INDIAN ROAD CONGRESS (IRC 45)

1. DESIGN OF WELL
FOUNDATION
R.UMABALACHANDRAN
M.TECH(SE)

2. Introduction:
 Well foundations are one of the types of deep foundations that
provide a solid and massive foundation typically for bridges
and heavy structures
 Caissons or well have been in use for foundations of bridges
and other structures since Roman and Mughal periods.
 “Caisson” means a box like structure, round or rectangular,
which is sunk from the surface of either land or water to some
desired depth

3. Types of Caisson
 Open caisson
 Box caisson
 Pneumatic caisson
Open Caisson:
 Open cassion is a box
opened both at top and
bottom.
 It is made up to either
timber, concrete or steel.
The open cassion is called
well.

4. Types of Caisson
Box caisson:
 It is open at the top and
closed at the bottom and
is made of timber,
reinforced concrete or
steel.
 This type of caisson is
used where bearing
stratum is available at
shallow depth.

5. Types of caisson
Pneumatic caisson
 Pneumatic caisson has its
lower end designed as a
working chamber in which
compressed air is forced to
prevent the entry of water
and thus excavation can be
done in dry conditions.

6. Shape of Well:
 Double-D
 Twin hexagonal
 Twin octagonal
Single circular
Twin circular
Dumb well

7. PROBLEM
1) A bridge pier in a sand deposit with external diameter d=8.5 and the
depth of well below scour level D=15m is subjected to the following
loads
 Vertical Load, W= 14000 kN
 Horizontal Load, H= 2000 kN
 Moment about base level, M= 42000 kN
 The value Ø of the sand = 30˚
 Wall friction, δ= 20˚
 Allowable bearing = 60t/m²
 kh/kv = m = 1
 Assume the weight of soil is 20 kN/m³
Check the lateral stability of the well under the above forces according to
IRC 45 (1972) recommendations.

8. Solution:
We check the four conditions for stability in the elastic state and also check the
ultimate state
 ELASTIC ANALYSIS
(1) H > (M/r)(1+µµ´)-µw
(2) H < (M/r)(1-µµ´)+µw
(3) Mm/I ≯γ(Kp – KA)
Where , γ - Submerged unit weight of soil
Kp – Passive earth pressure
Ka– Active earth pressure
(4) σ1, σ2 =
Where, σ- soil pressure
IB
MBW
2A
µ´P-
±

9. STEP-1:DETERMINE THE PARAMETER
 Projected dimension, L=0.9d = 0.9 X 8.5=7.65 m
 IV = LD³/12 = 7.65 X 15³ /12 = 2151m 4
where ,Iv – moment of inertia of vertical area of L * D
 IB = πd 4 /64 = π(8.5) 4 /64 = 256 m 4
where, IB - moment of inertia of rectangular or circle base
 α = d/ πD = 8.5/ π X 15 = 0.18
 µʹ= tan δ = tan 20 = 0.18
where, µʹ - Side friction
 µ = tan Ø = tan 30 = 0.58
where,µ - friction at base
 I = IB + MIv(1+2µ´α)
= 256+2151(1+2*0.36*0.18) = 2685.7
 r = ID/2mIV
= 2685.7*15/2*1*2151
r = 9.36m

10.  Kp = tan² (45+ø/2) = 3.69 (Passive earth pressure)
 Ka = tan² (45+ø/2) = 0.27 (Active earth pressure)
STEP-2:
Condition -1: Check whether the Base friction is safe or not
 H > (M/r)(1+µµ´)-µw
>
> -4182 kN
 2000kN > -4182kN (satisfied).
1400070.0)36.070.01(
36.9
42000


11. Condition -2:Direction of rotation is compatible
 H < (M/r)(1-µµ´)+µw
<
< 13,156 kN
 2000kN < 13,156 kN (satisfied).
Condition -3:Soil remains elastic
 Mm/I ≯γ(Kp – KA)
=1*42000/2685.7 ≯10(3.69-0.27)
 15.6 ≯34.2 (satisfied).
140070.0)36.070.01(
36.9
4200
×+×

12.  Condition – 4 :(M =external moment at base of well)
σ =
A=πd²/4 = 56.72m²
σ =218 ± 66 =284 and 152 kN/m²
 Being less than the value of 600 kN/m²,these value are acceptable .
BI
MBW
2A
µ´P-
±
r
M
P =
kNP 4487
36.9
42000
==
7.26852
5.842000
72.56
448736.014000
×
×
±
×
=

13. Ultimate Analysis:
The following two condition should be satisfied.
(1)
(2) Mr =0.7(Mb + Ms + Mf) and M < Mr
M = actual moment acting on the well
Step 1: Calculate stress at base in kN/m²
qu = 600 (ultimate bearing capacity)
 246 ≯ 600 (kN/m²) is satisfied.
Step 2: Calculate Mb (base moment)
Mb = (QB) (W tan ø )
uq
A
W
/>=
246
72.56
14000
==
A
W

14.  D/d = 15/18.5 =1.76
 Q = 0.53 *0.6 = 0.32 (refer the value of constant Q and also multiply the
shape factor 0.6 for circular base)
 Mb = 0.32*8.5*14000*0.70 = 26,656 kNm
Step 3: Calculate Ms (side earth pressure moment)
Ms = 0.1γD³ (KP –KA)L
= 0.1*10(15)³ *(3.69-0.27)*0.9*8.5
Ms = 88,300 kNm
Step 4: Calculate Mf (side friction moment) δ = 20˚
Mf = 0.11γ(KP-KA) B²D² sin δ
= 0.11*10*3.42*(8.5)²*(15)²*0.34
Mf = 20,793 kNm

15. Step 5: Find moment to be resisted at base
M = 42,000 kNm (given value)
Step 6: Estimate total resisiting moment
Mr = 0.7(Mb + Ms + Mf)
= 0.7(26,656+88,300+20,246)
Mr = 94,641 kNm
94,641 kNm > 42,000 kNm
 Mr > M (satisfied)
Step 7 : Final results
All condition are satisfied. Hence safe.

16. THANK YOU