PILE DRIVING ANALYSIS-SIMULATION OF HAMMERS,

CUSHIONS, PILES, AND SOIL

by
Lee Leon Lowery, Jr.
Assistant Research Engineer
T. J. Hirsch
Research Engineer
and
C. H. Samson, Jr.
Research Engineer

Research Report 33-9

Piling Behavior
Research Study No. 2-5-62-3'3

Sponsored by
The Texas Highway Department

in cooperation with the

U. S. Department of Transportation, Federal Highway Administration
Bureau of Public Roads

August 1967

TEXAS TRANSPORTATION INSTITUTE

Texas A&M University
College Station, Texas
 PREFACE
The information contained herein was developed on research study 2-5-62-33
entitled "Piling Behavior" which is a cooperative research study sponsored jointly
by the Texas Highway Department and the U. S. Department of Transportation,
Federal Highway Administration, Bureau of Public Roads. The broad objective of
this study is to fully develop the use of a computer solution of the wave equation so
that it may he used to predict driving stresses in piling and to estimate static load
hearing capacity of piling.
This report concerns itself with the following specific items in the work plan as
set forth in the study proposal:
. l. To determine the effect of .dynamic damping in concrete aU:d steel piling on
the impact longitudinal stress waves. This was accomplished by correlating theoreti-
cal stress waves with data obtained from full scale piles tested under controlled
conditions.
2. To study the dynamic load-deformation properties of cushioning materials
and their effect on the stress waves in piling. This was accomplished by correlating
theoretical stress waves with data from full scale pile tests under controlled condi-
tions. Theoretical results were compared with experimental data gathered for various
cushion materials.
3. To evaluate the true energy output of different pile driving hammers (single
acting steam hammers, double acting steam hammers, and open and closed end diesel
hammers) using the wave equation to analyze portions of data obtained by the Michi-
gan State Highway Commission and published in a report entitled "A Performance
Investigation of Pile Driving Hammers and Piles."
4. To determine a uniform basis of rating pile driver energy output applicable
to different type hammers.
5. To correlate the wave equation with suitable experimental test data.
During the course of investigation of the above items, the factors listed below
were also found to influence the wave equation results, and therefore were also in-
vestigated and are reported herein:
l. A study of the effect of ram elasticity on piling behavior.
2. A study of the influence of parameters used to describe soil behavior.
The information reported herein is necessary in order to understand the dynamic
behavior of piling and to properly simulate pile driving hammers, caphlocks and
cushion blocks, piles, and soils for wave equation analysis of piling behavior.
The opinions, findings· and conclusions expressed in this report are those of the
authors and not necessarily those of the Bureau of Public Roads.

iii
 LIST OF TABLES
Table Page
3.1 Effect of Breaking the Ram Into Segments When Ram Strikes a Cushion--------------------------- 5
3.2 Effect of Breaking Ram Into Segments When Ram Strikes a Steel AnviL ____________________________ 5
3.3 Summary of Belleville Cases Solved by Wave Equation _____________________ ---------------------- 7
3.4 Summary of Detroit Cases Solved by Wave Equation ______________________________________________ 8
3.5 Summary of Muskegon Cases Solved by Wave Equation ___________________________________________ 8
3.6 Effect of Cushion Stiffness on ENTHRU for BLTP-6; 10.0----------------------------------------- 9
3.7 Effect of Cushion Stiffness on FMAX for BLTP-6; 10.0------------------------------------------ 9
3.8 Effect of Cushion Stiffness on LIMSET for BLTP-6; 10.0 _________________________ ~--------------- 9
3.9 Data Reported in the Michigan Study ------------------------------------------------------------10
3.10 Summary of Results for Michigan Steam Hammers _______________________________________________ 12
3.11 Summary of Results for Michigan Diesel Hammers-------------------------------------------------14
3.12 Comparison of Energy Output Measured Experimentally With That Predicted by
Equation 3.6, for Diesel Hammers--------------------------------------------------------------16
3.13 Comparison of Measured Output With That Given by Equation 3.7, for Single Acting Steam Hammers ___ 16
3.14 Comparison of Measured Energy Output With That Predicted by Equation 3.11,
for Double Acting Steam Hammers--------------------------------------------------------------16
3.15 Summary of Hammer Properties and Operating Characteristics ______________________________________ 18
3.16 Effect of Removing Load Cell on ENTHRU, LIMSET, and Permanent Set of Pile _____________________ 19
3.17 Effect on ENTHRU Resulting From Removing the Load Cell Assembly _____________________________ 20
3.18 Effect of Cushion Stiffness on Maximum Point Displacement for Case BLTP-6; 10.0 and 57.9 __________ 20
3.19 Effect of Coefficient of Restitution on ENTHRU for Case BLTP-6; 10.0 and 57.9 ______________________ 20
3.20 Effect of Coefficient of Restitution on Maximum Point Displacement for Case BLTP-6; 10.0 and 57.9 _____ 21
3.21 Study of Various Hammers Driving the Same Pile_________________________________________________ 21
4.1 Suspended Pile Data--------------------------------------------------------------------------22
4.2 Dynamic Cushion Properties ___________________________________________________________________ 24
5.1 Dynamic Properties of New Cushion Blocks of Various Materials ___________________________________ 27
6.1 Comparison of Results Found by Using Elastic-Plastic vs Non-Linear Soil Resistance Curves ___________ 32
6.2 Influence of Soil Quake at Different Soil Resistances for Case BLTP-6; 57.9 With No Soil Damping _____ 33
6.3 Influence of Soil Damping on Different Soil Resistances for Case BLTP-6; 57.9
( Q = 0.1 for All Cases) ----------------------------------------------------------------------33

vi
 Pile Driving Analysis - Simulation of Hammers, Cushions,
Piles, and Soil

Chapter I
INTRODUCTION

General Background determine any additional information about the problem

being solved by referring to the original paper.
The problem of pile-driving analysis has been of
great interest to engineers for many years. Ever since
the first engineer proposed a method for predicting the Objectives
load carrying capacity of a pile, the whole subject of pile The objectives of this research were:
driving has become a much debated field in engineering.
In other areas new methods of analysis for structural l. To review and summarize Smith's original meth-
elements and systems are constantly being proposed with od of analysis and to derive a more general solution.
little or no resulting discussion. However, the proposal 2. To determine how the numerical solution is
of a new piling analysis is sure to stir much interest and affected by the elasticity of the ram.
often some rather heated discussions.
3. To determine the energy output of different type
Since over four-hundred pile-driving formulas have pile hammers.
been proposed, 1 not including the countless formula
modifications which are used, 2 many engineers resort to 4. To compare results given by the wave equation
the use of only one or two formulas regardless of the with those determined by laboratory experiments and
driving conditions encountered. 3 Although many of the field tests.
erroneous assumptions made in these formulas have been
widely discussed, 4 •5 the fact that they omit many signifi- 5. To illustrate the significance of the parameters
involved, including cushion stiffness and damping, ram
cant parameters which affect the problem seems to have
received less attention. However, when the driving for- velocity, material damping in the pile, soil damping and
quake, and to determine the quantitative effect of these
mulas omit parameters which change from case to case,
parameters where possible.
the engineer has no means of determining how signifi-
cant the parameter may be, nor can he tell in which 6. To show how the wave equation can be used to
direction or to what extent the change will vary the determine the dynamic or impact characteristics of the
results. Thus, to obtain an accurate solution obviously materials involved.
requires that fewer erroneous assumptions be made re-
garding the dynamic behavior of the materials and equip- 7. To determine the dynamic properties of the
ment used in pile driving, and that all significant param- cushion subjected to impact loading.
eters are included in the analysis. 8. To study the effect of internal damping in the
The first of these problems was solved when it was pile and its significance.
noted that pile driving is actually a case of longitudinal
impact, governed by the wave equation rather than by Literature Review
statics or rigid-body dynamics. 6 •7 However, since the
exact simulation and solution of the wave equation ap- The basic purpose of any pile driving formula is to
plied to piling are extremely complex for all but the permit the design of a functional yet economical foun-
dation. According to Chellis, 9 there are four basic types
simplest problems, many significant parameTers still had
to be neglected. of driving formulas:
l. Empirical formulas, which are based on statisti-
The second problem was solved by Smith8 who cal investigations of pile load tests,
proposed a numerical solution to the wave equation,
capable of including any of the known parameters in- 2. Static formulas, which are based on the side
volved in pile-driving analysis. This method of analysis frictional forces and point bearing force on the pile, as
was applicable to tapered, stepped, and composite piles, determined by soils investigations,
to nonlinear soil resistances and damping, to piles with 3. · Dynamic formulas, which assume that the dy-
cushions, followers, helm.ets, etc. In other words, it was namic soil resistance is equal to the static load capacity
a completely general method of analysis for the problem of the pile, and
of pile driving.
4. The wave equation, which assumes only those
It should be noted that much of the experimental material properties whose dynamic behavior is not com-
work used in this report was reported by other investi- pletely understood and has not yet been determined
gators. These cases are referenced, and the problem experimentally. Each of the preceding formulas has
number or name used herein will be the same as used advantages and disadvantages which have been widely
by the original reporter. This will enable the reader to noted10 •11 and need not be restated at this time.

