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Shear & Moment Diagram Calculator

Build a beam, add supports and loads, then get instant shear and bending moment graphs.

1) Beam Setup

Moments use Force×Length (kN·m or kip·ft). Distributed loads use Force/Length.

Enter downward loads with positive magnitudes, treat sagging bending moments as positive, and use counter-clockwise applied moments as positive.

2) Add Loads / Moments

TypeMagnitudePosition / SpanRemove

Positions are measured from the left end (\(x=0\)).

3) Visual Beam

Tap/click the beam to add the selected load at that position.
Shear Force Diagram \(V(x)\)
Bending Moment Diagram \(M(x)\)

Results

Practical Guide

Shear and Moment Diagram Calculator

This guide shows you how to build a beam, add supports and loads (including non-uniform distributed loads), and interpret the shear-force diagram \(V(x)\) and bending-moment diagram \(M(x)\) the calculator produces. You’ll also see when hand methods match the tool, what assumptions are baked in, and how to sanity-check results before you size members or sign off a design.

7–9 min read Updated 2025 Statics • Structures I/II

Quick Start

The calculator is visual, but it still follows classic statics rules. Use these steps to get correct reactions and clean diagrams on the first try.

  1. 1 Choose the beam type (Simply Supported or Cantilever), then enter the beam length \(L\). Confirm the length unit (m or ft).
  2. 2 Pick the force unit (kN or kip). The calculator automatically updates moment units to kN·m or kip·ft.
  3. 3 Add supports implicitly via beam type: Simply supported means a pin at \(x=0\) and a roller at \(x=L\); cantilever means a fixed support at \(x=0\).
  4. 4 Add loads in the left panel or by tapping the beam: point load \(P\), UDL \(w\), linearly varying load (LVdL) with end intensities \(w_a\) and \(w_b\), or a point moment \(M\). Enter positions/span limits in the same length unit as \(L\).
  5. 5 For distributed loads, make sure the span satisfies \(0 \le a < b \le L\). A UDL uses a single intensity \(w\). An LVdL uses two intensities \(w_a\) at \(a\) and \(w_b\) at \(b\), which covers triangular and trapezoidal loads.
  6. 6 Watch the diagrams update. The shear diagram \(V(x)\) should jump at point loads and slope under distributed loads. The moment diagram \(M(x)\) should slope where \(V\neq 0\) and curve under distributed loads.
  7. 7 Use Show Steps to see the equilibrium and piecewise build-up used internally, and compare with your own hand calc if needed.

Tip: If your diagram “looks wrong,” it often comes from a unit mismatch (e.g., kip with meters) or an invalid span where \(b \le a\).

Remember: This calculator targets statically determinate cases. If you add extra supports or constraints outside the provided beam types, the results will not represent a real structure.

Choosing Your Method

Engineers generate shear and moment diagrams in a few standard ways. The calculator mirrors these approaches but automates the bookkeeping. Here’s when each method is most useful.

Method A — Section Cuts + Equilibrium

The “intro statics” approach: cut the beam at a position \(x\), build a free-body diagram, then apply equilibrium to find \(V(x)\) and \(M(x)\).

  • Great for learning and for short beams with only a few loads.
  • Directly shows how jumps and slopes happen in \(V\) and \(M\).
  • Easy to validate with physical intuition.
  • Slow for many load regions.
  • Risk of sign mistakes when you have multiple point moments or LVdLs.
Cut at \(x\): \(\sum F_y=0 \Rightarrow V(x)\), \(\sum M=0 \Rightarrow M(x)\)

Method B — Singularity / Macaulay Functions

Represent loads with bracket functions, integrate to get \(V(x)\) and \(M(x)\), and solve constants using boundary conditions.

  • Compact for multiple point loads and moments.
  • Matches how many structural software tools work internally.
  • Easy to extend to deflection once you have \(M(x)\).
  • Step-function notation can be confusing at first.
  • Still requires careful constants and limits.
\(V(x)=R_A – \sum P_i\langle x-x_i\rangle^0 – \int w(x)\,dx\)

Method C — Numerical / Piecewise Integration (What this tool does)

Convert distributed loads to partial resultants for reactions, then numerically sample the beam from \(0\) to \(L\), accumulating shear and moment.

  • Fast for any mix of loads, including LVdLs.
  • Harder to “lose” a load region or mis-apply a centroid.
  • Ideal for iterative sizing or “what-if” checks.
  • Only as accurate as the model assumptions.
  • Does not solve indeterminate beams (requires stiffness methods/FEM).
\(\dfrac{dV}{dx}=-w(x),\quad \dfrac{dM}{dx}=V(x)\)

What Moves the Number

Small changes to key inputs can shift peak shear and moment a lot. These are the dominant “levers” you should think about during design.

