In the figure shown below, determine: 1) The final temperature if the normal stress at aluminium is Oa = -90 MPa and the initial temperature 20°C. 2) The final length of the aluminium member. Aluminum A=1800mm? E=73GPA a=23.2x10-6/°C Bronze A=1500mm? E=105GPA a=21.6x10-6/°C Gap=0.5mm 0.35m 0.45m
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- Narrow bars of aluminum are bonded to the two sides of a thick steel plate as shown. Initially, at T₁ = 70°F, all stresses are zero. Knowing that the temperature will be slowly raised to T₂ and then reduced to T₁, determine (a) the highest temperature T₂ that does not result in residual stresses, (b) the temperature T₂ that will result in a residual stress in the aluminum equal to 58 ksi. Assume aa = 12.8 x 10-6/°F for the aluminum and a = 6.5 × 10-6/°F for the steel. Further assume that the aluminum is elastoplastic with E = 10.9 × 106 psi and ay = 58 ksi. (Hint: Neglect the small stresses in the plate.) Fig. P2.121In the figure shown below, determine: 1) The final temperature if the normal stress at aluminium is Og = -90 MPa and the initial temperature %3D 20°C. 2) The final length of the aluminium member. Aluminum A=1800mm? Bronze A=1500mm? E=73GPA E=105GPA a=23.2x10-6/°C a=21.6x10-6/°C Gap=0.5mm 0.35m 0.45mIn the figure shown below, determine: 1) The final temperature if the normal stress at aluminium is oal = -90 MPa and the initial temperature 20°C. 2) The final length of the aluminium member. Aluminum A=1800mm? E=73GPA Bronze A=1500mm2 E=105GPA a=23.2x10-6/°C a=21.6x10-6/°C Gap=0.5mm 0.35m 0.45m
- Problem # 2. At room temperature (20°C) a 0.5-mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 140 °C, determine (a) the normal stress in the aluminum rod Ans: A -300 mm Aluminum A = 2000 mm² E = 75 GPa α = 23 x 16-6/°C 0.5 mm -250 mm Stainless steel A = 800 mm² B E = 190 GPa a = 17.3 x 10-6/°CThe following information about the o-e curve is given for a steel alloy. E = 0.001527 at o = 300 MPa and ɛ = 0.003054 for o = 600 MPa. (a) Draw the stress-strain diagram and calculate the E for this alloy.At room temperature (20°C) a 0.7-mm gap exists between the ends of the rods shown. At a later time when the temperature has reached 150°C, determine: (a.)The normal stress in the aluminum rod, (b.) The change in length of the aluminum rod. 0.7 mm 300 mm A Aluminum A =2,000 mm² E= 75 GPa α = 23 x 106/°C 250 mm B Stainless steel A = 800 mm² E = 190 GPa a = 17.3 x 10-6/°C
- 1.4-7 The data shown in the table below were obtained from a tensile test of high-strength steel. The test specimen had a diameter of 13 mm and a gage length of 50 mm (see figure for Prob. 1.4-3). At fracture, the elongation between the gage marks was 3.0 mm and the minimum diameter was 10.7 mm. Plot the conventional stress-strain curve for the steefor the steel and determine the proportional limit, modulus of elastics of elastic- ity (i.e., the slope of the initial part of the stress-strain,tress-strain curve), yield stress at 0.1% offset, ultimate stress, percent, elongation in 50 mm, and percent reduction in area. 'ess, percent area. TENSILE-TEST DATA FOR PROB. 1.4-7 Elongation (mm) 0.005 0.015 0.048 Load (kN) 5 10 30 50 0.084 60 0.099 64.5 0.109 67.0 0.119 68.0 0.137 69.0 0.160 70.0 0.229 72.0 0.259 76.0 0.330 84.0 0.584 92.0 0.853 100.0 1.288 112.0 2.814 113.0 FractureAn aluminum alloy [E = 67 GPa; ν = 0.33; α = 23.0 × 10–6/°C] plate is subjected to a tensile load P. The plate has a depth of d = 225 mm, a cross-sectional area of A = 5100 mm2, and a length of L = 4.1 m. The initial longitudinal normal strain in the plate is zero. After load P is applied and the temperature of the plate has been increased by ΔT = 63°C, the longitudinal normal strain in the plate is found to be 2900 με. Determine: (a) the magnitude of load P. (b) the change in plate depth Δd.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; a = 13.4 x 10-6/°F] bar with a width of 3 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; a = 10.1 x 10-6/°F] bar with a width of 2 in. and a thickness of 0.75 in. The supports at A and Care rigid. Determine the lowest temperature at which the two bars contact each other. (1) ↑ 3 in. 32 in. O 75.9°F O 146.5°F O 105.8°F O 122.3°F O 111.3°F 2 in. (2) 44 in. -0.04-in. gap
- Problem 1 An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown. Axial loads are applied at the positions indicated. Aluminum Steel A=800mm2 A=1000mm2 Bronze A=700mm? 4P 2P 500mm 600mm 700mm 1. What is the maximum value of P that will not exceed the axial stress bronze of 105 MPa? 2. What is the maximum value of P that will not exceed the axial stress in aluminum of 90 MPa? 3. If P=10KN, what is the axial force to be carried by the aluminum in KN? 4. If P is 5KN, what is the axial stress of steel?An aluminum alloy [E = 70 GPa; v = 0.33; a = 23.0×10-6/°C] bar is subjected to a tensile load P. The bar has a depth of d = 260 mm, a cross-sectional area of A = 14720 mm2, and a length of L = 5.5 m. The initial longitudinal normal strain in the bar is zero. After load P is applied and the temperature of the bar has been increased by AT = 46°C, the longitudinal normal strain is found to be 1680 µɛ. % D Calculate the change in bar depth d after the load P has been applied and the temperature has been increased. L P Answer: Ad = i mmAt 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; a = 12.7 x 10-6/°F] bar with a width of 3 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; a = 8.6 x 10-6/°F] bar with a width of 2 in. and a thickness of 0.75 in. The supports at A and C are rigid. Determine the lowest temperature at which the two bars contact each other. (1) 3 in. 32 in. 90.2°F O 69.9°F 139.2°F 103.5°F O 111.0°F B ↑ 2 in. ↓ 44 in. -0.04-in. gap