(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 .

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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

EBK MECHANICS OF MATERIALS

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