Explanation Given information: The rods are made from A992 steel. The modulus of elasticity of the steel is E . The diameter of each rod is d . The force is applied to the rods ( P ) is 100 kip . The allowable normal stress of the steel ( σ allow ) is 24 ksi . The length of the rods ( L ) is 2 ft . Calculation: Show the free-body diagram of joint A as in Figure 1. Refer Figure 1. Here, F A B is force of the rod AB , F A C is force of the rod AC , and F A D is force of the rod AD . Calculate the horizontal forces by applying the equation of equilibrium: Sum of horizontal forces is equal to 0. ∑ F x = 0 ( F A B × sin 30 ° ) − ( F A C × sin 30 ° ) = 0 sin 30 ° ( F A B − F A B ) = 0 F A B − F A B = 0 F A B = F A B Therefore, F = F A B = F A B Calculate the vertical forces by applying the equation of equilibrium: Sum of vertical forces is equal to 0. ∑ F y = 0 2 F × cos 30 ° + F A D − 100 = 0 1.732 F + F A D = 100 (1) Show the geometry of the deformation diagram of the rodsas in Figure 2. Refer to Figure 2. Here, δ F A D is deformation due to F A D and δ F is deformation due to force. Due to symmetry, the joint A will displace vertically. Calculate the forces of the rods ( F and F A D ) using the deformation: δ F = δ F A D cos 30 ° F L A E = F A D ( L cos 30 ° ) A E × cos 30 ° F = F A D ( cos 30 ° ) 2 F = 0

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Chapter 4.5, Problem 4.63P

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The required diameter to the nearest 18 in. of each rod (D) .

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Chapter 4.5, Problem 4.62P

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The required diameter to the nearest 1 8 in . of each rod ( D ) .

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The required diameter to the nearest 18 in. of each rod (D) .

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