ch-07

Introduction

Vertical or near-vertical slopes of soil are supported by retaining walls, cantileversheet-pile walls, sheet-pile bulkheads, braced cuts, and other, similar structures. Theproper design of those structures requires an estimation of lateral earth pressure,which is a function of several factors, such as (a) the type and amount of wall movement, (b) the shear strength parameters of the soil, (c) the unit weight of the soil,and (d) the drainage conditions in the backfill. Figure 7.1 sho'ws a retaining wall ofheight H. For similar types of backfill,

a. The wall may be restrained from moving (Figure 7.1a). The lateral earth pressure on the wall at any depth is called the at-rest earth pressure.

b. The wall may tilt away from the soil that is retained (Figure 7.1b). With sufficient wall tilt, a triangular soil wedge behind the wall will fail. The lateral pressure for this condition is referred to as active earth pressure.

c. The wall may be pushed into the soil that is retained (Figure 7.1c). With sufficient wall n10vement, a soil wedge will fail. The lateral pressure for this condition is referred to as passive earth pressure.

Height == H

tth !,\,:,:\~~",:,; :.i;::;').:':\y".i\?·:-' :.'.<).:.~-:;, l7"h(passive)..':,: /

~:::::.I S.l:':." 01:::: I- failure:./:.:,: wedge.:, ..:., .....:, :.::r:···I ~:'

\ \('/'.\.:·.i\(·;:::::···..:.;}/\ :"':. ,\ ~ l7"h(active)

\ ~;'::Soil\ :' failure\ ~wedge

Height = H' t!?::'\ ::;::

.::_....

:.:.:..~: ::~::·,;·if··:·; ::;~t~:~

Height = H !f!~

(a) (b) (c)

Figure 7.1 Nature of lateral earth pressure on a retaining wall

293

(7.1)

(7.2)

(~) = 0.001 for loosea sand to 0.04

for soft clay

(Th(passive)

(T~ == q + yz

(Th(active)

MlH

At any depth z below the ground surface, the vertical subsurface stress is

---~---------

IIIIII

(~) """ 0.01 for loose :P sand to 0.05 :

for soft clay IIIIIIII

~(~)p

Figure 7.2 Nature of variation of lateral earth pressure at a certain depth

s == c' + a' tan <p'

Lateral Earth Pressure at Rest

Figure 7.2 shows the nature of variation of the lateral pressure, (Fir, at a certain depthof the wall with the magnitude of wall movement.

In the sections that follow, we will discuss various relationships to determine theat-rest, active, and passive pressures on a retaining wall. It is assumed that the readerhas studied lateral earth pressure in the past, so this chapter will serve as a review.

where c' == cohesioncP' == effective angle of friction(F' == effective normal stress

Consider a vertical wall of height H, as shown in Figure 7.3, retaining a soil having aunit weight of y. A uniformly distributed load, q/unit area, is also applied at theground surface. The shear strength of the soil is

If the wall is at rest and is not allowed to move at all, either away from the soil massor into the soil mass (i.e., there is zero horizontal strain), the lateral pressure at adepth z is

where u = pore water pressureK o == coefficient of at-rest earth pressure

294 Chapter 7 Lateral Earth Pressure

7.2 Lateral Earth Pressure at Rest 295

I~~--Ko(q + I'H)

(b)

yc'1/

(a)

...'":::: -.- ..

H

Figure 7.3 At-rest earth pressure

For normally consolidated soil, the relation for K o (laky, 1944) is

(7.3)

Equation (7.3) is an empirical approximation.For normally consolidated clays, the coefficient of earth pressure at rest can be

approximated (Brooker and Ireland, 1965) as

(7.4)

where 4>' == drained peak friction angle

On the basis of Brooker and Ireland's experimental results, the value of K o fornormally consolidated clays may be approximately correlated with the plasticity index (PI) via the relationships

K o ==: 0.4 + 0.007 (PI) (for PI between 0 and 40) (7.5)

and

K o == 0.64 + 0.001 (PI) (for PI bet',veen 40 and 80) (7.6)

For overcol1solidated clays,

(7.7)


First unloading

__---"7'A

B------...__-

First reloading

o

Virgin loading

;,'1·.""····,,·,>~-'-,':I';

'r~

0-'o

Figure 7.4 Stress history for soil under K o condition

where OCR == overconsolidation ratio ·~.'I········,•;'1-·~7_ •

Mayne and Kulhawy (1982) analyzed the results of 171 different laboratory-tested soils and proposed the following general empirical relationship to estimate Ithe magnitude of K o for sand and clay:

III Figure 7.4, OCRmax is the value of OCR at point B.With a properly selected value of the at-rest earth pressure coefficient, Eq. (7.2)

can be used to determine the variation of lateral earth pressure with depth z. Figure 7.3b shows the variation of (Th with depth for the wall depicted in Figure 7.3a.Note that if the surcharge q == 0 and the pore water pressure u == 0, the pressurediagram will be a triangle. The total force, Po, per unit length of the wall given inFigure 7.3a can now be obtained from the area of the pressure diagram given in Figure 7.3b and is

K == (1 - sin ~')[ OCR + ~(1 - OCR )Jo 0/ OCR~a~sin4» 4 OCRmax

In this equation, OCR == present overconsolidation ratioOCRmax == maximllm overconsolidation ratio

Po == Pi + P2 == qKoH + ~yHZKo

where Pi == area of rectangle 1Pz == area of triangle 2

(7.8)

(7.9)

,·If-,~.

".\.'·1'.','.'•.'~,

II

II

J.

7.2 Lateral Earth Pressure at Rest 297

The location of the line of action of the resultant force, ~)' can be obtained by takingthe moment about the bottom of the wall. Thus,

-

z==~)

(7.10)

If the water table is located at a depth z < H., the at-rest pressure diagralTIshown in Figure 7.3b will have to be somewhat modified., as shown in Figure 7.5. Ifthe effective unit weight of soil below the water table equals y' (i.e., Ys,1t - YIl')' then

and

at z == H 2 , (Jh == Ko(J~ == Ko(q + yH I + y' H2 )

Note that in the preceding equations, u;) and (T/l are effective vertical and horizontalpressures, respectively. Determining the total pressure distribution on the wall requires adding the hydrostatic pressure u, which is zero from z == 0 to z == HI and isHzyw at z == Hz. The variation of a;1 and u with depth is shown in Figure 7.5b. Hence,the total force per unit length of the wall can be determined from the area of thepressure diagram. Specifically,

Po == A 1 + A z + A 3 + A 4 + As

where A == area of the pressure diagram

So

(7.11)

®

--+-~-u

®

CD

I~ ( + 'YH + 'Y 'H)l-l'wH2--Jo q 1 2

(b)

Ysatc'¢t ~

... ¢/.

Watertable

(a)

.' II]

