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«PILE DRIVING ANALYSIS BY THE WAVE EQUATION For technical assistance, contact: Dr. Lee L. Lowery, Jr., P.E. Department of Civil Engineering Texas A&M ...»

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Weights are denoted by WAM and internal springs (cushions and pile springs) are denoted by XKAM. Soil springs (external springs) are denoted by XKIM and soil dashpots are represented by SJ. The "top" weight of the system is denoted as WAM(l), and the adjacent masses are numbered sequentially to the point of the pile. Since there is no pile spring beneath the last pile weight, there will always be one less internal spring than the number of weights. The external springs are numbered according to the weight upon which they act. Hence, if WAM(5) is the first weight in the soil, its associated soil spring is XKIM(5). The soil spring beneath the point of the pile is denoted by a number one larger than that of the last pile weight. A similar notation applies to the soil dashpots.

To simulate a given system, the following information is required:

–  –  –

4. Soil characteristics a. length of pile embedment in the soil.

b. types of soil penetrated (soil profile) c. magnitude and distribution of the static resistance to penetration distributed along the side of the piled.

d. magnitude of the static resistance at the tip of the pile e. ultimate elastic displacement of the soil along the side of the pile and at the tip of the pile It should be recognized that the solution obtained with the program represents the results for one blow of the hammer at the specified soil embedment and soil resistance.

The Solution

The solution to an idealized pile driving problem is accomplished by a numerical technique proposed by Smith (24), which is based on concentrating the distributed mass of the pile into a series of relatively small weights, WAM(l) through WAM(MP), connected by weightless springs XKAM(l) through XKAM(MP-l), with the addition of soil resistance acting on the masses, as illustrated in Figure 22. Also, time was divided into small increments. The numerical solution to the wave equation is then applied by the repeated use of the following equations, developed by

Smith (21):

D(m,t) = D(m,t-l)+12*delt*V(m,t-l) Eq. 4.1 C(m,t) = D(m,t)-D(m+l,t) Eq. 4.2 F(m,t) = C(m,t)K(m) Eq. 4.3 R(m,t) = [D(m,t)-D'(m,t)]K'(m)[1+J(m)V(m,t-1)] Eq. 4.4 V(m,t) = V(m,t-l)+[F(m-l,t)-F(m,t)-R(m,t)]g*delt/W(m)Eq. 4.5 where m is the mass number; t denotes the time interval number; delt is the size of the time interval (sec); D(m,t) is the total displacement of mass number m during time interval t (in.);

V(m,t) is the velocity of mass m during time interval t(ft/sec); C(m,t) is the compression of the spring m during time interval t (in.); F(m,t) is the force exerted by spring number m between segment numbers (m) and (m+1) during time interval t(kips); and K(m) is the spring rate of element m(kip/in.).

Note that since certain parameters do not change with time, they are assigned a single subscript. The quantity R(m,t) is the total soil resistance acting on segment m(kip); K'(m) is the spring rate of the soil spring causing the external soil resistance force on mass m (kip/in.); D'(m, t) is the total inelastic soil displacement or soil inelastic yield during the time t at segment m (in.);

J(m) is a damping constant for the soil acting on segment number (m)(sec/ft); g is the gravitational acceleration (ft/sec^2); and W(m) is the weight of segment number m (kip).

The Numerical Solution The basic steps required for the numerical solution by the wave equation are outlined

below (see Figure 22 ):

1. The velocity of the top weight is set equal to the initial velocity of the pile driving ram at the instant of impact.

2. A short time interval )t is permitted to elapse (on the order of 1/5000 second).

3. The ram velocity is assumed to be uniform during this time interval and a new position of the ram is calculated.

4. Since the velocities of all other weights are zero, their displacements after the elapse of the first time interval will remain zero.

5. Because of the movement of the ram during the first time interval, the top spring is compressed and the resulting force may be calculated from the spring constant for that spring.

6. The force developed in the capblock acts between the ram and the helmet. This unbalanced force tends to reduce the downward velocity of the ram and to increase the velocity of the helmet from zero. New ram and helmet velocities are calculated, the other weight velocities still being zero.

