Foundations (Shallow and Deep) - Revision Notes
Foundations are the structural elements that transfer all building and infrastructure loads safely to the supporting soil or rock. In the Philippine context, foundation design must address diverse soil conditions — from the soft alluvial deposits of Metro Manila and Central Luzon to the volcanic soils of Mindanao — governed by NSCP 2015 (National Structural Code of the Philippines) and sound geotechnical engineering practice under RA 544 (Civil Engineering Law). This chapter covers the two major foundation categories tested in the PRC Civil Engineer Licensure Examination: shallow foundations (spread footings, combined footings, mat/raft) and deep foundations (piles, drilled shafts). Mastery of bearing capacity, pile capacity formulas, group effects, and negative skin friction is essential for the board exam.
1. Shallow Foundations — Types and Sizing
Formulas
A_req = P_service / q_a
Example
Column load P = 600 kN, q_a = 150 kPa → A_req = 600/150 = 4.0 m² → B = √4.0 = 2.0 m. Use 2.0 m × 2.0 m square footing.
Variables
A_req = required base area (m²); P_service = unfactored service load (kN); q_a = allowable bearing pressure (kPa = kN/m²)
Application
First-step sizing of any shallow footing. For a square footing, side B = √A_req.
q_net = q_gross − γ_soil × D_f
Example
P = 800 kN, B = 2.5 m, D_f = 1.2 m, γ = 18 kN/m³. q_gross = 800/6.25 = 128 kPa. q_net = 128 − 18(1.2) = 128 − 21.6 = 106.4 kPa.
Variables
q_net = net bearing pressure (kPa); q_gross = total applied bearing pressure (kPa); γ_soil = unit weight of overburden soil (kN/m³); D_f = depth of footing embedment (m)
Application
The net bearing pressure is what the soil 'sees' beyond the weight of the overburden already in place. Used when the allowable capacity is given as net (common in NSCP 2015 Section 304).
Exam tips
- In board exam problems, read carefully whether the given load is service (unfactored) or ultimate (factored). Most bearing capacity problems use service loads.
- For square footing: B = √(P/q_a). For rectangular footing: choose L/B ratio first, then solve — but square is default unless stated otherwise.
- When the problem mentions 'net allowable bearing capacity,' subtract the overburden pressure (γD_f) from the gross applied pressure before comparing.
- Mat foundation problems in the board exam often ask about the average contact pressure: q = (ΣP) / A_mat.
Key points
- Shallow foundations are used when competent bearing soil exists at or near the surface (typically Df/B ≤ 1, where Df = depth, B = width).
- Four main types: (1) Isolated (spread) footing — single column; (2) Combined footing — two columns sharing one footing; (3) Strap footing — two isolated footings connected by a strap beam; (4) Mat/raft — one large slab supporting the entire structure.
- Basic sizing rule: Required area A_req = P_service / q_a, where P_service is the unfactored (service) column load and q_a is the allowable bearing pressure.
- Switch from spread footings to a mat foundation when the total footing area would exceed approximately 50% of the building footprint, or when the soil is very weak/variable.
- After sizing, three checks are mandatory: (1) Bearing capacity, (2) Settlement (immediate + consolidation), (3) Structural design (shear and flexure per NSCP/ACI 318).
- Mat foundations also reduce differential settlement and are effective in resisting buoyancy in high water-table sites common in Metro Manila.
Definitions
Allowable Bearing Pressure (q_a)
The maximum net load per unit area that the soil can carry with adequate safety and tolerable settlement. It equals q_ultimate / FS, where FS ≈ 2.5 to 3.0.
Importance
Directly used to size the footing. Board exam problems almost always give q_a directly.
Mat/Raft Foundation
A continuous reinforced concrete slab extending under the entire structure, distributing column and wall loads over the full building footprint.
Importance
Used when spread footings would cover >50% of the footprint or in areas with very soft, variable soil. Reduces differential settlement.
