Chapter 9, Problem 53P

To determine

Minimum cross-sectional area required for each support.

Answer to Problem 53P

Solution: Minimum cross-sectional area required For A is 1.8×10-3 m²and B is 8×10-3 m².

Explanation of Solution

Given:Mass of beam, $m=2900\text{ kg}$

Factor of safety, $f=9$

Formula Used:

  ${\sigma }_{ut}=\frac{F}{A}$

Where, ${\sigma }_{ut}$ is the ultimate tensile stress, A is the area of cross section, F is the applied force.

Calculations: Taking moment about point A,

  $\begin{array}{l}{\displaystyle \sum M}=0\\ -F_B\times 20+(mg)(20+5)=0\\ F_B\times 20=(mg)(20+5)\\ F_B=(2900\times 9.81)(25)/20\\ F_B=35561N\end{array}$

For equilibrium condition, taking forces in y direction

  $\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_A+F_B-(9.81\times 2900)=0\\ F_A+35561-(9.81\times 2900)=0\\ F_A=-7112N\end{array}$

F_A is pointing in the opposite direction

To find the area, use stress formula, for support b which is having tensile force, use the tensile strength of wood which is equal to $40\times {10}^6{\text{N/m}}^{\text{2}}$ .

  $\begin{array}{l}{\sigma }_{ut}=\frac{F\times f}{A}\\ A=\frac{35561\times 9}{40\times 10^6}\end{array}$

  $A_B=8.0\times 10^{-3}{m}^2$

The required cross section area for B $8\times {10}^{-3}{\text{m}}^{\text{2}}$ .

To find the area of A, use stress formula, for support b which is having compressive force, we must use the compressive strength of wood. which is equal to $35\times {10}^6{\text{N/m}}^{\text{2}}$ .

  $\begin{array}{l}A=\frac{F\times f}{{\sigma }_{ut}}\\ A=\frac{7112\times 9}{35\times 10^6}\\ A_{}=1.82\times 10^{-3}{m}^2\end{array}$

The required cross section For A is $1.82\times 10^{-3}{m}^2$ .

Conclusion:Cross-sectional area required for A is $1.82\times 10^{-3}{m}^2$ and B is $8\times {10}^{-3}{\text{m}}^{\text{2}}$ .