CELE Geotechnical Engineering — Bearing Capacity of SoilsMisconception Buster
Misconception buster for Bearing Capacity of Soils. Every concept has a shadow — the subtly wrong version that looks right on first glance. Professional Regulation Commission (PRC) — Board of Civil Engineering builds CELE questions around those shadows. This page shows you the truth behind the traps.
Exam context
The Civil Engineer Licensure Examination is conducted by Professional Regulation Commission (PRC) — Board of Civil Engineering and is scheduled for May and November 2026. The Geotechnical Engineering subtest is marked as "Core" in the official pattern, and Bearing Capacity of Soils appears in position 9th of 11 in the CELE Geotechnical Engineering review rotation. Passing mark: 70% weighted average, no sub-test below 50%. Recent CELE 2026 papers have drawn roughly a meaningful share of questions from this subject.
Bearing Capacity of Soils - Misconception Buster
In the PRC Civil Engineer Licensure Examination, Geotechnical Engineering consistently yields some of the most avoidable errors. Bearing capacity problems appear every board exam cycle, yet reviewees repeatedly lose points on the same traps: omitting the surcharge term, misapplying shape factors, confusing gross with net bearing capacity, and ignoring water-table corrections. This guide targets those exact failure points. Understanding why you might be wrong is more powerful than memorizing formulas — it forces you to build correct mental models that survive the pressure of an actual exam. Work through every trap question honestly before reading the answer. If you get it wrong, that misconception was costing you marks.
Summary
The eight critical habits that separate high-scorers from average board exam takers in bearing capacity problems are: (1) Always distinguish gross from net allowable capacity — column load problems use the net form. (2) Never apply the 0.5 shape factor universally — strip = 0.5, square = 0.4, circular = 0.3 in front of γBN_γ. (3) Apply water-table corrections only to the affected term — γ' goes in the width term, not the surcharge term (unless the WT is above ground). (4) At φ = 0, use N_c = 5.7, N_q = 1, N_γ = 0 — never set all factors to zero. (5) For local shear in loose soils, reduce c and tan(φ) to 2/3 before finding N-factors — this is not just an FS adjustment. (6) Never omit the surcharge term qN_q — it is often the largest term for frictional soils. (7) Always divide q_u by FS (2.5–3) before comparing to applied pressure. (8) Bearing capacity adequacy does not eliminate the settlement check — both must always be performed. Master these eight rules and you eliminate the most common causes of wrong answers in Geotechnical Engineering bearing capacity problems on the PRC CE Licensure Examination.
Misconceptions
The allowable bearing capacity is simply q_u divided by FS — no further adjustment is needed.
Tags
- critical_error
- formula_confusion
- exam_trap
- net_vs_gross
Topic
Allowable Bearing Capacity — Gross vs Net
Severity
critical
Exam Impact
Board problems that ask for 'allowable column load' or 'net safe bearing capacity' require the net form. Using gross gives a numerically close but wrong answer that matches a distractor choice.
The Reality
There are TWO forms of allowable bearing capacity — gross and net. The GROSS allowable is q_a = q_u / FS. The NET allowable is q_a,net = (q_u − γD_f) / FS. The overburden pressure γD_f existed before the footing was constructed; it is replaced, not added, by the footing load. When a problem asks for the allowable COLUMN LOAD or NET safe load, you must use the net form. Using gross overestimates the safe load the soil can carry above its pre-existing state. In Example 2 of the reference: q_u = 388.5 kPa, γD_f = 18 kPa → q_u,net = 370.5 kPa → q_a,net = 123.5 kPa. The gross q_a = 129.5 kPa — a 5% overestimate that can tip a design into unsafe territory.
Trap Question
Question
A 2 m × 2 m square footing on saturated clay (c_u = 50 kPa, γ = 18 kN/m³) is placed at D_f = 1 m. Using Terzaghi factors (N_c = 5.7, N_q = 1, N_γ = 0) and FS = 3, what is the ALLOWABLE COLUMN LOAD the footing can carry?
Explanation
The 18 kPa overburden pressure was already present in the soil before construction. The column load is an ADDITIONAL stress above this pre-existing state. Therefore, only the net ultimate capacity (q_u − γD_f) is divided by FS to get the allowable NET pressure, which is then multiplied by area to get the safe column load.
