EGB375 Design of Concrete Structures
The Complete Exam Bible · S1 2027EGB375 Overview
- Queensland University of Technology
- S1 2027
- 12 chapters
- Engineering
EGB375 Design of Concrete Structures is a third-year civil-engineering unit at Queensland University of Technology, worth 12 credit points, taught in the Faculty of Engineering and graded on GPA (QUT's 7-point grade-point average).
- Assessed by Self-assessed Homework Reflection 10% · Design Project — Oral Poster Presentation 15% + 2 more
- Key terms Equivalent rectangular stress block, alpha2 and gamma, Neutral axis depth dn, Lever arm Z
- How to prepare Treat EGB375 as a set of repeatable design workflows rather than a reading unit: for each element…
- Most asked Where can I find past exam papers or practice for EGB375?
How EGB375 is assessed
| Component | Weight | Format |
|---|---|---|
| Self-assessed Homework Reflection | 10% | Single end-of-semester self-assessment form (student awards own grade 1-10 across ~11 weekly HW problems + 100-word justification), Week 13 |
| Design Project — Oral Poster Presentation | 15% | Group of 4, out of 15 marks (10 group presentation + 5 individual Q&A), Week 7 |
| Design Project — Written Report | 25% | Individual written report out of 25 marks, Week 11 |
| Final Exam | 50% | Closed-book, paper-based, out of 100, QUT Central Examinations; five A4 double-sided student-prepared note sheets permitted (handwritten, typed, or a mix); duration not stated |
What EGB375 covers
Twelve chapters walk EGB375 from the equivalent stress block through reinforced- and prestressed-concrete flexure, shear, serviceability, slabs, columns and footings — every element worked to AS3600-2018 for the QUT final.
Concrete & Steel Materials + Equivalent Rectangular Stress Block
AS3600 8.1.2/3.1.4 assumptions; alpha2 & gamma factors; fsy=500, ecu=0.003, Es=200GPa; N/L bars; bar areasThis 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.
- The AS3600-2018 (Clause 8.1.2) design assumptions: plane sections stay plane, concrete carries no tension, ultimate concrete strain ε<sub>cu</sub> = 0.003, and perfect steel-concrete bond
- Why the true parabolic concrete stress (peak 0.9 f′c) is replaced by the equivalent rectangular block of stress α<sub>2</sub> f′c over depth γ·d<sub>n</sub>
- The two block factors with units: α<sub>2</sub> = 0.85 − 0.0015 f′c ≥ 0.67 and γ = 0.97 − 0.0025 f′c ≥ 0.67 (f′c in MPa)
- The fixed material constants: f<sub>sy</sub> = 500 MPa, E<sub>s</sub> = 200 GPa and the yield strain ε<sub>sy</sub> = f<sub>sy</sub>/E<sub>s</sub> = 0.0025
- The 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)
- The two internal resultants C = α<sub>2</sub> f′c (γ·d<sub>n</sub>) b and T = A<sub>st</sub> f<sub>sy</sub>, and the neutral-axis depth from C = T
Singly Reinforced Beam — Ultimate Moment Capacity & Flexural Design
C=T equilibrium, dn, Z=d-gamma*dn/2, Mu=Ast*fsy*Z, ductility kuo<=0.36, phi=0.85, design-for-AstIn 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.
- 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
- Horizontal equilibrium C = T and the neutral-axis depth dn = A_st·f_sy / (α₂·f′c·γ·b)
- AS3600 stress-block factors α₂ = 0.85 − 0.0015 f′c ≥ 0.67 and γ = 0.97 − 0.0025 f′c ≥ 0.67
- Lever arm Z = d − γ·dn/2 and the moment couple M_u = A_st·f_sy·Z (N·mm → kN·m)
- 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
- Under-reinforced vs balanced (ku = 0.545) vs over-reinforced sections
Doubly Reinforced Beams & T-Beams
Cc+Cs=T quadratic for dn, compression-steel strain check; T-beam flange test Cf vs T, rectangular vs T actionThis 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.
