A Coulomb-type Solution for Backfills
under a Line of Inclined Surcharge
Venanzio R. Greco
Assistant Professor, Department of Structural Engineering, University of Calabria,
87030 Roges di Rende, Italy
e-mail venanziogreco@strutture.unical.it
ABSTRACT
The active thrust on retaining walls, when a line of surcharge acts on the
backfill is currently calculated using an analytical hybrid method, where
simultaneously the backfill soil is considered to be in the active state of
failure, to calculate the thrust in absence of the surcharge, and in the elastic
state, to calculate the thrust increment due to the surcharge.
The method of Coulomb permits us, however, to obtain a coherent solution
of the problem, through a numerical solution. However, with respect to a
numerical solution, an analytical one is clearly preferable if available.
Following Coulomb's approach, this paper gives the analytical solution of
the active thrust on retaining walls, when a line of inclined surcharge acts on
the wedge behind the wall. The position of the active thrust is also given in a
simple analytical form.
KEYWORDS: Active earth pressure, Retaining walls, Stability analysis,
Coulomb's method, Limit equilibrium, Effect of surcharges.
INTRODUCTION
The thrust exerted by a backfill, on which lines or strips of surcharge act, is globally due to the thrust wedge
weight and the surcharge on the backfill. In order to obtain an analytical solution, the problem is often
subdivided into two sub-problems, where the thrusts are calculated independently of one another and then
added (Bowles 1982).
In the first sub-problem, the surcharge is not considered and the thrust is calculated using a method for
assessing earth thrusts in active state, as the method of Coulomb (1773) or that of Rankine (1857).
In the second sub-problem, a thrust increment is calculated using the elasticity theory with reference to a
weightless semi-space subject to the surcharge (Cerruti, 1882; Boussinesq, 1885), by employing procedures
such as those suggested by Misra (1980) and Jarquio (1981). This approach is supported by the experimental
results of Spangler (1936) and Spangler and Mickle (1956), and the result interpretations due to Mindlin
(1936) and Terzaghi (1943). The theoretical formulas of elasticity were adapted to fit the test results, by
assuming the value 0.5 for Poisson's coefficient (whose effective value is very difficult to ascertain) and
modifying the numerical coefficients of the original elastic equations. From a practical point of view, this
approach is very attractive, the treatment of the mathematical model being very simple and permitting a
quick calculation (Kim and Barker, 2002). However, Steenfelt and Hansen (1983, 1984) assert that this
approach can only be used for unyielding walls, while structures in the active state of failure require a more
appropriate analysis, such as Coulomb's approach. This opinion seems to be confirmed by the study of
Georgiadis and Anagnostopoulos (1998), who, comparing experimental results on sheet pile walls with
predictions made using various methods, state that the values of lateral pressures and bending moments
determined via the elastic analysis are much larger than those determined experimentally. The results
obtained using numerical solutions of Coulomb's method are, on the contrary, sufficiently accurate.
The traditional approach to the thrust calculation in the presence of surcharge on the backfill, which
uncouples the two effects (wedge weight and surcharge), has notorious, clear inherent shortcomings. The
calculus of the thrust due to the surcharge is, in fact, based on the hypothesis that the backfill soil is in the
elastic state, while the thrust due to the wedge weight is computed supposing the same soil to be in the active
state of failure. In addition, the presence of the surcharge modifies the extension of the thrust wedge, which
differs from those relative to the thrust theories formulated in the absence of surcharges. The thrust due to the
surcharge is, finally, calculated with reference to a half-space, which, from a geometrical point of view,
differs sensibly from the real backfill shape. It is not surprising, therefore, that the traditional uncoupled
approach can obtain inaccurate previsions of lateral pressure. For these reasons, it is the opinion of the writer
that a coherent solution of the problem has to be found entirely in the approach of Coulomb. A numerical
solution of the problem of calculating the thrust, which includes both the effects of the backfill weight and
surcharge, is, in fact, possible and easy, using Coulomb's method. However, with respect to a numerical
solution, an analytical one is clearly preferable if available. Analytical solutions of active thrusts due to
backfills subject to vertical surcharge were given for distributed surcharges (Motta, 1993; Greco 2006b) and
lines of surcharge (Greco 2005, 2006a; Silvestri 2006). However, these studies are concerned with vertical
surcharges only.
This paper presents an analytical solution of the active thrust when a line of inclined surcharge acts on the
backfill. The study is developed in terms of limit equilibrium method, following Coulomb's approach.
Because the failure mechanism concerned with Coulomb's method is kinematically admissible, the thrust
calculated with this approach is an upper bound in the field of limit analysis (Davis and Selvadurai, 2002;
Yu, 2006). Finally, a well approximate analytical solution is also obtained for the evaluation of the point of
application of the thrust.
THRUST CALCULATION
Fig. 1 shows a schematic cross section of the thrust wedge A-B-C behind a retaining wall and the forces
acting on it. This wedge is limited by the topographic surface (supposed to be plane and inclined at an angle
 with respect to the horizontal) and by two failure planes: A-C inside the backfill and inclined at , and B-C
inclined at , separating wall and backfill. A force R acts on the plane A-C with inclination ' with respect to
the normal to the plane (' being the shear strength angle of the soil), while the thrust Pa acts on the plane BC with inclination  (the friction angle between soil and wall along B-C). On the backfill, at a horizontal
distance d from B, a force line F acts with inclination  with respect to the vertical. The wedge A-B-C is also
subject to its weight W. The backfill soil is assumed to be cohesionless and without pore pressures. All the
angles are positive if counterclockwise.
d
'
F
'


