book-chapter 20-sheet piled walls and braced cuts

7/27/2019 Book-Chapter 20-Sheet Piled Walls and Braced Cuts

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CHAPTER 20SHEET PILE WALLS AND BRACED CUTS

20.1 INTRODUCTIONSheet pile walls are retaining walls constructed to retain earth, water or any other fill material.These w alls are th inner in section as compared to masonry walls described in Chapter 19. Sheet pilewalls are generally used for the fol lowing:

1. Water front structures, fo r example, in bui lding wharfs , quays, an d piers2. Building diversion dams, such as cofferdams3. River bank protection4. Retaining th e sides of cuts made in earth

Sheet piles may be of timber, reinforced concrete or steel. Timber piling is used for shortspans and to resist light lateral loads. They are mostly used for temporary structures such as bracedsheeting in cuts. If used in permanent structures above th e water level, they require preservativetreatment and even then, their span of life is relatively short. Timber sheet piles are joined to eachother by tongue-and-groove joints as indicated in Fig. 20.1. Timber piles are not suitable fordriving in soils consisting of stones as the stones would dislodge th e joints.

Groove \ / Tongue

Figure 20.1 Timber pile wal l sect ion

881


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882 Chapte r 20

o~ ~ o " ~ a ~ T 3D _ _ o _ _ Q _ _ d b. _ o _ _ Q _ .a _ O - _ d

Figure 20.2 Rein fo rced concre te Sheet pile wall sect ion

( a ) S t r a i g h t sheet p i l i n g

(b ) S h a l l o w a r c h - w e b p i l i n g

( c ) A r c h - w e b p i l i n g

(d) Z-pileFigure 20.3 Sheet pile sect ions

Reinforced concre te shee t pi les a re precast concre te members, usua l ly with a tongue-and-groove jo int . Typica l sec t ion of pi les a re shown in Fig . 20.2. These pi les a re re la t ive ly heavy andbu lky . They d isplace la rge vo lum es of so l id dur ing dr iv ing . This la rge vo lume displacement of so i ltends to increase the d r iv in g resis tance . The design of pi les has to take in to accoun t the la rge dr iv ingstresses an d su i t a b le r e in fo r c e m e n t has to be pr ov ide d fo r th i s pu r pose .

Th e m o s t c o m m o n t y p e s o f pi les used a re steel sheet piles. Steel pi les possess severaladvantages over th e other types. Some of the im p o r ta n t a dva n ta ge s a r e :

1. T he y a r e r e s i s t a n t to h igh d r i v i n g stresses as deve loped in hard or rocky mater ia l2 . They a re l igh te r in sec t ion3. They may be used severa l t imes


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Sheet Pi le Wa lls and Braced Cuts 88 3

4. They can be used either below or above water and possess longer life5. Suitable joints wh ich do not deform during driving can be provided to have a continu ouswall6. The pile length can be increased either by welding or bolting

Steel sheet piles are available in the market in several shapes. Some of the typical pile sections areshown in Fig. 20.3. The archweb and Z-piles are used to resist large bending moments, as in anchored orcantilever walls. Where the bending moments are less, shallow-arch piles with corresponding smallersection moduli can be used. Straight-web sheet piles are used where the web will be subjected to tension, asin cellular cofferdams. Th e ball-and-socket type of joints, Fig. 20.3 (d), offer less driving resistance than thethumb-and-finger joints, Fig. 20.3 (c).

20.2 SHEET PILE STRUCTURESSteel sheet piles may con veniently be used in several civil engineering w orks. They m ay be used as:

1. Cantilever sheet piles2 . Anchored bulkheads3. Braced sheeting in cuts4. Single cell cofferd ams5. Cellular cofferdam s, circular type6. Cel lu lar cof ferdams (diaphragm)

Anchored bulkheads Fig. 20.4 (b) serve the same purpose as retaining walls. However, incontrast to retaining walls whose w eight always represent an appreciable fra ction of the weight ofthe sliding wedge, bulkheads consist of a single row of relatively light sheet piles of which thelower ends are driven into the earth and the upper ends are anchored by tie or anchor rods. Theanchor rods are held in place by anchors w hich are buried in the backfill at a considerable distancefrom the bulkhead.

Anchored bulkheads are widely used for dock and harbor structures. This constructionprovides a v ertical wall so that ships may tie up alongside, or to serve as a pier structure, which mayjet out into th e water. In these cases sheeting may be required to laterally support a fill on whichrailway lines, roads or warehouses may be constructed so that ship cargoes may be transferred toother areas. The use of an anchor rod tends to reduce the lateral deflection, the bending moment,an d the depth of the penetration of the pile.

Cantileve r sheet piles depend for their stability on an adequate em bedm ent into the soil belowthe dredge line. Since the piles are fixed only at the bottom and are free at the top, they are calledcantilever sheet piles. These piles ar e economical only fo r moderate wall heights, since th e requiredsection modulus increases rapidly with an increase in wall height, as the bending moment increaseswith the cube of the cantilevered height of the wall. The lateral deflection of this type of wall, becauseof the cantilever action, will be relatively large. Erosion and scour in front of the wall, i.e., loweringthe dredge line, should be controlled since stability of the wall depends primarily on the developedpassive pressure in front of the wall .20.3 FREE CANTILEVER SHEET PILE W A L L SWhen th e height of earth to be retained by sheet piling is small, th e piling acts as a cantilever. Theforces acting on sheet pile walls include:

1. The active earth pressure on the back of the wall which tries to push the w all away from thebackfill


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884 Chapte r 202. The passive pressure in f ron t of the wall below th e dredge line. Th e passive pressure resists

the movements of the wallThe active and passive pressure distributions on the wall are assumed hydrostatic. In the

design of the wal l , a l though th e Coulomb approach considering wal l f r ic t ion tends to be morerealist ic, the Rankine approach (with the angle of wall f r ic t ion 8 = 0) is normally used.

The pressu re due to water may be neglected if the water levels on both sides of the wall are thesame. If the diff erence in level is considerable, the effect of the differenc e on the pressure w ill haveto be considered. Effec t ive unit weights of soil should be considered in computing the active andpassive pressures.

VSheet pile

Backfi l l\Anchor rod

(a ) Cantilever sheet piles (b ) Anchored bulk head

Sheeting

(c ) Braced sheeting in cuts (d ) Single cell cofferdam

(e) Cellular cofferdamDiaphragms Granular fill

Tie-rods < \Outer sheetpile wall c

Inner sheetpile wall

(f ) Cellular cofferdam,diaphragm type (g) Double sheet pile walls

F i gur e 20.4 Use of sheet pi l es


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Sheet Pile Walls and Braced CutsP

a

885

//AXV/A//,

O'

Pp-PaD

(a) (b)Figure 20.5 Example i l lustrating earth pressure on cantilever sheet piling

General Principle of Design of Free Cant i lever S h e e t PilingThe action of the earth pressure against cantilever sheet piling can be best i l lustrated by a simplecase shown in Fig. 20.5 (a). In this case, the sheet piling is assumed to be perfectly rigid. When ahorizonta l force P is applied at the top of the pi l ing, th e upper portion of the piling til ts in thedirection of P and the lower portion moves in the opposite direction as shown by a dashed line inth e f igure . Thus the pi l ing rotates about a s ta t ionary point O'. The port ion above O' is subjected toa passive earth pressure from th e soil on the left side of the piles and an active pressure on the r ightside of the piling, whereas the lower po rtion O'g is subjected to a passive earth pressure on the rightside and an active pressure on the left side of the pi l ing. A t point O' th e piling does not move an dtherefore is subjected to equa l and opposite earth pressures (at-rest pressure from both sides) witha net pressure equal to zero. The net earth pressure (the difference between the pass ive and theactive) is represented by abO'c in Fig. 20.5 (b). For the purpose of design, the curve bO'c isreplaced by a straight line dc. Point d is located a t such a location on the l ine af that th e sheet pilingis in s ta tic equi l ibrium under th e action of force P and the earth pressures represented by the areasade and ecg . The posi t ion of point d can be determined by a trial and error method.

This discussion leads to the conclus ion tha t cant i lever sheet p i l ing derives it s stability frompassive earth pressure on both sides of the piling. However, the distribution of earth pressure isdifferent between sheet pi l ing in granu lar soi ls and sheet piling in cohesive soils. T he pressuredistribution is l ikely to change with t ime fo r sheet pilings in clay.

20.4 DEPTH OF EMBEDMENT OF CANTILEVER W A L L S IN SANDYSOILSCase 1: With Water Table at G re at DepthT he active pressure acting on the back of the wall tries to m o v e th e wall away from th e backf i l l . Ifth e depth of e m b e d m e n t is adequate th e wall rotates abou t a poin t O' s i tuated above th e bot tom ofth e wall as shown in Fig. 20.6 (a). T he t ypes of pressure that act on the wall when rotat ion is l ikelyto take place about O' are:

1. Active earth pressure at the back of wal l from th e surface of the backfill down to the poin tof rotat ion, O'. The pressure is des ignated as Pal.


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886 Chapter 202 . Passive earth pressure in f r on t of the wall f rom th e p o i n t o f r o ta t ion O ' to the dredge l ine.This pressure is des ignated as P ^ .3 . Active ear th pressure in f r on t of the wal l f r om th e p o i n t o f r o ta t ion to the bot tom of the

wal l . This pressure is designated a s P^.4. Pass ive ear th pressure a t the back of wall f rom the po int of rota t ion O' to the bottom of the

wall . This pressure is designated as P }2.The pressures act ing on the wa ll are show n in Fig. 20.6 (a) .If the passive and active pressures are algebraically com bined, the resultant pressure distr ibutionbelow the dredge line will be as given in Fig. 20.6 (b). The various notations used are:D = min imu m depth o f embedm ent wi th a f ac to r o f s a fe ty equa l to 1

K A = Rankine act ive ear th pressure coeff ic ientK p - Rankine pass ive ear th pressure coeff ic ient

K = Kp-KAp a = ef fect ive active earth pressure acting against the sheet pile at the dredge linep = effect ive pass ive ear th pressure a t the base of the pi le w all and act ing tow ards the

backf i l l = DK

/A\VA\VA\VA\\

O'

. . - Sand r":, . -

(a) (b)Figure 20.6 Pressure distribution on a can ti lever wall.


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Sheet Pile Wal ls and Braced Cuts 887

p' = effective passive earth pressure at the base of the sheet pile wall acting against thebackf i l l side of the wal l = p " p + yK D o

p" = effective passive earth pressure at level of O= /y y Q K + jH K pY = effective un i t weigh t of the soil assumed th e same below an d above dredge l ine

y Q = depth of point O below dredge line where the active and passive pressures areequa ly = height of point of application of the total active pressure P a above point Oh = heigh t of point G above th e base of the wall

D O = height of point O above the base of the wallExpression for y0A t point O, the passive pressure acting towards the r ight should equal the active pressure actingtowards th e left, that is

Therefore, 0(KP ~

Expression for hFor s tat ic equil ibr ium, the sum of al l the forces in the hor izontal d irection m ust equal zero. That is

pa - \PP V>-yQ) +\(P p +P'p)h =0

Solving fo r h,. pP(D-yJ-ipa=

Taking mom ents of al l the forces about the bottom of the p ile , and equating to zero,P(D 0+p) -ippxD0x-+(pp +p'p)xhx^ = 0

or 6Pa(D0+y)-ppDt+(pp+p'p)h2=0 (20.3)Therefore,

Subst i tut ing in Eq. (20.3) for p , p' and h and s im p l i fy ing ,


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888 Chap te r 20D 04 + C ,D 03 + C 2D < 2 + C3D0 + C 4=0 (20 .4)

where ,

The solution of Eq. (20.4) gives the depth D Q . The method of trial and error is generally adopted toso lve th i s equa tion . The m in im um dep th o f emb edmen t D with a fac tor of safety equal to 1 is thereforeD = D 0 + y Q (20.5)A min imum fac to r o f safety of 1 .5 to 2 may be obta ined by increas ing the m inim um depth Dby 20 to 40 percent .

