How to use this calculator

Choose SI or US units, then enter the span, the material stiffness E (or pick a material), the second moment of area I, and the loads. Results update as you type. The central point load and the uniform load are solved as two separate cases so you can see how each one behaves; the section on superposition below shows how to combine them.

Every input has its own unit menu. If you change a unit, the number converts so the physical value stays the same.

Formulas

For a prismatic beam of span L on a pin and a roller, with Young’s modulus E and second moment of area I:

Central point load P · maximum deflection at midspanδ_max = PL³48 EI

Central point load P · moment, reactions, end slopeM_max = PL4 R = P2 θ = PL²16 EI

Uniform load w · maximum deflection at midspanδ_max = 5 wL⁴384 EI

Uniform load w · moment, reactions, end slopeM_max = wL²8 R = wL2 θ = wL³24 EI

The deflection curves plotted above are, with x measured from the left support:

Point load, 0 ≤ x ≤ L/2 (mirror for the other half)δ(x) = Px (3L² − 4x²)48 EI

Uniform load, 0 ≤ x ≤ Lδ(x) = wx (L³ − 2Lx² + x³)24 EI

Symbol	Meaning	SI	US
L	Span between supports	m	ft or in
E	Young’s modulus of the material	Pa (GPa)	psi (ksi)
I	Second moment of area about the bending axis	m⁴ (mm⁴)	in⁴
P	Point load at midspan	N	lbf
w	Uniform load per unit length over the full span	N/m	lbf/ft
δ	Deflection, positive downward	m (mm)	in
M	Bending moment, positive sagging	N·m	lbf·ft
θ	Rotation at each support	rad	rad

Worked example

Steel beam, SI units (the calculator’s default values)

L = 2 m, E = 200 GPa = 200 × 10⁹ Pa, I = 800 cm⁴ = 8 × 10⁻⁶ m⁴, P = 1000 N, w = 500 N/m.

Flexural rigidity: EI = 200 × 10⁹ × 8 × 10⁻⁶ = 1.6 × 10⁶ N·m².
Point load: δ_max = 1000 × 2³ / (48 × 1.6 × 10⁶) = 1.042 × 10⁻⁴ m = 0.1042 mm. M_max = 1000 × 2 / 4 = 500 N·m.
Uniform load: δ_max = 5 × 500 × 2⁴ / (384 × 1.6 × 10⁶) = 6.510 × 10⁻⁵ m = 0.0651 mm. M_max = 500 × 2² / 8 = 250 N·m.
Span to deflection for the point load: 2 / 1.042 × 10⁻⁴ ≈ L/19,200, far stiffer than typical limits.

Combining both loads with superposition

Because the beam is linear elastic, results from separate load cases add. In this setup both maximum deflections and both maximum moments occur at midspan, so the combined values are simple sums: in the example above, 0.1042 + 0.0651 = 0.1693 mm and 500 + 250 = 750 N·m. For load cases whose peaks occur at different points, add the curves point by point instead of adding the maxima.

Typical deflection limits

Building codes limit deflection as a fraction of span. For example, the International Building Code (Table 1604.3) limits floor members to L/360 under live load and L/240 under total load. Compare those with the “span to deflection” result: a larger number after the slash means a stiffer beam.

Assumptions and limits

Ideal pin and roller supports with no rotational restraint.
Prismatic beam: E and I are constant along the span.
Linear elastic material and small deflections (Euler–Bernoulli beam theory).
Shear deformation is ignored, which slightly underestimates deflection for short, deep beams.
Self-weight is not included unless you add it to w.
No stress, lateral-torsional buckling, vibration, connection or code check is performed.

Frequently asked questions

Why is my deflection 1,000 times too big or too small?

Almost always a unit mix-up. 200 GPa is 200,000 MPa, and 1 cm⁴ is 10,000 mm⁴ or 10⁻⁸ m⁴. Check the unit menu next to E and I; the calculator converts for you once the units are right.

How do I include the beam’s own weight?

Add it to the uniform load. Weight per length is density × cross-sectional area × g. A steel section with A = 2,850 mm² weighs about 7,850 × 0.00285 × 9.81 ≈ 220 N/m.

Can I use this for a point load that is not at midspan?

Not yet. An off-center load moves the maximum deflection away from midspan and needs different formulas. This calculator handles the central load and the full-span uniform load only.

How do I get bending stress from the moment?

Use σ = M c / I, or σ = M / S with the elastic section modulus S. The moment of inertia calculator gives I, c and S for common shapes.

How were these results checked?

The closed-form results were compared with an independent solution that numerically integrates curvature M/EI twice and applies zero deflection at both supports, over many randomized inputs. The default example also matches the hand calculation above.

For education and preliminary checks only. Have a qualified engineer verify any design before it is built.