Deflection is a stiffness question, not a strength question
A beam can carry its load with stress to spare and still fail the job by sagging enough to crack the ceiling below it. That’s why deflection gets its own check: strength asks whether the beam breaks, deflection asks whether it moves too much while doing its work. For single-span beams the answer comes from closed-form formulas — no software, no iteration — and four load cases cover most of what walks in the door: a simply supported beam under a uniformly distributed load, a simply supported beam under a central point load, a cantilever under a distributed load, and a cantilever with a load at its free end.
Every one of those formulas is built from the same four ingredients. E is the material’s modulus of elasticity — how hard the stuff is to stretch. I is the second moment of area — how well the cross-section’s shape resists bending. L is the span or cantilever length, and the load enters as w (force per unit length) when it’s spread out or P (a single force) when it’s concentrated. To make the arithmetic honest, this guide follows one beam all the way through: a simply supported steel beam spanning 5,000 mm — about 16 ft 5 in — with E = 200,000 MPa, I = 1.0 × 10⁸ mm⁴, carrying w = 10 N/mm, which is 10 kN/m or roughly 685 lb per foot.
The four formulas
For a simply supported beam under a uniform load, the maximum deflection is δ = 5wL⁴ / (384EI), at midspan. Swap the uniform load for a single point load at the center and it becomes δ = PL³ / (48EI), still at midspan. A cantilever under a uniform load deflects δ = wL⁴ / (8EI) at its free end, and a cantilever with a point load at the tip deflects δ = PL³ / (3EI), also at the tip. These are the classical Euler–Bernoulli elastic solutions — the same expressions tabulated in Roark’s Formulas for Stress and Strain and in Gere & Timoshenko.
Before plugging in numbers, read the exponents, because they carry the engineering lesson. Distributed-load deflections grow with L to the fourth power: double the span of that uniformly loaded beam and, with everything else held fixed, the deflection multiplies by sixteen. Point-load deflections grow with L cubed — still brutal, at eight times per doubling. E and I sit in the denominator as a simple product, so a section with twice the I deflects exactly half as much. Span is the variable that punishes; stiffness only rescues linearly.
One beam, worked through
Now the example beam, digit by digit. Deflection: δ = 5 × 10 × 5000⁴ / (384 × 200,000 × 10⁸). The numerator is 5 × 10 × 6.25 × 10¹⁴ = 3.125 × 10¹⁶; the denominator is 7.68 × 10¹⁵; the quotient is 4.069 mm — about 0.16 inches of sag at midspan. Two companion numbers come along for free: the maximum bending moment M = wL² / 8 = 10 × 5000² / 8 = 3.125 × 10⁷ N·mm (31.25 kN·m, roughly 23,000 lb·ft), also at midspan, and the maximum shear V = wL / 2 = 25,000 N (about 5,600 lb) at each support.
Working the arithmetic once by hand is worth doing — after that, the beam deflection calculator evaluates all four load cases from the same inputs, reports deflection, moment, and shear together, and plots the elastic deflected shape along the member so you can see where the maximum actually lives instead of taking the formula’s word for it. The plot comes from the full deflection equation — for the uniform simply supported case, y(x) = wx(L³ − 2Lx² + x³) / (24EI) — of which the midspan formula above is just the peak value.
Point loads hit harder than the same load spread out
A comparison the formulas make easy: keep the example beam but gather its entire distributed load into one central point load, P = wL = 50,000 N. Now δ = PL³ / (48EI) = 50,000 × 5000³ / (48 × 200,000 × 10⁸) = 6.510 mm. Same beam, same total load, 60% more deflection — and the ratio is exact, because (1/48) ÷ (5/384) = 1.6 for any beam. Concentrating load where the beam is weakest costs stiffness you can compute before you commit to it.
Support conditions cost even more. Compare a tip-loaded cantilever with a center-loaded simply supported beam of the same length and load: PL³/(3EI) against PL³/(48EI) — a factor of sixteen, purely from losing the second support. The maximums also move: a simply supported beam bends hardest at midspan, but a cantilever takes both its maximum moment and its maximum shear at the fixed end (M = PL and V = P for the tip load), while the deflection peaks at the free end. Detailing a cantilever connection lightly because “the load is small” misreads where the formula puts the demand.
Units make or break the answer
The formulas have no unit conversions hiding inside them, which means they only work in a consistent set. The SI set the worked example uses is newtons and millimetres throughout: E in MPa (which is N/mm², so it matches), I in mm⁴, L in mm, w in N/mm, P in N — and deflection then falls out in mm, moment in N·mm, shear in N. The classic blunder is entering the span in metres or the load in kN while everything else is in N and mm; a factor of 1,000 in L enters the deflection formula raised to the fourth power, and the answer isn’t subtly wrong, it’s absurd — which is at least easy to catch.
The same formulas run just as happily in a consistent imperial set: E in psi, I in in⁴, L in inches, w in lb/in, P in lb, giving deflection in inches. What you cannot do is mix — feet for the span, psi for the modulus — without converting first. Whichever set you choose, convert every input into it before touching the formula, and check the output’s plausibility against the span: a hand-width of sag on a residential span is a unit error, not a design.
What the closed forms assume — and the recap
Three assumptions ride along with the elegance. First, the formulas solve for the applied load only: the beam’s self-weight isn’t included, so where it matters, add it to the distributed load w before computing. Second, they neglect shear deformation — a fine approximation for slender beams, which is the territory these formulas are meant for. Third, they’re elastic: they describe service behavior, not what happens near failure. Deflection results are graded against span-ratio limits — the timber design calculator, for instance, checks service deflection against L/360 under the NDS or L/300 under Eurocode 5 and reports the actual span-to-deflection ratio.
The example beam, end to end: a 5,000 mm (16 ft 5 in) simply supported steel beam with E = 200,000 MPa and I = 1.0 × 10⁸ mm⁴ under w = 10 N/mm deflects 5wL⁴/(384EI) = 4.069 mm at midspan — a span ratio of 5000/4.069 ≈ L/1229, comfortably stiffer than an L/360 target — while carrying a midspan moment of 31.25 kN·m and 25 kN of shear at each support. Gathered into a central point load, the same total load pushes the deflection to 6.510 mm, exactly 1.6 times worse. Four formulas, one consistent unit set, and a habit of checking where the maximum lives: that’s beam deflection done by hand.