Done as mini project m.tech. Manual Design of well foundation under a transmission tower/or under a bridge pier

1. DESIGN OF WELL FOUNDATION
Under the guidance of
Dr.G.YESURATNAM,Ph.D
PROFESSOR AND CHIEF ENGINEER
Presented By
Y. Bala Balaji
(19021D2004)
MASTER OF TECHNOLOGY
(STRUCTURAL ENGINEERING)
2019-2021 BATCH
UNIVERSITY COLLEGE OF ENGINEERING,JNTU
KAKINADA-533235

2. KEY CONTENT :
 ABSTRACT
 1.INTRODUCTION
 2.TYPES OF WELLS
 3.PARTS OF WELL FOUNDATION
 4.PROCEDURE WITH FIELD PHOTOS
 5.METHODOLOGY
 6.FOUNDATION DETAILS
 7.ESTIMATION OF WELL CAPACITY
 8.DESIGN OF WELL COMPONENTS
 9.REFERENCES

3. ABSTRACT
 Well foundations had their origin in India and have been used for
Hundred of years, for providing deep foundations, for important
buildings and structures. The techniques of sinking masonary wells
for drinking water is very ancient. Same technique was used for
construction of foundation wells in the earlier stage. Well foundations
were freely used during the Mughal period and many of the Mughal
monuments including the Taj Mahal have got well foundations .The
Mughal used well foundation for bridges across major river also .But
these bridges were washed away within a few years as the mechanism
of scour was not understood.
 In spite of the excellent development of
technology on well foundations there are still some areas where
engineers face difficulty while sinking of wells, some of which are
stated in this paper from the direct experience.
 Well foundations are quite appropriate
foundations for alluvial soils in rivers and creeks where maximum
depth of scour can be quite large. In India technology of well
foundation for design and construction is quite well developed

4. INTRODUCTION:
 FOUNDATION : It is the lowest part of a structure which is
constructed below the
 Ground level to transmit the weight of superstructure to the
subsoil. If D < B ,it is called as SHALLOW FOUNDATION.
 If D ≥ B, then it is called as DEEP FOUNDATION.
 Where ,
 D = depth of foundation B = width of
foundation
 1.SHALLOW FOUNDATION 2.DEEPFOUNDATION
 WALL FOUNDATION ► PILE FOUNDATION
 ISOLATED COLUMN ►UNDER REAMED PILE
 COMBINED FOUNDATION ► WELL FOUNDATION
 MAT or RAFT FOUNDATION
 STRIP , STRAP FOUNDATION

5. WELL OR CAISSON FOUNDATION :
Wells which are also known as caissons,
have been in use for foundations of bridges(roads , railways
and transmission line towers ) and other important structures
since the Roman and Moghal periods in India. Moughals in
particular used wells for the foundations of their monuments,
including Taj Mahal which is a standing testimony to the skill of
mankind in the earlier dates.
In modern days however ,one of the
earliest use in india is that for an aqueduct for the upper
Ganges Canal constructed in the earlier part of the 19th century
. With the advent of pneumatic sinking in 1850 A.D., and
discovery of better materials like reinforced concrete and
steel ,use of wells as foundations of bridges gained popularity.
Well foundations have been used for
most of the major bridges in india. Materials commonly used
for construction are reinforced concrete , brick or stone ,
masonry. Use of well or caisson foundations is equally popular
in the United States of America and other western countries.
The size of caisson used for the SanFrancisco Oakland
Bridge is 29.6*60.1 m in section in 74 m depth .

6. TYPES OF WELL FOUNDATION :
 • The cassion foundations are of three types.
 • Box Caissons
 • Open Caissons ( wells )
 • Pneumatic Caissons
 Box cassions are open at top and closed at the bottom
and they are made up of timber, reinforced concrete or
steel. They are built on land, then launched and floated to
pier site where they are sunk in position
 • Open cassion foundations are boxes of timber, metal,
reinforced concrete or masonry which is open at top and
bottom. They are used in constructing the foundations of
bridges and building
 • These are bit different from the above mentioned types
because the bottom of foundation is designed as working
chamber in which compressed air is forced to prevent the
entry of water and thus permit excavation in dry.

