FERS

Cantilever Beam Calculator

Calculate the reactions, shear force, bending moment, deflection and slope of a cantilever beam under any mix of point loads, distributed loads and moments — in SI or US units, with the full hand calculation written out. Every result is solved exactly and checked live against the FERS finite-element solver.

FERS Cloud engineering · last reviewed · how it is verified

Beam

E = 210.0 GPa

I = 3,893 cm⁴ · W_el = 324.4 cm³ · A = 39.13 cm²

Bending axis

Codes differ: several compare a cantilever's tip deflection with twice its length. Choose the ratio your project uses.

Advanced

For a connection or support that rotates under moment. The root turns by θ₀ = M/k_φ and the tip drops a further θ₀·L. Leave it off for a fully fixed end.

Loads

Positions from the fixed end · downward loads and clockwise moments are positive
  • 1. Point load

Results

Reaction at the fixed end, R
5.00 kN↑
Moment at the fixed end, M
12.5 kN·mhogging
Maximum shear |V|
5.00 kNat x = 0 m
Maximum moment |M|
12.5 kN·mat x = 0 m
Tip deflection δ
3.19 mmL/785
Tip rotation θ
0.00191 rad0.110°
Max bending stress σ = M/W
38.5 MPaσ/f_y = 0.164
Plastic moment M_pl = W_pl·f_y
86.2 kN·mM/M_pl = 0.145
W_el needed for σ ≤ f_y
≥ 53.2 cm³

Euler–Bernoulli beam theory: linear elastic, small deflections, shear deformation neglected. Lateral-torsional buckling, local buckling and connections are not checked. σ/f_y is an elastic stress ratio, not a design check. Check a steel beam to Eurocode 3 →

Checked by the FERS finite-element solver

✓ matches the closed form
QuantityClosed formFERS FEDifference
Reaction at the fixed end5.000 kN5.000 kN< 0.0001 %
Moment at the fixed end12.50 kN·m12.50 kN·m< 0.0001 %
Tip deflection3.185 mm3.185 mm< 0.0001 %
Tip rotation0.001911 rad0.001911 rad< 0.0001 %
Maximum deflection3.185 mm3.185 mm< 0.0001 %
Moment, worst FE station-2.500 kN·m-2.500 kN·m< 0.0001 %
Shear, worst FE station5.000 kN5.000 kN< 0.0001 %
Deflection, worst FE station3.185 mm3.185 mm< 0.0001 %

Precomputed with FERS 0.2.68 for this example; edit any input to re-run it live in your browser.

Downloadevery station, all four diagrams

Lighter sections that meet your limits

Screens a family of catalogue sections against your deflection limit and σ ≤ f_y, each with its own self-weight, and lists the lightest that meet both.

Diagrams

P = 5 kN @ 2.5 mR = 5 kNM = 12.5 kN·mL = 2.5 m
Draw moments
Bending moment M (kN·m) · sagging positive
-10-50012x (m)M = -12.5 kN·m

Hover or tap a diagram — or type x — to read every value at that point.

Step-by-step solution

Each load is solved on its own with the standard cantilever formula, and the results are added (superposition). Hand-calculation units: N and mm, so E is in N/mm² and w in N/mm.

  1. 1. Point load at the free end: P = 5 kN at a = 2.5 m

    Reaction R
    R = P = 5 kN = 5.00 kN
    Root moment M
    M = P·L = 5 kN × 2.5 m = 12.5 kN·m
    Tip rotation θ
    θ = P·L² / (2·E·I) = 5,000 N × (2,500 mm)² / (2 × 210,000 N/mm² × 3.893×10⁷ mm⁴) = 0.00191 rad
    Tip deflection δ
    δ = P·L³ / (3·E·I) = 5,000 N × (2,500 mm)³ / (3 × 210,000 N/mm² × 3.893×10⁷ mm⁴) = 3.19 mm

Add the rows (superposition)

Σ Reaction R
5.00 kN
Σ Root moment M
12.5 kN·m
Σ Tip rotation θ
0.00191 rad
Σ Tip deflection δ
3.19 mm
Bending stress
σ = M_max / W_el = 1.25×10⁷ N·mm / 324,420 mm³ = 38.5 MPa
Stress ratio
σ / f_y = 38.5 MPa / 235 MPa = 0.164
Span / deflection
L / δ_max = 2,500 mm / 3.185 mm = L/785

Cantilever beam formulas

The standard cases for a cantilever of length L, fixed at x = 0 and free at x = L, with bending stiffness E·I. R is the reaction and M_max the moment at the fixed end; θ_max and δ_max are the rotation and deflection of the free end. Each formula is checked against the calculator's exact solver in our test suite.

