Chapter 6, Problem 6.96P

To determine

The maximum contact force.

The height up to which the aluminum cylinder to be forged before the press stalls.

Answer to Problem 6.96P

The maximum contact force is $25\text{kN}$ .

The required height of the cylinder is $1.7\text{mm}$ .

Explanation of Solution

Given:

The power of motor is $p=40\text{kW}$ .

The number of strokes per minute is $N=40$ .

The stroke length of the crankshaft is $L=150\text{mm}$ .

The initial diameter of the cylinder is $d_o=10\text{mm}$ .

The initial height of the cylinder is $h_0=30\text{mm}$ .

Formula used:

The expression for the initial radius is given as,

  $r_o=\frac{d_o}{2}$

The expression for velocity is given as,

  $V=2LN$

The expression for power is given as,

  $p=FV$

Here, $F$ is the force.

The expression for volume consistency of the cylinder is given as,

  $\pi r_o^2{h}_o=\pi r^{2}h$

Here, $r$ is the final radius and $h$ is the final height of the cylinder.

The expression for relation of force in terms of yield strength is given as,

  $F=S_f\pi r^2\left(1+\frac{2\mu r}{3h}\right)$ …… (1)

Here, $S_f$ is the yield strength and $\mu$ is the coefficient of friction.

Calculation:

The initial radius of the cylinder can be calculated as,

  $\begin{array}{l}r=\frac{10\text{mm}}{2}\\ r=5\text{mm}\end{array}$

The velocity can be calculated as,

  $\begin{array}{l}V=2LN\\ V=2\times 150\text{mm}/\text{rev}\times 40\text{rev}/\min \times \frac{1\text{min}}{60\text{sec}}\\ V=200\text{mm}/\sec \times \frac{1\text{m}}{1000\text{mm}}\\ V=0.2\text{m}/\sec \end{array}$

The force from the relation of power can be calculated as,

  $\begin{array}{l}P=FV\\ 25\text{kW}=F\times 0.2\text{m}/\sec \\ F=\frac{25\text{kW}\times \frac{1\text{kN}\cdot \text{m}/\sec }{1\text{KW}}}{0.2\text{m}/\sec }\\ F=125\text{kN}\end{array}$

From volume consistency, relation between final radius and final height can be calculated as,

  $\begin{array}{l}\pi r_o^2{h}_o=\pi r^{2}h\\ r^2=\frac{r_o^2{h}_o}{h}\\ r^2=\frac{{\left(5\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\times 30\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}}{h}\\ h=\frac{7.5\times {10}^{-7}{\text{m}}^3}{r^2}\end{array}$ …… (2)

From the table 3.5, “Properties of various aluminum alloys at room temperature”, yield strength can be given as,

  $S_f=90\text{MPa}$

From equation (1) and (2)final radius of the cylinder can be calculated as,

  $\begin{array}{l}F=S_f\pi r^2\left(1+\frac{2\mu r}{3h}\right)\\ 125\text{KN}\times \frac{1000\text{N}}{1\text{kN}}=90\text{Mpa}\times \frac{{10}^6\text{Pa}}{1\text{MPa}}\times \pi r^2\left(1+\frac{2\mu r}{\frac{7.5\times {10}^{-7}{\text{m}}^3}{r^2}}\right)\\ 125000\text{N}=90\times {10}^6\text{Pa}\times \frac{1\text{N}/{\text{m}}^2}{1\text{Pa}}\times \pi r^2\left(1+\left(2.66\times {10}^6{\text{m}}^{-3}\right)\mu r^3\right)\\ \frac{125000\text{N}}{90\times {10}^6\text{N}/{\text{m}}^{\text{2}}\times \pi r^2}-1=2.66\times {10}^6\mu r^3{\text{m}}^{-3}\end{array}$

On further solving,

  $\begin{array}{l}\frac{4.418\times {10}^{-4}{\text{m}}^2}{r^2}-2.66\times {10}^6\mu r^3{\text{m}}^{-3}=1\\ r^2+2.66\times {10}^6\mu r^5{\text{m}}^{-3}=4.418\times {10}^{-4}{\text{m}}^2\\ r^2\left(1+2.66\times {10}^6\mu r^5{\text{m}}^{-3}\right)=4.418\times {10}^{-4}{\text{m}}^2\\ r=0.021\text{m}\end{array}$

Now from equation (2) height of the cylinder can be calculated as,

  $\begin{array}{l}h=\frac{7.5\times {10}^{-7}{\text{m}}^3}{r^2}\\ h=\frac{7.5\times {10}^{-7}{\text{m}}^3}{{\left(0.021\text{m}\right)}^2}\\ h=1.7\times {10}^{-3}\text{m}\times \frac{1\text{mm}}{1000\text{m}}\\ h=1.7\text{mm}\end{array}$

Conclusion:

Therefore, the maximum contact force is $25\text{kN}$ .

Therefore, the required height of the cylinder is $1.7\text{mm}$ .