Chapter 6, Problem 6.130P

To determine

Theexpression for wires drawing stress ${\sigma }_d$ in plane strain drawing of a flat sheet.

Answer to Problem 6.130P

The expression for wires drawing stress ${\sigma }_d$ in plane strain drawing of a flat sheet is ${\sigma }_d=S_y\left(1+\frac{\tan \alpha }{\mu }\right)\left[1-{\left(\frac{h-f}{h_0}\right)}^{\mu \cot \alpha }\right]$ .

Explanation of Solution

Formula used:

The expression flow stress is given as,

  $Y\text{'}={\sigma }_{\max }-{\sigma }_{\min }$ ...... (1)

Here, ${\sigma }_{\max }$ is the maximum stress generated and ${\sigma }_{\min }$ is the minimum stress generated.

Calculation:

The figure (1) shows the force analysis during the extrusion process,

  

Figure (1)

Here, $p$ is the pressure applied by the rollers, $\mu p$ is the frictional force, ${\sigma }_x$ is the tensile force, $h_0$ is the initial force and $h_f$ is the final force and $w$ is the width of the specimen.

Evaluate the forces in the x-direction,

  $\begin{array}{l}\left({\sigma }_x+d{\sigma }_x\right)\left(h+dh\right)w-{\sigma }_{x}hw+p\frac{wdx}{\cos \alpha }+\mu p\frac{dx}{\cos \alpha }=0\\ hw{\sigma }_x+hwd{\sigma }_x+dhw{\sigma }_x+d^{2}hw{\sigma }_x-{\sigma }_{x}hw+p\frac{wdx}{\cos \alpha }+\mu p\frac{dx}{\cos \alpha }=0\\ hwd{\sigma }_x+dhw{\sigma }_x+d^{2}hw{\sigma }_x+p\frac{wdx}{\cos \alpha }+\mu p\frac{dx}{\cos \alpha }=0\end{array}$

Ignoring the second order terms in the above equation and divide whole equation by $w$ .Therefore,

  $hd{\sigma }_x+dh{\sigma }_x+p\left(1+\frac{\mu }{\tan \alpha }\right)dh=0$ ...... (2)

Since positive pressure is indicated by negative stress,

  ${\sigma }_{\max }-{\sigma }_{\min }={\sigma }_x+p$

From equation (1),

  $\begin{array}{l}Y\text{'}={\sigma }_x+p\\ p=Y\text{'}-{\sigma }_x\end{array}$ ....... (3)

By equation (2) and (3),

  $\begin{array}{l}hd{\sigma }_x+dh{\sigma }_x+\left(Y\text{'}-{\sigma }_x\right)\left(1+\frac{\mu }{\tan \alpha }\right)dh=0\\ hd{\sigma }_x+dh{\sigma }_x+\left(Y\text{'}\left(1+\frac{\mu }{\tan \alpha }\right)dh-{\sigma }_x\left(1+\frac{\mu }{\tan \alpha }\right)dh\right)=0\\ hd{\sigma }_x+dh{\sigma }_x+\left(Y\text{'}\left(1+\mu \cot \alpha \right)dh-{\sigma }_x\left(1+\mu \cot \alpha \right)dh\right)=0\\ hd{\sigma }_x-{\sigma }_{x}dh\left[1-\left(1+\mu \cot \alpha \right)\right]+Y\text{'}\left(1+\mu \cot \alpha \right)dh=0\end{array}$

On further solving,

  $\begin{array}{l}hd{\sigma }_x-{\sigma }_{x}dh\mu \cot \alpha +Y\text{'}\left(1+\mu \cot \alpha \right)dh=0\\ hd{\sigma }_x=dh\left[{\sigma }_x\mu \cot \alpha +Y\text{'}\left(1+\mu \cot \alpha \right)\right]\\ \frac{dh}{h}=\frac{d{\sigma }_x}{{\sigma }_x\mu \cot \alpha -Y\text{'}\left(1+\mu \cot \alpha \right)}\end{array}$

Boundary conditions are as follows,

At $h=h_0$ , ${\sigma }_x=-{\sigma }_d$ and also at $h=h_f$ , ${\sigma }_x=0$ .

Integrating both sides in the limits,

  $\begin{array}{r}{\displaystyle \underset{h_0}{\overset{h_f}{\int }}\frac{dh}{h}={\displaystyle \underset{-{\sigma }_d}{\overset{0}{\int }}\frac{d{\sigma }_x}{{\sigma }_x\mu \cot \alpha -Y\text{'}\left(1+\mu \cot \alpha \right)}}}\\ {\sigma }_d=Y\text{'}\left(\frac{1+\mu \cot \alpha }{\mu \cot \alpha }\right)\left[1-{\left(\frac{h_f}{h_0}\right)}^{\mu \cot \alpha }\right]\\ {\sigma }_d=Y\text{'}\left(1+\frac{\tan \alpha }{\mu }\right)\left[1-{\left(\frac{h_f}{h_0}\right)}^{\mu \cot \alpha }\right]\end{array}$

Conclusions:

Therefore, the expression for wires drawing stress ${\sigma }_d$ in plane strain drawing of a flat sheet is ${\sigma }_d=S_y\left(1+\frac{\tan \alpha }{\mu }\right)\left[1-{\left(\frac{h-f}{h_0}\right)}^{\mu \cot \alpha }\right]$ .