7/16/2019 Lateral Earth Pressure http://slidepdf.com/reader/full/lateral-earth-pressure-563385d413d62 1/62 CHAPTER 11 LATERAL EARTH PRESSURE 11.1 INTRODUCTION Structures that are built to retain vertical or nearly vertical earth banks or any other material ar e called retaining walls. Retaining walls may be constructed of masonry or sheet piles. Some of the purposes for which retaining walls are used are shown in Fig. 11.1. Retaining walls m ay retain water also. The earth retained may be natural soil or fill. The principal types of retaining walls are given in Figs. 11.1 an d 11.2. Whatever may be the type of wall, all the walls listed above have to withstand lateral pressures either from earth or any other material on their faces. The pressures acting on the walls try to move th e walls from their position. The walls should be so designed as to keep them stable in their position. Gravity walls resist movement because of their heavy sections. They ar e built of mass concrete or stone or brick masonry. No reinforcement is required in these walls. Semi-gravity walls are not as heavy as gravity walls. A small amount of reinforcement is used for reducing the mass of concrete. The stems of cantilever walls ar e thinnerin section. The base slab is the cantilever portion. These walls ar e made of reinforced concrete. Counterfort walls ar e similar to cantilever walls except that th e stem of the walls span horizontally between vertical brackets known as counterforts. The counterforts are provided on the backfill side. Buttressed walls are similar to counterfort walls except the brackets or buttress walls are provided on the opposite side of the backfill. In al l these cases, th e backfill tries to move the wall from its position. The movement of the wall is partly resisted by the wall itself and partly by soil in front of the wall. Sheet pile walls are more flexible than th e other types. The earth pressure on these walls is dealt with in Chapter 20. There is another type of wall that is gaining popularity. This is mechanically stabilized reinforced earth retaining walls (MSE) which will be dealt with later on. This chapter deals with lateral earth pressures only. 419
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Struc tures that are bui l t to retain vertical or nearly vertical earth banks or any other material ar e
called retaining walls. Retaining walls may be constructed of masonry or sheet piles. Some of the
purposes for which retaining walls are used are shown in Fig. 11.1.
Retaining walls m ay retain water also. The earth retained may be natural soil or fill. The
principal types of retaining walls are given in Figs. 11.1 an d 11.2.
Whatever may be the type of wall, all the walls listed above have to withstand lateral
pressures either from earth or any other material on their faces. The pressures ac ting on the walls tryto move th e walls from their position. The walls should be so designed as to keep them stable in
their position. Gravity walls resist movement because of their heavy sections. They ar e buil t of
mass concrete or stone or brick m asonry. No reinforcement is required in these walls. Sem i-gravity
walls are not as heavy as gravity walls. A small amount of reinforcement is used for reducing the
mass of concrete. The stems of cantilever walls ar e th inner in section. The base slab is the cantilever
portion. These walls ar e made of reinforced concrete. Counterfort walls ar e s imilar to cantilever
walls except that th e stem of the walls span horizontally between vertical brackets known as
counterfor ts . The counterforts are provided on the backfil l side. Buttressed walls are similar to
counterfor t walls except the brackets or buttress walls are provided on the opposite side of the
backfill.
In al l these cases, th e backfill tries to move th e wall from it s position. The movemen t of the
wall is partly resisted by the wa ll itself and partly by soil in front of the wall.
Sheet pile walls ar e more flexible than th e other types. The earth pressure on these walls is
dealt with in Chapter 20. There is another type of wall that is gaining popularity. This is
mechanically stabilized reinforced earth retaining walls (MSE) which will be dealt with later on.
This chapter deals wi th lateral earth pressures only.
Table 11.1 C o e f f i c i e n t s o f e a r t h p r e s s u r e f o r a t r e s t condition
T y p e o f s o i l / K Q
L o o s e san d , sa tu ra te dDe n se san d , sa tu ra te d
D e n s e s a n d , dry (e =0.6) -
L o o s e s a n d , dry (e =0.8) -
Com pacte d c lay 9
Com pac te d c lay 31
O rg a n ic s i l ty c l a y , u n d i s tu rb e d (w { = 74%) 4 5
0.460.36
0.49
0.64
0.42
0.60
0.57
The value o f K Q depends upon the re lat ive dens i ty o f the sand and the p ro ce s s by which the
de p o s i t w as f o r m e d . I f th i s p ro ce s s do e s n o t in v o lv e a r t i f i c i a l tam p in g th e v a lu e o f K Q ranges f rom
a b o u t 0.40 fo r l o o s e s an d to 0 .6 for dense sand . Tamping th e layers m ay increase it to 0.8.
