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Chapter 5 · 3 hours

Flexible Retaining Structures and Coffer Dams

IOE past exam questions

Past questions and answers

12 questions set from this chapter, 2 of them more than once; 2 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 7 of 16 exams
  • Asked 7 times
  • 2081 Bhadra · 4 marks
  • 2079 Bhadra · 2 marks
  • 2076 Chaitra · 2 marks
  • 2075 Chaitra · 1+3 marks
  • 2074 Chaitra · 4 marks
  • 2076 Asoj · 1 mark
  • 2072 Chaitra · 2+2 marks

What is a cofferdam? Describe with neat sketches the different types of cofferdam, their uses and their relative merits and demerits.

Answer

A cofferdam is a temporary watertight enclosure built in water-bearing ground (river, lake, sea or wet soil) from which the water is pumped out so that foundations, piers or other structures can be built in dry conditions.

Types, sketches and uses

1. Earthen (and rockfill) embankment cofferdam

   ~~~ water ~~~\
                 \___________
                  /  clay core \
   ______________/______________\____

Used in shallow, slow rivers and low water depth (up to 3 m) for dams and bridge piers.

2. Single-wall (braced) sheet pile cofferdam

   ~~~ water   |sheet|====strut====|sheet|  water ~~~
               |pile |             |pile |
               |_____|  dry area   |_____|

A ring of interlocking sheet piles braced by wales and struts; used for piers and small foundations in water up to about 6-10 m deep.

3. Double-wall cofferdam

   | sheet pile |  sand/gravel fill  | sheet pile |
   |            |  (tied by rods)    |            |

Two parallel sheet-pile walls tied together with a gravel or sand fill between; used for deeper water (up to about 12 m).

4. Cellular cofferdam (circular, diaphragm or cloverleaf cells of steel sheet pile filled with sand/gravel)

    ()  ()  ()   (a row of circular cells joined by arcs,
   (  )(  )(  )   filled with granular soil)

Used for large works, dams, locks and deep water (up to about 25 m); self-supporting without internal bracing.

5. Crib (timber or concrete) and other special types: boxes or cribs filled with stones; used in rocky riverbeds.

Relative merits and demerits

TypeMeritsDemerits
Earth/rockfillCheap, simple, local materialNeeds large area; leaks; unsuitable for deep or fast water
Single-wall bracedEconomical, quick for small sitesStruts obstruct work; limited depth; leakage at joints
Double-wallStronger, good for greater depth, less leakageCostly, requires much material and space
CellularNo bracing, large clear working area, reusable, good in deep waterHigh cost, needs skilled work, needs rock/stiff base to avoid scour and slip, failure if a cell breaks
  • Most repeated · 3 of 16 exams
  • Asked 3 times
  • 2078 Kartik · 3 marks
  • 2074 Asoj · 3 marks
  • 2075 Asoj · 4 marks

In what respects does the design of a flexible retaining structure vary from that of a rigid retaining structure (in terms of stability and deformation analysis)?

Answer

PointRigid retaining structure (gravity/cantilever wall)Flexible retaining structure (sheet pile, anchored bulkhead)
BehaviourMoves as a rigid body (rotation or sliding); bending is smallBends (deflects) under the soil load; the deflection is part of the support
Earth pressureTriangular active pressure from Rankine/CoulombPressure redistributes due to arching and wall deflection; assumed simplified diagrams (net pressure)
Stability analysisOverturning, sliding, bearing capacity, tension at the baseDepth of embedment by moment equilibrium (free-earth/fixed-earth), anchor force, passive resistance in front
Structural designMostly mass or short stem designMaximum bending moment in the pile and the section modulus (for sheet piles); anchor and wale design
SupportWeight and base frictionPassive resistance of the embedded portion and anchors/struts
Deformation analysisSmall deformations; mostly checked for settlement and tiltDeflection and movement of retained ground must be checked; depends on pile stiffness EIEI and anchor level
  • 2078 Bhadra · 2 marks

Define cofferdam and explain the cellular cofferdam with the help of a neat sketch.

