23 - Deep Foundations
23 - Deep Foundations
23 Deep Foundations
23.1 OVERVIEW
23.1.1 Pile-Up
Putting structures on posts is not a new idea, but dates back at least 6000 years, to
a time when the European climate was warming after centuries of cold, drought,
and misery, and lake levels were rising. That was the time of ‘‘Ötzi,’’ the Ice Man
of the Italian Alps, whose discovery set off a crime investigation that must be the
ultimate in cold case files.
Prehistoric builders burned the ends of tree trunks to sharpen them, and probably
wielded rocks to drive the posts into soft ground. Today’s piles are driven with
pile drivers that range from drop hammers to steam or diesel operation.
The most common use of piles is to keep structures from sinking into the ground,
but piles also prevent bridge supports from being undercut by scour during
periods of rapid current and high water. The famous London Bridge—the one
that young children are taught is falling down, thereby fostering an unwarranted
disenchantment with civil engineering—was on wooden piles, and it did not fall
down; it was pushed down by river ice. It then was repaired and lasted for another
550 years. Somebody should write a song about that.
Modern piles are wood, steel, concrete, or composite if they are composed of more
than one material. An example of a composite pile is when a wood pile section is
used under a groundwater table where it is preserved by reducing conditions, and is
connected to a concrete section that extends through aerobic surroundings above
the water table.
Whereas piles are driven or can be jacked into the ground, piers are large-diameter
supports that are placed in pre-bored holes. Caissons are large tubes used for
construction below water, as in rivers, and may be pressurized to keep water out.
‘‘Caisson foundation’’ sometimes is applied to bored piers, and piers also are
sometimes called shafts or piles, leading to some deep confusion.
Figure 23.1
Deep,
intermediate, and
shallow
foundations.
Although end-bearing piles are long, slender columns, they are not free columns
and derive considerable lateral support against buckling from passive resistance of
the soil. Failure by buckling is virtually unknown unless the piles extend for long
distances through air or water. A different kind of bending of steel pile can occur
during driving, if the pile hits a glancing blow on a large boulder deep within
the soil. There have been rare occurrences where a driven steel pile has done a
1808 bend and emerged at the ground surface.
Vertical piles also may be designed to resist horizontal loading, particularly when
horizontal loading is not constant, as in wharves.
Although positive skin friction aids in support of a pile or pier foundation, it does
not peak out simultaneously with end support, particularly for long piles, because
of compression of the piles. That is, as a pile or pier is loaded, support initially
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Deep Foundations
Figure 23.2
Combined friction and end bearing: (a) ideal development, (b) negative skin
friction adds soil weight to the pile; (c) during loading, side friction develops
first and decreases, mobilizing end bearing.
derives from skin friction along the upper part, and then as additional load is
applied and the pile compresses, more friction is mobilized until finally the load
reaches the end of the pile or pier and starts building up end support. This is
illustrated in Fig. 23.2(c). It also means that with a factor of safety of 2 little or no
load may reach the lower end of the pile or pier.
Wood Piles
Timber piles were first, and still are commonly used depending on cost and
availability. Plain or untreated wood is highly susceptible to decay in aerobic wet
conditions, and most wood piles now are chemically treated. Older buildings
founded on untreated wood piles may suffer damage because of lowering of the
groundwater table. The length and bearing capacity of wood piles also is limited,
depending on the kind and size of the trees. In the U.S., southern pine is often
used and is limited to a length of about 80 ft (25 m).
Marine borers, both crustaceans and mollusks, are difficult to separate from their
appetites, and even wood piles whose surface has been impregnated with creosote
may last only 15 to 50 years, depending on climatic factors. The useful life can be
extended by periodic maintenance that involves caulking or coating damaged
areas.
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Deep Foundations
Steel H-Beams
Steel H-beam sections are especially useful to penetrate through loose or weath-
ered rock layers to obtain end bearing in solid rock. To minimize end damage
and bending they may be fitted with hardened steel shoes that are welded on at the
job site. Because of their small cross-sectional area, H-piles can be driven into
sand and gravel where it would be difficult to drive displacement piles such as
timber or concrete.
Steel H-beams also are used for trestle structures in which the piles serve both as
bearing piles and as braced columns.
Side friction on steel H-piles derives about one-half from soil-to-steel friction on
the outer surface of the flanges, and the other half from soil friction as soil
between the flanges moves with the pile. A driving shoe that is larger than the pile
oversizes the hole and reduces side friction, so H-piles intended for side friction
should not have such a shoe. However, the lower friction and side area are
advantageous for reducing negative skin friction.
When steel piles are exposed to the air or to alternate wetting and drying above a
groundwater table they are subject to corrosion, the same as any steel structure
under similar conditions. H-beam piles above ground can be protected from
corrosion by painting, or they may be encased in concrete for several feet above
and below the ground line.
Because handling and driving stresses are severe, steel reinforcement often is either
pre-tensioned or post-tensioned. Sufficient flexural strength is obtained with only
about one-fourth to one-third the usual level of prestressing for more orthodox
structural members. Prestressing also can help to prevent piles from breaking in
tension during driving, as compression waves going down the pile pass and add to
wave displacements echoing up from the bottom.
Whether prestressed or not, precast piles must be carefully handled and lifted by
slings under the one-third points to reduce bending stresses to a minimum.
Uniformly tapered precast piles are limited in length to about 40 ft (12 m) because
of the small cross-sectional area of the lower end. Parallel-sided piles with tapered
ends can be over 100 ft (30 m) long. Octagonal precast piles may be up to 36 in.
(0.9 m) in diameter, in which case a hole may be cast along the center axis to
reduce their weight.
As an illustration of the procedures that are involved, one type of tapered driven-
shell pile is made as follows:
1. A thin corrugated steel shell is closed at the bottom with a steel boot.
2. A steel mandrel or core is placed inside the shell.
3. The mandrel and shell are driven into the ground.
4. When a predetermined driving resistance has been reached, driving stops and
the mandrel is withdrawn.
5. The open lined hole is inspected for damage.
6. The shell is filled with concrete.
The driving mandrel protects the shell during driving. This type of shell can be
tapered uniformly from the bottom to top, or they may be step-tapered with
different-sized shells attached end to end. An advantage of the step-taper is that
the driving mandrel distributes driving energy along the length of the shell instead
of concentrating it at the bottom.