PAGE ONE
 Isaacs is thought to have first noted that the wave Immediately after the appearance of Smith's paper
equation is applicable to the problem of pile driving.1 2 in 1960, the Bridge Division of the Texas Highway
However, Fox13 was probably the first person to propose Department initiated a research project with the Texas
that an exact solution be used for pile-driving analysis. Transportation Institute to perform exhaustive studies
Shortly thereafter, Glanville, Grime, Fox, and Davies 14 of the behavior of piling by the wave equation. The first
published the first correlations between experimental report dealt with a computer program based on Smith's
studies and results determined by the exact solution to numerical solution. 24 This program was used to· deter-
the wave equation developed by Fox. Since this exact mine the driving stresses induced in a number of pre-
solution was extremely complex, they were forced to use stressed concrete piles which had failed during driving, 25
simplified boundary conditions including zero side fric- and later to check the conditions at similar sites at which
tional resistance, a perfectly elastic cushion block, and pile breakage due to excessive driving stresses might be
an elastic soil spring acting only at the tip o·f the pile. experienced. 26
However, even using these simplified boundary condi-
Forehand and Reese2 8 investigated the possibility
tions, they obtained reasonably accurate results.
of predicting the ultimate bearing capacity of piling
In 1940 Cummings 15 discussed several errors in- using the wave equation, but since complete data were
herent in dynamic pile-driving formulas and reviewed available for relatively few problems, they were unable
the previous work done using the wave equation. How- to draw many firm conclusions. They also studied the
ever, he also noted that even for the simplest problems, dynamic action of the soil during driving and recom-
"the complete solution includes long and complicated mended some values for the soil parameters used in the
mathematical expressions so that its use for a practical wave equation.
problem would involve laborious numerical calculations." In August, 1963 several extensions of Smith's method
A practical pile-driving ·problem usually involves were presented by the writers. 29 . Two simple cases for
side frictional soil resistance, soil damping constants, which "exact" solutions were known were compared
nonlinear cushion and capblock springs, and other fac- wifh Smith's numerical solution to indicate the method's
tors which prevent a direct solution of the resulting accuracy. A third section of the paper presented the
differential equation. However, in 1950 Smith 16 pro- results of a short parameter study which indicated how
posed a mathematical model and a corresponding nu- certain trends in pile driving might be determined and
merical method of analysis which accounted for the how to study the significance of various parameters.
effects of many of these parameters. He has continued The results for several theoretical and field test problems
to update this method and published various other were also compared.
works.17,18,19,20,21 In 1963 the writers30 published a study on the
methods employed in measuring dynamic stresses and
Smith's method of analysis did not really become displacements of piling during driving, and presented
popular until 1960 when he published a summary of the further experimental and theoretical comparisons "to
method's application to the problem of pile-driving analy- demonstrate that the computer solution of the wave
sis.22 In this paper he recommended a number of ma- equation offers a rational approach to· the problems
terial constants and the material behavior curves re- associated with the structural behavior of piling during
quired to account for the dynamic action of the soil, driving." This report was based on an earlier study
cushion, and pile material. dealing with driving prestressed concrete piles. 31
Smith's method of analyzing pile-driving problems An investigation by Hirsch 32 involved a study of the
received considerable interest, 23 and two immediate ap- variables which affected the behavior of concrete piles
plications of the wave equation were suggested: during driving. Over 2100 separate problems were
l. The immediate application of the wave equation, solved and the results were presented in the form of
using the most probable material properties to predict graphs for use by design engineers.
ultimate driving resistance ·and driving stresses. Later publications dealt with the dynamic load-
2. Its use to perform extensive parameter studies deformation properties of various pile cushion materials
in order to determine trends and to gain more insight and other dynamic properties of materials required to
into the behavior of pile driving, and also determine the simulate as closely as possible the actual behavior of
relative significance of these parameters. a pile during driving. 33 ·34 ·35 •36

Chapter II
A NUMERICAL METHOD OF ANALYSIS

The Basic Solution labor, and for most practical cases, required many sim-
plifying assumptions.
Since 1931, it has been realized that pile driving
involved theories of longitudinal impact rather than In 1950, Smith37 proposed an approximate solution
statics. However, the application of the wave equation based on concentrating the distributed mass of the pile,
to pile driving was restricted to very simple problems as shown in Figure 2.1a, into a series of small weights,
because the exact solution was complex, involved much W ( 1) thru W (MP) , connected by weightless springs

PAGE TWO
 J(m) is a damping constant for the soil acting on seg-
ment number (m) (sec/ft); g is the gravitational accel-
eration (ft/sec 2 ); and W(m) is the weight of segment
number m (lb) .
The solution is begun by initializing the time-
dependent parameters to zero and by giving the ram an
initial velocity. Then an incremental amount of time
At elapses during which the ram moves down an amount
given by Equation 2.1. The displacements D ( m,I) of
the other masses are computed in the same manner.
Equation 2.2 is then used to determine the com-
pressions C(m,I), after which the internal spring forces
acting between the masses are found from Equation 2.3
and the external soil forces R(m,I) are computed from
Equation 2.4.
SIDE Finally, a new velocity V(m,I) is determined for
FRICTIONAL
RESISTANCE each mass using Equation 2.5, after which another time
interval elapses. New displacements, compressions,
forces, and velocities are again computed using the same
equations and the cycle is repeated until the solution is
obtained. Smithau and others,~n.H give a detailed expla-
nation of this method of solution and the computer·pro-
NOTE' K (m)•INTERNAL
SPRING CONSTANT FOR
gramming required. The dynamic behavior of various
SEGMENT m. parameters will be discussed later.
K'!ml• SOIL SPRING
CONSTANT FOR Smith would have probably caused little interest had
he simply given a numerical solution for the wave equa-
tion. Instead he presented a simple, physical model,
(A) ACTUAL PILE (6) I OEALIZED PILE easily visualized, using parameters which are readily
understood. This and the simplicity of the equations
Figure 2.1. Idealization of a pile for purpose of aTUJJlysis. required for a solution doubtlessly account for much
of the wave equation's increasing popularity as a means
of studying the behavior of piling.
K{l) thru K(MP-1), with the addition of soil re-
sistance acting on the masses, as illustrated in Figure
2.1b. Time also was divided into small increments. This Modifications of the Original Solution
numerical solution was then applied by the repeated use Although the original method of analysis proposed
of the following equations, developed by Smith :~ 8 by Smith can be used to solve many of the problems
D(m,t) = D(m,t-1) + 12AtV(m,t-1) Eq. 2.1 given in this report, it has been greatly extended to in-
clude other idealizations. The major additions and
C(m,t) D(m,t) - D(m + 1,t) Eq. 2.2 changes are summarized here for reference only, and are
fully discussed in later chapters.
F(m,t) C(m,t)K(m) Eq. 2.3
l. The relationship between soil resistance to pene-
R(m,t) [D(m,t)-D' (m,t)] tration of the pile was originally limited to a series of
K' (m) [1+J(m)V(m,t-1)] Eq. 2.4 straight lines. The revised program allows the use of
any shape for this curve, as noted in Chapter VI.
V(m,t) = V(m,t-1)+[F(m,t)-R(m,t)]
gAt/W(m) Eq. 2.5 2. The elastic soil deformation "Q" and the soil
damping constant "J" were each limited to one value at
where m is the mass number, t denotes the time interval the point of the pile and a second value for side resist-
number, At is the size of the time interval (sec), D(m,t) ance. These parameters have been generalized to include
is the total displacement of mass number m during time different values at each pile segment.
interval number t(in.), V(m,t) is the velocity of mass 3. A new method by which internal damping in the
m during time interval t(ft/ sec), C (m,t) is the compres- pile can be accounted for is now included. This method
sion of spring m during time interval t(in.), F(m,t) is is explained in Chapter V.
the force exerted by spring number m between segment
numbers ( m) and ( m + t) during time interval t (lb) , 4. A second method is included to account for the
and K(m) is the spring rate of mass m (lb/in.). Note coefficient of restitution of the capblock or cushionblock.
that since certain parameters do not change with time,
they are assigned single rather than double subscripts. 5. For correlation with experimental data, it is now
possible to place forces directly on the head of the pile
The quantity R ( m,t) is the total soil resistance act- rather than having to calculate them from the hammer-
ing on segment m (lb/in.) ; K' (m) is the spring rate of cushion-anvil properties. This method was used exten-
the soil spring causing the external soil resistance force sively where the force vs time curve at the head of the
on mass m(lb/in.); D(m,t) is the total inelastic soil pile was known; since then the hammer, cushion, and
displacement or yielding during the tat segment m(in.); anvil properties did not influence the solution.
PAGE THREE
 6. The linear force vs compression curve for vari- parameters between specified limits in order to study
ous cushion materials used previously has been general- their influence on the solution, and to see if trends could
ized as noted in Chapter IV. be found.
7. The effect of gravity on the solution can now 9. For possible later use, several pile-driving for-
be accounted for. mulas were included in the computer program.
8. A special "parameter study" sub-program was 10. The soil resistance on the point segment now
written and included in the general program. This fea- uses two springs, one for the side friction acting on the
ture was used to vary specific parameters or groups of side of the pile and a second spring for point bearing.

Chapter III
PILE DRIVING HAMMERS

Ram Idealization divided into a series of weights and springs. However,

Smith42 suggests that since the ram is usually short no work has been done to determine how long the ram
in length, in many cases it can accurately be represented can be before its elasticity affects the accuracy of the
by a single weight having infinite stiffness. The exam- solution. The most common hammers in the above
ple illustrated in Figure 2.1 makes this assumption since class include drop, air, and steam hammers. Figures
K(l) represents the spring constant of only the cap 3.1 and 3.2 show how the ram may be idealized.
block, the elasticity of the ram having been neglected. In order to determine the significance of dividing
He also notes that where greater accuracy is desired, the ram into a number of segments, several ram lengths
or when the ram is long and slender, it can also be ranging from 2 to 10 ft were assumed, driving a 100-ft

W (I)
w (I)
K (I) K (I)
w (2) w (2)
~ I
I
K (2)

LONG RAM DIVIDED

lf
LONG RAM DIVIDED
INTO 'NR' SEGMENTS INTO 'NR' SEGMENTS
K (NR-1)
W (NR)
W (NR)
CUSHION BLOCK
K (NR) K (NR)
W (NR+I) TEEL ANVIL W (NR+I)
K (NR+I)
W (NR-t2) CUSHION BLOCK

*
K (NR .. I)
K (NR+2)

~
W (NR+2)
K (NR+2)
W (NR+3)
PILE PILE

K (MP-2)
W (MP-1)
K (MP-1) K (MP-1)
W (MP) W (MP)
K'(MP) K I (MP)

