CalcSteel · ToolsReal stiffness-method solverExact segment equations

Shear and Moment Diagram Calculator

SFD & BMD for the 8 classic beam cases or any custom loading (multiple loads, partial UDLs, applied moments) — with step-by-step equilibrium equations, CSV export, shareable links and steel profiles suggested from Mmax.

|V| max

30 kN

@ x = 6 m

M max (sagging)

45 kN·m

@ x = 3 m

M min (hogging)

-0 kN·m

@ x = 6 m

Reactions (kN)

R_A 30 · R_B 30

LOADING SKETCHw = 10 kN/mR_A = 30 kNR_B = 30 kNL = 6 m

Simply supported beam — uniformly distributed load

SFD · SHEAR FORCE V(x)[kN]030 kN-30 kNV = 0 @ x = 3 m
BMD · BENDING MOMENT M(x)[kN·m]045 kN·mx = 3 m

Segment equations — x in m, from the left end

0 m ≤ x ≤ 6 m

V(x) = 30 − 10·x [kN]

M(x) = 30·x − 5·x² [kN·m]

Profiles that resist this moment

Md = 45 kN·m → required Wx = Md / (fy/γa1) = 45 kN·m / (250/1.1) = 198 cm³

#1C 300x100x25x4.2517.4 kg/mWx = 204 cm³97% bendingδ ≈ 27.6 mm (L/218)NBR 8800 / AISC 360 check
#2U 300x100x4.7518.0 kg/mWx = 199 cm³99% bendingδ ≈ 28.2 mm (L/213)NBR 8800 / AISC 360 check
#3C 300x100x25x4.7519.4 kg/mWx = 226 cm³88% bendingδ ≈ 24.9 mm (L/241)NBR 8800 / AISC 360 check

Bending screen (Wx ≥ Md/(fy/γa1), NBR 8800 γa1 = 1.10 — AISC 360 φb = 0.90 is nearly identical); plastic Zx is valid for compact sections only. δ is the elastic deflection of THIS loading with E = 200 GPa and the section's Ix (loads taken at service value). LTB, shear, compactness and code deflection limits are verified on the profile page and in the 3D editor. "Open in 3D editor" recreates THIS beam — span, supports and every load — with the profile already assigned.

How to draw shear and moment diagrams (step by step)

Every shear force diagram (SFD) and bending moment diagram (BMD) is built with the same four-move routine. The calculator above automates it — and its step-by-step mode replays each move on the real numbers of the case you selected, so you can learn the method on the exact beam you care about.

  1. Find the support reactions. Draw the free body of the whole beam and apply global equilibrium: ΣFy = 0 and ΣM = 0. For a simply supported beam, take moments about one support to get the other reaction directly. (For a fixed-fixed beam this is not enough — see the FAQ on indeterminate beams.)
  2. Cut the beam region by region. A new region starts at every support, point load, and start/end of a distributed load. Inside one region nothing changes, so ONE pair of equations covers it.
  3. Write V(x) and M(x) for each region. Keep the left free body and sum: V(x) equals the upward forces to the left of the cut; M(x) equals their moments about the cut. That gives the exact polynomials this page prints under "Segment equations".
  4. Plot and close. Point loads and reactions create vertical jumps in V equal to their magnitude; distributed load makes V slope down at rate w. The moment curve is one degree higher than the shear in every region. Past the last support, both diagrams must return to zero — if they don't, a reaction is wrong. This "closure check" catches most exam mistakes.

Where the shear crosses zero, the bending moment is at a local peak — that is usually the section that governs the flexural design of the member.

The load–shear–moment relationships: dV/dx = −w and dM/dx = V

Two differential identities generate every rule of thumb used to sketch diagrams. With w(x) the downward load intensity:

dV/dx = −w(x) — the slope of the shear diagram equals minus the load intensity.

dM/dx = V(x) — the slope of the moment diagram equals the shear ordinate.

From them, the shape rules:

Loading on the regionShear diagram VMoment diagram M
No loadconstant (horizontal)straight line, slope = V
Uniform load wstraight line, slope −wparabola (2nd degree)
Triangular loadparabola (2nd degree)cubic (3rd degree)
Point load Pvertical jump of −Pkink (slope changes abruptly)
Applied/support momentno changevertical jump

Two consequences worth memorizing: M is extreme where V = 0 (because dM/dx = 0 there), and the change in M between two sections equals the area under the V diagram between them. You can verify both on any preset above — switch to step-by-step mode and watch the equations confirm it region by region.

Sign conventions (and why some BMDs look upside-down)

This calculator uses the classic strength-of-materials convention, the one used by Hibbeler, Beer & Johnston, and most design codes:

  • Positive shear V: the resultant of the forces on the left free body acts upward (the pair tends to rotate the element clockwise).
  • Positive moment M: sagging — the beam smiles, compression on the top fibre, tension on the bottom. Hogging moments (cantilever over a wall, moment over the middle support of an overhanging beam) are negative.
  • Applied loads P and w are entered as positive-down magnitudes; the solver handles the signs internally.

