Chapter 7, Problem 7.5.10P

A brass cube of 48 mm on each edge is comp ressed in two perpendicular directions by forces P = 160 kN (see figure).

(a) Calculate the change ...IV in the volume of the cube and the strain energy U stored in the cube. assuming E = 100 GPa and i’ = 0.34.

(b) Repeat part (a) if the cube is made of an alumim um alloy with E = 73 GPa and v = 0.33.

  

To determine

The change $\Delta V$ in the volume of the cube and strain energy U stored in the cube.

Answer to Problem 7.5.10P

The change $\Delta V$ in the volume of the cube is $-49.15{\text{mm}}^3$ .

The strain energy $U$ stored in the cube is $3.5195\text{J}$ .

Explanation of Solution

Given information:

Forces acting on the cube in the perpendicular directions is $-160\text{KN}$ , modulus of elasticity is $100\text{GP}$ , the length of edges is $48\text{mm}$ , and the Poisson’s ratio is $0.34$ .

  

     Figure (1)

Write the expression for the volumetric strain.

  $\frac{\Delta V}{V}={\epsilon }_x+{\epsilon }_y+{\epsilon }_z$ .....(I)

Here, the change in the volume of the cube is $\Delta V$ , the volume of cube is $V$ , the strain along x-axis is ${\epsilon }_x$ , the strain along y-axis is ${\epsilon }_y$ , and the strain along z-axis is ${\epsilon }_z$ .

Write the expression for the strain energy stored in the cube.

  $U=\frac{1}{2}{V}_o\left({\sigma }_X{\epsilon }_x+{\sigma }_Y{\epsilon }_y+{\sigma }_Z{\epsilon }_z\right)$ .....(II)

Here, the strain energy stored in the cube is $U$ .

Write the expression for the stresses.

  ${\sigma }_x=\frac{\text{Force}}{\text{Area}}$ .....(III)

Write the expression for the stresses along x-direction and y-direction.

  ${\sigma }_y=\frac{\text{Force}}{\text{Area}}$ .....(IV)

Here, the stress along x-axis is ${\sigma }_x$ , the stress along y-direction is ${\sigma }_y$ .

Write the expression for the strain along x-direction.

  ${\epsilon }_x=\frac{1}{E}\left({\sigma }_x-\nu {\sigma }_y\right)$ .....(V)

Here the Poisson’s ratio is $\nu$ .

Write the expression for the strain along y-direction.

  ${\epsilon }_y=\frac{1}{E}\left({\sigma }_y-\nu {\sigma }_x\right)$ .....(VI)

Write the expression for the strain along z-direction.

  ${\epsilon }_z=-\frac{\nu }{E}\left({\sigma }_x+{\sigma }_y\right)$ .....(VII)

Calculation:

Substitute $-160\times {10}^3\text{KN}$ for $\text{Force}$ , $0.048\text{m}\times 0.048\text{m}$ for $\text{Area}$ in Equation (III).

  $\begin{array}{c}{\sigma }_x=\frac{-160\times {10}^3\text{KN}}{0.048\text{m}\times 0.048\text{m}}\\ =-69.44\times {10}^6\text{kN}/{\text{m}}^2\left(\frac{1\text{MPa}}{{10}^6\text{kN}/{\text{m}}^2}\right)\\ =-69.44\text{MPa}\end{array}$

Substitute $-160\times {10}^3\text{KN}$ for $\text{Force}$ , $0.048\text{m}\times 0.048\text{m}$ for $\text{Area}$ in Equation (IV).

  $\begin{array}{c}{\sigma }_y=\frac{-160\times {10}^3\text{KN}}{0.048\text{m}\times 0.048\text{m}}\\ =-69.44\times {10}^6\text{kN}/{\text{m}}^2\left(\frac{1\text{MPa}}{{10}^6\text{kN}/{\text{m}}^2}\right)\\ =-69.44\text{MPa}\end{array}$

Substitute $100\times {10}^9\text{Pa}$ for $E$ , $0.34$ for $\nu$ , $-69.44\text{MPa}$ for ${\sigma }_x$ , and $-69.44\text{MPa}$ for ${\sigma }_y$ , in Equation (V).