8. BASED ON SHAPE OF WELL FOUNDATION :
 • Single Circular well foundation
 • Twin Circular well foundation
 • Dumb-well foundation
 • Double- D well foundation
 • Twin Hexagonal well foundation
 • Twin Octagonal well foundation
 • Rectangular well foundation

9. PARTS OF WELL FOUNDATION :
1. Well Cap :
It is a RCC slab. Its function is to transmit the forces from the piers to the body of well. It is
important that the dimension of well cap should be sufficient to accommodate pier.
2. Top Plug :
Top plug is situated below the well cap. This helps transferring the load through the granular
material into the steining.
3. Steining :
Steining is a wall that is built over a wedge shaped portion called well curb. The wall
should be designed such that it can be sunk under its own weight. Moreover, the
thickness should be enough to overcome the skin friction during sinking.
4. Well Curb:
As mentioned earlier, the well curb supports the steining. Well curb is made up of RCC with
steel cutting edge.
5. Cutting Edge :
The cutting edge is that part of the well which helps in penetrating the strata of soil. Thus
in case of hard strata such as gravels, the flat cutting edge is used whereas in case of
soft strata the sharp edged cutting edge is used.
6. Bottom Plug :
The bottom plug transmits the load to soil below and resists uplift forces. The thickness of
the plug varies from 0.5 – 1 times the inner diameter of the well.
7.Sand Filling :
Sand filling is the part that lies in between the top and bottom plug. Its main function is to
provide stability to well and reduce the tensile stress due to the bending moment along
with the distribution of load from the superstructure

10. 8. Intermediate Plug :
Wells resting on clayey strata, it is not preferable to fill the space inside the well completely
with sand. In such cases, sand filling is not done or sand is filled up to the scour level.
A concrete plug covering the filling is usually provided, known as intermediate plug.
Usually, thickness of intermediate plug is taken as 500 mm.

11. STEPS OF CONSTRUCTION:
 1.PLACING OF CUTTING EDGE IN LEVELLED PROPOSED GROUND
 2.CASING FOR WELL CURB

12. 3.FIXING OF REBAR :

13. 4.WELL STEINING,REINFORCING AND CURING:

14. 5.SINKING BY GRAB :

15. 6.GAUGING :
7.WELL PLUGGING :

17. 4.METHODOLOGY:
DETAILS OF TOWER:
RIVER DETAILS :
 REDUCED LEVEL (R.L) AT PROPOSED TOWER = 150 m
 REDUCED LEVEL (R.L) AT HIGH FLOW LEVEL (H.F.L) = 149.4 m
 REDUCED LEVEL (R.L) AT LOW FLOW LEVEL (L.W.L) = 141.4 m
 Width of River = 470 m
HYDRAULIC DATA :

18. CHARACTERISTICS OF SOIL
1. Poorly graded fine to medium sand (SP) upto 32m from G.L.
2. Average corrected SPT ‘N’ value = 10
3. Angle of friction = 32°
The legs of the tower are placed on the pedestal having height 2.5 m having cross- section
decreasing from 600x600 at bottom to 400x600 at top.
WELL CONFIGURATION :
Single well with outrigger arms supporting the tower legs.
FOUNDATION DETAILS:
5.1.PROPORTIONING OF FOUNDATION
The foundation shall be taken adequately below the minimum scour depth. The normal
depth of scour is estimated using Lacey’s formula as:
d = 0.473 (Q/f)1/3
= 0.473 ((1500+310)/0.68)1/3
= 6.614
where, d = normal depth of scour
Q = design discharge in cumecs, and
f = 0.68 Lacey’s silt factor.
IRC: 78- 2000 recommends that scour depth calculations for foundations may be made for a
discharge larger than the design discharge. Accordingly, 20 % increase in design
discharge has been assumed in scour depth calculations.
dmax = 1.27d = 8.4 m

