Consolidation and stress

This chapter covers the following:

• Consolidation
• Stress in soil mass

Consolidation of clay soils

Definitions

Consolidation

Consolidation is the time-dependent reduction in volume of saturated fine-grained soils caused by the dissipation of excess pore water pressure.

Settlement of fine-grained soils occurs in three stages:

• Immediate settlement occurs rapidly and is based on elasticity theory.
• Primary consolidation occurs due to the expulsion of water under pressure.
• Secondary compression (also known as “creep”) occurs as soil particles readjust and compress over time, beginning only after all excess pore pressure ($\Delta u$) has dissipated.

Consolidation refers to the primary consolidation phase in fine-grained soils (silts and clays). Because these soils have low permeability, excess pore water pressure ($\Delta u$) dissipates slowly, so settlement takes time.

Coarse-grained soils (sands and gravels) also settle, but they do so much faster because their high permeability allows pore water pressure to dissipate quickly.

Compaction vs. consolidation

Definitions

Compaction

Compaction is the process of increasing soil density by reducing air voids in an unsaturated soil.

Compaction and consolidation both increase soil density, but they happen in different ways: compaction reduces air voids in an unsaturated soil, while consolidation increases the density of a saturated soil over time by squeezing water out.

Stress history of clay soils

Definitions

Normally consolidated clay

A clay is normally consolidated if the present effective overburden pressure is the maximum pressure the soil has experienced.

Definitions

Overconsolidated clay

A clay is overconsolidated if the present effective overburden pressure is less than a past maximum pressure, called the preconsolidation pressure.

• Normally consolidated (NC) clay: The present effective overburden pressure is the maximum pressure the soil has been subjected to in the recent past.

• Overconsolidated (OC) clay: The present effective overburden pressure is less than what the soil has experienced in the past. The past maximum effective overburden pressure is called the preconsolidation pressure.

The stress history is summarized as:

• For normally consolidated soils:

  $${\sigma }_0^{\prime }={\sigma }_c^{\prime }$$

• For overconsolidated soils:

  $${\sigma }_0^{\prime }<{\sigma }_c^{\prime }$$

Where:

• ${\sigma }_0^{\prime }$ = Initial (present) effective overburden pressure
• ${\sigma }_c^{\prime }$ = Preconsolidation pressure

The overconsolidation ratio (OCR) is defined as:

$$OCR=\frac{{\sigma }_c^{\prime }}{{\sigma }_0^{\prime }}$$

Consolidation of normally consolidated (NC) clays

For a normally consolidated (NC) soil, primary consolidation settlement is calculated using the compression index $C_c$:

$$S_c=\left(\frac{C_c}{1+e_0}\right)H\log \left(\frac{{\sigma }_f^{\prime }}{{\sigma }_0^{\prime }}\right)$$

Where:

• $S_c$ = Primary consolidation settlement
• ${\sigma }_0^{\prime }$ = Initial effective overburden pressure
• ${\sigma }_f^{\prime }$ = Final effective overburden pressure (${\sigma }_0^{\prime }+\Delta \sigma$)
• $H$ = Thickness of soil layer
• $C_c$ = Compression index
• $e_0$ = Initial void ratio

Example: Primary consolidation settlement (NC clay)

A normally consolidated clay layer is $3\text{m}$ thick, has an initial void ratio $e_0=0.90$, and a compression index $C_c=0.30$. The initial effective overburden pressure is ${\sigma }_0^{\prime }=100\text{kPa}$, and a new structure increases the effective stress to ${\sigma }_f^{\prime }=200\text{kPa}$. Find the primary consolidation settlement.

$$S_c=\left(\frac{C_c}{1+e_0}\right)H\log \left(\frac{{\sigma }_f^{\prime }}{{\sigma }_0^{\prime }}\right)$$

$$S_c=\left(\frac{0.30}{1+0.90}\right)(3)\log \left(\frac{200}{100}\right)=(0.1579)(3)(0.3010)$$

Answer: $S_c\approx 0.1426\text{m}$ ($142.6\text{mm}$)

Consolidation of overconsolidated (OC) clay

Case I: ${\sigma }_f^{\prime }\le {\sigma }_c^{\prime }$

Use the recompression index $C_r$:

$$S_c=\left(\frac{C_r}{1+e_0}\right)H\log \left(\frac{{\sigma }_f^{\prime }}{{\sigma }_0^{\prime }}\right)$$

Case II: ${\sigma }_f^{\prime }>{\sigma }_c^{\prime }$

Use both $C_r$ and $C_c$:

$$S_c=\left(\frac{C_r}{1+e_0}\right)H\log \left(\frac{{\sigma }_c^{\prime }}{{\sigma }_0^{\prime }}\right)+\left(\frac{C_c}{1+e_0}\right)H\log \left(\frac{{\sigma }_f^{\prime }}{{\sigma }_c^{\prime }}\right)$$

The first term covers the stress increase from ${\sigma }_0^{\prime }$ up to the preconsolidation pressure ${\sigma }_c^{\prime }$ using the recompression index $C_r$, and the second term covers the remaining increase beyond ${\sigma }_c^{\prime }$ up to ${\sigma }_f^{\prime }$ using the compression index $C_c$. Always compare ${\sigma }_f^{\prime }$ to ${\sigma }_c^{\prime }$ first to decide whether Case I or Case II applies.

