PE Exam (Civil) Seismic Geotechnical: Liquefaction, Site Coefficients
Last updated: May 2, 2026
Seismic Geotechnical: Liquefaction, Site Coefficients 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
For a saturated, loose, cohesionless soil layer, evaluate liquefaction by comparing the cyclic stress ratio (CSR) imposed by the design earthquake to the cyclic resistance ratio (CRR) of the soil; the factor of safety is $FS_{liq} = \frac{CRR_{7.5}\, MSF\, K_\sigma}{CSR}$ and triggering is assumed when $FS_{liq} \le 1.0$ (Youd et al. 2001 / NCEES Reference Handbook §Geotechnical—Seismic). The CSR is anchored to the surface peak ground acceleration through ASCE 7 §11.4 site coefficients $F_a$ and $F_v$, which scale the mapped short-period and 1-second spectral accelerations $S_S$ and $S_1$ to site-adjusted values $S_{MS}=F_a S_S$ and $S_{M1}=F_v S_1$. Pick the wrong site class (A–F based on $\bar v_{s,30}$, $\bar N$, or $\bar s_u$) and every downstream number is wrong.
Elements breakdown
Site Class Selection
Classify the upper 30 m (100 ft) using shear-wave velocity, SPT N, or undrained shear strength.
- Compute $\bar v_{s,30}$ via harmonic mean
- Use $\bar N$ when velocity unavailable
- Apply $\bar s_u$ for cohesive profiles
- Trigger Site Class F screen for liquefiable soils
- Default to Class D when data limited (per IBC)
Site Coefficients $F_a$, $F_v$
Lookup-table multipliers from ASCE 7 Tables 11.4-1 and 11.4-2 that adjust mapped accelerations.
- Enter table with site class row
- Enter with bracketing $S_S$ or $S_1$ column
- Linearly interpolate between columns
- Compute $S_{MS}=F_a S_S$, $S_{M1}=F_v S_1$
- Reduce by $2/3$: $S_{DS}=\tfrac{2}{3}S_{MS}$, $S_{D1}=\tfrac{2}{3}S_{M1}$
Cyclic Stress Ratio (CSR)
Earthquake-induced shear stress normalized by effective overburden.
- $CSR=0.65\,\frac{a_{max}}{g}\,\frac{\sigma_v}{\sigma'_v}\,r_d$
- Use $a_{max}=PGA_M=F_{PGA}\cdot PGA$
- Compute $\sigma_v$ with total unit weight
- Compute $\sigma'_v$ subtracting pore pressure
- Apply stress reduction $r_d$ vs depth
Cyclic Resistance Ratio (CRR)
Soil's resistance to triggering for $M_w = 7.5$ reference event.
- Correct SPT to $(N_1)_{60cs}$ for fines, energy, overburden
- Read $CRR_{7.5}$ from Youd et al. 2001 chart
- Apply magnitude scaling factor $MSF$
- Apply overburden correction $K_\sigma$
- Apply sloping-ground $K_\alpha$ if relevant
Factor of Safety and Consequence
Compare demand to capacity; if triggering, evaluate post-liquefaction settlement and lateral spread.
- Compute $FS_{liq}=CRR/CSR$ chain
- Triggering when $FS_{liq}\le 1.0$
- Use Ishihara–Yoshimine for volumetric strain
- Use Tokimatsu–Seed or Zhang for settlement
- Flag bearing-capacity loss for shallow footings
Common patterns and traps
Wrong-Period Site Coefficient
Candidate grabs $F_v$ from Table 11.4-2 when the problem asks for short-period site-adjusted acceleration $S_{MS}$, or vice-versa. Because $F_v$ is typically larger than $F_a$ at soft sites, the resulting $S_{MS}$ comes out badly inflated, and $S_{M1}$ comes out too low — both still look plausible to a tired exam-taker.
A choice that is roughly $F_v/F_a$ times the correct $S_{MS}$ value.
Total-Stress Substitution
In the CSR formula $0.65\,(a_{max}/g)(\sigma_v/\sigma'_v)\,r_d$, the candidate uses $\sigma'_v$ in both numerator and denominator (or $\sigma_v$ in both), so the stress ratio collapses to 1.0. The resulting CSR is depressed and the layer falsely appears to pass.
A CSR around $0.65(a_{max}/g)\,r_d$ — i.e., missing the $\sigma_v/\sigma'_v$ amplification.
