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FE Civil Geotechnical Formulas

FE Civil geotechnical formulas organized by soil phase relationships, effective stress, compaction, seepage, consolidation, strength, earth pressure, and foundations.

What The FE Civil Exam Expects

The current NCEES FE Civil specification assigns 10 to 15 questions to geotechnical engineering within a 110-question, six-hour appointment. The listed subject matter includes index properties and classification, phase relationships, laboratory and field tests, effective stress, retaining structures, shear strength, bearing capacity, foundation types, consolidation and differential settlement, slope stability, and soil stabilization.

That range is broad, but the calculations are usually compact. Success depends on recognizing the model, locating the governing relationship in the supplied FE Reference Handbook, and maintaining unit discipline. The equations below are a study map. Always practice with the current handbook version available through MyNCEES because notation and organization may change.

Phase Relationships

Draw a three-phase diagram whenever weights and volumes are mixed. Put air and water in the void volume, and keep solids separate.

w=WwWs,e=VvVs,n=VvV=e1+eSr=VwVv,γ=WV,γd=WsV=γ1+wGs=γsγw,γsat=γwGs+e1+e,γ=γsatγw\begin{aligned} w &= \frac{W_w}{W_s}, & e &= \frac{V_v}{V_s}, & n &= \frac{V_v}{V}=\frac{e}{1+e} \\ S_r &= \frac{V_w}{V_v}, & \gamma &= \frac{W}{V}, & \gamma_d &= \frac{W_s}{V}=\frac{\gamma}{1+w} \\ G_s &= \frac{\gamma_s}{\gamma_w}, & \gamma_{sat} &= \gamma_w\frac{G_s+e}{1+e}, & \gamma^{\prime} &= \gamma_{sat}-\gamma_w \end{aligned}

Use water content and degree of saturation as decimals inside equations unless the equation explicitly uses percent. A reported water content of 18 percent is 0.180.18 in γd=γ/(1+w)\gamma_d=\gamma/(1+w).

Phase-Relationship Example

A moist soil has γ=118 pcf\gamma=118\ \text{pcf} and w=12%w=12\%. The dry unit weight and relative compaction are:

γd=1181.12=105.4 pcf,RC=105.4112(100%)=94.1%\gamma_d=\frac{118}{1.12}=105.4\ \text{pcf},\qquad RC=\frac{105.4}{112}(100\%)=94.1\%

The field dry unit weight and laboratory maximum dry unit weight must use compatible test methods, oversize corrections, and units. Relative compaction is not the same as relative density of granular soil.

Index Properties And Classification

For fine-grained soil, plasticity index and the USCS A-line are:

PI=LLPL,PIA=0.73(LL20)PI=LL-PL,\qquad PI_A=0.73(LL-20)

Whether a point plots above or below the A-line helps distinguish clay-like from silt-like behavior, subject to the classification rules.

For particle-size distributions:

Cu=D60D10,Cc=D302D10D60C_u=\frac{D_{60}}{D_{10}},\qquad C_c=\frac{D_{30}^2}{D_{10}D_{60}}

Do not classify using Cu and Cc alone. First determine the gravel, sand, and fines fractions and then apply the complete USCS decision path. Boundary symbols and dual symbols matter.

Total Stress, Pore Pressure, And Effective Stress

Vertical total stress, hydrostatic pore-water pressure, and effective vertical stress are:

σv=iγiHi,u=γwhp,σv=σvu\sigma_v=\sum_i\gamma_iH_i,\qquad u=\gamma_wh_p,\qquad \sigma_v^{\prime}=\sigma_v-u

Below a static water table, the effective-stress increment can also be calculated using submerged unit weight. Do not use submerged unit weight and then subtract the same pore pressure again.

Effective-Stress Example

Consider 6 ft of moist soil at 115 pcf above the water table and 8 ft of saturated soil at 125 pcf below it. At 14 ft depth:

σv=6(115)+8(125)=1,690 psf\sigma_v=6(115)+8(125)=1{,}690\ \text{psf}
u=62.4(8)=499 psf,σv=1,690499=1,191 psfu=62.4(8)=499\ \text{psf},\qquad \sigma_v^{\prime}=1{,}690-499=1{,}191\ \text{psf}

The same result is obtained from 6(115) + 8(125 - 62.4). The two paths are checks on each other.

Compaction

Relative compaction is:

RC=γd,fieldγd,max(100%)RC=\frac{\gamma_{d,field}}{\gamma_{d,max}}(100\%)

The corresponding moisture content should be considered because project specifications may require both density and moisture criteria.

Compaction raises dry unit weight by reducing air voids through mechanical effort. Consolidation is a time-dependent volume decrease caused primarily by drainage of pore water under sustained load. The two words are not interchangeable.

The zero-air-voids relationship provides a theoretical boundary, not an achievable compaction target. Field points should not plot above a correctly calculated zero-air-voids line.

Darcy Flow And Hydraulic Gradient

Hydraulic gradient, Darcy discharge, discharge velocity, and seepage velocity are:

i=ΔhL,Q=kiA,vd=QA,vs=vdni=\frac{\Delta h}{L},\qquad Q=kiA,\qquad v_d=\frac{Q}{A},\qquad v_s=\frac{v_d}{n}

Total head contains elevation head, pressure head, and velocity head. In most soil seepage problems, velocity head is negligible, but pressure head is not the same as total head.