Answer

A cofferdam is a temporary enclosure built in water or water-bearing ground, pumped dry to allow construction below water level.

A cellular cofferdam consists of a row of connected cells made of interlocking flat-web steel sheet piles, filled with sand, gravel or rock fill. Cells may be circular, diaphragm (arcs with cross walls) or cloverleaf.

   Plan:  ( )( )( )( )    Section:  | fill |
           circular cells            |      |  ~~~ water
        joined by connecting arcs    |______|    (dry on right)
  • The sheet piles act in tension (ring tension) and hold the fill; the filled cell behaves as a gravity wall resisting water pressure by its own weight.
  • It is self-supporting, so no internal bracing blocks the working area.
  • Stability is checked for sliding, overturning, interlock tension (hoop tension T=p rT = p\,r), shear along the centre line of the cell, and bearing.
  • Uses: piers, locks, dams, deep water work; reusable and quick.
  • 2079 Bhadra · 2 marks

Why is a sheet pile wall considered a flexible retaining structure?

Answer

A sheet pile wall is thin and long (slender), with low flexural rigidity EIEI compared to a gravity or cantilever wall. Under the earth pressure it bends and deflects, mostly in the cantilever or between supports, and it does not move as a rigid body. Its stability depends on bending and the passive resistance of the embedded part and anchors, rather than on its own weight. Therefore it is called a flexible retaining structure.

  • 2080 Baisakh · 1+3 marks

What is a sheet pile? Draw the clear deflected shapes for the following: (i) cantilever sheet pile, (ii) anchored bulkhead driven to shallow depth and (iii) anchored bulkhead driven to deeper depth.

Answer

A sheet pile is a slender, interlocking section of steel, timber or concrete driven into the ground to form a continuous wall that retains soil or water (e.g. in excavations, cofferdams and waterfronts).

Deflected shapes

 (i) Cantilever       (ii) Anchored,       (iii) Anchored,
                          shallow depth        deeper depth
  |  soil side         T o-->|                T o-->|
  |                          |                      |
  |\                         |\                     |  | \                        | \                    |  -|--\- dredge line        --|--\- dredge        --|--\- dredge
  |   )                      |   )                  |    )
  |  /  toe moves back       |  /  tip free         |   (  contra-
  | /   (pivot point)        | /   (rotates)        |    ) flexure
  |/                         |/                     |   / fixed toe
  • (i) Cantilever: the wall deflects toward the excavation and pivots about a point near the lower third of the embedded length; the toe moves backward.
  • (ii) Anchored bulkhead, shallow depth (free earth support): the pile is held at the anchor and rotates freely about the tip; single curvature, no fixity at the bottom.
  • (iii) Anchored bulkhead, deeper depth (fixed earth support): the toe is fixed in the soil, so the deflected curve has a point of contraflexure, with reverse curvature near the bottom.
  • 2078 Bhadra · 2 marks

Write down the differences between free earth and fixed earth support anchored pile.

Answer

PointFree earth supportFixed earth support
Embedment depthShallow (smaller)Deeper (larger)
Toe conditionPile tip free to rotate, no fixityPile tip is fixed (zero deflection and rotation)
Passive pressureOnly in front of the pile, up to the tip; the pile acts as simply supported at anchor and soilPassive pressure in front and a counter (reverse) passive force behind the toe
Bending momentLarger maximum momentSmaller maximum moment (negative moment near the toe)
Deflected shapeOne curve, no contraflexureHas a point of contraflexure
AnalysisSimple statics, moments about the anchorEquivalent beam (Blum) or other methods; more complex
  • 2080 Bhadra · 4 marks

Assume suitable condition to prove that use of anchor can reduce the depth of embedment of cantilever sheet pile wall.

Answer

Assumed condition: sheet pile wall retaining H=5H = 5 m of dry sand (ϕ=30∘\phi = 30^\circ, same on both sides), γ\gamma the same on both sides, passive resistance with FS=1FS = 1 for the comparison, anchor at 1 m below the top (free earth support).

