Reinforced Concrete Design / module 5
Shear, bond, anchorage, laps and torsion
The failures that are sudden. A truss you can derive, an empirical formula you cannot, and the reason a bar must be given length before it can be given a force.
What you will understand
- Explain why a diagonal crack forms, and why the failure is brittle
- Derive the variable-angle truss from equilibrium alone
- Choose a strut angle, and say what you are trading when you do
- Distinguish the empirical concrete term from the mechanical truss term
- Design links for a beam, and check the struts do not crush
- Derive the anchorage length from a force balance on one bar
- Curtail a bar correctly, accounting for both the shift and the anchorage
- Distinguish equilibrium torsion from compatibility torsion, and design for the first
Lessons
Why shear is the dangerous one
Diagonal tension, brittle failure, and why a beam without links gives no warning at all.
The variable-angle truss
Once the web has cracked, the beam is a truss. Derive it by cutting, and the design equations fall out.
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.
Torsion
The same truss, wrapped round a tube — and the one question you must answer before designing for it at all.
Module summary
- Shear failure is diagonal tension failure — concrete cracks perpendicular to the principal tension
- Without links, failure is sudden and without warning; this is why minimum links exist
- VRd,c is EMPIRICAL — a curve fit, with no mechanism behind it
- VRd,s is a MECHANISM — a truss, derivable by cutting and resolving
- Once links are needed, they carry the WHOLE shear; the concrete term is not added
- A flatter strut engages more links but pushes the struts closer to crushing
- Bond fails by splitting, so it depends on the concrete's TENSILE strength
- lb,rqd = (φ/4)(σsd/fbd) is pure equilibrium, not a code rule
- A bar stops at the theoretical cut-off PLUS the shift PLUS a full anchorage
Teaching basis and review status
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 notesCheck your understanding
Check yourself
What is physically happening when a beam 'fails in shear'?
Practice
A beam has bw = 300 mm, d = 450 mm, C30/37 (fcd = 17 N/mm²). At cot θ = 2.5, what is VRd,max, in kN?
Practice
For a beam with z = 405 mm, fywd = 434.8 N/mm² and cot θ = 2.5, what shear can H8 two-leg links at 175 mm centres carry, in kN? Take Asw = 100.5 mm².
Check yourself
A beam needs 250 kN of shear resistance. VRd,c is 95 kN. How much shear must the links be designed for?
Practice
An H20 bar in C30/37, good bond conditions, stressed to 435 N/mm² with fbd = 3.04 N/mm². What is lb,rqd, in mm?
Practice
A beam has z = 495 mm and is designed at cot θ = 2.5. What is the shift distance al that the bending moment diagram must be displaced by, in mm?
Check yourself
Two beams are identical except that one has d = 250 mm and the other d = 1000 mm. Which has the higher shear STRESS capacity vRd,c, all else equal?