- Why real beams get compression steel: ductility, extra moment, and reduced long-term deflection
- Writing equilibrium for a doubly-reinforced section as C_c + C_s = T, a quadratic in the neutral-axis depth d_n
- The compression-steel strain test: elastic (C_s = A_sc E_s epsilon_sc) versus yielded (C_s = A_sc f_sy)
- Moment capacity M_u = C_c(d - gamma d_n/2) + C_s(d - d_sc) and the design check phi M_u >= M*
- Effective flange width b_ef for a simply-supported T-beam
- The flange test C_f,test = alpha_2 f'c gamma t_f b_ef versus T — rectangular action or true-T action
Prestressed Concrete — Mechanics, Allowable Stresses & Magnel Method
eccentric fibre stresses (tension +ve), transfer/service stages, allowable fci/fti/fc/ft, Magnel 4 equations, parabolic tendonThis chapter of EGB375 Design of Concrete Structures at Queensland University of Technology covers the elastic mechanics of prestressed concrete (PSC) beams: how an eccentric tendon produces top- and bottom-fibre stresses, how those stresses must satisfy code limits at both the transfer and service stages, and how the Magnel method turns four stress checks into four inequalities that bracket the initial prestress force.
- Model an eccentric tendon as an axial force −P/A plus a moment P·e about the section centroid
- Write the top- and bottom-fibre elastic stresses with the correct AS3600 sign convention (tension +ve)
- Distinguish the transfer stage (full prestress, self-weight only) from the service stage (reduced prestress, full load)
- Compute the transfer moment Mₒ and the short-term service moment Mₛ = W_G + 0.7W_Q
- State the four allowable stresses: f_ci and f_ti at transfer, f_c and f_t at service
- Assemble the Magnel four inequalities and read the feasible prestress region on the Magnel diagram
PSC — Losses, Tendon Sizing & Ultimate Flexural Capacity
losses 10%/20%, jacking force & strand count, sigma_pu=fpb(1-k1k2/gamma), dn, Mu for fully/partially prestressedThis chapter of EGB375 Design of Concrete Structures at Queensland University of Technology closes the prestressed-concrete (PSC) design loop: from a required prestress force it sizes the tendon and then checks ultimate flexural strength to AS3600-2018.
- Why prestress is lost: immediate (~10%) vs time-dependent (~20%) losses, ~30% total
- Maximum jacking force per AS3600-2018 §17.3.4.6 — the 0.80 / 0.85 / 0.75 f_pb·A_p caps
- Jacking force P_j = P_i/(1 − immediate loss) and converting it to a strand count
- Total tendon area A_pt from the number of strands (always round UP)
- Ultimate tendon stress σ_pu = f_pb(1 − k1·k2/γ) and the reinforcement index k2
- AS3600 rectangular stress-block factors α₂ and γ as functions of f′c
Shear Design of RC & PSC Beams (AS3600 MCFT)
dv=max(0.72D,0.9d), ex mid-depth strain, theta_v=29+7000ex, Vu,max web crushing, Vuc=kv*bv*dv*sqrt(fc), Vus stirrups, min stirrups & torsion spacingShear 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.
- Why shear is the brittle mode, and the variable-angle truss (MCFT) idea
- The shear depth dv = max(0.72D, 0.9d) and where to cut V* and M*
- Mid-depth longitudinal strain ex = [M*/dv + |V*|] / (2 Es Ast) and its bounds
- The strut angle theta_v = 29 + 7000 ex (degrees)
- Web-crushing (maximum) capacity Vu,max and the phi = 0.75 check
- Concrete contribution Vuc = kv bv dv sqrt(f'c), factors kv and ks
Serviceability I — Crack Width & Spacing
shrinkage strain & residual stress, cracking stress fct.f=0.6sqrt(fc), tension chord hc,eff, Sr,max, Wcr<=w'maxA concrete beam that is strong enough at the ultimate limit state can still crack too widely in service — letting in moisture, corroding the steel and looking poor — so EGB375 Design of Concrete Structures at Queensland University of Technology teaches crack control to AS3600-2018 as a fixed workflow.
