Chapter 9, Problem 42P

A sign (mass 1700 kg) hangs from the bottom end of a vertical steel girder with a cross-sectional area of 0.012 m². (a) What is the stress within the girder? (b) What is the strain on the girder? (c) If the girder is 9.50 m long,how much is it lengthened? (Ignore the mass of the girder itself.)

To determine

Stress within the girder.

Answer to Problem 42P

Solution:Stress within the girder is found to be $13.88\times 10^{5}N/m^2$ .

Explanation of Solution

Given: Cross-sectional area, $A=0.012{\text{ m}}^{\text{2}}$

Mass of sign, $m=1700\text{ kg}$

Length of girder, $L_i=9.50\text{ m}$

Young’s Modulus of the steel, $Y=200\times {10}^9{\text{N/m}}^{\text{2}}$

Formula used: Stress can be obtained by:

  $\sigma =\frac{F}{A}$

Where, $\sigma$ is the stress, F is the applied force, A is the cross-sectional area

Calculation: Stress in girder when it is hanged is given by

  $\begin{array}{l}\sigma =\frac{F}{A}\\ \sigma =\frac{m\times g}{A}\\ \sigma =\frac{1700\times 9.8}{0.012}\\ \sigma =13.88\times 10^{5}N/m^2\end{array}$

Conclusion:Stress within the girder is found to be $13.88\times 10^{5}N/m^2$ .Part (b)

To determine

Strain in the girder.

Answer to Problem 42P

Solution: Strain value is $6.94\times 10^{-6}$

Explanation of Solution

Given Cross-sectional area, $A=0.012{\text{ m}}^{\text{2}}$

Mass of sign, $m=1700\text{ kg}$

Length of girder, $L_i=9.50\text{ m}$

Young’s Modulus of the steel, $Y=200\times {10}^9{\text{N/m}}^{\text{2}}$

Stress within the girder is found to be $13.88\times {10}^5{\text{N/m}}^2$

Formula Used:

Strain is given by

  $\frac{\sigma }{\epsilon }=E$

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

Calculation:

From Hooke’s law

  $\begin{array}{l}\frac{\sigma }{\epsilon }=Y\\ \frac{\sigma }{Y}=\epsilon \\ \epsilon =\frac{13.88\times 10^5}{200\times 10^9}\\ \epsilon =6.94\times 10^{-6}\end{array}$

Conclusion: Strain value is $6.94\times 10^{-6}$ .

To determine

Change in length of Girder.

Answer to Problem 42P

Solution: The girder is lengthened by $6.593\times 10^{-5}m$ .

Explanation of Solution

Given:Cross-sectional area, $A=0.012{\text{ m}}^{\text{2}}$

Mass of sign, $m=1700\text{ kg}$

Length of girder, $L_i=9.50\text{ m}$

Young’s Modulus of the steel, $Y=200\times {10}^9{\text{N/m}}^{\text{2}}$

Stress within the girder found to be $13.88\times {10}^5{\text{N/m}}^2$

Strain value is $6.94\times 10^{-6}$ .

Formula used:

Shortened length is given by

  $\epsilon =\frac{L_f-L_i}{L_i}$

Where, $\epsilon$ is the strain, $L_f$ is the final length, $L_i$ is the original length.

Calculation: Change in length is given by

  $\begin{array}{l}\epsilon =\frac{L_f-L_i}{L_i}\\ 6.94\times 10^{-6}=\frac{L_f-9.5}{9.5}\\ L_f=(6.94\times 10^{-6}\times 9.5)+9.5\\ L_f=9.5000659m\end{array}$

This is the final length after loading.

The column is shortened by

  $\begin{array}{l}{L}_f-L_i=9.5-9.5000659\\ L_i-L_f=6.593\times 10^{-5}m\end{array}$

Conclusion:The girder is lengthened by $6.593\times 10^{-5}m$ .