Strap Footing
Two isolated footings (often one near a property line) connected by a rigid strap beam that transfers eccentric moment, keeping both footings uniformly loaded.
Importance
Common in urban Philippine settings where a column is at the property boundary and an eccentric footing would be impractical.
Common mistakes
- Using factored (ultimate) loads instead of service loads when applying A_req = P/q_a. The allowable bearing pressure is a service-level concept.
- Forgetting to include the footing self-weight and soil overburden weight in the total load when q_a is expressed as gross (total) bearing pressure.
- Choosing a mat foundation without checking that spread footings would actually overlap — the 50% rule is a guideline, not an absolute trigger.
- Confusing N_c = 5.14 (Terzaghi's strip footing at surface) with N_c* = 9 (deep pile in clay) — these are completely different bearing capacity factors.
2. Deep Foundations — Pile Types and Capacity
Formulas
Q_u = Q_p + Q_s
Example
Q_p = 67.9 kN, Q_s = 814.5 kN → Q_u = 882.4 kN (see worked Example 1 below).
Variables
Q_u = ultimate pile capacity (kN); Q_p = end bearing capacity (kN); Q_s = total skin friction capacity (kN)
Application
Fundamental equation for all pile problems. Both components must be computed and summed.
Q_p = 9 × c_u × A_p [Clay — deep pile]
Example
D = 0.4 m pile, c_u = 60 kPa: A_p = π/4 × (0.4)² = 0.1257 m². Q_p = 9 × 60 × 0.1257 = 67.9 kN.
Variables
Q_p = end bearing (kN); c_u = undrained shear strength at pile tip (kPa); A_p = cross-sectional area of pile tip (m²); 9 = deep bearing capacity factor N_c* for piles in clay
Application
Used exclusively for piles (not shallow footings) in saturated clay under undrained (short-term) loading.
Q_s = α × c_u × (π × D) × L [Clay — α-method]
Example
α = 0.9, c_u = 60 kPa, D = 0.4 m, L = 12 m: Q_s = 0.9 × 60 × (π×0.4) × 12 = 0.9 × 60 × 1.257 × 12 = 814.5 kN.
Variables
Q_s = skin friction (kN); α = adhesion factor (0.5 to 1.0, dimensionless); c_u = undrained shear strength along shaft (kPa); π×D = pile perimeter (m); L = embedded pile length (m)
Application
Alpha method for skin friction in cohesive soils (clay). α decreases as c_u increases (stiff clays are less 'sticky' relatively).
Q_s = K × σ'_v × tan(δ) × (π × D) × L [Sand — β-method, simplified]
Example
K = 0.8, σ'_v(avg) = 60 kPa, δ = 28°, D = 0.4 m, L = 10 m: Q_s = 0.8 × 60 × tan(28°) × π×0.4 × 10 = 0.8 × 60 × 0.5317 × 1.257 × 10 = 320.6 kN.
Variables
K = lateral earth pressure coefficient (0.5–1.5); σ'_v = average effective vertical stress along pile shaft (kPa); δ = pile-soil friction angle (degrees); π×D = pile perimeter (m); L = pile length (m)
Application
Beta method for skin friction in cohesionless soils (sand/gravel). Note: K and δ depend on pile material and soil density.
Q_a = Q_u / FS
Example
Q_u = 882.4 kN, FS = 2.5 → Q_a = 882.4 / 2.5 = 353 kN.
Variables
Q_a = allowable pile capacity (kN); Q_u = ultimate pile capacity (kN); FS = factor of safety (typically 2.5 to 3.0 for piles)
Application
Applied after computing Q_u. Use FS = 2.5 unless the problem specifies otherwise.
Exam tips
- Memorize the sequence: (1) Compute A_p and perimeter (πD), (2) Q_p = 9c_u×A_p, (3) Q_s = α×c_u×(πD)×L, (4) Q_u = Q_p + Q_s, (5) Q_a = Q_u/FS.
