CELE Geotechnical Engineering — Bearing Capacity of SoilsDetailed Explanation
This is the "office hours" version of Bearing Capacity of Soils for the CELE 2026. No shortcuts, no hand-waving — just a full unpacking of why Professional Regulation Commission (PRC) — Board of Civil Engineering cares about each concept and how the Geotechnical Engineering section items tend to play out on exam day. Read this once, then hit the practice questions with real understanding.
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 Bearing Capacity of Soils is the 9th 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.
Bearing Capacity of Soils - Detailed Explanation
Bearing capacity of soils is one of the most consistently tested topics in the Geotechnical Engineering component of the PRC Civil Engineer Licensure Examination. Every foundation — whether a simple isolated footing under a residential column or a mat foundation supporting a high-rise — must be designed so that the soil beneath it does not fail in shear. The classical approach by Karl Terzaghi (1943) gives us a tractable, closed-form equation that the Board exam exploits repeatedly. This chapter builds the theory systematically: from the mechanics of shear failure, through Terzaghi's bearing-capacity equation and its factors, to shape corrections, the water-table adjustment, and the conversion from ultimate to allowable bearing pressure. Three fully worked board-style examples in SI units anchor the theory, and a bank of practice problems with complete solutions prepares you for the 5–8 items that typically appear on every examination. Read every pitfall note carefully — points are lost not because examinees do not know the formula, but because they misapply shape factors, forget the surcharge term, or confuse gross with net bearing capacity.
Concepts
Modes of Bearing-Capacity Failure
When a footing is loaded to failure, the soil beneath it fails in one of three characteristic modes, and identifying the correct mode determines which version of the bearing-capacity equations to use. **1. General Shear Failure** — occurs in dense sand or stiff, overconsolidated clay. A well-defined, continuous failure surface develops from the footing edge all the way to the ground surface. The load–settlement curve shows a clear peak (the ultimate load) followed by a sudden collapse. The soil bulges visibly on both sides of the footing. This is the idealized failure mode assumed in Terzaghi's standard equations. **2. Local Shear Failure** — occurs in medium-dense sand or medium-stiff clay. The failure surface is not continuous; it terminates within the soil mass and does not reach the surface. The load–settlement curve is gradual with no sharp peak. Terzaghi accounts for this by reducing the strength parameters: use c* = (2/3)c and tan φ* = (2/3)tan φ, then compute bearing-capacity factors with φ* instead of φ. **3. Punching Shear Failure** — occurs in very loose sand or soft, normally consolidated clay, and in deep footings. The footing simply punches straight down with very little lateral movement. This is the most unfavorable mode and is handled similarly to local shear (reduced parameters) in practice. For the Board exam, when the problem states 'dense sand' or 'stiff clay,' assume general shear. When it states 'loose sand,' 'soft clay,' or explicitly says 'local shear,' apply the (2/3) reduction.
Examples
Reducing c and tan φ by 2/3 is the Terzaghi correction for local shear. Note that c (not tan c) is reduced by 2/3, and tan φ (not φ itself) is reduced by 2/3.
Scenario
A footing rests on loose sand with c = 0, φ = 30°. Determine the modified parameters for local shear analysis.
Solution
c* = (2/3)(0) = 0 kPa tan φ* = (2/3) tan 30° = (2/3)(0.5774) = 0.3849 φ* = arctan(0.3849) = 21.05° ≈ 21° Use Nc, Nq, Nγ corresponding to φ* = 21° in the bearing-capacity factor table.
Applications
- Selecting the correct shear failure mode is the first decision in any bearing-capacity analysis.
- Geotechnical reports in the Philippines (required under NSCP 2015 Section 305) must identify the soil consistency and relative density to establish the governing failure mode.
- Pile design in Metro Manila's soft marine clay zones typically assumes punching-type behavior for end-bearing piles driven through compressible strata.
Misconceptions
- Many examinees reduce φ directly by 2/3 instead of reducing tan φ. The correct operation is: tan φ* = (2/3) tan φ.
- Not all loose-sand problems require local shear — always check if the problem explicitly states 'general shear' or gives the standard Nc, Nq, Nγ at the full φ.
Related Concepts
- Terzaghi's Bearing Capacity Equation
- Bearing Capacity Factors (Nc, Nq, Nγ)
- Shear strength parameters c and φ
Common Exam Questions
Example
Which failure mode applies to a footing on dense, well-graded sand? Answer: General shear failure.
Approach
Read the soil description. Dense = general shear. Loose/soft = local or punching shear. Apply 2/3 reduction only for local/punching.
Question Type
Multiple choice — identifying failure mode
Example
c = 20 kPa, φ = 35° → c* = 13.33 kPa; tan φ* = (2/3)(0.7002) = 0.4668; φ* ≈ 25°.
