Question

A tensile force of 9 kN is applied to the ends of a solid bar of 7.0 mm diameter. Under load, the diameter reduces to 5.00 mm. Assuming uniform deformation and volume constancy, (a) determine the engineering stress and strain; (b) determine the true stress and strain.

Answer 1

Step-by-step explanation:

Given

Force=9 kN

Diameter reduces from 7 mm to 5 mm

As volume remains constant therefore

`A_0L_0=A_fL_f`

`7^2* L_0=5^2* L_f`

`(L_0)/(L_f)=(25)/(49)`

Thus Engineering Strain
`\epsilon _E=(L_f-L_0)/(L_0)`

`=(49)/(25)-1=(49-25)/(25)=0.96`

Engineering stress
`=(Load)/(original\ Cross-section)`

`\sigma _E=(9* 10^3)/((\pi 7^2)/(4))=233.83 MPa`

(b)True stress

`\sigma _T=\sigma _E\left ( 1+\epsilon _E\right )`

`\sigma _T=233.83* (1+0.96)=458.30 MPa`

True strain

`\epsilon _T=ln\left ( 1+\epsilon _E\right )`

`\epsilon _T=ln\left ( 1.96\right )=0.672`