PE Exam (Civil) Shallow Foundations: Bearing Capacity (Terzaghi, Meyerhof, Hansen)
Last updated: May 2, 2026
Shallow Foundations: Bearing Capacity (Terzaghi, Meyerhof, Hansen) questions are one of the highest-leverage areas to study for the PE Exam (Civil). This guide breaks down the rule, the elements you need to recognize, the named traps that catch most students, and a memory aid that scales to test day. Read it once, then practice the same sub-topic adaptively in the app.
The rule
The ultimate bearing capacity $q_{ult}$ of a shallow footing is the sum of three contributions — cohesion, surcharge, and self-weight of the failure wedge — each scaled by a dimensionless bearing capacity factor ($N_c$, $N_q$, $N_\gamma$) that depends only on the friction angle $\phi'$. Terzaghi's general equation $q_{ult} = c'N_c s_c + qN_q + \tfrac{1}{2}\gamma B N_\gamma s_\gamma$ is the parent form; Meyerhof and Hansen extend it with explicit shape, depth, inclination, base, and ground factors so that rectangular, embedded, or eccentrically-loaded footings can be checked. Allowable bearing capacity is then $q_{all} = q_{ult}/FS$ with $FS = 3$ for typical static loads (NCEES Reference Handbook, Geotechnical §4.3).
Elements breakdown
Terzaghi's General Equation
Original closed-form bearing capacity equation for strip footings on homogeneous soil with rough base.
- Cohesion term: $c' N_c$
- Surcharge term at footing base: $q = \gamma D_f$
- Self-weight term: $\tfrac{1}{2}\gamma B N_\gamma$
- Shape factors $s_c, s_\gamma$ for square/circular
- Assumes general shear failure of dense soil
Meyerhof Modifications
Adds shape, depth, and load inclination factors and adopts curved $N_\gamma$.
- Shape: $s_c, s_q, s_\gamma$ from $B/L$ and $\phi$
- Depth: $d_c, d_q, d_\gamma$ from $D_f/B$
- Inclination: $i_c, i_q, i_\gamma$ from load angle
- Effective width $B' = B - 2e$ for eccentricity
- Best for medium-deep footings on sand
Hansen (and Vesić) Extensions
Adds base-tilt and ground-slope factors; widely used in modern practice.
- Base tilt factors $b_c, b_q, b_\gamma$
- Ground slope factors $g_c, g_q, g_\gamma$
- Most general form for sloped or tilted bases
- Vesić uses similar form, different $N_\gamma$
- Reduces to Terzaghi when factors equal 1
Bearing Capacity Factors
Tabulated functions of effective friction angle $\phi'$.
- $N_q = e^{\pi \tan\phi}\tan^2(45^{\circ}+\phi/2)$
- $N_c = (N_q - 1)\cot\phi$
- $N_\gamma$ varies by author (Meyerhof, Hansen, Vesić)
- For $\phi = 0$ (undrained clay): $N_c = 5.14, N_q = 1, N_\gamma = 0$
- Always read the same table the equation came from
Effective Stress and Water Table
Use effective unit weight when groundwater is within or near the failure zone.
- Water table at base: use $\gamma' = \gamma_{sat} - \gamma_w$ in $\tfrac{1}{2}\gamma B N_\gamma$
- Water table above base: reduce surcharge term too
- Water table below $B$ depth under footing: no reduction
- Always use effective $c'$ and $\phi'$ for drained analysis
- Use undrained $s_u$ with $\phi = 0$ for short-term clay
Allowable Capacity and FS
Convert $q_{ult}$ to a working design value.
- $q_{all} = q_{ult}/FS$
- Typical $FS = 3$ for static dead+live
- Net vs. gross: $q_{ult,net} = q_{ult} - \gamma D_f$
- Compare $q_{all}$ to applied $q_{applied} = P/A$
- LRFD alternative: $\phi q_n \ge \sum \gamma_i Q_i$
Common patterns and traps
The Drained–Undrained Swap
Distractor uses the wrong shear-strength parameters for the loading timeframe. Long-term settlement on clay needs effective $c', \phi'$, but short-term construction loading on saturated clay is undrained: $\phi = 0$, $s_u$, with $N_c = 5.14$. Picking the wrong column of the bearing factor table changes $q_{ult}$ by an order of magnitude.
A choice in the right ballpark if you assume drained behavior, but ~5× too high or too low when undrained behavior actually governs.
