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lys-0071 [83]
3 years ago
6

A specimen of some metal having a rectangular cross section 10.4 mm × 12.8 mm is pulled in tension with a force of 15900 N, whic

h produces only elastic deformation. Given that the elastic modulus of this metal is 79 GPa, calculate the resulting strain.
Engineering
1 answer:
marishachu [46]3 years ago
8 0

Answer:

\sigma=0.00151

Explanation:

We apply Hooke's Law as follows :

\sigma= E\varepsilon

where \sigma is the applied stress

#The applied stress is also equal to :

\sigma=\frac{F}{A_o}=\frac{F}{lw}

Where l\ and w are the cross-sectional dimensions.

=>We equate the two stress equations:

E\varepsilon=\frac{F}{lw}\\\\\varepsilon=\frac{F}{Elw}

#We substitute our values in the equation as:

\varepsilon=\frac{15900\ N}{(79\times 10^9\ N/m^2)(10.4\ m\times 10^{-3}\times 12.8\ m\times 10^{-3})}}\\\\\\=0.00151

Hence, the resulting strain is 0.00151

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