Reinforced Concrete Slabs - Detailed Explanation
Reinforced concrete slabs are among the most frequently tested structural elements in the PRC Civil Engineer Licensure Examination. A slab is a flat, wide, shallow flexural member that transfers floor and roof loads to the supporting beams, walls, or columns beneath it. Unlike beams, slabs are designed per unit width (typically a 1-metre strip), making the design process systematic and highly structured. This chapter covers the fundamental classification of slabs as one-way or two-way, the NSCP 2015 minimum thickness provisions for deflection control, the step-by-step flexural design of a 1-metre slab strip, shrinkage and temperature reinforcement, and an introduction to two-way slab analysis. Mastering this chapter requires a firm grasp of rectangular beam design (Chapter 2) and an understanding of how NSCP 2015 (aligned with ACI 318) governs slab detailing. Board exam problems on slabs typically test: (1) classification by span ratio, (2) minimum thickness calculations, (3) bar spacing from computed As, and (4) temperature steel — all of which are fully covered here with worked examples.
One-Way vs. Two-Way Slab Classification
Examples
Panel A: 6 m × 7 m slab supported on all four sides. Panel B: 3 m × 7 m slab supported on all four sides. Classify each panel.
Solution
Panel A: Ll/Ls = 7/6 = 1.167 < 2 → TWO-WAY slab.
Panel B: Ll/Ls = 7/3 = 2.333 ≥ 2 → ONE-WAY slab.
Panel A has nearly equal spans so both directions carry significant load — two-way behaviour. Panel B has one span more than twice the other, so almost all load is carried across the 3 m short span — one-way behaviour. Note that the long span of Panel B (7 m) still receives temperature/shrinkage steel but not designed for full flexural demand.
A 4 m × 4 m slab is supported on two opposite sides only (the 4 m sides). Classify the slab.
Solution
Because the slab is supported on only TWO opposite sides, it is ALWAYS classified as ONE-WAY, regardless of the span ratio (which here is 4/4 = 1.0 < 2).
The support condition overrides the span ratio rule. Load can only transfer to the two supported edges, so all flexural action is in one direction — by definition, one-way.
The most fundamental step in slab design is determining whether the slab behaves as one-way or two-way. This classification governs the entire analysis and design approach.
Definition and Basis:
A slab panel is rectangular with a short span Ls and a long span Ll. When the long-to-short span ratio Ll/Ls ≥ 2, the slab is classified as ONE-WAY because the bending resistance is provided almost entirely across the short span — the long span contributes negligibly. When Ll/Ls < 2, the slab is classified as TWO-WAY because bending occurs significantly in both directions and both spans must be designed.
Physical Intuition:
Imagine placing a uniformly distributed load on a rectangular plate. If one side is much longer (ratio ≥ 2), the plate deflects primarily like a beam spanning the short direction — hence 'one-way.' If the sides are nearly equal, the plate curves in both directions like a dish — hence 'two-way.'
Special Cases:
- A slab supported on only TWO opposite sides (e.g., a slab with free or unsupported edges on the long sides) is ALWAYS one-way, regardless of the span ratio.
- A cantilever slab is also always one-way.
Design Consequence:
- One-way slab: Design the main (flexural) steel along the short span; provide temperature/shrinkage steel along the long span.
- Two-way slab: Design main steel in BOTH directions; analysis uses coefficients, DDM, or EFM.
NSCP 2015 / ACI 318 Reference: Section 406.1 (ACI 318-14 Section 7.1) establishes these classification criteria for non-prestressed slabs.
Applications
- Preliminary design decision that determines which analysis tables or methods to use
- Determining the direction in which to place main flexural reinforcement
- Identifying which span to use for minimum thickness calculations
- Categorising slab systems in construction drawings and structural reports
Common misconceptions
- WRONG: Using Ls/Ll (short over long) instead of Ll/Ls — this gives a ratio always ≤ 1 and will lead to incorrect classification.
- WRONG: Assuming a nearly square slab (e.g., 4 m × 4.5 m) is one-way — ratio = 4.5/4 = 1.125 < 2, so it is two-way.
- WRONG: Forgetting that the support condition (two opposite sides only) automatically means one-way regardless of aspect ratio.
- WRONG: Placing main steel along the long span in a one-way slab — main steel always goes across the short span.
Related concepts
- Minimum slab thickness (directly tied to classification)
- Shrinkage and temperature steel (perpendicular to main steel in one-way slabs)
- Two-way slab analysis methods (DDM, EFM, Coefficient Method)
- Rectangular beam design (the 1-metre strip analogy)
Common exam questions
Example
A slab panel is 5 m × 9 m. Classify: Ll/Ls = 9/5 = 1.8 < 2 → two-way.
Approach
Compute Ll/Ls. Compare with 2. State one-way or two-way.
Example
A 4 m × 5 m slab rests on two parallel walls 4 m apart. Answer: one-way (only two supports).
Approach
Check if supported on all four sides. If only two sides, immediately answer one-way.
Example
3 m × 8 m one-way slab: main steel crosses the 3 m span; temperature steel runs parallel to the 3 m span (i.e., along the 8 m direction).
