(a) Explanation Given information: The size of the uniform rectangular cross section is 10 × 40 mm . The length ( L B E ) of link BE and ( L C F ) C F is 300 mm . The modulus of elasticity ( E ) of aluminium is 70 GPa . Calculation: Sketch the free body diagram of EFG as shown in Figure 1. Refer to Figure 1. Calculate the force along BE using the relation: Take a moment about F . ∑ M F = 0 − ( 400 × 2 F B E ) − ( 250 × 24 ) = 0 − 800 F B E = 6000 F B E = − 7.5 × 10 3 N Calculate the force along CF using the relation: Take a moment about E . ∑ M E = 0 ( 400 × 2 F C F ) − ( 650 × 24 ) = 0 − 800 F C F = 15 , 600 F C F = − 19.5 × 10 3 N Calculate the area of one link using the relation as follows: A = b t (1) Here, b is width of the cross section and t is the thickness of the cross section. Substitute 40 mm for b and 10 mm for t in Equation (1). A = 40 × 10 = 400 mm 2 ( 1 m 10 3 mm ) 2 = 400 × 10 − 6 m 2 Determine the deformation for B E using the relation: δ B E = F B E L E A (2) Substitute − 7.5 × 10 3 N for F B E , 300 mm for L , 70 GPa for E , and 400 × 10 − 6 m 2 for A in Equation (2). δ B E = − 7.5 × 10 3 × 300 mm ( 1 m 10 3 mm ) 70 GPa ( 10 9 Pa 1 GPa ) × 400 × 10 − 6 = − 7 (b) To determine Find the deflection at point F . (c) To determine Find the deflection at point G .

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ISBN: 9780100257061

Author: BEER

Publisher: YUZU

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Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial …

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Chapter 2, Problem 129RP

(a)

To determine

Find the deflection at point E.

(b)

To determine

Find the deflection at point F.

(c)

To determine

Find the deflection at point G.

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Chapter 2, Problem 128RP

Chapter 2, Problem 130RP

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Chapter 2 Solutions

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Find the deflection at point E .

Solution Summary: The author calculates the force along CF using the relation: Take a moment about F.

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