Approach
Compute c* = (2/3)c and φ* = arctan[(2/3)tan φ]. Then look up or use the given Nc, Nq, Nγ at φ*.
Question Type
Numerical — local shear parameters
Key Points To Remember
- General shear: dense/stiff soils, continuous failure surface, clear peak load — use standard Nc, Nq, Nγ.
- Local shear: medium soils, no surface heave — reduce c and tan φ by factor 2/3 before entering tables.
- Punching shear: very loose/soft soils or deep footings — treated like local shear.
- The exam usually specifies the mode or implies it through relative density/consistency.
Terzaghi's Bearing Capacity Equation
Terzaghi (1943) derived the ultimate bearing capacity of a shallow footing by analyzing the equilibrium of a rigid wedge of soil beneath the footing together with the Prandtl-type failure zones on either side. The result is the most widely used bearing-capacity formula in practice and on the PRC Board exam. **Strip (Continuous) Footing:** $$q_u = cN_c + qN_q + \tfrac{1}{2}\gamma B N_\gamma$$ **Square Footing:** $$q_u = 1.3cN_c + qN_q + 0.4\gamma B N_\gamma$$ **Circular Footing (diameter B):** $$q_u = 1.3cN_c + qN_q + 0.3\gamma B N_\gamma$$ where: - c = cohesion of the soil (kPa) - q = overburden pressure at footing base = γ·Df (kPa) - γ = unit weight of soil below the footing base (kN/m³) - B = width of footing (m) [diameter for circular] - Nc, Nq, Nγ = dimensionless bearing-capacity factors (functions of φ only) - Df = depth of footing below ground surface (m) **Physical meaning of each term:** 1. **cNc term** — contribution of soil cohesion (dominates in clays). 2. **qNq term** — contribution of the surcharge/overburden on the sides of the footing (the depth effect). This is why deeper footings carry more load. 3. **(1/2)γBNγ term** — contribution of the self-weight of the failure wedge below the footing (the width effect). Wider footings mobilize more soil. **Shape factors in Terzaghi's equation:** - Strip: coefficients 1.0 (on cNc) and 0.5 (on γBNγ) - Square: 1.3 and 0.4 - Circular: 1.3 and 0.3 Note: The Nq term carries a shape factor of 1.0 in Terzaghi's original formulation for all shapes.
Examples
Each of the three terms was computed separately before summing — a good exam habit that avoids arithmetic errors. The (1/2)γBNγ term = 0.5 × 18 × 2 × 8.34 = 150.12 kPa is often where points are lost if B is confused with area.
Scenario
BOARD-STYLE PROBLEM: A strip footing 2 m wide is placed at a depth of 1 m in a soil with c = 15 kPa, φ = 25°, γ = 18 kN/m³. Using Terzaghi factors Nc = 25.13, Nq = 12.72, Nγ = 8.34, determine: (a) the ultimate bearing capacity, and (b) the allowable bearing capacity with FS = 3.
Solution
Step 1 — Overburden at footing level: q = γDf = 18 × 1 = 18 kPa Step 2 — Apply strip footing equation: qu = cNc + qNq + (1/2)γBNγ qu = 15(25.13) + 18(12.72) + 0.5(18)(2)(8.34) qu = 376.95 + 228.96 + 150.12 qu = 756.03 kPa Step 3 — Allowable bearing capacity: qa = qu / FS = 756.03 / 3 = 252.0 kPa
For φ = 0, the γBNγ term vanishes entirely — only cohesion and surcharge contribute. The 1.3 shape factor on cNc is what distinguishes a square from a strip. Net capacity subtracts the overburden that was already there before the footing was placed.
Scenario
BOARD-STYLE PROBLEM: A 2 m × 2 m square footing rests at Df = 1 m on saturated clay with undrained cohesion cu = 50 kPa and γ = 18 kN/m³. Using φ = 0 parameters Nc = 5.7, Nq = 1, Nγ = 0, find qu and the net ultimate bearing capacity.
Solution
Step 1 — Overburden: q = 18 × 1 = 18 kPa Step 2 — Square footing on φ = 0 clay: qu = 1.3cNc + qNq + 0.4γBNγ qu = 1.3(50)(5.7) + 18(1) + 0.4(18)(2)(0) qu = 370.5 + 18 + 0 qu = 388.5 kPa Step 3 — Net ultimate bearing capacity: qu,net = qu − γDf = 388.5 − 18 = 370.5 kPa
Applications
- Sizing isolated column footings for residential and commercial buildings in the Philippines.
- Checking bearing capacity of raft foundations on Manila Bay soft clay (φ ≈ 0, cu controls).
