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7/21/2019 Active and Passive Pressure in Cohesive Soils
http://slidepdf.com/reader/full/active-and-passive-pressure-in-cohesive-soils 1/8
September,
1959
257
Nomograms
for
the Calculation of
Active
Pressure and PassiveResistance of
Cohesionless oils
By J. Rygol,B.Sc.(Eng.)
The wo extremes of pressure t o which a retaining
(i)
the active pressure of the soil on the back of a
wall resultingromlightmovement of the
wall away from the filling, and
(ii) the passive resistance of the soil on the front of
a wall to slight displacement of the wall towards
filling.
Theactive pressure is aminimum value, and is
attained when the wall yields b y moving away from
the filling, Fig.
1 .
The passive resistance is the
maximum pressure to which the wall is subjected
immediately before failure occurs by heaving up he
soil in front of the wall, Fig. 2.
For a cohesionless soil, assumingstraight rupture-
line through the foot
of
the wall, the expressions for
the total active thrust P a and the total passive resis-
tance
P,
can be written in the form
wall may be subjected are
:
P a
=
Q
y
H2
K,
. . . . .
1)
P p = Q y H z K p . .
.
. . (2)
where y is the equivalentdensity of soil given by
and , Figs. 1 and 2,
H verticalheight of wall
q surcharge oadperunitarea
y
unit weight of soil
K ,
coefficient of active pressure
K , coefficient of passive resistance
a)
S i g n C o n v e n t i o n
Similarly, the horizontal omponents of the ota l
pressures can be written in the form
P a h 4 y
H z
K a h
. . . . . 4)
P p h
=
4
y'
H2
K p h
.
. .
. . (5)
and the vertical pressure components
Pav
=
8 y'
H ZK,, . . . . . (6)
Pp,
=
y H2 K,, . . . . . (7)
Assuming linear istribution of earth pressures,
the horizontalomponents of active pressure and
passive resistance at depthH , measured on the vertical
from the top of the wall, are
P a h = YH q) K a h 8)
0 angle of internal friction of soil
6 angle of friction between soil and back of 'wall
a angle between ack of wall andhe vertical
p
angle of surcharge, between the upper surface
the coefficients K a h and K p h areiveny the
expression
of soil and the horizontal
COSY
0
-
)
1
ph
a
[I J
sin(
0
+ 8) in(0
-p)
cos (6 +a) cos (--S)
where in he above ormula the positive sign refers
to active pressure and he negative sign to passive
;r'.lr+?
b)
c
d l
E q u i l i b r i u m O f W e d g e
A c t i v e r e s s u r e
C o m p o n e n t s
H o r i z o n t a l A c t i v e P r e s s u r e
eh'=gr'H2
A CTIV EP RE S S URE
Fig. 1
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258
The Structural Engineer
resistance. The sign convention or a, /3 and S used
in formula
10)
s shown in Figs(
a)
and (2a) espectively
Note hat or hesameconditions of the wall and
backing, the signs
of
a and
p
for active pressure and
passiveesistancerenterchangeablenormula
(10).
Thus
+
and +p for active pressure become
--M. and -p for passive resistance.
The coefficients
K
and
K,
can be expressed in terms
of th e respective coefficients K h for horizontal thrust.
Thus
K,
= K a h
sec(a
+
8) . . . .
(11 )
K B V =
K a h tan(a + S) . .
,
. (12)
and
K, = K p h
sec( + 6 )
.
. . .
(13)
K,,= K p h tan(a
S) .
. .
.
(14)
Similarly, th e expressions or
P
and P, in erms of
Ph become
P a
=
Pah sec(a + S) . .
.
. (15)
Pa,= P a h t an (a + S) . . .
.
(16)
and
P,
= P p h
sec(a
+
8) .
.
.
. (17)
Pp,= P p h tan(
+
S) .
.
.
. (18)
Note that P,, and P,, are considered t o be positive
when directeddownwardsorupwards espectively.
