FE Civil Geotechnical Eng. practice problems

Dry unit weight relates to void ratio by $\gamma_d = G_s \gamma_w / (1+e)$. Solve for e: $e = G_s \gamma_w/\gamma_d - 1 = 2.68 \times 62.4/100 - 1 = 1.672 - 1 = 0.672$. The 1.672 distractor forgot to subtract 1 (that is the (1+e) value). The 0.401 distractor is the porosity $n = e/(1+e)$, not the void ratio. The 0.602 distractor comes from using a rounded or wrong unit weight of water.

1. Dry unit weight relation:
   $\gamma_d = \frac{G_s \gamma_w}{1+e}$
2. Solve for e:
   $e = \frac{G_s \gamma_w}{\gamma_d} - 1 = \frac{2.68 \times 62.4}{100} - 1$
3. Compute:
   $e = 1.672 - 1 = 0.672$

FE Reference Handbook 10.6, p. 265

Total unit weight is $\gamma = (G_s + Se)\gamma_w/(1+e)$. Here $Se = 0.60 \times 0.85 = 0.51$, so $\gamma = (2.72 + 0.51)(62.4)/(1.85) = 3.23 \times 62.4/1.85 = 201.6/1.85 = 108.9$ pcf. The 91.8 value is the dry unit weight $G_s\gamma_w/(1+e)$, dropping the Se term. The 120.9 value uses the saturated formula $(G_s+e)\gamma_w/(1+e)$, treating S as 1. The 98.7 value comes from using a partial Se contribution.

1. General total unit weight:
   $\gamma = \frac{(G_s + Se)\gamma_w}{1+e}$
2. Compute Se:
   $Se = 0.60 \times 0.85 = 0.51$
3. Substitute:
   $\gamma = \frac{(2.72 + 0.51)(62.4)}{1.85} = \frac{201.6}{1.85} = 108.9 \text{ pcf}$

FE Reference Handbook 10.6, p. 265

First $PI = LL - PL = 35 - 18 = 17$. The soil is fine-grained ($65\% > 50\%$ passes #200) with $LL = 35 < 50$ (low plasticity, L). The A-line at $LL = 35$ is $PI_A = 10.95$. Since $PI = 17 > 10.95$, the soil plots above the A-line, so it is a clay (C). Classification: CL. The ML option would require plotting below the A-line. The CH option requires $LL \geq 50$. The SC option is a coarse-grained soil with clay fines.

1. Plasticity index:
   $PI = LL - PL = 35 - 18 = 17$
2. $LL = 35 < 50$ → low plasticity (L).
3. A-line check:
   $PI_A = 0.73(35 - 20) = 10.95$
4. $PI = 17 > 10.95$ → above A-line → clay (C). Classification: **CL**.

FE Reference Handbook 10.6, p. 273

Coarse-grained, gravel (G) since more than half the coarse fraction is retained on No. 4. With < 5% fines, classify by gradation. $C_u = D_{60}/D_{10} = 8.0/0.40 = 20$ (gravel needs $\geq 4$, passes). $C_c = (D_{30})^2/(D_{10} D_{60}) = (2.4)^2/(0.40 \times 8.0) = 5.76/3.20 = 1.80$ (need 1–3, passes). Both pass → well-graded gravel, GW. The GP option would apply if a gradation criterion failed. The SW option misidentifies it as sand. The GM option requires more than 12% fines.

1. Coefficient of uniformity:
   $C_u = \frac{D_{60}}{D_{10}} = \frac{8.0}{0.40} = 20 \geq 4 \; \checkmark$
2. Coefficient of concavity:
   $C_c = \frac{(D_{30})^2}{D_{10} \times D_{60}} = \frac{5.76}{3.20} = 1.80$
3. $C_c = 1.80$ in the 1–3 range. Classification: **GW** (well-graded gravel).

FE Reference Handbook 10.6, p. 266

Total stress: $\sigma = 118(4) + 128(6) = 472 + 768 = 1{,}240$ psf. Pore pressure (6 ft below WT): $u = 62.4(6) = 374$ psf. Effective: $\sigma' = 1{,}240 - 374 = 866 \approx 865$ psf. Or use buoyant weight: $\sigma' = 118(4) + (128-62.4)(6) = 472 + 393.6 = 865.6$ psf. The 1,240 value is the total stress without subtracting pore pressure. The 394 value is only the buoyant contribution of the saturated zone, missing the moist sand overburden. The 766 value comes from an error in handling the pore pressure or zone thickness; it does not match either the correct method (865 psf) or the listed partial methods.

