Question

The assembly shown below is made of aluminum, which has an elastic modulus of 70 GPa. There are two members with cross-sectional areas of A1 125 mm2 and A2 75 mm2 a) Draw and label the axial force diagram for the system with the sign convention discussed in class b) Calculate the axial deformation of Members (1) and (2) in mm and state whether they are elongating or shortening. c) Evaluate the movement at point A, vA (in mm). Be sure to specify which direction it is moving. [Ans. to Check: VA 37.14 mm] 20 kN 10 m 15 m

Answer 1

Elastic modules (E) = 70 Gpa = 70000 mpa A_1 = 125 mm^2 A_2 = 75 mm^2 free body diagram of member (1) & (2) p_1 = 5 kN = 5000 N (tension) p_2 = 15 kN = -15000 N (compressive) L_1 = 10 m = 10,000 mm L_2 = 15 m = 15000 mm delta L_1 = p_1 L_1/A_1 E = 5000 times 10,000/125 times 70000 = 5.7 mm delta L_2 = p_2 L_2/A_2 E = -15000 times 15000/75 times 70000 = -42.8 mm \r\n delta L_1 = 5.7 mm (Elongation) delta L_2 = -42.8 mm (shortening) (c) movement of point (A) = v_A v_A = delta L_1 + delta L_2 = 42.8 - 5.7 = 37 mm