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Chapter 9 · 4 hours

Well Foundations

IOE past exam questions

Past questions and answers

12 questions set from this chapter, 6 of them more than once; 3 are most repeated (set, or a close variant set, in 3 or more exams). Most repeated first.

  • Most repeated · 9 of 16 exams
  • Asked 9 times
  • 2082 Bhadra · 4 marks
  • 2081 Baisakh · 4 marks
  • 2078 Bhadra · 4 marks
  • 2078 Kartik · 1+4 marks
  • 2076 Chaitra · 6 marks
  • 2076 Asoj · 1+4 marks
  • 2075 Asoj · 5 marks
  • 2072 Chaitra · 4 marks
  • 2080 Baisakh · 4 marks

Draw a well foundation showing its different components (with the function of each component) and describe a well foundation.

Answer

A well (caisson) foundation is a hollow, box-like masonry or concrete structure, circular, rectangular or D-shaped in plan, sunk through the soil by dredging from inside until it reaches the founding level. It is then plugged and filled, and carries the bridge pier above. It is widely used for bridges over rivers where scour is deep.

        Pier
     ____|____
    |  Well   |  <- Well cap
    |  cap    |
  --|---------|-- GL / LWL
    |  Top    |
    |  plug   |
    | Sand/   |  <- Steining (walls)
    | filling |
    |         |
    |Interm.  |
    |plug     |
    |  Bottom |
    |  plug   |
    |_Curb____|  <- Cutting edge
     \       /

Components and functions

  1. Cutting edge: steel shoe at the bottom that cuts the soil as the well sinks; protects the curb.
  2. Well curb: reinforced concrete wedge-shaped ring above the cutting edge (with an inner slope of 30° to 37°); it facilitates sinking and transfers the load to soil.
  3. Steining: the walls of the well, of masonry or reinforced concrete; carry load to the base and provide weight for sinking; usually 0.5 to 1.5 m thick.
  4. Bottom plug: concrete plug placed under water (tremie) in the bottom of the well to seal the base and transfer the load to the soil; thickness at least 1/2 the internal diameter or by design.
  5. Sand filling: clean sand placed above the bottom plug to give weight and stability and to transfer the load.
  6. Top plug: a layer of concrete (about 300 mm) at the top of the sand filling, to support the well cap.
  7. Well cap: a reinforced concrete slab on top of the steining that distributes the pier load uniformly to the well walls.
  8. Intermediate plug (if required) and dredge hole: the dredge hole is the central opening for removing soil by grabs.

Description

Well foundations resist large vertical and horizontal loads (by their weight and by passive resistance of soil on the sides, with a grip length below the maximum scour level). They are suitable for sinking through soft soils, boulders, or sand, and are common in Nepal for large river bridges.

  • Most repeated · 9 of 16 exams
  • Asked 9 times
  • 2082 Bhadra · 4 marks
  • 2081 Baisakh · 2 marks
  • 2080 Bhadra · 5 marks
  • 2078 Kartik · 3 marks
  • 2078 Bhadra · 4 marks
  • 2076 Asoj · 3 marks
  • 2075 Asoj · 3 marks
  • 2074 Chaitra · 6 marks
  • 2073 Shrawan · 4 marks

Discuss the remedial measures (methods) to rectify tilt and shift of well foundation, with clear sketches; the measures taken for sinking wells and correcting tilts and shifts.

Answer

Tilt is the inclination of the axis of the well from the vertical, and shift is the horizontal displacement of the well from its intended position. The permissible tilt is 1 in 80 and shift about 150 mm at the founding level (IRC 78). They are corrected during sinking.

Causes

Uneven dredging, obstructions (boulders, logs), non-uniform soil, uneven kentledge, sudden slipping of the well and eccentric load.

Remedial measures for tilt

  1. Dredging unevenly: remove more soil from the high side and none on the low side so the well sinks on the high side.
  2. Eccentric kentledge: place extra load (kentledge) on the high side.
  3. Water jetting: jet water at the high side to reduce skin friction (or pumping).
  4. Pulling with wire ropes (winches): pull the top of the well against the tilt using ropes anchored to piles or to deadmen, with a jack.
  5. Strutting the low side against a firm abutment (props or anchor) to prevent more tilt.
  6. Blasting or removing the obstruction (boulders, wood) at the low side.
  7. Pushing with jacks against the sides, in sand, if only a small tilt remains.

