9.3 Induced Stress Distribution & Immediate Elastic Settlement

Stress Distribution in Subsoils Under Surface Footings

When structural loads are applied to shallow foundations, vertical stress increases ($\Delta \sigma_z$) propagate downward and laterally through the underlying soil mass. Evaluating stress increments at various depths $z$ is essential for determining consolidation settlement in clay layers and immediate settlement in sand layers.

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Methods for Computing Vertical Stress Increments ($\Delta \sigma_z$)

1. The 2:1 (Vertical-to-Horizontal) Approximation Method

An empirical, widely used PE exam simplification that assumes surface stress spreads outward along a slope of 2 vertical to 1 horizontal ($1H:2V$).

                      2:1 Stress Spread Projection
                      
                      q0 (Surface Pressure)
                      ┌───────────────────┐ Footing Width B
                      └─────────┬─────────┘
                                │  z/2
                                ├───► z (Depth)
                                │
                      ┌─────────┴─────────────────┐
                      │  Spread Width = (B + z)   │
                      └───────────────────────────┘

• Rectangular Footing ($B \times L$ carrying total force $P = q_0 B L$): $$\Delta \sigma_z = \frac{P}{(B + z)(L + z)} = \frac{q_0 B L}{(B + z)(L + z)}$$
• Continuous Strip Footing (Width $B$): $$\Delta \sigma_z = \frac{q_0 B}{B + z}$$
• Square Footing ($B \times B$): $$\Delta \sigma_z = \frac{q_0 B^2}{(B + z)^2}$$

2. Boussinesq Theory for Elastic Half-Space

Boussinesq (1885) solved stress increments in a homogeneous, isotropic, semi-infinite linear elastic medium:

• Point Load $P$ at Radial Distance $r$ and Depth $z$:

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

• Corner of Uniformly Loaded Flexible Rectangular Area ($B \times L$):

  $$\Delta \sigma_z = q_0 \cdot I_m$$

  Where $I_m$ is Boussinesq corner influence factor determined as a function of parameters $m = B/z$ and $n = L/z$:

  $$I_m = \frac{1}{4\pi} \left[ \frac{2 m n \sqrt{m^2 + n^2 + 1}}{m^2 + n^2 + m^2 n^2 + 1} \left(\frac{m^2 + n^2 + 2}{m^2 + n^2 + 1}\right) + \arctan\left(\frac{2 m n \sqrt{m^2 + n^2 + 1}}{m^2 + n^2 - m^2 n^2 + 1}\right) \right]$$

Principle of Superposition for Interior/Exterior Stress Increments:

To calculate stress beneath the center of a rectangular area ($B \times L$), divide the area into four identical sub-rectangles of dimensions $(B/2) \times (L/2)$ meeting at the center:

$$\Delta \sigma_{z,center} = 4 \cdot q_0 \cdot I_m \left(m = \frac{B/2}{z}, n = \frac{L/2}{z}\right)$$

3. Westergaard Theory (Stratified Elastic Medium)

Westergaard (1938) assumed thin, rigid horizontal reinforcing sheets present within soil mass (preventing lateral strain, $\varepsilon_x = \varepsilon_y = 0$). Westergaard stresses are typically lower than Boussinesq values:

$$\Delta \sigma_{z,point} = \frac{P}{\pi z^2} \frac{1}{\left[ 1 + 2(r/z)^2 \right]^{3/2}}$$

4. Newmark's Influence Chart Method

Newmark (1942) developed circular influence charts where each grid unit represents a constant influence value (typically $I_{chart} = 0.005$). The chart is scaled such that length $AB$ equals depth $z$.

$$\Delta \sigma_z = I_{chart} \cdot N_{grid} \cdot q_0$$

Where $N_{grid}$ is the count of grid units covered by the foundation plan view.

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Immediate Elastic Settlement ($S_e$)

Immediate settlement occurs practically instantly upon load application in unsaturated cohesionless soils and saturated fine-grained clays without change in moisture content (undrained distortion).

                  Flexible vs Rigid Footing Settlement Profiles
                  
             Flexible Footing                       Rigid Footing
             Uniform Base Pressure q0               Non-Uniform Edge Pressure
             ┌───────────────────┐                  ┌───────────────────┐
             └─────────┬─────────┘                  └─────────┬─────────┘
             ┌─────────▼─────────┐                  ┌─────────▼─────────┐
             │                   │                  │                   │
             └───┐           ┌───┘                  └───────────────────┘
                 └───────────┘                      Uniform Settlement Se
               Dished Settlement Profile

Elasticity Equation for Flexible and Rigid Footings

$$S_e = q_0 B' \left( \frac{1 - \nu^2}{E_s} \right) I_s I_f$$

Where:

