A hollow steel [E = 30,000 ksi] tube (1) with an outside diameter of 3.75 in. and a wall thickness of 0.216 in. is faste

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A Hollow Steel E 30 000 Ksi Tube 1 With An Outside Diameter Of 3 75 In And A Wall Thickness Of 0 216 In Is Faste 1 (50.75 KiB) Viewed 19 times

A hollow steel [E = 30,000 ksi] tube (1) with an outside diameter of 3.75 in. and a wall thickness of 0.216 in. is fastened to a solid 2.50-in-diameter aluminum [E = 10,000 ksi] rod. The assembly is attached to unyielding supports at the left and right ends and is loaded as shown. Assume P=19 kips, Q=13 kips, L1=6 ft. L2=8ft, and L3=8 ft. Determine (a) the stresses in all parts of the axial structure. (b) the deflections of joints B and C. А B Q с D (1) (2) ) (3) P L L2 L3 Calculate the cross-sectional area of steel tube (1), A1, and the cross-sectional area of the aluminum rods (2) and (3). Answers: A1 = i i in.2 A2 = A3 = i in.2
Part 2 On a piece of paper, sketch a FBD of flange B. Write an equilibrium equation which relates F7, the force in steel tube (1), and F2. the force in aluminum rod (2). By convention, a tension force is positive, and a compression force is negative. Answer: O a) F1+F2 = 0 OC) F1+F2 = 38 kips d) F1-F2=-38 kips Ob) F1-F2 = 0 O e) F1-F2 = 38 kips eTextbook and Media Save for Later Attempts: unlimited Submit Answer Part 3 On a piece of paper, sketch a FBD of flange C. Write an equilibrium equation which relates F2, the force in aluminum rod (2), and F3 the force in aluminum rod (3). By convention, a tension force is positive, and a compression force is negative. Answer: OC) F2 + F3 = 26 kips b) F2-F3 = 0 O a) F2 +F3 =0 O e) F2-F3 = 26 kips O d) F2-F3 = -26 kips
Part 4 On a piece of paper, write a geometry-of-deformation equation in terms of deformations 81.82 and 83, which are the deformations of members (1).(2), and (3), respectively. Substitute force-deformation relationships into the geometry-of- deformation equation to obtain a compatibility equation. Choose the correct compatibility equation below where L1, L2, and L3 are the lengths of members (1). (2), and (3), respectively. Also, Est is the elastic modulus of steel, and Eal is the elastic modulus of aluminum. Answer: FIL F2L2 F3L3 a) + + 0 As Est AAEA AAEAI F F2 F3 d) = 0 Ast AAI ΑΛΙ Fi F2 F3 c) + + Ast AAL ΑΑΙ FIL F2L2 F3L3 b) = 0 Asi Est AAEAAAEAI =0
Part 5 Find the force in the steel tube (1),F2, and the forces F2 and F3, which are the forces in the aluminum rods. Use the correct sign for each force. By convention, a tension force is positive, and a compression force is negative. Answers: F1- = i kips F2 = kips F3 = = i kips e Textbook and Media Save for Later Attempts: unlimited Submit Answer Part 6 Find 0, 02, and 03, the normal stresses in members (1), (2), and (3), respectively. By convention, a tension stress is positive, and a compression stress is negative. Answers: 0 = ksi 02 = ksi 03 = ksi
Part 7 Determine di and 82, the deformations of members (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: ပ် = in. 82 = i in. e Textbook and Media Save for Later Attempts: unlimited Submit Answer Part 8 Determine the deflections of joints B and C. Report a positive value for a deflection to the right and a negative value for deflection to the left. Answers: Ug = PO i in. uc = Me in.