CELE Reinforced & Prestressed Concrete — Reinforced Concrete SlabsStudy Notes

The primary distinction between one-way and two-way slabs determines how loads are distributed and how the slab is analysed and designed. **One-Way Slab Action:** A rectangular slab supported on all four sides behaves as a one-way slab when the long-to-short span ratio satisfies: L_long / L_short ≥ 2 In a one-way slab, the majority of the load is carried in the short span direction. The slab bends like a series of parallel beams, with main reinforcement running perpendicular to the short span. One-way slabs supported on only two opposite sides (on beams along the long sides) are always classified as one-way regardless of the span ratio. **Two-Way Slab Action:** When the span ratio is less than 2 (L_long / L_short < 2), the slab is classified as two-way. Loads are distributed in both directions toward all four supports. The slab deflects and bends in both orthogonal directions, requiring reinforcement in both directions. Two-way slabs are more efficient than one-way slabs of the same thickness but require more complex analysis. **Practical Example:** For a rectangular panel measuring 5.0 m × 7.5 m, the span ratio is 7.5 / 5.0 = 1.5, which is less than 2.0, so this panel acts as a two-way slab. Conversely, a 3.0 m × 7.5 m panel has a ratio of 7.5 / 3.0 = 2.5, exceeding 2.0, and is designed as a one-way slab. This classification is critical because it dictates the design methodology, the location and direction of reinforcement, and the analysis technique (coefficient method, direct design method, or equivalent frame method).

To avoid explicit deflection calculations, NSCP 2015 (Table 8.3.1.1) specifies minimum slab thicknesses based on span length and support conditions. These minimums apply to non-prestressed slabs with f_y = 420 MPa. For other yield strengths, the minimum thickness is multiplied by a correction factor. **Minimum Thickness Equations (f_y = 420 MPa):** For simply supported one-way slabs: h_min = L / 20 For one-way slabs with one end continuous: h_min = L / 24 For one-way slabs with both ends continuous: h_min = L / 28 For cantilever slabs: h_min = L / 10 where L is the clear span in millimetres. **Adjustment for Different Yield Strength:** When f_y ≠ 420 MPa, apply the correction factor: h_adjusted = h_min × (0.4 + f_y / 700) This formula accounts for the reduced stiffness (higher steel strains) when lower-strength steel is used. **Physical Interpretation:** These limits ensure that steel strains remain manageable and deflections stay within acceptable limits (typically L/240 to L/360 under service loads). A stiffer slab (thicker) with lower steel stress produces smaller deflections. **Two-Way Slabs:** For two-way slabs, NSCP 2015 provides separate minimum thickness formulas depending on whether the slab is on stiff beams (coefficient method) or on columns (DDM/flat plate). Typical minimums range from L/30 to L/40 for the longer span in two-way systems on columns. **Common Steel Grades in the Philippines:** - Grade 415 (f_y = 415 MPa): Common deformed bars - Grade 275 (f_y = 275 MPa): Plain bars (less common in modern practice) - Grade 500 (f_y = 500 MPa): High-strength bars (sometimes imported)

One-way slabs are designed using the per-metre strip approach, where a 1 m wide section is treated as a rectangular beam with width b = 1000 mm. This simplifies the design process and allows direct application of beam flexural design principles. **Design Procedure:** 1. **Determine Load per Unit Area:** Calculate the factored load w_u (kN/m²) including self-weight, live load, and other applied loads, all multiplied by appropriate load factors (typically 1.4 for DL, 1.7 for LL per NSCP 2015). 