CELE Geotechnical Engineering — Foundations (Shallow and Deep)Detailed Explanation
Detailed explanations for CELE Geotechnical Engineering — Foundations (Shallow and Deep). This page treats you like a serious reviewer: we unpack the concepts thoroughly, show worked examples of how Professional Regulation Commission (PRC) — Board of Civil Engineering frames Foundations (Shallow and Deep) questions, and explain the underlying reasoning that gets you to the right answer every time.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Geotechnical Engineering section sits under a "Core" weighting, and Foundations (Shallow and Deep) is the 10th chapter in the 11-chapter CELE Geotechnical Engineering 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 Geotechnical Engineering.
Foundations (Shallow and Deep) - Detailed Explanation
Foundations are the structural elements that transfer all building loads — dead, live, wind, and seismic — safely into the ground. In the PRC Civil Engineer Licensure Examination, foundation problems consistently appear in the geotechnical engineering component, requiring both conceptual understanding and numerical proficiency. This chapter covers two broad foundation categories: (1) shallow foundations (spread footings, combined footings, strap footings, and mat/raft foundations), which distribute loads near the ground surface; and (2) deep foundations (driven piles, drilled shafts/bored piles), which transfer loads to competent strata at depth. Mastery of footing sizing, pile capacity computation (end bearing + skin friction), pile group efficiency, and the concept of negative skin friction is essential for exam success. All formulas are presented in SI units, consistent with NSCP 2015 practice.
Concepts
Shallow Foundations — Types and Sizing
A shallow foundation is one where the depth of embedment (Df) is small relative to its width (B), typically Df/B ≤ 1. Shallow foundations rely on the near-surface soil to resist the applied loads through bearing pressure. The four primary types are: (1) Isolated (Spread) Footing — supports a single column; most common and simplest to design. (2) Combined Footing — supports two or more columns, used when columns are closely spaced or near a property line. (3) Strap (Cantilever) Footing — an eccentrically loaded edge footing connected by a strap beam to an interior footing to eliminate net eccentricity. (4) Mat/Raft Foundation — a single slab covering the entire building footprint, used when individual footings would overlap (total footing area > ~50% of building footprint) or when the soil is weak and settlement must be minimized. The fundamental sizing equation for any shallow footing under concentric vertical load is: A_req = P_service / q_a, where P_service is the unfactored (service-level) column load in kN and q_a is the allowable net bearing capacity of the soil in kPa. For a square footing: B = √(A_req). For a rectangular footing: specify an aspect ratio (e.g., L = 1.5B) then solve for B. Note: the self-weight of the footing and the soil overburden above it are often accounted for by using net bearing pressure, or by subtracting an estimated footing+soil weight from P before dividing. Always clarify whether q_a is gross or net in the problem statement.
Examples
Always use SERVICE (unfactored) loads for footing sizing against allowable bearing capacity. Factored loads (1.2D + 1.6L) are used only for structural design of the footing section (bending, shear) per ACI 318/NSCP 2015 Sec. 405.
Scenario
A column carries a service dead load of 400 kN and a service live load of 200 kN. The allowable net bearing capacity of the soil is 150 kPa. Size a square isolated footing.
Solution
P_service = 400 + 200 = 600 kN. A_req = P_service / q_a = 600 / 150 = 4.0 m². B = √4.0 = 2.0 m. Use a 2.0 m × 2.0 m square footing.
If the ratio had exceeded 50%, a mat foundation would be the practical choice. This threshold is a common exam decision point.
Scenario
A building has 9 columns each carrying 500 kN service load on soil with q_a = 120 kPa. The building footprint is 12 m × 15 m = 180 m². Should isolated footings or a mat be used?
Solution
A per footing = 500/120 = 4.17 m². B = √4.17 = 2.04 m, use 2.1 m × 2.1 m. Total footing area = 9 × (2.1)² = 9 × 4.41 = 39.7 m². Ratio = 39.7/180 = 22%. Since 22% < 50%, isolated footings are still adequate.
Applications
- Sizing column footings in low-rise residential and commercial buildings in the Philippines.