Wrong Answer
518 kN (using q_a = q_u / FS = 388.5 / 3 = 129.5 kPa × 4 m²)
Correct Answer
494 kN (using q_a,net = (388.5 − 18) / 3 = 123.5 kPa × 4 m²)
Misconception Id
M1
Correct Vs Incorrect
Correct Approach
q_a,net = (q_u − γD_f) / FS = (388.5 − 18) / 3 = 123.5 kPa → Q = 123.5 × 4 = 494 kN (CORRECT)
Incorrect Approach
q_a = q_u / FS = 388.5 / 3 = 129.5 kPa → Q = 129.5 × 4 = 518 kN (WRONG — includes overburden already in place)
Why Students Believe It
The formula q_a = q_u / FS is the first thing taught and the easiest to remember. Most reviewers stop there. Students assume that dividing by a factor of safety already accounts for everything, including the weight of soil already present at the footing level.
The shape factor for the γBN_γ term is 0.5 for ALL footing shapes (strip, square, circular).
Tags
- formula_confusion
- shape_factor
- critical_error
- memorization_trap
Topic
Shape Factors — Terzaghi's Equations
Severity
critical
Exam Impact
Square and circular footings are the most common board-exam footing types. Applying 0.5 instead of 0.4 or 0.3 gives a higher (unconservative) q_u and a wrong numerical answer that matches a specific wrong-choice distractor.
The Reality
The coefficient in front of γBN_γ IS the shape factor and it varies by footing geometry: Strip = 0.5, Square = 0.4, Circular = 0.3. The full equations are: Strip: q_u = cN_c + qN_q + 0.5γBN_γ; Square: q_u = 1.3cN_c + qN_q + 0.4γBN_γ; Circular: q_u = 1.3cN_c + qN_q + 0.3γBN_γ. Using 0.5 for a square footing OVERESTIMATES the width contribution to bearing capacity by 25%.
Trap Question
Question
A 1.5 m square footing is placed at D_f = 1 m on c–φ soil: c = 20 kPa, φ = 30°, γ = 18 kN/m³. Terzaghi factors: N_c = 37.16, N_q = 22.46, N_γ = 19.13. Compute q_u.
Explanation
The 0.5 is NOT a universal constant — it is the shape factor specific to a strip footing. For a square footing, Terzaghi's equation explicitly writes 0.4 in front of γBN_γ. The difference here is about 51 kPa, which is large enough to select the wrong distractor choice on a board exam.
Wrong Answer
q_u = 1.3(20)(37.16) + 18(1)(22.46) + 0.5(18)(1.5)(19.13) = 966.16 + 404.28 + 258.26 = 1628.7 kPa (WRONG — used 0.5)
Correct Answer
q_u = 1.3(20)(37.16) + 18(1)(22.46) + 0.4(18)(1.5)(19.13) = 966.16 + 404.28 + 206.60 = 1577.0 kPa (CORRECT — used 0.4 for square)
Misconception Id
M2
Correct Vs Incorrect
Correct Approach
Square footing: q_u = 1.3cN_c + qN_q + 0.4γBN_γ — CORRECT. Strip: 0.5; Square: 0.4; Circular: 0.3. Memorize: 'S-S-C → 5-4-3' (Strip-Square-Circular → 0.5-0.4-0.3).
Incorrect Approach
Square footing: q_u = 1.3cN_c + qN_q + 0.5γBN_γ — WRONG. Using 0.5 for the width term of a square footing ignores the shape correction.
Why Students Believe It
Terzaghi's strip footing equation uses the coefficient 1/2 (= 0.5) in front of γBN_γ. Students memorize '0.5γBN_γ' and apply it universally, forgetting that the 0.5 is the strip-footing shape factor. The shape factor is embedded in the constant, not written separately.
When the water table is at the ground surface (or at footing level), simply replace γ with γ_sat everywhere in the formula.
Tags
- water_table
- unit_weight
- critical_error
- conceptual_gap
Topic
Water-Table Correction
Severity
critical
Exam Impact
Using γ_sat instead of γ' in the width term gives a q_u that is roughly 50% too high for submerged conditions. Water-table correction problems appear regularly in PRC board exams.
The Reality
In Terzaghi's equation q_u = cN_c + qN_q + ½γBN_γ: the term q = γD_f represents the overburden pressure AT the footing base; the ½γBN_γ term governs soil resistance in the FAILURE WEDGE below the footing. The water-table correction rules are: (1) Water table at or ABOVE footing base → use γ' (= γ_sat − 9.81) in the ½γBN_γ term; use γ_sat (or mixed) in computing q = γD_f depending on water table position relative to ground. (2) Water table more than B below footing base → no correction. (3) Water table between footing base and B below → linear interpolation. NEVER replace all γ with γ_sat — the terms are physically distinct.
