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## Homework Statement

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.

## Homework Equations

ε

_{x}= (σ

_{x}/E) + (-Vσ

_{y}/E) - (Vσ

_{z}/E)

ε

_{y}= (-vσ

_{x}/E) + (σ

_{y}/E) - (Vσ

_{z}/E)

ε

_{z}= (-vσ

_{x}/E) - (Vσ

_{y}/E) + (σ

_{z}/E)

## The Attempt at a Solution

(a) Area = (0.05)(0.0016) = 0.00008 m

^{2}

σ

_{x}= P/A = (2.75*10

^{3}) / 0.00008

σ

_{x}= 34.4*10

^{6}Pa

ε

_{x}= (34.4*10

^{6}) / (200*10

^{9})

ε

_{x}= 1.719*10

^{-4}

Δab = ε

_{x}(ab) = (1.719*10

^{-4}/ 0.05

Δab = 8.595*10

^{-6}

The answer should be 0.0358 mm. What am I doing wrong?