Civil Engineering Reference
In-Depth Information
1.
Solution A: Soil surface horizontal
δ
=
17.5°;
ψ
=
90°;
β
=
0°;
φ
′
=
35°.
{
}
2
sin
55
°
/ sin
90
°
K
a
=
sin
107 5
.
° +
sin
52 5
.
°
sin
35
°
/ sin
90
°
{
}
=
2
0 819
0 976
.
=
0 246
.
.
+
.
0 675
P
=
0 5
.
K H
γ
2
=
0 5 0 246 19 5
.
×
.
× × =
2
58 43
.
kN
a
a
This value is inclined at 17.5° to the normal to the back of the wall so that the total
horizontal active thrust, according to Coulomb, is 58.43
×
cos 17.5°
=
55.7 kN.
Note:
If
δ
had been assumed equal to 0°, the calculated value of total horizontal thrust
would have been the same as that obtained by the Rankine theory of Example
7.1.
2.
Solution B: Soil surface sloping at 35°
Substituting
φ
′
=
35°,
β
=
35°,
δ
=
17.5° and
ψ
=
90° into the formula gives
K
a
=
0.704. Hence
Total active thrust
=
0.5
×
0.704
×
19
×
5
2
=
167.2 kN
Total horizontal thrust
=
167.2
×
cos 17.5°
=
159.5 kN
Increase in horizontal thrust
=
159.5
−
55.7
=
104 kN
Example 7.6:
Coulomb active thrust; more than one soil
Determine the total horizontal active thrust acting on the back of the wall of Example
Solution:
Active pressure at the top of the wall,
P
a
0
=
.
Consider the upper soil layer:
For
φ
′
=
30°,
δ
=
φ
′
/2
=
15°,
β
=
0° and
ψ
=
90°, K
a
=
0.301
Hence active pressure at a depth of 3 m
=
0.301
×
16
×
3
=
14.5 kPa.
But this pressure acts at 15° to the horizontal (as
δ
=
15°).
Horizontal pressure at depth
=
3
m p
=
=
14 5
. cos
15
° =
14 0
.
kPa
.
a
3
Consider the lower soil layer:
For
φ
′
=
20°,
δ
=
φ
′
/2
=
10°,
β
=
0° and
ψ
=
90°, K
a
=
0.447
p
=
0 447 16 3
.
× × ×
cos
10
° =
21 1
.
kPa
a
3
p
=
[(
0 447 24 4 5
.
× ×
.
)
+
21 1
. ] cos
×
10
° =
68 3
.
kPa
a
7 5
.
These values are shown in brackets on the pressure diagram in Fig.
7.7b
.
7.6.2 The Culmann line construction
When the surface of the retained soil is irregular, Coulomb's analytical solution becomes difficult to apply
and it is generally simpler to make use of a graphical method proposed by Culmann in 1866, known as
