EGB375 · Design of Concrete Structures
Serviceability II — Steel Stresses, Section Inertia & Deflection
In 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. This chapter builds the serviceability chain to AS3600-2018: the cracking moment M_cr = (f_ct.f − σ_cs)·I_g/(D − d_n) decides whether to use the gross inertia I_g or the cracked inertia I_cr, tension stiffening blends them into the effective I_ef, and the 10M formula turns I_ef into a deflection that must satisfy span/250 (total) and span/500 (incremental).
What this chapter covers
- 011. 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
- 022. Deciding cracked vs uncracked by comparing the service moment M_s with M_cr
- 033. The transformed section and modular ratio n = E_s/E_c (≈ 6.6 for f′c = 32 MPa)
- 044. Gross inertia I_g = bD³/12 + bD(dn − D/2)² + (n−1)A_st(d − dn)² about the gross neutral axis
- 055. 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
- 066. Effective inertia (tension stiffening) I_ef = I_cr / [1 − (1 − I_cr/I_g)(M_cr/M_s)²] ≤ I_g
- 077. Cracked-section service steel stress σ_scr = M_s·n·(d − dn)/I_cr and the AS3600 stress-limit check
- 088. The 10M deflection Δ = 10·M·L²/(96·E_c·I_ef), service combos G+0.7Q and G+0.4Q, the multiplier k_cs = 2 − 1.2·A_sc/A_st ≥ 0.8, and the span/250 & span/500 limits
Cracking moment, effective inertia and short-term deflection
- +1Net cracking stress: f_ct.f − σ_cs = 3.39 − 0.4 = 2.99 MPa (shrinkage pre-tensions the surface, so the concrete cracks earlier than 3.39 MPa alone would suggest).
- +1Cracking moment: M_cr = (f_ct.f − σ_cs)·I_g/(D − dn) = 2.99 × 2.98 × 10⁹ / 235 = 37.9 × 10⁶ N·mm = 37.9 kN·m.
- +1Service moment (simply supported UDL, M = wL²/8 with L²/8 = 5.5²/8 = 3.78 m²): short-term W_s = G + 0.7Q = 14 + 7 = 21 kN/m → M_s = 21 × 3.78 = 79.4 kN·m. Since M_s = 79.4 > M_cr = 37.9, the section CRACKS → use I_ef.
- +1Effective inertia (tension stiffening): I_cr/I_g = 1.34/2.98 = 0.45; M_cr/M_s = 37.9/79.4 = 0.477. I_ef = I_cr / [1 − (1 − 0.45)(0.477)²] = 1.34 × 10⁹ / (1 − 0.125) = 1.34 × 10⁹ / 0.875 = 1.53 × 10⁹ mm⁴ (≤ I_g ✓).
- +1Short-term deflection (10M method, exact for a simply supported UDL): Δ_s = 10·M_s·L² / (96·E_c·I_ef) = 10 × 79.4 × 10⁶ × 5500² / (96 × 30 100 × 1.53 × 10⁹) = 5.4 mm (working in N and mm throughout).
- +1Limit check: span/250 = 5500/250 = 22 mm, and Δ_s = 5.4 mm ≤ 22 mm ✓ (short-term only). The total deflection Δ_T = Δ_s + k_cs·Δ_sus must also be checked against span/250, where k_cs = 2 − 1.2(A_sc/A_st) = 2.0 for this singly reinforced beam (A_sc = 0).
Key terms
- Cracking moment (M_cr)
- The bending moment at which the extreme tension fibre first reaches the net flexural tensile capacity: M_cr = (f_ct.f − σ_cs)·I_g/(D − dn), with f_ct.f = 0.6√f′c the characteristic flexural (modulus-of-rupture) strength and σ_cs the residual shrinkage stress. Below M_cr the section is uncracked (use I_g); above it, the section cracks (use I_cr, blended into I_ef).
- Modular ratio (n) & transformed section
- n = E_s/E_c replaces steel by an equivalent concrete area so ordinary section properties can be computed. For f′c = 32 MPa, E_c ≈ 30 100 MPa gives n ≈ 6.6. The gross section adds (n−1)A_st (the steel already displaces some counted concrete); the cracked section uses the full n·A_st.