Maximum Bending MomentT he m a x i m u m m o m e n t on section A B in Fig. 20.6(b) occurs at the poin t of zero shear. This pointoccurs below point O in the f igure . Let this poin t be represented by poin t C a t a depth y 0 belowp o i n t O . The net pressure (pass ive) of the t r iangle O C C ' m u s t balance the net ac t ive pressure P aact ing above the dredge l ine . The eq uat ion for P a is

-o r y 0 = -T r (20 .6)where 7= effect ive un i t w e i g h t of the soil. If the water table l ies above p oint O , 7 wi l l b e equa l to y b,the submerged uni t weight of the soi l .

Once the po in t o f ze ro shea r i s kno wn , the magni tud e o f the maxim um bend ing m om ent maybe obtained asMmax=Pa(y+yo o =?a(j +yo) ~ >K (20.7)

(20 .8)

The s ec t ion modulus Z s of the sheet pi le may be obta ined from the equat ionM--fh

w h e r e , /, = al lowable f lexural stress of the sheet pi le .


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Sheet Pile Walls an d Braced Cuts 889

Figure 20.7 Simplified method of determining D fo r cantilever sheet pile

Simplified MethodThe solut ion of the four th degree_equation is quite labor ious and the problem can be sim pl i f ied byassuming th e passive pressure p' (Fig. 20.6) as a concentrated force R acting at the foot of thepile. The simpl i f ied ar rangem ent is shown in Fig. 20.7.

For equi l ibr ium, th e moment s of the active pressure on the r ight an d passive resistance on theleft about the point of reaction R mus t balance .

=0

Now, P D = ^ an dTherefore, K pD 3 -KA(H + D )3 -0or KD 3 -3HD(H +D)K=0. (20.9)T he solution of Eq. (20.9) gives a value fo r D which is at least a guide to the required depth.The depth calculated should be increased by at least 20 percent to provide a factor of safety and toallow extra length to develop th e passive pressure R. An approximate depth of embedment may be

obtained from Table 20.1.Case 2: With Water Table Within the Backfil lFigure 20.8 gives th e pressure dist r ibut ion against th e wal l with a water table at a depth hl below th eground level. All the notations given in Fig. 20.8 are the same as those given in Fig. 20.6. In thiscase the soil above the water table has an effective uni t weight y and a saturated unit weight y satbelow th e water table. T he submerged unit weigh t is


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890 Chap te r 20Table 20.1 Approx imate penetrat ion (D ) of sheet piling

R e l a t i v e d e n s i t yV e r y looseLooseFi rmD e n s e

D e p t h , D2.0 H1.5 H1.0 H

0.75 HSource : T en g , 1969.

Yb=Ys a t ~ YH.)T he ac t iv e p r essu r e at the wate r t ab l e isPl=VhlKA

and pa at the dredge l ine isP a = YhiKA+ Y f c h2 K A = ( Y / i , + Y ^2 ) K

The o the r expr ess ions a rePP =

P P =Pa

pp(D-yo)-2Pah = ~~T he four th deg r ee equa t ion in t e rms of D g isD o4 + C, D^ + C2 D 02 + C 3 D o + C4 = 0 (20.10)

where ,


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Sheet Pi le Wa lls and Braced Cuts 891

Dredge line

Figure 20.8 Pressure distribution on a canti lever wall with a water table in thebackfi l l

The depth of embedment can be determined as in the previous case and also the maximumbending mom ent can be calculated. The depth D comp uted should be increased by 20 to 40 percent.Case 3: When the Cantilever is Free Standing with No Backfi l l (Fig. 20.9)The cantilever is subjected to a line load of P per unit length of wall. The expressions can bedeveloped on the same lines explained earlier for cantilever walls with backfill. The variousexpressions are

h = 2yDKwhere K=(Kp-KA)

The fourth degree equation in D isD4 + Cl D2 + C 2 D + C3 = 0 (20.11)

8Pwhere

C2=12PH


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892 C h a p t e r 20

Figure 20.9 Free stand ing can t i lever with no ba ckf i ll4P2

Equat ion (20.11) gives th e theoret ical depth D which shou ld be increased by 20 to 40 percent.Poin t C in Fig. 20.9 is the point of zero shear. Therefore,(20.12)

where 7 = effective uni t weigh t of the soilExample 20 .1Determine th e depth of embedmen t for the sheet-p il ing shown in Fig. Ex. 20.la by rigorousanalysis . Determine also th e min im um sec tion modu lu s . Assume an allowable f lexural stress/^ =17 5 MN / m 2. The soil has an effective un i t weig ht of 17 kN/m 3 and angle of internal fri ctio n of 30.Solution

F or 0= 30, K.= an 2 (45 - 0/2) = tan 2 30 = -

K =3, K = K a K. =3 2.67.p K ' p A 3The pressure d is tr ib ution along the sheet p i le is assumed as shown in Fig . Ex. 20. K b)p a = y HK A = 17x 6 x-= 34 k N / m 2


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Sheet Pi le Walls and Braced Cuts 89 3

From Eq (20.1)P 34 = o.75 m.7(Kp-KA) 17x2.67

=#+ 0=1*34x6+^x34x0.75= 102 + 12.75 = 1 14.75 kN/meter length of wall or say 115 kN/m.

= 17x6x3+ 17xDx2.67 = 306 + 45.4Z)p 'p = yH K p +W0(Kp -KA) = 17x6x 3+ 17x0.75x2.67 = 3 40 k N / m 2To find y

1_ H 1_ 2

= -x34x6x(2+0.75) +-x34x0.75x -xO.75 =286.92 2 3^ _ 286.9 286.9 ^^^Therefore, v =-=-=2.50m.P a 115

Now DQ can be found from Eq. (20.4), namelyD 0 4 + qrg + c 2D 0 2 + c3D0 + c4 = o

17x2.67 ' z yK 17x2.67(2 x2.50 x 17 x2.67 +340) = -189.9

= -310.4

(17x2.16x 115x2.50x340 +4x (115)2- 4 ( y K ) 2 (17X2.67) 2

Subst i tut ing fo r Cp C2, C3 an d C4, an d s impl i fy ing we h av eD04 +7.49D 03 - 20.3D2 -189.9DQ - 310.4 = 0

This equation when solved by the method of trial and error givesD0 *5 . 3 m

Depth of EmbedmentD = D0 + yQ = 5.3 + 0.75 = 6.05 m

Increasing D by 40%, w e haveD (design) = 1.4 x 6.05 = 8.47 m or say 8.5 m.


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894 Chap te r 20

H=6m

D = ?

Sandy = 1 7 k N /m 3

D

(a ) (b )Figure Ex . 20.1

( c ) Sec t io n m o du lu sFrom Eq. (20.6) (Th e p o in t of zero shear)

2 x 1 1 5yK V 17x2.67 = 2.25 m

M'"= 1 15(2.50 +2.25) - - (2.25) 3 x 17 x 2.676= 546.3 - 86.2 = 460 k N - m / mFrom Eq. (20.8)Sec t io n mo du lu s

460/, 175 X 1 0 3 -26.25X10-

2 m 3 / m o f w a l l

Example 20.2Fig . Ex. 20.2 show s a free s tan din g can ti lever sheet p ile with no ba ckfi l l dr iven into hom ogeneouss a n d . The fo l lo w ing da t a a re av a i l ab le :

H = 20 ft, P = 3000 Ib / f t of w a l l , 7 = 115 lb / f t 3 , 0 = 36.Dete r mine : (a ) the depth of p ene t r a t io n , D, and (b ) t he m a x i m u m b e n d i n g m o m e n t Mmax.


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Sheet Pile Wal l s and Braced Cuts 895

Solution

2 0 f t

P = 3000 Ib

y = 1 1 5 1 b / f t 30 = 36

Fig. Ex. 20.2

Kn = tan2 45+$. = tan2 45 + =3.85

KA== = 0.26A K p 3.85K =Kp - K A = 3.85 - 0.26 = 3.59The equation for D is (Eq 20.11)

where 8P 8x3000 = -58.133yK 115x3.5912PH 12x3000x20

115x3.59 = -11444x30002

(l 15x3.59)'Sub st i tu t ing and s impl i fy ing, we haveD^-58.133 D2 - 1744D -211.2 = 0From th e above equation D =13.5 f t .From Eq. (20.6)

= -211.2

12P^ = \2P_yK \yK2 x 3 0 0 0

115x3.59 = 3.81ft


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89 6 C h a p t e r 20F r o m E q . (20.12)

= 3000(20 + 3.81)-6

1 1 5x(3 .81 ) 3x3.596

= 71,430 - 3,806 = 67,624 I b - f t / f t of w a l lMmax=67'624]b'ft/ftfWal1

20.5 DEPTH OF EMBEDMENT OF CANTILEVER WALLS I NCOHESI VE SOILSCase 1 : When the Backf i l l is Cohesive Soi lT he pressure di s t r i but ion on a sheet pi le wal l is s h o w n in Fig. 20.10.

The act ive pressure pa at any depth z may be expressed as

w her eo ~ v = ver t ica l pressure , y zz = depth f rom the sur face of the b a c k f i l l .

T he pas s i ve p r e s su r e p } at any dep t h v b e l ow th e dredge l ine m ay be expressed as

T h e act ive pressure di s t r i but ion on the w al l f r om th e b ack f i l l s u r face to the d red ge l ine iss h o w n in Fig.20.10. Th e soil is supposed to be in t ens ion up to a depth of Z Q and the pre s s u re on thew al l is zero in th i s zone. T h e n e t pr es su r e d i s t r i b u t i on on the w a l l is s h o w n by t h e shaded t r i angle .A t t he dredge l i ne ( a t po i n t A )

(a) The act ive pressure p act ing towards the lef t isp a = YHKA~2cjK~Aw h e n 0 = 0 p a = y H - 2c = y H - qu (20.13a)w her e qu - u n c o n f in e d co m p ress iv e s t rength o f t h e clay soil = 2 c .

(b ) The pass ive pre ssure ac t in g towards the r ight a t the dredge l ine i sp} -2c s i nce ^ = 0

or P p = quT h e r e s u l t a n t of the pas s i ve and ac t i ve p r e s su r e s at the dredge l ine is

PP=Pa=Qu-(YH-ciu) =1qu-YH-p (20.13b)T h e r e su l t an t of the p a s s i v e an d act ive pressures at any dep t h y below th e dredge l ine is


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Sheet Pi le Wal ls and Braced Cuts 897

passive pressure, p = Yy + quactive pressure, pa = y (H + y) - quThe resultant pressure isP p - P a = p = ( r y + q u ) - \ - r ( H + y ) - q u ] = 2qu-rH ( 2 0 . 1 4 )Equa t ions (20.13b) and (20.14) indicate that the resultant pressure remains constant at

(2qu-yH) at all depths.If passive pressure is developed on the backfill side at the bottom of the pile (point B), thenp = Y (H + D) + qu acting towards the leftpa = yD-qu acting towards the right

The resultant isp / (20.15)

For static equilibrium , the sum of all the horizo ntal forces mu st be equal zero, that is,Pa-(2qu-YH}D + (2qu+2qu)h = 0

Simpl i fying,P a +2quh- 2quD + yHD = 0 , therefore,

D(2qu-yH}-Pa (20.16)Also, for equilibrium, the sum of the moments at any point should be zero. Taking momentsabout the base,

h2 (2q-yH)D2P a(y +D) +(2qu)- L-^-=0 (20.17)Subst i tu t ing fo r h in (Eq.20.17) and s impli fy ing,C1D2+C2D+C2=0 (20.18)where C { = (2 qu - yH )

__a

The depth computed from Eq. (20. 1 8) should be increased by 20 to 40 percent so that a factorof safety of 1.5 to 2.0 may be obtained. Alternativ ely the uncon fme d compressive strength qu m aybe divided by a factor of safety.


