Shear force and bending moment diagrams

Engineering · First- and second-year engineering · about 12 min

A shear force diagram (SFD) and a bending moment diagram (BMD) show what's happening inside a beam at every point along it. Engineers use them to find where a beam is working hardest, because that's where it will crack, bend too much, or need to be thicker.

The key idea

Shear force and bending moment are linked: the slope of the bending moment is the shear force.

\frac{dM}{dx} = V \qquad \frac{dV}{dx} = -w

So the bending moment is greatest where the shear force crosses zero, and the change in moment between two points equals the area under the shear force diagram between them.

Try it yourself

Change the loads and watch the shear force and bending moment diagrams redraw.

What shear force and bending moment mean

Imagine cutting the beam at a point.

The shear force $V$ is the net vertical force on one side of the cut: it's what tries to slide one part of the beam past the other.
The bending moment $M$ is the net moment on one side of the cut: it's what tries to bend the beam there.

You can use either side of the cut; both give the same answer if you keep the sign convention consistent.

Two parts side by side. Left: a simply supported horizontal beam with one downward force, and a thin dashed vertical line marking a cut to the left of the force. Right: only the piece of beam to the left of the cut, drawn larger. An upward dark blue arrow acts at its left support. On its cut right-hand face, one dark red arrow points straight down and one curved dark red arrow turns counterclockwise. — Cutting the beam shows the shear force and bending moment acting inside it.

Rules that make the diagrams fast

Point load: the SFD jumps by the size of the load; the BMD has a sharp corner there.
No load between two points: the SFD is flat; the BMD is a straight sloping line.
Uniformly distributed load: the SFD slopes in a straight line; the BMD is a curve (a parabola).
Simply supported ends: the bending moment is zero at a pin or roller.
Maximum moment: where the SFD crosses zero.

Results worth remembering

For a simply supported span $L$ :

Load	Maximum bending moment
Point load $P$ at mid-span	$\dfrac{PL}{4}$
UDL $w$ over the whole span	$\dfrac{wL^2}{8}$

They're handy for checking your diagram, but always work the actual case through.

Worked example: a point load off-centre

A simply supported beam is $6$ m long, with a $12$ kN point load at $2$ m from the left support $A$ . Draw the shear force and bending moment diagrams and find the maximum bending moment.

A horizontal beam 6 m long. Pin support at the left end A, roller support at the right end B. One downward force arrow at 2 m from A. Dimensions under the beam: 2 m and 4 m. — The beam in this example.

Show the solution

Reactions. Moments about $A$ : $B_y \times 6 = 12 \times 2$ , so $B_y = 4$ kN. Then $A_y = 12 - 4 = 8$ kN.

Shear force.

From $A$ to the load ( $0$ to $2$ m): $V = +8$ kN (flat).
At the load the SFD drops by $12$ kN.
From the load to $B$ ( $2$ to $6$ m): $V = 8 - 12 = -4$ kN (flat), and it closes back to zero at $B$ .

Bending moment. Zero at both supports. Under the load, from the left:

M = 8 \times 2 = 16 \text{ kN·m}

The SFD crosses zero at the load, so that's the maximum: $M_{max} = 16$ kN·m, with straight lines from $0$ at $A$ up to $16$ and back down to $0$ at $B$ .

Check from the right: $4 \times 4 = 16$ kN·m. It matches.

Three parts stacked and lined up vertically, all to the same horizontal scale. Top: the beam, 6 m long, with a pin at A, a roller at B and a downward force at 2 m from A. Middle: a shear force diagram with a horizontal axis. It is flat at +8 from 0 to 2 m, drops straight down at 2 m, then is flat at -4 from 2 m to 6 m. Bottom: a bending moment diagram made of two straight lines, rising from 0 at A to a peak of 16 at 2 m, then falling to 0 at B. — The finished diagrams. The maximum moment sits where the shear force crosses zero.

Common mistakes

Putting the maximum moment at the middle of the beam

It's where the shear force is zero, which is only the middle for symmetric loading.

Drawing a curve under a point load

Between point loads the moment changes linearly. Curves only appear under distributed loads.

Switching sign convention halfway through

Pick one (for example, sagging moment positive) and keep it for the whole beam. Checking from the other side catches the switch.

Check your understanding

1. Where on the beam is the bending moment largest?

2. What is the maximum bending moment for a $5$ kN/m UDL over a simply supported $4$ m span?

3. Between two point loads with nothing else on the beam, the shear force diagram is...

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