Reinforced Concrete Design / Shear, bond, anchorage, laps and torsion

Why shear is the dangerous one

Diagonal tension, brittle failure, and why a beam without links gives no warning at all.

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Checks needed for project use

A concrete-design reviewer must check the stated first-generation clauses, National Annex choices, detailing rules and factor sets. Numerical agreement alone does not establish code compliance.
Verify the first-generation shear model, axial-stress omissions, minimum-link exceptions and crack-direction diagrams. Assessment of actual cracks requires an engineer.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

Read the lesson register and outstanding review notes

Why this matters

Shear failure can be brittle and rapid. A suitably detailed under-reinforced beam can develop more ductile flexural behaviour, but neither visible cracks nor deflection guarantees warning or time to evacuate. This lesson explains the shear mechanisms and the role of reinforcement.

By the end of this lesson you should be able to

Explain shear failure as diagonal TENSION failure
Predict the orientation of shear cracks and say why they lean
Contrast the warning given by flexural and shear failure
Explain why minimum links are required even where calculation says none are needed

What you should already know

Shear force and bending moment diagrams (Structural Analysis Fundamentals, Module 5)
Principal stresses and Mohr's circle (Structural Analysis Fundamentals, Module 13)
Cracking and composite action (Module 1)

Diagonal tension and shear failure

Diagonal tension is one important mechanism in shear failure. Compression-strut crushing and reinforcement or anchorage failures must also be considered.

Take an element in the web of a beam, away from the extreme fibres. It carries a shear stress τ, and little or no direct stress. Put that on a Mohr's circle: a state of pure shear is a state of principal tension and principal compression of equal magnitude, at 45° to the shear.

So the element is being pulled apart along one diagonal and squeezed along the other. Concrete is roughly ten times weaker in tension than in compression. It therefore fails on the tension diagonal — and the crack runs perpendicular to that tension, which is to say along the compression diagonal, leaning at roughly 45° toward the nearer support.

What we call shear failure is diagonal tension failure. The concrete is not sliding; it is being pulled apart in a direction the beam's geometry disguises.

Once you see this, three things follow immediately, and none of them need memorising.

First, the cracks lean, and they lean the right way. Near a support the shear is large and the moment small, so the principal tension is close to 45°. Near midspan the shear falls to zero and the bending stress dominates, so the principal tension becomes horizontal and the cracks stand vertical. Walk along a cracked beam and you watch the cracks rotate from vertical at midspan to about 45° at the supports. That pattern is the principal stress trajectories made visible.

Second, links must cross the crack to be any use. A vertical link is not resisting a sliding action; it is stitching across a tension crack. Longitudinal bars contribute through chord forces and dowel action, and their ratio enters the concrete shear model. They do not replace properly anchored transverse reinforcement crossing the diagonal cracks.

Third, the failure is brittle. The concrete's tensile strength is reached, the crack forms, and if there is nothing crossing it the crack propagates instantly along its whole length. There is no yielding, no redistribution, no plateau. The beam splits.

The crack pattern is a map of the principal tension direction. Vertical at midspan where bending dominates; leaning at about 45° near the supports where shear does.

Predict first

In a simply supported beam under uniform load, the shear cracks near the left-hand support lean in a particular direction. Which way?

Why the concrete carries any shear at all

If concrete cracks in diagonal tension, how does an unreinforced web carry shear? Four mechanisms, and it is worth knowing them because they explain the shape of the empirical formula.

1.The uncracked compression zone. Above the neutral axis the concrete is in compression and has not cracked, so it can carry shear directly.
2.Aggregate interlock. A crack in concrete is rough, not smooth — it runs around the aggregate particles. The two faces catch on each other and transfer force across the crack.
3.Dowel action. The longitudinal bars cross the crack and resist its vertical displacement by bending, like dowels.
4.Residual tension across very fine cracks, before they open fully.

None of these is easy to quantify, and they interact. That is precisely why the design expression for shear without links is an empirical curve fit rather than a derivation — and knowing this tells you what the formula's terms are doing.

Shear resistance without shear reinforcement

v_{Rd,c} = C_{Rd,c}\,k\,\left(100\,\rho_l\,f_{ck}\right)^{1/3} \;\ge\; v_{min}

What it calculates: The shear stress a member can carry with no links at all.

$C_{Rd,c}$: Nationally determined coefficient, commonly 0.18/γ_C = 0.12 (—)
k: Size-effect factor, 1 + √(200/d), capped at 2.0 (—)
$\rho_l$: Tension reinforcement ratio A_sl/(b_w d), capped at 0.02 (—)
$f_{ck}$: Characteristic cylinder strength (N/mm²)
$v_{min}$: Lower bound, proportional to k to the power 3/2 and to √f_ck (N/mm²)

This assumes

EMPIRICAL — fitted to test data, with no derivation behind it
The tension steel is anchored past the section being checked, or it cannot provide dowel action
The coefficient is nationally determined and unverified in this course

In plain terms: Read the terms against the four mechanisms. ρ_l appears because more longitudinal steel means more dowel action, a smaller crack width and therefore better aggregate interlock, and a deeper compression zone. k appears because deep members are less efficient — a crack in a deep beam opens wider for the same rotation, so interlock is lost. f_ck appears as a cube root, not linearly, because tensile strength grows much more slowly than compressive strength. Nothing here is derived; every part of it is a curve through data.

Practice

A 300 mm wide beam with d = 550 mm has 3 H25 tension bars (1473 mm²) and C30/37 concrete. Taking C_Rd,c = 0.12, what is the shear resistance V_Rd,c, in kN?

Summary

Diagonal tension is an important shear failure mechanism; compression-strut crushing and anchorage also require checks
Cracks rotate from vertical at midspan to about 45° at supports, following the principal tension
Four mechanisms carry shear without links, none of them easy to quantify
V_Rd,c is therefore empirical: a curve fit, not a mechanism
Deep members are less efficient — the size effect is real and caught people out
Failure without links is brittle and unwarned, which is why minimum links are mandatory