The thinnest shells that can be used for this type of pile are about 18 in. (3 mm)
thick, but thicknesses of l.5 to 2 times this value are more common, depending on
the length and diameter of the pile.
A parallel-sided, dropped-in shell pile that is designed solely for tip resistance is
constructed by driving a heavy steel pipe, usually 16 in. (400 mm) in diameter and
½ in. (12.7 mm) thick, with the aid of a steel mandrel, then pulling out the
mandrel and inserting a thin metal shell such as a corrugated-metal culvert pipe,
filling it with concrete, and withdrawing the drive pipe.
Another alternative is to drive a steel pipe pile with either a closed end or an open
end. The closed-end pipe is filled with concrete. Sometimes a pipe is driven all in
one piece, or it may be made of several pieces of pipe either welded together or
fitted together with special internal sleeves. The open-end steel pipe is used to
obtain better penetration through weathered rock to find hard rock, and then
either flushed out with water and air or with a miniature orange-peel dredging
bucket. The open pipe then is given a few extra blows, pumped out, and filled with
concrete. By sealing into the rock, a pile installed in this manner in effect is a
small-diameter caisson and has a high bearing capacity.
1. A heavy steel pipe containing a mandrel or core is driven into the ground.
2. The mandrel is removed and a small charge of concrete is poured into the
pipe.
3. The mandrel is reintroduced and lowered to the top of the concrete.
4. The pipe is pulled upward 2 or 3 ft (23 to 1 m).
5. The mandrel is again driven with the pile-driving hammer, which forces
concrete out of the end of the pipe to form a bulb-shaped mass.
6. The mandrel is removed, and the pipe is filled with concrete up to the top.
7. The mandrel is reintroduced to hold the concrete down while the steel drive
pipe is pulled out of the ground.
23.2.11 Minipiles
Minipiles or micropiles are any small-diameter piles that are installed in
borings, jacked in, driven, or vibrated in. They are useful where there is little
headroom for installation, for example inside existing buildings. Hollow pipes
are used for higher loads and the ends pressure grouted. A simple adaptation used
for light underpinning is reinforcing steel bars that are jack-hammered into
the soil near a foundation, then bent back underneath a footing and encased in
concrete.
Diesel hammers (Fig. 23.3) have a self-contained power source. The ram acts as a
vertical piston that when dropped compresses an air-fuel mixture inside a cylinder,
causing it to explode and propel the ram back up again. Burned gases exhaust
Figure 23.3
Diesel hammer
puffing smoke as it
drives a steel pipe
pile.
through a port, additional fuel is injected into the cylinder, and the cycle repeats
automatically until the fuel is shut off. The ram also impacts an anvil that drives
the pile. Pile resistance is needed to keep a diesel hammer actuated, so if a pile
breaks or goes through very soft soil the hammer may stop.
The head of a precast concrete pile is protected by a metal drive cap or helmet.
As shown in Fig. 23.4, a cushion that usually consists of blocks or layers of
wood is used to reduce the impact of the hammer. The amount of energy lost
becomes apparent when the wood block smolders or catches on fire and must be
periodically replaced.
Figure 23.4
(a) Driver setup
showing ram,
wood
cushion, helmet
and pile.
(b) Single-acting
steam,
(c) double-acting,
(d) diesel, and
(e) vibratory
hammers. (After
Vesic, 1977.)
The most common driving formula also is one of the oldest, the Engineering
News formula that was suggested in 1888 by A. M. Wellington, the editor of that
magazine before it became Engineering News Record. The formula is based
on the principle of conservation of energy—energy in equals energy out—but
it also includes a generous and arbitrary energy loss factor. The formula is as
follows:
1 WH
Qp ¼ ð23:1Þ
6 S
where Qp ¼ allowable bearing capacity (tons);
W ¼ hammer weight (tons);
H ¼ height of fall (ft);
S ¼ penetration with one hammer blow (ft, inches in final formulas);
WH ¼ pile-driving energy per blow (ft-tons);
1/6 ¼ energy loss factor.
815WSI HSI
Qp ¼ for drop hammers ð23:2aÞ
SSI þ 25
815WSI HSI
Qp ¼ for single-acting steam hammers ð23:3aÞ
SSI þ 2:5
For double-acting hammers where power is added during the hammer stroke,
energy from the manufacturers’ specifications is substituted for WH. The
minimum hammer blow for piles that will carry 25 tons (220 kN) is 15,000 ft-lb
(20 kilojoules or kilonewton-meters).
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Deep Foundations
Figure 23.5
Engineering News
and Michigan
formulas applied
to results of pile
load tests from
several sources
(Spangler and
Mumma, 1958).
The Engineering News formula was developed when practically all piles were
of wood and were driven by drop hammers that now are considered light in
weight. The formula ignores pile length, weight, cross-section, taper, material,
and soil response, which can change with time. A graph comparing the formula
with over 100 load tests is shown in Fig. 23.5, where it will be seen that the
energy loss factor of 1/6 is valid to the extent that none of the piles had over
6 times the predicted capacity. The test piles included a wide variety of types,
lengths, sizes, soil conditions, and geographic locations. Included are 69 load
tests by the Michigan State Highway Commission at three sites, one having a
hard cohesive soil, another a soft cohesive soil, and the third a deep granular
deposit with interbedded organic materials. Pile types included H-section,
pipe, flute-tapered monotube, and step-tapered shell.
Despite obvious spread of the data in Fig. 23.5, according to these data the
probability of having a factor of safety less than 1.0 with the Engineering News
formula is less than 1 in 20, and will be further diminished for piles acting in
a group. The average factor of safety is about 2. Since the energy loss factor
nominally is 6, on the average about two-thirds of the driving energy is lost.
However, the wide range in factors of safety leaves room for improvement.
Example 23.1
The final penetration resistance for a pile driven with a 30,000 ft-lb (42 kJ or 42 kN-m) drop
hammer is 0.6 in./blow (15 mm/blow). Predict the bearing capacity according to the
Engineering News formula.
