Question

For the simple truss structure under the given load as shown below, determine the followings 1.5 ft 400 lb 6 ft 1.5 ft3ft 1.5 ft 6 ft 6 ft Support reactions obtained from the entire body Sign convention: negative if to the left or to downward o y component of the reaction at A, A, o x component of the reaction at A, A, o y component of the reaction at G, G,-- ° x component of the reaction at G, G lb lb Internal forces along members lb lb Sign convention: negative if in compression DE lb

Answer 1

Given truss structure under the given load as shown below. Consider free body diagram from the above figure. \r\n Now, Apply equilibrium condition of forces sigma F_y = 0 A_y + G_y = 400 lb A_y = 300 ln uparrow [because G_y = 100 lb uparrow] From the above eqn we consider and get the A_y. sigma M_A = 0 400 times 1.5 = G_y times 6 ft 600 = G_y times 6 G_y = 600/6 = 100 G_y = 100 At node 'A' AB = 302.25 downarrow A_x = AB sin 7 degrees = 302.25 times 0.12186 A_x = 36.83 rightarrow E_1 x = 36.83 leftarrow At node 'G' \r\n G_x = 0.85 G_B + 0.12 G_F G_y = G_B cos 40.83 degrees + GF cos 7 degrees 100 = 0.756 G_B + 0.992 G_F By solving of above equation we get G_B = 44.191 lb tension G_F = 67.531 lb tension At node 'F' FE sin 7