2. **Calculate Moment per Unit Width:** For a 1 m wide strip spanning length L: - Simply supported: M_u = w_u × L² / 8 - One end continuous: M_u = w_u × L² / 10 (at mid-span) - Both ends continuous: M_u = w_u × L² / 16 (at mid-span) - Cantilever: M_u = w_u × L² / 2 Express M_u in N·mm for direct use with d in mm. 3. **Calculate the Nominal Moment Capacity Factor (Moment Coefficient):** R_n = M_u / (φ × b × d²) where φ = 0.90 for flexure (NSCP 2015 Section 9.3.2.1) b = 1000 mm (per-metre width) d = effective depth in mm 4. **Find the Steel Ratio ρ:** ρ = (0.85 × f'_c / f_y) × [1 - √(1 - 2 × R_n / (0.85 × f'_c))] where f'_c is in MPa and f_y is in MPa. 5. **Check Against Code Limits:** - Minimum steel ratio (shrinkage/temperature): ρ_min = 0.0018 (for f_y ≥ 415 MPa) - Maximum steel ratio: ρ_max ≈ 0.75 × ρ_b (ensures ductile failure) Use ρ = max(ρ_calculated, ρ_min) and verify ρ ≤ ρ_max. 6. **Calculate Required Steel Area:** A_s = ρ × b × d = ρ × 1000 × d (in mm² per metre width) 7. **Convert to Bar Spacing:** Once A_s is determined (mm²/m), convert to bar spacing using: s = (A_bar × 1000) / A_s where A_bar is the cross-sectional area of one bar (mm²), and s is the centre-to-centre spacing in mm. 8. **Verify Spacing Limits:** - Main reinforcement: spacing ≤ min(3h, 450 mm) - Temperature/shrinkage steel: spacing ≤ min(5h, 450 mm) **Important Notes:** - The design is performed for a 1 metre strip (100 cm wide), but the result A_s is expressed per metre. This allows the same formula to apply regardless of total slab width. - All calculations follow ACI 318 / NSCP 2015 flexural design principles (see Chapter 2 notes on rectangular beams). - In many practical cases, the minimum steel ratio (temperature steel) governs the design, especially for lightly loaded or short-span slabs.

Reinforced concrete slabs undergo volumetric changes due to shrinkage during curing and thermal expansion/contraction from temperature variations. Uncontrolled shrinkage and temperature effects can cause random cracking in the slab, reducing durability and appearance. NSCP 2015 Section 7.10.4 requires minimum reinforcement perpendicular to the main flexural steel to control this cracking. **Purpose of Shrinkage/Temperature Steel:** - Limits crack widths to acceptable levels - Distributes cracks uniformly rather than concentrating them at weak points - Provides ductility and prevents sudden failure - Essential in thin slabs and slabs with low flexural demand **Required Steel Ratio:** The minimum ratio of shrinkage/temperature reinforcement is: ρ_temp = 0.0018 (for f_y = 415 – 420 MPa) ρ_temp = 0.0020 (for f_y = 275 MPa) These ratios apply to the full slab thickness h, not the effective depth d. **Calculation of Required Area:** For a 1 metre strip of slab with thickness h and width 1000 mm: A_s,temp = ρ_temp × b × h A_s,temp = ρ_temp × 1000 × h (mm²/m) Example: For a 200 mm slab with f_y = 415 MPa: A_s,temp = 0.0018 × 1000 × 200 = 360 mm²/m **Bar Spacing for Temperature Steel:** s_temp = (A_bar × 1000) / A_s,temp where A_bar is the bar area in mm² and s_temp is the spacing in mm. **Spacing Limits (NSCP 2015 Section 7.10.4.2):** Temperature/shrinkage reinforcement spacing must not exceed: - min(5h, 450 mm) This is a looser limit than main steel [min(3h, 450 mm)] because temperature steel is primarily for crack control, not flexural capacity. **Placement Considerations:** 1. Temperature steel is placed perpendicular to the main flexural steel 2. In one-way slabs, main steel spans the short direction; temperature steel runs parallel to the long side 3. Temperature steel is typically placed closer to the top and bottom surfaces (but still within cover requirements) in two-way systems 4. In two-way slabs, provide temperature/shrinkage steel in both directions **Interaction with Flexural Design:** In practice, the minimum flexural steel (governed by ρ_min = 0.0018 for f_y ≥ 415) is identical to the minimum temperature steel requirement. Thus: - If calculated flexural steel > ρ_min, place the calculated amount as main steel and provide separate temperature/shrinkage steel perpendicular to it - If calculated flexural steel < ρ_min, place ρ_min amount as both main steel and perpendicular direction (effectively a two-way minimum steel condition) **Common Misconception:** Temperature steel is NOT the same as flexural steel. Flexural steel is designed for applied moments in the primary bending direction; temperature steel is a minimum requirement to control cracking from non-mechanical causes (shrinkage, thermal) in the secondary direction.