A

B
zP
W
H



Pa


'
R
C
Figure 1. Cross section of thrust wedge and forces acting on it.
The active thrust Pa, relative to a given value of , is obtainable from the equilibrium
conditions of forces acting on the wedge A-B-C:
 A f ()  F g () for   
Pa ()  
otherwise
 A f ( )
(1)
sin (  )
1
,
A  H 2
2
sin 2 
(2)
where:
f () 
sin (  )
sin (  ' )
,
sin (  ) sin ( + '+  )
(3a)
sin (  ')
,
sin ( + '+  )
(3b)
g ( ) 
and  is the inclination angle, with respect to the horizontal, of the line connecting the point
of application of force line F with point C:

 H cot   d 
   tan 1 
.
2
 H  d tan  
(4)
For geometrical reasons connected with the wedge A-B-C, the functions f () and g() are
not considered outwit the range [, ]. In this range, f () is a unimodal function of , which
is zero for  = ' and  =  (Fig. 2) and attains its maximum for  = 0, whose value is given
in literature:
 P( P  Q)(1 QR)  P 

 0  ' tan 1 


1

(
P

Q
)
R


(5)
P  tan(') ,
(6a)
Q  tan(  ' ) ,
(6b)
R   cot(  ) .
(6c)
with:
The function g() is positive (and the load line F produces an increment of thrust) if   p,
while it is negative (and F determines a reduction of thrust) if   p, where
 p  ' .
(7)
Independently of the value of angle , the function g() increases strictly with  if:
        ,
(8)
Moreover, the tangential component of F is feasible, if it results (Fig. 1):
  '     ' .
(9)
f (), g()
g()
0
<
0

f ()
0



'
0

Figure 2. Schematic graphic of functions f() and g() in terms of angle  in the interval
[, ]
In the following, we assume that the value of  is admissible (Eq. 9) and the function g()
increases strictly with  (Eq. 8). Under these hypotheses, the function Af()+Fg() is
unimodal and attains its maximum for  = c, (Fig. 3a) with:
 C  C 2  C C
1
1
0
2
 c  ' tan 1 
C

2

and:




(10)
C0  Asin(   ' ) sin (  ) sin(   ' )  F sin 2 (  ' ) sin(    - )
(11a)
C1  A cos(  ' ) sin (  ) sin(   ' )  F sin(   ' ) cos(  ' ) sin(    - )
(11b)
C 2  Acos(  )sin (  )  Acos(  ' ) sin(   ) cos(  ' )  F sin (    ) cos 2 (  ' ) (11c)
It is possible to prove that c > 0 (Fig. 3a). Because g() is positive and increases with , if
the value of c given by Eq. 10 is higher than  or the value under root in Eq. 10 is negative,
the function Af()+Fg() attains its maximum in the feasible range for  =  (Figure 3b).
Pa()
A f(
)+F
Case c < 
Pa()
[ 0]
F g()
A f()+
Case c > 
g(
) [
0
]
A f()
A f()
'
m

0 c
'

a)
0
m


b)
Figure 3. Functions A f () and A f ()+F g () in terms of angle  for =0.
The function A f ()+F g () is zero for  = min where:
 N  N 2 4N N ) 
1
1
0
2 
 min  ' tan 1 

