CELE Reinforced & Prestressed Concrete — Reinforced Concrete Footings, Bond and DevelopmentDetailed Explanation
The Reinforced Concrete Footings, Bond and Development chapter rewards slow, careful thinking over quick pattern matching, especially on Professional Regulation Commission (PRC) — Board of Civil Engineering's scenario-based CELE items. This detailed explanation walks through the full derivation of every core idea, then links each one to a worked example pulled from recent CELE Reinforced & Prestressed Concrete papers.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Reinforced & Prestressed Concrete section sits under a "Core" weighting, and Reinforced Concrete Footings, Bond and Development is the 6th chapter in the 7-chapter CELE Reinforced & Prestressed Concrete rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Reinforced & Prestressed Concrete.
Reinforced Concrete Footings, Bond and Development - Detailed Explanation
Footings are the lowest structural elements of a building — they receive all column and wall loads and spread them safely onto the supporting soil. In the Philippines, where highly variable soil conditions (from soft coastal clays in Manila Bay areas to weathered volcanic rock in Visayas) demand careful foundation design, the civil engineer must master footing analysis and design thoroughly. This chapter covers the complete design sequence: (1) sizing the plan area from service loads and allowable bearing pressure, (2) checking two-way punching shear and one-way beam shear using factored loads, (3) flexural design for bottom reinforcement, and (4) verifying that bars are properly developed (anchored) to the concrete via adequate embedment length. These topics are perennial favorites in the PRC CE Board Examination, appearing in both the afternoon (design) and morning (theory) sessions. Mastery of these concepts is essential not only for passing the board exam but also for safe professional practice as a registered civil engineer under Republic Act 544.
Concepts
Footing Sizing — Plan Area from Service Loads
The first step in footing design is determining the required plan area. This is done using SERVICE (unfactored) loads because the allowable soil bearing capacity q_a is itself a service-level parameter established by geotechnical investigation. The fundamental relationship is: A_req = P_service / q_a, where P_service is the total service axial load from the column (dead + live, unfactored) in kN, and q_a is the allowable bearing capacity in kPa. For a square footing: side B = sqrt(A_req), rounded UP to the next practical dimension (typically to the nearest 50 mm or 100 mm). For a rectangular footing: choose dimensions L × B such that L × B >= A_req. An important practical refinement: the service load P_service should include the weight of the footing itself and the soil overburden above it (if significant). In many board-exam problems, these are either given or ignored. If the footing self-weight is to be estimated, a common approximation is to assume the footing weighs about 5–10% of the column load, or to compute it as W_ftg = gamma_conc × B × B × h_ftg. Once the plan area is fixed, the NET UPWARD FACTORED PRESSURE used for structural design is: q_u = P_u / A_footing, where P_u = 1.2D + 1.6L (NSCP 2015 / ACI 318-14 load combination). Note: q_u is the NET pressure — it does not include the weight of the footing and soil above, which cancel with the upward soil reaction when computing shear and moment. This simplification is standard in both NSCP 2015 and ACI 318.
Examples
Notice that the plan area is sized from the SERVICE load (1200 kN), not the factored load (1620 kN). Once the geometry is fixed, we switch to factored loads for all strength checks. This two-step process — service load for sizing, factored load for design — is a classic board-exam trap.
Scenario
A 400 mm square interior column carries a dead load of 750 kN and a live load of 450 kN. The allowable soil bearing capacity is q_a = 200 kPa. Size a square footing.
Solution
Step 1 — Service load: P_service = D + L = 750 + 450 = 1200 kN. Step 2 — Required area: A_req = P_service / q_a = 1200 / 200 = 6.00 m². Step 3 — Side dimension: B = sqrt(6.00) = 2.449 m → use B = 2.50 m (round up). Step 4 — Actual area: A = 2.50 × 2.50 = 6.25 m². Step 5 — Factored load: P_u = 1.2(750) + 1.6(450) = 900 + 720 = 1620 kN. Step 6 — Net factored pressure: q_u = P_u / A = 1620 / 6.25 = 259.2 kPa. Answer: Use 2.50 m × 2.50 m footing; q_u = 259.2 kPa for structural design.
Applications
- Isolated square and rectangular column footings for low- to mid-rise buildings.
- Wall footings (strip footings) — same concept but per unit length.
- Combined footings where two columns share one footing.
- Raft/mat foundations — pressure is distributed over the entire basement floor.
Misconceptions
- Using factored loads (P_u) to size the footing area — this overestimates the required area since q_a is a service-level value.
- Forgetting to round UP the footing dimension — using B = 2.449 m is non-conservative.
- Using q_u with the required area instead of the actual area — once you round up, recalculate q_u with the actual larger area.
- Confusing gross bearing pressure (includes footing weight) with net pressure (excludes self-weight) — most board problems use net pressure.
Related Concepts
- Soil bearing capacity (geotechnical engineering)
- NSCP 2015 load combinations
- Dead and live load definitions
- Combined and raft foundations
Common Exam Questions
Example
P_service = 900 kN, q_a = 150 kPa → A = 900/150 = 6.0 m² → B = 2.45 m → use 2.50 m.
Approach
Divide total service load by q_a to get A_req, then take the square root for a square footing. Round up.
