Chapter 2.9, Problem 91P

(a)

To determine

Find the change in length of the cube in the x direction.

Answer to Problem 91P

The change in length of the cube in the x direction is $\underset{\_}{0.0303\text{mm}}$.

Explanation of Solution

Given information:

The modulus of elasticity in x-direction $E_x$ is $50\text{GPa}$.

The modulus of elasticity y-direction $E_y$ is $15.2\text{GPa}$.

The modulus of elasticity in z-direction $E_z$ is $15.2\text{GPa}$.

The strain along xz axis is 0.254.

The strain along xy axis is 0.254

The strain along zy axis is 0.428.

Calculation:

Write the stress to strain Equation along x direction as follows:

${\epsilon }_x=\frac{{\sigma }_x}{E_x}-\frac{{\nu }_{yx}{\sigma }_y}{E_y}-\frac{{\nu }_{zx}{\sigma }_z}{E_z}$ (1)

Here, ${\sigma }_x,{\sigma }_y,{\sigma }_z$ is normal stress along x, y, z axis respectively.

Write the stress to strain Equation along y direction as follows:

${\epsilon }_y=-\frac{{\nu }_{xy}{\sigma }_x}{E_x}+\frac{{\sigma }_y}{E_y}-\frac{{\nu }_{zy}{\sigma }_z}{E_z}$ (2)

Write the stress to strain Equation along z direction as follows:

${\epsilon }_z=-\frac{{\nu }_{xz}{\sigma }_x}{E_x}-\frac{{\nu }_{yz}{\sigma }_y}{E_y}+\frac{{\sigma }_z}{E_z}$ (3)

Equate the stress Equation along xy axis.

$\frac{{\nu }_{xy}}{E_x}=\frac{{\nu }_{yx}}{E_y}$ (4)

Equate the stress Equation along yz axis.

$\frac{{\nu }_{yz}}{E_y}=\frac{{\nu }_{zy}}{E_z}$ (5)

Equate the stress Equation along zx axis.

$\frac{{\nu }_{zx}}{E_z}=\frac{{\nu }_{xz}}{E_x}$ (6)

Apply the constraint conditions as follows:

$\begin{array}{l}{\epsilon }_y=0\\ {\epsilon }_z=0\end{array}$

Substitute 0 for ${\epsilon }_y$ in Equation (2).

$\begin{array}{l}0=-\frac{{\nu }_{xy}{\sigma }_x}{E_x}+\frac{{\sigma }_y}{E_y}-\frac{{\nu }_{zy}{\sigma }_z}{E_z}\\ \frac{1}{E_y}{\sigma }_y-\frac{{\nu }_{zy}}{E_z}{\sigma }_z=\frac{{\nu }_{xy}}{E_x}{\sigma }_x\end{array}$ (7)

Substitute 0 for ${\epsilon }_z$ in Equation (3).

$\begin{array}{l}0=-\frac{{\nu }_{xz}{\sigma }_x}{E_x}-\frac{{\nu }_{yz}{\sigma }_y}{E_y}+\frac{{\sigma }_z}{E_z}\\ -\frac{{\nu }_{yz}}{E_y}{\sigma }_y+\frac{1}{E_z}{\sigma }_z=\frac{{\nu }_{xz}}{E_x}{\sigma }_x\end{array}$ (8)

Substitute $15.2\text{GPa}$ for $E_y$, 0.428 for ${\nu }_{zy}$, $15.2\text{GPa}$ for $E_z$, 0.254 for ${\nu }_{xz}$, and $50\text{GPa}$ for $E_x$ in Equation (7).

$\begin{array}{l}\frac{1}{15.2}{\sigma }_y-\frac{0.428}{15.2}{\sigma }_z=\frac{0.254}{50}{\sigma }_x\\ \frac{1}{15.2}{\sigma }_y-\frac{0.428}{15.2}{\sigma }_z=5.08\times {10}^{-3}{\sigma }_x\\ {\sigma }_y-0.428{\sigma }_z=0.077216{\sigma }_x\end{array}$

Substitute $15.2\text{GPa}$ for $E_y$, 0.428 for $v_{zy}$, $15.2\text{GPa}$ for $E_z$, 0.254 for ${\nu }_{xy}$, and $50\text{GPa}$ for $E_x$ in Equation (8).

$\begin{array}{l}-\frac{0.428}{15.2}{\sigma }_y+\frac{1}{15.2}{\sigma }_z=\frac{0.254}{50}{\sigma }_x\\ -\frac{0.428}{15.2}{\sigma }_y+\frac{1}{15.2}{\sigma }_z=5.08\times {10}^{-3}{\sigma }_x\\ -0.428{\sigma }_y+{\sigma }_z=0.077216{\sigma }_x\end{array}$

Solving Equation to get values,

${\sigma }_y={\sigma }_z=0.134993{\sigma }_x$

Apply the Equation (4) and (5) in Equation 1 as follows:

${\epsilon }_x=\frac{1}{E_x}{\sigma }_x-\frac{{\nu }_{xy}}{E_x}{\sigma }_y-\frac{{\nu }_{xz}}{E}{\sigma }_z$

Substitute 0.254 for ${\nu }_{xy}$, 0.254 for ${\nu }_{xz}$, $0.134993{\sigma }_x$ for ${\sigma }_y$, $0.134993{\sigma }_x$ for ${\sigma }_z$.

