Soil Mechanics

Consolidation and effective stress

Effective stress is defined by the Terzaghi principle as \sigma' = \sigma - u, where \sigma is total stress and u is pore water pressure. Only \sigma' acts on the soil skeleton and governs compression, consolidation and shear strength, so it is the quantity to track in every settlement calculation. In a fully saturated clay, a sudden increase in total stress is carried first by the pore water, so the excess pore pressure \Delta u equals the stress increment \Delta\sigma, and the effective stress stays at its former value at the instant of loading. As water drains, \Delta u dissipates and the load is transferred to the soil skeleton, so \sigma' rises gradually. This process is primary consolidation.

Changes in the groundwater table alter u without changing total stress, because the total vertical stress is set by the weight of the soil above. Raising the water table increases u and reduces \sigma', while lowering it increases \sigma' at the same depth. When the pore pressure rises or seepage forces bring \sigma' close to zero, the soil loses its strength and behaves like a liquid, which is liquefaction.

The overconsolidation ratio is \text{OCR} = \sigma'_p/\sigma'_0, where \sigma'_p is the preconsolidation pressure and \sigma'_0 is the present effective overburden stress. A past lower water table means the old effective stress was higher than today, which makes the soil overconsolidated and gives an OCR greater than 1.

For a normally consolidated clay, the one-dimensional primary settlement of a layer of thickness H and initial void ratio e_0 is S_c = \frac{C_c H}{1+e_0}\log_{10}\frac{\sigma'_f}{\sigma'_0}, where \sigma'_0 is the initial effective stress at mid-height and \sigma'_f is the final value.

When the final stress passes the preconsolidation pressure, settlement must be split into two parts: recompression along the swelling line with index C_s from p_0 to p_c, and virgin compression along the compression line with index C_c from p_c to p_0+\Delta p. The settlement is S_c = \frac{H}{1+e_0}\left[C_s\log\frac{p_c}{p_0} + C_c\log\frac{p_0+\Delta p}{p_c}\right], and the two indices must not be swapped.

The coefficient of consolidation is c_v = \frac{k(1+e_0)}{a_v\gamma_w} = \frac{k}{m_v\gamma_w}, where a_v = \Delta e/\Delta\sigma' and m_v = a_v/(1+e_0). A soil with high permeability k or low compressibility consolidates faster.

The time for consolidation follows from the time factor T_v = c_v t/H_{dr}^2, where H_{dr} is the longest drainage path. It is half the layer thickness when drainage occurs at both top and bottom, and the full thickness when drainage occurs at one face only. For U \le 60\%, T_v = \frac{\pi}{4}U^2, and for U > 60\%, T_v = 1.781 - 0.933\log_{10}(100-U). At U = 50\%, T_v \approx 0.197. For the same c_v and drainage path, the time is proportional to T_v, so the time ratio of two degrees of consolidation equals the ratio of their T_v values. For example, if 50% consolidation takes 10 years, 25% takes about 0.25 \times 10 = 2.5 years, because \frac{\pi}{4}(0.25)^2 / \frac{\pi}{4}(0.5)^2 = 0.25.

Beneath a point load, Boussinesq's theory gives the vertical stress increase on the load axis as \sigma_z = \frac{3P}{2\pi z^2}. This is used to decide how deep the site investigation must go, for example until \sigma_z falls to a specified fraction of the effective overburden stress. Below the water table, \sigma'_v = (\gamma_{sat}-\gamma_w)z, so the submerged unit weight must be used.

Soil phase relationships

A phase diagram separates a soil into solids, water and air. The basic indices are void ratio e = V_v/V_s, porosity n = e/(1+e), degree of saturation S = V_w/V_v and water content w = W_w/W_s. The three-phase relation wG = Se links water content, specific gravity G, degree of saturation and void ratio, and it is the quickest tool when one of these quantities is unknown.

The bulk, dry and saturated unit weights are \gamma = \frac{(G+Se)\gamma_w}{1+e}, \gamma_d = \frac{G\gamma_w}{1+e} and \gamma_{sat} = \frac{(G+e)\gamma_w}{1+e}. The dry unit weight is obtained from the bulk unit weight as \gamma_d = \gamma/(1+w), and the saturated unit weight can be found from the dry unit weight as \gamma_{sat} = \gamma_d + \frac{e\gamma_w}{1+e}.

Compaction expels air and water, but it does not change the volume of solids. The solid volume therefore stays constant, and a reduction in total volume reduces only the void volume, which lowers the void ratio.

For a cohesionless sand, the relative density is D_r = \frac{e_{max}-e}{e_{max}-e_{min}}. The limiting void ratios come from the maximum and minimum dry densities through e = G\gamma_w/\gamma_d - 1, so the maximum dry density corresponds to the minimum void ratio. A porosity given in the field must first be converted to a void ratio using e = n/(1-n).

When the volume of a soil does not change during wetting, the void ratio stays fixed, and the water content after wetting is w = Se/G. Raising the degree of saturation at constant void ratio therefore raises the water content in direct proportion.