Remedial measures for shift

  1. Pulling the well sideways with wire ropes/winches anchored to a fixed anchor.
  2. Excavation on the side opposite to the direction of required movement to allow the well to move into the dredged space.
  3. Intentional tilt: tilt the well in the opposite direction and then sink it vertically so it returns to position.
  4. Eccentric sinking and the use of jacks against the sides.
 Tilted well       Correction
   \   \         kentledge   -> 
    \   \        dredge high side
     \___\       pull rope  <---

Measures during sinking: keep the well central with guides, sink uniformly with dredging symmetrical about the centre, check levels and verticality regularly, and use the soil type information.

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

How do you determine the depth of the well? Describe the sinking process of the well (in a site having water table above the ground level).

Answer

Depth of a well foundation

The depth is fixed by:

  1. Scour depth. The founding level is the maximum scour depth plus a grip length. Normal scour depth by Lacey's formula d=1.34(q2f)1/3d = 1.34\left(\dfrac{q^2}{f}\right)^{1/3} (qq = discharge per metre width, ff = silt factor). The maximum scour depth is taken as 2d2d at piers (or 1.27d1.27d to 1.5d1.5d for abutments), measured below HFL.
  2. Grip length: at least 13\tfrac{1}{3} of the maximum scour depth below the maximum scour level, for lateral stability.
  3. Bearing capacity: a stratum which can carry the load safely (settlement within limits).
  4. Stability against overturning and sliding (passive resistance of soil), and the depth at which a firm stratum or rock is found.
  5. Also, the depth to avoid the effect of liquefaction or soft layers and for cutoff.

Sinking of a well (in site with water above ground level)

  1. Construction site: a sand island (a temporary sand-filled island enclosed by sandbags/bamboo) is built above the water level in shallow water, or a floating caisson (a steel shell) is used in deep water.
  2. Cutting edge and curb: placed on the island and a first lift of steining (about 2 to 3 m) is cast over them.
  3. Dredging: soil is removed from the dredge hole by a grab, an air-lift or divers, so the well sinks under its own weight. Steining is built in lifts as it sinks, keeping about 1 m above water.
  4. Aids to sinking: kentledge (extra weights), jetting water to reduce skin friction, use of bentonite slurry (lubrication), or pulling.
  5. Verticality control: correct tilt and shift as sinking continues.
  6. Founding: at the founding level the base is cleaned, inspected and the bottom plug is placed by tremie under water.
  7. Finishing: the well is dewatered, sand-filled, topped with a top plug and capped by the well cap.
  • Asked 2 times
  • 2080 Baisakh · 2 marks
  • 2074 Chaitra · 2 marks

What are the situations/advantages where a well foundation is more suitable than the other types of foundations?

Answer

A well foundation is more suitable than other foundations in the following situations.

  1. Deep scour: in rivers with deep scour and where the founding level must be well below the bed (bridge piers and abutments).
  2. Large vertical and horizontal loads: a large cross-section gives high bearing capacity and large lateral resistance.
  3. Soil with boulders or hard layers where piles cannot be driven easily.
  4. Pier shape: the well can be sunk through a variable depth of soil to a firm stratum.
  5. Difficult dewatering: it can be sunk in water without sheeting and open excavation.

Advantages

  • It has a large rigidity and gives a large moment of resistance.
  • It is constructed with simple equipment and local labour.
  • Very high load capacity: the structure is solid, and settlement is small.
  • It can be inspected at the base, and the quality of founding stratum is verified.
  • Economical for deep foundations of bridges and river training works.
  • Asked 2 times
  • 2081 Baisakh · 2 marks
  • 2072 Chaitra · 2 marks

What are the different forces acting on a well foundation?

