EGB375 · Design of Concrete Structures
Singly Reinforced Beam — Ultimate Moment Capacity & Flexural Design
In EGB375 Design of Concrete Structures at Queensland University of Technology, the singly reinforced beam is the foundation calculation for the whole unit: a concrete compression block balancing a steel tension force. This chapter derives the ultimate flexural capacity from horizontal equilibrium C = T — finding the neutral-axis depth dn, the lever arm Z = d − γ·dn/2 and the moment M_u = A_st·f_sy·Z — then enforces ductility with k_uo = dn/do ≤ 0.36 and checks design adequacy φM_u ≥ M* with φ = 0.85. Both analysis (find M_u) and design (find the steel A_st) are worked to AS3600-2018.
What this chapter covers
- 011. The two internal forces at ultimate: steel tension T = A_st·f_sy and the concrete stress block C = α₂·f′c·(γ·dn)·b
- 022. Horizontal equilibrium C = T and the neutral-axis depth dn = A_st·f_sy / (α₂·f′c·γ·b)
- 033. AS3600 stress-block factors α₂ = 0.85 − 0.0015 f′c ≥ 0.67 and γ = 0.97 − 0.0025 f′c ≥ 0.67
- 044. Lever arm Z = d − γ·dn/2 and the moment couple M_u = A_st·f_sy·Z (N·mm → kN·m)
- 055. Ductility gate: k_uo = dn/do ≤ 0.36 and the steel-yield strain check ε_st = 0.003(d−dn)/dn ≥ ε_sy = 0.0025
- 066. Under-reinforced vs balanced (ku = 0.545) vs over-reinforced sections
- 077. Design adequacy φM_u ≥ M* with the bending capacity-reduction factor φ = 0.85
- 088. Design-for-A_st: solving the quadratic M_u = f_sy·A_st(d − f_sy·A_st/(2α₂f′c·b)) then choosing bars and re-checking
Ultimate moment capacity of a singly reinforced beam
- +1Stress-block factors for f′c = 32 MPa: α₂ = 0.85 − 0.0015(32) = 0.802; γ = 0.97 − 0.0025(32) = 0.89 (both ≥ 0.67, so valid).
- +1Steel tension (bars assumed yielded): T = A_st·f_sy = 930 × 500 = 465 000 N = 465 kN.
- +1Neutral axis from equilibrium C = T: dn = T / (α₂·f′c·γ·b) = 465 000 / (0.802 × 32 × 0.89 × 250) = 465 000 / 5710 = 81.4 mm.
- +1Ductility check: k_uo = dn/do = 81.4 / 450 = 0.181 ≤ 0.36 ✓. Steel strain ε_st = 0.003(450 − 81.4)/81.4 = 0.0136 ≥ 0.0025 ✓, so the steel has yielded and T = A_st·f_sy holds.
- +1Lever arm: Z = d − γ·dn/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 = 192.4 × 10⁶ 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 idealisation of the curved concrete compression stress as a uniform stress α₂·f′c acting over a depth γ·dn from the compression face, where dn is the neutral-axis depth. The factors are α₂ = 0.85 − 0.0015 f′c ≥ 0.67 and γ = 0.97 − 0.0025 f′c ≥ 0.67.
- C = T equilibrium
- A beam carries no axial load, so the concrete compression resultant C = α₂·f′c·(γ·dn)·b must equal the steel tension T = A_st·f_sy. Setting them equal gives the neutral-axis depth dn = A_st·f_sy / (α₂·f′c·γ·b).
- Neutral-axis depth (dn)
- The depth from the compression face to the level of zero strain. Found from C = T. It sets both the depth of the compression block (γ·dn) and, via k_uo = dn/do, whether the section is adequately ductile.
- Lever arm (Z)
- The distance between the compression resultant C (acting at γ·dn/2 below the top fibre) and the tension resultant T (at the steel centroid, depth d): Z = d − γ·dn/2. The ultimate moment is the couple M_u = A_st·f_sy·Z.
- Ductility limit k_uo ≤ 0.36
- The AS3600 requirement that the neutral-axis parameter k_uo = dn/do (do = depth to the outermost tension bar) not exceed 0.36, ensuring an under-reinforced, ductile section that yields the steel and warns before failing. A balanced section has ku = 0.545; ku > 0.545 is over-reinforced and brittle.
- Steel-yield strain check
- From the linear strain diagram, ε_st = 0.003(d − dn)/dn must be ≥ ε_sy = f_sy/Es = 500/200000 = 0.0025 for the assumed tension T = A_st·f_sy to be valid (steel actually yielded).
- Capacity-reduction factor φ (bending)
- The strength-reduction factor applied to the nominal moment for a ductile flexural member: φ = 0.85. Design adequacy requires φM_u ≥ M*, where M* is the factored design moment.
- Design-for-A_st
- The inverse problem: given M*, find the steel. Substituting dn into Z leaves a quadratic M_u = f_sy·A_st(d − f_sy·A_st/(2α₂f′c·b)); set M_u = M*/φ, take the smaller root for A_st, choose bars (round the count up) and re-check k_uo and ε_st.
Singly Reinforced Beam — Ultimate Moment Capacity & Flexural Design FAQ
How do I know whether to use the singly reinforced formulae at all?
They apply when a rectangular section has tension steel only and is under-reinforced. After finding dn from C = T, check k_uo = dn/do ≤ 0.36 and ε_st = 0.003(d−dn)/dn ≥ 0.0025. If both pass, M_u = A_st·f_sy·Z is valid. If k_uo > 0.36 the section is over-reinforced — the steel has not yielded, so you must add compression steel (a doubly reinforced section) or enlarge the section instead.
What is the difference between d, do and dn, and why does it matter?
d is the effective depth to the centroid of the tension steel; do is the depth to the outermost tension bar layer; dn is the neutral-axis depth found from equilibrium. Use d in the lever arm Z = d − γ·dn/2 and do in the ductility ratio k_uo = dn/do. For a single bar layer d = do, but with two layers they differ, and mixing them up is a common mark-losing error.
Can AI help me with singly reinforced beam design in EGB375?
Yes — Sia is an AI tutor that can walk you through the method step by step: setting up C = T, solving for dn, running the k_uo ≤ 0.36 and steel-strain checks, and forming M_u = A_st·f_sy·Z, then checking φM_u ≥ M*. It is best used to build understanding and rehearse 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
Master this chapter first — it is the pattern every other reinforced-concrete topic reuses. Drill the fixed workflow until it is automatic: (1) compute the grade-dependent factors α₂ and γ; (2) T = A_st·f_sy; (3) dn from C = T; (4) the two ductility checks k_uo ≤ 0.36 and ε_st ≥ 0.0025; (5) Z = d − γ·dn/2; (6) M_u = T·Z, then φM_u with φ = 0.85. Work everything in N and mm and convert to kN·m only at the end. For design questions, practise inverting to the quadratic in A_st, taking the smaller root, rounding the bar count up and re-checking. Because the final exam is closed-book with five A4 double-sided student-prepared note sheets, put the stress-block factors, this six-step workflow and one worked analysis plus one worked design on your own sheets. Confirm the exam length and date on the QUT exam timetable, then pace by dividing the confirmed minutes across the 100 marks in proportion to each question's weight.
