Reinforced Concrete Footings, Bond and Development - Revision Notes
Footings are the structural elements that transfer column and wall loads to the supporting soil at a pressure that does not exceed the allowable bearing capacity. In the PRC Civil Engineer Licensure Examination, footing problems consistently appear and test three core competencies: (1) sizing the footing plan area using service loads and allowable soil pressure, (2) checking two modes of shear — two-way punching shear and one-way beam shear — using factored loads, and (3) ensuring reinforcing bars are properly developed (anchored) so they can reach yield before pulling out. This chapter also covers flexural design at the critical section (column face) and the computation of development lengths per NSCP 2015 (aligned with ACI 318). Mastery of these topics requires disciplined use of service vs. factored loads, correct identification of critical sections, and accurate application of the shear and development-length formulas.
1. Footing Sizing — Service Load and Allowable Soil Pressure
Formulas
A_req = P_service / q_a
Example
P_service = 1 200 kN, q_a = 200 kPa → A_req = 1 200/200 = 6.0 m² → B = √6.0 = 2.449 m → use 2.5 m × 2.5 m
Variables
A_req = required footing plan area (m²); P_service = total service load on footing (kN); q_a = allowable soil bearing pressure (kPa = kN/m²)
Application
First step in footing design — determines how large the footing must be to keep soil stress within safe limits.
q_u = P_u / A_footing
Example
P_u = 1 700 kN, A = 2.5 × 2.5 = 6.25 m² → q_u = 1 700/6.25 = 272 kPa
Variables
q_u = net upward factored soil pressure (kPa); P_u = factored column load (kN); A_footing = actual adopted footing area (m²)
Application
Used for all structural design checks: punching shear, one-way shear, and flexure.
Exam tips
- If the problem gives both dead load D and live load L separately, first compute P_service = D + L for sizing, then P_u = 1.2D + 1.6L for design.
- If the problem states 'factored load' directly, you can compute q_u immediately — no need to back-calculate service load.
- Always recompute q_u using the ADOPTED (rounded-up) footing area, not A_req.
Key points
- Footing plan area is sized using SERVICE loads (dead + live, unfactored), not factored loads.
- Allowable soil pressure q_a (kPa) is a geotechnical parameter from soil investigation; it accounts for safety already.
- For a square footing: A_req = P_service / q_a ; side B = √A_req ; always round UP to the next practical dimension (e.g., nearest 50 mm or 100 mm).
- For a rectangular footing: specify L and W such that L × W ≥ A_req and the aspect ratio L/W ≤ 2 (practical rule).
- The SELF-WEIGHT of the footing and backfill is often included in the service load for net bearing pressure calculations; exam problems usually state whether to include it.
- Once the plan area is fixed, the NET UPWARD FACTORED PRESSURE q_u = P_u / A_footing is used for ALL subsequent structural design checks.
- P_u is typically 1.2D + 1.6L (NSCP 2015 load combination 1) for gravity-dominated cases.
Definitions
Allowable Soil Bearing Pressure (q_a)
The maximum safe contact pressure between the footing base and the soil, determined from geotechnical analysis and including an appropriate factor of safety (typically FS = 3). It is a SERVICE-level limit.
Importance
Exam problems always state q_a; never apply load factors to q_a — it is already a safe allowable value.
Net Upward Factored Pressure (q_u)
The uniformly distributed upward soil reaction per unit area used in structural design, computed from the factored column load divided by the footing plan area.
Importance
This is the design pressure for shear and moment calculations. Confusing service and factored pressures is the #1 board-exam mistake.
Common mistakes
- Using factored load P_u to size the footing area (over-sizing) — footing AREA sizing uses SERVICE load.
- Using service load P_service for shear and moment checks (under-design) — shear and flexure use FACTORED load.
- Forgetting to round up the footing dimension after computing √A_req.
- Ignoring the self-weight of footing concrete and soil overburden when specified in the problem.
