12
Lateral Earth Pressure:
At-Rest,Rankine, and Coulomh
Retaining structurcssuchas retaining walls,bascmentwalls,and bulkheadsare commonly encountercdin foundation engineeringas thcy support slopesof earth masses.
Proper designand construction of thesestructuresrequire a thorough knowledgeof
thc lateral forcesthat act between the retaining structuresand the soil massesbeing
retaincd.These lateral forccs arc causedby lateral earth prcssure.This chapter is devotcd to the study oI the various earth pressuretheories.
12.1
At-Rest, Active, and Passive Pressures
Consider a massof soil shown in Figurc. l2.la. The massis bounded by a .frictionless
wall of height AB. A soil element located at a depth z is subjectedto a vertical effective prcssurerrj,and a horizontal efTectivepressureoj,. There arc no shearstresseson
thc vcrtical and horizontal planesof the soil element. Let us define the ratio of oj, to
a l , a s a n o n d i m e n s i o n aql u a n t i t v K , o r
C,r
(12.1)
K - -
(f,'
Now, three possiblecasesmay arise concerningthe retaining wall: and they are
described
Case 1. If the wall AB is static- that is, if it does not move either to the right or to
the left of its initial position - the soil masswill be in a state of stallc equilibrium. In
that case,rrj, is referred to as the ut-rest earth pressure,or
K: K,,:%
o',,
where K,, - at-rest earth pressure coefficient.
364
(r2.2)
12.1 At-Rest,Active, and Passive pressures
At-rest pressure
+l
'
I
:
t",,t;
-:.. :
I
'
A
C
'
I
c,,
H
A
l
Activepressure
Al, l+
I
r
K,,c',,=6',,
.,-{'Lo'ri're'
Pitssivepressurc
x l - , ' + o ' t a n0 '
(c,
Figure 12'1 Del\nitionof at-rcst.active,and passivepressures
(Note:WallAB is frictionless)
Case 2' If the frictionlesswall rotates sufficientlyabout its bottom to a position of
A'B (Figure l2.lb), then a triangular soil massABC' adjacentto the wall will reach
a state of plastic equilibrium and will fail sliding down the plane BC,. At this time,
the horizontal effective stress,oj,: o'u, will be ref'erred to as active pressare.Now.
K:Ku:4-4
a"
a"
(12.3)
where K,, : active earth pressure coefficient.
Case 3. If the frictionless wall rotates sufficiently about its bottom to a position
,4"8 (Figure 12.7c), then a triangular soil mass ABC" will reach a state of plastic
Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb
h
=
u
Passivepressure.oj,
At-rest pressure.oir
+
Wall tilt
Figure 12.2 Yariatictn ol thc magnitude of lateral earth pressure with wall tilt
Table 12.1 Typical Values of L.L,,lH and LL,,lH
Soil type
Loose sand
Dcnse sand
Soft clay
StilTclay
LL"IH
LLelH
0.(x)l 0.(D2
-0.00r
0.000-5
0.02
0.01
0.01
0.005
0.04
0.02
equilibriunt and will fail sliding upward along the plane BC". The horizontal effective stressat this time will be oi, : rr',, the so-calledpassivepressure.In this case,
K _ K ', :
oi.
a,u
, :
a,'
0,,
,
(r2.4)
where K,, : passiveearth pressure coefflcient
Figure 12.2showsthe nature of variation of lateral earth pressurewith the wall
t i l t . T y p i c a l v a l u e so f L , L , , l H( L L , , : A ' A i n F i g u r e 1 2 . 1 b )a n d L L p I H ( L L r : A " A i n
Figure 12.1c) for attaining the active and passivestatesin various soils are given in
Table 12.1.
AT.RESTLATERALEARTHPRESSURE
12.2
Earth Pressureat Rest
The fundamental concept of earth pressure at rest was discussedin the preceding
section. In order to define the earth pressure coefficient Kn at rest, we refer to Fig-
12.2 Earth Pressureat Rest
v
/r-
I
H
I!
367
t-
I
o'n= K,,Yz
It=t'+o'tan0'
Figure 12.3 Earth prcssLrrc
at rcst
urc 12.3,which showsa wall A Il rctaining a dry soil with a unit weight of 'l'hc wall
7.
i s s t a t i c .A t a d c p t h z ,
Vcrtical eflectivestrcss: o',,: yz,
HorizcrrrtalelTectivestress: oi, : K,,yz
So
K,, :
oi.
: at-rcst earth prcssurccoefficient
(f,,
For coarse-grainedsoils. the cocfficient of earth pressureat rest can bo estim a t e d b y u s i n gt h e c m p i r i c z rrl c l a t i o n s h i p( J a k y ,l 9 4 a )
K,:1-singl'
(1 2 . s )
where d' : drained friction angle.
While designinga wall that may be subjcctedto latcrerlearth pressureeltrcst.
one must take care in evaluatingthe value of K,,. Sherif, Fang,and Sherif (19g4),on
the basis of their laboratory tests,showed that Jaky's equation [<.trK,,
tEq. (12.-s)]
givesgood resultswhen the backlill is loose sand.However. for a densesand backfill.
Eq. (12.5) may grosslyunderestimatethe lateral earth prcssureat rest. This underestimation results becauseof the processof compaction of backlill. For this reason.
they recommended the designrelationship
K " : ( 1 * s i n@ ). l :
-
L /.I(mrnl
* r
ls.s
J
where y,1 : actual compacted dry unit weight of the sand behind the wall
: dry unit weight of the sand in the looseststate (Chapter
7,l1.rn;
2)
(12.6)
Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb
1
I
T l - . ' + o ' t a nQ
Unit weight
_
I
H
at-rcston a wall
Figure 12.4 Distributionol lateralcarthpressure
For finc-grained,normally consolidatcdsoils, Massarsch(1979) suggestedthe
following cquation for K,,:
I P r( % \ l
+ 0.421-ur]
K,,:0.44
(12.1)
For overconsolidatedclays,thc coeflicient clf carth pressureat rcst can be app r o x i m a t e da s
(r2.8)
: Krl,tnrtully.nn..lidor"d.;.\,m
Ko(overconsolidatcd)
'l'he
ovcrconsolidationratio was dellned in
whcre OCR : ovcrconsolidationratio.
Chapter 10as
OCR:
pressure,tri
Preconsolidatior-r
Presentelfective overburden pressure,oi,
( r 2.e)
Figure 12.4showsthe distribution of lateral earth pressureat rest on a wall of
height H retaining a dry soil having a unit weight of 7. The total force per unit length
of the wall, P,,.is equal to the area of the pressurediagram, so
(12.10)
P,,: lK,,yH2
12.3
Earth Pressureat Rest for Partially Submerged Soil
Figure 12.5ashows a wall of height FL The groundwater table is located at a depth
111below the ground surface, and there is no compensating water on the other side
of the wall. For z < Hr,the lateral earth pressureat rest can be given as oi,: K,,yz.
The variation of o'1,with depth is shown by triangle ACE in Figure 12.5a.However,
for z > H1 (i.e.,below the groundwatertable), the pressureon the wall is found from
the effective stressand pore water pressure components via the equation
effective vertical pressure : o',,: lHr + y'(z -
Ht)
Q2J'D
12.3 Earth Pressureat Restfor Partially Submerged Soil
I
H1
I
l
Saturatedunit weight
of soil = yr"r
'Ht)-l
( t
i
H1
I
I
I
l<--
where T' : 7,,,t
sure at rest is
Figure 12.5
Distribution of earth
pressureat rest for partially
submergedsoil
K,,(1lHt+y'H)+y,,,H2 +l
y?{r: the effectiveunit weight of soil. So the effectivelateral preso'1,: K,,o',,: K,,lyHt + y'(z -
Hr)l
(12.12)
The variation of cj, with depth is shown by CEG B in Figure 12.5a.Again the
lateral pressurefrom pore water is
u:
- H)
y,,,,(z
(12.13)
The variation of a with depth is shown in Figure 12.5b.
Hence, the total lateral pressurefrom earth and water at any depth z > l'1, is
equal to
o1:
Op -t U
: K,lyH, + y'(z - Hr)) t y,,,(z- H,)
(r2.14)
The force per unit length of the wall can be found from the sum of the areas of
the pressurediagramsin Figures 12.5aand 12.5band is equal to (Figure 12.5c)
(12.1s)
Area
ACE
Area
CEFB
Areas
EFG ancl IJK
370
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
Example12.1
Figure 12.6ashowsa l5-ft-high retainingwall. The wall is restrainedfrom yielding.Calculatethe lateralforce Puperunit lengthof the wall.Also,determinethe
location of the resultantforce,
u (lb/ft2)
o'/, (lb/ftl )
)und water table
Sand
c'- 0
Q'= 30'
Y,ur= 122.4lb/fir
Figure 12.6
Solution
K , , = I - S i n f ' : 1 - s i n 3 0: 0 . 5
AtZ:0:
a L = A ; c ' n = O ' ,u * - 0
At z : 10ft: ai = (10)(100)* 10001b/ft2
o'1,: Kor|* (0.5X1000): 500lb/ftz
u = O
At z : 15 ft: (r! * (10)(100)+ (5)(122.4* 62.4)= 13001b/ft2
o'1,: Koo'o: (0.5X1300): 6501b/ft2
u - (5)(y,,) = (5X62.4)= 312lblf*
The variations of oi and u with depth are shownin Figures 12.6band 1"2'6c.
:
+ Area$ + Area 4
*
Lateral force P^
" Area 1 Area 2
$
$
or
/t\
/r\
/t\
p,: (i
+ ( i )tsif:tzl
+ ( ; )(s)(1s0)
+ (sxs00)
)lroylsooy
\ L , /
\ L /
- 2500+ 2500+ 375+ ?80= 6155lblft
\ L /
12.4 Lateral Pressureon Retaining Walls from Surcharges-Based on Theory of Etasticity
371
The locationof the resultant,measuredfrom the bottomof the wall,is
z:
) momentof pressurediagramaboutC
or
/s\
/ \ \
l0\
(2soo)(s
T, ; t
+ txoor(|)+ (3?s)(,;,1
+ ( 7 8 0 ) ( ;I
\
J./
6155
12.4
= 4.7lft
r
Lateral Pressureon Retaining Wallsfrom
Surcharges-Basedon Theiry of Etasticity
Point-Load Surcharge
The equations for normal stressesinside a homogeneous,elastic,and isotropic medium produced from a point load on the surfacewerc given in chapter 9
[Eqs. (9. l0),
( 9 . 1l ) a n d 9 . 1 2 1 .
