How to Calculate Beam Load: Shear, Bending Moment, and Deflection Explained

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How to Calculate Beam Load: Shear, Bending Moment, and Deflection Explained

Every beam under load has to answer three structural questions: how much internal force is trying to shear it apart, how much it's being bent, and how far it physically deflects. Those three numbers — shear, bending moment, and deflection — are what beam sizing is actually built on, whether you're specifying a steel W-shape or sizing a timber joist for a deck.

The Three Numbers That Matter

• Maximum shear — the internal force trying to slide one section of the beam past the adjacent section. It governs connection design and, for wood, checks against splitting near supports.
• Maximum bending moment — the internal force trying to bend (rotate) the beam's cross-section. It governs whether the material yields or fractures, and drives the required section size.
• Maximum deflection — how far the beam physically sags under load. Even a beam strong enough to not break can still deflect more than a floor, roof, or code allows — deflection is frequently the controlling limit for longer spans, not strength.

Simply Supported vs. Cantilever Beams

A simply supported beam rests on a support (pin or roller) at each end and is free to rotate there — a joist spanning between two walls, for example. A cantilever beam is fixed rigidly at one end and completely free at the other — a balcony beam or a shelf bracket. The two configurations produce very different results for an identical load and span: a cantilever carrying the same uniform load develops four times the bending moment and five times the deflection of a simply supported beam of the same span, because the cantilever has no second support sharing the load.

Uniform Load vs. Point Load

A uniform load (measured in lb/ft) is distributed evenly along the beam's length — a floor's live load transferred to a joist, for example. A point load (measured in lb) is concentrated at a single location — a support post landing partway along a beam, or a piece of equipment. The same total weight produces a different bending moment and deflection depending on how it's distributed, which is why the two cases use separate formulas rather than one generic one.

The Standard Formulas

For a beam of span length L, load w (uniform, lb/ft) or P (point, lb), elastic modulus E, and moment of inertia I:

| Configuration | Max shear | Max bending moment | Max deflection | |---|---|---|---| | Simply supported, uniform load | wL / 2 | wL² / 8 | 5wL⁴ / (384EI) | | Simply supported, point load at center | P / 2 | PL / 4 | PL³ / (48EI) | | Cantilever, uniform load | wL | wL² / 2 | wL⁴ / (8EI) | | Cantilever, point load at free end | P | PL | PL³ / (3EI) |

Shear and moment scale directly with load, but deflection scales with the cube or fourth power of span length — doubling the span multiplies deflection by 8x (point load) or 16x (uniform load) if nothing else changes. This is why long spans need disproportionately deeper or stiffer sections, not just proportionally larger ones.

Elastic Modulus and Moment of Inertia: Material and Shape Both Matter

Elastic modulus (E) measures how stiff the material is — how much it resists stretching or compressing under stress, regardless of shape. Steel (≈29,000 ksi) is roughly 18 times stiffer than typical Douglas fir dimension lumber (≈1,600 ksi) and nearly 3 times stiffer than aluminum (≈10,000 ksi) — a big part of why steel beams can be so much shallower than wood beams for the same span and load.

Moment of inertia (I) measures how the cross-section's shape resists bending — a property of geometry, not material. For a rectangular wood section it's bh³/12 (base times height cubed, divided by 12) — notice that height is cubed while width isn't, which is why standing a 2×10 on edge is dramatically stiffer than laying it flat, even though it's the same piece of wood. For standard steel shapes, moment of inertia is published directly in structural steel manuals (a W12×50, for example, carries I=394 in⁴).

Both values feed into deflection identically — doubling either E or I cuts deflection in half — which is why increasing depth (which raises I with the cube of height for a rectangular section) is usually a far more efficient fix for a deflection problem than switching to a marginally stiffer material.

Sizing a Timber Beam: A Worked Example

Say you're checking a Douglas fir beam spanning 12 ft with a uniform load of 300 lb/ft, using a 2×10 (actual size 1.5"×9.25") standing on edge:

1. Moment of inertia: I = bh³/12 = 1.5 × 9.25³ / 12 ≈ 98.9 in⁴
2. Max shear: wL/2 = 300 × 12 / 2 = 1,800 lb
3. Max moment: wL²/8 = 300 × 12² / 8 = 5,400 ft·lb
4. Max deflection: using E ≈ 1,600 ksi for Douglas fir #2 and the formula above, plug in L (converted to inches), E, and I to get the deflection in inches — then compare it against the applicable deflection limit (commonly L/240 or L/360 depending on the application) to see whether the section is adequate.

Run these same four inputs through the calculator above to check shear, moment, and deflection together instead of by hand.

The Bottom Line

Shear, moment, and deflection are governed by simple, well-established formulas — the complexity is almost entirely in picking the right formula for your support condition and load type, and in choosing accurate E and I values for your actual material and section.

These formulas and this calculator are for preliminary, educational sizing only. Always follow manufacturer guidelines, the applicable building code, and industry design standards, and have final structural designs reviewed by a licensed engineer — incorrect beam sizing can result in structural failure, injury, or property damage.

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