Chapter 8, Problem 8.26P

Interpretation Introduction

(a)

Interpretation:

The amount of cold work for each of the given steps should be determined.

Concept Introduction:

The relationship between cold work and area is:

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

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

Answer to Problem 8.26P

Value of cold working for the diameter ranging from $\text{25 mm to 22 mm}$ is $\text{29}\text{.13}\%$.

Value of cold working for the diameter ranging between is $\text{22 mm to 20 mm}$

  $\text{21\%}$.

Explanation of Solution

Given information:

Steel bar having initial diameter of 25-mm is cold extruded to 22 mm, and then from 22 mm to a final diameter of 20 mm.

Initial diameter ${\text{d}}_0$ = $\text{25}\text{mm}$

Final diameter ${\text{d}}_{\text{f}}$ = $\text{22 mm}$

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

The value of area is:

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

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:

  $\%CW=\left(\frac{\left(\frac{\pi }{4}\right)\times {\left(25\right)}^2-\left(\frac{\pi }{4}\right)\times {\left(22\right)}^2}{\left(\frac{\pi }{4}\right)\times {\left(25\right)}^2}\right)\times 100$

  $\%CW=\left(\frac{{25}^2-{22}^2}{{25}^2}\right)\times 100$

  $\%CW=\left(\frac{625-484}{625}\right)\times 100$

  $\%CW=0.2913\times 100$

  $\%CW=\text{ 29}\text{.13}\%$

If the diameter is changing from $\text{22mm}\text{to}\text{20mm}$ cold work is calculated on the basis of this relationship:

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

The above equation gives dependency of diameter value on cold work

Initial Diameter ${\text{d}}_0$ = $\text{22mm}$

Final Diameter ${\text{d}}_{\text{f}}$ = $\text{20mm}$

The values of area based on diameter is:

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

On substituting the values in terms of diameter:

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

  $\%CW=\left(\frac{\left(\frac{\pi }{4}\right)\times {\left(22\right)}^2-\left(\frac{\pi }{4}\right)\times {\left(20\right)}^2}{\left(\frac{\pi }{4}\right)\times {\left(22\right)}^2}\right)\times 100$

  $\begin{array}{l}\%CW=\left(\frac{{22}^2-{20}^2}{{22}^2}\right)\times 100\\ \%CW=\left(\frac{484-400}{484}\right)\times 100\end{array}$

  $\begin{array}{l}\%CW=0\cdot 21\times 100\\ \%CW=\text{ 21}\%\end{array}$

Interpretation Introduction

(b)

Interpretation:

The comparison between the amounts of cold work for each step to the amount of cold work for direct conversion from 25 mm to 20 mm should be done.

Explanation of Solution

Given information:

Diameter is changing from $\text{25 mm}\text{to}\text{20 mm}$

Steel bar having initial diameter of 25-mm is cold extruded to 22 mm, and then from 22 mm to a final diameter of 20 mm.

Cold work is the measure of degree of deformation and its calculation is done based on the original and final value area and according to the below statement cold work is directly proportional to the change in area:

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

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

The above formula to be used for the calculation of percent cold working as given in the question the value of initial diameter that is $\text{25 mm}$ and reduced diameter that is final diameter for first case is $\text{20 mm}$

Initial Diameter ${\text{d}}_0$ = $\text{25 mm}$

Final Diameter ${\text{d}}_{\text{f}}$ = $\text{20 mm}$

On inserting the values of initial and final diameter we have

  $\begin{array}{l}\%CW=\left(\frac{\left(\frac{\pi }{4}\right)\times d_0^2-\left(\frac{\pi }{4}\right)\times d_f^2}{\left(\frac{\pi }{4}\right)\times d_0^2}\right)\times 100\\ \%CW=\left(\frac{\left(\frac{\pi }{4}\right)\times {\left(25\right)}^2-\left(\frac{\pi }{4}\right)\times {\left(20\right)}^2}{\left(\frac{\pi }{4}\right)\times {\left(25\right)}^2}\right)\times 10\end{array}$

Taking $\left(\frac{\text{π}}{\text{4}}\right)$ as common and solving:

  $\text{\%CW}=\left(\frac{{25}^2-{20}^2}{{25}^2}\right)\times 100$

On solving and taking square:

  $\%CW=\left(\frac{625-400}{625}\right)\times 100$

  $\%CW=\left(0.36\right)\times 100$

  $\%CW=36\%$

The value of cold work is $\text{36\%}$ for the diameter ranging from $\text{25}\text{mm}\text{to}\text{20}\text{mm}$

As calculated in the above part

Value of cold working for the diameter ranging from $\text{25 mm to 22 mm}$ is $\text{29}\text{.13}\%$

Value of cold working for the diameter ranging between is $\text{22 mm to 20 mm}$ is $\text{21\%}$

Sum ofthe cold work of diameter ranging from $\text{25 mm to 22 mm}$ and $\text{22 mm to 20 mm}$ is 50.13

Sum of the cold work of diameter ranging from $\text{25}\text{mm}\text{to}\text{20}\text{mm}$ is $\text{36\%}$

The amount of cold work is more for diameter ranging from $\text{25 mm to 22 mm}$ and $\text{22 mm to 20 mm}$ is 50.13 as compared to diameter ranging from $\text{25}\text{mm}\text{to}\text{20}\text{mm}$ is $\text{36\%}$.