Load caseReaction and momentFree-end rotation and deflection

Point load at the free end

A single load P at the tip — the classic cantilever.

R = P
M_max = P·L
θ_max = P·L² / (2·E·I)
δ_max = P·L³ / (3·E·I)
δ(x) = P·x²·(3L − x) / (6·E·I)

Point load at distance a

P at a from the fixed end; the beam beyond the load stays straight.

R = P
M_max = P·a
θ_max = P·a² / (2·E·I)
δ_max = P·a²·(3L − a) / (6·E·I)
x ≤ a: P·x²·(3a − x) / (6·E·I); x ≥ a: P·a²·(3x − a) / (6·E·I)

Uniformly distributed load

w over the full length.

R = w·L
M_max = w·L² / 2
θ_max = w·L³ / (6·E·I)
δ_max = w·L⁴ / (8·E·I)
δ(x) = w·x²·(6L² − 4L·x + x²) / (24·E·I)

Partial uniform load

w from a to b, measured from the fixed end.

R = w·(b − a)
M_max = w·(b² − a²) / 2
θ_max = w·(b³ − a³) / (6·E·I)
δ_max = w·[4L·(b³ − a³) − (b⁴ − a⁴)] / (24·E·I)

Triangular load, maximum at the fixed end

w at the support falling to zero at the tip.

R = w·L / 2
M_max = w·L² / 6
θ_max = w·L³ / (24·E·I)
δ_max = w·L⁴ / (30·E·I)
δ(x) = w·x²·(10L³ − 10L²·x + 5L·x² − x³) / (120·L·E·I)

Triangular load, maximum at the free end

Zero at the support rising to w at the tip.

R = w·L / 2
M_max = w·L² / 3
θ_max = w·L³ / (8·E·I)
δ_max = 11·w·L⁴ / (120·E·I)
δ(x) = w·x²·(20L³ − 10L²·x + x³) / (120·L·E·I)

Trapezoidal load

w₁ at the support varying linearly to w₂ at the tip.

R = (w₁ + w₂)·L / 2
M_max = L²·(w₁ + 2w₂) / 6
θ_max = L³·(w₁ + 3w₂) / (24·E·I)
δ_max = L⁴·(4w₁ + 11w₂) / (120·E·I)

Moment at the free end

A couple M at the tip bends the beam into a circular arc.

R = 0
M_max = M (constant)
θ_max = M·L / (E·I)
δ_max = M·L² / (2·E·I)
δ(x) = M·x² / (2·E·I)

Moment at distance a

A couple M at a from the fixed end.

R = 0
M_max = M
θ_max = M·a / (E·I)
δ_max = M·a·(2L − a) / (2·E·I)

P is a point load, w a distributed load (w₁ and w₂ at the two ends of a varying one), M an applied moment, a and b distances from the fixed end, E Young's modulus and I the second moment of area. Downward loads and clockwise moments are positive; for several loads, add the rows.

How to calculate the deflection of a cantilever beam

1. Reactions from equilibrium

A cantilever is fixed at one end and free at the other, so everything it carries goes back to the root. The vertical reaction R equals the sum of the loads, and the root moment M equals the sum of each load times its distance from the fixed end (plus any applied moments).

2. Shear and moment along the beam

Cut the beam at any x and look at the part beyond the cut: the shear V(x) is the sum of the loads on it, and the bending moment M(x) is the sum of their moments about the cut. For loads acting downward the moment is hogging along the whole length — tension on the top face — and largest at the root.

3. Slope and deflection by integration

Integrate E·I·v″ = M(x) twice, with v = 0 and v′ = 0 at the fixed end. Point loads and moments make the moment piecewise, which Macaulay's singularity functions handle in one expression; this calculator solves that expression exactly for any number of loads.

4. Superposition

Because the beam is linear elastic, the effect of several loads is the sum of their separate effects. That is how the step-by-step solution is built: one standard formula per load, then a sum — a second, independent route to the same answer as the exact solution.

Sign convention

x is measured from the fixed end. Downward loads and clockwise moments are positive; deflection is positive downward and rotation positive clockwise. Bending moment is positive when sagging, so a cantilever's root moment under downward loads is negative (hogging). Moment diagrams can be drawn on the tension side, as is usual in Europe, or with sagging up, as in most US textbooks.

Worked examples

Both are computed by the same solver as the calculator when this page is built, so the numbers below are exactly what the tool gives for these inputs.