The value of K Q may also be obtained on the basis of elastic theory. If a cylindrical sample of soil
is acted upon by vertical stress CT , and horizontal stress ah, the lateral strain e { m ay b e expressed as
(11.2)
where E = Y o u n g ' s m o d u l u s , n = P o i s s o n ' s rat io .
The lateral s t ra in e { = 0 when the ear th i s in the at res t cond it ion . For th is cond it ion , we may
write
ah V
o r — =— (11.3)
where ~T^~ = KQ, crv=yz (11.4)
A cco r d in g to Jaky (1944) , a g o o d ap p r o x im at io n fo r K 0 i s g iven b y Eq . (11.5) .
KQ=l-sin0 (11.5)
w h ic h fi ts most o f the exper imental data.
N u m e r ica l v a lu e s o f K Q for some so i l s are g iven in Table 11.1.
Example 11.1If a r e ta in in g w a l l 5 m high is res tra ined f r o m yie ld ing , what wi l l be the at- res t ear th pressure permeter length of the w a l l? G iv e n : th e b ack f i l l is co h e s io n le s s s o i l h av in g 0 = 30° and y = 18 k N / m
3.
A l s o d e te r m in e th e r e s u l tan t f o r ce f o r th e a t -r e s t c o n d i t i o n .
Solution
F ro m E q. (11.5)
K Q = l-sin^= l-sin30° =0.5
F ro m Eq. (1L i b ) , ah =KjH - 0.5 x 18 x 5 = 45 k N / m2
The problem now consists of determining the stresses associated with the states of plastic
equilibrium in the se mi-infinite mass and the orientation of the surface of sliding. The problem w as
solved by R ankine (1857).
The p lastic states w hich are produced by stretching or by com pressing a sem i-infinite mass ofsoil parallel to its surfa ce are called active an d passive Rankine states respectively. Th e orientation
of th e planes may be found by Mohr 's diagram.
Horizontal stretching or compressing of a semi-infinite mass to develop a state of plastic
equilibrium is only a concept. However, local states of plastic equilibrium in a soil mass can be
created by rotating a retaining wall about its base either away from th e backfill for an active state or
into the backfill for a passive state in the way shown in Figs. 1 1.3(c) and (d) respectively. In both
cases, the soil within wedge ABC will be in a state of plastic equ ilibrium and lin e AC represents the
rupture plane.
Mohr Circle for Active and Passive States of Equilibrium in Granular Soils
Point P { on the d-axis in Fig. 1 1.6(e) represents the state of stress on base AD of prismatic element
ABCD in Fig. 1 1.6(a). Since th e shear stress on AD is zero, th e vertical stress on the base
is a principal stress. OA an d OB are the two Mohr envelopes which satisfy th e Coulomb equation of
shear strength
j= c r t a n ^ (11.9)
Two circles C a an d C can be drawn passing through P l and at the same time tang ential to the M ohr
envelopes OA an d OB. Wh en the semi-infinite mass is stretched horizontally, the horizontal stress
on vertical faces AB an d CD (Fig. 1 1.6 a) at depth z is reduced to the minimum possible and this
stress is less than vertical stress ov.Mo hr circle Ca gives th e state of stress on the prismatic element
at depth z when the mass is in active failure. The intercepts OP l an d OP 2 are the major and minor
principal stresses respectively.
When the semi-infinite mass is compressed (Fig. 1 1.6 b), th e horizontal stress on the vertical
face of the prismatic element reaches the maximum value OP 3 and circle C is the Mohr circle
which gives that state of stress.
