8.3 Flexible Retaining Walls & Sheet Pile Bulkheads

Overview of Flexible Retaining Structures

Flexible retaining walls deflect under lateral earth pressure, mobilizing soil shear strength and inducing soil arching. Unlike rigid concrete gravity structures, flexible walls derive structural stability primarily from embedment depth into foundation soils and, for taller walls, upper anchor supports. Common structural forms include:

• Hot-Rolled Steel Sheet Piles: Interlocking Z-type or U-type steel profiles widely used in waterfront bulkheads, port quay walls, and deep excavations.
• Cold-Formed Sheet Piles: Lighter steel profiles utilized for shallow excavation support and temporary protection.
• Vinyl and Composite Sheet Piles: Corrosion-resistant synthetic profiles applied in marine environments.
• Timber Sheeting: Traditional temporary shoring method.

Flexible wall systems are broadly categorized into Cantilever Sheet Pile Walls and Anchored Sheet Pile Bulkheads.

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Cantilever Sheet Pile Walls

Cantilever sheet pile walls are un-anchored walls embedded into the ground below the excavation or dredge line. Stability depends entirely on passive soil resistance mobilized along the embedment depth ($D$).

          Wall Top
             |
             |   Retained Soil (Height H)
             |
-------------+------------- Dredge Line (z = 0)
             |   Passive Resistance
             |   (Depth D)
             v
         Pile Tip

1. Mechanics in Cohesionless Soils (Sand)

When lateral pressure pushes the retained height ($H$) outward, the wall rotates about a pivot point ($O$) located near the bottom tip. Active pressures act on the back face above point $O$, while massive passive pressures mobilize on the front face below the dredge line. Below point $O$, a small passive reaction mobilizes on the back face to maintain moment equilibrium.

To compute required embedment under simplified equilibrium:

1. Sum moments about the bottom rotation tip ($\sum M_O = 0$) to establish theoretical embedment depth $D_o$.
2. Increase $D_o$ by $20% \text{ to } 40%$ to provide an adequate safety factor: $D = (1.20 \text{ to } 1.40) \cdot D_o$
3. Determine the maximum bending moment ($M_{\text{max}}$) occurring at the depth below the dredge line where internal shear force equals zero ($V(z) = 0$).

2. Mechanics in Cohesive Soils (Clay under Undrained Conditions, $\phi = 0, c_u$)

For cohesive soils, the net passive resistance below the dredge line is constant: $p_{net} = 4c_u - \gamma H$. Theoretical embedment depth ($D_o$) is computed from:

$D_o = \frac{\gamma H^2}{8 c_u - 2 \gamma H}$

Limiting Height: A cantilever sheet pile in clay becomes unstable regardless of embedment depth when $\gamma H \ge 4 c_u$. Therefore, maximum retained height in clay is strictly bounded by $H_{\text{max}} < \frac{4 c_u}{\gamma}$. Cantilever walls in all soil types are generally limited to $H \le 5.0\text{ m}$.

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Anchored Sheet Pile Bulkheads

When retained heights exceed $5.0\text{ m}$, cantilever deflections and required sheet pile section sizes become prohibitive. Anchors (tie rods connected to deadman anchors or grouted tiebacks) are installed near the wall crest (typically at depth $a = 1.0 - 2.0\text{ m}$ below the top).

Anchored bulkheads are analyzed using two classical methods:

1. Free Earth Support (FES) Method

• Assumption: The sheet pile is relatively flexible or has minimal penetration. The embedded pile tip rotates freely without bottom bending moment fixity.
• Methodology:
  1. Take moments of all active and passive lateral forces about the anchor line ($\sum M_{\text{anchor}} = 0$) to solve directly for theoretical embedment depth $D_o$.
  2. Apply a safety factor of $1.20 - 1.30$ to theoretical embedment: $D = 1.20 D_o$.
  3. Sum horizontal forces ($\sum F_h = 0$) to calculate required anchor force per unit length of wall ($T$).

2. Fixed Earth Support (Blum's Equivalent Beam) Method

• Assumption: Deep embedment prevents rotation at the pile tip, creating a fixed-end boundary condition. An inflection point ($M = 0$) develops at depth $y_0$ below the dredge line.
• Methodology:
  • Position of inflection point: $y_0 \approx 0.10 H \text{ to } 0.25 H$ depending on sand friction angle.
  • The wall is split into an upper beam resting on the anchor and point of inflection, and a lower embedded cantilever beam.
  • Produces lower maximum flexural bending moments than FES, but requires $20% - 30%$ deeper penetration depths.