- For sand piles, the β-method formula Q_s = K×σ'_v×tan(δ)×(πD)×L mirrors the clay formula structure — same perimeter×length framework.
- The board exam often gives diameter in cm — always convert to meters before computing A_p and perimeter.
- If both end bearing and skin friction formulas are given in the problem stem, use them directly; do not derive from scratch.
Key points
- Deep foundations are used when weak or compressible soils near the surface require loads to be transferred to deeper, stronger strata.
- Most common type in PRC exams: driven piles (steel, precast concrete, timber) and bored piles (drilled shafts/caissons).
- Total ultimate pile capacity: Q_u = Q_p + Q_s (end bearing plus skin friction).
- End bearing Q_p acts at the pile tip (area A_p); skin friction Q_s acts along the pile shaft (perimeter × length).
- For clay: end bearing uses N_c* = 9 (deep failure mechanism, NOT 5.14); skin friction uses the α-method (adhesion factor).
- For sand: end bearing uses Meyerhof's bearing capacity factors with effective overburden; skin friction uses the β-method (lateral earth pressure × friction).
- Allowable pile capacity: Q_a = Q_u / FS, with FS = 2.5 to 3.0 for piles (higher uncertainty than shallow footings).
Definitions
End Bearing (Q_p)
The component of pile capacity developed at the pile tip, where stress is concentrated on the soil/rock beneath the tip area A_p. Dominant in piles bearing on hard rock or dense gravel.
Importance
Uses A_p (tip area, m²) — do NOT use perimeter here. Common board exam error is mixing up tip area and lateral surface area.
Skin Friction (Q_s)
The component of pile capacity developed by shear stress along the pile-soil interface over the entire embedded length. Dominant in long piles in clay or dense sand.
Importance
Uses perimeter × length (πDL for circular piles) — do NOT use tip area here. Skin friction usually dominates in clay piles.
Adhesion Factor (α)
A dimensionless empirical reduction factor (0.5 to 1.0) applied to the undrained shear strength c_u to obtain the unit skin friction in the α-method for clay. Accounts for remolding during pile driving.
Importance
Board exam problems always provide α. Remember: α × c_u = unit skin friction (kPa).
N_c* = 9
The deep bearing capacity factor for end bearing of piles in clay under undrained loading. Derived from cavity expansion theory — much higher than Terzaghi's N_c = 5.14 for shallow footings.
Importance
One of the most commonly tested values in the board exam. NEVER use 5.14 for pile end bearing in clay.
Common mistakes
- Using N_c = 5.14 (shallow footing) instead of N_c* = 9 (deep pile) for clay pile end bearing — this error cuts Q_p nearly in half.
- Computing skin friction with A_p (tip area) instead of πDL (lateral surface area) — always ask: 'Is this the tip or the shaft?'
- Forgetting to apply the factor of safety to get Q_a from Q_u — the problem asks for allowable load, not ultimate.
- For circular piles, forgetting to square the radius: A_p = π/4 × D² (not π × D).
- Mixing up gross and net pile capacity — in most exam problems, net capacity is the pile's own structural/geotechnical capacity without subtracting pile weight separately.
3. Pile Groups — Efficiency and Block Failure
Formulas
Q_group,ult = η × n × Q_single,ult
Example
3×3 group (n=9), η=0.80, Q_single=882.4 kN: Q_group,ult = 0.80 × 9 × 882.4 = 6,353 kN. Q_a,group = 6,353/2.5 = 2,541 kN.
Variables
Q_group,ult = ultimate group capacity (kN); η = group efficiency factor (dimensionless, ≤ 1.0); n = total number of piles in group; Q_single,ult = ultimate capacity of one pile (kN)
Application
Primary group capacity formula. The board exam always provides η — you do not need to compute it from spacing formulas.