Beam span \(L\)

For common cases, peak moment scales with \(L^2\). Doubling span can quadruple maximum bending moment, which directly impacts member size and deflection.

Load magnitude

Point loads raise shear abruptly. Distributed loads accumulate gradually. For LVdLs, the larger end intensity dominates peak moment location.

Load position \(x\)

Moving a point load toward midspan increases moment more than moving it toward a support. The calculator makes this visible by shifting the \(M(x)\) peak as you drag positions.

Distributed load span \([a,b]\)

A UDL over the full span yields a symmetric parabola in \(M(x)\). Shortening the loaded region shifts peaks toward the loaded center.

Load shape (UDL vs LVdL)

LVdLs create steeper shear curvature under the higher-intensity end. Triangular loads typically place the moment peak closer to the “heavy” side.

Support type

Cantilevers develop maximum moment at the fixed end. Simply supported beams usually peak near midspan. Choosing the right boundary condition matters more than any single load tweak.

Design intuition: Shear is “how much load is left to carry” at a cut. Moment is “how hard the loads try to rotate” about that cut.

Worked Examples

Example 1 — Simply Supported Beam with Point Load + UDL

  • Beam: Simply supported
  • Span: \(L=8\,\text{m}\)
  • Point load: \(P=18\,\text{kN}\) at \(x=3\,\text{m}\)
  • UDL: \(w=4\,\text{kN/m}\) from \(a=5\) to \(b=8\,\text{m}\)
1
Replace UDL with resultant: \(W = w(b-a)=4(8-5)=12\,\text{kN}\), located at \(\bar{x}=a+\frac{b-a}{2}=6.5\,\text{m}\).
2
Take moments about A to get \(R_B\): \(R_B L = P x_P + W\bar{x}\).
3
Compute: \(R_B = \dfrac{18(3)+12(6.5)}{8} = \dfrac{54+78}{8}=16.5\,\text{kN}\).
4
Vertical equilibrium: \(R_A = P+W-R_B = 18+12-16.5=13.5\,\text{kN}\).

Now build shear and moment:

5
For \(0<x<3\): \(V(x)=R_A=13.5\). \(M(x)=R_A x=13.5x\).
6
Jump at \(x=3\) by \(-P\): \(V=13.5-18=-4.5\). Moment stays continuous.
7
For \(3<x<5\): \(V(x)=-4.5\). \(M(x) = 13.5x – 18(x-3)\).
8
For \(5<x<8\) (UDL on): \(V(x)=-4.5 – 4(x-5)\). \(M(x)=M(5)+\int_5^x V(\xi)\,d\xi\).

Enter these inputs and you’ll see the shear line step down at \(x=3\), then slope downward from \(x=5\) to \(8\). The moment diagram is piecewise linear then curves (parabolic) where the UDL acts. The calculator’s peak \(M(x)\) should occur near the transition into the UDL region.

Example 2 — Cantilever with Linearly Varying Distributed Load (Triangular)

  • Beam: Cantilever fixed at left
  • Span: \(L=4\,\text{m}\)
  • LVdL: from \(a=0\) to \(b=4\,\text{m}\), \(w_a=0\), \(w_b=9\,\text{kN/m}\)
  • No other loads
1
Resultant of LVdL: \(W=\frac{w_a+w_b}{2}(b-a) =\frac{0+9}{2}(4)=18\,\text{kN}\).
2
Centroid for a triangular load with zero at \(a\): \(\bar{x}=a+\frac{2}{3}(b-a)=\frac{2}{3}(4)=2.667\,\text{m}\).
3
Fixed-end reactions: \(V_0=W=18\,\text{kN}\), \(M_0=W\bar{x}=18(2.667)=48.0\,\text{kN·m}\) (CCW).
4
Load function: \(w(x)=\dfrac{w_b-w_a}{L}x = \dfrac{9}{4}x\). Thus \(\dfrac{dV}{dx}=-w(x)\).

Integrate from the fixed end:

5
\(V(x)=V_0-\int_0^x \frac{9}{4}\xi\,d\xi =18-\frac{9}{8}x^2\).
6
\(M(x)=-M_0+\int_0^x V(\xi)\,d\xi =-48 + 18x -\frac{9}{24}x^3\).

In the calculator, set beam type to Cantilever and add an LVdL with \(w(a)=0\), \(w(b)=9\). You’ll see shear start at \(18\,\text{kN}\) and curve to zero at the free end; moment starts at \(-48\,\text{kN·m}\) (hogging by convention) and curves back to zero at \(x=L\). The peak magnitude is at the fixed support, exactly as expected for cantilevers.

Common Layouts & Variations

The calculator currently supports common statically determinate beams. Use this table as a quick reference for what each layout implies and where to expect maximums.