Question Type
Find required footing size
Example
D = 600 kN, L = 400 kN, B = 2.5 m → P_u = 1.2(600)+1.6(400) = 1360 kN → q_u = 1360/6.25 = 217.6 kPa.
Approach
Apply NSCP load combination to get P_u = 1.2D + 1.6L, then divide by the ACTUAL footing area (not A_req).
Question Type
Compute net factored pressure q_u
Key Points To Remember
- Use SERVICE load to size the footing plan area (divide by q_a).
- Use FACTORED load P_u and net q_u for all structural checks (shear, flexure).
- Round footing dimensions UP to ensure A >= A_req.
- q_u = P_u / A_footing (net upward pressure, factored).
- Include footing self-weight in service load if asked; it cancels in net q_u calculations.
- Square footing: B = sqrt(P_service / q_a), then round up.
- Factored load combination per NSCP 2015: U = 1.2D + 1.6L.
Two-Way (Punching) Shear in Footings
Two-way or punching shear is typically the GOVERNING shear failure mode for square or nearly square column footings. The column tends to punch through the footing slab along a truncated pyramid. Per NSCP 2015 Section 422.6 (and ACI 318-14 Section 22.6), the critical perimeter b_o is taken at d/2 from all column faces, where d is the effective depth of the footing. For a square column of side c (mm) with effective depth d (mm): b_o = 4(c + d) [perimeter of the critical square at d/2 from column faces] The concrete punching shear strength V_c is the MINIMUM of three expressions (NSCP 2015): (1) V_c = 0.17(1 + 2/β_c) λ√f'c · b_o · d (2) V_c = 0.083(α_s·d/b_o + 2) λ√f'c · b_o · d (3) V_c = 0.33 λ√f'c · b_o · d where β_c = ratio of long to short column dimension (= 1.0 for square columns), α_s = 40 for interior columns (30 for edge, 20 for corner), and λ = 1.0 for normal-weight concrete. For SQUARE interior columns: β_c = 1.0, so Expression (1) gives 0.17(1+2) = 0.51λ√f'c·b_o·d, which is LARGER than Expression (3). Thus Expression (3), V_c = 0.33λ√f'c·b_o·d, GOVERNS for square interior columns in most practical cases. The factored shear demand on the critical perimeter is the net upward force on the footing area OUTSIDE the critical perimeter: V_u = q_u × [A_footing − (c + d)²] [for a square column on a square footing] Strength check: φV_c ≥ V_u, where φ = 0.75 (shear). All forces in Newtons (N) and dimensions in mm when using the formula directly; convert V_c to kN for comparison with V_u in kN.
Examples
The large margin (2360 vs 1480 kN) shows that for well-proportioned footings, punching shear is usually satisfied. If φV_c < V_u, the footing depth must be increased — there is no practical way to add stirrups in footings for punching. Note how the demand is computed on the area OUTSIDE the critical perimeter, not the entire footing area.
Scenario
A 2.5 m × 2.5 m square footing supports a 400 mm square column. Effective depth d = 500 mm. Factored load P_u = 1700 kN, f'c = 28 MPa, normal-weight concrete. Check two-way (punching) shear.
Solution
Step 1 — Net factored pressure: q_u = 1700 / (2.5²) = 1700 / 6.25 = 272 kPa. Step 2 — Critical perimeter: b_o = 4(c + d) = 4(400 + 500) = 4(900) = 3600 mm. Step 3 — Critical area: (c + d)² = (900)² = 810,000 mm² = 0.81 m². Step 4 — Punching shear demand: V_u = q_u[A − (c+d)²] = 272 × (6.25 − 0.81) = 272 × 5.44 = 1479.7 kN ≈ 1480 kN. Step 5 — Concrete shear strength (Expression 3, governs for square column): V_c = 0.33(1.0)√28 × 3600 × 500 = 0.33 × 5.292 × 3600 × 500 = 3,147,288 N ≈ 3147 kN. Step 6 — Design strength: φV_c = 0.75 × 3147 = 2360 kN. Step 7 — Check: φV_c = 2360 kN > V_u = 1480 kN. → PUNCHING SHEAR IS ADEQUATE (OK).
Applications
- Checking adequacy of footing thickness (effective depth d) before detailing reinforcement.
- Two-way slab punching at column supports (same formulas apply).
- Determining if shear reinforcement (stud rails) is needed — required when φV_c < V_u.
- Comparing different footing depths to find the minimum d that satisfies punching.
Misconceptions
- Using b_o = 4c (column perimeter) instead of 4(c+d) — the critical perimeter is d/2 from the face, not at the face.
- Computing V_u as q_u × A_total — this ignores that the load directly under the column area is balanced by the column load itself.
- Forgetting to square (c+d): area inside perimeter = (c+d)², not 4(c+d).
- Applying φ = 0.90 (flexure) instead of φ = 0.75 (shear) to V_c.
- Using Expression (1) or (2) without checking which one governs — always take the MINIMUM.