Trap Question
Question
A strip footing B = 2 m, D_f = 1 m, c = 15 kPa, φ = 25° (N_c = 25.13, N_q = 12.72, N_γ = 8.34). Soil: γ = 18 kN/m³ above water table, γ_sat = 20 kN/m³. The water table rises to the footing base. Compute q_u.
Explanation
The width term ½γBN_γ captures resistance of soil in the failure wedge BELOW the footing. If the water table is at the footing base, this soil is submerged → use γ' = γ_sat − γ_w. The surcharge q uses the unit weight of soil ABOVE the water table (which is dry or moist, not submerged). Using γ_sat everywhere overestimates q_u by about 15% in this case.
Wrong Answer
q_u = 15(25.13) + (20×1)(12.72) + ½(20)(2)(8.34) = 376.95 + 254.40 + 166.80 = 798.2 kPa (WRONG — used γ_sat = 20 in width term)
Correct Answer
γ' = 20 − 9.81 = 10.19 kN/m³; q = 18(1) = 18 kPa (soil above WT is unsaturated). q_u = 15(25.13) + 18(12.72) + ½(10.19)(2)(8.34) = 376.95 + 228.96 + 85.09 = 691.0 kPa (CORRECT)
Misconception Id
M3
Correct Vs Incorrect
Correct Approach
γ' = 20 − 9.81 = 10.19 kN/m³. For q: water table at footing base, so q = γD_f uses the soil ABOVE the water table. Width term uses γ' = 10.19. q_u = cN_c + qN_q + ½(10.19)BN_γ — CORRECT.
Incorrect Approach
Water table at footing base, γ_sat = 20 kN/m³: q_u = cN_c + (20×1)N_q + ½(20)BN_γ — WRONG. Using γ_sat = 20 in the width term ignores buoyancy in the failure zone.
Why Students Believe It
Students think 'wet soil = saturated unit weight throughout.' They forget that the γ in different terms of the Terzaghi equation represents DIFFERENT physical quantities acting at DIFFERENT locations, so the water-table correction is applied selectively, not globally.
For purely cohesive soil (φ = 0 undrained), all three bearing-capacity factors (N_c, N_q, N_γ) are zero because φ = 0.
Tags
- phi_equals_zero
- N_factors
- critical_error
- conceptual_gap
Topic
Bearing-Capacity Factors for φ = 0 Clay
Severity
critical
Exam Impact
Setting N_c = 0 at φ = 0 makes q_u = γD_f only (just the overburden) — a catastrophically wrong and dangerously low answer. This is an immediate full-mark loss on any φ = 0 bearing capacity problem.
The Reality
At φ = 0: N_c = 5.7 (Terzaghi) or 5.14 (Meyerhof/general), N_q = 1, N_γ = 0. N_c is NOT zero — it represents the contribution of COHESION to bearing capacity, which is maximized at φ = 0 in a specific sense. Only N_γ = 0 (no frictional wedge resistance). The equation for a strip footing on φ = 0 clay becomes: q_u = 5.7c + q(1) + 0 = 5.7c_u + γD_f. This is why soft clay has measurable bearing capacity — entirely from cohesion and the surcharge.
Trap Question
Question
A 2 m × 2 m square footing sits on saturated clay: c_u = 60 kPa, φ = 0, γ = 17 kN/m³, D_f = 1.2 m. Using Terzaghi's method, compute q_u.
Explanation
N_c = 5.7 is a FIXED VALUE when φ = 0, not zero. It represents the maximum bearing contribution from cohesion alone. Only N_γ = 0 because there is no frictional failure wedge. The surcharge term (N_q = 1) is also retained. This is one of the most frequently tested scenarios in PRC board exams on bearing capacity.
Wrong Answer
Since φ = 0, all N factors = 0 → q_u = 0 kPa. (COMPLETELY WRONG)
Correct Answer
N_c = 5.7, N_q = 1, N_γ = 0 for φ = 0. q = 17(1.2) = 20.4 kPa. Square: q_u = 1.3(60)(5.7) + 20.4(1) + 0 = 444.6 + 20.4 = 465.0 kPa.
Misconception Id
M4
Correct Vs Incorrect
Correct Approach
φ = 0 → N_c = 5.7, N_q = 1, N_γ = 0 (Terzaghi). Strip: q_u = 5.7c_u + γD_f. Square: q_u = 1.3(5.7)c_u + γD_f = 7.41c_u + γD_f.
Incorrect Approach
φ = 0 → N_c = 0, N_q = 0, N_γ = 0 → q_u = 0 + 0 + 0 = 0. (COMPLETELY WRONG — soil would have zero bearing capacity, which is physically impossible for cohesive soil.)
