Chapter 6, Problem 6.77P

(a)

To determine

The work-done in open die forging of cylinder for no friction between the flat dies and the specimen.

Explanation of Solution

Given:

The initial thickness of the specimen is $h_o=10\text{mm}$ .

The initial radius of the specimen is $d_o=25\text{mm}$ .

The friction coefficient is $\mu =0$ .

Formula used:

The expression for the flow stress is given as,

  ${\sigma }_f=K{\epsilon }^n$ …… (1)

Here, ${\sigma }_f$ is the flow stress, $K$ is the strength coefficient, $\epsilon$ is the true strain, $n$ is the strain hardening coefficient.

The expression for the true strain is given as,

  $\epsilon =\ln \left(\frac{h_o}{h_f}\right)$

Here, $h_f$ is the final thickness.

The expression for the final radius by equating the volume is given as,

  $d_f^2=\frac{d_o^2{h}_o}{h_f}$

The expression for the forging force is given as,

  $F=p_a\frac{\pi d_f^2}{4}$

Here, $p_a$ is the average pressure.

The expression for the average pressure is given as,

  $p_a={\sigma }_f\left(1+\frac{\mu d_f}{3h_f}\right)$

The expression for final height for $10\%$ reduction in height in given as,

  $\begin{array}{l}\left(\frac{h_o-h_f}{h_o}\right)\times 100\%=10\%\\ h_f=0.9h_o\end{array}$

The expression for final height for $20\%$ reduction in height in given as,

  $\begin{array}{l}\left(\frac{h_o-h_f}{h_o}\right)\times 100\%=20\%\\ h_f=0.8h_o\end{array}$

The expression forfinal height for $30\%$ reduction in height in given as,

  $\begin{array}{l}\left(\frac{h_o-h_f}{h_o}\right)\times 100\%=30\%\\ h_f=0.7h_o\end{array}$

The expression for final height for $40\%$ reduction in height in given as,

  $\begin{array}{l}\left(\frac{h_o-h_f}{h_o}\right)\times 100\%=40\%\\ h_f=0.6h_o\end{array}$

The expression for final height for $50\%$ reduction in height in given as,

  $\begin{array}{l}\left(\frac{h_o-h_f}{h_o}\right)\times 100\%=50\%\\ h_f=0.5h_o\end{array}$

Calculation:

For $10\%$ reduction,

The final height can be calculated as,

  $\begin{array}{l}{h}_f=0.9h_o\\ h_f=0.9\times 10\text{mm}\\ h_f=9\text{mm}\end{array}$

The final radius can be calculated as,

  $\begin{array}{l}{d}_f^2=\frac{d_o^2{h}_o}{h_f}\\ d_f^2=\frac{{\left(25\text{mm}\right)}^2\times 10\text{mm}}{9\text{mm}}\\ d_f=26.35\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{9\text{mm}}\right)\\ \epsilon =0.1053\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=315\text{MPa}\times {\left(0.10536\right)}^{0.54}\\ {\sigma }_f=93.44\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed copper is,

  $\begin{array}{l}K=580\text{MPa}\\ n=0.34\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{\mu d_f}{3h_f}\right)\\ p_a=93.44\text{MPa}\times \left(1+\frac{0\times 26.35\text{mm}}{3\times 9\text{mm}}\right)\\ p_a=93.44\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\frac{\pi d_f^2}{4}\\ F=93.44\text{MPa}\times {\frac{3.14\times \left(26.35\text{mm}\right)}{4}}^2\\ F=50928\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.050928\text{MN}\end{array}$

For $20\%$ reduction,

The final height can be calculated as,

  $\begin{array}{l}{h}_f=0.8h_o\\ h_f=0.8\times 10\text{mm}\\ h_f=8\text{mm}\end{array}$

The final radius can be calculated as,

  $\begin{array}{l}{d}_f^2=\frac{d_o^2{h}_o}{h_f}\\ d_f^2=\frac{{\left(25\text{mm}\right)}^2\times 10\text{mm}}{8\text{mm}}\\ d_f=27.95\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{8\text{mm}}\right)\\ \epsilon =0.223\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=315\text{MPa}\times {\left(0.223\right)}^{0.54}\\ {\sigma }_f=140.135\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed copper is,

  $\begin{array}{l}K=580\text{MPa}\\ n=0.34\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{\mu d_f}{3h_f}\right)\\ p_a=140.135\text{MPa}\times \left(1+\frac{0\times 27.95\text{mm}}{3\times 8\text{mm}}\right)\\ p_a=140.135\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\frac{\pi d_f^2}{4}\\ F=140.135\text{MPa}\times {\frac{3.14\times \left(27.95\text{mm}\right)}{4}}^2\\ F=85936\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.085\text{MN}\end{array}$

