EGB375 · Design of Concrete Structures
Doubly Reinforced Beams & T-Beams
This chapter of the EGB375 Design of Concrete Structures (Queensland University of Technology) exam guide takes the equivalent rectangular stress block from singly-reinforced flexure and applies it to two real cross-sections: beams with steel in the compression zone as well as the tension zone, and T-beams cast monolithically with a slab. You learn to write horizontal equilibrium as the quadratic Cc + Cs = T for the neutral-axis depth, to test whether the compression steel has yielded, and to run the Cf-vs-T flange test that decides whether a T-beam behaves as a wide rectangle or a true T. Every factor and check follows AS3600-2018, the code that governs the whole unit.
What this chapter covers
- 01Why real beams get compression steel: ductility, extra moment, and reduced long-term deflection
- 02Writing equilibrium for a doubly-reinforced section as C_c + C_s = T, a quadratic in the neutral-axis depth d_n
- 03The compression-steel strain test: elastic (C_s = A_sc E_s epsilon_sc) versus yielded (C_s = A_sc f_sy)
- 04Moment capacity M_u = C_c(d - gamma d_n/2) + C_s(d - d_sc) and the design check phi M_u >= M*
- 05Effective flange width b_ef for a simply-supported T-beam
- 06The flange test C_f,test = alpha_2 f'c gamma t_f b_ef versus T — rectangular action or true-T action
- 07True T-beam equilibrium: flange overhangs (C_f) plus web strip (C_w), solved for d_n
- 08The two mandatory checks carried through both cases: tension yield (epsilon_st >= 0.0025) and ductility (k_uo = d_n/d_o <= 0.36)
- 09Exam traps: wrong compression-steel branch, skipping the flange test, mixing the two C_f forms, and dropping phi
Moment capacity of a doubly-reinforced beam
- +1Stress-block factors for f'c = 40 MPa: alpha_2 = 0.85 - 0.0015(40) = 0.79; gamma = 0.97 - 0.0025(40) = 0.87 (both >= 0.67, valid).
- +1Tension force (assume the tension steel yields): T = A_st f_sy = 2250 x 500 = 1 125 000 N = 1125 kN.
- +1Assume the compression steel is elastic. C_s = A_sc E_s (0.003)(d_n - 50)/d_n = 240 000(d_n - 50)/d_n N; concrete force C_c = alpha_2 f'c gamma b d_n = 0.79 x 40 x 0.87 x 350 d_n = 9622 d_n N.
- +2Set C_c + C_s = T and multiply by d_n: 9622 d_n^2 - 885 000 d_n - 12.0x10^6 = 0. Positive root d_n = [885 000 + sqrt(885 000^2 + 4 x 9622 x 12.0x10^6)]/(2 x 9622) = 104.0 mm.
- +1Verify the elastic assumption: epsilon_sc = 0.003(104.0 - 50)/104.0 = 0.00156 < 0.0025, so the compression steel is indeed elastic — the quadratic branch was correct.
- +2Mandatory checks: tension strain epsilon_st = 0.003(550 - 104.0)/104.0 = 0.0129 >= 0.0025 (tension steel yields, so T = A_st f_sy is valid); ductility k_uo = d_n/d_o = 104.0/550 = 0.189 <= 0.36 (section is ductile, taking d_o approx d for one tension layer).
- +1Resolve the forces: C_c = 9622 x 104.0 = 1000.4 kN; C_s = 240 000(104.0 - 50)/104.0 = 124.6 kN (sum = 1125.0 kN = T, equilibrium checks out).
- +2Moment about the tension steel: M_u = C_c(d - gamma d_n/2) + C_s(d - d_sc) = 1000.4 x (0.550 - 0.0452) + 124.6 x (0.550 - 0.050) = 1000.4 x 0.5048 + 124.6 x 0.500 = 505.0 + 62.3 = 567.3 kN.m.
- +1Apply the bending capacity-reduction factor phi = 0.85: phi M_u = 0.85 x 567.3 = 482.2 kN.m.
Key terms
- Doubly reinforced beam
- A beam with longitudinal steel in the compression zone (A_sc) as well as the tension zone (A_st). Equilibrium becomes C_c + C_s = T, a quadratic in the neutral-axis depth.
- Compression steel (A_sc)
- Reinforcement placed near the compression face at depth d_sc. It lowers the neutral-axis depth (improving ductility), restrains creep (reducing long-term deflection), and adds a modest amount of moment capacity.