Q_block = 9 × c_u × (L_g × B_g) + 2(L_g + B_g) × L_pile × c_u,avg [Block failure — clay]
Example
3×3 group with pile spacing 1.2 m (center-to-center), D = 0.4 m: L_g = B_g ≈ 2(1.2) + 0.4 = 2.8 m. Block end bearing = 9 × 60 × (2.8×2.8) = 4,234 kN. Block skin = 2(2.8+2.8) × 12 × 60 = 8,064 kN. Q_block = 12,298 kN — in this case, individual pile failure governs (6,353 < 12,298).
Variables
Q_block = block failure capacity (kN); L_g, B_g = length and width of pile group perimeter (m); L_pile = pile embedded length (m); c_u,avg = average undrained shear strength along pile length (kPa); first term = end bearing of block; second term = skin friction of block perimeter
Application
Compare with individual pile group capacity; the LOWER value governs. Critical for closely spaced piles in soft clay.
Exam tips
- Group problems almost always give you η directly. Just apply: Q_group,ult = η × n × Q_u,single.
- If the problem asks to 'check block failure,' set up the block as a rectangular pier with dimensions equal to the outer perimeter of the pile group.
- Remember: in the board exam, 'group efficiency' type questions are common — if η < 1, the group is less efficient than the sum of individual piles.
- For quick calculation: Q_a,group = (η × n × Q_u,single) / FS.
Key points
- A pile group consists of multiple piles connected at the top by a pile cap. The piles are typically arranged in a square or rectangular grid.
- Group capacity is NOT simply n × Q_single because adjacent piles create overlapping stress bulbs that reduce individual pile efficiency.
- Group efficiency factor η (eta, 0 < η ≤ 1) accounts for this reduction: Q_group,ult = η × n × Q_single,ult.
- In clay, a separate block failure check is mandatory: treat the entire pile group + enclosed soil as one large pier and compute its end bearing plus skin friction as a block.
- The GOVERNING (lower) value between individual pile failure and block failure controls the design.
- Typical pile spacing: 2.5D to 3D center-to-center. Closer spacing → lower η; wider spacing → η approaches 1.0.
- Allowable group load: Q_a,group = Q_group,ult / FS.
Definitions
Group Efficiency Factor (η)
A dimensionless factor (0 to 1) representing the ratio of actual group capacity to the sum of individual pile capacities. Caused by overlapping stress zones and soil disturbance during driving.
Importance
Directly multiplied into the group capacity formula. η = 1.0 would mean no group effect (ideal, rarely achieved in practice).
Block Failure
A failure mode where the pile group and the soil enclosed between piles act as a single large deep foundation block. The failure surface encloses the entire group, not individual piles.
Importance
Must be checked in addition to individual pile efficiency — whichever gives LOWER capacity governs. Frequently tested in board exams as a conceptual question.
Pile Cap
A thick reinforced concrete slab at the top of a pile group that distributes the column load to all piles and maintains their relative position. Designed structurally per NSCP/ACI 318.
Importance
The pile cap weight is often added to the total load on the pile group in practical design — watch for this in board exam problems.
Common mistakes
- Computing group capacity as simply n × Q_single without applying the efficiency factor η — this overestimates group capacity.
- Forgetting the block failure check in clay pile groups — the problem may specifically ask 'which failure governs?'
- Using the wrong n: a 3×3 group has n = 9 piles, not 3. Count all piles, not just the rows.
- Applying η to Q_a (allowable) instead of Q_u (ultimate) before dividing by FS — always apply η to ultimate, then divide.
4. Negative Skin Friction (Downdrag)
Formulas
Q_nsf = α_nsf × c_u × (π × D) × L_fill [NSF in clay]
Example
Pile through 5 m of settling fill: α=0.5, c_u=30 kPa, D=0.4 m → Q_nsf = 0.5 × 30 × (π×0.4) × 5 = 94.2 kN added to column load.
Variables
Q_nsf = negative skin friction force (kN) — acts as DOWNWARD load; α_nsf = adhesion factor for negative friction zone; c_u = undrained shear strength in settling layer (kPa); L_fill = thickness of settling fill/clay layer (m)
Application
The NSF force is added to the column load: Total pile load = P_column + Q_nsf. This total must be ≤ Q_a from the lower bearing zone.
