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**Unformatted text preview: **DEEP FOUNDATIONS 1 0.1 PILES
1. Uses: A. Bearing Piles (compression or downward loading)
B. Tension Piles (upward loading)
C. Laterally loaded piles
2. Types of Piles:
A. Driven
B. Drilled and cast-in-place 3. Static analysis of ultimate downward pile capacity: Qult = QEB + QSF
Where
QEB = end bearing contribution
031: = skin friction contribution
Note: It is not good practice to assume that both skin friction and end bearing act simultaneously.
This is not totally unrealistic as both skin friction and end bearing cannot be totally mobi-
lized unless some (and some times signiﬁcant) pile movement occurs. 10.2 PILES IN GRANULAR SOILS 1. End bearing (also referred to as point resistance) —-— the pile tip acts as a deeply
buried spread footing. The end bearing may be estimated by an equation in the form of the bearing capacity equation:
QEB = PT Nq AT
Where
PT = Effective vertical stress at pile tip
Nq = Bearing Capacity Factor
AT = Area of pile tip
2. Skin Friction — developed along the side of the pile
051: = (KHc) (P0) (tan 6) (S) (H) Where .
KHC = Ratio of horizontal to vertical stress on side of element when ele-
ment is in compression
P0 = Effective vertical stress over length of embedment, D
6 2 Friction angle between pile and soil
S = Surface area of pile/unit length
H = Length of pile (Refer to Figure 10.1.) [SOFT SOIL
(RESISTANCE IGNORED) PRESSURE DIAGRAM
KHCPO AND KHT Po (AI ULTIMATE LOAD CAPACITY IN COMPRESSION
H=HooD
out: = P1 Nq “T +2 IKHCXPOXTAN SIISI
H=Ho WHERE Gun :ULTIMATE LOAD CAPACITY IN COMPRESSDN
P1 = EFFECTIVE VERTICAL STRESS AT PILE TIP (SEE NOTE I)
Nq = BEARING CAPACITY FACTOR (SEE TABLE , FIGURE I (DNTINUED) A1- : AREA or PILE TIP
um; : RATIO or HORIZONTAL T0 VERTICAL EFFECTIVE STRESS 0N SIDE or ELEMENT wIIEN ELEMENT IS IN COMPRESSION.
P9 = EFFECTIVE VERTICAL STRESS OVER LENGTH OF EMBEDMENT, 0 (SEE NOTE II
8 = FRICTION ANGLE BETWEEN PILE AND SDIL (SEE TABLE,FIGURE I WNTINUEDI 4 S = SURFACE AREAOF PILE PER UNIT LENGTH
FOR CALCULATING 0°“ |USE F5 N 2 FOR TEMPORARY LOADS. 3 FOR PERMANENT LOADSISEE mTE 2) (B) ULTIMATE LOAD CAPACITY IN TENSION H=Hofo
Tun = 2 IKHTIIPOXTAN 8) IsIIIII
H=Ho
WHERE= TuII = ULTIMATE LOAD CAPACITY IN TENSION. PULLOUT
K,” 8 RATIO OF HORIZONTAL TO VERTICAL EFFECTIVE STRESS (N SIII OF ELEMENT WHEN ELEMENT IS IN TENSIW
lMI FOE CALCULATING To" , USE F5 = 3 ONTuis PLUS THE WEIGHT OF THE FILE (WPIJHUS To“ = 3-
(SEE NOTE 2) NOTE-I 3 EXPERIMENTAL AND'FIELD EVIDENCE INDICATE TNAT BEARING PRESSJRE AND SKIN FRICTIGI INCREASE
WITH VERTICAL EFFECTIVE STRESS Po UPTO A LIMITING DEPTH (I EMBEDMENT. DEPENDING ON
THE RELATIVE DENSITY (I: THE GRANULAR SOIL AND POSITION OF THE WATER TABLE. BEYOND THIS
LIMITING DEPTH (I08 3 TO 408 *I THEEIS VERY LITTLE INCREASE IN END BEARING.AND INCREASE
IN SIDE FRCTION IS DIRECTLY PROPORTIONAL TO THE SURFACE AREAOF THE PILE. THERE FORE .IF
