module 8 (lecture 33) pile foundations topics -...

NPTEL – ADVANCED FOUNDATION ENGINEERING-I

Module 8

(Lecture 33)

PILE FOUNDATIONS

Topics

1.1 PILE-DRIVING FORMULAS

1.2 NEGATIVE SKIN FRICTION

Clay Fill over Granular Soil

Granular Soil Fill over Clay

1.3 GROUP PILES

1.4 GROUP EFFICIENCY


PILE-DRIVING FORMULAS

To develop the desired load-carrying capacity, a point bearing pile must penetrate the dense soil layer sufficiently or have sufficient contact with a layer of rock. This requirement cannot always be satisfied by driving a pile to a predetermined depth because soil profiles vary. For that reason, several equations have been developed to calculate the ultimate capacity of a pile during driving. These dynamic equations are widely used in the field to determine whether the pile as reached satisfactory bearing value at the predetermined depth. One of the earliest of these dynamic equations-commonly referred to as the Engineering News Record (ENR) formula-is derived from the work-energy theory. That is,

Energy imparted by the hammer per blow = (pile resistance) (penetration per hammer blow)

According to the ENR formula, the pile resistance is the ultimate load 𝑄𝑄𝑢𝑢 , expressed as

𝑄𝑄𝑢𝑢 = 𝑊𝑊𝑅𝑅ℎ𝑆𝑆+𝐶𝐶

[8.118]

Where

𝑊𝑊𝑅𝑅 = 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑤𝑤 𝑟𝑟𝑟𝑟𝑟𝑟 (𝑜𝑜𝑜𝑜𝑟𝑟 𝑤𝑤𝑒𝑒𝑟𝑟𝑟𝑟𝑒𝑒𝑒𝑒𝑤𝑤, 𝑠𝑠𝑤𝑤𝑤𝑤 𝑡𝑡𝑟𝑟𝑡𝑡𝑒𝑒𝑤𝑤 𝐷𝐷. 4𝐴𝐴𝑒𝑒𝑒𝑒𝑤𝑤𝐴𝐴𝐴𝐴𝑤𝑤𝑒𝑒 𝐷𝐷)

ℎ = ℎ𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡 𝑜𝑜𝑜𝑜 𝑜𝑜𝑟𝑟𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑤𝑤 𝑟𝑟𝑟𝑟𝑟𝑟

𝑆𝑆 = 𝑒𝑒𝑤𝑤𝐴𝐴𝑤𝑤𝑡𝑡𝑟𝑟𝑟𝑟𝑡𝑡𝑤𝑤𝑜𝑜𝐴𝐴 𝑜𝑜𝑜𝑜 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤 𝑒𝑒𝑤𝑤𝑟𝑟 ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑤𝑤𝑟𝑟 𝑡𝑡𝑒𝑒𝑜𝑜𝑤𝑤

𝐶𝐶 = 𝑟𝑟 𝑐𝑐𝑜𝑜𝐴𝐴𝑠𝑠𝑡𝑡𝑟𝑟𝐴𝐴𝑡𝑡

The pile penetration, S, is usually based on the average value obtained from the last few driving blows. In the equations’ original form, the following values of C were recommended.

For drop hammers: C = 1 in. (if the units of S and h are in inches)

For steam hammers: C = 0.1 in. (if the units of S and h are in inches)

Also, a factor of safety, 𝐹𝐹𝑆𝑆 = 6, was recommended to estimate the allowable pile capacity. Note that, for single- and double-acting hammers, the term 𝑊𝑊𝑅𝑅ℎ can be replaced


by 𝐸𝐸𝐸𝐸𝐸𝐸 (where 𝐸𝐸 = ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑤𝑤𝑟𝑟 𝑤𝑤𝑜𝑜𝑜𝑜𝑤𝑤𝑐𝑐𝑤𝑤𝑤𝑤𝐴𝐴𝑐𝑐𝑖𝑖 𝑟𝑟𝐴𝐴𝐴𝐴 𝐸𝐸𝐸𝐸 = 𝑟𝑟𝑟𝑟𝑡𝑡𝑤𝑤𝐴𝐴 𝑤𝑤𝐴𝐴𝑤𝑤𝑟𝑟𝑤𝑤𝑖𝑖 𝑜𝑜𝑜𝑜 ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑤𝑤𝑟𝑟). Thus

𝑄𝑄𝑢𝑢 = 𝐸𝐸𝐸𝐸𝐸𝐸𝑆𝑆+𝐶𝐶

[8.119]

The ENR formula has been revised several times over the years, and other pile-driving formulas also have been suggested. Some of them are tabulated in table 1.

The maximum stress developed on a pile during the driving operation can be estimated from the pile-driving formulas presented in table 11. To illustrate, we use the modified ENR formula:

𝑄𝑄𝑢𝑢 = 𝐸𝐸𝑊𝑊𝑅𝑅ℎ𝑆𝑆+𝐶𝐶

𝑊𝑊𝑅𝑅+𝐴𝐴2𝑊𝑊𝑒𝑒

𝑊𝑊𝑅𝑅+𝑊𝑊𝑒𝑒

In this equation, S equals the average penetration per hammer blow, which can also be expressed as

𝑆𝑆 = 1𝑁𝑁

[8.120]

Where

𝑆𝑆 𝑤𝑤𝑠𝑠 𝑤𝑤𝐴𝐴 𝑤𝑤𝐴𝐴𝑐𝑐ℎ𝑤𝑤𝑠𝑠

𝑁𝑁 = 𝐴𝐴𝑢𝑢𝑟𝑟𝑡𝑡𝑤𝑤𝑟𝑟 𝑜𝑜𝑜𝑜 ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑤𝑤𝑟𝑟 𝑡𝑡𝑒𝑒𝑜𝑜𝑤𝑤𝑠𝑠 𝑒𝑒𝑤𝑤𝑟𝑟 𝑤𝑤𝐴𝐴𝑐𝑐ℎ 𝑜𝑜𝑜𝑜 𝑒𝑒𝑤𝑤𝐴𝐴𝑤𝑤𝑡𝑡𝑟𝑟𝑟𝑟𝑡𝑡𝑤𝑤𝑜𝑜𝐴𝐴

Table 11 Pile-Driving formulas

Name Formula

Modified ENR formula 𝑄𝑄𝑢𝑢 =

𝐸𝐸𝑊𝑊𝑅𝑅ℎ𝑆𝑆 + 𝐶𝐶

𝑊𝑊𝑅𝑅 + 𝐴𝐴2𝑊𝑊𝑒𝑒

𝑊𝑊𝑅𝑅 + 𝑊𝑊𝑒𝑒

Where

𝐸𝐸 = ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑤𝑤𝑟𝑟 𝑤𝑤𝑜𝑜𝑜𝑜𝑤𝑤𝑐𝑐𝑤𝑤𝑤𝑤𝐴𝐴𝑐𝑐𝑖𝑖

𝐶𝐶 = 0.1 𝑤𝑤𝐴𝐴. , 𝑤𝑤𝑜𝑜 𝑡𝑡ℎ𝑤𝑤 𝑢𝑢𝐴𝐴𝑡𝑡𝑠𝑠 𝑜𝑜𝑜𝑜 𝑆𝑆 𝑟𝑟𝐴𝐴𝐴𝐴 ℎ 𝑟𝑟𝑟𝑟𝑤𝑤 𝑤𝑤𝐴𝐴 𝑤𝑤𝐴𝐴𝑐𝑐ℎ𝑤𝑤𝑠𝑠

𝑊𝑊𝑒𝑒 = 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑤𝑤 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤

𝐴𝐴 =𝑐𝑐𝑜𝑜𝑤𝑤𝑜𝑜𝑜𝑜𝑤𝑤𝑐𝑐𝑤𝑤𝑤𝑤𝐴𝐴𝑡𝑡 𝑜𝑜𝑜𝑜 𝑟𝑟𝑤𝑤𝑠𝑠𝑡𝑡𝑤𝑤𝑡𝑡𝑢𝑢𝑡𝑡𝑤𝑤𝑜𝑜𝐴𝐴 𝑡𝑡𝑤𝑤𝑡𝑡𝑤𝑤𝑤𝑤𝑤𝑤𝐴𝐴 𝑡𝑡ℎ𝑤𝑤 𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟𝐴𝐴𝐴𝐴 𝑡𝑡ℎ𝑤𝑤 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤 𝑐𝑐𝑟𝑟𝑒𝑒

Typical values for E


Single- and double-acting hammers 0.7-0.85

Diesel hammers 0.8-0.9

Drop hammers 0.7-0.9

Typical values for n

Cast iron hammer and concrete pile (without cap) 0.4-0.5

Wood cushion on steel piles 0.3-0.4

Wooden pile 0.25-0.3

Michigan State Highway Commission formula (1965)

𝑄𝑄𝑢𝑢 =1.25𝐸𝐸𝐸𝐸𝐸𝐸𝑆𝑆 + 𝐶𝐶

𝑊𝑊𝑅𝑅 + 𝐴𝐴2𝑊𝑊𝑒𝑒

𝑊𝑊𝑅𝑅 + 𝑊𝑊𝑒𝑒

Where

𝐸𝐸𝐸𝐸 = 𝑟𝑟𝑟𝑟𝐴𝐴𝑢𝑢𝑜𝑜𝑟𝑟𝑐𝑐𝑡𝑡𝑢𝑢𝑟𝑟𝑤𝑤𝑟𝑟′𝑠𝑠𝑟𝑟𝑟𝑟𝑒𝑒𝑤𝑤𝑟𝑟𝑢𝑢𝑟𝑟 𝑟𝑟𝑟𝑟𝑡𝑡𝑤𝑤𝐴𝐴 ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑤𝑤𝑟𝑟 𝑤𝑤𝐴𝐴𝑤𝑤𝑟𝑟𝑤𝑤𝑖𝑖 (𝑒𝑒𝑡𝑡 −𝑤𝑤𝐴𝐴. )


𝐶𝐶 = 0.1 𝑤𝑤𝐴𝐴.

A factor of safety of 6 is recommended.

Danish formula (Olson and Flaate, 1967)

𝑄𝑄𝑢𝑢 =𝐸𝐸𝐸𝐸𝐸𝐸

𝑆𝑆 + �𝐸𝐸𝐸𝐸𝐸𝐸𝐿𝐿2𝐴𝐴𝑒𝑒𝐸𝐸𝑒𝑒

Where



𝐸𝐸𝐸𝐸 = 𝑟𝑟𝑟𝑟𝑡𝑡𝑤𝑤𝐴𝐴 ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑤𝑤𝑟𝑟 𝑤𝑤𝐴𝐴𝑤𝑤𝑟𝑟𝑤𝑤𝑖𝑖

𝐸𝐸𝑒𝑒 = 𝑟𝑟𝑜𝑜𝐴𝐴𝑢𝑢𝑒𝑒𝑢𝑢𝑠𝑠 𝑜𝑜𝑜𝑜 𝑤𝑤𝑒𝑒𝑟𝑟𝑠𝑠𝑡𝑡𝑤𝑤𝑐𝑐𝑤𝑤𝑡𝑡𝑖𝑖 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑤𝑤 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤 𝑟𝑟𝑟𝑟𝑡𝑡𝑤𝑤𝑟𝑟𝑤𝑤𝑟𝑟𝑒𝑒

𝐿𝐿 = 𝑒𝑒𝑤𝑤𝐴𝐴𝑤𝑤𝑡𝑡ℎ 𝑜𝑜𝑜𝑜 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤

𝐴𝐴𝑒𝑒 = 𝑟𝑟𝑟𝑟𝑤𝑤𝑟𝑟 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑤𝑤 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤 𝑐𝑐𝑟𝑟𝑜𝑜𝑠𝑠𝑠𝑠 𝑠𝑠𝑤𝑤𝑐𝑐𝑡𝑡𝑤𝑤𝑜𝑜𝐴𝐴

Pacific Coast Uniform Building

Code formula (International Conference of Building Officials, 1982)

𝑄𝑄𝑢𝑢 =(𝐸𝐸𝐸𝐸𝐸𝐸) �

𝑊𝑊𝑅𝑅 + 𝐴𝐴𝑊𝑊𝑒𝑒𝑊𝑊𝑅𝑅 + 𝑊𝑊𝑒𝑒

�

𝑆𝑆 + 𝑄𝑄𝑢𝑢𝐿𝐿𝐴𝐴𝐸𝐸𝑒𝑒

The value of n should be 0.25 for steel piles and 0.1 a for all other piles. A factor of safety of 4 is generally recommended.

Janbu’s formula (Janbu, 1953) 𝑄𝑄𝑢𝑢 =

𝐸𝐸𝐸𝐸𝐸𝐸𝐾𝐾′𝑢𝑢𝑆𝑆

Where

𝐾𝐾′𝑢𝑢 = 𝐶𝐶𝐴𝐴 �1 + �1 +𝜆𝜆′𝐶𝐶𝐴𝐴�

𝐶𝐶𝐴𝐴 = 0.75 + 0.14 �𝑊𝑊𝑒𝑒

𝑊𝑊𝑅𝑅�

𝜆𝜆′ = �𝐸𝐸𝐸𝐸𝐸𝐸𝐿𝐿𝐴𝐴𝑒𝑒𝐸𝐸𝑒𝑒𝑆𝑆2�

Gates formula (Gates, 1957)

𝑄𝑄𝑢𝑢 = 𝑟𝑟�𝐸𝐸𝐸𝐸𝐸𝐸(𝑡𝑡 − 𝑒𝑒𝑜𝑜𝑤𝑤 𝑆𝑆)

If 𝑄𝑄𝑢𝑢 is in 𝑘𝑘𝑤𝑤𝑒𝑒𝑠𝑠, then S is 𝑤𝑤𝐴𝐴 𝑤𝑤𝐴𝐴. 𝑟𝑟 = 27, 𝑡𝑡 = 1,𝑟𝑟𝐴𝐴𝐴𝐴 𝐸𝐸𝐸𝐸 is in 𝑘𝑘𝑤𝑤𝑒𝑒 − 𝑜𝑜𝑡𝑡.


If 𝑄𝑄𝑢𝑢 is in 𝑘𝑘𝑁𝑁, then S is in mm, 𝑟𝑟 = 104.5, 𝑡𝑡 = 2.4,𝑟𝑟𝐴𝐴𝐴𝐴 𝐸𝐸𝐸𝐸 is in 𝑘𝑘𝑁𝑁 −𝑟𝑟

𝐸𝐸 = 0.75 for drop hammer; 𝐸𝐸 = 0.85 for all other hammer

Use a factor of safety of 3.