- Why serviceability cracking matters and the master check Wcr <= w'max
- Shrinkage strain ecs as a looked-up value (AS3600 Table 3.1.7.2)
- Residual shrinkage stress scs = ecs Es (2.5 pw - 0.8 pcw)/(1 + 50 pw)
- Flexural tensile strength fct.f = 0.6 sqrt(f'c) and cracking stress scr = fct.f - scs
- The effective tension chord hc,eff = min[2.5(D-d), (D-dn)/3, D/2] and Ac,eff
- The effective reinforcement ratio rho_eff = Ast/Ac,eff and why it controls cracking
Serviceability II — Steel Stresses, Section Inertia & Deflection
Mcr, gross/cracked transformed section Ig/Icr, Ief tension stiffening, sigma_scr, 10M-method deflection, kcs, span/250 & span/500 limitsIn EGB375 Design of Concrete Structures at Queensland University of Technology, a beam that is strong enough at ultimate can still deflect or crack too much in service.
- The cracking moment M_cr = (f_ct.f − σ_cs)·I_g/(D − dn), with f_ct.f = 0.6√f′c, and why residual shrinkage stress σ_cs lowers it
- Deciding cracked vs uncracked by comparing the service moment M_s with M_cr
- The transformed section and modular ratio n = E_s/E_c (≈ 6.6 for f′c = 32 MPa)
- Gross inertia I_g = bD³/12 + bD(dn − D/2)² + (n−1)A_st(d − dn)² about the gross neutral axis
- Cracked neutral axis b·dn²/2 = n·A_st(d − dn) and cracked inertia I_cr = b·dn³/3 + n·A_st(d − dn)²
- Effective inertia (tension stiffening) I_ef = I_cr / [1 − (1 − I_cr/I_g)(M_cr/M_s)²] ≤ I_g
RC Slabs — One-Way & Two-Way + Deemed-to-Comply Deflection
aspect ratio Ly/Lx, span/depth Lef/d formula (k3,k4,Fd,ef), coefficient method Mx=bx*q*Lx^2, min & crack-control steel, corner meshIn EGB375 Design of Concrete Structures at Queensland University of Technology, a reinforced-concrete slab is designed as a wide, shallow beam on a 1 m strip.
- Classifying a slab by aspect ratio L_y/L_x — one-way (≥ 2) vs two-way (< 2), with L_x the shorter span, and designing a 1 m-wide strip (b = 1000 mm)
- Ultimate design pressure q* = 1.2 qG + 1.5 qQ (kPa), with qG including slab self-weight = Ds × 25 kN/m³
- The AS3600 9.4.4.1 deemed-to-comply span/depth rule Lef/d ≤ k3·k4[(Δ/Lef)(1000·Ec)/Fd,ef]^(1/3) and why Ec (MPa) is ×1000 for kPa units
- Support factors k3 = 1.0, k4 = 1.4 (SS one-way) or Table 9.4.4.2 for two-way, and the effective deflection load Fd,ef with kcs = 2.0 for slabs
- Total vs incremental deflection: Fd,ef(total) = (1+kcs)qG + (νs+kcs·ν1)qQ against Fd,ef(incr) = kcs·qG + (νs+kcs·ν1)qQ
- Main flexural steel on the 1 m strip (singly reinforced, φ = 0.85) and the three-way minimum-steel floor ρmin = 0.19(Ds/d)²(fct.f/fsy)
RC Column Sections — Interaction Diagram (Points A-D)
squash Nuo, decompression B, balanced C (dn=0.545d), pure bending D; plastic centroid; phi=0.6/0.65 compression, up to 0.85 bendingIn 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.
