Chapter 6, Problem 6.76P

(a)

To determine

The force vs. reduction in height curve 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 $\text{MN}$ )
$10$	$0.0509$
$20$	$0.085$
$30$	$0.1265$
$40$	$0.1792$
$50$	$0.2535$

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

  

Figure (1)

(b)

To determine

The force vs. reduction in height curve 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 $\text{MN}$ )
$10$	$0.05889$
$20$	$0.0959$
$30$	$0.1445$
$40$	$0.2113$
$50$	$0.3132$

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

  

Figure (2)