Lateral Earth Pressure: At-Rest,Rankine, and Coulomh

12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomh

Retainingstructurcs such as retainingwalls, bascment walls, and bulkheadsare com- monly encountercdin foundation engineeringas thcy support slopesof earth masses. Proper designand constructionof thesestructures require a thorough knowledgeof thc lateral forcesthat act betweenthe retaining structuresand the soil massesbeing retaincd.These lateral forccsarc causedby lateralearth prcssure.This chapteris de- votcd to the study oI the various earth pressuretheories.

12.1 At-Rest, Active, and Passive Pressures

Consider a massof soil shown in Figurc. l2.la. The massis bounded by a .frictionless wall of height AB. A soil element locatedat a depth z is subjectedto a vertical effective prcssurerrj, and a horizontal efTectivepressure oj,. There arc no shearstresses on thc vcrtical and horizontal planesof the soil element.Let us define the ratio of oj, to al, as a nondimensionalquantitv K, or

C,r K-- (12.1) (f,'

Now, three possiblecases may ariseconcerning the retainingwall: and they are described

Case 1. If the wall AB is static- that is, if it doesnot move either to the right or to the left of its initial position - the soil masswill be in a stateof stallcequilibrium. In that case,rrj, is referred to as the ut-restearth pressure,or

K: K,,:% (r2.2) o',,

where K,, - at-rest earth pressure coefficient.

364 12.1 At-Rest,Active, and Passive pressures

At-rest pressure Activepressure +l Al, l+ 'l A'AC' I c,, :

t",,t;Ir I H -:.. : K,,c',,=6',, I .,-{'Lo'ri're'

Pitssivepressurc

xl-,'+ o'tan0'

(c,

Figure 12'1 Del\nitionof at-rcst.active, and passive pressures (Note: Wall AB is frictionless)

Case 2' If the frictionlesswall rotates sufficientlyabout its bottom to a position of A'B (Figure l2.lb), then a triangular soil massABC' adjacentto the wall will reach a state of plastic equilibrium and will fail sliding down the plane BC,. At this time, the horizontal effective stress,oj,: o'u,will be ref'erred to as active pressare.Now.

K:Ku:4-4 (12.3) a" a" where K,, : active earth pressure coefficient.

Case 3. If the frictionless wall rotates sufficiently about its bottom to a position ,4"8 (Figure 12.7c), then a triangular soil mass ABC" will reach a state of plastic Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb

h = u

Passivepressure. oj,

At-rest pressure.oir

+ Wall tilt

Figure 12.2 Yariatictn ol thc magnitude of lateral earth pressure with wall tilt

Table 12.1 Typical Values of L.L,,lH and LL,,lH

Soil type LL"IH LLelH

Loose sand 0.(x)l 0.(D2 0.01 Dcnse sand 0.000-5-0.00r 0.005 Soft clay 0.02 0.04 StilTclay 0.01 0.02

equilibriunt and will fail sliding upward along the plane BC". The horizontal effective stressat this time will be oi, : rr',, the so-calledpassive pressure. In this case,

oi. a,u K_K,:' ,: , (r2.4) a,' 0,,

where K,, : passiveearth pressure coefflcient Figure 12.2shows the nature of variation of lateral earth pressurewith the wall tilt. Typicalvalues of L,L,,lH(LL,, : A'A in Figure12.1b) and LLpIH (LLr: A"A in Figure 12.1c) for attaining the active and passivestates in various soils are givenin Table 12.1.

AT.RESTLATERAL EARTH PRESSURE

12.2 Earth Pressureat Rest

The fundamental concept of earth pressure at rest was discussedin the preceding section. In order to define the earth pressure coefficient Kn at rest, we refer to Fig- 12.2 Earth Pressureat Rest 367

v /r- I! t- I I o'n= K,,Yz H

It=t'+o'tan0'

Figure 12.3 Earth prcssLrrcat rcst

urc 12.3,which 'l'hc showsa wall A Il rctaining a dry soil with a unit weight of 7. wall is static.At a dcpth z,

Vcrtical eflectivestrcss : o',,: yz,

HorizcrrrtalelTective stress : oi, : K,,yz So

oi. K,, : : at-rcst earth prcssurccoefficient (f,,-

For coarse-grainedsoils. the cocfficient of earth pressureat rest can bo esti- matedby usingthe cmpiriczrlrclationship (Jaky, l94a)

K,:1-singl' (12.s) where d' : drained friction angle. While designinga wall that may be subjcctedto latcrerlearth pressureelt rcst. one must take care in evaluatingthe value of K,,. Sherif, Fang,and Sherif (19g4),on the basisof their laboratory tests,showed that Jaky'sequation [<.trK,, tEq. (12.-s)] givesgood resultswhen the backlill is loosesand. However. for a densesand backfill. Eq. (12.5) may grosslyunderestimate the lateral earth prcssureat rest. This under- estimation resultsbecause of the processof compaction of backlill. For this reason. they recommendedthe designrelationship

* K": (1*sin @) .l: - r ls.s (12.6) L /.I(mrnl J

: where y,1 actual compacted dry unit weight of the sand behind the wall : 7,l1.rn; dry unit weight of the sandin the looseststate (Chapter 2) Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb

Tl - .'+ o'tanQ 1 Unit weight I _I H

Figure 12.4 Distributionol lateralcarth pressure at-rcst on a wall

For finc-grained,normally consolidatcdsoils, Massarsch (1979) suggested the following cquation for K,,: I Pr(%\l K,,:0.44+ 0.421-ur] (12.1)

For overconsolidatedclays, thc coeflicientclf carth pressureat rcst can be ap- proximatedas

Ko(overconsolidatcd): Krl,tnrtully.nn..lidor"d.;.\,m ( r2.8)

'l'he whcre OCR : ovcrconsolidationratio. ovcrconsolidationratio was dellned in Chapter 10as Preconsolidatior-rpressure, tri OCR: (r2.e) Presentelfective overburden pressure,oi,

Figure 12.4shows the distribution of lateral earth pressureat rest on a wall of height H retaining a dry soil havinga unit weight of 7. The total force per unit length of the wall, P,,.is equal to the area of the pressurediagram, so

P,,: lK,,yH2 (12.10)

12.3 Earth Pressureat Rest for Partially Submerged Soil

Figure 12.5ashows a wall of height FL The groundwater table is located at a depth 111below the ground surface, and there is no compensating water on the other side of the wall. For z < Hr,the lateral earth pressureat rest can be givenas oi,: K,,yz. The variation of o'1,with depth is shown by triangle ACE in Figure 12.5a.However, for z > H1 (i.e.,below the groundwatertable), the pressureon the wall is found from the effective stressand pore water pressure components via the equation - effectivevertical pressure: o',,: lHr + y'(z Ht) Q2J'D 12.3 Earth Pressureat Restfor Partially Submerged Soil I H1I

Saturatedunit weight l of soil = yr"r

(t 'Ht)-l i H1 I I Figure 12.5 I Distribution of earth l<-- K,,(1lHt+y'H)+y,,,H2 +l pressureat rest for partially submergedsoil

where T' : 7,,,t y?{r: the effectiveunit weight of soil. So the effectivelateral pressure at rest is o'1,: K,,o',,: K,,lyHt + y'(z - Hr)l (12.12) The variation of cj, with depth is shown by CEG B in Figure 12.5a.Again the lateral pressurefrom pore water is u: y,,,,(z- H) (12.13) The variation of a with depth is shown in Figure 12.5b. Hence, the total lateral pressurefrom earth and water at any depth z > l'1, is equal to

o1: Op -t U

: K,lyH, + y'(z - Hr)) t y,,,(z- H,) (r2.14)

The force per unit length of the wall can be found from the sum of the areas of the pressurediagrams in Figures 12.5aand 12.5band is equal to (Figure 12.5c) (12.1s)

Area Area Areas ACE CEFB EFG ancl IJK 370 Chapter 12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomb

Example12.1

Figure 12.6ashows a l5-ft-high retainingwall. The wall is restrainedfrom yielding.Calculate the lateralforce Puperunit lengthof the wall.Also, determine the location of the resultant force,

o'/, (lb/ftl ) u (lb/ft2)

)undwater table Sand c'- 0 Q'= 30' Y,ur= 122.4lb/fir

Figure 12.6

Solution K,,= I- Sinf' : 1 - sin30: 0.5

AtZ:0: aL=A; c'n=O',u*-0 At z : 10ft: ai = (10)(100)* 10001b/ft2 o'1,: Kor|* (0.5X1000): 500lb/ftz u=O At z : 15ft: (r! * (10)(100)+ (5)(122.4* 62.4)= 13001b/ft2 o'1,: Koo'o: (0.5X1300): 6501b/ft2 u - (5)(y,,)= (5X62.4)= 312lblf*

The variations of oi and u with depth are shownin Figures 12.6band 1"2'6c. : * + Area$ + Area 4 Lateral force P^"$ Area 1 Area 2 or $ /r\ /t\ /t\ p,: (i )lroylsooy+ (sxs00) + ( ; )(s)(1s0)+ ( i )tsif:tzl \L,/ \L/ \L/ - 2500+ 2500+ 375+ ?80= 6155lblft 12.4 Lateral Pressureon Retaining Walls from Surcharges-Based on Theory of Etasticity 371

The locationof the resultant,measured from the bottomof thewall, is z:) momentof pressurediagram about C

, l0\ /s \ /\\ (2soo)(sT;t + txoor(|)+ (3?s)(,;,1+ (780)(;I J./ \ = 4.7lft 6155 r

12.4 Lateral Pressureon Retaining Wallsfrom Surcharges-Basedon Theiry of Etasticity

Point-Load Surcharge The equationsfor normal stressesinside a homogeneous,elastic, and isotropic me- dium produced from a point load on the surfacewerc givenin chapter 9 [Eqs. (9.l0), (9.1l) and9.121. We now apply Eq. (9.10) to determinc the lateral pressureon a retaining wall causedby the concentratedpoint load Q placedat the surfaceof the backfill asshown in Figure r2.1a.lf the load Q is placedon the plane of the sectionshown, we can sub- : stitute y 0 in Eq. (9.10).Also, assumingthat ,p : 0.-5,we can writc , o /3x,2\ rrt- (t2.16) 2n\ /J / whereL : f7 + Z'z.Substitutin g x : mH and z : nH inroEq. ( 12.16),we have

, 3Q m2n tlr,: ^ - (12.n) " 2rHt (^t - ,,)"

The horizontal stressexpressed by Eq. (12.11)does not include the restrainingeffect of the wall. This expressionwas invesrigatedby Gerber (1929)and Spangler( l93g) with large-scaletests. on the basisof the experimentalfindings, Eq. (12.11)has been modified as follows to agreewith the real conditions:

For m > 0.4,

, 1.77Q mznz on: (12.l 8) nz @f +Ef For m * 0.4,

, 0.28Q n2 oa: (12.19) Hz (016 + ,rf Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb

q , r=mHI l*-*=,nn-l l..<+l Y +

l+"lYl H: l' lr ,f.

