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Beam deflection calculator

Find the maximum deflection and the deflected shape of a beam. Enter the span, supports and loads, then the section and E or EI. The deflection comes from Macaulay’s method, with the working and standard formulas to check against.

Units: kN, m and kN/m for loads, mm for the section and for support settlement, GPa for E. Beams are assumed to have a constant section. This is a learning and checking tool: check your own working and your course’s sign convention.

Start from an example Beam length L (m)

Supports

Support 1 Type Position (m) Settlement (mm, down)

Support 2 Type Position (m) Settlement (mm, down)

Loads

Uniform load 1Load (kN/m, down positive)From (m)To (m)

Section and material

What to use

Cross-sectionWidth b (mm)Depth h (mm)Young’s modulus E (GPa)

Significant figures Draw sagging moment below the beam (tension side)

Result

30.0 kNReaction at 0 m30.0 kNReaction at 6 m45.0 kN·mMax sagging moment at 3 m0 kN·mMax hogging moment at 0 m30.0 kNMax shear force at 0 m12.7 mmMax deflection downward at 3 m0.00675 radRotation at 0 m (clockwise)0.00675 radRotation at 6 m (anticlockwise)67.5 MPaMax bending stress2.25 MPaMax shear stress

Shear force, bending moment and deflection share the same horizontal scale as the beam. Loads and reactions act as drawn: downward loads are positive. Sagging moment is positive.

Send the stress to Mohr’s circle

At the critical section, the top or bottom fibre carries only bending stress. The neutral axis carries only shear stress.

Outer fibre: 67.5 MPa bending stress Neutral axis: 2.25 MPa shear stress

Working

Beam, supports and loads
Length L = 6 m. Loads are positive downward; reactions are positive upward; clockwise moments are positive.
Supports: pin or roller support at x = 0 m; pin or roller support at x = 6 m.
Load 1: Uniform load of 10 kN/m from x = 0 m to x = 6 m.
Support reactions
Take moments about the left support A (x = 0 m), clockwise positive. The right reaction R_B acts 6 m from A.
Load 1: resultant 10 × 6 = 60.0 kN acting at x = 3.00 m, so 60.0 × (3.00 − 0) = 180 kN·m
R_B × 6 = 180 kN·m, so R_B = 30.0 kN.
Vertical equilibrium: R_A + R_B = total downward load = 60.0 kN, so R_A = 30.0 kN.
Check: moments about B give R_A = 30.0 kN.
Shear force and bending moment
V is the sum of the vertical forces to the left of a section (up positive). M is the sum of the moments of those forces about the section (sagging positive). Where an arrow points, V jumps by the force; where a moment is applied, M jumps by the moment.
x (m) Shear force V (kN) Bending moment M (kN·m)
0 30.0 0
3 6.00 × 10⁻⁸ 45.0
6 −30.0 0
Maximum values
Maximum sagging moment: 45.0 kN·m at x = 3.00 m.
Maximum hogging moment: 0 kN·m at x = 0 m.
Largest shear force: 30.0 kN at x = 0 m.
There is no point of contraflexure.
A bending moment is largest where the shear force is zero or changes sign.
Deflection
Integrating M / EI twice (Macaulay’s method) and using y = 0 at every support gives the deflected shape.
EI = 13300 kN·m².
Rotation (slope) of the beam: 0.00675 rad clockwise at x = 0 m and 0.00675 rad anticlockwise at x = 6 m. A positive rotation is clockwise.
Largest deflection: 12.7 mm downward at x = 3.00 m.
Stresses
Section: Rectangle. I = 6.67 × 10⁷ mm⁴, c = 100 mm, section modulus Z = I / c = 6.67 × 10⁵ mm³.
Maximum bending stress σ = M c / I = 45.0 × 10⁶ × 100 / 6.67 × 10⁷ = 67.5 MPa at x = 3.00 m (outer fibre).
Maximum shear stress (at the neutral axis, where the shear force is largest) = 2.25 MPa.

x (m)	Shear force V (kN)	Bending moment M (kN·m)
0	30.0	0
3	6.00 × 10⁻⁸	45.0
6	−30.0	0

Practise beam questions

Press the button for a question with random numbers. Work it out on paper first, then type your answer.

How the beam calculator works

The calculator finds the support reactions, the internal forces along the beam and the deflected shape. The steps are:

Find the reactions from equilibrium: the vertical forces and the moments about a support must each add up to zero. Extra supports add compatibility conditions on the deflection, and so does any support settlement or imposed rotation.
Build the shear force V and the bending moment M along the beam, with jumps at point loads and applied moments.
Find where V is zero to locate the maximum bending moment.
Integrate M / EI twice for the slope and the deflection, using the support conditions (Macaulay’s method).
Use σ = M c / I and the shear stress formula for the chosen section.

Sign convention

Downward loads are positive and reactions act upward. A shear force is positive when the left part of the beam pushes up on the section. A bending moment is positive when the beam sags. Applied moments are positive when clockwise. Deflection is positive upward, so a sagging beam has a negative deflection.