2N2



(12)
N 0  F sin (  ' )sin  ,
(13a)
N 1  A sin (  ' )  F sin (  ') ,
(13b)
N 2  A cos(  ' )  F cos(  ' ) cos  .
(13c)
and:
The angle min can be less, equal to or higher than ', according to whether the value of  is
negative, zero or positive, respectively.
If A f (c)+F g (c) ≥ A f (0), we call m the value of  for which we have (Fig. 3):
A f ( m )  F g (  m )  A f ( 0 ) .
(14)
The value of m is given by:
 M  M 2 4M M
1
0
2
 1
 m  ' tan 
2M 2


1


 ,

(15)
where:
M 0  N 0  A f ( 0 ) sin (  ) sin(   ' ) ,
(16a)
M 1  N 1  A f ( 0 ) sin (      ) ,
(16b)
M 2  N 2  A f ( 0 ) cos(  ) cos(  ' ) .
(16c)
If, on the contrary, we have A f (c)+F g (c) < A f (0), we call m the value of  given by the
condition (Fig. 7):
A f ( c )  F g ( c )  A f ( m )
(17)
which remains given by Eq. 15, but with the following coefficients:
M 0  [ A f ( c )  F g ( c )]sin ( ) sin(  ' ) ,
(18a)
M 1 [ A f ( c)  F g ( c)]sin (  -)- Asin( -') ,
(18b)
M 2 [ A f ( c)  F g ( c)]cos( )cos(-')- Acos(-') .
(18c)
To solve the problem of finding the thrust in presence of an inclined surcharge line on the
backfill, we must determine the value of  maximizing the thrust Pa() in Eq. 1. From this
equation it emerges that Pa() follows the function Af()+Fg(), from min to , and the
function A f() for < < . To this end, we call crit the value of  maximizing Pa() and,
moreover, let Sa be the maximum of Pa():
S a  max{Pa ()} Pa ( crit ) .

(19)
The equation giving Sa is different according to the value of  and the angles , m, 0, c
and p which determine the reciprocal positions of the curves A f ()+F g () and A f ().
Case   0
Figs 3 and 4 show the cases  = 0 and  < 0, respectively, where A f ()+F g () ≥ A f () for
all the values of  in the interval [min, ]. Considering that Pa() follows the curve A f ()
for   m and the curve A f ()+F g () for  > m, we have (Figure 4b):
 if   m, then crit = 0 and
S a  A f ( 0 ) ;
(20a)
S a  A f ( )  F g ( ) ;
(20b)
S a  A f ( c )  F g ( c ) .
(20c)
 if m <  < c, then crit =  and
 if   c, then crit = c and
A f( 
)+F
g( )
[<
0]
Pa()
A f(c)+F g(c)
Sa
<0
<0
A
)
g(
+F
)

f(
A f(0)
A f()


min ' m
0 c
0
min ' m

c

b)
a)
Figure 4. Case <0: a) functions A f () and A f ()+F g () in terms of angle ; b) thrust Sa
in terms of angle .
Case >0 and p ≤ 0
When angle  is positive, a part of the curve A f ()+F g () is below the curve A f (). If,
moreover, p ≤ 0 we have the case shown in Fig. 5. As is evident, Sa is given by the same
Eqs (20) of the previous case.
Pa()
Sa
A f()+F g()
A f(0)
>0
p< 0
g
)+F
A f(
()
A f(c)+F g(c)
A f()


'
min p m
0 c
a)
'

min p m
0
c

b)
Figure 5. Case >0 and p <  0: a) functions A f() and A f()+F g() in terms of angle
; b) thrust Sa in terms of angle .
2.3. Case >0 and 0 < p ≤ c
Figure 6 reports the case  > 0 with 0 < p ≤ c. From this figure, it clearly emerges
 if   0, then crit = 0 and
S a  A f ( 0 ) ;
(21a)
S a  A f ( ) ;
(21b)
S a  A f ( )  F g ( ) ;
(21c)
 if 0 <   p, then crit =  and
 if p <  < c, then crit =  and
 if   c, then cri t = c and
S a  A f ( c )  F g ( c ) .
Pa()
Sa
A f()+Fg()
>0
0<P<c
(21d)
A f(0)
A f()
A f()+Fg()
A f(c)+Fg(c)
A f()