Answer:
2WH 2 15 ft-tons
Qp ¼ ¼ ¼ 19 tons, or
Sþ1 0:6 þ 1 ðin:Þ
815 42
¼ ¼ 860 kN
15 þ 25
Note that there actually is only one significant figure.
The coefficient or restitution is defined as the square root of the ratio of energy
out to energy in for the driving cushion, and therefore is dimensionless. The value
of e was established empirically and varies from 0.25 for timber pile and for a
concrete or steel pile with a soft wood cushion, to about 0.55 for a steel pile with
no cushion. A value of 0.5 is most commonly used to represent a common practice
of driving steel or concrete piles with an oak hardwood cushion. A special Micarta
cushion has e ¼ 0.8.
The Michigan formula gives a correction factor that is applied to the Engineering
News formula, and therefore still incorporates the driving energy loss factor of 6.
The correction factor is
Wr þ e2 Wp
Em ¼ 1:25 ð23:4Þ
Wr þ Wp
where Em ¼ a multiplier for the result from the EN formula and is dimensionless;
Wr ¼ weight of the hammer ram;
Wp ¼ weight of the pile;
e ¼ coefficient of restitution.
Results from the Michigan pile study in Fig. 23.5 show considerable improvement
over the EN formula by removing evaluations having a low factor of safety, so
this modification is recommended.
Example 23.2
The pile of the previous example is a steel pipe pile 40 ft (12 m) long, 12 in. (0.30 m) in
diameter, weighing 17.86 lb/ft (260 N/m). It is driven by a 10,000 lb (44.5 kN) single-acting
steam-driven ram operating on an oak cushion with e ¼ 0.5. Calculate Em.
Answer: The weight of the pile is Wp ¼ 40 ft 17.86 lb/ft ¼ 714 lb, or 12 m 260 N/
m ¼ 310 N.
Em ¼ 1:25½ð10,000 þ ð0:5Þ2 714=ð10,000 þ 714Þ ¼ 1:2, or
¼ 1:25½ð44,500 þ ð0:5Þ2 310=ð44,500 þ 310Þ ¼ 1:2
The wave equation treats the pile as a rod that is set into vibration by being hit on
the end with a hammer. The behavior can be demonstrated by suspending a pile
horizontally on cables, when instrumentation shows that a compression wave
travels down the pile from one end to the other; then the far end snaps back and
generates another compression wave that goes in the opposite direction. Tension
may develop instantaneously as two compression waves traveling in opposite
directions pass one another. The method received a considerable boost when some
long concrete bridge piles broke in load tests, and when pulled were found to have
broken in tension—an obvious conundrum because a hammer can push but not
pull.
Smith’s model is shown in Fig. 23.6. A pile is divided into an arbitrary series of
segments each 8 to 10 ft (2.5 to 3 m) long and having a mass W. Soil resistance to
driving is represented by individual R values along the sides or at the tip. Each
segment is connected to adjoining segments by an elastic connector represented
by a spring, K.
A ratio of tip to side resistance is assumed, and data for hammer energy and
elastic constants and mass of the pile are entered. Equations relate compression
and force in each connector spring to displacement, velocity, and accelerating
force of each segment. The equations are solved at discrete millisecond time
intervals starting when the hammer hits. This generates a record of forces and
displacements in time throughout the length of the pile, as shown in Fig. 23.7.
Time zero is when the ram hits the pile, and intercepts along the x time-zero axis
show the compression wave moving down the pile to the tip section D12. After
15 milliseconds the pile tip has moved downward 0.2 in., whereas the top of the
pile has moved down 0.86 in. and is bouncing back upward. There is a temporary
disconnect between the pile cap D2 and the upper end of the pile D3 at about
12.5 ms.
The pile-bearing capacity is obtained by summing all of the R’s that represent
both side friction and point bearing. The analysis is repeated with different
assumed R values to generate a graph relating pile capacity to blows per foot
(0.3 m), as shown in Fig. 23.8.
The wave equation addresses driving variables but it still does not take into
account time-related changes in the soil such as drainage of excess pore water
pressure and thixotropic hardening. These effects become obvious when driving is
stopped for a few hours and the pile has developed ‘‘set’’ or ‘‘freeze.’’ Re-driving a
test pile after a day or two then gives a new penetration resistance that is divided
Figure 23.6
Smith’s wave
equation model for
driven piles.
by the earlier resistance to give a ‘‘setup factor,’’ B/A in Fig. 23.8. The setup factor
varies from 1 for sands to 2 or more for clays. The setup factor reduces the
number of blows per foot required for a pile to reach a desired bearing capacity.
Because of the many equations and data entries required, the wave equation is
most conveniently solved with the aid of a computer, and programs are available.
Figure 23.7
Example of wave
equation analysis.
(From Smith,
1960, with
permission of the
American Society
of Civil
Engineers.)
depth of the pier with a square stem that slides up and down through a power
sleeve called a ‘‘kelly.’’ Holes may be bored dry, or to prevent caving they can be
kept full of mud. In clay soils this often is made on-site by mixing soil and water.
In this case a casing can be set to restrain the soil while the mud is being removed
by bailing. (This probably is how the name became ‘‘caisson.’’) Another
procedure is to leave the mud in the hole, lower the steel reinforcing cage, and
pour concrete through a canvas chute or tremie that extends to the bottom of
the hole. The fluid concrete because of its higher density then replaces the mud
from the bottom up.
Figure 23.8
Evaluating and
applying a setup
factor, B/A, to a
wave equation
analysis.
Figure 23.9
Expandable cutter
used for belling out
the bottom of
drilled piers to
increase the
end-bearing area.
(Photo courtesy of
Dr. Lyman C.
Reese.)
Augercast piles usually are smaller in diameter than drilled piers and more
expensive than driven piles, but are adapted for caving soils with difficult
groundwater conditions. Augercast piles do not require casing, and installation
does not generate the noise and tremors of pile driving. The pumping pressure
is monitored to ensure positive pressure as the auger is withdrawn, leaving a
continuous column of grout. A cardboard or plastic cylinder is added at the top
and filled with grout to the desired elevation, and reinforcing steel can then be
lowered into the fluid grout.
