Chapter 2.9, Problem 84P

(a)

To determine

Find the change in height and the change in volume of brass cylinder.

Answer to Problem 84P

The change in height of brass cylinder is $\underset{\_}{-0.0746\text{mm}}$.

The change in volume of brass cylinder is $\underset{\_}{-143.9{\text{mm}}^3}$.

Explanation of Solution

Given information:

The young’s modulus E is $105\text{GPa}$.

The poison ratio $\nu$ is 0.33.

The normal stress ${\sigma }_y$ along y axis is $-58\text{MPa}$.

Calculation:

Find the area of solid sphere as follows:

$A_0=\frac{\pi }{4}{d}_0^3$ (1)

Here, $d_0$ is diameter of the solid steel sphere.

Substitute $85\text{mm}$ for $d_0$ in Equation (1).

$\begin{array}{c}{A}_0=\frac{\pi }{4}{\left(85\right)}^2\\ =5.6745\times {10}^3{\text{mm}}^{\text{2}}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =5.6745\times {10}^{-3}{\text{m}}^{\text{2}}\end{array}$

Find the volume of solid sphere as follows:

$V_0=A_0{h}_0$ (2)

Here, $A_0$ is area of solid sphere and $h_0$ is height of solid sphere.

Substitute $5.6745\times {10}^{-3}{\text{m}}^{\text{2}}$ for $A_0$ and $135\text{mm}$ for $h_0$ in Equation (2).

$\begin{array}{c}{V}_0=5.6745\times {10}^{-3}\times 135\text{mm}{\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}^2\\ =5.6745\times {10}^{-3}\times 0.135\\ =766.06\times {10}^{-6}{\text{m}}^{\text{3}}\end{array}$

The normal stress $\left({\sigma }_x\right)$ along x axis is $0$.

The normal stress $\left({\sigma }_y\right)$ along y axis is $-58\times {10}^6\text{Pa}$

The normal stress $\left({\sigma }_z\right)$ along z axis is $0$

Write the expression of strain along y axis as follows:

${\epsilon }_y=\frac{1}{E}\left(-\nu {\sigma }_x+{\sigma }_y-\nu {\sigma }_z\right)$ (3)

Substitute 0 for ${\sigma }_x$ and 0 for ${\sigma }_y$ in Equation (3).

$\begin{array}{c}{\epsilon }_y=\frac{1}{E}\left(-\nu \left(0\right)+{\sigma }_y-\nu \left(0\right)\right)\\ =\frac{{\sigma }_y}{E}\end{array}$ (4)

Substitute $-58\times {10}^6\text{Pa}$ for ${\sigma }_y$ and $105\text{GPa}$ for E in Equation (4).

$\begin{array}{c}{\epsilon }_y=\frac{-58\times {10}^6}{105\text{GPa}\left(\frac{{10}^6\text{Pa}}{1\text{GPa}}\right)}\\ =\frac{-58\times {10}^6}{105\times {10}^6}\\ =-552.38\times {10}^{-6}\end{array}$

Determine the change in height of brass cylinder using the relation:

$\Delta h=h_0{\epsilon }_y$ (5)

Substitute $-552.38\times {10}^{-6}$ for ${\epsilon }_y$ and $135\text{mm}$ for $h_0$ in Equation (5).

$\begin{array}{c}\Delta h=-552.38\times {10}^{-6}\times 135\text{mm}\\ =-552.38\times {10}^{-6}\times 135\\ =-0.07457\\ =-0.0746\text{mm}\end{array}$

Thus, the change in height of brass cylinder is $\underset{\_}{-0.0746\text{mm}}$.

Find the value of dilatation e using the relation:

$e=\frac{1-2v}{E}\left({\sigma }_x+{\sigma }_y+{\sigma }_z\right)$ (6)

Substitute 0 for ${\sigma }_x$ and 0 for ${\sigma }_y$ in Equation (6).

$\begin{array}{c}e=\frac{1-2v}{E}\left(0+{\sigma }_y+0\right)\\ =\frac{\left(1-2v\right){\sigma }_y}{E}\end{array}$ (7)

Substitute 0.33 for $\nu$, $-58\times {10}^6\text{Pa}$ for ${\sigma }_y$ and $105\text{GPa}$ for E in Equation (7).