Example 1 — combined load on a steel cantilever (SI)

A 3 m IPE 240 cantilever in S235 carries a uniformly distributed load of 4 kN/m over its full length, a point load of 6 kN at the free end, and its own weight. The deflection is checked against 2L/250.

Reaction R
18.9 kN
Root moment M
37.4 kN·m
Tip deflection δ
11.9 mm
Span / deflection
2L/503
Bending stress σ
115 MPa
Limit 2L/250
24.0 mm ✓
Show the hand calculation

Step-by-step solution

Each load is solved on its own with the standard cantilever formula, and the results are added (superposition). Hand-calculation units: N and mm, so E is in N/mm² and w in N/mm.

Self-weight load
w = ρ·A·g = 7,850 kg/m³ × 39.13 cm² × 9.81 m/s² = 0.301 kN/m

  1. 1. Uniformly distributed load: w = 4 kN/m from 0 m to 3 m

    Reaction R
    R = w·L = 4 kN/m × 3 m = 12.0 kN
    Root moment M
    M = w·L² / 2 = 4 kN/m × (3 m)² / 2 = 18.0 kN·m
    Tip rotation θ
    θ = w·L³ / (6·E·I) = 4 N/mm × (3,000 mm)³ / (6 × 210,000 N/mm² × 3.893×10⁷ mm⁴) = 0.00220 rad
    Tip deflection δ
    δ = w·L⁴ / (8·E·I) = 4 N/mm × (3,000 mm)⁴ / (8 × 210,000 N/mm² × 3.893×10⁷ mm⁴) = 4.95 mm

  2. 2. Point load at the free end: P = 6 kN at a = 3 m

    Reaction R
    R = P = 6 kN = 6.00 kN
    Root moment M
    M = P·L = 6 kN × 3 m = 18.0 kN·m
    Tip rotation θ
    θ = P·L² / (2·E·I) = 6,000 N × (3,000 mm)² / (2 × 210,000 N/mm² × 3.893×10⁷ mm⁴) = 0.00330 rad
    Tip deflection δ
    δ = P·L³ / (3·E·I) = 6,000 N × (3,000 mm)³ / (3 × 210,000 N/mm² × 3.893×10⁷ mm⁴) = 6.61 mm

  3. 3. Self-weight: w = 0.3013 kN/m from 0 m to 3 m

    Reaction R
    R = w·L = 0.3013 kN/m × 3 m = 0.904 kN
    Root moment M
    M = w·L² / 2 = 0.3013 kN/m × (3 m)² / 2 = 1.36 kN·m
    Tip rotation θ
    θ = w·L³ / (6·E·I) = 0.3013 N/mm × (3,000 mm)³ / (6 × 210,000 N/mm² × 3.893×10⁷ mm⁴) = 0.000166 rad
    Tip deflection δ
    δ = w·L⁴ / (8·E·I) = 0.3013 N/mm × (3,000 mm)⁴ / (8 × 210,000 N/mm² × 3.893×10⁷ mm⁴) = 0.373 mm

Add the rows (superposition)

Σ Reaction R
18.9 kN
Σ Root moment M
37.4 kN·m
Σ Tip rotation θ
0.00567 rad
Σ Tip deflection δ
11.9 mm
Bending stress
σ = M_max / W_el = 3.736×10⁷ N·mm / 324,420 mm³ = 115 MPa
Stress ratio
σ / f_y = 115 MPa / 235 MPa = 0.490
Span / deflection
2L / δ_max = 6,000 mm / 11.93 mm = 2L/503
Allowed deflection
δ_allow = 2L / 250 = 6,000 mm / 250 = 24.0 mm

Example 2 — partial load and a tip moment (US)

A 10 ft W10×22 cantilever in ASTM A992 carries 0.6 kip/ft from 4 ft to the tip and a clockwise moment of 3 kip·ft at the free end. The deflection is checked against L/180.

Reaction R
3.60 kip
Root moment M
28.2 kip·ft
Tip deflection δ
0.422 in
Span / deflection
L/284
Bending stress σ
14.5 ksi
Limit L/180
0.667 in ✓
Show the hand calculation

Step-by-step solution

Each load is solved on its own with the standard cantilever formula, and the results are added (superposition). Hand-calculation units: inches, with E in ksi (psi for lb loads).