Forgot Shape Factor
Candidate plugs Terzaghi's strip-footing equation directly for a square or circular footing. The cohesion term should be multiplied by $s_c$ (about 1.3 for square) and the gamma term by $s_\gamma$ (about 0.6–0.8). Skipping these gives a $q_{ult}$ that is wrong in both directions for the two terms.
A numeric choice whose cohesion contribution is ~25% low and self-weight contribution ~25% high relative to the correct answer.
Water-Table Self-Weight Trap
When the water table sits at or above the footing base, the unit weight in $\tfrac{1}{2}\gamma B N_\gamma$ becomes $\gamma' = \gamma_{sat} - \gamma_w \approx 60 \text{ pcf}$, roughly half the moist value. Distractors are computed with full moist $\gamma$, overstating capacity by a noticeable margin.
The choice is ~10–25% higher than correct because the gamma term wasn't reduced for buoyancy.
Net vs. Gross Confusion
Gross $q_{ult}$ includes the surcharge that already exists at the footing level; net $q_{ult,net} = q_{ult} - \gamma D_f$ is what's available for the column load above existing grade. Mixing these gives an answer that's off by exactly $\gamma D_f$ — a small but tell-tale offset.
Two adjacent choices differing by exactly the surcharge $\gamma D_f$ — one is gross, one is net.
Eccentric Load Effective Width
For a footing with eccentricity $e$, Meyerhof requires using $B' = B - 2e$ (and $L' = L - 2e_L$) in the bearing equation. Skipping this and using full $B$ gives a non-conservative capacity that ignores the reduced contact area.
A choice computed with full $B$ rather than $B' = B - 2e$, producing a higher capacity than the correct effective-width result.
How it works
Picture a $B = 6 \text{ ft}$ square footing embedded $D_f = 4 \text{ ft}$ in a sand with $\phi' = 32^{\circ}$, $c' = 0$, $\gamma = 120 \text{ pcf}$, with the water table well below the influence zone. Because $c' = 0$, the cohesion term drops out and you only need $N_q$ and $N_\gamma$. From a Meyerhof table, $N_q \approx 23.2$ and $N_\gamma \approx 22.0$. Surcharge is $q = \gamma D_f = (120)(4) = 480 \text{ psf}$. With Meyerhof shape factors $s_q \approx 1.62$ and $s_\gamma \approx 0.6$ for a square footing, $q_{ult} = (480)(23.2)(1.62) + \tfrac{1}{2}(120)(6)(22.0)(0.6) \approx 18{,}040 + 4{,}752 = 22{,}792 \text{ psf}$. Apply $FS = 3$ to get $q_{all} \approx 7{,}600 \text{ psf}$. Notice every term carries pcf $\times$ ft = psf — if your units don't cancel to psf or ksf you've made an error before you ever opened the table.
Worked examples
For the Reyes Bridge Replacement Project, a continuous strip footing is to be founded in a homogeneous sand at the proposed Pier 3 location. The footing is $B = 5 \text{ ft}$ wide and embedded $D_f = 3 \text{ ft}$ below finished grade. The sand has a moist unit weight $\gamma = 118 \text{ pcf}$, effective cohesion $c' = 0$, and effective friction angle $\phi' = 34^{\circ}$. The groundwater table is well below the zone of influence and may be ignored. From Meyerhof's tables, $N_q = 29.4$ and $N_\gamma = 31.1$. Use Terzaghi's strip-footing form (no shape, depth, or inclination factors) to estimate the ultimate bearing capacity.
Most nearly, what is the ultimate bearing capacity $q_{ult}$?
- A $8{,}200 \text{ psf}$
- B $10{,}400 \text{ psf}$
- C $19{,}580 \text{ psf}$ ✓ Correct
- D $23{,}740 \text{ psf}$
Why C is correct: For a strip footing with $c' = 0$, $q_{ult} = qN_q + \tfrac{1}{2}\gamma B N_\gamma$. The surcharge is $q = \gamma D_f = (118)(3) = 354 \text{ psf}$. Then $q_{ult} = (354)(29.4) + \tfrac{1}{2}(118)(5)(31.1) = 10{,}408 + 9{,}174 = 19{,}582 \text{ psf}$. Units track: $\text{pcf} \cdot \text{ft} = \text{psf}$ in each term, so the sum is in $\text{psf}$, matching choice C.