Active State of Stress
From Mohr circle C a
Major principal stress = OP { = c r l = yz
Mino r principal stress = OP 2 = < 7 3
(7, + < J ~ , <J , — (Tonn—\J\J,—12
cr . — < T T < j,+
C T - ,From triangle 00,C,, — = — sin i1 J
2 2
( 1 + sin 0|"
\Therefore, pa = cr3 = — -=yzKA (U.ll)'V
where a, = y z, K A = coefficient o f earth pressure for the active state = ta n2
( a ) R e t a i n i n g w a l l (b ) P r essu r e d i s t r i bu t ion
O C T 3 0 On O]
(c ) Mohr diagram
Figure 11.11 R a n k i n e ' s a c t i v e p r e s s u r e f o r a sl op i n g c o h e s i o n l e s s b a c k f il l
n - < J v cos fi = yz cos /?cos fl=yz cos2
j3
T = a s i n / ? =
(11.20)
(1 1 . 2 1 )
A M o h r d i a g r a m c a n b e d r a w n a s s h o w n i n Fi g . 11.1 l (c ) . Here , length O A = yzcos/3 m a k e s
a n a n g l e ( 3 w i t h the (T-axis. OD =on - yzcos2/3 andAD = T= yzcosf} sin/3. OM is the M o h r envelope
m a k i n g a n a n g l e 0 w i t h the <7-axis. N o w Mo hr c i rcle C} c a n b e d r a w n pa s s i n g t h r o u g h p o i n t A a n d
at the s a me t i m e ta n g e n t i a l to e n v e l o p e O M . T h i s c i r c l e c u t s l i n e O A a t p o i n t B a n d t h e C T - a x i sa t E
andF.
N o w O B = th e l a t e r a l p r e s s u r e ol = pa in the ac t ive s ta te .
The p r i n c i p a l s t r e s s e s are
O F = C T j and O E = a 3
T h e f o l l o w i n g r e l a t i o n s h i p s c a n b e e x p r e ss e d w i t h r e f e re n c e t o t h e M o h r d i a g r a m .
An equation for P for a sloping backfill surface can be developed in the same way as for an active
case. The equation for P may be expressed as
(11.26)
n cos fi + Jcos2
fl- cos20
wher e , Kp=cos]3x /
co s /3 - ^cos2
j3- cos20
P acts at a heigh t H /3 above point A and paral lel to the s loping sur face.
(11.27)
Example 11.2
A canti lever retaining wall of 7 meter height (Fig. Ex. 11.2) retains sand. The properties of the sand
are: e - 0.5, 0 = 30° and G ^ = 2.7. Using Rankine ' s theory determine th e active earth pressure at the
base w h e n th e backf i l l is (i) dry, (ii)saturated an d (iii)submerged, an d also th e resultant active forcein each case. In addit ion determine the to tal water pressure under the submerged con dit ion.
A counterfort wal l of 10 m height re ta ins a non-cohes ive backfi l l . The void ra t io and angle of
i n t e rn a l fr ict ion of the backf i l l respect ive ly are 0.70and 30° in the loose state and they are 0.40and
40° in the dense s ta te . Calcu la te and compare ac t ive and passive earth pressures for both the cases.Take the specif ic gr av i t y of solids as 2.7.
Solution
(i ) In the loose state, e - 0.70 which gives
/""* -. r\ i— j
=_ L= __x9gj=156kNm3
dl + e 1 + 0.7
c j. ™ ° v l-sin0 1-sin 30° 1 1F o r 0 = 3 0 , K, -= = — ,and^0 = = 3
' A -i * ' 1 * O /"\ o O i TS1 + s i n 30 3 K,
M a x . pa =K AydH = - x 15.6 x 10 = 52 k N / m2
M a x . p = K pydH = 3 x 15.6 x 10 = 46 8 k N/ m2
(i i) In the dense state, e = 0.40, which g ives ,
Y =-22—x9.81 = 18.92 kN /m
3
d1 + 0.4
1-sin 40° 1For 0 = 40°, K=- —— = 0.217, Kp =-— = 4.6A 1 + s i n 4 0 ° p
K.f\
M a x . p f l =KAydH = 0 . 2 1 7 x 1 8 . 9 2 x 1 0 = 41.1 kN/ m 2
and Max . p = 4.6 x 18.92 x 10 = 870.3 kN/m2
Comment: The comparison of the resul ts indica tes tha t dens if ica t ion of soi l decreases the
act ive earth pressure and increases the pass ive earth pressure . This is advan tageous in the sense tha t
ac t ive earth pressure is a dis turb ing force and pass ive earth pressure is a res is t ing force .
