Chapter 7.3, Problem 7.86P

For the beam and loading shown, (a) write the equations of the shear and bending-moment curves, (b) determine the magnitude and location of the maximum bending moment.

To determine

To write the equations of the shear and bending moment curves.

Answer to Problem 7.86P

The equation for shear curve is $V=w_{0}L\left[\frac{1}{3}-\frac{x}{L}+\frac{1}{2}{\left(\frac{x}{L}\right)}^2\right]$ and the equation for bending moment curve is $M=w_0{L}^2\left[\frac{1}{3}\left(\frac{x}{L}\right)-\frac{1}{2}{\left(\frac{x}{L}\right)}^2+\frac{1}{6}{\left(\frac{x}{L}\right)}^3\right]$.

Explanation of Solution

Refer Fig P7.86 and figure 1.

Write the equation for the distributed load.

$w=w_0\left(1-\frac{x}{L}\right)$ (I)

Here, $w_0$ is the initial load, $w$ is the load, and $L$ is the length of the load.

Write the condition used to find the reaction force B.

$\Sigma M_A=0$ (II)

Here, $\Sigma M_A$ is the sum of the moments acting on A acting counter clockwise.

Write the condition used to find the reaction force A along y direction.

$\Sigma F_y=0$ (III)

Here, $\Sigma F_y$ is the sum of the forces acting along y direction.

Write the equation for change in shear moment with distance.

$\frac{dV}{dx}=-w$ (IV)

Here, $V$ is the shear moment.

Write the equation for change in bending moment with distance.

$\frac{dM}{dx}=V$ (V)

Here, $M$ is the bending moment.

Conclusion:

Use equation (I) to calculate the reaction force at B.

$\begin{array}{c}\Sigma M_A=0\\ \frac{L}{3}\left(\frac{1}{2}{w}_{0}L\right)-LB=0\\ B=\frac{w_{0}L}{6}\text{upwards}\end{array}$

Use equation (II) to calculate the reaction force at B.

$\begin{array}{c}\Sigma F_y=0\\ A_y-\frac{1}{2}{w}_{0}L+\frac{w_{0}L}{6}=0\\ A_y=\frac{w_{0}L}{3}\text{upwards}\end{array}$

Sketch the shear diagram.

Calculate the change in shear moment with distance using equation (IV) and thus find the equation for the shear moment curve.

$\begin{array}{c}\frac{dV}{dx}=-w\\ =V_A-{\displaystyle {\int }_0^x{w}_0\left(1-\frac{x}{L}\right)dx}\\ V=\left(\frac{w_{0}L}{3}\right)-w_{0}x+\frac{1}{2}\frac{w_0}{L}{x}^2\\ =w_{0}L\left[\frac{1}{3}-\frac{x}{L}+\frac{1}{2}{\left(\frac{x}{L}\right)}^2\right]\end{array}$

Sketch the bending moment diagram.

Calculate the change in bending moment with distance using equation (V) and thus find the equation for the bending moment curve.

$\begin{array}{c}\frac{dM}{dx}=V\\ M={\displaystyle {\int }_0^{x}Vdx}\\ =L{\displaystyle {\int }_0^{x/L}V\left(\frac{x}{L}\right)d\left(\frac{x}{L}\right)}\\ =w_0{L}^2{\displaystyle {\int }_0^{x/L}\left[\frac{1}{3}-\frac{x}{L}+\frac{1}{2}{\left(\frac{x}{L}\right)}^2\right]d\left(\frac{x}{L}\right)}\\ =w_0{L}^2\left[\frac{1}{3}\left(\frac{x}{L}\right)-\frac{1}{2}{\left(\frac{x}{L}\right)}^2+\frac{1}{6}{\left(\frac{x}{L}\right)}^3\right]\end{array}$

Therefore, the equation for shear curve is $V=w_{0}L\left[\frac{1}{3}-\frac{x}{L}+\frac{1}{2}{\left(\frac{x}{L}\right)}^2\right]$ and the equation for bending moment curve is $M=w_0{L}^2\left[\frac{1}{3}\left(\frac{x}{L}\right)-\frac{1}{2}{\left(\frac{x}{L}\right)}^2+\frac{1}{6}{\left(\frac{x}{L}\right)}^3\right]$.

To determine

The magnitude and location of the maximum bending moment.

Answer to Problem 7.86P

The magnitude of the maximum bending moment is $0.06415w_0{L}^2$ and the location is at $x=0.423L$.

Explanation of Solution

The maximum absolute value of the bending moment can be found using the equation for the bending moment curve.

The distance of the maximum bending moment is measured from the left hand side.

Conclusion:

At $x=L$,

$V=-\frac{w_{0}L}{6}$

If $V=0$ at ${\left(\frac{x}{L}\right)}^2-2\left(\frac{x}{L}\right)+\frac{2}{3}$, then,

$\frac{x}{L}=1-\sqrt{\frac{1}{3}}$

The magnitude of the maximum bending moment.

${M_{\max }|}_{\frac{x}{L}=1-\sqrt{\frac{1}{3}}}=0.06415w_0{L}^2$

According to figure 2, find the location of the maximum bending moment.

$x=0.423L$

Therefore, the magnitude of the maximum bending moment is $0.06415w_0{L}^2$ and the location is at $x=0.423L$.