A tensile test specimen has a gage length = 3.0 in anddiameter 0.75 in. Yielding occurs at a load of 38,000 lb. Thecorresponding gage length 3.01 03 in (neglect the 0.2 percentyield point). The maximum load of 54,000 lb is reached at a gagelength 3.453 in. Determine the tensile strength (round to thenearest whole Ksi)38

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Asked Oct 29, 2019
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A tensile test specimen has a gage length = 3.0 in and
diameter 0.75 in. Yielding occurs at a load of 38,000 lb. The
corresponding gage length 3.01 03 in (neglect the 0.2 percent
yield point). The maximum load of 54,000 lb is reached at a gage
length 3.453 in. Determine the tensile strength (round to the
nearest whole Ksi)
38
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A tensile test specimen has a gage length = 3.0 in and diameter 0.75 in. Yielding occurs at a load of 38,000 lb. The corresponding gage length 3.01 03 in (neglect the 0.2 percent yield point). The maximum load of 54,000 lb is reached at a gage length 3.453 in. Determine the tensile strength (round to the nearest whole Ksi) 38

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Expert Answer

Step 1

In order to determine the tensile strength of the specimen, the maximum load and area of cross section of the specimen are required.

Determine the area of cross-section, A of the specimen, where the diameter of the cross-section, is given.

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(d A = 4 (0.75 in) 2 4 0.4417 in2

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Step 2

Use the expression for Tensile Strength and substitute the provided and determined values.

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Step 3

Convert the value of tensile strength f...

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1 lb/in2 0.001 Ksi Therefore, 122254.94 lb/in2 (122254.94) ( 0.00 1) Ksi 122254.94 lb/in2 = 122.25 Ksi

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