Part 1 At a temperature of 60°F, a 0.04-in. gap exists between the ends of the two bars shown. Bar (1) is an aluminum alloy [E = 10,000 ksi; v = 0.32; α=12.5 x 10-6/°F] bar with a width of 3.0 in. and a thickness of 0.85 in. Bar (2) is a stainless steel [E 28,000 ksi; v = 0.12; α=9.6 x 10-6/°F] bar with a width of 1.5 in. and a thickness of 0.85 in. The supports at A and C are rigid. Assume h₁=3.0 in., h2=1.5 in., L₁=28 in., 42=50 in., and A =0.04 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 245°F. (c) the normal strain in the two bars at 245°F. (d) the change in width of the aluminum bar at a temperature of 245°F. A (1) h₁ h₂ B = Li L2 Determine the lowest temperature, T contact, at which the two bars contact each other. Answer: contact = °F Find a geometry-of-deformation relationship for the case in which the gap is closed. Express this relationship by entering the sum б₁ + S2, where S₁ is the axial deflection of Bar (1), and 2 is the axial deflection of Bar (2). Answer: 81 +82 = in. Part 5 Find the force in the Bar (1), F1, and the force in Bar (2), F2, at a temperature of 245°F. By convention, a tension force is positive and a compression force is negative. Answers: F1 F2 = = SHOW HINT LINK TO TEXT kips kips LINK TO VIDEO Attempts: 0 of 5 used SAVE FOR LATER SUBMIT ANSWER Part 6 Find 1 and 2, the normal stresses in Bars (1) and (2), respectively. By convention, a tension stress is positive and a compression stress is negative. Answers: 01 = 02= ksi ksi Part 7 Determine 1 and 2, 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: δι = in. 82 = in. SHOW HINT LINK TO TEXT Part 8 LINK TO VIDEO Attempts: 0 of 5 used SAVE FOR LATER SUBMIT ANSWER Determine &1 and 2, 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: με 82 = με Part 9 Determine €1,01 the component of the normal strain in Bar (1) due to the internal force F₁. This would be the normal strain in Bar 1 for an internal force F₁ with no effect of the temperature increase included. Also, determine the accompanying lateral strain due to the Poisson effect & lat 1,0, where the lateral strain here is the strain in the direction of the width of 3.0in. Answers: €1,σ = Elat 1,0 με = με ww SHOW HINT LINK TO TEXT LINK TO VIDEO Part 10 Determine lat 1,T the component of the lateral strain caused by the temperature change. Answer: Elat 1,T= the tolerance is +/-2% με Attempts: 0 of 5 used SAVE FOR LATER SUBMIT ANSWER Part 11 Determine the total lateral strain and the corresponding change in width of the aluminum bar at a temperature of 245°F. Answers: Elat 1 = Swidth με in.