Module 5 · Lesson 5.3
Bond, anchorage and laps
A bar can only be given a force over a length. Get that length wrong and the flexural design is worthless.
Why this matters
You can design the reinforcement in a beam perfectly and still have the beam fail, because a bar with the right area in the right place carries nothing at all unless it has been given enough length to develop its force. Anchorage failures are brittle, they happen at supports and laps rather than midspan, and they are invisible on a section drawing. Of all the topics in concrete design, this is the one where the calculation is easiest and the consequence of ignoring it is most severe.
By the end of this lesson you should be able to
- Explain bond as a splitting phenomenon, not a sliding one
- Derive the required anchorage length from equilibrium of one bar
- Explain why bond depends on the concrete's tensile strength
- Apply the α factors and know why their product is floored
- Calculate a lap length and say why laps are longer than anchorages
- Curtail a bar correctly, allowing for both shift and anchorage
What you should already know
- Bond and composite action (Module 1)
- The variable-angle truss and the shift rule (previous lesson)
Bond does not fail by sliding
The mental picture most people carry is of a bar being pulled through the concrete like a nail out of wood, resisted by friction along its surface. That picture predicts the wrong dependencies and the wrong failure mode.
A ribbed bar does not slide. Its ribs bear against the concrete, and because the ribs are inclined the bearing force has a radial component pushing outward, all around the bar. The concrete surrounding the bar therefore acts as a thick-walled cylinder under internal pressure — and it fails the way such a cylinder fails, by splitting along a radial crack to the nearest free surface.
Three consequences follow immediately, and all three are visible in the design rules:
- Bond strength depends on the concrete's tensile strength, not its compressive strength. That is what resists the splitting.
- Cover matters enormously. A bar near a face has a thin ring of concrete to split; a bar deep in a section has a thick one.
- Links around the bar help, because they confine the splitting crack in the same way hoops confine a barrel.
Every α factor in the anchorage equation is one of these three effects.
What it calculates: The stress that can be transferred between a bar and the concrete around it.
- 1.0 for good bond conditions, 0.7 for poor (—)
- 1.0 for bars up to 32 mm, (132 − φ)/100 above (—)
- Design TENSILE strength of the concrete (N/mm²)
This assumes
- Ribbed bars — plain round bars have far lower bond and are not covered
- EMPIRICAL: the 2.25 is a fitted coefficient
- Adequate cover and no splitting failure — the α factors handle departures
In plain terms: Note what is not in this equation: fck appears only through fctd. Doubling the concrete's compressive strength does very little for bond, because tensile strength grows roughly as the two-thirds power. And note η₁: a bar in the top of a deep pour has bleed water collecting under it, leaving a weak porous layer along its underside. That is a 30% penalty, decided entirely by casting position.
From first principles
The anchorage length, from one bar in equilibrium
We want to show: To find how far a bar must be embedded to develop a given stress, using nothing but a force balance.
A bar carries a force. That force has to be handed over to the concrete, and the only place the handover can happen is the bar's curved surface. So the question is a comparison of two areas: the bar's cross-section, over which the force is carried, and the bar's surface, over which it is delivered. The length needed is whatever makes the second big enough to supply the first. That is genuinely all there is to it, and the famous φ/4 in the answer is nothing more mysterious than the ratio of a circle's area to its circumference.
Worked example
Anchorage and lap for an H25 bar
Given
- H25 tension bar, fully stressed: σsd = fyd = 435 N/mm²
- C30/37 concrete
- Good bond conditions — the bar is in the bottom of the pour
- Straight bar, ordinary cover, no special confinement, so all α = 1.0
Find
The design anchorage length, and the lap length if all bars are lapped at one section.
Assumptions
- fctm is the two-thirds power of fck scaled by 0.30; fctk,0.05 = 0.7 fctm; fctd = fctk,0.05/γC
- γC = 1.5, nationally determined and requiring verification
Practice
An H16 bar in C30/37 with good bond conditions is stressed to 435 N/mm². Taking fbd = 3.04 N/mm², what is the basic anchorage length lb,rqd, in mm?
Practice
The same H16 bar is in the TOP of a 600 mm deep pour, so bond conditions are poor. What is the basic anchorage length now, in mm?
Practice
50% of bars are lapped at one section. What is the factor α₆ applied to the lap length?
Check yourself
Why does bond strength depend on the concrete's tensile strength rather than its compressive strength?
Summary
- Ribbed bars fail bond by SPLITTING the surrounding concrete, not by sliding
- So bond depends on tensile strength, and on cover and confinement
- lb,rqd = (φ/4)(σsd/fbd) is a pure force balance — φ/4 is area over circumference
- Anchorage length is proportional to bar diameter: large bars are hard to anchor
- Poor bond conditions cost 43% extra length, decided by casting position alone
- Laps are anchorages back to back, lengthened by α₆ when many bars lap together
- Staggering laps saves a third of the length and costs nothing
- A bar stops at the theoretical cut-off PLUS the shift PLUS a full anchorage
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