Related Concepts
- One-way (beam) shear in footings
- Two-way shear in flat slabs
- Effective depth d and cover requirements (NSCP 2015)
- Shear strength reduction factor φ
Common Exam Questions
Example
c=500 mm, d=600 mm, f'c=21 MPa, B=3m, P_u=2000 kN → b_o=4(1100)=4400 mm, V_c=0.33(1)√21(4400)(600)=3,978,400 N=3978 kN, φV_c=2984 kN; q_u=2000/9=222.2 kPa, V_u=222.2(9−1.21)=222.2(7.79)=1731 kN → OK.
Approach
1) Find b_o = 4(c+d). 2) Use V_c = 0.33λ√f'c·b_o·d for square interior column. 3) Multiply by φ=0.75. 4) Find V_u = q_u[A−(c+d)²]. 5) Compare.
Question Type
Compute φV_c for punching and compare with V_u
Example
This type requires trial-and-error: assume d, check φV_c vs V_u, adjust until φV_c just exceeds V_u.
Approach
Set φV_c = V_u, substitute expressions for both in terms of d (b_o and (c+d)² both contain d), and solve iteratively or algebraically.
Question Type
Find minimum effective depth d to satisfy punching shear
Key Points To Remember
- Critical perimeter b_o is at d/2 from column faces: b_o = 4(c + d) for square columns.
- Three V_c expressions — use the MINIMUM (governing) value.
- For square interior columns: Expression (3) V_c = 0.33λ√f'c · b_o · d governs.
- Punching shear DEMAND excludes the area inside the critical perimeter: V_u = q_u[A − (c+d)²].
- φ = 0.75 for shear strength checks.
- All dimensions in mm, forces in N for formula; convert to kN for final comparison.
- α_s = 40 (interior), 30 (edge), 20 (corner column).
One-Way (Beam) Shear in Footings
One-way shear treats the footing as a wide, shallow beam bending about an axis perpendicular to the direction of analysis. The critical section for one-way shear is located at a distance d from the FACE of the column (not the centerline), per NSCP 2015 Section 409.4.3 (ACI 318-14 Section 9.4.3). For a square footing of side B supporting a square column of side c, with effective depth d, the projecting cantilever length beyond the critical section is: ℓ_v = (B − c)/2 − d [from column face to critical section] The one-way shear demand at the critical section (entire footing width B resists): V_u = q_u × B × ℓ_v = q_u × B × [(B − c)/2 − d] The one-way shear strength of concrete (NSCP 2015 / ACI 318, no axial load, no shear reinforcement): V_c = 0.17 λ √f'c × B × d where B and d are in mm, f'c in MPa → V_c in Newtons. Convert to kN. Strength check: φV_c ≥ V_u, φ = 0.75. In most well-designed footings, punching shear controls over one-way shear. However, for RECTANGULAR footings that are elongated in one direction, one-way shear may govern in the SHORT direction. Both checks must ALWAYS be performed. Note: Footings are typically designed WITHOUT shear reinforcement (no stirrups). If φV_c < V_u, the solution is to INCREASE the footing depth d, not add stirrups (though NSCP technically permits shear reinforcement in footings).
Examples
The one-way shear demand (513 kN) is much less than φV_c (1113 kN), confirming that for compact square footings, one-way shear rarely governs. Note that ℓ_v is computed from the FACE of the column (not the edge of the footing), and measured inward by d to reach the critical section.
Scenario
Check one-way (beam) shear for a 3.0 m square footing with a 500 mm square column, d = 550 mm, P_u = 2200 kN, f'c = 28 MPa, normal-weight concrete.
Solution
Step 1 — Net factored pressure: q_u = 2200 / (3.0²) = 2200 / 9.0 = 244.4 kPa. Step 2 — Projecting length to critical section: ℓ_v = (B−c)/2 − d = (3000−500)/2 − 550 = 1250 − 550 = 700 mm = 0.70 m. Step 3 — One-way shear demand: V_u = q_u × B × ℓ_v = 244.4 × 3.0 × 0.70 = 513.2 kN. Step 4 — One-way shear strength: V_c = 0.17(1.0)√28 × 3000 × 550 = 0.17 × 5.292 × 1,650,000 = 1,484,154 N = 1484 kN. Step 5 — Design strength: φV_c = 0.75 × 1484 = 1113 kN. Step 6 — Check: φV_c = 1113 kN > V_u = 513 kN → ONE-WAY SHEAR IS ADEQUATE (OK).
Applications
- Checking whether the assumed footing thickness is sufficient without shear reinforcement.
- Designing wall footings (strip footings) — one-way shear is the primary shear concern.
- Rectangular footings where the long direction may have a more critical one-way shear condition.
Misconceptions
- Taking the critical section at the EDGE of the column (ℓ_v = (B−c)/2), forgetting to subtract d.
- Using V_c = 0.33λ√f'c·B·d (the punching formula coefficient) for one-way shear — one-way uses 0.17.
- Checking only punching shear and skipping one-way shear — both are required.
- Taking shear across half the footing width instead of the full width B.
Related Concepts
- Two-way punching shear
- Beam shear in rectangular concrete beams
- Effective depth and cover
- Wall footing design
Common Exam Questions
Example
B=2.5m, c=400mm, d=480mm, q_u=280kPa → ℓ_v=(2500−400)/2−480=1050−480=570mm → V_u=280×2.5×0.57=399kN; V_c=0.17√28(2500)(480)=1,081,344N=1081kN; φV_c=811kN>399kN ✓
Approach
1) Compute q_u = P_u/A. 2) Find ℓ_v = (B−c)/2 − d. 3) V_u = q_u × B × ℓ_v. 4) V_c = 0.17λ√f'c × B × d. 5) Check φV_c ≥ V_u.
