Chapter 6, Problem 6.86P

To determine

The work done will be same for two specimens.

Answer to Problem 6.86P

The work done associated with specimen A and specimen B is not same.

Explanation of Solution

Given:

The height of the specimen A is $h_{o1}=50\text{mm}$ .

The cross section of the specimen A is $a_1=6.25\times {10}^{-4}{\text{m}}^2$ .

The height of the specimen B is $h_{o2}=25\text{mm}$ .

The cross section of the specimen B is $a_2=0.0025{\text{m}}^2$ .

The flash thickness is $5\text{mm}$ .

The largest strain in finishing forging is $\epsilon =0.25$ .

Formula used:

The expression for the work done is given as,

  $w={\displaystyle \int Fd{h}_{o1}}$   ............. (1)

Here, $F$ is the force.

The expression for the force is given as,

  $F=p_{avg}A$   ............. (2)

Here, $p_{avg}$ is the average pressure and $A$ is the area.

The expression for the average pressure is given as,

  $$

  $p_{avg}={\sigma }_f\left(1+\frac{2\mu r_1}{3h_{o1}}\right)$   ............. (3)

Here, ${\sigma }_f$ is the flow stress and $r_1$ is the radius of specimen A and $\mu$ is the coefficient of friction.

The expression for volume is given as,

  $v=\pi r_1^2{h}_{o1}$   ............. (4)

Calculation:

Tre expression of specimen A radius from equation (4) can be calculated as,

  $\begin{array}{l}v=\pi r_1^2{h}_{o1}\\ r_1=\sqrt{\frac{v}{\pi h_{o1}}}\end{array}$   ............. (5)

Now from equation (3) and (5),

  $\begin{array}{l}{p}_{avg}={\sigma }_f\left(1+\frac{2\mu r_1}{3h_{o1}}\right)\\ p_{avg}={\sigma }_f\left(1+\frac{2\mu \left(\sqrt{\frac{v}{\pi h_{o1}}}\right)}{3h_{o1}}\right)\\ p_{avg}=\left({\sigma }_f\left(1+\left(\frac{2\mu }{3}\right)\left(\frac{v}{\pi }\right)\left(\frac{1}{h_1^{\frac{3}{2}}}\right)\right)\right)\end{array}$   ............. (6)

Substitute equation (6) in equation (2),

  $F=\left({\sigma }_f\left(1+\left(\frac{2\mu }{3}\right)\left(\frac{v}{\pi }\right)\left(\frac{1}{h_{o1}^{\frac{3}{2}}}\right)\right)\right)\left(\frac{v}{h_{o1}}\right)$   ............. (7)

Substitute equation (7) in equation (1),

  $\begin{array}{l}w={\displaystyle \int \left({\sigma }_f\left(1+\left(\frac{2\mu }{3}\right)\left(\frac{v}{\pi }\right)\left(\frac{1}{h_{o1}^{\frac{3}{2}}}\right)\right)\right)\left(\frac{v}{h_{o1}}\right)d{h}_{o1}}\\ w={\sigma }_{f}v{\displaystyle {\int }_{h_o}^{\frac{h_o}{2}}\left(\left(\left(\frac{1}{h_{o1}}\right)+\left(\frac{2\mu }{3}\right)\left(\frac{v}{\pi }\right)\left(\frac{1}{h_{o1}^{\frac{3}{2}}}\right)\right)\right)d{h}_{o1}}\\ w={\sigma }_{f}v\left({\left[\ln h_{o1}\right]}_{h_o}^{\frac{h_o}{2}}-\frac{3}{2}\left(\frac{2\mu }{\pi }\right){\left(\frac{1}{h_{o1}^{\frac{3}{2}}}\right)}_{h_o}^{\frac{h_o}{2}}\right)\end{array}$   ............. (8)

Work done by the specimen A can be calculated as,

  $w={\sigma }_{f}v\left(\left[\ln \frac{h_o}{2}-\ln h_o\right]-\frac{3}{2}\left(\frac{2\mu }{3}\right)\left(\frac{v}{\pi }\right)\left(\frac{1}{\frac{h_o^{\frac{3}{2}}}{2}-}\frac{1}{h_o^{\frac{3}{2}}}\right)\right)$   ............. (9)

Substitute $h_o=50\text{mm}$ and $v=6.25\times {10}^{-4}{\text{m}}^2$ in equation (9)

  $\begin{array}{l}w={\sigma }_f\left(6.25\times {10}^{-4}{\text{m}}^2\right)\left(\begin{array}{l}\left(\left(\frac{\ln 50\text{mm}\times \frac{1\text{m}}{1000\text{mm}}}{2}\right)-\left(\ln 50\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\right)-\\ \mu \left(\frac{6.25\times {10}^{-4}{\text{m}}^2}{\pi }\right)\left(\begin{array}{l}\frac{1}{{\left(\frac{50\text{mm}\times \frac{1\text{m}}{1000\text{mm}}}{2}\right)}^{\frac{3}{2}}}-\\ \frac{1}{{\left(50\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^{\frac{3}{2}}}\end{array}\right)\end{array}\right)\\ w=\left\{\begin{array}{l}{\sigma }_f\left(6.25\times {10}^{-4}{\text{m}}^2\right)\left(\left(\ln 0.025\text{m}\right)-\left(\ln 0.05\text{m}\right)\right)-\\ \mu \left(1.98\times {10}^{-4}{\text{m}}^2\right)\left(252.98{\text{m}}^{-1}-89.44{\text{m}}^{-1}\right)\end{array}\right\}\\ w={\sigma }_f\left(6.25\times {10}^{-4}{\text{m}}^2\right)\left(-0.6931\text{m}\right)-\mu \left(1.98\times {10}^{-4}{\text{m}}^2\right)\left(163.54{\text{m}}^{-1}\right)\\ w={\sigma }_f\left(-4.33\times {10}^{-4}{\text{m}}^3\right)-\mu \left(0.032\text{m}\right)\end{array}$

Work done by the specimen B can be calculated as,

Substitute $h_o=25\text{mm}$ and $v=0.0025{\text{m}}^2$ in equation (9)

  $\begin{array}{l}w={\sigma }_f\left(0.0025{\text{m}}^2\right)\left(\begin{array}{l}\left(\left(\frac{\ln 25\text{mm}\times \frac{1\text{m}}{1000\text{mm}}}{2}\right)-\left(\ln 25\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)\right)-\\ \mu \left(\frac{0.0025{\text{m}}^2}{\pi }\right)\left(\begin{array}{l}\frac{1}{{\left(\frac{25\text{mm}\times \frac{1\text{m}}{1000\text{mm}}}{2}\right)}^{\frac{3}{2}}}-\\ \frac{1}{{\left(25\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^{\frac{3}{2}}}\end{array}\right)\end{array}\right)\\ w=\left\{\begin{array}{l}{\sigma }_f\left(0.0025{\text{m}}^2\right)\left(\ln 0.0125\text{m}-\ln 0.025\text{m}\right)-\\ \mu \left(7.95\times {10}^{-4}{\text{m}}^2\right)\left(715.54{\text{m}}^{-1}-252.98{\text{m}}^{-1}\right)\end{array}\right\}\\ w={\sigma }_f\left(0.0025{\text{m}}^2\right)\left(-0.0693\text{m}\right)-\mu \left(0.3677\text{m}\right)\\ w={\sigma }_f\left(-1.7325\times {10}^{-4}{\text{m}}^3\right)-\mu \left(0.3677\text{m}\right)\end{array}$

Conclusion:

Therefore, work done associated with specimen A and specimen B is not same.