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Rowe's Moment Reduction Theory

Classical Free Earth Support theory overestimates maximum bending moments because it assumes rigid wall behavior. In reality, flexural wall deflection mobilizes soil arching, redistributing lateral pressures away from the flexible mid-span toward stiff anchor supports and embedded passive zones.

Rowe (1952, 1956) quantified moment reduction using the structural flexibility parameter ($\rho$):

$\rho = \frac{H^4}{E I} \quad \left(\text{expressed in } \frac{\text{m}^4}{\text{N}\cdot\text{m}^2} \text{ or } \frac{\text{ft}^4}{\text{lb}\cdot\text{in}^2}\right)$

where $H$ is total wall height ($H_{retained} + D$), $E$ is modulus of elasticity, and $I$ is moment of inertia of the sheet pile section.

 Moment Reduction Ratio (M_design / M_FES)
 1.0 |----------------\  (Rigid Wall)
     |                 \ 
 0.7 |                  \-- Medium Sand
     |                     \-- Dense Sand
 0.4 |                        \------------ (Very Flexible Wall)
     +------------------------------------
     -4.0        -3.5        -3.0    log10(rho)

• As wall flexibility increases (higher $\rho$), the design moment ratio ($M_{\text{design}} / M_{\text{FES}}$) decreases.
• In dense sand, design bending moments can be reduced by up to $30% \text{ to } 50%$ of theoretical FES moments, enabling selection of significantly lighter sheet pile sections.

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Design of Anchor Systems

Anchors provide critical horizontal restraint for bulkheads. Two main anchor configurations are used:

1. Deadman Anchors (Concrete Blocks or Continuous Walls)

Deadman anchors mobilize passive resistance in the upper backfill soil. The ultimate anchor capacity per unit length ($P_{ult}$) is:

$P_{ult} = P_p - P_a$

To ensure full passive capacity develops without interference, the deadman must be located completely behind the active Rankine failure surface originating from the base/dredge line of the wall:

$L_{\text{anchor}} \ge H \cot(45^\circ + \phi'/2) + D_{\text{deadman}} \cot(45^\circ + \phi'/2)$

2. Grouted Ground Anchors (Tiebacks)

Ground anchors consist of high-strength steel tendon bundles inserted into drilled boreholes, pressure-grouted in a competent soil or rock stratum:

• Unbonded Length ($L_u$): Extends beyond the potential slip surface to prevent anchor failure inside the active wedge.
• Bonded Length ($L_b$): Transfers tensile load to surrounding ground via grout skin friction ($\tau_{allow}$): $T_{\text{allow}} = \pi D_{\text{grout}} L_b \tau_{\text{allow}}$

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Worked Numerical Example: Free Earth Support Bulkhead Design

Problem Statement

An anchored sheet pile wall retains $H = 6.0\text{ m}$ of dense sand. An anchor line is located at depth $a = 1.20\text{ m}$ below the wall crest. Ground surface is horizontal, and groundwater is below the embedment depth. Soil properties:

• Unit weight $\gamma = 18.0\text{ kN/m}^3$, friction angle $\phi' = 34^circ$, $c' = 0$.
• Use Rankine's theory and the Free Earth Support method.

Calculate:

1. Theoretical embedment depth ($D_o$) and design embedment depth ($D = 1.20 D_o$).
2. Anchor force ($T$) per meter of wall.
3. Maximum flexural bending moment ($M_{\text{max}}$).