2. Two-Way (Punching) Shear
Formulas
b_o = 4(c + d) [square column]
Example
c = 400 mm, d = 500 mm → b_o = 4(400 + 500) = 3 600 mm
Variables
b_o = critical shear perimeter (mm); c = column side dimension (mm); d = effective depth of footing (mm)
Application
Defines the punching perimeter at d/2 from all four column faces.
V_c = 0.33λ√f'c · b_o · d (governing for square columns)
Example
f'c = 28 MPa, b_o = 3 600 mm, d = 500 mm → V_c = 0.33(1.0)√28 × 3 600 × 500 = 0.33 × 5.292 × 1 800 000 = 3 143 kN
Variables
V_c = nominal punching shear strength (N); λ = 1.0 for normal-weight concrete; f'c = concrete compressive strength (MPa); b_o = critical perimeter (mm); d = effective depth (mm)
Application
Nominal two-way shear capacity of the concrete footing.
V_u = q_u[A_footing − (c + d)²]
Example
q_u = 272 kPa, A = 6.25 m², (c+d)² = (0.9 m)² = 0.81 m² → V_u = 272(6.25 − 0.81) = 272 × 5.44 = 1 480 kN
Variables
V_u = factored punching shear demand (kN); q_u = net upward factored pressure (kPa); A_footing = total footing plan area (m²); (c + d)² = area inside critical perimeter (m², converted from mm²)
Application
Computes the upward force that the soil exerts on the footing area outside the punching perimeter — this is the force trying to punch the column through.
φV_c ≥ V_u (φ = 0.75)
Example
φV_c = 0.75 × 3 143 = 2 357 kN > V_u = 1 480 kN → SAFE
Variables
φ = strength reduction factor for shear = 0.75 per NSCP 2015
Application
Design adequacy check; if φV_c < V_u, increase footing depth d or increase f'c.
Exam tips
- Board exams almost always use square columns — memorize b_o = 4(c + d) and V_c = 0.33λ√f'c · b_o · d.
- Critical perimeter area = (c + d)² for a square column; this is the area to SUBTRACT from A_footing.
- To quickly check: if d > 0.3B, the footing is likely very deep — uncommon in practice but possible in exam problems.
- When asked which shear governs, compute both punching and one-way; report the one with the lower φV_c relative to V_u.
Key points
- Punching shear is the tendency of the column to punch through the footing like a cookie cutter. It governs for most square footings with compact columns.
- The critical perimeter b_o is taken at d/2 from ALL FACES of the column (or loaded area).
- For a square column of side c: b_o = 4(c + d), where d is the effective depth of the footing.
- NSCP 2015 Section 422.6 (ACI 318-19 Section 22.6) gives THREE expressions for V_c; the SMALLEST governs.
- For square or compact columns (β_c = 1), the expression V_c = 0.33λ√f'c · b_o · d typically governs and is tested most frequently on boards.
- The punching shear DEMAND V_u = q_u × [A_footing − (c + d)²] — i.e., the upward force on the area OUTSIDE the critical perimeter.
- Strength reduction factor φ = 0.75 for shear (NSCP 2015).
- Design requirement: φV_c ≥ V_u.
Definitions
Two-Way (Punching) Shear
A failure mode where the column punches through the footing along a truncated pyramid surface. The critical section is a closed perimeter at d/2 from the column face.
Importance
Usually the FIRST and most critical shear check for square footings. If it fails, you must increase depth d.
Critical Perimeter b_o
The perimeter of the critical section for two-way shear, measured at a distance d/2 from the faces of the column or other loaded area.
Importance
Fundamental parameter in punching shear calculations. For a square column of side c: b_o = 4(c + d).
β_c
The ratio of the long side to the short side of the column cross-section. For square columns β_c = 1. The NSCP/ACI punching shear formula with β_c governs only when the column is very elongated.
Importance
Exam problems with square columns bypass the β_c formula; use V_c = 0.33λ√f'c · b_o · d directly.
Common mistakes
- Placing the critical perimeter at d from the column face (beam shear location) instead of d/2 — punching is at d/2.
- Using the TOTAL footing area (not subtracting the interior area) for the punching demand V_u.
- Forgetting to convert (c + d) from mm to m when computing the area in m².