- Mat foundation design for high-rise buildings in Metro Manila on compressible alluvial deposits.
- Combined footing design at property boundaries (party walls in row houses).
- Strap footing design for eccentric loading at building edges adjacent to roads or adjacent structures.
Misconceptions
- Using factored loads (1.2D + 1.6L) for footing area sizing — WRONG. Use service loads only.
- Thinking q_a and q_ult are the same — q_a = q_ult / FS (FS ≈ 3 for bearing capacity).
- Assuming a mat always has higher bearing capacity than isolated footings — mats reduce contact pressure but capacity depends on soil, not foundation type.
- Using Nc* = 5.14 (for shallow footings) as end-bearing factor for deep piles in clay — for piles, Nc* = 9.
Related Concepts
- Ultimate bearing capacity (Terzaghi, Meyerhof equations)
- Settlement analysis (consolidation, immediate settlement)
- Structural design of footings (ACI 318 / NSCP 2015 — two-way shear, one-way shear, flexure)
- Soil bearing capacity from SPT (N-value correlations)
Common Exam Questions
Example
A column with DL = 350 kN, LL = 250 kN on q_a = 200 kPa. Find B of a square footing. → A = 600/200 = 3.0 m², B = 1.73 m → use 1.75 m or 1.8 m.
Approach
Identify P_service (sum of unfactored loads), identify q_a, compute A = P/q_a, then compute B (square) or use given L/B ratio (rectangular).
Question Type
Footing area/dimension calculation
Example
12 columns × 5.0 m² each = 60 m² total on a 100 m² footprint → 60% > 50% → use mat foundation.
Approach
Compute total footing area needed; compare to building footprint area; if ratio > 50%, recommend mat.
Question Type
Mat vs. isolated footing decision
Key Points To Remember
- Use A = P/q_a for footing area; P must be at service (unfactored) load level.
- Square footing: B = √A; Rectangular: set L/B ratio, solve for B.
- Switch to mat foundation when total isolated footing area exceeds ~50% of building footprint.
- Df/B ≤ 1 generally defines a shallow foundation.
- Mat foundations are preferred in weak, compressible soils (e.g., soft Manila Bay clay) to reduce differential settlement.
- Combined footing is used when two columns are close together or one column is at a property line.
- Strap footing uses a rigid beam to redistribute eccentric loads from an edge column.
Deep Foundations — Single Pile Capacity
A pile is a deep foundation element (typically L/D > 10) that transfers load to deeper, more competent soil or rock strata through two mechanisms: (1) End Bearing (Qp) — resistance developed at the pile tip due to the soil's or rock's bearing capacity; and (2) Skin Friction (Qs) — resistance developed along the shaft due to shear stress between pile surface and surrounding soil. The ultimate pile capacity is: Qu = Qp + Qs. The allowable capacity is: Qa = Qu / FS, where FS = 2.5 to 3.0 (PRC board exams commonly use FS = 2.5 or 3.0 — read the problem carefully). END BEARING: For clay (cohesive soil): Qp = Ap × qp = Ap × (cu × Nc*). For deep piles in clay, Nc* = 9 (this is critical — NOT 5.14 used for shallow footings). So Qp = 9 × cu × Ap. For sand (cohesionless): qp = q' × Nq*, where q' is the effective overburden pressure at the pile tip and Nq* is a bearing capacity factor for piles. SKIN FRICTION: Alpha (α) Method for Clay: fs = α × cu, where α = adhesion factor (ranges from 0.5 to 1.0; α decreases as cu increases). Qs = fs × (π × D) × L = α × cu × (π × D) × L. Beta (β) Method for Sand: fs = K × σ'v × tan(δ), where K = lateral earth pressure coefficient (0.5–1.5 depending on pile type), σ'v = average effective vertical stress, δ = pile-soil friction angle. Qs = fs × (π × D) × L. Important note on perimeter vs. area: skin friction uses the lateral surface area of the pile shaft = π × D × L (for circular pile), while end bearing uses the tip (cross-sectional) area Ap = π/4 × D².
