EGB375 · Design of Concrete Structures
Serviceability I — Crack Width & Spacing
A 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. You cannot stop concrete cracking, so instead you check that the likely crack width stays under an allowable limit, Wcr ≤ w'max, where the width is the maximum crack spacing multiplied by the mean strain the steel accumulates between cracks. This chapter builds that chain end to end: restrained-shrinkage stress, the cracking stress fct.f = 0.6√f'c, the effective tension chord hc,eff, the maximum crack spacing Sr,max, and the final width check.
What this chapter covers
- 01Why serviceability cracking matters and the master check Wcr <= w'max
- 02Shrinkage strain ecs as a looked-up value (AS3600 Table 3.1.7.2)
- 03Residual shrinkage stress scs = ecs Es (2.5 pw - 0.8 pcw)/(1 + 50 pw)
- 04Flexural tensile strength fct.f = 0.6 sqrt(f'c) and cracking stress scr = fct.f - scs
- 05The effective tension chord hc,eff = min[2.5(D-d), (D-dn)/3, D/2] and Ac,eff
- 06The effective reinforcement ratio rho_eff = Ast/Ac,eff and why it controls cracking
- 07Maximum crack spacing Sr,max = 3.4c + 0.3 k1 k2 db/rho_eff and the k1, k2 factors
- 08The bar-spacing validity limit 5(c + 0.5 db)
- 09The mean strain difference (esm - ecm), the shrinkage term and the 0.6 scr/Es floor
- 10Crack width Wcr = (esm - ecm) Sr,max and the allowable widths w'max
Worked example: full serviceability crack-width check
- +1Steel and moduli. Ast = 4 × 620 = 2480 mm^2; modular ratio n = Es/Ec = 200000/30100 = 6.65.
- +2Cracking stress. fct.f = 0.6√32 = 3.39 MPa. pw = Ast/(b·d) = 2480/(350×590) = 0.01201; scs = ecs·Es·(2.5×0.01201)/(1 + 50×0.01201) = 120×0.03003/1.601 = 2.25 MPa. So scr = 3.39 − 2.25 = 1.14 MPa (low, so the beam is expected to crack under M*).
- +2Cracked neutral axis. Solve b·dn^2/2 = n·Ast(d − dn): 175·dn^2 = 6.65×2480×(590 − dn) → 175·dn^2 + 16480·dn − 9723000 = 0 → dn = 193 mm.
- +2Cracked inertia and steel stress. Icr = b·dn^3/3 + n·Ast(d − dn)^2 = 350×193^3/3 + 16480×397^2 = 8.41×10^8 + 2.59×10^9 = 3.44×10^9 mm^4; sscr = M*·n·(d − dn)/Icr = (210×10^6 × 6.65 × 397)/(3.44×10^9) = 161 MPa.
- +2Tension chord. candidates 2.5(D − d) = 2.5(60) = 150; (D − dn)/3 = (650 − 193)/3 = 152; D/2 = 325 → hc,eff = 150 mm. Ac,eff = 150×350 = 52500 mm^2; rho_eff = 2480/52500 = 0.0472.
- +2Maximum crack spacing. Sr,max = 3.4c + 0.3·k1·k2·db/rho_eff = 3.4×40 + 0.3×0.8×0.5×28/0.0472 = 136 + 71 = 207 mm. Validity: bar spacing limit 5(c + 0.5db) = 5(40 + 14) = 270 mm, and the bars sit closer, so the formula applies.
- +3Mean strain difference. ne = (Es/Ec)(1 + phi_cc) = 6.65×3 = 19.9; fct = 0.5√32 = 2.83 MPa. Tension-stiffening term = 0.6·(fct/rho_eff)(1 + ne·rho_eff) = 0.6×(2.83/0.0472)(1 + 19.9×0.0472) = 69.7 MPa. (esm − ecm) = (161 − 69.7)/200000 + 0.0006 = 4.57×10^-4 + 6.0×10^-4 = 1.06×10^-3 (above the floor 0.6·sscr/Es = 4.83×10^-4).
- +2Crack width and verdict. Wcr = (esm − ecm)·Sr,max = 1.06×10^-3 × 207 = 0.22 mm ≤ w'max = 0.30 mm → crack control is satisfied.
Key terms
- Cracking stress scr
- The net tensile stress the concrete can reach before a load crack forms, scr = fct.f - scs in MPa, where fct.f = 0.6 sqrt(f'c) is the flexural tensile strength and scs is the residual shrinkage stress. Restrained shrinkage has already used up part of the tensile capacity, so scr is smaller than fct.f alone.