For $30\%$ reduction,

The final height can be calculated as,

  $\begin{array}{l}{h}_f=0.7h_o\\ h_f=0.7\times 10\text{mm}\\ h_f=7\text{mm}\end{array}$

The final radius can be calculated as,

  $\begin{array}{l}{d}_f^2=\frac{d_o^2{h}_o}{h_f}\\ d_f^2=\frac{{\left(25\text{mm}\right)}^2\times 10\text{mm}}{7\text{mm}}\\ d_f=29.88\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{7\text{mm}}\right)\\ \epsilon =0.3566\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=315\text{MPa}\times {\left(0.3566\right)}^{0.54}\\ {\sigma }_f=180.525\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed copper is,

  $\begin{array}{l}K=580\text{MPa}\\ n=0.34\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{\mu d_f}{3h_f}\right)\\ p_a=180.525\text{MPa}\times \left(1+\frac{0\times 29.88\text{mm}}{3\times 7\text{mm}}\right)\\ p_a=180.525\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\frac{\pi d_f^2}{4}\\ F=180.525\text{MPa}\times {\frac{3.14\times \left(29.88\text{mm}\right)}{4}}^2\\ F=126522\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.1265\text{MN}\end{array}$

For $40\%$ reduction,

The final height can be calculated as,

  $\begin{array}{l}{h}_f=0.6h_o\\ h_f=0.6\times 10\text{mm}\\ h_f=6\text{mm}\end{array}$

The final radius can be calculated as,

  $\begin{array}{l}{d}_f^2=\frac{d_o^2{h}_o}{h_f}\\ d_f^2=\frac{{\left(25\text{mm}\right)}^2\times 10\text{mm}}{6\text{mm}}\\ d_f=32.275\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{6\text{mm}}\right)\\ \epsilon =0.5108\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=315\text{MPa}\times {\left(0.5108\right)}^{0.54}\\ {\sigma }_f=219.17\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed copper is,

  $\begin{array}{l}K=580\text{MPa}\\ n=0.34\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{\mu d_f}{3h_f}\right)\\ p_a=219.17\text{MPa}\times \left(1+\frac{0\times 32.275\text{mm}}{3\times 6\text{mm}}\right)\\ p_a=219.17\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\frac{\pi d_f^2}{4}\\ F=219.17\text{MPa}\times {\frac{3.14\times \left(32.275\text{mm}\right)}{4}}^2\\ F=179218\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.1792\text{MN}\end{array}$

For $50\%$ reduction,

The final height can be calculated as,

  $\begin{array}{l}{h}_f=0.5h_o\\ h_f=0.5\times 10\text{mm}\\ h_f=5\text{mm}\end{array}$

The final radius can be calculated as,

  $\begin{array}{l}{d}_f^2=\frac{d_o^2{h}_o}{h_f}\\ d_f^2=\frac{{\left(25\text{mm}\right)}^2\times 10\text{mm}}{5\text{mm}}\\ d_f=35.35\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{5\text{mm}}\right)\\ \epsilon =0.69\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=315\text{MPa}\times {\left(0.69\right)}^{0.54}\\ {\sigma }_f=258.437\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed copper is,

  $\begin{array}{l}K=580\text{MPa}\\ n=0.34\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{\mu d_f}{3h_f}\right)\\ p_a=258.437\text{MPa}\times \left(1+\frac{0\times 35.35\text{mm}}{3\times 5\text{mm}}\right)\\ p_a=258.437\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\frac{\pi d_f^2}{4}\\ F=258.437\text{MPa}\times {\frac{3.14\times \left(35.35\text{mm}\right)}{4}}^2\\ F=253514\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.2535\text{MN}\end{array}$

For $\mu =0$ ,

Reduction (in $\%$ )	Forging force (in $\text{MN}$ )	Area under curve (in $\text{mm}\cdot \text{kg}$ )
$10$	$0.0509$	$0.509$
$20$	$0.085$	$0.72$
$30$	$0.1265$	$1.025$
$40$	$0.1792$	$1.5285$
$50$	$0.2535$	2.1635

Work done can be calculated by calculating the sum of area under the curve.

  $\text{Work}\text{done}=5.964\text{mm}\cdot \text{kg}$

The figure (1) shows the curve between the forging force and reduction in height,

  

Figure (1)

(b)

To determine

The work-done in open die forging of cylinder for $\mu =0.1$ between the flat dies and the specimen.

