Geotechnical Engineering

DHARM

CAISSONS AND WELL FOUNDATIONS 807

Allowable Transverse load,

Qa = q′max × L = 209 × 6 = 1254 kN

(The shape factor may be taken as unity since De > 6 m)

(b) Rotation about the base:

From Eq. 19.32,

q′max =

1

6

3

1

γ′()KK′ − D

paH

= (1/6) × 10{(3/2) – 1/3)}.

16

24

3

= 332 kN/m

Hence the allowable lateral load

Qa = q′max × L = 332 × 6 = 1992 kN

Since the actual horizontal load is given as 1000 kN, the well is safe against lateral load.

Example 19.5: A circular well has an external diameter of 7.5 m and is sunk into a sandy soil
to a depth of 20 m below the maximum scour level. The resultant horizontal force is 1800 kN.
The well is subjected to a moment of 36,000 kN.m about the maximum scour level due to the
lateral force. Determine whether the well is safe against lateral forces, assuming the well to
rotate (a) about a point above the base, and (b) about the base, Assume γ′ = 10 kN/m^3 , and φ =
36°. Use Terzaghi’s analysis, and a factor of safety of 2 against passive resistance.

With the notation of Example 19.4,

D = 20 m

H = M/Q =

36000

1800

= 20 m

H 1 = H + D = 20 + 20 = 40 m

Ka =

(sin )

(sin )

136

136

−°

+° = 0.2596

Kp =

(sinº)

(sin )

136

136

+

−°^ = 3.8518

Kp′ =

K

FS

p=3 8518

2

.

= 1.9259

(a) Centre of rotation above the base:

Using Eq. 19.30,

2 D 1 = 3923 HHDHD 11 ±− −^2 () 1

= 120 −−−14400 40 120 20()

= 120 – 102 = 18

∴ D 1 = 9 m

Using Eq. 19.28,

qmax = (1/2) γ′D(Kp′ – Ka) (D – 2D 1 )