- Applying φ = 0.90 (flexure factor) instead of φ = 0.75 (shear factor).
- Not checking all three NSCP punching shear expressions and assuming 0.33 always governs — it governs for square/compact columns but not for very elongated ones.
3. One-Way (Beam) Shear
Formulas
V_c = 0.17λ√f'c · B · d
Example
f'c = 28 MPa, B = 2 500 mm, d = 500 mm → V_c = 0.17(1.0)(5.292)(2 500)(500) = 1 124 kN
Variables
V_c = nominal one-way shear strength (N); λ = 1.0 for normal-weight concrete; f'c (MPa); B = full footing width (mm); d = effective depth (mm)
Application
One-way shear capacity of the footing acting as a wide beam.
L_crit = (B − c)/2 − d
Example
B = 2 500 mm, c = 400 mm, d = 500 mm → L_crit = (2 500 − 400)/2 − 500 = 1 050 − 500 = 550 mm
Variables
L_crit = distance from column face to critical shear section (mm); B = footing width (mm); c = column width (mm); d = effective depth (mm)
Application
Defines the overhang length whose upward soil pressure creates one-way shear at the critical section.
V_u = q_u × B × L_crit
Example
q_u = 272 kPa, B = 2.5 m, L_crit = 0.55 m → V_u = 272 × 2.5 × 0.55 = 374 kN
Variables
V_u = factored one-way shear demand (kN); q_u (kN/m²); B (m); L_crit (m)
Application
Upward soil force on the strip from the footing edge to the critical section.
Exam tips
- Quick memory aid: 0.33 for TWO-way (punching), 0.17 for ONE-way (beam shear).
- If the problem asks only for the 'critical section for beam shear,' the answer is: at distance d from the column face.
- When d is large relative to the cantilever, L_crit can become very small or even negative — meaning one-way shear is not critical (the critical section falls inside the column).
Key points
- One-way shear treats the footing as a wide beam spanning between column and footing edge.
- Critical section is taken at a distance d from the FACE OF THE COLUMN (not the center).
- For a square footing of side B with a square column of side c, the projecting cantilever length to the critical section is: L_crit = (B − c)/2 − d.
- The shear demand V_u = q_u × B × L_crit (force on the strip beyond the critical section).
- Nominal one-way shear capacity: V_c = 0.17λ√f'c · B · d (NSCP 2015 / ACI 318).
- Same strength reduction factor φ = 0.75.
- One-way shear usually does NOT govern for compact square footings but must always be checked.
- If one-way shear is critical, increase d (preferred) or increase f'c.
Definitions
One-Way (Beam) Shear
A shear failure mode where the footing acts as a wide, short cantilever beam. The failure plane is approximately vertical, extending across the full width B of the footing at a distance d from the column face.
Importance
Secondary shear check after punching shear. Critical for long, narrow footings or very shallow effective depths.
Common mistakes
- Placing the critical section at the column FACE (flexure location) instead of at distance d from the column face.
- Using the FULL cantilever length instead of the reduced length L_crit = (B−c)/2 − d.
- Forgetting to use the FULL footing width B (not B − c) for V_c.
- Using V_c = 0.33λ√f'c · Bd (punching coefficient) for one-way shear — one-way uses 0.17.
4. Flexural Design of Footings
Formulas
ℓ = (B − c)/2
Example
B = 2.5 m, c = 0.4 m → ℓ = (2.5 − 0.4)/2 = 1.05 m
Variables
ℓ = cantilevered projection from column face to footing edge (m or mm); B = footing width; c = column width
Application
Lever arm for moment calculation at the critical section.
M_u = q_u × B × ℓ²/2
Example
q_u = 272 kPa, B = 2.5 m, ℓ = 1.05 m → M_u = 272 × 2.5 × 1.05²/2 = 272 × 2.5 × 0.5513 = 375 kN·m
Variables
M_u = factored moment at column face (kN·m); q_u (kN/m²); B = footing width (m); ℓ = cantilever projection (m)
Application
Design moment for flexural reinforcement calculation.