0 IS GREATER THAN 20 8 , LIMIT Po AT THE FILE TIP TO THAT VALUE CORRESPGIDING TO D=208 NOTE>2= IF BUILIJNG LOADS AND SUBSURFACE CCNDTION ARE WELL DOCUMENTED IN THE OPIle (Y THE MINEERJI
LESSER FACTm CFSAFETY CAN E Di!) BUT NOT LESS THAN 2 Q PROVIDED FILE CAPACITY IS VERIFIED EY LOAD TEST AND SETTLEMENTS ARE ACCEPTABLE. {WP Figure 10.1
Load Carrying Capacity of Single Pile in Granular Soils
From DM—7.02 (Pages 7.2—193-195) BEARING CAPACITY FACTORS - Nq PILE TYPE
DRIVENSINGLE H-PILE DRIVEN SINGLE DISPLACEMENT
PILE DRIVEN SINGLE DISPLACEMENT
TAPERED PILE DRILLED PILE (LESS THAN
24" DIAMETER) PILE TYPE 8
m 0 STEEL TIMBER t LIMIT 4» To 28° IF .IETTING IS USED
In (A) IN CASE A BAILER oa GRAB BUCKET Is USED BELDw GROUNDWATER TABLE .CALCULATE END
BEARING BASED ON 4» NOT EXCEEDING 23°.
(3) FOR PIERS GREATER TNAN 24 -|NCH DIAMETER,SETTLEMENT NATNER THAN BEARING CAPACITY
USUALLY CONTROLS THE DESIGN. FOR ESTIMATING SETTLEMENTJAKE 50%0F TNE SETTLEMENT
FOR AN EQUIVALENT FOOTING RESTING 0N TNE SURFACE or COMPARABLE GRANULAR SOILS.
(CHAPTER 5 ,DM-‘l I). Figure 10.2
Load Carrying Capacity of Single File in Granular Soils BEARING STRATUM
DENSE SAND YB'GSPCF W30" I ‘ . ‘2' EFFECTIVE VERTlCAL STRESS,P°'FOR
FILE DESIGN son A 32"mmzmz CLOSED mo. 0mm PIPE PIL£.CONCRETE nun, rmo on" AND Ta" FOR
A aoroor LONG me. p. w occuas AT aoe.on zo'um BEARING STRATUM. 4H 30'. N. a 2: um «5.8- 20° KHT ' L0 AT - 1 x 0.57:0]: sr cmcuu. AREA I” s I x 1 - 3.14 sr/lt cu" H.535 x21x0.78+[(|.5 xxw-sg—lsﬁhmu 20x20:3!4)+(l.51l535xTAN 20. 5.344] a as.“ {$0.34 Hm] an“: m r. m... - at."
Tu" : IO “02”; )XTAN 201ml!" +1.0xl.535 mm 20:51 3.14 20.23 + 8.77
29.0 K W" .u1 A,“ news.“
Tm n3%Q—+3.5u12x Figure 10.3
Load Carrying Capacity of Single Pile in Granular soils 10.3 PILES IN COHESIVE SOILS rune cA/c FOR A RECTANGULAR FOUNDATION. WITH WIDTH.B,ANO LENGTH. LINE
EARING CAPACITY FACTOR IS "an = Ncc (No.2 B/LI BEARING CAPACITY FACTOR Nc 2 3
o --—_——- “Arm (f m 10 wmm mnmno" o 500 nooo I500 2000 2500 3000 z/aonz/zn
coucsmc,rsr WHENDED VALUES 0F ADHEan WNW!)-
WKTER HAS MEI-"FEST
m ULTIMATE LDAD
CAPACITY UNLESS
0anan IS WW. am Is mum LOAD 0H1. FILE WEBNT I5
BALM BY waem'
0F OVERNROEN AND IS not mam. so. ULTIMATE LOAD UNITY N TENSION
ULTIMATE LMD CAPAOTY IN WHIIW Tu" = CA 211 R2 ("m’ LIMITED 57 OTI'ER Fm, 5E TEXT. Figure 10.4
Ultimate Load Capacity of Single Pile or Pier in Cohesive Soils
From DM—7.02 (7.2-196) Table 10.1
Application of Pile Driving Resistance Formulas
From DM-7.02 (Page 72-203) - BASIC PILE DRIVING FORMULAS (SEE COMMENT IN SECTION 2)
FOR DROP HAMMER FOR SINGLE - ACTING HAMMER . FOR DOJBLE -ACTING UWENTIALNAWER on": 33': {usE wIIEN DRIVEN WEIGsz an" = $25“ gas: was" DRIVEN
ARE SMALLER THAN ' wEIGNTs APE SMALLER
sTRmNG wEIGRTs THEN STRIKING wEIGNTs. to
i
I 0”": SH can: 4'5:— IUSE WHEN DRIVEN WEIGHTS no": $00.! w° ARE LARGER TRAN
5 STRIKING VIEIGNTs. 2E
W {USE MEN DRIVEN WEIGHTS
0
50:“ w ARE LAmER THAN
STRIKING WEIGHTS. no" : ALLDwAaLE PILE LOAD IN POUNDS.
