9.4 Stresses in Soil Mass, Stress Distribution, and Immediate and Secondary Settlement

9.4 Stresses in Soil Mass, Stress Distribution, and Immediate and Secondary Settlement

The 2022 HGE table of specifications gives Stresses in Soil Mass 12% of the subject (6 items) and Settlement another 12% (6 items). Effective stress from self-weight and seepage is in Section 9.3. Primary consolidation is in Section 10.1. This section supplies the missing pieces:

• the stress increase caused by surface loads, which feeds every settlement calculation;
• immediate settlement;
• secondary settlement.

$S_{\text{total}} = S_{\text{immediate}} + S_{\text{primary}} + S_{\text{secondary}}$

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Point Loads: Boussinesq and Westergaard

For a vertical point load $Q$ on the surface of a homogeneous, isotropic, elastic half-space, Boussinesq gives the vertical stress increase at depth $z$ and radial distance $r$:

$\Delta\sigma_z = \frac{3Q}{2\pi z^2}\cdot\frac{1}{\left[1 + (r/z)^2\right]^{5/2}}$

Westergaard models soil reinforced by thin, rigid horizontal layers, as in stratified or varved clays. With Poisson's ratio taken as zero:

$\Delta\sigma_z = \frac{Q}{\pi z^2}\cdot\frac{1}{\left[1 + 2(r/z)^2\right]^{3/2}}$

Example. A column applies $Q = 500\text{ kN}$.

• Directly below at $z = 4\text{ m}$, Boussinesq gives $3(500)/(2\pi \cdot 16) = 14.9\text{ kPa}$.
• At $r = 3\text{ m}$ and the same depth, $r/z = 0.75$ and $[1.5625]^{2.5} = 3.052$, so $\Delta\sigma_z = 14.9/3.052 = 4.9\text{ kPa}$.
• Westergaard directly below gives $500/(\pi \cdot 16) = 9.9\text{ kPa}$, about two-thirds of the Boussinesq value.

Line load $q$ (kN/m) at horizontal distance $x$:

$\Delta\sigma_z = \frac{2qz^3}{\pi (x^2 + z^2)^2}$

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Approximate 2:1 Method

The load is assumed to spread at a slope of 2 vertical to 1 horizontal on each side. The loaded area therefore grows by $z$ in each plan dimension:

$\Delta\sigma_z = \frac{P}{(B + z)(L + z)} = \frac{qBL}{(B + z)(L + z)}$

For a strip footing, $\Delta\sigma_z = qB/(B + z)$.

Example. A $2 \times 3\text{ m}$ footing carries $q = 150\text{ kPa}$. At $z = 3\text{ m}$ below the base, $\Delta\sigma_z = 150(2)(3)/[(5)(6)] = 30\text{ kPa}$.

The 2:1 value is an average over the widened area. It is quick, but less accurate than the elastic solutions beneath the center at shallow depth.

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Uniformly Loaded Circular Area

Below the center of a flexible circular area of radius $R$ carrying uniform pressure $q$ (tanks, silos):

$\Delta\sigma_z = q\left[1 - \frac{1}{\left[1 + (R/z)^2\right]^{3/2}}\right]$

Example. An oil tank of radius 5 m applies $q = 100\text{ kPa}$. At $z = 5\text{ m}$, $R/z = 1$, so $\Delta\sigma_z = 100[1 - 1/2^{1.5}] = 100(1 - 0.3536) = 64.6\text{ kPa}$.

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Uniformly Loaded Rectangular Area

Below a corner of a flexible rectangle $B \times L$, $\Delta\sigma_z = qI$. The influence factor $I$ depends on $m = B/z$ and $n = L/z$. It is read from Fadum's chart or table, or computed from the integrated Boussinesq expression:

$I = \frac{1}{4\pi}\left[\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+m^2n^2+1}\cdot\frac{m^2+n^2+2}{m^2+n^2+1} + \tan^{-1}\frac{2mn\sqrt{m^2+n^2+1}}{m^2+n^2+1-m^2n^2}\right]$

When the denominator of the arctangent is negative, add $\pi$ to the angle.