The grout used in augercast piles consists of Portland cement, sand, fly ash, and
water. Fly ash is a byproduct from burning powdered coal in electric power
plants; it is a pozzolan, meaning that it reacts with lime liberated by the hydrating
cement to create additional cementitious compounds. The fine spherical particles
Figure 23.10
Continuous flight
auger and guide
for making
augercast piles.
After boring to full
depth, hoses carry
liquid grout to the
top of the
hollow-stemmed
auger for pumping
to the bottom of
the hole as the
auger is lifted and
confines the grout.
The horizontal bar
near the top is a
torque arm.
of fly ash reduce friction during pumping. Fluidizing agents and an expanding
agent such as aluminum powder, which reacts with alkali to give off hydrogen
bubbles, also may be added to help aid pumping and maintain positive pressure
through expansion until the grout sets.
Side friction depends in part on normal stress exerted by soil on a pile or pier,
which for the most part has seldom been measured except indirectly through pull
tests. The coefficient of friction can be compromised by several factors including
remolding.
Figure 23.11
Models for side
friction and end
bearing of piles
and piers. Other
things being
equal, side
support increases
as the length and
the diameter, and
end bearing as
the square of the
diameter.
Side friction has been shown to be increased if boring surfaces are rough or
grooved.
Temporary high pore water pressures have been measured in soil adjacent to
driven piles, and excess pore pressure also are likely to occur during ramming
of the bottom bulb of pedestal piles.
Temporary liquefaction was discovered in soil near Rammed Aggregate Piers,
described in the next chapter, and probably occurs near the bulb of pedestal
piles. Liquefaction also may occur in weak, saturated soil near tapered piles
and would aid driving, but those occurrences have not been verified.
Radial tension cracks have been conjectured to occur near driven piles to
explain rapid dissipation of pore water pressure, and have been confirmed in
the elastic zone near Rammed Aggregate Piers. Radial cracking may be an
important phenomenon for facilitating drainage of excess pore pressure near
displacement piles and piers.
Lateral stress directly influences side friction and to some extent end bearing of
deep foundations, but for the most part has not been measured except
indirectly from pullout resistance. Displacement from pile driving should
increase lateral stress, and lateral stress temporarily is decreased by borings and
reinstated by fluid pressure of concrete. The contact stress then can change as
load is applied. Lateral stress is the least known of the variables affecting pile
capacity.
Another complicating factor in Fig. 23.12 is a nearly complete loss of side friction
near the bottom of the pier. This can be expected if, as the base of the pier settles,
it carries a surrounding bulb of soil along with it.
Based on these and similar results, Reese and O’Neil recommend calculating side
friction of a pier in clay using an ‘‘alpha factor,’’ which represents a portion of the
undrained shear strength of the soil acting along a shaft length that excludes the
upper 5 ft (1.5 m) and the lower 5 ft (1.5 m) plus the length of a bell. The factor
experimentally was determined to have an average value of 0.55. Hence
Qs ¼ qs Ls D ð23:7Þ
where qs is the developed cohesion, Ls is the shaft length after deduction for end
effects, and D is the shaft diameter.
Figure 23.12
Development of
skin friction during
loading of a drilled
test pier in
expansive Houston
clay. (After Reese
et al., 1976, with
permission of the
American Society
of Civil Engineers.)
Example 23.3
Calculate the side friction for a 2.5 ft (0.76 m) diameter drilled pier 15 ft (4.6 m) long
penetrating clay with c ¼ 5 ton/ft2 (480 kPa). There is no bell.
Answer:
qs ¼ 0:55 5 ton=ft2 ¼ 2:75 ton=ft2
Qs ¼ 2:75 ton=ft2 ½ð15 5 5Þ 2:5 ft2 ¼ 108 tons, or
qs ¼ 5:5 480 kPa ¼ 2640 kPa or 2640 kN=m2
Qs ¼ 2640 kN=m2 ½ð4:6 1:5 1:5Þ 0:76 m2 ¼ 10:1 MN
Example 23.4
Calculate allowable total net bearing capacity of the 2.5 ft (0.75 m) diameter, 15 ft (4.6 m)
long pier of the previous example on the same overconsolidated clay.
Answer:
Qbnet ¼ ð7Þð5 tons=ft2 Þ ½ð1:25Þ2 ft2 ¼ 17 tons: Adding side resistance gives
Qnet ¼ 108 þ 7 ¼ 115 tons: With a factor of safety of 2,
Qa ¼ 52 tons, of which most is side friction, or
Qbnet ¼ ð7Þð480 kPaÞ ½ð0:38Þ2 m2 ¼ 0:49 MN
Qnet ¼ 10:1 þ 0:5 ¼ 10:6 MN; Qa ¼ 5:3 MN:
Example 23.5
Recalculate the previous example using eq. (23.10).
Answer:
L=B ¼ 15=2:5 ¼ 6:0ð4:6=0:75 ¼ 6Þ
qbnet ¼ 6½1 þ 0:2ð6Þ 5 ton=ft2 ¼ 66 tons=ft2 ; use 40 tons=ft2
Qbnet ¼ 40ð1:25 ftÞ2 ¼ 196 tons, compared with 170 tons
from the previous example:
Qnet ¼ 196 þ 67 ¼ 263 tons: With a factor of safety of 2,
Qa ¼ 130 tons instead of 118 tons, or
qbnet ¼ 6½1 þ 0:2ð6Þ 480 kPa ¼ 6:3 MPa; use 3:8 MPa:
Example 23.6
Determine the settlement correction factor to bearing capacity in the preceding
example.
Answer:
1 ¼ f0:0071 þ 0:0021ð15=2:5Þg ¼ 0:020; use 0:015
p 2
2 ¼ 1:125 10 kips=ft ¼ 3:56; use 0:15
Fr ¼ 2:5=½0:015 ð2:5Þ þ 0:15 ¼ 13:3 > 1:0; use 1:0, no reduction
where qsi is side friction per unit area of a section of pier passing through a soil
layer and zi0 is the vertical effective stress in soil at the middle of the layer. Beta is
evaluated from
p
¼ 1:5 0:35 zi , and 0:25 1:20 ð23:15Þ
Example 23.7
Calculate the maximum side friction on a 20 ft (6.1 m) long, 2.5 ft (0.76 m) diameter
straight pier in sand having a unit weight of 120 lb/ft3 (18.9 kN/m3). The ground-
water table is at 10 ft (3.05 m) depth.