- Foundation design reports submitted for DPWH and private projects must demonstrate qu/FS ≥ applied pressure.
Misconceptions
- Using L×B (area) instead of just B (width) in the γBNγ term — B is a linear dimension, not an area.
- Forgetting the 1.3 shape factor on the cNc term for square/circular footings.
- Using the net applied load instead of q = γDf for the surcharge term.
- Applying rectangular footing shape factors from Meyerhof to Terzaghi's equation — each method has its own set of factors.
Related Concepts
- Bearing Capacity Factors Nc, Nq, Nγ
- Overburden Pressure and Depth of Embedment
- Allowable Bearing Capacity and Factor of Safety
- Water-Table Correction
Common Exam Questions
Example
Given: square footing, B = 2.5 m, Df = 1.5 m, c = 0, φ = 35°, Nc = 57.75, Nq = 41.44, Nγ = 42.92, γ = 17 kN/m³ → qu = 0 + 1.5(17)(41.44) + 0.4(17)(2.5)(42.92) = 1056.7 + 728.0 = 1784.7 kPa
Approach
Identify footing shape → write the correct form of the equation → compute q = γDf → plug in Nc, Nq, Nγ from the given table → divide by FS for qa.
Question Type
Direct calculation — find qu and qa
Example
If P = 900 kN, FS = 3, use qu ≥ 3(900/B²) and solve for B.
Approach
Set qa = P/A = P/B² for square footing → express qu in terms of B → solve iteratively or by substitution.
Question Type
Back-calculate — find the required B
Key Points To Remember
- Memorize the three forms: strip (1.0, 0.5), square (1.3, 0.4), circular (1.3, 0.3) — these coefficients apply to the cNc and γBNγ terms respectively.
- The surcharge q = γDf — never use gross pressure from applied loads here; it is the overburden at footing level.
- B is the WIDTH (shorter dimension) for rectangular footings; for circular footings B is the diameter.
- Units: if c in kPa, γ in kN/m³, B in m → qu in kPa. Consistent SI units throughout.
- For φ = 0 (pure clay, undrained): Nc = 5.7, Nq = 1, Nγ = 0 (Terzaghi values).
- The formula applies to SHALLOW footings only: Df/B ≤ 1 is the classic Terzaghi criterion.
Bearing Capacity Factors Nc, Nq, and Nγ
The bearing-capacity factors are pure numbers that quantify how effectively the soil's friction angle amplifies each component of resistance. They depend solely on the internal friction angle φ and are obtained from tables or charts derived from the theoretical analysis. **Terzaghi's Original Factors (General Shear):** For φ = 0°: Nc = 5.7, Nq = 1.0, Nγ = 0.0 For φ = 10°: Nc = 9.61, Nq = 2.69, Nγ = 1.22 For φ = 20°: Nc = 17.69, Nq = 7.44, Nγ = 3.64 For φ = 25°: Nc = 25.13, Nq = 12.72, Nγ = 8.34 For φ = 30°: Nc = 37.16, Nq = 22.46, Nγ = 19.13 For φ = 35°: Nc = 57.75, Nq = 41.44, Nγ = 42.92 For φ = 40°: Nc = 95.67, Nq = 81.27, Nγ = 100.40 **Critical observations:** - All three factors INCREASE with φ — higher friction angle → larger bearing capacity. - Nγ = 0 when φ = 0 — the width term contributes nothing in pure clay under undrained conditions. - Nq = 1 when φ = 0 — the surcharge term contributes exactly q (no amplification). - The factors grow rapidly beyond φ = 30°, so the bearing capacity of dense sands can be very high. - Nc and Nq are related: Nq = eπ tan φ tan²(45 + φ/2) [Reissner, 1924]; Nc = (Nq − 1) cot φ [Prandtl]. - Nγ has multiple formulations; the Board exam will always provide the specific values to use. **Terzaghi's Local Shear Factors (N'c, N'q, N'γ):** These are Terzaghi's standard factors evaluated at the reduced angle φ* = arctan[(2/3)tan φ]. For φ = 30° → φ* ≈ 20.9° → use factors at ≈ 21°.
Examples
This exercise demonstrates that the formula gives Meyerhof-type factors, not Terzaghi's. The exam provides the specific values — use them as given without questioning.
Scenario
Given φ = 30°, verify the Nq value using the Reissner formula: Nq = e^(π tan φ) · tan²(45 + φ/2).
Solution
tan 30° = 0.5774 eπ·tan30° = e^(π × 0.5774) = e^1.8138 = 6.133 tan²(45 + 15) = tan²(60°) = (1.7321)² = 3.000 Nq = 6.133 × 3.000 = 18.40 Note: Terzaghi's tabulated value is 22.46 — the discrepancy arises because Terzaghi used a slightly different failure surface geometry. The Board exam provides Nq = 22.46 for φ = 30°; always use the given value.