- Why a column's strength is an interaction CURVE, not one number: axial force N and moment M act together
- Section notation (b, D, Ag, As, Ast, Asc, d, dsc) and the steel limits ρmin = 1%Ag, ρmax = 8%Ag
- The plastic centroid dpc = Σ(force·distance)/Σforce — the axis moments are taken about (= D/2 if symmetric)
- 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
- Point B — decompression: neutral axis at the tension steel (dn = d, T = 0), giving N and a modest M
- Point C — balanced: dn = 0.545 d from strain compatibility (εcu = 0.003 with εsy = 0.0025) — the maximum-moment nose
Slender Columns — Slenderness & Moment Magnification
r=0.3D, Le=kLu, short/slender limit Le/r, Euler-type Nc, magnifier delta_b=km/(1-N*/Nc), km=0.6-0.4(M1/M2), Mmax=delta*M2A short column fails by crushing at its section capacity, but a slender column bows sideways first, so the axial load acts at a deflection and adds a second-order (P–δ) moment.
- Why slender columns buckle: the P-delta second-order moment
- Radius of gyration r = 0.3D for a rectangular section (and why 0.289D rounds up)
- Effective length Le = kLu and the braced/unbraced k values (0.70, 0.85, 1.0; 1.2-2.2 sway)
- The slenderness ratio Le/r and evaluating it about each axis
- The short/slender limit: braced max{25, (38 - f'c/15) sqrt(1 + M1*/M2*)}; unbraced < 22
- The single- vs double-curvature sign convention on M1*/M2* (negative vs positive)
Shallow (Isolated Pad) Footing Design
qa,net bearing/size, punching shear fcv & Vuo at dom/2, flexure at 0.7asup, flexural shear at dv; four checks (service size, ultimate strength)An isolated pad footing spreads a single column load onto the soil, and in EGB375 Design of Concrete Structures at Queensland University of Technology it is designed to AS3600-2018 with four checks in two families: service (unfactored) loads fix the plan size so the soil is not overstressed, while factored ultimate loads verify the concrete strength.
- Why a footing is a thick upside-down cantilever, and the service-vs-ultimate split
- Net allowable soil pressure qa,net = qa - 25D and the size (bearing) check
- Ultimate net soil pressure Pnet = (1.2 NG + 1.5 NQ)/Apad
- Punching (two-way) shear at dom/2: perimeter u, Vuo = u dom fcv and phi = 0.7
- The punching stress fcv = 0.17(1 + 2/beta_h) sqrt(f'c) and its 0.34 sqrt(f'c) cap
- Flexure at 0.7 a_sup: the cantilever moment M* and the design steel
It is the reinforced- and prestressed-concrete design unit of the program, governed throughout by the Australian Standard AS3600-2018. It carries you from the equivalent rectangular stress block through the ultimate moment and shear capacity of beams, then serviceability, slabs, columns and isolated footings, always as a design calculation rather than pure theory.
This free guide mirrors how EGB375 is assessed at Queensland University of Technology and works every element end to end for the closed-book final that decides half the unit mark.
Worked example · free
Ultimate moment capacity of a singly reinforced beam (AS3600-2018)
- +2Stress-block factors (grade-dependent): alpha2 = 0.85 - 0.0015(32) = 0.802; gamma = 0.97 - 0.0025(32) = 0.89 (both >= 0.67, so valid).
- +1Steel tension (steel assumed yielded): T = Ast.fsy = 930 x 500 = 465 000 N = 465 kN.
- +2Neutral axis from horizontal equilibrium C = T: dn = T / (alpha2.f'c.gamma.b) = 465 000 / (0.802 x 32 x 0.89 x 250) = 465 000 / 5710 = 81.4 mm.
- +2Ductility gate: kuo = dn/do = 81.4/450 = 0.181 <= 0.36 (ductile); strain check eps_st = 0.003(450 - 81.4)/81.4 = 0.0136 >= 0.0025, so the steel has yielded and T = Ast.fsy holds.
- +1Lever arm of the internal couple: Z = d - gamma.dn/2 = 450 - 0.89 x 81.4/2 = 450 - 36.2 = 413.8 mm.
- +1Moment capacity: Mu = T.Z = 465 000 N x 413.8 mm = 1.924 x 10^8 N.mm = 192.4 kN.m.
- +1Design capacity for bending (phi = 0.85): phi.Mu = 0.85 x 192.4 = 163.5 kN.m, to be compared with the factored moment M*.