· Figure 3.1. Idealization for a long ram striking directly Figure 3.2. Idealization for a long ram striking directly
on· a cushion block. · on a steel anvil. ,
PAGE FOUR
 TABLE 3.1. EFFECT OF BREAKING THE RAM INTO occurs between two steel elements. One possible solu-
SEGMENTS WHEN RAM STRIKES A CUSHION
tion is to place the spring constant of the entire ram
Maximum Maximum Maximum between the weights representing the ram and anvil.
Length Compressive Tensile Point Also, the ram can be broken into a series of weights
Number of Pile Force Force Displace- and springs as is the pile.
of Ram Segments in Pile in Pile ment
Divisions (ft) (kip) (kip) (in.) To determine when the ram in this case should be
divided, a parameter study was run in which the ram
1 1.25 263.1 219.0 3.057 length varied between 6 and 10 ft and the anvil weight
2 1.25 262.6 218.8 3.058
10 1.25 262.9 218.5 3.059 from 1,000 to 2,000 lb. In each case, the ram diameter
was held constant and the ram was divided into equal
segment lengths as noted in Table 3.2. These variables
were picked because of their possible influence on the
pile with point resistance only. For this parameter solution.
study the total weight of the pile varied from 1,500 lb
to 10,000 lb, while the ultimate soil resistance ranged The pile used was again a 12H53 point bearing pile
from zero to 10,000 lb. The cushion was assumed to with a cushion of 2,000 kip/in. spring constant placed
have a stiffness of 2,000 kip/in. between the anvil and head of the pile. The soil parame-
ters used were RU[loint = 500 kip, Q = 0.1 in., and
Table 3.1 lists the results found for a typical prob- J = 0.15 sec/ft. These factors were held constant for
lem solved in this series, the problem consisting of a all problems listed in Table 3.2.
10-ft ram traveling at 20ft/sec, striking a cushion having
a stiffness of 2,000 kip/in. The pile used was a 100-ft The most obvious result shown by Table 3.2 is that
12H53 steel pile, driven by a 5,000-lb ram with an when the steel ram impacts directly on a steel anvil,
initial velocity of 12.4 ft/sec. · No pile cap was included dividing a long ram ( 6, 8 and 10 ft) into segments has
in the solution, the cushion being placed directly between a significant effect on the solution.
the hammer and the head of the pile. Since the ram
was divided into very short lengths, the pile was also Energy Output of Hammer
divided into short segments.
One of the most significant parameters involved in
As shown in Table 3.1, the solution is not changed pile driving is the velocity of the ram immediately before
to any extent, regardless of whether the ram is divided impact. This velocity is often used to determine the
into 1, 2, or 10 segments. The time interval Llt was maximum kinetic energy of the hammer and its energy
held constant in each case. output rating, and must be known or assumed before
the wave equation or dynamic formulas can be applied.
In certain hammers such as a diesel hammer, the
ram strikes directly on a steel anvil rather than on a Although the manufacturers of pile-driving equip-
cushion. This makes the choice of a spring rate be- ment furnish maximum energy ratings for their ham-
tween the ram and anvil difficult because the impact mers, these are usually downgraded by foundation ex-

TABLE 3.2. EFFECT OF BREAKING RAM INTO SEGMENTS WHEN RAM STRIKES A STEEL ANVIL

Length Maximum Compressive

Number of Each Force on Pile Maximum
Anvil Ram of Ram Ram At At At Point
Weight Length Divisions Segment Head Center Tip Displacement
(lb) (ft) (ft) (kip) (kip) (kip) (in.)

2000 10 1 10 513 513 884 0.207

2 5 437 438 774 0.159
5 2 373 373 674 0.124
10 1 375 375 678 0.125
8 1 8 478 478 833 0.183
4 2 359 359 648 0.117
8 1 360 360 651 0.118
6 1 6 430 430 763 0.155
3 2 344 344 621 0.110
6 1 342 342 616 0.109
1000 10 1 10 508 509 878 0.160
2 5 451 451 789 0.159
5 2 381 382 691 0.151
10 1 371 372 681 0.153
8 1 8 487 488 846 0.151
4 2 443 444 785 0.144
8 1 369 370 675 0.134
10 0.8 337 338 665 0.133
6 1 6 457 457 798 0.137
3 2 361 362 666 0.128
6 1 316 316 562 0.109
10 0.6 320 320 611 0.113

PAGE FIVE
perts for various reasons. A number of conditions such
as poor hammer condition, lack of lubrication, and wear
seriously reduce the energy output of a ·hammer. In
addition the energy of many hammers can be controlled
by regulating the steam pressure or diesel fuel. To
determine how much the rated energy of any given
hammer should be reduced is not a simple task.
Che1lis 48 discusses several reasons for this energy 1-E---- HAMMER BASE
reduction and recommends a number of possible effi- r----CUSHION
ciency factors for the commonly used hammers, based
on his observations and experience. ---HELMET

The Michigan Study of Pile Driving Hammers

In 1965 the Michigan State Highway Commission 44 ~~~~---ACCELEROMETER
completed an extensive research program designed to
obtain a better understanding of the complex problem ~~~2r--HELMET
of pile driving. Though a number of specific objectives
were given, one was of primary importance. As noted t - c - - - - PIPE ADAPTER
by Housel,4 5 "Hammer energy actually delivered to the
pile, as compared with the manufacturer's rated energy,
was the focal point of a major portion of this investiga-
tion of pile-driving hammers." In other words, they
hoped to determine the energy delivered to the pile and
to compare these values with the manufacturer's ratings.
---sOIL
The energy transmitted to the pile was termed
"ENTHRU" by the investigators 44 and was determined
by the summation
ENTHRU = X F .Ll S ~PIPE PILE CLOSED AT TIP
Where F, the average force on the top of the pile during
a short interval of time, was measured by a specially
designed load cell, and .Ll S, the incremental movement
of the head of the pile during this time interval, was Figure 3.3. Typical pile driving assembly (after refer-
found using displacement transducers and/ or reduced ence 44).
from accelerometer data. It should be pointed out that
ENTHRU is not the total energy output of the hammer
blow, but only a measure of that portion of the energy
~ } W(l} =RAM WEIGHT.

delivered below the load-cell assembly.

~-- K(l} =SPRING RATE OF

Since so many variables influence the value of

ENTHRU, :md since some of these variables were chang-
ing during the pile driving operation (e.g., condition of
the cushion, soil resistance, etc.) , the investigators were
w} .,. ';,;;~1,t
not able to determine the total energy output of the 1 -
r K(2): SPRING RATE OF
LOAD CELL.
hammer. As noted in the Michigan report :46 "Hammer
type and operation conditions; pile type, mass, rigidity,
and length; and the type and condition of cap blocks
were all factors that affected ENTHRU, but when, how,
and how much could not be ascertained· with any degree
of certainty." However, the wave equation can account
for each of these factors so that their effects can be
determined.
The Michigan report also noted that ENTHRU was
t1 W(3): WEIGHT OF 1/2 LOAD
CELL + UNIVERSAL
CAP + PIPE ADAPTER
( WHEN USED).

not actually a direct measurement of the hammer's effi-

ciency or energy output since the forces and displace-
ments were measured below the capblock, as shown in
Figure 3.3. Thus, ENTHRU was defined as "the amount
of work done on the load cell." SPRING RATE OF SOIL~
IN SHEAR ALONG K'(MP-1)
SEGMENT MP-1
The maximum displacement of the head of the pile
was also reported and was designated LIMSET. Oscillo-
graphic records of force vs time measured in the load
cell were also reported. ·Since force was measured only
at the load cell, the single maximum observed values
for each case will be called FMAX .. Figure 3.4. Idealization of a Vulcan hammer.
PAGE SIX
Problem Information
In selecting which of the Michigan pile problems to ~ } W(l) • RAM WEIGHT.

solve by the wave equation, it was decided to run at least

two problems for each hammer used at each of the three I_
r-- K(l). SPRING RATE OF
THERAM.
testing sites. As shown in Table 3.3, cases selected from
the Belleville site include two pile lengths for each of
four different hammers. Otherwise, the problems were
& } W(2) • ANVIL WEIGHT.

selected at random and the hammer energies determined

are not necessarily typical of the hammer's usual ,l,_ K(2)• SPRING RATE OF
~ CUSHION.
operating characteristics. Similarly, the Detroit and
Muskegon site problems are summarized in Tables 3.4
and 3.5. Figures 3.4 and 3.5 illustrate how these prob-
lems were idealized for purposes of analysis.
kid}
~~
W(3)• ~ti~~~ g~p +
1/2 LOAD CELL.

Even though the Michigan study is one of the most L _ K ( 3 ) • SPRING RATE OF
completely documented and fully reported research proj- 1· THE LOAD CELL.

t} t5lgHMr+1G~~J~~SAL
ects published concerning pile driving, certain informa-
tion was not reported which must be known in order to 4
apply the wave equation. This omission was not the W( ) •
CAP + PIPE ADAPTER
result of any failure in reporting the data, but was be- (WHEN USED).
cause this information was not required by the methods

F
of analysis used in the Michigan project and would have K(4) • SPRING RATE OF
been difficult to measure. Two examples are the lack FIRST PILE SEGMENT.
of information concerning the stiffness of the cushion W(5) • WEIGHT OF PILE
SEGMENT.
and the velocity of the ram at impact.

Preliminary Studies ~
} W(MP-1)• WEIGHT OF PILE
Since cushion-block information was not given, and SPRING RATE OF SOIL} SEGMENT.
IN SHEAR ALONG K'(MP-1)
because the cushion stiffness varies greatly during driv- SEGMENT MP ·I 1 - - - - - K(MP-1) • SPRING RATE OF
PILE SEGMENT.
ing, a broad parameter study was made using the first
W (MP) • WEIGHT OF FINAL
case mentioned in Table 3.3. In this study, the cushion SPRING RATE OF S111L~
}
PILE SEGMENT.
IN SHEAR ALC.NG K'(MP)
stiffness was varied by a factor of 50, from 540 kip;/in. SEGMENT MP SPRING RATE OF
up to 27,000 kip/in. Also studied was the effect of ~~E%-~E:.f~~~NT
varying the total soil resistance, RUT, using resistances
of 30, 90, and 150 kip and ram velocities of 8, 12, and
1 - - - - - K '(MP•Il
D MP.

16 ft/sec. Figure 3.5. Idealization of a diesel hammer.

TABLE 3.3 SUMMARY OF BELLEVILLE CASES SOLVED BY WAVE EQUATION

PILE INFORMATION
Total Embedded
Pile Length Length
I.D. Case* Hammer** Cushion Type (ft) (ft)

BLTP-6 10.0 V-1 Oak 12H53 32.5 10.0

57.9 V-1 Oak 12H53 72.5 57.9
BLTP-4 25.0
66.4 •
LB-312
LB-312
Micarta
Micarta
r-;n.1
Pipe
0.25
40.7
81.6
15.0
56.4
in.
wall
J
BRP-4

BLTP-5
20.0
50.0
15.0

66.0
M-DE30
M-DE30
D-D12

D-D12
Oak
Oak
German
Oak
German
Oak
r12H53
12H53

Pipe;n.}
0.179
in.
wall
40.0
60.0
40.0

80.0
20.0
50.0
5.0

50.0

*Case number indicates pile length below ground surface and not necessarily embeddment.
**Hammer designations are as follows:
V-1 Vulcan 1
V-.50C Vulcan 50C
V-80C Vulcan 80C
LB-312 Link Belt 312
LB-520 Link Belt 520
M-DE30 McKiernen-Terry DE-30
M-DE40 McKiernen-Terry DE-40
D-D12 Delmag D-12
D-D22 Delmag D-22
PAGE SEVEN
 TABLE 3.4. SUMMARY OF DETROIT CASES SOLVED BY WAVE EQUATION

PILE INFORMATION
Total Embedded
Pile Length Length
I. D. Case* Hammer* Cushion Type (ft) (ft)

DLTP-8 41.5 V-1 Oak 12H53 80.1 41.5

80.2 V-1 Oak 12H53 97.0 80.2

ri~J
DTP-5 20.0 V-50C Micarta 40.0 20.0
Pipe
0.179
79.0 V-50C Micarta in. 84.0 79.0
wall
DRP-3 40.0 LB-312 Micarta 12H53 80.0 40.0
LB-312 80.0

in)
60.0 Micarta 12H53 60.0
DTP-13 40.0 M-DE30 Oak
r12Pipe 45.0 40.0

80.7 M-DE30 Oak

l 0:179
In.
wall
90.7 80.7

DTP-15 20.0 D-D12 German 12H53 46.1 20.0

80.5 D-D12 Oak 12H53 86.1 80.5

*See Table 3.3 for notation.