Approach
Identify which is the short span. Main steel spans perpendicular to short span (i.e., across the short span).
Key points to remember
- One-way if Ll/Ls ≥ 2; two-way if Ll/Ls < 2 — ALWAYS use long divided by short, never the reverse.
- A slab supported on only two opposite sides is always one-way regardless of span ratio.
- In a one-way slab, main steel runs along the SHORT span; temperature steel runs along the LONG span.
- In a two-way slab, main steel runs in BOTH directions.
- The classification affects which design method (strip method vs. DDM/EFM) is used.
Flexural Design of One-Way Slabs (Per Metre Strip)
Examples
A one-way slab has h = 175 mm, d = 150 mm, b = 1000 mm, f'c = 28 MPa, fy = 415 MPa, Mu = 15 kN·m/m. Find the required As and bar spacing using 12 mm bars (Ab = 113 mm²).
Solution
Step 1: Convert Mu = 15 kN·m/m = 15 × 10⁶ N·mm/m.
Step 2: Rn = Mu/(φbd²) = (15 × 10⁶)/(0.90 × 1000 × 150²) = 15,000,000/20,250,000 = 0.741 MPa.
Step 3: ρ = (0.85 × 28/415) × [1 − √(1 − 2(0.741)/(0.85 × 28))]
= (0.05735) × [1 − √(1 − 1.482/23.8)]
= (0.05735) × [1 − √(1 − 0.06227)]
= (0.05735) × [1 − √(0.93773)]
= (0.05735) × [1 − 0.96836]
= (0.05735) × (0.03164)
= 0.001815.
Step 4: As,calc = ρ × b × d = 0.001815 × 1000 × 150 = 272 mm²/m.
As,min = 0.0018 × 1000 × 175 = 315 mm²/m.
Since 272 < 315, As,min GOVERNS → As = 315 mm²/m.
Step 5: Spacing with 12 mm bars (Ab = 113 mm²):
s = 113 × 1000/315 = 358.7 mm.
Step 6: Check max spacing: min(3 × 175, 450) = min(525, 450) = 450 mm.
Calculated s = 358.7 mm < 450 mm ✓
Adopt s = 350 mm (rounded down).
The minimum steel (temperature/shrinkage on full h) governed over the calculated flexural requirement — this is very common for thin slabs with low Mu. The maximum spacing check (450 mm) was satisfied. Final answer: 12 mm bars at 350 mm centre-to-centre.
Design the main steel for Mu = 22 kN·m/m, h = 180 mm, d = 155 mm, f'c = 28 MPa, fy = 415 MPa. Use 16 mm bars (Ab = 201 mm²).
Solution
Step 1: Mu = 22 × 10⁶ N·mm.
Step 2: Rn = 22 × 10⁶/(0.90 × 1000 × 155²) = 22,000,000/21,622,500 = 1.017 MPa.
Step 3: ρ = (0.85 × 28/415)[1 − √(1 − 2(1.017)/(0.85 × 28))]
= 0.05735 × [1 − √(1 − 2.034/23.8)]
= 0.05735 × [1 − √(0.91454)]
= 0.05735 × [1 − 0.95632]
= 0.05735 × 0.04368
= 0.002505.
Step 4: As,calc = 0.002505 × 1000 × 155 = 388.3 mm²/m.
As,min = 0.0018 × 1000 × 180 = 324 mm²/m.
As,calc = 388.3 > As,min = 324 → flexural As governs → As = 388.3 mm²/m.
Step 5: s = 201 × 1000/388.3 = 517.7 mm.
Step 6: Max spacing = min(3 × 180, 450) = 450 mm.
s_calc = 517.7 mm > 450 mm → SPACING LIMIT GOVERNS.
Use s = 450 mm, then actual As = 201 × 1000/450 = 446.7 mm²/m > 388.3 mm² ✓.
Adopt 16 mm bars at 450 mm c/c.
In this case, the required spacing (517 mm) exceeded the maximum permitted (450 mm). The spacing limit governed, so we adopt 450 mm — which provides more steel (447 mm²/m) than required (388 mm²/m). This is acceptable as it is conservative.
A one-way slab is designed as a series of parallel rectangular beams, each 1 m (1000 mm) wide. This unit-width approach simplifies the design and produces reinforcement expressed as mm²/m, which is then converted to a bar spacing.
Design Strip:
Set b = 1000 mm (unit width), determine effective depth d = h − cover − db/2 (typically cover = 20 mm for slabs), and proceed exactly as a rectangular beam design.
Step-by-Step Procedure:
Step 1: Factored Moment Mu
For a uniformly loaded simply supported slab: Mu = wu × L² / 8 (per metre, so wu in kN/m per metre width = kN/m²).
Step 2: Compute the nominal moment coefficient Rn
Rn = Mu / (φ × b × d²)
where φ = 0.90 (tension-controlled flexure), b = 1000 mm, Mu in N·mm.