19. DIMENSIONING OF THE WELL FOUNDATION
From the scour considerations minimum grip length for the well foundation = 0.33 * max. Scour
depth
= 2.772 m ≈ 3 m
However provide a grip length of 9 m as a conservative measure.
R.L. of base of well w.r.t. HFL = (149.4 -9 -8.4 )m = 131.1 m
Height of well w.r.t. NGL = ( 150 – 131.1 )m = 18.9 m
Thickness of well Steining :
As per IRC 78- 2000 ( Cl. 708.2.3 )
t = K*d*√l = 0.03*9*√18.9 = 1.174 m
t = Thickness of well steining
K = constant = 0.03 m
D = External diameter of well = 9m
l = depth of well w.r.t N.G.L = 18.9 m
Provide t = 1.5 m for sufficient sinking effect.
Internal Diameter of well = ( 9-1.5-1.5 ) =6 m (> 2 m) Hence OK.
As per IRC 78:2000
The thickness of well cap = 1.5 m
Thickness of top plug = 0.6 m (Because we are using well cap)
Height of well curb = 0.5 * internal diameter of well = (0.5*6 ) = 3.0 m
As per IRC 78 :2000 Projection ≥ 75 mm
Take 100 mm projection
Size of ISA cutting edge = 150*150*18 mm
After using AutoCAD for drawing well the length of outrigger for supporting
the pedestral is coming out to be = 13487.68 m

20. Taking , ( l/d ) = 5 13487.68 / d = 5
d = 2697.5 mm Take D = 2700 mm
The outriggers is tapered from 1 m to 2.7 m.
Total height of well = 18.22 m
Grip length = 9 m
External diameter of well = 9 m
Internal diameter of well = 6 m
Thickness of well steining = 1.5 m
Length of outrigger = 13.49 m
Max. Depth of outrigger = 2.7 m
Width of outrigger = 0.6 m

21. ESTIMATION OF WELL CAPACITY
Uplift capacity :
The safe uplift capacity, Qsafe, may be calculated as the submerged weight of the well,
conservatively ignoring the effect of side friction. Thus,
Qsafe = ((∏/4)*(92-62)*18.9*(25-10)+(4*0.5*(1+2.7)*13.49*0.6*(25-10)) = 10918.15 KN (> 1800 KN)
O.K.SAFE
Axial compression load capacity :
The effect of skin friction is conservatively ignored and the axial load capacity is taken as the base
resistance with a factor of safety of 3. The base resistance, Qsafe, is calculated as
Qsafe = (σ * (Nq-1)*Ap ) / F.O.S = 44582.9696 KN (> (1200*4 = 4800 KN )) HENCE OK
Where ,σ = effective over burden pressure at base (9*10 KN/m2)
Nq = bearing capacity factor
Ap = area of base of well ((∏/4)*9*9)
F.O.S = factor of safety
For soils at base of well, ϕ = 320 , hence Nq = 24.36
Lateral load capacity :
The lateral load acting on the well consists of two components :
Design lateral load corresponding to B.W.C. = 90.5*4 = 362 kN.
Lateral load due to water current force corresponding to H.F.L. acting on curved surface area of the
well. Intensity of water current pressure at HFL = 0.52KV2
Where, V = Velocity of the current at the point where the pressure intensity is being calculated
K = a constant having a value of 0.66 for circular piers.
P = 0.52*0.66*(√2*1.8)2 = 2.223 KN/m2

22. Water current (lateral) force = 0.5 x 8.4 x 2.223x 9 = 88.308 kN
Total lateral force = 359.4 + 88.308 = 450.3 kN
Ka = 0.31 and Kp = 3.25, for φ = 32°.
Let the total lateral is acting at a height ‘h’ above base of well.
450.3 x h = (88.308 x 15.2) + 362(18.9+2.5)
h = 20.184 m.
Elevation of resultant lateral load = 20.184 m.
Safe lateral capacity, Hsafe, may be computed as,
Where ,h = height of resultant lateral load above base = 20.184 m
D = grip length = 9 m
Kp = 3.25 Ka = 0.31
γ = submerged unit weight of soil = 10 KN/m3 F.O.S = 2.0
de = external diameter of well = 9 m
D1 = 56.42 or 4.10 (lower one is our answer)
Hsafe = ((( 0.5*10*2.94(9-2*4.10)*9) / 2 )*9 )
= 488.187 KN (>450.3 KN )
Stability check of outriggers :
Clear span of cantilever (lo) = 13.48 m.
Davg = (2.7 +1)/2 = 1.85 m or 1850 mm
Dav = 1775 mm b = 500 mm
a) 25b = 12500 mm
b) = 20281.69 mm
Take smaller value between (a) and (b) = 12500 mm
Lo< 15000 mm O.K

