Soil compressibility is the volume decrease of a soil mass under load. In saturated clay, most volume change comes from water being squeezed from the voids; in dry or partly saturated soil, air may also be expelled.
$S$ is settlement, $H$ is layer thickness, $e_0$ is initial void ratio, and $e_1$ is final void ratio.
A normally consolidated clay has never experienced an effective overburden pressure greater than its current effective overburden pressure. Settlement is computed along the virgin compression line.
$C_c$ is compression index, $P_0$ is the initial average effective stress at the middle of the clay layer, and $\Delta P$ is the average stress increase.
An overconsolidated clay has previously carried a larger effective stress than it carries now. The largest past effective stress is the preconsolidation pressure $P_c$.
$C_s$ is the swell or recompression index.
Secondary compression happens after primary consolidation, usually from long-term rearrangement of soil particles under essentially constant effective stress.
$C_\alpha$ is secondary compression index, $e_p$ is void ratio at the end of primary consolidation, $T_1$ is the time for completion of primary settlement, and $T_2$ is the final time considered.
Laboratory consolidation data are commonly expressed using compression index, coefficient of compressibility, coefficient of volume compressibility, and coefficient of consolidation.
The drainage path controls how fast consolidation happens. For double drainage, water escapes from both faces and $H_{dr}$ is half the clay thickness. For single drainage, water escapes from one face and $H_{dr}$ is the full clay thickness.
For a square footing on sand, the modulus of vertical subgrade reaction from a 0.3 m plate may be adjusted for footing width.
Given $P_1=100$ kN, $P_2=200$ kN, $e_1=1.46$, and $e_2=1.32$, compute the compression index.
Answer: $C_c=0.465$.
For an undisturbed clay specimen, $e_1=1.12$ at $P_1=90$ kPa and $e_2=0.90$ at $P_2=460$ kPa. Compute $a_v$.
Answer: $a_v=5.95\times10^{-4}$ m2/kN.
A 450 kN column load is transmitted through a square footing 1.5 m on a side. Assuming load spreads 2 vertical to 1 horizontal, evaluate the pressure 2.7 m below the footing.
Answer: $q_z=25.51$ kPa.
A 12 m soft clay layer has initial void ratio $e_0=0.90$. Under load, the void ratio decreases by one-third. Evaluate the reduction in thickness.
Answer: $S=1.895$ m.
A 10 m soft clay layer has initial void ratio $e_0=2.0$. Under compressive load, the void ratio decreases by one-half. Evaluate settlement.
Answer: $S=3.333$ m.
A 9.00 m soft clay layer has initial void ratio $e_0=1.0$. The void ratio decreases by one-fourth. Evaluate settlement.
Answer: $S=1.125$ m.
A building is on a clay layer over impervious rock. The clay layer is 6 m thick. In a laboratory test simulating the field condition, a 75 mm specimen has void ratio reduced from 0.92 to 0.88. Compute final settlement.
Answer: $S=0.125$ m or 125 mm.
Field data show water content $w=40\%$, liquidity index $LI=0.64$, and plastic limit $PL=20\%$. Compute primary compression index.
Answer: $C_c=0.371$.
Evaluate settlement for plastic clay with $P_0=200$ kPa, $\Delta P=150$ kPa, $H=10$ m, compression index $C_c=0.315$, and $e_0=1.5$.
Answer: $S_c=0.306$ m.
A soil profile is shown in the source PDF. Given saturated unit weight of sand $22.2$ kN/m3, saturated unit weight of clay $18.27$ kN/m3, $e_0=1.20$, $C_c=0.20$, structure stress increase $28$ kPa, and water table at the surface, compute construction settlement.
The layer thicknesses are embedded in the PDF figure, so use this workflow once the dimensions are read from the profile.
Use the clay thickness for $H$ and compute $P_0$ at the clay midpoint using effective unit weights because the water table is at the ground surface.
A 12 m high oil tank with unit weight $9.4$ kN/m3 is to be built on 3.6 m of sand over 16 m of clay, with the water table at the surface. Determine compression index, settlement under the center, and embedment depth to minimize settlement.
Some values for this CE Board problem are embedded in figures in the PDF. The governing steps are:
The embedment step balances removed soil pressure against tank pressure so the net stress increase on the clay is minimized.
A 3 m thick clay layer has $e_0=1.10$, $\Delta P=40$ kPa, $P_0=80$ kPa, $P_c=130$ kPa, and $C_s=0.06$. Compute primary consolidation settlement.
Answer: $S_c=0.0151$ m or 15.1 mm.
A soil profile has clay with $C_c=0.36$ and $e_0=1.2$. Lowering the water table by 1.5 m changes the effective pressure at the midpoint of clay from 83.13 kPa to 96.38 kPa. Determine settlement.
