Chapter 6, Problem 6.120P

To determine

The rolling force and the power requirement for single pass.

The rolling force and the power requirement for double pass.

Answer to Problem 6.120P

Rolling force is $57665.23\text{N}$ and the power required is $5150.1\text{W}$ for single pass.

Rolling force is $40796\text{N}$ and the power required is $2577.66\text{W}$ for double pass.

Explanation of Solution

Given:

The initial height is $h_o=1.5\text{mm}$ .

The final height is $h_f=0.9\text{mm}$ .

The radius of the rollers is $R=75\text{mm}$ .

The coefficient of friction is $\mu =0.1$ .

The width of the sheet is $w=25\text{mm}$ .

Velocity of the roll surface is $v=1\text{m}/\text{s}$ .

Formula used:

The expression for change in height for single pass is given as,

  $\Delta h_1=h_o-h_f$

The expression for change in height for double pass is given as,

  $\Delta h_2=\frac{h_o-h_f}{2}$

The expression for contact length for single passis given as,

  $L_1=\sqrt{R\Delta h_1}$

The expression for contact length is given as,

  $L_2=\sqrt{R\Delta h_2}$

The expression for rolling force is given as,

  $F_1={\sigma }_o{L}_{1}w$

The rolling force for double pass can be given as,

  $F_2={\sigma }_o{L}_{2}w$

Here, ${\sigma }_o$ is the roll separating forceis given as,

The expression for roll separatingstressis given as,

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

Here, strength coefficient is $K$ and strength hardening exponent is n.

The expression for the strain is given as,

  $\epsilon =\ln \left(\frac{h_0}{h_f}\right)$

The expression for the angular velocity is given as,

  $\omega =\frac{v}{R}$

The expression for the arm length is given as,

  $a_1=\frac{L_1}{2}$

The expression for the arm length in double pass is given as,

  $a_2=\frac{L_2}{2}$

The expression for the torque is given as,

  $T_1=F_1\times a$

The expression for the torque for double pass is given as,

  $T_2=F_2\times a$

The expression for the power is given as,

  $P_1=2\times T_1\times \omega$

The expression for the power for double pass is given as,

  $P_2=2\times T_2\times \omega$

Calculation:

The change in height can be calculated as,

  $\begin{array}{l}\Delta h_1=h_o-h_f\\ \Delta h_1=1.5\text{mm}-0.9\text{mm}\\ \Delta h_1=0.6\text{mm}\end{array}$

The contact length can be calculated as,

  $\begin{array}{l}{L}_1=\sqrt{R\Delta h_1}\\ L_1=\sqrt{75\text{mm}\times 0.6\text{mm}}\\ L_1=6.7\text{mm}\end{array}$

The strain can be calculated as,

  $\begin{array}{l}\epsilon =\ln \left(\frac{h_0}{h_f}\right)\\ \epsilon =\ln \left(\frac{1.5}{0.9}\right)\\ \epsilon =0.51\end{array}$

The roll separating stress can be calculated as,

  $\begin{array}{l}{\sigma }_o=\frac{K{\epsilon }^n}{n+1}\\ {\sigma }_o=\frac{580\text{MPa}\times {\left(0.51\right)}^{0.34}}{1+0.34}\\ {\sigma }_o=344.27\text{MPa}\end{array}$

Refer to table 2.2“Typical values of strength coefficient $K$ and strength hardening exponent $n$ ” for $85-15$ brass is,

  $\begin{array}{l}K=580\text{MPa}\\ n=0.34\end{array}$

The force for single pass can be calculated as,

  $\begin{array}{l}{F}_1={\sigma }_o{L}_{1}w\\ F_1=344.27\text{MPa}\times {10}^6\text{Pa}\times 6.7\times {10}^{-3}\text{m}\times 25\times {10}^{-3}\text{m}\\ F_1=57665.23\text{N}\end{array}$

The angular velocity can be calculated as,

  $\begin{array}{l}\omega =\frac{v}{R}\\ \omega =\frac{1\text{m}/\text{s}}{75\times {10}^3\text{m}}\\ \omega =13.33{\text{s}}^{\text{-1}}\end{array}$

The arm length can be calculated as,

  $\begin{array}{l}{a}_1=\frac{L_1}{2}\\ a_1=\frac{6.7\times {10}^{-3}\text{m}}{2}\\ a_1=3.35\times {10}^{-3}\text{m}\end{array}$

The torque can be calculated as,

  $\begin{array}{l}{T}_1=F_1\times a_1\\ T_1=57665.23\text{N}\times 3.35\times {10}^{-3}\text{m}\\ {\text{T}}_1\text{=193}\text{.178}\text{N}\cdot \text{m}\end{array}$

The power can be calculated as,

  $\begin{array}{l}{P}_1=2\times T_1\times \omega \\ P_1=2\times \text{193}\text{.178}\text{N}\cdot \text{m}\times 13.331/s\\ P_1=5150.1\text{W}\end{array}$

The change in height for double pass can be calculated as,

  $\begin{array}{l}\Delta h_2=\frac{h_o-h_f}{2}\\ \Delta h_2=\frac{1.5\text{mm}-0.9\text{mm}}{2}\\ \Delta h_2=0.3\text{mm}\end{array}$

The contact length for double pass can be given as,

  $\begin{array}{l}{L}_2=\sqrt{R\Delta h_2}\\ L_2=\sqrt{75\text{mm}\times 0.3\text{mm}}\\ L_2=4.74\times {10}^{-3}\text{m}\end{array}$

The arm length for the double pass can be calculated as,

  $\begin{array}{l}{a}_2=\frac{L_2}{2}\\ a_2=\frac{4.74\times {10}^{-3}\text{m}}{2}\\ a_2=2.37\times {10}^{-3}\text{m}\end{array}$

The rolling force for double pass can be given as,

  $\begin{array}{l}{F}_2={\sigma }_o{L}_{2}w\\ F_2=344.27\times {10}^6\text{Pa}\times 4.74\times {10}^{-3}\text{N}\times 25\times {10}^{-3}\text{m}\\ F_2=40796\text{N}\end{array}$

The torque for the double pass can be given as,

  $\begin{array}{l}{T}_2=F_2\times a_2\\ T_2=40796\text{N}\times \text{2}\text{.37}\times {\text{10}}^{-3}\text{m}\\ T_2=96.686\text{N}\text{.m}\end{array}$

The power required for double pass can be given as,

  $\begin{array}{l}{P}_2=2\times T_2\times \omega \\ P_2=2\times 96.686\text{N}\cdot \text{m}\times 13.33\text{1}/\text{s}\\ P_2=2577.66\text{W}\end{array}$

Conclusion:

Therefore, rolling force is $57665.23\text{N}$ and the power required is $5150.1\text{W}$ for single pass.

Therefore, rolling force is $40796\text{N}$ and the power required is $2577.66\text{W}$ for double pass.