Saturating a soil at constant volume means filling all of its void space with water. The weight of water to be added is W_{add} = (1-S)V_v\gamma_w, where V_v = eV/(1+e) for a total volume V.

For a fully saturated soil, the critical hydraulic gradient at which the effective stress becomes zero, a quick condition, is i_c = \frac{G-1}{1+e}. Since saturation gives e = wG, the water content and specific gravity are enough to find the critical gradient.

Shear strength of sands and clays

Shear strength is expressed by the Mohr-Coulomb criterion \tau_f = c' + \sigma'\tan\phi' in effective stress terms. For undrained clay analysed in total stress terms, the friction angle is zero, \phi_u = 0, and the strength equals c_u. Cohesionless sand has c' = 0, so in a drained direct shear test the friction angle follows from \phi' = \tan^{-1}(\tau_f/\sigma).

In a triaxial test on a cohesionless soil with zero cohesion, the failure envelope passes through the origin, and failure occurs when \sigma_1'/\sigma_3' = \tan^2(45^\circ + \phi'/2). For a circle in the \sigma-\tau plane, \sin\phi' = \frac{\sigma_1'-\sigma_3'}{\sigma_1'+\sigma_3'}, which equals the radius of the Mohr circle divided by the distance of its centre from the origin.

When drainage is blocked, the pore pressure at failure reduces the effective stress, so the failure condition must be checked in effective stress terms, with \sigma_1' = \sigma_1 - u and \sigma_3' = \sigma_3 - u. Ignoring this difference when interpreting a total-stress result gives an incorrect strength.

Dense sands and overconsolidated clays dilate, meaning their volume increases during drained shear, while loose sands and normally consolidated clays contract. The volumetric strain versus shear strain curve therefore rises for dilation and falls for compression.

The long-term stability of an excavated clay slope is checked with effective stress parameters from a consolidated drained test, because the excess pore pressures generated during excavation have dissipated. Short-term conditions are checked with undrained parameters from unconsolidated undrained tests or from in situ vane tests.

Elements along a potential slip surface are in different stress states. Near the toe the soil is stretched and is tested in triaxial extension. In the middle of the slip surface the stress path is close to simple shear, which is modelled by direct shear. Under the overlying mass the soil is compressed, which is modelled by triaxial compression.

The unconfined compression test is a special case of the unconsolidated undrained triaxial test with zero cell pressure, and the vane shear test measures the in situ undrained strength of saturated clay. Rapid construction on clay is analysed with undrained strength rather than consolidated drained parameters, because pore pressures do not dissipate during the short loading period.

Rankine earth pressure theory

Rankine's theory assumes a smooth vertical wall and a semi-infinite backfill in a limiting state of plastic equilibrium. The active coefficient is K_a = \tan^2(45^\circ - \phi'/2) and the passive coefficient is K_p = \tan^2(45^\circ + \phi'/2) = 1/K_a. In the passive state the horizontal effective stress reaches its maximum, so failure requires \sigma_h' = K_p\sigma_v' > \sigma_v', meaning the vertical effective stress is smaller than the horizontal one.

Because K_p exceeds 1 for any positive friction angle, the passive earth pressure can be larger than the vertical effective stress at the same point. The active pressure is the reverse case, with K_a < 1, so the statement that earth pressure is always smaller than the vertical effective stress is false.

In both the active and passive Rankine states, the failure planes make an angle of 45^\circ + \phi'/2 with the plane on which the major principal stress acts. For a friction angle of 30^\circ, this gives 60^\circ in both states. The angle 45^\circ - \phi'/2 is the angle that the passive failure plane makes with the horizontal, so mixing the two angles is a common error.

For layered backfill with cohesion, the active pressure at a depth is \sigma_a = K_a\sigma_v - 2c\sqrt{K_a}, where the vertical stress \sigma_v is accumulated layer by layer from the unit weights and any surcharge. The calculation is done separately for each layer, using that layer's friction angle and cohesion. If the result becomes negative near the top, a tension zone develops and is ignored in design.

Excavation stability

An unsupported excavation stands only if the soil has enough shear strength to carry its own weight over the cut face. The most important quantity is the maximum depth of an unsupported vertical face, which depends on cohesion, friction angle and unit weight. For a cohesive soil with internal friction, the Rankine-type estimate is H_c = \dfrac{4c}{\gamma}\tan\left(45^\circ + \dfrac{\phi}{2}\right), so the depth rises with cohesion and falls with unit weight. For saturated clay under undrained conditions (\phi_u = 0), the depth is H_{max} = N_s c_u / \gamma, where N_s is a stability number that depends on the slope geometry and must be taken from charts or from the configuration given. A vertical face in purely cohesive soil has a theoretical limit of 4c_u/\gamma. The factor of safety of a slope is the ratio of resisting moment to driving moment about the centre of a trial circle. For a purely cohesive soil, the resisting moment is c_u R^2\theta, where \theta is the arc angle in radians, and the driving moment is the weight times the horizontal lever arm of its centroid from the centre. In the long term, the excess pore pressure from excavation has dissipated, so the drained strength governs and the factor of safety is FS = (c' + \sigma' \tan\phi')/\tau, where \sigma' is the effective normal stress on the failure surface.