Why Students Believe It
Students know that bearing-capacity factors are functions of φ. When φ = 0, they assume N_c = 0 as well (since 'there is no friction'). This is wrong — cohesion still provides bearing capacity even when φ = 0.
Local shear failure just means using a smaller factor of safety — no change to c and φ is needed.
Tags
- local_shear
- two_thirds_reduction
- major_error
- formula_confusion
Topic
Local vs General Shear Failure
Severity
major
Exam Impact
Board problems that specify 'loose sand' or 'local shear' require the 2/3 reduction. Using full parameters overestimates q_u and gives a wrong final answer.
The Reality
For local shear failure (loose/compressible soils), Terzaghi requires reducing the shear strength parameters BEFORE entering the bearing-capacity equations: c* = (2/3)c and tan(φ*) = (2/3)tan(φ) → φ* = arctan(0.667 tan φ). These reduced parameters are used to look up MODIFIED bearing-capacity factors N'_c, N'_q, N'_γ (which are lower than the general shear factors). The factor of safety is still applied afterward. Simply reducing FS while using full c and φ is physically incorrect — it does not reflect the less-developed failure surface in loose soils.
Trap Question
Question
A strip footing on loose sand (φ = 30°, c = 0, γ = 16 kN/m³, B = 1.5 m, D_f = 1 m) is expected to undergo local shear failure. Which of the following correctly begins the solution?
Explanation
Terzaghi's local shear correction explicitly modifies the shear strength parameters to 2/3 of their values. This produces a less-developed, smaller failure wedge, which is physically what happens in loose compressible soils. The FS is applied after q_u is computed with the reduced parameters, not as a substitute for the correction.
Wrong Answer
Use φ = 30° and N_γ(30°) = 15.07, N_q(30°) = 18.40 directly — just apply FS = 2 at the end.
Correct Answer
Reduce: tan(φ*) = (2/3)tan(30°) = (2/3)(0.5774) = 0.3849 → φ* = 21.05°. Look up N'_q and N'_γ at φ* ≈ 21°, then compute q_u with reduced parameters.
Misconception Id
M5
Correct Vs Incorrect
Correct Approach
Local shear → c* = (2/3)c; tan(φ*) = (2/3)tan(28°) → φ* = arctan(0.3146) ≈ 17.5°. Use N'_c, N'_q, N'_γ at φ* = 17.5°. Then apply FS as usual.
Incorrect Approach
Loose sand, φ = 28°. Local shear → just use FS = 2 with full φ = 28° and standard N factors. q_u computed with N_c(28°), N_q(28°), N_γ(28°) — WRONG.
Why Students Believe It
Students conflate 'shear failure mode' with 'factor of safety selection.' They know FS decreases with uncertainty, so they assume local shear just means FS = 2 instead of 3. They do not realize that Terzaghi's local shear correction modifies the SOIL PARAMETERS before computing q_u.
The surcharge term q = γD_f can be ignored for shallow footings because D_f is small.
Tags
- surcharge_omission
- major_error
- N_q_term
- formula_incomplete
Topic
Terzaghi's Equation — Surcharge Term
Severity
major
Exam Impact
Omitting qN_q can reduce the computed q_u by 20–40% for typical φ values, causing selection of a much lower wrong answer on multiple-choice questions.
The Reality
The surcharge term qN_q = γD_f × N_q can be a MAJOR contributor, especially for frictional soils where N_q is large (e.g., N_q = 22.46 for φ = 30°). For a typical case: q = 18(1.2) = 21.6 kPa → qN_q = 21.6 × 22.46 = 485 kPa — this can exceed the cohesion term! Omitting it is not 'conservative' — it is simply wrong and drastically underestimates q_u for φ > 20°.
Trap Question
Question
Strip footing: B = 2 m, D_f = 1 m, c = 15 kPa, φ = 25°, γ = 18 kN/m³. N_c = 25.13, N_q = 12.72, N_γ = 8.34. A student computes q_u = 15(25.13) + ½(18)(2)(8.34) = 376.95 + 150.12 = 527.1 kPa. What did the student do wrong?
Explanation
qN_q = 18(1)(12.72) = 228.96 kPa was completely omitted. This is the SURCHARGE term reflecting the confining effect of soil at the footing depth. Omitting it reduces q_u by 30% in this case — a critical error. Every term in Terzaghi's equation must be included.
Wrong Answer
Nothing — the answer of 527.1 kPa is correct.