Two-way slabs, where L_long / L_short < 2, distribute loads to all four supports. The analysis is more complex than one-way systems and requires determining moments in both orthogonal directions. NSCP 2015 (based on ACI 318) provides two main methods: the Coefficient Method (or Method of Moments) for slabs on stiff beams, and the Direct Design Method (DDM) for flat plates and flat slabs on columns. **Method 1: Coefficient Method (for slabs supported on beams)** This method applies when the slab is supported on relatively stiff beams on all four sides. Moments are determined using tabulated coefficients that account for: - Aspect ratio λ = L_long / L_short - Support conditions (simply supported, continuous, etc.) - Relative stiffness of beams Moments are calculated as: M = C × w_u × L_s² where: - C = moment coefficient (from tables in NSCP 2015, Figure 8.10.1 or ACI 318) - w_u = factored load (kN/m²) - L_s = short span (m) The moment is distributed to column strips (width 0.5 × L_short on each side of centre-line) and middle strips (remaining width). Column strips carry a larger portion of the moment (typically 40–60% depending on configuration). Once moments are determined for each strip in each direction, each strip is designed as a one-way slab (per-metre analysis). **Method 2: Direct Design Method (DDM) for Flat Plates and Flat Slabs** The DDM applies to slabs supported directly on columns without beams, with or without drop panels or capitals. This is common in modern construction (flat plates). The method is more streamlined than coefficient method: 1. **Calculate Total Static Moment:** M_o = (w_u × L_2 × L_n²) / 8 where: - w_u = factored load (kN/m²) - L_2 = span perpendicular to the direction being analysed (m) - L_n = clear span in the direction of analysis, measured face-to-face of columns or capital edges (m) 2. **Distribute M_o to Negative and Positive Moments:** At an interior span: - Negative moment at support: M_neg = 0.65 × M_o - Positive moment at mid-span: M_pos = 0.35 × M_o At an exterior span (different factors for edge conditions). 3. **Further Distribute to Column and Middle Strips:** The negative moment at an interior support line is divided: - Column strip: 60% of the moment - Middle strips on either side: 40% total For positive moments: - Column strip: 60% - Middle strips: 40% 4. **Design the Strips:** Divide the moment by the strip width to get moment per unit width (kN·m/m), then design as a one-way slab. **Limitations and Applicability:** - **Coefficient Method:** Best for regularly proportioned slabs with beams of similar stiffness on all sides; complex for irregular layouts. - **DDM:** Limited to slabs with interior columns in a regular grid, rectangular bays, uniform load, and no abrupt changes in span length. Not applicable to edge columns with cantilever portions. - **Equivalent Frame Method (EFM):** More general; can handle irregular configurations, cantilevers, and varying beam stiffness. Not detailed here but is a refinement of DDM. **Minimum Thickness for Two-Way Slabs:** NSCP 2015 provides minimum thickness formulas for two-way slabs to avoid deflection checks: For slabs without interior beams (flat plates): h_min = (L_n / 33) to (L_n / 40) depending on support conditions where L_n is the clear span in the longer direction. For slabs on beams with relative stiffness parameter α_f ≥ 2: h_min = (L / 21) to (L / 30) depending on configuration These are more permissive than one-way slabs (which use L/20, L/24, L/28) because the two-way support reduces deflection.