The results of this study indicate the significance l) ENTHRU is nearly independent of the cushion
of the wave equation in helping to understand the many block stiffness used, since the cushion stiffness was in-
factors that affect pile-driving behavior. The solutions creased by a factor of 50 while influencing ENTHRU
for ENTHRU, FMAX, and LIMSET resulting from a only slightly,
change in the cushion stiffness, soil resistance, and ram
velocities are given in Tables 3.6, 3.7, and 3.8, respec· 2) FMAX is almost completely independent of the
driving resistance,
tively. Whereas before it could not be determined
"when, how, or how much," the results of this study 3) FMAX is almost linearly related to the ham-
indicate that in general for these particular problems, mer velocity, and

TABLE 3.5. SUMMARY OF MUSKEGON CASES SOLVED BY WAVE EQUATION

PILE INFORMATION
Total Embedded
Pile Length Length
J.D. Case* Hammer* Cushion Type (ft) (ft)

MLTP-2 20.0 V-1 Oak 45.0 20.0

rml Pipe

r
0.250
53.0 V-1 Oak in. 60.0 53.0
wall
MLTP-9 72.0

127.0
V-80C

V-80C
Micarta

Micarta
in,)
Pipe
0.250
in.
80.0

134.0
. 72.0

127.0
wall
MTP-12 30.5

70.8
LB-520

LB-520
Micarta

Micarta
I'l 'in.)
Pipe
0.250
in.
40.0

80.0
30.5

70.8
wall
MTP-11 69.5

150.0
M-DE40

M-DE40
Oak
and
Plywood rm.)
l
Pipe
0.250
in.
wall
80.0

165.0
69.5

150.0

MLTP-8 31.0

178.0
D-D22

D-D22
German
Oak
German
Oak
rm.)Pipe
0.250
in.
wall
40.0
185.0
185.0
31.0

178.0

*See Table 3.3 for notation.

PAGE EIGHT
TABLE 3.6. EFFECT OF CUSHION STIFFNESS ON Investigation of Steam Hammers
ENTHRU FOR BLTP-6; 10.0
Used in the Michigan Study
ENTHRU (kip ft)
As noted in Figure 3.4, the numerical solution to
Ram Cushion Stiffness (kip/in.)
Velocity RUT the wave equation uses a series of concentrated weights
(ft/sec) (kip) 540 1080 2700 27,000 and springs which closely represent the actual system
involved. Time is also divided into small intervals in
30 3.0 3.0 3.0 2.9 order to arrive at a solution.
8 90 3.1 3.2 3.3 2.9
150 3.0 3.2 3.3 3.0 As shown by Smith, 1 the wave equation can be used
to determine (among other quantities) the displacement
30 6.6 6.4 7.1 6.4
12 90 7.0 7.1 7.2 6.4 D(m,t) of any mass "m" at time "t", as well as the force
150 6.9 7.2 7.4 6.7 F(m,t) of any mass "m" at time "t." Thus the equa-
tion for ENTHRU at any point in the system can be
30 11.8 11.9 12.2 11.3
16 90 12.3 12.6 12.8 11.5 determined by simply letting the computer calculate the
150 12.4 12.9 13.2 11.4 equation previously mentioned:

ENTHRU = ~ F .!l S
or using the wave equation terms:
TABLE 3.7. EFFECT OF CUSHION STIFFNESS ON
FMAX FOR BLTP-6; 10.0 ENTHRU(m)
""[F(m,t)
~
+ F(m,t-1)
2.0 - ]
FMAX (kip) [D(m+1,t) - D(m+1;t-1)]
Ram Cushion Stiffness (kip/in.)
Velocity RUT
(ft/sec) (kip) 540 1080 2700 27,000 where ENTHRU(m) the work done on any weight
(m+1),
30 132 185 261 779
8 90 137 185 261 779 m the mass number, and
150 143 186 261 779
the time interval number.
30 198 278 391 1,169
12 90 205 278 391 1,169 For example, the Michigan report determined
150 215 279 391 1,169 ENTHRU (2) for the idealized system shown in Figure
30 264 371 522 1,558 3.4, since they recorded forces F ( 2,t) in the load cell
16 90 275 371 522 1,558 and displacements D (3,t) below the load cell. For the
150 288 371 522 1,558 system in Figure 3.5, ENTHR U ( 3) was determined.
Although it may not have been possible, ENTHRU
should actually have been measured directly under the
driving hammer ENTHRU(1), since ENTHRU(3) is
greatly influenced by several parameters, especially the
TABLE 3.8. EFFECT OF CUSHION STIFFNESS ON
LIMSET FOR BLTP-6; 10.0 type, condition, and coefficient of restitution of the
cushion, and the weights of the extra driving cap and
LIMSET (in.) load cell.
Ram Cushion Stiffness (kip/in.) As will be shown later, the coefficient of restitution
Velocity RUT
(ft/see) (kip) 540 1080 2700 27,000 alone can change ENTHRU (2) by 20%, simply by
changing e from 0.2 to 0.6. Nor is this variation in e
30 1.09 1.08 1.08 1.13 unlikely since cushion condition varied from new to
8 90 0.44 0.44 0.45 0.45 "badly burnt" and "chips added."
150 0.32 0.33 0.33 0.33
The wave equation was therefore used to analyze
30 2.21 2.14 2.19 2.25
12 90 0.80 0.82 0.84 0.84 the problems since what was needed was a method by
150 0.55 0.57 0.58 0.58 which the available data (ENTHRU, LIMSET, FMAX,
etc.) could be used to determine the actual hammer
30 3.62 3.59 3.63 3.68
16 90 1.30 1.31 1.32 1.34 energy involved, and also to compensate for the influence
150 0.85 0.87 0.88 0.90 of cushion stiffness, e, additional driving cap weights,
driving resistance encountered, etc.

4) FMAX consistently increases as the cushion Method Used to Correlate Theoretical

stiffness increases.
and Experimental Results
Thus for the first time, a number of trends may be
established for various pile driving situations by using In order to get the best possible correlation between
the wave equation. experimental and theoretical solutions, an iterative meth-
od was used. This approach was suggested by the pre-
In order to analyze other of the Michigan problems, liminary studies mentioned earlier. To demonstrate the
certain data given in the Michigan report were used. method, an example problem, BLTP-6;10.0, will be
This information is listed in Table 3.9. solved.
PAGE NINE
 TABLE 3.9. DATA REPORTED IN THE MICHIGAN STUDY

Manufacturer's
Maximum Estimated
Rated Permanent Static Soil
Driving Pile Hammer* Energy ENTHRU LIMSET Set Resistance
Location I.D. Case Type (ft lb) (ft lb) (in.) (in.) (kip)

Belleville BLTP-6 10.0 V-1 15,000 6,380 0.75 0.48 48

57.9 V-1 15,000 4,440 0.42 0.02 400
BLTP-4 25.0 LB-312 18,000 8,010 0.94 0.36 140
66.4 LB-312 18,000 11,200 0.92 0.02 690
BRP-4 20.0 M-DE-30 22,400 4,980 0.57 0.37 100
50.0 M-DE-30 22,400 4,470 0.41 0.12 320
BLTP-5 15.0 D-D12 22,500 9,040 1.86 1.43 80
60.0 D-D12 22,500 9,930 0.79 0.11 340
Detroit DLTP-8 41.5 V-1 15,000 5,760 1.22 1.00 60
80.2 V-1 15,000 4,540 0.54 0.50 360
DTP-5 20.0 V-50C 15,100 2.55 2.00 22
79.0 V-50C 15",100 11,420 0.82 0.09 235
DRP-3 40.0 LB-312 18,000 7,060 1.36 1.25 60
60.0 LB-312 18,000 6,620 1.41 0.77 76
DTP-13 40.0 M-DE30 22,400 9,100 2.21 2.00 30
80.7 M-DE30 22,400 9,480 1.12 O.G7 265
DTP-15 20.0 D-D12 22,500 10,100 2.07 2.00 40
80.5 D-D12 22,500 5,480 0.58 0.25 120
Muskegon MLTP-2 20.0 V-1 15,000 7,210 1.42 1.00 80
53.0 V-1 15,000 4,870 0.57 0.09 200
MLTP-9 72.0 V-80C 24,450 14,660 1.06 0.56 160
127.0 V-80C 24,450 . 13,110 1.03 0.23 470
MTP-12 30.5 LB-520 30,000 14,860 1.48 1.00 40
70.8 LB-520 30,000 13,140 1.02 0.77 156
MTP-11 69.5 M-DE40 32,000 16,760 1.16 0.67 160
150.0 M-DE40 32,000 17,900 1.41 0.05 500
MLTP-8 31.0 D-D22 39,700 25,500 2.35 1.25 40
178.0 D-D22 39,700 22,050 1.71 0.04 988

*See Table 3.3 for notation.

