.162 kN

(a) Normal stresses: F 21,162 N σ1 = 1 = = 41.5 MPa (T) A1 510 mm 2 F 9, 440 N σ2 = 2 = = 7.26 MPa (T) A2 1,300 mm 2 (b) Deflections of the rigid bar Calculate the elongation of member (1): FL (21,162 N)(1,500 mm) e1 = 1 1 = = 0.3112 mm A1 E1 (510 mm 2 )(200, 000 MPa) Since the pin at A is assumed to have a perfect connection, vA = e1 = 0.3112 mm. From Eq. (c), 1, 000 mm vD = v A = 1.176471(0.3112 mm) = 0.366 mm ↓ 850 mm

Ans. Ans.

Ans.

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5.38 The pin-connected structure shown in Fig. P5.38 consists of a rigid bar ABCD and two 1,500mm-long bars. Bar (1) is steel [E = 200 GPa] with a cross-sectional area of A1 = 510 mm2. Bar (2) is an aluminum alloy [E = 70 GPa] with a crosssectional area of A2 = 1,300 mm2. All bars are unstressed before the load P is applied; however, there is a 5-mm clearance in the pin connection at A. If a concentrated load of P = 200 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: ΣM C = (850 mm)F1 + (425 mm)F2 −(1, 000 mm)P = 0

(a)

Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL e1 = 1 1 e2 = 2 2 (b) A1 E1 A2 E2 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 = = 850 mm 425 mm 1, 000 mm

(c)

Since there is no clearance at pin B, the elongation of member (2) will equal the deflection of the rigid bar at B. However, there is a 5-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 elongation can be expressed: v A = e1 + 5 mm

(d)

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Equation (c) can be rewritten in terms of the member elongations as: e1 + 5 mm e2 = 850 mm 425 mm

(e)

Compatibility Equation Substitute the force-deformation relationships (b) into the geometry of deformation relationship (e) to derive the compatibility equation: F1 L1 FL 850 mm F2 L2 + 5 mm = = 2.0 2 2 (f) A1 E1 425 mm A2 E2 A2 E2 Solve the Equations Solve Eq. (f) for F2: ⎤ A E ⎡FL L A2 E2 (5 mm)A2 E2 F2 = 2 2 ⎢ 1 1 + 5 mm ⎥ = F1 1 + 2.0 L2 ⎣ A1 E1 2.0 L2 A1 E1 2.0 L2 ⎦ and substitute into Eq. (a) to solve for F1: (850 mm)F1 + (425 mm)F2 = (1, 000 mm)P ⎡ L A2 E2 (5 mm)A2 E2 ⎤ (850 mm)F1 + (425 mm) ⎢ F1 1 + ⎥ = (1, 000 mm)P 2.0 L2 ⎣ 2.0 L2 A1 E1 ⎦ L A2 E2 (5 mm)A2 E2 (850 mm)F1 + (425 mm)F1 1 = (1, 000 mm)P − (425 mm) 2.0 L2 A1 E1 2.0 L2 (5 mm)A2 E2 2.0 L2 L A2 E2 850 mm + (425 mm) 1 2.0 L2 A1 E1

(1, 000 mm)P − (425 mm) F1 =

The value of F1 is thus calculated as: (5 mm)(1,300 mm 2 )(70,000 N/mm 2 ) 2.0(1,500 mm) F1 = 1,500 mm 1,300 mm 2 70,000 N/mm 2 850 mm + (425 mm) 2.0(1,500 mm) 510 mm 2 200,000 N/mm 2 = 130,381 N = 130.381 kN (1,000 mm)(200,000 N) − (425 mm)

and backsubstituting into Eq. (a) gives F2: (1,000 mm)(200 kN) − (850 mm)F1 F2 = 425 mm (1,000 mm)(200 kN) − (850 mm)(130.381 kN) = 425 mm = 209.827 kN (a) Normal stresses: F 130,381 N σ1 = 1 = = 256 MPa (T) A1 510 mm 2

Ans.

