Beam Deflection Calculator

Select beam type and loading to see deflection curve, bending moment, and shear force diagrams.

Worked example — Beam Deflection Calculator

R_A	5 kN
R_B	5 kN
M_max	12.5 kN·m
δ_max	15.631 mm

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Beam Deflection Calculator

A beam’s stiffness is not a property of the beam. Load the same plank the same way and the answer changes by a factor of nearly fifty depending on nothing but how its ends are held — which is why a bookshelf sags in the middle and a balcony droops at its tip. Boundary conditions is the formal name for how the ends are held, and choosing them is most of what structural design is.

What each symbol means

EI: flexural rigidity — the material’s stiffness times the section’s second moment of area. This is the beam’s own contribution, and the only part of the answer the supports do not touch.
y′: the slope of the deflected shape. Whether a support permits it is the entire difference between a pin and a clamp, and that one distinction generates most of the dropdown.
δ_max: the largest deflection anywhere along the span — the headline number, and the one a floor’s occupants actually feel underfoot.
L/δ: span divided by peak deflection. Codes are written in this ratio rather than in millimetres, because a 10 mm sag is invisible over 12 m and alarming over 1 m.

Where the formula comes from

Write the governing equation once, because there is only one. For a beam carrying a distributed load w(x), EI·d⁴y/dx⁴ = w(x). Notice what is not in it: nothing about pins, clamps, props or overhangs. Integrate four times to recover y(x) and four constants of integration appear. Everything the support dropdown does is fix those four numbers.
So ask what a support is actually able to do, and the list is short. It can stop the end moving, stop it rotating, both, or neither. A pin holds the end in place but lets it turn freely: y = 0 with zero moment. A clamp holds it and refuses the rotation: y = 0 and y′ = 0. A free end withholds nothing — zero moment and zero shear. Every one of the eight entries in the dropdown is a pair drawn from that vocabulary, and nothing else.
Now price it. Leave everything else alone — the 1000 N/m spread over 5 m of steel this page opens with — and change only the ends. On two pins the beam sags 4.885 mm. Clamp both ends and the identical beam sags 0.977 mm: five times stiffer, exactly, because 5wL⁴/384EI has become wL⁴/384EI. Take one support away altogether and the cantilever drops 46.894 mm, forty-eight times the clamped figure. Nothing was added to the beam at any point.
Deflection is what you notice; moment is what breaks. A simply supported beam carries zero moment at its ends and wL²/8 at midspan — the page prints 3.125 kN·m. Clamp the ends and they become able to carry moment themselves, so the peak falls to wL²/12, 2.083 kN·m, and relocates from midspan out to the supports. The beam did not become stronger. The demand was redistributed onto material that had been doing nothing.
Which leaves the obvious question: why not clamp everything? Count the unknowns. Statics offers three equations, so a beam whose supports supply exactly three reactions is determinate — the reactions follow from equilibrium alone and nothing else can change them. A fourth reaction makes it indeterminate: stiffer, shallower, lighter, and from that moment on the beam has opinions about being moved. That is the bargain the propped case strikes, and the same bargain a continuous bridge deck strikes over every pier it crosses.

How to read what you see

Three panels stacked in causal order — the loaded beam with its supports drawn as they actually behave, then shear, then moment — and switching the support type redraws all three at once, which is the quickest way to see that the load never changed. Underneath sit the reactions, the peak moment, the peak deflection and a span-to-deflection ratio. That last line is the one a designer reads: a common serviceability limit is L/250, the simply supported case here clears it about fourfold, and the cantilever at roughly L/107 is the one configuration on this page that fails it.

Assumes: Euler–Bernoulli beam theory — plane sections stay plane, so shear deformation is ignored; deflections small enough that curvature can be written as y″; a linear-elastic material well below yield; and a section that does not change along the span. Above all, for this lesson: perfect supports. A pin with no friction, and a clamp with infinite rotational stiffness.
Breaks when: That last assumption is where the model parts company with the world, and the indeterminate cases are the ones that notice. A clamped end is only as fixed as whatever it is clamped to; real connections have finite rotational stiffness and land somewhere between the 4.885 mm and the 0.977 mm this page offers as separate choices. The deeper gap is that this page solves for a given load — and an indeterminate beam can be stressed by a given displacement, with no load at all. Let one support of the clamped beam settle three millimetres, or let both ends stay bolted while the steel warms in the afternoon sun, and moment and axial force appear that equilibrium cannot predict and that nothing on this page will draw. The determinate beam on two pins simply shrugs and moves. This is why a bridge deck sits on sliding bearings and meets the abutment through a toothed expansion joint: the engineer is buying back the movement that fixity took away, and paying for it in deflection.

Plane sections stay plane, until the beam gets short 🖖

Euler-Bernoulli beam theory assumes plane sections remain plane after bending. This holds well when beam length ≫ depth. Shear deformation, which the theory ignores, becomes significant in deep or short beams — for those, Timoshenko beam theory is used instead.

Three diagrams, one chain of cause and effect 🖖

The tool stacks three graphs because they are linked steps in a single story: the load creates a shear force, the shear builds up into a bending moment, and the moment curves the beam into its deflected shape. Each diagram is essentially the running total of the one above it. A handy consequence — the bending moment peaks exactly where the shear force crosses zero, and that is where the beam works hardest.

Galileo got the first beam problem wrong 🖖

Galileo posed the cantilever-strength problem in his 1638 Two New Sciences, the founding text of beam analysis — but he assumed every fibre of the cross-section pulled equally in tension. He missed that fibres on one side stretch while the other side compresses, pivoting about a neutral axis. His model overestimated bending strength by roughly a factor of 3, and it took nearly a century for Parent, and later Euler and Bernoulli, to correct it.

