A sample of aluminum is stretched by a 6 kip load. If the aluminum piece is 32 inches long, 2 in^2 in area, what is the deformation of the sample in inches? Answer to three decimal places. Type your answer...
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- A tensile test was performed on a metal specimen having a circular cross section with a diameter of 1 2 inch. The gage length (the length over which the elongation is measured) is 2 inches. For a load 13.5 kips, the elongation was 4.6610 3 inches. If the load is assumed to be within the linear elastic rang: of the material, determine the modulus of elasticity.The data in Table 1.5.3 were obtained from a tensile test of a metal specimen with a rectangular cross section of 0.2011in.2 in area and a gage length (the length over which the elongation is measured) of 2.000 inches. The specimen was not loaded to failure. a. Generate a table of stress and strain values. b. Plot these values and draw a best-fit line to obtain a stress-strain curve. c. Determine the modulus of elasticity from the slope of the linear portion of the curve. d. Estimate the value of the proportional limit. e. Use the 0.2 offset method to determine the yield stress.To stretch a 200 mm long tensile test specimen (a) of steel elastically by 0.08 mm. How many loads do I need to apply? (b) What is the maximum length of this sample without plastic deformation? How long can it be extended? (The modulus of elasticity of steel is 210 GPa, yield strength 580 MPa, tensile strength 920 MPa specified as.
- An aluminum alloy bar with a rectangular cross section that has a width of 12.5 mm, thickness of 6.25 mm, and a gauge length of 50 mm was tested in tension to fracture according to ASTM E-8 method. The load and deformation data were as shown in Table P4.6. Using a spreadsheet program, obtain the following: a. A plot of the stress-strain relationship. Label the axes and show units. b. A plot of the linear portion of the stress-strain relationship. Determine the modulus of elasticity using the best fit approach. c. Proportional limit. d. Yield stress at an offset strain of 0.002 m/m. e. Tangent modulus at a stress of 450 MPa. f. Secant modulus at a stress of 450 MPa. TABLE P4.6 Load (kN) AL (mm) Load (kN) AL (mm) 33.5 1.486 3.3 0.025 35.3 2.189 14.0 0.115 37.8 3.390 25.0 0.220 39.8 4.829 29.0 0.406 40.8 5.961 30.6 0.705 41.6 7.386 31.7 0.981 41.2 8.047 32.7 1.245A 19-mm reinforcing steel bar and a gauge length of 75 mm was subjected to tension, with the results shown in Table Using a computer spreadsheet program, plot the stress–strain relationship. From the graph, determine the Young’s modulus of the steel and the deformation corresponding to a 150-kN load.A high-yield-strength alloy steel bar with a rectangular cross section that has a width of 37.5 mm, a thickness of 6.25 mm, and a gauge length of 203 mm was tested in tension to rupture, according to ASTM E-8 method. The load and deformation data were as shown in Table Using a spreadsheet program, obtain the following:a. A plot of the stress–strain relationship. Label the axes and show units.b. A plot of the linear portion of the stress–strain relationship. Determine modulus of elasticity using the best-fit approach.c. Proportional limit.d. Yield stress.e. Ultimate strength.f. If the specimen is loaded to 155 kN only and then unloaded, what is the permanent deformation?g. In designing a typical structure made of this material, would you expect the stress applied in (f) safe? Why?
- A tensile test specimen of stainless steel alloy having a diameter of 0.495 in. and a gage length of 2.00 in. was tested to fracture. Stress and strain data obtained during the test are shown. Determine the modulus of elasticity. 120 Upper scale 90 60 Lower scale 30 0.040 0.004 0.060 0.006 0.120 0.012 0.020 0.100 0.010 0.0 0.080 0.008 0.0 0.002 Strain (in./in.) Stress (ksi)2.) Answer the following: a.) a 22 ft long solid steel rod is subjected to a load of 15,000 lb. This load causes the rod to stretch 0.435 in. The modulus of elasticity of the steel is 30,000,000 psi. Determine the diameter of the rod. b.) A solid circular titanium control rod must support an axial tension force of 12,000 lb. Assume that the modulus of elasticity of the titanium in 16,500,000 psi. If the rod must elongate no further than 0.35 in. when the normal stress of the rod is 68,000 psi, determine the minimum rod diameter AND the maximum rod length.1. The following data were obtained during a tension test of an aluminum alloy. The initial diameter of the test specimen was 0.505 in. and the gage length was 2.0 in. Load (lb) 0 2 310 4 640 6 950 9 290 11 600 12 600 Elongation (in.) 0.00220 0.00440 0.00660 0.00880 0.0110 0.0150 Load (lb) 14 000 14 400 14 500 14 600 14 800 14 600 13 600 Elongation (in.) 0.020 0.025 0.060 0.080 0.100 0.120 Fracture Plot the stress-strain diagram and determine the following mechanical properties: (a) proportional limit; (b) modulus of elasticity; (c) yield point; (d) yield strength at 0.2% offset; (e) ultimate strength; and (f) rupture strength.
- } 8) A tensile load of 3000 N is applied to a steel bar of cross-sectional area 500 mm?. If a tensile load was applied to an aluminum bar to achieve the same lateral (transvers) strain with the former (steel bar), what would be the load? Note: Original dimensions of the two bars are the same (Est = 2.1 x 105 MPa, Eat = 0.703 × 105 MPa, vg = 0.3, vat = 0.33) E = º , v = - Saterat Eaxial Fatuminum™ 3000 N A-500 mm² A=500 mm² Fatuminum=? 3000 N Steel Aluminum1. The following data were obtained during a tension test of an aluminum alloy. The initial diameter of the test specimen was 0.505 in., and the gage length was 2.0 in. Load ( Ib) Elongation (in.) Load ( Ib) Elongation (in.) 14000 0.020 2310 0.0022 14400 0.025 4640 0.0044 14 500 0.060 6950 0.0066 14600 0.080 9290 0.0088 14800 0.100 I1 600 0.0110 14600 0.120 13000 0.01 50 13600 Fracture Plot the stress-strain diagram and determine the following mechanical properties: a. proportional limit þ. modulus of elasticity S. vield stress at 0.2% offset d. ultimate stress e. nominal rupture stress.2. A steel bar, whose cross section is 0.55 inch by 4.05 inches, was tested in tension. An axial load of P = 30,500 lb. produced a deformation of 0.105 inch over a gauge length of 2.05 inches and a decrease of 0.0075 inch in the 0.55-inch thickness of the bar. Determine the lateral strain. * Your answer Determine the axial strain. Your answer Determine the Poisson's ratio v. * Your answer Determine the decrease in the 4.05-in. cross-sectional dimension (in inches). * Your answer