VI : WEIGHT or STRIKING PARTS OF HAMMER IN POUNDS. R s TNE EFFECTIVE HEIGHT or FALL IN FEET. E - TNE ACTUAL ENERGY DELIVERED BY HAMMER PER BLOW IN FOOT-POUNDS. s - AVERAGE NET PENETRATION IN INCHES PER eLow FOR THE LAST GINDEDRIVING. wo : DRIVEN WEIGHTS NOTE I RATIO or DRIVEN wEIGNTs T0 STRIKING wEIGRTs SHOULD NOT
ws s WEIGHTS or sTRIxINGRARTs 'ExcEEo s. IDDIFICATDNS OF BASIC PILE DRIVING FORMULAS A. FOR FILES DRIVEN 10 AND SEATED IN ROCK AS HIGH CAPACITY END-BEARING PILES=
MIVE T0 REFUSALIAPPROXIMATELY 4 TD 5 BLOWS RDR THE LAST QUARTER INCH W DRIVING). REDRIVE OPEN END PIPE PILES REPEATEDLY UNTIL RESISTANCE FOR REFUSAL IS REACHED
WITHIN I IN. OF ADDITIONAL PENETRATION. I. FILES DRIVEN THROUGH STIFF COMPRESSIBLE MATERIALS UNSUITABLE FOR PILE BEARING TO AN
UNDERLYING BEARING STRATUM I ADD BLDWS ATTAINED BEFORE REACHING BEARING STRATUM TO REQUIRED BLDWS ATTAINED IN
BEARING STRATUM (SEE EXAMPLE).
PILE EXAMPLE: REQUIRED LDADGAPAchYoE FILE aon=2smus
HAMMER ENERGY E ,,5°oon_m DDMPRESSIRLE w
7 STRATUM . —— (I *5
PENETRATDNIS) As PER BASIC FCRMULA =I/2”oR23Lows PER
INCH (24 BLOWS/FTI. 42 BLOWS/FT
REQUIRED BLOWS FOR PILE 24 9 I8 = 42 HMS/FT. C. FILES DRIVEN INTO IJMITED THIN EARING STRATUMI DRIVE TO PREDETERMINED TIP ELEVATION. “LE DETERMINE ALLOWABLE LOAD BY LOAD TEST. STRATUM
UNSUITABLE
FOR BEARING 5': sTRATUM I
TIFF CLAY sTRATUM INDDMRREssIaLE
jeuT UNSUITABLE FOR FONT BEARING 10.4 P. E. PROBLEMS 10.4.1 P. E. PROBLEM 1 —- ALLOWABLE LOAD ON PILE GROUP A pile group consists of 12 piles with a pile diameter of 1.0 ft and a pile length of 40 ft.