For a simple isotropic flow net:

q=kHNfNd,Δh=HNdq=kH\frac{N_f}{N_d},\qquad \Delta h=\frac{H}{N_d}

Critical hydraulic gradient is approximately:

ic=Gs11+e=γγwi_c=\frac{G_s-1}{1+e}=\frac{\gamma^{\prime}}{\gamma_w}

A simplified factor against heave or boiling may be expressed as ic/iexiti_c/i_{exit} when the model and geometry justify that comparison.

Consolidation And Settlement

One-dimensional primary consolidation settlement for a normally consolidated layer is commonly organized as:

S=H0Cc1+e0log10(σfσ0)S=\frac{H_0C_c}{1+e_0}\log_{10}\left(\frac{\sigma_f^{\prime}}{\sigma_0^{\prime}}\right)

For overconsolidated soil, use the recompression index through the preconsolidation range and the compression index beyond it.

Time rate is represented by:

Tv=cvtHdr2T_v=\frac{c_vt}{H_{dr}^2}

HdrH_{dr} is the maximum drainage path: layer thickness for single drainage and half the layer thickness for double drainage.

Settlement questions often require the stress increase at a representative depth, subdivision of thick layers, and a decision about whether immediate, primary, or secondary settlement is relevant. Do not infer acceptable settlement from bearing capacity alone.

Shear Strength

For drained effective-stress behavior and a simplified undrained total-stress idealization:

τf=c+σntanϕ,τf=su when ϕu=0\tau_f=c^{\prime}+\sigma_n^{\prime}\tan\phi^{\prime},\qquad \tau_f=s_u\ \text{when}\ \phi_u=0

The strength parameters must match the stress system. Combining effective normal stress with an undrained total-stress strength parameter is generally inconsistent unless the stated method specifically calls for it.

Peak, critical-state, and residual strengths represent different material histories and deformation levels. On the FE exam, the problem usually supplies the intended parameters, but you still need to recognize which normal stress belongs in the equation.

Lateral Earth Pressure

For level, cohesionless backfill and the assumptions of Rankine theory:

Ka=tan2(45ϕ2)=1sinϕ1+sinϕK_a=\tan^2\left(45^\circ-\frac{\phi^{\prime}}{2}\right)=\frac{1-\sin\phi^{\prime}}{1+\sin\phi^{\prime}}
Kp=tan2(45+ϕ2)=1+sinϕ1sinϕ,K01sinϕK_p=\tan^2\left(45^\circ+\frac{\phi^{\prime}}{2}\right)=\frac{1+\sin\phi^{\prime}}{1-\sin\phi^{\prime}},\qquad K_0\approx1-\sin\phi^{\prime}

The basic soil and surcharge resultants are:

Pa=12KaγH2  at  H3,Pq=KaqH  at  H2P_a=\frac{1}{2}K_a\gamma H^2\ \text{ at }\ \frac{H}{3},\qquad P_q=K_aqH\ \text{ at }\ \frac{H}{2}

Hydrostatic water pressure is another triangle and must be added separately when drainage is not assured.

The coefficient depends on wall movement. A restrained basement wall cannot ordinarily mobilize active conditions simply because active pressure is convenient.

Bearing Capacity And Foundations

A common strip-footing form before method-specific corrections is:

qu=cNc+qNq+12γBNγ,q=γDfq_u=cN_c+qN_q+\frac{1}{2}\gamma BN_{\gamma},\qquad q=\gamma D_f

Be clear whether the problem asks for ultimate, net ultimate, allowable gross, or allowable net pressure. Factor of safety applies to the defined resistance; it does not resolve settlement.

Foundation-type questions may be conceptual. Shallow foundations transfer load near the ground surface and depend on bearing and settlement performance. Deep foundations transfer load through shaft resistance, tip resistance, or both and may also be governed by group response, downdrag, uplift, or lateral loading.

Slope-Stability Relationships

A simple dry cohesionless infinite slope parallel to the ground surface reduces to:

FS=tanϕtanβFS=\frac{\tan\phi^{\prime}}{\tan\beta}

More general infinite-slope forms include cohesion, soil thickness, and pore pressure. The model is intended for a shallow translational surface approximately parallel to a long uniform slope; it is not a substitute for circular or noncircular stability analysis.

Water lowers effective normal stress and therefore frictional resistance. If two answer choices differ only by a groundwater assumption, sketch the pore-pressure condition before selecting the equation.

Unit And Calculator Checks

Use one coherent unit system inside each equation. In US customary calculations, remember that pcf times ft gives psf and psf times ft gives lb/ft of wall. In SI, kN/m3 times m gives kPa and kPa times m gives kN/m.

  • Check degree versus radian mode before trigonometric calculations.
  • Keep water content as a decimal unless percent is explicitly required.
  • Distinguish density from unit weight; mass density requires multiplication by gravitational acceleration to obtain unit weight.
  • Distinguish stress or pressure from resultant force.
  • Preserve intermediate values and round at the end.

How To Study These Formulas

Practice retrieval from the current FE Reference Handbook rather than memorizing an unofficial sheet in isolation. For every equation, write a one-line trigger and one-line limitation. For example: use Darcy's law for laminar flow through a saturated porous medium; do not treat hydraulic conductivity as independent of fluid and soil condition.

The official FE exam page is https://ncees.org/exams/fe-exam. Verify the current specification, handbook, calculator policy, and examinee requirements there before your appointment.