Examples
Note that skin friction (814.5 kN) vastly exceeds end bearing (67.9 kN) — typical for long piles in clay. This means even a small error in α significantly affects the answer. Always round D and compute Ap and perimeter carefully.
Scenario
A 0.4 m diameter, 12 m long concrete pile is driven into clay with cu = 60 kPa and adhesion factor α = 0.9. Compute Qu and Qa (FS = 2.5). Use Nc* = 9.
Solution
Ap = (π/4)(0.4)² = 0.1257 m². Qp = 9 × 60 × 0.1257 = 67.9 kN. Perimeter = π × 0.4 = 1.2566 m. Qs = 0.9 × 60 × 1.2566 × 12 = 814.5 kN. Qu = 67.9 + 814.5 = 882.4 kN. Qa = 882.4 / 2.5 = 353 kN.
This is Exercise 1 from the reference. Notice cu = 80 kPa is higher, but α = 0.7 < 0.9 (inverse relationship). Higher cu → lower α. The FS = 3 is more conservative, giving lower Qa.
Scenario
A 0.5 m diameter, 15 m long pile is driven into clay with cu = 80 kPa and α = 0.7. Find Qu and Qa (FS = 3).
Solution
Ap = (π/4)(0.5)² = 0.1963 m². Qp = 9 × 80 × 0.1963 = 141.3 kN. Perimeter = π × 0.5 = 1.5708 m. Qs = 0.7 × 80 × 1.5708 × 15 = 1319.5 kN. Qu = 141.3 + 1319.5 = 1460.8 kN. Qa = 1460.8 / 3 = 486.9 kN ≈ 487 kN.
This is Exercise 4 from the reference (β-method for sand). Key: tan(28°) = 0.5317. The β-method uses average σ'v over the pile length, not the σ'v at the tip.
Scenario
A pile in sand: K = 0.8, soil-pile friction angle δ = 28°, average effective vertical stress σ'v = 60 kPa, D = 0.4 m, L = 10 m. Find Qs.
Solution
fs = K × σ'v × tan(δ) = 0.8 × 60 × tan(28°) = 0.8 × 60 × 0.5317 = 25.5 kPa. Perimeter = π × 0.4 = 1.2566 m. Qs = fs × (π × D) × L = 25.5 × 1.2566 × 10 = 320.4 kN.
Applications
- Driven precast concrete piles for bridges and overpasses (common in DPWH projects).
- Bored piles (drilled shafts) for high-rise buildings in Metro Manila.
- Steel H-piles for pier foundations along Manila Bay and other coastal areas.
- Micro-piles for underpinning existing structures and slope stabilization.
- Timber piles for light structures in remote areas (historical use in the Philippines).
Misconceptions
- Using Nc* = 5.14 for deep pile end bearing in clay — use Nc* = 9 for piles.
- Using the lateral surface area (πDL) for end bearing — end bearing uses Ap = πD²/4.
- Adding skin friction and end bearing using factored loads — always use service loads for capacity check.
- Thinking skin friction acts upward for all cases — in negative skin friction, it acts downward (adds load).
- Using total stress for β-method (sand) — β-method requires effective stress σ'v.
Related Concepts
- Negative skin friction (downdrag)
- Pile group capacity and efficiency
- Settlement of pile foundations
- Pile load test (static and dynamic)
- NSCP 2015 Section 306 — deep foundation requirements
Common Exam Questions
Example
D = 0.3 m, L = 10 m, cu = 50 kPa, α = 0.8, FS = 2.5. Ap = 0.0707 m², Qp = 9(50)(0.0707) = 31.8 kN, P = 0.9425 m, Qs = 0.8(50)(0.9425)(10) = 377 kN, Qu = 408.8 kN, Qa = 163.5 kN.
Approach
Step 1: Compute Ap = (π/4)D². Step 2: Compute Qp = 9cu × Ap (clay). Step 3: Compute perimeter = πD. Step 4: Compute Qs = α × cu × πD × L. Step 5: Qu = Qp + Qs. Step 6: Qa = Qu / FS.
