Answer:n=0.973
Explanation:
Given
When True strain[tex]\left ( \epsilon _T_1\right )=0.171[/tex]
at [tex]\sigma _1=263.8 MPa[/tex]
When True stress[tex]\left ( \sigma _2\right )[/tex]=346.2 MPa
true strain [tex]\left ( \epsilon _T_2\right )[/tex]=0.226
We know
[tex]\sigma =k\epsilon ^n [/tex]
where [tex]\sigma [/tex]=True stress
[tex]\epsilon [/tex]=true strain
n=strain hardening exponent
k=constant
Substituting value
[tex]263.8=k\left ( 0.171\right )^n------1[/tex]
[tex]346.2=k\left ( 0.226\right )^n-----2[/tex]
Divide 1 & 2 to get
[tex]\frac{346.2}{263.8}=\left ( \frac{0.226}{0.171}\right )^n[/tex]
[tex]1.312=\left ( 1.3216\right )^n[/tex]
Taking Log both side
[tex]ln\left ( 1.312\right )=nln\left ( 1.3216\right )[/tex]
n=0.973