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σ2 =

F2 209,827 N = = 161.4 MPa (T) A2 1,300 mm 2

(b) Normal strains: F 130,381 N ε1 = 1 = = 1.278 × 10−3 mm/mm = 1, 278 με A1E1 (510 mm 2 )(200,000 N/mm 2 ) F 209,827 N ε2 = 2 = = 2.306 × 10−3 mm/mm = 2,310 με 2 2 A2 E2 (1,300 mm )(70,000 N/mm ) (c) Deflections of the rigid bar Calculate the elongation of member (2): FL (209,827 N)(1,500 mm) = 3.4587 mm e2 = 2 2 = A2 E2 (1,300 mm 2 )(70, 000 MPa) Since the pin at B is assumed to have a perfect connection, vB = e2 = 3.4587 mm. From Eq. (c), 1, 000 mm vD = vB = 2.35294(3.4587 mm) = 8.14 mm ↓ 425 mm

Ans.

Ans. Ans.

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 = π ⎡⎣(35 mm) 2 − (29 mm)2 ⎤⎦ = 1, 206.372 mm 2 A2 = π (15 mm) 2 = 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: ΣFy = − F1 − F2 − P = 0 (a) Force-Deformation Relationships: FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2 Geometry of Deformations: e2 = e1 + 2 mm

(b)

(c)

Compatibility Equation Substitute the force-deformation relationships (b) into the geometry of deformation relationship (c) 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: σ 2 L2 σ 1 L1 = + 2 mm E2 E1 and solve for σ2:

(e)

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σ 2 = σ1

L1 E2 E + (2 mm) 2 L2 E1 L2

600 mm 70, 000 MPa 70, 000 MPa + (2 mm) 602 mm 100, 000 MPa 602 mm Substitute σ1 = 340 MPa and solve for σ2: 70, 000 MPa N.G. σ 2 = 0.697674(340 MPa) + (2 mm) = 469.8 MPa ≥ 260 MPa 602 mm It is now evident that the stress in post (2) controls. Substitute σ2 = 260 MPa and solve for σ1: 70, 000 MPa ⎤ L2 E1 ⎡ σ 1 = ⎢σ 2 − (2 mm) 602 mm ⎥⎦ L1 E2 ⎣ = σ1

70, 000 MPa ⎤ 602 mm 100, 000 MPa ⎡ = ⎢ 260 MPa − (2 mm) 602 mm ⎥⎦ 600 mm 70, 000 MPa ⎣ OK = 39.333 MPa < 340 MPa

(a) Maximum load: Now that the stresses are known, the allowable forces F1 and F2 can be computed: F1 = σ 1 A1 = (39.333 MPa)(1,206.372 mm 2 ) = 47, 450 N = 47.450 kN F2 = σ 2 A2 = (260 MPa)(706.858 mm 2 ) = 183, 783 N = 183.783 kN 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: ΣFy = − 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: FL (−47, 451 N)(600 mm) e1 = 1 1 = = −0.2360 mm A1 E1 (1,206.372 mm 2 )(100, 000 N/mm 2 ) The displacement of cap B is 2 mm greater:

vB = −0.2360 mm − 2 mm = −2.2360 mm = 2.24 mm ↓

Ans.

(c) Normal strain in tube (1): The strain in bronze tube (1) is: σ −39.333 MPa ε1 = 1 = = −393 με E1 100, 000 MPa

Ans.

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5.40 A 7-m-long aluminum tube (1) is to be connected to a 3-m-long bronze pipe (2) at B. When put in place, however, a gap of 5 mm exists between the two members, as shown in Fig P5-40. Aluminum tube (1) has an elastic modulus of 70 GPa and a cross-sectional 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: ΣFy = F1 − F2 = 0 ∴ F1 = F2 (a) Force-Deformation Relationships: FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2

(b)

Geometry of Deformations: e1 + e2 = 5 mm

(c)

Compatibility Equation: Substitute Eqs. (b) into Eq. (c) to derive the compatibility equation: F1 L1 F2 L2 + = 5 mm A1 E1 A2 E2

(d)

Solve the Equations: Substitute Eq. (a) into Eq. (d) and solve for F1: ⎡ L L ⎤ F1 ⎢ 1 + 2 ⎥ = 5 mm ⎣ A1 E1 A2 E2 ⎦

F1 =

5 mm

7, 000 mm 3,000 mm + 2 2 (2,000 mm )(70, 000 N/mm ) (3, 600 mm 2 )(100, 000 N/mm 2 ) = 85, 714 N

(a) Normal stresses: F 85, 714 N σ1 = 1 = = 42.8570 MPa = 42.9 MPa (T) A1 2,000 mm 2 F 85, 714 N σ2 = 2 = = 23.8094 MPa = 23.8 MPa (T) A2 3,600 mm 2

(f)

Ans. Ans.