Applications
- Reading bearing-capacity factor tables from NSCP 2015 Appendix or geotechnical references in practice.
- Interpolating between table entries for non-standard φ values (e.g., φ = 27°).
- Sensitivity analysis: showing how much qu changes if the site investigation gives a φ range of ±2°.
Misconceptions
- Assuming Nc, Nq, Nγ values are the same for all bearing-capacity methods — Terzaghi, Meyerhof, and Hansen each have different factor tables.
- Interpolating Nγ linearly when it grows exponentially with φ — use the given table values and interpolate only when needed.
Related Concepts
- Internal Friction Angle φ from Lab/Field Tests
- Terzaghi's Bearing Capacity Equation
- Local Shear Failure and Reduced Parameters
Common Exam Questions
Example
Using Nc = 37.16, Nq = 22.46, Nγ = 19.13 (φ = 30°), compute qu for a 1.5 m strip at Df = 1.2 m, c = 10 kPa, γ = 17 kN/m³.
Approach
Problem provides Nc, Nq, Nγ directly. Simply plug into the correct equation form.
Question Type
Table look-up then computation
Key Points To Remember
- The Board exam ALWAYS provides the factor values in the problem — you do not need to derive them from formulas.
- φ = 0 special case: Nc = 5.7, Nq = 1, Nγ = 0 — memorize these three values.
- Factors increase sharply with φ; a 5° error in φ can change qu by 40–60%.
- For local shear, first compute φ*, then use the factor values at φ* (not at the original φ).
- Never mix Terzaghi factors with Meyerhof/Hansen shape and depth factors — use a consistent method.
Water-Table Correction
The unit weight γ that appears in both the surcharge term (q = γDf) and the width term (½γBNγ) must reflect whether the soil is submerged. Submergence reduces the effective unit weight from γ to γ' = γsat − γw, where γw = 9.81 kN/m³. The correction depends on the position of the water table relative to the footing base. **Case 1 — Water table at or above the ground surface (or at the base of the footing):** - Use γ' = γsat − γw in the ½γBNγ term. - Also use γ' in computing q = γ'Df if the water table is above the footing base. - This gives the most conservative (lowest) bearing capacity. **Case 2 — Water table at the base of the footing (depth Df):** - Surcharge: q = γDf (water table is AT the base, so the soil above is still moist; use γ for the upper Df). - Width term: use γ' = γsat − γw for the zone from the base downward. - qu = cNc + γDf·Nq + ½γ'BNγ **Case 3 — Water table at depth d below the footing base (0 ≤ d ≤ B):** - Interpolate the effective unit weight: γ_eff = γ' + (d/B)(γ − γ') [linear interpolation between γ' and γ] - Use γ_eff in the width term: ½γ_eff·B·Nγ **Case 4 — Water table deeper than B below the footing base:** - No correction needed; use γ throughout. **Practical note:** On the Board exam, water-table problems almost always place the table either at the surface or at the footing base (Cases 1 and 2) — the simplest corrections. Interpolation (Case 3) appears less frequently.
Examples
The main effect of the water table at the footing base is in the width (Nγ) term — γ drops from 18 to 10.19 kN/m³. The surcharge term actually increased slightly here because γsat (20) > γ (18). This counter-intuitive result is possible when γsat > γ — always compute both cases.
Scenario
BOARD-STYLE PROBLEM: Repeat the Example 1 strip footing (B = 2 m, Df = 1 m, c = 15 kPa, φ = 25°, γsat = 20 kN/m³, γw = 9.81 kN/m³, Nc = 25.13, Nq = 12.72, Nγ = 8.34) with the water table at the base of the footing.
Solution
Step 1 — Effective (submerged) unit weight: γ' = γsat − γw = 20 − 9.81 = 10.19 kN/m³ Step 2 — Surcharge (soil above footing base is moist, water table AT base): q = γsat × Df = 20 × 1 = 20 kPa (Conservative: use γsat above the water table if the soil is saturated throughout; use γmoist if the problem distinguishes — here γsat = 20 kN/m³ is used.) Step 3 — Strip footing equation with γ' in the width term: qu = cNc + qNq + (1/2)γ'BNγ qu = 15(25.13) + 20(12.72) + 0.5(10.19)(2)(8.34) qu = 376.95 + 254.40 + 85.19 qu = 716.5 kPa Compare with the dry case: 756.0 kPa — the water table reduced qu by ~39.5 kPa (~5%).
