A steel cylindrical sample was subjected to a tensile test. The yield load was 2100N. The maximum load was 3400N and the failure load was 2350N. The ultimate engineering stress was 497.4MPa. What was the initial diameter of the sample in mm?

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DeniceSandidge DeniceSandidge · Answer 1 · 2019-10-04T06:51:55+02:00

Answer:

initial diameter of the sample is 2.95 mm

Explanation:

given data

yield load = 2100 N

maximum load = 3400 N

failure load = 2350 N

ultimate engineering stress = 497.4 MPa = 497 × [tex]10^{6}[/tex] N/m²

to find out

What was the initial diameter of the sample in mm

solution

we will apply here ultimate engineering stress formula that is express as

ultimate engineering stress = [tex]\frac{Pmax}{A}[/tex] ...............1

here A is area and P max is maximum load applied

so area = [tex]\frac{\pi }{4} d^2[/tex]

here d is initial diameter

so put all value in equation 1

497 × [tex]10^{6}[/tex] = [tex]\frac{3400}{\frac{\pi }{4} d^2}[/tex]

solve it we get d

d = 2.95 × [tex]10^{-3}[/tex] m

so initial diameter of the sample is 2.95 mm