We now apply Eq. (9. 10) to determinc the lateral pressureon a retaining wall
causedby the concentratedpoint load Q placedat the surfaceof the backfill as shown
in Figure r2.1a.lf the load Q is placed on the plane of the sectionshown,we can substitute y : 0 in Eq. (9.10).Also, assumingthat ,p : 0.-5,we can writc
,
rrt-
where L : f7
o /3x,2\
2n\
/J /
(t2.16)
+ Z'z.Substitutin
g x : mH and z : nH inroEq. ( 12.16),we have
,
3Q
"
2rHt
tlr,:
m2n
^
-
(^t - ,,)"
(12.n)
The horizontal stressexpressedby Eq. (12.11)does not include the restrainingeffect
of the wall. This expressionwas invesrigatedby Gerber (1929) and Spangler( l93g)
with large-scaletests.on the basisof the experimentalfindings,Eq. (12.11)has been
modified as follows to agreewith the real conditions:
For m > 0.4,
o,n :
1.77Q
mznz
nz @f +Ef
(12.l 8)
For m * 0.4,
0.28Q
n2
,
oa:
H z ( 0 1 6+ , r f
(12.19)
Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb
q
, r = m HI
l*-*=,nn-l
l..<+l
Y
+
l+"
l
H
Y
l
:
ll r'
, f .
(b)
l.
rril-J+
t
nt:+l
t
l
.:
'
//
/ , .
loacl =
,, Strip
qlUrtI arcll
H
Figure 12.7
Latcral pressureon a retaining
wall due to a (a) point load,
( b ) l i n c l o a d ,a n d ( c ) s t r i p l o a d
Line-Load Surcharge
Figure 12.7bshowsthe distribution of lateral pressureagainstthe vertical back face
of the wall causedby a line-load surchargeplaced parallel to the crest.The modified
f o r m s o f t h e e q u a t i o n s[ s i m i l a rt o E q s . ( 1 2 . 1 8 )a n d ( 1 2 . 1 9 )f o r t h e c a s eo f p o i n t - l o a d
surcharge]for line-load surchargesare, respectively,
(r'h:
4q
rnzn
rrH (mz+ ,t)'
- ^ .{
m > u.41
llor
'
q
;'
(t2.20)
and
CL=
0.203q
n
H (oEl?r
where ? : load per unit length of the surcharge.
(for m o 0.4)
(r2.2r)
12.4 Lateral Pressure on Retaining Walls from Surcharges-Based on Theory of Elasticity
Strip-Load
373
Surcharge
Figure 12.7eshows a strip-load surcharge with an intensity of q per unit area located
at a distance rr , from a wall of height H. on the basis of the theory of elasticity, the
horizontal stressat a depth z on a retaining structure can be siven as
,',,:#@-sinBcosza)
(12.22)
The angles a and B are defined in Figure 72.1c.For actual soil behavior (from the
wall restraining effect). the precedingequation can be modified to
Zs
o'h: ;G
n
- sinp cos2a)
(12.23)
The nature of the distribution of oj, with depth is shown in Figure 12.lc.The force p
per unit length of the wall causedby the strip load akrne can be obtained by integration of oj, with limits of z from 0 to H.
Example12.2
Considerthe retainingwall shownin Figure12,8awhere H : l0 ft. A line load of
800 lb/ft is placedon the ground surfaceparallelto the crestat a distanceof 5 ft
from the backfaceof the wall.Determinethe increasein the lateralforceper unit
lengthof the wall causedby the line load.Use the modifiedequationgivenin Section12.4.
oi, (lb/lir)
,-Theorctical
shape
.{
".
800tb/ri
l+)u.+l
1l r T
I
Figure 12.8
I
I
2
l
t . ^ . .
' - - - ' - - - , OU.tr
' " I
t
: 4
c
a
O 6
t 4 | .\t.^)
H= l0ti
iI
:\6!16l
a
l
t ) \ 6 i' _ .
t "
I
5
,
r
,lr!,:
374
Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb
Solution
We are given l/ : 10 ft, q : 800lb/ft, and
*:4:0.5
1L)
> 0.4
So Eq. (12.20) will apply:
4q
.
on:
m2n
rrH Qrf;W
Now the following table can be prepared:
n = zn
0
0.2
0.4
0.6
0.8
1.0
4
q
m2n
1m2+ n2)2 o't llbllizl
,H
101.86
101.86
101.86
101.86
101.86
i01.86
0
0.595
0.595
0.403
0.252
0.16
0
60.61
60.61
41.05
25.67
16.3
Refer to the diagramin Figure 12.8b.
Area no,
rb/rt
(j)t'lr*'ut) = 6o'61
= r2r.2z
tbtrt
(|)rrlt*.ut + 60.61)
= 101.66
rb/rt
(])o,*o.ut + 41.0s)
= 66.72tbtrt
(i)trxrr.* + 2s,67)
= 41.e7tbtrl
(f;)elt t.u, + 16.3)
Toral= 392.18lb/ft
- 390lb/ft
RANKINE'SLATERALEARTHPRESSURE
12.5
Rankine's Theory of Active Pressure
The phrase plastic equilibrium in soil refers to the condition where every point in a
soil mass is on the verge of failure. Rankine ( 1857)investigated the stressconditions
in soil at a stateof plasticequilibrium. In this sectionand in section 12.6,wedeal with
Rankine's theory of earth pressure.
12.5 Rankine'sTheory of Active Pressure
+lAr!A ' A
T--*--
-2c',[9,
-
l-l
Unit weight of soil = T
lf=c'+o'tanQ'
T
e)I
I
a",
a
I
I
t
l-l
yzK,,- 2c'l K,,
(c)
'.h
(\
E
6
1
o '
*
q
\
\
\
r"o; \,'
6 ' , , N o r m a ls t r e s s
,,\, I
/,v./
(h)
Figure 12.9 Rankine's active carth pressure
Figure 12.9ashows a soil massthat is bounded by a frictionlesswall, AB, that
extends to an infinite depth. The vertical and horizontal effective principal stresses
on a soil element at a depth z are o'o and oi,, respectively.As we saw in Section 12.2,
if the wall ,48 is not allowed to move, then o'1,: K,,a',,.The stress condition in the
soil element can be represented by the Mohr's circle a in Figure 12.9b. However, if
the wall AB is allowed to move away from the soil mass gradually, the horizontal
principal stresswill decrease.Ultimately a statewill be reachedwhen the stresscondition in the soil element can be represented by the Mohr's circle b, the state of plastic equilibrium, and failure of the soil will occur. This situation represents Rankine's
active state, and the effective pressure oi on the vertical plane (which is a principal
plane) is Rankine's active earth pressure.We next derive oi in terms of y, z, c' , and $'
from Figure 12.8b
slnd
:
CD
AC
CD
AO+OC
376
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
But
CD : radius of the failure circle :
cu -
oo
AO : c' cot $'
and
oc:
o' I
o'..
2
"
So
a',, -
c'r,
2
s i nS ' :
*-
t -2,' ,a t t ' "
1v,co.(t, +"'
c'cos
<f' . q+4sinrp': ry
(r'
I - sind'
- 2 c ' cos
o',r: o',,,
* ,.r*
I *sinS'
(12.24)
But
U,,, -
vcrtical cf'fbctivcoverburden pressure
I - sin rf'
: t a n- '/( 4 5 - dt )' \
I *sin{'
and
c o sd '
l +,*d
- t a n/ ( 4 5- 6 ' \
, )
Substitutingthe precedingvaluesinto Eq. (12.24),we gel
*
cL: yztan'?(+s
+) zc'tun(+r 5)
t
(r2.2s)
The variation of o'owith depth is shown in Figure 12.9c.For cohesionlesssoils,
c':0and
^/
d'\
au-c',,tan'(4.5-t)
(12.26)
The ratio of oj, to oi, is called the coefficient ctf Rankine's active earth pressure
and is siven bv
12.6
Theory of Rankineb Passive Pressure
(ra
= t ? n ,' ([ 4, -5 K -o : - ,
oo
\
d'\
|
2/
377
(r2.27)
Again, from Figure 12.9b we can see that the failure planes in the soil make
-r (45 + g' l2)-degree
angleswith the direction of the major principal plane - that is,
the horizontal. These are called potential slip planes and are shown in Figure 12.9d.
It is important to realize that a similar equation for a" could be derived based
on the total stressshearstrengthparameters- that is, r, : c * o tan S. For this case,
oo: r2,""(* -
12.6
9)
* z,t^"(as
*
!)
(12.28)
Theory of Rankine3 Passive Pressure
Rankine'.spassivestatecan be explainedwith the aid of Figure 12.10.A B is a frictionl e s sw a l l t h a t e x t e n d st o a n i n f i n i t ed e p t h ( F i g u r e 1 2 . 1 0 a )T. h e i n i t i a l s t r e s sc o n d i t i o n
on a soil element is representedby the Mohr'.scircle a in Figure 12.l0b. If the wall is
gradualfyprzshedinto the soil mass,the cffectiveprincipal stressoi, will increase.Ultimately the wall will reach a situation where the stresscondition for the soil element
can be expressedby the Mohr\ circle b. At this time, failure of the soil will occur.
This situation is referred to as Rankine'spttssivestote.The lateral earth pressurerri,,
which is the maior principal stress,is called Rankine'spassiveearth pressure.From
Figure 12.10b,it can be shown that
o'p: o.'otunt(+s
.
*
+) ,r' ,^n(ot. +)
: wun?(+t.
+ zc'tan(t .
*)
+)
(12.2e)
The derivationis similarto that for Rankine'sactivestate.
Figure12.10c
showsthe variationof passive
pressure
with depth.For cohesionlesssoils(.' : 0),
d'\
"/
o'r: o',,tan'(4-5
*
t )
or
. +)
2 : *,: tan,(+s
(12.30)
Ko (the ratio of effectivestresses)in the precedingequationis referred to as the
coefficientof Rankine'spassiveearthpressure.
378
Chapter 12 LateralEarth Pressure: At-Rest,Rankine,and Coulomb
--+-lALl<A A '
1
1
I
1'
z
I
=-l=-2r'lE,
tzxr-------l
I
(c)
o',,
I
I
passiveearth pressure
Figure 12.70 Rankine'.s
'
The points D and D on the failure circle (see Figure 12.10b) correspondto the
+(45 - O'12)'
slip planes in the soil. For Rankine's passivestate, the slip planes make
degree angleswith the direction of the minor principal plane - that is, in the horizon'
tal direction. Figure 12.10dshows the distribution of slip planes in the soil mass.
12.7
Yielding of Wall of Limited Height
We learned in the preceding discussionthat sufficient movement of a frictionlesswall
extending to an infinite depth is necessaryto achieve a state of plastic equilibrium'
However, the distribution of lateral pressure against a wall of limited height is very
much influenced by the manner in which the wall actually yields. In most retaining
walls of limited height, movement may occur by simple translation or, more frequently, by rotation about the bottom.
r
12.7 Yielding of Wall of Limited Height
379
LLa
l**]-t,,+l
45
E
2
f'\
I
H
I
I
45-q
Figure 72.1? Rotationof frictionless
wall aboutthe bottom
For preliminary theoreticalanalysis,let us considera frictionlessretaining wall
representedby a plane AB as shown in Figure l2.l1a. If the wall ,4_Brotates sufficiently about its bottom to a position A' B, then a triangular soil mass,4BC, adjacent
to the wall will reach Rankine's activestate.Becausethe slip planesin Rankine'sactive statemake anglesof + (45 + O' 12)degreeswith the major principal plane,the soil
massin the stateof plasticequilibrium is bounded by the planeBC, ,which makes an
angle of (45 + 0'12) degreeswith the horizontal. The soil inside the zoneABC, undergoesthe same unit deformation in the horizontal direction everywhere,which is
equal to LL,,lLu. The lateral earth pressure on the wall at any depth z from the
ground surfacecan be calculatedby using Eq. (12.25).