Applications
- Foundation design in coastal Metro Manila and in flood-prone areas of Central Luzon where seasonal water tables fluctuate significantly.
- Design of footings for bridge abutments near river banks (e.g., Pasig River crossings) where the water table is near the ground surface year-round.
- Checking whether a footing designed in the dry season remains safe during the wet season when the water table rises.
Misconceptions
- Using γsat everywhere when the water table is at the footing base — only the width term should use γ'.
- Using γw = 10 kN/m³ when the problem does not specify — always use 9.81 kN/m³ unless told otherwise.
- Forgetting to apply the correction at all when the water table is between the surface and depth B — this is a common omission.
Related Concepts
- Effective Stress Principle
- Submerged (Buoyant) Unit Weight
- Terzaghi's Bearing Capacity Equation
Common Exam Questions
Example
Strip footing, B = 1.5 m, Df = 1 m, water table at ground surface, γsat = 19 kN/m³ → both q and width term use γ' = 9.19 kN/m³.
Approach
Determine which γ applies to the surcharge term and which to the width term based on water-table position. Compute γ' = γsat − 9.81.
Question Type
Compute qu with water table at a specified location
Key Points To Remember
- γ' = γsat − γw ≈ γsat − 9.81 kN/m³ (use 9.81, not 10, unless the problem specifies).
- Water table at ground surface or above footing base: both q and the width term use γ'.
- Water table exactly at footing base: q uses dry/moist γ above; width term uses γ' below.
- Water table more than B below footing base: NO correction; use natural γ.
- Submergence REDUCES qu — so ignoring a high water table is unconservative (dangerous).
Allowable Bearing Capacity and Net Bearing Capacity
The ultimate bearing capacity qu is the theoretical maximum pressure the soil can sustain before shear failure. In design, we must apply a factor of safety (FS) to account for uncertainties in soil properties, loading, and the model itself. **Gross Allowable Bearing Capacity:** $$q_a = \frac{q_u}{FS}$$ This is the maximum gross pressure (including the weight of the footing and overburden soil) that may be applied at the base of the footing. **Net Ultimate Bearing Capacity:** $$q_{u,net} = q_u - \gamma D_f$$ The net value represents the INCREASE in pressure at the footing base beyond the original overburden. Before the footing was constructed, the soil already carried the overburden pressure γDf. The net capacity measures only the additional capacity available for structural loading. **Net Allowable Bearing Capacity:** $$q_{a,net} = \frac{q_{u,net}}{FS} = \frac{q_u - \gamma D_f}{FS}$$ **Which to use?** - The Board exam usually asks for the **allowable column load Q**: use q_{a,net} × footing area = Q. - If asked for the **allowable soil pressure**: use q_a,net + γDf (gross). - The key distinction: the column load P is a NET additional force on the soil; the footing weight and overburden soil weight are sometimes added separately. **Typical FS values (per NSCP 2015 and geotechnical practice):** - FS = 3.0: normal conditions, uncertainties in soil data - FS = 2.5: well-characterized sites, large structures with thorough investigation - FS = 2.0: temporary structures or when settlements are controlled separately **Settlement vs. Shear Failure:** For many soils (especially clays), the allowable pressure is governed by the SETTLEMENT limit, not the shear failure. A footing may have a bearing capacity of 300 kPa but be limited to 75 kPa to keep settlement under 25 mm. The Board exam may present this as a 'which governs' question.
Examples
The 494 kN is the maximum column load the footing can safely support (in addition to the weight of the footing itself if that is separately accounted for). The net basis correctly credits the pre-existing overburden so we do not double-count it.
Scenario
BOARD-STYLE PROBLEM: A 2 m × 2 m square footing at Df = 1 m on clay (cu = 50 kPa, γ = 18 kN/m³, φ = 0) gives qu = 388.5 kPa (from the earlier example). Find: (a) net ultimate bearing capacity, (b) net allowable bearing capacity with FS = 3, and (c) the allowable column load.
Solution
(a) Net ultimate: qu,net = qu − γDf = 388.5 − 18(1) = 370.5 kPa (b) Net allowable: qa,net = qu,net / FS = 370.5 / 3 = 123.5 kPa (c) Allowable column load: Q = qa,net × A = 123.5 × (2 × 2) = 123.5 × 4 = 494 kN
Always round up to ensure the footing is adequate. A 3.0 m × 3.0 m footing with qa,net = 100 kPa can carry 900 kN > 800 kN — safe.
Scenario
A footing must support a column load of 800 kN. Using qa,net = 100 kPa, find the required square footing dimension B.
Solution
Q = qa,net × B² 800 = 100 × B² B² = 8 m² B = √8 = 2.83 m → Use B = 3.0 m (round up to the next practical size)
Applications
- Sizing all isolated footings for columns in reinforced concrete buildings per NSCP 2015.