Correct Answer
The student omitted the surcharge term qN_q. Correct: q_u = 15(25.13) + 18(1)(12.72) + ½(18)(2)(8.34) = 376.95 + 228.96 + 150.12 = 756.0 kPa.
Misconception Id
M6
Correct Vs Incorrect
Correct Approach
q = γD_f MUST always be included. q_u = cN_c + (γD_f)N_q + ½γBN_γ. Even for D_f = 1 m with γ = 18 kN/m³ and φ = 25° (N_q = 12.72): qN_q = 18(12.72) = 228.96 kPa — this is significant.
Incorrect Approach
Strip footing: q_u ≈ cN_c + ½γBN_γ (omitting the qN_q term) — WRONG. This treats the footing as if it is at the ground surface regardless of actual depth.
Why Students Believe It
Students see that D_f is often 1–2 m and γ is 17–20 kN/m³, giving q = 17–40 kPa — seemingly small compared to the cohesion and width terms. They drop it for 'simplicity' or because they forget the N_q term exists.
A higher friction angle φ always means higher bearing capacity, regardless of other conditions.
Tags
- failure_mode
- local_shear
- phi_misuse
- conceptual_gap
Topic
Failure Modes — General vs Local Shear
Severity
major
Exam Impact
Students may select the wrong failure mode or skip the 2/3 reduction for loose soils, leading to systematic overestimation of q_u in loose sand problems.
The Reality
In loose soils, a higher tabulated φ does NOT directly give a higher q_u because local shear governs and you must use φ* = arctan(0.667 tan φ) < φ. Additionally, a high φ in a cohesionless sand at shallow depth (small D_f) may still give lower q_u than a cohesive soil with c = 50 kPa at the same geometry. The full Terzaghi equation must be evaluated — no single parameter dominates universally. The governing failure mode must be identified first.
Trap Question
Question
Two soils, both with φ = 30°: Soil A is dense sand (general shear), Soil B is loose sand (local shear). For the same strip footing geometry, which soil gives a HIGHER computed q_u using Terzaghi's method?
Explanation
The failure mode determines which bearing-capacity factors to use. Same φ = 30° but different failure modes → different effective parameters → different q_u. Terzaghi explicitly provides for this distinction. Always identify the failure mode BEFORE selecting N factors.
Wrong Answer
Both give the same q_u because φ = 30° for both soils.
Correct Answer
Soil A (dense, general shear) gives the higher q_u. For Soil B: φ* = arctan(0.667 tan 30°) = arctan(0.385) ≈ 21°, giving much lower N' factors.
Misconception Id
M7
Correct Vs Incorrect
Correct Approach
Identify failure mode first. Loose sand → local shear → φ* = arctan(0.667 tan 35°) = arctan(0.467) ≈ 25°. Use N'_q(25°), N'_γ(25°) — significantly lower factors.
Incorrect Approach
Loose sand, φ = 35° → use N_q(35°) = 33.30, N_γ(35°) = 37.15 directly → very high q_u. (WRONG — local shear must be applied.)
Why Students Believe It
N_c, N_q, and N_γ all increase with φ — so students logically conclude that higher φ = higher q_u always. They forget that the FAILURE MODE changes (general → local shear) as the soil becomes looser, which REDUCES the effective parameters.
The bearing-capacity equation gives the ACTUAL stress applied to the footing — the design is safe as long as the applied load ÷ area < q_u.
Tags
- factor_of_safety
- allowable_vs_ultimate
- major_error
- design_concept
Topic
Ultimate vs Allowable Bearing Capacity — Factor of Safety
Severity
major
Exam Impact
Some board problems give q_u and ask if a specific load is 'safe' — using q_u directly instead of q_a = q_u/FS will incorrectly say loads are safe when they are not.
The Reality
q_u is the ULTIMATE capacity — the pressure at which shear failure occurs. The applied foundation pressure must be compared to q_ALLOWABLE (= q_u / FS), not to q_u. Applying the full column load P to get q = P/A and comparing q < q_u (without FS) is unsafe — the soil is being pushed to its failure stress, with no margin for uncertainty in soil properties, load estimation, or construction variability. The standard practice requires FS = 2.5 to 3.0. Additionally, the design must also check SETTLEMENT independently — sometimes settlement governs even when q_applied < q_allowable.
Trap Question
Question
A strip footing carries an applied foundation pressure of 400 kPa. The ultimate bearing capacity q_u = 756 kPa. A student says the design is safe. Is this correct?
