EGB375 · Design of Concrete Structures
RC Column Sections — Interaction Diagram (Points A-D)
In EGB375 Design of Concrete Structures at Queensland University of Technology, a reinforced-concrete column rarely carries pure axial load — it resists an axial force N and a bending moment M together, so its strength is a curve in (M, N) space called the interaction diagram, not a single value. This chapter builds that envelope from four anchor points — A squash load Nuo (M = 0), B decompression (T = 0), C balanced (dn = 0.545 d, the nose / maximum moment) and D pure bending (N = 0) — with moments taken about the plastic centroid. The nominal curve is then reduced by the AS3600-2018 capacity factor φ = 0.6 (or 0.65) in compression rising to 0.85 in bending, and a design action (M*, N*) is adequate only when it plots inside the factored curve.
What this chapter covers
- 011. Why a column's strength is an interaction CURVE, not one number: axial force N and moment M act together
- 022. Section notation (b, D, Ag, As, Ast, Asc, d, dsc) and the steel limits ρmin = 1%Ag, ρmax = 8%Ag
- 033. The plastic centroid dpc = Σ(force·distance)/Σforce — the axis moments are taken about (= D/2 if symmetric)
- 044. Point A — squash load Nuo = α·f′c·(Ag − As) + fsy·As, with α = 1.0 − 0.003 f′c bounded 0.72–0.85
- 055. Point B — decompression: neutral axis at the tension steel (dn = d, T = 0), giving N and a modest M
- 066. Point C — balanced: dn = 0.545 d from strain compatibility (εcu = 0.003 with εsy = 0.0025) — the maximum-moment nose
- 077. Point D — pure bending (N = 0): solve equilibrium for dn, then Mu, and check ductility kuo = dn/do ≤ 0.36
- 088. Stress-block factors α₂ = 0.85 − 0.0015 f′c ≥ 0.67 and γ = 0.97 − 0.0025 f′c ≥ 0.67, and the compression-steel yield check
- 099. Capacity-reduction factor φ: 0.6 (or 0.65 short column, Q/G ≥ 0.25) in compression, up to 0.85 in bending
- 1010. The design check: plot the factored (φM, φN) curve and confirm the action (M*, N*) lies inside it
Squash load (Point A) and balanced point (Point C) of a tied column
- +1Factors. Ag = b·D = 350 × 450 = 157 500 mm². α = 1.0 − 0.003(40) = 0.88 → capped at 0.85. α₂ = 0.85 − 0.0015(40) = 0.79; γ = 0.97 − 0.0025(40) = 0.87 (both ≥ 0.67 ✓). Symmetric → plastic centroid dpc = D/2 = 225 mm. Steel ratio ρ = 2480/157 500 = 1.57% (within 1–8% ✓).
- +1Point A — squash load. Nuo = α·f′c·(Ag − As) + fsy·As = 0.85 × 40 × (157 500 − 2480) + 500 × 2480 = 34 × 155 020 + 1 240 000 = 5 270 680 + 1 240 000 = 6 510 680 N = 6510.7 kN.
- +1Design squash. Point A is compression-controlled, so φ = 0.6: φNuo = 0.6 × 6510.7 = 3906.4 kN. (M = 0 at A.)
- +1Point C — neutral axis. The balanced point has the concrete at εcu = 0.003 as the tension steel reaches εsy = 0.0025, so by similar triangles dn = 0.003/(0.003 + 0.0025)·d = 0.545 × 405 = 220.7 mm. Block depth γ·dn = 0.87 × 220.7 = 192.0 mm.
- +1Concrete force. Cc = α₂·f′c·(γ·dn)·b = 0.79 × 40 × 192.0 × 350 = 2 123 900 N = 2123.9 kN.
- +1Steel forces (check the compression steel). Tension steel yields by definition: T = Ast·fsy = 1240 × 500 = 620 000 N = 620 kN. Compression steel strain εsc = 0.003(220.7 − 45)/220.7 = 0.00239 < 0.0025, so it is ELASTIC: σsc = Es·εsc = 200 000 × 0.00239 = 477.7 MPa, giving Cs = Asc·σsc = 1240 × 477.7 = 592 300 N = 592.3 kN.
- +1Axial force at C. Nub = Cc + Cs − T = 2123.9 + 592.3 − 620 = 2096.2 kN.
- +1Moment at C (about dpc = 225 mm). Mub = Cc(dpc − γ·dn/2) + Cs(dpc − dsc) + T(d − dpc) = 2 123 900 × (225 − 96.0) + 592 300 × (225 − 45) + 620 000 × (405 − 225) = 274.0 + 106.6 + 111.6 = 492.2 kN·m (this is the maximum-moment nose). Design values φ = 0.6: φNub = 1257.7 kN, φMub = 295.3 kN·m.
Key terms
- Interaction diagram
- The strength envelope of a column drawn in (M, N) space: the locus of every axial force N and moment M the section can carry simultaneously. It is built by joining four anchor points — A (squash), B (decompression), C (balanced), D (pure bending). A factored (φM, φN) version is the design curve.