Example 11.7
A wall of 8 m height retains sand having a density of 1.936 Mg/m3
and an angle of internal friction
of 34°. If the surface of the back fi l l s lopes up ward s a t 15° to the horizonta l , f ind the ac t ive thrus t per
un i t l eng th of the w a l l . U se Rank ine ' s cond i t ions .
Solution
There can be two solut ions : analy t ica l and graph ical . The a nalyt ica l solut ion can be obta ined f rom
If th e backfi l l material is cohesionless, the terms containi ng cohesion c in Eq. (11.42) reduce
to zero.
Example 11.8
A retaining wa ll has a vertical back and is 7.32 m high. The soil is sandy loam of unit weight17.3
kN/m3. It has a cohesion of 12 kN/m
2and 0 = 20°. Neglecting wall friction, determine the active
thrust on the wal l . The upper surface of the fill is horizontal.
Solution
(Refer to Fig. 11.14)
When the material exhibits cohesion, the pressure on the wall at a depth z is given by
(Eq. 11.30)
where K J_ iT= —= 0.49, IK 0.7v A
1-sin20 °
1 +s i n 20 °
When the depth is small the expression for z is negative because of the effect of cohesion up to a
theoretical depth z0. The soil is in tension and the soil draws away from th e wall .
-— I—-— Iy v Y
1 + sin ( f) i
where Kp = 7-7= 2.04, and JKP = 1.43- *p
2 x 1 2
Therefore Z Q = "TTT"x
1-43 = 1-98 m
The lateral pressure at the surface (z = 0) is
D = -2cJxT = -2 x 12 x 0.7 = -16.8 kN/m2
* u V •* »
The negative sign indicates tension.
The lateral pressure at the base of the wall (z = 7.32 m) is
pa =17.3 x7.32 x 0.49 -16.8 = 45.25 kN/m2
Theoretically the area of the upper triangle in Fig. 11.14(b) to the left of the pressure axis
represents a tensile force w hich should be subtracted from the compressive force on the lower part
of the wall below the depth Z Q . Since tension cannot be applied physica lly between the soil and the
wall, this tensile force is neglected. It is therefore com monly assumed that the active earth pressureis represented by the shaded area in Fig. 1 1 . 14(c). Th e total pressure on the wall is equal to the area
of the triangle in Fig. 11.14(c).
= -(17.3x7.32 x 0.49 - 2 x 12 x0.7) (7.32- 1.98) = 120.8 kN/m
A in Eq. (11.53) is the same as Rankine's. Th e effect of wall friction is frequentlyneglected where active pressures are concerned. Table 11.2 makes this clear. It is clear from this table
that K A decreases with an increase of 8 and the maxim um decrease is not more than 10 percent.
11.9 COULOMB'S E A R T H PRESSURE T H E O R Y FOR S A N D FOR
PASSIVE STATE
In Fig. 11.18, th e notat ions used are the same as in Fig. 11.17. As the wall moves into th e backf i l l ,
th e soil tr ies to move up on the pressure surface AB which is resisted by friction of the surface.
Shearing stress on this surface therefore acts downw ard. The passive earth pressure P is the
resultant of the normal pressure P and the shearing stress. The shearing force is rotated upward
with an angle 8 which is again the angle of wall f r ic t ion. In this case S is positive.
As the rupture takes place along assumed plane surface AC, the soil tries to move up the plane
which is resisted by the frictional force ac ting on that line. The shearing stress therefore, acts downw ard.
The reaction R makes an angle 0 with the normal and is rotated upw ards as shown in the figure.
The polygon of forces is shown in (b) of the Fig. 11.18. Proceeding in the same way as for
active earth pressure, we may write the following equations:
(11 .55 )
(11.56)
Different ia t ing Eq. (11.56) with respect to 0 an d setting th e derivative to zero, gives th e
min imum value of P as
22 sin
2a
.sm(#-/?)
