Statics: Relations between Loading, Shear and Moment

Section 8.5 Relations between Loading, Shear and Moment

Shear and bending moment diagrams are used to show how load on a beam affects the shear and bending moment within.

In this section, we will look at a completely general beam and loading, and use a free-body diagram and the principles of equlibrium to derive the fundamental mathematical relationships that exist between load, shear, and bending moment.

Then in the sections to follow, we will learn three methods to produce shear and bending moment diagrams: an equilibrium based approach, a graphical integration method, and a calculus-based analytical approach. All three methods are consistent with the equations derived in this section and understanding them all can be insightful.

Suppose that we have a simply supported beam upon which there is a load \(w(x)\) which is acts on the beam by some arbitrary function of \(x\text{,}\) as shown in Figure 8.5.1.

A simply supported beam with a distributed load that is a function of beam position. A segment is indicated with imaginary cuts at x and at x+delta x. — Figure 8.5.1. A simply supported beam with a distributed load that is a function of beam position \(w(x)\text{.}\)

For a thin section of this beam between \(x\) and \(x+\Delta x\text{,}\) we can assume that the distributed load is approximately constant and equal to \(w = w(x)\text{.}\) With this assumption, the distributed load is rectangular, and equivalent to a force \(W = w \Delta x\) acting at the centroid of the rectangle. The free-body diagram for this thin section is shown in Figure 8.5.2.

FBD of the small segment delta x, with shear and bending moments indicated on the "cut" surfaces. — Figure 8.5.2. A free-body diagram of a small section of the beam with a width of \(\Delta x\text{.}\)

Applying the force equilibrium in the vertical direction gives the following result:

\begin{align*} \sum F_y \amp = 0\\ V+w Delta x \amp = (V+\Delta V)\\ \Delta V \amp = w {\Delta x}\\ \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)\notag\\ \frac {dV}{dx}\amp = w(x)\tag{8.5.1} \end{align}

This differential equation tells us that at a given location \(x\text{,}\) the slope of the shear function at that point is the value of the loading directly above, \(w(x)\text{.}\)

Furthermore, if we separate the variables by multiplying both sides by \(dx\text{,}\) and then integrate between any two points along the beam we find that

\begin{equation} \Delta V=\int_a^b w(x) \ dx\tag{8.5.2} \end{equation}

In words, this equation says that the change in the shear \(\Delta V\) between any two points \(a\) and \(b\text{,}\) is the area under the loading curve between them.

Next, considering moment equilibrium of the FBD in Figure 8.5.2, and following the same procedure, we get:

\begin{align} \sum M \amp =0\notag\\ (M+\Delta M)\amp=\frac {\Delta x}{2}V+\frac {\Delta x}{2}(V+\Delta V)+M\notag\\ \Delta M\amp=(V+\frac{\Delta V}{2}) \Delta x\notag\\ \end{align}

Separating variables and taking limits as before, gives:

\begin{align} \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)\notag\\ \frac {dM}{dx} \amp =V(x)\tag{8.5.3} \end{align}

This equation tells us that at any point on the beam the slope of the moment function is equal to the value of the shear at that point.

Furthermore, if we multiply both sides by \(dx\text{,}\) we can integrate to find that

\begin{equation} \Delta M=\int_a^b V(x)\,dx\tag{8.5.4} \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.

Summary.

The internal shear force \(V(x)\text{,}\) bending moment \(M(x)\text{,}\) and loading \(w(x)\) at every point \(x\)are related to each other by the integral and differetial equations below.

\begin{equation*} \frac {dV}{dx}=w(x) \end{equation*}

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*} \Delta V=\int_a^b w(x)\ dx \end{equation*}

The change in the shear value between two points is the area under the loading function between those points.

\begin{equation*} \frac {dM}{dx}=V(x) \end{equation*}

The slope of the moment function at \(x\) is the value of the shear at the same position.

\begin{equation*} \Delta M=\int_a^b V(x)\ dx \end{equation*}

The change in the moment value between two points is the area under the shear curve between those points.

Remark 8.5.3. Sign Conventions.

It is sometimes convenient to assume that positive loads act down, and not up as we have assumed here. In that case simply subtitute \(-w(x)\) for \(w(x)\) into and equations (8.5.1) and (8.5.2) and use these variations instead:

\begin{align} \frac {dV}{dx}\amp = -w(x) \amp \Delta V\amp=-\int_a^b w(x) \ dx\tag{8.5.5} \end{align}

Thinking Deeper 8.5.4. What happens if we keep integrating?

As we have seen, integrating the load function once gives the shear force, and integrating the shear force gives the bending moment.

You may be wondering what happens if we keep integrating? Does this produce anything interesting?

In fact it does, as you will see in Mechanics of Materials. In the Statics world, all beams are assumed to be rigid, but in the real world they deform under load. The shape the beam takes on is described by a deflection curve which gives the actual displacement of the beam at any given point due to the applied loads

Integrating the bending moment \(M(x)\) (and dividing by the flexural rigidity, \(EI\)) gives the slope of the deflection curve, which represents the angle of the beam’s deformation at any given point.

Integrating that function once more (and dividing by \(EI\) again) gives the actual deflection function.

So, if you keep integrating the load function, you’re essentially moving through these stages of beam analysis, progressing from the applied forces to the resulting deformation: Load Shear Moment Slope Deflection. Each step provides valuable information about the beam’s behavior under the given loading conditions.

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