'
0 p m
min
c
'

min
0 p m
c

b)
a)
Figure 6. Case >0 and 0 < p <  c: a) functions A f() and A f()+F g() in terms of ;
b) thrust Sa in terms of .
2.4. Case >0 and p > c
This case is illustrated in Figure 7, where:
 if   0, then crit = 0 and
S a  A f ( 0 ) ;
(22a)
S a  A f ( ) ;
(22b)
S a  A f ( c )  F g ( c ) .
(22c)
 if 0   < m, then crit =  and
 if  > m, then crit = c and
Sa
Pa()
>0
p > c
A f(0)
A f()
A f()
A f(c)+Fg(c)


min
0
c
m p
'

min
0
c
m p

b)
A
a)
f( 
)+
Fg
(
)
'
Figure 7. Case >0 and p >  c: a) functions A f() and A f()+F g() in terms of ; b)
thrust Sa in terms of .
2.5. General formulation
We can recapitulate the previous four cases in the following way:

if   min0, m} then crit = 0 and
S a  A f ( 0 ) ;
(23a)
 if min0, m}<  < maxc,m} then crit =  and
S a  max{ A f (), A f ()  F g ()} ;
(23b)
 if   maxc,m}, then crit = c and
S a  A f ( c )  F g ( c ) .
(23c)
2.6. Analysis in terms of the distance d
Alternatively, we can express the thrust Sa in terms of the distance d instead of the angle .
To this end, we observe that the angles 0, m, p and c are independent of d, while the
angle  is a decreasing function of d, which is equal to  if d = 0, and converges to  when
ddiverges.
We introduce the following particular values of the distance d:
d0
the value of d given by the condition (d) = 0, i.e:
d 0 H
dp
(24a)
the value of d given by the condition (d) = p, i.e.
dP  H
dc
cos  sin(    0 )
sin  sin(  0  )
cos  sin(   P )
sin  sin(  P  )
(24b)
the value of d given by the condition (d) = c, i.e.
dc  H
cos  sin(  c )
sin  sin(c  )
(24c)
dm the value of d given by the condition (d) = m, i.e.
d m H
dmax = max{ d0, dm }.
cos  sin(    m )
sin  sin(  m  )
(24b)
(24e)
According to the value of d with respect to those of dc, dmax and dp, we have the following
cases (Fig. 8):
 if d  dmax, then Sa is unaffected by F and the wedge A-B-C is limited by a failure plane
inclined at crit = 0; the thrust Sa is hence given by Eq. 23a [Sa = A f (0)];
crit(d)
>c

0<<c
 if max{dm,dP}<d<dmax, then the failure plane A-C is inclined at an angle crit = 
depending on d; the thrust is given by Sa = A f ();
 if min{dm,dc}<d< min{max{dm,dP},dmax}, then the failure plane A-C is inclined at an angle
crit =  depending on d; the thrust is given by Sa = A f ()+F g ();
 if d  min{dc, dm} then both the thrust Sa and the failure angle crit depend only on F (they
do not depend on d) with crit = c and Sa given by Eq. 23c [Sa = A f (c)+F g (c)].
m<<0
<m
Case 0
and
case >0 with p<0
c
0
m
(
d)
p
a)
d
min
dc
d0
dp
dm
Case 0 with 0<p<c
crit(d)
>c
p<<c
0<<p
<0

c
(
d)
m
p
0
b) 
d
min
crit(d)
>p

dm
m<<p
dc
d0
dp
Case 0 with p>c
c<<m
<0
0<<c
p
m
(
d)
c
0
c)
min
d
dp
dm
dc
d0
Figure 8. Angles 0, c, m, P and  in terms of d: a) case  ≤ 0 and case  > 0 with
aP ≤ 0; b) case  > 0 with 0 < P < c; c) case >0 with P ≥c.
POINT OF APPLICATION OF THE THRUST
In addition to the thrust calculation, the design of the retaining walls requires the location of
the point of application of the thrust. The depth zS of this point below point B is obtainable
from the stress distribution along the thrust plane B-C. Such a distribution is calculated
under the hypothesis that the thrust wedge A-B-C also fails along planes starting from the
surface B-C (Huntington 1957). If D is a generic point of B-C at depth z below B, then the
thrust Sa(z) acting on the plane B-D can be calculated as in the previous section, by
substituting H with z. Consequently, the angles c,  and m, which before depended on H,
are now functions of z. In particular, the angle (z) is given by:

 z cot   d 
( z )   tan 1 
,
2
 z  d tan  
d
(25)
'
' 
F

B
F
z
H

Sa(z)
(z)
D

C
Figure 9. Thrust Sa(z) acting on the plane B-D and angle (z) of the line connecting the
points D and F.
The angles c(z) and m(z) are given by Eqs (10) to (11) and (15) to (18), respectively, by
substituting H with z. On the contrary, the angles 0 and p are constant in this case too.
As concerning the dependence of angles c(z), (z), m(z) on the depth z, the following
features could be proven (Figs 10a and 11a).
1.
The angle c(z) decreases when z increases, starting from the depth z, where it assumes
the value , and converges to 0 when z diverges; z being given by
z 
2.
3.
2 F sin  sin(     )
,
 sin(   ) sin(   )
(26)
The angle (z) is an increasing function of z, which is equal to  for z = 0, and converges
to  as z diverges.
The angle m(z) is equal to P for z = 0; moreover:
 If p < 0 then the coefficients M0, M1 and M2 are given by Eq. 16 (with z in lieu of
H). In this case, the function m(z), increases with z and converges to 0 when z
diverges.
 If p > 0 then the coefficients M0, M1 and M2 are calculated using Eq. 16 for z < zM
(upper zone) and with Eqs. (18) for z > zM (lower zone), where the value of zM is
given by the following iterative equation:
zM 
g ( c )
2F
sin 2 
 f ( 0 )  f ( c ) sin(   )
(27)
g(c) and f(c) being functions of zM. In the upper zone, m(z) increases with z and
converges to c when z converges to zM. For z = zM we have a jump from c to 0.
Below zM the angle m(z) increases until a depth where c = P = m and decreases
below it.
We also introduce the following specific values of the depth z below point B:
 z0
is the value of z given by the condition 0 =(z0), which furnishes:
z0 d
 zp
 zm
(28a)
the value of z given by the condition p =(zp), which gives:
zp d
 zc
sin(  0  ) sin 
sin(    0 ) cos
sin(  p  ) sin 
;
sin(    p ) cos
(28b)
the value of z given by the condition c (z)=(z)=c, whose solution is obtainable
through a trial and error procedure (Appendix 1).
the value of z obtainable by the condition m (z) =(z)=m, whose solution is
achievable by means of a trial and error procedure (Appendix 2).
Moreover let
sin(   )
1
.
A( z )   z 2
2
sin 2 
(29)
The angles c(z), (z), m(z), 0 and P in terms of z are shown in Figs 10a and 11. Figure
10a is referred to the case  ≤ 0 and the case  > 0 with P < 0. It is thus linked with Figs 3,
4, 5 and 8a. Depending on the z-value, the inclination crit of the failure plane differs as
follows:
 if z ≤ zm, then we have  < m and thus crit(z) = 0; consequently
S a ( z )  A( z ) f ( 0 ) ;
(30a)
 if z m < z < zc, then we have m < < c and hence crit(z) = (z) and
S a ( z )  A( z ) f [( z )] F g[( z )] ;
(30b)
 if z ≥ zc, then we have  ≥ c and thus crit (z) = c(z):
S a ( z )  A( z ) f [ c ( z )] F g[ c ( z )]
(30c)
Figure 10b shows a set of slip surfaces internal to the thrust wedge A-B-C. This figure
focalizes four different subwedges.
 Subwedge B-Dm-E with height zm, where the slip surfaces are parallel and inclined at an
angle 0; we call S1 the thrust acting on this subwedge.
 Subwedge E-Dm-F (corresponding to the jump on the value of crit(z) at depth zm), which
is not involved with internal slip surfaces and does not transfer thrust to the plane B-C
(S2 = 0);
 Subwedge Dm-Dc-F, where all the slip surfaces cross the point of application of the force
line F; S3 is the thrust exerted by this subwedge.
 Subwedge Dc-C-A-F, where the slip surface inclinations gradually decrease with the
depth; we call S4 the thrust on Dc-C.
Figure 11 concerns the case  > 0 with P > 0 and is linked to Figs 6, 7, 8b and 8c. Figure
11a shows the angles c(z), (z), m(z), 0 and p with different dotted lines, while solid
lines refer to the angle crit(z), which varies in the following way:
 if z ≤ z0, then we have  < 0, thus crit(z) = 0 and
S a ( z )  A( z ) f ( 0 ) ;
(31a)
 if z 0 < z < zP, then we have 0 < < P and hence crit(z) = (z); therefore:
S a ( z )  A( z ) f [( z )] ;
(31b)
 if zP < z < zc, then we have P < < c and therefore crit(z) = (z), with
S a ( z )  A( z ) f [( z )] F g[( z )] ;
(31c)
 if z ≥ zc, then we have  ≥ c and thus crit (z) = c(z), with:
S a ( z )  A( z ) f [ c ( z )] F g[ c ( z )]
(31d)
The slip surfaces internal to the thrust wedge A-B-C are displayed in Figure 11b. As can be
seen, there are four subwedges.
 Subwedge B-Dm-E, extended until depth z0, which transmits the thrust S1 to the wall; the
slip surfaces are parallel and inclined at an angle 0.
 Subwedge Dm-DP-F, where all the slip surfaces cross point F and the thrust S2 is
obtainable by Eq. 31b;
 Subwedge DP-Dc-F, which transmits the thrust S3 on the plane B-C; here also all the slip
surfaces cross the point F, but the thrust S3 is given by Eq. 31c;
 Subwedge Dc-C-A-F, where the slip surface inclinations gradually decrease with the
depth; the relative partial thrust is S4.
We also call z1, z2, z3 and z4 the depths of the points of application of the thrust S1, S2, S3 and
S4, respectively. If the wall is not sufficiently high (H < zc), then some of the previous thrusts
are not considered.
The depth zS of the point of application of the resultant Sa is, obviously, given by:
1
zS 
Sa
H