H

z

Figure 7.5 At-rest earth pressure with water table located at a depth z < H

Wall

b

c

a

z

movement ....~......-to left

----.J8X~

\\\\

(b)

~~~~~~- Rotation of wall~:-:<)/(:3:,-,-:':- f-:-::;-: about this point

(a)

Shearstress

Effective---__A- -----I__.....&.-_-----I a..-... normal

IT~ ITh KolT~ IT~ stress

H

The lateral earth pressure described in Section 7.2 involves walls that do not yield atall. However, if a wall tends to move away from the soil a distance ~x, as shown inFigure 7.6a, the soil pressure on the wall ~t any depth will decrease. For a wall that isfrictionless, the horizontal stress, Uln at depth z will equal Kou~(~KoYz) when Llx iszero. However, with Llx > 0, (Th will be less than Kou~.

The Mohr's circles corresponding to wall displacements of Llx == 0 and ~x > 0are shown as circles a and b, respectively, in Figure 7.6b. If the displacement of the wall,

Figure 7.6 Rankine active pressure

7.3 Rankine Active Earth Pressure

Active Pressure


.,..,. -

7.3 Rankine Active Earth Pressure 299

,.....-.. --(T~Ka-------·1 ~ -2c'~ J.(c)

II Pa

Eq. (7.16)

Figure 7.'6 (continued)

~x, continues to increase, the corresponding Mohr's circle eventually will just touch theMohr-Coulomb failllre envelope defined by the equation

s = c' + a' tan cp'

This circle, marked c in the figure, represents the failure condition in the soil mass;the horizontal stress then equals (T~, referred to as the Rankine active pressure. Theslip lines (failure planes) in the soil mass will then make angles of:± (45 + </J'/2) withthe horizontal, as shown in Figure 7.6a.

Equation (1.74) relates the principal stresses for a Mohr's circle that touchesthe Mohr-Coulomb failure envelope:

0"1 = 0"3 tan2( 45 + ~') + 2e' tan(45 + ~')

For the Mohr's circle c in Figure 7.6b,

Major principal stress, (T1 == (T~

and

Thus,

Minor principal stress, (T3 == (T~

~

O"~ = O"~ tan2( 45 + ~') + 2e' tan(45 + ~')(T~

(T~ == ------

tan2( 45 + ~')

2c'

(7.12)

(7.13)

(7.14)

(7.15)

(7.16)

Pa = iN(T~dz = iNyzKadz - iN2c'YiZadz '

== ~yH 2Ka - 2c'HYifa

or

where Ka == tan2 (45 - cP'j2) == Rankine active pressure coefficient

The variation of the active pressure with depth for the wall shown in Figure 7.6aisgiveninFigure7.6c.Notethat(T~== Oatz == Oand(T~ == yH atz == H.Thepressuredistribution shows that at z == 0 the active pressure equals - 2c'~, indicating atensile stress that decreases with depth and becomes zero at a depth z == zc' or

yzcKa - 2c'~ == 0

and

After the tensile crack appears, the force per unit length on the wall will becaused only by the pressure distribution between depths z == Zc and z == H, asshown by the hatched area in Figure 7.6c. This force may be expressed as

The depth Zc is usually referred to as the depth of tensile crack, because thetensile stress in the soil will eventually cause a crack along the soil-wall interface.Thus, the total Rankine active force per unit length of the wall before the tensilecrack occurs is

or

However, it is important to realize that the active earth pressure condition will bereached only if the wall is allowed to "yield" sufficiently. The necessary an10unt ofoutward displacement of the wall is about O.001H to O.004H for granular soil backfills and about O.01H to O.04H for cohesive soil backfills.


7.3 Rankine Active Earth Pressure 301

Note further that if the total stress shear strength parameters (c, ¢) were used,an equation similar to Eq. (7.12) could have been derived, namely,

(]" == (]" tan2(45 - ¢) -2c tan(45 - ¢)a () 2 2

7.4 Rankine Active Earth Pressure for Inclined Backfill 303

~~

y = 102 Ib/ft3

¢J{ == 30°, "10 ft c{ =0

z

CDT Water table 265.2

3401b/ft2

1-

Ysat == 1211b/ft3

·'.10 ft ¢2 = 36° ® +

1ci = 0

® CD

. o· 417.6 624Ib/ft2 Ib/ft2

(b)

Rankine Active Earth Pressure for Inclined Backfill

Granular Soil

If the backfill of a frictionless retaining wall is a granular soil (c' == 0) and rises at anangle a with respect to the horizontal (see Figure 7.8), the- active earth pressure coefficient may be expressed in the form

(7.17)

where <p' == angle of friction of soil

At any depth z, the Rankine active pressure may be expressed as

(7.18)

Figure 7.8 Notations for activepressure-Eqs. (7.17), (7.18), (7.19)

(7.19)

(7.20)(T~ == yzKa == yzK~ cos a

Also, the total force per unit length of the wall is

Note that, in this case, the direction of the resultant force Pa is inclined at an angle awith the horizontal and intersects the wall at a distance H /3 from the base of the wall.Table 7.1 presents the values of K a (active earth pressure) for varios values of a and 4>'.

c' -l/J' Soil

The preceding analysis can be extended to the case of an inclined backfill with ac'-c/J' soil. The details of the mathematical derivation are given by Mazindrani andGanjali (1997). As in Eq. (7.18), for this case

where

1 2cos2

a + 2( ~~ )cos cP' sin cP'K' == ---~ -1

a cos2

cP' [ 4cosZO'(COSZ 0'- COSZ4>') + 4(~~YCOSZ4>' + 8(~~ )COSZa sin 4>' cos 4>' ]


(7.21)

Some values of K~ are given in Table 7.2. For a problem of this type, the depth oftensile crack is given as

2c'z ==

c y1 + sin 4>'1 - sin <p'

(7.22)

7.4 Rankine Active Earth Pressure for Inclined Backfill 305

Table 7.1 Active Earth Pressure Coefficient K a [from Eq. (7.17)]

1J'(deg) ~

J.-a(deg) 28 30 32 34 36 38 40

0 0.361 0.333 0.307 0.283 0.260 0.238 0.2175 0.366 0.337 0.311 0.286 0.262 0.240 0.219

10 0.380 0.350 0.321 0.294 0.270 0.246 0.22515 0.409 0.373 0.341 0.311 0.283 0.258 0.23520 0.461 0.414 0.374 0.338 0.306 0.277 0.25025 0.573 0.494 0.434 0.385 0.343 0.307 0.275

Table 7.2 Values of K~

c'-

yz

ep' (deg) ll' (deg) 0.025 0.05 0.1 0.5

15 0 0.550 0.512 0.435 -0.1795 0.566 0.525 0.445 -0.184

10 0.621 0.571 0.477 -0.18615 0.776 0.683 0.546 -0.196

20 0 0.455 0.420 0.350 -0.2105 0.465 0.429 0.357 -0.212

10 0.497 0.456 0.377 -0.21815 0.567 0.514 0.417 -0.229

25 0 0.374 0.342 0.278 -0.2315 0.381 0.348 0.283 -0.233

10 0.402 0.366 0.296 -0.23915 0.443 0.401 0.321 -0.250

30 0 0.305 0.276 0.218 -0.2445 0.309 0.280 0.221 -0.246

10 0.323 0.292 0.230 -0.25215 0.350 0.315 0.246 -0.263


Figure 7.9 Calculation of Rankineactive force, c'-<// soil

i2.14 ill

L_1

5.36m

Coulomb's Active Earth Pressure