Key terms
- Equivalent rectangular stress block
- The AS3600 idealisation of the concrete compression zone as a uniform stress alpha2.f'c acting over a depth gamma.dn from the compression face, replacing the true curved stress distribution.
- alpha2 and gamma
- Grade-dependent stress-block factors: alpha2 = 0.85 - 0.0015 f'c >= 0.67 and gamma = 0.97 - 0.0025 f'c >= 0.67 (e.g. 0.802 and 0.89 at f'c = 32 MPa).
- Neutral axis depth dn
- The depth from the compression face to the level of zero strain, found from horizontal equilibrium C = T; dn = Ast.fsy / (alpha2.f'c.gamma.b) for a singly reinforced section.
- Lever arm Z
- The distance between the concrete compression resultant and the steel tension resultant, Z = d - gamma.dn/2; the moment capacity is the couple T.Z.
- Ductility ratio kuo
- kuo = dn/do; AS3600 requires kuo <= 0.36 so the section is under-reinforced and the steel yields (warns) before the concrete crushes.
- Capacity reduction factor phi
- The strength-reduction factor applied to nominal capacity: phi = 0.85 for bending, 0.75 for shear with N-type fitments, 0.7 for footing punching, 0.6 (0.65 short) for column compression.
- Prestress ultimate stress sigma_pu
- The tendon stress at ultimate, sigma_pu = fpb(1 - k1.k2/gamma), used with the neutral-axis depth to find the prestressed moment capacity Mu.
- Shear depth dv
- The effective shear (lever-arm) depth, dv = max(0.72D, 0.9d), at which V* and M* are taken and used in the MCFT-based shear equations.
- Strut angle theta_v
- The compression-strut inclination in the MCFT shear model, theta_v = 29 + 7000.eps_x, where eps_x is the mid-depth longitudinal strain.
- Effective second moment Ief
- The tension-stiffening inertia used for deflection, Ief = Icr / [1 - (1 - Icr/Ig)(Mcr/M*)^2] <= Ig, interpolating between cracked and gross behaviour.
- Interaction diagram
- The column strength envelope in (M, N) space defined by Points A (squash), B (decompression), C (balanced, dn = 0.545d) and D (pure bending); the design point (M*, N*) must lie inside the factored curve.
- Moment magnifier delta_b
- The slender-column factor delta_b = km/(1 - N*/Nc) >= 1.0 that amplifies the design moment to Mmax = delta_b.M2* once Le/r exceeds the short-column limit.
- Punching shear
- The two-way shear check on a footing or slab around a column, at a critical perimeter u a distance dom/2 from the face; require V*punch <= phi.Vuo with fcv <= 0.34.sqrt(f'c).
- AS3600-2018
- The Australian Standard for Concrete Structures that governs every design factor, capacity equation and detailing rule examined in EGB375.
EGB375 FAQ
Can AI help me study EGB375?
Yes. Sia is an AI study tutor that explains EGB375 topics step by step, so you can ask it to walk through the C = T to dn to ductility to Z to Mu flexure workflow, the MCFT shear procedure, or a column interaction check, and have it show each AS3600 step and unit conversion.
It is built to help you understand and rehearse the method on your own practice sections, not to hand you homework or exam answers, and no tool can promise a particular mark or grade.
Where can I find past exam papers or practice for EGB375?
Start inside Queensland University of Technology's own systems: previous exam problems, tutorial questions and homework solutions are posted on the EGB375 Canvas site, and the free unit workbook is the primary practice source. This guide adds ten worked, exam-style design problems with fresh numbers so you can practise the same skills without reproducing any real paper; use Sia to check your reasoning on each step.
What can Sia do that a textbook can't?
A textbook shows one worked example and then stops; Sia responds to your specific working, so when your dn or your stress-block factor is off it can pinpoint the line and explain why, and it will re-explain a step as many times and as many ways as you need.
It can generate a fresh practice section, walk the AS3600 check you are stuck on, and adapt to what you already know, all step by step and without ever promising an answer key or a grade.
Is EGB375 hard?
