In a tensile test, the force at the yield point was 1000 KN, at the ultimate point was 1500 KN and at the fracture point was 900 KN. The diameter of the specimen before the test was 10 mm. Find: a- The yield stress b- The tensile stress c- The fracture stress.

The magnitude of the shear stress to cause yielding in a torsion test and the magnitude of the normal stress to cause yielding in a uniaxial tensile test are usually equal". Is this true or false?

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39)A tensile test specimen has a gage length=3.0 inches and a diameter=  .75 inches. Yielding occurs at a load of 38000lbs. The corresponding gage length=3.0103 inches (neglect the .0.2% yield point). The maximum load of 54000lbs is reached at a gage length =3.453 inches. If fracture occurs at a gage length of 3.873inches, determine the percent elongation at fracture (Round to the nearest whole %)

For some metal alloy, a true stress 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 418 MPa (60630 psi) is applied if the original length is 500 mm (19.69in) ? Assume a value of 0.22 for the strain- hardening exponent, n

Question 4 Save Answer You do a series of tensile tests on plates of a magnesium alloy that have been subjected to prior cold rolling to true plastic strains of 0.1, 0.2 and 0.3. The resulting true stress-true strain curves are shown below (including a zoomed in version expanding on the small strain region). It is reasonable to approximate the the 0.2% offset yield strength of the magnesium as ✓ MPa for 0.2 and ✓ MPa for 0.3. Assuming the yield strength is ✓ MPa for 0.1 plastic strain, proportional to the square root of the prior true plastic strain results in a hardening coefficient of approximately k= MPa. Hence, we can predict that we need a prior plastic strain of approximately True Stress (MPa) True Stress (MPa) True Stress (MPa) 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 250 200 150 100 50 0 0.00 Ep = 0.1 0.01 Ep=0.2 0.01 Ep=0.3 0.01 0.02 True Strain 0.02 0.03 0.02 0.03 True Strain 0.03 0.04 0.04 0.04 True Stress (MPa) 0.05 0.000 0.001 True Stress (MPa) Ep=0.1 True…

Question-8. . A cylindrical metal specimen with a diameter of 16 mm and length of 88 mm is to be subjected to a tensile stress of 80 MPa; at this stress the resulting deformation is elastic. (a) If the elongation is less than 0.058 mm, which of the metals in the table can be considered as candidates? Why ? (b) If the maximum permissible diameter decrease is 0.0012 mm when the tensile stress of 68 MPa is applied, which of the metals satisfy the criterion in question (a) are the suitable candidates? Why ? Metal Shear Modulus / GPa Modulus of Elasticity / GPa 69 Poisson's Ratio Al 25 0.33 Cu 110 46 0.34 Mg 45 17 0.29 Ni 207 76 0.31 Ti 107 6.5 0.34 W 407 23.2 0.28

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Hello, the stress and cycles required for this question is Stress 1 = 420MPa at 7 cycles, Stress 2 = 360MPa at 65 cycles and Stress 3 = 130MPa at 325 cycles. Use the first three cycles (which shows how many cycles at that stress level to failure) to calculate when the specimen will fail if it underwent test of 7 cycles at 420MPa, then undergoes 360MPa for 65 cycles and finally 130MPa at 325 cycles. Will the specimen fail? Please show calculations of failure

3. A rod-shaped MgO specimen with round cross section is tested with the three-point bend method. What is the calculated minimum radius without fracture if the applied load is 5560 N, the flexural strength is 105 MPa, and the separation between the two support pins is 45 mm?