active and passive pressure in cohesive soils

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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



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



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



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



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.



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



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.



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.

active and passive pressure in cohesive soils

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