In a similar manner! if the frictionless wall,4B (Figure lz.rlb) rotates suffic i e n t l y i n t o t h e s o i l m a s s t o a p o s i t i o n A " B , l h e n t h e t r i a n g u l a r m a s s o fs o i l A B C ,
will reach Rankine's passivestate. The slip plane BC" bounding the soil wedge that
is at a state of plastic equilibrium will make an angle of (45 - g,12) degreeswith the
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
horizontal.Every point of the soil in the triangular zoneABC" will undergo the same
unit deformation in the horizontal direction, which is equal to LLpl Lr,. The passive
pressureon the wall at any depth z can be evaluatedby using F'q. (12.29).
12,8
Diagrams for Lateral Earth Pressure
Distribution against Retaining Walls
Backfill-Cohesionless
Soil with Horizontal Ground Surtace
Active Case Figure 12.12ashows a retaining wall with cohensionlesssoil backfill
that has a horizontal ground surface.The unit wcight and the angle of friction of the
soil are y and rf' , respectively.
For Rankine'sactivestate,the earth pressureat any clepthagainstthe retaining
wallcan be given by Eq. (12.25):
(lVote:c' : 0.)
o',,: K,,yz,
H
P,,
+
U
3
I
l+
(,yH+l
Failurewedge
v
a
H
H
tI t_
ll r 3y
l +
l-<-- K,,yH
--->l
(b)
soil backfill
distributionagainsta retainingwall for cohensionless
Figure 12.12 Pressrtre
with horizontalgroundsurface:(a) Rankinebactivestate;(b) Rankine'spassivestate
12.8 Diagrams for Lateral Earth PressureDistribution against Retaining Walls
381
Note that ai increaseslinearly with depth, and at the bottom of the wall, it is
o',,: K,1H
(12.31)
The total force per unit length of the wall is equal to the area of the pressurediagram, so
P,,: !K,,yHz
02.32)
Passive Case The lateral pressuredistribution againsta retaining wall of height H
for Rankine'.spassivestate is shown in Figure 12.12b.The lateral earth pressureat
a n y d e p t h z [ E q . ( 1 2 . 3 0 ) c, ' : 0 ] i s
o',, - K,,IH
(12.33)
The total force per unit length of thc wall is
P , ,- ) K g H t
1tz.z+1
Ba ckf i I I - Parti a Ily Subm erged Cohen si o n I ess
Soil Supporting a Surcharge
Active Case Figure l2.l3a shows a frictionless retaining wall of height 11 and a
backfill of cohensionlesssoil. The groundwater tablc is located at a depth of H, below the ground surface.and the backfill is supporting a surchargepressureof q per
unit area. From Eq. (12.27),the effectiveactive earth prcssure at any depth can be
given by
{r'r,-- K,,rr',,
(12.35)
where rr',,and o',, - the effective vertical pressure and lateral pressure, respectively.
Atz-0.
o.-o',,:q
(12.36)
and
A t d e p t hz :
o',,: K.q
(t2.31)
ol,: (q + vH,)
( 12.38)
Ht,
and
c',,: K,,(e+ yHt)
02.39)
a|,: (Q * yH, * y'Hz)
02.40)
o',,: Ko(e + yHt * y'H)
(12.4f)
At depthz: H,
and
wherel' : y,nt- 7,.. The variationof oi with depth is shownin Figure 72.13b.
382
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
Surcharge= r7
d
'
I
4r+;
I
1
1
v
a
ll
IT
I qK,,l<--
ln
|4 l
1,[l
N
l[]n
l l + + l
K,,(q+yHt+"{'H))
Y".Hz
(b)
(c)
l<___>l++l
Krlg + yHt)
Ka!',H2+\",H2
(d)
activeearth pressuredistributionagainsta retainingwall with
Figure 12.13 Rankine'.s
soilbackfillsupportinga surcharge
cohcsionless
partiallysubmerged
The lateral pressure on the wall from the pore water between z : 0 and H1 is
0, and for z ) H1, it increaseslinearly with depth (Figure L2.13c).At z : H,
u:
Y',,'H2
The total lateral pressure diagram (Figure 12.13d)is the sum of the pressurediagrams shown in Figures 12.13band 12.13c.The total active force per unit length of
the wall is the area of the total pressure diagram. Thus,
+ tr(x,y' + y,,)H3
Pu: K,,qH+ lx,yul * K,,yH1H2
(2.42)
12.8 Diagrams for Lateral Earth PressureDistribution against Betaining wails
IT
H,
II I
t
I
I
H
I
H.
II
I
I
Hl
K , , t y H1+ q t
II
\
II
n
H
t.\
H2
l
|<-+|+--------------- l
qKn Kr(.yH1+y'H2)
(b)
.
\
l<------------+l l<------------+l<-_-_____+l
"1,Hz
K,,(q+ yH11 Kry'Llz+\*Hz
(c)
(d)
Figure 12.14 Rankine'spassiveearth pressuredistributionagainsta retainingwall with
partially submergedcohesionless
soil backfillsupportinga surcharge
Passive Case Figure 72j4a shows the same retaining wall as was shown in Figure r2.l3a. Rankine'.spassivepressure at any depth against the wall can be given by
Eq. (12.30):
o', : Kr{r'r,
Using the preceding equation, we can determine the variation of o! with depth,
as shown in Figure 12.14b.The variation of the pressure on the wall from water with
depth is shown in Figure 72.74c.Figure12.14dshowsthe distribution of the total pressure ao with depth. The total lateral passiveforce per unit length of the wall is the area
of the diagram given in Figure 10.11d,or
po: KoeH + lxoyHl * KorHtHz+
+(Kfl, + y_)HZ
O2.43)
Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb
4-s + I2
, ?'---..----.-
t
I
+
?.
H
H
l+l
K,t\H
(b)
l<---------------l
l€l
2r'{K,
K,,yH - 2c'{l{,,
(d)
(c)
Figure 72.15 Rankine'sactiveearth pressuredistributionagainsta retainingwall with cohesivesoil backfill
Backfill-
Cohesive Soil with Horizontal
Backfill
Active Case Figure 12.75ashows a frictionless retaining wall with a cohesive soil
backfill. The active pressure against the wall at any depth below the ground surface
can be expressedas [Eq. (12.25)l
oL:
KoYz- 2f K-c'
The variation of Koyz with depth is shown in Figure 12.15b,and the variation
is not a function
depth is shown in Figure 12.75c.Note that 2ffic'
ot2{K-c'with
12.8 Diagrams for Lateral Earth PressureDistribution against Retaining Walts
385
of "1,'hence,Figure 12.15cis a rectangle.The variation of the net value of oj, with
depth is plotted in Figure 12.15d.Also note that, becauseof the effect of cohlsion,
ai is negative in the upper part of the retaining wall. The depth 2,,at which the active pressurebecomesequal to 0 can be found from Eq. (12.2-5)as
K,,y2,,, 2fK,c'' - 0
(12.44)
v{K
For the undrained condition - that is, rl.r: 0, K,, : tan245 :
(undrained cohesion)- from Eq. (12.28),
a
LLrt
l, and c : c,,
(12.4s)
v
s o . w i t h t i m e , t e n s i l ec r a c k sa t t h c s o i l - w a l li n t e r f a c ew i l l d e v e l o pu p t o a d e p t h 2 , , .
Thc total activeforce pcr unit length of thc wall cernbe found from the area of
t h e t o t a l p r e s s u r ed i a g r a m( F i g u r e l 2 . l - 5 c l )o, r
P,, : \ K,,yl12 - 2\,fK,,c''H
(12.46)
Forthe@:0condition,
P,,-lyII2-2c,,H
(1) 41\
For calculation of the total activc force, common practice is to take the tensile
cracks into account. Becauseno contact existsbetween the soil and the wall up to a
depth of z.(,after the devclopment of tensile cracks,only the active pressuredistrih u t i o n a g a i n s i l h e w a l l b e l w e ezn- 2 l l ( y V K , , ) a n c l 1 / ( F i g u r cl 2 . l . 5 d ) i s c o n s i d e r c d .
In this case,
p " = l ( K , y -Hz f f i t , )'\( o _ _ ? + )
v\/K,,/
.,r2
: $K"yHz- 2t/I{,c'H + 2:*
v
:,
$
(12.48)
Forthe@:0condition.
-2
P,: jyHz - 2cuH + 2!]t
(r2.4e)
386
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
t \
i \
H
\
6r
\
\
+l
*l z,'r.@l* K,,yH
(b)
(a)
Figure 12.16 Rankine'spassiveearth pressuredistribution againsta retaining wall with
cohesivesoil backfill
Passive Case Figure I2.l6a shows the same retaining wall with backfill similar to
that consideredin Figure 12.I5a.Rankine'spassivepressureagainstthe wall at depth
z can be given by lE.q. (12.29)l
o',,: Kryz+2t/Krc'
Atz:0,
o'p:2lKpc'
andat z:
(12.s0)
H,
o'o: KrlH + 2f Krc'
(12.s1)
The variationof o',,withdepthis shownin Figure|2.t6b. The passiveforceper
unit length of the wall can be found from the areaof the pressurediagramsas
ru: iKotHz + 2{\c'H
For the d : 0 condilron, Kr:
$2.s2)
1 and
Pr:
lyHz * 2c,,H
(12.s3)
12.8 Diagrams for Lateral Earth PressureDistribution against Retaining Walls
387
Example12.3
An 6 m high retainingwall is shownin Figure 12.17a.Determine
a. The Rankineactiveforceper unit lengthof the wall and the locationof
the resultant
b. The Rankinepassive
forceper unit lengthof the wall and the rocationof
the resultant
T
I
Y =l 5 k N / m r
0 ' =3 o '
6m
c'=0
(a)
70.2kN/m
T
t
l<___+l
2 3 . 4k N / r n 2
l.+346.-5
kN/m2-->l
(b)
Figure 12.17
Diagramsfor
determiningactive,
and passiveforces
Solution
a. Because
c' : 0, to determinetheactiveforce.we canuseEq. (12.27):
o|:
Koa| = K,TZ
l-sind'_ 1-sin36
:o)(t
l*sinS'
l*sin36
A t 3 : 0 , 0 ' o = 0 ;a t e = 6 m ,
a| = (0.26)(15)(6)- 23.4kN/m2
r*
The pressuredistribution diagramis shownin Figure 12.ffib.The active
force per unit lengthof the wall is as follows:
P" = i(6){23.4) : 70.2kN/m
Also,
6m
^ -" _
t
:2m
1
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
h. To determinethe passiveforce,we are giventhat c' : 0. So,from
E q .( 1 2 . 3 0 ) .
rr'o: Kocr|= KrTz
/<,:1+##-i+##:385
Arz:o,c'r-0;atz:6m,
on : (3.85)(lsX6):346'5 kN/m2
The pressuredistributiondiagramis shownin Figure 12.17c.The passive
force per unit lengthof the wall is
Pp : i(6x-r46s) : 103e.5kN/m
Also.
z:ry:2m
Exa mp l e1 2 .4
For the retainingwall shownin Figure 12.18a,determinethe force per unit width
of the wall for Rankine'sactivestate.Also lind the locationof the resultant.