The piles are placed in a 3 pile by 4 pile rectangular configuration with a pile spacing
of 2.0 ft center-to-center of piles. The piles are driven into clay that has the following
characteristics: Ground surface to 12 ft deep, cohesion = 1200 psf; 12 ft deep to 30 ft
deep, cohesion = 1600 psf; 30 ft deep to 40 ft deep, cohesion = 2000 psf. The clay layer
with a cohesion of 2000 psf also extends below the end of the piles. The unit weight of
the clay is approximately 120 pcf in all threeclay layers. A. Using a reasonable factor of safety, determine the allowable load on the pile group
considering that the piles act individually. B. Using a reasonable factor of safety, determine the allowable load on the piles con-
sidering group action. C. Which allowable load will you use? Brieﬂy justify your decision. 1’ diam. pile
Ground Surface / —IH—l||—l||——' l I
c = 1200 psf '5
7 = 120 pcf b ’
'5 12
.0.
I".
f l
5
c = 1600 psf 7 = 120 pcf 1;
fi 40’
‘0 18’
5
25.
D
.0.
E x
c = 2000 psf 7 = 120 pcf .5 10,
'1’.
g, V V Fig. P10.4.1 (a) Solution A. Determine individual capacity of each pile
Referring to Figure 10.4, Qult = C(NCS)J13R2 + cA ZnRZ Z = ﬂ = 40,useNcs = 9 E 1
Assuming piles are made of concrete,
0—12 ft: cA/c = 0.70, CA = 840 psf 12-30 ft: cA/c 0.57, CA 910psf
30-40 ftch/c = 0.46, CA = 920 psf Q“1t = (2000 psf) (9)31 (0.511)2
+ 2w (0.5) [ (840 psf) (12 ft) + (910 psi) (18 ft) + (920 psf) (10 ft) ]
= 14,1001b + (3.14) (35, 700) lb
= 14,1001b + 112,1001b = 126, 200 lb Using factor of safety of 2,
Qall = Quit/2 = 63 kip (single pile) Assuming each pile acts individually, the group capacity is:
Qau = 12 (63 kip) = 756 kip (12 piles) B. Determine group capacity Piles are 2’ on centers ©0007
OOQO4’ O O 0 OJ
L—sy—J Fig. P10.4.1 (b) Since piles are 2 ft on centers, very likely that group action will govern. Group efficiency, Eg, may be calculated using the following formula (found in
Bowles’ 2nd ed. of Foundation Analysis and Design.) 2(m+n — 2)s+4D Eg = mnnD
Where
m = number of columns (in plan View) of piles
n = number of rows
5 = pile spacing, center to center
D = pile diameter 2(4 + 3 — 2)2 + 4(1)
E = ———————————— x 100% = 63.7%
g (4) (3) (It) (1)
Therefore, considering group action, the allowable group capacity is 63.7% of 756
kip = 482 kip.
Which allowable capacity to use? Use reduced capacity of 482 kip because of group action; the piles are spaced too
closely to achieve 100% efficiency. The group just acts as if it were one large pile
with a surface area defined by the perimeter of the pile group. Note: To be 100% efficient, the minimum pile spacing may be calculated with the
following formula: 1.57Dmn — 2D
m+n - 2 For the given problem, the minimum pile spacing to achieve 100% efficiencywould
be: ' s = _ 1.57 (1 ft) (4) (3) — 2(1) _
S — W— — 10.4.2 P. E. PROBLEM 2 — MAXIMUM MOMENT IN PILE GROUP Situation: A structural foundation consists of 24 piles with a rigid concrete cap as shown below.
The foundation loads are a 600 ton concentric compression and a moment. The mo-
ment can be applied about axis x or y as shown below. The soil investigation report rec-
ommends individual pile capacities of 30 tons tension and 90 tons compression. Required: A. Determine the maximum moment that could be resisted about x or y shown below,
with the axial load. B. If the concrete cap is separated along the y-y axis and a rigid horizontal member
is added across both halves, determine the maximum moment that could be re-
sisted about the y-y axis with the axial load. ©——r O [email protected]’ 6" 0000 0000 0000 @000 0000
O y
[email protected]’—6" Fig. P10.4.2 (a) Load Per Pile, tons Solution A. Considering only the concentric compression load, the average load per pile = 600
tons/24 piles = 25 tons. Because the cap is rigid, moment will cause the pile cap to
rotate about the center line and the induced axial load in any pile will be a function
of distance. In fact, the load will be proportional to distance in a linear fashion. The
maximum moment will be governed by either the maximum allowable compressive or tensile load on a pile.
