(a) Explanation Given information: The modulus of elasticity E is 200 GPa . The yield stress ( σ Y ) is 250 MPa . The dimension a is 0.640 m . Calculation: Sketch the rigid bar ABC as shown in Figure 1. Refer to Figure 1. Take a moment about C , ∑ M C = 0 0.640 ( Q − P B E ) − 2.64 P A D = 0 Sketch the deformation as shown in Figure 2. Refer to Figure 2. The value of δ A = 2.64 θ . Find the δ B using the relation: δ B = a θ (1) Substitute 0.640 m for a in Equation (1). δ B = 0.640 θ Calculate the area ( A ) using the relation: A = 37.5 × 6 = 225 mm 2 ( 1 m 10 3 mm ) 2 = 225 × 10 − 6 m 2 Find the value of P A D using the relation: P A D = E A L A D δ A (2) Here, A is the cross sectional area, δ A is deflection at A , and L A D is length of rod AC . Substitute 200 GPa for E , 225 × 10 − 6 m 2 for A , 1.7 m for L A D in Equation (2). P A D = 200 GPa ( 10 9 Pa 1 GPa ) 225 × 10 − 6 1.7 δ A = 200 × 10 9 × 225 × 10 − 6 1.7 δ A = 26.47 × 10 6 δ A (3) Substitute 2.64 θ for δ A in Equation (3). P A D = 26.47 × 10 6 ( 2.64 θ ) = 69.88 × 10 6 θ Express the normal stress using the Equation as follows: σ A D = P A D A (4) Substitute 26.47 × 10 6 δ A for P A D and 225 × 10 − 6 m 2 for A in Equation (4). σ A D = 69.88 × 10 6 δ A 225 × 10 − 6 = 310.57 × 10 9 θ Find the force in the member BE , P B E using the relation: P B E = E A L B E δ B (5) Substitute 200 GPa for E , 225 × 10 − 6 m 2 for A , 1.0 m for L B E in Equation (5). P B E = 200 GPa ( 10 9 Pa 1 GPa ) ( 225 × 10 − 6 ) 1.0 δ B = 200 × 10 9 × ( 225 × 10 − 6 ) 1.0 δ B = 45 × 10 6 δ B Substitute 0.640 θ for δ B . P B E = 45 × 10 6 ( 0.640 θ ) = 28.80 × 10 6 θ Calculate the normal stress along BE using the relation: σ B E = P B E A (6) Substitute 28 (b) To determine Find the maximum deflection at point B .

7th Edition

ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill Education

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Chapter 2.13, Problem 113P

(a)

To determine

Find the normal stress in each link AD and BE.

(b)

To determine

Find the maximum deflection at point B.

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Chapter 2.13, Problem 112P

Chapter 2.13, Problem 114P

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Chapter 2 Solutions

Mechanics of Materials, 7th Edition

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The...Ch. 2.13 - Fig. P2.95 and P2.96 2.96 (a) For P = 13 kips and...Ch. 2.13 - 2.97 Knowing that the hole has a diameter of 9 mm,...Ch. 2.13 - For P = 100 kN, determine the minimum plate...Ch. 2.13 - Prob. 99P Ch. 2.13 - A centric axial force is applied to the steel bar...Ch. 2.13 - The cylindrical rod AB has a length L = 5 ft and a...Ch. 2.13 - Fig. P2.101 and P.102 2.102 The cylindrical rod AB...Ch. 2.13 - Rod AB is made of a mild steel that is assumed to...Ch. 2.13 - Prob. 104P Ch. 2.13 - Rod ABC consists of two cylindrical portions and...Ch. 2.13 - Prob. 106P Ch. 2.13 - Prob. 107P Ch. 2.13 - Prob. 108P Ch. 2.13 - Each cable has a cross-sectional area of 100 mm2...Ch. 2.13 - Prob. 110P Ch. 2.13 - Two tempered-steel bars, each 316 in. thick, are...Ch. 2.13 - Prob. 112P Ch. 2.13 - Prob. 113P Ch. 2.13 - Prob. 114P Ch. 2.13 - Prob. 115P Ch. 2.13 - Prob. 116P Ch. 2.13 - Prob. 117P Ch. 2.13 - Prob. 118P Ch. 2.13 - Prob. 119P Ch. 2.13 - For the composite bar in Prob. 2.111, determine...Ch. 2.13 - Prob. 121P Ch. 2.13 - Bar AB has a cross-sectional area of 1200 mm2 and...Ch. 2.13 - Bar AB has a cross-sectional area of 1200 mm2 and...Ch. 2 - The uniform wire ABC, of unstretched length 2l, is...Ch. 2 - The aluminum rod ABC (E = 10.1 106 psi), which...Ch. 2 - Two solid cylindrical rods are joined at B and...Ch. 2 - Prob. 127RP Ch. 2 - Prob. 128RP Ch. 2 - Prob. 129RP Ch. 2 - A 4-ft concrete post is reinforced with four steel...Ch. 2 - The steel rods BE and CD each have a 16-mm...Ch. 2 - Prob. 132RP Ch. 2 - Prob. 133RP Ch. 2 - The aluminum test specimen shown is subjected to...Ch. 2 - Prob. 135RP

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Find the normal stress in each link AD and BE .

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Find the maximum deflection at point B.

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Chapter 2 Solutions