  1. 1. Partial uniform load: w = 0.6 kip/ft from 4 ft to 10 ft

    Reaction R
    R = w·(b − a) = 0.6 kip/ft × (10 ft − 4 ft) = 3.60 kip
    Root moment M
    M = w·(b² − a²) / 2 = 0.6 kip/ft × ((10 ft)² − (4 ft)²) / 2 = 25.2 kip·ft
    Tip rotation θ
    θ = w·(b³ − a³) / (6·E·I) = 0.05 kip/in × ((120 in)³ − (48 in)³) / (6 × 29,000 ksi × 119 in⁴) = 0.00391 rad
    Tip deflection δ
    δ = w·[4L·(b³ − a³) − (b⁴ − a⁴)] / (24·E·I) = 0.05 kip/in × [4 × 120 in × ((120 in)³ − (48 in)³) − ((120 in)⁴ − (48 in)⁴)] / (24 × 29,000 ksi × 119 in⁴) = 0.347 in

  2. 2. Applied moment at the free end: M = 3 kip·ft at a = 10 ft

    Reaction R
    R = 0 = 0 = 0 kip
    Root moment M
    M = M = 3 kip·ft = 3.00 kip·ft
    Tip rotation θ
    θ = M·L / (E·I) = 36 kip·in × 120 in / (29,000 ksi × 119 in⁴) = 0.00125 rad
    Tip deflection δ
    δ = M·L² / (2·E·I) = 36 kip·in × (120 in)² / (2 × 29,000 ksi × 119 in⁴) = 0.0751 in

Add the rows (superposition)

Σ Reaction R
3.60 kip
Σ Root moment M
28.2 kip·ft
Σ Tip rotation θ
0.00516 rad
Σ Tip deflection δ
0.422 in
Bending stress
σ = M_max / W_el = 338.4 kip·in / 23.37 in³ = 14.5 ksi
Stress ratio
σ / f_y = 14.5 ksi / 50 ksi = 0.290
Span / deflection
L / δ_max = 120 in / 0.4218 in = L/284
Allowed deflection
δ_allow = L / 180 = 120 in / 180 = 0.667 in

Deflection limits for cantilevers

Deflection limits are set by codes and project specifications, not by beam theory. They are usually written as a span-to-deflection ratio — L/180, L/250 or L/360 are common — and sometimes as an absolute value.

For a cantilever the question is which span to use. Several codes take the span of a cantilever as twice its length, which is equivalent to comparing the tip deflection with 2L/ratio. Others use the cantilever length directly. The calculator offers both, and shows the ratio of your beam either way.

Deflection limits protect finishes, drainage, appearance and the feel of a structure. They do not replace a strength check: bending, shear and stability are verified separately — for steel members, for example, to EN 1993-1-1 or AISC 360.

Verified against finite-element analysis

Every case FERS's own beam-equation test suite checks for a cantilever, plus two combined cases, solved two ways: by the closed form and by the FERS finite-element engine. 5 m beam, E = 210 GPa, I = 13.21×10⁻⁶ m⁴. The difference column is the largest over the reactions, the tip and every finite-element station.

CaseR (kN) closed form / FEM (kN·m) closed form / FETip δ closed form / FEDifference
001 · 1 kN at the free end1.000 / 1.0005.000 / 5.00015.02 mm / 15.02 mm< 0.0001 %
002 · 1 kN at a = 3 m1.000 / 1.0003.000 / 3.0006.489 mm / 6.489 mm< 0.0001 %
003 · 1 kN/m over the full length5.000 / 5.00012.50 / 12.5028.16 mm / 28.16 mm< 0.0001 %
004 · 1 kN/m from 2.0 to 3.5 m1.500 / 1.5004.125 / 4.1258.463 mm / 8.463 mm< 0.0001 %
005 · triangle, 1 kN/m at the fixed end2.500 / 2.5004.167 / 4.1677.510 mm / 7.510 mm< 0.0001 %
005 · triangle, 1 kN/m at the free end2.500 / 2.5008.333 / 8.33320.65 mm / 20.65 mm< 0.0001 %
006 · triangle 1 → 0 kN/m over 0–3 m1.500 / 1.5001.500 / 1.5001.784 mm / 1.784 mm< 0.0001 %
006 · triangle 0 → 1 kN/m over 1.5–3.5 m1.000 / 1.0002.833 / 2.8335.958 mm / 5.958 mm< 0.0001 %
007 · 0.5 kN·m at the free end0 / 0.0000000000000.5000 / 0.50002.253 mm / 2.253 mm< 0.0001 %
Combined · 2 kN/m + 5 kN at the tip + 1 kN·m at 2 m15.00 / 15.0051.00 / 51.00134.3 mm / 134.3 mm< 0.0001 %
Combined · 3 → 1 kN/m, 2 kN up at 1.5 m8.000 / 8.00017.83 / 17.8339.53 mm / 39.53 mm< 0.0001 %

The same engine is validated against the NAFEMS benchmarks it can represent — see the NAFEMS validation and the benchmark suite.