  $\begin{array}{c}{\epsilon }_x=\frac{1}{100\times {10}^6\text{MPa}}\left(-69.44\text{MPa}-0.34\times \left(-69.44\text{MPa}\right)\right)\\ =\frac{-45.83\text{MPa}}{100\times {10}^3\text{MPa}}\\ =-485.3\times {10}^{-6}\end{array}$

Substitute $100\times {10}^9\text{Pa}$ for $E$ , $0.34$ for $\nu$ , $-69.44\text{MPa}$ for ${\sigma }_x$ , and $-69.44\text{MPa}$ for ${\sigma }_y$ , in Equation (VI).

  $\begin{array}{c}{\epsilon }_y=\frac{1}{100\times {10}^6\text{MPa}}\left(-69.44\text{MPa}-0.34\times \left(-69.44\text{MPa}\right)\right)\\ =\frac{-45.83\text{MPa}}{100\times {10}^3\text{MPa}}\\ =-485.3\times {10}^{-6}\end{array}$

Substitute $100\times {10}^9\text{Pa}$ for $E$ , $0.34$ for $\nu$ , $-69.44\text{MPa}$ for ${\sigma }_x$ , and $-69.44\text{MPa}$ for ${\sigma }_y$ in Equation (VII).

  $\begin{array}{c}{\epsilon }_z=-\frac{0.34}{100\times {10}^3\text{MPa}}\left(-69.44\text{MPa}+\left(-69.44\text{MPa}\right)\right)\\ =\frac{47.219\text{MPa}}{100\times {10}^3\text{MPa}}\\ =472.19\times {10}^{-6}\end{array}$

Substitute $-485.3\times {10}^{-6}$ for ${\epsilon }_x$ , $-485.3\times {10}^{-6}$ for ${\epsilon }_y$ , $472.19\times {10}^{-6}$ for ${\epsilon }_z$ , and $110,592{\text{mm}}^3$ for $V_o$ , in Equation (I).

  $\begin{array}{c}\Delta V=110592{\text{mm}}^3\times \left(-458.304-458.304+472.192\right)\times {10}^{-6}\\ =110592{\text{mm}}^3\times \left(444.416\right)\times {10}^{-6}\\ =-49.15{\text{mm}}^3\end{array}$

Substitute $-69.44\text{MPa}$ for ${\sigma }_x$ , and $-69.44\text{MPa}$ for ${\sigma }_y$

  $-485.3\times {10}^{-6}$

  $0\text{MPa}$ for ${\sigma }_Z$ , for ${\epsilon }_x$ , $-485.3\times {10}^{-6}$ for ${\epsilon }_y$ , $472.19\times {10}^{-6}$ for ${\epsilon }_z$ , and $110,592{\text{mm}}^3$ for $V_o$ , in Equation (II).

  $\begin{array}{c}U=\frac{1}{2}\left(110,592{\text{mm}}^3\right)\left(2\times 69.44\text{MPa}\times 485.3\times {10}^{-6}\right)\\ =3726{\text{mm}}^3\cdot \text{MPa}\left(\frac{1{\text{m}}^3}{{10}^9{\text{mm}}^3}\right)\left(\frac{1\text{kN}/{\text{m}}^2}{1\text{MPa}}\right)\\ =3.7\text{N}\cdot \text{m}\left(\frac{1\text{J}}{1\text{N}\cdot \text{m}}\right)\\ =3.7\text{J}\end{array}$

Conclusion:

The change $\Delta V$ in the volume of the cube is $-49.15{\text{mm}}^3$ .

The strain energy $U$ stored in the cube is $7.4\text{J}$ .

To determine

The change $\Delta V$ in the volume of the cube and the strain energy $U$ stored in the cube.

Answer to Problem 7.5.10P

The change $\Delta V$ in the volume of the cube is $-71.54{\text{mm}}^3$ .

The strain energy $U$ stored in the cube is $4.8943\text{J}$ .

Explanation of Solution

Given information:

Modulus of elasticity is $73\text{GP}$ , the length of edges is $48\text{mm}$ , and the Poisson’s ratio is $0.33$ .

Calculation:

Substitute $-160\times {10}^3\text{KN}$ for $\text{Force}$ , $0.048\text{m}\times 0.048\text{m}$ for $\text{Area}$ in Equation (III).

  $\begin{array}{c}{\sigma }_x=\frac{-160\times {10}^3\text{KN}}{0.048\text{m}\times 0.048\text{m}}\\ =-69.44\times {10}^6{\text{kN}/\text{m}}^2\left(\frac{1\text{MPa}}{{10}^6{\text{kN}/\text{m}}^2}\right)\\ =-69.44\text{MPa}\end{array}$

Substitute $-160\times {10}^3\text{KN}$ for $\text{Force}$ , $0.048\text{m}\times 0.048\text{m}$ for $\text{Area}$ in Equation (IV).