Since in nearly every case the condition of the in Table 3.9) ? By entering ENTHRU = 6,380 kip ft,
cushion is unknown, the first assumption must be for the and assuming RUT = 30 kip, project to the upper curve
cushion rate K(1). For illustrative purposes, assume where EINPUT is found to be 11,000 kip ft.
that K(1) = 180 kip/in. and that soil resistances of
30 and 90 kip were assumed. To further check the solution, determine the ram
velocity required for 11,000 kip ft of kinetic energy from:
The next step was to run the problem with various
hammer energies. As shown in Figure 3.6, for each
V= " I (EINPUT) (64.4)
= 11.9 ft/sec
energy input (EINPUT) the wave equation predicts a
corresponding theoretical value of ENTHRU. These
V Ram Weight
solutions are then used to plot the curves of Figure 3.6. Next, from Table 3.9, find the actual value of LIMSET
Also, since each solution predicts a value for LIMSET (determined experimentally) and enter this value of 0. 75
and initial ram velocity, it is possible to plot the curves in. and V = 11.9 ft/sec in Figure 3.7.
of Figure 3.7. Should the projection of these points intersect on
Returning to Figure 3.6, the question becomes what the RUT = 30 kip curve, then that assumption was
kinetic energy must the falling ram have had in order correct. However, this indicates a soil resistance of
to cause a value of ENTHRU = 6,380 kip ft (the meas- around 90 kip so that the RUT= 90 kip curve of Figure
ured experimental value reported by Michigan and listed 3.6 should probably have been used.
PAGE TE;:N
 20 lb (ENTHRU) of the 10,100 ft lb output reached the
load cell, the difference must have been lost in the helmet-
cushion-load cell assembly. Thus the efficiency of this
assembly must have been (6,380) X (100) /10,100 =
63 percent.
The ability to determine these efficiencies separately ·
is important since it indicates whether the driving ham-
mer or cushion-helmet assembly should be studied to
15 reduce energy losses during driving.
The preceding method was used to solve each of the
Michigan steam hammer cases listed in Tables 3.3, 3.4,
E INPUT Required To Give and 3.5.
ENTHRU Of 6380 ft lb

U,Q99_ft!~~-------
Correlation of Experimental
-
~

~
"-
____
____ J911.Q.Q_f_t_l_!l__ _________ ; (
and Theoretical Results
It is interesting to compare the final wave equation
10
1- I solution with the experimental results reported in the
:::>
ll.
z /i I
I
Michigan pile study. For the above case, comparisons
between the experimental results and those given by the
w I wave equation are shown in Figures 3.9 through 3.11.
'
I
These figures show the· experimental and theoretical
iI
I
forces and accelerations, displacements, and energy, vs.
I time, measured at the load cell. The correlations are
~: BELLEVILLE SITE reasonably accurate, especially during the first 0.01 sec,
::: CASE BLTP- 6; 10.0
0. although the reflected compressive wave seems to be
overestimated, as shown in Figure 3.9A at 0.014 sec.
~:
Ul, This did not greatly affect either the ENTHRU or dis-
i ENTHRU = 6380 It lb
~Determined Experimentally

I
5.0
'
I
I
''
I'
I 4.5
5 10
BELLEVILLE SITE
ENTHRU- (kip ft) CASE BL TP- 6; 10.0
4.0
Figure 3.6. EINPUT vs ENTHRU.

3.5
Returning to Figure 3.6, the new value of EINPUT
is found to be 10,100 ft lb, which gives a new ram ve-
locity of 11.4 ft/sec. Substituting this velocity into
Figure 3.7, the resulting value of RUT agrees closely 3.0
with the assumed value of 90 kip.
Since the ram velocity at impact is now known, the 1- 2.5
assumed cushion stiffness of 1080 kip/in. can be checked. w
Holding RUT =
90 kip and the initial ram velocity = en
~
11.4 ft/sec, and solving for the change in FMAX as the ::; 2.0
cushion stiffness varies, the curve of Figure 3.8 can be
drawn. The experimental value of FMAX reported in
the Michigan paper was 244 kip, which entered into
1.5
Figure 3.8 gives a value of K (1) = 900 kip/in. Since
this is close to the assumed value of 1080 kip/in., the
solution was considered to be satisfactory. However,
even in cases where the cushion stiffness was quite inac-
curate, ENTHRU was only slightly changed when a more LIM SET
accurate value of K ( 1) was used.
=0.75
i
i !
This solution now enables us to determine the energy !-;--RUT= 90 kip
I I
output of the hammer, and other quantities. Since this i ~RUT: 30 kip
hammer is rated at 15,000 ft lb and its actual output i I

was only 10,100 ft lb the hammer must have lost 3,900 0~----~--~----~----~----~----~--~
8 10 12 14 16 18 20
ft lb due to friction in the guides or from other causes. RAM VELOCITY (ft I sec)
Thus, the hammer efficiency is (10,100) X (100) /
15,000 =
67 percent. Furthermore, since only 6,380 ft Figure 3.7. Ram velocity vs LIMSET.
PAGE ELEVEN
 ---------------------------------------------------------------------------------------------,

BELLEVILLE SITE
experimentally to that predicted by the wave equation
CASE BLTP- 6; 10.0
is reasonably constant.

? Ram Velocity = 11.4 ft/sec Investigation of Diesel Hammers

.::::. 20,000 RUT = 90.0 kip
"-
;;;
Used in the Michigan Study
Because the diesel explosive force is much smaller
than the impact force, it was found to have little effect
~ 15,000 on the driving stresses. 41 However, if explosive pressure
(/)
is neglected, the ram velocity required to predict EN-
(/)
LIJ
THRU is much greater than that calculated from the
z
lJ..
free fall of the ram, even assuming 100 percent efficiency.
lJ..
i= 10,000 Therefore, it was necessary to run the diesel hammer
(/)
cases accounting for the explosive pressure in the
hammer.
z
0
:r During impact between the ram and anvil the force
g: 5,000 on the anvil will reach some maximum value and then
'-' decrease. Following this impact, the diesel explosion
occurs, exerting an explosive pressure and force between
the ram and anvil. This behavior has been studied and
reported by some of the hammer manufacturers. 71 In
200 400 600 800 order to simulate this action for wave-equation analysis,
FMAX (KIP ) the explosive force acting within the. diesel hammer is
assumed to behave as shown in Figure 3.12. The maxi-
Figure 3.8. Cushion stiffness vs FMAX. mum explosive force is held on the anvil for 0.01 sec
after which the force is tapered to zero at 0.0125 sec.
placement curves, although it may have caused the rather Actually, the explosive hammer force lasts considerably
large errors in the acceleration curve of Figure 3.9B. longer than this hut its magnitude is too small to he a
A summary of the results for the steam hammer significant factor in pushing the pile down except during
cases solved is given in Table 3.10. Listed are the the initial driving stages when little or no soil resistance
energy output of the hammer, the hammer efficiencies, is encountered. The magnitudes of explosive pressures
the ram velocity, and the total soil resistance, RUT, listed in Table 3.11 were obtained from the hammer
necessary to obtain correlation for each case. manufacturer or were assumed.
It should he noted that there was no way to deter- In previous solutions, it was an easy matter to solve
mine the soil damping or elasticity constants. Therefore, for the total energy of the ram at impact since only its
the constants recommended by Smith 47 were used. As kinetic energy, EINPUT, was involved. Now, since
shown by Forehand and Reese, 48 these constants affect explosive pressure is included, the total energy devel-
the resulting RUT values. Therefore, the theoretical oped includes both kinetic and explosive energy.
RUT values shown in Table 3.10 were not expected to This total energy, ENTOTL, is the sum of the energy
agree closely with the experimental values reported in transmitted to the anvil, ENTHRU1, and the kinetic re-
Table 3.9. However, it is interesting to note that in bound energy of the ram after impact, where ENTHRU1
several cases the ratio of the soil resistance determined is calculated by the same method as was used for EN-

TABLE 3.10. SUMMARY OF RESULTS FOR MICHIGAN STEAM HAMMERS

Ram- RUT
Ham- Cushion- (Theoretical)
mer** Helmet RUT
Effi- Assembly Ram (Experimental)
Driving Pile Hammer* EINPUT ENTHRUt ciency Efficiency Velocity RUT
Location I.D. Case Type (ft lb) (ft lb) ( o/o) ( o/o) (ft/sec) (kip) ( o/o)

Belleville BLTP-6 10.1 V-1 10,100 6,380 67 63 11.9 90 190

57.9 V-1 7,000 4,440 47 63 9.5 200 50
Detroit DLTP-8 41.5 V-1 9,700 5,760 65 60 11.2 50 83
80.2 V-1 7,200 4,540 48 63 9.6 120 33
DTP-5 20.0 V-50C 12,800 8,290 85 65 12.9 25 110
79.0 V-50C 15,600 11,420 103 73 14.2 300+ 128+
Muskeg an MLTP-2 20.0 V-1 12,200 7,210 81 59 12.5 50 62
53.0 V-1 7,700 4,870 51 63 10.0 1.50 75
MLTP-9 72.0 V-80C 19,700 14,660 81 74 12.6 175 109
127.0 V-80C 19,200 13,110 79 68 12.5 300 64
*See Table 3.3 for notation.
**Hammer efficiency computed on basis of the manufacturer's maximum rated output.
tNote: The problems were selected at random and the hammer energies determined are not necessarily typical of the
hammer's usual operating characteristics.
PAGE TWELVE
-
_ VI
<!> 200
0.8

BELLEVILLE SITE
...J CASE BLTP-6,10.0 .....,
...J 0.7
c
UJ
(.)
...J
0 100 _J 0.6 THEORETICAL - - - -
<( THEORETICAL w {'
EXPERIMENTAL ------0------
0 E XP ER IM ENTAL -----0----- u {
...J

I
0
<l 0.5
I.L 0
0 _J
¢
z
0
1-
<(
LL
0
1-
z
0.4

f
l
w
a:: ~ 0.3
UJ w
...J u
IJJ <l
_J
0-100 a. 0.2
~ (/)

0
BELLEVILLE SITE
0 50 100 150 CASE BLTP- 6,10.0

ELAPSED 4
TIME (SEC X 10- ) 0.1

4oor-----------------------------~
THEORETICAL
0
EXPERIMENTAL ·--0--· 100 150 200 250 300
ELAPSED TIME {SEC X 10-4)

a. 300 BELLEVILLE SITE Figure 3.10. Comparison of theoretical and experimental

.... CASE BLTP -6;10.0 lvad cell displacements.