Step 3: Compute the reinforcement ratio ρ
ρ = (0.85 × f'c / fy) × [1 − √(1 − 2Rn / (0.85 × f'c))]
Step 4: Check minimum reinforcement
For slabs, ACI 318 / NSCP 2015 uses the SHRINKAGE AND TEMPERATURE steel ratio as the minimum for the main steel:
As,min = ρtemp × b × h (using full thickness h, not d)
where ρtemp = 0.0018 for fy = 415–420 MPa, 0.0020 for fy ≤ 275 MPa.
Note: This differs from beams where As,min = 0.25√f'c/fy × bw × d. For slabs, the temperature steel minimum typically governs thin slabs.
Step 5: Compute required As
As = max(ρ × b × d, As,min) [mm²/m]
Step 6: Select bar size and compute spacing
s = Ab × 1000 / As
where Ab is the cross-sectional area of one bar (mm²).
Step 7: Check maximum spacing
Main steel: s ≤ min(3h, 450 mm)
Temperature steel: s ≤ min(5h, 450 mm)
Step 8: Adopt practical spacing
Round down s to the nearest 25 mm (e.g., 249 mm → adopt 225 mm or 200 mm).
Applications
- Design of floor slabs for residential, commercial, and industrial buildings
- Design of roof slabs and terrace decks
- Bridge deck slabs spanning between girders
- Precast slab panels
Common misconceptions
- WRONG: Using As,min = 0.25√f'c/fy × b × d (beam formula) for slabs — slabs use As,min = 0.0018bh (temperature steel on full h).
- WRONG: Using d (effective depth) instead of h (total thickness) in the As,min formula for slabs.
- WRONG: Forgetting to check maximum bar spacing — the computed spacing may exceed 3h or 450 mm.
- WRONG: Rounding the bar spacing UP — always round down to ensure As provided ≥ As required.
- WRONG: Not converting Mu from kN·m to N·mm (multiply by 10⁶) before computing Rn.
Related concepts
- Rectangular beam design — same Rn–ρ procedure, different b and minimum steel
- Shrinkage and temperature steel — provides As,min for main direction and transverse reinforcement
- Maximum bar spacing limits — NSCP 2015 Section 407.7
- Effective depth d computation — requires knowing cover and bar diameter
Common exam questions
Example
Mu = 18 kN·m/m, d = 145 mm, h = 170 mm, f'c = 21 MPa, fy = 415 MPa, b = 1000 mm.
Approach
Compute Rn → ρ → As,calc. Compare with As,min = 0.0018bh. Take the larger. Convert to spacing.
Example
As = 420 mm²/m, 12 mm bars (Ab = 113 mm²): s = 113×1000/420 = 269 mm. Max = min(3×175, 450) = 450 mm. Use 250 mm.
Approach
s = Ab × 1000 / As; check vs. max spacing min(3h, 450 mm).
Example
16 mm bars at 300 mm: As = 201×1000/300 = 670 mm²/m. Check φMn ≥ Mu.
Approach
Compute actual As from given spacing → compute φMn → compare with Mu.
Key points to remember
- Always design with b = 1000 mm (1 m strip) — this is the defining feature of slab design.
- Mu must be in N·mm when using Rn = Mu/(φbd²) with b and d in mm.
- As,min for slabs = 0.0018 × b × h (using FULL thickness h) for fy = 415–420 MPa.
- Bar spacing s = Ab × 1000 / As — the 1000 accounts for the 1 m (1000 mm) strip width.
- Maximum main steel spacing: min(3h, 450 mm); maximum temperature steel spacing: min(5h, 450 mm).
- Always round bar spacing DOWN (or to the nearest 25 mm below the calculated value) — never round up as that would reduce As.
Shrinkage and Temperature Reinforcement
Examples
A 175 mm thick one-way slab uses fy = 415 MPa. Find the required temperature steel per metre and the spacing of 10 mm bars (Ab = 78.5 mm²). Check against maximum spacing.
Solution
Step 1: ρtemp = 0.0018 (fy = 415 MPa).
Step 2: As,temp = 0.0018 × 1000 × 175 = 315 mm²/m.
Step 3: Spacing with 10 mm bars:
s = Ab × 1000/As = 78.5 × 1000/315 = 249.2 mm → adopt 225 mm.
Step 4: Maximum spacing = min(5 × 175, 450) = min(875, 450) = 450 mm.
s = 249.2 mm < 450 mm ✓
Adopt s = 225 mm c/c.
The 450 mm limit governs over 5h = 875 mm for this slab. The computed spacing of 249 mm is well within limits. Rounding down to 225 mm is standard practice. Temperature bars at 225 mm run perpendicular to the main flexural bars.
A 200 mm thick slab uses fy = 275 MPa. Find the temperature steel area and spacing of 12 mm bars (Ab = 113 mm²).
Solution
Step 1: ρtemp = 0.0020 (fy = 275 MPa).
Step 2: As,temp = 0.0020 × 1000 × 200 = 400 mm²/m.
Step 3: s = 113 × 1000/400 = 282.5 mm → adopt 275 mm.