  $\begin{array}{c}{\sigma }_y=\frac{-160\times {10}^3\text{KN}}{0.048\text{m}\times 0.048\text{m}}\\ =-69.44\times {10}^6{\text{kN}/\text{m}}^2\left(\frac{1\text{MPa}}{{10}^6{\text{kN}/\text{m}}^2}\right)\\ =-69.44\text{MPa}\end{array}$

Substitute $73\times {10}^9\text{Pa}$ for $E$ , $0.33$ for $\nu$ , $-69.44\text{MPa}$ for ${\sigma }_x$ , and $-69.44\text{MPa}$ for ${\sigma }_y$ , in Equation (V).

  $\begin{array}{c}{\epsilon }_x=\frac{1}{73\times {10}^6\text{MPa}}\left(-69.44\text{MPa}-0.33\times \left(-69.44\text{MPa}\right)\right)\\ =\frac{-46.523\text{MPa}}{73\times {10}^3\text{MPa}}\\ =-637.326\times {10}^{-6}\end{array}$

Substitute $73\times {10}^9\text{Pa}$ for $E$ , $0.33$ for $\nu$ , $-69.44\text{MPa}$ for ${\sigma }_x$ , and $-69.44\text{MPa}$ for ${\sigma }_y$ , in Equation (VI).

  $\begin{array}{c}{\epsilon }_x=\frac{1}{73\times {10}^6\text{MPa}}\left(-69.44\text{MPa}-0.33\times \left(-69.44\text{MPa}\right)\right)\\ =\frac{-46.523\text{MPa}}{73\times {10}^3\text{MPa}}\\ =-637.326\times {10}^{-6}\end{array}$

Substitute $73\times {10}^9\text{Pa}$ for $E$ , $0.33$ for $\nu$ , $-69.44\text{MPa}$ for ${\sigma }_x$ , and $-69.44\text{MPa}$ for ${\sigma }_y$ in Equation (VII).

  $\begin{array}{c}{\epsilon }_z=-\frac{0.33}{73\times {10}^3\text{MPa}}\left(-69.44\text{MPa}+\left(-69.44\text{MPa}\right)\right)\\ =\frac{44.8304\text{MPa}}{73\times {10}^3\text{MPa}}\\ =627.814\times {10}^{-6}\end{array}$

Substitute $-637.326\times {10}^{-6}$ for ${\epsilon }_x$ , $-637.326\times {10}^{-6}$ for ${\epsilon }_y$ , $627.814\times {10}^{-6}$ for ${\epsilon }_z$ , and $110,592{\text{mm}}^3$ for $V_o$ , in Equation (I).

  $\begin{array}{c}\Delta V=110,592{\text{mm}}^3\times \left(-637.326\times {10}^{-6}-637.326\times {10}^{-6}+627.814\times {10}^{-6}\right)\times {10}^{-6}\\ =110592{\text{mm}}^3\times 646\times {10}^{-6}\\ =-71.54{\text{mm}}^3\end{array}$

Substitute $-69.44\text{MPa}$ for ${\sigma }_x$ , and $-69.44\text{MPa}$ for ${\sigma }_y$

  $-485.3\times {10}^{-6}$ , $0\text{MPa}$ for ${\sigma }_Z$ , for $-637.326\times {10}^{-6}$ for ${\epsilon }_x$ , $-637.326\times {10}^{-6}$ for ${\epsilon }_y$ , $627.814\times {10}^{-6}$ for ${\epsilon }_z$ , and $110,592{\text{mm}}^3$ for $V_o$ in Equation (II).

  $\begin{array}{c}U=\frac{1}{2}\left(110592{\text{mm}}^3\right)\left(2\times -69.44\text{MPa}\times -637.326\times {10}^{-6}\right)\\ =4894{\text{mm}}^3\cdot \text{MPa}\left(\frac{1{\text{m}}^3}{{10}^9{\text{mm}}^3}\right)\left(\frac{1\text{kN}/{\text{m}}^2}{1\text{MPa}}\right)\\ =4.8943\text{N}\cdot \text{m}\left(\frac{1\text{J}}{1\text{N}\cdot \text{m}}\right)\\ =4.8943\text{J}\end{array}$

Conclusion:

The change $\Delta V$ in the volume of the cube is $-71.54{\text{mm}}^3$ .

The strain energy $U$ stored in the cube is $4.8943\text{J}$ .