Module 3 · Lesson 3.4
Relationships, overhangs and contraflexure
The calculus behind the shapes, and what happens when a beam hangs over its support.
Why this matters
Once you see that these three diagrams are linked by slopes and areas, you stop memorising shapes and start deducing them. It also gives you a fast way to check anyone's diagram, including your own.
By the end of this lesson you should be able to
- State how load, shear and moment relate to each other
- Use areas under the shear diagram to get moments
- Handle overhangs and find points of contraflexure
What it calculates: How the three diagrams are connected: the slope of the shear diagram is the load intensity, and the slope of the moment diagram is the shear.
- w
- Downward distributed load intensity (kN/m)
- V
- Shear force at the section (kN)
- M
- Bending moment at the section (kN·m)
- x
- Distance along the beam (m)
This assumes
- The beam is straight
- Loads act perpendicular to the beam
- No applied point moments at the section considered
In plain terms: Read it backwards and it becomes practical: the change in shear between two points is the load in between, and the change in moment is the area under the shear diagram between them. That is usually quicker than starting again from the reactions.
Predict first
The shear force at a section is zero. What does that tell you about the bending moment there?
An overhang is a length of beam projecting past its support. The overhang hogs, because the load out there tries to lift the beam off the far support and bends the beam the other way. Somewhere between the sagging span and the hogging overhang the moment must pass through zero. That point is called a point of contraflexure.
Practice
A simply supported beam spans 10.0 m and carries a UDL of 8 kN/m over its whole length. What is the maximum bending moment?
Practice
A cantilever 5.0 m long carries a UDL of 4 kN/m. What is the magnitude of the maximum bending moment?
Practice
A simply supported beam spans 12.0 m under a UDL of 6 kN/m. What is the maximum bending moment?
Practice
On a beam, the shear force diagram between two points encloses an area of 45 kN·m. If the bending moment at the first point is 20 kN·m, what is it at the second?
Summary
- dV/dx = −w and dM/dx = V
- Change in moment between two points = area under the shear diagram
- Overhangs hog; the main span sags
- Contraflexure is where the moment changes sign and the tension face swaps
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint