Chapter 2.9, Problem 70P

The block shown is made of a magnesium alloy for which E = 45 GPa and ν = 0.35, Knowing that ϭ_x = -180 MPa, determine (a) the magnitude of ϭ_y for which the change in the height of the block will be zero, (b) the corresponding change in the area of the face ABCD, (c) the corresponding change in the volume of the block.

Fig. P2.70

To determine

Find the magnitude of ${\sigma }_y$ for such that change in the height of the block will be zero.

Answer to Problem 70P

The magnitude of ${\sigma }_y$ is $\underset{\_}{-63.0\text{MPa}}$.

Explanation of Solution

Given information:

The poison’ ratio $\nu$ is 0.35.

The young’s modulus (E) is $45\text{GPa}$.

The normal stress, ${\sigma }_x$ is $-180\text{MPa}$.

Calculation:

Consider that the change in height of the block is zero.

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

Find the normal stress $\left({\sigma }_y\right)$ along y axis as follows:

${\sigma }_y=\nu {\sigma }_x$ (1)

Substitute 0.35 for $\nu$ and $-180\text{MPa}$ for ${\sigma }_x$ in Equation (1).

$\begin{array}{c}{\sigma }_y=0.35\times \left[-180\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)\right]\\ =0.35\times \left(-180\times {10}^6\right)\\ =-63\times {10}^6\text{Pa}\end{array}$

Thus, the magnitude of ${\sigma }_y$ is $\underset{\_}{-63.0\text{MPa}}$.

To determine

Find the corresponding change in the area of the face ABCD.

Answer to Problem 70P

The corresponding change in the area of the face ABCD is $\underset{\_}{-4.05{\text{mm}}^{\text{2}}}$.

Explanation of Solution

Calculation:

Consider the stress strain relationship for homogeneous, isotropic material.

Find the strain along z direction as follows:

Strain in z direction.

$\begin{array}{c}{\epsilon }_z=\frac{1}{E}\left[{\sigma }_z-\upsilon {\sigma }_x-\upsilon {\sigma }_y\right]\\ =-\frac{\upsilon }{E}\left({\sigma }_x+{\sigma }_y\right)\end{array}$ (2)

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

Substitute 0.35 for $\upsilon$, $45\text{GPa}$ for E, $-180\text{MPa}$ for ${\sigma }_x$, and $-63\times {10}^6\text{Pa}$ for ${\sigma }_y$ in Equation (2).

$\begin{array}{c}{\epsilon }_z=-\frac{0.35}{45\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)}\left(-180\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)-63\times {10}^6\right)\\ =-\frac{0.35}{45\times {10}^9}\left(-180\times {10}^6-63\times {10}^6\right)\\ =-7.7777\times {10}^{-12}\left(-243,000,000\right)\\ =1.890\times {10}^{-3}\end{array}$

Find the strain along x direction as follows:

$\begin{array}{c}{\epsilon }_x=\frac{1}{E}\left[{\sigma }_x-\upsilon {\sigma }_y-\upsilon {\sigma }_z\right]\\ =\frac{{\sigma }_x-\upsilon {\sigma }_y}{E}\end{array}$ (3)

Substitute 0.35 for $\upsilon$, $45\text{GPa}$ for E, $-180\text{MPa}$ for ${\sigma }_x$, and $-63\times {10}^6\text{Pa}$ for ${\sigma }_y$ in Equation (3).

$\begin{array}{c}{\epsilon }_x=\frac{-180\text{MPa}\left(\frac{{10}^6\text{Pa}}{1\text{MPa}}\right)-\left(0.35\right)\left(-63\times {10}^6\right)}{45\text{GPa}\left(\frac{{10}^9}{1\text{GPa}}\right)}\\ =\frac{-180\times {10}^6-\left(-22,050,000\right)}{45\times {10}^9}\\ =\frac{-157.95\times {10}^6}{45\times {10}^9}\\ =-3.510\times {10}^{-3}\end{array}$

Write the expression of area $A_0$ as shown follows:

$A_0=L_x{L}_z$

Write the expression of A as shown follows:

$\begin{array}{c}A=L_x\left(1+{\epsilon }_x\right)L_z\left(1+{\epsilon }_z\right)\\ =L_x{L}_z\left(1+{\epsilon }_x+{\epsilon }_z+{\epsilon }_x{\epsilon }_z\right)\end{array}$

Determine the change in area face ABCD as follows:

$\Delta A=A-A_0$ (4)

Substitute $L_x{L}_z\left(1+{\epsilon }_x+{\epsilon }_z+{\epsilon }_x{\epsilon }_z\right)$ for A and $L_x{L}_z$ for $A_0$ in Equation (4).