Navy-McKay formula 𝑄𝑄𝑢𝑢 =

𝐸𝐸𝐸𝐸𝐸𝐸

𝑆𝑆 �1 + 0.3𝑊𝑊𝑒𝑒𝑊𝑊𝑅𝑅

�

Use a factor of safety of 6.

Thus

𝑄𝑄𝑢𝑢 = 𝐸𝐸𝑊𝑊𝑅𝑅ℎ(1/𝑁𝑁)+0.1

𝑊𝑊𝑅𝑅+𝐴𝐴2𝑊𝑊𝑒𝑒

𝑊𝑊𝑅𝑅+𝑊𝑊𝑒𝑒 [8.121]

Different values of N may be assumed for a given hammer and pile and 𝑄𝑄𝑢𝑢 calculated. The driving stress can then be calculated for each value of N and 𝑄𝑄𝑢𝑢/𝐴𝐴𝑒𝑒 .

𝐴𝐴𝑒𝑒 = 100 𝑤𝑤𝐴𝐴2

The weight of the pile is

𝐴𝐴𝑒𝑒𝐿𝐿𝛾𝛾𝑐𝑐 = �100 𝑤𝑤𝐴𝐴2

144� (80 𝑜𝑜𝑡𝑡)(150 𝑒𝑒𝑡𝑡/𝑜𝑜𝑡𝑡3) = 8.33 𝑘𝑘𝑤𝑤𝑒𝑒

If the weight of the cap is 0.67 𝑘𝑘𝑤𝑤𝑒𝑒,

𝑊𝑊𝑒𝑒 = 8.33 + 0.67 = 9 𝑘𝑘𝑤𝑤𝑒𝑒

Again, from for an 11B3 hammer,

Rated energy= 19.2 𝑘𝑘𝑤𝑤𝑒𝑒 − 𝑜𝑜𝑡𝑡 = 𝐸𝐸𝐸𝐸 = 𝑊𝑊𝑅𝑅ℎ

Weight of ram= 5 𝑘𝑘𝑤𝑤𝑒𝑒

Assume that the hammer efficiency is 0.85 and that 𝐴𝐴 = 0.35. Substituting these values in equation (121) yields

𝑄𝑄𝑢𝑢 = �(0.85)(19.2×2)1𝑁𝑁+0.1

� �5+(0.35)2(9)5+9

� = 85.371𝑁𝑁+0.1

𝑘𝑘𝑤𝑤𝑒𝑒

Now the following table can be prepared:


𝑁𝑁 𝑄𝑄𝑢𝑢 (𝑘𝑘𝑤𝑤𝑒𝑒) 𝐴𝐴𝑒𝑒 (𝑤𝑤𝐴𝐴2) 𝑄𝑄𝑢𝑢/𝐴𝐴𝑒𝑒 (𝑘𝑘𝑤𝑤𝑒𝑒/𝑤𝑤𝐴𝐴2)

0 0 100 0

2 142.3 100 1.42

4 243.9 100 2.44

6 320.1 100 3.20

8 379.4 100 3.79

10 426.9 100 4.27

12 465.7 100 4.66

20 569.1 100 5.69

Both the number of hammer blows per inch and the stress can now be plotted in a graph, as shown in figure 8.47. If such a curve is prepared, the number of blows per inch of pile penetration corresponding to the allowable pile-driving stress can be easily determined.

Figure 8.47

Actual driving stresses in wooden piles are limited to about 0.7𝑜𝑜𝑢𝑢 . Similarly, for concrete and steel piles, driving stresses are limited to about 0.6𝑜𝑜′𝑐𝑐 𝑟𝑟𝐴𝐴𝐴𝐴 0.85𝑜𝑜𝑖𝑖 , respectively.


In most cases, wooden piles are driven with hammer energy of less than 45 𝑘𝑘𝑤𝑤𝑒𝑒 − 𝑜𝑜𝑡𝑡 (≈ 60 𝑘𝑘𝑁𝑁 ∙ 𝑟𝑟). Driving resistances are limited are limited mostly to 4-5 blows per inch of pile penetration. For concrete and steel piles, the usual N values are 6-8 and 12-14, respectively.

Example 11

A precast concrete pile 12 𝑤𝑤𝐴𝐴.× 12 𝑤𝑤𝐴𝐴. in cross sections in driven by a hammer. Given:

𝑀𝑀𝑟𝑟𝑒𝑒𝑤𝑤𝑟𝑟𝑢𝑢𝑟𝑟 𝑟𝑟𝑟𝑟𝑡𝑡𝑤𝑤𝐴𝐴 ℎ𝑟𝑟𝑟𝑟𝑟𝑟𝑤𝑤𝑟𝑟 𝑤𝑤𝐴𝐴𝑤𝑤𝑟𝑟𝑤𝑤𝑖𝑖 = 30 𝑘𝑘𝑤𝑤𝑒𝑒 − 𝑜𝑜𝑡𝑡

𝐸𝐸𝑟𝑟𝑟𝑟𝑟𝑟𝑤𝑤𝑟𝑟 𝑤𝑤𝑜𝑜𝑜𝑜𝑤𝑤𝑐𝑐𝑤𝑤𝑤𝑤𝐴𝐴𝑐𝑐𝑖𝑖 = 0.8

𝑊𝑊𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡 𝑜𝑜𝑜𝑜 𝑟𝑟𝑟𝑟𝑟𝑟 = 7.5 𝑘𝑘𝑤𝑤𝑒𝑒

𝑃𝑃𝑤𝑤𝑒𝑒𝑤𝑤 𝑒𝑒𝑤𝑤𝐴𝐴𝑤𝑤𝑡𝑡ℎ = 80 𝑜𝑜𝑡𝑡

𝐶𝐶𝑜𝑜𝑤𝑤𝑜𝑜𝑜𝑜𝑤𝑤𝑐𝑐𝑤𝑤𝑤𝑤𝐴𝐴𝑡𝑡 𝑜𝑜𝑜𝑜 𝑟𝑟𝑤𝑤𝑠𝑠𝑡𝑡𝑤𝑤𝑡𝑡𝑢𝑢𝑡𝑡𝑤𝑤𝑜𝑜𝐴𝐴 = 0.4

𝑊𝑊𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡 𝑜𝑜𝑜𝑜 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤 𝑐𝑐𝑟𝑟𝑒𝑒 = 550 𝑒𝑒𝑡𝑡

𝐸𝐸𝑒𝑒 = 3 × 106 𝑘𝑘𝑤𝑤𝑒𝑒/𝑤𝑤𝐴𝐴2

Number of blow for last 1 in. of penetration = 8

Estimate the allowable pile capacity by the

a. Modified ENR formula (use 𝐹𝐹𝑆𝑆 = 6) b. Danish formula (use 𝐹𝐹𝑆𝑆 = 4) c. Gates formula (use 𝐹𝐹𝑆𝑆 = 3)

Solution

Part a

𝑄𝑄𝑢𝑢 = (0.8)(30×12 𝑘𝑘𝑤𝑤𝑒𝑒−𝑤𝑤𝐴𝐴 .)18+0.1

× 7.5+(0.4)2(12.55)7.5+12.55

= 607 𝑘𝑘𝑤𝑤𝑒𝑒

𝑄𝑄𝑟𝑟𝑒𝑒𝑒𝑒 = 𝑄𝑄𝑢𝑢𝐹𝐹𝑆𝑆

= 6076≈ 101 𝑘𝑘𝑤𝑤𝑒𝑒

Part b

𝑄𝑄𝑢𝑢 = 𝐸𝐸𝐸𝐸𝐸𝐸

𝑆𝑆+�𝐸𝐸𝐸𝐸𝐸𝐸𝐿𝐿

2𝐴𝐴𝑒𝑒𝐸𝐸𝑒𝑒

Use 𝐸𝐸𝑒𝑒 = 3 × 106 𝑒𝑒𝑡𝑡/𝑤𝑤𝐴𝐴2.