0
d S a ( z)
1
z dz 
dz
Sa
4
S z
i
i
.
(32)
i 1
In the following, we call h1, h2, h3 and h4 the heights of the planes B-Dm, B-DP, B-Dc and BC, respectively, with
h1 = min{z0, zm, H},
(33a)
h2 = max{min{zP, H}, h1},
(33b)
h3 = max{min{zc, H},h2}
(33c)
h4 = max{h3, H}.
(33d)
The calculus of z1, z2, z3 and z4 is linked to the values of these heights.
Computing S1 and z1
For z  h1 the slip surfaces within the subwedge B-Dm-F are inclined at a constant angle
crit = o and the thrust Sa(z) is given by:
S a ( z )  A( z ) f ( 0 ) .
(34)
S1  S a (h1 )  A(h1 ) f ( 0 ) ,
(35)
2
z1  h1 .
3
(36)
Therefore the thrust S1 is:
and is applied at depth:
Computing S2 and z2
The thrust S2 is different from zero if h2 > h1 only. For h1 < z ≤ h2 the slip surface is inclined at
(z) and the thrust is given by:
S a ( z )  A( z ) f () ,
(37)
where f ( is expressible in terms of z:
f ( ) 
sin (  )
sin ( - ' )
d sin  z a1  d a 0

 f ( z) ,
sin (  ) sin ( + '+ - ) z cos  z c1  d c 0
(38)
with:
a1  cos  sin (  ') ,
(39a)
a 0  sin sin (  ' ) ,
(39b)
c1  cos  sin (') ,
(39c)
c 0  sin  sin (  '  ) ;
(39d)
Then Eq. 37 can be written in the form:
1 sin(   ) z a1  d a 0
S a ( z)  
d
z,
2 sin  cos  z c1  d c0
(40)
h a  d a0 
sin(   )  h2 a1  d a 0
1
S2  d

h2  1 1
h1  .
2 sin  cos   h2 c1  d c0
h1 c1  d c 0 
(41)
Therefore, the thrust S2 is:
and is applied at depth z2, with:
1 sin(   )
z2 S2  
d
2 sin  cos 

 1 a1
c a c a
c 0 c1 d (h1  h2 )
(c h  c d )  
2
2
(h2  h1 )  1 0 3 0 1 c 0 d 2 
 ln 1 2 0 

(c1 h1  c 0 d ) 
c1
 2 c1
 (c1 h2  c 0 d )(c1 h1  c 0 d )
(42)
Computing S3 and z3
For h2 < z ≤ h3 the slip surface is inclined at (z) and thus:
1 sin(   ) 2
S a ( z)  
z f ()  F g () .
2 sin 2 
(43)
Substituting Eq. 25 in the expression of g(, we have:
g () 
sin (  ') z b1  d b0

 g ( z) ,
sin ( + '+  ) z c1  d c0
(44)
with
b1  cos  sin (  ') ,
(45a)
b0  sin  sin (  ') .
(45b)
Then Eq. 43 can be written in the form:
S a ( z) 
with
e2 z 2  e1 z d  e0 d 2
,
z c1  d c0
(46)
1
sin(   )
e2   d
a1 ,
2 sin  cos 
(47a)
F
sin(  )
1
d
a0 
b1 ,
2
sin  cos 
d
(47b)
F
b0 .
d
(47c)
e1 
e0 
Therefore, the thrust S3 is:
e h  e1 d h3  e0 d 2 e2 h2  e1 d h2  e0 d 2
S3  2 3