23. DESIGN OF WELL COMPONENTS
Design of outriggers :
Leff = lo + (1.5/2) = 13.48 +0.75 = 14.24 m
LOAD (P) = 1200 KN
Taking ,
Overall depth (D) = 2700 mm Effective depth (d) =2500 mm Effective cover (dl) = 200 mm
Grade of concrete = M25 Grade of steel = Fe415
Self weight of beam
Volume of RCC = 0.5*(2.7+1)*14.24*0.6 = 15.08 m3
Weight of beam = 25 * 15.80 = 395.16 kN
FACTORED MOMENT (Mu) = 1.5*( 1200*(14.24-2*0.15)+(14.24/2)*395.16)
=1.5*19521.78*106 = 29282.67 *106 kN
=0.4791
= 13018.8 *106 Nmm
Since Mu > Mu, lim section is to be designed as doubly reinforced.
= 1.2 % = 15677.46 mm2
Ast,req = 19585.147 + 15677.46 = 35262.6 mm2
Provide 28 nos of 36ф bars and 6 nos. of 22ф bars.
Ast,prov. = 36888.58 ( > 35262.6 mm2)
Calculation of new d
36888.58* d = (((∏/4)*34*362)*2516) + (((∏/4)*6*222)* 2337)
 d = 2504.9 mm ≈ 2505 mm . OK d’ = 196 mm

24. Calculation of Asc :
ΔAst, prov= 36888.5-15677.4 = 21211.12 mm2 = 13693.165 mm2
= 353.308 N/mm2
Provide 22 nos. 36ф bars
Asc,prov. = 22393.2 mm2 (> 22283.9 mm2 ) O.K.
Pc =(100Ast/bd) = (100*22393.2)/(600*2505) = 1.48
Pt = 2.05 , = 0.8969
Pc > Pc* (hence beam is under reinforced ) O.K.
Side face reinforcement :
Side face reinforcement has to be provided because depth of the beam is more than 750 mm.
Minimum area = 0.001 x 600 x 2700 = 1620 mm 2
At a spacing not exceeding 300mm.
Provide 8 nos. of 12ф bar at each face at equal spaces
Design of shear reinforcement :
Shear at critical section
V = 0.5 x (2.7+1)x 13.48 x 25 x 0.6 +1200 kN = 1574.07 kN ; Vu = 1.5V =2361.105 kN
Mu = 1.5 ((395.16 x(13.48/2) ) + (1200 x 13.9)) = 29015.0676 kNm
tanβ = ((2.7-1)/13.48) = 0.126
( β = angle of sloping surface with the horizontal )
= 0.0732 N/mm2 ; = 2.45
From table 19 of IS 456 : 2000 , τc = 0.874 N/mm2

25. τc > τv . Hence section is safe in shear, and minimum shear reinforcement should be provided.
Minimum reinforcement should be provided as per the following formula
Where,
Asv = total cross sectional area of stirrups effective in shear.
sv = stirrup spacing along the length of the member.
b = breadth of the beam ( = 600 mm)
fy = Characteristic strength of stirrup reinforcement in N/mm2 . ( 415 N/mm2)
Provide 2-legged 10mm dia. Bars.
Asv = 157.08 mm2
Putting the values in above formula, we obtain Sv = 236.3 mm.
Give shear reinforcement at 230 mm c/c.
Development length :
Grade of Concrete = M25 Grade of steel = Fe415
The development length Ld is given by
Ld = (ϕσs /4 τbd )
Where, Ф = nominal diameter of the bar.
σs = Stress in bar at the section considered at design load (= 0.87fy), and
τbd = Design bond stress ( = 1.4 for M25)
Ld = (0.87*415ϕ)/(4*1.4) Ld = 64.47 ф
For 36mm dia. bar, Ld= 2320.92 mm ; For 22 mm dia. Bar Ld = 1418.34 mm

