Add supports. The beam is not restrained.
How to use the beam calculator
Each panel on the left holds one part of a beam problem. Open a panel, fill in its fields, and Apply to commit. The diagrams redraw as you type.
- Set the span. Open Span and enter the full length of the beam, including any overhang past a support. All positions are measured from the left end.
- Place the supports. Open Supports and place a pin, roller or fixed end at each restrained position. Click the beam to place one, or type the distance. Statically indeterminate arrangements are supported.
- Apply the loads. Point Loads takes a force at one station, Moments an applied couple, and Distributed Loads a run of load with an intensity at each end: equal for uniform, one of them zero for triangular. Downward is negative.
- Set the section. Open Section and enter E and I. Deflection always depends on them, as do the reactions of an indeterminate beam. Shear and moment on a determinate beam do not.
- Read the four diagrams. The beam, shear, moment and deflected shape share one x axis. Hover any station to read V, M and deflection at that point.
Reading the diagrams
The four drawings share one x axis because they are each other’s derivatives. The slope of the shear diagram is the distributed load, dV/dx = w, so a uniform load makes the shear fall in a straight line and a point load makes it jump by its own size. The slope of the moment diagram is the shear, dM/dx = V, so the moment peaks where the shear crosses zero.
Three checks follow from this and apply to any solution:
- The moment is zero at a free end, and at a pin or roller with nothing beyond it.
- The shear closes at zero at the far end of the beam.
- The area under the shear diagram between two stations equals the change in moment between them.
Sign conventions
- Loads: up is positive
- A downward load is entered negative: a 12 kN weight is −12, a 10 kN/m floor load is −10. The same rule applies to every input.
- Shear: positive where the left part pushes the right part up
- The shear at any station is the sum of the upward forces to the left of it, reactions included.
- Moment: positive sagging
- Tension in the bottom fibre. A hogging moment, over a middle support or at the root of a cantilever, is negative. The results report maximum sagging and maximum hogging separately, since they usually govern different parts of a design.
- Deflection: positive up
- A beam sagging under gravity has a negative deflection. The maximum is reported with its direction and its station.
Statically indeterminate beams
A beam in bending gives statics two equations: the vertical forces must balance, and the moments must balance. Two unknown reaction components can be found from those alone, as with a pin and a roller, or one fixed end. Beyond two, statics is not sufficient: a propped cantilever, a beam built in at both ends, or a continuous beam over three supports.
These are solved here by the stiffness method, which supplies the missing equations from the beam’s own resistance to curvature. One consequence is a genuine property of indeterminate structures: changing EI on a simply supported beam leaves the reactions unchanged, while changing it on a continuous beam moves them.
Deflection, EI, and serviceability
Deflection comes from Euler-Bernoulli theory, EI·d²v/dx² = M: small deflections, no shear deformation, and a section that stays constant along the beam. Only the product EI matters, so doubling the modulus and halving the second moment of area gives the same curve.
Strength and stiffness are separate checks, and a beam can pass one while failing the other. The span-to-deflection ratio in the results is compared against serviceability limits such as L/360 for a floor under live load, or L/250 in general. A slender beam often satisfies its bending check and still misses the deflection limit.
What this model does and does not include
The model is one straight prismatic member in bending. It covers point loads, applied couples and uniform, triangular or trapezoidal distributed loads; pins, rollers and fixed ends anywhere along the span; and any degree of static indeterminacy.
Not included: axial force and buckling, torsion, internal hinges, spring or settling supports, a section that varies along the beam, shear deformation in deep beams, plastic hinges and moment redistribution, and dynamic or moving loads. For a pin-jointed frame carrying only axial force, use the truss calculator.
Built-in examples
Every example has a closed-form answer, and most are asserted against it in the solver’s tests. Loading one and opening a panel shows the values that produced the diagrams.
- Simply supported, uniform load. wL²/8 at mid-span, with 5wL⁴/384EI of deflection.
- Simply supported, point load. PL/4 at mid-span, and PL³/48EI of sag under it.
- Cantilever, load at the tip. Hogging throughout: PL at the root, PL³/3EI at the tip.
- Propped cantilever. Indeterminate to degree 1: the prop takes 3wL/8, the fixed end wL²/8.
- Two-span continuous beam. Hogging wL²/8 over the middle support, which carries 1.25wL.
- Overhanging beam. The near support carries a negative reaction and holds the beam down.
- Fixed at both ends. Indeterminate to degree 2: wL²/12 hogging, wL²/24 sagging.
- Triangular load. Hydrostatic or soil pressure, peaking at 0.577 of the span.
Frequently asked questions
Is this beam calculator free?
Yes. It runs in the browser, requires no sign-up, and no input is uploaded.
Can it solve statically indeterminate beams?
Yes. A beam in bending gives statics two equations, vertical and moment equilibrium, so any arrangement with more than two reaction components is indeterminate: a propped cantilever, a beam fixed at both ends, a continuous beam over three or more supports. The solver uses the stiffness method with Euler-Bernoulli beam elements, so the additional reactions follow from the beam's stiffness. The reactions of an indeterminate beam therefore change with EI.
How does it calculate shear force and bending moment?
By cutting the beam at each station and summing everything to the left of the cut: shear is the sum of the upward forces, moment the sum of their moments about the cut. This is exact everywhere, including inside a distributed load.
What sign convention does it use?
Up is positive, so a downward load is entered as negative. Shear is positive where the left part of the beam pushes the right part up. Bending moment is positive sagging, meaning tension in the bottom fibre, so a hogging moment over a support is negative. Applied couples and moment reactions are positive counterclockwise. Deflection is positive up, so sag is negative.
How is the deflection calculated?
From the same finite element solve, using Euler-Bernoulli theory: EI·d²v/dx² = M, small deflections, no shear deformation. Nodal deflections under consistent element loads are exact, and each peak is located on the curve itself rather than at the nearest node. Slope is reported in milliradians, deflection in millimetres or inches.
What are E and I, and what should I use?
E is the elastic modulus of the material and I the second moment of area of the cross-section. Only their product affects the answer. Steel is about 200 GPa or 29 000 ksi, aluminium 69 GPa, structural timber 8 to 14 GPa, concrete 25 to 35 GPa. I comes from the section table for a rolled beam, or bd³/12 for a solid rectangle of breadth b and depth d.
What does span / deflection mean in the results?
The span divided by the largest deflection, written as L/n, which is the form serviceability limits take: L/360 under live load is a common floor limit, L/250 a common general one. A larger n is stiffer, so a beam at L/180 fails a limit of L/360.
Why is one of my reactions negative?
The support is holding the beam down rather than up. This occurs at the near support of a beam with a long overhang, where load on the tail lifts that end. If the real support cannot pull, such as a beam resting on a wall, the beam would lift off and the model no longer matches the structure.
Can it do an internal hinge, a spring support, or a varying section?
No. Every support is rigid, the section is constant along the beam, and there are no internal hinges, so a beam jointed mid-span cannot be entered. Axial force and torsion are outside the model: this is one straight member in bending.
How accurate is it, and how would I check?
Every built-in example has a closed-form answer and the solver is tested against it: wL²/8 and 5wL⁴/384EI for a simply supported uniform load, PL³/3EI for a cantilever tip, 3wL/8 for the prop of a propped cantilever, wL²/12 for a fixed end, wL²/8 hogging over the middle support of two equal spans. The results panel reports an equilibrium residual, which is what remains of shear and moment at the far end and should be zero to rounding.