Answer:
[tex]\sigma=0.00151[/tex]
Explanation:
We apply Hooke's Law as follows :
[tex]\sigma= E\varepsilon[/tex]
where [tex]\sigma[/tex] is the applied stress
#The applied stress is also equal to :
[tex]\sigma=\frac{F}{A_o}=\frac{F}{lw}[/tex]
Where [tex]l\[/tex] and [tex]w[/tex] are the cross-sectional dimensions.
=>We equate the two stress equations:
[tex]E\varepsilon=\frac{F}{lw}\\\\\varepsilon=\frac{F}{Elw}[/tex]
#We substitute our values in the equation as:
[tex]\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[/tex]
Hence, the resulting strain is 0.00151