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898 C h a p t e r 20

Dredge level

Figure 20.10 D e p t h of e m b e d m e n t of a cant i l eve r wall in cohes ive so i l

Limiting Height of WallEquat ion (20.14) indicates that when (2 qu - y//) = 0 the resultant pressure is zero. The wall will notbe stable. In order that th e wall may be stable, th e condi t ion that mu st be satisfied is

*->yHFwhere F = factor of safety.

(20.19)

Maximum Bending MomentAs per Fig. 20.10, th e m a x i m um b end ing m om en t m ay occur within th e depth (D-h) below th edredge line. Let this depth be ;y0 below the dredge l ine fo r zero shear. We may write,

or y0~~Th e expression fo r m a x i m u m b end ing m om en t is ,M = P (v + v ) m ax av^o J' ~

where p = 2qu- yHTh e sect ion m odu lus of the sheet pile may now be calculated as before.

(20.20a)

(20.20b)


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Sheet Pile Walls and Braced Cuts 899Case 2: When the Backfil l is Sand with Water Table at Great DepthFigure 20.11 gives a case where th e backfil l is sand with no water table within. T he fo l lowingrela t ionships may be writ ten as :

P a _ =p - 2qu-yH= 4c -yHp' = 2qu + yH =4c +yH

J _a o

h =A second degree eq uation in D can be developed as beforeC jD 2 + C2D + C3 = 0

where Cl = (2qu-yH)c, = - 2 P _

(20.21)

(20.22)

Hwhere y = , qu = 2cA n expression for comput ing maximum bending moment may be written asU -fff+y)---*ma x a v-r -^ o ^ ^, (20.23)

Sand

JL, Point ofzero shear

,= pFigure 20.11 Sheet pile wall embedded in clay with sand backfill.


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900

where 2(2qu-yH}

C h a p te r 20

(20.24)

Case 3: Cantilever Wall with Sand Backfi l l and Water Table Above DredgeLine [Fig. 20.12]The various expressions fo r this case may be developed as in the earlier cases. The variousre la t ionships may be written as

Pi = ^IKAP L = YhiKA +p = 2q-YH

K A = (yhj + Y bh 2) K A

h _(2ciu-(yhl+rbh2)]D-Pa

The expression for the second degree equation in D isC } D1 + C2 D + C3 = 0

where

(20.25)

(20.26)

(20.27)

Sand y, 0, c = 0

Sand y sat , 0, c =0

Clay /sat, 0-0, c

Figure 20.12 C a n t i l e v e r wall with sand backf i l l a n d wa te r t a b l e


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Sheet Pile Walls and Braced Cuts 901

C3 = -Eq (20.27) may be solved fo r D . T he depth computed should be increased by 20 to 40% to

obtain a factor of safety of 1.5 to 2.0.Case 4: Free-Standing Canti lever Sheet Pi le Wall Penetrating ClayFigure 20.13 shows a f reestanding canti lever wall penetra t ing clay. An express ion for D can bedeveloped as before. The various re la t ionships ar e given below.

ry __ /

The expression fo r D isC j D2 + C 2 D + C 3 = 0

where C j = 2quC 2 = -2P

(20.28)

The expression fo r h isft = 2 -f (20.29)T he m a x i m u m m o m e n t may be calculated per uni t length of wal l by u s in g th e expression

(20.30)

H

DC y

Point of zero shearClay ysat ,0

Figure 20.13 Free standing cantilever wall penetrat ing clay


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902 Chap t e r 20

where ~ depth to the point of zero shear. (20.31)

Examp l e20.3Solve Example 20.1, if the soil is clay having an unconf ined compressive strength of70 kN/m 2 and a uni t weight of 17 kN/m3 . Determine the maximum bending moment .SolutionThe pressure distribution is assumed as shown in Fig. Ex. 20.3.

For 0u =0, pa =yH-qu =17x6-70 = 32kN /m2

Figure Ex.20.3

1 1-^a(//-z0) = -x32=2x70- 1 7 x 6 =38 k N / m

of wall

pf =


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Sheet Pile Wal ls and Braced Cuts 903

y= (H-z0) = ^(6-4.12) = 0.63 mFor the determinat ion of h, equa te th e summat ion of all horizontal forces to zero, thus

or 30- 38 xD + -(38 + 242)^ = 0_ 3.8D-3Therefore h=-4For the determinat ion of D, taking moments of all the forces about th e base of the wall, w ehave

/

o 7or 30(D + 0.63)- 38 x+ (38 + 242)x = 02 6

Subst i tu t ing fo r h we have,O O J~)_13D + 1.89-1.9D2+4.7 =014

Simpl ifying, we haveD2-1.57D + 1.35 = 0Solving D = 2.2 m; Increasing D by 40%, w e have D = 1.4(2.2) = 3.1 m .

M a x i m u m bending momentFrom Eq. (20.20)

-- 2

y 0 = = = 0 . 7 9 mp 38y~= 0.63 m

= 42.6- 1 1.9 = 30.7kN-m/m of wall

Example 20.4Solve Exam ple 20.1 if the soil below th e dredge line is c lay having a cohesion of 35 kN/m 2 and thebackfill is sand having an angle of internal friction of 30.The u nit weight of both the soils may beassumed as 17 kN/m 3. Determine th e maximu m bending mom ent .


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904

SolutionRefer to Fig. Ex. 20.4

pap = _ x3 4 x6 = k N / m of wallp = 2qu -y H =2x2x35 -17x6 -38 kN/m2From Eq. (20.22)C, D2 + C2 D + C3 = 0wh ere Cl = (2 < y w - 7 H) = (2 x 2 x 35 - 17 x 6) = 38 k N / m 2 = p

C9 = - 2 P . = - 2 x 102 = - 204 kN

C/ >

f l (p

a+ 6 Therefore ,


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Deflected /Dredge shape of wall /level . ,

/ /A\V/A\V/A\V/A\V/A\V/A\V/A\V/

Passivewedge

/A\V/\\V/A\V/A\V/A\V/A\V/A\V/A\V/A' 4 ', .**r T" i fs

B

S AnchorActive 'wedge /

Figure 20.14 Cond itions for free-ea rth support of an anchore d bulkheadSimplifying th e equation,

> 2 + C3 = 0 (20.32)

Figure 20.15 Depth of embedment of an anchored bulkhead by the f ree-earthsupport method (method 1)


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910 Chap te r 20

whe re C { =

Y h = s ubm e rge d u n i t w e i g h t o f soilK - K p-K A

T he f o rc e i n t he a nc hor rod , Ta, i s f o u n d by s um m i ng t he hor i z on t a l fo rc e s a sTa=Pa~PP (20.33)

T h e m i n i m u m d e p t h o f e m b e d m e n t i sD=D0+y0 (20.34)

Increase the depth D by 20 to 40% to give a fac tor of safe ty of 1 .5 to 2.0.Maximum Bending MomentT h e m a xi m um t he ore t i c a l m om e n t in th i s case may be at a p o i n t C a n y depth hm below ground levelwhich l ies be tween h} an d H whe re th e shear is zero. T he depth hm m a y be determined from thee q u a t i o n

\Pi\-Ta (hm-\ \Yb(hm-h KA=Q (20.35)Onc e hm i s k n o w n th e m a x i m u m b e n d i n g m o m e n t can eas i ly be c a lc u la t e d .

Method 2: Depth of Embedment by Applying a Factor of Safety to K(a ) Granular Soil Both in the Backfill and Below the Dredge LineThe forces tha t are ac t ing on the sheet p i le wal l are as shown in F ig. 20.16. T he m a x i m um pa s s i vepressure tha t can be mob i l ized i s equal to the area of t r i a ng le ABC shown in the f igure . The pass ivepressure tha t has to be used i n t he c o m p u t a t i o n is the area o f f igure ABEF ( shaded) . T he t r i a ng leABC i s d ivided by a ver t ica l l ine EF s uc h t ha t

Area AB CArea ABEF =-- -P'Factor o r safe ty pT he wi d t h o f f i gur e ABEF a nd t he po i n t o f a ppl i c at i on o f P' c a n be c a lc u la t e d w i t hou t a ny

dif f icul ty .E q ui l i b r i um of the sys tem requi res tha t the sum of a l l the horizonta l forces an d m o m e n t sa bou t an y p o i n t , fo r i ns t a nc e , a bou t th e anchor rod, should b e e q ua l to zero.He nc e , P ' p +Ta-Pa=0 (20.36)

Paya-ph*=Q (20.37)

whe re ,


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Sheet Pi le Wa l l s and Brac ed Cuts 911

//\\V/ \\V/\\V/\\\ //\\V/\\\

Figure 20.16 Depth of e m b e d m e n t by f r e e - e a r t h support method (method 2)an d F s = assumed factor of safety.The tension in the anchor rod may be found from Eq. (20.36) and from Eq. (20.37) D can bedetermined.(b ) Depth of Embedment when th e Soil Below Dredge Line is Cohesive and theBackfill GranularFigure 20.17 shows th e pressure distribution.

The surcharge at the dredge line due to the backfil l may be written asq = yhl+ybh2 =yeH (20.38)

where h3 = depth of water above th e dredge line, y e effective equivalent unit weight of the soil, an dThe active earth pressure acting towards th e left at the dredge line is (when 0 = 0)

The passive pressure acting towards th e right is

The resultan t of the passive and active earth pressures is(20.39)


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912 C h a p te r 20

-PFigure 20.17 D ep th of e m b e d m e n t w h e n th e soil be l ow th e dredge l ine isc ohes i v eThe pressure rema ins consta nt with depth . Taking mom ents of al l the forces about the anchor

(20.40)

(20.41)

rod,

where ya = the d is tance of the an chor rod from Pa.Simpl i fy ing Eq. (15 .40) ,

whe re C, = 2h 3

or

The force in the an cho r rod is g iv en b y Eq . (20.33).It can be seen from Eq. (20.39) th at the wall wi l l be u nsta ble if2qu-q=04c - q = 0

Fo r allpractical purposes q - jH - ///, then Eq.(20.39) may be wr i t ten as4c - y# = 0

c 1or (20,42)


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Sheet Pile Walls and Braced Cuts 913Eq. (20.42) indicates th at the wa l l is unstab le i f the ra t io clyH is equal to 0.25. N s is termed isStability Number. The s tabi l i ty is a funct ion of the wal l h e i g h t //, but is re la t ively independent of themater ia l used in deve l op ing q. If the w al l adhes ion c a is t aken in to account th e s tab i li ty numb er N sbecomes

(20.43)

At pass ive fa i lure l + ca/c is appr oximate l y equa l to 1.25.The s tabi l i ty number for sheet pi le wal l s embedded in cohesive soi l s may be wr i t ten as

1.25c (20.44)

W h e n th e factor of safety F = 1 and = 0.25, N, = 0.30.s sThe s tab il i ty num ber N s required in de te r min ing th e depth of sheet pi le wal l s is thereforeN s = 0.30 x Fs (20.45)T he maximum bending moment occur s as per Eq. (20.35) a t depth hm w hich lies between hland H .