1. Total stress:
   $\sigma = 118(4) + 128(6) = 472 + 768 = 1{,}240 \text{ psf}$
2. Pore pressure (6 ft below WT):
   $u = 62.4 \times 6 = 374.4 \text{ psf}$
3. Effective stress:
   $\sigma' = 1{,}240 - 374 = 865 \text{ psf}$

FE Reference Handbook 10.6, p. 269

Under artesian pressure the pore pressure uses the full piezometric head, not just the depth. Head = 15 ft depth + 3 ft above surface = 18 ft. $u = \gamma_w h = 62.4 \times 18 = 1{,}123$ psf. Effective stress: $\sigma' = \sigma - u = 1{,}800 - 1{,}123 = 677$ psf. The 1,123 psf choice is the pore pressure itself. The 864 psf choice uses only the 15 ft depth ($u = 62.4 \times 15 = 936$, giving $1{,}800 - 936 = 864$), ignoring the artesian head. The 1,800 psf choice forgets to subtract pore pressure entirely.

1. Piezometric head:
   $h = 15 + 3 = 18 \text{ ft}$
2. Pore water pressure:
   $u = \gamma_w h = 62.4 \times 18 = 1{,}123 \text{ psf}$
3. Effective stress:
   $\sigma' = \sigma - u = 1{,}800 - 1{,}123 = 677 \text{ psf}$

FE Reference Handbook 10.6, p. 269

First $C_c = 0.009(50 - 10) = 0.36$. NC clay → Case 2: $\Delta H = H_0/(1+e_0) \times C_c \times \log((\sigma_0'+\Delta\sigma)/\sigma_0') = 15/2.20 \times 0.36 \times \log(2400/1200) = 6.818 \times 0.36 \times 0.3010 = 0.739$ ft = 8.9 in. The 19.5 value forgets to divide by $(1+e_0)$, giving $15 \times 0.36 \times 0.301 = 1.626$ ft. The 1.5 value uses $C_r = C_c/6 = 0.06$ instead of $C_c$. The 4.4 value halves the correct answer.

1. Compression index from correlation:
   $C_c = 0.009(50 - 10) = 0.36$
2. NC clay → Case 2:
   $\Delta H = \frac{H_0}{1+e_0} C_c \log \frac{\sigma_0' + \Delta\sigma}{\sigma_0'}$
3. Substitute:
   $\Delta H = \frac{15}{2.20} \times 0.36 \times \log \frac{2{,}400}{1{,}200}$
4. Compute:
   $\Delta H = 6.818 \times 0.36 \times 0.3010 = 0.739 \text{ ft} = 8.9 \text{ in.}$

FE Reference Handbook 10.6, p. 267

Because $c_v$ and $H_{dr}$ are fixed, time is proportional to the time factor: $t_2/t_1 = T_{v2}/T_{v1}$. So $t_{90} = t_{50} \times (T_{v90}/T_{v50}) = 40 \times (0.848/0.197) = 40 \times 4.305 = 172$ days. The 72 value scales by the ratio of U values (90/50 = 1.8) instead of the time-factor ratio. The 9 value inverts the ratio. The 340 value roughly doubles the correct answer.

1. Time is proportional to time factor:
   $\frac{t_{90}}{t_{50}} = \frac{T_{v,90}}{T_{v,50}}$
2. Solve:
   $t_{90} = 40 \times \frac{0.848}{0.197} = 40 \times 4.305 = 172 \text{ days}$

FE Reference Handbook 10.6, p. 268

For $c = 0$, $\sigma_1 = \sigma_3 \tan^2(45\degree + \phi/2)$. Here $45 + 18 = 63\degree$, $\tan 63\degree = 1.9626$, squared $= 3.852$. So $\sigma_1 = 2{,}500 \times 3.852 = 9{,}630 \approx 9{,}600$ psf. The 649 value uses $\tan^2(45 - \phi/2) = K_a$-style factor (0.2596) instead of the passive factor. The 4,910 value uses $\tan(63\degree)$ without squaring (1.9626 × 2,500). The 7,130 value reports the deviator stress $\sigma_1 - \sigma_3 = \sigma_3(\tan^2(45+\phi/2) - 1) = 2{,}500(2.852) = 7{,}130$ psf instead of the major principal stress.

1. Failure relation (c = 0):
   $\sigma_1 = \sigma_3 \tan^2\!\left(45\degree + \frac{\phi}{2}\right)$
2. Flow factor:
   $\tan^2(45\degree + 18\degree) = \tan^2(63\degree) = (1.9626)^2 = 3.852$
3. Compute:
   $\sigma_1 = 2{,}500 \times 3.852 = 9{,}630 \approx 9{,}600 \text{ psf}$

FE Reference Handbook 10.6, p. 268

Available strength: $\tau_f = c' + \sigma_N' \tan\phi' = 150 + 2{,}000 \tan 28\degree = 150 + 2{,}000(0.5317) = 150 + 1{,}063 = 1{,}213$ psf. FS = strength/mobilized = $1{,}213/1{,}000 = 1.21$. The 1.06 choice drops the cohesion term ($1{,}063/1{,}000$). The 0.82 choice inverts the ratio ($1{,}000/1{,}213$). The 2.13 choice forgets to multiply the normal stress by tan, adding $\sigma_N'$ directly.