It is a design-calculation unit, so the challenge is procedural rather than abstract: you must apply the correct AS3600 factors and checks to each element and carry units cleanly, which rewards steady practice over memorisation. Students who build fluency with the flexure, shear and serviceability workflows and keep every calculation in N and mm tend to find it manageable.
Is the EGB375 exam open or closed book?
It is a closed-book, paper-based final scheduled by QUT Central Examinations, marked out of 100 and worth 50% of the unit. You may bring five A4 double-sided student-prepared note sheets, so it is not open-book and there is no official formula sheet; the duration is not stated in the unit materials, so confirm it on the QUT exam timetable.
What is examined in EGB375, and is there a hurdle?
The final is design-computation heavy: ultimate moment and shear capacity for reinforced and prestressed members, serviceability (cracking and deflection), slab, column and footing design, and reading shear-force and bending-moment diagrams, all to AS3600-2018. No hurdle is stated on any single component in the unit materials, but the 50% exam is the largest single piece of the mark, so confirm the current rules on Canvas and the unit outline.
What GPA do I need for a good grade in EGB375?
EGB375 is graded on GPA, Queensland University of Technology's 7-point grade-point average, and each grade maps to a band: grade 7 (High Distinction), grade 6 (Distinction), grade 5 (Credit) and grade 4 (Pass) - you need a 4 to pass.
Your unit grade is built from the three assessment pieces (10% self-assessed homework reflection, 40% design project, 50% final exam), so a strong grade means scoring well on the design project and, above all, the 50% closed-book AS3600 final.
Sia can help you target those bands compliantly: it explains each flexure, shear, serviceability, slab, column and footing workflow step by step so you build the fluency the exam rewards - it never guarantees a grade, and no tool can promise you a particular GPA.
How many credit points is EGB375 and what do I need to know first?
EGB375 is a 12-credit-point third-year civil-engineering unit at Queensland University of Technology.
The unit materials do not list a specific prerequisite unit code, so treat the entry expectation as assumed knowledge rather than a named prerequisite: comfort with structural mechanics and analysis - drawing and reading shear-force and bending-moment diagrams for simply-supported and cantilever beams, and basic mechanics of materials - since EGB375 builds design methods on top of that (confirm the current prerequisites in the QUT unit outline).
The unit runs on the Canvas LMS, where lectures, the free unit workbook, tutorial and homework problems and previous exam papers are posted. The closed-book final is scheduled in the QUT Central Examination period, with a revision/study week before it, so plan your final revision into that study week.
Questions students ask in each EGB375 chapter
- 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. Chapter 01 · Concrete & Steel Materials + Equivalent Rectangular Stress Block
- 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. Chapter 02 · Singly Reinforced Beam — Ultimate Moment Capacity & Flexural Design
- 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. Chapter 03 · Doubly Reinforced Beams & T-Beams
- Why does a prestressed beam need checking at two different stages?
- Because the worst tension face swaps. At transfer the prestress is full but the load is only self-weight, so the large P·e moment can crack the top fibre. At service the load is full but the prestress has decayed to R·Pᵢ, so the bottom fibre is the one at risk. A single check would miss one of them, which is exactly why the Magnel method uses two inequalities per stage. Chapter 04 · Prestressed Concrete — Mechanics, Allowable Stresses & Magnel Method
- Why is the tendon stress at ultimate (σ_pu) less than the breaking stress f_pb?
- Because failure of the beam is governed by the concrete reaching its ultimate strain, not by the tendon snapping. At that point the bonded tendon has only developed part of its capacity, so AS3600 gives a reduced stress σ_pu = f_pb(1 − k1·k2/γ). The more heavily reinforced the section (larger reinforcement index k2), the lower σ_pu — the concrete crushes before the steel is fully stressed. Chapter 05 · PSC — Losses, Tendon Sizing & Ultimate Flexural Capacity
- 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. Chapter 06 · Shear Design of RC & PSC Beams (AS3600 MCFT)
- Why do we control crack width instead of trying to prevent cracking altogether?