R_n = M_u / (φ × B × d²) → ρ = (0.85f'c/f_y)[1 − √(1 − 2R_n/(0.85f'c))] → A_s = ρBd
Example
M_u = 375 kN·m = 375 × 10⁶ N·mm, φ = 0.90, B = 2 500 mm, d = 500 mm → R_n = 375×10⁶/(0.90 × 2 500 × 500²) = 375×10⁶/562 500 000 = 0.667 MPa
Variables
R_n = flexural resistance coefficient (MPa); φ = 0.90 for flexure; ρ = steel ratio; A_s = required steel area (mm²); B and d in mm, M_u in N·mm
Application
Standard beam flexural design procedure applied to the footing strip.
Exam tips
- Memorize the sequence: SIZE (service load) → q_u (factored) → PUNCHING shear → ONE-WAY shear → FLEXURE (column face) → DEVELOPMENT.
- For board exams that ask for M_u only, the formula M_u = q_u·B·ℓ²/2 is the fastest route.
- Minimum steel for footings is often taken as the slab temperature/shrinkage minimum: A_s,min = 0.0018 × b × h (for fy = 415 MPa).
Key points
- The critical section for MOMENT is at the FACE OF THE COLUMN (not d from the face).
- The footing cantilevers out from the column face; the net upward pressure q_u creates an upward moment.
- For a square footing: cantilevered projection ℓ = (B − c)/2.
- Factored moment at column face: M_u = q_u × B × ℓ²/2 (treating the full width B as the beam width).
- Once M_u is found, design A_s exactly as for a rectangular beam section: compute R_n = M_u/(φ × B × d²), then ρ = (0.85f'c/f_y)[1 − √(1 − 2R_n/(0.85f'c))], then A_s = ρ × B × d.
- Check: A_s ≥ A_s,min = (0.0018 × B × h) for Grade 60 / fy = 415 MPa deformed bars (temperature and shrinkage minimum for slabs, often used for footings).
- Bars are placed in TWO directions for square footings; for rectangular footings, a central band carries more steel in the short direction.
- The reinforcing bars must be DEVELOPED (anchored) within the footing — check ℓ_d available ≥ ℓ_d required.
Definitions
Critical Section for Moment
The section of maximum factored moment in a footing, located at the FACE of the column (for concrete columns) or at the middle of the base plate (for steel columns). This is where flexural reinforcement is designed.
Importance
Distinguished from the critical section for shear (which is at d from the face). Confusing these locations leads to errors in both M_u and V_u.
Cantilever Projection (ℓ)
The horizontal distance from the column face to the edge of the footing. The upward soil pressure over this length creates the bending moment at the critical section.
Importance
The moment M_u is proportional to ℓ²; larger footings with smaller columns have much larger moments.
Common mistakes
- Taking the critical moment section at d from the column face (beam shear location) — flexure is at the column FACE.
- Using only HALF the footing width (B/2) instead of the full width B in the moment formula.
- Forgetting to convert M_u from kN·m to N·mm before computing R_n (factor of 10⁶).
- Using φ = 0.75 (shear) instead of φ = 0.90 (flexure) in the moment check.
5. Development Length and Bond
Formulas
ℓ_d = (f_y · ψ_t · ψ_e)/(2.1 · λ · √f'c) · d_b [bars ≤ 20 mm, favorable conditions]
Example
d_b = 20 mm, f_y = 415 MPa, f'c = 35 MPa, ψ_t = 1.3 (top bar), ψ_e = 1.0, λ = 1.0 → ℓ_d = 415(1.3)(1.0)/[2.1(1.0)√35] × 20 = 539.5/(2.1 × 5.916) × 20 = 539.5/12.42 × 20 = 43.44 × 20 = 869 mm ≥ 300 mm ✓
Variables
ℓ_d = development length (mm); f_y = steel yield stress (MPa); ψ_t = casting position factor; ψ_e = coating factor; λ = lightweight factor (1.0 for NW); f'c (MPa); d_b = bar diameter (mm)
Application
Tension development length for smaller bars with adequate cover and bar spacing.