Question Type
Compute one-way shear demand and compare with φV_c
Key Points To Remember
- Critical section for one-way shear: distance d from the column face.
- Projecting length: ℓ_v = (B−c)/2 − d.
- V_c = 0.17λ√f'c × B × d (full footing width, no stirrups assumed).
- φ = 0.75 for shear.
- Both punching AND beam shear must be checked — do not skip either.
- If one-way shear fails: increase d (not add stirrups, in standard practice).
- For rectangular footings, check one-way shear in BOTH directions.
Flexural Design of Footings
After shear checks establish the required effective depth d, the bottom flexural reinforcement is designed to resist the bending moment caused by the net upward pressure acting on the cantilevered portion of the footing beyond the column face. CRITICAL SECTION FOR MOMENT: Per NSCP 2015 Section 415.4.2 (ACI 318-14 Section 13.2.7.1), the critical section for flexure in a footing is at the FACE of the column (for concrete columns), at the EDGE of the base plate (for steel columns), or at a point halfway between the center and edge of the wall (for masonry walls). For a square footing of side B with a square column of side c, the cantilever projection beyond the column face is: ℓ = (B − c) / 2 The ultimate bending moment at the column face (per unit width, then scaled to full width B): M_u = q_u × B × ℓ² / 2 = q_u × B × [(B−c)/2]² / 2 Note: This is the moment of the trapezoidal/rectangular pressure load on the cantilever strip of width B. Design the tension steel (bottom bars in both directions for square footings) using the standard rectangular beam formula: M_u = φ × A_s × f_y × (d − a/2) where a = A_s × f_y / (0.85 × f'c × B) and φ = 0.90 (tension-controlled flexure) Minimum reinforcement for footings (NSCP 2015 Section 407.6.1.1): A_s,min = 0.0018 × B × h [for Grade 415 deformed bars, one-way slabs / footings] A_s,min = 0.002 × B × h [for Grade 275 or smooth bars] where h = total footing thickness. Provide bars in BOTH perpendicular directions for square footings. For rectangular footings, the reinforcement in the LONG direction is uniform across the width; for the SHORT direction, a larger portion of the bars are concentrated in a central band of width equal to the SHORT dimension.
Examples
This moment acts uniformly across the full width B = 2.5 m of the footing. The factor of 1/2 comes from the triangular (uniform pressure on cantilever) moment formula: the pressure intensity q_u × B acts uniformly, and the moment arm from the resultant to the critical section is ℓ/2 ... wait, more precisely: M = (q_u × B × ℓ) × (ℓ/2) = q_u × B × ℓ²/2. The full width B is included because we are designing the entire footing cross-section at once, not a per-unit-width strip.
Scenario
Compute M_u at the column face for the 2.5 m × 2.5 m footing from Example 2 (q_u = 272 kPa, c = 400 mm).
Solution
Step 1 — Cantilever projection: ℓ = (B − c)/2 = (2500 − 400)/2 = 2100/2 = 1050 mm = 1.05 m. Step 2 — Factored moment at column face: M_u = q_u × B × ℓ²/2 = 272 × 2.5 × (1.05)²/2 = 272 × 2.5 × 0.5513 = 375.1 kN·m. Answer: M_u = 375.1 kN·m at the column face.
Applications
- Determining the required area of steel A_s for bottom flexural reinforcement.
- Checking if minimum steel governs (common in lightly loaded footings).
- Designing the bar layout and spacing for proper distribution.
- Verifying that selected bars can be developed within the available embedment length.
Misconceptions
- Taking the critical section at the column CENTERLINE — this grossly overestimates the moment.
- Using ℓ = B/2 (half the footing) instead of (B−c)/2 — this also overestimates the moment.
- Forgetting the 1/2 factor: M_u = q_u × B × ℓ (wrong) vs q_u × B × ℓ²/2 (correct).
- Using φ = 0.75 (shear) for the flexure calculation — flexure uses φ = 0.90.
- Forgetting to check minimum steel A_s,min = 0.0018Bh.
Related Concepts
- Rectangular beam flexural design
- Minimum reinforcement ratio
- Balanced and tension-controlled sections
- Development length (bars must be anchored within available embedment)
Common Exam Questions
Example
B=3m, c=0.5m, q_u=220kPa → ℓ=(3−0.5)/2=1.25m → M_u=220×3×(1.25)²/2=220×3×0.781=515.6 kN·m.
Approach
Identify ℓ = (B−c)/2, then apply M_u = q_u × B × ℓ²/2.
Question Type
Compute M_u at the column face
Example
M_u=375kN·m, φ=0.90, f_y=415MPa, d=500mm, B=2500mm, f'c=28MPa: trial a≈d/10=50mm → A_s=M_u/[φf_y(d−a/2)]=375×10⁶/[0.9×415×475]=375×10⁶/177,638≈2111mm², check a=2111×415/(0.85×28×2500)=14.7mm, re-solve: A_s=375×10⁶/[0.9×415×492.6]=2036mm².