- Concrete is weak in tension, so once the tensile stress from load, restrained shrinkage and temperature reaches the cracking stress the member cracks; that cannot be economically prevented in a reinforced section. AS3600 therefore controls the width of the cracks by detailing the reinforcement — more, smaller bars at modest cover raise the effective ratio rho_eff, which shortens the crack spacing Sr,max and narrows each crack so it stays under the allowable w'max for durability and appearance. Chapter 07 · Serviceability I — Crack Width & Spacing
- How do I decide whether to use the gross or the cracked section?
- Compute the cracking moment M_cr = (f_ct.f − σ_cs)·I_g/(D − dn) and compare it with the maximum service moment M_s. If M_s ≤ M_cr the section has not cracked, so use the gross inertia I_g (i.e. I_ef = I_g). If M_s > M_cr it has cracked: find the cracked inertia I_cr and blend it with I_g through the tension-stiffening formula I_ef = I_cr/[1 − (1 − I_cr/I_g)(M_cr/M_s)²]. Remember σ_cs (residual shrinkage stress) reduces M_cr, so the section cracks a little earlier than a shrinkage-free calculation predicts. Chapter 08 · Serviceability II — Steel Stresses, Section Inertia & Deflection
- How do I decide whether a slab is one-way or two-way, and why does it matter?
- Take L_x as the shorter effective span and form the aspect ratio L_y/L_x. If it is ≥ 2 the panel is one-way — it spans essentially in the short direction only, so you design a 1 m strip that way with k4 = 1.4 for a simply-supported case. If it is < 2 and the slab is supported on all four edges it is two-way, and you use the coefficient method Mx* = βx·q*·Lx² and My* = βy·q*·Lx² for the moments plus corner mesh at simply-supported corners. The classification changes which k4, which coefficients and which detailing rules apply, so always do it first. Chapter 09 · RC Slabs — One-Way & Two-Way + Deemed-to-Comply Deflection
- 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. Chapter 10 · RC Column Sections — Interaction Diagram (Points A-D)
- When is a column short enough to ignore buckling?
- When its slenderness ratio Le/r falls below the limiting slenderness. For a braced column that limit is max{25, (38 - f'c/15) sqrt(1 + M1*/M2*)}, and for an unbraced (sway) column it is 22. Below the limit the column is short and you design it straight from the Chapter 10 section (interaction) capacity for the actual moment; above it the column is slender and the end moment must be magnified for the P-delta second-order effect. Chapter 11 · Slender Columns — Slenderness & Moment Magnification
- Why are service loads used to size the footing but factored loads used for the strength checks?
- Bearing is a serviceability condition: the pad must not overstress the soil under the loads it actually sees day to day, which are the unfactored service loads NG + NQ. Once the plan area is fixed, the concrete itself (punching, flexure, one-way shear) is a strength condition, so it is checked against the factored ultimate pressure Pnet = (1.2 NG + 1.5 NQ)/Apad. Mixing the two up, for example sizing on factored loads, gives an oversized pad and is a common error. Chapter 12 · Shallow (Isolated Pad) Footing Design
How to study for the exam
Treat EGB375 as a set of repeatable design workflows rather than a reading unit: for each element, learn the sequence once - recompute the stress-block factors, balance the forces (C = T or Cc + Cs = T), find the neutral axis, prove ductility (kuo <= 0.36), then apply phi and check adequacy - and practise it until it is automatic.
Work every problem in newtons and millimetres and convert only at the end, because most lost marks are a wrong grade-dependent factor, a skipped ductility check, a mis-picked phi, or an N.mm to kN.m slip.
Because the final is closed-book with five A4 double-sided note sheets, build those sheets as you revise: put the factors, one solved example per element (flexure, shear, serviceability, slab, column, footing) and the key AS3600 limits where you can find them fast, then rehearse the ten practice problems in this guide against the clock once you confirm the exam length on the QUT exam timetable.
Use the QUT Central Examination period, with its revision/study week before the exam, as your dedicated window to consolidate those note sheets and work past exam papers and practice problems - the previous exam problems and tutorial and homework questions posted on Canvas are the best rehearsal for the real thing.
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