Chapter 6, Problem 6.74P

(a)

To determine

The force vs. reduction in height curve in open die forging of cylinder.

Explanation of Solution

Given:

The initial thickness of the specimen is $h_o=50\text{mm}$ .

The initial diameter 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,

  $r_f^2=\frac{r_o^2{h}_o}{h_f}$

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

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

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

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

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

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

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,

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

The expression for final height for $60\%$ reduction in height in given as,

  $\begin{array}{l}\left(\frac{h_o-h_f}{h_o}\right)\times 100\%=60\%\\ h_f=0.4h_o\end{array}$

The expression for final height for $60\%$ reduction in height in given as,

  $\begin{array}{l}\left(\frac{h_o-h_f}{h_o}\right)\times 100\%=70\%\\ h_f=0.3h_o\end{array}$

Calculation:

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

  $\begin{array}{l}K=315\text{MPa}\\ n=0.54\end{array}$

The flow stress can be calculated as,

  ${\sigma }_f=315\text{MPa}\times {\epsilon }^{0.54}$ …… (1)

The true strain can be calculated as,

  $\epsilon =\ln \left(\frac{50\text{mm}}{h_f}\right)$ …… (2)

Obtain the expression by substituting equation 2 in 1,

  ${\sigma }_f=315\text{MPa}\times {\left(\ln \left(\frac{50\text{mm}}{h_f}\right)\right)}^{0.54}$ …… (3)

The final radius can be calculated as,

  $\begin{array}{l}{r}_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 50\text{mm}}{h_f}\\ r_f^2=\frac{7812.5\text{mm}}{h_f}\end{array}$ …… (4)

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{h_f}\right)\right)}^{0.54}\times \left(1+\frac{2\times \mu \times 13.17\text{mm}}{3\times 45\text{mm}}\right)\end{array}$ …… (5)

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{h_f}\right)\right)}^{0.54}\times \left(1+\frac{2\times \mu \times 13.17\text{mm}}{3\times 45\text{mm}}\right)\pi r_f^2\end{array}$ …… (6)

For $10\%$ reduction,

The final height for $10\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=10\%\\ h_f=45\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{45\text{mm}}\\ r_f^2=173.61{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{45\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0\times 13.17\text{mm}}{3\times 45\text{mm}}\right)\times \pi \left(173.61\text{mm}\right)\\ F=49823\text{N}\left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.049\text{MN}\end{array}$

For $20\%$ reduction,

The final height for $20\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=20\%\\ h_f=40\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{40\text{mm}}\\ r_f^2=195.31{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{40\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0\times 13.17\text{mm}}{3\times 40\text{mm}}\right)\times \pi \left(195.31\text{mm}\right)\\ F=85042\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.085\text{MN}\end{array}$

For $30\%$ reduction,

The final height for $30\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=30\%\\ h_f=35\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{35\text{mm}}\\ r_f^2=223.21{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{35\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0\times 13.17\text{mm}}{3\times 35\text{mm}}\right)\times \pi \left(223.21\text{mm}\right)\\ F=125963.1\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.12\text{MN}\end{array}$

For $40\%$ reduction,

The final height for $40\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=40\%\\ h_f=30\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{30\text{mm}}\\ r_f^2=360.41{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{30\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0\times 13.17\text{mm}}{3\times 30\text{mm}}\right)\times \pi \left(260.41\text{mm}\right)\\ F=179295.26\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.17\text{MN}\end{array}$

For $50\%$ reduction,

The final height for $50\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=50\%\\ h_f=25\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{25\text{mm}}\\ r_f^2=360.41{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{25\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0\times 13.17\text{mm}}{3\times 25\text{mm}}\right)\times \pi \left(312.5\text{mm}\right)\\ F=253683.60\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.25\text{MN}\end{array}$

For $60\%$ reduction,

The final height for $50\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=60\%\\ h_f=20\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{20\text{mm}}\\ r_f^2=390.62{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{20\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0\times 13.17\text{mm}}{3\times 20\text{mm}}\right)\times \pi \left(390.62\text{mm}\right)\\ F=368733.58\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.36\text{MN}\end{array}$

For $70\%$ reduction,

The final height for $70\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=70\%\\ h_f=15\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{15\text{mm}}\\ r_f^2=520.83{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{15\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0\times 13.17\text{mm}}{3\times 15\text{mm}}\right)\times \pi \left(520.83\text{mm}\right)\\ F=569753.63\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.56\text{MN}\end{array}$

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

  

Figure (1)

To determine

The force vs. reduction in height curve in open die forging of cylinder.

