Question

The flow stress (in MPa) of a material is given by $\sigma =K{\in }^n,n=0.1$
where $\in$ is the true strain and K is strength coefficient. A rod made up of this material has a length of 200 mm. A compressive load is applied on the rod to reduce its length by 10 mm and then the load is removed. If the Young's modulus of elasticity of the material is 250 GPa and the final length of ro is 190.1362 mm, then the value of K is

• A. 241.19 MPa
• B. 150.82 MPa
• C. 200.33 MPa
• D. 260.19 MPa

Answer 1

The correct option is A 241.19 MPa

The rod is compressed by 10 mm by applying a compressive load.
We know, the true strain is given by,

$\Rightarrow {\in }_T=ln\left(\frac{190}{200}\right)=-0.0513$

${\sigma }_0=K{\in }^n=K\times {\left(0.0513\right)}^{0.1}$

${\in }_{\text{Elastic}}=\frac{{\sigma }_0}{E}$

${\left(\frac{\Delta L}{L}\right)}_{\text{Elastic}}=\frac{{\sigma }_0}{E}$

$\frac{0.1362}{190}=\frac{K\times {\left(0.0513\right)}^{0.1}}{250\times {10}^3}$