An alloy has a yield strength of 41 ksi, a tensile strength of 57 ksi, and an elastic modulus of 16 × 106 psi. A cylindrical specimen of this alloy 0.5 in. in diameter and 10 in. long is stressed in tension and found to elongate 0.3 in. On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why.
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- An aluminum alloy bar with a radius of 7 mm was subjected to tension until fracture and produced results shown in Table P4.3. a. Using a spreadsheet program, plot the stress–strain relationship. b. Calculate the modulus of elasticity of the aluminum alloy. c. Determine the proportional limit. d. What is the maximum load if the stress in the bar is not to exceed the proportional limit? e. Determine the 0.2% offset yield strength. f. Determine the tensile strength. g. Determine the percent of elongation at failure.arrow_forward1. 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.arrow_forwardAn aluminum alloy bar with a radius of 7 mm was subjected to tension until fracture and produced results shown in Table P4 a. Using a spreadsheet program, plot the stress–strain relationship. b. Calculate the modulus of elasticity of the aluminum alloy. c. Determine the proportional limit. d. What is the maximum load if the stress in the bar is not to exceed the proportional limit? e. Determine the 0.2% offset yield strength. f. Determine the tensile strength. g. Determine the percent of elongation at failure.arrow_forward
- ) Consider a cylindrical specimen of a steel alloy 10.0 mm in diameter and 75 mm long that is pulled in tension. (a) Determine its elongation when a load of 20,000 N is applied. (b) What is the modulus of elasticity? (c) What is the strength at a strain offset of 0.002? (d) What is the tensile strength?arrow_forwardS Figure P1.16 shows the stress-strain relations of metals A and B during ten- sion tests until fracture. Determine the following for the two metals (show all calculations and units): a. Proportional limit b. Yield stress at an offset strain of 0.002 m/m. c. Ultimate strength d. Modulus of resilience e. Toughness f. Which metal is more ductile? Why? 900 Metal A 600 Metal B 300 0.00 0.02 0.04 0.06 0.08 0.10 0.12 0.14 Strain, m/m FIGURE P1.16 Stress, MPaarrow_forward1arrow_forward
- A piece of Aluminum originally 213 mm long is pulled in tension with a stress of 844 MPa. If the deformation is entirely elastic, what will be the resultant elongation in mm? The modulus of elasticity of aluminum is 69 GPa. Round your answer to 2 significant figures.arrow_forwardA round steel alloy bar with a diameter of 19 mm and a gauge length of 76 mm was subjected to tension, with the results shown in Table P3.26. Using a computer spreadsheet program, plot the stress-strain relationship. From the graph, determine the Young's modulus of the steel alloy and the deformation corresponding to a 37 kN load. TABLE P3.26 Deformation, Load, kN mm 9 0.0286 18 0.0572 27 0.0859 36 0.1145 45 0.1431 54 0.1718arrow_forwardA brass alloy has a yield strength of 280 MPa, a tensile strength of 390 MPa,and an elastic modulus of 105 GPa. A cylindrical specimen of this alloy 12.7 mm in diameter and 250 mm long is stressed in tension and found to elongate 7.6 mm. On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If so, calculate the load. If not, explain why.arrow_forward
- 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 1.245arrow_forwardQ1/ Consider the brass alloy for which the stress-strain behavior is shown in the figure below. A cylindrical specimen of this material 10.0 mm in diameter and 101.6 mm long is pulled in tension with a force of 10,000 N. If it is known that this alloy has a value for Poisson's ratio of 0.35, compute (a) the specimen elongation, and (b) the reduction in specimen diameter. 500 70 Tensile strength 450 MPa (65,000 psi) 60 400 50 10 psi 300 MPa 40 40 30 200 Yield strength 30 200 250 MPa (36,000 psi) 20 100 20 10 100 10 0.005 0.10 0.20 0.30 0.40 Strain Stress (MPa) Stress (10 psi)arrow_forwardA round steel alloy bar with a diameter of 0.75 in. and a gauge length of 3.0 in. 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 alloy and the deformation corresponding to a 8,225-lb load.arrow_forward
- Materials Science And Engineering PropertiesCivil EngineeringISBN:9781111988609Author:Charles GilmorePublisher:Cengage Learning