Approach
Use M_u = φA_s f_y(d − a/2) with trial a, solve for A_s, iterate once.
Question Type
Find required A_s
Key Points To Remember
- Critical section for flexure is at the FACE of the column.
- Cantilever length ℓ = (B − c)/2.
- M_u = q_u × B × ℓ²/2.
- φ = 0.90 for flexure (tension-controlled sections).
- Minimum steel: A_s,min = 0.0018Bh (Grade 415 deformed bars).
- Provide bars in BOTH directions for square footings.
- For rectangular footings: short-direction bars are concentrated in a central band.
Development Length and Bond
Bond is the interface force between steel and concrete that allows them to act compositely. Development length ℓ_d is the MINIMUM embedment length required for a straight bar to reach its yield strength f_y through bond alone, without the bar pulling out. If ℓ_d cannot be provided, a STANDARD HOOK is used to reduce the required straight embedment. NSCP 2015 Section 425.5 (ACI 318-14 Section 25.4) provides two sets of simplified development length equations for straight deformed bars in tension: CASE A — Favorable spacing/cover conditions (clear spacing ≥ d_b, cover ≥ d_b, AND minimum stirrups/ties provided): Bars ≤ 20 mm: ℓ_d = [f_y ψ_t ψ_e / (2.1 λ √f'c)] × d_b Bars > 20 mm: ℓ_d = [f_y ψ_t ψ_e / (1.7 λ √f'c)] × d_b CASE B — Other (less favorable) conditions: Bars ≤ 20 mm: ℓ_d = [f_y ψ_t ψ_e / (1.4 λ √f'c)] × d_b Bars > 20 mm: ℓ_d = [f_y ψ_t ψ_e / (1.1 λ √f'c)] × d_b Minimum: ℓ_d ≥ 300 mm in all cases. MODIFICATION FACTORS: ψ_t = bar location factor: 1.3 for TOP bars (≥ 300 mm of fresh concrete cast below), 1.0 for other positions ψ_e = coating factor: 1.5 for epoxy-coated bars with cover < 3d_b or clear spacing < 6d_b; 1.2 for other epoxy-coated; 1.0 for uncoated/galvanized λ = lightweight concrete factor: 0.75 for all-lightweight, 0.85 for sand-lightweight, 1.0 for normal-weight Product ψ_t × ψ_e ≤ 1.7 For FOOTINGS specifically: the bars extend from the column face to within cover (typically 75 mm clear from the edge). The available development length is: ℓ_available = (B − c)/2 − 75 mm (side cover) If ℓ_available < ℓ_d, either: (a) increase footing size, (b) increase bar diameter to reduce ℓ_d/d_b ratio (wait — larger bars INCREASE ℓ_d), (c) use smaller bars at closer spacing, or (d) use standard hooks. For COMPRESSION development length (dowels from column into footing): ℓ_dc = [f_y ψ_r / (0.24 λ √f'c)] × d_b ≥ 200 mm, and ≥ 0.043 f_y d_b where ψ_r = 0.75 if bars are enclosed in ties/spirals, 1.0 otherwise
Examples
The 25 mm bar requires over 1.1 m of embedment. In a 2.5 m footing with a 400 mm column, the available length is (2500−400)/2 − 75 = 1050 − 75 = 975 mm, which is LESS than 1150 mm. This means the bars cannot be fully developed in straight embedment — standard hooks would be required, or the footing size must be increased.
Scenario
Find the tension development length for a 25 mm diameter deformed bar: f_y = 415 MPa, f'c = 28 MPa, uncoated, normal-weight concrete, bottom bar, favorable cover/spacing conditions.
Solution
Step 1 — Bar size: 25 mm > 20 mm → use the LARGER bar formula. Step 2 — Favorable conditions (Case A): coefficient = 1.7. Step 3 — Modification factors: ψ_t = 1.0 (bottom bar, not top), ψ_e = 1.0 (uncoated), λ = 1.0 (NW). Step 4 — Development length: ℓ_d = [f_y ψ_t ψ_e / (1.7 λ √f'c)] × d_b ℓ_d = [415 × 1.0 × 1.0 / (1.7 × 1.0 × √28)] × 25 ℓ_d = [415 / (1.7 × 5.292)] × 25 ℓ_d = [415 / 8.996] × 25 ℓ_d = 46.13 × 25 ℓ_d = 1153 mm ≈ 1150 mm Step 5 — Minimum: 1150 mm > 300 mm ✓. Answer: ℓ_d = 1150 mm (use 1200 mm for practical rounding).
The top bar factor ψ_t = 1.3 increases the required development length by 30% because fresh concrete bleeds and settles away from horizontal bars in the top portion, reducing bond quality. This is why 'top bars' always need more embedment than 'bottom bars' of the same size.
Scenario
Find ℓ_d for a 20 mm top bar: f_y = 415 MPa, f'c = 35 MPa, uncoated, normal-weight, favorable conditions.
