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Beam calculator

Add point loads, uniform loads, varying loads and moments to a beam with any supports. Get the support reactions, the shear force and bending moment diagrams, the deflection and the bending stress, with every step of the working shown.

Units: kN, m and kN/m for loads, mm for the section, 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)

Support 2 Type Position (m)

Loads

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

Point load 2Force (kN, down positive)Position (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

32.5 kNReaction at 0 m27.5 kNReaction at 8 m75.0 kN·mMax sagging moment at 3 m0 kN·mMax hogging moment at 0 m32.5 kNMax shear force at 0 m34.7 mmMax deflection downward at 3.88 m113 MPaMax bending stress2.44 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: 113 MPa bending stress Neutral axis: 2.44 MPa shear stress

Working

Beam, supports and loads
Length L = 8 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 = 8 m.
Load 1: Uniform load of 5 kN/m from x = 0 m to x = 8 m.
Load 2: point load of 20 kN at x = 3 m.
Support reactions
Take moments about the left support A (x = 0 m), clockwise positive. The right reaction R_B acts 8 m from A.
Load 1: resultant 5 × 8 = 40.0 kN acting at x = 4.00 m, so 40.0 × (4.00 − 0) = 160 kN·m
Load 2: 20 × (3 − 0) = 60.0 kN·m
R_B × 8 = 220 kN·m, so R_B = 27.5 kN.
Vertical equilibrium: R_A + R_B = total downward load = 60.0 kN, so R_A = 32.5 kN.
Check: moments about B give R_A = 32.5 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 32.5 0
3 17.5 → −2.50 75.0
8 −27.5 0
Maximum values
Maximum sagging moment: 75.0 kN·m at x = 3.00 m.
Maximum hogging moment: 0 kN·m at x = 0 m.
Largest shear force: 32.5 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.
Largest deflection: 34.7 mm downward at x = 3.88 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 = 75.0 × 10⁶ × 100 / 6.67 × 10⁷ = 113 MPa at x = 3.00 m (outer fibre).
Maximum shear stress (at the neutral axis, where the shear force is largest) = 2.44 MPa.

x (m)	Shear force V (kN)	Bending moment M (kN·m)
0	32.5	0
3	17.5 → −2.50	75.0
8	−27.5	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.
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 simply supported beam of span 8 m carries a uniform load of 5 kN/m over its whole length and a point load of 20 kN at 3 m from the left support. The section is a 100 mm by 200 mm rectangle and E = 200 GPa.

Beam, supports and loads
Length L = 8 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 = 8 m.
Load 1: Uniform load of 5 kN/m from x = 0 m to x = 8 m.
Load 2: point load of 20 kN at x = 3 m.
Support reactions
Take moments about the left support A (x = 0 m), clockwise positive. The right reaction R_B acts 8 m from A.
Load 1: resultant 5 × 8 = 40.0 kN acting at x = 4.00 m, so 40.0 × (4.00 − 0) = 160 kN·m
Load 2: 20 × (3 − 0) = 60.0 kN·m
R_B × 8 = 220 kN·m, so R_B = 27.5 kN.
Vertical equilibrium: R_A + R_B = total downward load = 60.0 kN, so R_A = 32.5 kN.
Check: moments about B give R_A = 32.5 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 32.5 0
3 17.5 → −2.50 75.0
8 −27.5 0
Maximum values
Maximum sagging moment: 75.0 kN·m at x = 3.00 m.
Maximum hogging moment: 0 kN·m at x = 0 m.
Largest shear force: 32.5 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.
Largest deflection: 34.7 mm downward at x = 3.88 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 = 75.0 × 10⁶ × 100 / 6.67 × 10⁷ = 113 MPa at x = 3.00 m (outer fibre).
Maximum shear stress (at the neutral axis, where the shear force is largest) = 2.44 MPa.

x (m)	Shear force V (kN)	Bending moment M (kN·m)
0	32.5	0
3	17.5 → −2.50	75.0
8	−27.5	0

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)

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, 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

Shear force and bending moment diagram calculator, Beam deflection calculator, Cantilever beam calculator, and the Mohr’s circle calculator for the stresses at a point on the beam.

Common questions

What kinds of beam can I calculate?

Any prismatic (constant section) beam on pin, roller or fixed supports: simply supported, overhanging, cantilever, propped cantilever, fixed at both ends, and continuous over several supports. Loads can be point loads, uniform loads, linearly varying (triangular or trapezoidal) loads and applied moments.

Where does the maximum bending moment occur?

The bending moment is largest where the shear force is zero or changes sign, which is usually under a point load or at a support. The calculator finds the exact position and also reports the largest hogging moment, the largest shear force and the points of contraflexure.

How is the bending stress found?

The bending stress is σ = M c / I, where M is the bending moment at the section, c is the distance from the neutral axis to the outer fibre and I is the second moment of area. Choose a section, or enter I and c yourself, and the calculator gives the largest stress along the beam.

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?

It is free to use, with no sign-up, no adverts and no personal tracking. The numbers you type stay in your browser.