Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)

Answer 1

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is $=21.29mm$

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

       A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as $\sigma_T$.

True Strain

     A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as $\epsilon_T$.

The mathematical relation between stress to strain on the plastic region of deformation is

              $\sigma _T =K\epsilon^n_T$

Where K is a constant

          n is known as the strain hardening exponent

           This constant K can be obtained as follows

                        $K = \frac{\sigma_T}{(\epsilon_T)^n}$

No substituting  $345MPa \ for \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and \ 0.22 \ for \ n$ from the question we have

                     $K = \frac{345}{(0.02)^{0.22}}$

                          $= 815.82MPa$

Making $\epsilon_T$ the subject from the equation above

              $\epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }$

$Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K \ and \ 0.22 \ for \ n$

       $\epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }$

            $=0.0443$

       

From the definition we mentioned instantaneous length and this can be  obtained mathematically as follows

           $l_i = l_o e^{\epsilon_T}$

Where

       $l_i$ is the instantaneous length

      $l_o$ is the original length

$Substituting \ 470mm \ for \ l_o \ and \ 0.0443 \ for \ \epsilon_T$

             $l_i = 470 * e^{0.0443}$

                $=491.28mm$

We can also obtain the elongated length mathematically as follows

            $Elongated \ Length =l_i - l_o$

$Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i$

          $Elongated \ Length = 491.28 - 470$

                                       $=21.29mm$