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Sheet Pile Walls

Sheet piles are interlocking sections driven to retain soil and water along waterfronts, cofferdams, and bulkheads. Cantilever sheet piles rely on passive resistance of the embedded length; anchored sheet piles add a tie rod near the top. The wall is designed against the soil's active pressure (driving) balanced by passive pressure (resisting), using Rankine coefficients.

For waterfront design, the governing water level on the retained side is the residual water level (RWL), taken above the low water level by two-thirds of the tidal range:

$$\text{RWL}=\text{LWL}+\tfrac{2}{3}\left(\text{HWL}-\text{LWL}\right)$$
$$K_a=\tan^2\!\left(45^\circ-\tfrac{\phi}{2}\right),\quad K_p=\tan^2\!\left(45^\circ+\tfrac{\phi}{2}\right),\quad P=\tfrac{1}{2}K\gamma H^2$$

★ Residual Water Level

For a sheet pile wall, the datum (MLLW) is 0.00 m, the high water level is +1.50 m, and the low water level is −0.30 m. Determine the residual water level (RWL).

$$\text{RWL}=\text{LWL}+\tfrac{2}{3}(\text{HWL}-\text{LWL})$$
$$=-0.30+\tfrac{2}{3}\big(1.50-(-0.30)\big)=-0.30+\tfrac{2}{3}(1.80)$$
$$=-0.30+1.20=+0.90\ \text{m}$$

Final answer: RWL = +0.90 m.

★ Active Earth Pressure Coefficient

A granular backfill has an angle of internal friction $\phi=30^\circ$. Determine the Rankine active earth pressure coefficient.

$$K_a=\tan^2\!\left(45^\circ-\tfrac{30^\circ}{2}\right)=\tan^2(30^\circ)$$
$$=\left(\tfrac{1}{\sqrt3}\right)^2=0.333$$

Final answer: $K_a = 0.333$.

★★ Active Thrust on the Wall

The backfill has $\gamma=18$ kN/m³, $\phi=30^\circ$ ($K_a=0.333$), and the retained height is 6 m. Determine the total active thrust per meter of wall and its point of application.

$$P_a=\tfrac{1}{2}K_a\gamma H^2=\tfrac{1}{2}(0.333)(18)(6)^2$$
$$=\tfrac{1}{2}(0.333)(18)(36)=108\ \text{kN/m}$$

The pressure is triangular, so the resultant acts at $H/3 = 2$ m above the base. Final answer: 108 kN/m at 2 m above the base.

★★ Passive Resistance on the Embedment

The same soil ($\gamma=18$ kN/m³, $\phi=30^\circ$) provides passive resistance over a 3 m embedment depth. Determine the passive thrust per meter.

$$K_p=\tan^2\!\left(45^\circ+\tfrac{30^\circ}{2}\right)=\tan^2(60^\circ)=3.0$$
$$P_p=\tfrac{1}{2}K_p\gamma d^2=\tfrac{1}{2}(3.0)(18)(3)^2=243\ \text{kN/m}$$

Final answer: 243 kN/m. Note $K_p=1/K_a$, so passive pressure greatly exceeds active.

★★★ Overturning Moment of the Active Thrust

Using the active thrust from the earlier item (108 kN/m acting 2 m above the base), determine the overturning moment per meter of wall taken about the base.

$$M_{OT}=P_a\times\bar{y}=108\times2=216\ \text{kN}\cdot\text{m/m}$$

This moment must be resisted by the passive pressure on the embedded length plus any anchor force. Final answer: 216 kN·m per meter.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: w57

MSTE - Ports and Harbors / Sheet Pile Walls / MSTE May 2019

Determine the elevation of the residual water level (RWL) for the sheet pile type if the following data are given:
Elev. of MLLW (datum) $= 0.00$ m
Elev. of HWL $= +1.26$ m
Elev. of LWL $= -0.23$ m

  1. +0.82 m
  2. +0.76 m
  3. +0.63 m
  4. +0.96 m
Tidal range: $h_r = 1.26 - (-0.23) = 1.49\text{ m}$.
The residual head is normally $\tfrac13$ to $\tfrac23$ of the tidal range; take $h_w = \tfrac23 h_r = \tfrac23(1.49) = 0.993\text{ m}$.
$RWL = -0.23 + 0.993 = \boxed{+0.763\text{ m}}$