T
1.j,,,^-^..-,,
urounq-
Jnt i""
L:'!'
.:
+
watertable
r
"T
3
rn ffl
;.ifri.;
il:r::ri
yI ;i;i!il
:
: .
y - 16kN/m3' ll.:'
"
Q = rtr
c ="
:
I
I
*
+
T,or
'-i,,=l8kN/nrr
;;.
0'= 3.5'
r.'= 0
t
3rn
I
i
3m
+
+l rol<--->l 19.67l<(b)
l+29.+3+l
+l
1 3 . 10 < -
3 6 . 1+ l
(d)
Figure 1Z 18 Retaining wall and pressurediagramsfor determining Rankine'sactive
earth pressure.(Note:Theunits of pressurein (b), (c), and (d) are kNlm')
12.8 Diagrams for Lateral Earth PressureDistribution against Retaining Walls
389
Solution
Giventhat c' : 0, we known that oL = Koa,,,.
For the upperlayerof the soil,Rankine'sactiveearth pressurecoefficientis
K q, , : Ku-\ ", , - l - s i n 3 o ' - l
| *sin30.
3
For the lower layer,
: o.2jr
Ko: Ko(z\:1 ;'l'11,
I -l_ sln -t-)"
At z = 0, oL: 0. At z : 3 m (just insidethe bottom of the upperlayer),aj, :
3 x 16 : 48 kN/m2.So
o',:
Kr,1yr',,:{ x 48 : 16kN/m2
Again,at z - 3m (in the lowerlayer),a',,: 3 x 16 : 48 kN/mz,and
ou: Kuplo',,:(0.271)X (48) : 13.0kN/m2
Ate=6m.
o", = 3 x 16 + 3(18* 9.81)= 72.-57kN/m2
t
7,,,
and
= (0.271)x (12.57)= 19.67kN/m2
o',,= Ku121o',,
The variation of oj, with depth is shownin Figure 12.lBb.
The lateralpressuresdue to the pore waterare asfollows:
At7=g'
u-0
Ate=3m: u=0
A t z : 6 m : u = 3 x T u , : 3 X 9 . 8 1= 2 9 . 4 3 k N / m 2
The variation of u with depth is shownin Figure r2.1&c,and that for ou (total active pressure)is shownin Figure 12.18d.Thus,
p ": (i x3 x1 6 +
:24+ 3e.0 + s4i1s:117.15kN/m
) 3 (1 3 .0+)(+X3X36.1)
The locationof the resultantcan be found by taking the &romentaboutthe
:
bottom of the wall:
+J). ,' o(i). 'z+(z
"(i)
117.15
: 1.78m
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
Example12.5
A frictionlessretaining wall is shownin Figure l2.Iga.Determine the activeforce,
Po,afterthe tensilecrackoccurs.
q = 15kN/m:
I
r
+
I
6 m
I
t
+
l
t
l
*
-6.64 kN/m2
l-l
T= 16.5kN/rn:
Q'=26"
c'= l0 kN/m2
I
I
--+l
l*lt.qz tN/rnr
(r)
rD,
diagram
distribution
wall;(b)activepressure
retaining
Figure12.19(a)Frictionless
Solution
Giventhat$' - 26o,wehave
I * s i n f ' : 1 - s i n 2 6:
u'3e
K' =
1 + r,"26
lTlin7
FromEq. \12.25).
oL: Koo'o'2c'{K
Atz:0,
oL: (0.39X1s)- (2X10)\639 :
-6.64kN/m2
Atz:6m,
- (2x10)1rc:39
: 31.e7
kNlmz
+ (6X16.5)l
oi = (0.3e)lr5
ij
this
diagram",{
From
r$lu'
distributio"otu**1';;*T;rttuure
Thepressure
j
* - =
z
o
6*z
r
I * 1.03m
I
'.$
.:
;
'i$
l
After the tensilecrackoccurs,
:79.215
: |1+.st112t.9't)
p": l(6 - z)(3r.97)
kN/m
:
rf
I
12.8 Diagrams for Lateral Earth pressure Distribution against Retaining
watts
Example12.6
A frictionlesrr:Tj.ning.Iall-isshownin Figure 12.20a.
Find the passiveresistance
(Po)on the backfilland the locationof the resultantpassive
force.
q = l0 kN/m2
I
I
I
1m
I
llJl
Y = l5 kN/mj
$'=26"
r" = 8 kN/ml
l++l+
1 . 5 3 . k6N / r n 2+ l
- 51 . 2k N / n r 2
(b)
Flgure 12.20 (a) Frictionlessretainingwall; (b) passivepressuredistribution
diagram
Solution
Giventhat$' : 26o,it followsthat
I + s i n d '_ 1 * s i n 2 6 _
"
*- _
) <A
" 1-sind' l-sin26"
FromEq. (12.29),
u'o: Ko{ro+ 2f4c'
=
At e 0, aL : 10kN/m2;thus,
o, = (2.56)(rA)+ 2\/736(S) : 25.e+ 25.6* 5t.2kN/m2
Again,at z = 4 m,c'o: (10+ 4 x15) = 70kNlm2.So
o', = (2.56)(70)
+ 2\A36(8) : 20+.S
kN/m2
Thepressure
distributiondiagramis shownin Figure1z.z0b.The
passive
resistanceperunit widthof thewallis asfollows:
i
$
Po= gLZ)(4) + t(4x1s3.6): 204.8+ 307.2=ku kN/m
locationof theresultantcar befoundby takingthemomentof thepressure1l*
dragramaboutthebottomof thewall.Thus.
(5r
,(:) * jr'rrurror(f
_
)
t : F : j m
392
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulamb
12,9
Rankine Active and Passive Pressure
with Sloping Backfill
In Sections12.-5through 12.8,we consideredretaining walls with vertical backs and
horizontal backfills. In some cases,however, the backfill may be continuously sloping at an angle a with the horizontal as shown in Figure 12.21for activepressurecase.
l n s u c h c i r s c s t. h e d i r e c t i o n o f R a n k i n e ' sa c t i v eo r p a s s i v ep r e s s u r e sa r e n o l o n g e r
horizontal. Rather, they are inclined at an angle a with the horizontal. If the backfill
',
: 0, then
is a granular soil with a drained friction angle <f and c'
o'rr: YzKu
where
K, : Rankine'sactivepressurecoefficient
= cosa
cosa-
Vcos2a-cos26'
cosd+ \,6P;:
(12.s4)
"*?
The activeforce per unit length of t h c w a l l c a n b c g i v e n a s
l
^
: :2 K,,vH'
P,,
'
(12.ss)
The finc of action of thc resultantactsat a distanceof H 13measuredfrom the bottom
''
of the wall. Table 12.2givesthe values of K,, for various combinations of a and f
In a similar manner, the Rankine passive earth pressure for a wall of height H
with a granular sloping backfill can be representedby the equation
P, : ltH2 K,,
(12.s6)
Frictionless
Figure 12.21
Frictionlessverticalretainingwall with
slopingbackfill
12.9 Ranking Active and Passive Pressure with Sloping Backfilt
Table 12.2 Valuesof K,, [Eq. (12.54)l
Q' (des) --+
I a (degl
U
.5
10
15
20
25
0.361
0.366
0.380
0.409
0.461
0.573
0.333
0.337
0.350
0.373
0.4t4
0.494
0.307
0.311
0.32r
0.34r
0.374
0.434
0.283
0.286
0.294
0 . 3 1I
0.338
0.385
0.260
0.262
0.270
0.283
0.306
0.343
0.238
0.240
0.246
0.258
0.277
0.307
0.217
0.219
0.225
0.235
0.250
0.275
4.204
4.136
3.937
3.61-5
3.lrig
2.676
4.599
Table 12.3 PassiveEarth PressureCoefficient,f,, IEq. (12.57)l
Q' (des) -->
J a (deg)
0
-5
10
1,5
20
25
2.770
2.715
2.551
2.281
t . 9 l1 3
1.434
3.000
2.943
2.775
2.502
2.132
1.664
3.2-55
3 . 19 6
3.022
2.740
l. _11)I
1.1394
3.537
3.476
3.295
3.003
2.612
2.135
3.u52
3.7uu
3.5gr3
3.293
2.8rJ6
2.394
A <)1
4.316
3.977
3.526
2.987
where
Kr, : cos a
cosrr + V!.rt'"
..f
d
cos., - \4;t:*?
(12.s7)
is the passiveearth pressurecoefficient.
As in the casc o[ the active forcc, the resultant force P,, is inclined at an anslc
'H13
a with the horizontal and intersectsthe wall at a distanceof
measuredfrom the
bottom of the wall. The valuesof K, (passiveearth pressurecoefficient)fbr various
valuesof a and (t' are given in Table 12.3.
c'-S'Soil
The preceding analysiscan be extended to the determination of the active and passive Rankine earth pressurefor an inclined backfill with a c'-rf' soil. The details of
the mathematicalderivation are given by Mazindrani and Ganjali (rggi).For a c'-$'
backfill, the active pressure is given by
o'" : yzK,, : yz.Ki cosa
:
whereKu Rankineactiveearth pressurecoefflcientand
K;:
K..
*r"
The passivepressureis givenby
o',, : TzKo : yzK", cosa
whereKo : Rankinepassiveearth pressurecoefficientand
A'
::!
K'i,:
'
cos.Y
( 12.58)
(r2.se)
(r2.60)
(12.61)
394
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
Also,
K " " ,K '' : , :
1
\
cos-@
c o s 2 a+ 2 ( { ) . o r 4 '
\yz /
sin rf '
t
'
o ( ) ' . o r ' a ' , B ( - ) . o r ' a s i n@ .' o , 4 ' l
\yz/
. i)
\yz/
"
'
(12.62)
{.