The extracted solution gives $P_1$ and $P_2$ at point A, the midpoint of the clay. The midpoint expression indicates a 2.4 m clay layer.
Answer: $S_c=0.0252$ m or 25.2 mm.
A profile is subjected to a surface load of 48 kPa. The clay has $P_c=160$ kPa, $LL=40\%$, $e_0=0.90$, and $C_s=\frac{1}{5}C_c$. The extracted solution gives $P_0=140.71$ kPa. Determine primary consolidation settlement for a 6 m clay layer.
Answer: $S_c=0.0706$ m or 70.6 mm.
A consolidation test gives $P_c=196$ kPa. The extracted solution computes present effective pressure as $P_0=97.43$ kPa. Compute OCR.
Answer: $OCR=2.01$.
A normally consolidated clay layer is 5 m thick. Primary settlement is 35 mm and requires 8 years. Given $C_\alpha=0.008$, $e_p=1.15$, estimate total settlement over 16 years.
Answer: Total settlement is 40.6 mm.
Primary consolidation settlement is 30 mm for a 5 m normally consolidated clay layer. Full primary settlement requires 8 years. If total expected settlement is 37.4 mm, $C_\alpha=0.008$, and $e_p=1.15$, find the time.
Answer: About 20 years.
A 3 m by 3 m square footing supports an axial load of 2250 kN on sand. If $K_v=45\times10^3$ kN/m3, estimate settlement.
Answer: $S=0.0184$ m or 18.4 mm.
Primary consolidation is the time-dependent compression that occurs as excess pore water pressure dissipates after a load is applied. Terzaghi's theory governs it through the time factor:
where $H_{dr}$ is the drainage path length (half the layer thickness for double drainage). Key time factors: $T_v=0.197$ for $U=50\%$, $T_v=0.848$ for $U=90\%$. Settlement magnitude comes from:
For overconsolidated clay with OCR $> 1$, use $C_s$ (swelling index) while $\sigma'_v < \sigma'_c$, then $C_c$ once the preconsolidation pressure is exceeded. $C_s \approx C_c/5$ to $C_c/10$. Secondary compression occurs after excess pore pressure is fully dissipated and is governed by the secondary compression index $C_\alpha$:
A consolidation test on a normally consolidated clay gives void ratio $e=1.24$ at effective stress $\sigma'_v=48$ kPa and $e=0.89$ at $\sigma'_v=192$ kPa. The clay layer is 3.0 m thick with initial void ratio $e_0=1.24$. Compute $C_c$ and estimate primary consolidation settlement when stress increases from 48 kPa to 192 kPa.
Answer: $C_c=0.581$, $S_c=469$ mm.
A 4.8 m thick clay layer is doubly drained. The coefficient of consolidation is $c_v=1.8\times10^{-3}$ cm²/s. Determine the time in days required to reach 50% and 90% average consolidation.
Double drainage: $H_{dr}=4.8/2=2.4$ m $=240$ cm.
At $U=50\%$, $T_v=0.197$:
At $U=90\%$, $T_v=0.848$:
Answer: $t_{50}=73.0$ days, $t_{90}=314$ days.
A normally consolidated clay layer is 6.0 m thick. Initial effective vertical stress at mid-depth is 80 kPa. A proposed building will increase stress at mid-depth by 45 kPa. The clay has initial void ratio $e_0=0.95$ and $C_c=0.40$. Compute expected primary consolidation settlement.
Answer: $S_c \approx 239$ mm.
A 4.0 m clay layer has $e_0=0.82$, $C_c=0.45$, $C_s=0.08$, and preconsolidation pressure $\sigma'_c=150$ kPa. Current effective stress at mid-depth is 90 kPa (OCR = 1.67). An applied load will increase mid-depth stress by 100 kPa to a final value of 190 kPa. Compute total primary settlement.
Since $\sigma'_{v,\text{final}}=190$ kPa $>\sigma'_c=150$ kPa, two stages are required.
Stage 1 — recompression (90 → 150 kPa), use $C_s$:
Stage 2 — virgin compression (150 → 190 kPa), use $C_c$:
Answer: $S_c=141$ mm (39 mm recompression + 102 mm virgin compression).
Primary consolidation of a 6.0 m clay layer is complete at $t_{100}=8$ years. At that time the void ratio is $e_p=0.72$. The secondary compression index $C_\alpha=0.018$. Estimate secondary compression from $t_{100}$ to 25 years and from $t_{100}$ to 50 years.
To $t=25$ years:
To $t=50$ years:
Answer: 31.1 mm by year 25, 50.0 mm by year 50 (after end of primary).