Question Type
Single pile ultimate and allowable capacity
Example
Given K, δ, average σ'v, D, L — compute fs, then Qs. Common PRC question: find the skin friction resistance only.
Approach
Compute fs = K × σ'v × tan(δ). Then Qs = fs × πD × L. If Qp is also required: Qp = Ap × q' × Nq*.
Question Type
Pile capacity in sand using β-method
Key Points To Remember
- Qu = Qp + Qs (always additive for single pile).
- For clay piles: Qp = 9 × cu × Ap (Nc* = 9, not 5.14).
- Skin friction in clay (α-method): Qs = α × cu × (πD) × L.
- Skin friction in sand (β-method): fs = K × σ'v × tan(δ).
- Qa = Qu / FS; FS = 2.5 to 3.0.
- Pile shaft surface area = πDL; pile tip area = (π/4)D².
- α decreases as undrained shear strength cu increases (stiff clays have lower α).
- End bearing in clay: Qu is often dominated by Qs (skin friction >> end bearing for long piles).
Pile Groups — Group Capacity and Efficiency
In practice, piles are almost always installed in groups (clusters) rather than as isolated single piles, because a single pile rarely has sufficient capacity for the column loads of a real structure. However, piles in a group do not behave as n independent piles — the overlapping stress zones in the soil reduce the effectiveness of each pile. This is quantified by the group efficiency factor η (eta): Q_group,ult = η × n × Q_single,ult, where n = total number of piles in the group (e.g., 3×3 = 9) and η ≤ 1.0. The group efficiency η is determined by empirical formulas such as the Converse-Labarre equation or given directly in exam problems. For CLAY soils, you must also check BLOCK FAILURE — the condition where the entire pile group + enclosed soil acts as one large 'block' foundation, failing as a unit. Block failure capacity: Q_block = cu × Nc* × (Lg × Bg) + 2cu × (Lg + Bg) × L, where Lg = group length (in plan), Bg = group width (in plan), L = pile length, and Nc* = 9. The governing (lower) value of individual pile group capacity vs. block failure capacity is used. In sand, block failure is rarely critical because sand has high shear strength and the block mechanism is less likely. PILE SPACING: Typical center-to-center pile spacing = 2.5D to 3.5D (D = pile diameter), with a minimum of 2.5D per most codes and NSCP 2015. Closer spacing increases group effect (lower η); wider spacing reduces group effect (η closer to 1.0). PILE CAP: The pile cap is the reinforced concrete slab connecting the pile heads and transferring the column load to the piles. Its design follows ACI 318 / NSCP 2015 provisions for punching shear and flexure.
Examples
Without the efficiency factor, one might incorrectly compute Q_group = 9 × 882.4 = 7942 kN, overestimating by 25%. The efficiency factor is the most commonly missed element in PRC pile group problems.
Scenario
A 3×3 group of 0.4 m diameter, 12 m long piles (from the single pile example: Q_single = 882.4 kN) has a group efficiency η = 0.80. Find the allowable group capacity (FS = 2.5).
Solution
n = 9 piles. Q_group,ult = η × n × Q_single = 0.80 × 9 × 882.4 = 6353 kN. Q_a,group = 6353 / 2.5 = 2541 kN.
This is Exercise 2 from the reference. Straightforward application of the group efficiency formula. PRC exams often give η directly and ask for Q_a,group.
Scenario
A 2×2 pile group has η = 0.85. Each single pile has Q_single,ult = 700 kN. Find the allowable group load (FS = 2.5).
Solution
n = 4. Q_group,ult = 0.85 × 4 × 700 = 2380 kN. Q_a,group = 2380 / 2.5 = 952 kN.
Applications
- Design of pile caps for building columns (2-pile, 4-pile, and 9-pile groups are most common).
- Bridge pier pile group design (DPWH bridge design standards).
- Offshore platform pile groups in petroleum industry.
- Pile group settlement analysis — group settles more than single pile due to deeper stress influence.
Misconceptions
- Multiplying single pile capacity by n without applying η — always apply group efficiency.
- Ignoring block failure check in clay groups — it can govern and give a lower capacity.
