Question

The assembly has the diameters and material makeup indicated. If it fits securely between its fixed supports when the temperature is T_1 = 70 degree F, determine the average normal stress in each material when the temperature reaches T_2 = 110 degree F. The coefficients of thermal expansion are alpha_al = 12.8 Times 10^-6, alpha_BR = 9.6 Times 10^-6 and alpha_st = 9.6 Times 10^-6 1/degree F and Young's moduli are E_al = 10.6 Times 10^3 ksi, E_BR = 15 Times 10^3 ksi and E_st = 38 Times 10^3 ksi The three suspender bars are made of the same material and have equal cross-sectional areas A and Young's modulus E. Determine the average normal stress in each bar if the rigid beam ACE is subjected to the force P. These stresses can be obtained in terms of P and A.

Answer 1

Sigma F_x = 0: F_A = F_B = F 3 delta_A/D = 0 - F(4)(12)/pi (6)^2(10.6) (10^6) + 12.8(10^-6) (110 - 70) (4) (12) = F(6) (12)/pi (4)^2(15)(10^6) + 9.60 (10^-6) (110 - 70) (6) (12) -F(3) (12)/pi (2)^2 (28)(10^6) + 9.60 (10^-6) (110) - 70) (3) (12) = 0 F = 277.69 kip sigma_ai = 277.69/pi(6)^2 = 2.46 ksi sigma_br = 277.69/pi (4)^2 = 5.52 ksi sigma_st = 277.69/pi(23)^2 = 22.1 ksi 4 + sigma M_A = 0: F_CD (d) + F_EF (2d) - P(d/2) = 0 F_CD + 2F_EF = p/2 + uparrow sigma F_y = 0 F_AB + F_CD + F_EF - P = 0 delta_C - delta_F/d = delta_A - delta_E/2d 2 delta_c = delta_A + delta_E 2 F_CD L/AE = F_AB L/AE + F_EFL/AE 2 F_CD - F_AB - F_EF = 0 rightarrow (3) Solving Eqs (1), (2) and (3) yields F_AB = 7P/12 F_CD = P/3 F_EF = P/12 sigma_AB = 7p/12 A sigma_CD = P/3 A sigma_EF = P/12A