The Rankine active earth pressure calculations discussed in the preceding sectionswere based on the assumption that the wall is frictionless. In 1776, Coulomb proposed a theory for calculating the lateral earth pressure on a retaining wall withgranular soil backfill. This theory takes wall friction into consideration.

To apply Coulomb's active earth pressure theory, let us consider a retainingwall with its back face inclined at an angle f3 with the horizontal, as shown in Figure 7.10a. The backfill is a granular soil that slopes at an angle a with the horizontal,Also, let 8 be the angle of friction between the soil and the wall (i.e., the angle ofwall friction).

7.5 Coulomb's Active Earth Pressure 307

w

{3-o

(b)

R

y

cP'c' = 0

------:;::::;;~~

/~ I ~"'./" I I 'T'.

, , r "-, 1 II I II I II I I

Activeforce

Pa(max)

H

:.:: ':: :''':~:: .~.~\: ~!/~:.~; ~:'~/:'::.: \;~7(?:ry/j~~.'..l} ~:··:i ~;:~ ~ ~.'

(a)

Wall movementaway from

... soil

Figure 7. 10 Coulomb's active pressure

Under active pressure, the wall will move away from the soil mass (to the leftin the figure). Coulomb assumed that, in such a case, the failure surface in the soilmass would be a plane (e.g., BC1 , BC2 , ..• ). So, to find the active force, consider apossible soil failure wedge ABC1• The forces acting on this wedge (per unit length atright angles to the cross section shown) are as follows:

1. The weight of the wedge, fV:2. The resultant, R, of the normal and resisting shear forces along the surface, Bel.

The force R will be inclined at an angle ¢' to the normal drawn to BC1.

3. The active force per unit length. of the wall, Pa, which will be inclined at an angle8 to the normal drawn to the back face of the walL

For equilibrium purposes, a force triangle can be drawn, as shown in Figure 7.1Gb.Note that 01 is the angle that Bel makes with the horizontaL Because the magnitudeof~ as well as the directions of all three forces, are known, the value of Pa can now bedetermined. Similarly, the active forces of other trial wedges, such as ABCz, ABC3, ••• ,

can be determined. The maximum value of Pa thus determined is Coulomb's activeforce (see top part of Figure 7.10), which may be expressed as

(7.23)


Table 7.3 Values of K a [Eq. (7.24)] for f3 == 90 0 and a == 0°

() (deg)

cjJ' (deg) 0 5 10 15 20 25

28 0.3610 0.3448 0.3330 0.3251 0.3203 0.318630 0.3333 0.3189 0.3085 0.3014 0.2973 0.295632 0.3073 0.2945 0.2853 0.2791 0.2755 0.274534 0.2827 0.2714 0.2633 0.2579 0.2549 0.254236 0.2596 0.2497 0.2426 0.2379 0.2354 0.235038 0.2379 0.2292 0.2230 0.2190 0.2169 0.216740 0.2174 0.2098 0.2045 0.2011 0.1994 0.199542 0.1982 0.1916 0.1870 0.1841 0.1828 0.1831

where

(7.24)

and H == height of the wall

The values of the active earth pressure coefficient, K a , for a vertical retainingwall (f3 == 90°) with horizontal backfill (a == 0°) are given in Table 7.3. Note that theline of action of the resultant force (Pa ) will act at a distance H /3 above the base ofthe wall and will be inclined at an angle 0 to the normal drawn to the back of the wall.

In the actual design of retaining walls, the value of the wall friction angle aisassumed to be between cp'/2 and ~4>'. The active earth pressure coefficients for various values of cP', a, and f3 with 8 == ~cP' and ~4>' are respectively given in Tables 7.4and 7.5. These coefficients are very useful design considerations.

J

Active Pressure for Wall Rotation about the Top:Braced Cut

In the preceding sections, we have seen that a retaining wall rotates about its bottom.(See Figure 7.11a.) With sufficient yielding of the wall, the lateral earth pressure isapproximately equal to that obtained by Rankine's theory or Coulomb's theory. Incontrast to retaining walls, braced cuts show a different type of wall yielding. (See Figure 7.11b.) In this case, deformation of the wall gradually increases with the depth ofexcavation. The variation of the amount of deformation depends on several factors,such as the type of soil, the depth of excavation, and the workmanship involved. However, with very little wall yielding at the top of the cut, the lateral earth pressure will beclose to the at-rest pressure. At the bottom of the wall, with a much larger degree of

1

7.6 Active Pressure for Wall Rotation about the Top: Braced Cut 309

Table 7.4 Values of K a [from Eg. (7.24)] for 0 == ~ q/

f3 (deg)

a (deg) cP' (deg) 90 85 80 75 70 65

0 28 0.3213 0.3588 0.4007 0.4481 0.5026 0.566229 0.3091 0.3467 0.3886 0.4362 0.4908 0.554730 0.2973 0.3349 0.3769 0.4245 0.4794 0.543531 0.2860 0.3235 0.3655 0.4133 0.4682 0.532632 0.2750 0.3125 0.3545 0.4023 0.4574 0.522033 0.2645 0.3019 0.3439 0.3917 0.4469 0.511734 0.2543 0.2916 0.3335 0.3813 0.4367 0.501735 0.2444 0.2816 0.3235 0.3713 0.4267 0.491936 0.2349 0.2719 0.3137 0.3615 0.4170 0.482437 0.2257 0.2626 0.3042 0.3520 0.4075 0.473238 0.2168 0.2535 0.2950 0.3427 0.3983 0.464139 0.2082 0.2447 0.2861 0.3337 0.3894 0.455340 0.1998 0.2361 0.2774 0.3249 0.3806 0.4468

41 0.1918 0.2278 0.2689 0.3164 0.3721 0.438442 0.1840 0.2197 0.2606 0.3080 0.3637 0.4302

5 28 0.3431 0.3845 0.4311 0.4843 0.5461 0.619029 0.3295 0.3709 0.4175 0.4707 0.5325 0.605630 0.3165 0.3578 0.4043 0.4575 0.5194 0.592631 0.3039 0.3451 0.3916 0.4447 0.5067 0.580032 0.2919 0.3329 0.3792 0.4324 0.4943 0.567733 0.2803 0.3211 0.3673 0.4204 0.4823 0.555834 0.2691 0.3097 0.3558 0.4088 0.4707 0.544335 0.2583 0.2987 0.3446 0.3975 0.4594 0.533036 0.2479 0.2881 0.3338 0.3866 0.4484 0.522137 0.2379 0.2778 0.3233 0.3759 0.4377 0.511538 0.2282 0.2679 0.3131 0.3656 0.4273 0.501239 0.2188 0.2582 0.3033 0.3556 0.4172 0.491140 0.2098 0.2489 0.2937 0.3458 0.4074 0.481341 0.2011 0.2398 0.2844 0.3363 0.3978 0.471842 0.1927 0.2311 0.2753 0.3271 0.3884 0.4625

10 28 0.3702 0.4164 0.4686 0.5287 0.5992 0.683429 0.3548 0.4007 0.4528 0.5128 0.5831 0.667230 0.3400 0.3857 0.4376 0.4974 0.5676 0.651631 0.3259 0.3713 0.4230 0.4826 0.5526 0.636532 0.3123 0.3575 0.4089 0.4683 0.5382 0.621933 0.2993 0.3442 0.3953 0.4545 0.5242 0.607834 0.2868 0.3314 0.3822 0.4412 0.5107 0.594235 0.2748 0.3190 0.3696 0.4283 0.4976 0.581036 0.2633 0.3072 0.3574 0.4158 0.4849 0.568237 0.2522 0.2957 0.3456 0.4037 0.4726 0.555838 0.2415 0.2846 0.3342 0.3920 0.4607 0.543739 0.2313 0.2740 0.3231 0.3807 0.4491 0.532140 0.2214 0.2636 0.3125 0.3697 0.4379 0.520741 0.2119 0.2537 0.3021 0.3590 0.4270 0.509742 0.2027 0.2441 0.2921 0.3487 0.4164 0.4990