26. 7.5. Deflection :The total deflection shall be taken as the sum of short-term deflection and the long term
deflection.
Short term deflection :
We have b=600 mm ;D = 2700 mm ; fck = 25MPa ; fy = 415 MPa ; W = 1200 kN ; l = 14237.68mm.
Unfactored moment M = 19521.78* 106 Nmm
Igr = bD3/12 = 9.84155 x 1011 mm4 ; fcr = 0.7√fck = 3.5 N/mm2 ; yt = D/2 = 1350 mm
Mcr =((fcr*Igr )/yt) = 2.6 x 109 Nmm
Ec = 5000 √fck= 25000 N/mm2 Es = 2 x 105 N/ mm2 m = 11
Let x be the depth of neutral axis, then taking moment of transformed section about N.A.
We get, x = 722.56 mm
Icr = (bx3/3) + (m-1)Asc(x-dl)2 + mAst(d-x)2 Icr = 13.57*1011 mm4
Ieff= 12.127 * 1011 (< Icr ) ,hence Ieff = 13.57*1011 mm4
Short Term Deflection
∆short term = (Wl3)/(3EIeff) = 34.02 mm
7.6. Deflection due to shrinkage :
αcs = k3 *фcs*l2 here, k3 = 0.5 ; фcs = (k4€cs)/D ; €cs = 0.0003 ; k4 = 0.72 x((Pt-Pc)/√Pt)
where, Pt = 2.45, Pc = 0.01 Putting the values we get αcs = 12.63 mm
Deflection due to creep :
Ece = (Ec/(1+Ɵ)) ; ( Ɵ= the creep coefficient. = 1.6 ) Ece = 9615.38
∆creep = ∆cc(perm) - ∆i(perm) Δcreep = 54.46 mm
TOTAL DEFLECTION :
Δtotal = 34.02+12.63+54.46 = 101.11 mm

27. Design of well cap :
Since no direct load is coming on well cap, minimum should be provided.
i.e. 0.12% of gross sectional area =(0.12/100) x 1500 x 1000 = 1800 mm2
Provide 18 mm. diameter bar at 300 c/c on top and bottom faces of the well cap at a clear
cover of 75 mm.
Ast provided = 2035.75 ( > 1800 mm2 ) HENCE O.K.
Design of well steining :
Lateral load acting on well = 362 + 88.308 kN = 450.3 kN
Distance of lateral load from base of well = 20.184 m.
The resultant earth pressure force at depth ‘y’ below M.S.L. is given by :
= 0.5 x γsub x (Kp-Ka) x y2 x De
Equating the lateral loads at depth y gives the location of zero shear (and max. moment
section).
450.3 = 0.5 x 10 x (3.25-0.31) x y2 x 9
y = 1.845 m
weight of well steining of 1.86 m height = (∏/4)x (92 - 6) x 1.845 x 25 = 1630.19 kN
Moment of lateral forces about section of zero shear
M = 450.3 * (18.9-9+1.845) = 5288.77 kNm
Total axial load at section of zero shear = Load from tower + Weight of pedestal+Weight of
outriggers + Weight of well cap + Weight of steining
= ( 4 x 1200) + (4 * 18.75) + (4 x 395.16) +
(∏/4 x 92 x 25 x 1.5 ) + 1630.19
P = 10471.476 KN

28. Area of cross section of steining = (∏/4) (92 – 62) = 35.34 m2
Ixx = Iyy = (∏/64)(94 – 64 ) = 258.44 m4
y =De = 4.5 m
The stresses in the steining
f1,2 =(P/A) ± (My/I)
putting the values we get.
f1,2 = 296.3 ± 92.089
f1 =0.388 MPa (< 8MPa )
f2 = 0.204 MPa ( > 0 )
Both the stresses f1 and f2 are compressive and significantly smaller than the allowable
stresses for M-25 grade concrete. Hence, the steining section is safe.
Reinforcement in well steining :
Provide vertical steel = 0.12 % of gross sectional area = (0.12/100)*(∏/4)*(90002-60002) =
42411.5 mm2
Provide vertical steel equally on both faces of steining.
Area of vertical steel on each face =(42411.5/2) = 21206 mm2
Provide 48 Nos. of equally spaced 25ф bars on the inner and outer faces of the
steining. Keep the vertical bars equally spaced.
Area of vertical steel provided = 23561.94 > 21206 mm2, Hence O.K.
Provide hoop steel at 0.04 % of the volume per unit height of steining.
Volume of hoop steel per ‘m’ height of steining. = (0.04/100)*(∏/4)* ( 92 – 62 ) x 1.0
= 0.014137107 m3 = 14137107 mm3