Answer: Arbitrarily divide the sand into two layers, zi ¼ 5 ft and 15 ft (1.5 and 4.6 m).
Respective values are 0.72 and 0.25. Respective qsi values are
where 0 is the effective soil unit weight, B is the pile diameter, and L the length of
the pile.
The bearing capacity factors depend on the friction angle of the sand, which
because of the difficulty of sampling for laboratory testing generally is evaluated
from in-situ tests. The most satisfactory option would be direct measurement, but
most common are penetration tests that are routinely conducted during site
evaluations.
Standard Penetration Test (SPT) results are in blows per foot of penetration of the
special sampler and are designated by N. As discussed in Chapter 26, several
correction factors have been proposed depending on the testing depth and on the
driving energy. The following formula was developed using uncorrected N values:
qo ¼ 0:6N 45 tons=ft2 , or
ð23:17Þ
qo ¼ 0:06 N 4:3 MN=m2
Example 23.8
Evaluate end-bearing and allowable bearing capacity for the 2.5 ft (0.76 m) diameter drilled
pier in the previous example if Nav ¼ 26 blows/ft at the bearing depth.
Answer:
q0 ¼ 0:6ð26Þ ¼ 15:6 tons=ft2
Qo ¼ ð1:25Þ2 ð15:6Þ ¼ 77 tons: Adding side friction,
Q ¼ 30 þ 77 ¼ 107 tons: Dividing by a factor of safety of 2 gives
Qa ¼ 53 tons, or
qo ¼ 0:06ð26Þ ¼ 1:6 MN=m2
Qo ¼ ð0:38Þ2 ð1:6Þ ¼ 730 kN
Q ¼ 264 þ 730 ¼ 994 MN
Qa ¼ 500 MN
Question: With the allowable bearing pressure, what percentages of the side and
base resistance theoretically will be mobilized?
Answer: 100% and about 30%. However, side friction may decrease when the pier
is loaded, which will increase the base percentage.
Critical parameters for side friction are the normal effective stress and the
developed friction angle. The normal stress following a concrete pour equals that
imposed by the fluid concrete, but it sometimes is assumed that in time soil pressure
will be relieved to its initial K0 condition, which is difficult to evaluate. Pier loading
can induce an increase in lateral stress because of rotation of the principal stress
directions, and dilation or compression of the soil. Finally, there is the developed
friction angle between the soil and the pier that because of the movement involved
will represent a residual strength without a dilatant component. A typical
assumption is that the angle of side friction is defined by tan ¼ 0.8 tan .
Example 23.9
Evaluate side friction using a soil mechanics approach for the pier in the preceding
example, the groundwater table being at a depth of 10 ft (3.05 m).
Answer: According to a criterion presented later in this book, an uncorrected N ¼ 26
corresponds to a friction angle in sand of about 358. An estimate of the developed side
friction is tan ¼ 0.8 tan , or ¼ 298.
Option 1: For a maximum value use K0 for fluid concrete of 1.0 and a unit weight of
150 lb/ft3. The average normal stresses at mid-depths in the layers above and below the
groundwater table and corresponding side friction are:
Above gwt: h ¼ 1:0ð5Þð150Þ ¼ 750lb=ft2 ; Ss ¼ 750ð2:5Þð10Þ tan 298 ¼ 32,600 lb
Below gwt: h ¼ 1:0½10ð150Þ þ 5ð150 62:4Þ
¼ 1940 lb=ft2 ; Ss ¼ 1940ð2:5Þð10Þ tan 29
¼ 84,400 lb
Total ¼ 58 tons
Option 2: For a minimum value use the soil normally consolidated Ko ¼ 1 – sin and the
unit weight of the soil.
Above gwt: h ¼ ð1 sin 35 Þð5Þð120Þ ¼ 256 lb=ft2 ; Ss ¼ 256ð2:5Þð10Þ tan 29
¼ 11,100 lb
The two options therefore bracket the answer obtained by the empirical method. The same
procedures can be followed using SI units. Obviously for a soil mechanics approach to be
valid the friction angle and lateral stress must be accurately evaluated.
For piers having a large base-bearing capacity four anchors may be required to
generate a sufficient side friction, in which case the reaction beams are assembled
as a letter H. Loads ordinarily are increased incrementally until they reach two
times the design capacity, to confirm a factor of safety of 2.
As test loads are applied, settlement is measured with dial gauges or transducers
on opposite sides of the pile or pier so that averages may be used to compensate
for bending. The gauges are supported on small independently supported beams
Figure 23.13
Load test of a
drilled pier using
hydraulic jacks
and two
independently
supported dial
gauges.
Sunshade is to
prevent thermal
warping from
affecting the
readings.
that are transverse to the pile row to avoid influences from ground heave. The
arrangement is shaded to reduce error from thermal expansion and warping of
gauge supports.
In the cyclic method, a load increment is applied, and the settlement is measured.
The load is released and the pile is permitted to rebound. Repeating this process
for several increments will permit data to be obtained for plotting a net or plastic-
settlement curve. The distinction between the two types of tests can be seen by
comparing the initial load curve starting at zero in Fig. 23.14 with the reloading
curve.
Figure 23.14
Pile load test and
failure criteria
based on
settlement, net
settlement, and
settlement per unit
of load.
The maximum test load seldom is sufficient to generate plunge, so many dif-
ferent criteria have been proposed to define acceptability without plunging
failure. No single criterion will be found that can apply to all load tests, and
a common procedure is to define failure on the basis of several criteria and
select the most relevant for a particular situation. For example, if settlement
cannot exceed a certain value, that settlement value becomes the criterion
for acceptability.
Some common criteria are as follows. In this connection ‘‘gross settlement’’ equals
total settlement, and ‘‘net settlement’’ is the gross settlement corrected for
rebound during unloading.