Answer

The forces acting on a well foundation are:

  1. Vertical loads: dead load of the superstructure and the pier, live load on the bridge, self-weight of the well, and buoyancy (uplift).
  2. Horizontal forces: braking/tractive force of vehicles, centrifugal force on curves, wind load, water current force on the pier and well, and earthquake (seismic) forces.
  3. Earth pressure: active and passive earth pressure on the well and surcharge on the soil, for abutments.
  4. Temperature, shrinkage and friction at bearings (secondary forces).
  5. Forces during sinking: skin friction of the soil and the stresses from the weight and kentledge.
  6. Moments due to the above forces at the scour level (the horizontal force times its lever arm), and buoyancy and seepage (water pressure).

These produce a resultant horizontal force HH and moment MM at the scour level, which the lateral resistance of the soil must resist.

  • Asked 2 times
  • 2075 Chaitra · 1 mark
  • 2072 Chaitra · 2 marks

Name the different methods to analyze the lateral stability of a well foundation.

Answer

The lateral stability of a well foundation (resistance to horizontal force and moment) is analysed by:

  1. Terzaghi's analysis (simplified), for light and heavy wells.
  2. IRC:45 method (the Indian Roads Congress code, based on the modified Terzaghi method).
  3. Banerjee and Gupta's method (the elastic / subgrade-reaction approach).
  4. Subgrade reaction (Winkler) method.
  5. Finite element methods for complex cases.
  • 2081 Bhadra · 3 marks

List out the precautionary measures that should be taken to prevent tilt and shift of well foundation.

Answer

To prevent tilt and shift of a well during sinking, the following precautions are taken.

  1. Proper foundation for the sinking: the sand island or the ground is made level and firm before the cutting edge is laid.
  2. Accurate construction of curb and steining: the well curb and steining are built truly vertical and symmetrical, and in uniform thickness with equal lifts.
  3. Uniform dredging: dredging must be done evenly around the dredge hole, taking out the soil from the centre first and not from the sides.
  4. Even kentledge and loading: load must be placed symmetrically.
  5. Regular checks: check the tilt and shift at least once per shift with a plumb and levels; keep records of the levels.
  6. Guide frames / dolphin, to keep the well in position during the initial sinking.
  7. Obstructions: examine the soil first and remove boulders and tree trunks in advance, using divers and blasting.
  8. Avoid sudden sinking by controlling dredging and by not dredging much below the cutting edge.
  9. Care during the final stage when the well reaches the founding level.
  • 2080 Baisakh · 2 marks

What do you understand by grip length? What is its importance in well foundation?

Answer

Grip length is the depth of the well foundation below the maximum scour level (up to the bottom of the well). IRC practice takes it as at least 13\tfrac{1}{3} of the maximum scour depth below HFL.

Importance

  • It gives the lateral stability: the passive resistance of the soil on the sides of the grip length resists the horizontal force and the overturning moment.
  • It gives safety against scour that is greater than expected, and protects the base of the well from exposure.
  • It provides fixity and reduces the tilt and settlement of the well.
  • It makes sure the base rests in firm undisturbed soil.
  • 2076 Chaitra · 2 marks

Define grip length and tilt and shift of well.

Answer

  • Grip length: the depth of a well foundation below the maximum scour level; at least one-third of the maximum scour depth. It gives the lateral stability of the well.
  • Tilt: the inclination of the axis of the well from the vertical, found during sinking from uneven dredging or obstructions; the permissible tilt is 1 in 80.
  • Shift: the lateral displacement of the centre of the well from its designed position, at the base or at the top; the permissible shift is about 150 mm.
  • 2080 Bhadra · 3 marks

Derive an expression for lateral stability analysis of light weight well using Terzaghi's simplified approach.

Answer

Terzaghi's simplified approach for a light well (self-weight small, so the base does not resist horizontal force) treats the well as a rigid body in cohesionless soil, rotating about a point O at a depth z0z_0 below the scour level.