$\begin{array}{c}e=\frac{\left(1-2\left(0.33\right)\right)-58\times {10}^6}{105\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)}\\ =-\frac{19,720,000}{105\times {10}^9}\\ =-187.81\times {10}^{-6}\end{array}$

Determine the change in volume of brass cylinder as follows:

$\Delta V=V_{0}e$ (8)

Substitute $-187.81\times {10}^{-6}$ for e and $766.66\times {10}^3{\text{mm}}^{\text{3}}$ for $V_0$ in Equation (8).

$\begin{array}{c}\Delta V=-187.81\times {10}^{-6}\times 766.66\times {10}^3\\ =-143.9{\text{mm}}^{\text{3}}\end{array}$

Hence, the change in volume of brass cylinder is $\underset{\_}{-143.9{\text{mm}}^3}$.

(b)

To determine

Find the change in height and the change in volume of brass cylinder when ${\sigma }_x={\sigma }_y={\sigma }_z=-70\text{MPa}$

Answer to Problem 84P

The change in height of brass cylinder when ${\sigma }_x={\sigma }_y={\sigma }_z=-70\text{MPa}$ is $\underset{\_}{-0.0306\text{mm}}$.

The change in volume of brass cylinder when ${\sigma }_x={\sigma }_y={\sigma }_z=-70\text{MPa}$ is $\underset{\_}{-521{\text{mm}}^3}$.

Explanation of Solution

Calculation:

Find the value of ${\sigma }_x+{\sigma }_y+{\sigma }_z$ as follows:

$\begin{array}{c}{\sigma }_x+{\sigma }_y+{\sigma }_z=-70\times {10}^6-70\times {10}^6-70\times {10}^6\\ =-210\times {10}^6\text{Pa}\end{array}$

Write the expression of strain along y axis as follows:

$\begin{array}{c}{\epsilon }_y=\frac{1}{E}\left(-\nu {\sigma }_x+{\sigma }_y-\nu {\sigma }_z\right)\\ =\frac{1-2\upsilon }{E}{\sigma }_y\end{array}$ (9)

Substitute 0.34 for $\nu$, $-70\times {10}^6\text{Pa}$ for ${\sigma }_y$ and $105\text{GPa}$ for E in Equation (9).

$\begin{array}{c}{\epsilon }_y=\frac{1-2\left(0.33\right)}{105\text{GPa}\left(\frac{{10}^6\text{Pa}}{1\text{GPa}}\right)}-70\times {10}^6\\ =\frac{1-2\left(0.33\right)}{105\times {10}^9}-70\times {10}^6\\ =-226.67\times {10}^{-6}\end{array}$

Determine the change in height of brass cylinder when ${\sigma }_x={\sigma }_y={\sigma }_z=-70\text{MPa}$ using the relation:

$\Delta h=h_0{\epsilon }_y$ (10)

Substitute $-226.67\times {10}^{-6}$ for ${\epsilon }_y$ and $135\text{mm}$ for $h_0$ in Equation (10).

$\begin{array}{c}\Delta h=-226.67\times {10}^{-6}\times 135\\ =-0.0306\text{mm}\end{array}$

Thus, the change in height of brass cylinder when ${\sigma }_x={\sigma }_y={\sigma }_z=-70\text{MPa}$ is $\underset{\_}{-0.0306\text{mm}}$.

Find the value of dilatation e using the relation:

$e=\frac{1-2v}{E}\left({\sigma }_x+{\sigma }_y+{\sigma }_z\right)$ (11)

Substitute 0.34 for $\nu$, $105\text{GPa}$ for E $-210\times {10}^6$ for $\left({\sigma }_x+{\sigma }_y+{\sigma }_z\right)$ in Equation (11).

$\begin{array}{c}e=\frac{1-2\left(0.33\right)}{105\times {10}^9}\left(-210\times {10}^6\right)\\ =-680\times {10}^{-6}\end{array}$

Determine the change in volume of brass cylinder when ${\sigma }_x={\sigma }_y={\sigma }_z=-70\text{MPa}$ as follows:

$\Delta V=V_{0}e$ (12)

Substitute $-680\times {10}^{-6}$ for e and $766.06\times {10}^{-6}{\text{m}}^{\text{3}}$ for $V_0$ in Equation (12).

$\begin{array}{c}\Delta V=-680\times {10}^{-6}\times 766.66\times {10}^3\\ =-521{\text{mm}}^{\text{3}}\end{array}$

Hence, the change in volume of brass cylinder when ${\sigma }_x={\sigma }_y={\sigma }_z=-70\text{MPa}$ is $\underset{\_}{-521{\text{mm}}^3}$.