Geotechnical Engineering

DHARM

STABILITY OF EARTH SLOPES 335

The shear resistance (stress) mobilised is:

τ =

cu

F

′+()tanσφn− ′

Total normal stress σn on the base of the slice = P/ls

∴τ =

1

F

[{(/)}tan]cPlu′+ s − φ′

Shearing force acting on the base of the slice, S = τls.

For equilibrium,

Sliding moment = Restoring moment

or ΣW. x = ΣS.r = Στ.ls.r

=

r

F

Σ[( )tan]cl′+ −ssP ul φ′

∴ F =

r

Wx

clssP ul

Σ

Σ[( )tan]′+ − φ′ ...(Eq. 9.29)

Let the normal effective force, P′ = (P – uls)
Resolving the forces vertically,
W = P cos α + S sin α ...(Eq. 9.30)
(The vertical components of Rn and Rn + 1 are taken to be equal and hence to be nullify-
ing each other, the error from this assumption being considered negligible). Here P = P′ + uls

S = 1/F (c′ls + P′ tan φ′)

Substituting these values of P and S in Eq. 9.30, we have:

W = (P′ + uls) cos α +

(tan)sincl P

F

′+′s φα′

= P

F

′+F ′ lus cF

HG

I

KJ

cosα tanφsinαα α++′{ cos ( / ) sin }

∴ P′ =

Wlus cF

F

−+′

+ ′

{ cos ( / ) sin }

cos tan sin

αα

α φα

...(Eq. 9.31)

Substituting this value of P′ for (P – uls) in Eq. 9.29, we get

F =

r

Wx

clssW ul clF

F

s

Σ

Σ

′+ − − ′

+

L

N

M

M

M

O

Q

P

P

P

′

′

cos sin tan

cos tan sin

ααφ

α φα

ej

ej

...(Eq. 9.32)

Here x = r sin α

b = ls cos α

ub

W

u

z

=

γ

= ru

Substituting these into Eq. 9.32,

F =

(^11)
Σ 1
Σ
W
cb W ru
sin F
{()tan}sectan tan
α
′+ − φ′ φαα

• L
  N
  M
  M
  M
  O
  Q
  P
  P
  P
  ′
  ej
  ...(Eq. 9.33)