15 28 0.4065 0.4585 0.5179 0.5868 0.6685 0.7670

(Continued)

yielding, the lateral earth pressure will be substantially lower than the Rankine activeearth pressure. As a result, the distribution of lateral earth pressure will vary substantially in comparison to the linear distribution assumed in the case of retaining walls.

The total lateral force per unit length of tIle wall, Pa, imposed on a wall may beevaluated theoretically by using Terzaghi's (1943) general wedge theory. (See Figure 7.12.) The failure surface is assumed to be the arc of a logarithmic spiral, defined as

where <p' ~ effective angle of friction of soil

In the figure, H is the height of the cut, and the unit weight, angle of friction,and cohesion of the soil are equal to y, cP', and c', respectively. Following are theforces per unit length of the cut acting on the trial failure wedge:

1. Weight of the wedge, W2. Resultant of the normal and shear forces along ab, R


Table 7.4 (Continued)

13 (deg)

a (deg) <{J' (deg) 90 85 80 75 70 65

29 0.3881 0.4397 0.4987 0.5672 0.6483 0.746330 0.3707 0.4219 0.4804 0.5484 0.6291 0.726531 0.3541 0.4049 0.4629 0.5305 0.6106 0.707632 0.3384 0.3887 0.4462 0.5133 0.5930 0.689533 0.3234 0.3732 0.4303 0.4969 0.5761 0.672134 0.3091 0.3583 0.4150 0.4811 0.5598 0.655435 0.2954 0.3442 0.4003 0.4659 0.5442 0.639336 0.2823 0.3306 0.3862 0.4513 0.5291 0.623837 0.2698 0.3175 0.3726 0.4373 0.5146 0.608938 0.2578 0.3050 0.3595 0.4237 0.5006 0.594539 0.2463 0.2929 0.3470 0.4106 0.4871 0.580540 0.2353 0.2813 0.3348 0.3980 0.4740 0.567141 0.2247 0.2702 0.3231 0.3858 0.4613 0.554142 0.2146 0.2594 0.3118 0.3740 0.4491 0.5415

20 28 0.4602 0.5205 0.5900 0.6714 0.7689 0.888029 0.4364 0.4958 0.5642 0.6445 0.7406 0.858130 0.4142 0.4728 0.5403 0.6195 0.7144 0.830331 0.3935 0.4513 0.5179 0.5961 0.6898 0.804332 0.3742 0.4311 0.4968 0.5741 0.6666 0.779933 0.3559 0.4121 0.4769 0.5532 0.6448 0.756934 0.3388 0.3941 0.4581 0.5335 0.6241 0.735135 0.3225 0.3771 0.4402 0.5148 0.6044 0.714436 0.3071 0.3609 0.4233 0.4969 0.5856 0.694737 0.2925 0.3455 0.4071 0.4799 0.5677 0.675938 0.2787 0.3308 0.3916 0.4636 0.5506 0.657939 0.2654 0.3168 0.3768 0.4480 0.5342 0.640740 0.2529 0.3034 0.3626 0.4331 0.5185 0.624241 0.2408 0.2906 0.3490 0.4187 0.5033 0.608342 0.2294 0.2784 0.3360 0.4049 0.4888 0.5930

(7.25)r == r e() tan cf/o

7.6 Active Pressure for Wall Rotation about the Top: Braced Cut 311

Table 7.5 Values of Ka [from Eq. (7.24)] for 8 == <p' /2

P (degl

a (deg) cP' (deg) 90 85 80 75 70 65

0 28 0.3264 0.3629 0.4034 0.4490 0.5011 0.561629 0.3137 0.3502 0.3907 0.4363 0.4886 0.549230 0.3014 0.3379 0.3784 0.4241 0.4764 0.537131 0.2896 0.3260 0.3665 0.4121 0.4645 0.525332 0.2782 0.3145 0.3549 0.4005 0.4529 0.513733 0.2671 0.3033 0.3436 0.3892 0.4415 0.502534 0.2564 0.2925 0.3327 0.3782 0.4305 0.491535 0.2461 0.2820 0.3221 0.3675 0.4197 0.480736 0.2362 0.2718 0.3118 0.3571 0.4092 0.470237 0.2265 0.2620 0.3017 0.3469 0.3990 0.459938 0.2172 0.2524 0.2920 0.3370 0.3890 0.449839 0.2081 0.2431 0.2825 0.3273 0.3792 0.440040 0.1994 0.2341 0.2732 0.3179 0.3696 0.430441 0.1909 0.2253 0.2642 0.3087 0.3602 0.420942 0.1828 0.2168 0.2554 0.2997 0.3511 0.4177

5 28 0.3477 0.3879 0.4327 0.4837 0.5425 0.611529 0.3337 0.3737 0.4185 0.4694 0.5282 0.597230 0.3202 0.3601 0.4048 0.4556 0.5144 0.583331 0.3072 0.3470 0.3915 0.4422 0.5009 0.569832 0.2946 0.3342 0.3787 0.4292 0.4878 0.556633 0.2825 0.3219 0.3662 0.4166 0.4750 0.543734 0.2709 0.3101 0.3541 0.4043 0;4626 0.531235 0.2596 0.2986 0.3424 0.3924 0.4505 0.519036 0.2488 0.2874 0.3310 0.3808 0.4387 0.507037 0.2383 0.2767 0.3199 0.3695 0.4272 0.495438 0.2282 0.2662 0.3092 0.3585 0.4160 0.484039 0.2185 0.2561 0.2988 0.3478 0.4050 0.472940 0.2090 0.2463 0.2887 0.3374 0.3944 0.462041 0.1999 0.2368 0.2788 0.3273 0.3840 0.451442 0.1911 0.2276 0.2693 0.3174 0.3738 0.4410

10 28 0.3743 0.4187 0.4688 0.5261 0.5928 0.671929 0.3584 0.4026 0.4525 0.5096 0.5761 0.654930 0.3432 0.3872 0.4368 0.4936 0.5599 0.638531 0.3286 0.3723 0.4217 0.4782 0.5442 0.622532 0.3145 0.3580 0.4071 0.4633 0.5290 0.607133 0.3011 0.3442 0.3930 0.4489 0.5143 0.592034 0.2881 0.3309 0.3793 0.4350 0.5000 0.577535 0.2757 0.3181 0.3662 0.4215 0.4862 0.563336 0.2637 0.3058 0.3534 0.4084 0.4727 0.549537 0.2522 0.2938 0.3411 0.3957 0.4597 0.536138 0.2412 0.2823 0.3292 0.3833 0.4470 0.523039 0.2305 0.2712 0.3176 0.3714 0.4346 0.510340 0.2202 0.2604 0.3064 0.3597 0.4226 0.497941 0.2103 0.2500 0.2956 0.3484 0.4109 0.485842 0.2007 0.2400 0.2850 0.3375 0.3995 0.4740

15 28 0.4095 0.4594 0.5159 0.5812 0.6579 0.7498

(Continued)

65

0.72840.70800.68840.66950.65130.63380.61680.60040.58460.56920.55430.53990.52580.51220.86130.83130.80340~7772

0.75240.72890.70660.68530.66490.64530.62660.60850.59120.57440.5582

70

0.63730.61750.59850.58030.56270.54580.52950.51380.49850.48380.46950.45570.44230.42930.75140.72320.69680.67200.64860.62640.60520.58510.56580.54740.52970.51270.49630.48050.4653

75

0.56110.54190.52350.50590.48890.47260.45690.44170.42710.41300.39930.38610.37330.36090.66080.63390.60870.58510.56280.54170.52160.50250.48420.46680.45000.43400.41850.40370.3894

II/I!T,IIII

II!

f3 (deg)

80

0.49640.47770.45980.44270.42620.41050.39530.38070.36670.35310.34010.32750.31530.30350.58440.55860.53450.51190.49060.47040.45130.43310.41570.39910.38330.36810.35350.33950.3261

.. ·r?</::··.~t<X>

(b)

85

0.44020.42200.40460.38800.37210.35680.34220.32820.31470.30170.28930.27730.26570.25460.51880.49400.47080.44910.42860.40930.39100.37360.35710.34130.32630.31200.29820.28510.2725

(a)

\\\\\\\\\\\,\\,\\\,

29 0.390830 0.373031 0.356032 0.339833 0.324434 0.309735 0.295636 0.282137 0.269238 0.256939 0.245040 0.233641 0.222742 0.212228 0.461429 0.437430 0.415031 0.394132 0.374433 0.355934 0.338435 0.321836 0.306137 0.291138 0.276939 0.263340 0.250441 0.238142 0.2263

l/J' (deg) 90

20

ll' (deg)


Figure 7.11 Nature of yielding of walls: (a) retaining wall; (b) braced cut


7.7 Rankine Passive Earth Pressure 313

a ~Arcofa

log spiral

y

cP'c'

Lateral pressure, o-~

Figure 7.12 Braced cutanalysis by generalwedge theory: wall rotation about top

3. Cohesive force along ab, C4. Adhesive force along ac, Co5. Pa' which is the force acting a distance naH from the bottom of the wall and is

