8.4 Temporary Support of Excavations

Mechanics of Soil Arching in Braced Cuts

Deep vertical excavations in urban environments require temporary support systems (braced cuts) to protect adjacent infrastructure. Braced cuts consist of vertical wall elements (soldier piles, sheet piles, secant piles, or diaphragm walls), horizontal beams (wales), and cross-excavation compression members (struts or tieback anchors).

The construction sequence of a braced cut creates a deformation pattern fundamentally different from conventional retaining walls:

1. Excavation proceeds to a shallow depth ($1.5 - 2.0\text{ m}$), and the top strut level is installed and preloaded.
2. As excavation continues deeper, the top strut prevents lateral movement at the top of the wall.
3. Lower unbraced wall segments yield inward toward the excavation before lower struts are placed.

Because upper wall movement is restricted while mid-to-lower wall sections deflect inward, soil arching occurs. Lateral earth pressure is transferred away from yielding middle soil zones upward to the rigid top strut and downward to lower struts. Consequently, classical Rankine or Coulomb triangular pressure distributions are invalid for braced excavations.

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Terzaghi and Peck Apparent Earth Pressure Envelopes (1967)

Terzaghi and Peck developed empirical apparent earth pressure envelopes based on field measurements of strut loads in deep excavations. These envelopes represent fictitious, envelope-bounding pressure diagrams designed to ensure all individual strut loads are safely enveloped.

   SAND (Dense to Loose)        SOFT-MEDIUM CLAY (Ns > 4)         STIFF CLAY (Ns <= 4)
  +---------------------+        +---------------------+        +--------------------+
  | p = 0.65 Ka gamma H |        | 0 to 0.25H: Linear  |        | 0 to 0.25H: Linear |
  |                     |        | 0.25H to 0.75H: max |        | 0.25H to 1.0H: max |
  | UNIFORM RECTANGLE   |        | 0.75H to 1.0H: lin  |        | (0.2 to 0.4) gamma H|
  +---------------------+        +---------------------+        +--------------------+

1. Braced Cuts in Sand

For loose to dense sand, the apparent earth pressure envelope is uniform rectangular over total height $H$:

$$p_a = 0.65 K_a \gamma H$$

where $K_a = \tan^2(45^\circ - \phi'/2)$. Total horizontal thrust per unit length of wall is:

$$P_a = 0.65 K_a \gamma H^2$$

2. Braced Cuts in Soft to Medium Clay (Stability Number $N_s > 4$)

The stability number ($N_s$) evaluates deep plastic yielding in cohesive soil:

$$N_s = \frac{\gamma H}{c_u}$$

When $N_s > 4$, the apparent pressure envelope is trapezoidal. Maximum pressure ($p_a$) is the larger of:

$$p_a = \gamma H \left( 1 - \frac{4 c_u}{\gamma H} \right) \quad \text{or} \quad p_a = 0.20 \gamma H \text{ to } 0.40 \gamma H$$

• Pressure increases linearly from zero at $z=0$ to $p_a$ at depth $z = 0.25 H$.
• Pressure remains uniform at $p_a$ from $z = 0.25 H$ to $z = 0.75 H$.
• Pressure decreases linearly from $p_a$ at $z = 0.75 H$ to zero at the bottom ($z = H$).

3. Braced Cuts in Stiff Clay ($N_s \le 4$)

For overconsolidated, stiff cohesive soils, the envelope is trapezoidal with maximum pressure:

$$p_a = 0.20 \gamma H \text{ to } 0.40 \gamma H \quad \text{(typically use } 0.30 \gamma H \text{ for design)}$$

• Pressure increases linearly from zero at $z=0$ to $p_a$ at $z = 0.25 H$.
• Pressure remains uniform at $p_a$ from $z = 0.25 H$ to $z = 1.0 H$.

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Structural Design of Braced Cut Components

1. Strut Load Calculation

Struts are structural compression members (steel wide-flange beams or structural steel pipes) spanning horizontally across the cut. Strut loads are calculated using tributary area methods or by treating the vertical sheeting as a continuous beam pinned at strut levels (the Hinge Method):

$$\text{Axial Strut Load: } P_{\text{strut}} = p_a \cdot S_h \cdot \Delta z$$

where $S_h$ is horizontal strut spacing, and $\Delta z$ is vertical tributary height.

Thermal Effects on Struts: Exposed steel struts expand under solar radiation. Thermal axial force increments must be added to structural loads: $\Delta P_{\text{temp}} = \alpha \cdot \Delta T \cdot E \cdot A$ where $\alpha = 1.2 \times 10^{-5} /^\circ\text{C}$ for structural steel.

2. Waler Design

Wales are continuous horizontal beams bolted to soldier piles or sheet piles to transfer wall reaction forces to struts. Wales are designed as continuous beams subjected to uniform line loads $w = p_a \cdot \Delta z$. Maximum bending moment is:

$$M_{\text{waler}} \approx \frac{w S_h^2}{10} \quad \text{or} \quad \frac{w S_h^2}{8}$$

3. Soldier Piles and Lagging

• Soldier Piles (H-Piles): Driven or pre-drilled vertically at $1.8 - 3.0\text{ m}$ spacing prior to excavation. Designed as vertical beams spanning between wale supports.
• Timber/Concrete Lagging: Installed horizontally between soldier pile flanges as excavation proceeds. Due to soil arching between soldier piles, design pressure on lagging is reduced to $50\% - 60\%$ of the apparent pressure envelope.