$\begin{array}{c}\Delta A=L_x{L}_z\left(1+{\epsilon }_x+{\epsilon }_z+{\epsilon }_x{\epsilon }_z\right)-\left(L_x{L}_z\right)\\ =L_x{L}_z\left({\epsilon }_x+{\epsilon }_z\right)\end{array}$ (5)

Substitute $100\text{mm}$ for $L_x$, $25\text{mm}$ for $L_z$, $-3.510\times {10}^{-3}$ for ${\epsilon }_x$, and $1.890\times {10}^{-3}$ for ${\epsilon }_z$ in Equation (5).

$\begin{array}{c}\Delta A=\left(100\times 25\right)\left(-3.510\times {10}^{-3}+1.890\times {10}^{-3}\right)\\ =2,500\left(-1.62\times {10}^{-3}\right)\\ =-4.05{\text{mm}}^{\text{2}}\end{array}$

Thus, the corresponding change in the area of the face ABCD is $\underset{\_}{-4.05{\text{mm}}^{\text{2}}}$.

To determine

Find the corresponding change in the volume of block.

Answer to Problem 70P

The corresponding change in the volume of block is $\underset{\_}{-162{\text{mm}}^3}$.

Explanation of Solution

Calculation:

Write the expression of area $V_0$ as shown follows:

$V_0=L_x{L}_z{L}_y$

Write the expression of V as shown follows:

$\begin{array}{c}V=L_x\left(1+{\epsilon }_x\right)L_y\left(1+{\epsilon }_y\right)L_z\left(1+{\epsilon }_z\right)\\ =L_x{L}_y{L}_z\left(1+{\epsilon }_x+{\epsilon }_y+{\epsilon }_z+{\epsilon }_x{\epsilon }_y+{\epsilon }_y{\epsilon }_z+{\epsilon }_z{\epsilon }_x+{\epsilon }_x{\epsilon }_y{\epsilon }_z\right)\end{array}$

Determine the change volume block as follows:

$\Delta V=V-V_0$ (6)

Substitute $L_x{L}_y{L}_z\left(1+{\epsilon }_x+{\epsilon }_y+{\epsilon }_z+{\epsilon }_x{\epsilon }_y+{\epsilon }_y{\epsilon }_z+{\epsilon }_z{\epsilon }_x+{\epsilon }_x{\epsilon }_y{\epsilon }_z\right)$ for V and $L_x{L}_z{L}_y$ for $V_0$ in Equation (6).

$\begin{array}{c}\Delta V=L_x{L}_y{L}_z\left(1+{\epsilon }_x+{\epsilon }_y+{\epsilon }_z+{\epsilon }_x{\epsilon }_y+{\epsilon }_y{\epsilon }_z+{\epsilon }_z{\epsilon }_x+{\epsilon }_x{\epsilon }_y{\epsilon }_z\right)-L_x{L}_z{L}_y\\ =L_x{L}_y{L}_z\left({\epsilon }_x+{\epsilon }_y+{\epsilon }_z+\text{small}\text{terms}\right)\end{array}$ (7)

Substitute $100\text{mm}$ for $L_x$, $40\text{mm}$ for $L_y$, $25\text{mm}$ for $L_z$, $-3.510\times {10}^{-3}$ for ${\epsilon }_x$, 0 for ${\epsilon }_y$, and $1.890\times {10}^{-3}$ for ${\epsilon }_z$ in Equation (7).

$\begin{array}{c}\Delta V=\left(100\times 40\times 25\right)\left(-3.510\times {10}^{-3}+0+1.890\times {10}^{-3}\right)\\ =100,000\left(-1.62\times {10}^{-3}\right)\\ =-162{\text{mm}}^{\text{3}}\end{array}$

Thus, the corresponding change in the volume of block is $\underset{\_}{-162{\text{mm}}^3}$.