(b)

rril-J+ nt:+l l. loacl = ttl,, Strip // qlUrtI arcll :,.'/.

Figure 12.7 Latcral pressureon a retaining wall due to a (a) point load, (b) linc load,and (c) strip load

Line-Load Surcharge Figure 12.7bshows the distribution of lateral pressureagainst the vertical back face of the wall causedby a line-load surchargeplaced parallel to the crest.The modified forms of the equations[similar to Eqs.(12.18) and (12.19)for the caseof point-load surcharge]for line-load surchargesare, respectively,

4q rnzn - ^ .{ (r'h: llor m > u.41 (t2.20) rrH (mz+ 'q ,t)' ;' and 0.203q n CL= H (oEl?r (for m o 0.4) (r2.2r)

where ? : load per unit length of the surcharge. 12.4 Lateral Pressureon Retaining Walls from Surcharges-Based on Theory of Elasticity 373

Strip-Load Surcharge Figure 12.7eshows a strip-load surchargewith an intensity of q per unit area located at a distance rr , from a wall of height H. on the basis of the theory of elasticity, the horizontal stressat a depth z on a retaining structurecan be siven as

,',,:#@-sinBcosza) (12.22)

The anglesa and B are defined in Figure 72.1c.For actual soil behavior (from the wall restrainingeffect). the precedingequation can be modified to

Zs o'h: - sinp cos2a) (12.23) ;Gn

The nature of the distribution of oj, with depth is shown in Figure 12.lc.The force p per unit length of the wall causedby the strip load akrne can be obtained by inte- gration of oj, with limits of z from 0 to H.

Example12.2

Considerthe retainingwall shownin Figure12,8a where H : l0 ft. A line load of 800lb/ft is placedon the groundsurface parallel to the crestat a distanceof 5 ft from the backface of the wall.Determine the increasein the lateralforce per unit lengthof the wall causedby the line load.Use the modifiedequation given in Sec- tion12.4.

oi, (lb/lir)

,-Theorctical .{ shape

". :\6!16l 800tb/ri I I I 2l l+)u.+l t.^.. OU.trI :4 '---'--- , ' " c t 1T a lr O6 t 4 | .\t.^) H= l0ti al

t )\ 6i'_. t" ,I i 5r I ,lr!,:

Figure 12.8 374 Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb

Solution We are given l/ : 10 ft, q : 800lb/ft, and

*:4:0.5 > 0.4 1L) So Eq. (12.20) will apply: 4q m2n on:. rrH Qrf;W

Now the following table can be prepared: n=nz4q m2n ,H 1m2+ n2)2 o't llbllizl 0 101.86 0 0 0.2 101.86 0.595 60.61 0.4 101.86 0.595 60.61 0.6 101.86 0.403 41.05 0.8 101.86 0.252 25.67 1.0 i01.86 0.16 16.3

Refer to the diagramin Figure 12.8b. Area no,

(j)t'lr*'ut) = 6o'61rb/rt (|)rrlt*.ut + 60.61)= r2r.2z tbtrt

(])o,*o.ut + 41.0s)= 101.66 rb/rt (i)trxrr.* + 2s,67)= 66.72tbtrt (f;)elt t.u,+ 16.3)= 41.e7tbtrl

Toral= 392.18lb/ft - 390lb/ft

RANKINE'SLATERAL EARTH PRESSURE

12.5 Rankine's Theory of Active Pressure

The phrase plastic equilibrium in soil refers to the condition where every point in a soil mass is on the verge of failure. Rankine ( 1857)investigated the stressconditions in soil at a stateof plasticequilibrium. In this sectionand in section12.6,we deal with Rankine's theory of earth pressure. 12.5 Rankine'sTheory of Active Pressure

+lAr!- -2c',[9, A'A - T--*-- Unit weight of soil = T l-l lf=c'+o'tanQ' T a", e)I I a I tI l-l yzK,,- 2c'l K,,

(c)

E 6 '.h 1 o'* q r"o; 6',, Normalstress \\ ,,\,\,' \ /,v./ I

(h)

Figure 12.9 Rankine's active carth pressure

Figure 12.9ashows a soil massthat is bounded by a frictionlesswall, AB, that extends to an infinite depth. The vertical and horizontal effective principal stresses on a soil element at a depth z areo'o and oi,, respectively.As we saw in Section 12.2, if the wall ,48 is not allowed to move, then o'1,: K,,a',,.The stresscondition in the soil element can be represented by the Mohr's circle a in Figure 12.9b.However, if the wall AB is allowed to move away from the soil mass gradually, the horizontal principal stresswill decrease.Ultimately a statewill be reachedwhen the stresscondition in the soil element can be represented by the Mohr's circle b, the state of plastic equilibrium, and failure of the soil will occur. This situation representsRankine's active state,and the effective pressure oi on the vertical plane (which is a principal plane) is Rankine's active earthpressure. We next derive oi in terms of y, z, c' , and $' from Figure 12.8b CD CD slnd : AC AO+OC 376 Chapter 12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomb

But

cu - oo CD : radius of the failure circle :

AO : c' cot$' and

oc: o' I o'.." 2 So

a',, - c'r, : 2 sinS' ', -* t2t-,att'" 1v,co.(t, +"'

c'cos' . q+4sin :

I - sind' cos(r' o',r: o',,, - 2c' (12.24) * ,.r* I *sinS' But

U,,, - vcrtical cf'fbctivcoverburden pressure - I sinrf' :tan'(45-t)-/ d'\ I *sin{' and cosd' / 6'\ - tan(45- l +,*d , ) Substitutingthe precedingvalues into Eq. (12.24),we gel

cL: yztan'?(+s- * +) zc'tun(+r 5) (r2.2s) t The variation of o'owith depth is shown in Figure 12.9c.For cohesionlesssoils, c':0and

^/ d'\ au-c',,tan'(4.5-t) (12.26)

The ratio of oj, to oi, is called the coefficientctf Rankine's active earth pressure and is siven bv 12.6 Theory of Rankineb Passive Pressure 377

(ra ,( ,- d'\ Ko:-,- =t?n'[ 45- | (r2.27) oo \ 2/

Again, from Figure 12.9bwe can see that the failure planes in the soil make -r (45 + g' l2)-degreeangles with the direction of the major principal plane - that is, the horizontal. These are called potential slip planesand are shown in Figure 12.9d. It is important to realize that a similar equation for a" could be derived based - on the total stressshear strength parameters that is, r, : c * o tan S. For this case,

oo: - * z,t^"(as* (12.28) r2,""(* 9) !)

12.6 Theory of Rankine3 Passive Pressure

Rankine'.spassive state can be explainedwith the aid of Figure 12.10.A B is a frictionlesswall that extendsto an infinitedepth (Figure12.10a). The initial stresscondition on a soil element is representedby the Mohr'.scircle a in Figure 12.l0b. If the wall is gradualfyprzshedinto thesoil mass,the cffectiveprincipal stressoi, will increase.Ul- timately the wall will reacha situation where the stresscondition for the soil element can be expressedby the Mohr\ circle b. At this time, failure of the soil will occur. This situation is referred to as Rankine'spttssive stote. The lateral earth pressurerri,, which is the maior principal stress,is called Rankine'spassive earth pressure. From Figure 12.10b,it can be shown that

o'p:o.'otunt(+s . * ,r' ,^n(ot. +) +)

: . + zc'tan( t . (12.2e) wun?(+t *) +)

The derivationis similarto thatfor Rankine'sactive state. Figure12.10c shows the variation of passivepressure with depth.For cohesionlesssoils (.' : 0), d'\ o'r: o',,tan'(4-5 "/ * t ) or : 2 *,: tan,(+s. +) (12.30)

Ko (the ratio of effectivestresses) in the precedingequation is referred to as the coefficientof Rankine'spassive earth pressure. 378 Chapter 12 LateralEarth Pressure: At-Rest,Rankine, and Coulomb

--+-lALl<- AA' 1' 1 1 z I I =-l=-- tzxr------l I 2r'lE, (c)

o',, I I

Figure 12.70 Rankine'.spassive earth pressure

' The points D and D on the failure circle (see Figure 12.1 0b) correspondto the +(45 - slip planes in the soil. For Rankine's passivestate, the slip planes make O'12)' degree angleswith the direction of the minor principal plane - that is, in the horizon' tal direction. Figure 12.10dshows the distribution of slip planes in the soil mass.

