Question

2A. A load of 85 kN is applied to the system of linkages at point B. All linkage components are made of Aluminum (E 70 GPa) Each linkage member has a cross- section as shown below. The pin at D is in single shear, and has a diameter of 40.0 mm. (a). Find the average shear stress in (b). Find the average normal stress (c) Find the extension/contraction (d) Find the average normal stress pin D at the midpoint of section CD of section CD at the midpoint of section CB (clear FBDs of all individual links are encouraged to solve for part d) 85 kN 10 mm 50 mm Linkage Cross-Sections

Answer 1

using method of joints analyzing joint 'B' F_BC cos 45 = F_AB F_BC sin 45 = 85 F_BC = 85 squareroot 2 F_AB = 85 squareroot 2 middot cos 45 F_AB = 85 KN Analyzing joint 'A' F_AB = F_CD cos 45 85/cos 45 = F_CD F_CA = 85 squareroot 2 Analyzing joint 'c' F_CA cos 45 + F_BC cos 45 = 0 F_CA = -F_BC shoes that F_CA = tensile F_BC = compressive F_CA sin 45 - F_BC sin 45 = F_CD F_CD = 2F_CA sin 45 + F_CA sin 45 F_CD = 2 F_CA sin 45 = 2 times 85 squareroot 2 times 1/squareroot 2 = 170 KN \r\n (a) tau_D = F_CD/A_CD tau_D = 170 times 10^3/pi/4 times (0.04)^2 tau_D = 135.28 MPa (b) sigma_CD = f_CD/A'CD sigma_CD = 170 times 10^3/50 times 10 times 10^-6 sigma_CD = 340 MPa (c) delta_CD = F_CD middot L_CD/A'CD middot E delta_CD = 170 times 10^3 times 2/50 times 10 times 10^-6 times 70 times 10^9 delta_c = 9.71 times 10^-3 m (d) sigma_BC = F_BC/A_BC = 85 squareroot 2/50 times 10 times 10^-6 = 240.380 MPa ACD = pi/4 d^2 alpha = 40 mm ACD = pi/4 times (0.04)^2 alpha = 0.040m