Ka=tan⁡2(45∘−15∘)=0.3333,Kp=tan⁡2(45∘+15∘)=3.0K_a = \tan^2 (45^\circ-15^\circ) = 0.3333,\qquad K_p = \tan^2(45^\circ+15^\circ) = 3.0

(a) Cantilever wall: the moment about the toe must balance:

12Kaγ(H+D)2(H+D)3=12KpγD2D3 ⇒ DH+D=(KaKp)1/3=0.4807\tfrac12K_a\gamma(H+D)^2\frac{(H+D)}{3} = \tfrac12K_p\gamma D^2\frac{D}{3}\ \Rightarrow\ \frac{D}{H+D} = \left(\frac{K_a}{K_p}\right)^{1/3} = 0.4807 Dcant=5×0.48071−0.4807=4.63 mD_{cant} = \frac{5\times0.4807}{1-0.4807} = 4.63\ \text{m}

(b) Anchored wall (free earth support): take moments about the anchor point (the anchor force TT drops out):

12Kaγ(H+D)2[23(H+D)−1]=12KpγD2[H+23D−1]\tfrac12K_a\gamma(H+D)^2\left[\tfrac23(H+D) - 1\right] = \tfrac12K_p\gamma D^2\left[H + \tfrac23D - 1\right]

Solving, Danch=1.90D_{anch} = 1.90 m, and the anchor force from ∑H=0\sum H = 0 is

T=12Kaγ(H+D)2−12KpγD2 (per unit γ)=7.94γ−5.43γ=2.51γT = \tfrac12K_a\gamma(H+D)^2 - \tfrac12K_p\gamma D^2 \ (\text{per unit }\gamma) = 7.94\gamma - 5.43\gamma = 2.51\gamma

(with γ=18\gamma = 18 kN/m3^3, T≈45T \approx 45 kN/m).

Comparison: Danchored=1.90D_{anchored} = 1.90 m ≪Dcantilever=4.63\ll D_{cantilever} = 4.63 m, about 59% less. The anchor supplies the horizontal reaction that the cantilever wall must get from the passive resistance of a deep embedment, so the embedment, and also the pile length and bending moment, are reduced.

  • 2078 Kartik · 1+3 marks

Enlist the uses of sheet pile. Write down the step by step procedure of analysis of anchored sheet pile by free earth support method in granular soil.

Answer

Uses of sheet piles

  • Retaining walls for waterfront structures (quay walls, bulkheads, jetties).
  • Temporary support of excavations (braced cuts and trenches).
  • Cofferdams for bridge piers and underwater foundations.
  • Cut-off walls under dams, seepage control, flood protection.
  • Protection against scour and for river training, and landfill containment.

Free earth support method (granular soil)

Assumption: the pile tip is free to rotate (no fixity); the active pressure acts on the full height (H+D)(H+D) on the back, passive pressure acts only in front; the wall is simply supported by the anchor and the passive resistance.

  1. Compute Ka=tan⁡2(45∘−ϕ/2)K_a = \tan^2(45^\circ-\phi/2) and Kp=tan⁡2(45∘+ϕ/2)K_p = \tan^2(45^\circ+\phi/2); use γ′\gamma' below water table and include surcharge and water pressure.
  2. Draw the active pressure diagram on the back from top to depth H+DH+D and the passive pressure diagram in front from the dredge line to depth DD (using Kp/FSK_p/FS if a factor of safety is applied to passive resistance).
  3. Find the active resultant PaP_a and its lever arm from the anchor level.
  4. Find the passive resultant PpP_p (in terms of DD) and its lever arm from the anchor.
  5. Take moments about the anchor point: Pa la=Pp lpP_a\, l_a = P_p\, l_p and solve for the embedment DD (trial and error).
  6. Find the anchor force per metre TT from horizontal equilibrium: T=Pa−PpT = P_a - P_p.
  7. Increase DD by 20-30% for the design depth, then find the maximum bending moment (at zero shear) and select the sheet pile section from Z=Mmax/σallowZ = M_{max}/\sigma_{allow}.
  • 2082 Bhadra · 4 marks

Find the depth of embedment for the cantilever sheet pile supporting 8 m deep excavation in cohesionless soil. Take γsat=20 kN/m3\gamma_{sat} = 20\ \text{kN/m}^3, γ=20 kN/m3\gamma = 20\ \text{kN/m}^3, ϕ=33∘\phi = 33^\circ and water table at 6.5 m from the ground surface.

