Answer:stress=79.56MPa
strain=[tex]3.8\times 10^{-4}[/tex]
Young Modulus=209.36 GPa
Explanation:
Given data
d=20 mm
Length[tex]\left ( L\right )[/tex]=80mm
[tex]\Delta {L}[/tex]=[tex]3.04\times 10^{-2}[/tex]mm
Load=[tex]25\times 10^{3}[/tex]N
[tex]\left ( i\right )[/tex]
Stress=[tex]\frac{Load\ applied}{cross-section}[/tex]
Stress=[tex]\frac{25\times 10^{3}}{314.2}[/tex]
Stress=79.56MPa
[tex]\left ( ii\right )[/tex]
Strain=[tex]\frac{Change\ in\ length}{Length}[/tex]
Strain=[tex]\frac{3.04\times 10^{-3}}{80}[/tex]
Strain=[tex]3.8\times 10^{-4}[/tex]
[tex]\left ( iii\right )[/tex]
young modulus[tex]\left ( E\right )[/tex]=[tex]\frac{Stress}{Strain}[/tex]
E=[tex]\frac{79.56\times 10^{6}}{3.8\times 10^{-4}}[/tex]
E=209.36GPa
