Chapter 6.13, Problem 61AAP

a)

To determine

The total percent of cold work has to be determined.

Answer to Problem 61AAP

The total percent of cold work is 39 %.

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)

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

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

Substitute 20% for $\%CR$, 2.80 mm for $d_f$ in equation (IV).

 $\begin{array}{l}20\%=\left|\frac{d_0{}^2-{\left(\text{2}\text{.80 mm}\right)}^2}{d_0{}^2}\right|\times 100\%\\ 0.2=\left|\frac{d_0{}^2-{\left(2.80\right)}^2}{d_0{}^2}\right|\\ 0.2d_0{}^2=d_0{}^2-7.84\\ d_0{}^2-0.2d_0{}^2=7.84\\ 0.8d_0{}^2=7.84\\ d_0=3.13\text{mm}\end{array}$

Substitute 3.13 mm for $d_i$ and 2.45 mm for $d_f$ in equation (III).

$\begin{array}{c}\%CR=\left|\frac{{\left(3.13\text{ mm}\right)}^2-{\left(2.45\text{ mm}\right)}^2}{{\left(3.13\text{ mm}\right)}^2}\right|\times 100\%\\ =\frac{9.797-6.003}{9.797}\times 100\%\\ =38.76\%\\ \simeq 39\%\end{array}$

Thus, the total percent of cold work that wire undergoes is 39%.

b)

To determine

Tensile strength, yield strength, and elongation have to be estimated.

Answer to Problem 61AAP

Ultimate tensile strength $\simeq 38\text{ ksi}$

Yield strength $\simeq 33\text{ ksi}$.

Elongation $\simeq 8\%$

Explanation of Solution

Refer the figure 6.44, to obtain a tensile strength, yield strength, and elongation for 40 percent of cold work as given as follows:

Ultimate tensile strength $\simeq 38\text{ ksi}$

Yield strength $\simeq 33\text{ ksi}$.

Elongation $\simeq 8\%$

Thus, the Ultimate tensile strength, Yield strength, and Elongation of the alloys is 38 ksi, 33 ksi, 8% respectively.