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- compare the effect of presence of a notch on ductile and brittle materials in terms of fracture behaviour3. A steel specimen is tested in tension. The specimen is 1.0 in. wide by 0.25 in. thick in the test region. By monitoring the load dial of the testing machine, it was found that the specimen yielded at a load of 12.5 kips and fractured at 17.5 kips. а. Determine the tensile stresses at yield and at fracture. b. Estimate how much increase in length would occur at 60% of the yield stress in a 2-in. gauge lengthDiscuss the principles of stress and deformation analysis for several kinds of stresses
- 2. Please estimate the number of cycles to failure of a steel specimen under tensile fatigue loading with the following parameters. The R ratio is 3, mean stress 200 MPa, yield strength 450 MPa, ultimate tensile strength 560 MPa, Young’s modulus 200 GPa, KIC = 140 MPa . Assume the initial crack length is 0.1 mm.A steel specimen is tested in tension. The specimen is 25 mm wide by 5 mm thick in the test region. By monitoring the load dial of the testing machine, it was found that the specimen yielded at a load of 55 kN and fractured at 78 kN.a. Determine the tensile stresses at yield and at fracture.b. Estimate how much elongation would occur at 60% of the yield stress in a 50-mm gauge length.A steel specimen is tested in tension. The specimen is 1.0 in. wide by 0.25 in. thick in the test region. By monitoring the load dial of the testing machine, it was found that the specimen yielded at a load of 12.5 kips and fractured at 17.5 kips.a. Determine the tensile stresses at yield and at fracture.b. Estimate how much increase in length would occur at 60% of the yield stress in a 2-in. gauge length.
- A steel specimen is tested in tension. The specimen is 1.0 in. wide by 0.25 in. thick in the test region. By monitoring the load dial of the testing machine, it was found that the specimen yielded at a load of 12.5 kips and fractured at 17.5 kips. a. Determine the tensile stresses at yield and at fracture. b. Estimate how much increase in length would occur at 60% of the yield stress in a 2-in. gauge length. Step-by-step solution: Step 1 of 4 Given that: Width of the specimen, b = 1 in Thickness of the specimen, t = 0.25 in Yield load on the specimen, Py = 12.5 kips Fracture load on the specimen, Pf = 17.5 kips Gauge length, L = 2 in Percentage of yield stress = 60%For the stress-strain diagram, which region or point represents fracture?For a specimen of a steel alloy with a plane strain fracture toughness of 80 MPa√m, fracture results at a stress of 510 MPa when the maximum (or critical) internal crack length is 6 mm. For the same alloy, will fracture occur at a stress level of 380 MPa when the maximum internal crack is 9.0 mm? Why or why not? Select the most appropriate answer based on your calculation. Select one: a. It will not fracture b. Not enough information c. It will fracture
- A steel specimen is tested in tension. The specimen is 25 mm wide by 12.5 mm thick in the test region. By monitoring the load dial of the testing machine, it was found that the specimen yielded at a load of 160 kN and fractured at 214 kN. a. Determine the tensile stress at yield and at fracture. b. If the original gauge length was 100 mm, estimate the gauge length when the specimen is stressed to 1/2 the yield stress.A tension test for a steel alloy results in the stress-strain diagram shown in Figure. Calculate the modulus of elasticity and the yield strength based on a 0.2% offset. Identify the graph the ultimate stress and fracture stress.Problem 4: A tensile test is carried out on a bar of a mild steel of diameter 2 cm. the bar yields under a load of 150 kN and breaks finally at a load of 70 kN. Estimate; 1-the tensile stress at the yield point 2-the ultimate tensile stress 3-the average stress at the breaking point, if the diameter of the fractured neck is 1 сm.