Some European and Brazilian schools plot the BMD on the tension side — positive moments drawn downward. The numbers are identical; only the plotting direction flips. If your textbook draws sagging moments below the axis, mentally mirror the BMD above (or compare the signed values at the labeled critical points, which never change).

On the diagrams above, positive regions are shaded green and negative regions red, so the convention is visible at a glance — a detail that matters when you check moment transfer at the supports of continuous or overhanging beams.

The 8 classic cases and their closed-form peaks

Every case in the preset bar has a textbook closed form. The calculator reproduces all of them exactly (that is part of its automated verification), and stays exact when you change L, P, w or the positions — which the formulas below cannot do for you once the loading gets mixed.

CaseMax shearMax momentWhere
Simple, P at centerP/2+PL/4midspan
Simple, P at aP·max(a, L−a)/L+P·a·(L−a)/Lunder the load
Simple, UDL wwL/2+wL²/8midspan
Simple, triangular 0→wwL/3 (at the steep end)+wL²/(9√3) ≈ 0.0642 wL²x = L/√3
Cantilever, P at tipP−PLat the wall
Cantilever, UDL wwL−wL²/2at the wall
Overhanging, UDL wdepends on b−w·c²/2 over the support (c = overhang)support B
Fixed-fixed, UDL wwL/2−wL²/12 at the ends, +wL²/24 at midspanends govern

The overhanging beam is the one students get wrong most often: the moment is negative over the interior support, positive in the span, and crosses zero at the point of contraflexure — the calculator marks that point explicitly on the BMD, since it is where you would splice reinforcement or check lateral bracing.

Beyond the classics: custom loadings and applied moments

Real beams rarely match a textbook card, so the Custom case builder accepts any combination the classic tables cannot: up to 6 point loads anywhere on the span, up to 4 distributed loads that may be partial (covering only part of the span) and trapezoidal (different intensities at each end), and up to 4 concentrated applied moments — on simply supported, cantilever, fixed-fixed or overhanging support layouts.

Two behaviors worth knowing:

  • Applied moments follow the clockwise-positive convention: a clockwise moment M₀ makes the BMD jump UP by +M₀ as you traverse it left to right, while the SFD does not change at all (a pure couple carries no net vertical force). The printed segment equations treat the moment exactly; the reactions come from the same stiffness-method engine as everything else.
  • Every result stays exact and auditable. The per-segment polynomials, the critical points, the step-by-step equilibrium narration and the CSV export all work for custom cases exactly as they do for the presets — including statically indeterminate layouts like fixed-fixed, which statics-only tools cannot solve at all.

When you are done, the Share link button encodes the whole configuration in the URL, so a colleague (or your students) can open the identical beam with one click — no account, no file.

From diagram to design: profiles that resist Mmax

A bending moment diagram is a means, not an end — the number you actually need is a steel section. Below the diagrams, this page closes that loop:

  1. Required section modulus. From the governing moment Md = max(|Mmax|, |Mmin|), the elastic requirement is Wx,req = Md / (fy/γa1) with γa1 = 1.10 per NBR 8800 (the AISC 360 LRFD factor φb = 0.90 lands within a few percent). Pick fy for your steel — A36, A572 Gr.50 or S355.
  2. Catalog sweep. The ~1,309-profile CalcSteel catalog (W/HP, channels, tubes — ASTM, EN and NBR series) is swept for the three lightest sections with Wx ≥ Wx,req, each shown with its weight and bending utilization, and each linking to its profile page where the full NBR 8800 / AISC 360 checks (LTB, shear, deflection) run on the real section properties.
  3. One click to a real model. "Open in 3D editor" recreates the exact beam you configured — span, supports, every point load, every distributed load — in the CalcSteel 3D structural editor, with the suggested profile already assigned, ready for load combinations, code checks and IFC/DXF export. The diagram you were studying becomes a design you can finish.

This is the difference between a formula card and a tool that sits inside a real structural product: the numbers keep living after the diagram.

Common mistakes (and how the diagrams expose them)

  • Diagram does not close. If V or M is not zero after the last support, a reaction is wrong. Recompute ΣM about a support. The step-by-step mode's final step performs this check for you.
  • Missing the jump at a point load. V must drop by exactly P at the load. If your jump differs, you mixed up the left/right free body.
  • Peak moment at the wrong place. The maximum of M is NOT always at midspan — for the off-center point load it sits under the load; for the triangular load at x = L/√3 ≈ 0.577L. Find where V = 0 first.
  • Sign error on the overhang. Hogging over the support is negative. If your BMD is all positive on an overhanging beam, the sign convention slipped.
  • Treating a fixed-fixed beam as determinate. Two clamped ends give 4 unknowns and only 2 equilibrium equations — statics alone cannot solve it. This page runs a real stiffness-method solve, which is why its fixed-fixed preset shows the correct −wL²/12 end moments instead of a wrong "wL²/8".
  • Forgetting units. V carries force units (kN, kip); M carries force×length (kN·m, kip·ft). The SI/imperial toggle above converts the equations' coefficients too, so you can hand equations to a colleague in either system without redoing algebra.