1. Available shear strength:
   $\tau_f = c' + \sigma_N' \tan \phi' = 150 + 2{,}000 \tan 28\degree$
2. Evaluate:
   $\tau_f = 150 + 2{,}000(0.5317) = 150 + 1{,}063 = 1{,}213 \text{ psf}$
3. Factor of safety:
   $FS = \frac{\tau_f}{\tau} = \frac{1{,}213}{1{,}000} = 1.21$

FE Reference Handbook 10.6, p. 269

A rigid wall uses the at-rest coefficient $K_0 = 1 - \sin\phi = 1 - \sin 32\degree = 1 - 0.530 = 0.470$. Total force: $P_0 = 0.5 K_0 \gamma H^2 = 0.5(0.470)(115)(16^2) = 0.5(0.470)(115)(256) = 6{,}919 \approx 6{,}920$ lb/ft. The 4,520 value uses the active coefficient $K_a = \tan^2(45-16) = 0.307$ instead of $K_0$. The 13,840 value forgets the 1/2 factor. The 3,460 value halves the answer an extra time.

1. At-rest coefficient:
   $K_0 = 1 - \sin 32\degree = 1 - 0.530 = 0.470$
2. Total at-rest force:
   $P_0 = \frac{1}{2} K_0 \gamma H^2 = \frac{1}{2}(0.470)(115)(256)$
3. Compute:
   $P_0 = 0.5 \times 0.470 \times 115 \times 256 \approx 6{,}920 \text{ lb/ft}$

FE Reference Handbook 10.6, p. 269

For $\phi_u = 0$, $K_a = 1$ and the active pressure is $\sigma_a = \gamma z - 2c_u$. Setting $\sigma_a = 0$: $z_c = 2c_u/\gamma = 2(400)/120 = 800/120 = 6.7$ ft. The 3.3 value drops the factor of 2 ($c_u/\gamma = 400/120$). The 13.3 value uses $4c_u/\gamma$, doubling the factor by mistake. The 0.3 value inverts the ratio ($\gamma/(2c_u)$-style error).

1. Active pressure (φ = 0, $K_a = 1$):
   $\sigma_a = \gamma z - 2 c_u$
2. Set to zero at the crack depth:
   $0 = \gamma z_c - 2 c_u \implies z_c = \frac{2 c_u}{\gamma}$
3. Compute:
   $z_c = \frac{2 \times 400}{120} = \frac{800}{120} = 6.7 \text{ ft}$

FE Reference Handbook 10.6, p. 269

Sliding resistance is base friction: $\Sigma F_R = \mu \Sigma V = 0.55 \times 12{,}000 = 6{,}600$ lb/ft. FS = resisting/driving = $6{,}600/4{,}800 = 1.38$. The 2.50 value divides $\Sigma V$ by $\Sigma F_D$ directly, forgetting to multiply by mu. The 0.73 value inverts the correct ratio. The 0.40 value divides driving by vertical without the friction term.

1. Sliding resistance:
   $\Sigma F_R = \mu \Sigma V = 0.55 \times 12{,}000 = 6{,}600 \text{ lb/ft}$
2. Factor of safety:
   $FS = \frac{\Sigma F_R}{\Sigma F_D} = \frac{6{,}600}{4{,}800} = 1.38$

FE Reference Handbook 10.6, p. 270

Check the middle third: $B/6 = 9/6 = 1.5$ ft. Since $e = 1.0 < 1.5$, the resultant is inside the middle third, so the full base is in compression and the trapezoidal formula applies. Minimum (heel) pressure: $q_{min} = (\Sigma V/B)(1 - 6e/B) = (9{,}000/9)(1 - 6(1.0)/9) = 1{,}000(1 - 0.667) = 1{,}000(0.333) = 333$ psf. The 1,000 psf option is the uniform pressure $\Sigma V/B$, ignoring eccentricity. The 1,667 psf option is the maximum (toe) pressure $q_{max} = 1{,}000(1 + 0.667)$. The 0 psf option wrongly assumes the resultant falls outside the middle third.

