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4 years ago

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A specimen of copper having a rectangular cross section 15.2 mm X19.1 mm (0.6 in. X 0.75 in.) is pulled in tension with 44500 N(

10000 lbf) force, producing only elastic deformation. Calculate the resulting strain.

1 answer:

vlabodo [156]4 years ago

7 0

Answer:

The elastic deformation is 0.00131.

Explanation:

Given that,

Force F = 44500 N

Cross section $A =15.2mm\times19.1\ mm$

We Calculate the stress

Using formula of stress

$\sigma=\dfrac{F}{A}$

Where, F = force

A = area of cross section

Put the value into the formula

$\sigma=\dfrac{44500}{15.2\times10^{-3}\times19.1\times10^{-3}}$

$\sigma=153.27\times10^{6}\ N/m^2$

We need to calculate the strain

Using formula of strain

$Y=\dfrac{\sigma}{\epsilon}$

$epsilon=\dfrac{\sigma}{Y}$

Where,

$\sigma$ =stress

Y = young modulus of copper

Put the value into the formula

$\epsilon=\dfrac{153.27\times10^{6}}{117\times10^{9}}$

$\epsilon =0.00131$

Hence, The elastic deformation is 0.00131.

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