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Engineering and Comp Sci Homework Help
Calculating Strain and Stress in a Steel Plate Under Tensile Load
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[QUOTE="fnicolep92, post: 4995249, member: 541076"] [h2]Homework Statement [/h2] A 2.75 KN tensile load is applied to at test coupon made from 1.6 mm flat steel plate (E=200 GPa, v=0.30). Determine the resulting change in the 50 mm gage length. [ATTACH=full]175854[/ATTACH] [h2]Homework Equations[/h2] ε[SUB]x[/SUB] = (σ[SUB]x[/SUB]/E) + (-Vσ[SUB]y[/SUB]/E) - (Vσ[SUB]z[/SUB]/E) ε[SUB]y[/SUB] = (-vσ[SUB]x[/SUB]/E) + (σ[SUB]y[/SUB]/E) - (Vσ[SUB]z[/SUB]/E) ε[SUB]z[/SUB] = (-vσ[SUB]x[/SUB]/E) - (Vσ[SUB]y[/SUB]/E) + (σ[SUB]z[/SUB]/E) [h2]The Attempt at a Solution[/h2] (a) Area = (0.05)(0.0016) = 0.00008 m[SUP]2[/SUP] σ[SUB]x[/SUB] = P/A = (2.75*10[SUP]3[/SUP]) / 0.00008 σ[SUB]x[/SUB] = 34.4*10[SUP]6[/SUP] Pa ε[SUB]x[/SUB] = (34.4*10[SUP]6[/SUP]) / (200*10[SUP]9[/SUP]) ε[SUB]x[/SUB] = 1.719*10[SUP]-4[/SUP] Δab = ε[SUB]x[/SUB](ab) = (1.719*10[SUP]-4[/SUP] / 0.05 Δab = 8.595*10[SUP]-6[/SUP] The answer should be 0.0358 mm. What am I doing wrong? [/QUOTE]
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Calculating Strain and Stress in a Steel Plate Under Tensile Load
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