Superposition. To find stress below any other point, divide the loaded area into rectangles that each have a corner at that point. Then add their influence factors, subtracting any that lie outside the loaded area.

Example. A $4 \times 6\text{ m}$ raft applies $q = 200\text{ kPa}$. Find the stress increase at $z = 4\text{ m}$ below the center.

• Split the raft into four $2 \times 3\text{ m}$ rectangles that meet at the center.
• For each, $m = 2/4 = 0.5$ and $n = 3/4 = 0.75$, which gives $I = 0.107$.
• So $\Delta\sigma_z = 4(200)(0.107) = 85.7\text{ kPa}$.
• The 2:1 method gives $200(24)/(8 \times 10) = 60\text{ kPa}$ at the same depth, an average rather than the center value.

Pressure bulbs. Under the center of a square footing, the stress increase falls to about $0.1q$ at a depth of about $2B$. Under a long strip it falls to about $0.1q$ at roughly $6B$. Soil investigation should reach these depths.

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Newmark's Influence Chart

Newmark's chart draws concentric circles and radial lines so that each element contributes the same stress, typically an influence value of $0.005$ (200 elements). To use it:

1. Draw the loaded area to a scale in which the chart's scale line equals the depth $z$.
2. Place the point of interest at the chart center.
3. Count the elements $N$ covered by the area.

$\Delta\sigma_z = q \times (\text{influence value}) \times N$

Example. With $N = 48$ elements, $q = 150\text{ kPa}$ and an influence value of 0.005, $\Delta\sigma_z = 150(0.005)(48) = 36\text{ kPa}$.

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Average Stress Increase in a Clay Layer

For consolidation, use the average stress increase across the clay layer. With values at the top ($t$), middle ($m$) and bottom ($b$), Simpson's rule gives:

$\Delta\sigma_{\text{av}} = \frac{\Delta\sigma_t + 4\Delta\sigma_m + \Delta\sigma_b}{6}$

Example. With 40, 25 and 15 kPa, $\Delta\sigma_{\text{av}} = (40 + 100 + 15)/6 = 25.8\text{ kPa}$.

This value goes into the primary consolidation equations of Section 10.1. Using only the mid-depth value is a common simplification.

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Immediate (Elastic) Settlement

Immediate settlement occurs as the load is applied, without volume change in saturated clay. It is the main settlement in sands. For a footing of width $B$ on an elastic layer:

$S_e = \frac{qB(1 - \mu^2)}{E_s}\,I_s$

Here $\mu$ is Poisson's ratio, $E_s$ is the soil's elastic modulus, and $I_s$ is an influence factor that depends on shape, rigidity and the point considered. For a square footing, $I_s$ is about $1.12$ at the center and $0.56$ at the corner of a flexible footing, about $0.95$ for its average, and about $0.82$ for a rigid footing.

Example. A rigid $2.5\text{ m}$ square footing applies $q = 180\text{ kPa}$ on sand with $E_s = 20\text{ MPa}$ and $\mu = 0.3$:

$S_e = \frac{180(2.5)(1 - 0.09)}{20{,}000}(0.82) = 0.0168\text{ m} = 16.8\text{ mm}$

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Secondary Compression

After excess pore pressure has dissipated, clays and organic soils keep compressing by creep of the soil skeleton:

$S_s = C_\alpha' H \log\frac{t_2}{t_1}, \qquad C_\alpha' = \frac{C_\alpha}{1 + e_p}$

Here:

• $C_\alpha$ is the secondary compression index, the change in void ratio per log cycle of time;
• $e_p$ is the void ratio at the end of primary consolidation;
• $t_1$ is the time at the end of primary consolidation.

Example. A 4 m clay layer has $C_\alpha = 0.02$ and $e_p = 1.1$, and primary consolidation ends at 2 years. Secondary settlement from 2 to 20 years is $(0.02/2.1)(4)\log 10 = 0.0381\text{ m} = 38.1\text{ mm}$.

Organic soils and peats have high $C_\alpha$, so secondary compression can exceed primary settlement in them.