$\begin{array}{c}{\epsilon }_x=\frac{1}{E_x}{\sigma }_x-\frac{0.254}{E_x}\left(0.134993{\sigma }_x\right)-\frac{0.254}{E}\left({\sigma }_z\right)\\ =\frac{1}{E_x}\left[1-\left(0.254\right)\left(0.134993\right)-\left(0.254\right)\left(0.134993\right)\right]{\sigma }_x\\ =\frac{1}{E_x}\left[0.93142{\sigma }_x\right]\end{array}$ (9)

Calculate the cross sectional area of cube as follows:

$A=a^2$ (10)

Here, a is the sides.

Substitute $40\text{mm}$ for a in Equation (10).

$\begin{array}{c}A={\left(40\right)}^2\\ =1,600{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{m}}\right)}^2\\ =1,600\times {10}^{-6}{\text{m}}^{\text{2}}\end{array}$

Find the normal stress along x axis as follows:

${\sigma }_x=\frac{P}{A}$ (11)

Here, P is the tensile load and A is the cross sectional area.

Substitute $65\text{kN}$ for P and $1,600\times {10}^{-6}{\text{m}}^{\text{2}}$ for A in Equation (11).

$\begin{array}{c}{\sigma }_x=\frac{65\text{kN}\left(\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\right)}{1,600\times {10}^{-6}{\text{m}}^{\text{2}}}\\ =\frac{65\times {10}^3}{1,600\times {10}^{-6}}\\ =40.625\times {10}^6\text{Pa}\end{array}$

Find the strain along x axis as follows:

Substitute $40.625\times {10}^6\text{Pa}$ for ${\sigma }_x$ and $50\text{GPa}$ for $E_x$ in Equation (9).

$\begin{array}{c}{\epsilon }_x=\frac{1}{50\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)}\left[0.93142\times 40.625\times {10}^6\right]\\ =\frac{1}{50\times {10}^9}\left[0.93142\times 40.625\times {10}^6\right]\\ =\frac{1}{50\times {10}^9}\left[37,838,937.5\right]\\ =756.78\times {10}^{-6}\end{array}$

Determine the change in length of the cube in the x direction using the relation:

${\delta }_x=L_x{\epsilon }_x$ (12)

Here, $L_x$ is side length and ${\epsilon }_x$ strain along x axis.

Substitute $40\text{mm}$ for $L_x$ and $756.78\times {10}^{-6}$ for ${\epsilon }_x$ in Equation (12).

$\begin{array}{c}{\delta }_x=40\times 756.78\times {10}^{-6}\\ =0.03027\\ =0.0303\text{mm}\end{array}$

Thus, the change in length of the cube in the x direction is $\underset{\_}{0.0303\text{mm}}$.

(b)

To determine

The stress values ${\sigma }_x,{\sigma }_y,{\sigma }_z$.

Answer to Problem 91P

The stress value ${\sigma }_x$ is $\underset{\_}{40.6\text{MPa}}$

The stress value ${\sigma }_y$ is $\underset{\_}{5.48\text{MPa}}$

The stress value ${\sigma }_z$ is $\underset{\_}{5.48\text{MPa}}$

Explanation of Solution

Calculation:

Find the normal stress along x axis as follows:

${\sigma }_x=\frac{P}{A}$ (13)

Here, P is the tensile load and A is the cross sectional area.

Substitute $65\text{kN}$ for P and $1,600\times {10}^{-6}{\text{m}}^{\text{2}}$ for A in Equation (13).

$\begin{array}{c}{\sigma }_x=\frac{65\text{kN}\left(\left(\frac{{10}^3\text{N}}{1\text{kN}}\right)\right)}{1,600\times {10}^{-6}{\text{m}}^{\text{2}}}\\ =\frac{65\times {10}^3}{1,600\times {10}^{-6}}\\ =40.625\times {10}^6\text{Pa}\end{array}$

Thus, the stress value ${\sigma }_x$ is $\underset{\_}{40.6\text{MPa}}$

Find the normal stress along y,z axis as follows:

${\sigma }_y={\sigma }_z=0.134993{\sigma }_x$

Substitute $40.625\times {10}^6\text{Pa}$ for ${\sigma }_x$.

$\begin{array}{c}{\sigma }_y={\sigma }_z=0.134993\times 40.625\times {10}^6\\ =5,484,090.625\text{Pa}\left(\frac{1\text{MPa}}{{10}^6\text{Pa}}\right)\\ =5.48\text{MPa}\end{array}$

Thus, the stress value ${\sigma }_y$ is $\underset{\_}{5.48\text{MPa}}$

Thus, the stress value ${\sigma }_z$ is $\underset{\_}{5.48\text{MPa}}$.