Why each wrong choice fails:
- A: Computes only the self-weight term $\tfrac{1}{2}\gamma B N_\gamma = 9{,}174 \text{ psf}$ and rounds — drops the surcharge term entirely. Common when candidates forget that embedment contributes capacity. (Net vs. Gross Confusion)
- B: Computes only the surcharge term $qN_q = 10{,}408 \text{ psf}$ and ignores the self-weight contribution from the footing's own width. The $\tfrac{1}{2}\gamma B N_\gamma$ piece is essential whenever $B > 0$.
- D: Applies square-footing shape factors ($s_q \approx 1.20$, $s_\gamma \approx 0.6$) inappropriately to a strip footing or double-counts surcharge. The problem specified strip, so shape factors equal 1. (Forgot Shape Factor)
At the Liu Civic Center site, a square spread footing $B = 7 \text{ ft}$ on a side will support a column on a saturated soft clay. The footing is embedded $D_f = 4 \text{ ft}$ below grade. Laboratory unconsolidated-undrained triaxial testing gives undrained shear strength $s_u = 1{,}200 \text{ psf}$. Saturated unit weight is $\gamma_{sat} = 122 \text{ pcf}$. The owner requires a factor of safety $FS = 3$ on ultimate bearing. For the short-term ($\phi = 0$) case, use $N_c = 5.14$, $N_q = 1.0$, $N_\gamma = 0$, with Meyerhof shape factor $s_c = 1 + 0.2(B/L) = 1.2$ and depth factor $d_c = 1 + 0.4(D_f/B) \approx 1.23$.
Most nearly, what is the allowable (net) bearing capacity?
- A $1{,}210 \text{ psf}$
- B $3{,}040 \text{ psf}$ ✓ Correct
- C $3{,}540 \text{ psf}$
- D $9{,}120 \text{ psf}$
Why B is correct: For undrained conditions with $\phi = 0$, $q_{ult} = c N_c s_c d_c + q N_q$ where $c = s_u = 1{,}200 \text{ psf}$ and $q = \gamma D_f = (122)(4) = 488 \text{ psf}$. So $q_{ult} = (1{,}200)(5.14)(1.2)(1.23) + (488)(1.0) = 9{,}104 + 488 = 9{,}592 \text{ psf}$ gross. Net: $q_{ult,net} = 9{,}592 - 488 = 9{,}104 \text{ psf}$. Allowable: $q_{all} = 9{,}104/3 \approx 3{,}035 \text{ psf}$, choice B.
Why each wrong choice fails:
- A: Divides only the cohesion term without shape or depth factors: $(1{,}200)(5.14)/3 \div 1.2 \approx 1{,}210 \text{ psf}$, or simply omits the corrections. Underestimates capacity by leaving out $s_c$ and $d_c$. (Forgot Shape Factor)
- C: Uses gross $q_{ult} = 9{,}592$ and divides by 3 to get $\approx 3{,}197 \text{ psf}$, then rounds — but adds the surcharge back in twice or misapplies $FS$. The exam form asked for net allowable, so the surcharge must be subtracted before dividing. (Net vs. Gross Confusion)
- D: Reports gross $q_{ult}$ (about $9{,}120 \text{ psf}$ after rounding) without applying the factor of safety. PE problems on allowable bearing always require the $FS$ division.
For the Okafor Warehouse expansion, a $B = 6 \text{ ft}$ wide continuous strip footing is embedded $D_f = 5 \text{ ft}$ in a uniform sand with $\phi' = 30^{\circ}$, $c' = 0$, $\gamma_{moist} = 115 \text{ pcf}$ above the water table, and $\gamma_{sat} = 125 \text{ pcf}$ below. The groundwater table sits exactly at the footing base. Bearing capacity factors are $N_q = 18.4$ and $N_\gamma = 15.7$. Take $\gamma_w = 62.4 \text{ pcf}$. Use Terzaghi's strip equation. Surcharge above the water table uses moist $\gamma$; the self-weight term beneath the footing must use buoyant unit weight.
Most nearly, what is the ultimate bearing capacity $q_{ult}$?
- A $10{,}580 \text{ psf}$
- B $13{,}530 \text{ psf}$ ✓ Correct
- C $16{,}160 \text{ psf}$
- D $22{,}160 \text{ psf}$
Why B is correct: Surcharge: $q = \gamma_{moist} D_f = (115)(5) = 575 \text{ psf}$, giving $qN_q = (575)(18.4) = 10{,}580 \text{ psf}$. Self-weight uses buoyant unit weight $\gamma' = 125 - 62.4 = 62.6 \text{ pcf}$, so $\tfrac{1}{2}\gamma' B N_\gamma = \tfrac{1}{2}(62.6)(6)(15.7) = 2{,}949 \text{ psf}$. Total $q_{ult} = 10{,}580 + 2{,}949 = 13{,}529 \text{ psf}$, choice B. Units check: $\text{pcf} \cdot \text{ft} = \text{psf}$ throughout.