�𝐸𝐸𝐸𝐸𝐸𝐸𝐿𝐿

2𝐴𝐴𝑒𝑒𝐸𝐸𝑒𝑒= �

(0.8)(30×12)(80×12)

2(12×12)�3×1061000 𝑘𝑘𝑤𝑤𝑒𝑒 /𝑤𝑤𝐴𝐴2�

= 0.566 𝑤𝑤𝐴𝐴.

𝑄𝑄𝑢𝑢 = (0.8)(30×12)18+0.566

≈ 417 𝑘𝑘𝑤𝑤𝑒𝑒

𝑄𝑄𝑟𝑟𝑒𝑒𝑒𝑒 = 4174≈ 104 𝑘𝑘𝑤𝑤𝑒𝑒

Part c

𝑄𝑄𝑢𝑢 = 𝑟𝑟�𝐸𝐸𝐸𝐸𝐸𝐸 (𝑡𝑡 − 𝑒𝑒𝑜𝑜𝑤𝑤 𝑆𝑆) = 27�(0.8)(30) [1 − 𝑒𝑒𝑜𝑜𝑤𝑤�18�] ≈ 252 𝑘𝑘𝑤𝑤𝑒𝑒

𝑄𝑄𝑟𝑟𝑒𝑒𝑒𝑒 = 2523

= 84 𝑘𝑘𝑤𝑤𝑒𝑒

NEGATIVE SKIN FRICTION

Negative skin friction is a downward drag force exerted on the pile by the soil surrounding it. This action can occur under conditions such as the following:

1. If a fill of clay soil is placed over a granular soil layer into which a pile is driven, the fill will gradually consolidate. This consolidation process will exert a downward drag force on the pile (figure 8.48a) during the period of consolidation.

2. If a fill of granular soil is placed over a layer of soft clay, as shown in figure 8. 48b, it will induce the process of consolidation in the clay layer and thus exert a downward drag on the pile.

3. Lowering of the water table will increase the vertical effective stress on the soil at any depth, which will induce consolidation settlement in clay. If a pile is located in the clay layer, it will be subjected to a downward drag force.

In some cases, the downward drag force may be excessive and cause foundation failure. This section outlines two tentative methods for the calculation of negative skin friction.


Figure 8.48 Negative skin friction

Clay Fill over Granular Soil (Figure 8.48a)

Similar to the 𝛽𝛽 method presented in section 12, the negative (downward) skin stress on the pile is

𝑜𝑜𝐴𝐴 = 𝐾𝐾′𝜎𝜎′ 𝑣𝑣 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿 [8.122]

Where

𝐾𝐾′ = 𝑤𝑤𝑟𝑟𝑟𝑟𝑡𝑡ℎ 𝑒𝑒𝑟𝑟𝑤𝑤𝑠𝑠𝑠𝑠𝑢𝑢𝑟𝑟𝑤𝑤 𝑐𝑐𝑜𝑜𝑤𝑤𝑜𝑜𝑜𝑜𝑤𝑤𝑐𝑐𝑤𝑤𝑤𝑤𝐴𝐴𝑡𝑡 = 𝐾𝐾𝑜𝑜 = 1 − 𝑠𝑠𝑤𝑤𝐴𝐴𝑠𝑠

𝜎𝜎′𝑣𝑣 = 𝑣𝑣𝑤𝑤𝑟𝑟𝑡𝑡𝑤𝑤𝑐𝑐𝑟𝑟𝑒𝑒 𝑤𝑤𝑜𝑜𝑜𝑜𝑤𝑤𝑐𝑐𝑡𝑡𝑤𝑤𝑣𝑣𝑤𝑤 𝑠𝑠𝑡𝑡𝑟𝑟𝑤𝑤𝑠𝑠𝑠𝑠 𝑟𝑟𝑡𝑡 𝑟𝑟𝐴𝐴𝑖𝑖 𝐴𝐴𝑤𝑤𝑒𝑒𝑡𝑡ℎ 𝑧𝑧 = 𝛾𝛾′𝑜𝑜𝑧𝑧

𝛾𝛾′𝑜𝑜 = 𝑤𝑤𝑜𝑜𝑜𝑜𝑤𝑤𝑐𝑐𝑡𝑡𝑤𝑤𝑣𝑣𝑤𝑤 𝑢𝑢𝐴𝐴𝑤𝑤𝑡𝑡 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡 𝑜𝑜𝑜𝑜 𝑜𝑜𝑤𝑤𝑒𝑒𝑒𝑒

𝛿𝛿 = 𝑠𝑠𝑜𝑜𝑤𝑤𝑒𝑒 − 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤 𝑜𝑜𝑟𝑟𝑤𝑤𝑐𝑐𝑡𝑡𝑤𝑤𝑜𝑜𝐴𝐴 𝑟𝑟𝐴𝐴𝑤𝑤𝑒𝑒𝑤𝑤 ≈ 0.5 − 0.7𝑠𝑠

Hence the total downward drag force, 𝑄𝑄𝐴𝐴 , on a pile is

𝑄𝑄𝐴𝐴 = ∫ (𝑒𝑒𝐾𝐾′𝛾𝛾′𝑜𝑜 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿)𝑧𝑧 𝐴𝐴𝑧𝑧 =𝑒𝑒𝐾𝐾′ 𝛾𝛾′𝑜𝑜 𝐸𝐸𝑜𝑜

2 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿

2𝐸𝐸𝑜𝑜

0 [8.123]

Where

𝐸𝐸𝑜𝑜 = ℎ𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑤𝑤 𝑜𝑜𝑤𝑤𝑒𝑒𝑒𝑒

If the fill is above the water table, the effective unit weight, 𝛾𝛾′𝑜𝑜 , should be replaced by the moist unit weight.