As indicated above, gross settlement may be limited to a specific value, for
example 0.5 or 1.0 in. (12 to 25 mm). This is the simplest criterion because it is
read directly from the load-settlement curve, as point H in Fig. 23.14, and is
based on the amount of settlement that can be tolerated by a structure instead
of depending on shifting pile-soil interactions.
Gross settlement can be defined as a function of rate, for example 0.01, 0.02, or
0.05 in./ton (0.03, 0.06, or 0.15 mm/kN). These criteria are shown by the slopes
of lines OA, OA0 , and OA00 in Fig. 23.14. A failure load is found by shifting
the selected line until it is tangent with the load-settlement curve. A factor of
safety of 2 then is applied. The steepest line, OA00 , probably defines a realistic
failure load for this pile. A limitation of this criterion is that the slope may
not be attained in the load test because of limits imposed by the jacks and
anchors.
The net settlement rate may be limited to an arbitrary value as in paragraph 1,
and a factor of safety of 2 applied. This in effect rotates the allowable
slope base line by an angle in Fig. 23.14 so that it is parallel to the rebound
curve, giving slopes OB, OB0 , and OB00 in the figure. The line then is shifted until
it is tangent to the load curve as in the preceding criterion. Net settle-
ment eliminates most of the elastic compression that depends on pile length
and material, but it should be recognized that some compression remains
after full removal of load because of residual side forces that act upward on the
lower part of the pile and downward on the upper part as the pile rebounds
elastically.
Net settlement may be limited to a specific value, for example 0.25 or 0.5 in.
(6 or 12 mm), in which case the failure load is obtained by parallel shifting of
the rebound curve so that it intersects the desired settlement value, and reading
where this line intersects the load curve, for example G in Fig. 23.14. A factor
of safety of 2 usually is applied.
Other criteria such as defining a ‘‘break in the curve,’’ or using a point defined by
tangents drawn on either side of a break, are more arbitrary and depend on the
scale of the plot and the maximum applied load. Another criterion is to define an
allowable settlement for a given load increment, in which case that load increment
must be defined.
The safe load for a pile or pier is one in which settlement is not excessive for the
intended use. A factor of safety is necessary to cover variable soil and pile
conditions and to a lesser extent unanticipated loading conditions. Since most
pile-soil systems gain strength with time, the eventual mean factor of safety
ordinarily will increase over that defined by the load test.
Figure 23.15
Schematic of the
Osterberg cell for
pushing upward on
a pile or pier.
to provide a base for strain measurements. The O-cell is particularly useful for
large-capacity piers that would require a massive reaction beam and jacking
capability. A special version of the device can be attached to the ends of driven
piles. However, the O-cell is not recovered after a test, which adds to the cost of
the test.
As can be seen from the data in Fig. 23.16, the maximum load depends on which
yields first, side friction or end bearing. In most cases side friction is the first to
peak out, and end bearing is estimated by extrapolation. The data in Fig. 23.16
indicate a failure load of about 7400 plus about 8000 tons (65 plus 55 MN). This is
one of the largest capacity piers ever tested.
With the O-cell side friction and bottom bearing are developed simultaneously,
whereas when a pile or pier is put into service with top-down loading, side friction
initially will carry all of the load prior to activation of bottom bearing. This
introduces a depth variable in the test, as side friction generally is higher at depth
because of higher normal stresses. If and when side friction is fully mobilized in
the test, indicated by side shear movement equaling and exceeding bottom
movement, it should make no difference whether shearing is acting upward or
downward. The predicted settlement then may equal bottom movement plus
elastic compression of the pile or pier, taking into account the cumulative nature
of side shear with depth.
Figure 23.16
O-cell data for a
9 ft diameter pier
50 ft deep at
Apalachicola
River, Florida.
(after Osterberg,
2004).
The high cost of pulling and replacing weak or broken piles or piers can result in
project abandonment, or in shifting to a slightly different footprint that will allow
installation of all new foundation elements. Far better is to get things right the
first time.
Side Friction
Side friction generally is not included in design of end bearing on rock, but is a
consideration for development of negative skin friction.
Rock Competence
The competence of rock commonly is estimated from the percentage of com-
petent core recovered during exploration drilling. This is the ‘‘rock quality
designation’’ or RQD. The modulus of a rock with weathered seams is mainly
a function of the thickness and spacing of the seams, which contain material
that may not be normally consolidated because of support from arching action.
The modulus therefore decreases rapidly and somewhat erratically depending
on the RQD, a 30 percent decrease in RQD causing about an 80 percent decrease
in modulus. Clues can be obtained from cross-hole seismic testing, and a modulus
can be more accurately evaluated with rock pressuremeter tests (Failmezger et al.
(2005)).
Experience in local areas often leads to bearing capacity criteria for particular
rocks and may become the basis for local building code requirements. Obviously,
caution must be used in applying such criteria to other areas and in particular to
other rock types, as even a rock name such as ‘‘shale’’ or ‘‘limestone’’ can cover a
wide range of strength properties.
The bearing capacity of a pile group therefore includes a factor for efficiency:
Qg ¼ EnQ ð23:20Þ
where Qg ¼ bearing capacity of the pile group;
E ¼ efficiency;
n ¼ number of piles in the group;
Q ¼ bearing capacity of individual pile.
Figure 23.17
Derivation and
example of the
Feld rule.
Substituting S ¼ 2.5D,
tanð=2Þ ¼ ð0:5DÞ=2:5D ¼ 0:20
a ¼ 22:6
22:6=360 ¼ 0:063 ¼ 1=16
Example 23.10
Calculate efficiency of a 9-pile group in Fig. 23.17 using the Feld rule.
Answer: There are 5 piles, and summing the reductions gives 16/16. This is the equivalent
of 5 – 1 ¼ 4 stand-alone piles, for an efficiency of 80%.
Question: Would eliminating the center pile increase the efficiency? Would it reduce the
load-carrying capacity?
Model tests summarized by Poulos and Davis (1980) indicate that, for a spacing of
2.5D in various size groups the efficiency varied from 0.7 to 0.9, averaging about
0.8. With larger spacings the efficiency is slightly higher. Pile group reduction
factors are appropriate for friction piles in clay but not for end-bearing or for
driven piles in sand, where the efficiency is 1.0 and may exceed 1.0 because of
densification of the sand.