Empirical relationships of $C_c$ and $C_r$

For all clays:

$$C_c=1.15(e_0-0.35),C_r=(0.05\text{ to }0.10)\cdot C_c$$

For undisturbed clays of low to moderate sensitivity:

$$C_c=0.009(LL-10)$$

Where:

• $LL$= Liquid limit

The compression index can also be found directly from two points on a consolidation test’s void ratio-effective stress curve:

$$C_c=\frac{e_1-e_2}{\log ({\sigma }_2^{\prime }/{\sigma }_1^{\prime })}$$

Example: Computing $C_c$ from consolidation test data

A consolidation test on a clay sample shows a void ratio $e_1=0.80$ at ${\sigma }_1^{\prime }=100\text{kPa}$, and $e_2=0.70$ at ${\sigma }_2^{\prime }=400\text{kPa}$. Find $C_c$.

$$C_c=\frac{e_1-e_2}{\log ({\sigma }_2^{\prime }/{\sigma }_1^{\prime })}=\frac{0.80-0.70}{\log (400/100)}=\frac{0.10}{0.602}$$

Answer: $C_c\approx 0.166$

Rate of consolidation

Degree of consolidation using settlement ratio:

$$U=\frac{S_t}{S_s}$$

Where:

• $S_t$ = Settlement at time $t$
• $S_s$ = Total settlement

Degree of consolidation using pore pressure:

$$U=\frac{u_i-u}{u_i}$$

Where:

• $u_i$ = Initial excess pore pressure
• $u$ = Excess pore pressure at time $t$

Average degree of consolidation as a function of time factor:

$$T_v=\frac{C_{v}t}{H_d^2}$$

Where:

• $T_v$ = time factor
• $C_v$ = coefficient of consolidation
• $t$ = time
• $H_d$ = length of the drainage path
• $H_d$ = $H/2$ for two-way drainage and $H$ for one-way drainage, where, $H$ is thickness of the layer.

Please refer to the FE Handbook for detailed variation of time factor with degree of consolidation.

Example: Time to reach a target degree of consolidation (double drainage)

A $4\text{m}$ thick clay layer drains at both the top and bottom (double drainage) and has a coefficient of consolidation $C_v=0.02{\text{m}}^2/\text{day}$. Using $T_v=0.848$ for $90\%$ consolidation (per the FE Handbook), find the time required.

Because the layer drains both ways, the drainage path is half the layer thickness: $H_d=H/2=4/2=2\text{m}$.

$$T_v=\frac{C_{v}t}{H_d^2}⟹t=\frac{T_v{H}_d^2}{C_v}$$

$$t=\frac{(0.848)(2{)}^2}{0.02}=\frac{3.392}{0.02}$$

Answer: $t=169.6\text{days}$

Watch out: settlement and strength formulas use effective stress, ${\sigma }^{\prime }=\sigma -u$, not total stress $\sigma$. Below the water table, always subtract the pore pressure $u=h_u{\gamma }_w$ before plugging into a consolidation or shear-strength equation.

Stress in soil mass

Stress definitions

Describing the state of stress at a point in a soil mass means finding the principal stresses and the orientation of the plane they act on. The quantities below set up that problem.

Where:

• $s$ = mean normal stress
• $t$ = maximum shear stress
• ${\sigma }_1$ = major principal stress
• ${\sigma }_3$ = minor principal stress
• ${\sigma }_x$, ${\sigma }_y$ = normal stresses on the x- and y-planes
• ${\tau }_{xy}$ = shear stress on the x-y plane
• $\theta$ = orientation angle between the plane of existing normal stress and the plane of the major principal stress

The mean normal stress and maximum shear stress are found from the principal stresses:

$$s=\frac{{\sigma }_1+{\sigma }_3}{2},t=\frac{{\sigma }_1-{\sigma }_3}{2}$$

When the state of stress is instead given in terms of ${\sigma }_x$, ${\sigma }_y$, and ${\tau }_{xy}$, the principal stresses are found from:

$${\sigma }_1,{\sigma }_3=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}^2}$$

Example: Principal stresses from ${\sigma }_x$, ${\sigma }_y$, and ${\tau }_{xy}$

At a point in a soil mass, ${\sigma }_x=120\text{kPa}$, ${\sigma }_y=60\text{kPa}$, and ${\tau }_{xy}=30\text{kPa}$. Find the major and minor principal stresses.

$${\sigma }_1,{\sigma }_3=\frac{120+60}{2}\pm \sqrt{{\left(\frac{120-60}{2}\right)}^2+30^2}=90\pm \sqrt{900+900}$$

$${\sigma }_1,{\sigma }_3=90\pm 42.4$$

Answer: ${\sigma }_1\approx 132.4\text{kPa}$, ${\sigma }_3\approx 47.6\text{kPa}$

Total normal stress

$${\sigma }_N=\frac{P}{A}$$

Where:

• $P$ = normal force
• $A$ = cross-sectional area over which force acts

Effective stress

Effective stress governs soil strength and deformation:

$${\sigma }^{\prime }=\sigma -u$$

$$u=h_u{\gamma }_w$$

Where:

• $h_u$ = uplift or pressure head

Shear stress

$$\tau =\frac{T}{A}$$

Where:

• $T$ = shearing force

Shear stress at failure

$${\tau }_F=c+{\sigma }_N\tan \varphi$$

Where:

• $c$ = cohesion
• $\varphi$ = angle of internal friction