1. Middle-third check:
   $\frac{B}{6} = \frac{9}{6} = 1.5 \text{ ft} > e = 1.0 \text{ ft} \; \checkmark$
2. Minimum pressure formula:
   $q_{min} = \frac{\Sigma V}{B}\left(1 - \frac{6e}{B}\right) = \frac{9{,}000}{9}\left(1 - \frac{6 \times 1.0}{9}\right)$
3. Compute:
   $q_{min} = 1{,}000(1 - 0.667) = 333 \text{ psf}$

FE Reference Handbook 10.6, p. 270

Square footing on φ = 0 clay: $q_{ult} = 1.3 c_u N_c + \gamma D_f = 1.3(60)(5.14) + 18(1.2) = 400.9 + 21.6 = 422$ kPa. The 401 value forgot the overburden term $\gamma D_f$. The 330 value dropped the 1.3 shape factor ($60 \times 5.14 + 21.6 = 330$). The 308 value dropped both the shape factor and overburden.

1. Square footing on φ = 0 clay:
   $q_{ult} = 1.3 c_u N_c + \gamma D_f$
2. Cohesion term:
   $1.3 \times 60 \times 5.14 = 400.9 \text{ kPa}$
3. Overburden term:
   $\gamma D_f = 18 \times 1.2 = 21.6 \text{ kPa}$
4. Total:
   $q_{ult} = 400.9 + 21.6 = 422 \text{ kPa}$

FE Reference Handbook 10.6, p. 270

Applied bearing pressure = load / area = $30{,}000 / 5 = 6{,}000$ psf (per foot of strip length). FS = $q_{ult}/q_{applied} = 18{,}000/6{,}000 = 3.0$. The 0.6 value compares $q_{ult}$ to the total line load $30{,}000$ instead of the contact pressure ($18{,}000/30{,}000$). The 1.5 value divides by the half-width $B/2 = 2.5$ ft instead of the full width ($q = 30{,}000/2.5 = 12{,}000$, giving $18{,}000/12{,}000 = 1.5$). The 6.0 value uses twice the width ($q = 30{,}000/10 = 3{,}000$, giving $18{,}000/3{,}000 = 6.0$).

1. Applied contact pressure:
   $q_{applied} = \frac{P}{B} = \frac{30{,}000}{5} = 6{,}000 \text{ psf}$
2. Factor of safety:
   $FS = \frac{q_{ult}}{q_{applied}} = \frac{18{,}000}{6{,}000} = 3.0$

FE Reference Handbook 10.6, p. 270

q = kH(Nf/Nd) = (1×10⁻⁴)(8)(5/10) = (1×10⁻⁴)(8)(0.5) = 4.0×10⁻⁴ m³/s per m. The 1.6×10⁻³ option inverts the ratio (uses Nd/Nf = 2). The 8.0×10⁻⁴ option drops the ratio (uses 1.0). The 2.0×10⁻⁴ option halves the head (uses H = 4).

1. Apply the flow-net formula:
   $q = kH\frac{N_f}{N_d} = (1\times10^{-4})(8)(0.5)$
2. Compute:
   $q = 4.0\times10^{-4}\text{ m}^3\text{/s per m}$

FE Reference Handbook 10.6, p. 298

Seepage parallel to a cohesionless slope multiplies the factor of safety by γ′/γ_sat ≈ 0.5, so a slope at FS ≈ 1.0 dry can drop to ≈ 0.5 and fail. The "unchanged" choice ignores the buoyancy/seepage effect on effective stress. The "roughly doubles" choice has the direction backwards. The "becomes infinite" choice is nonsensical.

1. Dry: FS = tan φ/tan β.
2. With seepage: FS = (γ′/γ_sat)(tan φ/tan β), and γ′/γ_sat ≈ 0.5.
3. So the factor of safety roughly halves.

FE Reference Handbook 10.6, p. 271

Dr = (emax − e)/(emax − emin) × 100 = (0.80 − 0.52)/(0.80 − 0.45) × 100 = 0.28/0.35 × 100 = 80%, a dense sand. The 20% choice uses (e − emin) in the numerator. The 65% choice miscomputes the denominator. The 35% choice inverts the numerator.

1. Apply the relative-density formula:
   $D_r = \frac{0.80 - 0.52}{0.80 - 0.45} \times 100$
2. Compute:
   $D_r = \frac{0.28}{0.35} \times 100 = 80\%$

FE Reference Handbook 10.6, p. 266

A pile bearing on sound rock acts essentially as an end-bearing pile: the strong tip support dominates, while the soft soils contribute little skin friction. The "skin friction in the soft soils" option describes a friction pile in deep uniform soil. The "equally by end bearing and negative skin friction" option invents a role for downdrag. The "buoyancy" option is not a real load-carrying mechanism here.

1. The soft soils provide little skin friction.
2. The pile tip rests on sound rock.
3. So load is carried mainly by end bearing.

FE Reference Handbook 10.6, Deep Foundations (FHWA/AASHTO)