Chapter 6, Problem 6.101P

To determine

The velocity of the strip leaving the rolls.

Answer to Problem 6.101P

The velocity of the strip leaving the rolls is $0.202\text{m}/\text{s}$ .

Explanation of Solution

Given:

The coefficient of friction is $\mu =0.3$ .

The radius of the roll is $R=300\text{mm}$ .

The speed of the roll is $N=100\text{rpm}$ .

The initial thickness of the strip is $t_1=25\text{mm}$ .

The final thickness of the strip is $t_2=20\text{mm}$ .

The width of the strip is $w=225\text{mm}$ .

Formula used:

The expression for length of arc of contact is given as,

  $L=\sqrt{R\left(t_1-t_2\right)}$

Here, $K$ is the strength coefficient, $\epsilon$ is the true strain and $n$ is the strain hardening

exponent.

The expression for true strain is given as,

  $\epsilon =\ln \left(\frac{t_1}{t_2}\right)$

The expression for average thickness is given as,

  $t_{avg}=\frac{t_1+t_2}{2}$

The expression for average flow stress is given as,

  ${\sigma }_f=\frac{K{\epsilon }^n}{n+1}$

Here, $K$ is the strength coefficient, $\epsilon$ is the strain and $n$ is the strain hardening coefficient.

The expression for roll force is given as,

  $F=Lw{\sigma }_f\left(1+\frac{\mu L}{2t_{avg}}\right)$

The expression for power of roll is given as,

  $p=\frac{\pi FLN}{60}$

The expression for velocity of strip is given as,

  $V=\frac{P}{F}$

Calculation:

The length of arc of contact can be calculated as,

  $\begin{array}{l}L=\sqrt{R\left(t_1-t_2\right)}\\ L=\sqrt{300\text{mm}\left(25\text{mm}-20\text{mm}\right)}\\ L=\sqrt{1500{\text{mm}}^2}\\ L=38.73\text{mm}\end{array}$

The true strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{t_1}{t_2}\right)\\ \epsilon =\ln \left(\frac{25\text{mm}}{20\text{mm}}\right)\\ \epsilon =\ln 1.25\\ \epsilon =0.223\end{array}$

The average thickness can be calculated as,

  $\begin{array}{l}{t}_{avg}=\frac{t_1+t_2}{2}\\ t_{avg}=\frac{25\text{mm}+20\text{mm}}{2}\\ t_{avg}=22.5\text{mm}\end{array}$

Refer to table 2.2 “The strength coefficient and strain hardening exponent”is given as,

  $\begin{array}{l}K=205\text{MPa}\\ n=0.\text{2}\end{array}$

The average flow stress can be calculated as,

  $\begin{array}{l}{\sigma }_f=\frac{K{\epsilon }^n}{n+1}\\ {\sigma }_f=\frac{205\text{MPa}\times {0.233}^{0.2}}{0.2+1}\\ {\sigma }_f=\frac{153.19\text{MPa}}{1.2}\\ {\sigma }_f=127\text{MPa}\end{array}$

The roll force can be calculated as,

  $\begin{array}{l}F=Lw{\sigma }_f\left(1+\frac{\mu L}{2t_{avg}}\right)\\ F=38.73\text{mm}\times 225\text{mm}\times 127\text{Mpa}\left(1+\frac{0.3\times 38.73\text{mm}}{2\times 22.5\text{mm}}\right)\\ F=8714.25{\text{mm}}^2\times \frac{1{\text{m}}^2}{{10}^6{\text{mm}}^2}\times 127\text{Mpa}\times \frac{1\text{MN}/{\text{m}}^2}{1\text{MPa}}\left(1.2582\right)\\ F=8.71\times {10}^{-3}{\text{m}}^2\times 127\text{MN}/{\text{m}}^2\left(1.2582\right)\end{array}$

On further solving as,

  $F=1.39\text{MN}$ .

The power of roll can be calculated as,

  $\begin{array}{l}p=\frac{\pi FLN}{60}\\ p=\frac{\pi \times 1.39\text{MN}\times \frac{{10}^6\text{N}}{1\text{MN}}\times 38.73\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\times 100\text{rpm}}{60}\\ p=\frac{4.3668\times {10}^6\text{N}\times 0.03873\text{m}\times \text{100}\text{rpm}}{60}\\ p=281.876\times {10}^3\text{N}\cdot \text{m}/\text{s}\times \frac{1\text{W}}{1\text{N}\cdot \text{m}/\text{s}}\end{array}$

On further solving as,

  $\begin{array}{l}p=281.876\times {10}^3\text{W}\times \frac{1\text{kW}}{{10}^3\text{W}}\\ p=281.876\text{kW}\end{array}$

The velocity of strip can be calculated as,

  $\begin{array}{l}V=\frac{P}{F}\\ V=\frac{281.876\times {10}^3\text{W}\times \frac{1\text{N}\cdot \text{m}/\text{s}}{\text{W}}}{1.39\text{MN}\times \frac{{10}^6\text{N}}{1\text{MN}}}\\ V=\frac{281.876\times {10}^3\text{N}\cdot \text{m}/\text{s}}{1.39\times {10}^6\text{N}}\\ V=0.202\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the velocity of the strip leaving the rolls is $0.202\text{m}/\text{s}$ .