Answer

Data and assumptions: excavation depth H=8H = 8 m (dredge level); γ=γsat=20\gamma = \gamma_{sat} = 20 kN/m3^3, ϕ=33∘\phi = 33^\circ; water table at 6.5 m below ground, taken at the same level on both sides (the excavation is flooded to the water table, no seepage and no net water pressure), so effective stresses with γ′=20−9.81=10.19\gamma' = 20 - 9.81 = 10.19 kN/m3^3 are used below the water table. Moments are taken about the pile tip (simplified cantilever method).

Coefficients

Ka=tan⁡2(45∘−16.5∘)=0.2948,Kp=tan⁡2(45∘+16.5∘)=3.3921K_a = \tan^2(45^\circ - 16.5^\circ) = 0.2948,\qquad K_p = \tan^2(45^\circ + 16.5^\circ) = 3.3921

Pressures

  • Active (back), depth zz: pa=Kaσv′p_a = K_a\sigma_v'; at 6.5 m =38.32= 38.32 kN/m2^2; below 6.5 m the effective stress increases with γ′\gamma'.
  • Passive (front), depth yy below the dredge line: pp=Kpγ′y=34.57 yp_p = K_p\gamma' y = 34.57\,y kN/m2^2.

Moment equilibrium about the tip: for a trial embedment DD,

Mactive(D)=∫08+Dpa (8+D−z) dz,Mpassive(D)=16Kpγ′D3M_{active}(D) = \int_0^{8+D} p_a\,(8+D-z)\,dz,\qquad M_{passive}(D) = \tfrac16K_p\gamma' D^3

Setting Mactive=MpassiveM_{active} = M_{passive} and solving by trial and error:

D=9.10 m:Mactive=4338,Mpassive=4338 kN m/mD = 9.10\ \text{m}:\quad M_{active} = 4338,\quad M_{passive} = 4338\ \text{kN m/m}

(Active force 699 kN/m, passive force 1430 kN/m at this depth.)

Design depth: increase by 20% for safety (practice 20-40%):

Ddesign=1.2×9.10=10.9 mD_{design} = 1.2\times9.10 = 10.9\ \text{m}

Answer: theoretical embedment D≈9.1D \approx 9.1 m (below the 8 m dredge level); provide ≈11\approx 11 m, i.e. a total pile length of about 19 m. For comparison, without the water table, D=6.4D = 6.4 m.

  • 2076 Asoj · 3 marks

An excavation of 5 m deep is to be carried out in sandy soil deposit having unit weight =22 kN/m3= 22\ \text{kN/m}^3 and angle of shearing resistance =33∘= 33^\circ. To support the soil, cantilever sheet pile walls are driven into the ground prior to excavation. Determine the depth of embedment needed for the sheet pile to retain the backfill. The water table is located below the base of the sheet pile.

Answer

Data: H=5H = 5 m, γ=22\gamma = 22 kN/m3^3, ϕ=33∘\phi = 33^\circ, no water above the pile tip. Cantilever sheet pile, simplified moment method (moments about the toe).

Ka=tan⁡2(45∘−16.5∘)=0.2948,Kp=tan⁡2(45∘+16.5∘)=3.3921K_a = \tan^2(45^\circ - 16.5^\circ) = 0.2948,\qquad K_p = \tan^2(45^\circ + 16.5^\circ) = 3.3921

Forces (with DD = embedment below the dredge line, γ\gamma same on both sides):

  • Active force on the back: Pa=12Kaγ(H+D)2P_a = \tfrac12K_a\gamma(H+D)^2, acting at (H+D)/3(H+D)/3 above the tip.
  • Passive force in front: Pp=12KpγD2P_p = \tfrac12K_p\gamma D^2, acting at D/3D/3 above the tip.