Figure 23.18
Bearing capacity
of a pile group.
where Qg is the net bearing capacity of the group, c is the undrained soil cohesion,
B and A are the plan dimensions of the group, and L is the pile length. Every large
pile group should be checked for this failure mode, particularly in soft soil with
short pile lengths.
Example 23.11
Find the maximum allowable net bearing capacity of a pile group with L ¼ 20 ft (6.1 m),
D ¼ 1 ft (0.3 m), A ¼ 30 ft (9.1 m), and B ¼ 20 ft (6.1 m), and compare with the bearing
capacity using the Feld rule. The soil is clay, c ¼ 500 lb/ft2 (23.9 kPa), and sub ¼ 60 lb/ft3
(9.4 kN/m3).
Single pile:
Qs ¼ cLðB=2Þ2 ¼ 500ð20Þð3:14Þð0:5Þ2 ¼ 7850 lb
Qbnet ¼ 7cB2 ¼ 7ð500Þð1Þ2 ¼ 3500 lb
Q ¼ 7850 þ 3500 ¼ 11,350 lb ¼ 5:68 tons ð50:5 kNÞ
Feld rule: Assume a spacing of 2.5D, or 2.5 ft center to center. The group will be
(30/2.5 – 1) ¼ 11 piles in the A direction and (20/2.5 – 1) ¼ 7 piles in the B direction, or
77 piles total. This will include:
4 corner piles each having an efficiency of 13/16, giving 4 13/16 ¼ 3.25 pile equivalencies.
2(7 – 2) þ 2(11 – 2) ¼ 45 edge piles, each with 9/16, which gives 25.3 equivalencies.
77 – 45 – 4 ¼ 29 center piles, each with 8/16, which gives 14.5 equivalencies for a total of
3.25 þ 25.3 þ 14.5 ¼ 43, or an efficiency of 43/77 ¼ 56%.
Q ¼ 0.56(77)(45) ¼ 1940 tons (17.3 MN) or approximately the same as the group bearing
capacity.
23.9.1 Overview
The larger the base area, the deeper the pressure bulb and the more soil will be
involved in consolidation settlement. A procedure suggested by Terzaghi and Peck
(1967) is widely used and is somewhat similar to that used for bearing capacity of
a pile group but with an important difference: the center of action is assumed
to be at two-thirds of the depth of the pile (Fig. 23.19). In other words, for
settlement calculations it is as if the pile group is replaced by a shallow foundation
having the same size and extending to a depth equal to two-thirds of the lengths
of the piles.
Fellenius (1991) suggested a rationale for this procedure, that settlement of soil
adjacent to the pile creates negative skin friction on the upper part of the pile.
As this is opposed by positive skin friction on the lower part, there exists a neutral
plane where there is no shearing stress and no shearing movement between the soil
and the pile. The load from the pile group therefore can be conceptually replaced
with a foundation at the elevation of the neutral plane. For friction piles the
neutral plane is at about the one-third point, and for end-bearing piles it is at
the bottom of the piles. The position of the neutral plane dictates the level of the
equivalent foundation.
Figure 23.19
Terzaghi-Peck equivalent foundation method for predicting settlement of a pile group. Negative
skin friction can be caused by surface loading of the soil and/or lowering of the groundwater
table.
Example 23.12
A 1000 ton (8.9 MN) surface load is supported by a 24 ft pile group 20 40 ft
(6.1 12.2 m), on soil having a cohesion of 400 lb/ft2 (19.2 kPa). Estimate the vertical
stress at a depth of 8 ft below the bottom of the piles.
Answer: A depth of 8 ft below the pile tips is 8 þ (24/3) ¼ 16 ft (4.9 m) below the neutral
plane. With a 2:1 slope this extends the base dimension 16/2 ¼ 8 ft in four directions, giving
an expanded area of (20 þ 16)(40 þ 16) ¼ 2016 ft2 (57.1 m2). The pressure at this level is
1000/2016 ¼ 0.5 ton/ft2 ¼ 1000 lb/ft2 (48 kPa).
Side shear can be reduced by using protective sheaths that can slip on the pile, or
coating the pile with soft mastic layers. Foundation elements of course must be
reinforced to withstand tensile forces.
Where structures are supported on piles, floors that normally would be in contact
with the soil also must be supported or they can be lifted by clay expansion. This
is called a ‘‘structural floor,’’ and is supported on beams connected to the pile.
It also is critical that a sufficient clear space be left under the floor to allow for
clay expansion.
A similar problem can occur as a result of freezing of soil to the surface of a pile or
pier, followed by frost heave of the soil. This mainly occurs in frost-susceptible
silty soils with access to water. In permafrost areas, foundations can be extended
into the permafrost where they may be presumed to remain permanently frozen,
subject to climatic influences.
Lateral forces can be provided for in design by combining vertical and nonvertical
or ‘‘battered’’ piles in the same pile group, the latter to act as braces. In some
applications, such as for bridges, lateral forces can be expected to come from one
direction, whereas in other situations, such as an earthquake, they may come from
any direction. Design with batter piles is simply a matter of drawing force
diagrams with forces parallel to the pile directions, or of calculating the horizontal
component of batter pile resistance and comparing to anticipated lateral loads.
If appropriately pinned and reinforced, batter pile may act in tension as well as in
compression.
Example 23.13
A bridge pier will exert a vertical force of 440 tons (3.9 MN). Select an array
of 50 ton (443 kN) piles with 70% efficiency due to group action, to support this load
plus a maximum anticipated lateral load of 44 tons (390 kN) caused by current, ice,
and debris.
Answer: Each pile capacity is 0.7 50 ¼ 35 tons (310 kN). The horizontal capacity of this
pile driven at 208 batter is 35 sin 208 ¼ 12 tons (106 kN), To meet the required 44 tons, four
piles in the group must be battered at that angle. The next step is to determine the vertical
capacity of the four battered piles, which is 4 35 cos 208 ¼ 131 tons (1165 kN). The required
number of vertical pile therefore is (440 – 121)/35 ¼ 8.8, so the pile group will have 9 vertical
piles plus 4 battered at 208.
