Question

Rod AC is made of mild the dimensions shown. A downward force downwards 0.5 mm. Use FEM steel with E 200 GPa, of two cylinders AB and BC with F is applied at B to move B 30-mm aiamete to find 250 mm 1. Force applied at B and the support reactions at A and C using Global Stiffness Equations 50-mmdiame Internal Forces in each of AB and BC using Element stiffness Equations each of AB and BC using Element Stiffness Equations 300 mm

Answer 1

K_1 = A_1 E_1/L_1 = pi/4 times 50^2 times 200 times 10^3/250 implies 540 pi times 10^3 /3 N/mm K_2 = A_2 E_2/L_2 = pi/4 times 50^2 times 200 times 10^3/300 implies 1250/3 pi times 10^3 a) Element shiftness matrix (K)_1 = K_1 matrix (1 1 -1 1) = pi times 10^3/3 matrix (540 -540 -540 540) (K)_2 = r_2 matrix (1 -1 -1 1) = pi/3 times 10^3 matrix (1250 -1250 -1250 150) Global shiftness (K) = (K)_1 + (K)_2 (K) = matrix (540 -540 0 -540 1790 -1250 0 -1250 1250) times pi times 10^3/3 Global displacement vector {v} = {u_1 u_2 u_3} Global load vector {F} = {R_1 F R_2} Global equilibrium equt (k) {v} = {F} pi/3 times 10^3 matrix (540 -540 0 -540 1790 -1250 0 -1250 1250) {u_1 u_2 u_3} = {R_1 F R_3} 1 & 3 are fixed u_1 = u_3 = 0 & u_2 = 0 5 mm \r\n equations (540 u_1 -