One-Way Slab Design (NSCP 2015 / ACI 318) Step by Step: Thickness, Moments, Main Bars, and Temperature Steel

By Engr. Ruel H. Cepeda, Structural Engineer

A slab is one-way when its longer-to-shorter span ratio, L/S, is 2.0 or greater — bending runs mainly across the short direction, and the slab designs as independent 1 m-wide strips. This article covers thickness, loads, ACI moment coefficients, main steel, temperature-and-shrinkage steel, bar spacing, and a shear check — then one full numeric example using a 3.5 m span and 125 mm slab.

Confirm the Classification: One-Way or Two-Way?

Check L/S first. When L/S ≥ 2, the short span carries essentially all the load and the one-way method here applies. When L/S < 2, both directions carry significant moment and a two-way method such as the Direct Design Method is required instead; see the sister site's two-way slab DDM guide on RHCES.

Step 1 — Minimum Thickness for Deflection Control

A slab can skip a deflection calculation if h meets a tabulated minimum ratio of span L, provided it is normal-weight concrete, uses Grade 420 bars, and does not support partitions likely to be damaged by deflection.

For fy other than 420 MPa, multiply the tabulated value by CF_fy = 0.4 + f_y/700 (fy in MPa). At fy = 415 MPa, CF_fy = 0.4 + 415/700 = 0.993 — a 0.7% reduction. Run any span through the free minimum beam depth calculator, which shares the same table family for beams and slabs.

Step 2 — Loads on the Slab

Load per square metre equals load per metre width of a 1 m design strip:

Sum the first three rows for dead load D; the fourth row is live load L for the governing occupancy.

Step 3 — Factored Design Load

Combine D and L with the governing strength load combination for gravity-only slabs:

w_u = 1.2D + 1.6L

w_u is per metre width of design strip (kN/m), numerically equal to the factored pressure in kPa since the strip is 1 m wide.

Step 4 — Approximate Moments via ACI Coefficients

ACI 318 permits approximate moment coefficients in lieu of a full elastic analysis, provided: (1) prismatic members, (2) uniform loads, (3) live load ≤ 3× dead load, (4) two or more spans, (5) adjacent spans differ by ≤ 20%. Moment = w_ul_n² ÷ coefficient below (l_n = clear span).

Spans of 3 m or less may alternatively take negative moment at all support faces as w_ul_n²/12; the worked example below has a 3.2 m clear span, so it uses the general /11 coefficient.

Step 5 — Main Reinforcement Design (R_n/ρ Method)

For each design moment, on a 1000 mm-wide strip:

R_n = M_u / (φ·b·d²)    ρ = (0.85f′_c/f_y)·[1 − √(1 − 2R_n/0.85f′_c)]    A_s = ρ·b·d

φ = 0.90, b = 1000 mm, d = h minus cover minus half the bar diameter. Compute ρ at each critical section, then compare against ρ_min below.

Step 6 — Minimum Steel Ratio and Temperature-Shrinkage Steel

For one-way slabs, minimum flexural ρ equals the temperature-and-shrinkage ratio — no separate, larger flexural ρ_min as for beams. Use whichever of computed ρ or ρ_min gives the larger A_s:

Temperature-and-shrinkage steel runs perpendicular to the main bars, sized on gross section: A_s,ts = ρ_min·b·h. Maximum spacing is the lesser of 3h or 450 mm for main bars, and the lesser of 5h or 450 mm for temperature-and-shrinkage bars. The main bars nearest the tension face must also satisfy the crack-control spacing of ACI 318 Section 24.3 (NSCP 2015 Section 424.3): s ≤ 380(280/f_s) − 2.5c_c and s ≤ 300(280/f_s), where f_s is the service-load bar stress (permitted to be taken as 2/3f_y) and c_c the clear cover.

Step 7 — Bar Spacing from Required Steel Area

Convert required A_s (mm²/m) to a center-to-center spacing s (mm) for a chosen bar area A_b:

s = 1000·A_b / A_s

Round s down to a practical value not exceeding the Step 6 cap. The free reinforcement calculator automates this conversion.

Step 8 — One-Way Shear Check

Design shear at a support face is V_u = w_ul_n/2, increased to 1.15w_ul_n/2 at the first interior support's exterior face. Compare against the concrete shear capacity of NSCP 2015 (ACI 318-14) for members without stirrups:

φV_c = φ·0.17λ√f′_c·b·d

φ = 0.75, λ = 1.0. Note that ACI 318-19 replaced this with V_c = 0.66λ_sλ(ρ_w)^1/3√f′_c·b·d for members without shear reinforcement, where λ_s = √(2/(1 + d/250)) ≤ 1.0 is the size-effect factor (equal to 1.0 for slabs with d ≤ 250 mm); it gives a lower value for lightly reinforced slabs, so it is worth checking both. Solid one-way slabs almost never need shear reinforcement, as the example below confirms.

Worked Example — 3.5 m Continuous One-Way Slab

Given: Interior span, both ends continuous, L = 3.5 m c/c of 300 mm supports, trial h = 125 mm. f′_c = 21 MPa; Grade 415 bars, 10 mm dia., 20 mm cover. Floor finish (assumed) 1.0 kPa; partition allowance (assumed) 1.0 kPa; live load L = 2.4 kPa (office).