20.7 DESIGN CHARTS FOR ANCHORED BULKHEADS IN SANDHager ty and Nofa l ( 1 992) provided a set of design char ts fo r de te r min ing

1 T he depth of embedment2. The tensile force in the anchor rod and3. The m a x i m u m m o m e n t in the sheet pi l ing

The char ts are appl icable to shee t p i l ing in sand and the analys is is based on the free-earthsuppor t method. T he assumptions made for the preparation of the design charts are:1. For act ive ear th pressure , Coulomb's theory is val id2. Logar i thmic fa i lure sur face below th e dredge l ine for the ana l ys i s of pass ive ear th pressure .3. The angle of fr ic t ion remains th e same above and below th e dredge l ine4. The ang l e of w al l friction be tw een th e pile and the soil is 0/2

The var ious symbols used in the charts are the same as given in Fig. 20.15w her e ,

ha = the depth of the anchor rod below the backfi l l sur facehl = t h e depth of the water table from th e backfi l l sur faceh2 - depth of the water above dredge l ineH = height of the sheet pi le w al l above the dredge l ineD = th e m i n i m u m d e p th of embedment r equ i r ed by the free-ear th suppor t methodT a = tens i le force in the anchor rod per uni t l ength of w a l lHager ty and Nofal developed th e cur ves g iven in Fig. 20 . 1 8 on the assumpt ion tha t th e w ate rtable is at the ground level , that is h{ = 0. Then they applied correction factors fo r /i , > 0. Thesecorrection factors are given in Fig. 20.19. The equa t ions fo r de te r min ing D , T a a nd M ( m a x ) a re


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914 Chap t e r 20

0.05

0.2 0.3Ancho r depth rat io , hJH

Figure 20.18 Genera l i zed (a ) depth of embedment, Gd, (b ) anchor fo rce G (f and (c)maximum moment G (a f ter Hager ty a nd Nofal, 1992)


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Sheet Pile Walls and Br aced Cuts 9151.181.161.14

e 1 . 1 0o1.081.061.04

0.01.08

1.061.04

0.94

0. 1 0.2 0.3

0.4

0.3

0.2

0.1

(a )

0.4 0.5

(c )

0. 1 0.2 0.3 0.4Anchor depth rat io h a/H

0.5

Figure 20.19 Correction factors fo r var iat ion of depth of water hr (a ) depthcor rect ion Cd, (b ) anchor fo rce c or rect ion Ct and (c) moment correct ion Cm (afterHagerty and Nofal , 1 9 9 2 )


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916 Chap t e r 20D = GdCdH (20.46 a)T a = GtCtjaH2 (20.46 b)

M ( m a x ) -GCn (20.46 c)w h e r e , G d = gene ra li zed non -d im ens iona l embedm ent = D /H for ft, =0G r = gene ra li zed non-d im ens iona l anchor fo rce = T a I (Y aH2) fo r hl~0

Gm - gene ra l i zed non-d imens iona l moment = M ( m a x ) / ya (#3) fo r / i j =0C rf , C r, C m = correc t ion fac tors for h { > 0Y fl = average effective u n i t w e i g h t of soil- /v h 2 + v h 2 + 2v / ? / z V /7 2' - ' m "l ^ I f e W2 Z'm 'M >V/7Y m = mo is t or dry u n i t we ight of soil above the water tableY ^ = s ubmerged u n i t w e i g h t of soil

The theore t ical d epth D as calculated by the use of des ign charts has to be increased by 20 to40% to give a fac tor of safety of 1.5 to 2.0 respect ive ly .

20.8 MOMENT REDUCTION FOR ANCHO RED SHEET PILE WALLSThe des ign o f anchored sheet pi l ing by the f ree-earth method is based on the as sumpt ion tha t th epi l ing is perfec t ly r igid and the earth pressu re dis t r ib ut io n is hydros tat ic , obeying c lass ical earthpressure theory . In real i ty , th e sheet pi l ing is rather f lexible and the earth pressure differscons iderably f rom th e hydros ta t i c d i s t r ibu t ion .

As such the bending moments M(max) calculated by the la teral earth pressure theories arehigher than th e actual values . Rowe (1952) sugges ted a procedure to reduce th e calculatedmom ents obtained by the/ree earth support method.Anchored Piling in Granular SoilsR o w e (1952) analyzed sheet pi l in g in granu lar soils and s tated that the fol low ing s ignif icant fac torsare required to be taken in the des ign

1. The re lat ive de ns i ty of the soil2 . The re lat ive flexibil i ty of the pi l ing wh ich is expressed as

H 4p=l09xlO-6 (20.47a)Elwhere ,

p = flexibil i ty n u m b e rH = the total he ight of the pi l in g in mEl = the modulus of e las t ic i ty and the moment of inert ia of the pi l ing (MN-m 2 ) pe r m of

wal lEq. (20.47a) may be expressed in Engl ish un i t s as

H4p = (20.47b)tAwhere , H is in ft , E is in lb/ in 2 and / is in in 4//if-of wall


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Sheet Pi le Walls and Braced Cuts 917Dense sand and gravel Loose sand

1.0

S 0 .6o'i04o d 0 .20

T= aH

D_L-4.0 -3.5 -3.0 -2.5Logp

-2.0

(a)

0.4 2.0

H

Logp = -2.6(working stress)

Logp = -2.0(y ield point of pi l ing)

1.0 1.5Stability n u mb e r

( b )Figure 20.20 Bending moment in anchored sheet piling by f ree-ear th support

method, (a) in granular soi ls, and (b) in cohesive soi ls (Rowe, 1952)Anchored Piling in Cohesive SoilsFor anchored piles in cohesive soils, th e most s ignificant factors ar e (Rowe, 1957)

1. The s tabi l i ty number


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918 Chap te r 20

( 2 0 - 4 8 )2. The relat ive height of pi l ing a

where,H = height of piling above the dredge line in metersy = effective un i t we ight of the soil above the dredge line = mo ist un i t weight above

water level and b u o y a n t un i t weight below water level, kN/m3c = th e cohesion of the soil below th e dredge line, kN/m 2

ca = adhesion between the soil and the sheet pile wall, kN/m2c-2- = 1 .25 fo r design purposesc

a = ratio between H and HM d = design moment

A f = maximum theoret ical momentiTlaXFig. 20.20 gives charts for computing design moments for pile walls in granular and cohesivesoils.

Example20.7Determine the depth of embedment and the force in the tie rod of the anchored bulkhead shown inFig. Ex. 20.7(a). The backfill above an d below th e dredge line is sand, having th e fol lowingproperties

G ^ = 2.67,y sa t = 18 kN/m 3 , jd = 13 kN/m 3 and 0 = 30Solve the problem b y the free-earth sup port method. A ssum e the backfill above the water

table remains dry.SolutionA s su m e th e soil above th e water table is dry

For ^=30, KA= , ,,=3.0

an d K =Kp-KA -3 = 2.67yb = y sat - yw = 18 - 9.81 = 8.19 kN/m 3.

where yw = 9.81 kN/m 3 .The pressure dis t r ibut ion along th e b u l k h e a d is as shown in Fig.Ex. 20.7(b)P j = YdhlKA =13x2x- = 8.67k N / m 2 at GW level3pa = p{+ybh2KA = 8.67 + 8.19 x3x-= 16.86 kN/m 2 at dredge l ine level


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Pa 16.86 = 0.77 mYbxK 8.19x2.671 - , - , 1 ,_

= - x 8.67 x 2 +8.67 x 3 + - (1 6.86 - 8.67)32 2+ - x 16.86 x 0.77 = 53.5 kN/ m ofwall2

_ L(a) (b)

Figure Ex. 20.7To f ind y, taking mo ments of areas about 0, we have53.5xy = -x8.67x2-+ 3 +0.77 + 8 . 6 7 x 3 ( 3 / 2 +0.77)

+ -(16.86-8.67)x 3(37 3+ 0.77) +-xl6.86x-x0.772 =122.6

W e have

N ow53.5 , ya =4 + 0.77-2.3= 2.47 m

p p =- = 10.93 D 2an d i ts d is tance from the anchor rod is

h4 = h3 + y0 + 2 7 3D0 = 4 + 0.77 + 2 / 3D0 = 4.77 + 0.67Z)0


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92 0 C h a p t e r 2 0Now, taki ng the mom ents of the forces abo ut the t ie rod, w e haveP \S -. , D \s hy * ' '/i

53.5 x 2.47 = 10.93D^ x (4.77 + 0.67D 0)S imp l i fy ing , w e haveD0 ~ 1.5 m, D = yQ + D Q = 0.77 + 1.5 = 2.27 m

D (des ign) = 1.4 x 2.27 = 3.18 mFor f inding the tens ion in the anchor rod, we have

Therefore, Ta=Pa-Pp= 53.5 -10.93(1.5)2 = 28.9 kN/m of wal l for the calculated depth D 0.

Examp l e 20.8Solve Example 20.7 b y a p p l y i n g F v = 2 to the passiv e earth pressure.SolutionRefer to Fig. E x. 20.8

The f o l l owing equa t ions m ay be wr i t t enp' =1YbKpD1. =- x8.19 x3 D 2 x - =6.14 D2P 2 F x 2 2pp=ybKpD= 8.19 x 3D-24.6DFG aPn D-hBC p D 1 n/1 ^r h = D(l-a)

D + h D + han -2 p 2Area ABEF = ap = ax24.6Dor 6.14 D 2 =a(D + h) x 12.3 DS ubs t i t u t i ng for h = D (1 - a) and s i mpl i f y i ngwe have2 2 - 4 a + l = 0Sol v ing th e equa t ion , w e g e t a = 0.3.N ow h = D (1-0.3) = 0.7 D and A G = D - 0. 7 D = 0. 3 DTaking moments of the area ABEF about the base of the pi le , and assuming ap = 1 in

Fig. E x. 20.8 w e h a v e- ( l ) x 0 . 3 Z > - X 0 . 3 Z 7 + 0.7D +(l)x0.7>x-Z, ~ j *


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ha=lm

Figure Ex. 20.8

s implifying wehave yp - 0.44Dy = 0.44 D

Now h, =h, +(D - y~ ) = 4 + (D - 0.44 D) = 4 +0.56Z)4 3 7p/From the active earth pressure diagram (Fig. Ex. 20.8) we havepl=ydhlKA=l3x2x-=$.61 kN/m 2

Pa=Pl+yb(h2+D) KA = 8.67 + = 1 6.86 +2.73D

(3+2 2

Taking mo men ts of active and passive forces about the tie rod, and sim plify ing , we have(a) for moments due to active forces = 0.89D3 +13.7D2 +66.7D + 104(b ) for mom ents due to passive forces = 6.14> 2(4 + 0.56D) = 24.56D2 + 3.44D3


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922 Chap te r 20Since the sum of the moments about th e anchor ro d should be zero, w e have0.89 D3 + 13.7 D2 + 66.7D + 104 = 24. 56 D2 + 3.44 D3Simpl i fy in g w e haveD 3 +4.26 D 2 - 26 . 16 D - 40.8 = 0By solv ing the equat ion we obtain D = 4.22 m with FS = 2.0Force in the anchor ro dT =P -P'a a p

where P a = 1.36 D2 + 16.86 D + 47 = 1.36 x (4.22)2 + 16.86 x 4.25 + 47 = 143 kNP'p = 6.14 D2 = 6.14 x (4.22)2 = 109 kNTherefore Ta = 143 - 109= 34 kN/m leng th of wal l .

