Module 5 · Lesson 5.2

The variable-angle truss

Once the web has cracked, the beam is a truss. Derive it by cutting, and the design equations fall out.

Why this matters

Most people meet the shear equations as two formulae to look up, with a mysterious angle θ that you are told to take as 45° or 22°. In fact both formulae come from cutting a truss and resolving forces — school statics, nothing more — and the choice of θ is a genuine engineering trade between steel and concrete that you make yourself. Derive it once and you will never need to look it up again.

By the end of this lesson you should be able to

Describe the truss analogy and identify each of its members in a real beam
Derive V_Rd,s by cutting along a crack and counting links
Derive V_Rd,max from the strut force
Explain the trade-off that choosing θ represents
State why the concrete and truss contributions are not added

What you should already know

Why shear is dangerous (previous lesson)
Method of sections for trusses (Structural Analysis Fundamentals, Module 6)

The beam becomes a truss

Once diagonal cracks have formed, the web is no longer a continuous plate. It is a series of concrete blocks separated by cracks, and force can only travel through the blocks — along the cracks' direction, never across them.

That is exactly a truss:

Top chord — the concrete compression zone.
Bottom chord — the longitudinal tension reinforcement.
Diagonal struts — the blocks of concrete between the cracks, in compression, inclined at angle θ to the horizontal.
Vertical ties — the links, in tension.

The striking part is that θ is not fixed. The cracks may form at 45°, but as load increases and the member deforms, the compression field can rotate to whatever angle the member needs, as long as the struts can carry it and the ties can hold it. The concrete has, in effect, a choice of load path — and design gets to pick.

From first principles

The variable-angle truss, from a cut

We want to show: To obtain both design equations for shear — the force the links carry, and the force that crushes the struts — using nothing but a section cut and resolution of forces.

Imagine cutting the beam's web along one of the diagonal cracks, at angle θ, and lifting away the part beyond the cut. What holds the two pieces together? Only the links that happen to cross the cut, each pulling with its yield force, and the concrete strut pushing along the crack. So the question 'how much shear can this beam carry?' becomes 'how many links does my cut cross?' — and that is a counting problem. Flatten the cut and it crosses more links. That is the entire idea, and everything below is arithmetic on it.

Try it

Choose the strut angle

Two curves: what the links can carry, which rises as the strut flattens, and what crushes the struts, which falls. The efficient design sits where they cross.

Web width b_w300 mm

Effective depth d550 mm

Concrete f_ck30 N/mm²

Link diameter8 mm, 2 legs

Link spacing s200 mm

cot θ2.5

Applied shear V_Ed350 kN

Strut angle θ: 21.8°
Lever arm z: 495 mm
Links carry V_Rd,s: 270 kN
Struts crush at V_Rd,max: 460 kN
Governing V_Rd: 270 kN
Governed by: links
Concrete alone V_Rd,c: 113 kN
ν₁: 0.528

Inadequate: V_Rd = 270 kN against 350 kN applied. Add link area or reduce the spacing.

Things worth trying

Flatten the strut towards cot θ = 2.5. The link curve rises and the strut curve falls — that crossing is the efficient design.
Narrow the web to 150 mm. The strut curve collapses and crushing governs everywhere. Now adding links achieves nothing at all.
Halve the link spacing. The link curve doubles and the strut curve does not move — the struts do not know how much steel is in the web.
Note V_Rd,c in the readout. Once links are needed it plays no part: the links carry the whole shear, not the excess.

Take care

The concrete term and the truss term are not added

This is the most-repeated error in shear design, and it follows directly from the derivation above.

The truss model assumes the web is fully cracked. In that state the four mechanisms behind V_Rd,c — uncracked compression zone, aggregate interlock, dowel action, residual tension — are no longer intact in the way the empirical fit assumed. The truss has already claimed the whole web.

So the rule is: if V_Ed ≤ V_Rd,c, provide minimum links and stop. If V_Ed > V_Rd,c, the links carry all of V_Ed, not the excess above V_Rd,c. Designing links for (V_Ed − V_Rd,c) will under-reinforce the beam, sometimes substantially.

This differs from older codes, including BS 8110, which did add the two. If you are checking an old calculation, expect to find the addition, and expect the answer to be less conservative than a modern one.

Worked example

Designing links for a beam

Given

Rectangular beam, b_w = 300 mm, d = 550 mm
C30/37 concrete: f_ck = 30 N/mm², f_cd = 17 N/mm²
Links of grade 500: f_ywd = 500/1.15 = 434.8 N/mm²
Tension steel 3 H25 = 1473 mm², fully anchored past the section
Design shear at the critical section, V_Ed = 350 kN

Find

The link size and spacing.

Assumptions

Vertical links
Lever arm z = 0.9d = 495 mm
cot θ limited to 2.5 — nationally determined, requires verification

The part everyone forgets: shear changes the bending reinforcement

The truss has an inclined compression field, and an inclined force has a horizontal component. That horizontal component has to go somewhere, and where it goes is into the chords.

The consequence is that the tension chord carries more force than bending alone would suggest:

Additional tension chord force from shear

Δ F_{t d} = 0.5 V_{E d} (cot θ - cot α)

What it calculates: The extra force the longitudinal tension steel must carry because of the inclined compression field.

$V_{E d}$: Design shear at the section (N)
$θ$: Strut angle
$α$: Link inclination — 90° for vertical links, so cot α = 0

This assumes

The same truss model as the shear derivation
Vertical links unless α is specified

In plain terms: For the beam above, ΔF_td = 0.5 × 350 × 2.5 = 437 kN — which needs an extra 1006 mm² of steel, comparable to the bending reinforcement itself. In practice this is handled by the equivalent and simpler shift rule: displace the bending moment diagram horizontally by a_l = z cot θ / 2, and design the bars for the shifted diagram. The two are the same statement. The total chord force need not exceed the value at the point of maximum moment.

Practice

A beam with b_w = 250 mm, d = 500 mm, C30/37 (f_cd = 17 N/mm²) is checked for strut crushing at cot θ = 2.5. What is V_Rd,max, in kN?

Practice

For a beam with z = 450 mm, f_ywd = 434.8 N/mm² and cot θ = 2.5, what link area per unit length A_sw/s is needed to carry V_Ed = 300 kN? Give the answer in mm²/mm.

Check yourself

A beam's links are found to be inadequate. The designer doubles the link area. What happens to V_Rd,max?

Summary

A cracked web behaves as a truss: chords, concrete struts, steel ties
V_Rd,s = (A_sw/s) z f_ywd cot θ — obtained by counting the links a cut crosses
V_Rd,max = b_w z ν₁ f_cd/(cot θ + tan θ) — obtained from the strut stress
Flatter struts engage more links but bring crushing closer; the two curves cross
Design strategy: flattest permitted angle first, steepen only if the struts crush
Once links are needed they carry the WHOLE shear — never add the concrete term
Adding links cannot fix a V_Rd,max failure; only more concrete can
The inclined compression field puts extra tension in the bottom chord — the shift rule

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This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint