EGB375 · Design of Concrete Structures
Concrete & Steel Materials + Equivalent Rectangular Stress Block
This chapter opens EGB375 Design of Concrete Structures at Queensland University of Technology (QUT) with the two material laws and the single idealisation that every later capacity calculation reuses. Reinforced concrete works because concrete carries compression and steel carries tension; AS3600-2018 lets you replace the curved concrete compressive stress with a simple equivalent rectangular stress block of stress α2 f′c acting over a depth γ·dn. Fix four constants once (εcu = 0.003, fsy = 500 MPa, Es = 200 GPa, εsy = 0.0025) and beam, slab, column and footing design all reduce to locating the neutral axis, writing force balance C = T, and taking moments.
What this chapter covers
- 01The AS3600-2018 (Clause 8.1.2) design assumptions: plane sections stay plane, concrete carries no tension, ultimate concrete strain εcu = 0.003, and perfect steel-concrete bond
- 02Why the true parabolic concrete stress (peak 0.9 f′c) is replaced by the equivalent rectangular block of stress α2 f′c over depth γ·dn
- 03The two block factors with units: α2 = 0.85 − 0.0015 f′c ≥ 0.67 and γ = 0.97 − 0.0025 f′c ≥ 0.67 (f′c in MPa)
- 04The fixed material constants: fsy = 500 MPa, Es = 200 GPa and the yield strain εsy = fsy/Es = 0.0025
- 05The elastic-perfectly-plastic steel model, N (normal) vs L (low) ductility bars, and standard single-bar areas (N12 = 110, N16 = 200, N20 = 310, N24 = 450 mm², and so on)
- 06The two internal resultants C = α2 f′c (γ·dn) b and T = Ast fsy, and the neutral-axis depth from C = T
- 07The ductility limit kuo = dn/do ≤ 0.36 and the strain-compatibility yield check εst ≥ 0.0025
- 08Applying the block to find the ultimate moment Mu = Ast fsy(d − γ·dn/2) and the design capacity φMu with φ = 0.85
- 09Keeping units straight (MPa × mm² = N; 1 kN·m = 10⁶ N·mm) and telling d, do and dn apart
Ultimate moment capacity of a singly reinforced beam using the stress block
- +1Block factors from f′c = 32 MPa: α2 = 0.85 − 0.0015(32) = 0.802; γ = 0.97 − 0.0025(32) = 0.89. Both exceed the 0.67 floor, so both stand.
- +1Steel area and tension force: A_st = 3 × 310 = 930 mm². Assume the steel has yielded, so T = A_st f_sy = 930 × 500 = 465 000 N = 465 kN.
- +2Neutral axis from C = T: d_n = A_st f_sy / (α2 f′c γ b) = 465 000 / (0.802 × 32 × 0.89 × 250) = 465 000 / 5710.5 = 81.4 mm.
- +1Ductility check: k_uo = d_n / d_o = 81.4 / 450 = 0.181 ≤ 0.36, so the section is under-reinforced and ductile.
- +1Yield check by strain compatibility: ε_st = 0.003(d − d_n)/d_n = 0.003(450 − 81.4)/81.4 = 0.0136 ≥ ε_sy = 0.0025, so the steel has yielded and T = 465 kN is valid.
- +1Lever arm: Z = d − γ·d_n/2 = 450 − 0.89 × 81.4 / 2 = 450 − 36.2 = 413.8 mm.
- +1Moment capacity: M_u = T·Z = 465 000 N × 413.8 mm = 1.924 × 10⁸ N·mm = 192.4 kN·m; design capacity φM_u = 0.85 × 192.4 = 163.5 kN·m.
Key terms
- Equivalent rectangular stress block
- The AS3600-2018 design idealisation that replaces the real curved concrete compressive stress with a uniform stress α2 f′c acting over a depth γ·d_n, chosen to keep the same resultant force and line of action so the moment can be found by simple statics.
