Foundation Engineering
ECIV 4352
Chapter (4) Bearing capacity of shallow foundation
Bearing capacity of layered soil
In practice the foundation may be based on layered soil profile, or in some times we have
to replace adequate thickness of weak soil by stronger one.
Calculating the ultimate bearing capacity for such case have some different as that will be
shown below.
qu  qb 
 D f  K s tan 1
2ca H

  1 H 2 1 
  1 H  q1
B
H
B


q1  c1 N c 1  q1 N q 1  0.5 1 BN  1  q   1 D f
, N q 1 and N  1 are given from the same tables but use   1 N c 1
: Ultimate bearing capacity. qu
: bearing capacity of bottom layer. qb


In this case qb  c 2 N c 2   q 2 N q 2   0.5 2 BN  2   q  D f  H  1
, N q  2  and N   2  are given from the same tables but use    2 N c 2 
: Adhesion ca
H: Thickness of top soil layer below bottom of foundation.
B: Width of foundation.
: Punching shear coefficient. K s
To find K s
To find ca
1- find q2 / q1
q1  c1 N c 1  0.5 1 BN  1
q 2  c2 N c 2   0.5 2 BN  2 
1- find q2 / q1
2- Go to figure 3.21 P.190 to find
c a / c1
2- Go to figure 3.21 P.190 to find K s
When the top layer height (H) is relatively large, the failure surface will occur in the top
layer and so qu = q1
Otherwise there will be punching shear failure in the top layer and then general shear
failure in the bottom layer.
For rectangular foundation:
 B  2c H 
 B  2 D f  k s tan  1 
qu  qb  1   a    1 H 2 1  1 

   1 H  q1
L  B 
L 
H 
B



qb  C 2 N c 2  Fcs 2    1 D f  H N q 2  Fqs2   0.5 2 BN  2  Fs 2 
q1  C1 N c 1 Fcs 1   1 D f N q 1 Fqs1  0.5 1 BN  1 Fs 1
Read the special cases from text book P.194 – 195
Example (1)
Rectangular footing 1.5m x 1m based on the shown soil profile; find the gross
allowable load that foundation can carry.
Special case III
B

 B  2c H 
qu  1  0.2 5.14C 2  1   a    1 D f  q1
L
L  B 


In this case q 2 / q1  C 2 / C1  48 / 120  0.4  ca / C1  0.9 from figure 3.22
 c a  0.9  120  108 KN / m 2 .
1 
1  2  108  1 


2
qu  1  0.2
  5.14  48  1 

  16.8  1  656.4 KN / m
1.5 
1

 1.5 

B
1 


2
q1  1  0.2 5.14C1   1 D f  1  0.2
  5.14  120  16.8  1  715.8 KN / m
L
1.5 


< q1  OK , otherwise use qu = q1 q u
656.4
 164.1KN / m 2
4
 164.1  1  1.5  246.15KN
q all 
Qall
Read Example 3.9 P.194
Bearing Capacity of foundation on top of slope
For continuous shallow foundation:
qu  CN cq  0.5BN q
For pure sand C = 0.00  qu  0.5BN q
For pure clay   0.00  qu  CN cq
b: Distance from edge of foundation to the
top of slope.
H: Height of top of slope.
: Angle of slope with horizontal
To find N cq :
1- Find stability number N s 
H
C
2- Find b/B and D f / B
3- Go to figure A shown below:
 If B < H: use curves with N s  0.00
 If B  H : use curves with calculated N s .
To find N q :
1- Find b/B and D f / B
2- Go to figure B shown below to find N q .
Fig.B Finding N
Fig.A Finding Ncq
Example 2)
For the shown soil profile determine the ultimate bearing capacity of the continuous foundation.
Solution
qu  0.5BN q C  0.00 
B=1.5m<H=6m  Use N s  0.00 .
Go to figure 3.25 P.197 using:
b/B=1, D f / B =1, and   30  .
N q  75
qu  0.5  16.8  1.5  75  945KN / m 2