1–4. Thickness, loads, factored load, clear span

h_min = L/28 = 3500/28 = 125.0 mm; CF_fy = 0.4+415/700 = 0.993 → h_min,adj = 125.0×0.993 = 124.1 mm ≤ 125 mm — OK.
Self-weight = 0.125×24 = 3.00 kPa; D = 3.00+1.00+1.00 = 5.00 kPa; L = 2.40 kPa.
w_u = 1.2(5.00)+1.6(2.40) = 6.00+3.84 = 9.84 kN/m.
l_n = 3.5−0.3 = 3.2 m.

5–6. Moments and effective depth

M_u⁺ (÷16) = 9.84×3.2²/16 = 9.84×10.24/16 = 6.30 kN·m/m.
M_u⁻ (÷11) = 9.84×10.24/11 = 9.16 kN·m/m.
d = 125−20−10/2 = 100 mm.

7–8. Main steel

Positive: R_n = 6.30×10&sup6;/(0.9×1000×100²) = 0.700 MPa; ρ = (0.85×21/415)[1−√(1−2(0.700)/17.85)] = 0.0430×0.0400 = 0.00172 < ρ_min (0.0018) → A_s⁺ = 0.0018×1000×100 = 180 mm²/m.
Negative: R_n = 9.16×10&sup6;/(0.9×1000×100²) = 1.018 MPa; ρ = 0.0430×[1−√(1−2(1.018)/17.85)] = 0.0430×0.0588 = 0.00253 > ρ_min → A_s⁻ = 0.00253×1000×100 ≈ 253 mm²/m.

9–10. Spacing and temperature-shrinkage steel

10 mm bar: A_b = 78.5 mm². Main cap = min(3h,450) = 375 mm.
Positive: s = 1000×78.5/180 = 436 > 375 cap → 10 mm@350 o.c. (224 > 180 req'd).
Negative: s = 1000×78.5/253 = 310 < 375 cap → 10 mm@300 o.c. (262 > 253 req'd).
T&S: A_s,ts = 0.0018×1000×125 = 225 mm²/m; cap = min(5h,450) = 450 mm; s = 1000×78.5/225 = 349 < cap → 10 mm@300 o.c. transverse (262 > 225 req'd).
Crack control (Step 6): with f_s = 2/3f_y = 277 MPa the limit is 300(280/277) = 303 mm, which the 300 mm top bars meet. For the 350 mm bottom bars, the service moment M⁺ = (5.00+2.40)×10.24/16 = 4.74 kN·m/m on a cracked section with A_s = 224 mm² gives f_s ≈ 225 MPa, so s ≤ 300(280/225) = 373 mm — 350 mm OK.
(If ρ_min = 0.0020 is applied to Grade 415 instead, A_s⁺ = 200 and A_s,ts = 250 mm²/m; the same 10 mm@350 and 10 mm@300 selections still satisfy them.)

11. Shear check

V_u = 9.84×3.2/2 = 15.74 kN/m; worst case 1.15×15.74 = 18.11 kN/m.
φV_c = 0.75×0.17×√21×1000×100 = 58.4 kN. 18.11 << 58.4 kN — no stirrups needed.
ACI 318-19 cross-check: λ_s = √(2/(1+100/250)) = 1.20 → 1.0; ρ_w = 262/(1000×100) = 0.00262; V_c = 0.66×1.0×1.0×(0.00262)^1/3×√21×1000×100 = 0.66×0.138×4.58×10&sup5; = 41.7 kN; φV_c = 31.3 kN > 18.11 kN — still OK.

Assumptions & Limitations

• Applies to nonprestressed, solid one-way slabs only (L/S ≥ 2) — not two-way, ribbed, or prestressed slabs.
• The ACI moment coefficients require all five applicability conditions in Step 4; otherwise use a proper elastic analysis.
• Floor finish and partition allowance in the example are assumed values; confirm against the actual finish schedule.
• ρ_min pairs Grade 415 with 0.0018 (treating it as Grade 420) and Grade 275 with 0.0020, common in Philippine practice; read literally, the NSCP 2015 / ACI 318 minimum-reinforcement table gives 0.0020 for any fy below 420 MPa, which does not change the bar selections in this example. Verify the fy-based formula for other grades.
• Support width (300 mm) for clear span is an assumption; use the actual width from the framing plan.
• NSCP 2015 mirrors these ACI 318 values closely; confirm against the code edition adopted by the building official of record.

Frequently Asked Questions

How do I know if my slab panel is one-way or two-way before starting this procedure?

Measure the two span directions, L (longer) and S (shorter), and compute L/S. If ≥ 2, follow this article. If < 2, a two-way method such as the Direct Design Method applies instead — see the RHCES two-way slab DDM guide.

Can I use the ACI approximate moment coefficients if my slab has only two spans, or spans that are not quite equal?

Two spans are fine — the table has a /9 divisor for that case. Unequal spans are fine up to a 20% difference; beyond that, or with live load exceeding three times dead load, run a full elastic analysis instead.

Why did the minimum steel ratio (ρ_min) govern the positive-moment steel in the worked example instead of the computed value?

Thin, lightly loaded slabs are often controlled by ρ_min at midspan, since demand moment is small relative to capacity while the temperature-and-shrinkage minimum is a fixed fraction of gross area. It is also why the 3h/450 mm and 5h/450 mm caps often govern spacing more than computed A_s, as happened above.

Getting slab thickness, moments, and steel right the first time avoids formwork rework and rebar call-backs on site. Try the minimum beam depth calculator and the reinforcement calculator, or browse all free web tools. Offline spreadsheet versions are on the download page for use without internet.