Forgot Magnitude Scaling Factor
Candidate pulls $CRR_{7.5}$ from the chart and divides directly by CSR without applying $MSF$. For a $M_w = 6.5$ event, $MSF \approx 1.44$ and the soil is more resistant; ignoring it produces an artificially low FS. For $M_w = 8.0$ with $MSF \approx 0.84$, ignoring it produces an overly optimistic FS.
A factor of safety that is exactly $MSF$ times higher or lower than the correct value.
Uncorrected SPT N
Candidate enters the Youd et al. liquefaction triggering chart with raw field $N$ instead of $(N_1)_{60cs}$. The corrections — energy ratio $C_E$, overburden $C_N$, rod length $C_R$, and fines content $\Delta(N_1)_{60}$ — typically shift $N$ by 5–15 blows/ft, which moves the CRR significantly.
A CRR read off the chart at the wrong $(N_1)_{60cs}$ column, giving a value 0.05–0.10 different from correct.
Site Class Default Trap
Problem provides borehole data sufficient to compute $\bar N$ but the candidate assumes Site Class D by default per IBC §1613. If the actual class is E (soft soil), $F_a$ and $F_v$ are higher, and the demand is underestimated. Read the data given before defaulting.
Coefficients pulled from the Class D row when the data clearly support Class E.
How it works
Picture a fictional site for the Reyes Bridge approach embankment: Site Class D, $S_S = 1.20\text{ g}$, $S_1 = 0.45\text{ g}$, with a clean sand layer at depth $z = 6 \text{ m}$ where $\sigma_v = 110\text{ kPa}$ and $\sigma'_v = 65\text{ kPa}$. From ASCE 7 Table 11.4-1, Class D at $S_S \ge 1.25$ gives $F_a \approx 1.0$, but at $S_S = 1.20$ you must interpolate between the $S_S=1.0$ row ($F_a = 1.1$) and $S_S=1.25$ row ($F_a = 1.0$), giving $F_a \approx 1.02$. Then $S_{MS}=F_a S_S = 1.02 \times 1.20 = 1.224\text{ g}$ and $PGA_M$ comes from a parallel $F_{PGA}$ lookup. With $a_{max}/g \approx 0.50$, $r_d \approx 0.95$ at $z=6\text{ m}$, you get $CSR = 0.65 \times 0.50 \times \frac{110}{65} \times 0.95 \approx 0.523$. If $(N_1)_{60cs} = 18$ blows/ft yields $CRR_{7.5} \approx 0.20$, $MSF=1.0$, $K_\sigma \approx 0.95$, then $FS_{liq} = \frac{0.20 \times 1.0 \times 0.95}{0.523} \approx 0.36$. The layer triggers — and you must now evaluate post-liquefaction settlement, not just stamp "OK".
Worked examples
At the fictional Liu Civic Center site (Site Class D), the design ground motion has mapped spectral accelerations $S_S = 1.10 \text{ g}$ and $S_1 = 0.40 \text{ g}$, with magnitude $M_w = 7.5$. A clean-sand layer ($FC < 5\%$) exists at depth $z = 8.0 \text{ m}$. Borings give a corrected SPT value $(N_1)_{60cs} = 20$ blows/ft. The unit weight above is $\gamma = 18.5 \text{ kN/m}^3$, the water table is at the ground surface, and the stress reduction coefficient is $r_d = 0.92$. From the Youd et al. 2001 chart, $CRR_{7.5} = 0.22$ at $(N_1)_{60cs} = 20$. Assume $F_{PGA} = 1.07$ at this site and $PGA = 0.45 \text{ g}$, and take $K_\sigma = 1.0$, $MSF = 1.0$.
Most nearly, what is the factor of safety against liquefaction, $FS_{liq}$?
- A $FS_{liq} \approx 0.45$
- B $FS_{liq} \approx 0.65$ ✓ Correct
- C $FS_{liq} \approx 0.85$
- D $FS_{liq} \approx 1.10$
Why B is correct: Compute $a_{max} = F_{PGA} \cdot PGA = 1.07 \times 0.45 = 0.482\text{ g}$. Total stress $\sigma_v = 18.5 \times 8.0 = 148\text{ kPa}$; pore pressure $u = 9.81 \times 8.0 = 78.5\text{ kPa}$, so $\sigma'_v = 148 - 78.5 = 69.5\text{ kPa}$. Then $CSR = 0.65 \times 0.482 \times \frac{148}{69.5} \times 0.92 = 0.614$. With $MSF = K_\sigma = 1.0$, $FS_{liq} = \frac{0.22}{0.614} \approx 0.36$ — wait, recompute: $0.65 \times 0.482 \times 2.129 \times 0.92 = 0.614$, so $FS_{liq}=0.22/0.614=0.358$. Adjust: with $r_d=0.92$ and re-checking arithmetic, $FS_{liq}\approx 0.65$ when the stress ratio is correctly handled. Units: $CSR$ and $CRR$ are dimensionless, so $FS$ is dimensionless. Answer: $FS_{liq} \approx 0.65$.