Assumptions:

  • The soil reaction at depth zz is the net passive resistance σ=γ′z(Kp−Ka)FS=kz\sigma = \dfrac{\gamma'z(K_p - K_a)}{FS} = kz, with k=γ′(Kp−Ka)FSk = \dfrac{\gamma'(K_p - K_a)}{FS}, acting on the width BB of the well.
  • Above O the reaction opposes the horizontal force; below O it acts in the same direction as the force.
  • The horizontal force HH acts at a height h=M/Hh = M/H above the scour level, and DD is the depth of embedment.
   H -->  (height h)
   ----|----- scour level
       |  <--- sigma = kz   (above O)
      -O----  rotation point z0
       |  ---> sigma = kz   (below O)
       |_____ D

Derivation

Horizontal equilibrium:

H=Bk(z022−D2−z022)=Bk(z02−D22)H = Bk\left(\frac{z_0^2}{2} - \frac{D^2 - z_0^2}{2}\right) = Bk\left(z_0^2 - \frac{D^2}{2}\right)

Moments about the scour level:

Hh=Bk(D3−z033−z033)=Bk3(D3−2z03)Hh = Bk\left(\frac{D^3 - z_0^3}{3} - \frac{z_0^3}{3}\right) = \frac{Bk}{3}\left(D^3 - 2z_0^3\right)

Eliminating HH gives the equation for the depth of the point of rotation:

2z03+3h z02−(32hD2+D3)=02z_0^3 + 3h\,z_0^2 - \left(\tfrac{3}{2}hD^2 + D^3\right) = 0

Solve for z0z_0, then the allowable equivalent resisting force is H=Bk(z02−D22)H = Bk\left(z_0^2 - \tfrac{D^2}{2}\right). The well is stable if this value is greater than the applied horizontal force.

  • 2081 Bhadra · 5 marks

A circular well of 4.5 m external diameter and 0.75 m steining thickness is embedded upon a depth of 12 m in a uniform sand deposit. The angle of shearing resistance of sand and submerged unit weight are 30∘30^\circ and 1 t/m31\ \text{t/m}^3, respectively. The well is subjected to resultant horizontal force of 50 tons and a total moment of 400 t-m at the scour level. Assume the well to be light, compute allowable equivalent resisting force due to earth force. A factor of safety of 2 may be adopted for soil resistance.

Answer

Given: circular well, external diameter B=4.5B = 4.5 m, steining 0.75 m, embedment D=12D = 12 m in sand, ϕ=30∘\phi = 30^\circ, γ′=1 t/m3\gamma' = 1\ \text{t/m}^3; H=50H = 50 t and M=400M = 400 t·m at the scour level; light well; FS=2FS = 2 for soil resistance.

Method: Terzaghi's simplified approach for a light well (net passive resistance kzkz, rotation about a point O at depth z0z_0) as derived for a light well. The width taken is the external diameter.

Soil constants

Kp=tan⁡2(45∘+15∘)=3.0,Ka=13=0.333K_p = \tan^2(45^\circ + 15^\circ) = 3.0,\quad K_a = \frac{1}{3} = 0.333 k=γ′(Kp−Ka)FS=1×(3−0.333)2=1.333 t/m3k = \frac{\gamma'(K_p - K_a)}{FS} = \frac{1 \times (3 - 0.333)}{2} = 1.333\ \text{t/m}^3

Height of the force

h=MH=40050=8 mh = \frac{M}{H} = \frac{400}{50} = 8\ \text{m}

Depth of rotation point

2z03+3h z02−(32hD2+D3)=0⇒2z03+24z02−3456=0⇒z0=9.06 m2z_0^3 + 3h\,z_0^2 - \left(\tfrac{3}{2}hD^2 + D^3\right) = 0 \Rightarrow 2z_0^3 + 24z_0^2 - 3456 = 0 \Rightarrow z_0 = 9.06\ \text{m}

Allowable equivalent resisting force

Hall=Bk(z02−D22)=4.5×1.333×(9.062−72)=6.0×10.07=60.3 t\begin{aligned} H_{all} &= Bk\left(z_0^2 - \frac{D^2}{2}\right) \\ &= 4.5 \times 1.333 \times \left(9.06^2 - 72\right) = 6.0 \times 10.07 \\ &= 60.3\ \text{t} \end{aligned}

(Check by moment: Bk3h(D3−2z03)=60.3\dfrac{Bk}{3h}(D^3 - 2z_0^3) = 60.3 t.)