Granular Soil Fill over Clay (figure 8.48b)

In this case, the evidence indicates that the negative skin stress on the pile may exist from 𝑧𝑧 = 0 𝑡𝑡𝑜𝑜 𝑧𝑧 = 𝐿𝐿1, which is referred to as the neutral depth (see Vesic, 1977, pp. 25-26, for discussion). The neutral depth may be given as (Bowles, 1982):

𝐿𝐿1 = (𝐿𝐿−𝐸𝐸𝑜𝑜)𝐿𝐿1

�𝐿𝐿−𝐸𝐸𝑜𝑜2

+ 𝛾𝛾′𝑜𝑜𝐸𝐸𝑜𝑜𝛾𝛾′

� − 2𝛾𝛾′𝑜𝑜𝐸𝐸𝑜𝑜𝛾𝛾′

[8.124]

Where

𝛾𝛾′𝑜𝑜 𝑟𝑟𝐴𝐴𝐴𝐴 𝛾𝛾′ =𝑤𝑤𝑜𝑜𝑜𝑜𝑤𝑤𝑐𝑐𝑡𝑡𝑤𝑤𝑣𝑣𝑤𝑤 𝑢𝑢𝐴𝐴𝑤𝑤𝑡𝑡 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡𝑠𝑠 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑤𝑤 𝑜𝑜𝑤𝑤𝑒𝑒𝑒𝑒 𝑟𝑟𝐴𝐴𝐴𝐴 𝑡𝑡ℎ𝑤𝑤 𝑢𝑢𝐴𝐴𝐴𝐴𝑤𝑤𝑟𝑟𝑒𝑒𝑖𝑖𝑤𝑤𝐴𝐴𝑤𝑤 𝑐𝑐𝑒𝑒𝑟𝑟𝑖𝑖 𝑒𝑒𝑟𝑟𝑖𝑖𝑤𝑤𝑟𝑟, 𝑟𝑟𝑤𝑤𝑠𝑠𝑒𝑒𝑤𝑤𝑐𝑐𝑡𝑡𝑤𝑤𝑣𝑣𝑤𝑤𝑒𝑒𝑖𝑖

For end-bearing piles, the neutral depth may be assumed to be located at the pile tip (that is, 𝐿𝐿1 = 𝐿𝐿 − 𝐸𝐸𝑜𝑜).

Once the value of 𝐿𝐿1 is determined, the downward drag force is obtained in the following manner. The unit negative skin friction at any depth from 𝑧𝑧 = 0 𝑡𝑡𝑜𝑜 𝑧𝑧 = 𝐿𝐿1 is

𝑜𝑜𝐴𝐴 = 𝐾𝐾′𝜎𝜎′𝑣𝑣 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿 [8.125]

Where

𝐾𝐾′ = 𝐾𝐾𝑜𝑜 = 1 − 𝑠𝑠𝑤𝑤𝐴𝐴𝑠𝑠

𝜎𝜎′𝑣𝑣 = 𝛾𝛾′𝑜𝑜𝐸𝐸𝑜𝑜 + 𝛾𝛾′𝑧𝑧

𝛿𝛿 = 0.5 − 0.7𝑠𝑠

𝑄𝑄𝐴𝐴 = ∫ 𝑒𝑒𝑜𝑜𝐴𝐴 𝐴𝐴𝑧𝑧 = ∫ 𝑒𝑒𝐾𝐾′ �𝛾𝛾′𝑜𝑜𝐸𝐸𝑜𝑜𝛾𝛾′𝑧𝑧� 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿 𝐴𝐴𝑧𝑧𝐿𝐿1

0𝐿𝐿1

0

= (𝑒𝑒𝐾𝐾′𝛾𝛾′𝑜𝑜𝐸𝐸𝑜𝑜 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿)𝐿𝐿1 + 𝐿𝐿12𝑒𝑒𝐾𝐾′ 𝛾𝛾′ 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿

2 [8.126]

If the soil and the fill are above the water table, the effective unit weights should be replaced by moist unit weights. In some cases, the piles can be coated with bitumen in the downdrag zone to avoid this problem. Baligh et al. (1978) summarized the results of several field tests that were conducted to evaluate the effectiveness of bitumen coating in reducing the negative skin friction. Their results are presented in table 12.

A limited number of case studies on negative skin friction is available in the literature. Bjerrum et al. (1969) reported monitoring of downdrag force on a test pile at Sorenga in the harbor of Oslo, Norway (noted as pile G in the original paper). This was also discussed by Wong and The (1995) in terms of the pile being driven to bedrock at 40 m. Figure 8.49 a shows the soil profile and the pile. Wong and The (1995) estimated the following:


Figure 8.49 Negative skin friction on a pile in the harbor of Oslo, Norway [based on Bjerrum et al., (1969); and Wong and The (1995)]

𝐹𝐹𝑤𝑤𝑒𝑒𝑒𝑒:𝑀𝑀𝑜𝑜𝑤𝑤𝑠𝑠𝑡𝑡 𝑢𝑢𝐴𝐴𝑤𝑤𝑡𝑡 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡, 𝛾𝛾𝑜𝑜 = 16 𝑘𝑘𝑁𝑁/𝑟𝑟3

𝑆𝑆𝑟𝑟𝑡𝑡𝑢𝑢𝑟𝑟𝑟𝑟𝑡𝑡𝑤𝑤𝐴𝐴 𝑢𝑢𝐴𝐴𝑤𝑤𝑡𝑡 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡, 𝛾𝛾𝑠𝑠𝑟𝑟𝑡𝑡 (𝑜𝑜) = 18.5 𝑘𝑘𝑁𝑁/𝑟𝑟3

So

𝛾𝛾′𝑜𝑜 = 18.5 − 9.81 = 8.69 𝑘𝑘𝑁𝑁/𝑟𝑟3

𝐸𝐸𝑜𝑜 = 13 𝑟𝑟

𝐶𝐶𝑒𝑒𝑟𝑟𝑖𝑖:𝐾𝐾′ 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿 ≈ 0.22

𝑆𝑆𝑟𝑟𝑡𝑡𝑢𝑢𝑟𝑟𝑟𝑟𝑡𝑡𝑤𝑤𝐴𝐴 𝑤𝑤𝑜𝑜𝑜𝑜𝑤𝑤𝑐𝑐𝑡𝑡𝑤𝑤𝑣𝑣𝑤𝑤 𝑢𝑢𝐴𝐴𝑤𝑤𝑡𝑡 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡, 𝛾𝛾′ = 19 − 9.81 = 9.19 𝑘𝑘𝑁𝑁/𝑟𝑟3

𝑃𝑃𝑤𝑤𝑒𝑒𝑤𝑤: 𝐿𝐿 = 40𝑟𝑟

𝐷𝐷𝑤𝑤𝑟𝑟𝑟𝑟𝑤𝑤𝑡𝑡𝑤𝑤𝑟𝑟,𝐷𝐷 = 500𝑟𝑟

Thus, the maximum downdrag force on the pile can be estimated from equation. (126). Since it is a point bearing pile, the magnitude of 𝐿𝐿1 = 27 𝑟𝑟, so

𝑄𝑄𝐴𝐴 = (𝑒𝑒)(𝐾𝐾′ 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿)[𝛾𝛾𝑜𝑜 × 2 + (13 − 2)𝛾𝛾′𝑜𝑜 ](𝐿𝐿1) + 𝐿𝐿12𝑒𝑒𝛾𝛾′ (𝐾𝐾′ 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿)

2


Or

𝑄𝑄𝐴𝐴 = (𝜋𝜋 × 0.5)(0.22)[(16 × 2) + (8.69 × 11)](27) + (27)2(𝜋𝜋×0.5)(9.19)(0.22)2

= 2348 𝑘𝑘𝑁𝑁

The measured value of maximum 𝑄𝑄𝐴𝐴 was about 2500 𝑘𝑘𝑁𝑁 (figure 8. 49b), which is in good agreement with the calculated value.