- Applying block failure check to sand groups — rarely applicable; group efficiency governs in sand.
- Assuming η = 1.0 when not stated — a conservative assumption; use η < 1 unless stated or computed.
- Using factored loads when checking against pile group capacity — use service loads.
Related Concepts
- Pile cap design (ACI 318 / NSCP 2015)
- Group settlement
- Converse-Labarre equation for group efficiency
- Block failure mechanism in clay
- Pile spacing requirements (NSCP 2015 Sec. 306)
Common Exam Questions
Example
3×3 group, η = 0.75, Q_single = 500 kN, FS = 2.5. Q_group = 0.75 × 9 × 500 = 3375 kN. Q_a = 3375/2.5 = 1350 kN.
Approach
Identify n (total pile count), η, Q_single,ult. Compute Q_group = η × n × Q_single. Apply FS for Q_a,group.
Question Type
Group ultimate and allowable capacity
Example
If block failure gives 3000 kN and individual pile group gives 3375 kN, the governing capacity is 3000 kN.
Approach
Compute Q_block using block dimensions (based on pile layout and spacing). Compare to Q_group = η×n×Q_single. Use the smaller value.
Question Type
Block failure check for clay
Key Points To Remember
- Q_group = η × n × Q_single (group efficiency formula).
- η ≤ 1.0; do not use η > 1 even if some formulas theoretically allow it.
- For clay: always check block failure capacity separately; use the lower value.
- Minimum pile spacing: 2.5D center-to-center (NSCP 2015).
- Block failure: Q_block = 9cu(Lg×Bg) + 2cu(Lg+Bg)L for clay pile group.
- In sand, block failure is rarely critical; group efficiency governs.
- Group efficiency generally decreases as pile spacing decreases.
- Allowable group capacity: Q_a,group = Q_ult,group / FS.
Negative Skin Friction (Downdrag)
Negative skin friction (NSF), also called downdrag, is a phenomenon that occurs when the surrounding soil settles MORE than the pile itself. Instead of the pile skin resistance acting upward (resisting the applied load), the settling soil drags the pile downward — adding to the axial load in the pile rather than relieving it. This is the OPPOSITE of normal (positive) skin friction. WHEN DOES IT OCCUR? Negative skin friction typically occurs: (1) When piles are driven through a recently placed fill (embankment or reclaimed land) that is still consolidating — common in Philippines coastal reclamation projects. (2) When piles pass through a soft compressible clay layer (e.g., soft marine clay along Manila Bay and Laguna Lake areas) that is undergoing consolidation under its own weight or due to surface loads. (3) When lowering of the groundwater table causes increased effective stress and consolidation of compressible layers. EFFECT ON PILE CAPACITY: The pile must now carry the applied structural load PLUS the downdrag force. The effective capacity is reduced: Q_net = Q_total,ult − Q_NSF, where Q_NSF (downdrag force) = α × cu × πD × L_NSF (the length of pile within the settling zone). The NEUTRAL POINT is the depth where the pile and surrounding soil have equal settlement — above this point, NSF acts; below it, positive skin friction acts. DESIGN IMPLICATION: For piles in consolidating soils, NSF can be 20–40% of the pile's capacity, which is why it must be explicitly accounted for in design.
Examples
This 94.2 kN acts DOWNWARD on the pile, adding to the structural load. If the pile allowable capacity was 300 kN, the effective load capacity for the structure is reduced to 300 − 94.2 = 205.8 kN. This is critical in reclaimed land sites.
Scenario
A pile passes through a 5 m thick soft clay fill (cu = 30 kPa, α = 0.5) and is embedded into dense sand below. Pile diameter D = 0.4 m. Estimate the downdrag force.
Solution
Q_NSF = α × cu × πD × L_NSF = 0.5 × 30 × π × 0.4 × 5 = 0.5 × 30 × 1.2566 × 5 = 94.2 kN.
Applications
- Pile design in reclaimed coastal areas (e.g., Bay Area Manila, Mactan reclamation projects).
- Bridge piles on embankment approaches where fill is newly placed.
- Building piles near rivers or coastal zones with active sedimentation and consolidation.