It is convenient t o express Kah in terms of:
Ph
p =
tan
z ~
coefficient of internal friction
o l
soil
Y = tan S wall roughness
a
= ta n a inclination of wall to the vertical
b = tan p inclination
of
the surcharge to
the horizon tal
This gives
K a h
=
ph
1 +
w
l2
-I
.
. . .
(19)
Table 1
Densities of cohesionless mate rials Ib/fta
Material
-- -
--
Gravel
Coarse and med iim sands
:
Fine and silty sands
( granites and ;hales.
basalts and dolerites
Rock
{
limeston?s and
sandatones .
Broker?rick ..
. .
ha h
..
. .
Ashes . . .
.
draineg above
Densit when
Ground Water
Level
ym
100-1
25
105-130
110-135
100-130
110-140
80-1
20
60-430
70-1 10
40-60
submerged below
Density when
Ground Water
Level Yb
~ _ _
60-80
60-80
60-80
70-100
40-80
40-80
20-40
40-60
20-30
The sign convention or a,
b
and
r
follows directly
from he s i p s of the respective values of the angles
a, p and S, as shown in Figs. 1 and 2.
The densities of cohesionless
soils
as recommended
by TheCode of Practice Earth Retaining Structu res
1
are given in Table 1 .
The unit weight of the soil varies with the amount
of
moisture content, ndwithhe position of the
groundwater able. In he case of a dry backing
thedensity is
yd.
Where thebacking is moist but
is above ground water level
GWL)
the moist density
y m
should besubstituted for
y
inequation
(3).
The
density of waterlogged acking below the round
waterevels yb, the submergedensity. I t is
usual to assume that abovegroundwater evel he
soil is ompletely saturated.The aturateddensity
y S of a soil may be taken as ts submergeddensity
plus thedensity of thewatercontained
(ys
= y b
+
62.5
lb/ft3).
It is
apparent hat below theground
water level there is the pressure risingrom the
submergeddensity of the soil (acting at an angle S
with henormal o he wall) togetherwith he n-
dependenthydrostaticpressure normal to he wall).
A rough guide for the value of 0 (or p) for cohesion-
less soils may be obtained from Table 2.
p r
9
c
d)
. S i g no n v e n t i o n E q u i l i b r i u m of Wedge
P a s s i v e R e s i s t a n c e
Horizontal P a s s ~ v e
C o m p o n e n t s
R e s i s t a n c e
Pph=+$H'K,h
P A S S IV E R ES
I
STANCE
Fig.
2
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September,
I959
259
Table 2
Typical values
of 4
and p for cohesionless materials
1
d
35 -45
35 -40
30 -35
30 --35
35 -45
~
35 -45
Materials
-
______
Sandy gravel .
Compact sand . .
Loose sand. .
Shale filling . .
. .
Rock filling
Ashes or broken brick
. .
. .
. .
. . . .
. . . .
P
0.700-1 .OOO
0.700-0.839
0.577-0.700
0.577-0.700
0.700--1.000
0.700-1.000
ar
oractive pressure it seems easonable to assume
a value of
6
between 0 nd 0. he Code of Practice
Earth RetainingStructures recommends tha t in
the absence of definite tes t data 6 should be taken as
20"
(r
=
0.364) for walls of concrete or brick, as
30"
(r
= 0.577)
or steel piling coated with tar or bitumen,
and as 15" (r
= 0.268)
for uncoated steel piling.
Table 3 gives theresults of tests on the values of
wall roughness Y carried out in the Franzius Institute
of Hannover.2
Inall cases where the tructure or the backing
behind it isubjectedoontinuous ibration,
6
shouldbe takenas zero.
It
should also be taken as
zero in cases where there may be a tendency for the
structureo moveownwardswith theacking
material (e.g. in n xcavation where the heeting
does notpenetrate oany appreciable depth below
the bottom).