Solution
Step 1 — Bar size: 20 mm → ≤ 20 mm formula. Step 2 — Case A coefficient: 2.1 (favorable). Step 3 — ψ_t = 1.3 (TOP BAR — more than 300 mm of concrete cast below), ψ_e = 1.0, λ = 1.0. Step 4 — Check cap: ψ_t × ψ_e = 1.3 × 1.0 = 1.3 ≤ 1.7 ✓. Step 5 — Development length: ℓ_d = [415 × 1.3 × 1.0 / (2.1 × 1.0 × √35)] × 20 ℓ_d = [539.5 / (2.1 × 5.916)] × 20 ℓ_d = [539.5 / 12.424] × 20 ℓ_d = 43.43 × 20 ℓ_d = 869 mm ≈ 870 mm Step 6 — Minimum: 870 mm > 300 mm ✓. Answer: ℓ_d ≈ 870 mm.
Applications
- Checking if bottom flexural bars in footings can be developed within the available cantilever length.
- Designing lap splices for column vertical reinforcement (compression development).
- Providing standard hooks where straight embedment is insufficient.
- Detailing dowels projecting from footing into column for force transfer.
Misconceptions
- Using coefficient 2.1 for bars LARGER than 20 mm — the higher coefficient is for 20 mm AND SMALLER.
- Forgetting the 300 mm minimum floor — short calculations sometimes yield less than 300 mm.
- Applying ψ_t = 1.3 to bottom footing bars — this factor is only for bars with ≥300 mm of fresh concrete BELOW them.
- Forgetting to check the ψ_t × ψ_e ≤ 1.7 cap.
- Confusing development length (anchoring yield) with lap splice length (different formulas).
- Using the tension development formula for compression dowels — compression has a separate formula.
Related Concepts
- Standard hooks and equivalent embedment
- Lap splice length
- Compression development length (column dowels)
- Bar cutoff points in beams
- Bond stress distribution
Common Exam Questions
Example
db=16mm (≤20mm), f_y=415, f'c=28, top bar, uncoated, NW, Case A: ℓ_d=[415×1.3/(2.1×√28)]×16=[539.5/11.11]×16=48.6×16=778mm.
Approach
1) Identify bar size vs 20 mm threshold. 2) Check cover/spacing for Case A or B. 3) Determine all ψ factors. 4) Apply formula. 5) Compare with 300 mm minimum.
Question Type
Compute ℓ_d for given bar size and material properties
Example
B=2.4m, c=400mm, cover=75mm → ℓ_avail=(2400−400)/2−75=1000−75=925mm. If ℓ_d=1153mm > 925mm → hooks required.
Approach
Compute ℓ_available = (B−c)/2 − cover. Compute ℓ_d. Compare. If ℓ_available < ℓ_d, bars need hooks.
Question Type
Check if bars can be developed in available length
Key Points To Remember
- Development length ℓ_d = minimum embedment for bar to yield without pullout.
- Case A (favorable): coefficient 2.1 (≤20 mm) or 1.7 (>20 mm).
- Case B (unfavorable): coefficient 1.4 (≤20 mm) or 1.1 (>20 mm).
- ψ_t = 1.3 for top bars; ψ_e = 1.0 for uncoated bars; λ = 1.0 for NW concrete.
- Product ψ_t × ψ_e ≤ 1.7 (cap).
- Minimum ℓ_d = 300 mm always.
- Available length in footing = (B−c)/2 − cover (typically 75 mm).
- If available < required: use standard hook or smaller bars.
Practice Problems
Service load (900 kN) governs footing sizing because q_a is a service-level parameter. After rounding up to 2.5 m, the actual area (6.25 m²) is larger than the required area (6.0 m²), which is conservative for bearing but means q_u = 192 kPa (not 192.5 kPa). Always recalculate q_u using the ACTUAL area, not A_req.
Problem
PROBLEM 1 (Footing Sizing): A 450 mm square interior column carries a dead load of 600 kN and a live load of 300 kN. The allowable bearing capacity of the soil is q_a = 150 kPa. (a) Determine the required square footing dimension. (b) Compute the net factored upward pressure q_u for structural design.
Solution
(a) Service load: P_service = D + L = 600 + 300 = 900 kN. Required area: A_req = P_service / q_a = 900 / 150 = 6.00 m². Side dimension: B = √6.00 = 2.449 m → use B = 2.50 m. (b) Factored load: P_u = 1.2D + 1.6L = 1.2(600) + 1.6(300) = 720 + 480 = 1200 kN. Actual footing area: A = 2.50 × 2.50 = 6.25 m². Net factored pressure: q_u = P_u / A = 1200 / 6.25 = 192.0 kPa. Answer: Use 2.50 m × 2.50 m footing; q_u = 192.0 kPa.
The key steps are: b_o at d/2 from column faces, V_c with the 0.33 coefficient (governs for square), and V_u from the pressure on the area OUTSIDE the critical perimeter. The margin here is about 38%, showing that the footing depth is more than adequate for punching shear.
Problem
PROBLEM 2 (Punching Shear): A 3.0 m × 3.0 m square footing supports a 500 mm square column. The effective depth is d = 600 mm. Factored column load P_u = 2500 kN. f'c = 21 MPa, normal-weight concrete. (a) Compute q_u. (b) Find b_o. (c) Determine φV_c for two-way shear. (d) Find V_u. (e) Is punching shear adequate?