ℓ_d = (f_y · ψ_t · ψ_e)/(1.7 · λ · √f'c) · d_b [bars > 20 mm, favorable conditions]
Example
d_b = 25 mm, f_y = 415 MPa, f'c = 28 MPa, ψ_t = ψ_e = 1.0, λ = 1.0 → ℓ_d = 415(1.0)(1.0)/[1.7(1.0)√28] × 25 = 415/(1.7 × 5.292) × 25 = 415/8.996 × 25 = 46.13 × 25 = 1 153 mm ≈ 1 150 mm
Variables
Same as above; coefficient 1.7 applies to bars larger than 20 mm diameter.
Application
Tension development length for larger bars (25 mm, 28 mm, 32 mm, 36 mm) with adequate cover and spacing.
ℓ_d,hook = (0.24 · ψ_e · f_y)/(λ · √f'c) · d_b [standard 90° hook]
Example
d_b = 25 mm, f_y = 415 MPa, f'c = 28 MPa → ℓ_dh = 0.24(1.0)(415)/(1.0 × √28) × 25 = 99.6/5.292 × 25 = 18.82 × 25 = 470 mm; apply 0.7 factor if adequate cover → 0.7 × 470 = 329 mm
Variables
ℓ_d,hook = hook development length (mm); ψ_e = 1.0 for uncoated bars; other symbols same as above
Application
When straight embedment length is insufficient (e.g., at column-footing interface), standard hooks provide anchorage in shorter length.
Exam tips
- In footings, main bars are BOTTOM bars (ψ_t = 1.0) and usually UNCOATED (ψ_e = 1.0); this simplifies the formula.
- Quick check: for f_y = 415 MPa, f'c = 28 MPa, uncoated bottom bar > 20 mm → ℓ_d/d_b ≈ 415/(1.7 × 5.292) = 46.1 → ℓ_d ≈ 46 × d_b.
- Bar cutoff in footings: bars must extend to within cover of the footing edge, so available ℓ_d = ℓ − cover_end.
- When asked 'is the bar adequately developed?': compute ℓ_d required and compare with ℓ_available = (B − c)/2 − end cover.
Key points
- Development length ℓ_d is the minimum bar embedment required for the bar to reach its yield stress f_y through bond with concrete. Shorter embedment = bar pulls out before yielding.
- NSCP 2015 Section 425 (ACI 318 Chapter 25) gives ℓ_d formulas with MODIFICATION FACTORS ψ_t, ψ_e, ψ_s, λ.
- Simplified NSCP/ACI formulas (favorable spacing/cover conditions): ≤ 20 mm bars: ℓ_d = (f_y · ψ_t · ψ_e)/(2.1 · λ · √f'c) · d_b ; Larger bars (> 20 mm): ℓ_d = (f_y · ψ_t · ψ_e)/(1.7 · λ · √f'c) · d_b
- Less-favorable conditions (tight spacing or small cover): coefficients change from 2.1 to 1.4 (≤ 20 mm) and from 1.7 to 1.1 (> 20 mm).
- Minimum ℓ_d = 300 mm regardless of calculation.
- Modification factors: ψ_t = 1.3 for top bars (more than 300 mm of fresh concrete below the bar during casting); ψ_e = 1.5 for epoxy-coated bars with cover < 3d_b or spacing < 6d_b; ψ_e = 1.2 for other epoxy conditions; ψ_s = 0.8 for bars ≤ 20 mm (some references); λ = 1.0 for normal-weight concrete.
- In footings, bars are BOTTOM bars (ψ_t = 1.0) and typically uncoated (ψ_e = 1.0), so ψ_t · ψ_e = 1.0.
- Available development length = ℓ - cover (from column face to bar end minus end cover). Must be ≥ ℓ_d.
- Standard hooks provide anchorage equivalent to ℓ_dh (hook development length) when straight embedment is insufficient.
Definitions
Development Length (ℓ_d)
The minimum length of bar embedment in concrete required to develop the full yield force T = A_s · f_y through bond stress between bar and concrete. A bar with embedment less than ℓ_d will pull out before yielding.