Step 4: Max spacing = min(5 × 200, 450) = min(1000, 450) = 450 mm.
s = 282.5 mm < 450 mm ✓
Adopt s = 275 mm c/c.
Grade 40 (fy = 275 MPa) bars require a higher steel ratio (0.0020 vs. 0.0018) because the lower-strength bars are less effective per unit area in restraining cracks. More steel area is needed to compensate.
In one-way slabs, the main flexural steel runs across the short span. In the perpendicular direction (the long span), there is no significant bending. However, concrete shrinks as it cures and expands/contracts with temperature changes. Without steel in the transverse direction, uncontrolled cracks would form. NSCP 2015 (Section 407.6.1 / ACI 318-14 Section 24.4.3) therefore mandates minimum transverse reinforcement called shrinkage and temperature (S&T) steel.
Minimum Ratios (NSCP 2015):
- fy = 275 MPa (Grade 40) deformed bars: ρtemp = 0.0020
- fy = 415–420 MPa (Grade 60) deformed bars or welded wire fabric: ρtemp = 0.0018
- fy > 420 MPa: ρtemp = max(0.0014, 0.0018 × 420/fy) — less common in board exams
Formula:
As,temp = ρtemp × b × h (using FULL thickness h, per metre length of slab)
Important Distinction:
In the main (short-span) direction, this same formula As,min = 0.0018 × b × h serves as the minimum steel requirement for the flexural reinforcement when the computed As is less than this value.
Maximum Spacing:
For temperature steel: s ≤ min(5h, 450 mm)
For main flexural steel: s ≤ min(3h, 450 mm)
Physical Basis:
The factor 0.0018 was empirically calibrated to provide enough distributed reinforcement to keep crack widths acceptably small (typically ≤ 0.3–0.4 mm) under restrained shrinkage conditions. The 5h spacing limit ensures no large unrestrained areas exist.
S&T Steel in Two-Way Slabs:
In two-way slabs, main steel is provided in both directions and typically exceeds the minimum S&T requirement. S&T steel as a separate entity is mainly a concern in one-way slabs.
Applications
- Providing transverse bars in one-way slab construction
- Setting minimum reinforcement in the main direction when flexural demand is very low
- Controlling crack widths due to volumetric changes (temperature cycles, concrete curing shrinkage)
- Slab-on-grade construction where temperature swings are significant
Common misconceptions
- WRONG: Using ρtemp = 0.0018 for fy = 275 MPa — Grade 40 bars need 0.0020.
- WRONG: Computing As,temp using d (effective depth) instead of h (full thickness).
- WRONG: Applying max spacing of min(3h, 450) for temperature bars — temperature bars use min(5h, 450).
- WRONG: Omitting temperature bars entirely in a one-way slab — they are MANDATORY per NSCP.
Related concepts
- Minimum flexural reinforcement for slabs (same formula as temperature steel)
- Maximum bar spacing limits (Section 407.7, NSCP 2015)
- Crack control in reinforced concrete (ACI 318 Chapter 24)
- One-way slab flexural design (temperature As,min may govern main steel)
Common exam questions
Example
200 mm slab, fy = 420, 12 mm bars: As = 0.0018×1000×200 = 360 mm²/m; s = 113×1000/360 = 314 mm < 450 mm ✓.
Approach
As,temp = ρtemp × 1000 × h; s = Ab × 1000/As; check min(5h, 450).
Example
If As,calc = 280 mm²/m and As,temp = 315 mm²/m → temperature minimum governs → use 315 mm²/m.
Approach
Compare As,calc (from Rn–ρ) with As,temp. Use the larger.
Example
Multiple choice: fy = 275 MPa → ρtemp = 0.0020 (not 0.0018).
Approach
Match fy to the NSCP table: fy ≤ 275 → 0.0020; fy = 415–420 → 0.0018.
Key points to remember
- ρtemp = 0.0018 for fy = 415–420 MPa; ρtemp = 0.0020 for fy = 275 MPa.
- As,temp = ρtemp × b × h — use FULL thickness h, not effective depth d.
- For a 1 m strip: As,temp = ρtemp × 1000 × h [mm²/m].
- Maximum spacing for temperature steel: min(5h, 450 mm).
- Maximum spacing for main steel: min(3h, 450 mm).
- The shrinkage/temperature As,min also acts as the minimum for main flexural steel in slabs — this frequently governs for lightly loaded slabs.
Two-Way Slab Analysis: Direct Design Method and Coefficient Method
Examples
A flat plate floor system has panels 6 m × 7 m (L1 = 6 m in design direction, L2 = 7 m transverse). Clear span Ln = 5.5 m. Factored load wu = 12 kN/m². Compute the total static moment Mo and distribute it for an interior span.
Solution
Step 1: Mo = wu × L2 × Ln² / 8 = 12 × 7 × 5.5² / 8 = 12 × 7 × 30.25 / 8 = 2541/8 = 317.6 kN·m.
Step 2 (Interior span):
Negative moment = 0.65 × 317.6 = 206.4 kN·m
Positive moment = 0.35 × 317.6 = 111.2 kN·m.