5-
I io*
25'
30
35'
40
OD 5 O
IOD
e
Graph
l
For passive resistance
,
according to Terzaghi, the
angle of wall friction 6 shouldnotbe takengreater
than
3 . When
there
is a
considerable friction between
Table 3
Wall roughness
Material
Wall roughness y
6
14
3 -20
31
17
17
26
27
31
27
294
318
88
20 -14
17
14
214
178
17
6
6
114
timber
0.30
steel I concrete
brick
-
Gravel or sand
(rough) 0 60
(smooth)
0.30
Sand, coarse, dry
moist
0.48
0
50
(rendered)
0.60
0.52
(rough)
Sand, fine, dry
moist
~ ~
oam and clay
Clay, moist
0.56
0-61
0 -15
--
0.35-0.2
(coarse) 0 30
(smooth) 0.20
~ ~~
Silty sands
(rendered)
0.39
0.3
(rough)
0.10
Silty clay
0-30
(coarse) 0 20
(smooth) 0.10
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260
The Structural
Engineer
and
A, =
pa (2 +
pa
.
.
.
.
. (24)
The new auxiliary parameter t is given by
t =
( 1
+ At) t o . . . . . 25)
where
t a = ( p + r )
p - b ) . . . . (26)
and
a2
At
=
. . . .
(27)
(1
+
ab) ( l r )
W
Le t
a = t a n
b - t a n p
p = t a n
J
r - t a n J
t h e n
P ~ ~ = ~ ~ H ' K , , ,
pah'r H Kah
where
Kah'(l L h ~ ah
t
- ( l + A t ) t o
ta=(per ) (p-b)
R e a d :
K:h- f rom th I r N o m o g r a m
A t
-
f r o m N o m o g r a m 3
d , - f r o m N o m o g r a m 4
Coefficient of ActivePressure
K o 8 h
Nomogram
1
the wall and he filling, and 6 increases beyond
the sliding urfaceunder passive resistance annot
any more be assumed to be straight.
0
The expressions for
K
a h and K p h can be written in
the form
l
l
= [JP + p2) JJ2
A
nomographic solution s offeredfor quick evaluation
of K o a h ,K o p h , At and AK. K o a h may be read
from Nomogram
1 ,
K o p h from Nomogram
2,
At from
Nomogram 3 and A, from Nomogram 4. Note
that the factors At and AK are equal to zero
if
a
= 0.
Hence, it may be concluded that
KOah
and
K o p h
are
the coefficientsof active pressure and passive resistance
for a vertical wall, and hen, A , and A, may be
looked upon as correctionactorsccounting for
the inclination of the wall,
a.
Graph
1
gives thevalues of tan
8
for 8 ranging
from 0 to 45 .
The following procedure for evaluating
K a h
and
K ph therefore applies :
1.00
0.90
Oa0
c1
10.70
0 .60 -
0 . 5 0 -
040
0 3 0
0.20
0.10.
0
K e y
L e t
a - t a n a
U- t a n
9
b = t a n p
r =tan
t h e n
P = -FH'K
ph
2 ph
Pph=TH
K ~ , , = ( I + & \ K ; ~
ph
where
t =(I + A t )
to
( p
+r)(p b)
R e a d .
$h f r o m h l rN o m o g r a m
t
. - f r o m N o m o g r a m 3
A , - f r o mN o m o g r a m
4
i
t
Coefficient of PassiveResistance Koph
Nomogram 2
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September,
I9 9
261
Case
I-Vertical Wall
1 .
Evaluate
2. From Nomogram l read off
K a h
= KOah
3.
From Nomogram
2
read off
K p h = K o p h
t = t o = p +
Y ) p - - b )
Case
11-Inclined Wall
1 .
2.
3.
4.
5 .
Note
For he given values of
a , b
and
r ,
read off
At
from Nomogram 3
Evaluate
and then compute
For the given values of p and
a
read off
A
K
from Nomogram
4.