Consider moment about y-y axis.
Case I: Assume compression governs. Draw load distribution across one line of piles (will
multiply by‘4 later). Set maximum compression load as 90 tons, then determine
load in other piles using knowledge that average load in all piles is 25 tons. Case 11: .
Assume tension governs. Set maximum tensile load as 30 tons. —1OO - I
U‘
o O
l 0"
O
- 100 - Fig. P10.4.2 (b) Note that Case I exceeds allowable tension capacity, Case 11 will govern. To determine the maximum moment, do not consider the axial load. Then the load
diagram is: —100 — ~55
_5Q __ —33 —11 O ll 33 Load Per Pile, tons
01
o
| 55 100 — Fig. P10.4.2 (c) The maximum moment about the y-y axis: M = 4 {(55 tons) 5 (3.5 ft) + (33 tons) 3 (3.5 ft) + (11 tons) (3.5 ft)}
= 5,390 ton-ft Consider Moment about x-x axis.
Using a similar procedure, you may do for practice:
M = 3,849 ton-ft If the concrete pile cap is separated along the y-y axis and a rigid horizontal mem—
ber is added across both halves, the maximum moment that could be resisted should still be 5 ,390 ton-ft. 10.4.3 P. E. PROBLEM 3 — DESIGN LOAD ON PILES IN DENSE SAND A pile is driven into a dense sand layer. After a few days, its capacity is load tested.
Curve A in the figure shows the load settlement relation obtained from the load test.
Because of the fact that the dense sand is overlain by several strata of soft compressible
materials, it may not be advisable to count on that portion of the pile capacity devel-
oped through skin friction. To determine the contribution of the soft compressible soils,
a second pile is driven into the ground in such a way that its point ends up three feet
above the dense sand. The second pile is also load tested. The results of the load test
on the second pile are given by curve B of the figure. Use a factor of safety of 2.0. Estimate the design load that may be used on piles driven into the dense sand. Load in tons O 20 4O 60 80 Settlement, in. Fig. P10.4.3 (a) Solution Pile A — resistance mostly due to end bearing in the sand and small part skin friction
in the soft compressible soils. Pile B — resistance due only to skin friction in the soft compressible soils. // Soft Compressive
Soil its " Dense ‘Sand".-'- ‘ '.' Fig. P10.4.3 (b)
Decide what is allowable settlement. This will depend on type of structure. Say 1 in.
settlement is allowable.
At] in. settlement: Ultimate capacity of piles: Pile A. . . . 56 tons
Pile B . . . . 10 tons To neglect capacity of skin friction, then capacity due to end bearing only is:
56 — 10 = 46 tons
For factor of safety of 2.0, then allowable pile capacity is 23 tons. 10.4.4 P. E. PROBLEM 4 — MAXIMUM POINT LOAD IN A TYPICAL FILE A storage warehouse that is being designed to carry a uniform distributed live load of
1200 psf will have 2-ft thick reinforced concrete floor. The entire structure will be sup-
ported by 18-in.-diameter circular concrete piles that span at 10 ft center to center in a square grid pattern. The piles penetrate 9 ft of recent granular fill that weighs 110 pcf and 25 ft of very soft
normally consolidated clay that weighs 95 pcf. The unconfined compressive strength is
800 psf. The piles are point bearing in a thick dense sand layer that is beneath the soft clay layer. Special precautions consisting of the installation of permanent sleeve to be taken dur-
ing construction to minimize the transmission of shear forces from the recent granular fill to the piles. Requirements: '
A. Compute the maximum point load that is likely to be developed in a typical pile. B. If subsequent consolidation tests on the very soft clay indicates that the clay is over
consolidated rather than normally consolidated as assumed in requirement A.