Examp l e 20.9Solve Example 20.7, if the backfi l l is sand wi th 0 = 30 and the soil below th e dredge line is clayh av ing c = 20 kN/m 2 . For both th e soils, assume G v = 2.67.SolutionThe pressure dist r ibut ion along th e bulkhead is as shown in Fig. Ex. 20.9.

p, = 8.67 kN/m 2 as in Ex. 20.7, pa = 16.86 kN/m 2

= - x 8.67 x 2 +8.67 x 3 + - (l 6.86 - 8.67) x 3-47.0 kN /mh, = 1 m

h, = 2 m

= 5mh - , = 4 m

D

= 3 m

c = 0

HHHiVHHP C : HClayc = 20 kN/m2

= 2< -

Figure E x . 20.9


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S he e t P i l e W a l l s a n d B r a c e d C u t s 9 2 3To d e t e r mi n e ya, t a ke mome n t s a bout the tie rod.

Paxya =-x8.67x2 -x2-l + 8 . 6 7 x 3 x 2 . 5

+ -(16.86- 8.67) x 3 x 3= 104.82 V '_ _ 1048 _ 104,8Therefore ^ ~~~~ ~ ~ 223m

Now, < ? = Ydh { + rA= 13 x 2 + 8.19x 3 = 50.6kN/ m 2p = 2x2x20- 50.6 = 29.4 k N / m 2

Therefore, P=p*D= 29A D kN/m

a n d 4Taking moments of forces about the t ie rod, we ha ve

or 47x2.23-29.4D 4 + = 02or 1.47D 2 +11.76>-10.48 = 0 or D = 0.8 m.D = 0.8 m is obtained w ith a factor of safety equal to one. It should be increased by 20 to 40

percent to increase the factor of safety from 1.5 to 2.0.For a factor of safety of 2, the depth ofembed ment should be a t leas t 1 .12 m. How ever the suggested depth (design) = 2 m.H e n c e P for D (design) = 2 m isPp = 29 A x 2 = 58.8 kN/ m l e n gt h of w a l l

The tension in the t ie rod i sTa=Pa-Pp=41-59 = -1 2 kN/m of wal l .

This indicates that the t ie rod wi l l not be in tension under a design depth of 2 m. Howeverthere i s tension for the ca lcula ted depth D = 0.8 m.

Example 20.10Determine the depth of embed ment for the sheet pi le given in Fig. E x a m p l e 20.7 using the designchar ts given in Sect ion 20.7.Given: H = 5 m , hl = 2 m , h 2 ~ 3 m , ha = 1 m , h 3 = 4 m, 0 = 30


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924 Chapter 20Solution

= 1= 0.2H 5From Fig. 20.18For haIH = 0.2,and 0 = 30 we haveGd = 0.26, Gt = 0.084, Gm =0.024From Fig. 20.19 for 0 = 30,hJH- 0.2 and / z , / H = 0.4 we haveCd= 1.173, C,= 1.073, Cm= 1.036Now from Eq. (20.46 a)D=GdCdH = Q.26x 1.173x5= 1 . 5 2 mD (design) = 1.4 x 1.52 = 2.13 m.

where ya - H1V l ^ V 9 V ^ = 1 1 . 2 7 k N / m 252

Substi tut ing and s i m p l i f y i n gT a = 0.084 x 1.073x 11.27 x 52 = 25.4 k NM =G C Y H 3m ax m m 'a

= 0.024 x 1.036x 11.27x53 = 35.03kN-m/mofwallThe va l ues of D (design) and T a from Ex. 20.7 ar eD (design) = 3.18 rnT a = 28.9 kNThe design chart gives less by 33% in the valu e of D (design) and 12% in the valu e of Ta.

Example20.11Refer to Example 20.7. Determine for the pi le (a ) the bending moment Mmax, and (b) the reducedm o m e n t b y Rowe's method.SolutionRefer to Fig.Ex. 20.7. The f o l l o w i n g data ar e a v a i l a b l e

H = 5 m, hl - 2 m, H 2 = 3 m, / z 3 = 4 mY rf = 1 3 k N/ m 3 , Y 6 = 8.19 k N / m 3 and 0 = 30The m a x i m um be n d i n g m o m e n t o c c ur s at a depth hw f rom ground leve l where th e shear iszero. The equation which gives th e v a l ue of h is-p.h.-T +p.(h -h.)+-v,~ "1 1 a "1 v m \' 9 "


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where p l = 8.67kN/m 2/ i j = 2 m , T a = 28.9 kN/mSubs t i tu t ing-x8.67x2-28.9+8.67(/z -2) +-x8.19x-(/z -2) 2=02 m 2 3 mSimp l i fy ing , we have^+2.35^-23.5 = 0or hm~ 3.81 mTaking mom ents abo ut the poin t of zero shear ,

MmiiX=-\PA -f*! +Ta(hm-ha)-pl(hm~2hl) ~YbKA(hm-hf

= --x8.67x2 3.81--X2 + 28.9(3.81-1)- 8.67 (3'81~2)l8.19x-(3.81-2)32 3 2 6 3= - 21.41 + 81.2 - 14.2 - 2.70 42.8kN-m/m

From Ex.20.7 > (design) = 3 . 1 8 m , H = 5mTherefore #=8.18m

H4From Eq. (20.47a) p = 109 x 10"6 ElAssume E = 20.7x 10 4 M N / m 2F or a section P z - 27, / = 25.2 x 10~ 5 m 4/m

. . 1 0 . 9 x l O ~ 5 x ( 8 . 1 8 ) 4 ^or^ 1rt .Subs t i tu t ing p- ; ^ = 9.356 x l O ~ 3& 2.07 x!05x 25.2 x l O - 5logp = log y= -2.0287 or say 2.00Assum ing the sand backfi l l is loose, we have from Fig. 20.20 (a)

M d = 0.32 for i0g p = _ 2 . 0 0m axTherefore M d (design) = 0.32 x 42.8 = 13.7 kN-m/m

20.9 A N C H O R A G E OF B U L K H E A D SSheet pi le walls are many t imes t ied to some kind of an chors through t ie rods to g ive them greaterstability as shown in Fig.20.21. The types of ancho rage that are normally used are a lso sh own inthe same f igure .


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926 Chapte r 20A nchor s such as anchor wal l s an d anchor pla tes which depend fo r the i r res is tance ent i re ly o n

pass ive ear th pressure must be g i ven such d i mens i on s t hat th e anchor pul l does n o t exceed a certainfract ion of the pu l l r equ i r ed to pr oduce f a i l u r e . Th e ra t io between th e t ens ion in the a nchor T an dt he ma x i mu m pu l l w h i ch t he a nchor ca n s t a nd i s ca l l ed the fa c to r o f s a fe ty o f t he a nchor .

The types of anchorages given in Fig . 20.21 are :1. Deadmen, anchor plates, anchor beams etc.: D e a d m e n ar e shor t concre te blocks or

con t i nu ous conc r e t e bea ms d e r i v i ng t he i r r e s i st a nce f r om pa s s i ve earth pressure . T h i s typeis sui table when i t can be ins ta l led below the level of the or igina l grou nd sur face .

2 . Anchor block supported by battered piles: F ig (20.21b) s hows a n a nchor block supportedb y two bat tered pi les . T h e force Ta exer ted by the t ie rod t ends to i nduce compres s ion inpile P j a nd t ens i o n i n p i l e P9. Thi s t ype i s empl oyed w he r e firm soil is at great depth.

3. Sheet piles: Shor t sheet pi les are dr iven to form a con t i nuous w a l l w h i ch de r i ves it sres i s tance from pass ive ear th pressure in the sa me ma nne r as d e a d m e n .

4. Existing structures: Th e rods can be connec t ed to hea vy founda t i ons such as bu i l d i ngs ,c r a ne founda t i ons e t c .

Orig ina lg round B ac kf i l l Or ig ina lg round Backf i l l

Concre te cas t aga in s torig inal soi l

Sand and gravelcompac ted in l aye rs

(a )Precast concrete

Final ground

O ri g i n a lg round

Backf i l l

C o n t i n u o u ssheet p i les

(c )

C o m p .Tensionpile

Ta = ancho r p u l l

Pairs of sheet piles driven togreater depth at f req u en t i n t e rv a l sas vert ica l support

(d )Figure 20.21 Types of anchorage : (a) deadman; (b ) braced piles; (c ) sheet piles; (d)large st ruc ture (a f ter Teng , 1969)


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Locat ion of Anc hora geThe minimum distance between the sheet pile wall and the anchor blocks is determined by thefailure wedges of the sheet pile (un der free-earth support condition) and deadmen. The anchoragedoes not serve any purpose if it is located w ithin the failure wedge ABC shown in Fig. 20.22a.

If th e fa i lure wedges of the sheet pile and the anchor interfere with each other, th e location ofthe anchor as shown in Fig.20.22b reduces its capacity. Full capacity of the anchorage will beavailable if it is located in the shaded area shown in Fig. 20.22c. In this case1. The active sliding wedge of the backfil l does not interfere with the passive sliding w edge ofth e deadman.2. The deadman is located below the slope line starting from the bottom o f the sheet pile andmaking an angle 0 with the horizontal, 0 being the angle of internal friction of the soil.

Capaci ty of Deadman (After Teng,1969)A series of deadmen (anchor beams, anchor blocks or anchor plates) are normally placed atintervals parallel to the sheet pile walls. These anchor blocks may b e constructed near the groundsurface or at great depths, and in short lengths or in one continu ous beam. The holding cap acity ofthese anchorages is discussed below.Continuous An chor Beam Near Ground Surface (Teng,1969)If the length of the beam is considerably greater than its depth, it is called^ a continuou s deadman.Fig. 20.23(a) shows a deadman. If the depth to the top of the deadman, h, is less than about one-third to one-half of H (where H is depth to the bottom of the deadman) , the capacity may becalculated by assuming that the top of the deadman extends to the ground surface. The ultimatecapacity of a deadman may be obtained from (per uni t length)F or granular soil (c = 0)

or Tu=^yH2(Kp-KA) (20.49)

For clay soil (0 = 0)

-=~ (20-50)where qu = unconfmed compressive strength of soil,

Y = effective uni t weight of soil, an dK p, KA = Ran kine's active and passive earth pressure coefficients.It may be noted here that the active earth pressure is assumed to be zero at a depth = 2c/y

which is the depth of the tension cracks. It is likely that the magnitude and distribution of earthpressure may chang e slowly with time. For lack of sufficient data on this, the design of deadmen incohesive soils should be made with a conservative factor of safety.


