This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

Pile foundations act in a variable combination of two methods of support. In one case the piles are driven into the soil to such a depth that the frictional resistance of the soil to further penetration of the pile is greater than any load which will be placed on the pile. As the soil becomes more and more soft, the frictional resistance furnished by the soil is less and less;

Fig. 51. Cast-Iron Pile-Shoe.

and it then becomes necessary that the pile shall penetrate to a strata of much greater density, into which it will penetrate but little if any. Under such conditions, the structure rests on a series of columns (the piles) which are supported by the hard subsoil, and whose action as columns is very greatly assisted by the density of the very soft soil through which the piles have passed. It practically makes but little difference which of these methods of support exists in any particular case. The piles are driven until the resistance furnished by each pile is as high as is desired. The resistance against the sinking of a pile depends on such a very large variety of conditions, that attempts to develop a formula for that resistance based on a theoretical computation taking in all these various factors, are practically useless. There are so many elements of the total resistance which are so large, and also so very uncertain, that they entirely overshadow the few elements which can be precisely calculated. Most formulas for pile-driving are based on the general proposition that the resistance of the pile, multiplied by its motion during the last blow, equals the weight of the hammer multiplied by the distance through which it falls. To express this algebraically:

If R = Resistance of pile; s = Penetration of pile during last blow; w = Weight of hammer; h = Height of fall of hammer; then, according to the above principle, we have:

Rs = wh.

Practically, such a formula is considerably modified, owing to the fact that the resistance of a pile (or its penetration for any blow) depends considerably on the time which has elapsed since the previous blow. This practically means that it is far easier to drive the pile, provided the blows are delivered very rapidly; and also that when a pause is made in the driving for a few minutes or for an hour, the penetration is very much less (and the apparent resistance very much greater), on account of the earth settling around the pile during the interval. The most commonly used formula for pile-driving is known as the Engineering News formula, which, when used for ordinary hammer-driving, is as follows:

R = 2wh..(4) s + 1

This formula is fundamentally the same as the formula given above, except that,

R = Safe load, in pounds; s = Penetration, in inches, w = Weight of hammer, in pounds; h = Height of fall of hammer, in feet.

In the above equation, s is considered a free-falling hammer (not retarded by hammer rope) striking a pile having a sound head. If a friction-clutch driver is used, so that the hammer is retarded by the rope attached to it, the penetration of the pile is commonly assumed to be just one-half what it would have been had no rope been attached (that is, had it been free-falling).

Also, the quantity s is arbitrarily increased by 1, to allow for the influence of the settling of the earth during ordinary hammer pile-driving, and a factor of safety of 6 for safe load has been used. In spite of the extreme simplicity of this formula compared with that of others which have attempted to allow for all possible modifying causes, this formula has been found to give very good results. When computing the bearing power of a pile, the penetration of the pile during the last blow is determined by averaging the total penetration during the last five blows.

The pile-driving specifications adopted by the American Railway Engineering & Maintenance of Way Association, require that,

"All piles shall be driven to a firm bearing satisfactory to the Engineer, or until five blows of a hammer weighing 3,000 pounds, falling 15 feet (or a hammer and fall producing the same mechanical effect), are required to drive a pile one-half (1/2) inch per blow, except in soft bottom, when special instructions will be given."

This is equivalent to saying (applying the Engineering News formula) that the piles must have a bearing power of 60,000 pounds.

The total penetration during the last five blows was 14 inches for a pile driven with a 3,000-pound hammer. During these blows the average drop of the hammer was 24 feet. How much is the safe load?

s + 1 (1/5 x 14) +1 3.7

It is required (if possible) to drive piles with a 3,000-pound hammer until the indicated resistance is 70,000 pounds. What should be the average penetration during the last five blows when the fall is 25 feet?

70.000 = 2 wh = 2 X 3,000 X 25 = 150,000 s+l s + 1 s + 1

150,000 - 1 = 2.14 - 1 = 1.14inches. S = 70,000

200. The last problem suggests a possible impracticability, for it may readily happen that when the pile has been driven to its full length its indicated resistance is still far less than that desired. In some cases, such piles would merely be left as they are, and additional piles would be driven beside them, in the endeavor to obtain as much total resistance over the whole foundation as is desired.

The above formula applies only to the drop-hammer method of driving piles, in which a weight of 2,500 to 3,000 pounds is raised and dropped on the pile.

When the steam pile-driver is used, the blows are very rapid, about 55 to 65 per minute. On account of this rapidity the soil does not have time to settle between the successive blows, and the penetration of the pile is much more rapid, while of course the resistance after the driving is finished is just as great as is secured by any other method. On this account, the above formula is modified so that the arbitrary quantity added to s is changed from one to 0.1, and the formulabecomes:

R = 2 wh ..(5) s = 0.1

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