Table 12.4 Variationof K'1,and K'i,*
c'
yz
a
(deg)
1.0
0.025
0.05
0.-stt8t3
1.6984
0.6069
t.6477
0.673rJ
r.4841
1.0000
1.0000
0.-5504
1;7637
0.-5658
I . 71 5 6
0.6206
1.-5641
0.1762
1.2506
0 . 51 2 1
1.828'l
0.5252
r.7830
0.570'7
1.640u
0.6u34
t.3702
0.4353
1.9590
0.4449
L9169
0.4'769
1.7882
0.5464
1.5608
-0.1785
3.0016
-0.1804
2.9709
-0.1861
2.8799
(\.1962
2.732t
-0.9459
4.3048
-0.9-518
4.2782
-0.9696
4.1993
- 1.0000
4.0718
0.4903
2.0396
0.5015
1.9940
0.5394
L8539
0.6241
1.6024
0.45-53
2 . 1 lI 0
0.4650
2.0669
0.4974
1.9323
0.5666
1.6962
0.4203
2.1824
0.4287
2.1396
0.4564
2.0097
0.-5137
1.7856
0.3502
2.3252
0.3,565
2.2846
0.3'767
2.1622
0.4165
1.95-56
-0.2099
3.4678
-0.2t19
3.4353
-0.2180
3.3392
-0.228t
3.1831
-0.9101
4.8959
-0.9155
4.8669
-0.9320
4;7812
-0.9599
4.6422
0.4059
2.4639
0.4133
2.4r95
0.4376
2.2854
0.4860
2.0575
0.3140
2.5424
0.3805
2.4989
0.4015
2.3680
0.4428
2.1474
0.3422
2.6209
0.3478
2.5782
0.3660
2.4502
0.4011
2.235'7
0.2'784
2.7779
0.2826
2.1367
0.2960
2.6]35
0.3211
2.4090
-0.2312
4.0336
-0.2332
3.9986
-0.2394
3.8950
-0.2503
3.7264
-0.8683
s.6033
-0.8733
5.5713
-0.8884
5.4765
-0.9140
5.3228
K';IK;
u <b': 15"
0
0
-5
-5
l0
I0
1-5
15
b
,b':20"
0
o
-5
5
It)
10
1-s
15
K'i,
K';,
K":,
K';,
K""
K';,
c 6' :25"
U
0
5
5
l0
10
t5
15
(continued)
12.9 Ranking Active and Passivepressure with Stoping Backfitl
Table 12.4 gives the variation of K'/ and Ki with at,c,llz, and 0, .
For the active case,the depth of the tensile crack can be given as
2c'
(r2.63)
I
Table 12.4 (continued\
c'
yz
a
(des)
K:tK;
0.0
0.05
0.1
0.2756
-0.2440
4.7321
*0.2460
4.6935
-0.2522
4.5794
-0.2628
4.3936
-0.8214
6.4641
-0.tt260
6.4282
,0.8399
6.3218
-0.8635
6.1489
*0.7'/01
'7.5321
1.0
d $' :30"
0
0
5
-5
l0
l0
15
1-5
U
0
5
5
10
10
15
15
0
0
5
5
10
10
15
15
0.3045
3.0866
0.3090
3.0416
0.3233
2.9070
0.3502
2.6836
0.2795
3.1288
0.2919
2.9961
0.3150
2.7766
0.2179
3.3464
0.2207
3.3030
0.2297
3.1737
0.2462
2.9608
0.2710
3.6902
0.2746
3.6473
0.2861
3.4953
0.3073
3.2546
0.2450
3.7862
0.2481
3.1378
0.2581
3.5933
0.2764
3.3555
0.2r t39
3.8823
0.2217
3.8342
0.2303
3.6912
0.2459
3.4559
0 . 16 6 9
4.0744
0.1688
4.0271
0.1749
3.8866
0.1860
3.6559
-0.2496
5.6172
-0.2-515
5.-s678
-0.2575
s.4393
-0.2678
5.2300
-0.7872
7.3694
-0.8089
7.1715
0.2174
4.-5989
0.2200
4.5445
0.2282
4.3826
0.2429
4.1168
0 . 19 4 1
4.7061
0.1964
4.6521
0.2034
4.4913
0.216I
4.2275
0.1708
4.8134
0.1727
4.7597
0.r787
4.5999
0.1895
4.3380
0.1242
5.0278
0.1255
4.9747
0.1296
4.8168
0.r370
4.5584
-0.2489
6.7434
-0.2507
6.693-5
-0.2564
6.5454
-0.2662
6.304\
-0.7152
8.8879
-0.7190
8.8400
-0.7308
8.6980
-0.7507
8.4669
J.t t3z
Q' :35'
e
f
0.3333
3.0000
0.3385
2.9543
0.3549
2.8176
0.386r
2.5900
-0.7744
'7.4911
Q' :40'
Ki,
K';
K';
K';
Ki
K';
K';
K';,
* After Mazindrani and Ganjali (1997)
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
Example12.7
on page392.Giventhat H = 6.1m, a - 5o,f = 16'5kNlm',
Refer to Figure 12.21:
10
determinethe Rankine activeforce Pnon the retaining
c'
kN/m2,
20o,
6'
crack
occurs.
tensile
wall after the
Solution
From Eq. (12.63),the depth of tensilecrackis
2c'
1 * sind'
1* sinf'
(2X10)/T + stn20
1 6 . -Vs l - s i n 2 0
So
Atz:0:
o',:0
Atz:6.1 m: o'u: yzK'lcosa
l0
c'
,," : (ra5x6t,: o'l
From Table 12.4,forat = 5" andc'lyz: 0.1,the magnitudeof Kii : 0.3565.So
5") - 35.t5kN/m2
oi, : (16.5)(6.1
)(0.3565)(cos
Hence,
r":
)rn
- r.t3)(3s.i5)
- 2,,)(35.75)
- 78.tkN/m
=
lrc.t
THEORY
COULOMB'SEARTHPRESSURE
Morc than 200 ycars ago, Coulomb (1116) presenteda theory for active and passive
earth pressurcsagainst retaining wetlls.In this thcory, Coulomb assumedthat the
failurc surfacc is a planc. The wull .f'rictionwas tzrkeninto consideration.The following sectionsdiscussthe generalprinciplesof the derivation of Coulomb'searth pressure theory for a cohesionlessbackfill (shear strength defined by the equation 11=
c' tan $').
12.10
Coulombb Active Pressure
Let AB (Figure 12.22a)be the back face of a retaining wall supporting a granular
soil, the surfaceof which is constantlysloping at an angle a with the horizontal. BC
is a trial failure surface.In the stability considerationof the probable failure wedge
ABC, the following forces are involved (per unit length of the wall):
12.10 Coulomb'sActive Pressure
iI
90-0+cr
9 0 + e + 6B + 0
W
397
4,,-,6
II
+
t1
\ r
-Y
'
r90+e-
D
l
p-0
I
\
\(a)
(b)
Figure 12.22 C.<tulomb'.s
activcprcssurc:
(a) trial lailurewcdgc;(b) lirrcepolygon
l. tr44
the weight of the soil wedge.
2. F, Lheresultant of the shcar and normal forces on the surfaceof failure. BC.
T h i s i s i n c l i n e da t a n a n g l c o f { ' t o t h e n o r m a l d r a w n t o t h c p l a n e B C .
3. P,,,the active force per unit length o1'thewall. The clirectiono1 P,,is inclined at
a n a n g l e6 t o t h e n o r m a l d r a w n t o t h e l ' a c eo f t h e w a l l t h a t s u p p o r t st h e s o i l .6 i s
t h c a n g l eo f f r i c t i o n b e t w e c nt h e s o i l a n d t h c w a l l .
The force triangle lor the wedgc is shown in Figure l2.z2b. From the law of
sincs.we havc
s i n ( 9 0 + 0 + 6 -F + 4 t ' )
P,,:
J,,
sin(B
4t')
sin(B- rf')
W
s i n ( 9 0 + 9 + 6B
- +O')
(12.64)
(r2.6s)
The pre.cedingcquation cernbe written in the form
cos(0- B)cos(0* a)sin(Be"- )tn.lcos2Flsin(B
* a)sin(90+ g + D -
r/')
B + O'))
rt oot
where 7 : unit weight of the backfill.The valuesof y, H, 0, a,6, ,and 6 are constants.
and B is the only variable. To determine the critical value of 6 for maximum p.,. we
have
dP." : 0
dp
(r2.61)
398
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
Table 12.5 Valuesof K, [Eq. (12.69)]for 0 : 0o,a : 0o
6 6eg) --+
10
.l-d' (deg)
28
30
32
34
36
38
40
42
0.3448
0.3189
0.2945
0.21t4
0.249'7
0.2292
0.2089
0.1916
0.3610
t r.-r-t-tJ
0.3073
0.2827
0.2596
0.2379
0.2t74
0.1982
0.3330
0.3085
0.2853
0.2633
0.2426
0.2230
0.2045
0.1870
15
0.3257
0.3014
0.2791
0.2579
0.2379
0.2190
0.2011
0.1841
0.3203
0.2973
0.2755
0.2549
0.2354
0.2169
0.1994
0.1828
0.3186
0.2956
0.2745
0.2542
0.2350
0.2167
0.1995
0.1831
After solvingEq. (12.67),when the relationship of B is substituted into Eq.
(12.66),we obtain Coulomb's active earth pressureas
Po: lKoyH2
( 12.68)
whereK,,is Coulomb'sactivecarth pressurecoefficientand is givenby
6,:
, " o t ' ( d ' , l f ) ,, , , , , ' , - . - u
ol)sin(q' t)
gcos(6
+ 0)l
cos2
' L I * .,/sin(9-+
l'
c o s ( 6 + d ) c o s ( O- d ) J
02.69)
\
Note that when d - 0', g - 0", and 6 : 0", Coulomb's activeearth pressurecoeflicient becomesequal to (1 - sin rb')l(1 + sin {'), which is the same as Rankine's
earth pressurecoefficientgiven earlier in this chapter.
The variation of the valuesof K,, for retaining walls with a vertical back (0 0')
anclhorizontal backfill (a : 0") is given in Table 12.5.From this table, note that for
a given value of @', the effect of wall friction is to reduce somewhat the active earth
pressure coefficient.
12.11
GraphicSolution for Coulomb'sActive Earth Pressure
An expedient method for creating a graphic solution of Coulomb's earth pressure
theory was given by Culmann (1875). Culmann'ssolution canbe used for any wall friction, regardlessof irregularity of backfill and surcharges.Hence, it provides a powerful technique for estimating lateral earth pressure. The steps in Culmann's solution
of active pressure with granular backfill (c' : 0) are described next, with reference
to Figure I2.23a:
1. Draw the features of the retaining wall and the backfill to a convenient scale.
2. Determine the value of ry'(degrees) : 90 0 6, where 0 : the inclination of
the back face of the retaining wall with the vertical, and 6 : angle of wall friction.
12.11 Graphic Solution for Coulombb Active Earth Pressure
(b)
Figure 12.23 Culmann's solution for active carth rrressure
3. Draw a line BD that makes an angle d' with the horizontal.
4. Draw a line BE that makes an angle r/ with line B D.
To consider some trial failure wedges,draw lines BCt, BC2, BC., . . ., 8C,,.
6. F i n d t h e a r e a so f A B C t , A B C 2 , A B C j , . . . , A B C , , .
7. Determine the weight of soil, w, per unit length of the retaining wall in each of
the trial failure wedgesas follows:
W1: (Area of ABC,) x (7) x (1)
W2: (Areaof ABC) x (7) x (1)
: (Rr"u of ABC.) x (7) x (1)
%
W,: (Area of ABC,) x (7) x (1)
8 . A d o p t a c o n v e n i e n tl o a d s c a l ea n d p l o t t h e w e i g h t sW , , W r , W 3 , . , W n d e t e r mined from step 7 on line BD. (Note: Bc1 : W1, Bcz: W2, Bc1 : Wt, . . .,
Bc,,: Wo)
9. Draw cp\, c2c'2,
c{\, . . ., c,,c'nparallel
to the line BE. (Note: c\, c2,cj, . . . , c,,,
are located on lines BCt, BC2, BCr, . . ., BC,, respectively.)