Table 12 Summary of Case Studies of Bitumen-Coated Piles(After Baligh et al. (1978))

Downward drag Test loadings

Case number 1 2 3 4 5 6 7

Soil type Fill, sand, and clay

Fill and silty clay

Fill and clay

Sand and silty clay

Silty clay

Silty clay

Sand fill, clay, and peat

Pile type

Pile cross section (mm)

Cast-in-place concrete

𝐷𝐷 = 530

Steel pipe

𝐷𝐷 = 300

Steel pipe

𝐷𝐷= 500

Steel pile

𝐷𝐷= 760

6 RC piles

300× 300

6 RC piles

300× 300

Precast concrete

380× 450

Length in contact with settling soil (m)

25 26 40 25 7-17 9-16 24

Installation method

Predriven casing

Enlarged tip and slurry

Enlarge tip and casing

Driving Driving Driving Driving

Bitumen coating

Type (pen 25° 𝐶𝐶)

Coating thickness (mm)

20/30

10

80/100

1.2

80/10

1.2

60/70

1.5

60/70

1.2

80-100 RC-0 cutback

1.2

43 special grade

10


Measured shaft resistance

Uncoated pile (ton)

Coated pile (ton)

Coating effectiveness (%)

70-80

5-7

92

120

10

92

300

15

95

180

3

98

31-40

10-33

30-80

31-40

20-42

30-80

160

Predicted downdrag

Coated pile (ton)

Coating Effectiveness (%)

0.1

100

2-11

91-98

5

98

0-23

87-100

Example 12

Refer to figure 8. 48a; 𝐸𝐸𝑜𝑜 = 3 𝑟𝑟. The pile is circular in cross section with a diameter of 0.5 m. For the fill that is above the water table, 𝛾𝛾𝑜𝑜 = 17.2 𝑘𝑘𝑁𝑁/𝑟𝑟3 𝑟𝑟𝐴𝐴𝐴𝐴 𝑠𝑠 = 36°. Determine the total drag force. Use 𝛿𝛿 = 0.7 𝑠𝑠.

Solution

From equation. (123),

𝑄𝑄𝐴𝐴 =𝑒𝑒𝐾𝐾′ 𝛾𝛾𝑜𝑜 𝐸𝐸𝑜𝑜

2 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿

2

𝑒𝑒 = 𝜋𝜋(0.5) = 1.57 𝑟𝑟

𝐾𝐾′ = 1 − 𝑠𝑠𝑤𝑤𝐴𝐴 𝑠𝑠 = 1 − 𝑠𝑠𝑤𝑤𝐴𝐴 36° = 0.41

𝛿𝛿 = (0.7)(36) = 25.2°

𝑄𝑄𝐴𝐴 = (1.57)(0.41)(17.2)(3)2 𝑡𝑡𝑟𝑟𝐴𝐴 25.22

= 23.4 𝑘𝑘𝑁𝑁


Example 13

Refer to figure 8. 48b. Here, 𝐸𝐸𝑜𝑜 = 2 𝑟𝑟,𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤 𝐴𝐴𝑤𝑤𝑟𝑟𝑟𝑟𝑤𝑤𝑡𝑡𝑤𝑤𝑟𝑟 = 0.305 𝑟𝑟, 𝛾𝛾𝑜𝑜 = 16.5 𝑘𝑘𝑁𝑁/𝑟𝑟3, 𝑠𝑠𝑐𝑐𝑒𝑒𝑟𝑟𝑖𝑖 = 34°, 𝛾𝛾𝑠𝑠𝑟𝑟𝑡𝑡 = 17.2 𝑘𝑘𝑁𝑁/𝑟𝑟3, 𝑟𝑟𝐴𝐴𝐴𝐴 𝐿𝐿 = 20 𝑟𝑟. The water table coincides with the top of the clay layer. Determine the downward drag force. Assume 𝛿𝛿 = 0.6𝑠𝑠𝑐𝑐𝑒𝑒𝑟𝑟𝑖𝑖 .

Solution

The depth of the neutral plane in given in equation (124) as

𝐿𝐿1 = 𝐿𝐿−𝐸𝐸𝑜𝑜𝐿𝐿1

�𝐿𝐿−𝐸𝐸𝑜𝑜2

+ 𝛾𝛾𝑜𝑜𝐸𝐸𝑜𝑜𝛾𝛾′� − 2𝛾𝛾𝑜𝑜𝐸𝐸𝑜𝑜

𝛾𝛾′

Note that 𝛾𝛾′𝑜𝑜 in equation (124) has been replaced by 𝛾𝛾𝑜𝑜 because the fill is above the water table, so

𝐿𝐿1 = (20−2)𝐿𝐿1

�(20−2)2

+ (16.5)(2)(17.2−9.81)

� − (2)(16.5)(2)(17.2−9.81)

𝐿𝐿1 = 242.4𝐿𝐿1

− 8.93; 𝐿𝐿1 = 11.75 𝑟𝑟

Now, referring to equation (126), we have

𝑄𝑄𝐴𝐴 = (𝑒𝑒𝐾𝐾′𝛾𝛾𝑜𝑜𝐸𝐸𝑜𝑜 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿)𝐿𝐿1 + 𝐿𝐿12𝐾𝐾′ 𝛾𝛾′ 𝑡𝑡𝑟𝑟𝐴𝐴 𝛿𝛿

2

𝑒𝑒 = 𝜋𝜋(0.305) = 0.958 𝑟𝑟

𝐾𝐾′ = 1 − 𝑠𝑠𝑤𝑤𝐴𝐴 34° = 0.44

𝑄𝑄𝐴𝐴 = (0.958)(0.44)(16.5)(2)[𝑡𝑡𝑟𝑟𝐴𝐴(0.6 × 34)](11.75)

+ (11.75)2(0.958)(0.44)(17.2−9.81)[𝑡𝑡𝑟𝑟𝐴𝐴 (0.6×34)]2

= 60.78 + 79.97 = 140.75 𝑘𝑘𝑁𝑁

GROUP PILES

GROUP EFFICIENCY

In many cases, piles are used in groups, as shown in figure 8.50, to transmit the structural load to the soil. A pile cap is constructed over group piles. The pile cap can be contact with the ground, as in most cases (figure 8.50a), or well above the ground, as in the case of offshore platforms (figure 8.50b).


Figure 8.50 Pile groups

Determining the load-bearing capacity of group piles is extremely complicated and has not yet been fully resolved. When the piles are placed close to each other, a reasonable assumption is that the stresses transmitted by the piles to the soil will overlap (figure 8. 50c), reducing the load-bearing capacity of the piles. Ideally, the piles in a group should be spaced so that the load-bearing capacity of the group should not be less than the sum of the bearing capacity of the individual piles. In practice, the minimum center-to-center pile spacing, 𝐴𝐴, 𝑤𝑤𝑠𝑠 2.5 𝐷𝐷, and in ordinary situations, is actually about 3 − 3.5𝐷𝐷.