Moment about the tip

12Kaγ(H+D)2H+D3=12KpγD2D3 ⇒ DH+D=(KaKp)1/3=(0.29483.3921)1/3=0.4429\tfrac12K_a\gamma(H+D)^2\frac{H+D}{3} = \tfrac12K_p\gamma D^2\frac{D}{3} \ \Rightarrow\ \frac{D}{H+D} = \left(\frac{K_a}{K_p}\right)^{1/3} = \left(\frac{0.2948}{3.3921}\right)^{1/3} = 0.4429 D=0.4429×51−0.4429=3.98 mD = \frac{0.4429\times5}{1-0.4429} = 3.98\ \text{m}

Design depth: since the method assumes full passive resistance and ignores the extra tip reaction, increase by 20%:

Ddesign=1.2×3.98=4.77 mD_{design} = 1.2\times3.98 = 4.77\ \text{m}

Answer: theoretical embedment D=3.98D = 3.98 m; adopt D≈4.8D \approx 4.8 m (total pile length about 9.8 m).

  • 2074 Asoj · 5 marks

A cantilever sheet pile wall is driven into sand deposit having friction angle 35∘35^\circ and bulk unit weight 22 kN/m322\ \text{kN/m}^3. One side of the sheet pile was backfilled to 3 m height. The backfill material is cohesionless sand having ϕ=32∘\phi = 32^\circ and bulk unit weight of 18 kN/m318\ \text{kN/m}^3. Using the simplified method determine the depth of penetration needed for the sheet pile to retain the backfill. Provide a safety factor of 2 for the passive resistance. The water table is below the base of the sheet pile.

Answer

Data: deposit (in which the pile is driven): ϕ2=35∘\phi_2 = 35^\circ, γ2=22\gamma_2 = 22 kN/m3^3. Backfill 3 m high on one side: ϕ1=32∘\phi_1 = 32^\circ, γ1=18\gamma_1 = 18 kN/m3^3. FS=2FS = 2 on passive resistance. Water table below the pile tip. Simplified method: moments about the pile tip.

Coefficients

Ka1=tan⁡2(45∘−16∘)=0.3073,Ka2=tan⁡2(45∘−17.5∘)=0.2710,Kp=tan⁡2(45∘+17.5∘)=3.6902K_{a1} = \tan^2(45^\circ-16^\circ) = 0.3073,\quad K_{a2} = \tan^2(45^\circ-17.5^\circ) = 0.2710,\quad K_p = \tan^2(45^\circ+17.5^\circ) = 3.6902

Active pressures (back, depth zz from the top of the backfill; the dredge line is at z=3z=3)

  • Backfill (0≤z≤30 \le z \le 3): p=Ka1γ1zp = K_{a1}\gamma_1 z, giving 16.59 kN/m2^2 at 3 m.
  • Below 3 m (in the deposit): p=Ka2 [18×3+22(z−3)]=14.63+5.962(z−3)p = K_{a2}\,[18\times3 + 22(z-3)] = 14.63 + 5.962(z-3) kN/m2^2.

Passive pressure (front, depth yy below the dredge line)

pp=Kpγ2yFS=3.690×22 y2=40.59 y kN/m2p_p = \frac{K_p\gamma_2 y}{FS} = \frac{3.690\times22\,y}{2} = 40.59\,y\ \text{kN/m}^2

Moments about the tip (embedment DD): the active moment is

Ma(D)=∫03+Dpa (3+D−z) dzM_a(D) = \int_0^{3+D} p_a\,(3+D-z)\,dz

and the passive moment is Mp=16 (Kpγ2/FS) D3=6.765 D3M_p = \tfrac16\,(K_p\gamma_2/FS)\,D^3 = 6.765\,D^3. Solving Ma=MpM_a = M_p by trial and error gives

D=3.10 m:Ma=202.4,Mp=202.4 kN m/mD = 3.10\ \text{m}:\qquad M_a = 202.4,\quad M_p = 202.4\ \text{kN m/m}

(active force 99.0 kN/m, passive force 195.6 kN/m).

