In 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=1800mm2 Bronze A=1500mm2 E=105GPA a=21.6×10-6/°C E=73GPA a=23.2x10-6/°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 0.45mAt 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
- Strain = 600 Stress = Strain = 500 Stress = 400 500 300 400 300 200 200 100 100 0.000 0.002 0.004 0.006 Strain 0.00 0.04 0.08 0.12 0.16 0.20 Strain Stress (MPa)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 = 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. gapAt 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
- 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 = 14.4 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 = 9.6 × 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. 105.3°F 75.3°F O 147.3°F 86.6°F 113.4°F B ↑ 2 in. ↓ (2) 44 in. 0.04-in. gapTwo triangular wedges are glued together as shown in the figure. The stress acting normal to the interface, o, is MPa. 100 MPa 100 MPa on 45% 100 MPa -100 MPaProblem # 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/°C
- Calculate Young's modulus for an iron crystal when tension is applied along [2 -2 1] direction. S11=7.6 (TPa)-', S12 = - 2.8 (TPa)-', S44 = 8.6 (TPa) -!The 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 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 = 12.3 × 106/°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 = 8.9 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. O 80.1°F O 118.6°F O 150.7°F O 132.9°F O 110.9°F B 2 in. 44 in. 0.04-in. gap