Question

Using the method of joints, determine the force in each member of the truss shown. The load P = 3.6 kN. (Round the final answers to two decimal places.) 0.75 m. 0.4 m 1.4 m The force in member AC (FAc) is The force in member BC (Fc) is3.2 KN. (Compression) The force in member AB (FAB) is KN. (Tension) kN. (Tension)

Answer 1

Calculate angle made by the member AB with the horizontal theta = tan^-1 90.4/0.75) = 28.07 degree calculate angle made by the member BC with the horizontal phi = tan^-1 (1/0.75) = 53.13 degree consider the joint B consider equilibrium of forces sigma F_x = 0 F_AB cos theta + F_BC cos phi = 0 F_AB cos 28.07 + F_BC cos 53.13 = 0 F_AB = -0.6738 F_BC sigma F_y = 0 F_AB sin theta - F_BC sin phi - p = 0 -0.6738 (sin 28.07 - F_BC sin 53.13 - 3.6 = 0 F_BC = -3.22 KN F_AB = -0.6738 F_BC Rightarrow - 0.6738 times (-3.22) F_AB = 2.17 KN sigma F_y = 0 F_AC + F_BC sin phi = 0 F_AC - 3.22 sin 53.13 = 0 F_AC = 2.58 KN