Shear and Moment Diagrams

NCEES FE Civil spec: Mechanics of Materials, A. Shear and moment diagrams. FE Reference Handbook: Mechanics of Materials chapter, beams (load, shear and moment relationships).

Shear and moment diagrams show how internal shear force V and bending moment M vary along a beam. They tell you where the beam is most highly stressed, so they come before every bending stress, shear stress and design question. On the FE Civil exam you rarely draw a full diagram; you need its key values, usually the maximum moment and where it occurs.

What the NCEES spec covers

Section A covers shear and moment diagrams for statically determinate beams: simply supported, cantilever and overhanging beams with point loads, uniform loads, linearly varying loads and applied couples.

Key concepts

Sign convention. The common beam convention takes shear as positive when it would rotate a segment clockwise, and moment as positive when it bends the beam concave up (sagging).
Load-shear relation. The slope of the shear diagram equals the distributed load intensity (with downward load negative). A point load causes a jump in V equal to the load.
Shear-moment relation. The slope of the moment diagram equals V, so the change in M between two points equals the area under the shear diagram. A concentrated couple causes a jump in M.
Maximum moment occurs where V crosses zero (or at a support or point of jump in an overhang).
Shape rules. No load gives constant V and linear M. A uniform load gives linear V and parabolic M. A triangular load gives parabolic V and cubic M.

Core formulas

\frac{d V}{d x} = w (x) \frac{d M}{d x} = V

V_{2} - V_{1} = \int_{x_{1}}^{x_{2}} w d x M_{2} - M_{1} = \int_{x_{1}}^{x_{2}} V d x

Standard maxima worth knowing on sight:

Simple span, UDL: M_{m a x} = \frac{w L ^{2}}{8} Simple span, center load: M_{m a x} = \frac{P L}{4} Cantilever, UDL: M_{m a x} = \frac{w L ^{2}}{2}

Where it is in the FE Reference Handbook

The Mechanics of Materials chapter states the load-shear-moment relationships and the beam sign convention, and includes tables of beam slopes and deflections for simply supported and cantilever beams. The Civil Engineering chapter also has beam formulas for common support and load cases.

Worked example

A simply supported beam spans 8 m from A to B. It carries a uniform load of 5 kN/m over the full span and a 20 kN point load 2 m from A. Find the reactions, the shear at key points, and the maximum moment.

Step 1: reactions. The uniform load totals 40 kN at 4 m.

\sum M_{A} = 8 R_{B} - 40 (4) - 20 (2) = 0 \Rightarrow R_{B} = 25 kN R_{A} = 60 - 25 = 35 kN

Step 2: shear diagram.

V = +35 kN at A, falling at 5 kN/m to 35 − 10 = 25 kN just left of the point load.
The 20 kN load drops V to +5 kN just right of x = 2 m.
V keeps falling at 5 kN/m and crosses zero at x = 2 + 5/5 = 3 m.
V reaches 5 − 5(5) = −25 kN at B, which matches the reaction. ✓

Step 3: maximum moment at x = 3 m, where V = 0:

M (3) = 35 (3) - \frac{5 ( 3 ) ^{2}}{2} - 20 (1) = 105 - 22.5 - 20 = 62.5 kN \cdotp m

Check with areas: from A to 2 m the shear area is (35 + 25)/2 × 2 = 60 kN·m, and from 2 m to 3 m it is (5)(1)/2 = 2.5 kN·m, for a total of 62.5 kN·m. ✓

Common mistakes

Taking the maximum moment at the point load or midspan without finding where V = 0.
Forgetting the jump in V under a point load, or the jump in M under an applied couple.
Mixing sign conventions between the left and right segments of a cut.
For overhangs, missing the negative moment over the interior support, which can govern.
Using wL²/8 when the load does not cover the full span.

Draw the diagrams for any load case with the shear and moment diagram calculator, and see the worked shear moment diagram example.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A simply supported beam of span L = 8 m carries a uniformly distributed load w = 10 kN/m. What is the maximum bending moment? See the worked solution or practice the Beam Analysis deck
For the same beam - simply supported, L = 8 m, w = 10 kN/m - what is the maximum shear force? See the worked solution or practice the Beam Analysis deck
A 3 m cantilever carries a uniformly distributed load of 8 kN/m along its whole length. What is the bending moment at the fixed support? See the worked solution or practice the Beam Analysis deck
A simply supported beam spans 10 m. A 40 kN point load acts 4 m from the left support A. What is the maximum bending moment? See the worked solution or practice the Beam Analysis deck
A simply supported beam carries a single 20 kN downward point load at midspan. Neglect self-weight. What is the shear force magnitude immediately to the left of the load? See the worked solution or practice the Beam Analysis deck

Related Calculators and Examples

Frequently Asked Questions

Where does the maximum bending moment occur?

Where the shear force crosses zero, because the slope of the moment diagram equals the shear. Also check supports of overhanging beams and points where the moment jumps due to an applied couple.

How are load, shear and moment related?

The slope of the shear diagram equals the distributed load intensity, and the slope of the moment diagram equals the shear. The change in moment between two points equals the area under the shear diagram.

What is the maximum moment for a simply supported beam with a uniform load?

It is w L^2 / 8 at midspan, where w is the load per unit length and L is the span.

Back to all Mechanics of Materials topics