12.7 Yielding of Wall of Limited Height

We learned in the preceding discussionthat sufficient movement of a frictionlesswall extending to an infinite depth is necessaryto achieve a state of plastic equilibrium' However, the distribution of lateral pressure against a wall of limited height is very much influenced by the manner in which the wall actually yields. In most retaining walls of limited height, movement may occur by simple translation or, more fre- quently, by rotation about the bottom. r 12.7 Yielding of Wall of Limited Height 379

LLa l**]-t,,+l

45 E 2 f'\ I H I I

45-q

Figure 72.1? Rotationof frictionlesswall about the bottom

For preliminary theoreticalanalysis, let us considera frictionlessretaining wall representedby a plane AB as shown in Figure l2.l1a. If the wall ,4_Brotates sufficiently about its bottom to a position A' B, then a triangular soil mass,4BC, adjacent to the wall will reach Rankine'sactive state. Because the slip planesin Rankine'sac- + tive statemake anglesof (45 + O' 12)degrees with the major principal plane,the soil massin the stateof plasticequilibrium is bounded by the planeBC, ,which makesan angle of (45 + 0'12) degreeswith the horizontal. The soil inside the zoneABC, un- dergoesthe same unit deformation in the horizontal direction everywhere,which is equal to LL,,lLu. The lateral earth pressure on the wall at any depth z from the ground surfacecan be calculatedby using Eq. (12.25). In a similar manner! if the frictionlesswall,4B (Figure lz.rlb) rotates suffi- cientlyintothesoilmasstoapositionA"B, lhenthetriangularmassofsoil ABC, will reach Rankine's passivestate. The slip plane BC" bounding the soil wedge that is at a stateof plasticequilibrium will make an angleof (45 - g,12) degreeswith the Chapter 12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomb

horizontal.Every point of the soil in the triangular zoneABC" will undergo the same unit deformation in the horizontal direction, which is equal to LLplLr,. The passive pressureon the wall at any depth z can be evaluatedby using F'q. (12.29).

12,8 Diagrams for Lateral Earth Pressure Distribution against Retaining Walls

Backfill-Cohesionless Soil with Horizontal Ground Surtace Active Case Figure 12.12ashows a retaining wall with cohensionlesssoil backfill that has a horizontal ground surface.The unit wcight and the angleof friction of the soil are y and rf' , respectively. For Rankine'sactive state, the earth pressureat any clepthagainst the retaining wallcan be given by Eq. (12.25): o',,: K,,yz, (lVote:c' : 0.)

H +P,, U 3 I l+ (,yH+l

Failurewedge v H a H Itt_ lry l3 l+ --->l l-<-- K,,yH (b)

Figure 12.12 Pressrtredistribution against a retainingwall for cohensionlesssoil backfill with horizontalground surface: (a) Rankinebactive state; (b) Rankine'spassive state 12.8 Diagrams for Lateral Earth PressureDistribution against Retaining Walls 381

Note that ai increaseslinearly with depth, and at the bottom of the wall, it is o',,: K,1H (12.31)

The total force per unit length of the wall is equal to the area of the pressurediagram, so

P,,: !K,,yHz 02.32)

Passive Case The lateral pressuredistribution againsta retaining wall of height H for Rankine'.spassive state is shown in Figure 12.12b.The lateral earth pressureat any depth z [Eq. (12.30),c' : 0] is o',, - K,,IH (12.33)

The total force per unit length of thc wall is - P,, )K gHt 1tz.z+1

Ba ckf i I I - Parti a Ily Subm erged Cohen si o n I ess Soil Supporting a Surcharge Active Case Figure l2.l3a shows a frictionless retaining wall of height 11 and a backfill of cohensionlesssoil. The groundwater tablc is located at a depth of H, below the ground surface.and the backfill is supporting a surchargepressure of q per unit area. From Eq. (12.27),the effectiveactive earth prcssureat any depth can be given by

{r'r,-- K,,rr',, (12.35)

where rr',,and o',,- the effective vertical pressure and lateral pressure, respectively. Atz-0.

o.-o',,:q (12.36) and

o',,: K.q (t2.31)

Atdepthz: Ht,

ol,: (q + vH,) ( 12.38) and

c',,: K,,(e+ yHt) 02.39) At depthz: H,

a|,: (Q* yH, * y'Hz) 02.40) and o',,: Ko(e + yHt * y'H) (12.4f) - wherel' : y,nt 7,.. The variationof oi with depthis shownin Figure 72.13b. 382 Chapter 12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomb

Surcharge= r7 d'I 4r+; I 11

v a ll IT lnI qK,,l<-- |4 l l[]n1,[l N ll++l l<___>l++l K,,(q+yHt+"{'H)) Y".Hz Krlg + yHt) Ka!',H2+\",H2 (b) (c) (d)

Figure 12.13 Rankine'.sactive earth pressure distribution against a retainingwall with partiallysubmerged cohcsionless soil backfill supporting a surcharge

The lateral pressure on the wall from the pore water between z : 0 and H1 is : 0, and for z ) H1, it increaseslinearly with depth (Figure L2.13c).At z H,

u: Y',,'H2

The total lateral pressurediagram (Figure 12.13d)is the sum of the pressurediagrams shown in Figures 12.13band 12.13c.The total active force per unit length of the wall is the area of the total pressure diagram. Thus,

Pu: K,,qH+ lx,yul * K,,yH1H2+ tr(x,y' + y,,)H3 (2.42) 12.8 Diagrams for Lateral Earth PressureDistribution against Betaining wails

T IH, I I I H t I H. II II

I Hl I K,,tyH1 + qt I \ I H H2 I nt.\ I l.\ |<-+|+------l l<------+l l<------+l<-_-_____+l qKn Kr(.yH1+y'H2) "1,Hz K,,(q+ yH11 Kry'Llz+\*Hz (b) (c) (d)

Figure 12.14 Rankine'spassive earth pressure distribution against a retainingwall with partiallysubmerged cohesionless soil backfillsupporting a surcharge

Passive Case Figure 72j4a shows the same retaining wall as was shown in Fig- ure r2.l3a. Rankine'.spassive pressure at any depth against the wall can be given by Eq. (12.30): o', : Kr{r'r, Using the preceding equation, we can determine the variation of o! with depth, as shown in Figure 12.14b.The variation of the pressure on the wall from water with depth is shown in Figure 72.74c.Figure12.14d shows the distribution of the total pressure ao with depth. The total lateral passiveforce per unit length of the wall is the area of the diagram givenin Figure 10.11d,or po: KoeH + lxoyHl * KorHtHz+ +(Kfl, + y_)HZ O2.43) Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb

4s+-2I , ?'---..----.-t +I ?. H

l+l l€l l<------l K,t\H 2r'{K, K,,yH - 2c'{l{,, (b) (c) (d)

Figure 72.15 Rankine'sactive earth pressuredistribution against a retainingwall with cohesivesoil backfill

Backfill- Cohesive Soil with Horizontal Backfill Active Case Figure 12.75ashows a frictionless retaining wall with a cohesive soil backfill. The active pressure against the wall at any depth below the ground surface can be expressedas [Eq. (12.25)l

oL: KoYz - 2f K-c'

The variation of Koyz with depth is shown in Figure 12.15b,and the variation ot2{K-c'with depth is shown in Figure 12.75c.Note that 2ffic' is not a function 12.8 Diagrams for Lateral Earth PressureDistribution against Retaining Walts 385

of "1,'hence,Figure 12.15cis a rectangle.The variation of the net value of oj, with depth is plotted in Figure 12.15d.Also note that, becauseof the effect of cohlsion, ai is negativein the upper part of the retaining wall. The depth 2,,at which the active pressurebecomes equal to 0 can be found from Eq. (12.2-5)as K,,y2,,, 2fK,c'' - 0

v{K (12.44)

For the undrained condition - that is, rl.r: 0, K,, : tan245 : l, and c : c,, (undrained cohesion)- from Eq. (12.28),

a- LLrt (12.4s) v

so. with time, tensilecracks at thc soil-wallinterface will developup to a depth 2,,. Thc total activeforce pcr unit length of thc wall cernbe found from the area of the total pressurediagram (Figure l2.l-5cl),or

P,,: \ K,,yl12 - 2\,fK,,c''H (12.46) Forthe@:0condition,

P,,-lyII2-2c,,H (1) 41\

For calculationof the total activc force, common practiceis to take the tensile cracksinto account.Because no contact existsbetween the soil and the wall up to a depth of z.(,after the devclopment of tensile cracks,only the active pressuredistri- - hutionagainsilhewallbelweenz 2ll(yVK,,)ancl 1/(Figurcl2.l.5d)isconsidercd. In this case,

p"=l(K,yH -zffit,)('\ o __?+) v\/K,,/ .,r2 : - $K"yHz 2t/I{,c'H + 2:*v$ :, (12.48)

Forthe@:0condition.

-2 P,: jyHz - 2cuH+ 2!]t (r2.4e) 386 Chapter 12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomb

t\ i\ H \

\ 6r \

*l z,'r.@l* K,,yH+l

(a) (b)

Figure 12.16 Rankine'spassive earth pressuredistribution againsta retaining wall with cohesivesoil backfill

Passive Case Figure I2.l6a showsthe same retaining wall with backfill similar to that consideredin Figure 12.I5a.Rankine's passive pressure against the wall at depth z can be given by lE.q. (12.29)l

o',,: Kryz+2t/Krc'

Atz:0,

o'p:2lKpc' (12.s0)

andat z: H,

o'o: KrlH + 2f Krc' (12.s1)

The variationof o',,withdepth is shownin Figure|2.t6b. The passiveforce per unit lengthof the wall canbe found from the areaof the pressurediagrams as

ru: iKotHz + 2{\c'H $2.s2)

For the d : 0 condilron, Kr: 1 and

Pr: lyHz * 2c,,H (12.s3) 12.8 Diagrams for Lateral Earth PressureDistribution against Retaining Walls 387

Example12.3

An 6 m high retainingwall is shownin Figure 12.17a.Determine a. The Rankineactive force per unit lengthof the wall andthe locationof the resultant b. The Rankinepassive force per unit lengthof the wall andthe rocationof the resultant T I Y=l5 kN/mr 6m 0'=3o' c'=0

(a)

70.2kN/m T t l<___+l Figure12.17 23.4kN/rn2 Diagramsfor l.+346.-5 kN/m2-->l determiningactive, (b) and passiveforces

Solution a. Becausec' : 0, to determinethe active force. we can useEq. (12.27): o|: Koa| = K,TZ l-sind'_ 1-sin36 ^-_ :o)(t " l*sinS' l*sin36 At3:0,0'o= 0;ate = 6m,

a| = (0.26)(15)(6)- 23.4kN/m2 r*t The pressuredistribution diagram is shownin Figure12.ffib.The active forceper unit lengthof the wall is as follows: P" = i(6){23.4) : 70.2kN/m Also, 6m:2m 1 Chapter 12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomb

h. To determinethe passiveforce, we are giventhat c' : 0. So,from Eq.( 12.30). rr'o: Kocr|= KrTz /<,:1+##-i+##:385