Step-by-Step Solution

1. Earth Pressure Coefficients

$K_a = \tan^2(45^circ - 34^circ/2) = \tan^2(28^circ) = 0.2827$ $K_p = \tan^2(45^circ + 34^circ/2) = \tan^2(62^circ) = 3.537$ $\text{Net Passive Coefficient: } K_{net} = K_p - K_a = 3.537 - 0.2827 = 3.2543$

2. Active Thrust Force Above Dredge Line

Active pressure at dredge line ($z = 6.0\text{ m}$): $\sigma_{a,dredge}' = K_a \gamma H = 0.2827 \times 18.0 \times 6.0 = 30.53\text{ kPa}$ Total active force above dredge line: $P_a = \frac{1}{2} \sigma_{a,dredge}' H = \frac{1}{2} \times 30.53 \times 6.0 = 91.59\text{ kN/m}$ Point of application below wall top: $y_a = \frac{2}{3} H = \frac{2}{3} \times 6.0 = 4.00\text{ m}$ (or $2.00\text{ m}$ above dredge line). Distance from anchor line ($a = 1.20\text{ m}$): $l_a = y_a - a = 4.00 - 1.20 = 2.80\text{ m}$

3. Moments About Anchor Line ($\sum M_{\text{anchor}} = 0$)

Active pressures below dredge line and net passive pressure over depth $D_o$:

• Net passive force over depth $D_o$: $P_{net} = \frac{1}{2} (K_{net} \gamma D_o) D_o = \frac{1}{2} \times 3.2543 \times 18.0 \times D_o^2 = 29.29 D_o^2\text{ kN/m}$
• Moment arm of $P_{net}$ from anchor line: $l_p = (H - a) + \frac{2}{3} D_o = 4.80 + 0.6667 D_o$
• Active pressure increment below dredge line force: $P_{a,sub} = \sigma_{a,dredge}' D_o + \frac{1}{2} K_a \gamma D_o^2 = 30.53 D_o + 2.544 D_o^2$
• Moment arm of active pressure rectangle below dredge line: $l_{a1} = 4.80 + 0.50 D_o$
• Moment arm of active triangle below dredge line: $l_{a2} = 4.80 + 0.6667 D_o$

Equating resisting moment of net passive force to active overturning moments: $M_{\text{passive}} = 29.29 D_o^2 (4.80 + 0.6667 D_o) = 140.59 D_o^2 + 19.53 D_o^3$ $M_{\text{active}} = (91.59 \times 2.80) + 30.53 D_o (4.80 + 0.50 D_o) + 2.544 D_o^2 (4.80 + 0.6667 D_o)$ $M_{\text{active}} = 256.45 + 146.54 D_o + 27.48 D_o^2 + 1.70 D_o^3$

Setting $M_{\text{passive}} = M_{\text{active}}$ and gathering terms: $17.83 D_o^3 + 113.11 D_o^2 - 146.54 D_o - 256.45 = 0$

Solving trial values for $D_o$:

• Try $D_o = 2.0\text{ m}$: $17.83(8) + 113.11(4) - 146.54(2) - 256.45 = 142.64 + 452.44 - 293.08 - 256.45 = 45.55 \approx 0$
• Refined solution: $D_o = 1.90\text{ m}$

Design Embedment Depth: $D = 1.20 \times D_o = 1.20 \times 1.90 = 2.28\text{ m} \quad \text{(Use } D = 2.30\text{ m)}$

4. Calculate Anchor Force ($T$)

Summing horizontal forces ($\sum F_h = 0$): $T = P_a + P_{a,sub} - P_{net}$ $P_{a,sub} = (30.53 \times 1.90) + (2.544 \times 1.90^2) = 58.01 + 9.18 = 67.19\text{ kN/m}$ $P_{net} = 29.29 \times (1.90)^2 = 105.74\text{ kN/m}$ $T = 91.59 + 67.19 - 105.74 = 53.04\text{ kN/m}$

5. Maximum Flexural Bending Moment ($M_{\text{max}}$)

Maximum moment occurs where shear force $V(z) = 0$ below the anchor ($z > a$): $V(z) = T - \frac{1}{2} K_a \gamma (z)^2 = 0$ $53.04 = \frac{1}{2} \times 0.2827 \times 18.0 \times z^2 = 2.544 z^2$ $z = \sqrt{\frac{53.04}{2.544}} = \sqrt{20.85} = 4.57\text{ m below wall top}$

$\text{Distance below anchor: } x_m = 4.57 - 1.20 = 3.37\text{ m}$ $M_{\text{max}} = T \cdot x_m - \frac{1}{2} K_a \gamma (4.57)^2 \cdot \left(\frac{4.57}{3}\right)$ $M_{\text{max}} = (53.04 \times 3.37) - (2.544 \times 20.89 \times 1.523) = 178.74 - 80.93 = 97.81\text{ kN}\cdot\text{m/m}$