Chapter 10, Problem 22P

(a)

To determine

The stress in the column.

Answer to Problem 22P

The stress is $2.8\times {10}^7\text{Pa}$.

Explanation of Solution

The cross sectional area is $25{\text{cm}}^{\text{2}}$, load is $7.0\times {10}^4\text{N}$, Young’s modulus is $60\text{GPa}$, and the compressive strength is $200\text{MPa}$.

Write the equation for stress.

$S=\frac{F}{A}$

Here, $S$ is the stress, $F$ is the applied force, and $A$ is the area of cross section.

Conclusion:

Substitute $7.0\times {10}^4\text{N}$ for $F$ and $25{\text{cm}}^{\text{2}}$ for $A$ in the above equation to find $S$.

$\begin{array}{c}S=\frac{7.0\times {10}^4\text{N}}{25{\text{cm}}^{\text{2}}\left(\frac{{10}^{-4}{\text{m}}^{\text{2}}}{1{\text{cm}}^{\text{2}}}\right)}\\ =2.8\times {10}^7\text{N}\end{array}$

Therefore, the stress is $2.8\times {10}^7\text{Pa}$.

(b)

To determine

The strain in the column.

Answer to Problem 22P

The strain is $4.7\times {10}^{-4}$.

Explanation of Solution

The cross sectional area is $25{\text{cm}}^{\text{2}}$, load is $7.0\times {10}^4\text{N}$, Young’s modulus is $60\text{GPa}$, and the compressive strength is $200\text{MPa}$.

Write the equation for strain.

$S_t=\frac{F}{YA}$

Here, $S_t$ is the strain and $Y$ is the young’s modulus.

Conclusion:

Substitute $7.0\times {10}^4\text{N}$ for $F$, $60\text{GPa}$ for $Y$, and $25{\text{cm}}^{\text{2}}$ for $A$ in the above equation to find $S$.

$\begin{array}{c}S=\frac{7.0\times {10}^4\text{N}}{\left(60\text{GPa}\left(\frac{{\text{10}}^{\text{9}}\text{Pa}}{\text{1}\text{GPa}}\right)\right)\left(25{\text{cm}}^{\text{2}}\left(\frac{{10}^{-4}{\text{m}}^{\text{2}}}{1{\text{cm}}^{\text{2}}}\right)\right)}\\ =4.7\times {10}^{-4}\text{N}\end{array}$

Therefore, the strain is $4.7\times {10}^{-4}$.

(c)

To determine

The elongation due to the supporting load.

Answer to Problem 22P

The elongation is $9.3\times {10}^{-4}\text{m}$.

Explanation of Solution

The height of column is $2.0\text{m}$, cross sectional area is $25{\text{cm}}^{\text{2}}$, load is $7.0\times {10}^4\text{N}$, Young’s modulus is $60\text{GPa}$, and the compressive strength is $200\text{MPa}$.

Write the equation to find the elongation due to the supporting load.

$\Delta L=\frac{FL}{YA}$

Conclusion:

Substitute $7.0\times {10}^4\text{N}$ for $F$, $2.0\text{m}$ for $L$, $60\text{GPa}$ for $Y$, and $25{\text{cm}}^{\text{2}}$ for $A$ in the above equation to find $\Delta L$.

$\begin{array}{c}\Delta L=\frac{\left(7.0\times {10}^4\text{N}\right)\left(2.0\text{m}\right)}{\left(60\text{GPa}\left(\frac{{\text{10}}^{\text{9}}\text{Pa}}{\text{1}\text{GPa}}\right)\right)\left(25{\text{cm}}^{\text{2}}\left(\frac{{10}^{-4}{\text{m}}^{\text{2}}}{1{\text{cm}}^{\text{2}}}\right)\right)}\\ =9.3\times {10}^{-4}\text{m}\end{array}$

Therefore, the strain is $9.37\times {10}^{-4}\text{m}$.

(d)

To determine

The maximum weight that can be supported by the load.

Answer to Problem 22P

The maximum load is $5.0\times {10}^5\text{N}$.

Explanation of Solution

The height of column is $2.0\text{m}$, cross sectional area is $25{\text{cm}}^{\text{2}}$, load is $7.0\times {10}^4\text{N}$, Young’s modulus is $60\text{GPa}$, and the compressive strength is $200\text{MPa}$.

When the column supports the maximum possible weight, the compressive stress will be equal to the stress.

Write the equation for the maximum weight that can be supported by the load.

$W_{\max }=S_{com}A$

Here, $W_{\max }$ is the maximum weight and $S_{com}$ is the compressive strength.

Conclusion:

Substitute $200\text{MPa}$ for $S_{com}$ and $25{\text{cm}}^{\text{2}}$ for $A$ in the above equation to find $W_{\max }$.

$\begin{array}{c}{W}_{\max }=\left(200\text{MPa}\left(\frac{{\text{10}}^{\text{6}}\text{Pa}}{\text{1}\text{MPa}}\right)\right)\left(25{\text{cm}}^{\text{2}}\left(\frac{{10}^{-4}{\text{m}}^{\text{2}}}{1{\text{cm}}^{\text{2}}}\right)\right)\\ =5.0\times {10}^5\text{N}\end{array}$

Therefore, the maximum load is $5.0\times {10}^5\text{N}$.