Arz:o,c'r-0; atz:6m, on : (3.85)(lsX6):346'5 kN/m2 The pressuredistribution diagram is shownin Figure 12.17c.The passive force per unit lengthof the wall is

Pp : i(6x-r46s) : 103e.5kN/m Also. z:ry:2m

Example12.4

For the retainingwall shownin Figure 12.18a,determine the forceper unit width of the wall for Rankine'sactive state. Also lind the locationof the resultant.

y - 16kN/m3 : . T 1.j,,,urounq-^-^..-,, 'll.:' " : Jnt i"" Q = rtr L:'!' watertable I c =" + .: : r I = 3"T rn ffl T,or'-i,, l8 kN/nrr* + ;.ifri.; 3.5';;. I il:r::ri 0'= y ;i;i!il r.'= 0

t 3rn I i 3m+ +l rol<- l+29.+3+l +l 13.01<- 36.1+l --->l 19.67l<- (b) (d)

Figure 1Z 18 Retaining wall and pressurediagrams for determining Rankine'sactive earth pressure.(Note:The units of pressurein (b), (c), and (d) arekNlm') 12.8 Diagrams for Lateral Earth PressureDistribution against Retaining Walls 389

Solution Giventhat c' : 0, we known thatoL = Koa,,,.For the upperlayer of the soil,Rank- ine'sactive earth pressurecoefficient is

K,,:K-,,-l-sin3o'-lq u\" | *sin30. 3 For the lower layer,

Ko: Ko(z\:1 ;'l'11, : o.2jr I -l_ sln -t-)" = At z 0, oL: 0. At z : 3 m (just insidethe bottomof the upperlayer), aj, : 3 x 16 : 48 kN/m2.So

o',: Kr,1yr',,:{ x 48 : 16kN/m2

Again,at z - 3m (in the lowerlayer), a',, : 3 x 16 : 48 kN/mz,and

ou: Kuplo',,:(0.271) X (48): 13.0kN/m2 Ate=6m.

o",= 3 x 16+ 3(18* 9.81)= 72.-57kN/m2 t 7,,, and

o',,= Ku121o',,= (0.271) x (12.57)= 19.67kN/m2

The variationof oj, with depthis shownin Figure 12.lBb. The lateralpressures due to the pore waterare asfollows: At7=g' u-0 Ate=3m: u=0

Atz : 6m: u = 3 x Tu,: 3 X 9.81= 29.43kN/m2 The variationof u with depth is shownin Figure r2.1&c,and that for ou (total active pressure)is shownin Figure 12.18d.Thus, p": (ix3x16)+ 3(13.0)+ (+X3X36.1) :24+3e.0 + s4i1s:117.15kN/m

The locationof the resultantcan be found by taking the &romentabout the bottom of the wall: : z+(z+ . ,' o(i). '- J) "(i) 117.15 : 1.78m Chapter 12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomb

Example12.5

A frictionlessretaining wall is shownin Figure l2.Iga.Determine the activeforce, Po,after the tensilecrack occurs.

q = 15kN/m: -6.64 kN/m2 ++t*rtll l-l I I T= 16.5kN/rn: 6m Q'=26" I c'= l0 kN/m2 I I --+l l*lt.qz tN/rnr

(r) rD, Figure12.19 (a) Frictionless retaining wall; (b) active pressure distribution diagram

Solution Giventhat $' - 26o,we have I * sinf' 1- sin26 K' = : : u'3e lTlin7 1 + r,"26 FromEq. \12.25). oL: Koo'o'2c'{K Atz:0, oL: (0.39X1s)- (2X10)\639 : -6.64kN/m2 Atz:6m, oi = (0.3e)lr5+ (6X16.5)l- (2x10)1rc:39: 31.e7 kNlmz ij Thepressure distributio" otu**1';;*T;rttuure r$lu' Fromthis diagram",{

*-= j I z 6*z '.$ .: or; I * 1.03m 'i$ l After the tensilecrack occurs, : p": l(6 - z)(3r.97): |1+.st112t.9't) :79.215 kN/m rf I 12.8 pressure Diagrams for Lateral Earth Distribution against Retainingwatts

Example12.6

A frictionlesrr:Tj.ning.Iall-is shown in Figure 12.20a.Find the passiveresistance (Po)on the backfilland the locationof the resultantpassive force.

q = l0 kN/m2

I llJl Y = l5 kN/mj I $'=26" 1m = I r" 8 kN/ml I l++l+ 1.53.6kN/rn2 +l -51.2 kN/nr2 (b) Flgure 12.20 (a) Frictionlessretaining wall; (b) passivepressure distribution diagram Solution Giventhat $' : 26o,it followsthat I + sind' 1 * sin26" *- _ _ _ )

12,9 Rankine Active and Passive Pressure with Sloping Backfill

In Sections12.-5 through 12.8,we consideredretaining walls with vertical backsand horizontal backfills. In some cases,however, the backfill may be continuously sloping at an anglea with the horizontal asshown in Figure 12.21for activepressure case. ln such cirscs.the directionof Rankine'sactive or passivepressures are no longer horizontal. Rather, they are inclined at an angle a with the horizontal. If the backfill ', : is a granular soil with a drained friction angle

o'rr: YzKu

where

K, : Rankine'sactive pressure coefficient

cosa- Vcos2a-cos26' = cosa (12.s4) cosd+ \,6P;: "*?

The activeforce per unit length of thc wall can bc given as l^ P,,'2: : K,,vH' (12.ss)

action of thc resultantacts at a distanceof H 13measured from the bottom The finc of '' of the wall. Table 12.2gives the valuesof K,, for various combinationsof a and f In a similar manner, the Rankine passive earth pressure for a wall of height H with a granular sloping backfill can be representedby the equation P, : ltH2 K,, (12.s6)

Frictionless

Figure 12.21 Frictionlessvertical retaining wall with slopingbackfill 12.9 Ranking Active and Passive Pressure with Sloping Backfilt

Table 12.2 Valuesof K,, [Eq. (12.54)l Q' (des) --+ I a (degl

U 0.361 0.333 0.307 0.283 0.260 0.238 0.217 .5 0.366 0.337 0.311 0.286 0.262 0.240 0.219 10 0.380 0.350 0.32r 0.294 0.270 0.246 0.225 15 0.409 0.373 0.34r 0.31I 0.283 0.258 0.235 20 0.461 0.4t4 0.374 0.338 0.306 0.277 0.250 25 0.573 0.494 0.434 0.385 0.343 0.307 0.275

Table 12.3 PassiveEarth PressureCoefficient, f,, IEq. (12.57)l Q' (des) --> J a (deg)

0 2.770 3.000 3.2-55 3.537 3.u52 4.204 4.599 -5 2.715 2.943 3.196 3.476 3.7uu 4.136 A <)1 10 2.551 2.775 3.022 3.295 3.5gr3 3.937 4.316 1,5 2.281 2.502 2.740 3.003 3.293 3.61-5 3.977 20 t.9l13 2.132 l. _11)I 2.612 2.8rJ6 3.lrig 3.526 25 1.434 1.664 1.1394 2.135 2.394 2.676 2.987 where

cosrr + V!.rt'" ..f d Kr, : cos a (12.s7) cos., - \4;t:*? is the passiveearth pressurecoefficient. As in the casco[ the active forcc, the resultantforce P,,'H13is inclined at an anslc a with the horizontal and intersectsthe wall at a distanceof measuredfrom the bottom of the wall. The valuesof K, (passiveearth pressurecoefficient) fbr various valuesof a and (t' are given in Table 12.3. c'-S'Soil The precedinganalysis can be extendedto the determination of the activeand passive Rankine earth pressurefor an inclined backfill with a c'-rf' soil. The details of the mathematicalderivation are given by Mazindrani and Ganjali (rggi).For a c'-$' backfill, the active pressure is given by o'" : yzK,, : yz.Ki cosa ( 12.58) whereKu : Rankineactive earth pressurecoefflcient and K.. K;: (r2.se) *r" The passivepressure is givenby : : o',, TzKo yzK", cosa (r2.60) whereKo : Rankinepassive earth pressurecoefficient and A' K'i,: ::! (12.61) ' cos.Y 394 Chapter 12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomb

Also, 1 K"",K':,: ' \ cos-@ ' cos2a+ 2({).or4' sin rf \yz / ' t " o( )'.or' a' , B( -).or' a sin@' .o, 4'l {. \yz/ \yz/ .i)' (12.62)

Table 12.4 Variationof K'1,and K'i,* c' yz

a (deg) K';IK; 0.025 0.05 1.0 u

c 6' :25" -0.8683 U 0.4059 0.3140 0.3422 0.2'784 -0.2312 0 2.4639 2.5424 2.6209 2.7779 4.0336 s.6033 -0.8733 5 0.4133 0.3805 0.3478 0.2826 -0.2332 5 2.4r95 2.4989 2.5782 2.1367 3.9986 5.5713 -0.8884 l0 0.4376 0.4015 0.3660 0.2960 -0.2394 10 2.2854 2.3680 2.4502 2.6]35 3.8950 5.4765 -0.9140 t5 0.4860 0.4428 0.4011 0.3211 -0.2503 15 2.0575 2.1474 2.235'7 2.4090 3.7264 5.3228 (continued) 12.9 Ranking Active and Passivepressure with Stoping Backfitl

Table 12.4 gives the variation of K'/ and Ki with at,c, llz, and 0, . For the active case,the depth of the tensile crack can be given as

