Answer:
Explanation:
Given
Force=9 kN
Diameter reduces from 7 mm to 5 mm
As volume remains constant therefore
[tex]A_0L_0=A_fL_f[/tex]
[tex]7^2\times L_0=5^2\times L_f[/tex]
[tex]\frac{L_0}{L_f}=\frac{25}{49}[/tex]
Thus Engineering Strain [tex]\epsilon _E=\frac{L_f-L_0}{L_0}[/tex]
[tex]=\frac{49}{25}-1=\frac{49-25}{25}=0.96[/tex]
Engineering stress[tex]=\frac{Load}{original\ Cross-section}[/tex]
[tex]\sigma _E=\frac{9\times 10^3}{\frac{\pi 7^2}{4}}=233.83 MPa[/tex]
(b)True stress
[tex]\sigma _T=\sigma _E\left ( 1+\epsilon _E\right )[/tex]
[tex]\sigma _T=233.83\times (1+0.96)=458.30 MPa[/tex]
True strain
[tex]\epsilon _T=ln\left ( 1+\epsilon _E\right )[/tex]
[tex]\epsilon _T=ln\left ( 1.96\right )=0.672[/tex]
