Chapter 5.1, Problem 6P

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. P5.6

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 D by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ D_y\left(L\right)-wa\left(L-\frac{a}{2}\right)-wa\times \frac{a}{2}=0\\ D_y\left(L\right)-waL+\frac{w{a}^2}{2}-\frac{w{a}^2}{2}=0\\ D_y=wa\end{array}$

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

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

Substitute wa for $D_y$.

$\begin{array}{l}{A}_y-wa-wa+wa=0\\ A_y=wa\end{array}$

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 as in Figure 2.

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

$\begin{array}{c}{\displaystyle \sum F_y=0}\\ wa-wx-V=0\\ V=wa-wx\\ =w\left(a-x\right)\end{array}$

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

$\begin{array}{c}{\displaystyle \sum M_x=0}\\ -wax+wx\times \frac{x}{2}+M=0\\ M=wax-\frac{w{x}^2}{2}\\ =w\left(ax-\frac{x^2}{2}\right)\end{array}$

Show the free-body diagram of the section 2-2 as in Figure 3.

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ wa-wa-V=0\\ V=0\end{array}$

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

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ -wax+wa\left(x-\frac{a}{2}\right)+M=0\\ -wax+wax-\frac{w{a}^2}{2}+M=0\\ M=\frac{w{a}^2}{2}\end{array}$

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

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ wa-w\left(L-x\right)+V=0\\ wa-wL+wx+V=0\\ V=w\left(L-x-a)\right)\end{array}$

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

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ wa\left(L-x\right)-w\left(L-x\right)\times \frac{\left(L-x\right)}{2}-M=0\\ M=wa\left(L-x\right)-\frac{w{\left(L-x\right)}^2}{2}\\ =w\left[a\left(L-x\right)-\frac{{\left(L-x\right)}^2}{2}\right]\end{array}$

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)
A	wa	0
B (1-1)	0	$\frac{w{a}^2}{2}$
B (2-2)	0	$\frac{w{a}^2}{2}$
C (2-2)	0	$\frac{w{a}^2}{2}$
C (3-3)	0	$\frac{w{a}^2}{2}$
D	–wa	0

Plot the shear force and bending moment diagrams as in Figure 5.

To determine

The equations of the shear and bending-moment curves.

Answer to Problem 6P

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

For section AB;

$\underset{\_}{V=w\left(a-x\right);M=w\left(ax-\frac{x^2}{2}\right)}$.

For section BC;

$\underset{\_}{V=0;M=\frac{w{a}^2}{2}}$.

For section CD:

$\underset{\_}{V=w\left(L-x-a\right);M=w\left[a\left(L-x\right)-\frac{{\left(L-x\right)}^2}{2}\right]}$

Explanation of Solution

Determine the reactions of the beam.

Show the free-body diagram of the entire beam as in Figure 6.

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

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ D_y\left(L\right)-wa\left(L-\frac{a}{2}\right)-wa\times \frac{a}{2}=0\\ D_y\left(L\right)-waL+\frac{w{a}^2}{2}-\frac{w{a}^2}{2}=0\\ D_y=wa\end{array}$

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

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

Substitute wa for $D_y$.

$\begin{array}{l}{A}_y-wa-wa+wa=0\\ A_y=wa\end{array}$

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 as in Figure 7.

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

$\begin{array}{c}{\displaystyle \sum F_y=0}\\ wa-wx-V=0\\ V=wa-wx\\ =w\left(a-x\right)\end{array}$

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

$\begin{array}{c}{\displaystyle \sum M_x=0}\\ -wax+wx\times \frac{x}{2}+M=0\\ M=wax-\frac{w{x}^2}{2}\\ =w\left(ax-\frac{x^2}{2}\right)\end{array}$

Show the free-body diagram of the section 2-2 as in Figure 8.

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ wa-wa-V=0\\ V=0\end{array}$

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

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ -wax+wa\left(x-\frac{a}{2}\right)+M=0\\ -wax+wax-\frac{w{a}^2}{2}+M=0\\ M=\frac{w{a}^2}{2}\end{array}$

Show the free-body diagram of the section 3-3 as in Figure 9.

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

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ wa-w\left(L-x\right)+V=0\\ wa-wL+wx+V=0\\ V=w\left(L-x-a)\right)\end{array}$

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

$\begin{array}{l}{\displaystyle \sum M_x=0}\\ wa\left(L-x\right)-w\left(L-x\right)\times \frac{\left(L-x\right)}{2}-M=0\\ M=wa\left(L-x\right)-\frac{w{\left(L-x\right)}^2}{2}\\ =w\left[a\left(L-x\right)-\frac{{\left(L-x\right)}^2}{2}\right]\end{array}$

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

For section AB;

$\underset{\_}{V=w\left(a-x\right);M=w\left(ax-\frac{x^2}{2}\right)}$.

For section BC;

$\underset{\_}{V=0;M=\frac{w{a}^2}{2}}$.

For section CD:

$\underset{\_}{V=w\left(L-x-a\right);M=w\left[a\left(L-x\right)-\frac{{\left(L-x\right)}^2}{2}\right]}$