Chapter 6, Problem 6.125P

To determine

The expression for extrusion pressure.

Explanation of Solution

The figure (1) demonstrate the forces on the specimen in a wire drawing process.

  

Figure (1)

The expression for the equilibrium in x direction can be given as,

  $\left({\sigma }_x+d{\sigma }_x\right)\frac{\pi }{4}{\left(D+dD\right)}^2-{\sigma }_x\frac{\pi }{4}{D}^2+p\pi Ddx\sin \alpha +\mu p\pi Ddx\cos \alpha =0$

Here, pressure along the die contact is $p$ , $\mu$ is the coefficient of friction, $D_o$ is the initial diameter, $D_f$ is the final diameter.

After expanding the above equation,

  $\begin{array}{l}\left({\sigma }_x+d{\sigma }_x\right)\frac{\pi }{4}\left(D^2+{\left(dD\right)}^2+2D\left(dD\right)\right)-{\sigma }_x\frac{\pi }{4}{D}^2+p\pi Ddx\sin \alpha +\mu p\pi Ddx\cos \alpha =0\\ \left\{\begin{array}{l}\left(\frac{\pi }{4}{\sigma }_x{D}^2+\frac{\pi }{4}{\sigma }_x{\left(dD\right)}^2+2{\sigma }_{x}D\frac{\pi }{4}\left(dD\right)\right)\\ +\left(\frac{\pi }{4}d{\sigma }_x{D}^2+\frac{\pi }{4}d{\sigma }_x{\left(dD\right)}^2+2d{\sigma }_{x}D\frac{\pi }{4}\left(dD\right)\right)\\ -{\sigma }_x\frac{\pi }{4}{D}^2+p\pi Ddx\left(\sin \alpha +\mu \cos \alpha \right)\end{array}\right\}=0\\ \frac{\pi }{4}d{\sigma }_x{D}^2+2d{\sigma }_{x}D\frac{\pi }{4}\left(dD\right)+p\pi Ddx\left(\sin \alpha +\mu \cos \alpha \right)=0\\ \frac{1}{4}d{\sigma }_x{D}^2+2d{\sigma }_{x}D\frac{1}{4}\left(dD\right)+pDdx\left(\sin \alpha +\mu \cos \alpha \right)=0\end{array}$

Further solving the above equation,

  $\begin{array}{l}d{\sigma }_{x}D+2d{\sigma }_x\left(dD\right)+4pdx\left(\sin \alpha +\mu \cos \alpha \right)=0\\ d{\sigma }_{x}D+2d{\sigma }_x\left(dD\right)+4p\frac{dD}{2\sin \alpha }\left(\sin \alpha +\mu \cos \alpha \right)=0\\ d{\sigma }_{x}D+2d{\sigma }_x\left(dD\right)+2pdD\left(1+\mu \frac{1}{\tan \alpha }\right)=0\end{array}$ …… (1)

Expression for pressure $p$ can be given as,

  $p=S_y-{\sigma }_x$

Substituting the value of $p$ in the equation (1),

  $\begin{array}{l}d{\sigma }_{x}D+2d{\sigma }_x\left(dD\right)+2\left(S_y-{\sigma }_x\right)dD\left(1+\mu \frac{1}{\tan \alpha }\right)=0\\ d{\sigma }_{x}D+2d{\sigma }_x\left(dD\right)+\left(2S_{y}dD\left(1+\mu \cot \alpha \right)-2{\sigma }_{x}dD\left(1+\mu \cot \alpha \right)\right)=0\\ d{\sigma }_{x}D+\left(2S_{y}dD\left(1+\mu \cot \alpha \right)-2{\sigma }_{x}dD\left(\mu \cot \alpha \right)\right)=0\\ \left(\frac{d{\sigma }_x}{\left(-2S_y\left(1+\mu \cot \alpha \right)+2{\sigma }_x\left(\mu \cot \alpha \right)\right)}\right)=\frac{dD}{D}\end{array}$ …… (2)

Integration of equation (2) for boundary condition $D=D_0$ at ${\sigma }_x={\sigma }_d$ and $D=D_f$ at ${\sigma }_x=0$ can be given as,

  $\begin{array}{l}{\displaystyle {\int }_{-{\sigma }_d}^0\left(\frac{d{\sigma }_x}{\left(-2S_y\left(1+\mu \cot \alpha \right)+2{\sigma }_x\left(\mu \cot \alpha \right)\right)}\right)}={\displaystyle {\int }_{D_o}^{D_f}\frac{dD}{D}}\\ {\left(\frac{\ln -\left(2S_y\left(1+\mu \cot \alpha \right)+2{\sigma }_x\left(\mu \cot \alpha \right)\right)}{2\left(\mu \cot \alpha \right)}\right)}_{-{\sigma }_d}^0=\ln \left(\frac{D_f}{D_o}\right)\\ \ln \left(\frac{-2S_y\left(1+\mu \cot \alpha \right)}{\left(-2S_y\left(1+\mu \cot \alpha \right)+2{\sigma }_x\left(\mu \cot \alpha \right)\right)}\right)=\ln {\left(\frac{D_f}{D_o}\right)}^{2\left(\mu \cot \alpha \right)}\end{array}$

Simplifying the above equation,

  $\begin{array}{l}{\sigma }_d=S_y\left(\frac{1+\mu \cot \alpha }{\mu \cot \alpha }\right)\left(1-{\left(\frac{D_f}{D_o}\right)}^{2\left(\mu \cot \alpha \right)}\right)\\ {\sigma }_d=S_y\left(1+\frac{\tan \alpha }{\mu }\right)\left(1-{\left(\frac{D_f}{D_o}\right)}^{2\left(\mu \cot \alpha \right)}\right)\end{array}$ …… (3)

The expression for the ratio of area,

  ${\left(\frac{D_f}{D_o}\right)}^2=\left(\frac{A_f}{A_o}\right)$

Substituting the above value in the equation (3),

  ${\sigma }_d=S_y\left(1+\frac{\tan \alpha }{\mu }\right)\left(1-{\left(\frac{A_f}{A_o}\right)}^{\left(\mu \cot \alpha \right)}\right)$

For, $p={\sigma }_d$

  $p=S_y\left(1+\frac{\tan \alpha }{\mu }\right)\left(1-{\left(\frac{A_f}{A_o}\right)}^{\left(\mu \cot \alpha \right)}\right)$

Conclusion:

Therefore, the equation is $p=S_y\left(1+\frac{\tan \alpha }{\mu }\right)\left(1-{\left(\frac{A_f}{A_o}\right)}^{\left(\mu \cot \alpha \right)}\right)$ .