Shear and Moment Diagram Calculator

How to use this calculator

1. Set the span between supportsA is at 0 and B is at the far end. Both are treated as pinned or roller supports, which carry vertical force but no moment.
2. Place the distributed loadSet the start and end positions. For a UDL across the whole beam, use 0 and the full span. Set the intensity to zero to omit it.
3. Add point loads by positionDistance is measured from support A. Leave a load at zero to skip it. All loads are taken as acting downward.
4. Read the diagramsThe shear diagram jumps vertically at each point load and slopes under a distributed load. The moment diagram peaks where shear crosses zero.

Why shear crossing zero locates the maximum moment

The relationship between load, shear and moment is a chain of derivatives, and understanding it removes most of the guesswork from this topic.

The rate of change of bending moment along the beam equals the shear force at that point. In symbols, dM/dx = V. And the rate of change of shear equals the intensity of the distributed load, dV/dx = −w.

That first relationship is the useful one. A maximum or minimum of any function occurs where its derivative is zero, so the bending moment is stationary exactly where the shear force is zero. Finding the maximum moment therefore reduces to finding where the shear diagram crosses the axis.

The second relationship explains the shapes. Where there is no distributed load, dV/dx = 0, so shear is constant and the shear diagram is horizontal. Since dM/dx = V is then constant, the moment diagram is a straight sloping line.

Under a uniformly distributed load, shear falls linearly and the moment diagram becomes a parabola. Under a triangular load the shear is parabolic and the moment is cubic. Each integration raises the order by one.

At a point load the shear jumps discontinuously by the size of the load — a finite force over zero length. The moment diagram has no jump but changes slope abruptly, producing the characteristic kink.

This also explains why the moment is zero at both supports of a simply supported beam. A pin and a roller cannot resist rotation, so they cannot transmit moment. Any calculation producing a non-zero moment at either support has an error in it, which makes this the fastest check available.

Worked example: 6 m span, 20 kN point load, 5 kN/m throughout

Total load: the point load of 20 kN plus the UDL of 5 kN/m over 6 m, which is 30 kN. Total 50 kN.

Take moments about A. The point load contributes 20 × 2 = 40 kN·m. The UDL acts through its centroid at mid-span, contributing 30 × 3 = 90 kN·m. Total 130 kN·m.

So R_B × 6 = 130, giving R_B = 21.67 kN.

Vertical equilibrium: R_A = 50 − 21.67 = 28.33 kN.

Now trace the shear. Just right of A it is +28.33 kN. The UDL pulls it down at 5 kN per metre, so just left of the point load at x = 2 it has fallen to 28.33 − 10 = 18.33 kN.

At x = 2 the 20 kN point load drops the shear vertically to 18.33 − 20 = −1.67 kN. It has crossed zero — at the point load itself.

Beyond that it continues falling at 5 kN/m, reaching 28.33 − 20 − 30 = −21.67 kN at B, which equals −R_B exactly. That closure is another useful check.

Since shear crosses zero at x = 2, that is where the moment peaks.

Moment at x = 2: R_A × 2 less the UDL to the left of that section. The UDL over the first 2 m totals 10 kN acting at 1 m, so its moment about the section is 10 × 1 = 10 kN·m.

M = 28.33 × 2 − 10 = 56.67 − 10 = 46.67 kN·m.

That is the maximum bending moment, and it occurs at the point load rather than at mid-span — which is exactly why locating the shear zero matters instead of assuming the centre.

Standard cases worth memorising

The first two are worth knowing by heart because they cover most preliminary sizing. Note that wL²/8 and PL/4 are not interchangeable — a UDL of total weight W = wL produces a maximum moment of WL/8, exactly half the WL/4 produced by the same total weight applied as a single point load at mid-span. Spreading a load halves the peak moment, which is the whole reason distributed loading is preferable.

What this covers, and what a real design needs

A simply supported beam is statically determinate: two unknown reactions and two useful equilibrium equations. That is why the analysis needs no material properties, no section size and no assumptions about stiffness. The answers here are exact statics.

Other support arrangements are not all so simple. A cantilever is also determinate, with a fixed end carrying both force and moment. But a propped cantilever, a fixed-ended beam or a continuous beam over three or more supports is statically indeterminate — there are more unknown reactions than equations, and the distribution of forces depends on the relative stiffness of the members. Those need moment distribution, slope-deflection or a stiffness matrix approach.

Overhanging beams, where the span extends past a support, produce negative bending moment over the support — hogging rather than sagging. That matters because the tension face flips to the top, and reinforcement has to follow it.

What this calculator gives you is the bending moment and shear force. Design needs more.

Section sizing requires the section modulus, from bending stress σ = M ÷ Z. Deflection is a separate calculation and is frequently the governing criterion rather than strength — a beam can be perfectly safe and still unacceptably bouncy, and serviceability limits on deflection are what usually decide the size in practice.

Shear capacity, lateral torsional buckling for steel, and web crippling at supports are all separate checks. Reinforced concrete needs the moment converted into a reinforcement area, with separate checks for shear links and crack width.

Load factors and combinations under the governing code apply before any of this. The loads entered here should already be factored if you are designing to limit states.

And every real design must be checked and signed by a competent engineer. This tool is for learning the analysis and for checking a hand calculation — it is not a design tool, and nothing structural should be built from its output.

Assumptions and limitations

• Simply supported beams only — not cantilevers, overhangs or continuous beams
• Statically determinate analysis, so no material or section properties are needed or used
• All loads act vertically downward; no axial loads, moments or inclined loads
• Handles up to three point loads and one uniformly distributed load
• Does not calculate deflection, which frequently governs the design instead of strength
• Does not size a section, check buckling, or produce reinforcement
• A learning and checking tool — not a design tool, and not a substitute for a competent engineer

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