Civil Engineering Reference
In-Depth Information
Solution:
′
1
1
−
sin
sin
φ
φ
1
−
sin
sin
25
°
K
a
=
′
=
°
=
0 41
.
+
1
+
25
−
′
p
=
K h
γ
2
c K
a
a
a
Pressure distribution:
P
= − × ×
0
(
2 4
)
0 41
.
=−
5 1
.
kPa
0
m
P
=
(
0 41 18 10
.
× ×
)
− × ×
(
2 4
)
0 41
.
=
68 7
.
kPa
10
m
Tension crack:
′
′
=
2
c
φ
tan
.
h
c
=
45
°+
0 70
m
γ
2
γ
= 18kN/m
3
c' = 4 kPa
σ
' = 25°
10m
Fig. 7.12
Example
7.4.
Theoretical distribution:
Distribution used for design:
-5.1 kPa
h
c
= 0.7 m
0.7 x 9.81 = 6.9 kPa
68.7kPa
68.7 kPa
Fig. 7.13
Pressure distributions
(not to scale)
.
7.5.3 The occurrence of tensile cracks
A tension zone, and therefore tensile cracking, can only occur when the soil exhibits cohesive strength.
Gravels, sands and most silts generally operate in a drained state and, having no cohesion, do not experi-
ence tensile cracking.
Clays, when undrained, can have substantial values of c
u
but, when fully drained, almost invariably have
effective cohesive intercepts that are either zero or, have a small enough value to be considered
negligible.
It is therefore apparent that tensile cracks can only occur in clays and are only important in undrained
conditions. The value of h
c
, as determined from the formula derived above, is seen to become smaller as