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928 Chapter 20

S l i d i n g sur faceS l i d i n g sur face

O re pi le

Soft c lay

Anchorage subjected too ther hor i zon ta l fo rces Tw o sl iding wedges interferewi th each o ther(b)

Deadm an l oca ted in th i sarea has fu l l capac i ty

Figure 20.22 Locat ion of deadmen: (a ) o f f e r s n o res is tance; (b ) ef f ic iency great lyimpai red; (c ) full cap ac i ty , (a f ter Teng, 19 69)Short Deadman Near Ground Sur f ace in Granula r Soil (F ig . 20.23b)If the leng th of a deadm an is shor ter than 5h (h = heigh t o f deadm an) there wi l l be an end ef fect wi thr egards to th e h o ld ing ca p ac i ty o f th e anch o r . Th e eq u a t i o n su gges ted b y Teng f o r co mp u t i ng th eu l t ima te t ens i l e cap ac i ty T u is

T -where

hh

LHP ,'

h e igh t o f d e a d m a ndepth to the top of deadmanl eng th o f deadmandepth to the b o t to m of the dead m an from th e gro u nd su r faceto tal pass ive an d act iv e ear th pressures per u n i t l eng th

(20.51)


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Ground surface

Anchor pul l H ^Deadman Activewedge

Granular soilCohesive soil

Passivewedgea Active

* *" * /wedge

Deadmanh i' HHT1

Footing

(c )

Figure 20.23 Capaci ty of deadmen: (a ) continuous deadmen near ground surface(hlH < 1/3 ~ 1/2 ); (b) short deadmen near ground surface; (c ) deadmen a t greatdepth below ground sur face (after Teng, 1969)

KoYK0

a, KA

coefficient of earth pressures at-rest, taken equal to 0.4effective unit weight of soilRankine 's coefficients of passive an d active earth pressuresangle of internal friction

Anchor Capacity of Short Deadman in Cohes ive Soil Near Ground SurfaceIn cohesive soils, the second term of Eq. (20.51) sh ould be replaced by the cohesive resistance

T u = L(P -Pc) + qulf- (20.52)where q = unconfmed comp ressive strength of soil.Deadman at Grea t DepthThe ultimate capacity of a deadman at great depth below the ground surface as shown inFig. (20.23c) is approximately equal to the bearing capacity of a footing whose base is located at adepth ( h + ft/2), corresp ond ing to the mid heig ht of the deadman (Terzaghi, 1943).


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930 C hap t e r 20Ultimate Lateral Resistance of Vertical Anchor Plates in SandT he load-displacement behavior of horiz on tal ly loaded vert ical anchor pla tes w as ana lyzed byGhaly (1997). He mad e use of 128 pub lished f ield and laboratory tes t resul ts and presentedequa t ions fo r com pu t ing the fo l lowing :

1. Ultimate horizontal res is tance T u of s ingle anchors2. Horizon ta l d i sp lacem ent u at any load level T

The equa t ions area

nr C AyA Htan0 A

T = 2.2

whereAHhY

0.3

(20.53)

(20.54)

area of anchor pla te = hLdepth f rom the groun d surface to the bot tom of the pla teheight of plate and L = w i d t heffective uni t weigh t of the sandangle of f r ic t ion

16

12--

General equat ion -

Square plate

4 8 12 16 20Geomet ry factor , A

Figure 20.24 Rela t ionsh ip of pu l lou t -c apac i t y f ac to r ve r sus geomet r y f ac t o r (a f te rG ha l y , 1997)


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Sheet Pile Walls and Braced Cuts1.0

931

0.00.00 0.02 0.04 0.06 0.08Displacement ratio, u/ H 0.10

Figure 20.25 Relationship o f load ratio versus displacement ratio [from datareported by Das and Seeley, 1 9 7 5 ) ] (after Ghaly , 1 9 9 7 )C A = 5.5 for a rectangular plate= 5.4 for a general equ ation

= 3.3 for a square plate,a = 0.31 for a rectangu lar plate

0.28 for a general equation0.39 for a square plate

The equations d eveloped are v alid for relative depth ratios (Hlh) < 5. Figs 20.24 and 20.25give relationships in non-dimensional form fo r computing T u an d u respectively. Non-d imensionalplots for computing Tu fo r square an d rectangu lar plates are also given in Fig. 20.24.

20.10 BRACED CUTSGenera l Considera t ionsShallow excavations can be made w ithou t supporting the surro unding material if there is adequatespace to establish slopes at which the material can stand. The steepest slopes that can be used in agiven locality are best determined b y experience. M any bu ilding sites extend to the edges of theproperty lines. Under these circumstances, the sides of the excavation have to be made vertical andmust usually be supported by bracings.

Common methods of bracing the sides when the depth of excavation does not exceed about3 m are shown in Figs 20.26(a) an d (b). T he practice is to drive vertical timber planks known assheeting along the sides of the ex cavation. The sheeting is held in place by means of horizontal


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932 C h a p te r 20beams cal led wales tha t in tu rn are com m on ly suppor ted by hor izon ta l struts extending from s ide tos ide of the excavat ion . The s truts are usua l l y of t imber for widths not exceeding about 2 m. Forgreater w idth s metal pipes cal led trench braces a re com m on ly used .

When the excavat ion depth exceeds about 5 to 6 m, the use of vert ical t imber sheet ing wil lb e c o m e u n e c o n o m i c a l . A c c o r d i n g to one procedure , steel sheet piles are used around th e boun daryof the excavat ion . As the soi l is removed f rom the enclosure , wales and s truts are inserted. Thewales are comm only o f steel and the s t ruts may be of s teel or wood. The process con tinues unt i l th e

Steelsheet pi les m^ WaleH a r d w o o d b l o c k

(a )

WedgeWaleHardwood b lock

Ground l eve lwhi l e t i ebackis instal ledL a g g i n g -

Final groundlevel \

G r o u t or concreteSpacer

Steelanchor ro dBel l

(c)

Figure 20.26 C r o ss sec t io n s , t h r o u g h t yp ica l b rac ing in d eep e xc a va t i on , ( a ) sidesr e t a i n ed by s tee l shee t p i l es , (b ) s ides re ta ined by H pi les a nd lagg ing , (c) one ofsev e r a l t i eb ack sys t em s fo r s u p p o r t i n g ver t ica l s ides o f open cu t. s eve ra l se ts o fa n c h o r s m a y b e used , a t d i f f e r en t e l ev a t i o n s ( P eck , 1969)


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excavat ion is complete . In most types of soil, i t may be possible to el iminate sheet pi les and toreplace them with a series o f / / p i l e s spaced 1.5 to 2.5 m apart . The //piles, k n o w n as soldier pilesor soldier beams, are driven with their f langes para l le l to the sides of the excavation as shown inFig. 20.26(b). As the soil next to the piles is removed horizonta l boards known as lagging areintroduced as shown in the f igure and are wedged against the soi l outs ide the cut . As the genera ldepth of excavat ion advances f rom one level to another, wales and s truts are inserted in the samem anner as for s tee l sheet ing.

If th e w i d t h of a deep excavat ion is too great to permit economical use of struts across th eent i re excavat ion, tiebacks are often used as an a l ternat ive to cross-bracings as shown inFig. 20.26(c) . Incl ined holes are dri l led into th e soil outside th e sheet ing or H piles. Tensilere in fo rcement is then inserted and concreted into th e hole. Each tieback is usual ly prestressedbefore the depth of excavat ion is increased.Example 20 . 12Fig. Ex. 20.12 gives an anchor pla te f ixed vert ica l ly in medium dense sand with the bot tom of theplate at a depth of 3 ft below the ground surface. The size of the plate is 2 x 12 ft. Determine theul t ima te la tera l resis tance of the plate. The soi l parameters are 7= 115 lb/f t3 , 0 = 38.

SolutionFor all pract ica l purposes if L/h_> 5, the pla te may be considered as a long beam. In this caseL /h = 12/2 = 6 > 5. If the depth h to the top of the plate is less than abo ut 1/3 to 1/2 of// (where H isth e depth to the bottom of the pla te) , th e lateral capacity may be calculated using Eq. (20.49). In thiscase h lH = 1/3, As such

where K = t a n 2 45 + = 4.204p 2K A = = 0.2384.204T = -x 1 15 x32 (4.204 -0.238) = 2051 lb/ft

Example 20 .13Solve Example 20.12 i f 0 = 0 and c = 300 lb/ft2. All the other data remain th e same.


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934 Chap te r 20SolutionU se Eq. (20.50)

T =4cH- = 4x300x3-2x30(h = 2035 I b / f tY 11 5

Examp le 20.14Solve th e problem in Example 20 .12 for a plate length of 6 f t. All the other data remain th e same.SolutionUse Eq. (20.51) for a shorter length of plate

where (P - P a) = 2051 Ib / f t from Ex. 20 .12K o =1 - sin 0 = 1 - sin 38 = 0.384ta n 0= t an 38 = 0.78JK^ =V4.204 = 2.05, ^K ~ A =V0.238 -0.488

subst i tu t ingTu = 6 x 2 0 5 1 + -x0.384xl 15(2.05 + 0.488) x 3 3 x 0.78 = 12,306 + 787 = 13,093 Ib

Examp le 20.15Solve Example 20.13 if the pla te length L =6 ft . All the other data remain th e same.SolutionUse Eq. (20.52)

T u = L(Pp-Pu) +qulPwhere Pp-Pu = 2035 Ib/f t f rom Ex. 20.13

qu = 2 x 300 = 600 Ib / f t2Subst i tu t ingT u = 6 x 2035 + 600 x 3 2 = 12,210 + 5,400 = 17,610 Ib

Example 20.16Solve the problem in Exam ple 20.14 using Eq. (20.53). All the other data remain the same.SolutionUse Eq. (20.53)

5.4 Y AH H '28t a n 0 A

where A = 2 x 6 = 12 sq ft , H = 3 ft, C A = 5.4 and a = 0.28for a general equation


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Sheet Pile Wal ls and Braced Cuts 935ta n 0 = tan 38 = 0.78Subs t i tu t i ng

0.285 . 4 X H 5 X 1 2 X 3 3j =0.78 12

20.11 LATERAL EARTH PRESSURE DISTRIBUTION ON BRACED-CUTSSince most open cuts ar e excavated in s tages with in th e boundaries of sheet pile walls or wallsconsist ing of soldier piles and lagg ing , and since struts are inserted progressiv ely as the excavationprecedes, the wa lls are l ikely to deform as sho wn in Fig. 20.27. Lit t le inw ard movem ent can occurat the top of the cut after th e first s trut is inserted . The pat tern of deformat ion differs so greatly fromthat required fo r Rankine 's s ta te that th e dis t r ibut ion of earth pressure associa ted with reta iningwalls is not a satisfactory basis for design (Peck et al, 1974). The pressures against the upperportion of the wal l s are sub s tant ia l ly greater than those indicated by the equa t ion .

p* vfo r Rankine 's condit ion

(20.55)

wherepv - vertical pressure,0 = fr ict ion angle

Apparent Pressure DiagramsPeck (1969) presented pressure d is t r ibu t ion d iagrams on braced cu ts . These diagrams are based ona weal th of in form at ion col lected by actua l measurem ents in the field. Peck called these pressured iagrams apparent pressure envelopes which represent f icti tious pressure d is t r ibut ions fo rest imat ing s t rut loads in a sys tem of loading. Figure 20.28 gives the apparent pressu re d is t r ibut iondiagrams as proposed by Peck.Deep Cuts in SandThe apparent pressure diagram for sand given in Fig. 20.28 was developed by Peck (1969) after agreat deal of s tudy of actual pressure measurem ents on braced cuts used for subw ays .