- Shrinkage strain ecs
- The design (30-year) shrinkage strain of the concrete, read from AS3600-2018 Table 3.1.7.2 as a function of f'c, the hypothetical thickness th = 2Ag/ue and the climate/exposure zone. It is a looked-up value (typically of order 0.0005-0.0007), not computed from a closed formula; confirm the value on AS3600 and the unit materials.
- Residual shrinkage stress scs
- The tensile stress locked into the concrete at the reinforcement level by restrained shrinkage, scs = ecs Es (2.5 pw - 0.8 pcw)/(1 + 50 pw) in MPa, with pw = Ast/(b d) and pcw = Asc/(b d) the tension- and compression-steel ratios and Es = 200 000 MPa.
- Flexural tensile strength fct.f
- The modulus of rupture of the concrete, fct.f = 0.6 sqrt(f'c) in MPa (f'c in MPa). It governs whether a section cracks and the cracking moment. The lower mean/direct strength fct = 0.5 sqrt(f'c) is used instead inside the tension-stiffening strain term; the two must not be swapped.
- Effective tension chord hc,eff
- The depth of the band of concrete around the bars that shares tension with the steel, hc,eff = min[2.5(D - d), (D - dn)/3, D/2] in mm, with dn the cracked-section neutral-axis depth. Its area Ac,eff = hc,eff b sets the effective reinforcement ratio.
- Effective reinforcement ratio rho_eff
- The steel ratio of the tension chord, rho_eff = Ast/Ac,eff (dimensionless). It is far larger than the gross ratio pw because the chord concentrates the steel into a thin effective area, and a larger rho_eff gives closer, finer cracks.
- Maximum crack spacing Sr,max
- The largest expected spacing between cracks, Sr,max = 3.4c + 0.3 k1 k2 db/rho_eff in mm, where c is clear cover, db the bar diameter, k1 = 0.8 deformed / 1.6 plain and k2 = 0.5 bending / 1.0 pure tension. It is valid only while the bar spacing stays <= 5(c + 0.5 db).
- Crack width Wcr and allowable w'max
- The characteristic crack width Wcr = (esm - ecm) Sr,max in mm, the product of the mean strain difference and the spacing; it must satisfy Wcr <= w'max. AS3600 Table C2.3.3.1 sets w'max by purpose, about 0.3 mm for corrosion/durability and roughly 0.2-0.7 mm for appearance by exposure.
Serviceability I — Crack Width & Spacing FAQ
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.
Which tensile strength do I use, 0.6 sqrt(f'c) or 0.5 sqrt(f'c)?
Both, in different places. Use the flexural tensile strength fct.f = 0.6 sqrt(f'c) to decide whether the section cracks (and for the cracking moment in the deflection chapter), but use the lower mean/direct tensile strength fct = 0.5 sqrt(f'c) inside the tension-stiffening term of the mean strain difference (esm - ecm). They are different numbers for the same concrete, and swapping them is a classic lost mark.
Can AI help me with crack width and spacing in EGB375?
Yes, for learning and rehearsal. Sia is an AI tutor that walks the crack-control chain with you step by step — cracking stress, the tension chord hc,eff, the spacing Sr,max, the mean strain difference and the final Wcr <= w'max check — and helps you spot slips such as dropping the shrinkage term or using the wrong area in rho_eff. It supports your study rather than replacing it: it will not sit the closed-book exam for you, and no tool can promise a particular mark or grade. Always confirm code clauses and table values against AS3600 and your unit materials.
Studying with AI? Sia — free AI civil engineering tutor works through EGB375 step by step.
Exam move
Learn crack control as one ordered chain rather than a set of loose formulae: cracking stress scr = fct.f - scs, then the cracked-section dn, Icr and steel stress sscr, then the tension chord hc,eff = min[2.5(D-d), (D-dn)/3, D/2] and rho_eff = Ast/Ac,eff, then the spacing Sr,max = 3.4c + 0.3 k1 k2 db/rho_eff, then the mean strain difference (esm - ecm) with its +ecs shrinkage term and its 0.6 sscr/Es floor, and finally the width Wcr = (esm - ecm) Sr,max compared with w'max. Rehearse the chain end to end on a couple of sections until the order is automatic, keeping every stress in MPa, every length in mm and rho_eff as a decimal so the numbers stay clean, and take particular care to keep fct.f = 0.6sqrt(f'c) apart from fct = 0.5sqrt(f'c). Because the final exam is closed-book and paper-based, condense the whole chain, the k1/k2 factors and one fully worked crack-width check onto your permitted note sheets, and pair it with the deflection workflow of the next chapter since the two share the cracked section. The exam is marked out of 100 and worth 50% of the unit; once you confirm its duration on the QUT exam timetable, divide those minutes by 100 to set a consistent minutes-per-mark rate and spend on a crack-control question in proportion to its marks.