Explanation
q_u is the FAILURE pressure, not the design pressure. An applied pressure of 400 kPa is already 400/756 = 0.53 of the failure load — meaning the actual factor of safety is only 756/400 = 1.89, which is below the minimum required FS of 2.5–3. This represents an unsafe condition despite the applied pressure being 'less than q_u'.
Wrong Answer
Yes — 400 kPa < 756 kPa, so the design is safe.
Correct Answer
No. q_a = 756 / 3 = 252 kPa (using FS = 3). Since 400 > 252, the design is UNSAFE.
Misconception Id
M8
Correct Vs Incorrect
Correct Approach
q_a = q_u / FS = 756 / 3 = 252 kPa. Applied = 400 kPa > 252 kPa → UNSAFE. The design fails with FS = 3.
Incorrect Approach
q_u = 756 kPa. Applied pressure = 400 kPa. Since 400 < 756, the design is safe. (WRONG — no factor of safety applied.)
Why Students Believe It
Students confuse bearing CAPACITY with bearing PRESSURE. They think q_u is some kind of applied load limit and compare gross applied pressure directly to q_u without using a factor of safety.
Settlement analysis is only needed for very soft or highly compressible soils — if bearing capacity is satisfied, settlement is automatically fine.
Tags
- settlement
- design_completeness
- major_error
- conceptual_gap
Topic
Settlement vs Bearing Capacity — Governing Criterion
Severity
major
Exam Impact
Board exam questions may present a scenario where q_applied < q_allowable but ask whether the design is adequate — the correct answer is 'not necessarily, settlement must also be checked.' Students who skip this reasoning lose marks.
The Reality
Settlement and bearing capacity are INDEPENDENT design checks. A footing can have adequate bearing capacity (no shear failure) yet fail in service due to excessive settlement — particularly in clay soils where consolidation settlement can reach tens of centimeters over years. The reference explicitly states: 'Settlement often governs over shear bearing.' In practice, footings on soft clay are often governed by tolerable settlement limits (typically 25 mm for isolated footings per general practice), not by q_u. Both checks MUST be performed in every foundation design.
Trap Question
Question
A foundation engineer computes q_u = 300 kPa for a footing on soft clay. With FS = 3, q_a = 100 kPa. The column load produces q_applied = 80 kPa. The engineer concludes the design is complete and safe. What critical check is missing?
Explanation
Soft clay consolidates slowly under sustained loading, causing long-term settlement independent of shear failure. A tolerable settlement limit (e.g., 25 mm for isolated footings) must be verified. This is often the controlling design criterion for footings on compressible soils.
Wrong Answer
The design is complete — q_applied (80 kPa) < q_a (100 kPa), so the footing is safe.
Correct Answer
Settlement analysis is missing. The footing on soft clay may experience excessive consolidation settlement even though shear failure will not occur. The design is incomplete without a settlement check.
Misconception Id
M9
Correct Vs Incorrect
Correct Approach
Step 1: Check shear: q_applied = 120 kPa < q_a = 150 kPa ✓. Step 2: Check settlement independently (especially for clay). If settlement > tolerable → redesign (larger footing, deeper placement, or piled foundation).
Incorrect Approach
Computed q_a = 150 kPa, applied q = 120 kPa < 150 kPa → design is adequate. Done. (WRONG — settlement not checked.)
Why Students Believe It
The Terzaghi bearing capacity equation clearly addresses shear failure, and students equate 'no shear failure' with 'adequate performance.' Settlement is a separate topic in textbooks, so students treat it as optional or secondary.
The bearing-capacity factors N_c, N_q, and N_γ are constants that students must memorize for all values of φ.
Tags
- N_factors
- exam_strategy
- minor_error
- memorization
Topic
Bearing-Capacity Factors — When and What to Memorize
Severity
minor
Exam Impact
Misremembering N factors is rarely the source of error since the exam provides them. The real risk is not knowing HOW to plug them into the correct formula — which is covered in other misconceptions.
The Reality
N_c, N_q, and N_γ are functions of φ from semi-empirical tables or charts — they are NOT universal constants to be memorized for all φ values. In PRC board exams, these values are ALWAYS GIVEN in the problem. What you must memorize are the special-case values: (1) φ = 0: N_c = 5.7, N_q = 1, N_γ = 0 (Terzaghi). (2) The correct FORMULA with the right coefficients for each footing shape. (3) The fact that N factors increase with φ (so higher φ → higher all three N values). Focus on knowing HOW to use the factors in the formula, not on memorizing tables.
Trap Question
Question
A board exam problem states: 'φ = 25°, N_c = 25.13, N_q = 12.72, N_γ = 8.34.' A student second-guesses the given N_q = 12.72 and uses N_q = 10 (memorized incorrectly). The correct q_u should be 756 kPa. What does the student compute?