• $q_0 =$ Net applied foundation contact pressure (kPa or psf)
• $B' =$ Characteristic footprint dimension:
  • For settlement at center of flexible rectangular footing: $B' = B/2$
  • For settlement at corner of flexible rectangular footing: $B' = B$
• $\nu =$ Poisson's ratio of soil (0.20-0.30 for sand, 0.40-0.50 for undrained clay)
• $E_s =$ Soil Young's modulus of elasticity (kPa or psf)
• $I_s =$ Shape influence factor (tabulated based on $L/B$ ratio)
• $I_f =$ Embedment depth factor (Fox factor, $I_f \le 1.0$)

Typical values of shape factor $I_s$ (flexible center vs rigid):

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Schmertmann's Method for Immediate Settlement in Granular Soil

Schmertmann (1970, 1978) established an empirical method integrating vertical strain influence factors $I_z$ over depth intervals beneath footings based on CPT or SPT profiles.

$$S_e = C_1 C_2 (q_0 - \sigma_{v0}') \sum_{i=1}^{n} \frac{I_{z,i}}{E_{s,i}} \Delta z_i$$

Where:

• $C_1 = 1 - 0.5 \left( \frac{\sigma_{v0}'}{q_0 - \sigma_{v0}'} \right) \ge 0.5 =$ Embedment depth correction factor
• $C_2 = 1 + 0.2 \log_{10}\left(\frac{t}{0.1}\right) =$ Creep correction factor for time $t$ in years
• $q_0 - \sigma_{v0}' =$ Net foundation pressure increase at base level
• $I_{z,i} =$ Vertical strain influence factor at midpoint of layer $i$
• $E_{s,i} =$ Equivalent elastic modulus of layer $i$:
  • $E_s = 2.5 q_c$ (for square/circular footings)
  • $E_s = 3.5 q_c$ (for continuous strip footings, $L/B \ge 10$)

Strain Influence Factor $I_z$ Peak Characteristics:

                  Schmertmann Strain Influence Factor Iz Curves
                  
             Depth z        Iz Value
               0.0 ─────────┼──────► Iz_initial (0.1 for Square, 0.2 for Strip)
               0.5B ────────┼──────► Iz_peak (Square)
               1.0B ────────┼──────► Iz_peak (Strip)
               2.0B ────────┼──────► 0.0 (Square Limit)
               4.0B ────────┼──────► 0.0 (Strip Limit)

• Peak Strain Influence Factor Value ($I_{z,p}$): $$I_{z,p} = 0.5 + 0.1 \sqrt{\frac{q_0 - \sigma_{v0}'}{\sigma_{vp}'}}$$ Where $\sigma_{vp}'$ is initial vertical effective stress at the depth of $I_{z,p}$ peak ($0.5 B$ for square, $1.0 B$ for strip).

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Comprehensive Worked Calculation Example

Problem Statement

A square footing measuring $3.0\text{ m} \times 3.0\text{ m}$ is founded at depth $D_f = 1.5\text{ m}$ and carries a uniform gross pressure $q_0 = 180\text{ kPa}$. Soil unit weight is $\gamma = 18.5\text{ kN/m}^3$. Groundwater is deep.

1. Calculate vertical stress increment $\Delta \sigma_z$ at depth $z = 3.0\text{ m}$ below the center of the footing using the 2:1 method.
2. Calculate immediate elastic settlement $S_e$ at the center of the footing assuming a flexible footprint resting on an elastic layer ($E_s = 30\text{ MPa}, \nu = 0.30$, shape factor $I_s = 1.12$, embedment factor $I_f = 0.90$).

Solution Steps

Step 1: Compute 2:1 Stress Increment $\Delta \sigma_z$ at $z = 3.0\text{ m}$

Total column force $P = q_0 \cdot (B \times L) = 180 \times (3.0 \times 3.0) = 1620\text{ kN}$. At depth $z = 3.0\text{ m}$:

$$B_z = B + z = 3.0 + 3.0 = 6.0\text{ m}$$ $$L_z = L + z = 3.0 + 3.0 = 6.0\text{ m}$$ $$\Delta \sigma_z = \frac{P}{(B + z)(L + z)} = \frac{1620}{6.0 \times 6.0} = \frac{1620}{36} = 45.0\text{ kPa}$$

Step 2: Compute Net Foundation Pressure $q_{net}$

$$\sigma_{v0}' = \gamma D_f = 18.5 \times 1.5 = 27.75\text{ kPa}$$ $$q_{net} = q_0 - \sigma_{v0}' = 180 - 27.75 = 152.25\text{ kPa}$$

Step 3: Compute Immediate Elastic Settlement $S_e$

For flexible center calculation, $B' = B / 2 = 3.0 / 2 = 1.5\text{ m}$:

$$S_e = q_{net} B' \left( \frac{1 - \nu^2}{E_s} \right) I_s I_f$$ $$S_e = 152.25 \times 1.5 \times \left( \frac{1 - 0.30^2}{30,000} \right) \times 1.12 \times 0.90$$ $$S_e = 228.375 \times \left( \frac{0.91}{30,000} \right) \times 1.008$$ $$S_e = 228.375 \times 0.000030333 \times 1.008 = 0.006983\text{ m} = 6.98\text{ mm}$$