Active and Passive Pressure in Cohesive Soils

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7/21/2019 Active and Passive Pressure in Cohesive Soils

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September,

Nomograms

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

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.

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

the wall, the expressions for

the total active thrust P a and the total passive resis-

can be written in the form

wall may be subjected are

. . . . .

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

unit weight of soil

coefficient of active pressure

K , coefficient of passive resistance

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

. . . . . 4)

. . (5)

and the vertical pressure components

H ZK,, . . . . . (6)

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

angle of surcharge, between the upper surface

the coefficients K a h and K p h areiveny the

expression

of soil and the horizontal

+ 8) in(0

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+?

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

The Structural Engineer

resistance. The sign convention or a, /3 and S used

in formula

s shown in Figs(

and (2a) espectively

Note hat or hesameconditions of the wall and

backing, the signs

for active pressure and

passiveesistancerenterchangeablenormula

and +p for active pressure become

--M. and -p for passive resistance.

The coefficients

can be expressed in terms

of th e respective coefficients K h for horizontal thrust.

= K a h

8) . . . .

K B V =

K a h tan(a + S) . .

. (12)

K, = K p h

sec( + 6 )

K,,= K p h tan(a

Similarly, th e expressions or

and P, in erms of

Ph become

Pah sec(a + S) . .

. (15)

Pa,= P a h t an (a + S) . . .

= P p h

. (17)

Pp,= P p h tan(

. (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:

coefficient of internal friction

Y = tan S wall roughness

= ta n a inclination of wall to the vertical

b = tan p inclination

the surcharge to

the horizon tal

This gives

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

limeston?s and

sandatones .

Broker?rick ..

Ashes . . .

draineg above

Densit when

Ground Water

105-130

110-135

100-130

110-140

60-430

70-1 10

submerged below

Density when

Ground Water

Level Yb

70-100

The sign convention or a,

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

as recommended

by TheCode of Practice Earth Retaining Structu res

are given in Table 1 .

The unit weight of the soil varies with the amount

moisture content, ndwithhe position of the

groundwater able. In he case of a dry backing

thedensity is

Where thebacking is moist but

is above ground water level

the moist density

should besubstituted for

inequation

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

lb/ft3).

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.

. 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

STANCE

September,

Table 2

Typical values

and p for cohesionless materials

35 -45

35 -40

30 -35

30 --35

35 -45

Materials

______

Sandy gravel .

Compact sand . .

Loose sand. .

Shale filling . .

Rock filling

Ashes or broken brick

. . . .

0.700-1 .OOO

0.700-0.839

0.577-0.700

0.700--1.000

0.700-1.000

oractive pressure it seems easonable to assume

a value of

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

0.364) for walls of concrete or brick, as

= 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,

shouldbe takenas zero.

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).

OD 5 O

For passive resistance

according to Terzaghi, the

angle of wall friction 6 shouldnotbe takengreater

3 . When

considerable friction between

Table 3

Wall roughness

Material

Wall roughness y

20 -14

timber

steel I concrete

Gravel or sand

(rough) 0 60

(smooth)

Sand, coarse, dry

(rendered)

(rough)

Sand, fine, dry

oam and clay

Clay, moist

0.35-0.2

(coarse) 0 30

(smooth) 0.20

Silty sands

(rendered)

(rough)

Silty clay

(coarse) 0 20

(smooth) 0.10

The Structural

Engineer

pa (2 +

. (24)

The new auxiliary parameter t is given by

+ At) t o . . . . . 25)

t a = ( p + r )

p - b ) . . . . (26)

. . . .

ab) ( l r )

a = t a n

b - t a n p

p = t a n

r - t a n J

t h e n

P ~ ~ = ~ ~ H ' K , , ,

pah'r H Kah

Kah'(l L h ~ ah

- ( 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

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

the wall and he filling, and 6 increases beyond

the sliding urfaceunder passive resistance annot

any more be assumed to be straight.

The expressions for

a h and K p h can be written in

the form

= [JP + p2) JJ2

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

K o p h from Nomogram

At from

Nomogram 3 and A, from Nomogram 4. Note

that the factors At and AK are equal to zero

Hence, it may be concluded that

K o p h

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,

gives thevalues of tan

for 8 ranging

from 0 to 45 .

The following procedure for evaluating

K ph therefore applies :

0 .60 -

0 . 5 0 -

a - t a n a

U- t a n

b = t a n p

r =tan

t h e n

P = -FH'K

Pph=TH

K ~ , , = ( I + & \ K ; ~

t =(I + A t )

+r)(p b)

R e a d .

$h f r o m h l rN o m o g r a m

. - 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

Coefficient of PassiveResistance Koph

Nomogram 2

September,

I-Vertical Wall

Evaluate

2. From Nomogram l read off

= KOah

From Nomogram

read off

K p h = K o p h

t = t o = p +

Y ) p - - b )

11-Inclined Wall

For he given values of

read off

from Nomogram 3

Evaluate

and then compute

For the given values of p and

read off

from Nomogram

From Nomogram

r ) p - - )

+ A , ) t o

Ka h =

A K ) K o a h

From Nomogram

read off K o p h

K p h = ( 1 + A K ) K o p h

that because of the sign convention for a and

b the values of

At and

or passive resistance

are not the same as for active pressure. I t is apparent

from Figs. 1 and 2 that

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.

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

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

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

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 =

Angle of wall friction

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

p + r )

(0.577+0.364) (0.577-0.176) = 0.377

From Nomogram

K o a h

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

Example

The Strl4ctlrrtrl Engineer

Total horizontal active thrust :

Groundurfaceloping p =

Wallnclined

-15 , a = - .268

internal friction G= 35", p

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

-12 read 6 = +la

Example 2

p + ~ ) p-b) = 0.913 X 0.613

From Nomogram

At = 0.070

t = ( l

At) to =

1 -070 X 0.560

From Nomogram 1

KOah = 0.251

From Nomogram 4 : A K= - -340

K a h = (1 +

KOah = 0.660 X 0.251 = 0.166

Equivalent density of backing

2q 2 x 240

y' = y + - = 1204- 140 lb/ft.'

