Chapter 6, Problem 6.79P

(a)

To determine

The force vs. reduction in height curve in open die forging of cylinder for $\mu =0.2$ 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 diameter of the specimen is $d_o=25\text{mm}$ .

The friction coefficient is $\mu =0.2$ .

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,

  $r_f^2=\frac{r_o^2{h}_o}{h_f}$

The expression for the forging force is given as,

  $F=p_a\pi r_f^2$

Here, $p_a$ is the average pressure.

The expression for the average pressure is given as,

  $p_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)$

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 for final 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,

  $\begin{array}{l}{r}_f^2=\frac{r_o^2{h}_o}{h_f}\\ r_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 10\text{mm}}{9\text{mm}}\\ r_f=13.17\text{mm}\end{array}$

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,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=1400\text{MPa}\times {\left(0.10536\right)}^{0.015}\\ {\sigma }_f=1353.53\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed Ti-6Al-4V is,

  $\begin{array}{l}K=14000\text{MPa}\\ n=0.015\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=1353.53\text{MPa}\times \left(1+\frac{2\times 0.2\times 13.17\text{mm}}{3\times 9\text{mm}}\right)\\ p_a=1617.61\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=1617.61\text{MPa}\times 3.14\times {\left(13.17\text{mm}\right)}^2\\ F=880998.2\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.88\text{MN}\end{array}$

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,

  $\begin{array}{l}{r}_f^2=\frac{r_o^2{h}_o}{h_f}\\ r_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 10\text{mm}}{8\text{mm}}\\ r_f=13.97\text{mm}\end{array}$

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,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=1400\text{MPa}\times {\left(0.22314\right)}^{0.015}\\ {\sigma }_f=1368.85\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed Ti-6Al-4V is,

  $\begin{array}{l}K=14000\text{MPa}\\ n=0.015\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=1368.85\text{MPa}\times \left(1+\frac{2\times 0.2\times 13.97\text{mm}}{3\times 8\text{mm}}\right)\\ p_a=1652.15\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=1652.15\text{MPa}\times 3.14\times {\left(13.97\text{mm}\right)}^2\\ F=1012446.15\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=1.033\text{MN}\end{array}$

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,

  $\begin{array}{l}{r}_f^2=\frac{r_o^2{h}_o}{h_f}\\ r_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 10\text{mm}}{7\text{mm}}\\ r_f=14.94\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{7\text{mm}}\right)\\ \epsilon =0.356\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=1400\text{MPa}\times {\left(0.356\right)}^{0.015}\\ {\sigma }_f=1378.47\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed Ti-6Al-4V is,

  $\begin{array}{l}K=14000\text{MPa}\\ n=0.015\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=1378.47\text{MPa}\times \left(1+\frac{2\times 0.2\times 14.94\text{mm}}{3\times 7\text{mm}}\right)\\ p_a=1770.74\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=1770.74\text{MPa}\times 3.14\times {\left(14.97\text{mm}\right)}^2\\ F=1241039.604\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=1.241\text{MN}\end{array}$

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,

  $\begin{array}{l}{r}_f^2=\frac{r_o^2{h}_o}{h_f}\\ r_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 10\text{mm}}{6\text{mm}}\\ r_f=16.13\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{6\text{mm}}\right)\\ \epsilon =0.51\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=1400\text{MPa}\times {\left(0.51\right)}^{0.015}\\ {\sigma }_f=1385.96\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed Ti-6Al-4V is,

  $\begin{array}{l}K=14000\text{MPa}\\ n=0.015\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=1385.96\text{MPa}\times \left(1+\frac{2\times 0.2\times 16.13\text{mm}}{3\times 6\text{mm}}\right)\\ p_a=1882.749\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=1882.749\text{MPa}\times 3.14\times {\left(16.13\text{mm}\right)}^2\\ F=1538122.08\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=1.538\text{MN}\end{array}$

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,

  $\begin{array}{l}{r}_f^2=\frac{r_o^2{h}_o}{h_f}\\ r_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 10\text{mm}}{5\text{mm}}\\ r_f=17.67\text{mm}\end{array}$

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,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=1400\text{MPa}\times {\left(0.69\right)}^{0.015}\\ {\sigma }_f=1392.32\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed Ti-6Al-4V is,

  $\begin{array}{l}K=14000\text{MPa}\\ n=0.015\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=1392.32\text{MPa}\times \left(1+\frac{2\times 0.2\times 17.67\text{mm}}{3\times 5\text{mm}}\right)\\ p_a=2048.381\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=2048.381\text{MPa}\times 3.14\times {\left(17.67\text{mm}\right)}^2\\ F=2008230.344\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=2.00\text{MN}\end{array}$

For $\mu =0.2$

Reduction (in $\%$ )	Forging force (in $\text{MN}$ )
$10$	$0.88$
$20$	$1.033$
$30$	$1.24$
$40$	$1.53$
$50$	$2.00$

The plot between forging force and reduction in height is shown in figure (1) below,

  

  Figure (1)

(b)

To determine

The force vs. reduction in height curve in open die forging of cylinder for $\mu =0.4$ 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 diameter of the specimen is $d_o=25\text{mm}$ .