- Gross inertia (I_g)
- The second moment of area of the uncracked transformed section (all concrete active): I_g = bD³/12 + bD(dn − D/2)² + (n−1)A_st(d − dn)², taken about the gross neutral axis dn found from first moments of area. Used for M_cr and while the section is uncracked.
- Cracked inertia (I_cr)
- The second moment of area with the tension concrete below the neutral axis ignored: first find the cracked neutral axis from b·dn²/2 = n·A_st(d − dn), then I_cr = b·dn³/3 + n·A_st(d − dn)². The cracked neutral axis is higher (smaller dn) than the gross one, and I_cr < I_g.
- Effective inertia (I_ef) — tension stiffening
- Between cracks the concrete still carries some tension, so the real stiffness lies between I_cr and I_g: I_ef = I_cr / [1 − (1 − I_cr/I_g)(M_cr/M_s)²] ≤ I_g. As the service moment M_s rises above M_cr, I_ef drifts from I_g toward I_cr.
- Cracked steel stress (σ_scr)
- The service tensile stress in the bars from cracked-section bending: σ_scr = M_s·n·(d − dn)/I_cr, using the cracked neutral axis and I_cr. It is checked against the AS3600 Table 8.6.2.2 stress limit (by bar diameter/spacing) that guards crack width.
- 10M deflection method
- For a simply supported member under a uniformly distributed load, the midspan deflection is Δ = 10·M·L²/(96·E_c·I_ef) — the exact 5wL⁴/384E_cI rewritten with M = wL²/8. Short-term uses M_s (from W_s = G + 0.7Q); sustained uses M_sus (from W_sus = G + 0.4Q).
- Creep-shrinkage multiplier (k_cs) & span limits
- k_cs magnifies the sustained deflection for long-term creep and shrinkage: k_cs = 2 − 1.2(A_sc/A_st) ≥ 0.8 for beams (2.0 for slabs, or when A_sc = 0). The total ΔT = Δs + k_cs·Δsus must be ≤ span/250, and the incremental Δinc = (Δs − ΔG) + k_cs·Δsus ≤ span/500.
Serviceability II — Steel Stresses, Section Inertia & Deflection FAQ
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.
Why are there two different neutral-axis depths, and which one goes where?
The gross (uncracked) neutral axis sits near mid-depth because the whole concrete section acts; the cracked neutral axis is higher (a smaller dn) because the tension concrete below it is ignored. Use the gross dn in the cracking moment (with D − dn measured to the extreme tension fibre) and the cracked dn in both I_cr and the steel stress σ_scr. Mixing them up — or using (n−1)A_st for the cracked section instead of the full n·A_st — is a common mark-losing slip.
Can AI help me with serviceability and deflection in EGB375?
Yes — Sia is an AI tutor that can walk you through the method step by step: forming M_cr and the cracked/uncracked decision, setting up the gross and cracked transformed sections, blending them into I_ef, and running the 10M deflection with k_cs against the span/250 and span/500 limits. 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, E_c/n values 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 chain and drill it until each link is automatic: (1) f_ct.f = 0.6√f′c and M_cr = (f_ct.f − σ_cs)·I_g/(D − dn); (2) compare M_s with M_cr to decide cracked or uncracked; (3) build I_g (gross, with (n−1)A_st) and I_cr (cracked, with the full n·A_st and its own higher neutral axis); (4) blend to I_ef with the tension-stiffening formula; (5) service steel stress σ_scr = M_s·n·(d − dn)/I_cr; (6) 10M deflections Δs (from G + 0.7Q) and Δsus (from G + 0.4Q); (7) k_cs = 2 − 1.2·A_sc/A_st ≥ 0.8 and the two checks ΔT ≤ span/250, Δinc ≤ span/500. Work everything in N and mm and convert to kN·m or mm only at the end. Because the final exam is closed-book with five A4 double-sided student-prepared note sheets, put this chain, the transformed-section relations and one worked deflection 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.