inclined at an angle Dto the horizontal

The adhesive force is

(7.26)

(7.27)

where ca' == unit adhesion

A detailed outline for the evaluation of Pa is beyond the scope of this text;those interested should check a soil mechanics text for more information (e.g., Das,1998). Kim and Preber (1969) provided tabulated values of Pa/~yH2 determined byusing the principles of general wedge theory, and these values are given in Table 7.6.In developing the theoretical values for that table, it was assumed that

ca' tan 8

c' tan 4>'

Passive Pressure

Rankine Passive Earth Pressure

Figure 7.13a shows a vertical frictionless retaining wall with a horizontal backfill. Atdepth z, the effective vertical pressure on a soil element is (J"0' == yz. Initially, if thewall does not yield at all, the lateral stress at that depth will be O"h' == KoO"o'. Thisstate of stress is illustrated by the Mohr's circle a in Figure 7.13b. Now, if the wall ispushed into the soil mass by an amount ~x, as shown in Figure 7.13a, the verticalstress at depth z will stay the same; however, the horizontal stress will increase. Thus,ah will be greater than Koa0'. The state of stress can now be represented by theMohr's circle b in Figure 7.13b. If the wall moves farther inward (i.e., dx is increasedstill more), the stresses at depth z will ultimately reach the state represented by


Table 7.6 Pa/[(1)yH 2] versus </J',D,na, and c'/yH*

"a = 0.3 na = 0.4 na = 0.5 na = 0.6c'/y H c'/y H c'/y H c'/y H

t!J', in (j', indegrees degrees 0 0.1 0.2 0 0.1 0.2 0 0.1 0.2 0 0.1 0.2

(1 ) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11 ) (12) (13) (14)

0 0 0.952 0.558 0.164 0.652 0.192 0.782 0.230 0.978 0.288

5 0 0.787 0.431 0.076 0.899 0.495 0.092 1.050 0.580 0.110 1.261 0.697 0.134

5 0.756 0.345 -0.066 0.863 0.399 -0.064 1.006 0.474 -0.058 1.209 0.573 -0.063

10 0 0.653 0.334 0.015 0.734 0.378 0.021 0.840 0.434 0.027 0.983 0.507 0.032

5 0.623 0.274 -0.074 0.700 0.312 -0.077 0.799 0.358 -0.082 0.933 0.420 -0.093

10 0.610 0.242 -0.125 0.685 0.277 -0.131 0.783 0.324 -0.135 0.916 0.380 -0.156

15 0 0.542 0.254 -0.033 0.602 0.285 -0.033 0.679 0.322 -0.034 0.778 0.370 -0.039

5 0.518 0.214 -0.089 0.575 0.240 -0.094 0.646 0.270 -0.106 0.739 0.310 -0.118

10 0.505 0.187 -0.131 0.559 0.210 -0.140 0.629 0.238 -0.153 0.719 0.273 -0.174

15 0.499 0.169 -0.161 0.554 0.191 -0.171 0.623 0.218 -0.187 0.714 0.251 -0.212

20 0 0.499 0.191 -0.067 0.495 0.210 -0.074 0.551 0.236 -0.080 0.622 0.266 -0.090

5 0.430 0.160 -0.110 0.473 0.179 -0.116 0.526 0.200 -0.126 0.593 0.225 -0.142

10 0.419 0.140 -0.139 0.460 0.156 -0.149 0.511 0.173 -0.165 0.575 0.196 -0.184

15 0.413 0.122 -0.169 0.454 0.137 -0.179 0.504 0.154 -0.195 0.568 0.174 -0.219

20 0.413 0.113 -0.188 0.454 0.124 -0.206 0.504 0.140 -0.223 0.569 0.160 -0.250

25 0 0.371 0.138 -0.095 0.405 0.150 -0.104 0.447 0.167 -0.112 0.499 0.187 -0.125

5 0.356 0.116 -0.125 0.389 0.128 -0.132 0.428 0.141 - 0.146 0.477 0.158 -0.162

10 0.347 0.099 -0.149 0.378 0.110 -0.158 0.416 0.122 -0.173 0.464 0.136 -0.192

15 0.342 0.085 -0.172 0.373 0.095 -0.182 0.410 0.106 -0.198 0.457 0.118 -0.221

20 0.341 0.074 -0.193 0.372 0.083 -0.205 0.409 0.093 -0.222 0.456 0.104 -0.248

25 0.344 0.065 -0.215 0.375 0.074 -0.228 0.413 0.083 -0.247 0.461 0.093 -0.275

30 0 0.304 0.093 -0.117 0.330 0.103 -0.124 0.361 0.113 -0.136 0.400 0.125 -0.150

5 0.293 0.078 -0.137 0.318 0.086 -0.145 0.347 0.094 -0.159 0.384 0.105 -0.175

10 0.286 0.066 -0.154 0.310 0.073 -0.164 0.339 0.080 -0.179 0.374 0.088 -0.198

15 0.282 0.056 -0.171 0.306 0.060 -0.185 0.334 0.067 -0.199 0.368 0.074 -0.220

20 0.281 0.047 -0.188 0.305 0.051 -0.204 0.332 0.056 -0.220 0.367 0.062 -0.242

25 0.284 0.036 -0.211 0.307 0.042 -0.223 0.335 0.047 -0.241 0.370 0.051 -0.267

30 0.289 0.029 -0.230 0.313 0.033 -0.246 0.341 0.038 -0.265 0.377 0.042 -0.294

35 0 0.247 0.059 -0.129 0.267 0.064 -0.139 0.290 0.069 -0.151 0.318 0.076 -0.165

5 0.239 0.047 -0.145 0.258 0.052 -0.154 0.280 0.057 -0.167 0.307 0.062 -0.183

10 0.234 0.038 -0.157 0.252 0.041 -0.170 0.273 0.046 -0.182 0.300 0.050 -0.200

15 0.231 0.030 -0.170 0.249 0.033 -0.183 0.270 0.035 -0.199 0.296 0.039 -0.218

20 0.231 0.022 -0.187 0.248 0.025 -0.198 0.269 0.027 -0.215 0.295 0.030 -0.235

25 0.232 0.015 -0.202 0.250 0.016 -0.218 0.271 0.019 -0.234 0.297 0.020 -0.256

30 0.236 0.006 -0.224 0.254 0.008 -0.238 0.276 0.011 -0.255 0.302 0.011 -0.281

35 0.243 0 -0.243 0.262 0.001 -0.260 0.284 0.002 -0.279 0.312 0.002 -0.307

40 0 0.198 0.030 -0.138 0.213 0.032 -0.148 0.230 0.036 -0.159 0.252 0.038 -0.175

5 0.192 0.021 -0.150 0.206 0.024 -0.158 0.223 0.026 -0.171 0.244 0.029 -0.186

10 0.189 0.015 -0.158 0.202 0.016 -0.170 0.219 0.018 -0.182 0.238 0.020 -0.199

15 0.187 0.008 -0.171 0.200 0.010 -0.180 0.216 0.011 -0.195 0.236 0.012 -0.212

20 0.187 0.003 -0.181 0.200 0.003 -0.195 0.216 0.004 -0.208 0.235 0.004 -0.227

25 0.188 -0.005 0.202 -0.003 0.218 -0.003 0.237 -0.003

30 0.192 -0.010 0.205 -0.010 0.222 -0.011 0.241 -0.012

35 0.197 -0.018 0.211 -0.018 0.228 -0.018 0.248 -0.020



na == 0.3 na == 0.4 na == 0.5 na == 0.6e'ly H e'ly H e'ly H e'/y H

l/J', in S', indegrees degrees 0 0.1 0.2 0 0.1 0.2 0 0.1 0.2 0 0.1 0.2

(1 ) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11 ) (12) (13) (14)

40 0.205 -0.025 0.220 -0.025 0.237 -0.027 0.259 -0.030

45 0 0.156 0.007 -0.142 0.167 O.OOg -0.150 O.IRO 0.009 -0.162 0.196 0.010 -0.177

5 0.152 0.002 -0.148 0.163 0.002 -0.158 0.175 0.002 -0.170 0.190 0.003 -O.U·)5

10 0.150 -0.003 0.160 -0.004 0.172 -0.003 0.187 -0.004

15 0.148 -0.009 0.159 -0.008 0.171 -0.009 0.185 -0.010

20 0.149 -0.013 0.159 -0.014 0.171 -0.014 0.185 -0.016

25 0.150 -0.018 0.160 -0.020 0.173 -0.020 0.187 -0.022

30 0.153 -0.025 0.164 -0.026 0.176 -0.026 0.190 -OJ)29

35 0.158 -0.030 0.168 - 0.031 0.181 -0.034 0.196 -0.037

40 0.164 -0.038 0.175 -0.040 0.188 -0.042 0.204 -0.045

45 0.173 -0.046 0.184 -0.048 0.198 -0.052 0.215 -0.057

*After Kim and Preber (1969)

Mohr's circle c. Note that this Mohr's circle touches the Mohr-Coulomb failure envelope, which implies that the soil behind the wall will fail by being pushed upward.The horizontal stress, a~, at this point is referred to as the Rankine passive pressure,or '- ,

(J"o - (Ip.