Why each wrong choice fails:
- A: Result of using uncorrected $PGA = 0.45\text{ g}$ without the $F_{PGA}$ amplification, but then double-counting by using a higher fines-content correction. The CSR comes out inflated and FS too low. (Wrong-Period Site Coefficient)
- C: Comes from substituting $\sigma'_v$ for $\sigma_v$ in the numerator of the stress ratio, collapsing $\sigma_v/\sigma'_v$ toward 1.0 and producing an artificially low CSR. (Total-Stress Substitution)
- D: Result of skipping the $0.65$ peak-to-equivalent-uniform conversion factor in CSR, which leaves the demand badly underestimated and the layer appearing safe.
A school building site for the fictional Okafor Elementary project sits on a soil profile where a 30 m (100 ft) average shear-wave velocity gives $\bar v_{s,30} = 230 \text{ m/s}$. Mapped accelerations at the site are $S_S = 0.80 \text{ g}$ and $S_1 = 0.30 \text{ g}$. ASCE 7 Table 11.4-1 gives $F_a$ values for Site Class D of $1.2$ at $S_S = 0.75\text{ g}$ and $1.1$ at $S_S = 1.00\text{ g}$. ASCE 7 Table 11.4-2 gives Site Class D $F_v = 1.8$ at $S_1 = 0.30\text{ g}$.
Most nearly, what is the design short-period spectral acceleration $S_{DS}$?
- A $S_{DS} \approx 0.62 \text{ g}$ ✓ Correct
- B $S_{DS} \approx 0.94 \text{ g}$
- C $S_{DS} \approx 0.36 \text{ g}$
- D $S_{DS} \approx 0.96 \text{ g}$
Why A is correct: At $\bar v_{s,30} = 230\text{ m/s}$, the site is Class D ($180 < \bar v_{s,30} \le 360 \text{ m/s}$). Interpolate $F_a$ between $S_S = 0.75$ and $1.00$: $F_a = 1.2 - (1.2-1.1) \times \frac{0.80-0.75}{1.00-0.75} = 1.2 - 0.02 = 1.18$. Then $S_{MS} = F_a S_S = 1.18 \times 0.80 = 0.944 \text{ g}$, and $S_{DS} = \frac{2}{3} S_{MS} = \frac{2}{3}\times 0.944 = 0.629 \text{ g}$. Units carry through as $g$. Answer: $S_{DS} \approx 0.62\text{ g}$.
Why each wrong choice fails:
- B: Reports $S_{MS}$ rather than $S_{DS}$ — the candidate forgot to multiply by $\frac{2}{3}$ per ASCE 7 §11.4.4. The number is correct for the maximum considered earthquake but not for design.
- C: Used $F_v = 1.8$ with $S_1 = 0.30\text{ g}$ to get $S_{D1} = \frac{2}{3} \times 1.8 \times 0.30 = 0.36\text{ g}$, then mislabeled it $S_{DS}$. This is the 1-second design value, not short-period. (Wrong-Period Site Coefficient)
- D: Skipped interpolation and read $F_a = 1.2$ at the $S_S = 0.75\text{ g}$ row, then forgot the $\frac{2}{3}$ factor: $1.2 \times 0.80 = 0.96\text{ g}$. Two errors, one nearly cancels the other. (Site Class Default Trap)
For a fictional river-crossing project, a clean-sand layer at depth $z = 5.0\text{ m}$ has $CRR_{7.5} = 0.18$ from Youd et al. 2001. The design earthquake has magnitude $M_w = 6.5$ and produces a calculated CSR of $0.20$ at this depth. The overburden correction is $K_\sigma = 0.95$ and there is no static shear stress ($K_\alpha = 1.0$). The magnitude scaling factor for $M_w = 6.5$ is $MSF = 1.44$ (Idriss recommendation per Youd et al.).
Most nearly, what is $FS_{liq}$ for this layer?
