Shear and Moment Diagram Calculator

Load, shear and bending moment are three levels of the same story, linked by calculus:

$\frac{dV}{dx} = -w(x), \qquad \frac{dM}{dx} = V(x)$

Symbols and units:

• $w$ — distributed load intensity, newtons per metre (N/m), kN/m in practice
• $V$ — internal shear force, N or kN
• $M$ — internal bending moment, newton-metres (N·m), kN·m in practice
• $x$ — distance along the beam, metres (m)

Integrating gives the two rules that make diagrams quick:

• the change in shear between two points equals minus the area under the load diagram
• the change in moment equals the area under the shear diagram

The immediate consequence: the bending moment is maximum where the shear crosses zero, because that is where $dM/dx = 0$. Standard results follow directly — $M_{\max} = PL/4$ for a central point load on a simply supported span, $wL^2/8$ for a uniform load on the same span, and $wL^2/2$ at the wall of a cantilever.

The assumption people forget: these relations hold for statically determinate beams with small deflections.

The bending moment produces a stress that varies linearly through the depth:

$\sigma = \frac{M c}{I}$

• $\sigma$ — bending stress, pascals (Pa), tension on one face and compression on the other
• $M$ — bending moment at that section, N·m
• $c$ — distance from the neutral axis to the extreme fibre, m
• $I$ — second moment of area, m⁴; for a rectangle $I = bh^3/12$, so depth dominates

The ratio $S = I/c$ is the section modulus in m³, reducing the check to $\sigma = M/S$.

Stiffness is a separate question, governed by $EI$ in N·m²:

$\delta_{\text{cantilever, UDL}} = \frac{wL^4}{8EI}, \qquad \delta_{\text{simply supported, central }P} = \frac{PL^3}{48EI}$

Deflection scales with $L^3$ or $L^4$, which is why long spans fail serviceability checks long before they fail on stress.

The assumption people forget: $\sigma = Mc/I$ assumes a linear-elastic, initially straight, symmetric beam in pure bending — it is not valid past yield.

• Looking for $M_{\max}$ at midspan by habit — it sits where the shear crosses zero, which is midspan only for symmetric loading.
• Losing the minus sign in $dV/dx = -w$ — shear falls, not rises, under a downward load.
• Mixing kN·m and N·m in $\sigma = Mc/I$ — $40$ kN·m is $4.0 \times 10^4$ N·m.
• Using the full depth as $c$ — it is the half-depth for a symmetric section, so $0.15$ m for a $300$ mm beam.
• Writing $I = bh^3/3$ — that is about the base, not the neutral axis; bending uses $bh^3/12$.
• Confusing the shear and moment diagrams — the moment is the area under the shear plot, not a rescaled copy of it.
• Forgetting the fixed-end moment on a cantilever — the wall carries both a reaction force and a moment.