Chapter 5.1, Problem 2P

5.1 through 5.6 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves.

Fig. P5.2

To determine

To draw: The shear and bending-moment diagrams.

Explanation of Solution

Determine the reactions of the beam.

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

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

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ C_y\left(L\right)-P\left(a\right)=0\\ C_y=\frac{Pa}{L}\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-P+C_y=0\end{array}$

Substitute $\frac{wL}{2}$ for $B_y$.

$\begin{array}{c}{A}_y-P+\frac{Pa}{L}=0\\ A_y=P-\frac{Pa}{L}\\ =P\left(\frac{L-a}{L}\right)\\ =\frac{Pb}{L}\end{array}$

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

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

Show the free-body diagram of the section 1-1 and 2-2 as in Figure 2.

Section 1-1:

Determine the shear force at the section by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \frac{Pb}{L}-V=0\\ V=\frac{Pb}{L}\end{array}$

Determine the moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ -\frac{Pb}{L}\left(x\right)+M=0\\ M=\frac{Pbx}{L}\end{array}$

Section 2-2:

Determine the shear force at the section by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \frac{Pb}{L}-V=0\\ V=\frac{Pb}{L}\end{array}$

Determine the moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ -\frac{Pb}{L}\left(x\right)+M=0\\ M=\frac{Pbx}{L}\end{array}$

Show the free-body diagram of the section 3-3 and 4-4 as in Figure 3.

Section 3-3:

Determine the shear force at the section by resolving the vertical component of forces.

$\begin{array}{c}{\displaystyle \sum F_y=0}\\ \frac{Pb}{L}-V-P=0\\ V=\frac{Pb}{L}-P\\ =P\left(\frac{b}{L}-1\right)\end{array}$

$\begin{array}{c}V=P\left(\frac{b-L}{L}\right)\\ =-P\left(\frac{L-b}{L}\right)\\ =-\frac{Pa}{L}\end{array}$

Determine the moment at the section by taking moment about the section.

$\begin{array}{c}{\displaystyle \sum M_x=0}\\ -\frac{Pb}{L}\left(x\right)+M+P\left(x-a\right)=0\\ M=\frac{Pbx}{L}-P\left(x-a\right)\\ =\frac{Pbx}{L}-Px+Pa\end{array}$

When the section 3-3 is at point B, $x=a$.

$\begin{array}{c}{M}_B=\frac{Pb\left(a\right)}{L}-P\left(a\right)+Pa\\ =\frac{Pab}{L}\end{array}$

Section 4-4:

Determine the shear force at the section by resolving the vertical component of forces.

$\begin{array}{c}{\displaystyle \sum F_y=0}\\ \frac{Pb}{L}-V-P=0\\ V=\frac{Pb}{L}-P\\ =P\left(\frac{b}{L}-1\right)\end{array}$

$\begin{array}{c}V=P\left(\frac{b-L}{L}\right)\\ =-P\left(\frac{L-b}{L}\right)\\ =-\frac{Pa}{L}\end{array}$

Determine the moment at the section by taking moment about the section.

$\begin{array}{c}{\displaystyle \sum M_x=0}\\ -\frac{Pb}{L}\left(x\right)+M+P\left(x-a\right)=0\\ M=\frac{Pbx}{L}-P\left(x-a\right)\\ =\frac{Pbx}{L}-Px+Pa\\ =\frac{Pbx-PLx+PLa}{L}\end{array}$

$\begin{array}{c}M=\frac{Px\left(b-L\right)+PLa}{L}\\ =\frac{-Pxa+PLa}{L}\\ =\frac{Pa\left(L-x\right)}{L}\end{array}$

Shear force and bending moment values:

Show the calculated shear force and bending moment values as in Table 1.

Location (x)	Shear force (V)	Bending Moment (M)
1-1	$\frac{Pb}{L}$	0
2-2	$\frac{Pb}{L}$	$\frac{Pab}{L}$
3-3	$-\frac{Pa}{L}$	$\frac{Pab}{L}$
4-4	$-\frac{Pa}{L}$	0

Plot the shear force and bending moment diagrams as in Figure 4.

To determine

The equations of the shear and bending-moment curves.