Worked example

A beam of length 6 m is fixed at both ends. It carries a 30 kN point load 2 m from the left end. The flexural rigidity is EI = 12 000 kN·m². Find the end moments and the maximum deflection.

Beam, supports and loads
Length L = 6 m. Loads are positive downward; reactions are positive upward; clockwise moments are positive.
Supports: fixed support at x = 0 m; fixed support at x = 6 m.
Load 1: point load of 30 kN at x = 2 m.
Support reactions
This beam has more unknown reactions than the two equations of equilibrium can find, so it is statically indeterminate. The extra equations come from the shape of the beam: the deflection is zero at every support and the slope is zero at every fixed support. Solving all of these together (Macaulay’s method) gives:
Support 1 at x = 0 m: R = 22.2 kN, reaction moment 26.7 kN·m anticlockwise
Support 2 at x = 6 m: R = 7.78 kN, reaction moment 13.3 kN·m clockwise
Check: the reactions add up to 30.0 kN, and the total downward load is 30.0 kN.
Shear force and bending moment
V is the sum of the vertical forces to the left of a section (up positive). M is the sum of the moments of those forces about the section (sagging positive). Where an arrow points, V jumps by the force; where a moment is applied, M jumps by the moment.
x (m) Shear force V (kN) Bending moment M (kN·m)
0 22.2 −26.7
2 22.2 → −7.78 17.8
6 −7.78 −13.3
Maximum values
Maximum sagging moment: 17.8 kN·m at x = 2.00 m.
Maximum hogging moment: −26.7 kN·m at x = 0 m.
Largest shear force: 22.2 kN at x = 0 m.
The bending moment changes sign (point of contraflexure) at x = 1.20 m and x = 4.29 m.
A bending moment is largest where the shear force is zero or changes sign.
Deflection
Integrating M / EI twice (Macaulay’s method) and using y = 0 at every support gives the deflected shape.
EI = 12000 kN·m².
Rotation (slope) of the beam: zero at x = 0 m and zero at x = 6 m. A positive rotation is clockwise.
Largest deflection: 2.18 mm downward at x = 2.57 m.

x (m)	Shear force V (kN)	Bending moment M (kN·m)
0	22.2	−26.7
2	22.2 → −7.78	17.8
6	−7.78	−13.3

Standard beam formulas

These are the cases found in most textbooks and data sheets. The last column is the calculator’s own answer for a beam of length 4 m, so you can check it against the formulas.

Case	Reactions	Maximum bending moment	Maximum deflection	Example: moment, deflection in kN·m³
Simply supported, central point load P	P/2 at each end	PL/4 at midspan	PL³/(48EI)	10 kN·m, 13.3/EI (P = 10 kN, L = 4 m)
Simply supported, uniform load w	wL/2 at each end	wL²/8 at midspan	5wL⁴/(384EI)	10 kN·m, 16.7/EI (w = 5 kN/m, L = 4 m)
Cantilever, end point load P	P	PL at the fixed end	PL³/(3EI)	40 kN·m, 213/EI (P = 10 kN, L = 4 m)
Cantilever, uniform load w	wL	wL²/2 at the fixed end	wL⁴/(8EI)	40 kN·m, 160/EI (w = 5 kN/m, L = 4 m)
Fixed at both ends, central point load P	P/2 at each end	PL/8 at the ends and at midspan	PL³/(192EI)	5 kN·m, 3.33/EI (P = 10 kN, L = 4 m)
Fixed at both ends, uniform load w	wL/2 at each end	wL²/12 at the ends, wL²/24 at midspan	wL⁴/(384EI)	6.67 kN·m, 3.33/EI (w = 5 kN/m, L = 4 m)
Propped cantilever (fixed left, pin right), uniform load w	5wL/8 at the fixed end, 3wL/8 at the prop	wL²/8 at the fixed end	wL⁴/(185EI) approximately	10 kN·m, 6.93/EI (w = 5 kN/m, L = 4 m)

Off-centre loads, overhangs and fixed ends

Point loads away from the centre, a beam with overhangs, a fixed-fixed beam and a propped cantilever. W is the load, a and b are its distances from the left and right ends, and L is the span.