Rutledge based his procedure on lateral pull-tests using a 1.5-inch diameter steel
auger, and significantly, signposts that have endured a hurricane generally are
bent instead of being ripped out of the ground. The assumed stress distribution is
shown at the right in Fig. 23.20, which also includes Rutledge’s nomograph that
can be used for a solution. A factor of safety is incorporated into the soil strength
data, and a deeper embedment is required in sands because of the dependence of
frictional strength on overburden pressure. It is recommended that SI units be
converted to ft-lb-sec for use of this method.
Example 23.14
Determine the embedment depth in soil having an unconfined compressive strength
of 2000 lb/ft2 for a 20 ft long, 1.5 ft diameter pole supporting a sign having an area of 20 ft2,
to resist a wind of 100 mph.
Figure 23.20
Nomograph for
calculating
required
embedment depth
for pole
structures.
(Redrawn from
Rutledge, 1947.)
Figure 23.21
Wind pressure as
a function of
velocity. Arrows
show pressures
from 199 mph
(320 km/hr) wind.
(Drawn from data
of Farr, 1980.)
Answer: With a factor of safety of 2 the allowable horizontal stress is 1000 lb/ft2, which is
entered on the left scale. The wind force is 1.5 20 16 þ 20 25 ¼ 980 lb, which is entered
on the P scale. A line through these two points reads 0.97 on the C scale. Then draw a line
through 0.97 and 18 in. on the b scale and read 0.68 on the L scale. A horizontal line from
0.68 intersects the H ¼ 10 ft mid-height to give 5 ft embedment.
Problems
23.1. What are friction piles? End-bearing piles? Compaction piles? Describe the
soil requirements for each. What distinguishes between a pile, a pier, and a
caisson?
23.2. What are the principal materials in deep foundation piles?
23.3. What are batter or spur piles and why are they used?
23.9. A pile is driven by a drop hammer weighing 17.8 kM (4000 lb). The height
of fall is 3 m (10 ft) and the average penetration of the last few blows is
1.27 mm (0.05 in.). What is the bearing capacity of the piles according to
the Engineering News formula?
23.10. The pile of Problem 23.15 is concrete, 254 254 mm (10 10 in.) 9.1 m
(30 ft) long. The driving cushion is oak. What is the bearing capacity by
the Michigan formula?
23.11. A pile is driven by a single-acting steam hammer which weighs 8.9 kM
(2000 lb). The height of fall is 1.2 m (4 ft) and the average penetration
under the last few blows is 8.4 mm (0.33 in). What is the bearing capacity
of the pile by the Engineering News formula?
23.12. Explain how driving forces can induce tension in a pile.
23.13. A second test pile is driven under the same conditions as in Fig. 23.8 until
the driving resistance is 50 blows/0.3 m (ft). Two weeks later the pile is
re-driven with the same hammer, and 16 blows were found to be necessary
to advance the pile 25 m (1 in.). What is the setup factor? How is this
used in design?
23.14. As a professional engineer you review a foundation design with piers
appropriately belled out to rest on a top of a layer of stiff clay, and above
the clay is a saturated sand. Is this design acceptable? Why (not)? What is
meant by ‘‘buildable?’’
23.15. An 0.9 m (3 ft) diameter straight bored pier is founded at a depth of 10.4 m
(34 ft) in clay which has an average unconfined compressive strength of
47.9 kPa (1000 lb/ft2). Calculate end bearing, side friction, and the allow-
able bearing capacity with a factor of safety of 2.0.
23.16. Check the previous answer for settlement.
23.17. The pier of Problem 23.15 was installed and tested, and settled excessively
at 180% of the design load. Revise the design with a longer pier that will
provide a factor of safety of 2.2 and explain why you require 2.2 instead of
the target 2.0.
23.18. Design a pier that will carry 20 tons in end bearing on sand with an
average blow count of 12 blow/ft and check for settlement. The factor of
safety will be 2.0.
23.19. Make a preliminary estimate of side friction and end bearing of a 15 ft, 1 ft
diameter pile driven in soil with an unconfined compressive strength of
800 lb/ft2. What assumptions are made in your analysis?
23.20. A 10 ft, 0.5 ft diameter pipe pile is driven in sand having an internal
friction angle of 308. Calculate side friction assuming that K ¼ 1.0,
¼ 120 lb/ft3, and contact friction is 80% of the internal friction.
23.21. A group of 24 piles 0.3 m (1 ft) in diameter is driven in four rows of six
piles each. They are spaced 1.1 m (3.5 ft) center to center. If each individual
pile has a bearing capacity of 214 kN (24 tons), determine the bearing
capacity of the group and indicate the criterion.
23.22. If the piles in Problem 23.21 are driven 6.1 m (20 ft) into cohesive soil
having an unconfined compression strength of 19.2 kPa (400 lb/in.2), check
to determine stability of the whole group.
23.23. What area should be used to determine the average vertical stress in soil at
the bottom of the piles in Problems 23.21 and 23.22?
23.24. Calculate vertical soil stress imposed by the pile group in Problems 23.21
and 23.22 at a depth 10 ft below the bottom. How would this value be used
to predict settlement?
23.25. Sketch the arrangement for a pile load test, indicating jacks, reinforced
anchor piles, and dial gauges. What loading sequence should be used for a
pile with a design load of 120 tons?
23.26. What are the failure loads and working loads of the pile in Fig. 23.14
according to the 29 mm/N (0.01 in./ton) gross settlement and net
settlement criteria? Are these reasonable criteria for this test?
23.27. Deep foundation elements are to be installed in a soil with the
groundwater table at a depth of 8 ft. Soil samples show mottled gray
and brown colors to a depth of 18 ft. What water table depth should be
used in design? How will this affect a design?
23.28. Give reasons why the natural groundwater table generally is lowered in
cities.
23.29. Calculate the maximum tensile and net uplift forces on a 0.3 m (12 in.)
diameter end-bearing pile extending through 6.1 m (20 ft) of expansive
clay, one-third of which is below the permanent water table. Assume
c0 ¼ 14 kPa (2 lb/in.2).
23.30. Design a vertical and batter 6-pile system to sustain bidirectional
lateral loads equal to 30% of the vertical load, all piles being equal in
length.
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