For Mohr's circle c in Figure 7.13b, the major principal stress is (T~, and the minor principal stress is (T~. Substituting these quantities into Eq. (1.74) yields

(7.28)

Now, let

(7.29)

Then, from Eq. (7.28), we have

(7.30)

',','1;.·.:,:'.,'·,'.. ~.

-~ ---

)~)

I

':1,·,.:'.···.·,·... •f ~.~

'i

I1 ~

tj

~I;~>

Effectivenormalstress

l'

yc'

45 - 1'/2

(b)

H

z

Rotation aboutthis point (a)

Figure 7.13 Rankine passive pressure

Shear stress

Direction of--...~~ wall movement

Chapter 7 Lateral Earth Pressure316

Equation (7.30) produces Figure 7.13c, the passive pressure diagram for thewall shown in Figure 7.13a. Note that at z == 0,

·,:,'1····,·.'·;.·

;

<:,~ --.....,

a-~ == 0 and a-' == 2c'~p p

andatz == H,

(J"~ = yH and (J"~ = yHKp + 2c'YK;

The passive force per unit length of the wall can be determined from the areaof the pressure diagram, or

Pp = hH2K p + 2c'HYK; (7.31)

The approximate magnitudes of the wall movements, ~x, required to develop.failure under passive conditions are as follows:

·.,·.".··'·;.:·.I·~··.~

I

I

f.1

H


114-4I--Kp 'YH + 2c'~---...~1

(c)

Soil type

Dense sandLoose sandStiff claySoft clay

Figure 7. 13 (continued)

Wall movement forpassive condition, ax

O.OOSHO.01HO.OIHO.OSH

If the backfill behind the wall is a granular soil (i.e., c' == 0), then, from Eq. (7.31),the passive force per unit length of the wall will be

1P == -",H2K

P 2' P(7.32)

The passive force on a frictionless inclined retaining wall (see Figure 7.14) with a horizontal granular backfill (c' == 0) can also be expressed by Eq. (7.32). The variation

H

& .. •• •

'" .. ".,," .... " .... :

.' Friction lesswall

yc' == 0¢'

Figure 7.14 Passive force on africtionless inclined retaining wall


Table 7.7 Variation of K p [see Eq. (7.32) and Figure 7.14]*

(J (deg)

~' (deg) 30 25 20 15 10 5 0

20 1.70 1.69 1.72 1.77 1.83 1.92 2.0421 1.74 1.73 1.76 1.81 1.89 1.99 2.12

22 1.77 1.77 1.80 1.87 1.95 2.06 2.20

23 1.81 1.81 1.85 1.92 2.01 2.13 2.28

24 1.84 1.85 1.90- 1.97 2.07 2.21 2.37

25 1.88 1.89 1.95 2.03 2.14 2.28 2.46

26 1.91 1.93 1.99 2.09 2.21 2.36 2.5627 1.95 1.98 2.05 2.15 2.28 2.45 2.6628 1.99 2.02 2.10 2.21 2.35 2.54 2.7729 2.03 2.07 2.15 2.27 2.43 2.63 2.8830 2.07 2.11 2.21 2.34 2.51 2.73 3.00

31 2.11 2.16 2.27 2.41 2.60 2.83 3.1232 2.15 2.21 2.33 2.48 2.68 2.93 3.25

33 2.20 2.26 2.39 2.56 2.77 3.04 3.3934 2.24 2.32 2.45 2.64 2.87 3.16 3.5335 2.29 2.37 2.52 2.72 2.97 3.28 3.6836 2.33 2.43 2.59 2.80 3.07 3.41 3.8437 2.38 2.49 2.66 2.89 3.18 3.55 4.0138 2.43 2.55 2.73 2.98 3.29 3.69 4.1939 2.48 2.61 2.81 3.07 3.41 3.84 4.3840 2.53 2.67 2.89 3.17 3.53 4.00 4.5941 2.59 2.74 2.97 3.27 3.66 4.16 4.8042 2.64 2.80 3.05 3.38 3.80 4.34 5.0343 2.70 2.88 3.14 3.49 3.94 4.52 5.2744 2.76 2.94 3.23 3.61 4.09 4.72 5.5345 2.82 3.02 3.32 3.73 4.25 4.92 5.80

*Based on Zhu and Qian, 2000

of K p for this case, with wall inclination () and effective soil friction angle 4J'., isgiven in Table 7.7 (Zhu and Qian, 2000).


r ) "

r == 100 Ib/ft3

<Pi == 30°z 8ft c{ = 0

~ Water tableCD

-

4ftYsat == 1201b/ft3

@ +<Pz = 26°

1 C2 == 210 Ib/ft2 CD3309.8 249.6Ib/ft2 Ib/ft2


Rankine Passive Earth Pressure: Inclined Backfill

Granular Soil

For a frictionless vertical retaining wall (Figure 7.8) with a granular backfill

(e' == 0), the Rankine passive pressure at any depth can be determined in amanner similar to that done in the case of active pressure in Section 7.4. Thepressure is

(]" , == yzKp p

and the passive force is

cos a + Vcos2 a - cos2 4>'where K p == cos a ---------;:==================-

cos a- V cos2 a - cos2 <p'

(7.33)

(7.34)

(7.35)

As in the case of the active force, the resultant force, Pp , is inclined at an angle Q'

with the horizontal and intersects the wall at a distance H /3 from the bottom of thewall. TIle values of K p (the passive earth pressure coefficient) for various values of aand 4J' are given in Table 7.8.

C' -CP' Soil

If the backfill of the frictionless vertical retaining wall is a c' - <p' soil (see Figure 7.8),then (Mazindrani and Ganjali, 1997)

(T~ == 'YzKp == 'YzK~ cos a

where

(7.36)

1K' =---<..

P cosz 4J'+

2cos2

a + 2( ~~ )cos <P' sin <P'

( C')2 (C' )4cos2a (cos2

a - cos2 <P') + 4 yz cos2 <P' + 8 yz cos2a sin <P' cos <P'

-1

(7.37)

The variation of K~ with <p',a, and c'/yz is given in Table 7.9 (Mazindrani and Ganjali, 1997).

Table 7.8 Passive Earth Pressure Coefficient, K p [from Eq. 7.35)]

cP' (deg) ~

ta (deg) 28 30 32 34 36 38 40

0 2.770 3.000 3.255 3.537 3.852 4.204 4.5995 2.715 2.943 3.196 3.476 3.788 4.136 4.527

10 2.551 2.775 3.022 3.295 3.598 3.937 4.31615 2.284 2.502 2.740 3.003 3.293 3.615 3.97720 1.918 2.132 2.362 2.612 2.886 3.189 3.52625 1.434 1.664 1.894 2.135 2.394 2.676 2.987

7.9 Coulomb's Passive Earth Pressure 321

Table 7.9 Values of K~

c' /yz

<P' (deg) a (deg) 0.025 0.050 0.100 0.500

15 0 1.764 1.829 1.959 3.0025 1.716 1.783 1.917 2.971

10 1.564 1.641 1.788 2.88015 1.251 1.370 1.561 2.732

20 0 2.111 2.182 2.325 3.4685 2.067 2.140 2.285 3.435

10 1.932 2.010 2.162 3.33915 1.696 1.786 1.956 3.183

25 0 2.542 2.621 2.778 4.0345 2.499 2.578 2.737 3.999

10 2.368 2.450 2.614 3.89515 2.147 2.236 2.409 3.726

30 0 3.087 3.173 3.346 4.7325 3.042 3.129 3.303 4.674

10 2.907 2.996 3.174 4.57915 2.684 2.777 2.961 4.394

Coulomb's Passive Earth Pressure

Coulomb (1776) also presented an analysis for determining the passive earth pressure (i.e., when the wall moves into the soil mass) for walls possessing friction(8 == angle of wall friction) and retaining a granular backfill material similar to thatdiscussed in Section 7.5.

To understalld the determination of Coulomb's passive force, Pp , consider thewall shown in Figure 7.16a.As in the case of active pressure, Coulomb assumed that thepotential failure surface in soil is a plane. For a trial failure wedge of soil, such as ABCl ,

the forces per unit length of the wall acting on the wedge are

1. The weight of the wedge, W2. The resultant, R, of the normal and shear forces on the plane Bel' and3. The passive force, Pp

Figure 7.16b shows the force triangle at equilibrium for the trial wedge ABCl .

From this force triangle, the value of Pp can be determined, because the direction of allthree forces and the magnitude of one force are known.

Similar force triangles for several trial wedges, such as ABCl , ABCz, ABC3 , ..• ,