Importance
Critical for structural safety. Board exams frequently ask for ℓ_d computation and comparison with available anchorage length.
Casting Position Factor (ψ_t)
Modification factor that accounts for reduced bond quality for bars with more than 300 mm of fresh concrete cast below them ('top bars'). ψ_t = 1.3 for top bars, 1.0 for all others.
Importance
Always check if the bar is a top bar (ψ_t = 1.3) — this increases ℓ_d by 30%.
Coating Factor (ψ_e)
Modification factor for epoxy-coated bars, which have reduced bond. ψ_e = 1.5 or 1.2 for epoxy-coated; 1.0 for uncoated bars.
Importance
Most footing bars are uncoated (ψ_e = 1.0); exam problems will explicitly state if bars are epoxy-coated.
Standard Hook
A bar end bent at 90° (with 12d_b extension) or 180° (with 4d_b extension) that mechanically anchors the bar in concrete. Used when straight development length ℓ_d exceeds the available embedment.
Importance
Hooks allow shorter embedment — a common design solution at footing-column interfaces.
Common mistakes
- Using coefficient 2.1 (≤ 20 mm) for a 25 mm bar — coefficient 1.7 applies to bars larger than 20 mm.
- Forgetting to apply ψ_t = 1.3 for top bars — increases ℓ_d by 30%.
- Not applying the 300 mm minimum floor on ℓ_d.
- Comparing ℓ_d with the FULL cantilever ℓ instead of ℓ minus end cover.
- Confusing ℓ_d (straight development) with ℓ_dh (hook development) — these are different formulas.
Connections
- Footing flexural design uses the same rectangular beam theory (Whitney stress block, R_n, ρ) as beams — the footing is just a very wide, short cantilever beam.
- The shear strength formulas (0.33 and 0.17 × λ√f'c) are the same concrete shear expressions used for beams and slabs; the difference is only the critical perimeter or section definition.
- Development length theory underpins ALL reinforced concrete elements — columns, beams, slabs, walls, and footings all require bar development; same formulas, same ψ factors apply.
- Load combinations (1.2D + 1.6L) from NSCP 2015 Section 405 apply universally to all RC elements including footings — the factored load P_u used in footing design follows the same LRFD load factors as beams and columns.
- Soil-structure interaction: the footing connects structural engineering (RC design) with geotechnical engineering (bearing capacity, settlement). Allowable bearing pressure q_a from soil mechanics directly governs footing sizing.
- Transfer of force from column to footing through bearing and dowels connects column design to footing design — dowel bars must also satisfy development length requirements at the column-footing interface.
- Prestressed concrete footings (uncommon but possible) would use the same plan-area sizing but different flexural and shear approaches — the bond/development concept shifts to transfer length and development length for prestressing strand.
- Retaining wall footings follow the same two shear checks and flexural design as isolated footings, adding overturning and sliding stability checks from geotechnical analysis.
Quick review questions
A column carries a service dead load of 700 kN and live load of 500 kN. The allowable soil bearing pressure is 180 kPa. What is the minimum required area for a square footing, and what size should be adopted?
Answer
A_req = (700 + 500)/180 = 1 200/180 = 6.667 m². B = √6.667 = 2.582 m → Adopt 2.6 m × 2.6 m (A = 6.76 m²).
Footing area sizing uses the SERVICE load (unfactored D + L) divided by the allowable soil pressure. After computing the theoretical B, always round UP to a practical dimension. A 2.6 m × 2.6 m footing provides A = 6.76 m² ≥ 6.667 m² required.
For the footing in Q1 with P_u = 1.2(700) + 1.6(500) = 1 640 kN, compute q_u using the adopted 2.6 m × 2.6 m footing.
Answer
q_u = P_u/A = 1 640/(2.6 × 2.6) = 1 640/6.76 = 242.6 kPa ≈ 243 kPa.
q_u must be computed using the ADOPTED (rounded-up) footing area, not A_req. This ensures the actual soil pressure is modeled correctly for design.