- Evaluation of pile settlements in areas with lowering groundwater table (over-extraction of groundwater in Metro Manila aquifers).
Misconceptions
- Thinking NSF helps the pile by adding more friction — NSF acts downward, ADDING to load, NOT capacity.
- Applying NSF to piles in non-consolidating soils — NSF only occurs in settling/consolidating ground.
- Using NSF for all soft clay layers — NSF only applies in the zone where soil settles more than the pile.
- Ignoring NSF in reclaimed land sites — extremely important for coastal Philippine construction projects.
Related Concepts
- Consolidation settlement
- Neutral point of a pile
- Pile design in reclaimed land (NSCP 2015 Sec. 306)
- Positive skin friction vs. negative skin friction
- Groundwater lowering effects on effective stress
Common Exam Questions
Example
A pile is driven through a 6 m thick compressible clay layer that is consolidating. Determine if NSF applies and estimate the downdrag force.
Approach
Look for keywords: recently placed fill, soft compressible clay, consolidating soil, lowering of water table. NSF occurs when soil settles more than pile.
Question Type
Identifying when NSF occurs
Example
Pile Qu = 800 kN (below settling zone). Q_NSF = 120 kN. Net Qu for structural use = 800 − 120 = 680 kN. Qa = 680/2.5 = 272 kN.
Approach
Compute total pile capacity (Qp + Qs for pile in bearing stratum). Compute Q_NSF for consolidating layer. Net capacity = Q_total − Q_NSF.
Question Type
Adjusted pile capacity with NSF
Key Points To Remember
- NSF occurs when soil settles MORE than the pile — soil drags pile downward.
- NSF ADDS to the applied load; it REDUCES net pile capacity.
- Common scenarios: piles through fill, soft clay under consolidation, lowering of groundwater.
- Downdrag force: Q_NSF = α × cu × πD × L_NSF (clay layer) or β-method for sand fill.
- The neutral point separates the zone of NSF (above) from positive skin friction (below).
- Philippines context: reclaimed coastal areas, soft alluvial deposits in Metro Manila.
- Design must ensure pile tip capacity alone (below neutral point) is sufficient to carry structural load + NSF.
- NSF is a load effect, not a resistance — never add it to the capacity side of the equation.
Practice Problems
This is Exercise 3 from the reference. Always round up to the next practical dimension (typically to the nearest 0.1 m or 0.05 m in practice). Use service loads (NOT factored loads) for bearing capacity checks. Factored loads are only used for structural design of the footing section.
Problem
PROBLEM 1 (Spread Footing Sizing): A square isolated footing supports a column with a dead load of 500 kN and a live load of 350 kN. The allowable net bearing capacity of the soil is 180 kPa. Determine the required side dimension B of the square footing.
Solution
P_service = DL + LL = 500 + 350 = 850 kN. A_req = P_service / q_a = 850 / 180 = 4.722 m². B = √4.722 = 2.173 m. Round up to B = 2.2 m (use 2.2 m × 2.2 m).
This is Exercise 1. Key observations: (1) Skin friction (1319.5 kN) is about 9.3× the end bearing (141.4 kN) — typical for long clay piles. (2) Nc* = 9 is used for deep piles. (3) FS = 3 (not 2.5) — read the problem carefully. (4) Perimeter = πD, NOT πD²/4.
Problem
PROBLEM 2 (Single Pile Capacity in Clay): A 0.5 m diameter, 15 m long concrete pile is driven into clay with undrained shear strength cu = 80 kPa and adhesion factor α = 0.7. Compute: (a) end bearing capacity Qp, (b) skin friction capacity Qs, (c) ultimate capacity Qu, and (d) allowable capacity Qa with FS = 3.
Solution
(a) Ap = (π/4)(0.5)² = (π/4)(0.25) = 0.19635 m². Qp = 9 × cu × Ap = 9 × 80 × 0.19635 = 141.4 kN. (b) Perimeter = π × D = π × 0.5 = 1.5708 m. Qs = α × cu × (πD) × L = 0.7 × 80 × 1.5708 × 15 = 1319.5 kN. (c) Qu = Qp + Qs = 141.4 + 1319.5 = 1460.9 kN. (d) Qa = Qu / FS = 1460.9 / 3 = 486.97 kN ≈ 487 kN.
