In the truss shown, members AC and AD consist of rods made of the same metal all
ID: 1826033 • Letter: I
Question
In the truss shown, members AC and AD consist of rods made of the same metal alloy. Knowing that AC is of 1-in. diameter and that the ultimate load for that rod is 75 kips, determine (a) the factor of safety for AC, (6) the required diameter of AD if It is desired that the both rods have the same factor of safety. The free body diagram of the member is shown below The free body diagram at joint C is shown below Consider the equilibrium condition at joint C Sum of all vertical forces is zero V = 0 -(FAC times sin theta)+10 = 0 -[FAD times 5/11. 18]+10 = 0 FAC = 22. 37 kips Find the stress in the member AC sigma AC = FAC/AAC sigma AC = FAC/pidAC2/4 (1) Here, Stress in the member is sigma Cross-sectional area of member AC is AAC Diameter of the member AC is dAc Substitute 22. 37 kips for FAC, 1 in for dAC in equation (1) sigma AC = 22. 37/pi(1)2/4 = 28. 497 kips/in. 2 Also we have, sigma = F/A = Fall/F. S/pid2/4 (2) Calculate the factor of safety by substituting 75 for Fall, 1 in for d and 28. 497 kips/in. 2 for sigma AC equation(2) 28. 497 = 75/F. S/pi(1)2/4 F. S. = 75/28. 497 times 0. 785 Consider the equilibrium condition at joint D Sum of all vertical forces is zero V = 0 -(FAD times sin theta) + 10 = 0 -(FAD times 5/20. 61) + 10 = 0 FAD = 41. 15kips Find the stress in the member AD sigma AD = FAD/AAD sigma AD = FAD/pidAD2/4 (2) Here, Cross-sectional area of member AD is AAD Diameter of the member AD is dAD Substitute 28. 497kips/in2. for sigma and 41. 15 kips for FAD in equation (2) 28. 497 = 41. 15/pidAD2/4 dAD = 1. 356in. Therefore the diameter of the member AD is 1. 356in.Explanation / Answer
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