.
c1 h3  c0 d
c1 h2  c 0 d
2
2
(48)
and is applied at depth z3 given by:
S3 z3 
1 e2
2
2
(h3  h2 )
2 c1
c 0 e2  c1 c 0 e1  c1 e0
2

2
c1
3

c 0 c1 d (h2  h3 )
c h c d 
d 
 ln 1 3 0 
c1 h2  c 0 d 
 (c1 h2  c 0 d )(c1 h3  c 0 d )
(49)
2
Computing S4 and z4
For h3 < z  h4 we have crit = c(z), thus:
S a ( z )  A( z ) f [ c ( z )]  F g[ c ( z )] .
(50)
The thrust S4 is therefore:
S 4  A(h4 ) f [ c (h4 )]  A(h3 ) f [ c (h3 )]  F{ g[ c (h4 )]  g[ c (h3 )]}
(51)
As concerning the depth of the point of application of the thrust S4, since the angle c is a
complicated function of z, a simple but accurate analytical solution can be obtained by
approximating the thrust Sa(z) in the interval [h2, h3] with a second degree polynomial which
verifies the conditions:
S a (h3 )  S1  S 2  S 3 .
(52a)
S a ( h4 )  S 1  S 2  S 3  S 4
(52b)
1
h  (h3  h4 ) .
2
(52c)
   c (h )
(52d)
S  A(h) f (  )  F gf (  )
(52e)
Under this assumption, we have:
5h4  h3 2 S

(h4  h3 )
6
3 S4
z4 
(53)
A
F
p

m
0 c

B
crit
m(z)
zp
z1
(z)
zm
E
F
S1
zm
Dm
z3
z0
S3
z4
z
c(z)
zc
zc
Dc
m(z)
S4
(z)
a)
b) H
z
C
z
Figure 10. Case  ≤ 0 and case  > 0 with p <  0: a): Angles 0, m, c ,  and crit versus
the depth z below the point B; the angle crit is reported with solid curves ; b) Slip surfaces
internal to the thrust wedge and thrusts acting on the three parts in which the thrust plane BC is divided.
F

0
'
p
c

B
crit
(z)
z1
z0
zp
z
zc
zM
z*
z3
zp
m(z)
S1
z2
z0
A
E=F
Dm
S2
DP
S3
z4
Dc
zc
c(z)
S4
a)
(z)
z
b)
H
z
C
Figure 11. Case  > 0 with p >  0: (a) Angles 0, m, c ,  and crit versus the depth z
below the point B; the angle crit is reported with solid curves ; (b) Slip surfaces internal to
the thrust wedge and thrusts acting on the four parts in which the thrust plane BC is divided.
A NUMERICAL APPLICATION
We apply the method presented in the previous sections to a problem having the following data:
 = 95°,  = 10°, '=36°,  = 18°,  = 19 kN/m3, H = 7m, F = 50kN/m, d = 1
 = -5°.
With these data, we have A = 467.3 kN/m and (Eqs (6) to (9))

P=0.488,
Q=1.664,
R=0.424,
0=60.0°;
C0=-170.1,
C1 = -79.72,
C2 = -345.3,
c = 62.9°;
P = 36° - 5° = 31°
Since  is negative and P < 0, the problem is referred to the graph of Figure 4, where clearly
A f (c)+F g (c) is greater than A f (0). Therefore, the angle m must be calculated using Eqs
(23) to (25)
M 0 = -55.52,
M 1 = 284.2,
m = 50.0°;
M 2 = 245.9,
Eq. 5 gives:  = 86.9°. As a result   maxc,m}, thus we use Eq. 43 to calculate Sa, with
crit = c = 62.9°, with f(c) = 0.302 and g(c) = 0.530,
Sa = A f(c) + F g(c) = 167.7 kN/m
Alternatively, using the distances, we have
d0 = 5.19 m;
dm = 7.62 m;
dmax =max{d0, d m}= 7.62 m;
dc= 4.61 m;
dP=17.32 m.
Since d  min{dc, dm}= 4.61 m, the thrust Sa is calculated using Eq. 43.
Concerning zS, we have:
z0 = 1.35 m,
zP = 0.40 m.
The values zm and zc are obtained via trial and error procedures, because m and c are functions
of zm and zc, respectively. From these we have found:
zm = 0.42 m
zc = 3.23 m.
As a consequence,
h1 = 0.42 m,
S1 = 0.52 kN/m
z1 = 0.28 m
h2 = 0.42 m,
S2 = 0
h3 = 3.23 m,
S3 = 60.28 kN/m
z3 =1.55 m
h4 = 7.00 m,
S4 =106.87 kN/m
z4 =5.39 m
and hence: zS = 4.00 m
CONCLUSIONS
The paper has presented a simple analytical solution to calculate the active thrust acting on
retaining walls subject to a line of surcharge with generic inclination. The effect of the change on
the thrust acting on a wall depends on the distance d of the surcharge line from the wall (Wu
1975). Synthesizing, we can compare the value of d with some fixed values, dmax, dP, dm and dc,
depending on wedge geometry, soil geotechnical parameters and intensity and inclination of
surcharge line.


