Answer:(a)[tex]45,300.24 lb/in/^2[/tex]
(b)0.255
Explanation:
Given
Strain hardening exponent(n)=0.22
Strength coefficient(k)[tex]=54000 lb/in/^2[/tex]
and we know
[tex]\sigma =k\left ( \epsilon \right )^n[/tex]
where
[tex]sigma =true\ stress[/tex]
[tex]\epsilon =true\ strain[/tex]
(a)True strain=0.45
[tex]\sigma =54000\times 0.45^{0.22}[/tex]
[tex]\sigma =45,300.24 lb/in^2[/tex]
(b)true stress[tex]=40,000 lb/in^2[/tex]
[tex]40000=54000\times \epsilon ^{0.22}[/tex]
[tex]\epsilon ^{0.22}=0.7407[/tex]
[tex]\epsilon =0.7407^{4.5454}=0.255[/tex]