7. A second time interval is permitted to elapse.

8. Assuming that the new ram and helmet velocities are uniform during the second elapsed time interval, their new displacements are calculated. These new displacements result in new spring compressions in the first and second springs from which new spring forces may be computed.

This results in unbalanced forces on the first three weights and new velocities for these weights may be determined. This procedure is continued until maximum stresses and displacements have been found.

It should be emphasized that the results of this procedure are for a single blow of the ram with the pile at a specified embedment in the soil. To determine the number of blows of the ram required to attain one foot of penetration at this pile embedment, it is assumed that the calculated permanent set per blow of the ram will be uniform during the one foot penetration. Hence, the reciprocal of the permanent set per blow is used to predict the number of blows per foot of penetration at the given embedment.

There is no limitation to the number of time intervals which can elapse during the computer solution. However, the significant results are generally obtained after a relatively few number of intervals have elapsed. The following equation may be used to determine an estimate of the number of iterations which will normally be adequate for determining the solution to a problem.

NSTOP = 30Lp/Lmin

where NSTOP = maximum number of iterations, Lp = length of pile, and Lmin = length of shortest pile segment used in the analysis. This is usually greater than necessary, so the program incorporates an automatic shut-off which can be used to shorten the running time should the user desire.

Idealization of Hammers The program is formulated to handle drop hammers, steam hammers and diesel hammers.

The techniques presented in this section are general in scope and are presented for illustration purposes.

Figures 23 through 25 describe the idealization for the following cases:

1. Case I - Steam Hammer with ram, capblock, helmet, cushion and pile (Figure 23),

2. Case II - Steam Hammer with ram, capblock, helmet, and pile, (Figure 24 ), and

3. Case III - Diesel Hammer with ram, anvil, capblock, helmet and pile (Figure 25).

The Ram The idealization of the ram of a pile driver depends upon its construction. Drop hammers and steam hammers are usually constructed so that the ram impacts directly on a cushion (the capblock), whereas the ram of a diesel hammer impacts directly on an anvil. Rams which impact directly on a cushion can be represented accurately by a single concentrated weight of infinite stiffness (i.e. the ram is assumed rigid). Thus, according to Figure 23, WAM(l) is equal to the weight of the ram and XKAM(l) represents the capblock spring. However, a ram which impacts directly on an anvil must be represented by at least one concentrated weight and a weightless spring (Figure 25), since every weight must be separated from its neighboring weight by a spring.

The concentrated weight of the ram is WAM(l) and the associated spring constant is calculated by

XKAM(l) = K(RAM) = pie(DT)(DB)E/(4L)

where XKAM(l) = spring constant of ram for the diesel hammer to be inserted between the ram and anvil weights (kips/in), DT is the diameter of the top of the ram (in), DB is the diameter of the base of the ram, which comes in contact with the anvil (in), E is the modulus of elasticity of the ram material (ksi), and L is the length of the ram (in).

It has been found that the diameter of contact between the ram and anvil is usually around 1/10 of the full ram diameter, such that this equation becomes XKAM(l) = XKAM(RAM) = pi*E(DR)^2/(40L) where DR is diameter of the ram.

–  –  –

The idealization of the helmet and the anvil is similar to that of the ram in that they are ordinarily short bodies which can each be represented with sufficient accuracy by single rigid weights. The anvil is represented by WAM(2) and is considered a rigid weight, since it is relatively short. Appendix C shows the idealization and pertinent information for common hammers.

Ram Velocity at Impact The initial ram velocity, VELMI, of the ram for specific hammer types can be calculated as


–  –  –

where VELMI = initial ram velocity (ft/sec) he = equivalent stroke derived from bounce chamber pressure gage(ft) (he = E/WAM(l); E = Indicated Ram Energy) Work done on the pile by the diesel explosive force is automatically accounted for by including an explosive pressure as later shown. In the hammer idealization, note that the elements of the pile hammer are physically separated, i.e., the ram is capable of transmitting compressive force to the anvil but not tension. The same is true between the anvil and helmet and the helmet and the head of the pile. The program contains provisions for eliminating the ability of various elements to transmit these tensile forces. The mechanics of this provision are explained in the following sections.

