Chapter 5.2, Problem 48P

To determine

The maximum normal stress due to bending on section a-a.

Answer to Problem 48P

The maximum normal stress due to bending on section a-a is $\underset{\_}{139.2\text{MPa}}$.

Explanation of Solution

Show the free-body diagram of the entire beam as in Figure 1.

Determine the vertical reaction at point A by taking moment about point B.

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ 30\left(0.8\right)+50\left(1.6\right)+50\left(2.4\right)+30\left(3.2\right)-A_y\left(4\right)=0\\ 24+80+120+96-4A_y=0\\ A_y=80\text{kN}\end{array}$

Determine the vertical reaction at point B by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-30-50-50-30+B_y=0\\ 80-160+B_y=0\\ B_y=80\text{kN}\end{array}$

Determine the horizontal direction at point A by resolving the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ A_x=0\end{array}$

Shear force:

Show the calculation of shear force as follows;

$V_A=+80\text{kN}$

$\begin{array}{c}{V}_{0.8m}-V_A=0\\ V_{0.8m}^{Left}=V_A\\ =80\text{kN}\end{array}$

$\begin{array}{c}{V}_{0.8}^{Right}=80-30\\ =50\text{kN}\end{array}$

$\begin{array}{c}{V}_{1.6m}-V_{0.8m}=-30\\ V_{1.6m}^{Left}=-30+V_{0.8}\\ =-30+80\\ =50\text{kN}\end{array}$

$\begin{array}{c}{V}_{1.6m}^{Right}=50-50\\ =0\end{array}$

$\begin{array}{c}{V}_{2.4m}-V_{1.6m}=-50\\ V_{2.4m}^{Left}=-50+V_{1.6m}\\ =-50+50\\ =0\end{array}$

$\begin{array}{c}{V}_{2.4m}^{Right}=0-50\\ =-50\text{kN}\end{array}$

$\begin{array}{c}{V}_{3.2m}-V_{2.4m}=-50\\ V_{3.2}^{Left}=-50+V_{2.4m}\\ =-50+0\\ =-50\text{kN}\end{array}$

$\begin{array}{c}{V}_{3.2}^{Right}=-50-30\\ =-80\text{kN}\end{array}$

$\begin{array}{c}{V}_B-V_{3.2m}=-30\\ V_B=-30+V_{3.2m}\\ =-30-50\\ =-80\text{kN}\end{array}$

Bending moment:

Show the calculation of bending moment as follows;

$M_A=0$

$\begin{array}{c}{M}_{0.8m}-M_A=80\times 0.8\\ M_{0.8m}=64+M_A\\ =64\text{kN-m}\end{array}$

$\begin{array}{c}{M}_{1.6m}-M_{0.8m}=50\times 0.8\\ M_{1.6m}=40+M_{0.8m}\\ =40+64\\ =104\text{kN-m}\end{array}$

$\begin{array}{c}{M}_{a-a}-M_{1.6m}=0\\ M_{a-a}=0+M_{1.6m}\\ =0+104\\ =104\text{kN-m}\end{array}$

$\begin{array}{c}{M}_{2.4m}-M_{1.6m}=0\\ M_{2.4m}=M_{1.6m}\\ =104\text{kN-m}\end{array}$

$\begin{array}{c}{M}_{3.2m}-M_{2.4m}=-50\times 0.8\\ M_{3.2m}=-40+M_{2.4m}\\ =-40+104\\ =64\text{kN-m}\end{array}$

$M_B=0$

The bending moment at the section a-a is $\left|M_{a-a}\right|=104\text{kN}\cdot \text{m}$.

Refer to Appendix C “Properties of Rolled-Steel Sections” in the textbook.

The section modulus for $\text{W}310\times 52$ section is $S=747\times {10}^3{\text{mm}}^3$.

Determine the maximum normal stress $\left({\sigma }_m\right)$ using the equation.

${\sigma }_m=\frac{\left|M\right|}{S}$

Substitute $104\text{kN}\cdot \text{m}$ for M and $747\times {10}^3{\text{mm}}^3$ for S.

$\begin{array}{c}{\sigma }_m=\frac{104\text{kN}\cdot \text{m}\times \frac{1,000\text{N}}{1\text{kN}}}{747\times {10}^3{\text{mm}}^3\times {\left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^3}\\ =139.2\times {10}^6\text{Pa}\times \frac{\text{1}\text{MPa}}{{\text{10}}^{\text{6}}\text{Pa}}\\ =139.2\text{MPa}\end{array}$

Therefore, the maximum normal stress due to bending on section a-a is $\underset{\_}{139.2\text{MPa}}$.