The efficiency of the load-bearing capacity of a group pile may be defined as

𝜂𝜂 = 𝑄𝑄𝑤𝑤(𝑢𝑢 )

𝛴𝛴𝑄𝑄𝑢𝑢 [8.127]

Where

𝜂𝜂 = 𝑤𝑤𝑟𝑟𝑜𝑜𝑢𝑢𝑒𝑒 𝑤𝑤𝑜𝑜𝑜𝑜𝑤𝑤𝑐𝑐𝑤𝑤𝑤𝑤𝐴𝐴𝑐𝑐𝑖𝑖


𝑄𝑄𝑤𝑤(𝑢𝑢) = 𝑢𝑢𝑒𝑒𝑡𝑡𝑤𝑤𝑟𝑟𝑟𝑟𝑡𝑡𝑤𝑤 𝑒𝑒𝑜𝑜𝑟𝑟𝐴𝐴 − 𝑡𝑡𝑤𝑤𝑟𝑟𝑟𝑟𝑤𝑤𝐴𝐴𝑤𝑤 𝑐𝑐𝑟𝑟𝑒𝑒𝑟𝑟𝑐𝑐𝑤𝑤𝑡𝑡𝑖𝑖 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑤𝑤 𝑤𝑤𝑟𝑟𝑜𝑜𝑢𝑢𝑒𝑒 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤

𝑄𝑄𝑢𝑢 = 𝑢𝑢𝑒𝑒𝑡𝑡𝑤𝑤𝑟𝑟𝑟𝑟𝑡𝑡𝑤𝑤 𝑒𝑒𝑜𝑜𝑟𝑟𝐴𝐴 − 𝑡𝑡𝑤𝑤𝑟𝑟𝑟𝑟𝑤𝑤𝐴𝐴𝑤𝑤 𝑐𝑐𝑟𝑟𝑒𝑒𝑟𝑟𝑐𝑐𝑤𝑤𝑡𝑡𝑖𝑖 𝑜𝑜𝑜𝑜 𝑤𝑤𝑟𝑟𝑐𝑐ℎ 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤 𝑤𝑤𝑤𝑤𝑡𝑡ℎ𝑜𝑜𝑢𝑢𝑡𝑡 𝑡𝑡ℎ𝑤𝑤 𝑤𝑤𝑟𝑟𝑜𝑜𝑢𝑢𝑒𝑒 𝑤𝑤𝑜𝑜𝑜𝑜𝑤𝑤𝑐𝑐𝑡𝑡

Many structural engineers use a simplified analysis to obtain the group efficiency for friction piles, particularly in sand. This type of analysis can be explained with the aid of figure 8. 50a. Depending on their spacing within the group, the piles may act in one of two ways: (1) as a block with dimensions 𝐿𝐿𝑤𝑤 × 𝐵𝐵𝑤𝑤 × 𝐿𝐿, or (2) as individual piles. If the piles act as a block, the frictional capacity is 𝑜𝑜𝑟𝑟𝑣𝑣𝑒𝑒𝑤𝑤𝐿𝐿 ≈ 𝑄𝑄𝑤𝑤(𝑢𝑢). [Note: 𝑒𝑒𝑤𝑤 = perimeter of the cross section of block= 2(𝐴𝐴1 + 𝐴𝐴2 − 2)𝐴𝐴 + 4 𝐷𝐷,𝑟𝑟𝐴𝐴𝐴𝐴 𝑜𝑜𝑟𝑟𝑣𝑣 = average unit frictional resistance.] Similarly, for each pile acting individually, 𝑄𝑄𝑢𝑢 ≈ 𝑒𝑒𝐿𝐿𝑜𝑜𝑟𝑟𝑣𝑣 . (Note: 𝑒𝑒 = perimeter of the cross section of each pile.) Thus


𝛴𝛴 𝑄𝑄𝑢𝑢= 𝑜𝑜𝑟𝑟𝑣𝑣 [2(𝐴𝐴1+𝐴𝐴2−2)𝐴𝐴+4𝐷𝐷]𝐿𝐿

𝐴𝐴1𝐴𝐴2𝑒𝑒𝐿𝐿𝑜𝑜𝑟𝑟𝑣𝑣= 2(𝐴𝐴1+𝐴𝐴2−2)𝐴𝐴+4𝐷𝐷

𝑒𝑒𝐴𝐴1𝐴𝐴2 [8.128]

Hence

𝑄𝑄𝑤𝑤(𝑢𝑢) = �2(𝐴𝐴1+𝐴𝐴2−2)𝐴𝐴+4𝐷𝐷𝑒𝑒𝐴𝐴1𝐴𝐴2

� 𝛴𝛴 𝑄𝑄𝑢𝑢 [8.129]

From equation (129), if the center-to-center spacing, 𝐴𝐴, s large enough, 𝜂𝜂 > 1. In that case, the piles will behave as individual piles. Thus, in practice, if 𝜂𝜂 < 1,

𝑄𝑄𝑤𝑤(𝑢𝑢) = 𝜂𝜂 𝛴𝛴 𝑄𝑄𝑢𝑢

And, if 𝜂𝜂 ≥ 1,

𝑄𝑄𝑤𝑤(𝑢𝑢) = 𝛴𝛴 𝑄𝑄𝑢𝑢

There are several other equations like equation (129) for the group efficiency of friction piles. Some of these are given in table 13.

Feld (1943) suggested a method by which the load capacity of individual piles (friction) in a group embedded in sand could be assigned. According to this method, the ultimate capacity of a pile is reduced by one-sixteenth by each adjacent diagonal or row pile. The technique can be explained by referring to figure 8.51, which shows the plan of a group pile. For pile type A, there are eight adjacent piles; for pile type B, there are five adjacent piles; and for pile type C, there are three adjacent piles. Now the following table can be prepared:


Figure 8.51 Feld’s method for estimation of group capacity of friction piles

Table 13 Equations for Group Efficiency of Friction Piles

Converse-Labarre equation 𝜂𝜂 = 1 − �

(𝐴𝐴1 − 1)𝐴𝐴2 + (𝐴𝐴2 − 1)𝐴𝐴1

90𝐴𝐴1𝐴𝐴2� 𝜃𝜃

𝑤𝑤ℎ𝑤𝑤𝑟𝑟𝑤𝑤 𝜃𝜃 (𝐴𝐴𝑤𝑤𝑤𝑤) = 𝑡𝑡𝑟𝑟𝐴𝐴−1(𝐷𝐷/𝐴𝐴)

Los Angles Group Action equation 𝜂𝜂 = 1 −

𝐷𝐷𝜋𝜋𝐴𝐴𝐴𝐴1𝐴𝐴2

[𝐴𝐴1(𝐴𝐴2 − 1)] + 𝐴𝐴2(𝐴𝐴1 − 1)

+ √2(𝐴𝐴1 − 1)(𝐴𝐴2 − 1)]

Seiler and Keeney equation (Seiler and Keeney, 1944) 𝜂𝜂 = �1 − �

111𝐴𝐴7(𝐴𝐴2 − 1)

� �𝐴𝐴1 + 𝐴𝐴2 − 2𝐴𝐴1 + 𝐴𝐴2 − 1

�� +0.3

𝐴𝐴1 + 𝐴𝐴2

𝑤𝑤ℎ𝑤𝑤𝑟𝑟𝑤𝑤 𝐴𝐴 𝑤𝑤𝑠𝑠 𝑤𝑤𝐴𝐴 𝑜𝑜𝑡𝑡

Pile type No. of Piles

No. of adjacent piles/pile

Reduction factor for each pile

Ultimate capacity

A 1 8 1 −

816

0.5𝑄𝑄𝑢𝑢

B 4 5 1 −

516

2.75𝑄𝑄𝑢𝑢

C 4 3 1 −

316

3.25𝑄𝑄𝑢𝑢


𝛴𝛴 6.5 𝑄𝑄𝑢𝑢= 𝑄𝑄𝑤𝑤(𝑢𝑢)

