Skip to main content
Logo image

Engineering Statics Open and Interactive

Daniel W. Baker, William Haynes

Section 8.8 Integration Method

In Section 8.5 we saw that loading, shear and bending moments are related by integral and differential equations, and in Subsection 8.7.2 used this knowledge to draw shear and bending moment diagram using a graphical approach. This method is straightforward and fast for concentrated forces and uniformly distributed loads, because they produce triangular and rectangular areas that are easy to calculate but it falls short when the areas are more complicated.
🔗
For example, a uniformly varying load is a first degree linear function of \(x\text{,}\)
\begin{equation*} w(x) = w_0 x \end{equation*}
It integrates to a second degree parabolic shear function, and a third degree cubic moment function. Finding the zero crossings and areas under these curves with geometery is troublesome. More complex loading functions are even more difficult.
🔗
When the loading becomes complex it is better to use the fundamental relations (8.5.2) and (8.5.4) to find the shear and bending moment functions directly
\begin{align*} \Delta V\amp =\int_a^b w(x)\ dx \amp \Delta M\amp=\int_a^b V(x)\ dx \end{align*}
Rather than finding areas and slopes using geometry, we will directly integrate the load function \(w(x)\) once to find the shear function \(v(x)\text{,}\) then integrate that result to find the moment function \(m(x)\text{.}\)
🔗
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 applying 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.
🔗

Subsection 8.8.1 Determining Loading Functions

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.
🔗
When determining equations for the loading segments from the load diagram, consider the following.
  • No load.
    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.
    🔗
    🔗
  • Point Load.
    A point load is a concentrated force acting at a single point which causes a jump in the shear diagram.
    🔗
    🔗
  • Uniformly Distributed Load.
    A uniformly distributed load is constant over the segment and results in a linear slope, either a triangle or a trapezoid, on the shear diagram. The loading function on such sections is
    \begin{equation*} w(x) = C \qquad V(x) = Cx + b\text{.} \end{equation*}
    The constant value is negative if the load points down, and positive if it points upward.
    🔗
    🔗
  • Uniformly Varying Load.
    In this case the loading function is a straight, sloping lie forming a triangle or trapezoidal shape. The resulting shear function is parabolic. The general form of these functions are
    \begin{equation*} w(x) = m x + b \qquad V(x) = \frac{mx^2}{2} + bx + c\text{.} \end{equation*}
    The slope \(m\text{,}\) intercept \(b\text{,}\) and constant \(c\) must be determined from the situation, and will depend on whether you are writing a global or local equation.
    🔗
    🔗
  • Arbitrary Load.
    The loading function will be a given function of \(x\text{.}\)
    \begin{equation*} w(x) = f(x)\text{,} \end{equation*}
    and the shear and moment functions are found by integration.
    \begin{equation*} V(x) = \int f(x) dx \qquad M(x) = \int V(x) dx \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. In other words, for downward pointing loads, the scalar values are negative.
    🔗
    🔗
🔗
This interactive compares the local and global equations for a beam segment with a uniformly varying load.
🔗

Instructions.

Notice as you switch between global and local coordinate systems the \(y\) intercept (\(b\) vs. \(b'\)) and the \(x\) coordinate of point \(P\) change, but the formula of the line gives the same result either way.
🔗
🔗
Figure 8.8.1. Global vs. Local coordinate systems.
🔗
🔗

Subsection 8.8.2 Application of the Integration Method

You can either use 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 \(V(x)\) 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 shear and bending moments, recall from calculus that the local maximum/minimum points of a function occur at the endpoints and where the function’s first derivative is equal to zero.
    1. For shear, evaluate the shear function \(V(x)\) at the ends and where ever the load function crosses the \(x\) axis.
      🔗
      🔗
    2. For bending moments, find the roots of the shear function by solving \(V(x)=0\text{,}\) then evaluate the moment function \(M(x)\) at these points, and also at the endpoints.
      🔗
      🔗
    The critical values we are looking for are the points where the magnitudes of the shear and bending moment are maximum. The direction of the internal forces is not usually significant.
    🔗
    🔗
🔗

Example 8.8.2. Uniformly Distributed Load.

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 \(L\text{.}\)
🔗
Simply supported beam carrying a uniformly distributed load w over its entire length L.
Answer.
\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 and points down, 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 (8.5.2) and (8.5.4) we have.
🔗
\begin{align*} \Delta V \amp = - \int_0^x w(x) \ dx\\ V(x) - \cancelto{wL/2}{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\\ M(x)- \cancelto{0}{M(0)} \amp = \int_0^x w\left(\frac{L}{2} - x \right) \ dx\\ M(x) \amp = \frac{w}{2} (Lx - x^2) \end{align*}
🔗
🔗
🔗

Example 8.8.3. Piecewise Function.

