Geotechnical Engineering

DHARM

382 GEOTECHNICAL ENGINEERING

(ii) for point A 2 ,

θ 0 = tan.

.

−^118 =°.

12

56 31 = 0.9828 rad.

σ 1 =

180

0 9828 0 8321

π

(. +. ) = 104 kN/m^2

τmax =

(^180) 0 8321
π
.(. ) = 47.68 kN/m^2
(iii) for point A 3 ,
θ 1 = tan.
.
−^106 =°.
12
26 565 = 0.464 rad.
θ 2 = tan
.
.
− (^1) =°.
24
12
63 435 = 1.107 rad.
θ 0 = (θ 2 – θ 1 ) = (1.107 – 0.464) rad. = 0.643 rad.
σ 1 = (180/π) (0.643 + 0.600) = 71.22 kN/m^2
τmax =
(^18006)
π
×. = 34.38 kN/m^2
Absolute maximum shear stress = q/π = 180/π = 57.3 kN/m^2
This occurs at a depth B/2 or 0.9 m below the centre of the footing.
Example 10.6: A circular area on the surface of an elastic mass of great extent carries a
uniformly distributed load of 120 kN/m^2. The radius of the circle is 3 m. Compute the intensity
of vertical pressure at a point 5 metres beneath the centre of the circle using Boussinesq’s
method.
(S.V.U.—B.E., (Part-Time)—Apr., 1982)
Radius ‘a’ of the loaded area = 3 m
q = 120 kN/m^2
z = 5 m
z = q
az
1 1
1 232
−

• L
  N
  M
  M
  O
  Q
  P
  {(/)}/ P
  = 120 1
  1
  135232
  −


• 
  

  L
  N
  M
  M
  O
  Q
  P
  {(/)}/ P
  = 120 1
  1
  34 25^32
  −
  L
  N
  M
  O
  Q
  P
  (/)/
  = 44.3 kN/m^2.
  Example 10.7: A raft of size 4 m-square carries a load of 200 kN/m^2. Determine the vertical
  stress increment at a point 4 m below the centre of the loaded area using Boussinesq’s theory.
  Compare the result with that obtained by the equivalent point load method and with that
  obtained by dividing the area into four equal parts the load from each of which is assumed to
  act through its centre.