## Section8.8Integration Method

In Section 8.6 we learned that loading, shear and bending moments are related by integral and differential equations, and used this knowledge to draw shear and bending moment diagram suing a graphical approach. This method is easy and fast in cases when you can easily calculate the areas under the loading and shear curves without integration. Beams consisting of point and uniformly distributed loads only do not require the use of the calculus method.

However, there are times that the graphical technique falls short when the areas are more complicated than rectangles or triangles. For example, a uniformly varying load, which is a firs degree linear function of $$x\text{,}$$ integrates to a second degree parabolic shear function, and a third degree cubic moment function. To use the graphical method you would need to find the area under the parabolic shear curve to compute the cubic moment. When the loading becomes more complex it is better to use perform the integration directly.

We will use fundamental equations (8.6.2) and (8.6.4) to find the shear and bending moment functions

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

but instead of finding areas and slopes using geometry, we will integrate the load function $$w(x)$$ to find the $$\Delta V\text{,}$$ then integrate that result to find the $$\Delta M\text{.}$$

These results are the change in shear and moment over a segment; to find the actual shear and moment functions $$V(x)$$ and $$M(x)$$ for the entire beam we will need to find initial values for each segment. This is equivalent to using boundary conditions to find the constant of integration when solving a differential equation. The initial values come from either the final value of the previous segment or from point loads or point moments. Because of the requirement for these segment starting values, no segment can be computed in isolation from the other segments. Physically this means that the shear and moment along a beam are not just due to the loading in one segment, but are related to the loading on the rest of the beam as well.

Before you can find shear and bending moment functions with integration you must know the equation for the load on each segment of the beam. These equations may be given in the problem statement if you're lucky, or you may have to determine them from a loading diagram.

When determining equations for loading segments, you may choose either global equations, where all segments use the same origin, usually at the left end of the beam, or local equations, where each segment uses its own origin, usually at the left end of the segment. Often local equations are easier because you can simply use the variable $$x$$ in your equations as opposed to “$$x$$ + constant”, and you do not have to project the $$y$$-intercept values back to an axis system which is not adjacent to the segment. See interactive Figure 8.8.1 to explore the difference between local and global equations.

Whenever there is no load at all on a segment there will be no change in the shear on the segment. On such sections the loading function is

\begin{equation*} w(x) = 0\text{.} \end{equation*}

Note that this can only occur when the weight of the beam itself is neglected.

A uniformly distributed load is constant over the segment and results in a rectangular shear diagram. The loading function on such sections is

\begin{equation*} w(x) = C \text{.} \end{equation*}

The constant value is negative if the load points down, and positive if it points upward.

In this case the loading function varies linearly, and will produce a triangular or trapezoidal shear function. The loading function is a straight line.

\begin{equation*} w(x) = m x + b\text{.} \end{equation*}

The slope $$m$$ and intercept $$b$$ must be determined from the statement, and will depend on whether you are writing a global or local equation.

The loading function will be a given function of $$x\text{.}$$

\begin{equation*} w(x) = f(x)\text{.} \end{equation*}

Most gravitational distributed loads are drawn with the arrows pointing down and resting on the beam. If you slide these along their line of action so that their tails are on the beam, the tips define the loading equation.

This interactive compares the local and global equations for a beam segment with a uniformly varying load.

### Subsection8.8.2Application of the Calculus Method

You can either us this method from the start or use the graphical method until you need areas of shapes more complicated than rectangles and triangles.

1. You will need to have solved the loading segment to the left of your desired segment.

2. Write an equation for the loading $$w$$ in the segment using either local or global coordinates.

3. Integrate the loading equation $$w(x)$$ to find the change in the shear $$\Delta V$$ and include the shear value at the beginning of your loading segment including the influence of any point loads at that location, which is equivalent to the integration constant.

4. Integrate the shear equation $$w(x)$$ to find the change in the bending moment $$\Delta M$$ and include the moment value at the beginning of your loading segment including the influence of any point couple-moments at that location, equivalent to the integration constant.

5. To find maximum values

1. For shear, use $$V(x)$$ equation.

2. For moment, use $$M(x)\text{.}$$ If the value of the maximum moment is in the middle of a loading segment (often where the shear value is zero), set the $$V(x)=0\text{,}$$ solve for $$x$$ and then put that distance value into the moment $$M(x)$$ equation.

###### Example8.8.2.Example.

Use the integration method to find the equations for shear and moment as a function of $$x\text{,}$$ for a simply supported beam carrying a uniformly distributed load $$w$$ over its entire length $$L\text{.}$$ \begin{align*} V(x) \amp = w\left( \frac{L}{2} -x\right ) \amp M(x) \amp = \frac{w}{2} (Lx - x^2) \end{align*}
Solution.

This beam has only one load section, and on that section the load is constant so,

\begin{equation*} w(x)= w\text{.} \end{equation*}

There is a pinned connection at $$x=0$$ which provides a vertical force and no concentrated moment, so the initial conditions there are $$V(0) = wL/2\text{,}$$ and $$M(0) = 0\text{.}$$

Integrating equations (8.6.2) and (8.6.4) we have.

\begin{align*} \Delta V \amp = - \int_0^x w(x) \ dx\\ V(x) - V_0 \amp = -w x \\ V(x)\amp = \frac{wL}{2} - wx\\ \amp = w\left(\frac{L}{2} - x\right) \end{align*} \begin{align*} \Delta M \amp = \int_0^x V(x) \ dx\\ V(x)- \cancelto{0}{M_0} \amp = \int_0^x w\left(\frac{L}{2} - x \right) \ dx \\ V(x) \amp = \frac{w}{2} (Lx - x^2) \end{align*} 