6 + a = a)
(a ) Forces on the sliding wedge (b) Polygon of forces
Figure 11.18 Conditions for fai lure under passive state
where K is called the passive earth pressure coefficient.
Kp =
si n2
asin(a (11.58)
Eq . (11.58) is valid for both positive and negative va lues of ft an d 8.
The total normal component of the passive earth pressure P on the back of the wall is
(11.59)< • - - /,
For a smooth vertical wall with a horizontal backf i l l, w e have
Nt (11.60)
Eq . (11.60) is Rank ine 's passive earth pressure coefficient. We can see from Eqs. (11.53) an d
(11.60) that
1Kp =" l< y j-. iy
Coulomb sliding wedge theory of plane surfaces of failure is valid with respect to passive
pressure, i.e., to the resistance of no n-cohesive soils only. If wall friction is zero for a vertical wall
an d horizontal backf i l l , th e value of Kp may be calculated using Eq. (11.59). If wall friction is
considered in conjunction with plane surfaces of failure, much to o high, .and therefore unsafe
values of earth resistance will be obtained, especially in the case of high friction angles 0. Forexample for 0= 8 = 40°, and for plane surfaces of failure, Kp = 92.3, whereas fo r curved surfaces of
failure K p = 17.5. However, if S is smaller than 0/2, the difference between the real surface of
sliding and Coulomb 's plane surface is very small and we can compute the corresponding passive
earth pressure coefficient by means of Eq. (11.57). If S is greater than 0/2, th e values of Kp should
be obtained by analyz ing curved surfaces of failure.
11.10 ACTIVE PRESSURE B Y CULMANN'S METHOD FOR
COHESIONLESS SOILS
Without Surcharge Line Load
Culmann's (1875) method is the same as the trial wedge method. In Culmann's method, the force
polygons are constructed directly on the 0-line AE taking AE as the load line. The procedure is asfollows:
In Fig. 11.19(a) AB is the retaining wall drawn to a suitable scale. The various steps in the
construct ion of the pressure locus are:
1. Draw 0 -line AE at an angle 0 to the horizontal.
2. Lay off on AE distances, AV , A1, A2, A3, etc. to a suitable scale to represent th e weights of
(e) The weights of the wedges in (d ) above per meter length of wal l may be determined by
mul t ip ly ing the areas by the un it weight of the soil. The resul ts are tabulated below:
Wedge
BAc^
BAc2
BAc3
Weigh t , kN
115
230
345
Wed g e
BAc4
BAc5
Weigh t , kN
460
575
(f ) The w e igh t s of the wedges BAc }, BAc2, etc. are respect ively plot ted are Bdv Bd2, etc. on
the 0-line.
(g) Lines are drawn paral lel to the pressure l ine from points d{, d2, d 3 etc. to meetrespect ively the t r ial r u p tu r e l ines Bcr Bc2, Bc^ etc. at points e}, e2, ey etc.
(h ) The pres sure locus is drawn pass ing through points e\ , e2, ey etc.
(i) Line zz is drawn t angent ia l to the pressure locus at a point at which zz is parallel to the 0
l ine. This po int coincides w ith the poin t ey
(j ) e3d ^ gives th e act ive ear th pressure when conver ted to force units .
Pa = 180 kN per meter length of wal l ,
(k ) Bc3 is the cr i t ical rupture plane.
11.11 LATERAL PRESSURES B Y T H E O R Y OF ELASTICITY FOR
S U R C H A R G E LOAD S ON THE SURFAC E OF BACK FILL
The surcharges on the surface of a backfill parallel to a retaining wall may be any one of the
The body of soil mass BAe]dl (Fig. 1 1.23b) is acted on by the follow ing forces:
1. The weight W j of the soil mass acting through th e center of gravity of the mass having a
lever arm /2 with respect to Or the center of the spiral.
2. The passive earth pressure /^acting on the vertical section eld } having a lever arm /3.
3. The passive earth pressure Pj acting on the surface AB at an angle S to the normal and at a
height H/3 above A h av in g a lever ar m l { .