...J
...J
w
u
0
<l
0 7000
...J
z ,0'0-0-0-0-0-0-0-0-0-0-0
0'0
w 6000 :.e~0-0'
u
0:: ):J'
0 f/ THEORETICAL --
LL
5000
•~l EXPERIMENTAL ---0---
0
p(
0 0~----~~~~~~~--~~~~ >if
50 100 150 200 ·y~
I

ELAPSED TIME (SEC X 10-4) ~ 4000 l

Figure 3.9. Comparison of theoretical and experimental
load cell forces and accelerations.
--
:::>
~ 3000 1/J
I
I
I
I
I
I

BELLEVILLE SITE
CASE BLTP-6;10.0
I
1- I
z
w

THRU at the load cell, and the kinetic rebound energy 2000
remaining in the ram after impact is given by WV2 /64.4,
where W is the weight of the ram and V is the rebound
velocity of the ram determined by the wave equation. 1000
The efficiencies and initial ram velocities noted in
Table 3.11 were found by plotting ENTHRU and EN-
THRU1 vs the initial ram velocity as shown in Figure 0~~~7----7.~--~=---~~~~--~~
50 100 150 200 250 300
3.13. Plotting the values of LIMSET vs ram velocity as 4
ELAPSED TIME (SEC X I0- 1
in Figure 3.14 then gives the total soil resistance pre-
dicted by the wave equation. This procedure was used
on all diesel hammer cases, and the results are sum- Figure 3.11. Compa.rison of theoretical and experimental
marized in Table 3.11. values of ENTHRU.
PAGE THIRTEEN
 TABLE 3.11. SUMMARY OF RESU!,TS FOR MICHIGAN DIESEL HAMMERS
Ram- RUT
Ham- Cushion Ram (Theoretical)
Explosive mer* Assembly Velocity
Force on Ef- Ef- at RUT
Driving· Pile Hammer Anvil ENTOTL ENTHRU** ficiency ficiency Impact RUT (Experimental)
Location I. D. Case Type (kip) (ft lb) (ft lb) (%) (%) (ft/sec) (kip) (%)

Belleville BLTP-4 25.0 LB-312 98.0 10,630 8,010 59 75 8.2 70 50

66.4 16,030 11,200 89 70 6.4 250 36
BRP-4 20.0 M-DE30 98.0 9,450 4,980 42 53 9.8 100 100
50.0 9,100 4,470 41 49 10.6 200 63
BLTP-5 15.0 D-D12 93.7 13,000 9,040 58 69 12.8 40 50
60.0 14,730 9,930 66 67 15.0 400 118
Detroit DRP-3 40.0 LB-312 98.0 9,270 7,060 52 76 9.8 45 75
60.0 13,900 6,620 77 48 5.2 60 79
DTP-13 40.0 M-DE30 98.0 14,390 9,100 64 63 13.7 35 117
80.7 15,280 9,480 68 62 15.1 120 45
DTP-15 20.0 D-D12 93.7 15,270 10,100 68 66 15.2 45 112
80.5 9,430 5,480 42 58 11.6 110 92
Muskegon MTP-12 30.5 LB-520 98.0 22,140 14,860 74 67 16.4 75 187
70.8 21,260 13,140 71 62 14.4 70 45
MTP-11 69.5 M-DE40 138.0 32,800 16,760 102 50 20:6 150 94
150.0 36,8.50 17,900 115 49 21.5 250 50
MLP-8 31.0 D-D22 158.7 31,600 25,500 80 81 17.8 70 175
178.0 27,300 22,050 69 81 17.1 300 30
*Hammer efficiency based on manufacturer's maximum rated energy.
**Note: The problems were selected at random and the hammer energies determined are not necessarily typical of the
hammer's usual operating characteristics.

Determination of Hammer Energy Output as the kinetic energy of the falling ram plus the explo-
Diesel Hammers sive energy found by thermodynamics. Other manufac-
At present the manufactures of diesel hammers ar- turers simply give the energy output per blow as the
rive at the energy delivered per blow by two different product of the weight of the ram-piston WR and the
methods. One manufacturer feels that "Since the amount length of the stroke h, or the equivalent stroke in the
of (diesel) fuel injected per blow is constant, the com- case of closed-end diesel hammers.
pression pressure is constant, and the temperature con- The energy ratings given by these two methods dif-
stant, the energy delivered to the piling is also con- fer considerably since the ram stroke h varies greatly
stant."09 The energy output per blow is thus computed thereby causing much controversy as to which, if either,
35r-------------------------------------------,

HAMMER EFFICIENCY • ENTOTL X 100 • 13,000 X 100 = 59 %

_. ~
1-
30 ERATED 22,000
u.
>
z 0.. CUSHION EFFICIENCY= ENTHRU XIOO 9040 X 100 • 69 %
<t g ENTOTL 13,000
a >- 25
z (l)
a:
<t MAXIMUM IMPACT FORCE L&J
ON THE ANVIL CAUSED z
::!' L&J
<t BY THE FALLING RAM
a: 20
z
w
w
3:
1- 15
w
Ill
w
u
a: 10
~ BELLEVILLE SITE
CASE BLTP-5; 15.0
IDEALIZED DIESEL EXPLOSIVE
FORCE ON THE ANVIL AND RAM

RAM VELOCITY AT
IMPACT •12.8 FT /SEC

0 25 50 75 100 125 150

0~----------~-A--------~------------~----~
4 8 12 16 20
TIME (SEC X I0- 1 RAM VELOCITY AT IMPACT CFT /SEC I
Figure 3.12. Typical force vs time curve for a diesel Figure 3.13. ENTHRUI an.d ENTHRU vs ram velocity
hammer. determined by wave equation analysis.
PAGE FOURTEEN
 4-0r-------------~--------------------------~
remammg fall of the ram.i 0 Therefore the velocity of
the ram at the exhaust ports is essentially the same as
at impact, and the kinetic energy at impact can be closely
3.5
approximated by:
Ek = Wn (h - d) Eq. 3.3
3.0
where Wn the ram weight,
h the total observed stroke of the ram,
and d the distance the ram moves after closing
~ 2.5
the exhaust ports and impacts with the
l-
UI
anvil.
"'~ 2.0
..J
The total amount of explosive energy Ee(totall is
dependent upon the amount of diesel fuel injected, com-
pression pressure and temperature and therefore may
1.5 vary somewhat.
BELLEVILLE SITE Unfortunately, the wave equation must be used in
CASE BLTP-5;15.0
each case to determine the exact magnitude of Ee since
1.0
it not only depends on the hammer characteristics but
also on the characteristics of the anvil, helmet, cushion,
pile, and soil resistance. However, values of Ee deter-
.5 mined by the wave equation for several typical pile prob-
lems indicates that it is usually small in portion to the
total explosive energy output per blow, and furthermore,
that it is on the same order of magnitude as W n X d.
8 12 16 20
RAM VELOCITY AT IMPACT (FT /SEC)
Thus, Eq. 3.1 can be simplified by assuming:
Figure 3.14. LIMSET vs ram velocity detennined by Ee = WR X d Eq. 3.4
wave equation analysis.
Substituting Eqs. 3.3 and 3.4 into Eq. 3.1 gives:

method is correct and what energy output should be used £total= E~r + E" = Wn (h- d) + Wn d Eq. 3.5
in dynamic pile analysis. so that:
In conventional single acting steam hammers the Etotal = Wn h Eq. 3.6
steam pressure or energy is used to raise the ram for The results given by this equation are compared with the
each blow. The magnitude of the steam force is too actual values found by the wave equation in Table 3.12.
small to force the pile downward and consequently it Nnte that the results are relatively constant, the average
works only pn the ram to restore its potential energy, efficiency being 100 7r .
Wn x h, for the next blow. In a diesel hammer on the
otherhand, the diesel explosive pressure used to raise the Steam Hammers
ram is, for a short time at least, relatively large. Again using the wave equation in conjunction with
the Michigan report, Tables 3.13 and 3.14 suggest effi-
While this explosive force works on the ram to ciency ratings of 601ft, for the single-acting steam ham-
restore its potential energy W R x h, the initially large mers, and 87% for the double-acting hammer, based on
explosive pressure also- does some useful work on the an energy output given by:
pile given by :

where F
E~ = s F ds
the explosive force, and
Eq. 3.1
Etotal = Wn h
In order to determine an equivalent ram stroke for
the double-acting hammers, the internal steam pressure
Eq. 3.7

above the ram which is forcing it down must be taken

ds the infinitesimal distance through which
into consideration. The manufacturers of such hammers
the force acts.
state that the maximum steam pressure or force should
Since the total energy output is the sum of the not exceed the weight of the housing or casing, or the
kinetic energy at impact plus the work done by the housing may be lifted off the pile. Thus the maximum
explosive force. downward force on the ram is limited to the total weight
Etotal = Ek + E. Eq. 3.2 of the ram and housing.

where Etotal the total energy output per blow, Since these forces both act on the ram as it falls
through the actual ram stroke h, they add kinetic energy
Ek the kinetic energy of the ram at the tn the ram, which is given by:
instant of impact,
and E. = the diesel explosive energy which does
Etotal = Wn h + Fn h Eq. 3.8
useful work on the pile. where Wn the ram weight,
It has been noted that after the ram passes the Fn = a steam force not exceeding the weight
exhaust ports, the energy required to compress the air· and h = the observed or actual ram stroke.
fuel mixture is nearly identical to that gained by the Since the actual steam pressure is not always applied at
PAGE FIFTEEN
TABLE 3.12. COMPARISON OF ENERGY OUTPUT MEASURED EXPERIMENTALLY WITH THAT PREDICTED
BY EQUATION 3.6, FOR DIESEL HAMMERS

Ram Wt. Observed Wn X h ENTOTL

Pile ENTOTL WR Ram Stroke Etotnl
Hammer. I. D. Case (ft lb) (lb) h (ft) (ft lb) Etotal

LB-312 BLTP-4 25.0 10,630 3,857 3.3* 12,800 .83

66.4 16,030 3,857 3.6* 13,900 1.15
DRP-3 40.0 9,270 3,857 2.9* 11,000 .84
60.0 13,900 3,857 3.0* 11,600 1.20
DE-30 BRP-4 20.0 9,450** 2,800 6.6 18,500
50.0 9,100** 2,800 6.9 19,300
DTP-13 40.0 14,390 2,800 .5.2 14,600 .99
80.7 15,280 2,800 7.0 19,600 .78
D-12 BLTP-5 15.0 13,000 2,750 4.9 13,500 .96
60.0 14,730 2,7.50 6.1 16,800 .88
DTP-15 20.0 15,270 2,750 6.0 16,500 .93
80.5 9,430** 2,750 7.0 19,300
LB-520 MTP-12 30.5 22,140 5,070 3.7* 18,500 1.20
70.8 21,260 5,070 4.5* 22,750 .93
DE-40 MTP-11 69.5 32,800 4,000 7.6 30,400 1.08
150.0 . 36,850 4,000 8.2 32,800 1.12
D-22 MLTP-8 31.0 31,600 4,850 5.6 27,200 1.16
178.0 27,300 4,850 5.5 26,700 1.02
Avg. = 1.00
*Equivalent stroke derived from bounce chamber pressures.
**Experimental results for these cases appear to be quite inaccurate.