Worked example

Simply supported beam, 6 m span, UDL of 10 kN/m (the default case)

Given

  • Span L = 6.00 m
  • UDL w = 10.0 kN/m (down)
  • Supports: pin at A (x = 0), roller at B (x = 6 m)
  1. 1. Support reactions (symmetry)

    R_A = R_B = wL/2 = 10 × 6 / 2

    30.0 kN each

  2. 2. Shear equation — single region, cut at x

    V(x) = R_A − w·x = 30 − 10x → V = 0 at x = 3.00 m

    Vmax = ±30.0 kN at the supports

  3. 3. Moment equation — moments about the cut

    M(x) = R_A·x − w·x²/2 = 30x − 5x²

    M(3) = 90 − 45 = 45.0 kN·m

  4. 4. Check against the closed form and the FEM engine

    wL²/8 = 10 × 6² / 8 = 45 — stiffness solver reports the same V, M at all 61 stations

    exact match

Result

Vmax = 30.0 kN · Mmax = 45.0 kN·m @ x = 3.00 m · diagrams close at both ends

Frequently asked questions

What is a shear and moment diagram?

They are plots of the internal shear force V(x) and internal bending moment M(x) along a beam. The SFD shows the transverse force a cross-section must transmit at every position x; the BMD shows the bending moment. Together they identify the critical sections where the beam must be checked or designed.

Where is the maximum bending moment on a beam?

At a point where the shear force equals zero, or at a support/point of discontinuity. Because dM/dx = V, the moment has a local extreme wherever the shear diagram crosses zero — under the load for a point load, at midspan for a symmetric UDL, at x = L/√3 for a triangular load, and at the wall for a cantilever.

Why does the shear diagram jump at a point load?

Equilibrium of an infinitesimal slice at the load requires the internal shear just right of the load to differ from the shear just left of it by exactly the applied force. So every point load (and every support reaction) creates a vertical step in the SFD equal to its magnitude.

What do dV/dx = −w and dM/dx = V mean in practice?

They are slope rules: at any x, the slope of the shear diagram equals minus the distributed load intensity, and the slope of the moment diagram equals the shear value. They also give the area rule: the change in moment between two sections equals the area under the shear diagram between them.

Can this calculator handle statically indeterminate beams?

Yes — the fixed-fixed preset is indeterminate and is solved with a real stiffness-method (FEM) engine, the same approach used in CalcSteel’s 3D structural editor. For a UDL it reproduces the exact closed form: end moments of −wL²/12 and a midspan moment of +wL²/24.

Why is the bending moment negative on a cantilever?

A tip load bends a cantilever concave-down (hogging): tension on the top fibre. Under the sagging-positive convention, hogging moments are negative, so the cantilever BMD runs from −PL at the wall up to zero at the free tip.

Can I use the exported PNG in my class notes or website?

Yes. The PNG export is free and contains the loading sketch, the SFD and the BMD with all critical values labeled. You may embed it in lecture slides, homework solutions or articles — a credit link back to this calculator is appreciated but not required.

Can I analyze several loads at once, or an applied moment?

Yes — the Custom case builder accepts up to 6 point loads, 4 partial or trapezoidal distributed loads and 4 concentrated applied moments on simply supported, cantilever, fixed-fixed or overhanging layouts. A clockwise applied moment M₀ produces a vertical jump of +M₀ in the BMD and no change in the SFD; the segment equations and the step-by-step mode handle it exactly.

Can I export the V(x) and M(x) values or share a case with someone?

Both. The CSV button downloads every station of the shear and moment diagrams (about 160 points plus one row per side of each discontinuity) in your active unit system, ready for Excel or Python. The Share-link button copies a URL that encodes the full case — preset or custom loads — so anyone opening it sees the identical beam, no account needed.

How do I pick a steel profile from the maximum moment?

Compute the required elastic section modulus Wx = Md/(fy/γa1) with Md the governing moment and γa1 = 1.10 (NBR 8800). This page does it automatically: it sweeps the 1,309-profile catalog, lists the three lightest sections that pass, links each to its full NBR 8800/AISC 360 verification page, and can open the beam in the CalcSteel 3D editor with the profile already assigned.

Reviewed by Eng. Rilis Reis · Structural Engineer — CalcSteel·Updated