Queensland University of Technology · FACULTY OF ENGINEERING

EGB375 · Design of Concrete Structures

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Chapter 9 of 12 · EGB375

RC Slabs — One-Way & Two-Way + Deemed-to-Comply Deflection

In 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. This chapter shows how to classify a panel by its aspect ratio L_y/L_x (one-way if ≥ 2, two-way if < 2), size the thickness with the AS3600 deemed-to-comply span/depth rule L_ef/d ≤ k_3·k_4[(Δ/L_ef)(1000·E_c)/F_d,ef]^(1/3), find the moments from the coefficient method M_x* = β_x q* L_x², and detail the minimum, crack-control and corner-mesh reinforcement to AS3600-2018.

In this chapter

What this chapter covers

  • 011. 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)
  • 022. Ultimate design pressure q* = 1.2 qG + 1.5 qQ (kPa), with qG including slab self-weight = Ds × 25 kN/m³
  • 033. 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
  • 044. 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
  • 055. Total vs incremental deflection: Fd,ef(total) = (1+kcs)qG + (νs+kcs·ν1)qQ against Fd,ef(incr) = kcs·qG + (νs+kcs·ν1)qQ
  • 066. 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)
  • 077. Two-way coefficient method Mx* = βx·q*·Lx² and My* = βy·q*·Lx² (both use the short span) — short-span moment governs and gets the larger effective depth
  • 088. Detailing: secondary crack-control steel (0.00175 / 0.0035 · Ds/d) and corner mesh (0.5 Ast, top & bottom, over 0.2L each way) for simply-supported two-way panels
Worked example · free

Deemed-to-comply depth check for a one-way slab

Q [8 marks]. A simply-supported one-way RC floor slab spans L_x = 3.6 m between beams (the panel is 3.6 m × 9.0 m, so L_y/L_x = 2.5). It carries a superimposed dead load qG(super) = 1.0 kPa and imposed load qQ = 2.0 kPa. Concrete f′c = 32 MPa (Ec = 30 100 MPa), cover 20 mm, N12 main bars (db = 12 mm). A trial thickness Ds = 180 mm is proposed. Is it adequate for total-deflection control (Δ/Lef = 1/250)? [8 marks]
  • +1Classify: L_y/L_x = 9.0/3.6 = 2.5 ≥ 2 → one-way. Design a 1 m strip spanning L_x, so k3 = 1.0 and k4 = 1.4 (simply-supported one-way slab).
  • +1Self-weight and dead load: g_sw = Ds × 25 = 0.180 × 25 = 4.5 kPa, so total qG = 4.5 + 1.0 = 5.5 kPa; qQ = 2.0 kPa.
  • +2Effective deflection load (total): Fd,ef = (1 + kcs)qG + (νs + kcs·ν1)qQ with kcs = 2.0 (slab), νs = 0.7, ν1 = 0.4 → (1+2)×5.5 + (0.7 + 2×0.4)×2.0 = 3×5.5 + 1.5×2.0 = 16.5 + 3.0 = 19.5 kPa.
  • +1Span/depth bracket: (Δ/Lef)(1000·Ec)/Fd,ef = (1/250)×(1000×30 100)/19.5 = 0.004×30 100 000/19.5 = 120 400/19.5 = 6174. Cube root = 18.35.
  • +1Limiting ratio: Lef/d ≤ k3·k4×18.35 = 1.0×1.4×18.35 = 25.7.
  • +1Required depth: with Lef = 3600 mm, d ≥ Lef/25.7 = 3600/25.7 = 140 mm.
  • +1Provided depth and verdict: d = Ds − cover − db/2 = 180 − 20 − 6 = 154 mm ≥ 140 mm ✓ → the 180 mm slab is adequate for total-deflection control, so no full deflection calculation is needed.
Required d = 140 mm; provided d = 154 mm ≥ 140 mm, so Ds = 180 mm is adequate (deemed to comply for total deflection). The effective deflection load is Fd,ef = 19.5 kPa and the limiting span/depth ratio is Lef/d = 25.7.
Sia tip — Keep Ec in MPa but multiply by 1000 so the cube-root bracket is dimensionless against Fd,ef in kPa — dropping the 1000 makes Lef/d come out ten times too small. The rule gives a ceiling on Lef/d, i.e. a floor on d; compute d_req = Lef ÷ (k3·k4·bracket), add cover + db/2 back to get the required Ds, and round up. If the trial slab fails, thicken it and re-run, because the self-weight in Fd,ef rises too.
Glossary

