Geotechnical Engineering

DHARM

806 GEOTECHNICAL ENGINEERING

horizontal force of 1000 kN acting at a height of 8 m above the scour level. Determine the total
allowable equivalent resisting force due to earth pressure, assuming that (a) the well rotates
about a point above the base, and (b) the well rotates about the base. Assume γ′ = 10 kN/m^3 ,
φ = 30°, and factor of safety against passive resistance = 2. Use Terzaghi’s Approach:

D = 16 m H = 8 m φ = 30°

∴ Ka =

(sin )

(sin )

130

130

1

3

−°

+°

=

Kp =

(sin )

(sin )

130

130

3

+°

−°

=

Total height above base + H 1 = 16 + 8 = 24

Modified passive pressure coefficient,

Kp′ =

K

FS

p

.

==.

3

2

150

(a) Rotation about a point above the base:

From Eq. 19.30:

2 D 1 = 3923 HHDHD 11 ±− −^2 () 1

= 3 24×± × −× ×−9 24^2 2 16 3 24 16()

=^72 ±−5184 1792

P

8m

16 m

Scour level

D 1

P

H=8m

D=16m

H = 24 m 1

(a) Well with lateral load (b) Pressure distribution

Fig. 19.24 Cylindrical well with lateral load (Ex. 19.4)

2 D 1 = 72 – 58.24 (rejecting the +ve sign, as it leads to a value for D 1 > D)

= 13.76

D 1 = 6.88 m

From Eq. 19.28,

q′max =

1

2

γ′DK().()pa′ −K D− 2 D 1

= (1/2) × 10 × 16{(3/2) – (1/3)} (16 – 13.76)

= 209.1 kN/m