Introduction
The Beam Calculator’s outputs — moment, shear, deflection, required section size — come from a small set of classical beam formulas. They look intimidating as symbols, but each one answers a plain physical question: how much does the beam bend, where does it bend most, and how stiff must it be? This guide derives every formula the calculator uses, shows where each term comes from, and works two full examples in metric and imperial with cross-checked arithmetic.
If you haven’t run your numbers yet, the Beam Calculator applies all of these automatically — this guide is for understanding what it’s doing and sanity-checking the results. For the condensed version with a single worked example, see the complete guide.
The Setup: Simply Supported Beam, Uniform Load
The standard case is a beam resting on two supports (a “simply supported” beam) carrying a uniform load w — force per unit length, in lb/ft or kN/m — over a span L. Deck beams, floor girders, and headers all reduce to this case when joists or framing bear evenly along the beam. Everything that follows assumes static loads and small deflections, the standard assumptions of elementary beam theory. If you haven’t measured your span and built your load yet, the estimating guide covers that fieldwork first.
Reactions: where the load goes
By symmetry, each support carries half the total load:
R = wL / 2
The reactions are also the maximum shear forces, acting at the supports. For a 400 lb/ft load over 12 ft: total load = 4,800 lb, so each reaction is 2,400 lb (10.7 kN). Metric: 5.83 kN/m × 3.66 m = 21.3 kN total; each reaction 10.7 kN. Cross-check: 2,400 × 4.44822 = 10,676 N ✓.
Bending moment: why wL²/8
Moment at any point is the sum of force × distance to one side. At midspan, the left reaction (wL/2) acts at distance L/2, giving (wL/2)(L/2) = wL²/4 — but the uniform load on that half-span also pulls down, contributing (wL/2)(L/4) = wL²/8 in the opposite sense. Subtracting:
M_max = wL²/8 (at midspan)
Moment is zero at the supports and follows a parabolic curve between them. For our 400 lb/ft, 12-ft beam: M = 400 × 144/8 = 7,200 lb-ft (86,400 lb-in). Metric: 5.83 × 3.66²/8 = 9.76 kN·m. Cross-check: 7,200 × 1.35582 = 9,762 N·m ✓.
Shear force: why wL/2
Shear is the vertical force trying to slice the beam. It equals the reaction at the support and decreases linearly to zero at midspan (for symmetric uniform load):
V_max = wL/2 (at the supports)
Shear rarely controls sawn-lumber or LVL sizing — bending and deflection get there first — but it matters at notches, holes, and connections, and it controls short, deep, heavily loaded beams. Our example: 2,400 lb (10.7 kN) at each support.
Deflection: why L⁴
Deflection comes from integrating the curvature (M/EI) along the beam. For uniform load, the result is:
Δ_max = 5wL⁴ / (384EI) (at midspan)
Three things to notice. First, span appears to the fourth power — double the span, sixteen times the sag. Second, stiffness appears as the product EI: E (modulus of elasticity, a material property) times I (moment of inertia, a shape property). Typical E values run from ~1.4×10⁶ psi for common sawn lumber to ~2.0×10⁶ psi for LVL and 29×10⁶ psi for steel — the material guide tabulates them. Third, the 5/384 constant is specific to this load case; other load patterns have their own constants, tabulated below.
Point Loads: PL/4 and PL³/48EI
A concentrated load P at midspan — a post from the floor above, a girder framing into the side — produces:
M_max = PL/4 · V_max = P/2 · Δ_max = PL³/(48EI)
Compare with the uniform case: a point load P equal to the total uniform load (P = wL) gives M = wL²/4 — twice the uniform-load moment. This is why unidentified point loads are dangerous: the same total weight, concentrated, doubles the bending demand. If a load sits off-center at distance a from the left support (span L), the moment under the load is M = Pab/L where b = L − a, and each reaction is found by static equilibrium (R_left = Pb/L).
Worked point-load example
A 3,000 lb (13.3 kN) point load at the center of a 10-ft (3.05 m) header: M = 3,000 × 10/4 = 7,500 lb-ft. Metric: 13.3 × 3.05/4 = 10.2 kN·m. Cross-check: 7,500 × 1.35582 = 10,169 N·m ✓. Shear at each support: 1,500 lb (6.67 kN).