H,= I f t .

P a h = 8

K a h = 4

X 140 X 24'

=67001b.

Total active thrust :

P a h sec(a+6)

6710 lb.

Example 3

p = + 5 ,

+ 0.087

= +17 ,

+ 0-306

Backing : density bove GWL

= Y m = 110 lb/ft.3

density below GWT,

70 ib/ftB3

taken as equal to ground water level.

so06 X 0-613

From Nomogram 1

K a h =

K o a h

Horizontal active pressures:

atdepth 0 0 PO= 0

at depth 11-1

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

70 X 10 X 0-249

= 301 + 174

475 lb/ft.2

Horizontal active pressure thrusts :

1656 lb

1656 + 3880

5536 lb.

Total active pressure thrust :

= Pah sec 6

= 5790

Example

September, 1959

H,= O f t .

H,=lOft.

surcharge load =Q =ZOO lb/ft2

6. Example 4

Example

friction, varies

from 0 ft.-l0 ft. a

+ 30°, a

Wall inclination, wall friction, and angle of internal

25 = 35 ,

from 10 t.-20 ft. a =

15 , a =

p. = 0.577

+ 17 ,

from 20t.-30 ft.

= 25 ,

+ 15 ,

+ 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

H2 = H3

10 ft.

(a) coefficients of active earth pressurehorizontal

component) :

(1) from 0-10

to = 1 -064 X

-524 = 0.558

1 -375 X 0.558

from Nomogram

from Nomogram 1

KOah = 0.228

from Nomogram 4

1 -97 X

0 -450

to= 0.883

0.401 = 0.354

from Nomogram 3

t = 1 0074 X 0-354 = 0-380

from Nomogram 1 :

K o a h =

from Nomogram 4 : A K=

(2) from 10 ft.-20 ft .

Kah = 1

X 0.320 =

t = to

0 -734 X 0.290 = 0.213

from 20 ft.-30

Nomogram 1

(b) horizontal ear th pressures rom urchargeoad:

0.450 = 90 lb/ft.2

= 200 X 0.426

85 lb/ft.2

p3* = 200 X 0.410

82 lb/ft.2

(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

= y H1 Kah

100 X 10

426 lb/ft.2

lb/ft.2

p'2 = 100 X 20

lb/ft.Z

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.

from backing material

450 x 10 = 2250 lb.

x 10 = 6390 lb.

10250 lb.

(iii) resultants

P3 + P*3

2250 + 900 = 3150 lb.

850 = 7240 lb.

= 10250

11070 lb.

(iv) total

21460 lb.

(e) vertical active thru sts :

P1 3150 X 1.1917 = 3755 lb.

P2 = 7240

0.6249 = 4525 lb.

P3 = 11070

0.268 = 2970 lb.

3755 + 4525 + 2970

11250 lb.

f ) total resultant active thrust :

$14602

112502

24200 lb.

The Structural

Engrnoer

C a l m l a t i o m

Passive Resis tance.

Exa mp le 5

=- .087

= 0.466

= + 0.141

y = 110 lb/ft.3

.l f tt

7. Example 5

t = t o

= (p+?,) p-b )

= (0.466

0-141)(0.466 k 0.087

= 0.336

From Nomogram 2

K o p h = 3-65

Horizontal pressure a t foot of wall :

12050 lhlft.2

Total horizontal passive resistance :

P p h =

H p p h

X 30 X 12050

180750

Total passive resistance :

= P p h sec(sct6)

= 180750 X l

*0098=

182500 ib.

Exa mp le 6

= 0, = 0

= 0, -- 0

= l oo 1bp.3

= 200 lbIft.2

= 30°, p == 0.577

20 ft.

b. / f t 2

p,=5 500

b . / f t z

Fig. 8.

Example

0 -5 7 7

0.577 =

from Nomogram 3 : A t

from Nomogram

4 : A K

=- 193

from Xomogram 2 : K o p h = 3.10

t = 1 -031 X 0.334 = 0 -344

(1 + AK) K o p h

resultant earth pressures :

at depth 0-0

500 Ib / f t . ?

P o + H

=: 500 + 1 0 0 x 20 A 2 -50

Il>/ft.z

tota l horizontal component

passive resistance :

Total passive resistance

60000 X 1.0154 = 60900 lb.

References

Earth Retaining tructures,

Civil Engineering

Code o f

Taschenbuch f.ir Bauingenieure, dited by

Practice

2, Inst.Struct. Engs.

(1957)

Schleicher,Springer-Verlag.

Review

Structural Mechanics

Morgan and D. T. Williams

(London

Pitman, 1958) in.

54 in., 427 plus

viipp. Price 301-

Building, architectural and surveyor students have

commonly

much slendererknowledge and under-

standing

mathematicshan is the casewith

engineering tudents, ndts for them hat the

authors

writtenhis approach book-an

approachohelementaryheory

structures

using theminimum

mathematicsand employing

graphicalmethodswherever possible. The

chapters deal withorces, their moments

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

sections. Simple beam design is next considered as

flitched beams an d simple reinforced-concrete

beams.

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

few fullyworked-outnumericalexamplesare given

in the text throughout

book and these are supple-

mented y ther exercises to be tackled by the

student.

book is clearly llustratedand hould

prove qllite suitable for its purpose.

L. A. B.

Active and Passive Pressure in Cohesive Soils

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