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Basal Stability (Heave) in Cohesive Soils

In deep excavations in soft clay, high overburden stresses ($\gamma H + q$) outside the cut push soft clay downward, driving the base of the excavation to heave upward inside the cut.

 Outside Excavation          Inside Excavation
 Ground Surface               Bottom of Cut
   |                            |
   | Overburden                 | Heave Resistance
   v gamma H + q                v cu Nc
  ================================================
           <-- Deep Plastic Shear Flow -->

Terzaghi's Factor of Safety against basal heave ($\text{FS}_{heave}$) is:

$$\text{FS}_{heave} = \frac{c_u N_c}{\gamma H + q - \frac{c_u H}{B_1}}$$

where $N_c$ is the bearing capacity factor (typically $N_c = 5.14$ for strip cuts, increasing up to $7.5$ for deep square cuts based on Bjerrum and Eide charts), and $B_1$ is failure zone width ($B/\sqrt{2}$ or wall penetration depth).

If $\text{FS}_{heave} < 1.50$, soldier piles or sheet piles must be driven deeper into rigid underlying stratum, or ground improvement (jet grouting, deep soil mixing) applied below the excavation floor.

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Seepage, Piping, and Bottom Blowout

When excavating below the water table, upward hydraulic gradients ($i$) develop at the bottom of the cut. If the upward seepage force exceeds effective soil weight, fluidization (quicksand condition or piping) occurs:

$$\ni_{\text{crit}} = \frac{\gamma'}{\gamma_w} = \frac{\gamma_{sat} - \gamma_w}{\gamma_w}$$ $$\text{FS}_{piping} = \frac{i_{\text{crit}}}{i_{\text{exit}}} \ge 1.50 - 2.00$$

To prevent bottom piping and hydrodynamic instability, structural wall cutoffs (secant piles or slurry diaphragm walls) are driven deep into impermeable clay or bedrock to lengthen the seepage path.

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Advanced Excavation Support Wall Systems

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Worked Numerical Example: Braced Cut Strut Forces

Problem Statement

A $9.0\text{ m}$ deep braced excavation in dense sand is supported by sheet piling and three levels of horizontal struts placed at depths $z_1 = 1.0\text{ m}$, $z_2 = 4.0\text{ m}$, and $z_3 = 7.0\text{ m}$. Horizontal strut spacing $S_h = 3.0\text{ m}$. Soil properties:

• Unit weight $\gamma = 19.0\text{ kN/m}^3$, friction angle $\phi' = 36^\circ$, $c' = 0$.
• Water table is deep below the base.

Calculate:

1. Terzaghi and Peck uniform apparent earth pressure ($p_a$).
2. Strut compressive load at each level ($P_{strut1}, P_{strut2}, P_{strut3}$) using the Tributary Area Method.

Step-by-Step Solution

1. Calculate Active Earth Pressure Coefficient and Apparent Pressure

$$K_a = \tan^2(45^\circ - 36^\circ/2) = \tan^2(27^\circ) = 0.2596$$ $$p_a = 0.65 K_a \gamma H = 0.65 \times 0.2596 \times 19.0 \times 9.0 = 28.87\text{ kPa}$$

2. Determine Vertical Tributary Heights ($\Delta z_i$)

• Strut 1 (at $z_1 = 1.0\text{ m}$):

  • Top boundary: $z = 0\text{ m}$
  • Midpoint to Strut 2: $z = 1.0 + \frac{4.0 - 1.0}{2} = 2.50\text{ m}$
  • Tributary height: $\Delta z_1 = 2.50 - 0 = 2.50\text{ m}$

• Strut 2 (at $z_2 = 4.0\text{ m}$):

  • Upper midpoint: $z = 2.50\text{ m}$
  • Midpoint to Strut 3: $z = 4.0 + \frac{7.0 - 4.0}{2} = 5.50\text{ m}$
  • Tributary height: $\Delta z_2 = 5.50 - 2.50 = 3.00\text{ m}$

• Strut 3 (at $z_3 = 7.0\text{ m}$):

  • Upper midpoint: $z = 5.50\text{ m}$
  • Bottom boundary: $z = 9.0\text{ m}$
  • Tributary height: $\Delta z_3 = 9.0 - 5.50 = 3.50\text{ m}$

3. Compute Axial Load Per Strut ($P = p_a \cdot S_h \cdot \Delta z_i$)

• Strut Level 1:

  $$P_{\text{strut1}} = 28.87\text{ kPa} \times 3.0\text{ m} \times 2.50\text{ m} = 216.53\text{ kN}$$

• Strut Level 2:

  $$P_{\text{strut2}} = 28.87\text{ kPa} \times 3.0\text{ m} \times 3.00\text{ m} = 259.83\text{ kN}$$

• Strut Level 3:

  $$P_{\text{strut3}} = 28.87\text{ kPa} \times 3.0\text{ m} \times 3.50\text{ m} = 303.14\text{ kN}$$