Exam tips
- Key trigger for NSF problems: 'pile driven through recently filled area,' 'soft clay undergoing consolidation,' or 'lowering of water table.' Watch for these phrases.
- The conceptual question 'what is the effect of negative skin friction on pile capacity?' — Answer: it REDUCES the effective capacity by adding to the applied load.
- NSF value is computed exactly like positive skin friction (same formula), but it is added to the structural load, not to Q_u.
- In a two-zone problem: Zone 1 (fill) = NSF zone (adds load); Zone 2 (bearing stratum) = positive friction + end bearing (resists load).
Key points
- Negative skin friction (NSF) occurs when the surrounding soil settles MORE than the pile itself — the settling soil drags downward on the pile shaft, adding load instead of providing resistance.
- This is the OPPOSITE of normal skin friction: it acts downward as an additional load, reducing the net capacity of the pile.
- Most critical in: (1) Piles driven through recently placed fills over soft clay, (2) Areas with ongoing consolidation settlements, (3) Lowering of groundwater table causing increased effective stress and settlement.
- The neutral point is where pile and soil settlements are equal — above it, NSF acts downward (drag zone); below it, positive skin friction acts upward (resistant zone).
- NSF must be added to the structural load when checking pile structural capacity and geotechnical capacity of the lower bearing zone.
- In Manila Bay reclamation areas and soft clay zones of Metro Manila (common board exam context), NSF is a critical design consideration.
- Mitigation: bitumen coating on pile shaft, preloading the fill before driving, or using batter piles.
Definitions
Negative Skin Friction (NSF) / Downdrag
A downward-acting skin friction force on a pile shaft caused by consolidating or settling surrounding soil moving downward relative to the pile. It adds to the applied load, reducing the effective pile capacity.
Importance
Critical concept in Philippine soft clay and reclamation fill areas. Board exams test understanding of whether NSF adds to or subtracts from pile capacity (it adds to the LOAD, reduces the net safety margin).
Neutral Point
The depth at which pile and soil settlements are equal. Above the neutral point: soil settles more → NSF (downward drag). Below: pile settles more → positive skin friction (upward resistance).
Importance
Defines the transition between drag zone and resistance zone. Understanding this concept is essential for answering conceptual board exam questions.
Common mistakes
- Thinking NSF helps the pile — it does NOT. It ADDS to the downward load, reducing the available capacity from the bearing zone.
- Forgetting to add Q_nsf to the structural (column) load when checking pile capacity in the lower bearing zone.
- Confusing NSF with ordinary skin friction — NSF acts in the same direction as the applied load (both downward); positive skin friction acts upward (resists the load).
- Assuming NSF only occurs in fills — it also occurs during groundwater lowering or on piles through natural soft clay layers undergoing primary consolidation.
Connections
- Bearing Capacity (Terzaghi, Meyerhof) → Provides q_ultimate for computing q_a used in shallow footing sizing. The N_c factors differ significantly between shallow footings (5.14–9.0 depending on shape/depth) and deep piles (N_c* = 9).
- Consolidation Settlement → Drives negative skin friction problems; consolidating clay layers create the differential settlement condition that causes downdrag on piles.
- Effective Stress Principle → Used in the β-method (sand pile skin friction): f_s = K × σ'_v × tan(δ). Effective vertical stress σ'_v depends on depth, groundwater level, and unit weights.
- Soil Classification (USCS/AASHTO) → Determines which pile capacity method to use: α-method for clay (cohesive), β-method for sand (cohesionless). Correct soil identification is prerequisite to correct formula selection.
- Structural Design (NSCP/ACI 318) → After geotechnical sizing, footings must be checked for wide-beam shear (one-way), punching shear (two-way), and bending moment — all governed by NSCP 2015 Section 406 / ACI 318 Chapter 13.