10. Draw a smooth curve through points c1,c2,c\, . . ., c',. This curve is calleclthe
Culmann line.
11. Draw a tangent B'D' to the smooth curve drawn in step I0. B,D,is parallel to
line BD. Let c'"bethe point of tangency.
12. Draw a line coc'oparallel
to the line BE.
13. Determinethe activeforce per unit lengthof wall as
P, : (Lengthof coc) x (Load scale)
14. Draw aline Bc'oC,,.ABC, is the desired failure wedge.
Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb
Note that the constructionprocedure entails,in essence,drawing a number of
force polygons for a number of trial wedges and flnding the maximum value of the
active force that the wall can be subjected to. For example, Figure 12.23bshows the
force polygon for the failure wedge,4BC, (similar to that in Figure 12.22b),in which
l4z:
P,, :
F B -
weight of the failure wedge of soil .48C,,
active force on the wall
the resultant of the shear and normal forces acting along BC,,
LC,,BF (the angle that the failure wedge makes with the horizontal)
Thc force triangle (Figure 12.23b)is simply rotated in Figure I2.23a and is rep, cp\, '..,
r e s e n t e db y t h e t r i a n g l e B c , , c i ,S. i m i l a r l y ,t h e f o r c e t r i a n g l e sB c l c !, B c 2 c 2B
B c , , c ' ,c, o r r e s p o n dt o t h e t r i a l w e d g c sA B C t , A B C 2 , A B C ; , . . . , A B C , , .
The prcccding graphic pKtcedurc is given in a step-by-stepmanner only to facilitatc basic understanding.These problems can be easily and cffectivelysolvedby
thc use ol computcr programs.
Thc Culmann solution provides us with only the magnitude of the activeforce
per unit length of the retaining wall - not with the point of application of the resultant. Thc analytic procedure usedto find the point of applicationof the resultantcan
bc teclious.For l.hisreason.an apprcximatc method, which does not sacrif,cemuch
accuracy,can be used.This n-rethodis demonstratedin Figure 12.24,in whichABC
is the failurc wedge determined by Culmzrnn\ method. O is the center of gravity of
the wedgeABC. lf trline OO' is drawn parallel to the surfaceof sliding,BC, the point
of intersectionof this line with the back face of the wall will give the point of application of P,,.Thus, P,,actsat O' inclined at angle6 with the normal drawn to the back
face o[ thc wall.
a
()
----/o'
p
B
Figure 12.24 Approximate method for finding the point of application of the resultant
active fbrce
12.11 Graphic Solution for Coulomb's Active Earth Pressure
401
Example12.8
A 15-ft-high retaining wall with a granular soil backfill is shown in Figure 12.25.
Giventhat y : 100lblft3,6' : 35',0 : 5o,and6 = 10'.determinethe activethrust
per foot lengtho[ the wall.
Solution
For this problem,,lt : 90 * 6 - 6 : 90o* 5" * 10" : 75'. The graphicconstruction is shownin Figure12.25.The weightsof the wedgesconsideredare asfollows:
Wedge
Weight (lbl
ABCl
ABC2
ABC3
ABC4
ABCs
= 3,916
'(4.38X17.S8)(100)
- 6,106
+ t; (2.36X18.56)l(100)
3,e16
: 8,2e5
+ l+(2.24)(1e.s4)l(100)
6,106
: 10,486
+ ti Q.rr)(20.77)l(100)
8,29s
= 12,67
+ [, (1.e7)(22,2U](100)
10,486
5
-l2.sri
2.s ft | 2..5li I
I.el<------>l
tcl+
cr
tcs
17.-5
II
v=90-0-6=7s"
I
Weight ( 1000lb)
|-_-_1
0=5"
2 3 4
Length(fi)
5
Fl--_-
1 2 3 4 5
Figure 12.25 Culmannssolutionfor determiningactivethrust per unit lengthof wall
Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb
In Figure12'25
rq : z,gtota
Bc'z: 6'106lb
tU
aq : S,Z9S
Bq : 10,4861b
nct : lZ,OlSrc
The activethrust per unit lengthof the wall is 4,090lb.
12,12
Active Force on Retaining Walls
with Earthquake Forces
Coulomb's analysisfor activeforce on retaining walls discusscdin Section 12.I 0 can
be convenicntly extendcd to include earthquarkcforces.To do so, let us considera
retaining wall of height H with a sloping grttnulur bucklillas shown in Figure 12.26a.
Lct the unit weight and thc friction trngleof the granular soil retained by the wall be
equaf to y and$' , respectivcly.Also, let 6 be the anglc of friction betweenthe soil and
the wirll. AtlC is a trial ferilurewedge.The forces acting on the wcdge are as follows:
1 . W e i g h t o f t h e s o i l i n t h e w e d g c ,W
2. Resultant of thc shear and normal forceson the failurc surfaceBC, F
v
q
6
(a)
Figure
12.26
Active force on a retaining wall with earthquake
(b)
forces
403
12.12 Active Force on Retaining Walls with EarthquakeForces
3. Active force per unit length of the wall, P,,,.
4. Horizontal inertial force, k,,W
Vertical inertial force, k,,W
Note that
Horizontal component of earthquake acceleration
(12.70)
g
t . -
Vertical component of earthquake acceleration
(t2.11)
where g : accelerationdue to gravity.
The force polygon demonstrating these forces is shown in Figure 12.26b.The
dynamic activeforce on the wall is the maximum value of P,,,,exerted by any wedge.
This value can be exprcssedas
p,": LyHr(I * k,,)K1,
(t2.12)
where
- a *
@')sin(@'
c o s 2 g c o s B c o s+
( d0 +
0 + p)cos(0*
p) 'l'oI t
ot)
) I
(t2.13)
and
F:,""-'(#?)
(12.74)
Note that with no inertia forccs from earthquakes.B is equal to 0. Hence, K',,: K,,
as given in Eq. (12.69).Equations (12.72) and (12.73)are generallyreferred to as the
Mononobe-Okabe equation.s(Mononobe. 1929,Okabe, 1926).The variation of Kj,
with 0 - 0' and k,, : 0 is given in Table 12.6.
Consideringthe activeforce relation given by Eqs. (12.72)through (12.74),we
find that the term sin (@' - q - B) in Eq. (12.13) has two important implications.
First, if rp' B < 0 (i.e.,negative),no real solution of Ki is possible.Physically,
"
this implies that an equilibrium condition will nr,tterisl. Hence, for stability, the limiting slope of the backfill may be given as
cr=o'-E
(t2.ts)
Table 12.6 Valuesof rKi,[Eq. (12.73)lwith 0 : 0" and k, : g
6'@eg)
kh
6 {deg}
a (deg)
28
30
35
40
45
0.1
0.2
0.3
0.4
0.5
0.427
0.-508
0.611
0.753
1.00-5
0.39"7
0.473
0..569
0.697
0.890
0.328
0.396
0.478
0.581
0.'716
0.268
0.382
0.400
0.488
0.-596
0.2t7
0.2'70
0.334
0.409
0.500
0.1
0.2
0.3
0.4
0.5
0.451
0.-554
0.69i)
0.942
0.423
0 . 5l 4
0.635
0.825
0.341
0.421
0.522
0.653
0.8-55
0.282
0.349
0.431
0.-s35
0.673
0.227
0.285
0.356
0.442
0.551
0.1
0.2
0.3
0.4
0.5
0.497
0.623
0.n56
0.457
0.570
0.74tt
0.371
0.461
0.58-s
0.7u0
0.21)r)
0.37,5
0.4'72
0.604
0.1309
0.238
0.303
0.383
0.4u6
0.624
0.1
0.2
0.3
0.4
0.5
,bl2
0.396
0.4u5
0.604
0.778
l.ll5
0.368
0.452
0.563
0.7Itr
0.912
0.306
0.380
0.474
0.-599
0.771
0.2.53
0.3I9
0.402
0.508
0.64u
0.207
0.267
0.340
0.433
0.-552
0.1
0.2
0.3
0.4
0.-5
,bl2
0.42tt
0.537
0.699
t.025
0.396
0.491
0.640
0.881
0.326
0.4t2
0.526
0.690
0.962
0.26t3
0.342
0.438
0.56u
0.752
0.2Itt
0.2n3
0.361
('t.4'75
0.620
0.1
0.2
0.3
0.4
0.-5
tf l2
0.472
0.66
I
0.90u
0.433
0.562
0.7130
0.352
0.454
0.602
0.u-57
0.285
0.371
0.4tt7
0.6.56
0.944
0.230
0.303
0.400
0.531
0.722
0.393
0.366
0.306
0.256
0.212
0.486
0.612
0.t301
t.177
0.454
0.572
0.740
1.023
0.3u4
0.4u6
0.622
0.819
o.326
0.416
0.-s33
0.693
0.276
0.357
0.462
0.600
0.427
0.39-5
0.327
0.27|
0.224
0.541
0.714
1.073
0.-501
0.6-55
0.921
0.4u
1
0.-541
0.722
1.034
0.35t)
0.455
0.600
0.812
0.294
0.386
0.509
0.679
0.472
0.434
0.3-54
0.290
0.237
0.625
0.942
0.570
0.rJ07
0.463
0.624
0.909
0.381
0.509
0.699
t.037
0.317
0.423
0.573
0.800
0.1
l0
2 ,
;-')E
0.2
0.3
0.4
0.-5
0.1
2 ,
;-)E
0.2
0.3
0.4
0.5
0.1
0.2
0.3
0.4
0.5
2 ,
;-)E
t0
12.12 Active Force on Retaining Walls with Earthquake Forces
405
For no earthquake condition. F : 0'; for stability, Eq. (12.75) gives the familiar
relation
a=0'
(12.76)
Second,for horizontalbackfill,a : 0', for stability,
B=o'
(12.77)
Because F : tan-t[kt,l(I - k,)), for stability, combining Eqs. (12.74) and (12.77) results in
k,, = (I
k,,)tanS'
(12.18a)
Hence.the criticalvalueof the horizontalacceleration
canbe definedas
knk : (l - k,)tan S'
(12.18b)
where kr,(.,): critical of horizontal acceleration(Figure 12.21).