Step 3 (Column strip width): min(L1/2, L2/2) = min(6/2, 7/2) = min(3, 3.5) = 3 m on each side → total column strip width = 2 × 3 = 6 m.
For interior negative moment, column strip typically takes 75% (if no beams):
Column strip negative = 0.75 × 206.4 = 154.8 kN·m (over 6 m width)
Middle strip negative = 0.25 × 206.4 = 51.6 kN·m.
The total static moment Mo = 317.6 kN·m is the equivalent simply-supported moment for the equivalent beam of width L2 spanning Ln. The 65%–35% split allocates more moment to the supports (negative) than to midspan (positive) for interior spans — consistent with an interior continuous bay. The column strip carries more moment because the column regions are stiffer.
Classify a 6 m × 7 m two-way slab (all four sides supported) and identify which method is appropriate if the slab rests directly on columns with no beams.
Solution
Ll/Ls = 7/6 = 1.167 < 2 → TWO-WAY slab.
Supported directly on columns (no beams) → FLAT PLATE → use Direct Design Method (DDM) or Equivalent Frame Method (EFM).
The Coefficient Method is NOT appropriate here because it applies to slabs on stiff beams, not flat plates.
The Coefficient Method assumes significant beam stiffness to redistribute moments through beams. A flat plate has no beams, so DDM or EFM must be used. This distinction is frequently tested in the board exam.
When Ll/Ls < 2, the slab must be designed as two-way. Two primary methods are used in Philippine practice and the board exam:
METHOD 1: Coefficient Method (For Slabs on Stiff Beams)
Used when the slab panel is supported on relatively stiff beams or walls on all four sides. Moments are computed as:
Ma or Mb = C × wu × Ls² (or Ll²)
where C is a tabulated coefficient that depends on the edge conditions (simply supported, continuous, or fixed) and the aspect ratio m = Ls/Ll. The coefficients Ca and Cb for the short and long spans, respectively, are obtained from NSCP tables. This is the simpler method, common in older Philippine practice and still appears in boards.
METHOD 2: Direct Design Method (DDM) — For Flat Plates/Flat Slabs
Applies to slabs supported directly on columns (no beams), with at least three spans in each direction, regular rectangular panels, and uniform loads. The procedure:
Step 1: Total Static Moment Mo
Mo = (wu × L2 × Ln²) / 8
where:
- wu = factored uniform load [kN/m²]
- L2 = width of the strip (transverse span centre-to-centre of columns) [m]
- Ln = clear span in the direction being designed [m]
Step 2: Distribute Mo to Negative and Positive Moments
For an interior span:
- Negative moment (at supports) = 0.65 Mo
- Positive moment (at midspan) = 0.35 Mo
For an end span:
- Interior negative = 0.70 Mo
- Positive = 0.52 Mo
- Exterior negative = 0.26 Mo (with spandrel beam) or 0.30 Mo (without)
Step 3: Distribute to Column Strip and Middle Strip
The column strip (width = min(L1/2, L2/2) centred on the column line) receives a larger share of the moment than the middle strip. Distribution percentages are given in NSCP tables and depend on the relative beam stiffness.
Step 4: Design Each Strip
Each strip is designed as a rectangular beam of known width and moment.
Limitations of DDM:
- Must have ≥ 3 spans in each direction
- Spans must not vary by more than 1/3
- Loads must be gravity only (no lateral loads governing)
- Ratio of live to dead load ≤ 2
Equivalent Frame Method (EFM): A more rigorous approach for irregular geometry or non-uniform loads; typically not fully worked out in board exams but may be tested conceptually.
Applications
- Design of flat plate and flat slab floor systems in multi-storey buildings
- Design of two-way slabs on beams in warehouses and industrial facilities
- Punching shear check at column supports in flat plates (related to DDM)
- Determination of reinforcement layout in column and middle strips
Common misconceptions
- WRONG: Using L1 (design span) instead of L2 (transverse span) in the Mo formula.
- WRONG: Using total span L instead of CLEAR span Ln in the Mo formula — always use the clear span.
- WRONG: Applying DDM moment distribution factors (0.65/0.35) to the Coefficient Method — these are separate methods with different coefficients.
- WRONG: Assuming DDM applies to any two-way slab — DDM has specific applicability conditions (≥3 spans, regular, gravity only).
- WRONG: Confusing the column strip width (min(L1/2, L2/2) × 2) with the total panel width.
Related concepts
- One-way vs. two-way classification (prerequisite for choosing the analysis method)
- Flat plate and flat slab systems (relevant structural systems for DDM)
- Punching shear at column supports (critical check for flat plates)
- Equivalent Frame Method (more rigorous alternative to DDM)
Common exam questions
Example
wu = 15 kN/m², L2 = 5 m, Ln = 4.6 m: Mo = 15×5×4.6²/8 = 15×5×21.16/8 = 198.4 kN·m.
Approach
Identify wu, L2 (transverse span), Ln (clear span). Apply Mo = wu × L2 × Ln²/8.
Example
Mo = 200 kN·m, interior span: M(−) = 130 kN·m; M(+) = 70 kN·m.
