1. a test specimen in a tensile test has a gage length of 2.0 in and an area = 0.5 in2. during the test the specimen yields under a load of 32,000 lb. the corresponding gage length = 2.0083 in. this is the 0.2 percent yield point. the maximum load of 60,000 lb is reached at a gage length = 2.60 in. determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength. (d) if fracture occurs at a gage length of 2.92 in, determine the percent elongation. (e) if the specimen necked to an area = 0.25 in2, determine the percent reduction in area.

(a) 64000 Pa. (b) 29.7 MPa.  (c) 120000 Pa. (d) 46%. (e) 50%

Explanation:

(a) The value of yield strength (Y) is calculated by dividing the maximum load (L) with the area (A). Where the maximum load (L) is 60000 lb and area is 0.5 in^2, thus:

Y = L/A = 32000/0.5 = 64000 Pa

(b) Modulus of Elasticity is calculated by dividing the stress with the elongation. Thus: E = stress/elongation

However, we need to estimate the 0.2% offset. Therefore we have:

elongation = (2.0083 - 2)/2 - 0.2% = 0.00415 - 0.002 = 0.00215

E = 64000/0.00215 = 29.7 MPa

(c) The tensile stress is calculated by dividing the maximum load with the area. Thus:

Tensile stress = maximum load / area = 60000/0.5 = 120000 Pa

(d) The percent elongation = (2.92-2)/2 * 100% = 46%

(e) The percent reduction in area = (0.5 - 0.25)/0.5 * 100% = 50%

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