Chapter 6.13, Problem 42AAP

To determine

Total percent of cold work done

Answer to Problem 42AAP

• • The total percent of cold work done is $\text{80}\text{.8}\text{\%}$.

Explanation of Solution

Write the expression for the initial cross sectional area of the wire.

  $A_0=\frac{\pi }{4}{d}_0{}^2$                                                                                                 (I)

Here, initial diameter of the wire is $d_0$.

Write the expression for the final cross sectional area of the wire.

  $A_f=\frac{\pi }{4}{d}_f{}^2$                                                                                                (II)

Here, final diameter of the wire is $d_f$.

Write the expression for the percentage of cold reduction:

  $\%CR=\left|\frac{A_0-A_f}{A_0}\right|\times 100\%$                                                                            (III)

Here, initial thickness of the sheet is $t_0$ and final thickness is $t_f$.

Conclusion:

Substitute equation (I) and (II) in equation (III).

$\%CR=\left|\frac{\frac{\pi }{4}{d}_0{}^2-\frac{\pi }{4}{d}_f{}^2}{\frac{\pi }{4}{d}_0{}^2}\right|\times 100\%$

$\%CR=\left|\frac{d_0{}^2-d_f{}^2}{d_0{}^2}\right|\times 100\%$ . (III)

Substitute 25% for $\%CR$ and 1.10 mm for $d_f$ in Equation (III)

  $\begin{array}{c}25\%=\left|\frac{d_0^2-1.10{\text{ mm}}^2}{d_o{}^2}\right|\times 100\%\\ 0.25d_o{}^2=d_0^2-1.10{\text{ mm}}^2\\ 1.10{\text{ mm}}^2=\left(1-0.25\right)d_0^2\\ d_0=1.211\end{array}$

Substitute $1.211\text{ mm}$ for $A_0$ in equation (I).

$\begin{array}{l}{A}_0=\frac{\pi }{4}{\left(1.211\text{ mm}\right)}^2\\ A_0=1.152{\text{ mm}}^2\end{array}$

Substitute $0.900\text{ mm}$ for $A_i$ in equation (II).

$\begin{array}{l}{A}_f=\frac{\pi }{4}{\left(0.900\text{ mm}\right)}^2\\ A_f=0.6362{\text{ mm}}^2\\ \end{array}$

Substitute $1.152{\text{ mm}}^2$ for $A_0$ and $0.6362{\text{ mm}}^2$ for $A_f$ in equation (III).

$\begin{array}{c}\%CR=\left|\frac{1.152{\text{ mm}}^2-0.6362{\text{ mm}}^2}{1.152{\text{ mm}}^2}\right|\times 100\%\\ =0.4477\times 100\%\\ \simeq 44.77\%\end{array}$

Thus, the total percent of cold reduction is 44.77%.