Approach
Interior span: 0.65Mo (−ve) and 0.35Mo (+ve). End span: 0.70Mo (int. −ve), 0.52Mo (+ve), 0.26–0.30Mo (ext. −ve).
Example
5 m × 6 m two-way slab on 300 mm × 600 mm beams → Coefficient Method.
Approach
If on stiff beams → Coefficient Method. If on columns (flat plate/slab) → DDM or EFM.
Key points to remember
- DDM total static moment: Mo = wu × L2 × Ln² / 8 — memorise this formula exactly.
- L2 is the TRANSVERSE span (perpendicular to the direction being analysed); Ln is the CLEAR span in the design direction.
- Interior span moment split: 0.65Mo negative, 0.35Mo positive.
- End span moment split: 0.70Mo (interior negative), 0.52Mo (positive), 0.26–0.30Mo (exterior negative).
- Column strip receives the larger share of the moment; middle strip receives the remainder.
- DDM requires at least 3 spans, regular panels, and gravity loads with LL/DL ≤ 2.
Practice problems
PROBLEM 1: A one-way slab spans 4.8 m between simple supports. The main reinforcement is Grade 60 bars (fy = 415 MPa). (a) Find the minimum slab thickness per NSCP 2015. (b) If h = 200 mm and d = 170 mm, find the temperature steel area per metre (fy = 415 MPa) and the spacing of 10 mm bars (Ab = 78.5 mm²).
Solution
(a) Minimum Thickness:
— Support condition: simply supported → divisor = 20.
— fy correction factor: (0.4 + 415/700) = (0.4 + 0.5929) = 0.9929.
— h_min = (4800/20) × 0.9929 = 240 × 0.9929 = 238.3 mm → round up → h_min = 240 mm.
(b) Temperature Steel (h = 200 mm, fy = 415 MPa):
— ρtemp = 0.0018.
— As,temp = 0.0018 × 1000 × 200 = 360 mm²/m.
— Spacing: s = 78.5 × 1000/360 = 218.1 mm → adopt s = 200 mm.
— Max spacing check: min(5 × 200, 450) = min(1000, 450) = 450 mm. s = 218 mm < 450 mm ✓.
— Final answer: 10 mm bars at 200 mm c/c.
Part (a) shows that for a 4.8 m simply-supported span, the minimum thickness is effectively L/20 corrected for fy = 415 MPa (very close to fy = 420). The 240 mm result means the designer should use at least a 240 mm thick slab to skip deflection calculations. Part (b) demonstrates that for a 200 mm slab, the required temperature steel (360 mm²/m) is met comfortably by 10 mm bars at 200 mm spacing. The 450 mm max spacing limit governs over the 5h limit for this slab thickness.
PROBLEM 2: A one-way continuous slab (both ends continuous) has a span of 5.0 m, h = 200 mm, d = 170 mm, b = 1000 mm, f'c = 21 MPa, fy = 415 MPa. The factored moment is Mu = 28 kN·m/m. (a) Find the required main steel area As. (b) Determine the spacing of 16 mm bars (Ab = 201 mm²). (c) Check the maximum spacing limit.
Solution
(a) Main Steel:
Step 1: Rn = Mu/(φbd²) = (28 × 10⁶)/(0.90 × 1000 × 170²) = 28,000,000/26,010,000 = 1.076 MPa.
Step 2: ρ = (0.85 × f'c/fy)[1 − √(1 − 2Rn/(0.85f'c))]
= (0.85 × 21/415)[1 − √(1 − 2(1.076)/(0.85 × 21))]
= (0.04301)[1 − √(1 − 2.152/17.85)]
= (0.04301)[1 − √(1 − 0.12057)]
= (0.04301)[1 − √0.87943]
= (0.04301)[1 − 0.93779]
= (0.04301)(0.06221)
= 0.002676.
Step 3: As,calc = 0.002676 × 1000 × 170 = 454.9 mm²/m.
Step 4: As,min = 0.0018 × 1000 × 200 = 360 mm²/m.
As,calc = 454.9 > As,min = 360 → flexural steel governs → As = 455 mm²/m.
(b) Spacing:
s = 201 × 1000/455 = 441.8 mm → adopt s = 425 mm.
(c) Max spacing check:
min(3h, 450) = min(3 × 200, 450) = min(600, 450) = 450 mm.
Adopted s = 425 mm < 450 mm ✓.
Final: 16 mm bars at 425 mm c/c.
For this more heavily loaded slab, the flexural requirement (455 mm²/m) exceeds the temperature minimum (360 mm²/m), so the flexural As governs. The computed spacing of 442 mm was close to the 450 mm limit — using 425 mm rounds down safely. Note that for both-ends-continuous slabs, h_min = (5000/28)(0.4 + 415/700) = 178.6 × 0.993 = 177 mm, so h = 200 mm satisfies the minimum thickness.
PROBLEM 3: A flat plate floor system has square panels of 7 m × 7 m (centre-to-centre of columns). Columns are 500 mm square. The factored uniform load wu = 14 kN/m². (a) Classify the slab. (b) Compute the total static moment Mo for the design strip in one direction. (c) Distribute Mo to positive and negative moments for an interior span. (d) Determine the column strip width.