From Nomogram
1
read
off
KOah
Then
t o
= p
+
r ) p - - )
t =
( 1
+ A , ) t o
Ka h =
( 1
-k
A K ) K o a h
From Nomogram
2
read off K o p h
Then
K p h = ( 1 + A K ) K o p h
that because of the sign convention for a and
b the values of
t ,
At and
AUK
or passive resistance
are not the same as for active pressure. I t is apparent
from Figs. 1 and 2 that
+ a
and
b
for active pressure
become -a and -b for passive resistance, and vice
versa. N s o , in general, different values for the wall
roughness ? n a y be taken for active pressure and
passive resistance.
The examples which follow illustrate the application
of the Nomograms.
1
O O l
060
N o m o g r a m
-
for a c t l v e p r c r s u r c
N o m o g r o r n 2 - f o r o s s i v cc s ~ s t a n c c
06 \
10 50 o r s i g n c o n v e n t i o n o r a 5 c e :
-o.40 N o m o g r a m
- N o m o g r a m
I - t o r a c t i v e
2- t o r pass ivc
I ,
e y
p r e s s u r e
r e s i s t o n c c
Correction Factor -- A k
Nomogram 4
Examples
A . Calculations of ActivePressure
Example
1 .
Ground surface sloping 8 = lo",b = tan p = 0.176
Wall vertical a =
O , a = O
Angle of internal friction 0 = 30°, p = tan 0 =
0.577
Angle of wall friction
6
=
20°, r
= tan 6 = 0.364
Wall height H = 30 ft.
Backing : dry y
=
110 lb/ft.3
Surcharge load : none
H - 3 0 t l
For p = -10" read p= +10"
Fig. 3. Example 1
t
= to
=
p + r )
p-b)
=
(0.577+0.364) (0.577-0.176) = 0.377
From Nomogram
1 :
K a h
=
K o a h
=
0.320
Horizontal pressure a t foot of wall :
Total horizontal active thrust :
Total active thrust :
Pah = y H K o h
==
110 X 30 X 0.320 = 1056 lb/ft.'
Pah = 4 H Pall = X 30 X 105615840 Ih.
P a = Pah sec(a+6) = 15840 X 1.064 = 16860
lb.
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262
Example
2
The Strl4ctlrrtrl Engineer
Total horizontal active thrust :
Groundurfaceloping p =
+
5 O ,
b = +
0.087
Wallnclined
a=
-15 , a = - .268
Angle
of
internal friction G= 35", p
=
0.700
Angle of wall friction 6= +12 , Y
=
+ 0.213
Wall height H
=
24 ft.
Backing : moist, above GWL y = 120 lb/ft.3
Surcharge load q
=
240 lb/ft.2
P< 6
7 0 0
For 6
=
-12 read 6 = +la
Fig.
4.
Example 2
to
=
p + ~ ) p-b) = 0.913 X 0.613
=
0.560
From Nomogram
3 :
At = 0.070
t = ( l
+
At) to =
1 -070 X 0.560
=
0-600
From Nomogram 1
:
KOah = 0.251
From Nomogram 4 : A K= - -340
K a h = (1 +
A,)
KOah = 0.660 X 0.251 = 0.166
Equivalent density of backing
:
2q 2 x 240
H
24
y' = y + - = 1204- 140 lb/ft.'
H,= I f t .
1
P a h = 8
y H2
K a h = 4
X 140 X 24'
X
0.166
=67001b.
Total active thrust :
P a
=
P a h sec(a+6)
=
6700
X 1
a0014
=
6710 lb.
Example 3
0,
a = O
p = + 5 ,
b
=
+ 0.087
=
35 ,
p =
0.700
6
= +17 ,
Y
=
+ 0-306
Backing : density bove GWL
yl
= Y m = 110 lb/ft.3
density below GWT,
y
y b =
70 ib/ftB3
TWL
taken as equal to ground water level.
to= l
so06 X 0-613
=
0.617
From Nomogram 1
:
K a h =
K o a h
=
0.249
Horizontal active pressures:
atdepth 0 0 PO= 0
at depth 11-1
l
p1= y1 H1 K a h
=
110 X l 1 X 0.249
=
301 lb/ft.2
at de pth 21-21 pz= p1 + yz H2
K a h
=
301
+
70 X 10 X 0-249
= 301 + 174
=
475 lb/ft.2
Horizontal active pressure thrusts :
p l =
1 P1
-
x
301
-
1656 lb
2 2
P a h
= P1
+
P2 =
1656 + 3880
=
5536 lb.
