Chapter 2, Problem 2.61P

(a)

To determine

The work done in frictionless compression of solid cylinder of 1100-O Aluminum.

Answer to Problem 2.61P

The work done in the frictionless compression of the cylinder is $W=1562\text{Nm}$

Explanation of Solution

Calculation:

The work done in frictionless compression of solid cylinder can be calculated by multiplying the specific energy and the volume of the cylinder.

For calculating the value of specific energy, the value of true strain is also required. The reduction in the height of the cylinder is 75%, thus the final height of the cylinder is 10 mm.

The value of true strain can be calculated as

  $\begin{array}{l}\epsilon =\ln \left(\frac{40}{10}\right)\\ \epsilon =1.386\end{array}$

The value of bulk modulus and n for the 1100-O Al are

  $\begin{array}{l}K=180MPa\\ n=0.20\end{array}$

The specific energy can be calculated as

  $\begin{array}{l}u=\frac{K{\epsilon }^{n+1}}{n+1}\\ u=\frac{\left(180\right){\left(1.386\right)}^{1.2}}{1.2}\\ u=222MN/m^3\end{array}$

Now, the volume of the cylinder is

  $\begin{array}{l}V=\pi r^{2}l\\ V=\pi {\left(0.0075\right)}^2\left(0.04\right)\\ V=7.069\times {10}^{-6}{m}^3\end{array}$

Now, the work done is

  $\begin{array}{l}W=u\times V\\ W=222\times 7.069\text{Nm}\\ W=1562\text{Nm}\end{array}$

Conclusion: Thus, the work done in the frictionless compression of the cylinder is $W=1562\text{Nm}$

(b)

To determine

The work done in frictionless compression of solid cylinder of annealed copper.

Answer to Problem 2.61P

The work done in the frictionless compression of the cylinder is $W=2391\text{Nm}$

Explanation of Solution

Calculation:

The work done in frictionless compression of solid cylinder can be calculated by multiplying the specific energy and the volume of the cylinder.

For calculating the value of specific energy, the value of true strain is also required. The reduction in the height of the cylinder is 75%, thus the final height of the cylinder is 10 mm.

The value of true strain can be calculated as

  $\begin{array}{l}\epsilon =\ln \left(\frac{40}{10}\right)\\ \epsilon =1.386\end{array}$

The value of bulk modulus and n for the annealed copper are

  $\begin{array}{l}K=315MPa\\ n=0.54\end{array}$

The specific energy can be calculated as

  $\begin{array}{l}u=\frac{K{\epsilon }^{n+1}}{n+1}\\ u=\frac{\left(315\right){\left(1.386\right)}^{1.54}}{1.54}\\ u=338MN/m^3\end{array}$

Now, the volume of the cylinder is

  $\begin{array}{l}V=\pi r^{2}l\\ V=\pi {\left(0.0075\right)}^2\left(0.04\right)\\ V=7.069\times {10}^{-6}{m}^3\end{array}$

Now, the work done is

  $\begin{array}{l}W=u\times V\\ W=338\times 7.069\text{Nm}\\ W=2391\text{Nm}\end{array}$

Conclusion: Thus, the work done in the frictionless compression of the cylinder is $W=2391\text{Nm}$

(c)

To determine

The work done in frictionless compression of solid cylinder of annealed 304 stainless steels.

Answer to Problem 2.61P

The work done in the frictionless compression of the cylinder is $W=10,808\text{Nm}$

Explanation of Solution

Calculation:

The work done in frictionless compression of solid cylinder can be calculated by multiplying the specific energy and the volume of the cylinder.

For calculating the value of specific energy, the value of true strain is also required. The reduction in the height of the cylinder is 75%, thus the final height of the cylinder is 10 mm.

The value of true strain can be calculated as

  $\begin{array}{l}\epsilon =\ln \left(\frac{40}{10}\right)\\ \epsilon =1.386\end{array}$

The value of bulk modulus and n for the annealed 304 stainless steel are

  $\begin{array}{l}K=1300MPa\\ n=0.30\end{array}$

The specific energy can be calculated as

  $\begin{array}{l}u=\frac{K{\epsilon }^{n+1}}{n+1}\\ u=\frac{\left(1300\right){\left(1.386\right)}^{1.3}}{1.3}\\ u=1529MN/m^3\end{array}$

Now, the volume of the cylinder is

  $\begin{array}{l}V=\pi r^{2}l\\ V=\pi {\left(0.0075\right)}^2\left(0.04\right)\\ V=7.069\times {10}^{-6}{m}^3\end{array}$

Now, the work done is

  $\begin{array}{l}W=u\times V\\ W=1529\times 7.069\text{Nm}\\ W=10,808\text{Nm}\end{array}$

Conclusion: Thus, the work done in the frictionless compression of the cylinder is $W=10,808\text{Nm}$

(d)

To determine

The work done in frictionless compression of solid cylinder of annealed 70-30 brass.

Answer to Problem 2.61P

The work done in the frictionless compression of the cylinder is $W=6908\text{Nm}$

Explanation of Solution

Calculation:

The work done in frictionless compression of solid cylinder can be calculated by multiplying the specific energy and the volume of the cylinder.

For calculating the value of specific energy, the value of true strain is also required. The reduction in the height of the cylinder is 75%, thus the final height of the cylinder is 10 mm.

The value of true strain can be calculated as

  $\begin{array}{l}\epsilon =\ln \left(\frac{40}{10}\right)\\ \epsilon =1.386\end{array}$

The value of bulk modulus and n for the annealed 70-30 brass are

  $\begin{array}{l}K=895MPa\\ n=0.49\end{array}$

The specific energy can be calculated as

  $\begin{array}{l}u=\frac{K{\epsilon }^{n+1}}{n+1}\\ u=\frac{\left(895\right){\left(1.386\right)}^{1.49}}{1.49}\\ u=977MN/m^3\end{array}$

Now, the volume of the cylinder is

  $\begin{array}{l}V=\pi r^{2}l\\ V=\pi {\left(0.0075\right)}^2\left(0.04\right)\\ V=7.069\times {10}^{-6}{m}^3\end{array}$

Now, the work done is

  $\begin{array}{l}W=u\times V\\ W=977\times 7.069\text{Nm}\\ W=6908\text{Nm}\end{array}$

Conclusion: Thus, the work done in the frictionless compression of the cylinder is $W=6908\text{Nm}$