Consider a \(\m{20}\) beam loaded as as shown. Use the integration method to derive the shear and bending moment as functions of \(x\) and plot them.
🔗
Answer.
On the range \(0 < x <= 10\)
\begin{align*} V(x) \amp= -12 x + 75\\ M(x) \amp= - 6 x^2 + 75 x \end{align*}
🔗
On the range \(10 <x <= 20\)
\begin{align*} V(x) \amp = - \frac{9}{20} (x-10) ^2 + 9 (x-10) - 45\\ M(x)\amp= - \frac{3}{20} (x-10) ^3 + \dfrac{9}{2} (x-10) ^2-45 (10-x) + 150 \end{align*}
🔗
🔗
Solution.
  1. Reactions.
    Find the reactions using the methods of Section 7.8.
    🔗
    \begin{align*} \Sigma M_A \amp= 0\\ - W_1 x_1 + W_2 x_2 + R_B (\m{20}) \amp =0\\ R_B (\m{20}) \amp = \kN{120} \cdot \m{5} - \kN{45} \cdot \m{13.33}\\ R_B \amp= 0\\ \\ \Sigma F_y \amp=0\\ R_A - W_1 + W_2 + R_B \amp =0\\ R_A \amp= \kN{120} - \kN{45}\\ R_A \amp= \kN{75} \end{align*}
    🔗
    🔗
  2. Segment 1.
    On the first half of the beam, for the range \(x = 0\) up to and including \(x = \m{10}\text{,}\) the loading is constant and pointing down, so the loading function on this segment is
    \begin{equation*} w_1(x) = - \kNperm{12}\text{.} \end{equation*}
    🔗
    Integrating twice gives the shear and bending moment functions on this segment
    \begin{align*} V_1(x) \amp= -12 x + C_0\\ M_1(x) \amp= - 6 x^2 + C_0 x + C_1 \end{align*}
    🔗
    Evaluating the boundary conditions, \(V(0) = R_A\) and \(M(0) = 0\) we learn that
    \begin{gather*} C_0 = \kN{75}.\\ C_1 = 0 \end{gather*}
    🔗
    🔗
  3. Segment 2.
    To simplify the analysis of the second half of the beams we will use local coordinates by defining
    \begin{equation*} x' = (x-10)\text{.} \end{equation*}
    In local coordinates, the function is valid over the range \(0 < x' <= 10\text{.}\)
    🔗
    The load over this segment starts at \(\kNperm{9}\) but decreases to zero after \(\m{10}\text{,}\) so the slope is
    \begin{equation*} m = \dfrac{\kNperm{-9}}{\m{10}} = -\kNperm{\dfrac{9}{10}}^2\text{.} \end{equation*}
    In terms of \(x'\) the load the loading function is
    \begin{equation*} w_2(x') = 9 - \frac{9}{10} x'\text{.} \end{equation*}
    Integrating twice gives the shear and moment functions.
    \begin{align*} V_2(x') \amp = 9 x' - \frac{9}{20} x'^2 + C_2\\ M_2(x')\amp=\dfrac{9}{2} x'^2 - \frac{3}{20} x'^3 + C_2 x' + C_3 \end{align*}
    🔗
    The constants are evaluated by matching the functions at the boundary \(x=\m{10}\text{,}\) where
    \begin{gather*} V_1(10) = -\kN{45}\\ M_1(10) = \kNm{150} \end{gather*}
    Evaluating \(V_2(x')\) and \(M_2(x')\) at \(x' = 0\) we find that
    \begin{align*} C_2 \amp= -\kN{45}\\ C_3 \amp= \kNm{150} \end{align*}
    🔗
    🔗
  4. Results.
    Combining results we get the final version of shear and bending moment functions in terms of \(x\text{.}\)
    🔗
    On the range \(0 < x <= 10\)
    \begin{align*} V(x) \amp= -12 x + 75\\ M(x) \amp= - 6 x^2 + 75 x \end{align*}
    🔗
    On the range \(10 <x <= 20\)
    \begin{align*} V(x) \amp = - \frac{9}{20} (x-10) ^2 + 9 (x-10) - 45\\ M(x)\amp= - \frac{3}{20} (x-10) ^3 + \dfrac{9}{2} (x-10) ^2-45 (10-x) + 150 \end{align*}
    🔗
    🔗
🔗
A plot of the two functions is shown below. Notice how the two halves of each function joins continuously at \(x =\m{10}\text{.}\)
🔗
🔗
🔗
🔗
🔗
PrevTopNext
Feedback