Chapter 8, Problem 8.25P

Interpretation Introduction

Interpretation:

The percent cold work (% CW) for the given two cases should be determined.

Concept Introduction:

The general relationship between the percentage cold working and area is based on the formula:

  $\begin{array}{l}\%CW=\left(\frac{A_0-A_f}{A_0}\right)\times 100\\ A_0=\left(\frac{\pi }{4}{d}_0^2\right)\\ A_f=\left(\frac{\pi }{4}{d}_f^2\right)\end{array}$

For the above equation,

  $\%CW$ is the percentage of cold working.

  ${\text{A}}_{\text{0}}$ is the initial area of copper bar.

  $A_f$ is the final area of coper bar.

Answer to Problem 8.25P

The value of percent cold working for 2-inch diameter reduced to 1.5 in. and then to 1 in. diameter is $43.7\text{5 \%}$.

The value of percent cold working for 2inches to 1-inch diameter is $7\text{5 \%}$.

Explanation of Solution

Given Information:

A copper rod is reduced from 2 inches diameter to 1.5-inchdiameter and on further reduction attaining the value of final diameter as 1 inch.

For the next case, the diameter is reducing in a single step that is from a value of 2 inches to 1 inch

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

As,

  $\begin{array}{l}{A}_0=\left(\frac{\pi }{4}\right)\times d_0^2\\ A_f=\left(\frac{\pi }{4}\right)\times d_f^2\end{array}$

On substituting the value of area in the above formula of percentage cold working,

  $\%CW=\left(\frac{\left(\frac{\pi }{4}\right)\times d_0^2-\left(\frac{\pi }{4}\right)\times d_f^2}{\left(\frac{\pi }{4}\right)\times d_0^2}\right)\times 100$

Now, on substituting the value of initial diameter and final diameter,

  $\%CW=\left(\frac{\left(\frac{\pi }{4}\right)\times {\left(2\right)}^2-\left(\frac{\pi }{4}\right)\times {\left(1.5\right)}^2}{\left(\frac{\pi }{4}\right)\times {\left(2\right)}^2}\right)\times 100$

  $\%CW=\left(\frac{2^2-1\cdot 5^2}{2^2}\right)\times 100$

  $\begin{array}{l}\%CW=\left(\frac{4-2\cdot 25}{4}\right)\times 100\\ \%CW=\left(\frac{1\cdot 75}{4}\right)\times 100\end{array}$

  $\text{\%CW}\text{=}43\cdot 75\%$

Calculation of percent cold working when the initial diameter of 2 inches is further reduced to 1 inch

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

  $\begin{array}{l}{A}_0=\left(\frac{\pi }{4}\right)\times d_0^2\\ A_f=\left(\frac{\pi }{4}\right)\times d_f^2\end{array}$

As given,

Initial diameter ${\text{d}}_{\text{0}}$ = 2 inches

Reducing diameter $\left(\text{final}\text{diameter}\right)$

  ${\text{d}}_{\text{f}}$ = $\text{1}\text{inch}$

On substituting the values of diameter in the above equation of cold working,

  $\%CW=\left(\frac{\left(\frac{\pi }{4}\right)\times d_0^2-\left(\frac{\pi }{4}\right)\times d_f^2}{\left(\frac{\pi }{4}\right)\times d_0^2}\right)\times 100$

  $\%CW=\left(\frac{\left(\frac{\pi }{4}\right)\times {\left(2\right)}^2-\left(\frac{\pi }{4}\right)\times {\left(1\right)}^2}{\left(\frac{\pi }{4}\right)\times {\left(2\right)}^2}\right)\times 100$

  $\begin{array}{l}\%CW=\left(\frac{2^2-1^2}{2^2}\right)\times 100\\ \%CW=\left(\frac{4-1}{4}\right)\times 100\\ \%CW=\left(\frac{3}{4}\right)\times 100\\ \%CW=75\%\\ \end{array}$

For third case, calculation of percent cold working when the initial diameter of 2-inch is reduced to 1-inch

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

  $\begin{array}{l}{A}_0=\left(\frac{\pi }{4}\right)\times d_0^2\\ A_f=\left(\frac{\pi }{4}\right)\times d_f^2\end{array}$

As given:

Initial diameter ${\text{d}}_{\text{0}}$ = 2inches

Reducing diameter $\left(\text{final}\text{diameter}\right)$

  ${\text{d}}_{\text{f}}$ = $\text{1}\text{inch}$

On substituting the values of diameter in the above equation of cold working,

  $\%CW=\left(\frac{\left(\frac{\pi }{4}\right)\times d_0^2-\left(\frac{\pi }{4}\right)\times d_f^2}{\left(\frac{\pi }{4}\right)\times d_0^2}\right)\times 100$

  $\%CW=\left(\frac{\left(\frac{\pi }{4}\right)\times {\left(2\right)}^2-\left(\frac{\pi }{4}\right)\times {\left(1\right)}^2}{\left(\frac{\pi }{4}\right)\times {\left(2\right)}^2}\right)\times 100$

  $\begin{array}{l}\%CW=\left(\frac{2^2-1^2}{2^2}\right)\\ \%CW=\left(\frac{4-1}{4}\right)\times 100\\ \%CW=\left(\frac{3}{4}\right)\times 100\\ \%CW=75\%\end{array}$

Conclusion

The value of percent cold working for a diameter change of 2inch to 1.5 inch is 43.75%.

The value of percent cold working for a diameter change of 2inch to 1 inch is 75%.