Shear and moment diagrams

Why it matters

A beam carries load by bending, and the internal forces change along its length. To check a beam you need to know where the internal bending moment is largest (that is where bending stress peaks) and where the shear force is largest (that is where shear and bearing problems appear). Shear force and bending moment diagrams, often called V and M diagrams, show both at a glance. Drawing them is a core skill in every structural and mechanical engineering course.

Internal forces in a beam

Cut a beam at a distance x from the left end and look at the left-hand piece. To keep it in equilibrium, the cut face must carry:

• a shear force V, acting across the beam, and
• a bending moment M, trying to rotate the piece.

A common sign convention (used here): V is the sum of the upward forces to the left of the cut, and M is positive when it makes the beam sag (concave up, compression on top, tension on the bottom). Some textbooks use different conventions; always state yours.

For a beam with point loads, you get V and M at any x from the left-hand piece alone:

V(x) = (upward reactions left of x) − (downward loads left of x)
M(x) = Σ (each force left of x) × (its distance to the cut)

Worked example 1: one point load

A simply supported beam spans 6 m with a 12 kN load at 2 m from the left support. From the first lesson, RA = 8 kN and RB = 4 kN.

0 < x < 2:   V = +8 kN            M = 8·x            (M = 16 kN·m at x = 2)
2 < x < 6:   V = 8 − 12 = −4 kN   M = 8·x − 12·(x − 2)
                                    at x = 6:  48 − 48 = 0

The V diagram is a step: +8 kN, dropping by 12 kN at the load, to −4 kN, then back to zero at the right support. The M diagram is a triangle: zero at both supports, peaking at 16 kN·m under the load.

The relationships that make sketching fast

For distributed load w (kN/m, downward positive), shear V and moment M:

dV/dx = −w        dM/dx = V

So:

1. With no distributed load, V is constant and M is a straight line.
2. A point load makes V jump by the size of the load.
3. Under a uniform load, V is a sloping straight line and M is a parabola.
4. M is at its maximum where V crosses zero. This is the most useful rule: find where V = 0 and you have found the peak moment.
5. The change in M between two points equals the area of the V diagram between them. In example 1, the area of the V diagram from 0 to 2 m is 8 × 2 = 16 kN·m, matching M at x = 2.

At a simple support with no applied moment, M = 0.

Worked example 2: a uniform load

The same 6 m beam carries a uniform load w = 5 kN/m along its whole length. The total load is 30 kN, so by symmetry RA = RB = 15 kN.

V(x) = 15 − 5·x                  zero at x = 3 m (mid-span)
M(x) = 15·x − 5·x²/2
M(3) = 45 − 22.5 = 22.5 kN·m

That matches the standard result for a uniformly loaded simply supported beam:

M_max = w·L² / 8 = 5 · 6² / 8 = 22.5 kN·m

And for a single central point load P: M_max = P·L/4.

Drawing the diagrams step by step

1. Find the reactions.
2. Mark the key points: supports, point loads, starts and ends of distributed loads.
3. Walk from left to right, building V: jump at each point load, slope under distributed loads.
4. Build M from the area under V, starting from zero at a simple support.
5. Check that M returns to zero at the right-hand simple support. If it does not, there is an arithmetic error.
6. Label the maximum values and where they occur.

Common mistakes

• Wrong reactions. Every diagram after that is wrong. Check ΣFy = 0 and moments about a second point.
• Missing the jump at a point load in the V diagram.
• Looking for M_max at the load every time. With several loads or distributed loads, find where V crosses zero.
• Using w·L²/8 for other cases. It applies only to a uniform load over a simply supported span. Cantilevers and continuous beams have different results.

Recap

• V(x) and M(x) come from equilibrium of the piece on one side of a cut.
• dM/dx = V: M peaks where V crosses zero.
• Point loads make V jump; uniform loads make V slope and M curve.
• Standard results: P·L/4 for a central point load, w·L²/8 for a full uniform load.

Try it

On the Do tab, finish the bending moment function for a simply supported beam with point loads and find the peak moment numerically.