- Checking adequacy of existing footings when a building undergoes additional loading (renovation or added floors).
- Reporting 'allowable bearing capacity' in geotechnical investigation reports for DPWH-funded infrastructure projects.
Misconceptions
- Dividing qu,net by the GROSS applied pressure (P/A + γDf) instead of the net applied pressure — this underestimates FS.
- Forgetting to subtract footing weight from Q when the problem asks for the NET column load.
- Using FS = 2 for permanent structures without justification — standard practice requires FS ≥ 2.5 to 3.
Related Concepts
- Terzaghi's Bearing Capacity Equation
- Settlement Analysis
- Factor of Safety in Foundation Design
Common Exam Questions
Example
3 m square, Df = 1.5 m, qu = 600 kPa, γ = 18 kN/m³, FS = 3 → Q = [(600 − 27)/3] × 9 = 191 × 9 = 1719 kN
Approach
Compute qu using the appropriate equation → subtract γDf for net → divide by FS → multiply by footing area.
Question Type
Find the allowable column load
Example
P = 500 kN, B = 2 m, Df = 1 m, qu = 400 kPa, γ = 17 kN/m³ → net qu = 383 kPa; applied net = 500/4 = 125 kPa; FS = 383/125 = 3.06
Approach
Compute applied pressure = P/A → compute qu,net → FS = qu,net / applied net pressure.
Question Type
Back-calculate FS given the applied load and footing size
Key Points To Remember
- qa = qu/FS (gross); qa,net = (qu − γDf)/FS (net).
- Allowable column load Q = qa,net × B² (for square footing of width B).
- FS is typically 3 for normal conditions in the Philippines.
- Net capacity is ALWAYS less than gross capacity since we subtract γDf.
- Settlement may control over bearing capacity — especially for clays and silts.
Practice Problems
Note how the Nq and Nγ terms dominate for φ = 30° (dense sand) while the cohesion term contributes only 371.6 kPa out of 1073.9 kPa total. For clean sand (c = 0), qu would be 702.3 kPa — still substantial due to friction. The net allowable of 351 kPa is typical of dense granular soil, confirming that footings on dense sand can safely carry very high column loads.
Problem
PROBLEM 1 (Strip Footing — φ soil): A strip footing B = 1.5 m is placed at Df = 1.2 m in a soil with c = 10 kPa, φ = 30°, γ = 17 kN/m³. Terzaghi bearing-capacity factors at φ = 30°: Nc = 37.16, Nq = 22.46, Nγ = 19.13. Determine: (a) the ultimate bearing capacity, and (b) the net allowable bearing capacity with FS = 3.
Solution
Step 1 — Surcharge: q = γDf = 17 × 1.2 = 20.4 kPa Step 2 — Ultimate bearing capacity (strip): qu = cNc + qNq + (1/2)γBNγ qu = 10(37.16) + 20.4(22.46) + 0.5(17)(1.5)(19.13) qu = 371.60 + 458.18 + 244.16 qu = 1,073.9 kPa Step 3 — Net ultimate: qu,net = 1073.9 − 17(1.2) = 1073.9 − 20.4 = 1053.5 kPa Step 4 — Net allowable: qa,net = 1053.5 / 3 = 351.2 kPa
For φ = 0 clay, only the cNc and qNq terms survive. The 1.3 shape factor for the square footing increases the cohesion contribution by 30% compared to a strip. The net ultimate capacity 555.75 kPa exactly equals 1.3 × 75 × 5.7 — the surcharge (27 kPa) cancels out in the net computation, which is mathematically consistent.
Problem
PROBLEM 2 (Square Footing on Clay — φ = 0): A 2.5 m square footing at Df = 1.5 m supports a column on saturated clay with cu = 75 kPa, γ = 18 kN/m³. Using Nc = 5.7, Nq = 1, Nγ = 0 (Terzaghi, φ = 0), and FS = 3, find: (a) qu, (b) qu,net, (c) qa,net, and (d) the maximum allowable column load.
Solution
Step 1 — Surcharge: q = 18 × 1.5 = 27 kPa Step 2 — Square footing on φ = 0 clay: qu = 1.3cNc + qNq + 0.4γBNγ qu = 1.3(75)(5.7) + 27(1) + 0 qu = 555.75 + 27 + 0 qu = 582.75 kPa Step 3 — Net ultimate: qu,net = 582.75 − 27 = 555.75 kPa Step 4 — Net allowable: qa,net = 555.75 / 3 = 185.25 kPa Step 5 — Allowable column load: Q = qa,net × A = 185.25 × (2.5)² = 185.25 × 6.25 = 1,157.8 kN ≈ 1,158 kN
The water table at the footing base affects ONLY the width term (½γBNγ). The surcharge term is unchanged because the soil above the footing base remains moist. This is the most common water-table scenario on Board exams — the two-line answer: q stays the same; replace γ with γ' in the width term only.