above, how would this information affect your approach to the determination of
the maximum point load in a typical pile, explain the basis of your answer. Solution A. Determine maximum load likely to develop in a pile. The structural loads are due
to:
1. Floor weight
2. Distributed live load
3. Weight of the pile
Tributary area to a single pile is 10 ft x 10 ft = 100 ft2- f 2’ Thick R.C. Floor O'Q_ 7=150pcf ’ Granular fill \
' ’y = 110 pcf ' , ' 25 Soft Normally
Consolidated clay ///////7/
7 = 95 pcf
/////// é / \\
L oooa_ on 0005‘ °agaooeoo
009 o o o 00 p 000 OF‘DOO p oDense sgidolayerco 9 a0 Op
0 9 5‘ D C Fig. P10.4.4 (21) Load = 2ft x 150 pcf x 100 ft2 + 1,200 psf x 100 ft2 2 + n(1'gft) x 34 ft x 150pcf = 30,0001b + 120,0001b + 9,0001b
= 159 kip Sleeving of piles through the fill will minimize transmission of forces, however,
consolidation of the clay will induce downdrag forces in the pile. For the clay, since qu = 800 psf, then c = 1/2 qu = 400 psf. The downdrag force will
be equal to the cohesion of the clay times the surface area of pile in the contact with the clay.
Downdrag = C x A = 400 psf >< It(l.5 ft)(25 ft) = 47 kip.
Therefore, total maximum load = 159 + 47 kip = 206 kip If the clay is over-consolidated, the consolidation of the clay layer under load will
be much less and the downdrag forces should be insignificant. Then the maximum load would be 159 kip. 10.5 CAISSONS Where deep foundations are required and end bearing is desired, caissons, which are
drilled-and-belled shafts as shown in Figure 10.5, may be used. The bearing capacity of soils at the base level of a caisson may be determined by the
end bearing equation. Figure 10.5 Caisson The belled portion of the caisson must be hand cleaned, and OSHA requires that the
caisson excavation be shored and a fresh air supply be provided. These regulations have
made caissons uneconomical in some situations. Caissons provide excellent uplift capacity. The uplift capacity is the lesser of either: A. Shear resistance of a cylinder of soil extending from the bottom of the bell of the
caisson upward Figure 10.6 Shear Resistance of Cylinder 01' B. The weight of the soil and concrete contained in the frustum of an inverted cone
extending from the bottom of the bell upward. Figure 10.7 Inverted Cone 10.6 EXAMPLE PROBLEM 10.6.1 EXAMPLE PROBLEM 1 — UPLIFT CAPACITY OF A CAISSON Determine the uplift capacity of a belled caisson founded at a depth of 30 ft. Soil properties: y = 110 pcf
c = 200 psf
(1) = 35°
Caisson dimensions: 30 in. shaft
48 in. bell
Solution Find average shear strength of soil at depth of 15 ft.
5 = c + oh tan (1)
= c +K0 0V tan (1) ; assume KO = 0.5
= 200 psf + 0.5 (110 pcf) (15 ft) tan 35°
= 778 psf
Use: 5 = 750 psf
1. Shear failure: Straight cylindrical failure 48 in. bell
4 ft diameter of cylinder Surface area of cylinder = red X h = 311 (4 ft) (30) = 377 ft2 Uplift capacity in shear = 750 psf x 377 ft2 + (weight of caisson and soil)
= 283 k + 47 k (you verify the 47 k)
= 330 kip i 7— -\:- i
/- AO‘E: ‘ %
w <5 Fig. P10.6.1 (a) 2. Friction failure: Frustum of a cone Fig. P10.6.1 (b) Volume of frustum = 1212— h (D2 + Dd + d2) d = 4ft, h = 30ft, D = d+2htan¢
D = 4ft + (2) (30ft) tan 35° D = 4ft+42ft 46ft V = f—Z (30 ft) [(46 £02 + (46 ft) (4 ft) + (4 £02] = 18, 200 ft3
If entire frustum were soil with y = 110 pcf, the weight of frustrum
= 18200 ft3 x 110 pcf / (10001b/kip) = 2,002 kip. Net addition and weight of concrete in shaft 2
= n ft) 30 ft x (150 — 110 pcf) = 5.9 kip
(Say 6 kip) Neglect concrete in belled portion, then weight of frustum = 2002 kip + 6 kip
Uplift capacity in friction failure = 2008 kip 3. Governing capacity Shear governs, uplift capacity is 330 kip Note that no factors of safety have been applied to this example. ...

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