4. The resul tant reaction force Fl on the curved surface Ae{ and passing through the center
Determinat ion of the Force />1 Graph ica l ly
The directions of all the forces men tioned above except that of Fl are known . In order to determine
the direction of F, combine the weight W { and the force Pel which gives the resultant /? , (Fig.
1 1.23c). This resultant passes through the point of intersection nl of W { andPel in Fig. 1 1.23b and
intersects force P { at point n2. Equilibrium requires that force F { pass through the same point.According to the property of the spiral, it must pass through the same point. According to the
property of the spiral, it must pass throu gh the center Ol of the spiral also. Hence, the direction of
Fj is know n and the polygon of forces shown in Fig. 1 1 .23c can be completed. Thus we obtain the
intensity of the force P } required to produce a slip along surface Aelcl .
Determinat ion of /* , by Moments
Force Pl can be calculated by taking moments of all the forces about the center O { of the spiral.
Equi l ibr ium of the system requires that the sum of the moments of all the forces must be equal to
zero. Since the direction of F l is now known and since it passes through Ol , it has no mom ent. The
sum of the moments of all the other forces may be written as
P 1 / 1 + W 1 / 2 + J P 1 / 3 = 0 (11.72)
Therefore,P\
=-7( 2
+ P) (11.73)l
i
Pl is thus obtained for an assumed failure surface Ae^c^. The next step consists in repeating
the investigation for more trial surfaces passing through A which intersect line BD at points e2,e3etc. The values of Pr P 2 P 3 etc so obtained may be plotted as ordinates dl d { , d2 d'2 etc., as shown
in Fig. 1 1 .23b and a smooth cu rve C is obtained by join ing points d { , d'2 etc. Slip occurs along the
surface corresponding to the min imu m value P which is represented by the ordinate dd'. The
corresponding failure surface is s h o wn as Ae c in Fig. 1 1.23b.
11.13 COEFFICIENTS OF PASSIVE EARTH PRESSURE TABLES
A N D G RAPHS
Concept of Coulomb 's Formula
Coulomb (1776) computed the passive earth pressure of ideal sand on the simplifying assumption
that the entire surface of s liding consists of a plane through the lower edge A of contact face AB as
shown in Fig. 11.24a. Line AC represents an arbitrary plane section through this lower edge. Th e
forces acting on this wedge and the polygon of forces are shown in the figure. The basic equation
fo r computing the passive earth pressure coefficient may be developed as follows:
From Table 11.3 K 'p for a curved surface of failure (Caquot an d Kerisel. 1948) for 0 = 40°
an d 8 =20° is 10.38.
From Eq. (11.77)
p = -y H2
K ' = - x 18.5 x102x 10.38
p2
p2
= 9 6 02k N /m
Comments
For S = $2, th e reduction in the passive earth pressure due to a curved surface of failure is
10,828-9602Reduct ion = — — x 100 = 11.32%
Example 11.16
For the data given in Example 11.15, determine the reduction in passive earth pressure for a
curved surface of failure if 8 = 30°.
Solution
For a plane surface of failure P from Eq. (11.76) is
P = -xl8.5x!02x — = 22,431 kN/m
p2 s i n 9 0 ° c o s 3 0 °
where, K = 21 from Fig.11.24 for § = 30° and < /> = 40 °
From Table 11.3 for 8 = 30° and < /» = 40°
K .f =10.38 +17.50
From Eq(l 1.77)
P =-x18.5x!02x 13.94 =12,895 kN/m
p2
o A .• • • 22,431-12,895 „ „ _ _ ,
Reduction in passive pressure = = 42.5%22,431
It is clear from th e above calculations, that th e soil resistance under a passive state gives
highly erroneous values for plane surfaces of failure with an increase in the v alue of S. This errorcould lead to an unsafe condition because the computed values of P would become higher than
the actual soil resistance.
11.14 LATERAL EARTH PRESSURE ON RETAINING WALLS
DURING EARTHQUAKES
Ground motions during an earthquake tend to increase the earth pressure above the static earth
pressure. Retaining walls with horizontal backfills designed with a factor of safety of 1.5 for static
loading a re expected to with stand horizontal accelerations up to 0.2g.For larger accelerations, and
fo r walls with sloping backf i l l , additional allowances should b e made for the earthquake forces.