if d  dmax then Sa is unaffected by F and the thrust Sa is hence calculated using the traditional
formulation of Mueller-Breslau (1906): Sa = A f (0);
if min{dc, dm}<d< dmax, then the failure plane A-C is inclined at an angle crit =  depending
on d; the thrust Sa is given:



if max{dm,dP}<d<dmax, by Sa = A f ()
if min{dm,dc}<d< min{max{dm,dP},dmax}, by Sa = A f ()+F g ();
if d  min{dc, dm} then both the thrust Sa and the failure angle crit depend only on F (they do
not depend on d); with crit = c and Sa = A f (c)+F g (c).
Concerning the position of the point of application of the thrust, the thrust wedge is divided in
four sub-wedges, characterized by different inclinations of internal planes of failure. For each
sub-wedge, the relative thrust and the position of its point of application are also obtained
analytically with a rigorous procedure for z < zc and simplified but accurate for z > zc.
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APPENDIX 1
The depth zc is obtained by the following iterative procedure.
1.
2.
A trial value is assumed for zc (for example zc = H).
The following parameters are calculated:
Ac 
1
2 sin(   )
 zc
2
sin 2 
C0  Ac sin(   ' )sin (  )sin(   ' )  F sin 2 (  ' ) sin(    - )
(54)
(55a)
C1  Ac cos(  ' )sin (  )sin(   ' )  F sin(   ' )cos(  ' )sin(    - )
C 2  Ac cos(  )sin (  )  Ac cos(  ' ) sin(   ) cos(  ' )
 F sin (    ) cos 2 (  ' )
C  C 2 C C 
1
1
0
2 
 c  ' tan 

C2



1
3.
5.
(55c)
(56)
A new value zc' is calculated for zc as
zc '  d
4.
(55b)
sin(  c  ) sin 
sin(    c ) cos 
(57)
If zc' is sufficiently close to zc
 then the final value of zc is the average between the trial value and the result [(zc + zc')/2];
 otherwise a new value is assumed for zc and the procedure restarts from step 2. The
assumption zc = zc' for starting a new iteration is generally ineffective; starting from
(n zc+ zc')/(n+1), with n>0, the convergence is slower but more effective.
End
APPENDIX 2
The depth zm is obtained by the following iterative procedure.
1.
2.
A trial value is assumed for zm (for example zm = z0).
The following parameters are calculated:
Am 
3.
4.
1
2 sin(   )
 zm
2
sin 2 
(58)
M 0  F sin (  ' )sin   Am f ( 0 ) sin (  ) sin(   ' )
(59a)
M 1  Am sin (  ' )  F sin (  ')  Am f ( 0 ) sin (      )
(59b)
M 2  Am cos(  ' )  F cos(  ' ) cos   Am f ( 0 ) cos(  ) cos(  ' )
(59c)
If M12 + 4 M0 M1 < 0 then the coefficients M0, M1 and M2 are recalculated as follows:
M 0  [ Am f ( c )  F g ( c )] sin ( ) sin(  ' ) ,
(60a)
M 1  [ Am f ( c )  Fg ( c )] sin (     - ) - Am sin( -' )
(60b)
M 2  [ Am f ( c )  F g ( c )] cos( ) cos(-' )- Am cos(-' ) .
(60c)
The value of m is calculated as
 M  M 2  4M M
1
0
2
 m  ' tan 1  1

2M2

5.
7.
(61)
A new value zm' is calculated for zm as
zm '  d
6.




sin(  m  ) sin 
sin(    m ) cos 
(62)
If zm' is sufficiently close to zm

then the final value of zm is equal to (zm + zm')/2;

otherwise a new value is assumed for zm and the procedure restarts from step 2. The
assumption zm = zm' for starting a new iteration is generally effective.
End
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