Answer to Problem 6.74P

The force vs. reduction in height curve in open die forging of cylinder.

Explanation of Solution

Calculation:

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

  $\begin{array}{l}K=315\text{MPa}\\ n=0.54\end{array}$

The flow stress can be calculated as,

  ${\sigma }_f=315\text{MPa}\times {\epsilon }^{0.54}$ …… (1)

The true strain can be calculated as,

  $\epsilon =\ln \left(\frac{50\text{mm}}{h_f}\right)$ ...... (2)

Obtain the expression by substituting equation 2 in 1,

  ${\sigma }_f=315\text{MPa}\times {\left(\ln \left(\frac{50\text{mm}}{h_f}\right)\right)}^{0.54}$ ...... (3)

The final radius can be calculated as,

  $\begin{array}{l}{r}_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 50\text{mm}}{h_f}\\ r_f^2=\frac{7812.5\text{mm}}{h_f}\end{array}$ ...... (4)

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{h_f}\right)\right)}^{0.54}\times \left(1+\frac{2\times \mu \times 13.17\text{mm}}{3\times 45\text{mm}}\right)\end{array}$ ...... (5)

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{h_f}\right)\right)}^{0.54}\times \left(1+\frac{2\times \mu \times 13.17\text{mm}}{3\times 45\text{mm}}\right)\pi r_f^2\end{array}$ ...... (6)

For $10\%$ reduction,

The final height for $10\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=10\%\\ h_f=45\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{45\text{mm}}\\ r_f^2=173.61{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.25$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{45\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.25\times 13.17\text{mm}}{3\times 45\text{mm}}\right)\times \pi \left(173.61\text{mm}\right)\\ F=52214\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.052\text{MN}\end{array}$

For $20\%$ reduction,

The final height for $20\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=20\%\\ h_f=40\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{40\text{mm}}\\ r_f^2=195.31{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.25$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{40\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.25\times 13.17\text{mm}}{3\times 40\text{mm}}\right)\times \pi \left(195.31\text{mm}\right)\\ F=59590.24\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.059\text{MN}\end{array}$

For $30\%$ reduction,

The final height for $30\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=30\%\\ h_f=35\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{35\text{mm}}\\ r_f^2=223.21{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.25$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{35\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.25\times 13.17\text{mm}}{3\times 35\text{mm}}\right)\times \pi \left(223.21\text{mm}\right)\\ F=134780.55\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.134\text{MN}\end{array}$

For $40\%$ reduction,

The final height for $40\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=40\%\\ h_f=30\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{30\text{mm}}\\ r_f^2=260.41{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.25$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{30\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0\times 13.17\text{mm}}{3\times 30\text{mm}}\right)\times \pi \left(260.41\text{mm}\right)\\ F=195761.25\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.195\text{MN}\end{array}$

For $50\%$ reduction,

The final height for $50\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=50\%\\ h_f=25\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{25\text{mm}}\\ r_f^2=312.5{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.25$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{25\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.25\times 13.17\text{mm}}{3\times 25\text{mm}}\right)\times \pi \left(312.5\text{mm}\right)\\ F=283597.53\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.28\text{MN}\end{array}$

For $60\%$ reduction,

The final height for $60\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=60\%\\ h_f=20\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{20\text{mm}}\\ r_f^2=390.62{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.25$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{20\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.25\times 13.17\text{mm}}{3\times 20\text{mm}}\right)\times \pi \left(390.62\text{mm}\right)\\ F=429205.88\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.42\text{MN}\end{array}$

For $70\%$ reduction,

The final height for $70\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=70\%\\ h_f=15\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{15\text{mm}}\\ r_f^2=520.83{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.25$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{15\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.25\times 13.17\text{mm}}{3\times 15\text{mm}}\right)\times \pi \left(520.83\text{mm}\right)\\ F=694387.24\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.69\text{MN}\end{array}$

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

  

Figure (2)

To determine

The force vs. reduction in height curve in open die forging of cylinder.