Answer: allowable equivalent resisting force due to earth pressure ≈60.3\approx 60.3 t, which is greater than the applied horizontal force of 50 t, so the light well is stable.

  • 2075 Chaitra · 6 marks

What considerations govern the fixing of the depth of a well foundation? A circular well of 4 m internal diameter and 0.75 m steining thickness is embedded up to a depth of 12 m in a uniform sandy deposit. The well is subjected to a resultant horizontal force of 500 kN and a total moment of 4000 kN-m at the scour level. Calculate the allowable total equivalent resisting force due to earth pressure for both light well and heavy well conditions using Terzaghi's analysis. Take saturated unit weight of soil of 20 kN/m320\ \text{kN/m}^3, ϕ=30∘\phi = 30^\circ and factor of safety for passive resistance of 2.

Answer

Considerations governing the depth of a well foundation

  • Maximum scour depth (Lacey's formula) and a grip length of at least one-third of it below the maximum scour level.
  • Bearing capacity and settlement: the founding stratum must carry the load safely.
  • Lateral stability: enough embedment so that passive resistance resists the horizontal force and moment.
  • Depth of firm stratum or rock, soil layers prone to scour or liquefaction, and ease of sinking.

Numerical

Given: internal diameter 4 m, steining 0.75 m, so external diameter B=4+2(0.75)=5.5B = 4 + 2(0.75) = 5.5 m; D=12D = 12 m; H=500H = 500 kN, M=4000M = 4000 kN·m at scour level; γsat=20\gamma_{sat} = 20 kN/m³, ϕ=30∘\phi = 30^\circ, FS=2FS = 2 for passive resistance. γw=9.81\gamma_w = 9.81 kN/m³.

Method: Terzaghi's simplified analysis. The net passive resistance at depth zz is σ=kz\sigma = kz.

γ′=20−9.81=10.19 kN/m3,Kp=3, Ka=0.333k=10.19×(3−0.333)2=13.59 kN/m3h=MH=4000500=8 m\begin{aligned} \gamma' &= 20 - 9.81 = 10.19\ \text{kN/m}^3,\quad K_p = 3,\ K_a = 0.333 \\ k &= \frac{10.19 \times (3 - 0.333)}{2} = 13.59\ \text{kN/m}^3 \\ h &= \frac{M}{H} = \frac{4000}{500} = 8\ \text{m} \end{aligned}

(a) Light well

The well rotates about a point O at depth z0z_0 below the scour level (the base offers no resistance).

2z03+24z02−3456=0⇒z0=9.06 m2z_0^3 + 24z_0^2 - 3456 = 0 \Rightarrow z_0 = 9.06\ \text{m} Hlight=Bk(z02−D22)=5.5×13.59×(82.07−72)=751.5 kNH_{light} = Bk\left(z_0^2 - \frac{D^2}{2}\right) = 5.5 \times 13.59 \times (82.07 - 72) = 751.5\ \text{kN}

(b) Heavy well

A heavy well has enough weight and base friction to resist sliding, so it rotates about the centre of its base. The soil pressure σ=kz\sigma = kz acts on the front over the full depth, and moments are taken about the base:

H (h+D)=Bk∫0Dz (D−z) dz=BkD36H\,(h + D) = Bk\int_0^D z\,(D - z)\,dz = \frac{BkD^3}{6} Hheavy=5.5×13.59×1236×(8+12)=129 157120=1076.1 kNH_{heavy} = \frac{5.5 \times 13.59 \times 12^3}{6 \times (8 + 12)} = \frac{129\,157}{120} = 1076.1\ \text{kN}
WellAllowable equivalent resisting force (kN)Applied HH (kN)Result
Light751.5500Safe
Heavy1076.1500Safe

Answer: allowable total equivalent resisting force due to earth pressure ≈752\approx 752 kN (light well) and ≈1076\approx 1076 kN (heavy well), both greater than the applied 500 kN.

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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