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Part 1 At a temperature of 60°F, a 0.03-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; a = 12.5 x 10-6/°F] bar with a width of 3.0 in and a thickness of 0.65 in. Bar (2) is a stainless steel [E = 28.000 ksi; v= 0.12; a = 9.6 x 10-6/°F] bar with a width of 2.0 in and a thickness of 0.65 in. The supports at A and Care rigid. Assume h2=3.0 in., h2=2.0 in., L2=27 in., L2=48 in., and 4 = 0.03 in. Determine (a) the lowest temperature at which the two bars contact each other. (b) the normal stress in the two bars at a temperature of 240°F. (c) the normal strain in the two bars at 240°F. (d) the change in width of the aluminum bar at a temperature of 240°F. (1) (2) h h2 B B L L2 Determine the lowest temperature, Tcontact, at which the two bars contact each other. Answer: Tcontact i °F
Part 2 From the result obtained for Part 1, we know that the gap will be closed at 240°F, making this a statically indeterminate axial configuration. On a piece of paper, sketch a FBD at joint B for the case in which the gap is closed. Write an equilibrium equation which relates F1, the force in bar (1), and F2, the force in bar (2). By convention, a tension force is positive, and a compression force is negative. Answer: O a) F1+F2 = 0 OC) L1F1 + L3F2 = 0 O d) L1F1-L2F2 = 0 O b)-F1+F2 = 0 e Textbook and Media Save for Later Attempts: unlimited Submit Answer Part 3 Find a geometry-of-deformation relationship for the case in which the gap is closed. Express this relationship by entering the sum 81 +82, where is the axial deflection of Bar (1), and is the axial deflection of Bar (2) Answer: 8 + 8 = PR in.
Part 4 Substitute force-temperature-deformation relationships into the geometry-of-deformation equation to obtain a compatibility equation. Choose the correct compatibility equation, where E,a, L, and A are the elastic modulus, coefficient of thermal expansion, length, and bar cross-sectional area, respectively, with subscript"1" referring to Bar (1) and subscript"2" referring to Bar (2). Also, At = 240°F -60°F, and dşap = 0.03 in., the original gap distance. Answer: FIL F2L2 + a LAT - a2L2AT = dgap A E A2E2 FIL F2L2 d) + a LAT + + a2L2AT = dgap AE A₂ E2 FIL F2L2 a) + a LAT + + a2LAT = 0 ALE A2E2 FL F2L2 b) +aL AT + + a_LAT AE A₂E2 = = dgap
Part 5 Find the force in the Bar (1),F1, and the force in Bar (2),F2, at a temperature of 240°F. By convention, a tension force is positive and a compression force is negative. Answers: F1 = kips F2 = kips e Textbook and Media Save for Later Attempts: unlimited Submit Answer Part 6 Find o and 02, the normal stresses in Bars (1) and (2), respectively. By convention, a tension stress is positive and a compression stress is negative. Answers: 01 = i ksi 02 = i ksi
Part 7 Determine d, and 82, the deformations of Bars (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: 8. i in. = 82 = in. e Textbook and Media Save for Later Attempts: unlimited Submit Answer Part 8 Determine ej and €, the normal strains in Bars (1) and (2), respectively. By convention, a normal strain is positive for a member that elongates, and it is negative for a member that is shortened. Answers: EL με E2 - i με
Part 9 Determine €1,0, the component of the normal strain in Bar (1) due to the internal force Fị This would be the normal strain in Bar 1 for an internal force Fi with no effect of the temperature increase included. Also, determine the accompanying lateral strain due to the Poisson effect Elatl.0, where the lateral strain here is the strain in the direction of the width of 3.0in. Answers: E1,6 = με Elatl. = με eTextbook and Media Save for Later Attempts: unlimited Submit Answer Part 10 Determine Elat1,7, the component of the lateral strain caused by the temperature change. Answer: Elat),T i με
Part 11 Determine the total lateral strain and the corresponding change in width of the aluminum bar at a temperature of 240°F. Answers: . Elati = με dwidth in. 11