Compute the density of lead and molybdenum.
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Compute the density of lead and molybdenum.
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- c). Interpret the following diagram? The effect of carbon and heat treatment on the properties of plain carbon steels. 160 140 Tensile 120 strength 100 Impact energy | 100 Annealed 80 80 Normalized 60 60 Annealed 40 `Yield strength 20-% Elongation Annealed 20 Normalized- 0.2 0.4 0.6 0.8 1.0 Weight percent carbon Figure-4 Yield and tensile strength (ksi) Normalized % Elongation or impact energy (ft • Ib)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=α=12.5 x 10-6/°F] bar with a width of 2.5 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; α=α=9.6 x 10-6/°F] bar with a width of 1.7 in. and a thickness of 0.75 in. The supports at A and C are rigid. Assume h1=2.5 in., h2=1.7 in., L1=31 in., L2=46 in., and Δ=Δ= 0.04 in. (A) Determine the lowest temperature, Tcontact, at which the two bars contact each other. (B) Find a geometry-of-deformation relationship for the case in which the gap is closed. Express this relationship by entering the sum δ1+δ2, where δ1 is the axial deflection of Bar (1), and δ2 is the axial deflection of Bar (2). δ1+δ2= _____in. (C) Find the force in the Bar (1), F1, and the force in Bar (2), F2, at a temperature of 225oF. By convention, a tension force is positive and a compression force is negative. IN KIPS (D) Find σ1 and σ2,…Compute the volume change of a solid copper cube, 40 mm on each edge, when subjected to a pressure of 20MPa. The bulk of modulus for copper is 125GPa. Please solve this problem with complete and detailed solution. Thank you
- 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 = 12.5 x 10-6/°F] bar with a width of 3.0 in. and a thickness of 0.75 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 0.75 in. The supports at A and C are rigid. 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 250°F. (c) the normal strain in the two bars at 250°F. (d) the change in width of the aluminum bar at a temperature of 250°F. (1) 3.0 in. 32 in. 2.0 in. B ↓ (2) 44 in. 0.04-in. gap Determine the lowest temperature, Tcontact, at which the two bars contact each other.At a temperature of 60°F, a 0.02-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=α=12.5 x 10-6/°F] bar with a width of 2.8 in. and a thickness of 0.85 in. Bar (2) is a stainless steel [E = 28,000 ksi; v = 0.12; α=α=9.6 x 10-6/°F] bar with a width of 1.6 in. and a thickness of 0.85 in. The supports at A and C are rigid. Assume h1=2.8 in., h2=1.6 in., L1=26 in., L2=40 in., and Δ=Δ= 0.02 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 225°F.(c) the normal strain in the two bars at 225°F.(d) the change in width of the aluminum bar at a temperature of 225°F.Why does the iron–carbon phase diagram go only to 6.7% carbon?
- At a temperature of 60°F, a 0.04 in. gap exists between the ends of the two bars shown in the figure. Bar (1) is an aluminum alloy [E=10000 ksi; v=0.32; a= 12.5 x10^-6/,°F] bar with a width of 3 in. and a thickness of 0.75 in. Bar (2) is a stainless steel [E= 28000 ksi; v=0.12; a=9.6x10^-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 normal stress in bar (1) at a temperature of 380°F. Answer in ksiAt 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. gapWhat is a ceramic which consists of layered silicates.
- 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. 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 = 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. gapDetermine ratio of axial deformation of steel to aluminum