Q = 500 N T= 300 N 20° F- 600 N 30° P= 400 N
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For the given system of concurrent force system, the x-component of the 500N-force is?
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- Calculate the change in dimensions in the bar.F = 12000 N; h= 0.01 m; w= 0.03 m; L = 0.8 m; E = 60 GPa; v = 0.3Suppose that the following rod system is subjected to a specific amount of load that makes the displacements of nodes 2 and 3 to be q2=2.18 mm and q3=0.32 mm. The system is made of a material with Young's modulus of E=150 GPa. Suppose that the vector B (strain-displacement vector) for node 1 is B 1 = 14.9 × [ - 1 . 5 2 - 0 . 5 ] 1/m. Workout the stress of node 1 in GPa. aAt 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; α=α=12.5 x 10-6/°F] bar with a width of 3.0 in. and a thickness of 0.70 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 1.9 in. and a thickness of 0.70 in. The supports at A and C are rigid. Assume h1=3.0 in., h2=1.9 in., L1=27 in., L2=47 in., and Δ=Δ= 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 260°F.(c) the normal strain in the two bars at 260°F.(d) the change in width of the aluminum bar at a temperature of 260°F.
- 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; α=α=12.5 x 10-6/°F] bar with a width of 3.0 in. and a thickness of 0.70 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 1.9 in. and a thickness of 0.70 in. The supports at A and C are rigid. Assume h1=3.0 in., h2=1.9 in., L1=27 in., L2=47 in., and Δ=Δ= 0.03 in. Determine:At a temperature of 60°F, a 0.05-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 2.9 in. and a thickness of 0.80 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 1.9 in. and a thickness of 0.80 in. The supports at A and C are rigid. Assume h1=2.9 in., h2=1.9 in., L1=28 in., L2=42 in., and Δ=Δ= 0.05 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 255°F.(c) the normal strain in the two bars at 255°F.(d) the change in width of the aluminum bar at a temperature of 255°F. Determine the lowest temperature, Tcontact, at which the two bars contact each other.The stepped shaft in the figure transmits 19.2 kW of power at 180 d/d (rpm). Determine the required shaft diameter using the maximum strain energy (Von Mises) hypothesis. Take (L=0.8 m, F=6kN, τem=72 MPa, σem= 108 MPa)
- 5. A round aluminum sleeve with an outer diameter of 300 mm and an inner diameter of 200 mm is pressed onto a round steel sleeve with an inner diameter of 100 mm. The radial interference between the two sleeves is 6 = 0.25 mm. Let Ex = 72 GPa, VA = 0.33, E, = 200 GPa and v; = 0.29. Determine the following: (19 points) The interference pressure, p. in MPa between the two sleeves The change in the inner radius in the aluminum sleeve, u, in millimeters If the allowable tangential stress, do, in the aluminum is 180 MPa, determine the safety factor in the aluminum with respect to failure due to tangential stress. You may ignore the radial stres 7The two cylindrical rod segments shown in the figure below (Figure 1) are fixed to the rigid walls such that there is a gap of 0.01 in. between them when T1=75 degrees F . Each rod has a diameter of 1.5 in. Take αal=13(10^-6)/∘F ,Eal=10(10^3)ksi, (σY)al=40ksi, αcu=9.4(10^−6)/∘F, (σY)cu=50ksi, and Ecu=15(10^3)ksi. a) What larger temperature T2 is required in order to just close the gap? b) Determine the magnitude of the average normal stress in each rod if T2=350∘FAnswer no. 17. Derive with the formula below, show the cancellation, explain the step by step and DRAW a free body diagram/figure. refer with this: PE1 + KE1 + U1 + WF1 + Q1 + W1 = PE2 + KE2 + U2 + WF2 + Q2 + W2 + E losses.
- [Note: Dont attempt this question second time if u have already provided the solution one time ] The stepped shaft in the figure transmits 19.2 kW of power at 180 rpm. Determine the required shaft diameter using the maximum strain energy (Von Mises) hypothesis. Take L=0.8 m, F=6kN, τem=72 MPa, σem= 108 MPa.The figure shows a sign hanging from a beam anchored to a wall by two cables. In an element of the beam, the stress tensor has been measured, resulting in the following values σxx = 2250 [psi], σyy = 1175 [psi] and ???=−820 [psi]. Q) It is known that σcreep= 2000 [psi]. Is it possible that the beam could fail? Apply the von mises and tresca criteriaThe two cylindrical rod segments shown in the figure below (Figure 1) are fixed to the rigid walls such that there is a gap of 0.01 in.between them when T1=65∘F. Each rod has a diameter of 1.5 in. Take αal=13(10−6)/∘F, Eal=10(103)ksi, (σY)al=40ksi, αcu=9.4(10−6)/∘F, (σY)cu=50ksi, and Ecu=15(103)ksi. 1) Part A: What larger temperature T2 is required in order to just close the gap? 2) Part B: Determine the magnitude of the average normal stress in each rod if T2=270∘F.