EGB375 · Design of Concrete Structures
Shear Design of RC & PSC Beams (AS3600 MCFT)
Shear is the brittle failure mode of a concrete beam — it can fail with little warning — so EGB375 Design of Concrete Structures at Queensland University of Technology teaches the AS3600-2018 shear method as a fixed, defensible workflow rather than a single formula. The 2018 code models a cracked web as a variable-angle truss (modified compression field theory): the concrete forms diagonal compression struts whose angle is not fixed at 45° but varies with how hard the longitudinal steel is working. This chapter builds that chain for both reinforced (RC) and prestressed (PSC) beams — shear depth, mid-depth strain, strut angle, the web-crushing cap, the concrete contribution and the stirrup design.
What this chapter covers
- 01Why shear is the brittle mode, and the variable-angle truss (MCFT) idea
- 02The shear depth dv = max(0.72D, 0.9d) and where to cut V* and M*
- 03Mid-depth longitudinal strain ex = [M*/dv + |V*|] / (2 Es Ast) and its bounds
- 04The strut angle theta_v = 29 + 7000 ex (degrees)
- 05Web-crushing (maximum) capacity Vu,max and the phi = 0.75 check
- 06Concrete contribution Vuc = kv bv dv sqrt(f'c), factors kv and ks
- 07The decision: minimum stirrups only vs designed stirrups
- 08Stirrup design Vus, Asv/s = Vus tan(theta_v)/(fsy,f dv), and minimum stirrups
- 09Detailing: minimum steel and maximum spacing / torsion limits
- 10The PSC extension terms to the mid-depth strain
Worked example: designing the shear stirrups at a support
- +1Shear depth. dv = max(0.72×700, 0.9×640) = max(504, 576) = 576 mm.
- +1Section and moment. Cut at dv from the support face: x = 300/2 + 576 = 726 mm. M* = V*·x = 520 × 0.726 = 377.5 kN·m (≥ V*·dv = 299.5 kN·m, so it is consistent).
- +2Mid-depth strain. ex = [M*/dv + |V*|]/(2·Es·Ast) = [377.5×10^6/576 + 520×10^3] / (2×200000×4500) = (655382 + 520000)/1.8×10^9 = 6.53×10^-4 (within −0.2 to 3.0×10^-3).
- +1Strut angle. theta_v = 29 + 7000×0.000653 = 29 + 4.57 = 33.6°.
- +2Web-crushing check. sin(theta_v)cos(theta_v) = 0.5·sin(67.1°) = 0.461. Vu,max = 0.55×0.9×32×350×576×0.461 = 1471 kN; phi·Vu,max = 0.75×1471 = 1103 kN ≥ 520 kN, so the web does not crush.
- +2Concrete contribution. kv = 0.4/(1 + 1500×0.000653) = 0.202. Vuc = 0.202×350×576×√32 = 230 kN. D = 700 ≥ 650 so ks = 0.5; ks·phi·Vuc = 0.5×0.75×230 = 86.4 kN < 520 kN → design stirrups.
- +2Stirrup area. Vus = (V* − phi·Vuc)/phi = (520 − 0.75×230)/0.75 = 463 kN. Asv/s = Vus·tan(theta_v)/(fsy,f·dv) = 463000×0.663/(500×576) = 1.07 mm^2/mm; Asv,min/s = 0.08×350×√32/500 = 0.32 mm^2/mm (smaller, so the design governs).
- +1Choose bars. Two-leg N12: Asv = 2×110 = 220 mm^2; s = 220/1.07 = 206 mm → use 2-leg N12 @ 200 mm (provided 220/200 = 1.10 ≥ 1.07, and 200 mm < the maximum spacing 0.5D = 350 mm and < 300 mm).
Key terms
- Shear depth dv
- The internal lever arm used only for shear, dv = max(0.72D, 0.9d) in mm. For a typical beam the 0.9d term usually governs. It is neither the effective depth d nor the overall depth D.
- Mid-depth strain ex
- The longitudinal strain at mid-depth of the web, ex = [M*/dv + |V*|]/(2 Es Ast) for a cracked RC section (dimensionless). It measures how hard the tension steel is working and is bounded between -0.2x10^-3 and 3.0x10^-3.