can be constructed, and the corresponding values of Pp can be determined. The top partof Figure 7.16a shows the nature of variation of the Pp values for different wedges. Theminimum value of Pp in this diagram is Coulombs passive force, mathematically expressed as

(7.38)

322 Chapter 7 Lateral Earth Pressure IPassiveforce

Wall movement--.....~~ toward

the soil

I

I

yc'= 0

¢/

"" /I............ .-/

---t--~-;fI

Pp(min)

H

Figure 7. 16 Coulomb's passive pressure

where

The values of the passive pressure coefficient, K p , for various values of 0/' and 5 aregiven in Table 7.10 (f3 == 90°,lr == 0°).

Note that the resultant passive force, Pp , will act at a distance HI3 from thebottom of the wall and will be inclined at an angle 0 to the normal drawn to the backface of the wall.

7.10 Comments on the Failure Surface Assumption for Coulomb's Pressure Calculations 323

Table 7.10 Values of Kp [from Eq. (7.39)] for f3 == 90° and a == 0°

S (degJ

l/J' (deg) 0 5 10 15 20

15 1.698 1.900 2.130 2.405 2.73520 2.040 2.313 2.636 3.030 3.52525 2.464 2.830 3.286 3.855 4.59730 3.000 3.506 4.143 4.977 6.10535 3.690 4.390 5.310 6.854 8.32440 4.600 5.590 6.946 8.870 11.772

Comments on the Failure Surface Assumptionfor Coulomb's Pressure Calculations

Coulomb's pressure calculation methods for active and passive pressure have beendiscussed in Sections 7.5 and 7.9. The fundamental assumption in these analyses isthe acceptance of plane failure surface. However, for walls with friction, this assumption does not hold in practice. The nature of actual failure surface in the soilmass for active and passive pressure is shown in Figure 7.17a and b, respectively (fora vertical wall with a horizontal backfill). Note that the failure surface BC is curvedand that the failure surface CD is a plane.

Although the actual failure surface in soil for the case of active pressure issomewhat different from that assumed in the calculation of the Coulomb pressure,the results are not greatly different. However, in the case of passive pressure, as thevalue of 0 increases, Coulomb's method of calculation gives increasingly erroneousvalues of Pp • This factor of error could lead to an unsafe condition because the values of Pp would become higher than the soil resistance.

Several studies have been conducted to determine the passive force Pp , assuming that the curved portion Be in Figure 7.17b is an arc of a circle, an ellipse,or a logarithmic spiral. The results of three of these studies are described brieflynext.

Analysis of Caquot and Kerisel

Caquot and Kerisel (1948) developed the chart shown in Figure 7.18 for estimatingthe value of the passive pressure coefficient K p with a curved failure surface in granular soil (c' == 0), such as that shown in Figure 7.17b. In their solution, the portionBC of the failure surface was assumed to be an arc of a logarithmic spiral. In usingFigure 7.18, the following points should be kept in mind:

1. The curves are for <p' == ~.

2. If B/</>' is less than unity, then

K p (8) == RKp (8=4/)


A

T!H:3

(a)

(b)

D

I

'r.·.I",.·.·..~.::' . .

Figure 7.17 Nature of failure surface in soil with wall friction: (a) active pressure;(b) passive pressure

where R == reduction factor

The reduction factor R is given in Table 7.11.3. The passive pressure is

Analysis of Shields and Tolunay

Shields and Tolunay (1973) analyzed the problem of passive pressure for a verticalwall with a horizontal granular soil backfill (c' == 0). This analysis was done by considering the.stability of the wedge ABCC' (see Figure 7.17b), using the method ofslices. From Figure 7.17b, the passive force per unit length of the wall can be expressed as

1P == -K 'VH 2

p 2 pI

I

7. 10 Comments on the Failure Surface Assumption for Coulomb's Pressure Calculations 325

403020

¢' (deg)

10

0.2I--------t-------f---------f-----+--F----J'-----+--R

30 1-----------+---------+----------+------#--1----1------4----1

90 r---------,.-------r--------,--------.,.1.0801--------f---------f-------f----------+-J70 0.8

60 l-------I----------I----------t-------------r---r---i

0.650 l-------I--------I--------t--------f-I----+t

401-------f---------f--------+------F---JL--,,f-----I0.4

200

T8

-0.2

Kp 10 H _ -9 18 -0.47

6 up = KpyH

5-0.6

4

3

-0.8

2

Figure 7.18 Caquot and Kerisel's (1948) passive pressure coefficient, K p , for granular soil

The values of the passive earth pressure coefficient, K p , obtained by Shields andTolunay are given in Table 7.12.

Analysis of Zhu and Qian

Zhu and Qian (2000) determined the passive earth pressure coefficient K p by usingtriangular.slices within the framework of the limit equilibrium method as shown inFigure 7.19 for a vertical retaining wall with a granular soil backfill. According to thisanaiysis,

1P == -K 'VH2

p 2 pi

(7.40)

yc' = 0<p'

'4'5'" . :j,/I '.' :. : ' :',''. .:.;.. '¥" /2 .,' .. " ',,'

Rankinepassive zone

Triangularslices

.- -44

• 0 ••

~t__

Figure 7. 19 Passive earth pressure by method of triangular slices

where K p = 1] tanz(45 + ~')

The variation of TJ with 4J' and S/4J' is shown in Figure 7.20.


Table 7.11 Reduction Factor, R, for Use in Conjunction with Figure 7.18

S/~

t/J' (deg) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0

10 0.978 0.962 0.946 0.929 0.912 0.898 0.880 0.86415 0.961 0.934 0.907 0.881 0.854 0.830 0.803 0.77520 0.939 0.901 0.862 0.824 0.787 0.752 0.716 0.678 Y,j

.. '

25 0.912 0.860 0.808 0.759 0.711 0.666 0.620 0.57430 0.878 0.811 0.746 0.686 0.627 0.574 0.520 0.46735 0.836 0.752 0.674 0.603 0.536 0.475 0.417 0.36240 0.783 0.682 0.592 0.512 0.439 0.375 0.316 0.26245 0.718 0.600 0.500 0.414 0.339 0.276 0.221 0.174

iJ.--IJI

Table 7.12 Values of K p from the Method of Slices I~f:;:tD (deg)

1

4>' (deg) 0 5 10 15 20 25 30 35 40 45 -".~r20 2.04 2.26 2.43 2.55 2.70 I25 2.46 2.77 3.03 3.23 3.39 3.6330 3.00 3.43 3.80 4.13 4.40 4.64 5.0335 3.69 4.29 4.84 5.34 5.80 6.21 6.59 7.2540 4.69 5.44 6.26 7.05 7.80 8.51 9.18 9.83 11.0345 5.83 7.06 8.30 9.55 10.80 12.04 13.26 14.46 15.60 18.01

7.10 Comments on the Failure Surface Assumption for Coulomb's Pressure Calculations 327

6l------+-----+------t------.-r-------lII

51----+-----+-----+-----+-~---1

'rJ 4 I-----+-----+-------+----r-----+-----~