Solution
(a) q_u = P_u / A = 2500 / (3.0²) = 2500 / 9.0 = 277.8 kPa. (b) b_o = 4(c + d) = 4(500 + 600) = 4(1100) = 4400 mm. (c) V_c = 0.33 λ √f'c × b_o × d [governing expression for square interior column] V_c = 0.33 (1.0) √21 × 4400 × 600 V_c = 0.33 × 4.583 × 4400 × 600 V_c = 0.33 × 4.583 × 2,640,000 V_c = 3,992,285 N ≈ 3992 kN φV_c = 0.75 × 3992 = 2994 kN (d) Area inside critical perimeter: (c + d)² = (1100)² = 1,210,000 mm² = 1.21 m². V_u = q_u [A − (c+d)²] = 277.8 × (9.0 − 1.21) = 277.8 × 7.79 = 2164.1 kN ≈ 2164 kN. (e) φV_c = 2994 kN > V_u = 2164 kN → PUNCHING SHEAR IS ADEQUATE. ✓
One-way shear critical section is at d from the face (not the edge). Using ℓ_v = (B−c)/2 without subtracting d is a common error that overestimates the demand. Here, ℓ_v = 650 mm (not 1250 mm). The one-way shear demand (542 kN) is far less than the punching demand (2164 kN), confirming punching governs.
Problem
PROBLEM 3 (One-Way Shear): Using the same footing as Problem 2 (B=3.0m, c=500mm, d=600mm, q_u=277.8 kPa, f'c=21 MPa), check one-way (beam) shear adequacy.
Solution
Step 1 — Projecting length to critical section: ℓ_v = (B − c)/2 − d = (3000 − 500)/2 − 600 = 1250 − 600 = 650 mm = 0.65 m. Step 2 — One-way shear demand: V_u = q_u × B × ℓ_v = 277.8 × 3.0 × 0.65 = 541.7 kN. Step 3 — One-way concrete shear strength: V_c = 0.17 λ √f'c × B × d = 0.17 (1.0) √21 × 3000 × 600 V_c = 0.17 × 4.583 × 1,800,000 = 1,402,398 N ≈ 1402 kN φV_c = 0.75 × 1402 = 1052 kN Step 4 — Check: φV_c = 1052 kN > V_u = 542 kN → ONE-WAY SHEAR IS ADEQUATE. ✓ Summary: Both punching (φV_c=2994 kN > V_u=2164 kN) and one-way (φV_c=1052 kN > V_u=542 kN) shear are satisfied. The footing depth d=600mm is adequate.
This problem illustrates an important phenomenon: for lightly loaded or deeply proportioned footings, minimum reinforcement often governs over the calculated required steel. Always check A_s,min = 0.0018Bh for Grade 415 deformed bars. The total footing depth h is used (not d) in the minimum steel formula.
Problem
PROBLEM 4 (Flexure): For the footing in Problem 2 (B=3.0m, c=500mm, d=600mm, q_u=277.8 kPa), (a) compute M_u at the column face, and (b) find the required A_s if f_y=415 MPa, f'c=21 MPa.
Solution
(a) Cantilever projection to column face: ℓ = (B − c)/2 = (3000 − 500)/2 = 1250 mm = 1.25 m. M_u = q_u × B × ℓ²/2 = 277.8 × 3.0 × (1.25)²/2 = 277.8 × 3.0 × 0.7813 = 651.1 kN·m. (b) Trial: assume a = d/10 = 600/10 = 60 mm. A_s = M_u / [φ f_y (d − a/2)] = 651.1×10⁶ / [0.90 × 415 × (600 − 30)] A_s = 651.1×10⁶ / [0.90 × 415 × 570] A_s = 651.1×10⁶ / [212,895] A_s = 3058 mm² Check a: a = A_s f_y / (0.85 f'c B) = 3058 × 415 / (0.85 × 21 × 3000) = 1,269,070 / 53,550 = 23.7 mm. Re-iterate with a = 23.7 mm: A_s = 651.1×10⁶ / [0.90 × 415 × (600 − 11.85)] A_s = 651.1×10⁶ / [0.90 × 415 × 588.15] A_s = 651.1×10⁶ / [219,512] A_s = 2966 mm² Check a again: a = 2966 × 415 / 53,550 = 22.99 mm ≈ 23 mm (converged). Check minimum: A_s,min = 0.0018 × B × h = 0.0018 × 3000 × (600+75+db/2). Assuming h ≈ 700 mm: A_s,min = 0.0018 × 3000 × 700 = 3780 mm². Since A_s,min = 3780 mm² > A_s,req = 2966 mm², MINIMUM STEEL GOVERNS. Use A_s = 3780 mm². Answer: M_u = 651.1 kN·m; A_s = 3780 mm² (minimum governs).
The 20 mm bar sits at the boundary — the '≤ 20 mm' category uses the favorable coefficient 2.1, giving a shorter ℓ_d (862 mm) than a 22 mm bar would (which would use 1.7 and require about 1050 mm). This boundary behavior is a classic board-exam setup. The 38 mm margin, while technically adequate, is tight for construction practice.
Problem
PROBLEM 5 (Development Length): Column dowels consist of 20 mm deformed bars (f_y = 415 MPa, f'c = 21 MPa, normal-weight concrete, uncoated, bottom bars, favorable conditions). The available embedment length in the footing is 900 mm. (a) Compute ℓ_d. (b) Are the bars adequately developed?
