A flat stainless steel test piece type 304 of 0.700 in width and 0.125 in thickn

Question

A flat stainless steel test piece type 304 of 0.700 in width and 0.125 in thickness with 8 in length center has a calibrated distance of 2 in. This stainless steel test piece is deformed until reaching a distance in the section calibrated from 2.002 in (consider the width of the specimen in the calibrated section of 30% of the total width of the specimen) a charge of 735 lbs.

Calculate the conventional unit strain, the corresponding percentage of deformation and the Elastic Modulus of the material in [GPa].

Explanation / Answer

given width = w = 0.7 in
thickness, t = 0.125 in
length, l = 8 in

now dl = 0.002 in
so unit strain = dl/(l/8) = 0.002/2 = 0.001
percent deformation = 0.001*100 = 0.1 %

now loading , P = 735 lbs
stress = P/A
A = 0.3*wl [ considering just 30 percent of width]
stress = sigma = 735/0.3*0.7*0.125 = 28000 lbs/sq in

so, Modulus of elasticity = stress/unit strain = 28000/0.001 = 28000000lbs/in^2 = 0.028 lbs/in^2
now 1 lbs = 0.453592 kg
1 in = 0.0254 m

so Y = 0.122*0.453592/0.0254^2 *10^12 = 19.68 GPa

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