Location of Line of Action of Resultant Force, Pu,
Seedand Whitman (1970)proposeda simple procedure to determine the location of
the line of action of the resultant, P,,".Their method is as follows:
1. Let
Pur: P,,+ 4P,,,.
where f, : Coulomb's active force as deterntinedfrom Eq. (12.68)
LPu": additional active forcc causedby the earthquake effect
2. Calculate P,, IEq. ( l2.6lJ)].
3. Cafculate P,,"lBcl.(12.72)1.
4. Calculate LPu"- Pu"- Pu.
0.5
0
S o i l f r i c t i o na n g l e ,Q ' ( d e g )
Figure 12.27 Criticalvaluesof horizontalacceleration(Eq. 12.78b)
(12.7e)
Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb
ll
l
I
H
3
I
Y
Figure 12.28 Locati<lnof thc linc of actionofP,,,,
5. According to Figure 12.28.P,,will act at a distanccof H/3 from the baseof the
wall. Also, AP,,,,will zrctat a distanceof 0.6H from thc baseof the wall.
6. Calculate thc location of P,,,,as
/ H \
4 , {; l
+ 4P,,,,(0.611)
( 12.80)
Pu"
whcrc ? - distanceof thc line of action of P,,,,from the baseof the wall.
Note that thc line o1 action of P,,,.will be inclined zrtan angle of 6 to the normal
drawn to the back l'nccof the retaining wall. It is very important to realize that this
mcthod of determining P,,,,is approximate and does not actually model the soil
dynamics.
E xa mp l e1 2 .9
kN/m3,d' = 30o,6=
soil backfill.7 = 15.-5
For a retainingwall with a cohesionless
:
:
:
:
15', 0 0o,rv 0", H 4 m, k.u 0, and kn 0.2.DetermineP,". Also determine
the locationof the resultantline of actionof Po"- that is, ?.
$olution
To determinePn",we useEq. (12.72):
p"": !ryH,(l- k,,)K'"
We are giventhat $' = 30"and 5: 15",so
6
6:t
;i
;
q
12.13
Puufor c'-cf' Soil Backfill
407
Also, 0 * 0o,d = 0o,kh: 0.2.From thesevaluesand Table 12.6,wefind that the
magnitudeof Ki is equalto 0.452.Hence,
- 0x0,4s2)
p"": i(1s.5x4),(1
: 56.05
kN/m
We now locatethe resultantline of action.From Bq. (12.68),
Po* lK,,yH2
'
For f - 30oand 6 : 15o,K" : 0.3014(Table12.5),so
= 37.37kNtm
P, : +(0.3014X15.5X4)'z
Hence,APo": 56.05- 37.3'1= 18.68kN/m. From Eq. (12.80),
/a\
(37.37)
,"(+) + ^P,"(0.6H)
[r,)
1 . *
Pu"
12.13
. (18.68X2.4)
- 1.69m
56.05
r
P",for c'-O' Soil Backfill
The Mononobe-Okabc cquation for estimating P,,,.frtr cohesionlessbackfill des c r i b e di n S c c t i o n 1 2 . 1 2c a na l s o b e e x t c n d e dr < 'ct ' - < f 's o i l ( P r a k e r s a
h n d S a r a n ,1 9 6 6 ;
S a r a ne r n dP r a k a s h ,1 9 6 1 3F) .i g u r e 1 2 . 2 9s h o w sa r e t a i n i n gw a l l o f h e i g h tH w i t h a h o r i z o n t a l c ' - r f' b a c k f i l l . ' l ' h ed e p t h o [ t e n s i l cc r a c k t h a t m a y d e v e l o pi n a c ' - r ! ' s c l i lw a s
g i v c n i n E q . ( 1 2 . 4 4 )a s
2c'
Y\FK,
where K,, : tan2(45- ,b'l2).
T
.
II
f
I
j .
a
'
t:'
H
Il ,
:i.
'
lI '{
I
Figure 12.29 Trtal failure wedge behind a retaining wall with a c'-{" backfill
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
Refering to Figure 72.29 the forces acting on the soil wedge (per unit length of
wall)
are as follows:
the
1.
2.
3.
4.
5.
6.
The weight of the wedge ABCDE, W
Resultant of the shear and normal forces on the failure surface CD, F
Active force, P,,"
Horizontal inertia force, kr,W
Cohesiveforce along CD, C - c(CD)
Adhesive force along BC, C' - .( BC )
It is important to realize that the following two assumptionshave been made:
1. The vertical inertia force (k,,trV)has bccn takcn to be zero.
2. The unit udhesionalong the soil-wall interface (BC) has been taken to be
equal to the cohesion(c) of the soil.
Considering these forces,we can shclwthat
P , , ,-, y ( F I - Z , , ) t N ' , , , c ' ( H
2,,)N',,,
(12.81)
where
:gu_':99 r ,q4 !9!r
(12.82)
s i n ( a '+ 5 )
N",,
s (, lr ' ) + k r , s i n (+i q 5 ' ) ]
[ ( n + 0 . - 5 ) ( t a n*0 t a ni ) + n 2 1 a n g ] [ c o +
( 12.83)
s i n ( 4 ' +6 )
in which
(t2.84)
11'-0+i+(b'
n :
2,,
H - 2.,,
( 12.8s)
The values of lVj,, and Nj,, can bc dctcrmincd by optimizing each coefficientseparately. Thus, Eq. (12.81)givesthe upper bound of P,,".
For the static condition, ki, : 0. Thus,
P , , " : y ( H - Z , , ) t N u -- c ' ( H -
Zu)Nu,
(12.86)
The relationships for N,,. and N., can be determined by substituting k7,: 0 into
E q s . ( 1 2 . 8 2 )a n d ( 1 2 . 8 3 ) .H e n c e ,
1V,,- Nl,.
/t
cos4' sec0 * cos@'secr
sin(4'+ 6)
0.5)(tan0 * tani) * n2tan g]cos(l+ @')
s i n ( 4 '+ 6 )
(12.81)
(12.88)
N,r, and tr with $' and 0 are shown in Figures 12.30 through
The variations of -Ay'o,.,
12.33.
12.13 Pu"for c'-g, Soil Backfitt
a
r
409
3.0
'6
o
2.5
E
2.0
I.5
1 . 0i
20
25
30
3-5
40
15
0'(deg)
Figure 12.30 Yariation of N,,, : N i , , w i t h t [ ' a n d 0 ( b a s e c l
on prakash and Saran, 1966, and
Saran and Prakash, l96lt)
I.0
5 0.u
a
.
E
9
:
(.,.t)
o
/ o.t:
c
?
d
I!
n )
".4
0 '
0
l0
|5
20
25
Q'(deg)
Figure 12.31 Yariation
with @, and 0 (n:
_o^f
_{:,,
and Saran and Prakash.196g)
30
3-5
40
45
0.2) (based on prakash and Saran, 1966,
410
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
l.(-)
0.ti
;
t
I
:
:
]
n=o
z
'5
0.6
E
e=20'
t5'
t0"
5'
3 0.4
E
0.2
-s
0
r0
-2(P
-'4
15
20
2.s
30
0'(dcg)
Figure 12.32 Yarialion of N,,, with r!' and 0 (rr : 0) (based on Prakash and Saran, 1966,
and Saran and Prakash.196iii)
2.0
t.9
|0"
Lu
:'t.i
1.1
1.6
l" l.s
ozo
{
t.4
1.3
1.2
20
0"
/11"
)'
; l (0"
l
.20
'0'
|0'
2f
-0'
10"
20"
o.ro{
l.l
I
0.05'
0
1
0 2 0 3 0 4 0
Angleof internalfiiction,$'
Figure 12.33 Yariation of tr with k,,, rf
Saran and Prakash. 1968)
5
0
', and 0 (based on Prakash and Saran, 1966,and
12.14 Coulomb's PassivePressure
411
Example12.10
For a retainingwall, the following are given:
H : 28 ft
c' : 21.}lblftz
g - *10"
y:1181b/ft3
6' : 20"
kn: 0.1
Determinethe magnitudeof the activeforce,Po".
$olution
From Eq. (12.44),
'o:
2c'
,rKo:
2c
17
.:--?T
r t a n [ 4 5- T
:
)
(2)(210)
-/
-( l r s ) t a n---70\
( 4- ;s
- 5'08ft
)
From Eq. (12.85),
n:
'"
5'08
*
: o'22 - o'2
17 - zr, 28 - 5.oB
From Eqs. (12.81),(12.87),and (12.88),
P,,"* l(H * z,)2(IN,,7)* c'(H - zu)Nu,
For 0 : I0", O' : 20",kt, = 0.1,and n o 0.2.
Nn,: 1..67 (Figure 12.30)
: 0.375 (Figure12.31)
A4,r,
),:
1.17 (Figure12.33)
Thus,
p,, : (118X28- s.0S)11..17
x 0.375)- (210X28* s.08X1.67)
: 1901"60lb/ft
12.14
Coulomb's PassivePressure
Figure 12.34ashows a retaining wall with a sloping cohensionlessbackfill similar to
that consideredin Figure 12.22a.The force polygon for equilibrium of th ewedgeABC
for the passivestate is shown in Figure 12.34b.P,, is the notation for the passiveforce.
Other notations used are the same as those for the activecase (Section 12.10).In a
procedure similar to the one that we followed in the activecase[Eq. (12.68)],we get
Pr:
\KryHz
(2.8g)
412
Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb
Y
....'
(b)
Figure 12.34 Coulomb'spassivepressure:(a) trial failurewedge;(b) force polygon
where K, : Coulomb's passiveearth pressure coefficient, or
cosz({'+ o)
Ko=
(12.e0)
For a frictionlesswall with the verticalbackfacesupportinggranularsoil backfrllwithahorizontalsurface(thatis,0:0o,a:0o,and5:0'),Eq.(12.90)yields
Kr:
l+sind'
-/
d'\
: t a n - [ o t* ,
r
sina'
/
12.15 PassiveForce on Retaining Walls with EarthquakeForces
413
Table 12.7 Valuesof Ku [Eq. 12.90]for 0 : 0', a : 0'
6 (deg) -+
I q5'(deg)
l5
20
25
30
-1f
40
15
1.698
2.040
2.464
3.000
3.690
4.600
1.90t)
1.,) t -)
2.rJ30
3.-506
4.39t)
5..59t)
2.130
2.636
3.286
4.143
- s . l30
6.946
2.405
3.030
3.85.5
4.9'77
6.8-54
8.870
2.135
3.525
4.591
6.105
8.324
rt.772
This relationship is the same as that obtained for the passiveearth pressurecoefficient in Rankine'.scase,given by Eq. (12.30).
The variation of K,, with 95'ancl 6 (for 0 : 0' and a : 0") is given inTable 12.1.
We can see frclm this table that for given value of f ', the value of Kn increaseswith
the wall friction.
12.15
PassiveForce on Retaining Walls
with Earthquake Forces
Figurc 12.3-5
shows the failurc wedge analysisfor a passiveforce againsta retaining
wall of height H with a granular backlill and earthquake forces.As in Figure 12.25,
the failure surfaceis assumedto be a plane. P1,"is the passiveforce. All other notations in Figure 12.35are the samc as those in Figure 12.26.Following a procedure
similar to that used in Section 12.12,(atLerKarrila, 1962) we obtain
Pr,,: )tH2(l - k,)K'p
(r2.e1)
el
H
Figure 12.35
Passiveforce on a
retaining wall with
earthquake forces
414
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
5
i z 3
0'(cleg)
Figure 12.36
Variationol Ki, with kl,for
k,, - tt : 0 : 6 - 0
where
K',,:
cos2(4'+o-B)
,)
sin(D+<i')sin(d'
"-p)
cos(6-d+B)cos(a-9)
(12.e2)
'/ k,, \
i n w h i c h B - t a n( ; - " , I
l(/. /
\ I
Figure 12.36shows a plot of Ki, with @' for various valuesof k, (for k,,: a =
0:6:0).