- RA 544 (Civil Engineering Law) → Registered Civil Engineers are legally responsible for foundation design safety in the Philippines. Foundation failures (e.g., due to NSF in reclamation areas) are professional liability issues.
- NSCP 2015 Section 304 → Governs allowable bearing pressures for presumptive values and the requirements for foundation investigations. Board exams reference NSCP provisions on minimum footing depth and bearing capacity.
- Pile Load Testing → Static load tests and dynamic pile testing (PDA) are used to verify computed pile capacities. Understanding why we use FS = 2.5–3.0 for piles (higher uncertainty than footings) connects to reliability theory.
Quick review questions
A column carries a service load of 850 kN on soil with q_a = 180 kPa. What is the required area and minimum side length of a square footing?
Answer
A_req = 4.72 m²; B = 2.17 m → Use 2.2 m × 2.2 m
A_req = P/q_a = 850/180 = 4.722 m². For a square footing, B = √4.722 = 2.173 m. Round up to the next practical dimension: 2.2 m × 2.2 m (area = 4.84 m² > 4.72 m², adequate).
A 0.5 m-diameter, 15 m-long pile is driven in clay with c_u = 80 kPa and α = 0.7. Find Q_u and Q_a (FS = 3).
Answer
Q_p = 141.4 kN; Q_s = 1,319.5 kN; Q_u = 1,460.9 kN; Q_a = 487.0 kN
A_p = π/4 × (0.5)² = 0.1963 m². Q_p = 9 × 80 × 0.1963 = 141.4 kN. Perimeter = π × 0.5 = 1.5708 m. Q_s = 0.7 × 80 × 1.5708 × 15 = 1,319.5 kN. Q_u = 141.4 + 1,319.5 = 1,460.9 kN. Q_a = 1,460.9 / 3 = 487.0 kN.
A 2×2 pile group has group efficiency η = 0.85. Each pile has Q_u = 700 kN. Find the allowable group load (FS = 2.5).
Answer
Q_a,group = 952 kN
n = 4 piles. Q_group,ult = η × n × Q_u = 0.85 × 4 × 700 = 2,380 kN. Q_a,group = 2,380 / 2.5 = 952 kN.
What bearing capacity factor N_c* is used for end bearing of a pile in clay, and how does it differ from the N_c used for shallow footings?
Answer
N_c* = 9 for deep piles in clay; N_c ≈ 5.14 (Terzaghi) or 5.71 for shallow footings.
For deep piles, the failure mechanism around the tip resembles cavity expansion, giving N_c* ≈ 9 (Meyerhof, Skempton). For shallow footings under undrained loading, Terzaghi gives N_c = 5.7 (strip) and Skempton gives 6.17 (square) or 5.14 (general formula). Using 5.14 for a pile end bearing UNDERESTIMATES Q_p by ~43%.
Define negative skin friction and state one common condition that causes it.
Answer
Negative skin friction (NSF) is a downward-acting drag force on a pile shaft from settling surrounding soil, adding to the applied pile load. Common cause: pile driven through recently placed fill over soft consolidating clay.
When surrounding soil settles faster than the pile, friction reverses direction — pulling the pile DOWN instead of holding it up. This is an additional load, not a resistance. It is common in Metro Manila reclamation and low-lying areas with thick soft clay deposits.
A pile in sand has K = 0.8, average σ'_v = 60 kPa, δ = 28°, D = 0.4 m, and L = 10 m. Find Q_s.
Answer
Q_s = 320.6 kN
Q_s = K × σ'_v × tan(δ) × (π × D) × L = 0.8 × 60 × tan(28°) × (π × 0.4) × 10. tan(28°) = 0.5317. Q_s = 0.8 × 60 × 0.5317 × 1.2566 × 10 = 0.8 × 60 × 0.5317 × 12.566 = 320.6 kN.
When should a mat foundation be preferred over individual spread footings?
