Chapter 2, Problem 2.91P

To determine

The load required for each step.

Answer to Problem 2.91P

The load required in intermediate step is $P_1=-410.64\text{kN}$ and the load required in final step is $P_2=-1139.43\text{kN}$ .

Explanation of Solution

Given:

The initial height of the cylinder is $h_0=100\text{mm}$ .

The intermediate height of the cylinder is $h_1=70\text{mm}$ .

The final height of the cylinder is $h_2=30\text{mm}$ .

The initial diameter of the cylinder is $d_0=50\text{mm}$ .

Formula used:

The expression for the intermediate diameter is given as,

  $d_1=d_0\sqrt{\frac{h_0}{h_1}}$

The expression for the final diameter is given as,

  $d_2=d_0\sqrt{\frac{h_0}{h_2}}$

The expression for intermediate cross-sectional area is given as,

  $A_1=\frac{\pi }{4}{d}_1^2$

The expression for final cross-sectional area is given as,

  $A_2=\frac{\pi }{4}{d}_2^2$

The expression for true stress at intermediate step is given as,

  ${\sigma }_1=K{\epsilon }_1^n$

The expression for true stress at finalstep is given as,

  ${\sigma }_2=K{\epsilon }_2^n$

The load required at intermediate step is given as,

  $P_1={\sigma }_1{A}_1$

The load required at final step is given as,

  $P_2={\sigma }_2{A}_2$

Calculation:

The true strain for intermediate step calculated is, ${\epsilon }_1=-0.3566$ .

The true strain for final step calculated is, $\epsilon =-0.847$ .

Refer to table 2.2 for aluminium $1100-O$ the following values can be obtained $K=180\text{MPa}$ and $n=0.2$

The intermediate step diameter can be calculated as,

  $\begin{array}{l}{d}_1=d_0\sqrt{\frac{h_0}{h_1}}\\ d_1=50\text{mm}\times \sqrt{\frac{100\text{mm}}{70\text{mm}}}\\ d_1=59.76\text{mm}\end{array}$

The final step diameter can be calculated as,

  $\begin{array}{l}{d}_2=d_0\sqrt{\frac{h_0}{h_2}}\\ d_2=50\text{mm}\times \sqrt{\frac{100\text{mm}}{30\text{mm}}}\\ d_2=91.28\text{mm}\end{array}$

The intermediate step cross-sectional area can be calculated as,

  $\begin{array}{l}{A}_1=\frac{\pi }{4}{d}_1^2\\ A_1=\frac{\pi }{4}{\left(59.76\text{mm}\right)}^2\\ A_1=2804{\text{mm}}^2\end{array}$ .

The final step cross-sectional area can be calculated as,

  $\begin{array}{l}{A}_2=\frac{\pi }{4}{d}_2^2\\ A_2=\frac{\pi }{4}{\left(91.28\text{mm}\right)}^2\\ A_2=6543.96{\text{mm}}^2\end{array}$ .

The true stress at final step can be calculated as,

  $\begin{array}{l}{\sigma }_2=K{\epsilon }_2^n\\ {\sigma }_2=180\text{MPa}{\left(-0.847\right)}^{0.2}\\ {\sigma }_2=-174.12\text{MPa}\end{array}$ .

The true stress at intermediate step can be calculated as,

  $\begin{array}{l}{\sigma }_1=K{\epsilon }_1^n\\ {\sigma }_1=180\text{MPa}{\left(-0.3566\right)}^{0.2}\\ {\sigma }_1=-146.45\text{MPa}\end{array}$ .

The load required for intermediate step can be calculated as,

  $\begin{array}{l}{P}_1={\sigma }_1{A}_1\\ P_1=-146.45\text{MPa}\left(\frac{\text{N}}{{\text{mm}}^2\times \text{MPa}}\right)\times 2804{\text{mm}}^2\\ P_1=-410.64\text{kN}\end{array}$ .

The load required for final step can be calculated as,

  $\begin{array}{l}{P}_2={\sigma }_2{A}_2\\ P_2=-174.12\text{MPa}\left(\frac{\text{N}}{{\text{mm}}^2\times \text{MPa}}\right)\times 6543.96{\text{mm}}^2\\ P_2=-1139.43\text{kN}\end{array}$ .

Conclusion:

Therefore, the load required in intermediate step is $P_1=-410.64\text{kN}$ and the load required in final step is $P_2=-1139.43\text{kN}$ .