Total active pressure thrust :
P a
= Pah sec 6
=
5536
X 1
-0457
= 5790
lb.
Fig.
5.
Example
3
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September, 1959
263
H,= O f t .
H,=lOft.
surcharge load =Q =ZOO lb/ft2
Fig.
6. Example 4
Example
4
friction, varies
:
from 0 ft.-l0 ft. a
=
+ 30°, a
= +
0.577
Wall inclination, wall friction, and angle of internal
25 = 35 ,
p
=
0.700
6
=
+
20 ,
Y
=
+
0.364
from 10 t.-20 ft. a =
+
15 , a =
+
0.268
0
=
30
p. = 0.577
6 =
+ 17 ,
Y
= +
0.306
from 20t.-30 ft.
a =
0,
a =
0
= 25 ,
p =
0.466
p.
=
+ 15 ,
Y
=
+ 0.268
Angle of surcharge p = + lo , b = + 0 176
Backing : density constant y
=
100 lb/ft.3
Surcharge oad q = 200 lb/ft.2
H1
=
H2 = H3
=
10 ft.
(a) coefficients of active earth pressurehorizontal
component) :
(1) from 0-10
ft.
to = 1 -064 X
0
-524 = 0.558
t =
1 -375 X 0.558
=
0.768
from Nomogram
3
:
A t
=
0.375
from Nomogram 1
:
KOah = 0.228
from Nomogram 4
:
A K
= +
0 -97
K a h
=
1 -97 X
0.228
=
0 -450
to= 0.883
X
0.401 = 0.354
from Nomogram 3
:
A ,
=
0.074
t = 1 0074 X 0-354 = 0-380
from Nomogram 1 :
K o a h =
0.320
from Nomogram 4 : A K=
+
0
-333
(2) from 10 ft.-20 ft .
Kah = 1
-333
X 0.320 =
0
-426
t = to
=
0 -734 X 0.290 = 0.213
(3)
from 20 ft.-30
ft.
from
Nomogram 1
:
Ka h
=
KOah
=
0.410
(b) horizontal ear th pressures rom urchargeoad:
pi*
=
q
K a h
= 200
X
0.450 = 90 lb/ft.2
p2*
= 200 X 0.426
=
85 lb/ft.2
p3* = 200 X 0.410
=
82 lb/ft.2
1
(c)horizontal ear th pressures from
backing
material:
at depth 10-10.
p1 = H1 K a h
= 100 X 10 X 0 e450 = 450 lb/ft.2
p'l
= y H1 Kah
2
=
100 X 10
X 0
-426
=
426 lb/ft.2
p2
=
100
X
20
X
0.426
=
852
lb/ft.2
p'2 = 100 X 20
X
0.410
= 820
lb/ft.Z
p3
= 100
X
30 X 0.410 = 1230 lb/ft.2
at depth 20-20
at depth 30-30
(d) horizontal active thrusts :
(i) from surcharge load
PI* = pi* H1
=
90 X 10
=
900 lb.
P2* = p2*H2 = 85 X 10
=
850 lb.
P3* = p3*H3 = 82 X 10 = 820 lb.
(ii)
from backing material
-
+
450 x 10 = 2250 lb.
2
26 +
852
x 10 = 6390 lb.
2
+
x 10
=
10250 lb.
2
(iii) resultants
ah
P1
= P1
+
P*l
P2
= P2
+ P*2
P3
=
P3 + P*3
=
2250 + 900 = 3150 lb.
ah
=
6390
+
850 = 7240 lb.
ah
= 10250
+
820
=
11070 lb.
(iv) total
P a h
=
3150
+
7240
+
11070
=
21460 lb.