2c' (r2.63) I

Table 12.4 (continued\

c' yz a (des) K:tK; 0.0 0.05 0.1 1.0 d $' :30" 0 0.3333 0.3045 0.2756 0.2179 -0.2440 -0.8214 0 3.0000 3.0866 J.t t 3z 3.3464 4.7321 6.4641 5 0.3385 0.3090 0.2795 0.2207 *0.2460 -0.tt260 -5 2.9543 3.0416 3.1 288 3.3030 4.6935 6.4282 l0 0.3549 0.3233 0.2919 0.2297 -0.2522 ,0.8399 l0 2.8176 2.9070 2.9961 3.1737 4.5794 6.3218 15 0.386r 0.3502 0.3150 0.2462 -0.2628 -0.8635 1-5 2.5900 2.6836 2.7766 2.9608 4.3936 6.1489

e Q' :35' U 0.2710 0.2450 0.2r t39 0.1669 -0.2496 *0.7'/01 '7.5321 0 3.6902 3.7862 3.8823 4.0744 5.6172 5 0.2746 0.2481 0.2217 0.1688 -0.2-515 -0.7744 '7.4911 5 3.6473 3.1378 3.8342 4.0271 5.-s678 10 0.2861 0.2581 0.2303 0.1749 -0.2575 -0.7872 10 3.4953 3.5933 3.6912 3.8866 s.4393 7.3694 15 0.3073 0.2764 0.2459 0.1860 -0.2678 -0.8089 15 3.2546 3.3555 3.4559 3.6559 5.2300 7.1715 f Q' :40' 0 Ki, 0.2174 0.1941 0.1708 0.1242 -0.2489 -0.7152 0 K'; 4.-5989 4.7061 4.8134 5.0278 6.7434 8.8879 5 K'; 0.2200 0.1964 0.1727 0.1255 -0.2507 -0.7190 5 K'; 4.5445 4.6521 4.7597 4.9747 6.693-5 8.8400 10 Ki 0.2282 0.2034 0.r787 0.1296 -0.2564 -0.7308 10 K'; 4.3826 4.4913 4.5999 4.8168 6.5454 8.6980 15 K'; 0.2429 0.216I 0.1895 0.r370 -0.2662 -0.7507 15 K';, 4.1168 4.2275 4.3380 4.5584 6.304\ 8.4669

* After Mazindrani and Ganjali (1997) Chapter 12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomb

Example12.7

Refer to Figure 12.21-on page392. Given that H = 6.1m, a - 5o,f = 16'5kNlm', 6' - 20o,c' : 10 kN/m2,determine the Rankine activeforce Pnon the retaining wall after the tensilecrack occurs. Solution From Eq. (12.63),the depth of tensilecrack is

2c' 1 * sind' (2X10)/T + stn20 1* sinf' 16.-sVl-sin20 So Atz:0: o',:0 Atz:6.1 m: o'u: yzK'lcosa c' : l0 : o'l ,," (ra5x6t, FromTable 12.4,forat = 5" andc'lyz: 0.1,the magnitudeof Kii : 0.3565.So

oi, : (16.5)(6.1)(0.3565)(cos 5") - 35.t5kN/m2 Hence, r": - = - r.t3)(3s.i5)- 78.t kN/m )rn 2,,)(35.75)lrc.t

COULOMB'SEARTH PRESSURE THEORY

Morc than 200 ycars ago, Coulomb (1116)presented a theory for activeand passive earth pressurcsagainst retaining wetlls.In this thcory, Coulomb assumedthat the failurc surfaccis a planc. The wull .f'rictionwas tzrken into consideration.The following sectionsdiscuss the generalprinciples of the derivation of Coulomb'searth pressure theory for a cohesionlessbackfill (shearstrength defined by the equation 11= c' tan $').

12.10 Coulombb Active Pressure

Let AB (Figure 12.22a)be the back face of a retaining wall supporting a granular soil, the surfaceof which is constantlysloping at an angle a with the horizontal. BC is a trial failure surface.In the stability considerationof the probable failure wedge ABC, the following forces are involved (per unit length of the wall): 12.10 Coulomb'sActive Pressure 397

90-0+cr 90+e+6B+0 4,,-,6 i W I I t1 +I \rr90+e- -Y'l I p-0

D\ \-

(a) (b)

Figure 12.22C.

l. tr44the weight of the soil wedge. 2. F, Lheresultant of the shcar and normal forceson the surfaceof failure. BC. This is inclinedat an anglcof {' to the normal drawn to thc plane BC. 3. P,,,the active force per unit length o1'thewall. The clirectiono1 P,,is inclined at an angle6 to the normal drawn to the l'aceof the wall that supportsthe soil.6 is thc angleof friction betwecnthe soil and thc wall.

The force triangle lor the wedgc is shown in Figure l2.z2b. From the law of sincs.we havc

J,, (12.64) sin(90+0+6-F+4t') sin(B 4t')

sin(B- rf') P,,: W (r2.6s) sin(90+9+6-B+O') The pre.cedingcquation cernbe written in the form - - cos(0 B)cos(0* a)sin(B- r/') e" * - rt oot )tn.lcos2Flsin(B a)sin(90+ g + D B + O')) : where 7 unit weight of the backfill.The valuesof y, H, 0, a,6, ,and 6 are constants. and B is the only variable.To determine the critical value of 6 for maximum p.,. we have dP. ":0 (r2.61) dp 398 Chapter 12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomb

: Table 12.5 Valuesof K, [Eq. (12.69)]for 0 : 0o,a 0o 6 6eg) --+

.l-d' (deg) 10 15 28 0.3610 0.3448 0.3330 0.3257 0.3203 0.3186 30 t r.-r-t-tJ 0.3189 0.3085 0.3014 0.2973 0.2956 32 0.3073 0.2945 0.2853 0.2791 0.2755 0.2745 34 0.2827 0.21t4 0.2633 0.2579 0.2549 0.2542 36 0.2596 0.249'7 0.2426 0.2379 0.2354 0.2350 38 0.2379 0.2292 0.2230 0.2190 0.2169 0.2167 40 0.2t74 0.2089 0.2045 0.2011 0.1994 0.1995 42 0.1982 0.1916 0.1870 0.1841 0.1828 0.1831

After solvingEq. (12.67),when the relationship of B is substitutedinto Eq. (12.66),we obtain Coulomb'sactive earth pressureas

Po: lKoyH2 ( 12.68)

whereK,, is Coulomb'sactive carth pressurecoefficient and is givenby

6,: ,"ot'(d',lf),,,,,,', - .-u 02.69) gcos(6 ol)sin(q' t) cos2 + 0)l'L I * .,/sin(9-+ - l' \ cos(6+ d)cos(O d) J

Note that when d - 0', g - 0", and 6 : 0", Coulomb'sactive earth pressureco- eflicient becomesequal to (1 - sin rb')l(1+ sin {'), which is the same as Rankine's earth pressurecoefficient given earlier in this chapter. - The variation of the valuesof K,,for retainingwalls with a vertical back (0 0') anclhorizontal backfill (a : 0") is given in Table 12.5.From this table, note that for a given value of @', the effect of wall friction is to reduce somewhat the active earth pressure coefficient.

12.11 GraphicSolution for Coulomb'sActive Earth Pressure

An expedient method for creating a graphic solution of Coulomb's earth pressure theory was given by Culmann (1875). Culmann'ssolution canbe used for any wall friction, regardlessof irregularity of backfill and surcharges.Hence, it provides a power- ful technique for estimating lateral earth pressure.The steps in Culmann's solution of active pressure with granular backfill (c' : 0) are described next, with reference to Figure I2.23a:

1. Draw the features of the retaining wall and the backfill to a convenient scale. - - 2. Determine the value of ry'(degrees) : 90 0 6, where 0 : the inclination of the back face of the retaining wall with the vertical, and 6 : angle of wall friction. 12.11 Graphic Solution for Coulombb Active Earth Pressure

(b)

Figure 12.23 Culmann's solution for active carth rrressure

3. Draw a line BD that makes an angle d' with the horizontal. 4. Draw a line BE that makes an angle r/ with line B D. To considersome trial failure wedges,draw lines BCt, BC2, BC., . . ., 8C,,. 6. Find the areasof ABCt, ABC2, ABCj, . . ., ABC,,. 7. Determine the weight of soil, w, per unit length of the retaining wall in each of the trial failure wedgesas follows: W1 : (Area of ABC,) x (7) x (1) W2: (Areaof ABC) x (7) x (1) : (Rr"u of ABC.) x (7) x (1) %

W,: (Area of ABC,) x (7) x (1) 8. Adopt a convenientload scaleand plot the weightsW,,Wr,W3,. ,Wndeter- mined from step 7 on line BD. (Note: Bc1 : W1,Bcz: W2,Bc1 : Wt, . . ., Bc,,: Wo) 9. Draw cp\, c2c'2,c{\, . . ., c,,c'nparallelto the line BE. (Note:c\, c2,cj, . . . , c,,, are located on lines BCt, BC2, BCr, . . ., BC,, respectively.) 10. Draw a smooth curve through points c1,c2, c\, . . ., c',.This curve is calleclthe Culmann line. 11. Draw a tangent B'D' to the smooth curve drawn in step I0. B,D,is parallel to line BD. Let c'"bethe point of tangency. 12. Draw a line coc'oparallelto the line BE. 13. Determinethe activeforce per unit lengthof wall as P, : (Lengthof coc) x (Load scale) 14. Draw aline Bc'oC,,.ABC, is the desired failure wedge. Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb

Note that the constructionprocedure entails,in essence,drawing a number of force polygons for a number of trial wedges and flnding the maximum value of the active force that the wall can be subjected to. For example, Figure 12.23bshows the force polygon for the failure wedge,4BC, (similar to that in Figure 12.22b),in which

l4z: weight of the failure wedge of soil .48C,, P,,: active force on the wall F - the resultantof the shearand normal forcesacting along BC,, B - LC,,BF (the angle that the failure wedge makes with the horizontal)

Thc force triangle (Figure 12.23b)is simply rotated in Figure I2.23aand is representedby the triangleBc,,ci,. Similarly, the forcetriangles Bclc! , Bc2c2,Bcp\, '.., Bc,,c',,correspond to the trial wedgcsABCt, ABC2, ABC;,. . ., ABC,,. The prcccding graphic pKtcedurc is given in a step-by-stepmanner only to fa- cilitatc basicunderstanding. These problems can be easilyand cffectivelysolved by thc use ol computcr programs. Thc Culmann solution providesus with only the magnitudeof the activeforce per unit length of the retaining wall - not with the point of applicationof the resultant. Thc analyticprocedure used to find the point of applicationof the resultantcan bc teclious.For l.hisreason. an apprcximatc method, which does not sacrif,cemuch accuracy,can be used.This n-rethodis demonstratedin Figure 12.24,in whichABC is the failurc wedge determined by Culmzrnn\ method. O is the center of gravity of the wedgeABC. lf trline OO' is drawn parallel to the surfaceof sliding,BC, the point of intersectionof this line with the back face of the wall will give the point of application of P,,.Thus, P,,acts at O' inclinedat angle6 with the normal drawn to the back face o[ thc wall.