Problem
PROBLEM 3 (Water Table Effect): A strip footing B = 2 m, Df = 1.5 m rests on sand with c = 0, φ = 30°, γmoist = 17 kN/m³, γsat = 20 kN/m³. Nc = 37.16, Nq = 22.46, Nγ = 19.13. Compute qu for two cases: (a) water table well below the footing (no correction), and (b) water table at the base of the footing. Take γw = 9.81 kN/m³.
Solution
CASE (a) — No water table correction: q = 17 × 1.5 = 25.5 kPa qu = 0 + 25.5(22.46) + 0.5(17)(2)(19.13) qu = 572.73 + 325.21 qu = 897.9 kPa CASE (b) — Water table at footing base: Surcharge (soil above base is moist): q = γmoist × Df = 17 × 1.5 = 25.5 kPa (same) Effective unit weight below footing: γ' = 20 − 9.81 = 10.19 kN/m³ qu = 0 + 25.5(22.46) + 0.5(10.19)(2)(19.13) qu = 572.73 + 195.12 qu = 767.9 kPa Reduction due to water table: 897.9 − 767.9 = 130.0 kPa (about 14.5% reduction)
This problem illustrates why it is critical to correctly identify the failure mode. Using general shear factors on loose sand would overestimate qu by nearly 3×, a catastrophic design error. Always read the problem carefully: 'loose sand,' 'soft clay,' or 'local shear' are keywords for applying the 2/3 reduction.
Problem
PROBLEM 4 (Local Shear Failure): A 1.8 m strip footing at Df = 1.0 m rests on loose, fine sand with c = 0, φ = 30°, γ = 16 kN/m³. Terzaghi factors at φ = 21°: Nc = 18.92, Nq = 8.26, Nγ = 5.84. Determine qu using local shear parameters.
Solution
Step 1 — Reduce parameters for local shear: c* = (2/3)(0) = 0 kPa tan φ* = (2/3) tan 30° = (2/3)(0.5774) = 0.3849 → φ* = arctan(0.3849) ≈ 21° Step 2 — Surcharge: q = 16 × 1.0 = 16 kPa Step 3 — Strip footing using local shear factors (at φ* = 21°): qu = c*Nc + qNq + (1/2)γBNγ qu = 0(18.92) + 16(8.26) + 0.5(16)(1.8)(5.84) qu = 0 + 132.16 + 84.10 qu = 216.3 kPa For comparison, general shear at φ = 30°: qu = 0 + 16(22.46) + 0.5(16)(1.8)(19.13) = 359.36 + 275.47 = 634.8 kPa Local shear qu is only 34% of the general shear value — a dramatic reduction for loose sand.
Back-calculation problems require solving a nonlinear equation in B — iteration (trial and error) is the standard approach. Always verify your answer at the end. Note that rounding up B from 1.55 to 1.6 m provides a small additional reserve (1278 kN vs. 1200 kN required).
Problem
PROBLEM 5 (Size a Footing — Back-Calculation): A column carries a service load of 1,200 kN. The footing will be square at Df = 1.5 m in soil with c = 0, φ = 35°, γ = 17 kN/m³. Terzaghi factors at φ = 35°: Nc = 57.75, Nq = 41.44, Nγ = 42.92. Using FS = 3, determine the required footing width B.
Solution
Step 1 — Express qu in terms of B (square footing, c = 0): q = 17 × 1.5 = 25.5 kPa qu = 0 + 25.5(41.44) + 0.4(17)(B)(42.92) qu = 1056.72 + 291.86B Step 2 — Net allowable: qa,net = (qu − γDf) / FS = (1056.72 + 291.86B − 25.5) / 3 qa,net = (1031.22 + 291.86B) / 3 Step 3 — Applied net pressure: Applied net pressure = P / B² = 1200 / B² Step 4 — Set demand ≤ capacity: 1200 / B² = (1031.22 + 291.86B) / 3 3600 = B²(1031.22 + 291.86B) 3600 = 1031.22B² + 291.86B³ Step 5 — Solve by trial (iterative): Try B = 1.5 m: 1031.22(2.25) + 291.86(3.375) = 2320.2 + 985.0 = 3305.2 (< 3600 — too small) Try B = 1.6 m: 1031.22(2.56) + 291.86(4.096) = 2639.9 + 1195.5 = 3835.4 (> 3600 — OK) Try B = 1.55 m: 1031.22(2.4025) + 291.86(3.724) = 2477.6 + 1087.0 = 3564.6 (≈ 3600 — very close) Use B = 1.6 m (round up for safety) → adopt B = 1.6 m square footing. Verification at B = 1.6 m: qu = 1056.72 + 291.86(1.6) = 1056.72 + 467.0 = 1523.7 kPa qu,net = 1523.7 − 25.5 = 1498.2 kPa qa,net = 1498.2 / 3 = 499.4 kPa Capacity = 499.4 × (1.6)² = 499.4 × 2.56 = 1278.5 kN > 1200 kN ✓
Exam Preparation Tips
- MEMORIZE THE THREE FORMULA FORMS: Strip (1.0, 0.5), Square (1.3, 0.4), Circular (1.3, 0.3) for the (cNc, γBNγ) coefficients. The Nq term always has coefficient 1.0 in Terzaghi's method. This alone will solve 60% of bearing-capacity Board problems.
