(a) Explanation Given information: The magnitude of force Q is 260 kN . The modulus of elasticity E is 200 GPa . The yield stress ( σ Y ) is 250 MPa . Calculation: Refer to problem 2.113. The normal stress in each link and the deflection of point B after loading as follows: σ A D = 250 × 10 6 Pa σ B E = 124.3 × 10 6 Pa δ B = 621.53 × 10 − 6 m Refer to problem P2.113 applies to the loading. Q ′ = 317.06 × 10 6 θ ′ Find the slope ( θ ′ ) using the relation: θ ′ = Q ′ 317.06 × 10 6 (1) Substitute 260 kN for Q ′ in Equation (1). θ ′ = 260 kN ( 10 3 N 1 kN ) 317.06 × 10 6 = 260 × 10 3 317.06 × 10 6 = 820.03 × 10 − 6 Find the stress along AD ( σ ′ A D ) using the relation: σ ′ A D = 310.6 × 10 9 θ ′ (2) Substitute 820.03 × 10 − 6 for θ ′ in Equation (2). σ ′ A D = 310.6 × 10 9 ( 820.03 × 10 − 6 ) = 254.70 × 10 6 Pa Find the stress along BE ( σ ′ B E ) using the relation: σ ′ B E = 128 × 10 9 θ ′ (3) Substitute 820.03 × 10 − 6 for θ ′ in Equation (3). σ ′ B E = 128 × 10 9 ( 820.03 × 10 − 6 ) = 104 (b) To determine Find the deflection at point B .

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Chapter 2.13, Problem 115P

(a)

To determine

Find the residual stress in link AD and BE.

(b)

To determine

Find the deflection at point B.

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Chapter 2.13, Problem 114P

Chapter 2.13, Problem 116P

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Find the residual stress in link AD and BE .

Solution Summary: The author calculates the residual stress in link AD and BE. The magnitude of force Q is 260kN.

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Find the deflection at point B.

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Chapter 2 Solutions