Explanation
Always trust the values given in the problem. The PRC board exam consistently provides N_c, N_q, and N_γ. Using a memorized (and wrong) value instead of the provided value is a self-inflicted error. The only N values to know by heart are the φ = 0 special case: N_c = 5.7, N_q = 1, N_γ = 0.
Wrong Answer
q_u = 15(25.13) + 18(1)(10) + ½(18)(2)(8.34) = 376.95 + 180 + 150.12 = 707.1 kPa (WRONG — used wrong N_q)
Correct Answer
Use the GIVEN N_q = 12.72: q_u = 15(25.13) + 18(1)(12.72) + ½(18)(2)(8.34) = 376.95 + 228.96 + 150.12 = 756.0 kPa
Misconception Id
M10
Correct Vs Incorrect
Correct Approach
Always use the N values PROVIDED in the problem. Know the special cases (φ = 0) and the formula structure cold. Do not waste study time memorizing the full N-factor table.
Incorrect Approach
Trying to recall N_q for φ = 25° from memory as 10 (wrong) instead of using the provided value of 12.72 — resulting in a wrong final answer when the problem provides the correct value.
Why Students Believe It
Some review books list a few N values and students try to memorize them all. They become anxious when a board problem gives different values or uses a different φ. They do not realize the exam almost always provides these values in the problem statement.
The width B in the bearing-capacity formula refers to the LONGER dimension of a rectangular footing.
Tags
- footing_dimensions
- minor_error
- notation_confusion
- rectangular_footing
Topic
Footing Geometry — Definition of B
Severity
minor
Exam Impact
For square footings (B = L), this error does not occur. For rectangular footings, it overestimates the width contribution. A board problem with a 1 m × 2 m footing would give a 100% error in B if 2 m is used instead of 1 m.
The Reality
In geotechnical engineering and Terzaghi's bearing capacity equation, B is ALWAYS the SHORTER dimension (least lateral dimension) of the footing. For a 1.5 m × 3 m rectangular footing, B = 1.5 m (shorter side), L = 3 m (longer side). This convention is consistent with Meyerhof's shape factors and all geotechnical references. Using the longer dimension overestimates the width term, inflating q_u.
Trap Question
Question
A rectangular footing 1.2 m × 2.4 m is analyzed using Terzaghi's strip footing equation as a conservative approximation. What value of B should be used in the ½γBN_γ term?
Explanation
Geotechnical convention universally defines B as the short side of a rectangular footing. This is consistent with Terzaghi's original derivation, where B controls the depth of the failure wedge below the footing. Using the long dimension would model a deeper failure mechanism than actually exists.
Wrong Answer
B = 2.4 m (the longer dimension, since 'width' usually means the larger side).
Correct Answer
B = 1.2 m (the SHORTER dimension). In geotechnical engineering, B is always the least lateral dimension.
Misconception Id
M11
Correct Vs Incorrect
Correct Approach
B = 1.5 m (shorter side), L = 3 m. ½γBN_γ = ½(18)(1.5)(8.34) — CORRECT. B = shorter dimension always.
Incorrect Approach
Footing: 1.5 m × 3 m. B = 3 m (longer side) → ½γBN_γ = ½(18)(3)(8.34) — WRONG. Uses B = 3 m.
Why Students Believe It
In everyday language, 'width' often means the larger dimension. Students also associate 'B' with breadth = larger side. In structural design, B sometimes refers to the overall dimension.
Terzaghi's bearing capacity equation applies to all footing conditions including inclined loads, eccentric loads, and footings on slopes.
Tags
- terzaghi_limitations
- eccentricity
- minor_error
- scope_of_equation
Topic
Limitations of Terzaghi's Equation — Eccentric and Inclined Loads
Severity
minor
Exam Impact
Board problems involving eccentric loading require computing effective dimensions B' = B − 2e_B and L' = L − 2e_L before applying the bearing capacity equation. Ignoring eccentricity overestimates the safe load.
The Reality
Terzaghi's equation is valid ONLY for: centrally applied vertical loads, horizontal ground, and strip/square/circular footings on level ground. For inclined loads, eccentric loads, footings on slopes, or rectangular footings (general case), MEYERHOF's general bearing capacity equation with inclination factors (i_c, i_q, i_γ), eccentricity corrections (effective B' = B − 2e), and slope factors must be used. In PRC board exams, the type of equation required is usually specified — but students who do not know this distinction may apply Terzaghi to inclined-load or eccentric problems incorrectly.
