Geotechnical Engineering

DHARM

464 GEOTECHNICAL ENGINEERING

()/

()/

σσ

σσ

13

13

2

2

−

+

= sin φ

or (σ 1 – σ 3 ) = (σ 1 + σ 3 ) sin φ ...(Eq. 13.13)

OG = OC. cosβ = [(σ 1 + σ 3 )/2] cos β

CG = OC. sin β = [(σ 1 + σ 3 )/2] sin β

FG = GE = CF^22 CG 132 13

2

2

2

−= − −RS +

T

U

V

W

{(σσ) / } ()σσsinβ

= [(σ 1 + σ 3 )/2] sin^22 φβ−sin , using Eq. 13.13.

Now, σv = OG + GE =

()

cos

()

sin sin

σσ

β

σσ

(^131322) φβ
22

• 
  


• 
  


• 
  

  −
  σv =
  ()
  (cos sin sin )
  σσ
  (^13) βφβ^22
  2

  
  


• 
  

  +−
  or σv =
  ()
  (cos cos cos )
  σσ
  (^13) ββφ 22
  2

  
  


• 
  

  +− ...(Eq. 13.14)
  σl = OG – FG =
  σσ
  β
  σσ
  (^131322) φβ
  22
  F +
  HG
  I
  KJ
  −
  F +
  HG
  I
  KJ
  cos sin −sin
  σl =
  σσ
  (^13) βφβ 22
  2
  F +
  HG
  I
  KJ
  (cos −−sin sin )
  or σl =
  σσ
  (^13) ββφ 22
  2
  F +
  HG
  I
  KJ
  (cos −−cos cos ) ...(Eq. 13.15)
  σ
  σ
  l
  v
  = K =
  cos cos cos
  cos cos cos
  ββφ
  ββφ
  −−
  +−
  22
  22 ...(Eq. 13.16)
  K is known as the ‘Conjugate ratio’.
  Using Eq. 13.12,
  σl = γz. cosβ.
  cos cos cos
  cos cos cos
  ββφ
  ββφ
  −−
  +−
  F
  H
  GG
  I
  K
  JJ
  22
  22
  ...(Eq. 13.17)
  If σl is defined as Ka. γz as usual,
  Ka = cos
  cos cos cos
  cos cos cos
  β
  ββφ
  ββφ
  −−
  +−
  F
  H
  GG
  I
  K
  JJ
  22
  22 ...(Eq. 13.18)
  Ka is the ‘Rankine’s Coefficient, of active earth pressure for the case inclined surcharge—
  sloping backfill.
  The distribution of pressure with the height of the wall is linear, the pressure distribu-
  tion diagram being triangular as shown in Fig. 13.12 (c). The total active thrust Pa per unit
  length of the wall acts at (1/3)H above the base of the wall and is equal to^12 Kaγ.H^2 ; it acts
  parallel to the surface of the fill.