This is Exercise 4. The β-method is used for piles in sand (cohesionless soil). Note that tan(28°) = 0.5317. The average σ'v (not the σ'v at the tip) is used in this method. This is a common PRC board exam format — providing K, δ, and average σ'v and asking for Qs.
Problem
PROBLEM 3 (Single Pile Capacity in Sand — β-method): A 0.4 m diameter, 10 m long pile is driven into sand. Given: K = 0.8 (lateral earth pressure coefficient), δ = 28° (pile-soil friction angle), average effective vertical stress σ'v = 60 kPa. Compute the skin friction capacity Qs.
Solution
fs = K × σ'v × tan(δ) = 0.8 × 60 × tan(28°) = 0.8 × 60 × 0.5317 = 25.52 kPa. Perimeter = πD = π × 0.4 = 1.2566 m. Qs = fs × (πD) × L = 25.52 × 1.2566 × 10 = 320.7 kN.
This is Exercise 2. The critical step is applying η before multiplying by n. Without η: Q_group = 4 × 700 = 2800 kN (incorrect, overestimates by 17.6%). The group efficiency accounts for the overlapping stress zones that reduce individual pile effectiveness.
Problem
PROBLEM 4 (Pile Group Capacity): A 2×2 pile group has a group efficiency η = 0.85. Each pile has an ultimate single-pile capacity of 700 kN. Compute: (a) the group ultimate capacity, and (b) the allowable group capacity (FS = 2.5).
Solution
(a) n = 2 × 2 = 4 piles. Q_group,ult = η × n × Q_single,ult = 0.85 × 4 × 700 = 2380 kN. (b) Q_a,group = Q_group,ult / FS = 2380 / 2.5 = 952 kN.
This problem combines single pile calculation with group efficiency. Step-by-step: compute Ap and perimeter separately, compute Qp and Qs separately, sum for Qu, then apply η and n for the group, then divide by FS. This multi-step format is typical of PRC board problems.
Problem
PROBLEM 5 (Combined Single Pile and Group): A 3×3 group of 0.3 m diameter, 10 m piles is driven in clay with cu = 50 kPa, α = 0.8, Nc* = 9. Group efficiency η = 0.75, FS = 2.5. Find: (a) Qu for a single pile, (b) Q_group,ult, (c) Q_a,group.
Solution
(a) Ap = (π/4)(0.3)² = 0.07069 m². Qp = 9 × 50 × 0.07069 = 31.8 kN. Perimeter = π × 0.3 = 0.9425 m. Qs = 0.8 × 50 × 0.9425 × 10 = 377.0 kN. Qu,single = 31.8 + 377.0 = 408.8 kN. (b) n = 9. Q_group,ult = 0.75 × 9 × 408.8 = 2759.4 kN. (c) Q_a,group = 2759.4 / 2.5 = 1103.8 kN ≈ 1104 kN.
The 70.7 kN downdrag force acts DOWNWARD on the pile, effectively adding 70.7 kN to the applied structural load. If the pile was designed for an allowable capacity of 300 kN, the actual available capacity for supporting the structure is only 300 − 70.7 = 229.3 kN. Always identify the length of the consolidating zone (L_NSF), not the total pile length.
Problem
PROBLEM 6 (Negative Skin Friction): A 0.45 m diameter pile is driven through a 4 m thick soft clay layer (cu = 25 kPa, α = 0.5) that is currently undergoing consolidation. Below the soft clay, the pile bears on dense sand. Compute the downdrag (negative skin friction) force on the pile.
Solution
Q_NSF = α × cu × (πD) × L_NSF = 0.5 × 25 × (π × 0.45) × 4 = 0.5 × 25 × 1.4137 × 4 = 70.7 kN.