Answer: required depth of penetration D≈3.10D \approx 3.10 m below the 3 m backfill (dredge) level. Since FS=2FS = 2 is already applied to the passive resistance, no further increase is needed; the total pile length is about 6.1 m.

  • 2081 Baisakh · 3+1 marks

Determine the depth of embedment of the free earth support anchored sheet pile shown in figure below. Also, find the tensile force (T) per meter acting on the anchor. Assume necessary conditions. [Figure: anchored sheet pile with 8.0 m height above dredge level, anchor (tie rod) T about 2.5 m below top; soil above water level γ=16 kN/m3\gamma = 16\ \text{kN/m}^3, ϕ=35∘\phi = 35^\circ for about 3.0 m; below that γsat=19 kN/m3\gamma_{sat} = 19\ \text{kN/m}^3, ϕ=35∘\phi = 35^\circ. Dimensions as read from the scan.]

Answer

Data and assumptions (figure values as read): height above dredge level H=8.0H = 8.0 m; anchor (tie rod) at 2.5 m below the top; soil above the water table: γ=16\gamma = 16 kN/m3^3, ϕ=35∘\phi = 35^\circ for the top 3.0 m; below: γsat=19\gamma_{sat} = 19 kN/m3^3, ϕ=35∘\phi = 35^\circ. Water level is taken the same on both sides (no net water pressure), so γ′=19−9.81=9.19\gamma' = 19 - 9.81 = 9.19 kN/m3^3 is used below the water table. Free earth support method, passive resistance with FS=1FS = 1 for the theoretical depth.

Ka=tan⁡2(45∘−17.5∘)=0.2710,Kp=tan⁡2(45∘+17.5∘)=3.6902K_a = \tan^2(45^\circ-17.5^\circ) = 0.2710,\qquad K_p = \tan^2(45^\circ+17.5^\circ) = 3.6902

Active pressure (back, depth zz): pa=Kaσv′p_a = K_a\sigma_v'

  • z=3z = 3 m: σv′=48\sigma_v' = 48, pa=13.01p_a = 13.01 kN/m2^2.
  • z=8z = 8 m (dredge): σv′=48+9.19×5=93.95\sigma_v' = 48 + 9.19\times5 = 93.95, pa=25.46p_a = 25.46 kN/m2^2.
  • Below the dredge line: the effective stress increases by γ′\gamma' per metre.

Passive pressure (front): pp=Kpγ′y=33.91 yp_p = K_p\gamma' y = 33.91\,y kN/m2^2, yy measured below the dredge line.

Moment about the anchor (level 2.5 m): for a trial embedment DD,

∑Mactive=∫08+Dpa (z−2.5) dz,∑Mpassive=∫0Dpp (8+y−2.5) dy\sum M_{active} = \int_0^{8+D} p_a\,(z-2.5)\,dz,\qquad \sum M_{passive} = \int_0^{D} p_p\,(8 + y - 2.5)\,dy

Equating and solving by trial and error:

D=2.57 mD = 2.57\ \text{m}

At this depth: active force Pa=189.3P_a = 189.3 kN/m, passive force Pp=112.0P_p = 112.0 kN/m.

Design embedment: increase by 30% (free earth support): Ddesign=1.3×2.57=3.3D_{design} = 1.3\times2.57 = 3.3 m.

Tension in the anchor from horizontal equilibrium:

T=Pa−Pp=189.3−112.0=77.3 kN/mT = P_a - P_p = 189.3 - 112.0 = 77.3\ \text{kN/m}

Answer: depth of embedment D≈2.57D \approx 2.57 m (adopt about 3.3 m); anchor force T≈77T \approx 77 kN per metre run.

Questions from Old Question Collection (CE 602) (IOE BCE exam papers from 2072 Chaitra to 2082 Bhadra (last two scans cut off)). Answers are written for this site; check them against your class notes.

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