TABLE 3.13. COMPARISON OF MEASURED OUTPUT WITH THAT GIVEN BY EQUATION 3.7, FOR SINGLE
ACTING STEAM HAMMERS

Pile Hammer EINPUT* WR h** Etotal EINPUT

I.D. Case Type (ft lb) (lb) (ft) (ft lb) Etotal

BLTP-6 10.0 V-1 10,100 5,000 3 15,000 0.67

57.9 V-1 7,000 .5,000 3 15,000 0.47
DLTP-8 41.5 V-1 9,700 5,000 3 15,000 0.65
80.2 V-1 7,200 5,000 3 15,000 0.48
MLTP-2 20.0 V-1 12,200 5,000 3 15,000 0.81
53.0 V-1 7,700 5,000 3 15,000 0.51
Avg. = 0.60

*EINPUT found by wave equation and listed in Table 3.10.

**The observed ram stroke h or equivalent ram stroke h. was given in the Michigan report text.

TABLE 3.14. COMPARISON OF MEASURED ENERGY OUTPUT WITH THAT PREDICTED BY EQUATION 3.11,
FOR DOUBLE ACTING STEAM HAMMERS

Pile Hammer EINPUT* WR h.** Etotnl EINPUT

I.D. Case Type (ft lb) (lb) (ft) (ft lb) Etotal
DTP-5 20.0 V-50C 12,800 5,000 3.02 15,100 0.85
79.0 V-50C 15,600 5,000 3.02 15,100 1.03
MLTP-9 72.0 V-80C 19,700 8,000 3.0.5 24,450 0.81
127.0 V-80C 19,200 8,000 3.05 24,450 0.79
Avg. = 0.87

*EINPUT found by wave equation and listed in Table 3.10.

**The observed ram stroke h or equivalent ram stroke h. was given in the Michigan report text.
PAGE SIXTEEN
 the rated maximum, the actual steam force can he ex- where WR = ram weight
pressed as:
VR initial ram velocity
FR = (
_P ) WH Eq. 3.9 h equivalent stroke derived from bounce
Prated chamber pressure gage
where W H is the housing weight, p is the operating d distance from anvil to exhaust ports
pressure, and Prated is the maximum rated steam pressure. e efficiency of closed end diesel hammers,
approximately 1007< when energy is
The total energy output is then given by computed by this method.
Double-Acting Steam. Hammers
Etotal = WR h + r=p WH h Eq. 3.10
E WR h. (e)
L_:rated
V y2g h. (e)
This can he reduced in terms of Eq. 3. 7 by using where W R ram weight
an equivalent stroke h. which will give the same energy h. = equivalent ram stroke
output as Eq. 3.10.
Thus:
Etotal = WR he Eq. 3.ll = h L1 + _P_
Prated
X WH
Wn
Setting Eqs. 3.10 and 3.ll equal yields
h actual or physical ram stroke
WR he WR h + 1-----..L_ WH h p
Prated
operating steam pressure
maximum steam pressure recommended
l]>rated
by manufacturer
WH weight of hammer housing
= h II WR + _P_
Prated
WH e efficiency of double-acting steam ham-
mers, approximately 85 7< by this method.
Single-Acting Steam Hammers
or solving for the equivalent stroke:
E WR h (e)

h, = h [ 1 + p;:..,- X ::j Eq. 3.12

VR
where WR
h
V2g h (e)
ram weight
ram stroke
Conclusions e efficiency of single-acting steam hammers,
normally recommended around 75 7<· to
The preceding discussion has shown that it is possi-
ble to determine reasonable values of hammer energy 85%. 43 In this study of the Michigan
output simply by taking the product of the ram weight data, a figure of 60 7<· was found. The
and its observed or equivalent stroke, and applying an writers feel the 60% figure is unusually
efficiency factor listed in Tables 3.12 thru 3.14. This low and would not recommend it as a typ-
ical value.
method of energy rating can be applied to all types of
impact pile drivers with reasonable accuracy. A summary of the properties and operating character-
istics of the various hammers is given in Table 3.15.
A brief summary of this simple procedure for arriv-
ing at hammer energies and initial ram velocities is as
follows: Effects of the Experimental
Open End Diesel Hammers
Measuring Devices
E WR h (e) Another example of the application of the wave
V 2g (h-d) (e) equation to the Michigan pile study is the solution of
each of the previous problems, but excluding any effects
ram weight of the experimental apparatus. When the question was
initial ram velocity first raised as to how the elasticity of the load cell and
observed total stroke of ram the additional weight of the load cell and extra driving
Distance from anvil to exhaust ports cap might affect the results, it was decided to drive a
e Belleville H pile to refusal with a Delmag D-12 hammer
efficiency of open end diesel hammers,
with the load cell and extra driving cap removed. The
approximately 100% when energy is data recorded for this pile were then compared with the
computed by this method.
data for similar piles which were driven by the same
Closed End Diesel Hammers hammer hut which included the extra driving cap and
load cell.
E* WR h. (e)
VR =
V 2g (h.-d) (e) The only data obtainable for the noninstrumented
pile were the blow count and rate of penetration at vari-
*Note: For the Link Belt Hammers, this energy can be ous depths, since there was no way to measure the forces,
read directly from the manufacturer's chart using bounce displacements, ENTHRU, etc. It is also possible that a
chamber pressure gage. pipe pile might have been affected differently than the
PAGE SEVENTEEN
 TABLE 3.15. SUMMARY OF HAMMER PROPERTIES AND OPERATING CHARACTERISTICS
Hammer Hammer Maximum Ram Casing Anvil Maximum d Rated Explosive Cap
Manu- Type Rated Weight Weight Weight or Equiva- (ft) Steam Pressure Block
facturer (ft lb) (lb) (lb) (lb) lent Pressure (lb)
Stroke (psi)
(ft)

Vulcan #1 15,000 5,000 4,700 3.00

014 42,000 14,000 13,500 3.00
50C 15,100 5,000 6,800 3.02 120
soc 24,450 8,000 9,885 3.06 120
140C 36,000 14,000 13,984 2.58 140
Link Belt 312 18,000 3,857 1188 4.66 0.50 98,000 5 Micarta
disks
1" X 10%"
dia.
520 30,000 5,070 1179 5.93 0.83 98,000
MKT Corp DE20 16,000 2,000 640 8.00 0.92 46,300 nylon disk
2" X 9"
dia.
DE30 22,400 2,800 77.5 8.00 1.04 98,000 nylon disk
2·" x 19"
dia.
DE40 32,000 4,000 1350 8.00 1.17 138,000 nylon disk
2" X 24"
dia.
Delmag D-12 22,500 2,750 754 8.19 1.25 93,700 15" X 15"
X 5"
German
Oak
D-22 39,700 4,850 1147 8.19 1.48 158,700 15" X 15"
X 5"
German
Oak

H-pile tested, and that the soil conditions of the Detroit These effects are easily determined by the wave
or Muskegon sites could be of influence. Furthermore, equation, simply by omitting the weights and springs
only one hammer was studied (the Delmag D-12) and corresponding to the extra driving cap and load cell
the effect on the other hammers could be different. shown in Figures 3.4 and 3.5. The modified idealiza-
Obviously, these questions cannot be completely an- tions are shown in Figures 3.15 and 3.16.
swered experimentally since this would mean that every
time the hammer, pile type, driving location, or any
other parameter changed, a similar noninstrumented pile ~ } Wlll=RAM WEIGHT.

would also have to be driven under identical conditions.

r--.
I_ - . K (I) • SPRING RATE OF
THE RAM.

~ } W(ll=RAM WEIGHT. & } W(2}• ANVIL WEIGHT.

~K(Il L--
=r-- . =SPRING RATE OF
CUSHION. ~
K(2} =SPRING RATE OF
CUSHION.

}
W(3l = ~ ~13tiJG ~P
PIPE ADAPTER
(WHEN USED}.
+

K(3} = SPRING RATE OF

FIRST PILE SEGMENT.
W{4} = WEIGHT OF PILE
SEGMENT.

~ .

} W{MP-11• WEIGHT OF PILE

SPRING RATE OF SOIL} SEGMENT.
SPRING RATE OF SOIL~ IN SHEAR ALONG K'(MP-1)
IN SHEAR ALONG K'(MP-1) SEGMENT MP-1 :>------- K{MP-1}• SPRING RATE OF
SEGMENT MP -I PILE SEGMENT.

} W {MP} = WEIGHT OF FINAL

SPRING RATE OF SOIL} PILE SEGMENT.
IN SHEAR ALONG K'(MP)·--~
SEGMENT MP SPRING RATE OF
K'IMP + t} SOIL IN BEARING
U
BENEATH SEGMENT
MP.

Figure 3.15. Idealization of a Vulcan hammer without Figure 3.16. Idealization of a diesel hammer without
measuring devices. measuring devices.

PAGE EIGHTEEN
 TABLE 3.16. EFFECT OF REMOVING LOAD CELL ON ENTHRU, LIMSET, AND PERMANENT SET OF PILE

ENTHRU LIMSET PERMANENT SET

(kip ft) (in.) (in.)
Ram With Without With Without With Without
Velocity Load Load Load Load Load Load
Case (ft/sec) Cell Cell Cell Cell Cell Cell

8 1.5 1.6 0.27 0.34 0.23 0.25

DTP-15, 12 3.3 3.6 0.53 0.67 0.57 0.57
8(}.5 16 5.8 6.5 1.02 1.03 0.94 0.97
20 9.1 10.1 1.54 1.54 1.43 1.47
8 3.1 3.8 0.62 0.71 0.51 0.62
DLTP-8, 12 7.1 8.5 1.15 1.32 1.06 1.29
80.2 16 12.5 15.1 1.91 2.10 1.82 2.15
20 19.5 23.6 2.70 3.08 2.65 3.13