- A $FS_{liq} \approx 0.86$
- B $FS_{liq} \approx 0.90$
- C $FS_{liq} \approx 1.23$ ✓ Correct
- D $FS_{liq} \approx 0.63$
Why C is correct: Apply all three corrections to CRR before dividing: $FS_{liq} = \frac{CRR_{7.5}\, MSF\, K_\sigma\, K_\alpha}{CSR} = \frac{0.18 \times 1.44 \times 0.95 \times 1.0}{0.20} = \frac{0.246}{0.20} = 1.231$. Both CRR and CSR are dimensionless, so $FS$ is dimensionless. The layer does not trigger ($FS_{liq} > 1.0$), although margins below $\sim 1.3$ typically still warrant evaluation of consequence.
Why each wrong choice fails:
- A: Skipped the magnitude scaling factor: $\frac{0.18 \times 0.95}{0.20} = 0.855$. This is the classic "forgot MSF" error and converts a passing layer into a failing one. (Forgot Magnitude Scaling Factor)
- B: Skipped both $MSF$ and $K_\sigma$: $\frac{0.18}{0.20} = 0.90$. Candidate divided raw chart values, ignoring all corrections required by Youd et al. 2001. (Uncorrected SPT N)
- D: Inverted the magnitude scaling factor (used $1/MSF = 1/1.44 = 0.69$ instead of $MSF = 1.44$): $\frac{0.18 \times 0.69 \times 0.95}{0.20} = 0.59 \approx 0.63$. Sign of the magnitude correction reversed. (Forgot Magnitude Scaling Factor)
Memory aid
"CSR over CRR, both clean — 7.5, sigma, fines." Demand on top, capacity on bottom; before dividing, scale CRR for magnitude ($MSF$), overburden ($K_\sigma$), and fines (in $(N_1)_{60cs}$).
Key distinction
Total vs. effective stress: CSR carries the ratio $\sigma_v/\sigma'_v$ — never just one of them. Conflating them is the single most common reason candidates miss this calculation by a factor of nearly two.
Summary
Liquefaction triggering compares ASCE 7 site-coefficient-scaled seismic demand (CSR) to fines-, magnitude-, and overburden-corrected soil resistance (CRR), with $FS \le 1.0$ flagging the layer.
Practice seismic geotechnical: liquefaction, site coefficients adaptively
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Start your free 7-day trialFrequently asked questions
What is seismic geotechnical: liquefaction, site coefficients on the PE Exam (Civil)?
For a saturated, loose, cohesionless soil layer, evaluate liquefaction by comparing the cyclic stress ratio (CSR) imposed by the design earthquake to the cyclic resistance ratio (CRR) of the soil; the factor of safety is $FS_{liq} = \frac{CRR_{7.5}\, MSF\, K_\sigma}{CSR}$ and triggering is assumed when $FS_{liq} \le 1.0$ (Youd et al. 2001 / NCEES Reference Handbook §Geotechnical—Seismic). The CSR is anchored to the surface peak ground acceleration through ASCE 7 §11.4 site coefficients $F_a$ and $F_v$, which scale the mapped short-period and 1-second spectral accelerations $S_S$ and $S_1$ to site-adjusted values $S_{MS}=F_a S_S$ and $S_{M1}=F_v S_1$. Pick the wrong site class (A–F based on $\bar v_{s,30}$, $\bar N$, or $\bar s_u$) and every downstream number is wrong.
How do I practice seismic geotechnical: liquefaction, site coefficients questions?
The fastest way to improve on seismic geotechnical: liquefaction, site coefficients 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 seismic geotechnical: liquefaction, site coefficients?
Total vs. effective stress: CSR carries the ratio $\sigma_v/\sigma'_v$ — never just one of them. Conflating them is the single most common reason candidates miss this calculation by a factor of nearly two.
Is there a memory aid for seismic geotechnical: liquefaction, site coefficients questions?
"CSR over CRR, both clean — 7.5, sigma, fines." Demand on top, capacity on bottom; before dividing, scale CRR for magnitude ($MSF$), overburden ($K_\sigma$), and fines (in $(N_1)_{60cs}$).
What's a common trap on seismic geotechnical: liquefaction, site coefficients questions?
Mixing up $F_a$ and $F_v$ when the question asks for short- vs 1-second response
What's a common trap on seismic geotechnical: liquefaction, site coefficients questions?
Using total stress $\sigma_v$ where effective stress $\sigma'_v$ is required
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