Preparing for the PRC Civil Engineer Licensure Examination requires not only understanding the theory but also recognizing common errors that appear repeatedly in board exams. This section highlights mistakes that cost points and strategies to avoid them. **Pitfall 1: Incorrect Slab Classification (One-Way vs. Two-Way)** Error: Using L_short / L_long instead of L_long / L_short, or misremembering the threshold. Correct Approach: - Always compute the aspect ratio as: Ratio = Longer span / Shorter span - If Ratio ≥ 2.0 → ONE-WAY slab - If Ratio < 2.0 → TWO-WAY slab - A slab supported on only two opposite sides is always ONE-WAY, regardless of ratio Example: A 5 m × 8 m panel: Ratio = 8 / 5 = 1.6 < 2.0 → Two-way (not one-way) **Pitfall 2: Forgetting the f_y Correction Factor for Minimum Thickness** Error: Using h_min = L / 20 (or other base value) directly without adjusting for f_y ≠ 420 MPa. Correct Approach: - If f_y = 420 MPa: use base value directly (no adjustment) - If f_y ≠ 420 MPa: multiply by (0.4 + f_y / 700) Example: - f_y = 415 MPa: Factor = 0.4 + 415/700 = 0.9929 (slightly less than 1.0, so h is slightly reduced) - f_y = 275 MPa: Factor = 0.4 + 275/700 = 0.7929 (much less than 1.0, so h is significantly reduced) A simply supported 4 m slab with f_y = 275 MPa: h_min = (4000 / 20) × 0.7929 = 158.6 mm, not 200 mm **Pitfall 3: Confusing Minimum Steel Ratio with Main Flexural Steel** Error: Not checking whether calculated ρ exceeds ρ_min, and therefore placing inadequate steel. Correct Approach: 1. Calculate ρ from R_n 2. Compare with ρ_min = 0.0018 (for f_y ≥ 415) or 0.0020 (for f_y = 275) 3. **Use the larger value**: A_s = max(ρ_calculated, ρ_min) × b × d 4. In many thin, lightly loaded slabs, ρ_min governs and becomes the main steel 5. Place additional temperature/shrinkage steel perpendicular (also at A_s,temp ≥ ρ_min × b × h) Boardroom note: An examiner might ask "Why did you use 315 mm²/m instead of the 272 mm²/m you calculated?" The correct answer is: "Because the minimum steel ratio (0.0018 × 1000 × 175 = 315 mm²/m) exceeds the calculated design steel, so the minimum governs to ensure ductility and crack control." **Pitfall 4: Wrong Spacing Calculation or Limits** Error 1: Reversing the numerator and denominator in s = (A_bar × 1000) / A_s (results in very large spacings). Error 2: Using the wrong limit [min(3h, 450) for temperature steel instead of min(5h, 450)]. Error 3: Computing spacing but not verifying it against code limits. Correct Approach: - Main flexural steel: s_main = (A_bar × 1000) / A_s, then check s_main ≤ min(3h, 450 mm) - Temperature/shrinkage steel: s_temp = (A_bar × 1000) / A_s,temp, then check s_temp ≤ min(5h, 450 mm) - If calculated spacing exceeds limit, select a smaller bar diameter or use closer spacing Example: A 200 mm slab with 12 mm main bars: A_s = 400 mm²/m s = (113.1 × 1000) / 400 = 283 mm Max = min(3 × 200, 450) = min(600, 450) = 450 mm 283 mm < 450 mm ✓ Acceptable But if A_s = 600 mm²/m: s = (113.1 × 1000) / 600 = 189 mm ✓ Still OK If A_s = 1000 mm²/m: s = 113 mm (very close spacing; might switch to larger bars) **Pitfall 5: Per-Metre Strip Confusion** Error: Treating the 1 m strip as the entire slab width, or miscalculating area for a slab wider than 1 m. Correct Approach: - Design is always performed for a 1 metre (1000 mm) wide strip, regardless of total slab width - All formulas use b = 1000 mm for this reference strip - Result A_s is in mm²/m, meaning: for every metre of slab width, you need this much steel area - Spacing s (mm c/c) tells you how far apart the bars are spaced along the 1 m width Example: If A_s = 400 mm²/m and you use 12 mm bars: s = 283 mm means: in a 1 m width, place bars 283 mm apart, which fits approximately 3–4 bars per metre Actual number of bars per metre ≈ 1000 / 283 ≈ 3.5 bars/m ✓ **Pitfall 6: DDM Applied Outside Its Scope** Error: Attempting