Explanation of Solution

Calculation:

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

  $\begin{array}{l}K=315\text{MPa}\\ n=0.54\end{array}$

The flow stress can be calculated as,

  ${\sigma }_f=315\text{MPa}\times {\epsilon }^{0.54}$ ...... (1)

The true strain can be calculated as,

  $\epsilon =\ln \left(\frac{50\text{mm}}{h_f}\right)$ ...... (2)

Obtain the expression by substituting equation 2 in 1,

  ${\sigma }_f=315\text{MPa}\times {\left(\ln \left(\frac{50\text{mm}}{h_f}\right)\right)}^{0.54}$ ...... (3)

The final radius can be calculated as,

  $\begin{array}{l}{r}_f^2=\frac{{\left(12.5\text{mm}\right)}^2\times 50\text{mm}}{h_f}\\ r_f^2=\frac{7812.5\text{mm}}{h_f}\end{array}$ ...... (4)

The average pressure can be calculated as,

  $\begin{array}{l}{p}_a={\sigma }_f\left(1+\frac{2\mu r_f}{3h_f}\right)\\ p_a=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{h_f}\right)\right)}^{0.54}\times \left(1+\frac{2\times \mu \times 13.17\text{mm}}{3\times h_f}\right)\end{array}$ ...... (5)

The forging force can be calculated as,

  $\begin{array}{l}F=p_a\pi r_f^2\\ F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{h_f}\right)\right)}^{0.54}\times \left(1+\frac{2\times \mu \times 13.17\text{mm}}{3\times h_f}\right)\pi r_f^2\end{array}$ ...... (6)

For $10\%$ reduction,

The final height for $10\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=10\%\\ h_f=45\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{45\text{mm}}\\ r_f^2=173.61{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.5$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{45\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.5\times 13.17\text{mm}}{3\times 45\text{mm}}\right)\times \pi \left(173.61\text{mm}\right)\\ F=54656.24\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.054\text{MN}\end{array}$

For $20\%$ reduction,

The final height for $20\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=20\%\\ h_f=40\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{40\text{mm}}\\ r_f^2=195.31{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.5$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{40\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.5\times 13.17\text{mm}}{3\times 40\text{mm}}\right)\times \pi \left(195.31\text{mm}\right)\\ F=94822.72\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.094\text{MN}\end{array}$

For $30\%$ reduction,

The final height for $30\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=30\%\\ h_f=35\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{35\text{mm}}\\ r_f^2=223.21{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.5$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{35\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.5\times 13.17\text{mm}}{3\times 35\text{mm}}\right)\times \pi \left(223.21\text{mm}\right)\\ F=143723.93\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.14\text{MN}\end{array}$

For $40\%$ reduction,

The final height for $40\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=40\%\\ h_f=30\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{30\text{mm}}\\ r_f^2=260.41{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.5$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{30\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.5\times 13.17\text{mm}}{3\times 30\text{mm}}\right)\times \pi \left(260.41\text{mm}\right)\\ F=211389.11\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.21\text{MN}\end{array}$

For $50\%$ reduction,

The final height for $50\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=50\%\\ h_f=25\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{25\text{mm}}\\ r_f^2=312.5{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.5$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{25\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.5\times 13.17\text{mm}}{3\times 25\text{mm}}\right)\times \pi \left(312.5\text{mm}\right)\\ F=313451.46\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.31\text{MN}\end{array}$

For $60\%$ reduction,

The final height for $60\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=60\%\\ h_f=20\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{20\text{mm}}\\ r_f^2=390.62{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.5$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{20\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.5\times 13.17\text{mm}}{3\times 20\text{mm}}\right)\times \pi \left(390.62\text{mm}\right)\\ F=489678.19\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.48\text{MN}\end{array}$

For $70\%$ reduction,

The final height for $70\%$ reduction in height can be calculated as,

  $\begin{array}{c}\left(\frac{50\text{mm}-h_f}{50\text{mm}}\right)\times 100\%=70\%\\ h_f=15\text{mm}\end{array}$

The final radius can be calculated by substituting the values in equation 4,

  $\begin{array}{l}{r}_f^2=\frac{7812.5{\text{mm}}^3}{15\text{mm}}\\ r_f^2=520.83{\text{mm}}^{\text{2}}\end{array}$

The forging force can be calculated at $\mu =0.5$ by substituting the values in equation 6,

  $\begin{array}{l}F=315\text{MPa}{\left(\ln \left(\frac{50\text{mm}}{15\text{mm}}\right)\right)}^{0.54}\times \left(1+\frac{2\times 0.25\times 13.17\text{mm}}{3\times 15\text{mm}}\right)\times \pi \left(520.83\text{mm}\right)\\ F=819020.84\text{N}\times \left(\frac{1\text{MN}}{{10}^6\text{N}}\right)\\ F=0.81\text{MN}\end{array}$

The plot between forging force and reduction in height is shown in figure (3) below,

  

Figure (3)