ConfigurationTypical LoadsWhere \(|V|_{max}\) HappensWhere \(|M|_{max}\) HappensNotes / Use Cases
Simply Supported (Pin–Roller)Point loads, UDLs, LVdLs, point momentsNear supports or at large point loadsOften near midspan or under heavy load regionFloor beams, simple bridge girders, purlins.
Cantilever (Fixed–Free)End point loads, triangular wind/snow, equipment momentsAt fixed endAt fixed endBalconies, sign brackets, projecting canopy beams.
Partial UDLLive load over a room bay, storage regionJust inside load start/endUnder loaded patchCommon for mixed occupancy floor loading.
Triangular LVdLSoil/water pressure, wind uplift varying with heightAt higher-intensity endShifted toward heavier sideModel trapezoids by using nonzero \(w_a\) and \(w_b\).
Point Moment AddedApplied couple from connection eccentricityShear unchanged at that pointMoment jumps by \(\pm M\)Useful for bracketed loads or torsion-to-bending effects.
  • Make sure all loads fit within \(0\le x\le L\).
  • Expect \(V(x)\) to be piecewise constant between point loads.
  • Expect \(M(x)\) to be continuous unless a point moment is applied.
  • Symmetric loading on simply supported beams should yield symmetric \(M(x)\).
  • Cantilever shear and moment must be zero at the free end.
  • Check that the moment diagram’s slope matches the shear diagram.

Specs, Logistics & Sanity Checks

The diagrams are only step one. Before you finalize sizing, verify that the load model and boundary conditions match your real structure.

Model Assumptions

  • Beam is prismatic (constant \(E\) and \(I\)) for the statics portion.
  • Loads act in a single plane and are applied quasi-statically.
  • Supports are ideal pins/rollers or a fully fixed end for cantilevers.
  • Small deflections: geometry does not change under load.

Load Checklist

  • Separate dead, live, wind, seismic, and equipment loads.
  • Convert area loads to line loads using tributary width.
  • For LVdLs, confirm which end is heavier and why.
  • Include self-weight if it matters for moment (long spans).

Diagram Sanity

  • \(\int_0^L w(x)\,dx\) should equal total drop in \(V(x)\).
  • \(M(L)=0\) for simply supported; \(M(L)=0\) and \(V(L)=0\) for cantilevers.
  • Max moment location often aligns with \(V(x)=0\).
  • Units: \(V\) in kN/kip; \(M\) in kN·m/kip·ft.

Safety note: Use code-based load combinations (ASCE 7, ACI, AISC, Eurocode, etc.) for final member design. This calculator gives line-load statics; it does not choose governing combinations or check limit states.

Best practice: Run at least one hand check on a critical case. If your hand equilibrium and the calculator disagree, fix the model before proceeding.

Frequently Asked Questions

What’s the difference between shear force and bending moment?
Shear \(V(x)\) is the internal vertical force at a cut; it balances the net applied load to one side. Bending moment \(M(x)\) is the internal couple that balances rotational effects of those loads. They’re linked by \(\frac{dM}{dx}=V\).
Why does the shear diagram jump at point loads?
A point load is an impulse in the load function. Since \(\frac{dV}{dx}=-w(x)\), an impulse in \(w\) produces an instantaneous change in \(V\), so the diagram steps up or down by the load magnitude.
How are non-uniform distributed loads handled?
The calculator supports linearly varying distributed loads (LVdLs). You enter \(w(a)\) and \(w(b)\) over a span \([a,b]\). Internally it computes the equivalent resultant \(W=\frac{w_a+w_b}{2}(b-a)\) at centroid \(\bar{x}=a+\frac{(b-a)(2w_b+w_a)}{3(w_a+w_b)}\), then samples \(V(x)\) and \(M(x)\) piecewise.
Why is the maximum moment often where shear crosses zero?
Because \(\frac{dM}{dx}=V\). At a local maximum or minimum of \(M(x)\), the slope is zero, so \(V(x)=0\) at that location (except where a point moment causes a discontinuity).
Can this calculator solve continuous or multi-span beams?
Not yet. Continuous beams are statically indeterminate and require stiffness methods, slope-deflection, moment distribution, or FEM. This tool is for determinate cases (simple span or cantilever) where reactions come from equilibrium alone.
What sign convention does the calculator use?
Downward loads are entered as positive magnitudes. Positive shear is upward on the left cut face, and positive bending moment is sagging (tension at the bottom fiber). Applied counter-clockwise point moments are positive and appear as jumps in \(M(x)\).
How should I use these diagrams for design?
Use \(|V|_{max}\) to check shear capacity and \(|M|_{max}\) for flexural capacity. Then verify deflection using your material code equations (e.g., \(EI\) methods). Always apply factored load combinations from your governing design standard.
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