() a

----/o'

Figure 12.24 Approximate method for finding the point of application of the resultant active fbrce 12.11 Graphic Solution for Coulomb's Active Earth Pressure 401

Example12.8

A 15-ft-high retaining wall with a granular soil backfill is shown in Figure 12.25. Giventhat y : 100lblft3, 6' : 35',0 : 5o,and6 = 10'.determine the activethrust per foot lengtho[ the wall.

Solution For this problem,,lt : 90 * 6 - 6 : 90o* 5" * 10" : 75'. The graphicconstruction is shownin Figure12.25. The weightsof the wedgesconsidered are as follows:

Wedge Weight (lbl ABCl '(4.38X17.S8)(100)= 3,916 ABC2 3,e16+ t; (2.36X18.56)l(100)- 6,106 ABC3 6,106+ l+(2.24)(1e.s4)l(100) : 8,2e5 ABC4 8,29s+ ti Q.rr)(20.77)l(100): 10,486 ABCs 10,486+ [, (1.e7)(22,2U](100)= 12,67 5

2.sri 2.sft 2..5li -l I.el<------>l| I cr tcl+ tcs

17.-5

I I v=90-0-6=7s" Weight ( 1000lb) |-_-_- 12345 Length(fi) 0=5" Fl--_- 12345

Figure 12.25 Culmannssolution for determiningactive thrust per unit lengthof wall Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb

In Figure12'25 rq : z,gtota Bc'z: 6'106lb aq : S,Z9StU Bq : 10,4861b nct : lZ,OlSrc The activethrust per unit lengthof the wall is 4,090lb.

12,12 Active Force on Retaining Walls with Earthquake Forces

Coulomb's analysisfor activeforce on retaining walls discusscdin Section 12.I 0 can be convenicntly extendcd to include earthquarkcforces. To do so, let us considera retaining wall of height H with a slopinggrttnulur bucklillas shown in Figure 12.26a. Lct the unit weight and thc friction trngleof the granular soil retainedby the wall be equafto y and$' , respectivcly.Also, let 6 be the anglcof friction betweenthe soil and the wirll. AtlC is a trial ferilurewedge. The forcesacting on the wcdge are as follows: 1. Weightof the soil in the wedgc,W 2. Resultantof thc shearand normal forceson the failurc surfaceBC, F

v q 6

(a) (b)

Figure 12.26 Active force on a retaining wall with earthquake forces 12.12 Active Force on Retaining Walls with EarthquakeForces 403

3. Active force per unit length of the wall, P,,,. 4. Horizontal inertial force, k,,W Vertical inertial force, k,,W

Note that

Horizontal component of earthquakeacceleration (12.70) g

Vertical component of earthquakeacceleration t.- (t2.11)

where g : accelerationdue to gravity. The force polygon demonstratingthese forcesis shown in Figure 12.26b.The dynamic activeforce on the wall is the maximum value of P,,,,exerted by any wedge. This value can be exprcssedas

p,": LyHr(I * k,,)K1, (t2.12)

where

'l'o @')sin(@'- a * p t cos2gcosBcos(d+ 0 + ) I 0 + p)cos(0* ot) )I

(t2.13) and

F:,""-'(#?) (12.74)

Note that with no inertia forccs from earthquakes.B is equal to 0. Hence, K',,: K,, as givenin Eq. (12.69).Equations (12.72)and (12.73)are generallyreferred to as the Mononobe-Okabe equation.s(Mononobe. 1929,Okabe, 1926).The variation of Kj, with 0 - 0' and k,, : 0 is given in Table 12.6. Consideringthe activeforce relation givenby Eqs. (12.72)through (12.74),we - - find that the term sin (@' q B) in Eq. (12.13)has two important implications. - - First, if rp' B < 0 (i.e.,negative), no real solution of Ki is possible.Physically, " this implies that an equilibrium condition will nr,tterisl. Hence, for stability, the lim- iting slope of the backfill may be given as cr=o'-E (t2.ts) g Table 12.6 Valuesof rKi,[Eq. (12.73)lwith 0 : 0" and k, : 6'@eg)

kh 6 {deg} a (deg) 28 30 35 40 45 0.1 0.427 0.39"7 0.328 0.268 0.2t7 0.2 0.-508 0.473 0.396 0.382 0.2'70 0.3 0.611 0..569 0.478 0.400 0.334 0.4 0.753 0.697 0.581 0.488 0.409 0.5 1.00-5 0.890 0.'716 0.-596 0.500 0.1 0.451 0.423 0.341 0.282 0.227 0.2 0.-554 0.5l4 0.421 0.349 0.285 0.3 0.69i) 0.635 0.522 0.431 0.356 0.4 0.942 0.825 0.653 0.-s35 0.442 0.5 0.8-55 0.673 0.551 0.1 0.497 0.457 0.371 0.21)r) 0.238 0.2 0.623 0.570 0.461 0.37,5 0.303 0.3 0.n56 0.74tt 0.58-s 0.4'72 0.383 0.4 0.7u0 0.604 0.4u6 0.5 0.1309 0.624 0.1 ,bl2 0.396 0.368 0.306 0.2.53 0.207 0.2 0.4u5 0.452 0.380 0.3I9 0.267 0.3 0.604 0.563 0.474 0.402 0.340 0.4 0.778 0.7Itr 0.-599 0.508 0.433 0.5 l.ll5 0.912 0.771 0.64u 0.-552 0.1 ,bl2 0.42tt 0.396 0.326 0.26t3 0.2Itt 0.2 0.537 0.491 0.4t2 0.342 0.2n3 0.3 0.699 0.640 0.526 0.438 0.361 0.4 t.025 0.881 0.690 0.56u ('t.4'75 0.-5 0.962 0.752 0.620 0.1 tf l2 l0 0.472 0.433 0.352 0.285 0.230 0.2 0.6I6 0.562 0.454 0.371 0.303 0.3 0.90u 0.7130 0.602 0.4tt7 0.400 0.4 0.u-57 0.6.56 0.531 0.-5 0.944 0.722 2, 0.1 0.393 0.366 0.306 0.256 0.212 ;E-') 0.2 0.486 0.454 0.3u4 o.326 0.276 0.3 0.612 0.572 0.4u6 0.416 0.357 0.4 0.t301 0.740 0.622 0.-s33 0.462 0.-5 t.177 1.023 0.819 0.693 0.600 2, 0.1 0.427 0.39-5 0.327 0.27| 0.224 ;E-) 0.2 0.541 0.-501 0.41u 0.35t) 0.294 0.3 0.714 0.6-55 0.-541 0.455 0.386 0.4 1.073 0.921 0.722 0.600 0.509 0.5 1.034 0.812 0.679 2, 0.1 t0 0.472 0.434 0.3-54 0.290 0.237 ;E-) 0.2 0.625 0.570 0.463 0.381 0.317 0.3 0.942 0.rJ07 0.624 0.509 0.423 0.4 0.909 0.699 0.573 0.5 t.037 0.800 12.12 Active Force on Retaining Walls with EarthquakeForces 405

For no earthquake condition. F : 0'; for stability, Eq. (12.75) gives the familiar relation a=0' (12.76) Second,for horizontalbackfill, a : 0', for stability, B=o' (12.77) - Because F : tan-t[kt,l(I k,)), for stability, combining Eqs. (12.74) and (12.77) re- sultsin

k,,= (I k,,)tanS' (12.18a) Hence.the criticalvalue of the horizontalacceleration can be definedas - knk : (l k,)tan S' (12.18b) where kr,(.,): critical of horizontal acceleration(Figure 12.21).

Location of Line of Action of Resultant Force, Pu, Seedand Whitman (1970)proposed a simple procedureto determine the location of the line of action of the resultant, P,,".Their method is as follows: 1. Let

Pur: P,,+ 4P,,,. (12.7e)

where f, : Coulomb's active force as deterntinedfrom Eq. (12.68) LPu": additional active forcc causedby the earthquakeeffect

2. CalculateP,, IEq. ( l2.6lJ)]. 3. CafculateP,,"lBcl. (12.72)1. 4. Calculate LPu"- Pu"- Pu.

0.5

Soil frictionangle, Q'(deg) Figure 12.27 Criticalvalues of horizontalacceleration (Eq. 12.78b) Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb

l I H 3 I ll Y Figure 12.28 Locati

5. According to Figure 12.28.P,, will act at a distanccof H/3 from the baseof the wall. Also, AP,,,,will zrctat a distanceof 0.6H from thc baseof the wall. 6. Calculatethc location of P,,,,as / H\ 4,{;l + 4P,,,,(0.611) ( 12.80) Pu"

whcrc ? - distanceof thc line of action of P,,,,from the baseof the wall. Note that thc line o1action of P,,,.will be inclined zrtan angleof 6 to the normal drawn to the back l'nccof the retaining wall. It is very important to realize that this mcthod of determining P,,,,is approximate and does not actually model the soil dynamics.