- Strut angle theta_v
- The inclination of the diagonal concrete compression struts, theta_v = 29 + 7000 ex in degrees. A more strained web (larger ex) cracks at a flatter angle, so a diagonal crack crosses more stirrups.
- Vu,max (web crushing)
- The maximum shear before the diagonal struts crush the web, Vu,max = 0.55[0.9 f'c bv dv (sin theta_v cos theta_v)] in N. The design shear must satisfy V* <= phi Vu,max with phi = 0.75; if not, no stirrups help and the web must be enlarged.
- Vuc (concrete contribution)
- The shear carried by the concrete alone, Vuc = kv bv dv sqrt(f'c) in N (with sqrt(f'c) capped at f'c = 64). With at least minimum stirrups, kv = 0.4/(1 + 1500 ex).
- Size factor ks
- A depth-dependent factor applied in the stirrup decision ks phi Vuc: ks = 1.0 for D <= 300 mm, (1000 - D)/700 for 300 < D < 650 mm, and 0.5 for D >= 650 mm.
- Vus and Asv/s
- The stirrup shear contribution Vus = (V* - phi Vuc)/phi, converted to a required area per unit length Asv/s = Vus tan(theta_v)/(fsy,f dv) in mm^2/mm. fsy,f = 500 MPa for N (deformed) fitments, 250 MPa for R (plain) fitments.
- Minimum stirrups and max spacing
- Asv,min/s = 0.08 bv sqrt(f'c)/fsy,f in mm^2/mm; the spacing is capped at s <= min(300 mm, 0.5D, 0.12 un), where un is the closed-stirrup centreline perimeter. This applies even when only minimum stirrups are required.
Shear Design of RC & PSC Beams (AS3600 MCFT) FAQ
Why does the strut angle depend on the strain instead of being fixed at 45 degrees?
The AS3600-2018 method is based on modified compression field theory, which recognises that a more heavily strained web cracks at a flatter angle. theta_v = 29 + 7000 ex captures this: the flatter the strut, the more stirrups a diagonal crack crosses, so the calculation adapts the concrete and stirrup shares to how hard the longitudinal steel is actually working. A fixed 45 degrees would be a cruder, less economical assumption.
What is the difference between dv, d and D, and why does shear use dv?
D is the overall section depth and d is the effective depth to the tension-steel centroid. dv = max(0.72D, 0.9d) is a shear-only internal lever arm representing the truss depth. Using d or D in place of dv is a common error; every shear equation in this chapter (the strain, the web-crush cap, Vuc and the stirrup area) uses dv.
Can AI help me with shear design in EGB375?
Yes, for learning and rehearsal. Sia is an AI tutor that walks the dv -> ex -> theta_v -> Vu,max -> Vuc -> Vus workflow with you step by step, checks your reasoning on practice sections, and helps you interrogate why each AS3600 check is there. It supports your study rather than replacing it: it will not sit the closed-book exam for you, and no tool can promise a particular mark or grade. Always confirm code clauses against AS3600 and your unit materials.
Studying with AI? Sia — free AI civil engineering tutor works through EGB375 step by step.
Exam move
Treat shear as a fixed workflow, not a lookup: dv = max(0.72D, 0.9d), then the mid-depth strain ex, then the strut angle theta_v = 29 + 7000 ex, then the web-crushing cap Vu,max (check V* <= phi Vu,max first, because if it fails no stirrups help), then the concrete Vuc, then the ks phi Vuc decision, and only then the stirrup area Asv/s and the minimum-steel and maximum-spacing checks. Rehearse the chain end to end on a couple of sections until the order is automatic, keeping every force in newtons and every length in millimetres so ex comes out as a clean decimal times 10^-4. Because the final exam is closed-book and paper-based, condense the whole chain, the AS3600 factors (phi = 0.75 for N-type fitments) and one fully worked stirrup design onto your permitted note sheets. The exam is marked out of 100 and worth 50% of the unit; once you confirm its duration on the QUT exam timetable, divide those minutes by 100 to set a consistent minutes-per-mark rate and spend on a shear question in proportion to its marks.
