The physics
A beam carries loads across a span by bending. At any cross-section, the part of the beam on one side pushes and pulls on the other with three internal forces: a normal (axial) force N along the beam, a shear force V across it, and a bending moment M. Cutting the beam and writing equilibrium for one side gives them everywhere, and they are linked to the distributed load w (downward per metre):
So the moment is largest where the shear crosses zero. A point load makes the shear jump by its size, and an applied moment makes the bending moment jump. Bending curves the beam. In Euler–Bernoulli beam theory the curvature is proportional to the moment:
Integrating twice gives the slope θ and the deflection v. The bending stress is largest at the top and bottom of the section, σ = N/A ± Mc/I.
| Symbol | Meaning | SI unit |
|---|---|---|
| x | Position along the beam, from the left end | m |
| N | Normal (axial) force, tension positive | kN |
| V | Shear force: the upward resultant on the part left of the cut | kN |
| M | Bending moment, sagging (concave up) positive | kN·m |
| w | Distributed load, downward positive | kN/m |
| v, θ | Deflection (up positive) and slope (counterclockwise positive) | mm, mrad |
| E | Young's modulus of the material | GPa |
| I, A | Second moment of area and area of the cross-section | cm⁴, cm² |
| c | Distance from the neutral axis to the outer fibre | mm |
Supports, joints, and stability
- A fixed support (clamp) stops movement and rotation, so it provides two forces and a moment. A pin stops movement but lets the beam rotate (two forces). A roller stops only vertical movement (one force).
- A hinge is an internal joint. It passes forces from one part of the beam to the next but no bending moment, so M = 0 there and the slope can change suddenly.
- A straight beam in its plane has three equilibrium equations, and each hinge adds one (M = 0). If the supports provide exactly that many reactions, the beam is statically determinate: statics alone gives the reactions. With more, it is indeterminate, and the reactions also depend on how the beam deforms. This tool solves both kinds the same way. With too few, or badly arranged, restraints the beam is a mechanism and cannot carry load. For example, a hinge between a pin and a roller lets the beam fold. The tool detects this and says why.
Reading the diagrams
- The diagrams follow the equilibrium rules: the shear jumps at point loads and reactions and slopes under distributed loads, and the moment jumps at applied moments. A free end and a pin or roller at an end carry no moment, and a hinge carries none either.
- Use the probe on any diagram to read values. The largest and smallest values are labelled.
- The deflected shape in the scene is exaggerated so that it is visible. The scale factor is shown, and Deflection scale → True scale shows it as it really is. Span / deflection compares with serviceability limits such as L/250 or L/360 used in design codes.
- The equilibrium check confirms that the loads and the computed reactions balance, in forces and moments.
Solving by hand: the moment equation and the load equation
Equations (1) and (2) can be combined in two ways. The moment equation is second order:
It needs M(x) first. For a statically determinate beam, equilibrium gives the reactions and then M(x) directly, and integrating twice leaves only two constants, fixed by the support conditions. Differentiating twice more gives the fourth-order load equation:
It starts from the loads alone. Integrating four times gives V, M, θ, and vwith four constants, and the conditions at the ends (for example v = 0 and M = 0 at a pin,v = 0 and θ = 0 at a fixed support, V = 0 and M = 0 at a free end) determine them together with any unknown reactions at interior supports. That makes it the natural method for statically indeterminate beams.
Both are written with singularity (Macaulay) functions ⟨x − a⟩ⁿ, which are zero for x < a and (x − a)ⁿ beyond it, so one expression holds along the whole beam however many loads it carries. A point force at a is ⟨x − a⟩⁻¹ and a couple ⟨x − a⟩⁻². At a hinge the slope may jump, which adds a term Δθ⟨x − a⟩ with the condition M = 0 there.
The step-by-step solution below the simulation works the current beam through either method. It recommends the moment equation for determinate beams and the load equation for indeterminate ones. It offers the moment equation up to twice indeterminate in bending; beyond that, the extra unknown reactions make it impractical by hand.
Diagrams accompany the steps. A free-body diagram replaces each support by its unknown reactions. The reactions are then found as by hand: each time from an equation with a single unknown left (moments about a support, or about a hinge for the part on one side of it), with an unused equation as a check. For the moment equation, the beam is cut once in every segment between loads and supports. The part to the left of the cut, with the shear force V(x) and bending moment M(x) at the cut, gives both by equilibrium. Further sketches show the sign convention, a short element dx of the beam, where each boundary condition holds, and the final V, M, and v.
With Symbols selected, the same solution is written in symbols instead of numbers: L for the length, a, b, … for positions (or L/2, L/3, … where they fall), P, w, M₀ for the loads, and EI. The result is the general deflection equation, also given segment by segment, which shows how each load contributes. For example, a simply supported beam with a point load P at a gives RA = P(L − a)/L and a deflection under the load of −Pa²(L − a)²/(3EIL).
Method and validation
The beam is solved with the finite element method using cubic Euler–Bernoulli beam elements. Distributed loads are converted to their consistent (work-equivalent) nodal loads. For a beam with constant EI, this gives the exact deflection and slope at every node for any loading. Nodes sit at every support, joint, and load point, and elements are at most L/160 long, so the curves between nodes are exact to drawing accuracy. Hinges get a separate rotation on each side. The reactions follow from the stiffness equations at the supports. The diagrams N, V, and M are then computed exactly from equilibrium of the beam to the left of each cut, so their jumps sit exactly at the loads. The axial problem is solved separately with bar elements, which splits axial loads between several pins or fixed supports according to their stiffness.
Automated tests compare the results with handbook formulas (Roark; Gere & Goodno), agreeing to floating-point round-off (about eight significant digits). They cover simply supported beams with point, uniform, and triangular loads and applied moments, cantilevers, overhangs, a Gerber beam with a hinge, fixed–fixed and propped cantilevers, and two-span continuous beams. They also check how axial loads are shared and detect mechanisms and sliding. For random beams, the tests check global equilibrium and that M = EIv″ along the beam.
The step-by-step solution is computed independently, in closed form. Automated tests check it against the finite-element solution for every preset and 300 random beams. Where both methods apply, the moment equation and the load equation must also agree with each other to 10⁻⁹.
Assumptions and limitations
- Linear-elastic material and small deflections. Stresses beyond the yield strength are not detected.
- Euler–Bernoulli theory: plane sections stay plane and shear deformation is neglected. This is accurate for slender beams (span more than about 10× the depth).
- A straight, prismatic beam (constant section and material) loaded in one plane. Axial forces do not change the bending stiffness, so buckling and other second-order effects are not included.
- Supports are rigid. Shear stresses use elementary formulas (for the I-beam, the average over the web), and the I-section ignores the root radii of rolled profiles.