Question

Draw FBDs as required Clearly state all assumptions made towards solving the problem 1) In the structure shown, 큭in dia. pin is used at A and jin dia, pins are used at B and D If the thickness of member BD is ^in, the ultimate shearing stress for the pins at A, B&D are 5ksi and the ultimate normal stress for the link BD is 8ksi and the overall factor of safety of 3 is desired. Determine the allowable horizontal force P that can be applied at C Double shear C 5 (in) 30 Single shear 9 4 (in) 12(in)

Answer 1

"BE/4 = 5/12 + 4 therefore BE = 1.25 sigma M_A = 0 p(5 "") - F_BD cos 30 degree (4) - F_BD sin 30 degree (1.25) = 0 therefore F_BD = 1.223 p sigma F_x = 0 A_x - F_BD sin 30 degree + p = 0 therefore A_X = -0.723 p sigma Fy = 0 Rightarrow Ay + F_BD cos 30 degree = 0 therefore Ay = 1.05 p therefore reaction at a = squareroot A_x^2 + A^2_y = 1.282 p stress at pin - A tou = R_A/A 5 ksi/3 = 1.282 p/pi/4 (3/4)^2 Rightarrow p = 0.574^k normal stress in BD sigma = F_BD/A 8 ksi/3 = 1.223 p/1 "" times 1/2 "" Rightarrow p = 1.09 ^k therefore allowable horizontal force p is the smallest of the three values p = 0.268 ^k (or) 268 lb"