Cantilevers: The Overhang Case
Deck beams often cantilever past the outer posts. For a cantilever of length L with uniform load w (fixed at the wall, free at the end):
M_max = wL²/2 (at the fixed end) · V_max = wL · Δ_max = wL⁴/(8EI) (at the free end)
Note the moment is four times the simply-supported value for the same w and L — cantilevers are demanding, which is why codes limit overhangs (commonly to about one-quarter of the adjacent span for deck joists; confirm locally). A point load P at the free end gives M = PL, V = P, Δ = PL³/(3EI).
From Moment to Beam Size: S = M/Fb
The bending stress in a beam is σ = M/S, where S is the section modulus. Setting stress equal to the material’s allowable bending stress Fb and solving gives the required size:
S_required = M / Fb
For a rectangular section, S = bd²/6 and I = bd³/12 — depth d dominates both. Using our 86,400 lb-in moment with #2 Southern Pine (Fb ≈ 1,000 psi): S_required = 86.4 in³. A 2×12 (actual 1.5 × 11.25 in) provides S = 1.5 × 11.25²/6 = 31.6 in³, so three plies give 94.9 in³ ≥ 86.4 ✓ — the same conclusion as the complete guide’s worked example.
Deflection check with real E and I
Taking E = 1.6×10⁶ psi (11.0 GPa) and I = 3 × 178 = 534 in⁴ for the 3-ply 2×12, with w = 400/12 = 33.33 lb/in and L = 144 in: Δ = 5 × 33.33 × 144⁴/(384 × 1.6×10⁶ × 534) = 0.22 in (5.6 mm), under the L/360 limit of 0.40 in (10.2 mm) ✓. Metric cross-check: 5 × 5,830 × 3.66⁴/(384 × 11.0×10⁹ × 2.22×10⁻⁴) = 0.00555 m = 5.55 mm ✓.
Formula Reference Table
| Load case | Max moment | Max shear | Max deflection |
|---|---|---|---|
| Uniform load, simply supported | wL²/8 | wL/2 | 5wL⁴/(384EI) |
| Point load at center, simply supported | PL/4 | P/2 | PL³/(48EI) |
| Point load off-center (a from left) | Pab/L | max(Pb/L, Pa/L) | — (see references) |
| Uniform load, cantilever | wL²/2 | wL | wL⁴/(8EI) |
| Point load at free end, cantilever | PL | P | PL³/(3EI) |
What the Formulas Don’t Cover
These closed-form formulas assume prismatic (constant cross-section) beams, static loads, and supports that behave as idealized pins or fixed ends. They don’t account for load duration (wood is weaker under sustained load), moisture, lateral-torsional buckling of deep narrow beams without bracing, notches and holes, or continuous spans over three or more supports (which need different analysis). Real designs apply adjustment factors for all of these — which is exactly where a structural engineer earns their fee. The common mistakes guide shows how each of these assumptions gets violated in practice.
Related Guides
- Complete Beam Sizing Guide — the full method with a single start-to-finish worked example.
- Estimating Guide — getting span, tributary width, and loads right before you calculate.
- Beam Material Guide — E and Fb values across lumber, LVL, PSL, and steel.
- Beam Calculator FAQ — quick answers on moments, deflection limits, and conversions.
- Beam Sizing Checklist — every verification step in order.
Get Your Number
Five formulas carry the whole calculation: reactions wL/2, moment wL²/8, shear wL/2, deflection 5wL⁴/384EI, and the sizing rule S = M/Fb — with point-load and cantilever variants in the table above. The span’s fourth-power grip on deflection is the insight that matters most: when a beam looks marginal, shortening the span beats upsizing the beam nearly every time. Plug your values into the Beam Calculator and verify the outputs against the hand calculations here.
Disclaimer: These formulas are planning tools based on idealized beam theory. Real beam design applies adjustment factors for load duration, moisture, stability, and connections, and must satisfy local building codes. Verify selections with published span tables or a structural engineer, and check your local code before building.
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