- Block stress factor α2
- The dimensionless factor that sets the magnitude of the block stress: α2 = 0.85 − 0.0015 f′c, with a lower limit of 0.67 (f′c in MPa). It falls as concrete strength rises.
- Block depth factor γ
- The dimensionless factor that sets the block depth a = γ·d_n, where γ = 0.97 − 0.0025 f′c ≥ 0.67 (f′c in MPa). It also falls as concrete strength rises.
- Neutral-axis depth d_n
- The depth from the extreme compression fibre to the level of zero strain, found from horizontal equilibrium C = T. Concrete above it is in compression; below it the concrete is assumed to carry no tension.
- Ductility ratio k_uo
- The neutral-axis depth divided by the depth to the outermost tension bar, k_uo = d_n/d_o. AS3600 caps it at 0.36 to force a ductile, steel-yielding (under-reinforced) failure rather than a sudden concrete-crushing one.
- Yield strain ε_sy
- The strain at which reinforcing steel starts to yield: ε_sy = f_sy/E_s = 500 MPa / 200 000 MPa = 0.0025 (dimensionless). A tension bar strained to at least 0.0025 carries exactly its yield stress of 500 MPa.
- Effective depth d
- The distance from the extreme compression fibre to the centroid of the tension reinforcement; it sets the internal lever arm. It differs from d_o, the depth to the outermost bar layer, only when the steel is in more than one layer.
Concrete & Steel Materials + Equivalent Rectangular Stress Block FAQ
Why is the concrete assumed to carry no tension in these calculations?
Concrete is weak and unreliable in tension, so once the section cracks the concrete below the neutral axis is ignored and all tension is assigned to the steel. This is one of the AS3600-2018 (Clause 8.1.2) design assumptions, and it is what lets you write the tension force simply as T = A_st f_sy for a yielded bar.
How do I know whether the tension steel has actually yielded?
Check strain compatibility. From the linear strain diagram, ε_st = 0.003(d − d_n)/d_n; if this is at least the yield strain ε_sy = 0.0025 the steel has yielded and carries 500 MPa. Equivalently the section is under-reinforced when k_u is below the balanced value of 0.545, and the AS3600 ductility rule tightens this to k_uo = d_n/d_o ≤ 0.36.
Can AI help me with the equivalent rectangular stress block in EGB375?
Yes. Sia is an AI tutor that walks through the method step by step — how to read α2 and γ off f′c, set up C = T to find the neutral axis, run the ductility and yield checks, and take moments for M_u — using practice numbers so you learn the procedure. It supports your own working and revision; it does not sit your closed-book exam for you or promise any particular mark or grade, and you should always confirm code clauses and factors against AS3600 and the current unit materials.
Studying with AI? Sia — free AI civil engineering tutor works through EGB375 step by step.
Exam move
Treat this chapter as the foundation you drill until it is automatic, because almost every reinforced- and prestressed-concrete question in the unit begins here. Put the two block-factor equations, the four material constants and the standard bar areas on your permitted note sheets, then practise the full loop on varied numbers: read α2 and γ from f′c, write C = T to find d_n, report both the ductility check (k_uo ≤ 0.36) and the yield check (ε_st ≥ 0.0025), then take the moment M_u = A_st f_sy(d − γ·d_n/2) and apply φ = 0.85. The final is a closed-book, paper-based exam worth 50% of the unit, marked out of 100, with five A4 double-sided student-prepared note sheets (handwritten, typed, or a mix) permitted, held in the end-of-Semester-1 central examination period (around June for a Semester-1 offering — confirm the exact date and the paper duration on the QUT exam timetable). Once you have the duration, budget your time as (duration in minutes) ÷ 100 per mark and keep that same rate across every question. The marks that are most often dropped are unit slips between N·mm and kN·m and skipped ductility or yield checks, so carry units on every line and never leave the two checks unstated.