Answer
When the total area of individual spread footings would exceed approximately 50% of the building footprint, or when the soil is very weak or variable, causing unacceptable differential settlements.
If footings would overlap or nearly touch, it is more economical and structurally sound to use one continuous mat. Mat foundations also help resist hydrostatic uplift in high water-table areas (common in Metro Manila), reduce differential settlement, and are used when soil variability makes individual footing design unreliable.
For a 3×3 pile group, what does 'block failure' mean and why must it be checked separately?
Answer
Block failure occurs when the entire pile group plus enclosed soil acts as one large deep pier. It must be checked because it may give a LOWER capacity than the individual pile efficiency formula — the lower value governs.
When piles are closely spaced in clay, the failure surface may pass around the outside of the entire group rather than around individual piles. The block has a much larger perimeter skin area but also a much larger base area. Compare Q_block vs. η×n×Q_single — use whichever is smaller for design.
Diagrams
Foundations — Complete Chapter Mind Map
A hierarchical mind map showing all major topics in the Foundations chapter: shallow foundations (types, sizing, when to use mat), deep foundations (pile types, capacity components, formulas), pile groups (efficiency, block failure), negative skin friction, and key constants. Use this as a quick-review overview before studying individual sections.
Step-by-Step Pile Capacity Calculation in Clay
A detailed flowchart showing the complete step-by-step procedure for computing ultimate and allowable pile capacity in clay. Starting from raw inputs (D, L, c_u, α, FS), it guides through computing A_p and perimeter, end bearing Q_p using N_c*=9, skin friction Q_s using the α-method, summing to Q_u, and applying FS to get Q_a. Follow this flowchart during exam problem-solving to avoid missing steps.
Foundation Type Selection Flowchart
A decision flowchart guiding the selection of appropriate foundation type based on soil conditions, footing area vs building footprint, number of columns, and presence of soft clay. Also routes deep foundation problems to the correct skin friction method (alpha for clay, beta for sand, layered analysis for mixed profiles) and flags when negative skin friction must be checked.
Pile Group Capacity Check — Efficiency vs Block Failure
A flowchart illustrating the two-check procedure for pile group capacity in clay: (1) Individual pile efficiency method and (2) Block failure method. The lower of the two capacities governs the design. This is a frequently missed step in board exam problems where students only compute the efficiency formula without checking block failure.
Negative Skin Friction — Pile-Soil Interaction States
A state diagram showing the progression of negative skin friction development from pile installation through fill placement, consolidation, and steady-state load transfer. Distinguishes the NSF zone (above neutral point, drag acting downward) from the positive friction zone (below neutral point, resistance acting upward). Helps visualize why NSF is a load addition, not a capacity reduction.
Exam strategy
For PRC board exam foundation problems, follow this structured attack: (1) IDENTIFY foundation type (shallow vs. deep) and soil type (clay vs. sand) from the problem stem — this determines which formulas apply. (2) For shallow footings: immediately apply A = P/q_a; for square footing B = √A. (3) For piles in clay: memorize the sequence A_p = πD²/4 → Q_p = 9c_uA_p → perimeter = πD → Q_s = α×c_u×(πD)×L → Q_u = Q_p + Q_s → Q_a = Q_u/FS. (4) For pile groups: Q_group,ult = η×n×Q_single; also mentally check if block failure is asked. (5) For NSF: NSF ADDS to the load (never reduces Q_u directly). (6) Watch the critical differentiators: N_c*=9 for piles (NOT 5.14), perimeter for skin friction (NOT area), η applied to Q_u (NOT Q_a). (7) Always write down given values and draw a quick sketch of the pile with Q_p at tip and Q_s along shaft — this prevents formula confusion under exam pressure. (8) Manage time: shallow footing problems are quick (2–3 min); pile capacity problems require 5–7 min. Flag and return to group/NSF problems if pressed for time. (9) Cross-check units: loads in kN, areas in m², pressures in kPa — never mix kN and MN.