Flexibleshee t ingRig idshee t ing

Deflect iona t mud l ine

(a ) (b )

y//////////////////////

(c )Figure 20.27 Typical pattern of deformation of vert ical walls (a) anchoredbulkhead, (b ) braced cut, and (c) t ieback cut (Peck et al., 1974)


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936 Chap te r 20The pressure d iagram given in Fig. 20 .28(b) is applicable to both loose and dense sands. The

s t ru ts are to be designed based on this apparent pressure d is t r ibut ion. T he most probable va lue ofany i nd iv idua l s t ru t load is a b ou t 25 percent lower than th e m a x i m u m (Peck, 1969). I t may be notedhere tha t th is apparen t pressure d is t r ib ut ion diagram is based on the assum pt ion tha t the water tab leis below the bottom of the cut .The pressure pa i s un i f o rm w i th respect to depth. The expression for p is

pa = 0.65yHKA (20.56)where ,

K A = ta n 2 (45 - 0/2)y = u n i t w e i g h t of sand

Cuts in Saturated ClayPeck (1969) developed two apparent pressure d iagrams, one for soft to medium c lay and the otherfo r stiff f is sured c lay . H e c lass i f ied these c lays on the basis of non-dimens ional fac tors ( s tab i l i tyn u m b e r N J as fol lows .St i f f F issured c lay

N = (20.57a)

So f t to Medium c lay

N,=c (20.57b)whe re y = un i t we igh t of clay, c - undra ined cohes ion (0 = 0)

Sand

H

0.65 yH tarT (45 -0/2)

(b )

(i ) Stiff f issured clay (i i) Soft to m e d i u m c l ayyH yHc ~ c

0.25 H

0.50 H

0.25 H

0. 2 yHto 0. 4 yH(c )

(ii)

yH-4c

0.25 H

0.75 H

(d )Figure 20.28 Apparent pressure diagram for calculating loads in struts of braced

cuts: (a ) ske t ch of wa l l of cut, (b ) diagram fo r cuts in dry or moist sand , (c ) diagramfo r clays if jHIc is less t han 4 (d) diagram fo r clays if yH/c is greater than 4 where cis th e average undra ined shear ing s t reng th of the soi l (Peck, 1969)


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0 0.1 0.2 0.3 0.4Values of yH

0.5 0 0.1 0.2 0.3 0.4

0.5 H

l.OH

London1 6 mA.J Oslo4 m

Hous ton

(a )

Hu mbl e B ldg ( 16 m )A 50 0 Jefferson Bldg ( 1 0 m ) One Shell Plaza ( 1 8 m )

(b )Figure 20.29 Maximum apparent pressures for cuts in stiff clays: (a) f issured claysin London and Oslo, (b) st i f f sl ickensided c lays in Houston (Peck, 1969)

The pressure diagrams fo r these tw o types of clays ar e given in Fig. 20.28(c) and (d)respectively. The apparent pressure diagram fo r soft to medium clay (Fig.20.28(d)) ha s been foundto be conservative fo r estimating loads fo r design supports. Fig. 20.28(c) shows th e apparentpressure diagram fo r stiff-fissured clays. Most stiff clays ar e weak an d contain fissures. Lowerpressures should be used only when th e results of observations on similar cuts in the vicinity soindicate. Otherwise a lower limit fo r pa = 0.3 yH should be taken. Fig.20.29 gives a comparison ofmeasured and computed pressures distr ibution fo r cuts in London, Oslo an d Houston clays.Cuts in Stratified SoilsIt is very rare to find uniform deposits of sand or clay to a great depth. Many times layers of sandan d clays overlying one another th e other ar e found in nature. Even th e simplest of these conditionsdoes no t lend itself to vigorous calculations of lateral earth pressures by any of the methods

H

Sand

Clay720=0

Figure 20.30 Cuts in stratif ied soils


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938 Chap te r 20available. Based on field experience, empirical or semi-empirical procedures for es t imat ingapparent pressure diagram s may be jus t i f ied. Peck (1969) proposed th e fol lowing uni t pressure forexcavations in layered soils (sand and clay) with sand overlying as shown in Fig. 20.30.

When layers of sand and soft clay ar e encounte red , th e pressure distribution shown inFig. 20.28(d) may be used if the unconf med compress ive s t rength qu is subst i tuted by the averagequ and the un it we igh t of soil 7 by the average value 7 (Peck, 1969). The express ions for qu an d7are

13u = "77 t X i KS hi2tan 0 +h2n


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Sheet Pile Walls an d Braced Cuts 939Case 1: Formation of Full Plastic Failure Zone Below th e Bottom of Cut.Figure 20.31 (a) is a ver t ica l sec t ion throug h a long cut of wi d t h B and dep th //in sa tura ted cohesivesoil (0 = 0). The soil below the bottom of the cut is un i fo rm up to a considerable depth for theformat ion of a ful l plastic failure zone. The undrained cohesive strength of soil is c. The w e igh t o fthe blocks of c lay on e i ther s ide of the cut tends to displace the un de rlying c lay tow ard theexcavat ion. I f the underlying c lay experiences a bear ing capaci ty fa i lure , the bot tom of theexcavation heaves and the earth pressure against th e bracing increases considerably.

The anchorage load block of soil a b c d in Fig. 20.31 (a) of w i d t h B ( a ssumed) at the levelof the bottom of the cut per unit length may be expressed as

(20.60)

(20.61)

Q = yHB-cH= B HThe vertical pressure q per uni t length of a horizonta l , ba , is

B B

(a )

/xx\

H

'/

s

D\' >

- t ^

2fi, ~= fi

^ t ^

s /WN /WN /WX

11 '1

(b )

/ / / / / / / / / / / / / / / / / / / / / ^ ^ ^ ^Hard s tra tum

Figure 20.31 Stability of braced cut: (a ) heave of bottom of t imbered cut in softclay if no hard stratum interferes with f low of clay, (b) as before, if clay rests atshallow depth below bottom of cut on hard stratum (after Terzaghi, 1943)


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940 Chapter 20The bearing capaci ty qu per un it area at level ab isqu = N cc = 5.1c (20.62)

where N =5.1The factor of safety against heaving isp = q = 5'7c

q H y-4 (20.63)BBecause of the geometrical condi t ion, it has been found that th e width B cannot exceed0. 7 B . Sub st i tut ing th is value fo r 5,F - = , ., (20.64)Q.1BThis indicates that th e width of the fa i lure slip is equa l to B V 2 = 0.7B .

Case 2: When the Formation of Full Plastic Zone is Restricted by thePresence of a Hard LayerIf a hard layer is located at a depth D below the bottom of the cut (which is less than 0.75), th efailure of the bot tom occurs as shown in Fig.20.31(b). The width of the strip which ca n sink is alsoequal to D.

Replacing 0.75 by D in Eq. (20.64), the factor of safety is represented b y5.7cF = u -

H Y- (20-65)D

For a cut in soft clay with a co nstant value of cu below the bottom of the cut, D in Eq. (20.65)becomes large, and Fs approaches th e valueF _ - _ 5-7S~~JT~~N^ (20'66)

where #5 = (20.67)Cu

is termed th e stability number. The stabi l i ty number is a useful indicator of potential soilmovements . The soi l movement is smaller fo r smaller values of N s.The analysis discussed so far is for long cuts . For short cuts, square, circular or rectangular,th e factor of safety against heave can be found in the same way as for foot ings.

20.13 BJERRUM AND EIDE (1956) METHOD OF ANALYSISThe m ethod o f analysis discussed earlier gives reliable results provided the w idth of the braced cutis larger than th e depth of the excavat ion an d that th e braced cut is very long. In the cases where th ebraced c uts are rectang ular, square or circular in plan or the depth of excav ation exceeds the widthof th e cut,th e fol lowing analysis should be used.


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Circular or square -= 1.0A-/ 1 I

H qm my0 S/0 NH

0 1

Figure 20.32 Stability of bottom excavat ion (af ter Bjerrum and Eide, 1956)In this analysis th e braced cut is visual ized as a deep foot ing whose depth and horizontaldime nsion s are identical to those at the bottom of the braced cut. This deep foot ing would fail in anidentical manner to the bottom braced cut failed by heave. The theory of Skempton for computing

N c (bearing capaci ty factor) for different shapes of footing is made use of. Figure 20.32 givesvalues of N c as a function of H/B for long, circula r or square footings. For rectang ular footings, thevalue of N c may be computed by the expressionN (sq) (20.68)

wherelength of excavationwidth of excavation

The factor of safety for bottom heave may be expressed asF cN .

whereyti + q

= effective unit weight of the soil above the bottom of the excavation= un iform surcharg e load (Fig. 20.32)

Example 20.17A long t rench is excavated in medium dense sand for the foundat ion of a mult is torey bui lding.The sides of the trench are supported with sheet pile walls fixed in place by struts and wales asshown in Fig. Ex. 20.17. The soil properties are:

7= 18.5 kN/m 3, c = 0 and 0 = 38 Determ ine: (a) The pressure distribution on the walls with respect to depth.

(b) Strut loads. The struts are placed ho rizon tally at distances L = 4 m center to center.(c) The maximu m bending moment fo r determining th e pile wall section.(d) The ma ximu m bending momen ts for determin ing the section of the wales.

Solution(a ) For a braced cut in sand use the apparent pressure envelope given in Fig. 20.28 b. Theequat ion fo r pa is

pa =0.65 y H K A =0.65 x 18.5 x 8 ta n 2 (45 - 38/2) = 23 kN/m 2


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942 Chap t e r 20pa = 23 k N / n r T

H

D

(a ) Sect ion 3 m

pa = 0.65 yH K A(b ) Pressure en velope

-3 m-8.33 kN

1 . 3 3 m 1 . 3 3 m i .67C

23 kN \/ 2 3 kNSection Dfi, "Section B\E

(c) Shear force distribution

*-! m-^38.33 k N

Figure Ex. 20.17Fig. Ex. 20.17b shows th e pressure envelope ,

(b ) St rut loadsT h e reac t ions at the e n d s of st ruts A, B a n d C a r e represented b y RA,RB an d Rc respectivelyFor reac t ion RA, t a ke mome nt s a b ou t Bfl. x3 = 4x23x-or R. == 61.33 kNA 2 3RB1 = 23 x 4-61.33 = 30.67 kND ue t o t he s y m m e t r y of the load d i s t r i b u t i o n ,RBl =RB2 = 30.67 kN, and RA =Rc = 61.33 kN.N ow the st rut loads are ( for L - 4 m )St rut A,PA = 61.33 x 4-245 kNSt rut B, PB = (R Bl +RB2 ) x 4 = 61.34 x 4-245 kNSt rut C,Pc = 245 kN


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Sheet Pile Walls and Braced Cuts 94 3

(c) M om ent of the pile wall sectionTo determine moments at different poin ts it is necessary to draw a diagra m show ing the shearforce dis tr ibut ion.Consider sections D B { an d B 2E of the w all in Fig. Ex. 20 . 17(b). The d istribution of the shearforces are shown in Fig. 20.17(c) along with the points of zero shear.The mome n t s at different points may be determined as fol lows

1M A = ~ x 1 x 23 = 1 1.5 k N - m21Mc = - x 1 x 23 = 1 1.5 kN- m1M m = - x 1.33 x30.67 = 20.4 kN- m

M n = ~ x 1.33 x 30.67 = 20.4 kN - mT he ma x i mum mome n t A /ma x = 20.4 k N - m . A suitable section of sheet pile can be determinedas per standard practice.

(d ) M a x i m u m m o m e n t fo r walesThe bending moment equa t ion fo r wales is

RL2

where R = ma xim um strut load = 245 kNL = spacing of struts = 4 m

245 x 42

M m a x =-= 490kN-mA suitable section for the wales can be determined as per standard practice.Example 20.18Fig. Ex. 20.18a gives th e section of a long braced cut . T he sides are supported by steel sheet pilewalls with stru ts and wales. Th e soil excavated at the site is stiff clay with the following properties

c = 80 0 lb/ft2, 0 = 0, y = 1 1 5 lb/ft3Determine: (a) The earth pressure distribution envelope.