29. Volume of hoop steel required on each face = (14137107 / 2) = 7068583.5 mm3
Total cross-sectional area of hoop bars required on each face per meter height of steining =
= (7068553.5/(∏*(9000-75-5))) = 252.24 mm2
Provide 12ф hoops in the form of closed rings on both the inner as well as the outer face
of the well steining @ 250 mm c/c.
Area provided = 452.39 mm2 > 252.24 mm2 O.K
Design of well curb :
Provide normal steel at 72 kg/m3 in the well curb.
Vol. of concrete in well curb = (area of well curb * (∏*d)) = 62.26 m3
Total weight of steel in well curb = 72 x 62.26 =4482.72 kg
Consider the following arrangement of steel in well curb.
(i) 40 nos. of 25ф hoops of average dia. = 7.25 m.
Weight provided = 3510.6491 kg.
(ii) 20 mm ф triangular at 280mm c/c.
Total no. of rings = 80
Average Length of one ring = (1.6-(2*0.075))+(3-(2*0.075))+ (3.332-2.075)
= 5.557 m
Weight of 80 stirrups = 80 *((∏/4)*0.0202) *5.557*7850 = 1096.35 kg.
Total weight of steel provided in well curb = 3510.65 + 1096.35 kg
= 4607Kg (> 4482.72 kg) Hence O.K

30. REINFORCEMENT DETAILS :
OUTRIGGERS:
Provide 28 nos of 36ф bars and 6 nos. of 22ф bars. (Ast)
Provide 22 nos. 36ф bars (Asc)
SIDE FACE REINFORCEMENT:
Provide 8 nos. of 12ф bar at each face at equal spaces
SHEAR REINFORCEMENT:
Provide 2-legged 10mm dia. Bars reinforced at 230 mm c/c.
WELL CAP :
Provide 18 mm. diameter bar at 300 c/c on top and bottom faces of the well cap at
a clear cover of 75 mm.
WELL STEINING :
Provide 48 Nos. of equally spaced 25ф bars on the inner and outer faces of the
steining. Keep the vertical bars equally spaced.
Provide 12ф hoops in the form of closed rings on both the inner as well as the
outer face of the well steining @ 250 mm c/c.
WELL CURB :
(1.)40 nos. of 25ф hoops of average dia. = 7.25 m.
(2.) 20 mm ф triangular at 280mm c/c.

31. REFERENCE :
 1. IRC : 6-2000. Standard Specification and Code of Practice for Road Bridges,
Section II, Loads and Stresses (Fourth Revision). The Indian Roads Congress, New
Delhi, 2000, 29 pp.
 2. IRC : 78-2000. Standard Specification and Code of Practice for Road Bridges,
Section VII, Foundation and Substructure (Second Revision). The Indian Roads
Congress, New Delhi, 2000, 97 pp.
 3. IS Indian Standard. Code of Practice for Design and Construction of Well
Foundation. IS, Bureau of Indian Standard, 1967, IS 3955 -1967.
 4. IS Indian Standard. Plain and Reinforced Concrete- Code of Practice. IS, Bureau
of Indian Standard, 2000, IS 456: 2000.
 5. Pillai, S Unnikrishna and Menon, Devdas. Reinforced Concrete Design, 3rd edn.,
McGraw-Hill, New Delhi, 2010.
 6. Punmia B.C., Jain A.K. and Jain A.K. Soil Mechanics, 16th edn., Laxmi
Publications, New Delhi, 2005.
 7. SP Special Publication. Handbook on Concrete Reinforcement and detaiing.
Bureau Of Indian Standards, 1987, SP34: 1987

32. THANK YOU