Shear Force in Beams Explained with a Worked Example

When a beam carries load, two internal actions develop inside it. One is bending moment, which makes the beam sag. The other is shear force, the internal force that tries to slide one part of the beam vertically past the adjacent part. Shear is usually greatest close to the supports, where the load is being handed over to the columns or walls, and it is behind some of the most sudden and brittle failures in concrete structures.

This article explains shear intuitively, sets out how shear force is calculated and how it relates to bending moment, works through a complete example, explains how shear stress is distributed within a section, and covers how shear failures look and the mistakes that cause them.

The intuition: slicing the beam

Imagine a beam made of a stack of vertical slices, like a loaf of bread, sitting on two supports with weight on top. Each slice is pushed down by the load on it and must be held up by its neighbours. The slices nearest the supports have to pass on the most load, because everything between them and the middle of the span is being transferred through them to the support. If the slices were not glued together, they would slide past each other vertically. The force each cut has to transmit to prevent that sliding is the shear force at that point.

That picture explains why shear is greatest at the supports and why it often falls to zero somewhere in the span. At midspan of a symmetrically loaded beam, each half supports itself equally from its own support, so nothing needs to cross the centreline vertically.

How shear force is calculated

The shear force at any section along a beam equals the algebraic sum of all vertical forces, loads and reactions, acting on one side of that section. Working from the left end, you add the upward reaction, subtract each downward load as you pass it, and the running total at any point is the shear force there. Engineers commonly use the convention that shear is positive when the left-hand part tends to move up relative to the right-hand part.

Key results follow directly:

For a simply supported beam with a uniformly distributed load w over span L, the shear at each support is V = w·L/2, falling linearly to zero at midspan.
For a central point load P, the shear is V = P/2, constant from each support to the load, changing sign under it.
A point load produces a sudden step in the shear diagram equal to its magnitude; a distributed load produces a sloping line.

The relationship with bending moment is the most useful fact in beam analysis: shear force is the rate of change of bending moment along the beam, V = dM/dx. So wherever the shear force passes through zero, the bending moment reaches a maximum or minimum. Finding the zero-shear point is how engineers locate the maximum moment. The uniformly distributed load versus point load comparison shows how each load type shapes both diagrams.

A worked example

Take a 6 m simply supported beam carrying a uniformly distributed load of 10 kN/m over its full length plus a point load of 30 kN located 2 m from the left support.

Reactions. Total load is 10 × 6 + 30 = 90 kN. Taking moments about the left support: R_right × 6 = (60 × 3) + (30 × 2) = 240, so R_right = 40 kN and R_left = 90 − 40 = 50 kN.

Shear along the beam. At the left support the shear is +50 kN. Moving right, the UDL reduces it by 10 kN per metre, so just before the point load at 2 m it is 50 − 20 = +30 kN. The 30 kN point load drops it immediately to 0. Beyond that the UDL continues to reduce it, reaching 0 − 40 = −40 kN at the right support, which matches the right reaction. The check closing to the reaction confirms the arithmetic.

Maximum moment. The shear passes through zero at the point load, 2 m from the left, so the bending moment peaks there: M = 50 × 2 − 10 × 2² / 2 = 100 − 20 = 80 kNm. Drawing both diagrams together is covered step by step in our article on shear force and bending moment diagrams, and the moment itself in our article on bending moment.

Shear stress inside the section

The shear force is a total; inside the cross-section it is carried as shear stress, which is not uniform. The general expression is τ = V·Q / (I·b), where Q is the first moment of the area beyond the level being considered, I the moment of inertia and b the width at that level. For a rectangular section, the stress is zero at the top and bottom faces and greatest at the neutral axis, where it equals 1.5·V/A, half as much again as the simple average.

In an I-beam, almost all of the shear is carried by the web, because that is where the section is narrow and the stress concentrated, while the flanges carry the bending. A quick hand check divides the shear force by the web area. This division of labour is why webs are thin but must be checked for shear buckling when they are very deep, and why web openings for services near supports need careful design.

Shear failure in concrete

Shear combines with bending to create diagonal tension in the web of a beam, inclined at roughly 45 degrees near the supports. Concrete is weak in tension, so a reinforced concrete beam that lacks adequate shear reinforcement can crack diagonally from near the support towards the load and fail suddenly, with little of the warning sagging that typically precedes a bending failure. This brittleness is why codes are deliberately cautious about shear.

The remedy is shear reinforcement: links, also called stirrups, wrapped around the main bars at close spacing, especially near supports. In the classic truss model, the links act as vertical ties and the concrete between diagonal cracks acts as inclined struts. Our guide to rebar reinforcement shows how links are detailed. In flat slabs, the equivalent problem is punching shear around columns, where the slab can fail by a cone punching through, which is why flat slab column heads, drop panels or shear studs are used.

Diagonal cracks near the supports of a beam are therefore taken seriously in inspections, as described in our guide to types of concrete cracks.

Common mistakes

Checking only midspan. Bending often governs at midspan but shear governs near supports and under heavy point loads. Both must be checked.
Sign errors. Use one convention consistently and confirm that the shear at the far end equals the reaction there.
Cutting openings near supports. Coring through a beam web close to a support removes material exactly where shear stress is highest.
Omitting links in short, heavily loaded beams. Transfer beams and beams with loads near supports are often shear-critical.
Confusing shear force with shear stress. Shear force is the total in kN; shear stress is its intensity in N/mm², and varies across the section.

Frequently asked questions

What is shear force in simple terms?

Shear force is the internal force that resists one part of a beam sliding vertically past the adjacent part. At any section it equals the sum of the vertical loads and reactions on one side of that section, and it is usually largest near the supports.

Where is shear force maximum in a beam?

For a simply supported beam under typical gravity loads, shear force is maximum at or near the supports. Under a heavy point load close to a support, the shear between that load and the support can be particularly high.

How are shear force and bending moment related?

Shear force is the rate of change of bending moment along the beam. Where shear force is zero, the bending moment is at a maximum or minimum, which is why engineers locate the zero-shear point to find the critical bending moment.

What does a shear crack look like?

In a concrete beam, shear cracks are usually diagonal, inclined at around 45 degrees, starting near the support and running towards the load. They differ from vertical flexural cracks near midspan and should be assessed promptly by an engineer.

What resists shear in a reinforced concrete beam?

Shear is resisted by the concrete itself, by the dowel action of the main bars and aggregate interlock across cracks, and above all by shear reinforcement in the form of links or stirrups placed at regular spacing along the beam.

Seeing diagonal cracks in a beam, or planning openings through one? Talk to our consultancy or get in touch.