Why each wrong choice fails:
- A: Reports only the surcharge term $qN_q = 10{,}580 \text{ psf}$ and drops the self-weight contribution. The footing has finite width, so $\tfrac{1}{2}\gamma' B N_\gamma$ must be included.
- C: Uses the full moist $\gamma = 115 \text{ pcf}$ in the self-weight term: $\tfrac{1}{2}(115)(6)(15.7) = 5{,}417 \text{ psf}$, then $10{,}580 + 5{,}417 \approx 16{,}000 \text{ psf}$. Ignores the water-table reduction to $\gamma'$ in the failure zone below the footing. (Water-Table Self-Weight Trap)
- D: Uses moist $\gamma$ for self-weight AND adds Meyerhof shape factors as if the footing were square ($s_q \approx 1.5, s_\gamma \approx 0.6$), inflating the surcharge term. Strip footings get $s = 1$ and the buoyancy reduction must apply. (Forgot Shape Factor)
Memory aid
Read the equation as 'C-Q-G': Cohesion + surcharge (Q) + self-weight (Gamma). Multiply each by N, then by shape × depth × inclination factors, divide by FS=3.
Key distinction
Drained ($c', \phi'$, long-term) vs. undrained ($s_u$, $\phi = 0$, short-term) analysis — the wrong choice swaps $N_c = 5.14$ for a $\phi$-dependent value or vice versa, and the bearing capacity changes by a factor of 5 or more.
Summary
Bearing capacity is C-Q-G times bearing factors times shape/depth/inclination corrections, divided by FS — and the friction angle decides whether you're in the drained or undrained world.
Practice shallow foundations: bearing capacity (terzaghi, meyerhof, hansen) adaptively
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Start your free 7-day trialFrequently asked questions
What is shallow foundations: bearing capacity (terzaghi, meyerhof, hansen) on the PE Exam (Civil)?
The ultimate bearing capacity $q_{ult}$ of a shallow footing is the sum of three contributions — cohesion, surcharge, and self-weight of the failure wedge — each scaled by a dimensionless bearing capacity factor ($N_c$, $N_q$, $N_\gamma$) that depends only on the friction angle $\phi'$. Terzaghi's general equation $q_{ult} = c'N_c s_c + qN_q + \tfrac{1}{2}\gamma B N_\gamma s_\gamma$ is the parent form; Meyerhof and Hansen extend it with explicit shape, depth, inclination, base, and ground factors so that rectangular, embedded, or eccentrically-loaded footings can be checked. Allowable bearing capacity is then $q_{all} = q_{ult}/FS$ with $FS = 3$ for typical static loads (NCEES Reference Handbook, Geotechnical §4.3).
How do I practice shallow foundations: bearing capacity (terzaghi, meyerhof, hansen) questions?
The fastest way to improve on shallow foundations: bearing capacity (terzaghi, meyerhof, hansen) is targeted, adaptive practice — working questions that focus on your specific weak spots within this sub-topic, getting immediate feedback, and revisiting items you missed on a spaced-repetition schedule. Neureto's adaptive engine does this automatically across the PE Exam (Civil); start a free 7-day trial to see your sub-topic mastery climb in real time.
What's the most important distinction to remember for shallow foundations: bearing capacity (terzaghi, meyerhof, hansen)?
Drained ($c', \phi'$, long-term) vs. undrained ($s_u$, $\phi = 0$, short-term) analysis — the wrong choice swaps $N_c = 5.14$ for a $\phi$-dependent value or vice versa, and the bearing capacity changes by a factor of 5 or more.
Is there a memory aid for shallow foundations: bearing capacity (terzaghi, meyerhof, hansen) questions?
Read the equation as 'C-Q-G': Cohesion + surcharge (Q) + self-weight (Gamma). Multiply each by N, then by shape × depth × inclination factors, divide by FS=3.
What's a common trap on shallow foundations: bearing capacity (terzaghi, meyerhof, hansen) questions?
Forgetting shape factors on square or circular footings
What's a common trap on shallow foundations: bearing capacity (terzaghi, meyerhof, hansen) questions?
Mixing $\phi = 0$ undrained with drained bearing factors
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