Shear Force and Bending Moment Calculator for Beams

Shear Force and Bending Moment Calculator

Find support reactions, maximum shear force, and peak bending moment for a simply supported or cantilever beam under a single point load, a uniformly distributed load, or both combined by superposition.

🎯Real Beam Presets

📝Beam and Load Inputs

Distance between supports, or the free length for a cantilever.

Measured from the left support. For a cantilever, from the fixed end.

Max bending moment 0 kN·m at critical section
Max shear force 0 kN at critical section
Left reaction 0 kN upward support
Right reaction 0 kN upward support

🔢Statics Snapshot

PPoint load kN
wUDL kN per m
LSpan metres
a, bLoad arms

📊Shear and Moment at Key Points

SectionPosition x (m)Shear V (kN)Moment M (kN·m)Note
Enter values above to generate the shear and moment table.

📐Standard Load Case Formulas

Beam and LoadLeft / Fixed ReactionRight ReactionMax ShearMax MomentMoment Location
SS, point P at centerP / 2P / 2P / 2P×L / 4Mid-span
SS, point P at a (b=L–a)P×b / LP×a / Lmax(R_left, R_right)P×a×b / LUnder the load
SS, UDL w over spanw×L / 2w×L / 2w×L / 2w×L² / 8Mid-span
Cantilever, point P at tipP (at fixed end)PP×LFixed end
Cantilever, UDL w fullw×L (at fixed end)w×Lw×L² / 2Fixed end
Cantilever, point P at aP (at fixed end)PP×aFixed end

🗂Beam Case Comparison Grid

ScenarioSupportLoadSpan LPeak VPeak MTypical Use
Center point loadSimply supported20 kN @ mid6 m10 kN30 kN·mTransfer beam under column
Off-center loadSimply supported25 kN @ 2 m6 m16.67 kN33.33 kN·mMachine base support
Uniform floorSimply supported5 kN/m UDL4 m10 kN10 kN·mResidential floor joist
Bridge girderSimply supported12 kN/m UDL10 m60 kN150 kN·mShort-span deck girder
Balcony cantileverCantilever8 kN @ tip2.5 m8 kN20 kN·mProjecting balcony beam
Canopy cantileverCantilever4 kN/m UDL3 m12 kN18 kN·mRoof canopy overhang

Full Formula Breakdown

EquilibriumSum of vertical forces equals zero and sum of moments equals zero. Reactions carry the total applied load.
SS point loadWith load P at distance a from the left support and b = L – a: R_left = P×b / L and R_right = P×a / L.
SS point momentMaximum bending moment sits under the load: M_max = P×a×b / L. At mid-span this reduces to P×L / 4.
SS UDLFor w over the full span: R_left = R_right = w×L / 2. Max shear is w×L / 2 at each support.
SS UDL momentMaximum moment is at mid-span: M_max = w×L² / 8, where shear passes through zero.
CantileverFixed end reaction equals the total load. Point P at tip gives M = P×L; UDL w gives M = w×L² / 2.
SuperpositionPoint + UDL reactions add: R_left and R_right sum the two cases. Peak shear and moment are combined from both effects.

📋Sign Convention and Reference Guide

QuantitySymbolPositive SenseWhere It PeaksUnits
Shear forceVUpward left of sectionAt or near supportskN
Bending momentMSagging (concave up)Where shear crosses zerokN·m
Support reactionRUpward on the beamUnder heavier load sidekN
Point loadPDownward gravity loadApplied at one sectionkN
Distributed loadwDownward per metreSpread along the spankN/m

💡Practical Beam Tips

Shear tip: On a simply supported beam the largest shear force lives at a support, so check both reactions and design connections for the greater value.
Moment tip: The peak bending moment occurs where the shear diagram crosses zero. For a symmetric UDL that point is mid-span, giving w×L² / 8.

Next, you specify the type of beam. For most applications, it’s a simple “simply supported” beam that sits on two separate supports (e.g., bridge deck, floor joist). In contrast, cantilevered beams has a single fixed end with the other end being free (think diving board or balcony). That shifts all the physics.

Instead of the max bending moment being in the center as before, it’s now at the fixed end, where you see the beams gets thick and heavy right next to the wall. The calculator’s internal logic takes this all into account. However, it has to knows what kind of beam you want, or else you will use wrong formula for the beam.

Beam Types and Loads

Sometimes it is not just how much load but where you put it. If I have a simply supported beam with a heavy point load in the center, then the shear are evenly distributed. But if I slide that same load toward either support it has a dramatic effect on reaction forces. The near support bear most of the load and far support is relaxed. Novices often overlook that what happens here is asymmetrical. Designing for average loads fail to take both ends into account. Instead, the tool size the connection for each end specifically. A weak anchor on the heavily loaded end won’t necessarily break the beam; instead, it will fail first.

Uniform distributed loads (UDLs) are called distributed loads because they consist of weights, like furniture and people, spread across the structure. While point loads forms distinct peaks on shear diagrams, distributed loads will produce smooth curves on bending moment diagram. For a simply supported beam, the maximum moment from a UDL is always located at the midpoint of the span, making it much easier to check designs around this area. It’s safe to assume the middle is the key spot here.

Adding point loads to UDLs makes things tricky. Manual calculations becomes almost impossible without some sort of automation. This is why combined mode exists to handle these load combinations. They also mess up calculations.

If you mix feet with newtons or meters with kilograms, you end up either building the structure too light and running a high risk of it collapsing, or you overbuild it and waste money. Most structural codes operate in terms of kilonewtons and meters, which means keeping them all in kilonewtons and meters will keep things on track. Switching will automatically convert for you, although it’s still your job to check your units, since a misplaced decimal point can double your estimated bending moment, and that sort of mistake tends not to show itself until halfway through construction.

Critical points is where the shear force intersects zero; it’s also the point on the beam where the bending moment is maximum. Knowing this statics relationship can help you identify the danger spots without having to draw out a complete diagram each time. For instance, if you have an equal load across a span and you know the shear goes to zero in the middle of that span, you’ll want to put as much steel (or large timbers) there as possible. It’s simple math, but get these figures correct and you would of stability instead of collapse.

Shear Force and Bending Moment Calculator for Beams