- ALWAYS COMPUTE q = γDf FIRST: The surcharge term is the most frequently omitted in rushed exam conditions. Write it as the very first step of every bearing-capacity solution.
- KNOW THE φ = 0 SPECIAL CASE BY HEART: Nc = 5.7, Nq = 1, Nγ = 0. For a square footing on clay: qu = 1.3(c)(5.7) + γDf = 7.41c + γDf. Recognizing this immediately saves 2–3 minutes per item.
- WATER TABLE RULE-OF-THUMB: Water table at or above footing base → replace γ with γ' = γsat − 9.81 in the width term. Water table more than B below → no change. These two extremes cover most exam scenarios.
- DISTINGUISH NET vs. GROSS: When asked for 'allowable column load,' use qa,net × Area. When asked for 'allowable bearing pressure at the footing base,' use qa,gross = qu/FS (gross).
- LOCAL SHEAR TRIGGER WORDS: 'loose sand,' 'soft clay,' 'local shear failure,' 'punching shear' → reduce c by 2/3 and tan φ by 2/3. NEVER reduce φ directly.
- CHECK UNITS THROUGHOUT: c in kPa, γ in kN/m³, B in m → qu in kPa. If γ is given in kN/m³ and B in m, the formula is dimensionally consistent. A result of 700 kPa for dense sand or 350 kPa for clay is in the right ballpark — sanity-check your answer.
- USE THE FACTOR VALUES PROVIDED IN THE PROBLEM: Board exam problems always supply Nc, Nq, Nγ. Do not use your own table values — use exactly what the problem states.
- FACTOR OF SAFETY CLARITY: FS = 3 is the default for the Board exam unless stated otherwise. Some problems give FS = 2.5 for 'carefully characterized sites' — read carefully.
- PRACTICE BACK-CALCULATION PROBLEMS: At least one Board item per examination requires finding B given P and qa,net. Set up the cubic equation and solve by trial. Practice three iterations from a reasonable starting guess (B = 1.5 to 2.0 m is usually a good start for typical column loads of 500–2000 kN).
- SETTLEMENT GOVERNS STATEMENT: When the problem states that settlement governs, report the settlement-limited allowable pressure as the controlling value — do not use the higher bearing-capacity-based value.
- DEPTH REQUIREMENT (Terzaghi Shallow Foundation): Terzaghi's equation applies for Df/B ≤ 1 (approximately). For deeper foundations, Meyerhof or Hansen depth factors should be used — but the Board exam at this level stays within Terzaghi's range.
In summary
The bearing capacity of soils — built on Terzaghi's three-term equation — is one of the most algebraically straightforward yet conceptually rich topics in the PRC Civil Engineer Licensure Examination. Mastery requires four things: (1) knowing the correct form of the equation for each footing shape by heart; (2) applying the surcharge, water-table, and local-shear corrections accurately; (3) converting ultimate capacity to allowable pressure and then to column load using the net basis; and (4) recognizing when settlement, not shear, governs the design. The five practice problems in this chapter span every scenario that appears on the Board exam: strip and square footings, pure clay (φ = 0), dense and loose sands, water-table correction, and back-calculation of footing size. Work through all of them with pencil and paper, checking every intermediate value. The flowchart and mind-map visual aids provide a rapid review tool for the exam morning. Under the practice of civil engineering in the Philippines (RA 544), every licensed engineer is expected to apply these principles correctly in the field — not just answer multiple-choice questions. Understand the physics: soil fails because the applied pressure exceeds the combined resistance of cohesion, confinement from depth, and the frictional weight of the failure wedge. Keep that physical picture in mind, and the equations will always make sense.
Previous chapter
Lateral Earth Pressure and Retaining Structures
Next chapter
Foundations (Shallow and Deep)
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.