Groundwater flow and Darcy's law

Darcy's law states that the discharge velocity is proportional to the hydraulic gradient, v = ki, where k is the coefficient of permeability and i = \Delta h / L is the gradient. The total head is the sum of elevation head, pressure head and velocity head, and in seepage problems the velocity head is usually neglected. In a stratified deposit with flow in series, the same discharge passes through each layer, so the total head loss is shared among the layers, and the equivalent resistance is the sum of L/k for each layer. The discharge is then Q = A\,\Delta h / \sum (L/k). For flow parallel to the layers, the equivalent permeability is the thickness-weighted arithmetic mean. For anisotropic soil, the governing Laplace-type equation has different coefficients in the horizontal and vertical directions, so the flow net can be made isotropic by scaling the vertical dimension by \sqrt{k_H/k_V}, while horizontal dimensions remain unchanged. A flocculated fine-grained compacted soil has randomly oriented particles, and its permeability is approximately isotropic.

Coulomb earth pressure with wall friction

Coulomb's theory assumes a planar failure wedge behind a wall and includes the friction between the wall and the backfill, \delta. The wall thrust is inclined at \delta from the normal to the wall, so the thrust has both horizontal and vertical components. When the wall is vertical and there is no adhesion between wall and wedge, the thrust is horizontal. For equilibrium of the failure wedge, the four forces, namely the wall thrust P, the weight W, the cohesive force C along the failure plane and the reaction R on the failure plane, must close into a single force polygon. The reaction R is inclined at the friction angle from the normal to the failure plane, and the cohesive force is parallel to the failure plane. Direction of wall friction depends on the wall movement. When a rough wall moves toward the backfill, the soil is pushed upward relative to the wall, so the wall friction acts downward on the wall, which is the passive case. Most classical theories such as Rankine and Coulomb represent a surcharge as a uniform stress added to the lateral pressure, and they do not use the true elastic stress distribution under the surcharge.

Soil consistency limits

Atterberg limits describe how a fine-grained soil changes with water content. The liquid limit w_L is the water content at the boundary between liquid and plastic states, the plastic limit w_P is the boundary between plastic and semi-solid states, and the plasticity index is I_P = w_L - w_P. The liquidity index I_L = (w - w_P)/I_P shows whether the natural water content lies near the plastic limit (I_L near 0) or the liquid limit (I_L near 1). The toughness index I_T = I_P / I_F relates plasticity to the flow index I_F, which is the slope of the flow curve. The Casagrande plasticity chart separates inorganic clays from silts with the A-line I_P = 0.73(w_L - 20), and the U-line I_P = 0.9(w_L - 8) gives an upper bound for natural soils. Inorganic clays at a given liquid limit lie between these two lines. If the natural water content, plasticity index and liquidity index are known, the plastic limit follows from w_P = w - I_L I_P.

USCS soil classification

The Unified Soil Classification System uses a two-letter symbol. The first letter gives the main soil group: G for gravel, S for sand, M for silt, C for clay, O for organic soil, and Pt for peat. The second letter describes gradation or plasticity. For sands and gravels, W means well graded and P means poorly graded. For fine soils, L means low compressibility and H means high compressibility, and the boundary is a liquid limit of 50 percent. Thus MH is inorganic silt with a liquid limit of 50 percent or more, while CL and CH are the low- and high-plasticity inorganic clays. The hydraulic conductivity of a soil generally decreases as particle size falls and as fines and plasticity increase. Well-graded sand has the highest conductivity among these groups, poorly graded sand is lower, silts are lower still, and high-plasticity clay has the lowest conductivity because of its very small pores.

Soil sampler area ratio

The area ratio of a soil sampler measures how much the tube wall displaces the soil during sampling. It is defined as A_r = \dfrac{D_o^2 - D_i^2}{D_i^2} \times 100\%, where D_o is the outer diameter of the cutting edge and D_i is the internal diameter of the tube. A low area ratio is needed for undisturbed sampling, typically below about 10 percent for thin-walled tubes such as the Shelby tube. A higher area ratio causes more disturbance and compression of the sample. The relation can be inverted to find the inner diameter from a target area ratio, giving D_i = D_o / \sqrt{1 + A_r}, with A_r as a fraction.

Boussinesq vertical stress under point load

Boussinesq's solution gives the vertical stress increase in a linearly elastic, isotropic, semi-infinite half-space under a concentrated surface load P. At a point with radial distance r and depth z, the vertical stress is \sigma_z = \dfrac{3P z^3}{2\pi (r^2+z^2)^{5/2}}. On the line of action of the load, r = 0, and the expression reduces to \sigma_z = \dfrac{3P}{2\pi z^2}, so the stress varies inversely with the square of depth. The ratio of stresses at two depths on this line is therefore the inverse square of the ratio of depths. For a fixed radial distance, the depth at which the vertical stress is maximum is found by setting d\sigma_z/dz = 0. This gives 3(r^2+z^2) = 5z^2, so the locus of maximum vertical stress is z^2 = \dfrac{3}{2} r^2.