Beam Load Calculator
Simply-supported beam analysis: support reactions, shear force & bending moment diagrams, and deflection estimates for point, uniform, and triangular loads — in metric or imperial units.
Calculator
Simply supported beam (pin left, roller right). Enter loads, pick a material and section, get reactions, shear/moment values, diagrams, and deflection.
1 · Span
2 · Point loads (up to 4)
3 · Distributed loads
4 · Stiffness (for deflection)
Results
Support reactions
Shear force & bending moment
| Quantity | Value |
|---|
Diagrams
Load diagram
Shear force diagram
Bending moment diagram
Deflection (exaggerated)
Deflection
Stiffness guide: –
Common serviceability guides: L/360 for floors, L/240 for roofs (stiffness only — not a strength check).
How it works
Equilibrium first: for a simply supported span L, reactions come from force and moment balance. A point load P at distance a from the left gives R1 = P·b/L, R2 = P·a/L (with b = L − a). A uniform load w gives R1 = R2 = w·L/2. A triangular load of max intensity w₀ gives R1 = w₀·L/6 and R2 = w₀·L/3 on the max-intensity side.
Shear V(x) and moment M(x) are built by superposition — each load's contribution is added. Moment peaks where shear crosses zero. Deflection uses the Euler–Bernoulli relation with your E and I inputs (closed-form curves, e.g. δ = P·L³/(48·E·I) for a center point load, δ = 5·w·L⁴/(384·E·I) for uniform load).
Frequently asked questions
What is a beam load calculator?
A beam load calculator computes how a simply supported beam responds to loads: the support reactions (R1, R2), the shear force and bending moment at any point along the span, and the estimated deflection. This tool handles point loads, a uniform distributed load, and a triangular distributed load, and draws the shear, moment, and deflection diagrams.
Worked example: 4 m beam with a 10 kN load at the center — what are the reactions and maximum moment?
Each support carries half the load: R1 = R2 = 5 kN. Shear is +5 kN on the left half and −5 kN on the right half. The maximum bending moment occurs under the load: Mmax = P·L/4 = 10 × 4 / 4 = 10 kN·m. With E = 200 GPa and I = 20×10⁻⁶ m⁴, the midspan deflection is δ = P·L³/(48·E·I) = 3.3333 mm downward.
How do you calculate support reactions for a point load that is not at the center?
Use moment equilibrium: the reaction nearer the load takes the larger share. For an 8 kN load on a 5 m beam, 1.5 m from the left support: R1 = P·b/L = 8 × 3.5 / 5 = 5.6 kN and R2 = P·a/L = 8 × 1.5 / 5 = 2.4 kN. The two reactions always add up to the total load (8 kN). The maximum moment sits under the load: M = P·a·b/L = 8.400 kN·m.
Worked example: 6 m beam carrying a uniform load of 2 kN/m — what are the key values?
Each reaction is half the total load: R1 = R2 = w·L/2 = 6 kN. Shear varies linearly from +6 kN at the left support to −6 kN at the right, crossing zero at midspan. Bending moment follows a parabola peaking at midspan: Mmax = w·L²/8 = 9 kN·m. Maximum deflection is also at midspan: δ = 5·w·L⁴/(384·E·I) = 296.58 mm (E = 13.1 GPa, I = 8.69×10⁻⁶ m⁴).
How does a triangular distributed load work? Example: 3 m beam, zero at the left, 6 kN/m at the right.
The total load is the triangle's area: ½ × 6 × 3 = 9 kN, acting 1 m from the right (one-third of the span). Reactions: R1 (left) = 3 kN, R2 (right) = 6 kN. Maximum moment is 3.4641 kN·m at 1.732 m from the left (x = L/√3). Peak deflection is about 0.00652·w₀·L⁴/(E·I), located slightly past midspan at 0.519·L.
Can I combine several point loads? Example: 8 m beam with 5 kN at 2 m and 5 kN at 6 m.
Yes — this tool uses superposition, adding each load's effect. Here R1 = R2 = 5 kN by symmetry. The bending moment rises linearly to 10 kN·m under each load and stays constant between the two loads (shear is zero there), then falls back to zero at the right support.
Imperial example: 12 ft beam with a 2000 lb center load on a 2×8 Douglas-fir joist?
Reactions: R1 = R2 = 4448 lb. Maximum moment at center: Mmax = P·L/4 = 500 lb·ft (6000 lb·in). With E = 1,900,000 psi and I = 47.635 in⁴, midspan deflection is δ = P·L³/(48·E·I) = 1.375 in downward — a span-to-deflection ratio of about L/105.
What is the L/360 deflection rule and does the example beam pass it?
Many building codes limit deflection to span/360 for floors (a serviceability guide, not a strength check). For the 4 m example beam, δ = 3.3333 mm, so L/δ = 1200 — well above 360, so it passes the L/360 guide comfortably. The tool also checks the looser L/240 guide. These are stiffness guidelines only and do not replace structural design by a licensed engineer.
How do I read the shear force diagram for the 4 m / 10 kN center-load example?
Shear starts at +5 kN just right of the left support and stays constant across the left half. At the 10 kN load (midspan) it drops vertically by 10 kN to −5 kN, then stays constant to the right support. The point where shear crosses zero is where the bending moment is largest — here, exactly at midspan.
Why is the bending moment zero at the supports?
A simple (pin or roller) support can push up or down but cannot resist rotation, so it provides no fixing moment. The bending moment diagram therefore starts at zero at the left support, rises to its peak where shear is zero, and returns to zero at the right support.
What elastic modulus (E) values should I use for wood and steel?
Typical values built into this tool's presets: structural steel E = 200 GPa (29,000,000 psi); aluminum 6061 E = 68.9 GPa; Douglas-fir–larch E = 13.1 GPa (1,900,000 psi); LVL engineered lumber E = 13.8 GPa (2,000,000 psi); SPF softwood E = 9.65 GPa (1,400,000 psi). Always use the graded value from your lumber stamp or steel supplier for real decisions — presets are starting points for education.
How do I find the moment of inertia (I) of my beam?
For a rectangular section, I = b·h³/12 using the ACTUAL dressed dimensions. The tool includes presets, e.g. a nominal 2×8 (1.5 × 7.25 in) has I = 47.635 in⁴, and 38 × 184 mm has I = 1972.4 cm⁴. For steel I-beams, read I from the manufacturer's section tables (AISC). Enter I in cm⁴ (metric) or in⁴ (imperial).
Does this calculator work for cantilever beams?
No — this tool models a simply supported beam (pin at one end, roller at the other), the most common textbook case. A cantilever (fixed at one end, free at the other) has completely different formulas: for a tip load P, the fixed-end moment is P·L and tip deflection is P·L³/(3·E·I). Do not apply simply-supported results to a cantilever.
Can I use these results to size a real beam for construction?
No. This is a simplified educational model: it assumes ideal simple supports, small deflections, linear-elastic material, and static loads — it ignores self-weight unless you add it, load combinations, lateral buckling, shear deformation, connections, and code safety factors. Never make structural engineering decisions from it; consult a licensed structural engineer for anything that carries real loads.
Free tool by NRTechStudio