Murphy (1960) show s that when subjected to a horizontal acceleration at the base, failure occurs in
th e soil mass along a plane inclined at 35° from th e horizontal . The analysis of Mononobe (1929)considers a soil wedge subjected to vertical and horiz onta l accelerations to behave as a rigid body
sliding over a plane slip surface.
The current practice for earthquake design of retaining w alls is generally based on design rules
suggested by Seed and Whitman (1970). Richards et al. (1979) discuss the design and behavior of
gravity retaining walls with unsaturated cohesionless backfill. Most of the papers make use of the
popular Mononobe-Okabe equations as a starting p oint fo r their ow n analysis. They follow generally the
pseudoplastic approach fo r solving th e problem. Solutions are available fo r both the active an d passive
cases with as granular backfill materials. Thou gh solutions for (c-0)soils have been presented by some
investigators (Prakash and Saran, 1966, Saran and Prakash, 1968), their findings have not yet been
confirmed, and as such the solutions fo r (c-0)soils have no t been taken up in this chapter.
Earthquake Ef fect on Active Pressure with Granular Backfill
The Mononobe-Okabe method (1929, 1926) for dynamic lateral pressure on retaining walls is a
straight forward extension of the Coulomb sliding wedge theory. The forces that act on a wedge
under the active state are shown in Fig. 11.25
In Fig. 11.25 AC in the sliding surface of failure of wedge ABC having a weight W with
inertial components kv W an d khW. The equat ion for the total active thrust Pae acting on the wall AB
under dynam ic force condi t ions as per the analys is of Mononobe-Okabe is
(11.79)
in which
K. =•Ae
co s / /cos2
<9cos(#+ 0+77 ) 1 +cos(8+ 9+ /7)cos(/?- 9]
(11.80)
Figure 11.25 Act i ve f o rce on a retaining wall with ear thquake fo rces
Effect of Wall Lateral Displacement on the Design of Retaining Wall
It is the usu al practice of some designers to ignore the inert ia forces of the mass of the gravity
re ta ining wall in seismic design. Richards and Elms (1979) have shown that this approach is
unconserva t ive since it is the weight of the wall which provides most of the resistance to lateralm o v e m e n t . Taking into account all the seismic forces acting on the wall and a t the base they have
developed an expression for the weight of the wall W w u n d e r th e equi l ibrium condi t ion as(for
failing by s l iding)
Ww=±yH2(l-kv)KAeCIE (11.88)
in which ,
cos(S + 6 > ) - sin(£ + 6 > ) ta n S1E
( l - & v ) ( t a n £ - t a n 7 7 ) ( 1 1 . 8 9 )
where W w = w e i g h t of retaining wall (Fig. 11.25)8 = angle of fr iction between the wall and soil
Eq. (11.89) is considerably affected by 8. If the wall inert ia factor is neglected, a designer will
have to go to an exorbitant expense to design gravity walls.
It is clear that tolerable displacem ent of gravity walls has to be considered in the design. The
we igh t of the retaining wall is therefore required to be determined to limit the displacement to the
tolerable limit . The procedure is as follows
1. Set the tolerable displacement Ad
2. Determine the design value of kh by making use of the following equation (Richards et al., 1979)
0.2 A ,2
^
where A a, AV = acceleration coefficients used in the Applied Technology Counci l (ATC) Bui lding
Code (1978) for various regions of the United States. M is in inches.
3. Us ing th e values of kh calculated above, and assuming k v - 0, calculate K Ae from E q (11.80)
4. Using the value of K Ae , calculate the w eight , W w, of the re ta ining wal l by m aking use of
Eqs . (11.88) and (11.89)
5. Ap ply a suitable factor of safety, say, 1.5 to W w .
Passive Pressure During Earthquakes
Eq. (11.79) gives an expression for computing seismic active thrust which is based on the well
k n o w n Mononobe-Okabe analys is for a plane surface of fa i lure . The corresponding express ion fo r
2. Fig.11.33 shows th e effect of 0 on K pg . The f igure shows th e difference between
Mononobe-Okabe and log spiral values of K versus kh with 8=( 2/30) an d kv = 0. It is also
clear from th e f i gu re th e di ffe rence between the two approaches is greatest fo r kh - 0 and
decreases with an increase in the value of kh.