- Point A — squash load Nuo
- The pure-compression corner (M = 0): the whole section is uniformly squashed with concrete at α·f′c and all bars at yield. Nuo = α·f′c·(Ag − As) + fsy·As, with α = 1.0 − 0.003 f′c bounded 0.72 ≤ α ≤ 0.85. It is the maximum axial capacity — the top of the diagram.
- Point B — decompression
- The state where the neutral axis sits at the tension-steel level (dn = d), so that steel is at zero strain and carries no force (T = 0). The section is still wholly in compression: N = Cc + Cs, with a modest moment from the off-centre stress block.
- Point C — balanced (dn = 0.545 d)
- The state where the concrete reaches εcu = 0.003 at the same instant the tension steel reaches yield εsy = 0.0025. Similar triangles fix dn = 0.003/(0.003 + 0.0025)·d = 0.545 d directly. It is the nose of the interaction diagram — the point of maximum moment capacity (Nub, Mub).
- Point D — pure bending
- The N = 0 corner, where the column behaves as a doubly reinforced beam. Axial equilibrium T = Cc + Cs fixes a (small) neutral axis by solving a quadratic in dn, then Mu = Cc(d − γ·dn/2) + Cs(d − dsc). Ductility must satisfy kuo = dn/do ≤ 0.36.
- Plastic centroid (dpc)
- The line of action of the squash load — the force-weighted centroid of the fully-plastic section, dpc = Σ(force·distance)/Σforce. It is the axis all interaction-diagram moments are referenced to. For a symmetric column it is simply mid-depth, dpc = D/2.
- Equivalent rectangular stress block (α₂, γ)
- The AS3600-2018 idealisation of the concrete compression as a uniform stress α₂·f′c over a depth γ·dn from the compression face, where α₂ = 0.85 − 0.0015 f′c ≥ 0.67 and γ = 0.97 − 0.0025 f′c ≥ 0.67. Both vary with concrete strength — they are not fixed at 0.85.
- Capacity-reduction factor φ (columns)
- The AS3600 Table 2.2.2 strength-reduction factor: φ = 0.6 for compression-controlled points A, B, C (0.65 for a short column with Q/G ≥ 0.25), rising toward pure bending to 0.65 ≤ φ = 1.24 − 13·kuo/12 ≤ 0.85. Design requires the action (M*, N*) to plot inside the factored (φM, φN) curve.
RC Column Sections — Interaction Diagram (Points A-D) FAQ
Why does a column need an interaction diagram instead of a single capacity?
Because a column carries axial force N and bending moment M at the same time, and the two interact: adding axial compression increases the moment capacity below the balanced point but decreases it above the balanced point. There is no single 'strength', only a curve of the (M, N) combinations the section can carry. Computing the four anchor points A–D and joining them (then applying φ) is the exam-efficient way to draw that curve and test whether a design action lies safely inside it.
How do I know whether the compression steel has yielded at Points B, C or D?
Use strain compatibility from the linear strain diagram: εsc = 0.003(dn − dsc)/dn. If εsc ≥ εsy = 0.0025 the compression steel has yielded and σsc = fsy = 500 MPa; if εsc < 0.0025 it is still elastic and you must use σsc = Es·εsc (with Es = 200 000 MPa), which is smaller. Assuming yield when the steel is actually elastic overstates Cs and the capacity — it is one of the most common column-question errors, so always compute εsc before choosing the stress.
Can AI help me with RC column interaction diagrams in EGB375?
Yes — Sia is an AI tutor that can walk you through the method step by step: computing the plastic centroid, deriving each of the four points (squash, decompression, balanced dn = 0.545 d, pure bending), running the compression-steel yield check, applying the right φ, and testing whether (M*, N*) plots inside the factored curve. It is best used to build understanding and rehearse your own practice sections, and to explain why each AS3600 check exists. It will not sit the closed-book exam for you, and no tool can promise a particular mark or grade — always confirm factors and clauses against the unit outline and AS3600.
Studying with AI? Sia — free AI civil engineering tutor works through EGB375 step by step.
Exam move
Treat this as a fixed four-point routine and drill it until it is automatic: (1) compute α, α₂ and γ from f′c and locate the plastic centroid dpc; (2) Point A squash Nuo with the net area (Ag − As); (3) Point B at dn = d with T = 0; (4) Point C balanced at dn = 0.545 d — the maximum-moment nose; (5) Point D pure bending at N = 0 by solving the quadratic, then check kuo = dn/do ≤ 0.36; (6) apply φ (0.6 or 0.65 in compression, up to 0.85 in bending) and test whether (M*, N*) lies inside the factored curve. At every point, run the compression-steel strain check before choosing σsc, and keep forces in N and lengths in mm so moments land in N·mm (÷10⁶ for kN·m). Because the final is closed-book with five A4 double-sided student-prepared note sheets, put the four point-formulae, the α₂/γ expressions, this six-step workflow and one fully worked column on your own sheets. The exam is out of 100 marks and worth 50%; its duration is not fixed in advance, so confirm it on the QUT exam timetable for the ~June end-of-Semester-1 exam period (this unit next runs in Semester 1, 2027), then pace at a single rate of (confirmed exam minutes ÷ 100) minutes per mark and hold that rate across every question.
