Section 8.6 Relation Between Loading, Shear and Moment
Suppose that we have a simply supported beam upon which there is an applied load \(w(x)\) which is distributed on the beam by some function of position, \(x\text{,}\) as shown in Figure 8.6.1.
If we select a small section of this beam from \(x\) to \(x+\Delta x\) to look at closely, we have the free-body diagram shown in Figure 8.6.2.
Since \(\Delta x\) is infinitely narrow, we can assume that the distributed load over this small distance is constant and equal to the value at \(x\text{,}\) and call it \(w\text{.}\)
Applying the force equilibrium in the vertical direction gives the following result:
\begin{align*}
\sum F_y \amp = 0\\
V+w (\Delta x)-(V+\Delta V) \amp = 0\\
\frac {\Delta V}{\Delta x} \amp = w
\end{align*}
Taking the limit of both sides as \(\Delta x\) approaches 0, we get this important result
\begin{align*}
\lim_{\Delta x \to\ 0}\left( \frac {\Delta V}{\Delta x} \right) \amp= \lim_{\Delta x \to\ 0}\left( w \right)\\
\frac {dV}{dx}\amp = w
\end{align*}
This equation tells us that, at a given location \(x\text{,}\) the slope of the shear function \(V(x)\) there is the value of the loading directly above, \(w(x)\text{.}\) Furthermore, if we multiply both sides by \(dx\text{,}\) we can integrate to find that
\begin{equation*}
\Delta V=\int w(x) \ dx
\end{equation*}
In words, this equation says that over a given distance, the change in the shear \(V\) between two points is the area under the loading curve between them.
Now looking at the internal bending moments on the FBD in Figure 8.6.2, when we apply moment equilibrium about the centroid of the element, and take the limit similarly,
\begin{align*}
\sum M \amp =0\\
-\frac {\Delta x}{2}V-\frac {\Delta x}{2}(V+\Delta V)-M+(M+\Delta M)\amp=0\\
\frac {\Delta M}{\Delta x}\amp=\frac{1}{2}(2V+\Delta V)\\
\lim_{\Delta x \to\ 0}\left( \frac {\Delta M}{\Delta x}\right)\amp= \lim_{\Delta x \to\ 0}\left(V+\frac{\Delta V}{2}\right)\\
\frac {dM}{dx} \amp =V
\end{align*}
This final equation tells us that, the slope of the moment diagram is the value of the shear. Furthermore, if we multiply both sides by \(dx\text{,}\) we can integrate to find that
\begin{equation*}
\Delta M=\int V\ dx
\end{equation*}
In words, this equation says that over a given segment, the change in the moment value is the area under the shear curve.
Hence, the functional relationships between the internal shear force \(V(x)\text{,}\) internal bending moment \(M(x)\) at a point \(x\text{,}\) and the value of the loading at that point \(w(x)\) are simply the derivatives and integrals that you learned in Calculus I. These relationships are summarized below.
The slope of the shear function at \(x\) is the value of the loading function at the same position. An upward load is considered a positive load.
\begin{equation}
\frac {dV}{dx}=w(x)\tag{8.6.1}
\end{equation}
The change in the shear value between two points is the area under the loading function between those points.
\begin{equation}
\Delta V=\int_a^b w(x)\ dx\tag{8.6.2}
\end{equation}
The slope of the moment function at \(x\) is the value of the shear at the same position.
\begin{equation}
\frac {dM}{dx}=V(x)\tag{8.6.3}
\end{equation}
The change in the moment value between two points is the area under the shear curve between those points.
\begin{equation}
\Delta M=\int_a^b V(x)\ dx\tag{8.6.4}
\end{equation}
Shear and bending moment diagrams show the effect of the load on the internal forces within the beam and are a graphical representation of equations (8.6.1)–(8.6.4). The diagrams are made up of jumps, slopes and areas as a result of the load.
- Jumps are vertical changes in shear and moment diagrams.
- Slopes are gradual changes in shear and moment diagrams. Positive slopes go up and to the right.
- Areas are “areas” under the loading and shear curves, i.e. integration. The area under the loading curve is actually the force, and the area under the shear curve is actually the bending moment.
Diagram |
Jumps |
Slopes |
Areas |
Shear |
Concentrated forces cause the shear diagram to jump by the same amount. Upward loads cause upward jumps.
Concentrated moments on the beam have no effect on the shear diagram.
|
The slope of the shear diagram at a point is equal to the value of the distributed load above that point. A downward distributed load will give the shear diagram a negative slope. |
The change in the shear between two points is equal to the corresponding area under the loading curve. |
Moment |
Concentrated moments cause jumps on the moment curve. Counterclockwise moments cause downward jumps and vice-versa. |
The slope of the moment diagram at a point is equal to the value of shear at that point. A positive shear causes a positive slope on the moment diagram and vice-versa. |
Change in the moment between two points is equal to the corresponding area under a shear curve. |
You can use the interactive below to explore how changes to concentrated load \(P\) and distributed load \(w\) affects the slopes, jumps, and areas of the resulting shear and bending moment diagrams.
Instructions.
Try moving the red dots to change the load and see the effects on the diagrams.
