Skip to main content
Logo image

Engineering Statics Open and Interactive

Daniel W. Baker, William Haynes

Section 8.7 Graphical Method

Once you have a firm grasp on the relations between load, shear and bending moments, you can quickly draw shear and bending moment diagrams using a graphical approach. This technique is really a graphical integration process; rather than using calculus, the integrals are found using simple geometry.

Subsection 8.7.1 Relationship Review

Shear and bending moment diagrams are governed by equations (8.5.1) through (8.5.4) and must be consistent with them.
\begin{align*} \dfrac{dV}{dx} \amp = w(x) \amp \dfrac{dM}{dx} \amp = V(x)\\ \Delta V \amp = \int_a^b w(x)\,dx \amp \Delta M \amp= \int_a^b V(x)\,dx \end{align*}
From top to bottom, the shear function is determined from the area under the load, and the moment function is determined from the area under the shear function. At the same time, from bottom to top, differentiating the moment function gives the shear function, and differentiating the shear function gives the load.
Shear and bending moment are piecewise functions made up of a series of jumps, slopes and areas. Jumps are vertical changes in shear and moment diagrams caused by concentrated forces and moments. Slopes are gradual changes in the functions. Areas are the ‘areas’ under the loading and shear curves, i.e. the integral. The area under the loading curve is actually a force, and the area under the shear curve is actually a bending moment.
The relationship between load, shear, and bending moment are summarized below.
Shear Diagram
Jumps Concentrated forces cause the shear curve to jump by the amount of the force.
Upward loads cause upward jumps.
Slopes The slope of the shear diagram at a point is equal to the value of the distributed
load above that point.
Downward distributed loads create negative slopes on the shear diagram.
Areas The area under the loading curve between two points is equal to
the corresponding change in value of the shear.
Moment Diagram
Jumps Concentrated moments cause the moment curve to jump by the amount of the
moment.
Counterclockwise moments cause downward jumps.
Slopes The slope of the moment diagram at a point is equal to the value of the
shear at that point.
Positive shears cause positive slopes on the moment diagram.
Areas The area under the shear curve between two points is equal to
the corresponding change in the value of the moment.
You can use the interactive below to explore how changes to concentrated load \(P\) and distributed load \(w\) affect 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.
Figure 8.7.1. Building Blocks for Shear and Moment Diagrams

Subsection 8.7.2 Graphical Method Procedure

You can draw shear and bending moments efficiently and accurately using this procedure
  1. First, determine the reaction forces and moments by drawing a free-body diagram of the entire beam and applying the equilibrium equations. Double check that your reactions are correct.
  2. Establish the shear graph with a horizontal axis below the beam and a vertical axis to represent shear. Positive shears will be plotted above the \(x\) axis and negative below.
  3. Make vertical lines at all the “interesting points”, i.e. points where concentrated forces or moments act on the beam and at the beginning and end of any distributed loads. This divides the beam into segments between vertical lines.
  4. Draw the shear diagram by starting with a dot at \(x = 0\text{,}\) \(V = 0\) then proceeding from left to right until you reach the end of the beam. Choose and label a scale which keeps the diagram a reasonable size.
    1. Whenever you encounter a concentrated force, jump up or down by that value
    2. Whenever you encounter a concentrated moment, do not jump.
    3. Whenever you encounter a distributed load, move up or down by the “area” under the loading curve over the length of the segment, according to equation (8.5.2). The “area” is actually a force.
    4. The slope of the curve at each point \(x\) is given by (8.5.1). Distributed loads cause the shear diagram to have a slope equal to value of the distributed load at that point. For unloaded segments of the beam, the slope is zero, i.e. the shear curve is horizontal. For segments with uniformly distributed load, the slope is constant. Downward loads cause downward slopes.
    5. The shear diagram should start and end at \(V = 0\text{.}\) If it doesn’t, recheck your work.
  5. Add another interesting point wherever the shear diagram crosses the \(x\)-axis, and determine the \(x\) position of the zero crossing.
  6. After you have completed the shear diagram, calculate the area under the shear curve for each segment. Areas above the axis are positive, areas below the axis are negative. The areas represent moments and the sum of the areas plus the values of any concentrated moments should add to zero. If they don’t, then recheck your work.
  7. Establish the moment graph beneath the shear diagram with a horizontal axis below the shear diagram and a vertical axis to represent moment. Positive moments will be plotted above the \(x\) axis and negative below.
  8. Draw and label dots on the moment diagram by starting with a dot at \(x = 0, M = 0\) then proceed from left to right placing dots until you reach the end of the beam. As you move over each segment move up or down from the current value by the “area” under the shear curve for that segment and place a dot on the graph. In this step, you are applying (8.5.4).
    1. Positive areas cause the moment to increase, negative areas cause it to decrease.
    2. If you encounter a concentrated moment, jump straight up or down by the amount of the moment and place a dot. Clockwise moments cause upward jumps and counter-clockwise moments cause downward jumps.
    3. When you reach the end of the beam you should return to \(M = 0\text{.}\) If you don’t, then recheck your work.
  9. Connect the dots with correctly shaped lines. Segments under constant shear are straight lines, segments under changing shear are curves. The general curvature of the lines can be determined by considering equation (8.5.3).

Example 8.7.2. Graphical Method: Combined Loads.

Draw a neat, accurate, labeled shear and bending moment diagram for the beam and loading shown using the method of this section. Distance units are in feet.
Answer.
Solution.
Click the buttons below to step through the solution.

Example 8.7.3. Graphical Method: Concentrated Moments.

Draw a neat, accurate, labeled shear and bending moment diagram for the \(\m{15}\) long beam loaded as shown using the method of this section.
Answer.
Solution.
Click the buttons below to step through the solution.

Remark 8.7.4. Tips for Drawing Good Shear and Bending Moment Diagrams.

  1. Work neatly and professionally, and take pride in your work. Use a straightedge!
  2. Draw a labeled free-body diagram of the entire structure, and your work to find the reactions off to the side. For this free-body diagram, you may replace the distributed loads with equivalent concentrated loads.
  3. Next draw a careful sketch of the beam showing its reactions and the actual distributed loads. Distributed loads must be shown here, because their distributed nature is significant.
  4. Draw the shear diagram directly below the load diagram and the moment diagram beneath that. It is convenient to draw these graphs on graph paper. Indicate the scale used each function.
  5. Downward forces make the shear digram move down, counterclockwise concentrated moments make the moment curve jump down.
  6. Show the correct shape and curvature for each curve segment: zero, constant slope, polynomial. Changes in curve shapes should align with the load which causes them.
  7. Indicate the values of shear and moment at maximums, minimums and points of inflection.
  8. Include any other work needed to justify your results.
PrevTopNext
Feedback