Exam Preparation Tips
- MEMORIZE the key formula set: A = P/q_a (footing); Qu = Qp + Qs; Qp = 9cu×Ap (clay piles); Qs = α×cu×πD×L (clay); Q_group = η×n×Q_single; Qa = Qu/FS. These cover ~80% of PRC foundation problems.
- KNOW YOUR Nc* VALUES: Nc* = 5.14 (general shear, strip, at surface) and Nc* = 5.7 (Terzaghi, strip) are for SHALLOW foundations. For DEEP piles in clay, Nc* = 9. This is the single most common error in PRC pile problems.
- DISTINGUISH AREAS: Pile tip uses Ap = (π/4)D² for end bearing. Pile shaft uses perimeter = πD (times L) for skin friction. Never mix these two.
- SERVICE LOADS for sizing, FACTORED LOADS for structural design: Footing area = P_service/q_a. Bending/shear design uses Pu = 1.2DL + 1.6LL per NSCP 2015 Sec. 405.
- PILE GROUP EFFICIENCY: Never compute pile group capacity as simply n × Q_single. Always apply the efficiency factor η. If η is not given, check if the problem asks you to compute it (Converse-Labarre equation) or state η = given value.
- NEGATIVE SKIN FRICTION IDENTIFICATION: Look for keywords — 'recently placed fill', 'soft clay undergoing consolidation', 'lowering of water table', 'reclaimed land'. NSF adds to applied load, reducing net capacity.
- BLOCK FAILURE IN CLAY GROUPS: For a pile group in clay, after computing individual group capacity, check block failure (entire group + soil acts as one block). The LOWER of the two governs. This step is often omitted, leading to unconservative answers.
- UNIT CONSISTENCY: D in meters → Ap in m², perimeter in m. cu in kPa → Qp and Qs in kN (because kPa × m² = kN). Always verify units at each step.
- FACTOR OF SAFETY: FS = 2.5 and FS = 3 are both commonly used in PRC problems. Read the problem statement carefully — do not default to 2.5 if FS = 3 is stated.
- PRACTICE THE COMPLETE SOLUTION PATHWAY: PRC board problems often ask for intermediate results (Qp, Qs, Qu, Qa as separate parts). Practice writing out each step clearly — partial credit and following the solution path is important.
- MAT FOUNDATION DECISION RULE: If total isolated footing area > 50% of building footprint → recommend mat/raft. This is a frequently tested conceptual question.
- REVIEW NSCP 2015 SECTION 306 (Deep Foundations): Minimum pile diameter, minimum pile spacing (2.5D), minimum number of piles per cap, and load test requirements are codified in NSCP 2015 and can appear as code-based questions.
- DRAW A FREE BODY DIAGRAM: For every pile problem, sketch the pile showing Qp at tip (upward), Qs along shaft (upward for positive, downward for NSF), and applied load at top (downward). This prevents sign errors.
- ALPHA (α) vs BETA (β) METHOD: α-method (adhesion) → clay soils, uses cu. β-method → sand soils, uses effective stress σ'v and friction angle δ. Distinguish by soil type in the problem.
In summary
Foundations — both shallow and deep — represent one of the most consistently tested areas in the PRC Civil Engineer Licensure Examination. This chapter has covered the essential computational framework: shallow footing sizing using A = P/q_a, single pile capacity as Qu = Qp + Qs (with Nc* = 9 for clay piles), pile group capacity using Q_group = η × n × Q_single (with mandatory block failure check for clay), and the design implications of negative skin friction (downdrag) in consolidating soils. To succeed in the PRC CE board exam, you must internalize three critical distinctions: (1) Nc* = 9 for deep piles vs. 5.14/5.7 for shallow footings, (2) pile tip area (πD²/4) vs. pile shaft perimeter (πD) — these are the most common computational errors, and (3) service loads for capacity design vs. factored loads for structural design. The Philippine context — reclaimed coastal areas, soft alluvial deposits in Metro Manila, and active DPWH infrastructure projects — makes foundation engineering particularly relevant and practically important. Mastery of these concepts not only prepares you for licensure examination success but builds the foundation (pun intended) for sound engineering practice as a licensed Filipino Civil Engineer under RA 544.
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