Example12.9

For a retainingwall with a cohesionlesssoil backfill.7 = 15.-5kN/m3, d' = 30o,6= 15', 0 : 0o,rv - 0",H : 4 m, k.u: 0, and kn : 0.2.Determine P,". Also determine the locationof the resultantline of actionof Po"- that is, ?. $olution ;i To determinePn", we useEq. (12.72): ; q p"": !ryH,(l- k,,)K'" We are giventhat $' = 30"and 5: 15",so 6:t6 12.13 Puufor c'-cf' Soil Backfill 407

Also, 0 * 0o,d = 0o,kh: 0.2.From thesevalues and Table 12.6,wefind that the magnitudeof Ki is equalto 0.452.Hence, p"": i(1s.5x4),(1- 0x0,4s2): 56.05kN/m We now locatethe resultantline of action.From Bq. (12.68),

Po* lK,,yH2 ' For f - 30oand 6 : 15o,K" : 0.3014(Table 12.5), so P, : +(0.3014X15.5X4)'z= 37.37 kNtm Hence,APo" : 56.05- 37.3'1= 18.68kN/m. From Eq. (12.80), /a\ ,"(+) + ^P,"(0.6H)(37.37) . (18.68X2.4) [r,) 1.* - 1.69m r Pu" 56.05

12.13 P",for c'-O' Soil Backfill

The Mononobe-Okabc cquation for estimating P,,,.frtr cohesionlessbackfill describedin Scction12.12 can also be extcndedr<'t c'-

Figure 12.29 Trtal failure wedge behind a retaining wall with a c'-{" backfill Chapter 12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomb

Refering to Figure 72.29the forces acting on the soil wedge (per unit length of the wall) are as follows:

1. The weight of the wedge ABCDE, W 2. Resultant of the shear and normal forces on the failure surface CD, F 3. Active force, P,," 4. Horizontal inertia force, kr,W 5. Cohesiveforce along CD, C - c(CD) 6. Adhesive force along BC, C' - .( BC ) It is important to realizethat the following two assumptionshave been made:

1. The vertical inertia force (k,,trV)has bccn takcn to be zero. 2. The unit udhesionalong the soil-wallinterface (BC) has been taken to be equal to the cohesion(c) of the soil.

Consideringthese forces, we can shclwthat

P,,,,- y(FI - Z,,)tN',,, c'(H 2,,)N',,, ( 12.81)

where r ,q4 :gu_':99 !9!r (12.82) sin(a'+ 5)

[(n + 0.-5)(tan0* tan i) + n21ang][cos(l+,r') + kr,sin(i+ q5')] N",, ( 12.83) sin(4'+6)

in which 11'-0+i+(b' (t2.84)

2,, n: ( 12.8s) H - 2.,,

The valuesof lVj,,and Nj,, can bc dctcrmincd by optimizing each coefficientsepa- rately.Thus, Eq. (12.81)gives the upper bound of P,,". For the static condition, ki, : 0. Thus,

P,,": y(H - Z,,)tNu-- c'(H - Zu)Nu, (12.86)

The relationshipsfor N,,. and N., can be determined by substituting k7,: 0 into Eqs. (12.82)and (12.83).Hence,

cos4' sec0 * cos@' sec r 1V,,- Nl,. (12.81) sin(4'+ 6)

0.5)(tan0 * tani) * n2tan g]cos(l + @') (12.88) /t sin(4'+ 6)

The variations of -Ay'o,.,N,r, and tr with $' and 0 are shown in Figures 12.30 through 12.33. 12.13 Pu"for c'-g, SoilBackfitt 409

a r 3.0 '6

o 2.5

E 2.0

I.5

1.0i

20 25 30 3-5 40 15 0'(deg)

Figure 12.30 Yariation of N,,, : Ni,, with t[' prakash and0 (basecl on and Saran, 1966, and Saran and Prakash,l96lt)

I.0

5 0.u a - .9: E (.,.t) o

/ o.t: c

? d n) I! ".4

0' 0 l0 |5 20 25 30 3-5 40 45 Q'(deg)

Figure 12.31 Yariation with and 0 (n: prakash _o^f_{:,, @, 0.2) (based on and Saran, 1966, and Saranand Prakash.196g) 410 Chapter 12 Lateral Earth Pressure: At-Rest,Rankine,and Coulomb

l.(-); t: I n=o :] 0.ti

z '5 0.6 E

e=20' t5' 3 0.4 t0" E 5'

0.2

-2(P -'4 0 -s r0 15 20 2.s 30 0' (dcg)

Figure 12.32 Yarialion of N,,, with r!' and 0 (rr : 0) (based on Prakash and Saran, 1966, and Saran and Prakash.196iii)

2.0

t.9 |0" Lu 200" 1.1 :'t /11")' 1.6 .i ;l(l0" .20 l" l.s ozo '0' { |0' t.4 2f -0' 1.3 10" 20" 1.2 o.ro { l.l 0.05' I 01020304050 Angleof internalfiiction, $' ', Figure 12.33 Yariation of tr with k,,, rf and 0 (based on Prakash and Saran, 1966,and Saranand Prakash.1968) 12.14 Coulomb's PassivePressure 411

Example12.10

For a retainingwall, the following are given: H : 28 ft c' : 21.}lblftz g - *10" y:1181b/ft3 6' : 20" kn: 0.1 Determinethe magnitudeof the activeforce, Po". $olution From Eq. (12.44), 2c' 2c (2)(210) 'o: : -/ - 5'08ft ,rKo: 17 .:--?T - ---70\ rtan[45- (lrs)tan(4s-; T ) ) From Eq. (12.85), '" 5'08 n: * : o'22 - o'2 17 - zr, 28 - 5.oB From Eqs. (12.81),(12.87), and (12.88),

P,,"* l(H * z,)2(IN,,7)* c'(H - zu)Nu, For 0 : I0", O' : 20",kt, = 0.1,and n o 0.2. Nn,: 1..67 (Figure 12.30) A4,r,: 0.375 (Figure12.31) ),: 1.17 (Figure12.33) Thus, p,, : (118X28- s.0S)11..17x 0.375) - (210X28* s.08X1.67) : 1901"60lb/ft

12.14 Coulomb's PassivePressure

Figure 12.34ashows a retaining wall with a sloping cohensionlessbackfill similar to that consideredin Figure 12.22a.The forcepolygon for equilibrium of th ewedgeABC for the passivestate is shown in Figure 12.34b.P,, is the notation for the passiveforce. Other notations used are the same as those for the activecase (Section 12.10).In a procedure similar to the one that we followed in the activecase [Eq. (12.68)],we get

Pr: \KryHz (2.8g) 412 Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb

Y ....'

(b)

Figure 12.34 Coulomb'spassive pressure: (a) trial failurewedge; (b) forcepolygon

where K, : Coulomb's passiveearth pressure coefficient, or

cosz({'+ o) Ko= (12.e0)

For a frictionlesswall with the verticalback face supporting granular soil back- frllwithahorizontalsurface(thatis,0:0o,a:0o,and5:0'),Eq.(12.90)yields l+sind' -/ d'\ Kr: : tan-[ot*, r - sina' / 12.15 PassiveForce on Retaining Walls with EarthquakeForces 413

Table 12.7 Valuesof Ku [Eq. 12.90]for 0 : 0', a : 0' 6 (deg) -+

I q5'(deg) 15 l5 1.698 1.90t) 2.130 2.405 2.135 20 2.040 1.,) t -) 2.636 3.030 3.525 25 2.464 2.rJ30 3.286 3.85.5 4.591 30 3.000 3.-506 4.143 4.9'77 6.105 -1f 3.690 4.39t) -s.3l0 6.8-54 8.324 40 4.600 5..59t) 6.946 8.870 rt.772

This relationship is the same as that obtained for the passiveearth pressurecoefficient in Rankine'.scase, given by Eq. (12.30). The variation of K,, with 95'ancl 6 (for 0 : 0' and a : 0") is given inTable 12.1. ', We can see frclm this table that for given value of f the value of Kn increaseswith the wall friction.

12.15 PassiveForce on Retaining Walls with Earthquake Forces

Figurc 12.3-5shows the failurc wedge analysisfor a passiveforce againsta retaining wall of height H with a granular backlill and earthquake forces.As in Figure 12.25, the failure surfaceis assumedto be a plane. P1,"is the passiveforce. All other notations in Figure 12.35are the samc as those in Figure 12.26.Following a procedure similar to that used in Section 12.12,(atLer Karrila, 1962) we obtain

- Pr,,: )tH2(l k,)K'p (r2.e1)

Figure 12.35 Passiveforce on a retaining wall with earthquake forces 414 Chapter 12 Lateral Earth Pressure: At-Rest,Rankine, and Coulomb

iz3

Figure 12.36 Variationol Ki, with kl,for 0'(cleg) k,,- tt : 0 : 6 - 0

where cos2(4'+o-B) K',,: sin(D+

12.16 Summary and General Comments

This chapter covers the general topics of lateral earth pressure, including the following: 1". At-rest earth pressure 2. Active earth pressure- Rankine's and Coulomb's Problems 415

3. Passiveearth pressure- Rankine's and Coulomb's 4. Pressureon retaining wall due to surcharge(based on the theory of elasticity) 5. Active and passiveearth pressure,which includesearthquake forces. This is an extensionof Coulomb's theory

For design,it is important to realizethat the lateral activepressure on a retaining wall can be calculatedusing Rankine's theory only when the wall movessalj?- cientlyoutward by rotation about the toe of the footing or by deflectionof the wall. If sufficientwall movement cannot occur (or is not allowedto occur) then the lateral earth pressurewill be greater than the Rankine activcpressure and sometimesmay be closer to the at-rest earth pressure.Hence, proper selectionof the lateral earth pressurecoefficient is crucialfor safeand proper design.It is a generalpractice to assume a value for the soil friction angle (@') of the backfill in order to calculatethe Rankine activepressure distribution, ignoring the contribution of the cohesion(c'). The general range of ry''uscd for the designof retaining walls is given in the follow- ins table:

Soil friction Soil type angle, d' (deg) Soft clay 0-l-5 Compactedclay 20-30 Dry sand anclgravcl 30-40 Silty sancl 20-30

In Section 12.5,we saw thzrtthe lateral earth pressurcon a retaining wall is greatly increasedin the prescnceof a water table above the baseof the wall. Most retaining wallsare not designcdto withstandfull hydrostaticpressure; hence, it is important that adequatedrainage facilities are provided to ensurethat thc backllll soil does not bccome fully saturated.This can be achievedby providing weepholesat regular intervalsalong the length of the wall.