Answer to Problem 2P

The equation of shear force and bending-moment curves is:

For section AB;

$\underset{\_}{V=\frac{Pb}{L};M=\frac{Pbx}{L}}$.

For section BC;

$\underset{\_}{V=-\frac{Pa}{L};M=\frac{Pa\left(L-x\right)}{L}}$.

Explanation of Solution

Determine the reactions of the beam.

Show the free-body diagram of the entire beam as in Figure 5.

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

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ C_y\left(L\right)-P\left(a\right)=0\\ C_y=\frac{Pa}{L}\end{array}$

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ A_y-P+C_y=0\end{array}$

Substitute $\frac{wL}{2}$ for $B_y$.

$\begin{array}{c}{A}_y-P+\frac{Pa}{L}=0\\ A_y=P-\frac{Pa}{L}\\ =P\left(\frac{L-a}{L}\right)\\ =\frac{Pb}{L}\end{array}$

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

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

Show the free-body diagram of the section 1-1 and 2-2 as in Figure 6.

Section 1-1:

Determine the shear force at the section by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \frac{Pb}{L}-V=0\\ V=\frac{Pb}{L}\end{array}$

Determine the moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ -\frac{Pb}{L}\left(x\right)+M=0\\ M=\frac{Pbx}{L}\end{array}$

Section 2-2:

Determine the shear force at the section by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ \frac{Pb}{L}-V=0\\ V=\frac{Pb}{L}\end{array}$

Determine the moment at the section by taking moment about the section.

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ -\frac{Pb}{L}\left(x\right)+M=0\\ M=\frac{Pbx}{L}\end{array}$

Show the free-body diagram of the section 3-3 and 4-4 as in Figure 7.

Section 3-3:

Determine the shear force at the section by resolving the vertical component of forces.

$\begin{array}{c}{\displaystyle \sum F_y=0}\\ \frac{Pb}{L}-V-P=0\\ V=\frac{Pb}{L}-P\\ =P\left(\frac{b}{L}-1\right)\end{array}$

$\begin{array}{c}V=P\left(\frac{b-L}{L}\right)\\ =-P\left(\frac{L-b}{L}\right)\\ =-\frac{Pa}{L}\end{array}$

Determine the moment at the section by taking moment about the section.

$\begin{array}{c}{\displaystyle \sum M_x=0}\\ -\frac{Pb}{L}\left(x\right)+M+P\left(x-a\right)=0\\ M=\frac{Pbx}{L}-P\left(x-a\right)\\ =\frac{Pbx}{L}-Px+Pa\end{array}$

When the section 3-3 is at point B, $x=a$.

$\begin{array}{c}{M}_B=\frac{Pb\left(a\right)}{L}-P\left(a\right)+Pa\\ =\frac{Pab}{L}\end{array}$

Section 4-4:

Determine the shear force at the section by resolving the vertical component of forces.

$\begin{array}{c}{\displaystyle \sum F_y=0}\\ \frac{Pb}{L}-V-P=0\\ V=\frac{Pb}{L}-P\\ =P\left(\frac{b}{L}-1\right)\end{array}$

$\begin{array}{c}V=P\left(\frac{b-L}{L}\right)\\ =-P\left(\frac{L-b}{L}\right)\\ =-\frac{Pa}{L}\end{array}$

Determine the moment at the section by taking moment about the section.

$\begin{array}{c}{\displaystyle \sum M_x=0}\\ -\frac{Pb}{L}\left(x\right)+M+P\left(x-a\right)=0\\ M=\frac{Pbx}{L}-P\left(x-a\right)\\ =\frac{Pbx}{L}-Px+Pa\\ =\frac{Pbx-PLx+PLa}{L}\end{array}$

$\begin{array}{c}M=\frac{Px\left(b-L\right)+PLa}{L}\\ =\frac{-Pxa+PLa}{L}\\ =\frac{Pa\left(L-x\right)}{L}\end{array}$

Therefore, the equation of shear force and bending-moment curves is:

For section AB;

$\underset{\_}{V=\frac{Pb}{L};M=\frac{Pbx}{L}}$.

For section BC;

$\underset{\_}{V=-\frac{Pa}{L};M=\frac{Pa\left(L-x\right)}{L}}$.