Solution
(a) Classification:
Ll/Ls = 7/7 = 1.0 < 2 → TWO-WAY slab (flat plate, as it is on columns with no beams).
(b) Total Static Moment Mo:
— Clear span: Ln = 7000 − 500 = 6500 mm = 6.5 m.
— Transverse span: L2 = 7 m (same in both directions for square panels).
— Mo = wu × L2 × Ln² / 8 = 14 × 7 × 6.5² / 8 = 14 × 7 × 42.25 / 8 = 4140.5/8 = 517.6 kN·m.
(c) Moment Distribution (Interior Span):
— Negative moment = 0.65 × Mo = 0.65 × 517.6 = 336.4 kN·m.
— Positive moment = 0.35 × Mo = 0.35 × 517.6 = 181.2 kN·m.
(d) Column Strip Width:
— On each side of column line: min(L1/2, L2/2) = min(7/2, 7/2) = 3.5 m.
— Total column strip width = 2 × 3.5 = 7.0 m (i.e., the full panel width for a square panel).
Note: In practice, the column strip takes the full panel width for square flat plates, and moment is further distributed between column half-strips.
For a square flat plate, the column strip spans the full panel width (7 m). The total static moment Mo = 517.6 kN·m is the benchmark moment used for all further distribution. The interior span distribution (0.65/0.35) reflects the continuity at both supports. This problem structure — computing Mo and distributing it — is a classic DDM board exam question.
PROBLEM 4: A rectangular slab panel is 4 m × 9 m. (a) Classify the slab. (b) If it is a one-way slab (simply supported along the 9 m edges), determine the minimum slab thickness for fy = 420 MPa. (c) If h = 180 mm, d = 155 mm, f'c = 28 MPa, fy = 420 MPa, and Mu = 10 kN·m/m, find As for the main and temperature steel, and the spacing of 12 mm bars (Ab = 113 mm²) for each.
Solution
(a) Classification:
Ll/Ls = 9/4 = 2.25 ≥ 2 → ONE-WAY slab.
(b) Minimum Thickness (simply supported, fy = 420 MPa):
Factor = (0.4 + 420/700) = (0.4 + 0.60) = 1.0.
h_min = (4000/20) × 1.0 = 200 mm.
(c) Design (h = 180 mm given — note this is less than h_min = 200 mm, so in practice we would increase h; however, the problem asks us to design with h = 180 mm):
MAIN STEEL:
Rn = (10 × 10⁶)/(0.90 × 1000 × 155²) = 10,000,000/21,622,500 = 0.4625 MPa.
ρ = (0.85 × 28/420)[1 − √(1 − 2(0.4625)/(0.85 × 28))]
= (0.05667)[1 − √(1 − 0.925/23.8)]
= (0.05667)[1 − √(1 − 0.03887)]
= (0.05667)[1 − √0.96113]
= (0.05667)[1 − 0.98040]
= (0.05667)(0.01960) = 0.001111.
As,calc = 0.001111 × 1000 × 155 = 172.2 mm²/m.
As,min (temperature) = 0.0018 × 1000 × 180 = 324 mm²/m.
Since 172 < 324 → minimum governs → As,main = 324 mm²/m.
Spacing (main): s = 113 × 1000/324 = 348.8 mm → adopt 325 mm.
Max main spacing: min(3 × 180, 450) = 450 mm. 325 mm < 450 mm ✓.
TEMPERATURE STEEL:
As,temp = 0.0018 × 1000 × 180 = 324 mm²/m (same value, same formula).
Spacing (temp): s = 113 × 1000/324 = 348.8 mm → adopt 325 mm.
Max temp spacing: min(5 × 180, 450) = min(900, 450) = 450 mm. 325 mm < 450 mm ✓.
This problem highlights a critical concept: the minimum temperature steel (0.0018bh = 324 mm²/m) governs both the main steel (since flexural As = 172 mm²/m is less) and the temperature steel itself. Both sets of bars end up with the same requirement of 324 mm²/m, which is a common outcome for lightly loaded slabs. In practice, h would be increased to 200 mm to satisfy the minimum thickness requirement.
Exam preparation tips
- MEMORISE the four divisors for minimum thickness: 20 (simple), 24 (one-end continuous), 28 (both-end continuous), 10 (cantilever). A mnemonic: 'Simple-20, One-24, Both-28, Cantilevered-10' — note that 10 is the smallest divisor, giving the THICKEST slab.
- ALWAYS apply the fy correction factor (0.4 + fy/700) unless explicitly told fy = 420 MPa. Board problems frequently use fy = 415 or fy = 275 as a deliberate trap.
- Keep UNITS consistent: Mu must be in N·mm (not kN·m) when substituting into Rn = Mu/(φbd²) with b and d in mm. Multiply kN·m by 10⁶ to get N·mm.