40°

31----+-----+------+--+---~-+-----1

35°

Figure 7.20 Variation of 1]

with oj<P and <P'

1.00.80.4 0.6

8/<//

0.21......-=:~-..L.....----'-----......L....-----I---- .......o


7.1 In Figure 7.5a, let H l == 3 m, H 2 == 3 ill, q == 0, y == 16.5 kN/m3, Ysat ==

19 kN/m3, c' == 0, and <P' == 30°. Determine the at-rest lateral earth force per me

ter length of the wall. Also, find the location of the resultant force. Use Eq. (7.3).7.2 In Figure 7.5a, let Hi == 5 m, H2 == 0, q == 0, and Y == 18 kN/m3

. The backfill isan overconsolidated clay with a plasticity index of 26. If the overconsolidationratio is 2.0, determine the at-rest lateral earth force per meter length of thewall. Also, find the location of the resultant. Use Eqs. (7.5) and (7.7).

7.3 Redo Problem 7.2, assuming that the surcharge q == 50 kN/m2•

7.4 Use Eq. (7.3), Figure 7.5, and the following values to determine the at-rest lat~

eral earth force per unit length of the wall: H == 10 ft, Hi == 4 ft, Hz == 6 it,Y == 105Ib/ft3

, Ysat == 122Ib/ft3, </J' == 30°, c' == 0, and q == 300 Ib/ft2. Find the

location of the resultant force.7.5 A vertical retaining wall (Figure 7.6a) is 5.5 m high with a horizontal backfill.

For the backfill, Y == 18.7 kN/m3, </J' == 24°, and c' == 10 kN/m2

• Determine theRankine active force per unit length of the wall after a tensile crack appears.,

7.6 In Figure 7.6a, let the height of the retaining wall be H == 21 ft. The backfill isa saturated clay with 4J == 0°, C == 630 Ib/ft2

, and Ysat == 113 Ib/ft3.

3. Determine the Rankine active pressure distribution diagram behind the walLb. Determine the depth of the tensile crack, Zc.

c. Estimate the Rankine active force per foot length of the wall after the tensile crack appears.

Problems 329

z

H

Groundwatertable

Figure Pl.l

7.7 In Figure P7.7, let HI == 8.2 ft, Hz == 14.8 ft, 1'1 == 107Ib/ft3, q == 0, c/Ji == 34°,

ci == O,l'z == 140 Ib/ft3, cP2 == 25°, and C2 == 209Ib/ft2

. Determine the Rankineac~ive force per unit length of the wall.

7.8 For the retaining wall of Figure 7.8, H == 6 m, c/J' == 34°, ex == 10°, I' ==17 kN/m3

, and c' == O.a. Determine the intensity of the Rankine active force at z == 2, 4, and 6 m.b. Determine the Rankine active force per meter length of the wall and also

the location and direction of the resultant force.7.9 In Figure 7.8, H == 22 ft, I' == 115 Ib/ft3

, 4>' == 25°, c' == 250 Ib/ft2, and ex == 10°.

Calculate the Rankine active force per unit length of the wall after the tensile--~ crack appears.

<7.10)In Figure 7.10a, let H = 6 m, Y = 16.5 kN/m3, 1>' = 30°, 8 = 20°, c' = 0,

V a = 10°, and f3 = 80°. Determine the Coulomb's active force per meterlength of the wall and the location and direction of the resultant force.

7.11 In Figure 7.10a, let H == 12 ft, 'Y == 105Ib/ft3, 4>' == 30°, c' == 0, and f3 == 85°.

Determine the Coulomb's active force per unit length of the wall and the location and direction of the resultant force for the following cases:a. a == 10° and 8 == 20°b. a == 20° and <5 == 15°

7.12 A retaining wall is shown in Figure P7.12. If the wall rotates about its top, determine the n1agnitude of the active force per unit length of the wall forna === 0.3, 0.4, and 0.5. Assume unit adhesion, c~ == c' (tan 8/tan c/J').

7.13 With regard to Problem 7.6,a. Draw the Rankine passive pressure distribution diagram behind the wall.b. Estimate the Rankine passive force per foot length of the wall and also the

location of the resultant force.7.14 Repeat Problem 7.5 for the Rankine passive case.7.15 ~epeat Problem 7.7 for the Rankine passive case.7.16 For the retaining wall described in Problem 7.10, determine the Coulomb's

passive force per meter length of the wall and the location and direction ofthe resultant force.

330 Chapter 7 Lateral Earth Pressure I

8m

".:. y = 17.5 kN/m3

~':::: cP' = 25°:::. 8 = 15°

~t-~ :...... c' = 14kN/m2!Y ~:'~:

Fa {;

I

Figure P7.12

7.17 For the retaining wall with a granular backfill shown in Figure 7.18, letH == 6 ffi,a == +10°,4>' == 36°,1 == 15 kN/m3

, and 8/4>' == 0.4. Calculate thepassive force per unit length of the wall, assuming a curved failure surfacein the soil.

7.18 In Figure 7.17b, which shows a vertical retaining wall with a horizontal backfill, let H == 13.5 ft,1 == 105Ib/ft3,c!J' == 35°, and 8 == 10°. Based on Shieldsand Tolunay's work (see Table 7.12), what would be the passive force per meter length of the wall?

7.19 Solve Problem 7.18, using Eq. (7.40) and Figure 7.20.

Brooker, E. W., and Ireland, H. O. (1965). "Earth Pressure at Rest Related to Stress HistoryCanadian Geotechnical Journal, Vol. 2, No.1, pp. 1-15.

Caquot, A., and Kerisel, J. (1948). Tables for Calculation of Passive Pressure, Active Pressureand Bearing Capacity ofFoundations, Gauthier-Villars, Paris, France.

Coulomb, C. A. (1776). Essai sur une Application des Regles de Maximis et Minimum aquelques Problemes de Statique Relatifs al'Architecture, Mem.Acad. Roy. des Sciences,Paris, Vol. 3, p. 38.

Das, B. M. (1998). Principles of Geotechnical Engineering, 4th ed., PWS Publishing Company, Boston.

Jaky, J. (1944). "The Coefficient of Earth Pressure at Rest," Journal for the Society ofHungarian Architects and Engineers, October, pp. 355-358.

Kim, J. S., and Preber, T. (1969). "Earth Pressure against Braced Excavations," Journal of theSoil Mechanics and Foundations Division, ASCE, VoL 96, No.6, pp. 1581-1584.

Mayne, ~ W., and Kulhawy, F. H. (1982). "Ko-OCR Relationships in Soil," Journal of the Geotechnical Engineering Division, ASCE, Vol. 108, No. GT6, pp. 851-872.

Mazindrani, Z. H., and Ganjali, M. H. (1997). "Lateral Earth Pressure Problem of CohesiveBackfill with Inclined Surface," Journal of Geotechnical and Geoenvironmental Engineering, ASCE, Vol. 123, No.2, pp.110-112.

Shields, D. H.,-and Tolunay, A. Z. (1973). "Passive Pressure Coefficients by Method of Slices,"Journal of the Soil Mechanics and Foundations Division, ASCE, Vol. 99, No. SM12,1043-1053.

Terzaghi, K. (1943). Theoretical Soil Mechanics, Wiley, New York.Zhu, D. Y., and Qian, Q. (2000). "Determination of Passive Earth Pressure Coefficient by the

Method of Triangular Slices," Canadian Geotechnical Journal, Vol. 37, No.2, pp. 485-491.

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