Frequently asked questions

How do you calculate the deflection of a cantilever beam?

Integrate the bending moment twice: E·I·v″ = M(x), with zero deflection and zero slope at the fixed end. For a point load P at the free end this gives δ = P·L³/(3·E·I); for a uniform load w it gives δ = w·L⁴/(8·E·I). With several loads, add the deflection of each load on its own (superposition) — the step-by-step solution above does exactly that with your numbers.

Where are the maximum bending moment and the maximum deflection on a cantilever?

When all loads act in the same direction, the maximum bending moment is at the fixed end (a hogging moment: tension on top) and the maximum deflection is at the free end. With loads in opposite directions either can move along the beam; the calculator finds the true maximum and reports where it occurs.

What is the maximum moment of a cantilever with a uniformly distributed load?

M = w·L²/2 at the fixed end, with a reaction R = w·L. A cantilever carries its whole load back to one support, so its root moment is four times that of a simply supported beam of the same length and load (w·L²/8).

Why does a cantilever deflect so much more than a simply supported beam?

Under a point load at the end, a cantilever deflects P·L³/(3·E·I); the same load at midspan of a simply supported beam of the same length deflects P·L³/(48·E·I) — sixteen times less. The cantilever has one support to hold both the load and the rotation, so all of its curvature adds up towards the tip.

How does the length of a cantilever affect its deflection?

Strongly. Tip deflection grows with L³ under an end load and with L⁴ under a distributed load, so doubling the length multiplies the deflection by 8 or 16. Stiffness grows only linearly with E and I, which is why cantilever length usually governs.

What is a typical deflection limit for a cantilever?

It depends on the code and the use. Many codes express limits as a span/deflection ratio such as L/180 or L/250, and several take the span of a cantilever as TWICE its length, which doubles the allowed tip deflection. The calculator lets you pick the ratio and whether it applies to L or 2L; it does not choose one for you.

How far can a beam cantilever?

There is no single answer: it is set by whichever governs first — the bending stress at the root, the tip deflection your project allows, vibration, or the connection that has to carry the root moment. To find it for a given section, increase L here until the stress or the deflection reaches your limit.

Is a cantilever beam statically determinate?

Yes. The fixed support provides exactly three reactions — a horizontal force, a vertical force and a moment — and a planar beam has three equilibrium equations, so the reactions follow from statics alone. A propped cantilever, with an extra support at the free end, is statically indeterminate.

Does the calculator include self-weight and shear deformation?

Self-weight is optional: tick the box and it is added as a uniform load ρ·A·g from the material density and section area. Shear deformation is neglected, as in Euler–Bernoulli theory. At a span-to-depth ratio of 10 it would add under 1% to the deflection of a solid rectangle and about 3% to that of an I-beam, whose thin web carries the shear; it grows quickly for shorter, deeper cantilevers.

How accurate is it?

The closed-form solution is exact for Euler–Bernoulli beam theory. It is then checked against the FERS finite-element solver — the same engine as the full FERS app, running in your browser — and the panel shows the difference for every quantity, typically below one part in a billion.

Can I export the results to Excel?

Yes. The CSV download lists the shear, bending moment, rotation and deflection at every station along the beam, with both sides of each jump, and opens directly in Excel, Google Sheets or Numbers. The diagrams download as SVG or PNG for a report or a drawing.

How do I find the section size a cantilever needs?

Set a deflection limit and choose a material with a yield strength. The results then show the second moment of area your loads need to meet the limit and the section modulus that keeps the bending stress below f_y, and the catalogue screen lists the lightest sections in a family that meet both, each solved with its own self-weight. It checks deflection and elastic bending stress only: lateral-torsional buckling, shear, local buckling and the connection need their own checks.

Is this cantilever calculator free?

Yes. Everything on this page runs in your browser with no installation and no sign-up: reactions, shear and moment diagrams, peak and tip deflection, stresses, the step-by-step solution, the finite-element check and the CSV and image downloads. The deflection and rotation curves, on the page and in the downloads, are free on your first 3 days of use; after that they, and the PDF report, need a free account.

References

  • W. C. Young, R. G. Budynas, A. M. Sadegh — Roark's Formulas for Stress and Strain, Table 8.1 (cantilevered beams).
  • J. M. Gere, B. J. Goodno — Mechanics of Materials, Appendix: deflections and slopes of cantilever beams.
  • AISC Steel Construction Manual, Table 3-23: shears, moments and deflections.
  • EN 1990, Annex A1.4 (serviceability) and EN 1993-1-1 (design of steel structures).

Written and maintained by FERS Cloud engineering. Last reviewed .

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