Explanation of Solution

Given:

The initial thickness of the specimen is $h_o=10\text{mm}$ .

The initial radius of the specimen is $d_o=25\text{mm}$ .

The friction coefficient is $\mu =0.1$ .

Formula used:

The expression for the flow stress is given as,

  ${\sigma }_f=K{\epsilon }^n$ …… (1)

Here, ${\sigma }_f$ is the flow stress, $K$ is the strength coefficient, $\epsilon$ is the true strain, $n$ is the strain hardening coefficient.

The expression for the true strain is given as,

  $\epsilon =\ln \left(\frac{h_o}{h_f}\right)$

Here, $h_f$ is the final thickness.

The expression for the final radius by equating the volume is given as,

  $d_f^2=\frac{d_o^2{h}_o}{h_f}$

The expression for the forging force is given as,

  $F=p_a\frac{\pi d_f^2}{4}$

Here, $p_a$ is the average pressure.

The expression for the average pressure is given as,

  $p_a={\sigma }_f\left(1+\frac{\mu d_f}{3h_f}\right)$

The expression for final height for $10\%$ reduction in height in given as,

  $\begin{array}{l}\left(\frac{h_o-h_f}{h_o}\right)\times 100\%=10\%\\ h_f=0.9h_o\end{array}$

The expression for final height for $20\%$ reduction in height in given as,

  $\begin{array}{l}\left(\frac{h_o-h_f}{h_o}\right)\times 100\%=20\%\\ h_f=0.8h_o\end{array}$

The expression forfinal height for $30\%$ reduction in height in given as,

  $\begin{array}{l}\left(\frac{h_o-h_f}{h_o}\right)\times 100\%=30\%\\ h_f=0.7h_o\end{array}$

The expression for final height for $40\%$ reduction in height in given as,

  $\begin{array}{l}\left(\frac{h_o-h_f}{h_o}\right)\times 100\%=40\%\\ h_f=0.6h_o\end{array}$

The expression for final height for $50\%$ reduction in height in given as,

  $\begin{array}{l}\left(\frac{h_o-h_f}{h_o}\right)\times 100\%=50\%\\ h_f=0.5h_o\end{array}$

Calculation:

For $10\%$ reduction,

The final height can be calculated as,

  $\begin{array}{l}{h}_f=0.9h_o\\ h_f=0.9\times 10\text{mm}\\ h_f=9\text{mm}\end{array}$

The final radius can be calculated as,

  $\begin{array}{l}{d}_f^2=\frac{d_o^2{h}_o}{h_f}\\ d_f^2=\frac{{\left(25\text{mm}\right)}^2\times 10\text{mm}}{9\text{mm}}\\ d_f=26.35\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{9\text{mm}}\right)\\ \epsilon =0.1053\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=315\text{MPa}\times {\left(0.10536\right)}^{0.54}\\ {\sigma }_f=93.44\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed copper is,

  $\begin{array}{l}K=580\text{MPa}\\ n=0.34\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{\mu d_f}{3h_f}\right)\\ p_a=93.44\text{MPa}\times \left(1+\frac{0.1\times 26.35\text{mm}}{3\times 9\text{mm}}\right)\\ p_a=102.56\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\frac{\pi d_f^2}{4}\\ F=102.56\text{MPa}\times {\frac{3.14\times \left(26.35\text{mm}\right)}{4}}^2\\ F=58898\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.05889\text{MN}\end{array}$

For $20\%$ reduction,

The final height can be calculated as,

  $\begin{array}{l}{h}_f=0.8h_o\\ h_f=0.8\times 10\text{mm}\\ h_f=8\text{mm}\end{array}$

The final radius can be calculated as,

  $\begin{array}{l}{d}_f^2=\frac{d_o^2{h}_o}{h_f}\\ d_f^2=\frac{{\left(25\text{mm}\right)}^2\times 10\text{mm}}{8\text{mm}}\\ d_f=27.95\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{8\text{mm}}\right)\\ \epsilon =0.223\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=315\text{MPa}\times {\left(0.223\right)}^{0.54}\\ {\sigma }_f=140.135\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed copper is,

  $\begin{array}{l}K=580\text{MPa}\\ n=0.34\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{\mu d_f}{3h_f}\right)\\ p_a=140.135\text{MPa}\times \left(1+\frac{0.1\times 27.95\text{mm}}{3\times 8\text{mm}}\right)\\ p_a=156.454\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\frac{\pi d_f^2}{4}\\ F=156.454\text{MPa}\times {\frac{3.14\times \left(27.95\text{mm}\right)}{4}}^2\\ F=95943.4\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.0959\text{MN}\end{array}$