Idealization of Cushions

The primary purpose of cushion material in a pile driver assembly is to limit impact stresses in both the pile and the ram. An inherent disadvantage to the use of cushions is that much of the available impact energy may be absorbed as the result of nonlinear load-deformation characteristics. The idealization of the cushion material consists of specifying a spring constant for the load-deformation characteristics and a coefficient of restitution for the energy absorbing characteristics.

It has been shown that a cushion material can be adequately described if its loaddeformation behavior is represented by two straight lines of different slope (Figure 26). The slope of the loading line is denoted as the spring constant of the cushion. The slope of the unloading line is determined from the cushion spring constant and coefficient of restitution of the cushion material such that the area enclosed by the two lines is proportional to the energy absorbed by the material. Appendix C gives values of cushion material constants. The spring constant of a

cushion can be calculated using its modulus of elasticity from Appendix C:

–  –  –

It should be noted that an exact description of the behavior of the cushion during driving is difficult because of cushion deterioration through heating, compaction, and wear during use. For this reason, further refinement in the idealization of the cushion does not seem warranted.

Average values for well compacted yet acceptable cushions were determined by field studies and are presented in Appendix C.

Idealization of the Pile The idealization of the pile is handled by breaking the pile into discrete segments. Each segment is represented by its weight and a spring representing its total stiffness.

Pile Segment Length To calculate the concentrated weights and spring constants for a pile, it is necessary to establish a criterion for segmenting the pile into discrete weights and springs. Piles should be divided into segments not to exceed approximately 10 feet in length. Sufficient solution accuracy is obtained if the pile is broken into approximately 10-foot lengths. A slight increase in solution accuracy is possible by using segment lengths of less than 10 feet, however, this usually is not justified because of the increase in solution time and the relative accuracy of the input data.

Further, it is desirable that the lengths of all the segments in the hammer-pile system be approximately equal.

Two different cases arise in segmenting a pile. One case is that of a pile with a uniform cross-section. The length of the pile may be such that it can be divided into an integer number of 10-foot segments. For the later case, the pile should be divided into an integer number of segments, the length of which are close to 10 feet. For example, a pile with a total length of 313.5 feet could be divided into 33 segments of 9.5 feet per segment or 31 segments of 10.113 feet per segment. Comparable solution accuracy would be obtained with either division scheme.

A second case arises when dealing with piles of nonuniform cross-section, where variations in cross-section do not occur at 10-foot intervals. For example, a pipe pile may have an 8-foot, 1-inch wall section, a 20-foot 1-inch wall section, and a 150-foot, 1-inch wall section for its makeup. Since it is desirable to have approximately equal segment lengths, the shortest segment length required by the cross-sectional variations will be used as a basis for dividing the other sections of the pile. Hence, the 8-foot section is considered to be the base segment length.

The 20-foot section can be divided into two segments of 10 feet each or three segments of 6.667 feet each. The 150-foot section can be divided into either 18 segments 8.333 feet each or 19 segments 7.895 feet each.

Pile Segment Weight

The weight of a segment of the pile can be calculated from:

WAM(I) = [A(I)][L(I)][d] where WAM(I) = weight of the I-th segment in the system (kips) A(I) = cross-sectional area of the I-th segment (in^2) L(I) = length of the I-th segment(in) d = unit density of the material (kips/in^3) Pile Segment Springs

The spring constant associated with a segment in a pile can be calculated from:

XKAM(I) = [A(I)][E]/[L(I)] where XKAM(I) = spring constant associated with the I-th spring in the system (kips) A(I) = cross-sectional area of the I-th spring (in^2) E = modulus of elasticity of the material (ksi)

–  –  –

Note that the number of springs in the pile equals one less than the number of corresponding weights.

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