The shear strength of a metal is usually (a) greater than or (b) less than its tensile strength? (b). Answer. Problems A tensile test uses a test specimen that has a gage length of 50 mm and an area = 200 mm2. During the test the specimen yields under a load of 98,000 N. The corresponding gage length = 50.23 mm. This is the 0.2 percent yield point. The maximum load of 168,000 N is reached at a gage length = 64.2 mm. Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength. (d) If fracture occurs at a gage length of 67.3 mm, determine the percent elongation. (e) If the specimen necked to an area = 92 mm2, determine the percent reduction in area. Solution: (a) Y = 98,000/200 = 490 MPa. (b) s = E e Subtracting the 0.2% offset, e = (50.23 - 50.0)/50.0 - 0.002 = 0.0026 E = s/e = 490/0.0026 = 188.5 x 103 MPa. (c) TS = 168,000/200 = 840 MPa. (d) EL = (67.3 - 50)/50 = 17.3/50 = 0.346 = 34.6% (e) AR = (200 - 92)/200 = 0.54 = 54% Online Problems Online 1: Three methods of solid solution strengthening. Grain size reduction: as we reduce the size of grains we make it harder for defects to move due to the increased numbers of grain boundaries. However, we bring ourselves closer to the tensile strength and strain at failure since defects will tend to aggregate at boundaries. Allloying : by adding atoms we may distort the regular order of the matrix. These “impurity” local matrix distortions make it harder for defects to move thus increasing I nthe form “Strengh” of yield and tensile strengths. Changing the atomic in attraction through alloying would affect the modulus of elasticity. Cold Work: by deforming a metal we reshape grains and add energy into the grain structure; the distorted grains are not easily distorted further hence the increases in the for of “strength” increased yield strength.
The shear strength of a metal is usually (a) greater than or (b) less than its tensile strength? Answer. (b). Problems A tensile test uses a test specimen that has a gage length of 50 mm and an area = 200 mm2. During the test the specimen yields under a load of 98,000 N. The corresponding gage length = 50.23 mm. This is the 0.2 percent yield point. The maximum load of 168,000 N is reached at a gage length = 64.2 mm. Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength. (d) If fracture occurs at a gage length of 67.3 mm, determine the percent elongation. (e) If the specimen necked to an area = 92 mm2, determine the percent reduction in area. Solution: (a) Y = 98,000/200 = 490 MPa. (b) s = E e Subtracting the 0.2% offset, e = (50.23 - 50.0)/50.0 - 0.002 = 0.0026 E = s/e = 490/0.0026 = 188.5 x 103 MPa. (c) TS = 168,000/200 = 840 MPa. (d) EL = (67.3 - 50)/50 = 17.3/50 = 0.346 = 34.6% (e) AR = (200 - 92)/200 = 0.54 = 54% Online Problems Online 1: Three methods of solid solution strengthening. Grain size reduction: as we reduce the size of grains we make it harder for defects to move due to the increased numbers of grain boundaries. However, we bring ourselves closer to the tensile strength and strain at failure since defects will tend to aggregate at boundaries. Allloying : by adding “impurity” atoms we may distort the regular order of the matrix. These local matrix distortions make it harder for defects to move thus increasing “Strengh” I nthe form of yield and tensile strengths. Changing the atomic in attraction through alloying would affect the modulus of elasticity. Cold Work: by deforming a metal we reshape grains and add energy into the grain structure; the distorted grains are not easily distorted further hence the “strength” increases in the for of increased yield strength.
MSE-hmwk2-soln.docx