Answer Happy

Part 1 Three rods of different materials are connected and placed between rigid supports at A and D, as shown. Properties for each of the three rods are given in the accompanying table. The bars are initially unstressed when the structure is assembled at 20°C. After the temperature has been increased to 80°C, determine (a) the normal stresses in the three rods. (b) the force exerted on the rigid supports. (c) the deflections of joints B and C relative to rigid support A. Aluminum (1) Cast Iron (2) Bronze (3) L1 = 440 mm L2 = 250 mm L3 = 330 mm A1 = 1100 mm A2 = 2900 mm2 Az = 900 mm E1 = 70 GPa E2 = 155 GPS E3 = 100 GP a1 = 22.5 x 10-6/°C 02 = 13.5 x 10-6/°C az = 17.0x 10-6/°C Aluminum (1) Cast Iron (2) Bronze (3) B C D L L2 L3 On a piece of paper, sketch a FBD at joint B. Write an equilibrium equation which relates F1, the force in rod (1), and F2, the force in rod (2). By convention, a tension force is positive, and a compression force is negative. Answer: O d) L1F1-L2F2 = 0 e) F1+F2 = (a L1 - a2L2)/(E1-E2) O a) F1 +F2 = 0 Ob) -F1+F2 = 0 O c) L1F1 +L2F2 = 0
Part 2 On a piece of paper, sketch a FBD at joint C. Write an equilibrium equation which relates F2, the force in rod (2), and F3, the force in rod (3). By convention, a tension force is positive, and a compression force is negative. Answer: O a) L2F2 +L3F3 = 0 Oc) F2+F3 = 0 O b) L2F2-L3F3 = 0 e) F2+F3 = (azL2-a3L3)/(E2-E3) d) -F2+F3 = 0 Save for Later Attempts: 0 of 3 used Submit Answer Part 3 On a piece of paper, write a geometry-of-deformation equation in terms of deformations 81, 82, and 83, which are the deformations of rods (1), (2), and (3), respectively. Choose the correct geometry-of-deformation equation below. Answer: O c) 81-02-03 = 0 O d) 81 +83 - 82 = 0 O a) 81 +82 + 63 = L1 +L2 +L3 O e) 81 = 82 = 63 Ob) 81 +82 +83 = 0 Save for Later Attempts: 0 of 3 used Submit Answer
Part 4 Substitute force-temperature-deformation relationships into the geometry-of-deformation equation to derive the compatibility equation. Then, find the force in rod (1),F1, the force in rod (2), F2, and the force in rod (3), F3. Be sure to use the correct sign for each force. By convention, a tension force is positive and a compression force is negative. Answer: F1 = i F2 = 2 2 2 F3 = i Save for Later Attempts: 0 of 3 used Submit Answer Part 5 Find 01, 02, and 03, the normal stresses in rods (1), (2), and (3), respectively. By convention, a tension stress is positive, and a compression stress is negative. Answers: 01 = MPa 02 = i MPa 03 = i MPa Save for Later Attempts: 0 of 3 used Submit Answer
Part 6 Determine the magnitudes of the reaction forces at supports A and B. Note that the reaction forces are equal in magnitude to the forces exerted on the rigid supports. Since the magnitudes are required, be sure to enter positive values here. Answers: RA| = i KN |RB| = i kN Save for Later Attempts: 0 of 3 used Submit Answer Part 7 Determine 81 and 82, the deformations of rods (1) and (2), respectively. By convention, a deformation is positive for a member that elongates, and it is negative for a member that is shortened. Answers: 81 = i mm 82= i mm Save for Later Attempts: 0 of 3 used Submit Answer
Part 8 Determine the deflections of joints B and Crelative to rigid support A. Report a positive value for a deflection to the right and a negative value for deflection to the left. Answers: UB = mm uc = mm Save for Later Attempts: 0 of 3 used Submit Answer