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Rigid bar ABC is supported by bronze rod (1) and aluminum rod (2), as shown. A concentrated load Pis applied to the free end of aluminum rod (3). Bronze rod (1) has an elastic modulus of E1 = 15,000 ksi and a diameter of d1 = 0.55 in. Aluminum rod (2) has an elastic modulus of E2 = 10,000 ksi and a diameter of d2 = 0.85in. Aluminum rod (3) has a diameter of d3 = 1.05in. The yield strength of the bronze is 48 ksi and the yield strength of the aluminum is 40 ksi. Assume a = 2.5 ft, b = 1.5 ft, L1 = 6 ft. L2 = 8 ft, and L3= 3 ft. (a) Determine the magnitude of load P that can safely be applied to the structure if a minimum factor of safety of 2.0 is required. (b) Determine the deflection of point D for the load determined in part (a). (c) The pin used at B has an ultimate shear strength of 59 ksi. If a factor of safety of 3.0 is required for this double shear pin connection, determine the minimum pin diameter that can be used at B. Aluminum (2) Bronze (1) L2 L1 a b А B с Aluminum L3 D Р

For a factor of safety of 2.0, calculate the allowable stresses in the bronze and the aluminum rods. Answers: Gallow.Br = Me ksi Callow, Al = IN ksi eTextbook and Media Save for Later Attempts: unlimited Submit Answer Part 3 On a piece of paper, sketch a free body diagram (FBD) cut through rod (3) and also a FBD of the rigid bar. Using these, find a relationship between the force in rod 1 (F1) and the applied load, P, a relationship between the force in rod 2 (F2) and the applied load, P, and also a relationship between the force in rod 3 (F3) and the applied load, P. Answers: P/F1 = i P/F2 = i P/F3 -

Based on the allowable stress for the bronze, calculate the maximum allowable force that can be supported by rod (1), and the corresponding maximum value for the applied load P. Answers: Fmax 1 i kips Pmax, 1 PO i kips e Textbook and Media Save for Later Attempts: unlimited Submit Answer Part 5 Based on the allowable stress for the aluminum, calculate the maximum allowable force that can be supported by rod (2), and the corresponding maximum value for the applied load P. Answers: Fmax 2 = PE i kips Pmax 2 i kips

Part 6 Based on the allowable stress for the aluminum, calculate the maximum allowable force that can be supported by rod (3), and the corresponding maximum value for the applied load P. Answers: Fmax, 3 = PE kips Pmax 3" PE kips e Textbook and Media Save for Later Attempts: unlimited Submit Answer Part 7 What is the magnitude of load P that can safely be applied to the structure for a minimum factor of safety of 2.0? Answer: P = i kips

Determine the deformations in rods (1), (2), and (3) when load Pequals Pmax from Part 7 Answers: 8 = i in. 5) = PE in. 83 = in. e Textbook and Media Save for Later Attempts: unlimited Submit Answer Part 9 Using the results from Part 8, determine the deflections of Joint B and Point D. Answers: VB- i in. VD i . in.

The pin used at Bhas an ultimate shear strength of 59 ksi. If a factor of safety of 3.0 is required, determine the allowable shear stress in this pin. Answer: Tallow i ksi e Textbook and Media Save for Later Attempts: unlimited Submit Answer Part 11 The pin used at B is a double shear pin connection. Determine the minimum pin cross-sectional area, and the corresponding minimum pin diameter that can be used at B. Answer: Amin i in 2 dpin . in.