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(b) The deflection of flange B is equal to the elongation of bronze pipe (2): FL (85, 714 N)(3,000 mm) e2 = 2 2 = = 0.714 mm ↑ A2 E2 (3,600 mm 2 )(100,000 N/mm 2 )

Ans.

In its final position, flange B is located 3,000.714 mm away from support A.

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5.41 The assembly shown in Fig. P5.41 consists of a steel rod (1) [E1 = 30,000 ksi; A1 = 1.25 in.2], a rigid bearing plate B that is securely fastened to rod (1), and a bronze post (2) [E2 = 15,000 ksi; A2 = 3.75 in.2]. The yield strengths of the steel and bronze are 62 ksi and 75 ksi, respectively. A clearance of 0.08 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 Elongation 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. F L (65 kips)(12 ft)(12 in./ft) e1 = 1 1 = = 0.2496 in. > 0.08 in. gap ∴ contact will occur A1 E1 (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: ΣFy = F1 − F2 − P = 0 (a) Force-Deformation Relationships: FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2 Geometry of Deformations: e1 + e2 = 0.08 in.

(b)

(c)

Compatibility Equation: Substitute Eqs. (b) into Eq. (c) to derive the compatibility equation: F1 L1 F2 L2 + = 0.08 in. A1 E1 A2 E2

(d)

Solve the Equations: Solve Eq. (d) for F1: ⎡ F L ⎤ AE AE L A E F1 = ⎢0.08 in. − 2 2 ⎥ 1 1 = (0.08 in.) 1 1 − F2 2 1 1 A2 E2 ⎦ L1 L1 L1 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.

and substitute the resulting expression into Eq. (a) to determine an expression for F2: AE L A E (0.08 in.) 1 1 − F2 2 1 1 − F2 − 65 kips = 0 L1 L1 A2 E2 ⎡ L A E ⎤ AE F2 ⎢1 + 2 1 1 ⎥ = (0.08 in.) 1 1 − 65 kips L1 ⎣ L1 A2 E2 ⎦ AE (0.08 in.) 1 1 − 65 kips L1 F2 = L A E 1+ 2 1 1 L1 A2 E2 The axial force in post (2) is thus: (1.25 in.2 )(30, 000 ksi) (0.08 in.) − 65 kips (144 in.) F2 = = −39.75 kips 24 in. 1.25 in.2 30, 000 ksi 1+ 144 in. 3.75 in.2 15, 000 ksi and the axial force in rod (1) is: F1 = F2 + P = −39.75 kips + 65 kips = 25.25 kips

(a) Normal stresses: F 25.25 kips σ1 = 1 = = 20.20 ksi = 20.2 ksi (T) A1 1.25 in.2 F −39.75 kips = −10.60 ksi = 10.60 ksi (C) σ2 = 2 = A2 3.75 in.2 (b) Factors of safety: 62 ksi FS1 = = 3.07 20.2 ksi

FS2 =

75 ksi = 7.08 10.60 ksi

(c) Displacement of plate B: The displacement of plate B is equal to the elongation of rod (1). FL (25.25 kips)(144 in.) e1 = 1 1 = = 0.09696 in. ↓ A1 E1 (1.25 in.2 )(30,000 ksi)

Ans. Ans.

Ans.

Ans.

