Chapter 9, Problem 40P

A marble column of cross-sectional area 1.4 m² supports a mass of 25,000 kg. (a) Whais the stress within the column? (b) What is the strain?

To determine

Stress of the column.

Answer to Problem 40P

Solution:Stress in the column found to be $1.75\times {10}^5{\text{N/m}}^{\text{2}}$ .

Explanation of Solution

Given: Cross sectional area, $A=1.{\text{4 m}}^{\text{2}}$

Applied mass, $m=25000\text{ kg}$

Young’s Modulus of the Marble, $Y=5\times {10}^{10}\text{Pa}$

Formula used:

Stress is given by

  $\sigma =\frac{F}{A}$ Where, $\sigma$ is the stress, F is the applied force, A is the cross-sectional area.

Calculation:

Stress is given by

  $\begin{array}{l}\sigma =\frac{F}{A}\\ \sigma =\frac{m\times g}{A}\\ \sigma =\frac{25000\times 9.8}{1.4}\\ \sigma =1.75\times 10^{5}N/m^2\end{array}$

Conclusion: a) Stress in the column found to be $1.75\times 10^{5}N/m^2$ .

To determine

Strainwithin the column.

Answer to Problem 40P

Solution: Strain value is $3.5\times {10}^{-6}$

Explanation of Solution

Given:Cross sectional area, $A=1.{\text{4 m}}^{\text{2}}$

Applied mass, $m=25000\text{ kg}$

Young’s Modulus of the Marble, $Y=5\times {10}^{10}\text{Pa}$

Stress $\sigma =1.75\times {10}^5{\text{N/m}}^{\text{2}}$

Formula used:

  $\frac{\sigma }{\epsilon }=Y$

Where, $\epsilon$ is the strain, Y is modulus of elasticity

Calculation:

From Hooke’s law

  $\begin{array}{l}\frac{\sigma }{\epsilon }=Y\\ \frac{\sigma }{Y}=\epsilon \\ \epsilon =\frac{1.75\times 10^5}{5\times 10^{10}}\\ \epsilon =3.5\times 10^{-6}\end{array}$

Conclusion:Strain value is $3.5\times 10^{-6}$ .