(No. of piles)(𝑄𝑄𝑢𝑢 ) (reduction factor)

𝑄𝑄𝑢𝑢 = 𝑢𝑢𝑒𝑒𝑡𝑡𝑤𝑤𝑟𝑟𝑟𝑟𝑡𝑡𝑤𝑤 𝑐𝑐𝑟𝑟𝑒𝑒𝑟𝑟𝑐𝑐𝑤𝑤𝑡𝑡𝑖𝑖 𝑜𝑜𝑜𝑜𝑟𝑟 𝑟𝑟𝐴𝐴 𝑤𝑤𝑠𝑠𝑜𝑜𝑒𝑒𝑟𝑟𝑡𝑡𝑤𝑤𝐴𝐴 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤𝑠𝑠

Hence


𝛴𝛴 𝑄𝑄𝑢𝑢= 6.5 𝑄𝑄𝑢𝑢

9𝑄𝑄𝑢𝑢= 72%

Figure 8.52 shows a comparison of field test results in clay with the theoretical group efficiency calculated from the Converse-Labarre equation (table 13). Reported by Brand et al. (1972), these tests were conducted in soil for which the details are given in figure 8. 7 from chapter 3. Other test details include

𝐿𝐿𝑤𝑤𝐴𝐴𝑤𝑤𝑡𝑡ℎ 𝑜𝑜𝑜𝑜 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤𝑠𝑠 = 6 𝑟𝑟

𝐷𝐷𝑤𝑤𝑟𝑟𝑟𝑟𝑤𝑤𝑡𝑡𝑤𝑤𝑟𝑟 𝑜𝑜𝑜𝑜 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤𝑠𝑠 = 150 𝑟𝑟𝑟𝑟

𝑃𝑃𝑤𝑤𝑒𝑒𝑤𝑤 𝑤𝑤𝑟𝑟𝑜𝑜𝑢𝑢𝑒𝑒𝑠𝑠 𝑡𝑡𝑤𝑤𝑠𝑠𝑡𝑡𝑤𝑤𝐴𝐴 = 2 × 2

𝐿𝐿𝑜𝑜𝑐𝑐𝑟𝑟𝑡𝑡𝑤𝑤𝑜𝑜𝐴𝐴 𝑜𝑜𝑜𝑜 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤 ℎ𝑤𝑤𝑟𝑟𝐴𝐴 = 1.5 𝑟𝑟 𝑡𝑡𝑤𝑤𝑒𝑒𝑜𝑜𝑤𝑤 𝑡𝑡ℎ𝑤𝑤 𝑤𝑤𝑟𝑟𝑜𝑜𝑢𝑢𝐴𝐴𝐴𝐴 𝑠𝑠𝑢𝑢𝑟𝑟𝑜𝑜𝑟𝑟𝑐𝑐𝑤𝑤

Figure 8.52 Variation of group efficiency with 𝐴𝐴/𝐷𝐷 (after Brand et al., 1972)


Pile tests were conducted with and without a cap (free-standing group). Note that for 𝐴𝐴/𝐷𝐷 ≥ 2, the magnitude of 𝜂𝜂 was greater than 1.0. Also for similar values of 𝐴𝐴/𝐷𝐷 the group efficiency was greater with the pile cap than without the cap. Figure 8.53 shows the pile group settlement at various stages of the load test.

Figure 8.53 Variation of group pile settlement at various stages of load (after Brand et al., 1972)

Figure 8.54 Variation of efficiency of pile group in sand (based on Kishida and Meyerhof, 1965)


Figure 8.55 Behavior of low-set ad high-set pile groups in terms of average skin friction (based on Liu et al., 1985)


Figure 8. 55 (Continued)

Figure 8.54 shows the variation of group efficiency (𝜂𝜂) 𝑜𝑜𝑜𝑜𝑟𝑟 𝑟𝑟 3 × 3 group pile in sand (Kishida and Meyerhof, 1965). It can be seen that, for loose and medium sands, the magnitude of group efficiency is larger than one. This is primarily due to the densification of sand surrounding the pile.

Liu et al. (1985) reported the results of field tests on 58 pile groups and 23 single piles embedded in granular soil. Test details include

𝑃𝑃𝑤𝑤𝑒𝑒𝑤𝑤 𝑒𝑒𝑤𝑤𝐴𝐴𝑤𝑤𝑡𝑡ℎ, 𝐿𝐿 = 8𝐷𝐷 − 23𝐷𝐷

𝑃𝑃𝑤𝑤𝑒𝑒𝑤𝑤 𝐴𝐴𝑤𝑤𝑟𝑟𝑟𝑟𝑤𝑤𝑡𝑡𝑤𝑤𝑟𝑟,𝐷𝐷 = 125 𝑟𝑟𝑟𝑟 − 330 𝑟𝑟𝑟𝑟

𝑇𝑇𝑖𝑖𝑒𝑒𝑤𝑤 𝑜𝑜𝑜𝑜 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤 𝑤𝑤𝐴𝐴𝑠𝑠𝑡𝑡𝑟𝑟𝑒𝑒𝑒𝑒𝑟𝑟𝑡𝑡𝑤𝑤𝑜𝑜𝐴𝐴 = 𝑡𝑡𝑜𝑜𝑟𝑟𝑤𝑤𝐴𝐴

𝑆𝑆𝑒𝑒𝑟𝑟𝑐𝑐𝑤𝑤𝐴𝐴𝑤𝑤 𝑜𝑜𝑜𝑜 𝑒𝑒𝑤𝑤𝑒𝑒𝑤𝑤𝑠𝑠 𝑤𝑤𝐴𝐴 𝑤𝑤𝑟𝑟𝑜𝑜𝑢𝑢𝑒𝑒,𝐴𝐴 = 2𝐷𝐷 − 6𝐷𝐷

Figure 8. 55 shows the behavior of 3 × 3 pile groups with low-set and high-set pile caps in terms of average skin friction, 𝑜𝑜𝑟𝑟𝑣𝑣 . Figure 8.56 shows the variation of average skin friction based on the location of a pile in the group.

Figure 8.56 Average skin friction based on pile location (based on Liu et al., 1985)


Based on eh experimental observations of the behavior of group piles in sand to date, the following general conclusions may be drawn.

1. For driven group piles in sand with 𝐴𝐴 ≥ 3𝐷𝐷,𝑄𝑄𝑤𝑤(𝑢𝑢) may be taken to be 𝛴𝛴 𝑄𝑄𝑢𝑢 , which includes the frictional and the point bearing capacities of individual piles.

2. For bored group piles in sand at conventional spacings (𝐴𝐴 ≈ 3𝐷𝐷),𝑄𝑄𝑤𝑤(𝑢𝑢) may be taken to be 2

3 𝑡𝑡𝑜𝑜 34 𝑡𝑡𝑤𝑤𝑟𝑟𝑤𝑤𝑠𝑠 𝛴𝛴 𝑄𝑄𝑢𝑢 (frictional and point bearing capacities of individual

piles).

module 8 (lecture 33) pile foundations topics -...

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