Explanation Given information: The posts are made of aluminum. The modulus of elasticity of the aluminum ( E ) is 70 GPa . The length of the posts B ( L B ) is 124.7 mm . The length of the posts A and C ( L A and L C ) is 125 mm . The force is applied to the posts ( P ) is 100 kip . The cross-sectional of each post ( A ) is 400 mm 2 . The uniformly distributed load act at top and bottom of the beam ( w ) is 800 kN/m . Calculation: Convert the uniformly distributed load ( w ) to point load ( P ): P = 800 kN/m × ( 200 mm × 10 − 3 m 1 m ) = 160 kN Show the free-body diagram of joint A as in Figure 1. Refer to Figure 1. Here, F A is force of the post A , F B is force of the post B , and F C is force of the post C . Calculate the moment about point B by applying the equation of equilibrium: Sum of moments about point B is equal to 0. ∑ M B = 0 − ( F A × 100 ) + ( F C × 100 ) = 0 − 100 F A + 100 F C = 0 d ( − F A + F C ) = 0 − F A + F C = 0 F A = F C Therefore, F = F A = F C Calculate the vertical forces by applying the equation of equilibrium: Sum of vertical forces is equal to 0. ∑ F y = 0 2 F + F B − 160 = 0 2 F + F B = 160 (1) Show the geometry of the deformation diagram of the postsas in Figure 2. Refer Figure 2. Here, δ A is deformation at A and δ B is deformation at B . Calculate the forces of the posts ( F and F B ) using the deformation: δ A = δ B + 0.3 F L A A E = F B L B A E + 0.3 (2) Substitute 125 mm for L A , 124.7 mm for L B , 400 mm 2 for A , and 70 GPa for E in Equation (2). F × 125 400 × ( 70 GPa × 10 3 N/mm 2 1 GPa ) = F B × 124.7 400 × ( 70 GPa × 10 3 N/mm 2 1 GPa ) + 0.3 4.46428 × 10 − 6 F = 4.45357 × 10 − 6 F B + 0

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Chapter 4.5, Problem 4.64P

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Theaverage normal stress in each post (σA, σB, and σC) .

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Chapter 4.5, Problem 4.63P

Chapter 4.5, Problem 4.65P

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Theaverage normal stress in each post ( σ A , σ B , and σ C ) .

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