Example 11.17
A grav ity reta ining wall is required to be designed for seismic condit ions for the act ive s ta te. The
following data a re given:
Height of wall = 8 m 0 = 0 ° , 0=0, 0 = 30 ° , & = 15°, £ , = 0, kh = 0.25 and y = 19kN/m3. Determine
P ae and the approximate point of application. What is the additional active pressure caused by the
earthquake?
Solution
From Eq. (11.79)
Pae=\rH2(l-kv)KAe=^yH^KAe, since *y = 0
For 0 = 30°,5 = 15° and kh = 0.25, w e have from Fig. 1 1.26 a
KAe = 0.5.Therefore
pag = - ? - 1 9 x 82x 0 .5 = 3 0 4 k N / m
1 9
From Eq. (11.14) Pa=-y H2K A
2° -
2where K A =ta n
2 (45° - ^2 ) = tan230° = 0.33
Therefore Pa = - x 19 x82x0.33 = 202.7 kN/m
&Pae = th e add it ional pressure due to the earthquake = 304 - 202.7 = 101.3 kN /m
For all practic al purpo ses, the point of application of P ae may be taken as equal to H/2 abovethe base of the w all or 4 m above the base in this c ase.
Example 11.18
For the wall g iven in Example 11.17, dete rmine th e total passive pressure P e under se i smic
condit ions. What is the addi t ional pressure due to the earthquake?
Solut ion
From Eq. (11.91),
Pae = rH*(l-kv)Kpe =7H*Kpe, since *v = 0
From Fig 1 1.32, (from M -O curves) , Kpe = 4.25 for 0 = 30°, an d 8= 15 °
shear strength parameters for the soil are c = 600 lb/ft2, and 0 = 22°, and the uni t weight
is 110 lb/ft3. Using Rankine theory, determine the total active thrust behind th e wall and the
total passive resistance in front of the w all . Assu me the w ater table is at a great depth .
11.13 For the retaining w all given in Fig. Prob. 11.12, assum e the w ater table is at a depth of 10 ftbelow the backfi l l surface. The saturated uni t weig ht of the soil is 120 lb/ft
3. The soil above
th e GWT is also saturated. Compute the resultant active and passive thrusts per uni t length
of the wall .
11.14 A retaining wall has a vertical back face and is 8 m high. The backfill has the following
properties:
cohesion c = 15 kN/m2, 0 = 25°, y = 18.5 kN/m
3
The water table is at great depth. The backfill surface is horizontal. Draw the pressure
distribution diagram and determine the magnitude and the point of application of the
resultant active thru st.
11.15 For the retaining wall given in Prob. 11.14, the water table is at a depth of 3 m below the
backfill surface. Determine the magnitude of the resultant active thrust.
11.16 For the retaining wall given in Prob. 11.15, compute th e magni tude of the resultant activethrust, if the backfill surface carries a surcharge load of 30 kN/m
2.
11.17 A smooth retaining wall is 4 m high and supports a cohesive backfill with a unit weight of
17 kN/m3. The shear strength parameters of the soil ar e cohesion =10 kPa and 0 = 10°.
Calculate the total active thrust acting against the wall and the depth to the point of zero
lateral pressure.
11.18 A rigid retaining wall is subjec ted to passive earth pressure. Determine th e passive earth
pressure distribution and the magnitude and point of application of the resultant thrust by
Rankine theory.
Given: Height of wall = 10 m; depth of water table from ground surface = 3 m;
c - 2 0 kN/m2, 0 = 20° and y sat = 19.5 kN/m
3. T he backfill carries a uniform surcharge of
20 kN/m2.
Assume the soil above the water table is saturated.11.19 Fig. Prob. 11.19 gives a retaining wall with a vertical back face and a sloping backfill . A ll
th e other data ar e given in the figure. Determine th e magnitude an d point of application of