Trap Question
Question
A 2 m × 2 m square footing carries an eccentric vertical load with eccentricity e = 0.25 m. What effective width B' should be used in the bearing capacity computation?
Explanation
Eccentric loading shifts the resultant force away from the footing centroid, creating a non-uniform pressure distribution. The Meyerhof effective area method replaces B with B' = B − 2e (and L with L' = L − 2e_L if biaxial eccentricity). Terzaghi's equation, which assumes central loading, must be modified for eccentric conditions.
Wrong Answer
B' = 2 m — eccentricity does not change the footing dimension used in the bearing capacity equation.
Correct Answer
B' = B − 2e = 2 − 2(0.25) = 1.5 m. This reduced effective width is used in the bearing capacity formula.
Misconception Id
M12
Correct Vs Incorrect
Correct Approach
Effective width B' = B − 2e = 2 − 2(0.2) = 1.6 m. Use B' in the bearing capacity equation (Meyerhof's approach). The allowable load is then q_a × B' × L'.
Incorrect Approach
Footing with eccentric load e = 0.2 m on a 2 m square footing. Use B = 2 m in Terzaghi's equation as normal. (WRONG — eccentricity reduces the effective bearing area.)
Why Students Believe It
Terzaghi's equation is the most prominent formula in review books, so students apply it universally. They are unaware that Terzaghi's original formulation assumed: (1) vertical, centrally applied load; (2) horizontal ground surface; (3) footing base at or above the general ground surface level.
Quick Self Check
For a square footing, the coefficient is 0.4 (not 0.5). The value 0.5 applies to strip footings only. Circular footings use 0.3. Remember: Strip-Square-Circular = 0.5-0.4-0.3.
Statement
For a square footing, the coefficient in front of γBN_γ in Terzaghi's equation is 0.5.
At φ = 0, N_c = 5.7 (Terzaghi) and N_q = 1. Only N_γ = 0. N_c represents cohesion contribution, which exists even at φ = 0. Setting N_c = 0 would give zero capacity, which is physically impossible.
Statement
When φ = 0 for a saturated clay, the bearing-capacity factor N_c = 0 because there is no frictional resistance.
The net form subtracts the pre-existing overburden pressure γD_f from q_u before dividing by FS. This gives the additional net pressure the footing can safely carry above the existing soil pressure. It is used when computing allowable column loads.
Statement
The net allowable bearing capacity is computed as q_a,net = (q_u − γD_f) / FS.
When the water table is at the footing base: the soil ABOVE the footing (used in q = γD_f) may still be moist/unsaturated (use γ or γ_sat depending on actual condition above the footing); the soil BELOW the footing base (in the failure wedge) is submerged, so γ' = γ_sat − γ_w is used in the ½γBN_γ term. The two terms use different unit weights.
Statement
If the water table rises to the level of the footing base, γ_sat should be used in BOTH the surcharge term q = γD_f and the width term ½γBN_γ.
Terzaghi's local shear correction explicitly reduces both c and tan(φ) to 2/3 of their actual values. These reduced parameters are used to determine modified N-factors (N', N'_q, N'_γ) for local shear conditions in loose or compressible soils.
Statement
For local shear failure, the correction involves using c* = (2/3)c and tan(φ*) = (2/3)tan(φ) to get reduced shear parameters.
The applied pressure must be less than q_ALLOWABLE = q_u/FS (with FS = 2.5 to 3), not less than q_u directly. Additionally, settlement must be checked independently — bearing capacity adequacy does NOT guarantee acceptable settlements, especially in compressible soils.
Statement
If the applied foundation pressure is less than q_u (ultimate bearing capacity), the footing design is complete and safe.
By universal geotechnical convention, B = shorter (least) lateral dimension of the footing, and L = longer lateral dimension. This is consistent across Terzaghi, Meyerhof, and all standard geotechnical references.
Statement
In Terzaghi's bearing capacity equation, B always refers to the shorter plan dimension of the footing.
qN_q = γD_f × N_q can be enormous. For D_f = 1 m, γ = 18 kN/m³, φ = 30° (N_q = 22.46): qN_q = 18 × 22.46 = 404 kPa. This is often the largest term. Omitting it causes 20–40% error in q_u. The surcharge term must NEVER be omitted.
Statement
The surcharge term qN_q in Terzaghi's equation can be neglected for footings at shallow depths (D_f ≤ 1 m) without significant error.
Previous chapter
Lateral Earth Pressure and Retaining Structures
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Foundations (Shallow and Deep)
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