∴ vB = 0.0970 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. A1 =

π

⎡⎣(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. A2 = A3 =

π 4

(2.00 in.) 2 = 3.1416 in.2

Equilibrium: Consider a FBD of flange B. Sum forces in the horizontal direction to obtain: ΣFx = − F1 + F2 − 34 kips = 0 (a)

Consider a FBD of flange C. Sum forces in the horizontal direction to obtain: ΣFx = − F2 + F3 + 26 kips = 0 (b) Force-Deformation Relationships: FL FL FL e1 = 1 1 e2 = 2 2 e3 = 3 3 A1 E1 A2 E2 A3 E3

(c)

Geometry of Deformations: e1 + e2 + e3 = 0 Compatibility Equation: Substitute Eqs. (c) into Eq. (d) to derive the compatibility equation: F1 L1 F2 L2 F3 L3 + + =0 A1 E1 A2 E2 A3 E3 Solve the Equations: Solve Eq. (a) for F1: F1 = F2 − 34 kips

(d)

(e)

(f)

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Solve Eq. (b) for F3: F3 = F2 − 26 kips

(g)

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: ⎡ L L ⎤ (34 kips)L1 (26 kips)L3 L F2 ⎢ 1 + 2 + 3 ⎥ = + A1 E1 A3 E3 ⎣ A1 E1 A2 E2 A3 E3 ⎦ and solve for F2: (34 kips)L1 (26 kips)L3 + A1 E1 A3 E3 F2 = L L1 L + 2 + 3 A1 E1 A2 E2 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: F −17.6773 kips = −7.9325 ksi = 7.93 ksi (C) σ1 = 1 = A1 2.2285 in.2

Ans.

σ2 =

F2 16.3227 kips = = 5.1957 ksi = 5.20 ksi (T) A2 3.1416 in.2

Ans.

σ3 =

F3 −9.6773 kips = = −3.0834 ksi = 3.08 ksi (C) A3 3.1416 in.2

Ans.

(b) Joint deflections: The deflection of flange B is equal to the elongation (i.e., contraction in this instance) of member (1): FL (−17.6773 kips)(48 in.) uB = e1 = 1 1 = = −0.012692 in. = 0.01269 in. ← Ans. A1 E1 (2.2285 in.2 )(30, 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.

The deflection of flange C is equal to the sum of the elongations in members (1) and (2). The elongation of member (2) is: FL (16.3227 kips)(60 in.) e2 = 2 2 = = 0.031174 in. A2 E2 (3.1416 in.2 )(10, 000 ksi) And thus, the deflection of flange C is: uC = e1 + e2 = −0.012692 in. + 0.031174 in. = 0.018482 in. = 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: ΣM A = −(36 in.)F1 + (84 in.)F2 − (98 in.)P = 0 (a) Force-Deformation Relationships The relationship between the internal force and the deformation of an axial member can be stated for members (1) and (2): FL FL e1 = 1 1 e2 = 2 2 A1 E1 A2 E2 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 v = C = D (c) 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 elongation 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. (c) can be rewritten in terms of the member elongations as: e e − 1 = 2 36 in. 84 in.

(b)

(d)

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Compatibility Equation Substitute the force-deformation relationships (b) into the geometry of deformation relationship (d) 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: ΣM A = −(36 in.)F1 + (84 in.)F2 − (98 in.)P = 0

(36 in.)F2

(f)

36 in. L2 A1 E1 + (84 in.)F2 = (98 in.)P 84 in. L1 A2 E2

⎡ ⎤ 3 L2 A1 E1 + (84 in.) ⎥ = (98 in.)P F2 ⎢ (36 in.) 7 L1 A2 E2 ⎣ ⎦

(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: F −13.7241 kips = 10.98 ksi (C) σ1 = 1 = A1 1.25 in.2 F 25.6183 kips σ2 = 2 = = 12.81 ksi (T) A2 2.00 in.2

Ans. Ans.

(b) Deflections of the rigid bar Calculate the elongation of one of the axial members, say member (2): FL (25.6183 kips)(40 in.) e2 = 2 2 = = 0.051237 in. (h) 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 elongation of member (2); therefore, vC = e2 = 0.051237 in. (downward). From similar triangles [Eq. (c)], the deflection of the rigid bar at D is therefore: 98 in. 98 in. (0.051237 in.) = 0.0598 in. ↓ vD = vC = Ans. 84 in. 84 in.

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