A 2.5 m × 2.5 m footing supports a 350 mm square column with effective depth d = 450 mm and f'c = 21 MPa. What is φV_c for punching shear?
Answer
b_o = 4(350 + 450) = 3 200 mm. V_c = 0.33(1.0)√21 × 3 200 × 450 = 0.33(4.583)(1 440 000) = 2 179 kN. φV_c = 0.75 × 2 179 = 1 634 kN.
For a square column, b_o = 4(c + d). The punching V_c uses the coefficient 0.33. φ = 0.75 for shear. Always convert final answer to kN (divide N result by 1 000).
Using the footing in Q3 with q_u = 272 kPa (hypothetically), what is the punching shear demand V_u?
Answer
V_u = q_u[A_footing − (c + d)²] = 272[6.25 − (0.35 + 0.45)²] = 272[6.25 − 0.64] = 272 × 5.61 = 1 526 kN.
(c + d) = 0.35 + 0.45 = 0.80 m; (c + d)² = 0.64 m². The area inside the critical perimeter is subtracted from the total footing area to get the area over which upward pressure acts to cause punching.
For a 3.0 m square footing with a 500 mm square column, d = 550 mm, q_u = 250 kPa, and f'c = 28 MPa, check one-way shear adequacy.
Answer
L_crit = (3 000 − 500)/2 − 550 = 1 250 − 550 = 700 mm = 0.70 m. V_u = 250 × 3.0 × 0.70 = 525 kN. V_c = 0.17(1.0)√28 × 3 000 × 550 = 0.17(5.292)(1 650 000) = 1 484 kN. φV_c = 0.75 × 1 484 = 1 113 kN > 525 kN → ONE-WAY SHEAR OK.
The critical section for one-way shear is at distance d from the column face. L_crit is the strip length beyond this section. V_u is the upward force on this strip. One-way V_c uses coefficient 0.17 and the full footing width B = 3 000 mm.
Compute M_u at the column face for a 2.5 m footing with 400 mm square column and q_u = 272 kPa.
Answer
ℓ = (2 500 − 400)/2 = 1 050 mm = 1.05 m. M_u = q_u × B × ℓ²/2 = 272 × 2.5 × 1.05²/2 = 272 × 2.5 × 0.5513 = 375 kN·m.
Critical section for moment is at the COLUMN FACE. ℓ is the cantilever projection. The formula M_u = q_u·B·ℓ²/2 treats the footing as a uniformly loaded cantilever slab of width B.
Find the tension development length for a 32 mm bar with f_y = 415 MPa, f'c = 28 MPa, normal-weight concrete, uncoated bottom bar, favorable spacing/cover.
Answer
32 mm > 20 mm → use 1.7 coefficient. ℓ_d = [415(1.0)(1.0)/(1.7 × 1.0 × √28)] × 32 = [415/8.996] × 32 = 46.13 × 32 = 1 476 mm ≈ 1 480 mm ≥ 300 mm ✓.
For bars larger than 20 mm with favorable cover and spacing, the coefficient is 1.7. ψ_t = 1.0 (bottom bar, not top bar), ψ_e = 1.0 (uncoated), λ = 1.0 (normal-weight). The result 1 476 mm exceeds the 300 mm minimum, so 1 480 mm controls.
A 20 mm top bar (ψ_t = 1.3) in a beam with f'c = 35 MPa, f_y = 415 MPa, uncoated, normal-weight, favorable conditions. Find ℓ_d.
Answer
20 mm ≤ 20 mm → use 2.1 coefficient. ℓ_d = [415(1.3)(1.0)/(2.1 × 1.0 × √35)] × 20 = [539.5/(2.1 × 5.916)] × 20 = [539.5/12.42] × 20 = 43.44 × 20 = 869 mm ≥ 300 mm ✓.
The casting position factor ψ_t = 1.3 applies because this is a top bar (more than 300 mm of concrete below it). This increases ℓ_d by 30% compared to a bottom bar. The 20 mm bar uses the 2.1 coefficient (favorable conditions).
Why is the coefficient 2.1 (not 1.7) used for bars ≤ 20 mm in the development length formula?