(e) vertical active thru sts :
P1 3150 X 1.1917 = 3755 lb.
P2 = 7240
X
0.6249 = 4525 lb.
P3 = 11070
X
0.268 = 2970 lb.
av
av
av
Pa, =
3755 + 4525 + 2970
=
11250 lb.
f ) total resultant active thrust :
p a =
J R L
=
$14602
+
112502
=
24200 lb.
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264
The Structural
Engrnoer
B
C a l m l a t i o m
of
Passive Resis tance.
Exa mp le 5
a
=
0,
a = 0
p
=-
0,
h
=- .087
Z =
25 ,
p
= 0.466
6 =
+
8 ,
= + 0.141
Q = 0
y = 110 lb/ft.3
H
=
30
ft.
I \
.l f tt
Fig.
7. Example 5
t = t o
= (p+?,) p-b )
= (0.466
+
0-141)(0.466 k 0.087
= 0.336
From Nomogram 2
:
K p h
=
K o p h = 3-65
Horizontal pressure a t foot of wall :
p p h
y H
K p h
= 110
X
30
X
3-65
=
12050 lhlft.2
Total horizontal passive resistance :
P p h =
4
H p p h
= $
X 30 X 12050
=
180750
lb.
Total passive resistance :
P,
= P p h sec(sct6)
= 180750 X l
*0098=
182500 ib.
Exa mp le 6
=
-
oo,
Q = -
.176
p
= 0, = 0
6
= 0, -- 0
= l oo 1bp.3
q
= 200 lbIft.2
= 30°, p == 0.577
H
=
20 ft.
q=200
b. / f t 2
p,=5 500
b . / f t z
Fig. 8.
Example
6
to =
0 -5 7 7
X
0.577 =
0-334
from Nomogram 3 : A t
=
0.031
from Nomogram
4 : A K
=- 193
from Xomogram 2 : K o p h = 3.10
t = 1 -031 X 0.334 = 0 -344
K p h
=
(1 + AK) K o p h
=
0 807
X
3-10
=
2.50
resultant earth pressures :
at depth 0-0
Po =
c1
K p h
= 200
x 2
-50 =
500 Ib / f t . ?
at
depth
20-20
p1
=
P o + H
K p h
=: 500 + 1 0 0 x 20 A 2 -50
=
5500
Il>/ft.z
tota l horizontal component
o f
passive resistance :
H
Total passive resistance
P p
=
60000 X 1.0154 = 60900 lb.
References
1.
Earth Retaining tructures,
Civil Engineering
Code o f
2.
Taschenbuch f.ir Bauingenieure, dited by
Prof.
Dr.
1
Practice
No.
2, Inst.Struct. Engs.
(1957)
Schleicher,Springer-Verlag.
Book
Review
Structural Mechanics
by W.
Morgan and D. T. Williams
(London
:
Pitman, 1958) in.
x
54 in., 427 plus
viipp. Price 301-
Building, architectural and surveyor students have
commonly
a
much slendererknowledge and under-
standing
of
mathematicshan is the casewith
engineering tudents, ndts for them hat the
authors
have
writtenhis approach book-an
approachohelementaryheory
of
structures
using theminimum
of
mathematicsand employing
graphicalmethodswherever possible. The
first
six
chapters deal withorces, their moments
and
resultants,
followed by woon forces in simple rames. After
thi s come hapterson tress, train nd lasticity,
bending moment and shearing force, and the properties
of
sections. Simple beam design is next considered as
well
as
flitched beams an d simple reinforced-concrete
beams.
T w o
chaptersrcevotedo deflection
problems, and matters deal t with finally are
:
axially-
loadedcolumns, iveted and boltedconnections, the
addition of direct an d bendingtresses, andhe
stability of darns andearth-retaining walls. Quite
a
few fullyworked-outnumericalexamplesare given
in the text throughout
the
book and these are supple-
mented y ther exercises to be tackled by the
student.
The
book is clearly llustratedand hould
prove qllite suitable for its purpose.
L. A. B.