- For the minimum As in slabs, use 0.0018bh (fy = 415–420) or 0.0020bh (fy = 275) — use FULL thickness h, NOT effective depth d. This is different from beams.
- ALWAYS check maximum bar spacing AFTER computing s from As. Fail to check this and you may provide spacing that violates NSCP — a common source of board exam point deduction.
- In the DDM formula Mo = wu × L2 × Ln²/8: L2 is the TRANSVERSE span (width of design strip), Ln is the CLEAR span in the design direction. Do not confuse these or use centre-to-centre spans for Ln.
- For classification, always compute Ll/Ls as LONG over SHORT. If you get a ratio < 1, you divided backwards — recompute.
- Remember that both the main AND temperature steel formulas use As = ρ × b × h (full h). The distinction is that main steel uses ρ from Rn or the minimum 0.0018, while temperature steel always uses exactly 0.0018 (fy = 415–420) regardless of Mu.
- Practice the bar spacing formula: s = Ab × 1000/As. The '1000' is the strip width in mm. Verify your answer by back-computing As = Ab × 1000/s and confirming it ≥ required As.
- For DDM moment distribution: the interior span split (0.65/0.35) and end span split (0.70/0.52/0.26) are fixed percentages in NSCP — write them on your scratch paper at the start of the exam to avoid recall errors.
Diagrams
Reinforced Concrete Slabs — Chapter Overview
A comprehensive mind map showing all major topics in the Reinforced Concrete Slabs chapter: classification criteria, minimum thickness formulas, flexural design procedure, temperature steel requirements, and two-way slab analysis methods. Use this as a study overview before diving into individual concepts.
Slab Design Decision Flowchart — From Classification to Reinforcement
A step-by-step flowchart guiding the designer through the complete slab design process: classification → minimum thickness check → setting up the 1-metre strip → computing Rn and ρ → checking minimum steel → determining bar spacing → verifying maximum spacing limits. Includes branching for two-way slabs toward DDM. Use this flowchart as a checklist during board exam problems.
Minimum Thickness Selection — NSCP 2015 One-Way Slab
A decision flowchart for selecting the correct minimum slab thickness per NSCP 2015. The chart identifies the four support conditions (each with its divisor), computes the base thickness, applies the fy correction factor (critical for fy ≠ 420 MPa), and rounds up to a practical value. The cantilever is highlighted in yellow to flag it as the most restrictive condition (smallest divisor = thickest slab).
Direct Design Method — Total Static Moment and Distribution
A flowchart illustrating the Direct Design Method (DDM) for two-way flat plates: computing the total static moment Mo, distributing it to positive and negative moments based on interior vs. end span conditions, subdividing into column and middle strips, and designing each strip independently. The key formula Mo = wu × L2 × Ln²/8 and the distribution percentages (0.65/0.35 for interior spans) are highlighted.
DDM Moment Distribution — Interior Span of Flat Plate
A pie chart showing the NSCP/ACI DDM moment distribution for an interior span: 65% of the total static moment Mo is assigned to the negative (support) moments and 35% to the positive (midspan) moment. This reflects the higher moment demand at continuous supports compared to midspan in interior bays.
Slab Design Procedure — Step-by-Step Timeline
A timeline-style visual guide showing all nine steps of the one-way slab design procedure in chronological order. This serves as a quick reference checklist for board exam problem-solving, ensuring no critical step (like the fy correction in Step 2 or the spacing check in Step 8) is overlooked.
In summary
Reinforced concrete slab design is a systematic, formula-driven process that rewards careful attention to code provisions and unit consistency. The key competencies tested in the PRC Civil Engineer Licensure Examination can be summarised as follows:
Classification: Always compute Ll/Ls (long over short). A ratio ≥ 2 gives a one-way slab; support on only two sides is always one-way. This single decision determines your entire analysis path.
Minimum Thickness: The four NSCP divisors (20, 24, 28, 10) must be memorised along with the fy correction factor (0.4 + fy/700). Forgetting to apply this factor when fy ≠ 420 MPa is the single most common board exam error in this topic.
Flexural Design: Design a 1 m strip (b = 1000 mm) using the Rn–ρ–As procedure from rectangular beam design. Always compare computed As with As,min = 0.0018bh; for lightly loaded slabs, the minimum steel almost always governs.
Bar Spacing: Convert As to spacing using s = Ab × 1000/As, then verify against maximum spacing limits: min(3h, 450 mm) for main steel and min(5h, 450 mm) for temperature steel.
Two-Way Slabs: For DDM problems, memorise Mo = wu × L2 × Ln²/8 and the interior span distribution (0.65/0.35). Know the difference between the Coefficient Method (slabs on stiff beams) and DDM/EFM (flat plates/slabs on columns).
Mastery of this chapter not only prepares you for direct slab problems in the board exam but also strengthens your understanding of the broader NSCP 2015 framework — as the same φ factors, material constants, and design philosophy that govern slabs apply throughout the entire reinforced concrete design subject. Study the worked examples carefully, practise the exercises, and verify your process against the flowcharts provided. With consistent practice, slab problems become among the most straightforward and reliably solved items in the civil engineering licensure examination.