Problems solved in full

A 5 m simply supported steel beam carrying 10 kN at midspan 5 steps
A 5 m steel beam, simply supported, carrying 10 kN at midspan. Find the reactions, the bending moment and the sag — then check it against the limit that actually governs the design.
1. Symmetry does the statics. The load sits at midspan, so each support carries half of it, and no equation is needed beyond that observation.
  
  $\sum M = 0 ⟹ R_{A} = R_{B} = \frac{P}{2} = 5 kN$
2. The maximum moment is at the load, where each reaction has acted over half the span. PL/4 is worth memorising: it is the simply-supported point-load case that every other case gets compared against.
  
  $M_{m a x} = R_{A} \cdot \frac{L}{2} = \frac{P L}{4} = \frac{10 \times 5}{4} = 12.5 kN m$
3. Deflection needs the beam's stiffness, not just the load. Two properties enter — the material through E and the shape through I — and they enter only as their product, which is why a stiffer alloy and a deeper section are interchangeable here.
  
  $δ_{m a x} = \frac{P L ^{3}}{48 E I} = \frac{1 0 ^{4} \times 5 ^{3}}{48 \times 200 \times 1 0 ^{9} \times 8.33 \times 1 0 ^{- 6}}$
4. Substituting gives 15.631 mm. Note where the span sits: cubed. Deflection is far more sensitive to length than to anything else in the expression.
  
  $= \frac{1.25 \times 1 0 ^{6}}{7.997 \times 1 0 ^{7}} = 0.015631 m = 15.631 mm$
5. Now the check that decides the design. Serviceability limits are commonly quoted as span over 360, which for 5 m is 13.89 mm.
  
  $\frac{L}{360} = \frac{5000}{360} = 13.89 mm < 15.631 mm$
Answer

The tool prints 5 kN at each support, 12.5 kN·m of moment, and 15.631 mm of sag. That last figure fails the L/360 serviceability limit of 13.89 mm — while the beam is nowhere near yielding. This is the thing structural courses lead with and intuition resists: beams are usually sized by how much they move, not by how close they are to breaking. The cube on the span is why. Take the same beam to 10 m and the sag does not double, it goes up eightfold to 125 mm, and no amount of strength margin helps.

$δ = \frac{P L ^{3}}{48 E I} = 15.631 mm > \frac{L}{360} = 13.89 mm$
A beam sagging 15.631 mm against a serviceability limit of 13.89 mm 6 steps
The panel says this beam sags 15.631 mm and that the serviceability limit L/360 is 13.89 mm — over by an eighth. The question a builder asks next is how much bigger the beam has to be, and the answer is far smaller than the overshoot suggests.
1. Take the ratio of the two numbers already on the panel. The beam is 1.1254 times more flexible than the limit allows.
  
  $\frac{δ _{m a x}}{L /360} = \frac{15.631}{13.89} = 1.1254$
2. Deflection is inversely proportional to the second moment of area, and for a rectangle that quantity carries the depth cubed while the width enters only once. The two dimensions are not interchangeable.
  
  $I = \frac{b d ^{3}}{12} ⟹ δ \propto \frac{1}{I} \propto d^{- 3}$
3. That consequence is worth stating on its own before using it: double the depth and the sag falls by a factor of eight, not two.
  
  $d \to 2 d : δ \to \frac{δ}{2 ^{3}} = \frac{δ}{8}$
4. So invert the cube. Dividing the deflection by 1.1254 needs the depth multiplied by the cube root of the same number.
  
  $(\frac{d}{d ^{'}})^{3} = \frac{1}{1.1254} ⟹ \frac{d ^{'}}{d} = 1.125 4^{1/3} = 1.0402$
5. Four per cent. A 300 mm joist becomes 312 mm, and the beam that failed passes.
6. Now price the alternative. Fixing it with width instead means the whole 1.1254 — three times the extra material for the same millimetres saved.
  
  $\frac{d ^{'}}{d} - 1 = 0.0402, \frac{b ^{'}}{b} - 1 = 0.1254$
Answer

A 4% deeper beam fixes a 12.5% overshoot; a 12.5% wider one does the same job for three times the added material. That is the entire reason floor joists are tall rectangles standing on edge rather than squares or planks laid flat: a millimetre of depth is worth three of width, and the cube is why the same board feels like a diving board one way up and like a floor the other.

$δ \propto d^{- 3} : δ \times 1.1254 ⟺ d \times 1.0402 ⟺ b \times 1.1254$

References (2)

The lesson — where the support-case coefficients (5/384, 1/384, 1/8) are tabulated: W. C. Young, R. G. Budynas and A. M. Sadegh, Roark's Formulas for Stress and Strain, 8th edition, ch. 8 — beams; flexure of straight bars. McGraw-Hill, 2012. ISBN 978-0-07-174247-4.
The shear correction this tool's first block says Euler–Bernoulli leaves out: S. P. Timoshenko, "On the correction for shear of the differential equation for transverse vibrations of prismatic bars." Philosophical Magazine Series 6, 41, 744–746, 1921.

By the 42Math team · Updated 2026-07-18

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Midpoint load - A point load at mid-span costs 15.631 mm of sag and 12.5 kN·m of moment, each support carrying 5 kN.
Cantilever tip - Free at one end, the tip drops 27.011 mm — a span-to-deflection ratio of L/111.07.
UDL bridge - Spread out as a distributed load instead, the beam sags 78.156 mm, the worst of these four presets.
Off-centre load - Off-centre is gentler than mid-span: 13.437 mm, with the reactions splitting 6.67 kN against 3.33 kN.