- Neutral-axis depth (d_n)
- Depth from the extreme compression fibre to the neutral axis, where the linear strain diagram is zero. The equivalent rectangular stress block has depth a = gamma d_n.
- Equivalent rectangular stress block
- AS3600 idealisation of the concrete compression stress as a uniform block of magnitude alpha_2 f'c over depth gamma d_n, with alpha_2 = 0.85 - 0.0015 f'c >= 0.67 and gamma = 0.97 - 0.0025 f'c >= 0.67.
- Effective flange width (b_ef)
- The width of slab that acts effectively as the compression flange of a T-beam; for a simply-supported beam b_ef = b_w + 0.2L or b_w + S_b depending on the span-to-spacing check.
- Flange test (C_f,test vs T)
- Assume the neutral axis is at the flange bottom (d_n = t_f) and compute C_f,test = alpha_2 f'c gamma t_f b_ef. If C_f,test >= T the section acts as a rectangle of width b_ef; if C_f,test < T it is a true T-beam.
- Ductility limit (k_uo)
- k_uo = d_n/d_o, where d_o is the depth to the outermost tension bar. AS3600 requires k_uo <= 0.36 so the section is under-reinforced and fails in a ductile (tension-controlled) manner.
- Compression-steel strain check
- epsilon_sc = 0.003(d_n - d_sc)/d_n. If epsilon_sc < 0.0025 the compression steel is elastic (C_s = A_sc E_s epsilon_sc); if epsilon_sc >= 0.0025 it has yielded (C_s = A_sc f_sy).
Doubly Reinforced Beams & T-Beams FAQ
What is the difference between rectangular and true T-beam action?
It depends on whether the compression block stays inside the flange. Run the flange test: with the neutral axis at the flange bottom (d_n = t_f), compute C_f,test = alpha_2 f'c gamma t_f b_ef and compare it with the steel force T = A_st f_sy. If C_f,test is greater than or equal to T the block fits within the flange, so you design a plain rectangular section of width b_ef. If C_f,test is less than T the block reaches into the web, and you must split the compression into the flange overhangs (C_f) and the web strip (C_w) and solve C_f + C_w = T.
Why doesn't adding compression steel greatly increase the moment capacity?
Because the concrete already supplies most of the compression force, so the extra couple from the compression steel is usually small — often only a few percent of M_u. The real benefits are elsewhere: compression steel lowers the neutral-axis depth d_n (which lowers k_uo and improves ductility) and restrains long-term creep (which reduces deflection). If an exam asks why a beam is doubly reinforced, lead with ductility and deflection, not moment.
Can AI help me with doubly reinforced beams and T-beams in EGB375?
Yes, as a study aid. Sia, the AskSia tutor, can explain the method step by step — how to set up the C_c + C_s = T quadratic, when the compression steel is elastic versus yielded, and how to run the C_f-vs-T flange test — and it can check your reasoning on a practice problem. Because EGB375's final is a closed-book invigilated paper, use it to build genuine understanding rather than to obtain answers: it will not sit your exam or guarantee a grade, and it should never be used during an assessment. Confirm all factors against AS3600-2018 and your unit outline.
Studying with AI? Sia — free AI civil engineering tutor works through EGB375 step by step.
Exam move
Put the two decision lines on one of your five permitted A4 note sheets: for a doubly-reinforced section write C_c + C_s = T and the elastic-compression-steel quadratic; for a T-beam write the flange test C_f,test = alpha_2 f'c gamma t_f b_ef versus T. Practise until you instinctively state which equation governs before touching numbers, because that framing secures the method marks even if the arithmetic slips. Every solution must show the same skeleton — stress-block factors, equilibrium, the solved d_n, the compression-steel strain check, the tension-yield and ductility checks (epsilon_st >= 0.0025 and k_uo <= 0.36), the moment, then phi M_u with phi = 0.85 — since marks are spread across those steps, not saved for the final number. The final exam is closed-book, paper-based and marked out of 100 (worth 50% of the unit); the duration is not fixed in the unit materials, so confirm it on the QUT exam timetable and then budget your time strictly in proportion to the marks on each part. Keep your units in SI throughout and divide N.mm by ten to the sixth for kN.m so a T-beam capacity never comes out a thousand times too large.