Case	Reactions	Bending moment	Deflection	Calculator (W = 10 kN, a = 1 m, L = 4 m)
Simply supported, point load W at a from the left support and b from the right	W·b/L at the left, W·a/L at the right	W·a·b/L under the load	W·a²·b²/(3EI·L) under the load	reactions 7.5 and 2.5 kN, moment 7.5 kN·m, deflection 7.5/EI
Simply supported with equal overhangs a, a load W on each end	W at each support	W·a, hogging, constant between the supports	W·a·L²/(8EI) upward at the centre, L being the span between the supports	reactions 10 kN, moment 10 kN·m, centre deflection 20/EI upward
Fixed at both ends, point load W at a from the left end and b from the right	W·b²(3a + b)/L³ at the left, W·a²(a + 3b)/L³ at the right	Hogging W·a·b²/L² at the left end and W·a²·b/L² at the right end; sagging 2W·a²·b²/L³ under the load	W·a³·b³/(3EI·L³) under the load	reactions 8.44 and 1.56 kN, end moments 5.63 and 1.88 kN·m, moment under the load 2.81 kN·m, deflection 1.41/EI
Propped cantilever (fixed left, pin right), point load W at a from the fixed end	Pin: W·a²(3L − a)/(2L³). Fixed end: W minus that	Fixed-end moment W·a·b(L + b)/(2L²)	W·a³·b²(3L + b)/(12EI·L³) under the load	pin reaction 0.859 kN, fixed-end moment 6.56 kN·m, deflection 1.76/EI
Cantilever, point load W at a from the fixed end	W at the fixed end	W·a at the fixed end, zero beyond the load	W·a²(3L − a)/(6EI) at the free end	reaction 10 kN, moment 10 kN·m, tip deflection 18.3/EI

Support settlement and imposed rotation

A support that sinks, or a fixed end that is turned, creates forces without any load. These are the stiffness results behind the slope-deflection and stiffness methods. Give a support a settlement or a fixed support a rotation, and enter EI.

Case	Force or reaction	Moment	Calculator (EI = 12 000 kN·m², L = 4 m)
Propped cantilever, the prop settles by Δ	3EI·Δ/L³	3EI·Δ/L² at the fixed end	Δ = 10 mm: reaction 5.63 kN, moment 22.5 kN·m
Fixed at both ends, one end settles by Δ	12EI·Δ/L³	6EI·Δ/L² at each end	Δ = 10 mm: reaction 22.5 kN, moment 45 kN·m
Propped cantilever, the fixed end is rotated by θ	3EI·θ/L² at the prop	3EI·θ/L at the rotated end	θ = 0.01 rad: reaction 22.5 kN, moment 90 kN·m
Fixed at both ends, one end is rotated by θ	6EI·θ/L² at each end	4EI·θ/L at the rotated end, 2EI·θ/L at the far end	θ = 0.01 rad: reaction 45 kN, moments 120 and 60 kN·m
End couple M on a pin end: the rotation of that end	Far end pinned: θ = M·L/(3EI), stiffness 3EI/L	Far end fixed: θ = M·L/(4EI), stiffness 4EI/L	M = 10 kN·m: 0.00111 rad (far end pinned), 8.33 × 10⁻⁴ rad (far end fixed)

Which formulas does this use?

Simply supported beam: uniform load, Simply supported beam: central point load, Cantilever with a tip load, Bending stress in a rectangular beam, Bending stress, Second moment of area, rectangle, Deflection beneath an eccentric point load, Cantilever tip displacement from an interior point load, Fixed-ended beam midspan deflection under uniform load, Prop reaction of a cantilever under full-span uniform load, Interior support moment of a two-span continuous beam.

Related calculators

Beam calculator, Cantilever beam calculator, Shear force and bending moment diagram calculator, the Mohr’s circle calculator for the stresses at a point on the beam, and the shaft torsion calculator for shafts that twist.

Common questions

How do I find the rotation at the end of a beam?

Enter EI. The results then list the rotation of the beam at each end in radians, clockwise positive. For a simply supported beam with a moment M at one end, the rotation of that end is ML/(3EI), and ML/(4EI) when the far end is fixed. This is where the stiffness factors 3EI/L and 4EI/L come from.

How is the beam deflection found?

The calculator uses Macaulay’s method. It writes the bending moment along the beam, integrates M / EI once to get the slope and again to get the deflection, and fixes the constants from the supports: the deflection is zero at every support and the slope is zero at a fixed end.

What do I enter for EI?

EI is the flexural rigidity: Young’s modulus E times the second moment of area I. Choose a section and E and the calculator works it out, or type EI in kN·m² if you have it. If you give neither, the deflection is shown as a multiple of 1/EI.

What are the standard deflection formulas?

Simply supported with a central point load: PL³/(48EI). Simply supported with a uniform load: 5wL⁴/(384EI). Cantilever with an end load: PL³/(3EI). Cantilever with a uniform load: wL⁴/(8EI). The table on this page checks them against the calculator.

How do you find the reactions of a beam?

Take moments about one support to find the other reaction, then use vertical equilibrium for the remaining one. For a distributed load, replace it by its resultant acting at the centroid of the load. A cantilever has one fixed support, which supplies a vertical reaction and a reaction moment. A beam with extra supports is statically indeterminate, and the extra equations come from the deflection being zero at each support.

What is the sign convention used here?

Downward loads are positive and reactions act upward. A shear force is positive when the left part of the beam pushes up on the section. A bending moment is positive when it makes the beam sag, so the top is in compression. Deflection is positive upward, so a sagging beam has a negative deflection. Check this against your own course notes before comparing answers.

Is the calculator free, and does it keep my numbers?

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