For $30\%$ reduction,

The final height can be calculated as,

  $\begin{array}{l}{h}_f=0.7h_o\\ h_f=0.7\times 10\text{mm}\\ h_f=7\text{mm}\end{array}$

The final radius can be calculated as,

  $\begin{array}{l}{d}_f^2=\frac{d_o^2{h}_o}{h_f}\\ d_f^2=\frac{{\left(25\text{mm}\right)}^2\times 10\text{mm}}{7\text{mm}}\\ d_f=29.88\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{7\text{mm}}\right)\\ \epsilon =0.3566\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=315\text{MPa}\times {\left(0.3566\right)}^{0.54}\\ {\sigma }_f=180.525\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed copper is,

  $\begin{array}{l}K=580\text{MPa}\\ n=0.34\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{\mu d_f}{3h_f}\right)\\ p_a=180.525\text{MPa}\times \left(1+\frac{0.1\times 29.88\text{mm}}{3\times 7\text{mm}}\right)\\ p_a=206.211\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\frac{\pi d_f^2}{4}\\ F=206.211\text{MPa}\times {\frac{3.14\times \left(29.88\text{mm}\right)}{4}}^2\\ F=144524\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.1445\text{MN}\end{array}$

For $40\%$ reduction,

The final height can be calculated as,

  $\begin{array}{l}{h}_f=0.6h_o\\ h_f=0.6\times 10\text{mm}\\ h_f=6\text{mm}\end{array}$

The final radius can be calculated as,

  $\begin{array}{l}{d}_f^2=\frac{d_o^2{h}_o}{h_f}\\ d_f^2=\frac{{\left(25\text{mm}\right)}^2\times 10\text{mm}}{6\text{mm}}\\ d_f=32.275\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{6\text{mm}}\right)\\ \epsilon =0.5108\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=315\text{MPa}\times {\left(0.5108\right)}^{0.54}\\ {\sigma }_f=219.17\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed copper is,

  $\begin{array}{l}K=580\text{MPa}\\ n=0.34\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{\mu d_f}{3h_f}\right)\\ p_a=219.17\text{MPa}\times \left(1+\frac{0.1\times 32.275\text{mm}}{3\times 6\text{mm}}\right)\\ p_a=258.47\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\frac{\pi d_f^2}{4}\\ F=258.47\text{MPa}\times {\frac{3.14\times \left(32.275\text{mm}\right)}{4}}^2\\ F=211354\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.2113\text{MN}\end{array}$

For $50\%$ reduction,

The final height can be calculated as,

  $\begin{array}{l}{h}_f=0.5h_o\\ h_f=0.5\times 10\text{mm}\\ h_f=5\text{mm}\end{array}$

The final radius can be calculated as,

  $\begin{array}{l}{d}_f^2=\frac{d_o^2{h}_o}{h_f}\\ d_f^2=\frac{{\left(25\text{mm}\right)}^2\times 10\text{mm}}{5\text{mm}}\\ d_f=35.35\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{5\text{mm}}\right)\\ \epsilon =0.69\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=315\text{MPa}\times {\left(0.69\right)}^{0.54}\\ {\sigma }_f=258.437\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed copper is,

  $\begin{array}{l}K=580\text{MPa}\\ n=0.34\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{\mu d_f}{3h_f}\right)\\ p_a=258.437\text{MPa}\times \left(1+\frac{0.1\times 35.35\text{mm}}{3\times 5\text{mm}}\right)\\ p_a=319.34\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\frac{\pi d_f^2}{4}\\ F=319.34\text{MPa}\times {\frac{3.14\times \left(35.35\text{mm}\right)}{4}}^2\\ F=313256\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.3132\text{MN}\end{array}$

For $\mu =0.1$

Reduction (in $\%$ )	Forging force (in $\text{MN}$ )	Area under curve (in $\text{mm}\cdot \text{kg}$ )
$10$	$0.05889$	$0.294$
$20$	$0.0959$	$0.774$
$30$	$0.1445$	$1.202$
$40$	$0.2113$	$1.779$
$50$	$0.3132$	$2.6225$

Work done can be calculated by calculating the sum of area under the curve.

  $\text{Work}\text{done}=6.6715\text{mm}\cdot \text{kg}$

The figure (2) shows the curve between the forging force and reduction in height,

  

Figure (2)