DHARM
LATERAL EARTH PRESSURE AND STABILITY OF RETAINING WALLS 487
But the ratio
()
()
bc
bd
−
−
may be transformed as follows:
()
()
bc
bd
−
−
=
bbd
bd
db
db db
−
−
= −
−
=
+
1
1
1
1
/
/ /
...(Eq. 13.49)
From triangles ABE and ABD, by the application of the sine rule, one may obtain d/b as
follows:
d/b = d
AB
AB
b
. sin( )
sin
.sin( )
sin( )
sin( ).sin( )
sin( ).sin( )
= +−
+
= +−
−+
φδ
ψ
φβ
αβ
φδ φβ
αδ αβ
...(Eq. 13.50)
Hence, substituting this in Eq. 13.49.
()
()
bc
bd
−
− = –
1
1 +
+−
−+
sin( ).sin( )
sin( ).sin( )
φδ φβ
αδ αβ
...(Eq. 13.51)
Substituting this in Eq. 13.48,
x =
H
sin
.sin( )
sin( )
.
sin( ).sin( )
sin( ).sin( )
α
φα
αδ φδ φβ
αδ αβ
+
−
+ +−
−+
1
1
...(Eq. 13.52)
Since Pa =
1
2
γψx^2 .sin from Eq. 13.46, one obtains
Pa =
1
2
1
1
2
2
2
2
2
γαδ
α
φα
αδ φδ φβ
αδ αβ
.sin( ).
sin
.sin ( )
sin ( )
.
sin( ).sin( )
sin( ).sin( )
− +
−
+
+−
−+
L
N
M
M
M
M
M
O
Q
P
P
P
P
P
H
or Pa =
1
2
1
2
2
2
.. 2
sin ( )
sin .sin( )
sin( ).sin( )
sin( ).sin( )
γ αφ
ααδ
φδ φβ
αδ αβ
H +
−+
+−
−+
L
N
M
M
O
Q
P
P
which is the same as Eq. 13.33 obtained previously from Coulomb’s theory.
Another form for Pa is as follows (Taylor, 1948):
Pa =
1
2
γ^2 ααφ
αδ φδ φβ
αβ
H
cos .sin( )
sin( ) sin( ).sin( )
sin( )
ec −
−+ +−
+
L
N
M
M
M
M
M
O
Q
P
P
P
P
P
...(Eq. 13.53)
It is interesting to note that when α = 90° and δ = φ, both Eqs. 13.33 and 13.53 reduce to
the corresponding value obtained by using Eq. 13.17 of Rankine’s theory. (Also the form for Pa
expressed in Eq. 13.53 may be derived by considering the equality of the side ratios x/a and
CD/AD, and those in the similar triangles BJC and BCD, JG being parallel to CD, and substi-
tuting in Eq. 13.46 for Pa)