Chapter 5.2, Problem 62P

The beam AB supports two concentrated loads P and Q. The normal stress due to bending on the bottom edge of the beam is +55 MPa at D and +37.5 MPa at F. (a) Draw the shear and bending-moment diagrams for the beam. (b) Determine the maximum normal stress due to bending that occurs in the beam.

Fig. P5.62

To determine

Draw the shear and bending-moment diagrams for the beam.

Explanation of Solution

Given information:

The normal stress due to bending at the point D is ${\sigma }_{D\left(\text{Bottom}\text{edge}\right)}=+55\text{MPa}$.

The normal stress due to bending at the point F is ${\sigma }_{F\left(\text{Bottom}\text{edge}\right)}=+37.5\text{MPa}$

Determine the section modulus (S) of the rectangular beam section using the equation.

$S=\frac{1}{6}b{h}^2$

Here, the width of the beam is b and the depth of the beam is h.

Substitute 24 mm for b and 60 mm for h.

$\begin{array}{c}S=\frac{1}{6}\times 24\times {60}^2\\ =14,400{\text{mm}}^3\end{array}$

Determine the bending moment at point D $\left(M_D\right)$ using the relation.

$M_D={\sigma }_{D}S$

Here, the normal stress at point D is ${\sigma }_D$.

Substitute 55 MPa for ${\sigma }_D$ and $14400{\text{mm}}^3$ for S.

$\begin{array}{c}{M}_D=55\text{MPa}\times \frac{{10}^6\text{Pa}}{1\text{MPa}}\times 14400{\text{mm}}^3\times {\left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^3\\ =792\text{N-m}\end{array}$

Determine the bending moment at point F $\left(M_F\right)$ using the relation.

$M_F={\sigma }_{F}S$

Here, the normal stress at point F is ${\sigma }_F$.

Substitute 37.5 MPa for ${\sigma }_D$ and $14,400{\text{mm}}^3$ for S.

$\begin{array}{c}{M}_F=37.5\text{MPa}\times \frac{{10}^6\text{Pa}}{1\text{MPa}}\times 14,400{\text{mm}}^3\times {\left(\frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)}^3\\ =540\text{N-m}\end{array}$

Show the free-body diagram of the region FB as in Figure 1.

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

$\begin{array}{l}{\displaystyle \sum M_F=0}\\ -540+B_y\left(0.3\right)=0\\ 0.3B_y=540\\ B_y=1,800\text{N}\end{array}$

Show the free body diagram of the region DEFB as in Figure 2.

Determine the magnitude of the load Q by taking moment about the point D.

$\begin{array}{l}{\displaystyle \sum M_D=0}\\ -M_D+B_y\left(0.8\right)-Q\left(0.3\right)=0\\ -792+0.8\left(1,800\right)-0.3Q=0\\ Q=2,160\text{N}\end{array}$

Show the free body diagram of the entire beam as in Figure 3.

Determine the magnitude of the load P by taking moment about the point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -P\left(0.2\right)-2,160\left(0.7\right)+1,800\left(1.2\right)=0\\ -0.2P-1,512+2,160=0\\ P=3,240\text{N}\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-P-2,160+1,800=0\\ A_y-3,240-360=0\\ A_y=3,600\text{N}\end{array}$

Shear force:

Show the calculation of shear force as follows;

$V_A=3,600\text{N}$

$\begin{array}{c}{V}_C-V_A=0\\ V_C^{Left}=V_A\\ =3,600\text{N}\end{array}$

$\begin{array}{c}{V}_C^{Right}=3,600-3,240\\ =360\text{N}\end{array}$

$\begin{array}{c}{V}_E-V_C=0\\ V_E^{Left}=0+V_C\\ =0+360\\ =360\text{N}\end{array}$

$\begin{array}{c}{V}_E^{Right}=360-2160\\ =-1,800\text{N}\end{array}$

$\begin{array}{c}{V}_B-V_E=0\\ V_B=V_E\\ =-1,800\text{N}\end{array}$

Show the calculated shear force values as in Table 1.

Location (x) m	Shear force (V) N
A	3600
C (Left)	3600
C (Right)	360
E (Left)	360
E (Right)	–1800
B	–1800

Plot the shear force diagram as in Figure 4.

Bending moment:

Show the calculation of the bending moment as follows;

$M_A=0$

$\begin{array}{c}{M}_C-M_A=3600\times 0.2\\ M_C=720+M_A\\ =720+0\\ =720\text{N-m}\end{array}$

$\begin{array}{c}{M}_E-M_C=360\times 0.5\\ M_E=180+M_C\\ =180+720\\ =900\text{N}\end{array}$

$M_B=0$

Show the calculated bending moment values as in Table 2.

Location (x) m	Bending moment (M) N-m
A	0
C	720
E	900
B	0

Plot the bending moment diagram as in Figure 5.

Refer to Figure 5;

The maximum absolute bending moment is $\left|M_{\max }\right|=900\text{N-m}$.

To determine

The maximum normal stress due to bending.

Answer to Problem 62P

The maximum normal stress due to bending is $\underset{\_}{62.5\text{MPa}}$.

Explanation of Solution

Given information:

Determine the section modulus (S) of the rectangular beam section using the equation.

$S=\frac{1}{6}b{h}^2$

Here, the width of the beam is b and the depth of the beam is h.

Substitute 24 mm for b and 60 mm for h.

$\begin{array}{c}S=\frac{1}{6}\times 24\times {60}^2\\ =14400{\text{mm}}^3\end{array}$

The maximum absolute bending moment is $\left|M_{\max }\right|=900\text{N-m}$.

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

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

Substitute $900\text{N-m}$ for M and $14400{\text{mm}}^3$ for S.

$\begin{array}{c}{\sigma }_m=\frac{900\text{N-m}}{14400{\text{mm}}^3\times {\left(\frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)}^3}\\ =62.5\times {10}^6\text{Pa}\times \frac{\text{1}\text{MPa}}{{\text{10}}^6\text{Pa}}\\ =62.5\text{MPa}\end{array}$

Therefore, the maximum normal stress due to bending is $\underset{\_}{62.5\text{MPa}}$.