Answer
Smaller bars have a higher surface-area-to-cross-sectional-area ratio, providing better relative bond per unit of tensile force. Therefore, less embedment length is needed to develop f_y, resulting in a higher (more favorable) coefficient in the denominator of the formula, which gives a shorter ℓ_d.
The development length formula ℓ_d = (f_y·ψ·...)/(k·λ·√f'c)·d_b has k in the denominator. A larger k means shorter ℓ_d. Smaller bars bond more efficiently, so k = 2.1 (shorter ℓ_d required) vs k = 1.7 for larger bars (longer ℓ_d required).
In a square footing with a 25 mm main bar and the available embedment from column face to bar end (minus 75 mm end cover) is 900 mm, while ℓ_d required = 1 153 mm. What design solution is appropriate?
Answer
Since ℓ_d,required (1 153 mm) > ℓ_available (900 mm), the bar cannot be developed straight. Provide a STANDARD 90° HOOK at the bar end. Compute ℓ_dh and verify it fits within the 900 mm available embedment.
When the available straight embedment is less than ℓ_d, a standard hook converts the anchorage to bearing of the bent portion, reducing the required embedment length to ℓ_dh. For 25 mm bars, ℓ_dh ≈ 470 mm (before modification factors), which would fit in 900 mm.
Diagrams
RC Footing Design — Chapter Overview Mind Map
A hierarchical mind map showing all five major topics in RC footing design: sizing, punching shear, one-way shear, flexure, and development length — with key formulas and parameters at each branch. Use this as a quick visual reference before attempting board exam problems.
Step-by-Step RC Footing Design Procedure
A decision-tree flowchart showing the complete six-step footing design sequence used in board exams. The flowchart clearly shows which step uses service load vs factored load, and the decision points for shear adequacy checks. The feedback loop shows what to do when a shear check fails.
Selecting the Correct Development Length Formula
A decision flowchart guiding the selection of the correct development length formula coefficient based on bar size (≤20 mm vs >20 mm), cover/spacing conditions (favorable vs less-favorable), and casting position (top bar vs bottom bar). Ends with the 300 mm minimum check. Essential for eliminating coefficient-selection errors on the board exam.
Critical Sections for Punching, One-Way Shear, and Flexure
A linear flowchart showing the spatial relationship between the three critical sections in a footing cross-section from edge to edge: (1) bar end / available development length, (2) column face for moment, (3) distance d for one-way shear, and (4) distance d/2 for punching shear. This diagram directly addresses the most common board-exam confusion about where each critical section is located.
Exam strategy
For PRC board exam footing problems, follow this disciplined FIVE-STEP sequence and never skip a step: (1) SIZE — compute A_req = P_service/q_a, select B (round up), this step uses SERVICE load only. (2) PRESSURE — compute q_u = P_u/A_adopted using FACTORED load and ADOPTED area. (3) PUNCHING SHEAR — compute b_o = 4(c+d), V_c = 0.33λ√f'c·b_o·d, V_u = q_u[A−(c+d)²], check φV_c ≥ V_u with φ = 0.75. (4) ONE-WAY SHEAR — compute L_crit = (B−c)/2 − d, V_c = 0.17λ√f'c·B·d, V_u = q_u·B·L_crit, check φV_c ≥ V_u. (5) FLEXURE and DEVELOPMENT — M_u = q_u·B·ℓ²/2 at column face, design A_s, then verify ℓ_available ≥ ℓ_d. Memory aids: SIZE=service, DESIGN=factored; PUNCHING critical at d/2 (0.33), ONE-WAY critical at d from face (0.17); MOMENT critical at column face (no d offset). For development length: ≤20 mm bar uses 2.1, >20 mm bar uses 1.7 (favorable conditions); always check ψ_t (=1.3 for top bars) and apply 300 mm floor. Budget approximately 8–12 minutes for a complete footing design problem; if time-pressed, focus on whichever check the question specifically asks for rather than working all five steps. Watch for the classic trick of mixing service and factored loads — identify which load type is given before computing anything.