Key terms

Aspect ratio L_y/L_x
The ratio of the longer effective span L_y to the shorter L_x. If L_y/L_x ≥ 2 the slab is one-way (it sheds load across the short span only); if L_y/L_x < 2 it is two-way (supported on four sides and bending in both directions). L_x is always the shorter span.
Deemed-to-comply span/depth rule
The AS3600-2018 clause 9.4.4.1 shortcut that satisfies deflection without a full calculation by limiting Lef/d ≤ k3·k4[(Δ/Lef)(1000·Ec)/Fd,ef]^(1/3). Ec is in MPa (×1000 to match Fd,ef in kPa) and Δ/Lef is the deflection limit (1/250 total, 1/500 or 1/1000 incremental).
Effective span Lef
The span used in the deemed-to-comply check, Lef = min(Ln + Ds, L), where Ln is the clear span, Ds the slab thickness and L the centre-to-centre span. The effective depth is d = Ds − cover − db/2.
Effective deflection load Fd,ef
The sustained serviceability load driving the span/depth rule. For total deflection Fd,ef = (1+kcs)qG + (νs+kcs·ν1)qQ; for incremental Fd,ef = kcs·qG + (νs+kcs·ν1)qQ, with kcs = 2.0 for slabs, νs = 0.7 and ν1 = 0.4. Distinct from the ultimate load q* = 1.2 qG + 1.5 qQ.
Factors k3 and k4
Support-condition multipliers in the span/depth rule. k3 = 1.0 for one-way and for two-way slabs supported on four sides; k4 = 1.4 for a simply-supported one-way slab, and for two-way slabs is read from AS3600 Table 9.4.4.2 by edge condition and L_y/L_x (e.g. 3.60 for four edges continuous down to 2.25 for four edges discontinuous at L_y/L_x = 1.0).
Coefficient method (Mx* = βx·q*·Lx²)
The AS3600 simplified method for two-way slab moments: Mx* = βx·q*·Lx² and My* = βy·q*·Lx², where both directions use the short span Lx squared. βx (short) and βy (long) come from Table 6.10.3.2(A) by edge condition and L_y/L_x; the short-span moment is the largest.
Minimum reinforcement ρmin
The AS3600 shrinkage/temperature floor on main-direction steel: ρmin = 0.19(Ds/d)²(fct.f/fsy) for two-way slabs (0.20 for one-way), where fct.f = 0.6√f′c. The provided steel must be the largest of the bending requirement ρBM, this ρmin, and 75% of the secondary crack-control steel.
Corner mesh
Reinforcement required at the corners of a simply-supported two-way panel, which tends to curl up. It is placed top and bottom over a length 0.2L each way, with an area of 0.5·Ast of the span steel.
FAQ

RC Slabs — One-Way & Two-Way + Deemed-to-Comply Deflection FAQ

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.

Why does the short span carry the larger moment and get the larger effective depth?

In a two-way panel the stiffer, shorter span attracts more of the load, so βx > βy and the short-span moment Mx* is the largest. To resist the biggest moment most efficiently you give that steel the greatest lever arm, so the short-span bars are placed in the outer layer with the larger effective depth dx, while the long-span bars sit inside with a slightly smaller dy. Placing the long span outside instead is a common mark-losing error.

Can AI help me with RC slab design in EGB375?

Yes — Sia is an AI tutor that can walk you through the slab workflow step by step: classifying by L_y/L_x, forming q*, applying the deemed-to-comply span/depth rule (keeping Ec in MPa but ×1000 against Fd,ef in kPa), pulling the coefficient-method moments and detailing the minimum, crack-control and corner-mesh steel. It is best used to build understanding and rehearse practice panels 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, coefficients and clauses against the unit outline and AS3600.

Studying with AI? Sia — free AI civil engineering tutor works through EGB375 step by step.

Study strategy

Exam move

Treat slabs as a fixed pipeline and drill it until it is automatic: (1) classify by L_y/L_x with L_x the shorter span; (2) build q* = 1.2 qG + 1.5 qQ, remembering to add the slab self-weight Ds × 25 into qG; (3) size or check the thickness with the deemed-to-comply rule Lef/d ≤ k3·k4[(Δ/Lef)(1000·Ec)/Fd,ef]^(1/3), taking care with the ×1000 unit factor and the total-vs-incremental form of Fd,ef; (4) for two-way panels read βx and βy and compute Mx* = βx·q*·Lx² and My* = βy·q*·Lx² (both on the short span); (5) design the main steel on a 1 m strip and enforce the minimum floor ρmin = 0.19(Ds/d)²(fct.f/fsy); (6) add secondary crack-control steel and, for simply-supported two-way corners, corner mesh of 0.5 Ast over 0.2L each way. Because the final exam is closed-book with five A4 double-sided student-prepared note sheets, put this workflow, the k3/k4 and β values and one worked deemed-to-comply check 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.

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