Beam Moment At X⚠ unverified
Mechanical / Beams · Compute the internal bending moment at a given distance along a beam
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| M | M | N*m | 1.0 | Reference bending moment at the start point |
| V | V | N | 1.0 | Shear force at the start point |
| w | w | N/m | 1.0 | Uniformly distributed load per unit length |
| x | x | m | 1.0 | Distance from the start point at which the moment is evaluated |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | M | N*m | Internal bending moment at ``x``, in newton-metres (N*m) |
The science & history
Understanding the Parameters
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Reference moment $M$ — the bending moment already present at the starting station ($x=0$). It is the constant of integration: the moment diagram is built relative to this known value (often zero at a simple support or free end).
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Reference shear $V$ — the shear force at the start. It sets the initial slope of the moment diagram, because $\mathrm{d}M/\mathrm{d}x = V$: where shear is large the moment climbs steeply; where shear is zero the moment is momentarily flat (a local maximum or minimum).
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Distributed load $w$ — a uniform load along the span (self-weight, snow, pressure). It bends the moment diagram into a parabola: as you move along, the accumulating load steadily reduces the shear, so the moment's slope keeps decreasing — the $-wx^2/2$ term.
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Distance $x$ — how far along the beam from the reference point you evaluate. Sweeping $x$ traces the whole moment diagram.
Derivation (Approaching a Proof)
The relations between load, shear, and moment come from equilibrium of an infinitesimal beam element $\mathrm{d}x$. Vertical equilibrium and moment equilibrium of that slice give the two fundamental differential relations of beam statics:
$$\frac{\mathrm{d}V}{\mathrm{d}x} = -w, \qquad \frac{\mathrm{d}M}{\mathrm{d}x} = V.$$
Integrate the first from the reference point, where the shear is $V_0 \equiv V$:
$$V(x) = V - w x.$$
Integrate the second, using $M(0) = M$ and substituting $V(x)$:
$$M(x) = M + \int_0^x V(\xi)\,\mathrm{d}\xi = M + \int_0^x (V - w\xi)\,\mathrm{d}\xi = M + V x - \frac{w x^2}{2}.$$
So the formula is the double integration of the load, with the reference moment and shear as the two integration constants. The peak moment occurs where $V(x) = 0$, i.e. $x = V/w$ — the classic rule that the maximum bending moment sits where the shear diagram crosses zero.
Dimensional check. $[M x\text{-term}] = \text{N}\cdot\text{m}$, $[V x] = \text{N}\cdot\text{m}$, $[w x^2] = (\text{N/m})\,\text{m}^2 = \text{N}\cdot\text{m}$. All terms N·m. ✓
History and Development
The differential relations $\mathrm{d}V/\mathrm{d}x = -w$ and $\mathrm{d}M/\mathrm{d}x = V$ and the graphical shear/moment diagrams built from them were systematised in the 19th century as strength-of-materials matured with railway and bridge engineering (Rankine, Mohr). They remain the first analysis every structural engineer learns, and underpin how finite-element post-processors report beam internal forces.
Related Concepts: Beam Bending Stress, Continuous Beam Moment, Section Modulus, Shear Deflection, Rectangular Moment of Inertia
Notes: Assumes a single uniform load segment and a straight, prismatic beam. For point loads or varying distributed loads, apply the formula piecewise between load discontinuities, carrying the end-of-segment $M$ and $V$ forward as the next segment's reference values.