to use DDM for an irregular grid, a cantilever slab, or non-uniform loading. Correct Approach: - DDM (Direct Design Method) is restricted to: * Regular rectangular grid of columns * Uniform load on all panels * Similar bay dimensions (longer span / shorter span ≤ 1.5 typically) * No abrupt changes in floor level * No cantilevers beyond the column grid - If any restriction is violated, use the Equivalent Frame Method (EFM), which is more general **Pitfall 7: Miscalculating M_u from Span Formulas** Error: Using the wrong formula for moment based on support conditions, or forgetting to convert to N·mm. Correct Approach: For a 1 m wide strip with factored load w_u (kN/m²) and span L (m): - Simply supported: M_u = w_u × L² / 8 - One end continuous: M_u = w_u × L² / 10 (at mid-span) - Both ends continuous: M_u = w_u × L² / 16 (at mid-span) - Cantilever: M_u = w_u × L² / 2 Result is in kN·m. Convert to N·mm by multiplying by 10⁶ for use with d in mm. Example: w_u = 12 kN/m², L = 4.5 m, simply supported: M_u = 12 × 4.5² / 8 = 12 × 20.25 / 8 = 30.375 kN·m M_u = 30.375 × 10⁶ N·mm = 30.375 × 10⁶ N·mm **Pitfall 8: Effective Depth (d) Miscalculation** Error: Using d = h instead of d = h – cover – bar radius, or using inconsistent assumptions about bar placement. Correct Approach: d = h – clear cover – half bar diameter For a typical slab: - h = total thickness (given or designed) - Clear cover = 25 mm (to bottom main bars in normal environments) or 35–40 mm (exterior / high exposure) - Bar radius = diameter / 2 Example: 200 mm slab, 12 mm bottom bars, 25 mm cover, 10 mm top temperature steel For bottom (main) steel: d = 200 – 25 – 6 = 169 mm For top (temperature) steel: d ≈ 200 – 35 – 5 = 160 mm (approximation; exact value depends on bar placement) **Pitfall 9: Mixing ACI 318 and NSCP 2015 Provisions** Error: Using different φ factors, steel ratios, or formulas from different standards, or misremembering a coefficient. Correct Approach: - Philippine exams expect NSCP 2015 (which is largely based on ACI 318-14 with local modifications) - Key factors: * φ_flexure = 0.90 (not 0.85) * ρ_min = 0.0018 (for f_y ≥ 415 MPa), 0.0020 (for f_y = 275 MPa) * f_y typically 275 or 415 MPa (not 500 MPa in most board problems) * Spacing limits: main ≤ min(3h, 450 mm); temperature ≤ min(5h, 450 mm) **Pitfall 10: Not Showing Working or Rounding Errors** Error: Jumping to answers without clear steps, or rounding intermediate results too aggressively. Correct Approach: - Show all calculation steps on the exam board - Carry sufficient decimal places through intermediate steps - Round only the final answer and practical design dimensions (spacing, thickness) - Example: ρ = 0.00364 (not 0.004), then later round spacing to 260 mm (not 265 mm) - Examiners can give partial credit if your method is correct even if final answer is slightly off due to rounding **Exam Strategy Tips:** 1. **Read the problem carefully.** Identify whether the slab is one-way or two-way, what support condition, what load, and what steel grade. This classification drives the entire solution path. 2. **Sketch the slab.** Draw a simple diagram showing span, support, bar layout. This prevents misunderstanding and can impress the examiner. 3. **State assumptions.** If cover, bar size, or other data are not given, state your assumption (e.g., "Assuming 25 mm cover to main bars"). 4. **Check your answer.** After calculating A_s, verify spacing against limits. After finding thickness, check if it is reasonable for the span and load. 5. **Explain minimum steel.** If minimum steel governs, explicitly state why: "A_s,calc = 270 mm²/m, but A_s,min = 315 mm²/m (minimum required by NSCP 2015 Section 7.10.4), so provide 315 mm²/m." 6. **Practice with similar problems** before the exam. The exam will likely present a one-way slab design, a two-way slab classification, and a minimum thickness question.