(b) Strut loads.(c) The ma xi mum mome n t of the sheet pile section.

The struts are placed 1 2 ft apart center to center horizontally.Solution(a) The s tabil i ty num ber N s from Eq. (20.57a) is

= c 800


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944 Chap t e r 20D

H

Jflk. ^90^

2S ft Clavc - 800 lb/ f t20= 0y= 115 lb /f t3

/ v /7

iv5 f t A|

7.5 ft 57.5 ft

C ,5 f t ,

S^ /W xi?


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St r u t loads are:P A - 5243 x 12 = 62,916 Ib = 62.92 kipsP B = 2 x 2848 x 12 = 68,352 Ib = 68.35 kipsPc = 62.92 kips

(c) MomentsThe shear force diagram is shown in Fig. 20.18c for sections D B { an d B 2 EMoment at A =-x 5x 690 x - = 2,875 Ib-ft/ ft ofwall2 3M o m e n t at m =2848 x 3.3 - 863 x 3.3 x = 4699 Ib-ft /f t2Because of symmetrical loadingMoment a t A = Mom ent a t C = 2875 Ib-ft /f t of wallMoment a t m = Moment a t n = 4699 Ib-ft /f t of wal lHence , the maximum m oment = 4699 Ib-ft /f t of wall .The sect ion mo dulus and the requi red sheet pi le sect ion can be determined in the usual way.

20.14 PIPING FAILURES IN SAND CUTSSheet piling is used for cuts in sand and the excavation must be dewatered by pumping from th ebot tom of the excavat ion. Suffic ient penetra t ion below the bot tom of the cut must be provided toreduce the amount of seepage and to avoid the danger of piping.

Piping is a phenomenon of wa te r ru sh ing up through pipe-shaped channels due to largeu p w a r d seepage pressure. When piping takes place, the weight of the soil is counteracted by theu p w a r d hydr aul ic pressure and as such there i s no contact pressure be tween the gra ins a t the bot tomof th e excavation. Therefore, it offers no lateral support to the sheet piling and as a result th e sheetpiling may collapse. Further the soil will become very loose and may not have any bearing power.It is therefore, essential to avoid piping. For fur ther discussions on piping, see Chapter 4 on Soi lPermeabili ty and Seepage. Piping can be reduced by increasing the depth of penetration of sheetpiles below th e bot tom of the cut .

20 .15 PROBLEM S20.1 Figure Prob. 20.1 shows a cant i lever sheet pi le wal l penetra t ing m edium dense sand withth e following properties of the soil:

Y = 1151b/f t3 , 0 = 3 8 ,A ll th e other data are given in the f igure .Determine: (a) the depth of embedment for design, and (b) the maximum theore t ica lm o m e n t of the sheet pile.

20.2 Figure Prob. 20.2 shows a sheet pi le penetra t ing med ium dense sand wi t h the fol lowingdata :h { = 6 ft, h 2 = 18 ft, y s a t = 120 lb / f t3, 0 = 38,Determine: (a) the depth of embedment for design, and (b) the maximum theore t ica lmoment of the sheet pile. The sand above th e water table is sa tura ted.


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946 Chap t e r 20

y s a t = 1 2 0 1 b / f t 3

S and

y s a t = 1 2 0 1 b / f t j ^ gA . _ o o o a: ^ t0-^8 Mc = 0

S and

D = ?

y ( mo i s t ) = 170= 34, c = 0_ T ^ .^ _..., _ .^ _ .

y sa t = 19.5 k N/m 30 = 34c = 0

F i g u r e P r o b . 2 0 . 120.3

F i g u r e Prob. 20.2 F i g u r e Pr ob . 20 .3Figure Prob. 20 .3 shows a sheet p i l e p e n e t ra t i n g loose t o m e d i u m d e n s e s a n d w i t h th efo l l o wi n g d a t a :/ z , = 2 m , h2 = 4 m, 7 ( m o i s t ) = 1 7 k N / m 3 , y s a t = 19.5 kN/m 3 , 0 = 34,D e t e rmi n e : ( a ) t h e d e p t h o f e mb e d me n t , a n d (b ) t h e ma x i mu m b e n d i n g mo me n t o f t h esheet pile.

20.4 Solve Problem 20 .3 for the water t ab l e a t grea t dep th. Ass um e y = 1 7 k N / m 3 . All the otherd a t a re ma i n th e s a me .20.5 Figure Problem 20.5 shows f rees ta nd ing cant i lever wal l wi th n o b a c k f i l l . T h e sheet pilep e n e t ra te s me d i u m d e n s e s a n d w i t h th e f o l lo w i n g d a t a :

H = 5 m , P = 2 0 k N / m , y = 17.5 k N / m 3 , 0 = 36.D e t e rmi n e : ( a ) t h e d e p t h o f e mb e d me n t , a n d (b ) t h e ma x i mu m mo me n t

20.6 Solve Pr ob 20.5 with th e f o l lo w i n g d a t a :H = 20 f t , P = 3000 Ib / f t , 7 = 1 1 5 lb / f t 3 , 0 = 3 6

20.7 Figure Prob. 20 .7 shows a sheet p i le wa l l pen et ra t ing c lay soi l and the back f i l l i s a lso c lay .T h e f o l lo w i n g d a t a a re g i v e n .H = 5 m , c = 30 k N / m 2 , y = 17.5 k N / m 3D e t e rmi n e : ( a ) th e d e p t h o f p e n e t ra t io n , a n d (b ) t h e ma x i mu m b e n d i n g mo m e n t .20.8 Figure Prob. 20 .8 has s a n d backf i l l an d clay be low th e d re d g e l i n e . T he propert ies of theBa c kf i l l are:0 = 3 2 , y= 17.5 kN/m 3 ;Determine the dep th of penet ra t ion of p i le .

2 0 . 9 S o lv e P r o b . 2 0 . 8 w i t h th e w a t e r t a b l e a b o v e d r e d g e l i n e : G i v e n h{ = 3 m , h2 = 3 m ,y sa t = 1 8 k N / m 3 , where h{ = d e p t h of water t ab le be low backf i l l surface a n d h2 = (H - h{).T h e soi l above th e w a t e r t a b le is a lso sa tura ted . All the other da ta remain th e sa m e .

2 0 .1 0 Fi g u re Prob. 20.10 s h o w s a f r ee - s t a n d i n g sheet p i le pen et ra t ing c lay . T h e f o l lo w i n g d a t a a reavai lable :H = 5 m , P = 5 0 k N / m , c = 35 k N / m 2 , y = 17.5 k N / m 3 ,D e t e rmi n e : ( a ) t h e d e p t h o f e mb e d me n t , a n d (b ) t h e ma x i mu m b e n d i n g mo me n t .


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Sheet Pile Walls an d Braced Cuts 947,

{I/a

z

'

ii*H

)

P = 20 kN/m

H =

7 = 17.5 kN/m 30 = 36C=0 D

1

5 m

7

Clayc = 30 kN/m 37 = 1 7 . 5 kN/m 3

Clay

Sand7 = 1 7 . 5 kN/m 3= 32

Clay

Figure Prob. 20.5 Figure Prob. 20.7 Figure Prob. 20.8

20.11 Solve Prob. 20.10 with the f o l lo wi n g data:H = 15 ft , P = 3000 Ib/f t , c = 300 lb / f t2 , 7= 100 lb /f t3 .

20.12 Figure Prob. 20.12 shows a n anchored sheet pi le wal l fo r wh i c h th e f o l lo wi n g data a regivenH = 8 m, ha = 1.5 m, hl = 3 m, y sa t = 19.5 kN/m 3 , 0 = 38Determine: th e force in the tie rod.Solve th e problem by the free-earth support method.

= 5m

D

= 50 kN/m

Clayc = 35 kN/m27 = 17.5 kN/m 3

ha- 1.5 m_ TH = 8 m

Sand

7 sa t= 19.5 kN/m 3j = 3 m

Sand "7 sa t = 19.5 kN/m 30 = 38, c = 0

; = 6 .5 m= 5 m

Sand

Figure Prob. 20.10 Figure Prob. 20.12


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948 Chapte r 20

20.15

20.13 Solve the Prob. 20.12 with the follow ing data: H=24 ft , hl = 9 ft , h2 = 15 ft, and ha = 6 ft .Soil properties: 0 = 32, y sat = 120 lb/ft3 . The soil above th e water table is also saturated.20.14 Figure Prob. 20.14 gives an anchored sheet pile wall penetr ating clay. The backfill is

sand. The fol lowin g data ar e given:H = 24 ft, h =6 ft , h, = 9 ft ,' a ' I 'For sand: d ) = 36, y . = 120 lb/ft3,' 'SalThe soil above the water table is also saturated.For clay: 0 = 0, c =600 lb/ft2Determine: (a) the depth of em bedm ent, (b) the force in the tie rod, and (c) the maximu mmoment .Solve Prob. 20.14 with the fol lowing data:H=8m, h a= 1.5m, h} - 3 mFo r sand: y =19.5 kN/m 3, (b ~ 36 ' SU l 'The sand above the WT is also saturated.For clay: c = 30 kN/m 2Solve Prob. 20.12 by the use of design charts given in Section 20.7.Refer to Prob. 20.12. Determine for the pile (a) the bending moment Mmax, and (b) thereduced moment by Rowe's method.

20.18 Figure Prob. 20.18 gives a rigid anchor plate fixed vertically in medium dense sand withthe bottom of the plate at a depth of 6 ft below the ground surface. The height (h ) an dlength (L) of the plate are 3 ft and 18 ft respect ively. The soil properties are:y= 120 lb/ft3 , and 0= 38. Determine th e ult ima te lateral resis tance per unit length of theplate.

20.1620.17

ha = 6 ft \i, = 9 ftV

= 24 ft Sandy s a t= 12 0 lb/ft30 = 36, c =0

ClayD

Clay0 = 0,c = 600 lb/ft2

Y= 120 lb/ft

L = 1 8 f tFigure Prob. 20.14 Figure Prob. 20.18

20.19 Solve Prob. 20.18 for a plate of length = 9 ft. All the other data remain the same.20.20 Solve Prob. 20.18 for the plate in clay (0 = 0) having c = 40 0 lb/ft2 . All the other data

remain the same.


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Sandysat = 18.5 kN/m30 = 38, c = 0

Fig. Prob. 20.23

20.21 Solve Prob. 20.20 for a plate of length 6 ft. All the other data remain the same.20.22 Solve th e prob. 20.19 by Eq (20.53). All the other determine th e same.20.23 Figure Prob. 20.23 shows a braced cut in medium dense sand. Given 7= 18.5 kN/m 3, c =0and 0 = 38.

(a ) Draw th e pressure envelope, (b ) determine th e strut loads, and (c) determine th emaximum m oment of the sheet pile section.Th e struts ar e placed laterally at 4 m center to center.

20.24 Figure Prob. 20.24 shows the section of a braced cut in clay. Given: c = 65 0 lb/ft2,7= 115 lb/ft3.(a ) Draw th e earth pressure envelope, (b ) determine th e strut loads, and (c) determine th emaximum moment of the sheet pile section.Assume that the struts are placed laterally at 12 ft center to center.

Fig. Prob. 20.24


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book-chapter 20-sheet piled walls and braced cuts

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