The friction coefficient is $\mu =0.4$ .

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,

  $r_f^2=\frac{r_o^2{h}_o}{h_f}$

The expression for the forging force is given as,

  $F=p_a\pi r_f^2$

Here, $p_a$ is the average pressure.

The expression for the average pressure is given as,

  $p_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)$

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 for final 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,

  $\begin{array}{l}{r}_f^2=\frac{r_o^2{h}_o}{h_f}\\ r_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 10\text{mm}}{9\text{mm}}\\ r_f=13.17\text{mm}\end{array}$

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,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=650\text{MPa}\times {\left(0.10536\right)}^{0.064}\\ {\sigma }_f=562.81\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed Ti-6Al-4V is,

  $\begin{array}{l}K=650\text{MPa}\\ n=0.064\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=562.81\text{MPa}\times \left(1+\frac{2\times 0.4\times 13.17\text{mm}}{3\times 9\text{mm}}\right)\\ p_a=782.43\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=782.43\text{MPa}\times 3.14\times {\left(13.17\text{mm}\right)}^2\\ F=426135.02\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.42\text{MN}\end{array}$

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,

  $\begin{array}{l}{r}_f^2=\frac{r_o^2{h}_o}{h_f}\\ r_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 10\text{mm}}{8\text{mm}}\\ r_f=13.97\text{mm}\end{array}$

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,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=650\text{MPa}\times {\left(0.22314\right)}^{0.064}\\ {\sigma }_f=590.479\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed Ti-6Al-4V is,

  $\begin{array}{l}K=14000\text{MPa}\\ n=0.015\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=590.479\text{MPa}\times \left(1+\frac{2\times 0.4\times 13.97\text{mm}}{3\times 8\text{mm}}\right)\\ p_a=865.445\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=865.44\text{MPa}\times 3.14\times {\left(13.97\text{mm}\right)}^2\\ F=530349.456\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.53\text{MN}\end{array}$

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,

  $\begin{array}{l}{r}_f^2=\frac{r_o^2{h}_o}{h_f}\\ r_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 10\text{mm}}{7\text{mm}}\\ r_f=14.94\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{7\text{mm}}\right)\\ \epsilon =0.356\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=650\text{MPa}\times {\left(0.356\right)}^{0.064}\\ {\sigma }_f=608.42\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed Ti-6Al-4V is,

  $\begin{array}{l}K=650\text{MPa}\\ n=0.064\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=608.42\text{MPa}\times \left(1+\frac{2\times 0.4\times 14.94\text{mm}}{3\times 7\text{mm}}\right)\\ p_a=954.69\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=954.69\text{MPa}\times 3.14\times {\left(14.97\text{mm}\right)}^2\\ F=671793.22\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.67\text{MN}\end{array}$

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,

  $\begin{array}{l}{r}_f^2=\frac{r_o^2{h}_o}{h_f}\\ r_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 10\text{mm}}{6\text{mm}}\\ r_f=16.13\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_o}{h_f}\right)\\ \epsilon =\ln \left(\frac{10\text{mm}}{6\text{mm}}\right)\\ \epsilon =0.51\end{array}$

The flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=650\text{MPa}\times {\left(0.51\right)}^{0.064}\\ {\sigma }_f=622.58\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed Ti-6Al-4V is,

  $\begin{array}{l}K=650\text{MPa}\\ n=0.064\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=622.58\text{MPa}\times \left(1+\frac{2\times 0.4\times 16.13\text{mm}}{3\times 6\text{mm}}\right)\\ p_a=1068.90\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=1068.90\text{MPa}\times 3.14\times {\left(16.13\text{mm}\right)}^2\\ F=873244.25\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.87\text{MN}\end{array}$

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,

  $\begin{array}{l}{r}_f^2=\frac{r_o^2{h}_o}{h_f}\\ r_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 10\text{mm}}{5\text{mm}}\\ r_f=17.67\text{mm}\end{array}$

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,

  $\begin{array}{l}{\sigma }_f=K{\epsilon }^n\\ {\sigma }_f=650\text{MPa}\times {\left(0.69\right)}^{0.064}\\ {\sigma }_f=634.74\text{MPa}\end{array}$

Refer to table 2.2 “Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for annealed Ti-6Al-4V is,

  $\begin{array}{l}K=650\text{MPa}\\ n=0.064\end{array}$

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=634.74\text{MPa}\times \left(1+\frac{2\times 0.4\times 17.67\text{mm}}{3\times 5\text{mm}}\right)\\ p_a=1232.91\text{MPa}\end{array}$

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=1232.918\text{MPa}\times 3.14\times {\left(17.67\text{mm}\right)}^2\\ F=1208752.21\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=1.2\text{MN}\end{array}$

For $\mu =0.4$

Reduction (in $\%$ )	Forging force (in $\text{MN}$ )
$10$	$0.42$
$20$	$0.52$
$30$	$0.67$
$40$	$0.87$
$50$	$1.20$

The plot between forging force and reduction in height is shown in figure (2) below,

  

  Figure (2)