Solution
(a) Bar size = 20 mm → use the ≤ 20 mm formula. Case A (favorable): coefficient = 2.1. ψ_t = 1.0 (bottom bar), ψ_e = 1.0 (uncoated), λ = 1.0 (NW). ℓ_d = [f_y ψ_t ψ_e / (2.1 λ √f'c)] × d_b ℓ_d = [415 × 1.0 × 1.0 / (2.1 × 1.0 × √21)] × 20 ℓ_d = [415 / (2.1 × 4.583)] × 20 ℓ_d = [415 / 9.624] × 20 ℓ_d = 43.12 × 20 ℓ_d = 862 mm Check minimum: 862 mm > 300 mm ✓. (b) Available length = 900 mm > ℓ_d = 862 mm → BARS ARE ADEQUATELY DEVELOPED. ✓ (Margin: 900 − 862 = 38 mm — barely adequate; consider using larger footing or standard hooks if construction tolerances are tight.)
Exam Preparation Tips
- MASTER THE SERVICE vs FACTORED LOAD DISTINCTION: The single most common source of error in board exam footing problems is using factored loads for footing sizing (should be service) or using service loads for shear/flexure design (should be factored). Write this rule at the top of your scratch paper on exam day.
- MEMORIZE THE THREE CRITICAL SECTION LOCATIONS: (1) Punching shear — at d/2 from column face; (2) One-way shear — at d from column face; (3) Flexure — AT the column face. These three locations are tested separately and often appear as fill-in-the-blank or multiple-choice questions.
- KNOW THE COEFFICIENTS BY HEART: Punching V_c uses 0.33λ√f'c; One-way V_c uses 0.17λ√f'c. Do not mix these up. A useful memory trick: 0.33 ≈ 1/3 (two-way punching), 0.17 ≈ 1/6 (one-way beam) — the one-way case is roughly half the two-way case.
- FOR DEVELOPMENT LENGTH: Remember the 20 mm threshold. Bars ≤ 20 mm: coefficient 2.1 (Case A) or 1.4 (Case B). Bars > 20 mm: coefficient 1.7 (Case A) or 1.1 (Case B). The 300 mm minimum always applies. Top bars get ψ_t = 1.3 (30% penalty for inferior bond).
- PRACTICE UNIT CONSISTENCY: When computing V_c = 0.33√f'c·b_o·d, use f'c in MPa, b_o and d in mm → V_c comes out in Newtons. Divide by 1000 to get kN. Writing out units prevents costly errors on the board exam.
- PUNCHING DEMAND TRAP: V_u = q_u × [A_total − (c+d)²], NOT q_u × A_total. The area inside the critical perimeter is EXCLUDED. This subtraction is frequently omitted by examinees under time pressure.
- ALWAYS CHECK BOTH SHEAR MODES: In PRC board problems, if only one shear check is performed, the answer will match one of the distractors but the correct answer requires both. Never assume one governs without checking both.
- MINIMUM STEEL IN FOOTINGS: A_s,min = 0.0018Bh for Grade 415 deformed bars. Note it uses h (total thickness), not d (effective depth). For a footing with d = 500 mm and 75 mm cover + 12 mm bar, h ≈ 590 mm. Minimum steel often governs in lightly loaded footings.
- AVAILABLE DEVELOPMENT LENGTH IN FOOTINGS: ℓ_available = (B−c)/2 − side cover (75 mm for footings per NSCP 2015). If ℓ_available < ℓ_d, state that standard 90° hooks are required. This is a common design decision question in board exams.
- TIME MANAGEMENT: A complete footing design (sizing → punching → one-way → flexure → development) takes about 15–20 minutes. In the afternoon board exam (structural design), allocate time accordingly. Practice the sequence until it becomes automatic.
In summary
Reinforced concrete footing design is a systematic, multi-check process that tests a civil engineer's ability to correctly identify critical sections, apply appropriate formulas, and maintain the crucial distinction between service and factored loads. For the PRC Civil Engineer Board Examination, mastery of this chapter requires four core competencies: (1) sizing the plan area from service loads using q_a; (2) performing two-way punching shear checks using b_o = 4(c+d) and V_c = 0.33λ√f'c·b_o·d; (3) performing one-way beam shear checks with the critical section at d from the column face using V_c = 0.17λ√f'c·B·d; and (4) computing development lengths using the correct coefficient (2.1 or 1.7 for Case A favorable conditions, with the 300 mm minimum always applying). The most commonly tested board-exam traps are: using factored loads for footing sizing, computing punching demand on the gross area (forgetting to subtract the interior area), using the wrong critical section location, and neglecting minimum flexural reinforcement. By following the step-by-step design procedure diagrammed in this chapter — sizing, punching shear, one-way shear, flexure, development — examinees can solve complex footing problems systematically and confidently within the time constraints of the board examination. Practice the five worked problems repeatedly until the sequence is instinctive, and you will be well-prepared for this high-value topic in the PRC CE Licensure Examination.
Ready to practise for the CELE 2026?
Super Tutor's AI review plan adapts to your weak areas and builds a weekly practice schedule around your target CELE exam date.