12.16
Summary and General Comments
This chapter covers the general topics of lateral earth pressure, including the
following:
1". At-rest earth pressure
2. Active earth pressure- Rankine's and Coulomb's
Problems
415
3. Passiveearth pressure- Rankine's and Coulomb's
4. Pressureon retaining wall due to surcharge(basedon the theory of elasticity)
5. Active and passiveearth pressure,which includesearthquake forces.This is an
extensionof Coulomb's theory
For design,it is important to realize that the lateral activepressureon a retaining wall can be calculated using Rankine's theory only when the wall moves salj?ciently outward by rotation about the toe of the footing or by deflection of the wall.
If sufficientwall movement cannot occur (or is not allowed to occur) then the lateral
earth pressurewill be greater than the Rankine activc pressureand sometimesmay
be closer to the at-rest earth pressure.Hence, proper selectionof the lateral earth
pressurecoefficientis crucial for safeand proper design.It is a generalpracticeto assume a value for the soil friction angle (@') of the backfill in order to calculatethe
Rankine activepressuredistribution, ignoring the contribution of the cohesion(c').
The general range of ry''uscd for the design of retaining walls is given in the followins table:
Soil friction
angle, d' (deg)
Soil type
Soft clay
Compacted clay
Dry sand anclgravcl
Silty sancl
0-l-5
20-30
30-40
20-30
In Section 12.5,we saw thzrtthe lateral earth pressurc on a retaining wall is
greatly increasedin the prescnceof a water table above the base of the wall. Most
retaining walls are not designcdto withstand full hydrostaticpressure;hence,it is important that adequatedrainagefacilitiesare provided to ensure that thc backllll soil
does not bccome fully saturated.This can be achievedby providing weepholesat
regular intervals along the length of the wall.
Problems
12.7-12.6 Assumingthat the wallshownin Figure12.37is restrainedfrom yielding,find the magnitudeand locationof the resultantlateralforccper unit
width of the wall.
6',
12.1
12.2
12.3
12.4
t2.5
12.6
10ft
12fr
18ft
3m
4.5m
5.-5m
I t0 tb/fc
98 lb/fc
100lb/ft3
17.6kN/mr
19.95kN/m3
17.ttkN/m3
32'
28'
40"
36'
42"
37'
72.7 Consider a 5-m-high retaining wall that has a vertical back face with a horizontal backfrll. A vertical point load of 10 kN is placed on the ground surface
at a distance of 2 m from the wall. Calculate the increase in the lateral force
on the wall for the section that contains the point load. Plot the variation of
416
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
Sand
Unit weight = y (or dcnsily = p)
a
H
I
I
c'=0
6 (angleof wall fiiction) = $
Figure 12.37
thc pressureincreasewith depth. Use the modified equation given in Section 12.4.
Assume that the retaining wall shown in Figure 12.37is frictionless.
12.8-l2,ll
For each problem. determine the Rankine active force per unit length of the
wall, the variation of activeearth pressurcwith depth, and the location of
the resultant.
Problem
H
d'(degl
l2.x
12.9
12.10
l2.ll
1 5[ l
lu fl
4m
5m
.10
32
36
40
y
I 0 5l b / f r l
100Ib/ftl
Iu kN/rnl
l7 kN/mr
12.72-12J4 A retaining wall is shown in Figure 12.38.For each problem, determine the Rankine activeforce, P,,,per unit length of the wall and the location of the resultant.
o\
" fz
12.t2
12.t3
12.14
l0 ft
20 ft
6m
5fr
6ft
3m
l0-5lb/ft''
122lb ltit3
n0 rbifc
l26lb/fc
1 5 . 5k N / m r
1 9 . 0k N / m 3
(deg)
q
30
34
30
30
34
36
0
300rb/fc
Sttrcharge= q
I
lffi
tffi
t r t
A
I
Ht
I
|
V
lffi
sant
It
Qr
''' =tt
:
v
+i
(degl
Ground water table
Sand
y2 (saturatedunit weight)
Q:
c'2=0
Figure 12.38
l-5kN/m2
Problems
417
12.15 A 15-ft-high retaining wall with a vertical back face retains a homogeneous
saturated soft clay. The saturated unit weight of the clay is 122 lb lft3. Laboratory testsshowedthat the undrained shear strength c,,of the clay is equal to
350 Ib/ft2.
a. Make the necessarycalculationsand draw the variation of Rankine'sactive pressureon the wall with depth.
b. Find the depth up to which a tensile crack can occur.
c. Determine the total activeforce per unit length of the wall before the tensile crack occurs.
d. Determine the total activeforcc per unit length of the wall after the tensile
crack occurs.Also find the location of the resultant.
1 2 . 1 6 R e d o P r o b l e m 1 2 . 1 -a5s s u m i n gt h a t t h e b a c k f i l l i s s u p p o r t i n ga s u r c h a r g eo f
200tbtf(.
12.17 A 5-m-high retaining wall with zrvertical back face has a r:'-{' soil for backfill. For the backfill, y : 19 kN/m3, c' - 26 kN/m2, and r/,' : 16".Considering
the cxistenceof thc tcnsile crack, dctcrmine the active lorce P, on the wall
for Rankineb activestatc.
12.18 For thc retaining wall shown in Figurc 12.39,deterrnir.rcthe activeforce P,
f o r R a n k i n e ' .sst a t e .A l s o , l i n d t h c p o s i t i o no f t h e r c s u l t a n t .A s s u m c t h a t t h e
tensile crack exists.
p : 2 1 0 0 k g / m r , Q - 0 " , ( : : ( : ! t: 3 0 . 2k N / m r
1 2 . 1 9 R e p e a tP r o b l c m l 2 . l B u s i n gt h e f o l l o w i n gv a l u c s :
p - 1 9 5 0k g i m 3 ,d ' - l t i " . t ' ' : 1 9 . 4k N / m :
12.20-12.23 Assume that the rctaining wall shown in Figure 12.37is frictionless.
For each problcm, cleterminethc Rankine passivcf'orceper unit length of
t h e w a l l , t h e v t r r i a t i o no l l a t er a l p r c s s u r ew i t h d c p t h . a n c lt h e l o c a t i o no f t h e
resultant.
Problem
t2.20
t2.21
12.22
t2.23
H
d'(deg)
lJlr
l0 fr
.5nt
4m
-l+
36
3-5
30
y
ll0lb/li'
1 0 5l b / l ' r r
l4 kN/rn'
l5 kN/nr''
1 2 . 2 4 F o r t h e r e t a i n i n gw a l l d e s c r i b e di n P r o b l e m 1 2 . 1 2c, l e t e r m i n et h e R a n k i n e
pnssivcforce per unit lcngth of the wall and the location of the resultant.
II
6.5rn
I
I
Clay
( ,(
a,a
d e n s i t y= P
Figure 12.39
418
Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb
I
' l ' r
i
l l
'
'
t
*l
t
"
!*
Sand
0- 10":
H
= p)
Unitwcight= y (ordensity
r"=0
0' = 36'
6 (wallll'iction)
I
,
',
Figure 12.40
12.25 For the reteriningwall describedin Problcm 12.13,detcrmine the Rankine
passiveforcc per unit lcngth of the wall and the location of the resultant.
12.26 A retaining wall is shown in Figurc 12.40.The hcight of the wall is 5 m, and
t h e u n i t w e i g h t o f t h c s a n db a c k l i l l i s l t t k N i m r . U s i n g C o u l o m b ' se q u a t i o n ,
calculatc the activeforcc P,,on thc wall for the following valuesof the angle
of wall lriction:
a. 5 - llJ'
b. 6:24"
C o m m e n t o n t h e d i r e c t i o na n d l o c a t i o no f t h e r e s u l t t r n t .
12.27 Ref erring to Figure I 2.41, dctermine Coulombh activc force P,,per unit
lcngth of thc wall for thc following cascs:
a . I t : l - 5f t , B - [ J - 5 o ,:n l , H r - 2 0 t t . y : l 2 t { l b / f t r , q 5 '- 3 8 ' , 6 : 2 0 "
. : 2, Ht - 22lt,y : ll6lblf(,0' : 34',5 : lJ'
b . H - 1 t 3f t . B : 9 0 o n
:
5
.
5
[
J
0
" , r : l , H r : 6 . 5 m , 7 : l 6 l 3 t ) k g / m r , d ': 3 0 ' , 6 : 3 0 "
r
r
i
,
p
c. H
g
r
a
p
h
i
cc o n s t r u c t i o np r o c e d u r c .
Usc Culmann\
.
I
I
I
* - - - - - J
.
t
l
Cllhcsionlesssoil
U n i t w e i g h t- y ( r l r d e n s i t y= p )
t ' =0
H
0'
6 ( a n g l eo f w a l l f r i c t i o n )
R z
Y,/'
Figure 12.41
References
419
12.28Refer toFigure 72.26.GiventhatH - 6 m,0 - 0",a:0o, y: 15kN/mr.
6' : 35",6 - 2136' , kn - 0.3, and ft,,: 0, determine the activeforce P,,,,per
unit length of the retaining wall.
L2.29 Refer to Problem 12.28.Determine the location of the point of intersection
of the resultant force P,,"with the back face of the retaining wall.
12.30 Repeat Problem 12.28with the following Values:H - 10 ft. f, : 10",a : 10',
7 : 1 1 0 l b l f t 3 , O '- 3 0 ' , 6 : 1 0 " k. n : 0 . 2 5 , a n d k , : 0 .
1 2 . 3 1R e f e r t o F i g u r e 1 2 . 2 9 . G i v e n t h a t L-I 6 m , 0 - 1 0 " ,< b ': 1 5 " , c ' : 2 0 k N / m 2 ,
y : 1 9 k N i m 3 , a n d / c 7:, 0 . t 5 , u s i n gt h e m e t h o d c l e s c r i b e d
i n S e c t i o n1 2 . 1 3 ,
determine Pn,,.Assume that the depth of tcnsile crack is zero.
1 2 . 3 2R e p e a t P r o b l e m1 2 . 3 1 w i t h t h e f o l l o w i n g V a l u eH
s: l0 ft,0 - -5',6' -20".
c' : 200lblft2, y : 100lb/ftr, anLlkr, : 0.2-5.
References
C o t . t l c l n t uC
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Gr-:ntrrn, E. ( 1929).Untarstrchtmganiiber dic Dntc'kvcrteilung int Orliclt hclustetcnSrrarl,Te chn i s c h eH o c h s c h u l eZ. u r i c h .
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'l'hcory
Orrrsl,, S. (1926). "General
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Civil Engineer.r',Tokyo. Vol. 12, No. l.
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