Problems

12.7-12.6 Assumingthat the wallshownin Figure12.37 is restrainedfrom yielding,find the magnitudeand location of the resultantlateral forcc per unit widthof the wall. 6', 12.1 10ft I t0 tb/fc 32' 12.2 12fr 98 lb/fc 28' 12.3 18ft 100lb/ft3 40" 12.4 3 m 17.6kN/mr 36' t2.5 4.5m 19.95kN/m3 42" 12.6 5.-5m 17.ttkN/m3 37'

72.7 Consider a 5-m-high retaining wall that has a vertical back face with a horizontal backfrll. A vertical point load of 10 kN is placed on the ground surface at a distance of 2 m from the wall. Calculate the increase in the lateral force on the wall for the section that contains the point load. Plot the variation of 416 Chapter 12 Lateral Earth Pressure:At-Rest, Rankine, and Coulomb

Sand

Unit weight = y (or dcnsily = p) H a c'=0 I 6 (angleof wall fiiction) = $ I Figure 12.37

thc pressureincrease with depth. Use the modified equation givenin Sec- tion 12.4. 12.8-l2,ll Assume that the retaining wall shown in Figure 12.37is frictionless. For each problem. determine the Rankine activeforce per unit length of the wall, the variation of activeearth pressurcwith depth, and the location of the resultant. Problem H d'(degl y l2.x 15[l .10 I05lb/frl 12.9 lu fl 32 100Ib/ftl 12.10 4 m 36 Iu kN/rnl l2.ll 5 m 40 l7 kN/mr

12.72-12J4 A retaining wall is shown in Figure 12.38.For eachproblem, determine the Rankine activeforce, P,,,per unit length of the wall and the location of the resultant. o\ +i "fz (degl (deg) q 12.t2 l0 ft 5fr l0-5lb/ft'' 122lb ltit3 30 30 0 12.t3 20 ft 6ft n0 rbifc l26lb/fc 34 34 300rb/fc 12.14 6m 3m 15.5kN/mr 19.0kN/m3 30 36 l-5kN/m2

Sttrcharge= q I trt A I sant Ht It I Qr lffi ''' =tt | VvGround watertable : tffi Sand y2 (saturatedunit weight) Q: c'2=0

lffi Figure 12.38 Problems 417

12.15 A 15-ft-high retaining wall with a vertical back face retains a homogeneous saturated soft clay. The saturated unit weight of the clay is 122 lb lft3. Labora- tory testsshowed that the undrained shearstrength c,, of the clay is equal to 350 Ib/ft2. a. Make the necessarycalculations and draw the variation of Rankine'sactive pressureon the wall with depth. b. Find the depth up to which a tensilecrack can occur. c. Determine the total activeforce per unit length of the wall before the tensile crack occurs. d. Determine the total activeforcc per unit length of the wall after the tensile crack occurs.Also find the location of the resultant. 12.16 Redo Problem 12.1-5assuming that the backfillissupporting a surchargeof 200tbtf(. 12.17 A 5-m-high retaining wall with zrvertical back face has a r:'-{' soil for backfill. For the backfill, y : 19 kN/m3, c' - 26 kN/m2, and r/,' : 16".Considering the cxistenceof thc tcnsile crack, dctcrmine the activelorce P, on the wall for Rankineb activestatc. 12.18 For thc retaining wall shown in Figurc 12.39,deterrnir.rc the activeforce P, for Rankine'.sstate. Also, lind thc positionof the rcsultant.Assumc that the tensilecrack exists. p :2100 kg/mr,Q - 0",(: : (:!t: 30.2kN/mr 12.19Repeat Problcm l2.lB usingthe followingvalucs: p - 1950kgim3, d' - lti". t'' : 19.4kN/m: 12.20-12.23 Assume that the rctaining wall shown in Figure 12.37is frictionless. For each problcm, cleterminethc Rankine passivcf'orce per unit length of the wall, the vtrriationol lateral prcssurewith dcpth.ancl the locationof the resultant.

Problem H d'(deg) y

t2.20 lJlr -l+ ll0lb/li' t2.21 l0 fr 36 105lb/l'rr 12.22 .5nt 3-5 l4 kN/rn' t2.23 4m 30 l5 kN/nr''

12.24For the retainingwall describedin Problem 12.12,cletermine the Rankine pnssivcforce per unit lcngth of the wall and the location of the resultant.

II Clay 6.5rn (,( a,a I density= P I Figure 12.39 418 Chapter 12 Lateral Earth Pressure:At-Rest, Rankine,and Coulomb

I 'l'r i 'tll 't" *l !* Sand 0- 10": H I Unitwcight = y (ordensity = p) , r"=0 ', 0' = 36' 6 (wallll'iction)

Figure 12.40

12.25 For the reteriningwall describedin Problcm 12.13,detcrmine the Rankine passiveforcc per unit lcngth of the wall and the location of the resultant. 12.26 A retaining wall is shown in Figurc 12.40.The hcight of the wall is 5 m, and the unit weightof thc sandbacklill is ltt kNimr. UsingCoulomb's equation, calculatcthe activeforcc P,,on thc wall for the following valuesof the angle of wall lriction: a. 5 - llJ' b. 6:24" Commenton the directionand locationof the resulttrnt. 12.27 Ref erring to Figure I 2.41, dctermine Coulombh activc force P,,per unit lcngth of thc wall for thc following cascs: a. It : l-5ft,B - [J-5o,n: l, Hr - 20tt.y : l2t{ lb/ftr, q5'- 38', 6 : 20" b. H - 1t3ft.B : 90o.n : 2, Ht - 22lt,y : ll6lblf(,0' : 34',5 : lJ' : c. H :-5.5 rri,p - [J0",r: l, Hr: 6.5m, 7 : l6l3t)kg/mr,d': 30',6 30" Usc Culmann\ graphicconstruction procedurc.

.I I I *-----J .tl

Cllhcsionlesssoil

Unit weight- y (rlrdensity = p) t'=0 H 0' 6 (angleof wall friction)

Rz Y,/'

Figure 12.41 References 419

12.28 Refer toFigure 72.26.GiventhatH - 6 m,0 - 0",a:0o, y: 15kN/mr. 6' : 35",6 - 2136' , kn - 0.3,and ft,,: 0, determine the activeforce P,,,,per unit length of the retaining wall. L2.29 Refer to Problem 12.28.Determine the location of the point of intersection of the resultantforce P,,"with the back face of the retaining wall. 12.30 Repeat Problem 12.28with the following Values:H - 10ft. f, : 10",a : 10', 7 : 110lblft3,O'- 30',6 : 10".kn:0.25, andk, : 0. 12.31RefertoFigure 12.29.GiventhatLI - 6m,0 - 10",

References

Cot.tlclntu,C. A. ( 1776)."Essai sur unc Applicationcles Riglcs dc Maxin'risct Minimisa qucl- qucs Probldmesdc Statique,relatif.s a I'Architecture," Ment. Rov.das Sclcrrccs,Paris. Vol. 3. 3li. Cut.vnNN, C. ( lu75). Dic gruphischcStutik, Meycr and Zeller. Zwich. Gr-:ntrrn, E. ( 1929).Untarstrchtmgan iiber dic Dntc'kvcrteilung int Orliclt hclustetcnSrrarl, Te ch- nischeHochschule. Zurich. Jnrv, J. (1944)."Thc Coefficicntol'F,arth Prcssurc at Rcsl,"./orrlrul o.lthc Stx'ictvo.f' Iltrngur- iun Architet'ts und Engincers, Vol. 7. 3-5-53-5IJ. K,q,ptrn,J. P. (1962). "E,arthquake ResistarrtDesign ol Rctaining Walls." I'rocacding.t,2ntl Eurthqrutkt Symposiunt,Univcrsity ol' Roorkcc, Iloorkcc, India. Mnssnpsc'u, K. R. (1979). "Latcral Earth Prcssurein Nornrally C-\rnsoliclatcclClay." f'rrr- ceadingso.f the Sevcnth Ettntpcun ()on.larcnceon Soil Mcchunit's urulfituntlution Engi- neering,Brighton. England. Vrl. 2. 24-5-250. MnzrNnn,qNt,z.H.. and ci,lN:nr-r.M. H. (IgL)l)."Latcral Earth Pressur-cProble m ol'c'shc- sivcBacktill with Inclined Sur[acc.",/r.rurrtulof'(icotcthnitu! untl GutcnvintnntantulEn- gincarinS4,ASCIE, Vol. 123,No. 2, I l0 I 12. 'l'hcory Orrrsl,, S. (1926). "General ol'E,arth Pressurc.".lournul o.fthc.lultanesa Sot:ict1'o.f Civil Engineer.r',Tokyo. Vol. 12,No. l. Pnnra.su, S., and Sannr. S. (1966)."Static and Dynamic Earth PressurcBehind Retain- ing Walls," Proceedings,3rdSymposium on Earthquakc Engineering,Roorkcc. lndia. YoL1.277-288. R,,rNrtNE,W. M. J. (18-57)."On Stability on Loose Earth." Pltiktsophic'l'runsuctionsof'Royul Society,London, ParI 1.9,2'/ . SRnaN,S., and Pnnrastl. S.( 1968)."Dimcnsionless Parameters for Staticand DynanricEarth Pressurefor Retainir.rgWalls." Indian Geotechnical.lounnl, Vol. 7. No. 3. 29-5 310. Sep.o.H. B., and WutrvRN, R. V. ( 1970)."Design of Earth RctainingStructures for Dynamic Loads," Proceedings,Specialty Conference on Lateral Stressesin thc Ground and Dc- sign of Earth Retaining Structurcs,ASCE, 103-14'7. SHl,ntr',M. A., FnNc, Y. S.,and SHERrp,R. I. (1984)."K,, and K,, Bchind Rotating and Non- Yielding Walls," Journal of GeotechnicalEngineering, ASCE, Vol. 110, No. CT1. 41--56. SeeNclr,n, M. G. (1938)."Horizontal Pressureson RetainingWalls Due to ConcentratedSur- face Loads," Iowa State University EngineeringExperiment Station,Bulletin, No. 140.