Although no problems were solved which involved -only about lO percent. This effect can also be seen
H piles driven by a Delmag D-12 hammer at the Belle- in Table 3.18, in which the cushion stiffness varies
ville site, a similar pile was driven at Detroit for which greatly, while the displacement of the pile point changes
a wave equation solution was obtained. less than lO percent.
The results for this problem with the load cell as- However, if a different cushion is used, the coeffi-
sembly included and excluded are given in Table 3.16. cient of restitution will probably change too. Since the
This agrees with the Michigan study conclusion that for coefficient of restitution of the cushion may affect EN-
case DLTP-15 ;80.5 the permanent set per blow including THRU, a number of cases were solved with "e" ranging
the load cell agrees with that found when the load cell from 0.2 to 0.6. As shown in Tables 3.19 and 3.20, an
is excluded. The corresponding values for ENTHRU increase in "e" from 0.2 to 0.6 normally increases
do not agree nearly so· well. ENTHRU from 18 to 20 percent, while increasing the
permanent set from 6 to l l percent. Thus, for the case
The results for a similar problem solved at the shown, the coefficient of restitution of the cushion has
Detroit site, DLTP-8;80.2, do not agree with this con- a greater influence on rate of penetration and ENTHRU
clusion. This pile was also an H-pile, was embedded than does its stiffness. This same effect was noted in
to within 0.3 ft of the first H-pile, and also had 55 kip the other solutions, and the cases shown in Tables 3.19
soil resistance. However, DLTP-8;80.2 differs from the and 3.20 are typical of the results found in other cases.
Michigan test pile in that this pile was ll ft longer, and
was driven by a Vulcan-1 hammer rather than the Del- As was noted in Table 3.7, any increase in cushion
mag D-12. As shown in the lower half of Table 3.16, stiffness also increases the driving stress. Thus, accord-
ENTHRU, LIMSET, and the permanent set per blow all ing to the wave equation, increasing the cushion stiff-
show large changes when the measuring devices are ness to increase the rate of penetration {for example by
omitted. This might be overlooked if only the experi- not replacing the cushion until it has been beaten to a
mental results for case DLTP-15;80.5 were known. fraction of its original height or by omitting the cushion
entirely) is both inefficient and poor practice because
Table 3.17 shows how ENTHRU increases when the of the high stresses induced in the pile. It would be
load cell assembly is removed. better to use a cushion having a high coefficient of resti-
tution and a low cushion stiffness in order to increase
ENTHRU and to limit the driving stress.
E"jfects of Cushion Properties on Driving
This suggests that a long micarta cushion having
Although the general effects of cushioning materials a relatively low spring rate, and a high coefficient of
on pile driving are discussed in Chapter IV, the follow- restitution might be very effective.
ing discussion is given since it deals with the Michigan
pile study.
Comparison of Various Hammers
As. previously noted, the Michigan report states that Driving the Sam{! Pile
the cushion properties influence the values of ENTHRU
significantly, although "how, when, or how much" One of the objectives of the Michigan pile study
ENTHRU was affected could not be determined. It was was to determine just how effective the various hammers
thought that ENTHRU could be increased by using a actually were during driving. Therefore, every attempt
more resistant cushion block, in the case of the Vulcan l was made to equalize any variables which would affect
and McKiernan-Terry DE-30 hammers. Although this the results, such as choosing the driving location to give
conclusion seems reasonable, results given by the wave comparable driving conditions. However, it would be
equation did not seem to agree. For example, as seen impossible to test several hammers without having some
in Table 3.6, ENTHRU does not always increase with variations occur, perhaps in the soil resistance or ham-
increasing cushion stiffness, and furthermore, the maxi- mer condition. Since the wave equation does not have
mum increase in ENTHRU noted here is relatively small this limitation, it can be used to advantage here.
PAGE NINETEEN
 TABLE 3.17. EFFECT ON ENTHRU RESULTING FROM REMOVING THE LOAD CELL ASSEMBLY

ENTHRU
(ft lb) Increase
With Without ENTHRU
Driving Pile Load Load in
Location I.D. Case Cell Cell ( o/o)

Belleville BLTP-6 10.0 6380 7500 18

.57.9 4440 5300 19
BLTP-4 25.0 8010 8800 10
66.4 11200 12000 8
BRP-4 20.0 4980 5750 15
50.0 4470 6450 44
BLTP-5 15.0 9040 10750 19
60.0 9930 12300 24
Detroit DLTP-8 41.5 5760 6900 21
80.2 4540 5400 19
DTP-5 20.0 8290 10000 23
79.0 11420 12700 12
DRP-3 40.0 7060 7600 13
60.0 6620 7200 11
DTP-13 40.0 9100 10850 13
80.7 9480 11400 20
DTP-15 20.0 10100 11500 14
80..5 5480 6600 20
Muskegon MLTP-2 20.0 7210 8800 23
53.0 4870 5700 17
MLTP-9 72.0 14660 17000 16
127.0 13110 16000 22
MTP-12 30.5 14860 17000 14
70.8 13140 15000 14
MTP-11 69.5 16760 22000 31
150.0 17900 25300 41
MLTP-8 31.0 25500 31000 22
178.0 22050 26600 21

TABLE 3.18. EFFECT OF CUSHION STIFFNESS ON MAXIMUM POINT DISPLACEMENT FOR CASES BLTP-6;
10.0 AND 57.9

Maximum Point Displacement (in.)

Ram Cushion Stiffness (kip/in.) Maximum
Pile RUT Velocity Change
I.D. (kip) (ft/sec) 540 1080 2700 27,000 ( o/o)

BLTP-6; 10.0 30 12 2.20 2.14 2.22 2.26 5

16 3..54 3.47 3.52 3.70 6
20 4.66 4.93 5.00 5.01 7
BLTP-6; 57.9 150 12 0.45 0.48 0.38 0.48 6
16 0.72 0.76 0.76 0.79 9
20 1.06 1.10 1.11 1.15 8

TABLE 3.19. EFFECT OF COEFFICIENT OF RESTITUTION ON ENTHRU FOR CASE BLTP-6; 10.0 AND 57.9

Ram ENTHRU (kip ft) Maximum

Pile RUT Velocity Change
I. D. (kip) (ft/sec) e = 0.2 e = 0.4 e = 0.6 ( o/o)

BLTP-6; 10.0 30 12 6.0 6.5 7.3 18

16 10.5 11.8 12.8 18
20 16.5 17.4 20.0 17
BLTP-6; 57.9 150 12 6.7 7.2 8.2 18
16 11.6 12.7 14 ..5 20
20 18.2 19.7 22.4 19

PAGE TWENTY
 TABLE 3.20. EFFECT OF COEFFICIENT OF RESTITUTION ON MAXIMUM POINT DISPLACEMENT FOR CASE
BLTP-6; 10.0 AND 57.9

Ram Maximum Point Displacement (in.) Maximum

Pile RUT Velocity Change
I.D. (kip) (ft/sec) e = 0.2 e = 0.4 e = 0.6 (%)

BLTP-6; 10.0 30 12 2.13 2.14 2.36 10

16 3.38 3.47 3.58 6
20 4.73 4.93 5.17 8
BLTP-6; 57.9 150 12 0.46 0.48 0.50 8
16 0.73 0.76 0.81 10
20 1.05 1.10 1.18 11

TABLE 3.21. STUDY OF VARIOUS HAMMERS DRIVING THE SAME PILE

Permanent
Maximum Set of
Ram Explosive Point Pile Per
Velocity Force Displacement Blow Blows
Hammer (ft/sec) (kip) (in.) (in.) Per Inch

Vulcan-! 10.0 0 0.125 0.025 8

Vulcan-50C 14.5 0 0.284 0.184 3
Vukan-80C 12.5 0 0.360 0.260 2
Link Belt 312 7.0 98.0 0.119 0.019 8
Link Belt 520 16.0 98.0 0.357 0.257 3
McKiernen-Terry DE-30 13.0 98.0 0.139 0.039 7
McKiernen-Terry DE-40 21.0 138.0 0.592 0.492 1
Delmag D-12 15.0 93.7 0.173 0.073 5
Delmag D-22 17.5 158.7 0.473 0.373 2

As an example of such a comparison, Case BLTP- example, the ram velocity at impact must be known, as
6;57.9 is used, with the load cell and extra helmet well as the dynamic behavior of the cushion, the diesel
omitted, and with a soil resistance of 300 kips. This explosive pressure in the hammer, and the length of time
pile was then analyzed by the wave equation to deter- it exerts a force on the pile. Since the above data were
mine its penetration per blow when driven by each of not directly measured in the Michigan research pro-
the hammers listed in Table 3.10. In each case, the soil gram, they were being calculated from the previous data
and pile parameters were held constant. Thus, for ex- reported. The ram velocities at impact and explosive
ample, even though the values of the soil damping con- forces on the pile for the diesel hammers were based on
stant or quake may not be exact, they remained constant the results given in Table 3.11, assuming the explosive
for each problem while experimental results would vary force to be acting as shown in Figure 3.12. The Vulcan
unless Q and J did not change at each new driving hammer properties were based on Table 3.10.
location.
The results of driving this pile with the eight differ-
Again, certain quantities had to be known for each ent hammers are listed in Table 3.21 in the form of per-
hammer before the wave equation could be applied. For manent set of the pile per blow and blows per inch.

Chapter IV
CHARACTERISTIC CUSHION PROPERTIES

Introduction cushion assemblies using wood, while the Micarta as-

semblies have an average efficiency of 66 percent. As
Although a pile cushion serves several purposes, shown in Table 3.7, an increase in cushion stiffness will
its primary function is to limit impact stresses in both also cause an increase in impact . stresses which might
the pile and hammer. 00 In general, it has been found damage the pile or hammer during driving. This in-
that a wood or rope cushion is more effective in reducing crease in stress is particularly important when driving
the driving stresses than one of a relatively stiff material concrete or prestressed concrete piles.
such as Micarta. However, a stiffer cushion is usually
more durable and transmits a greater percentage of the
hammer's energy to the pile. Dynamic Stress-Strain Curves
For example, the results given in Tables 3.10 and In order to apply the wave equation to pile driving,
3.11 give an overall average efficiency of 52 percent for Smith" 1 assumes that the cushion's stress-strain curve is
PAGE TWENTY-ONE
 B

SLOPE= K
(J)
(J) 1 - - - - - - GUIDE RAIL
LLI
a::
t-
(/)

1=:::::}-----l-+----- CUSHION BLOCK

c
PEDESTAL
STRAIN
FLOOR SLAB
Figure 4.1. Stress-strain curve for a cushion block
(after reference 51). FIXED BASE
-+-----!~,-------~CONCRETE PILE

Figure 4.3. Cushion test stand.

a series of straight lines as shown in Figure 4.1. Even

though this curve might be sufficiently accurate to pre-
dict maximum compressive stresses in the pile, the shape
of the stress wave often disagrees with that of the actual
SUPPORTING stress wave. 52 This discrepancy was at first thought to
CABLES AREA=Ap
be the result of inaccurate soil data, since very little was
known concerning the soil behavior during driving. It
was therefore decided to suspend several test piles hori-

r4 '"'"'"
~~BLOCK
AREA= Ac
zontally above the ground 53 as shown in Figure 4.2 to
eliminate the effects o.f soil resistance.
Table 4.1 lists the pertinent information concerning
these piles. The cushion was then hit by the ram and
the resulting strains were measured at six points along
the pile. Displacements and accelerations of both the
ram and the head of the pile were also measured. How-
Figure 4.2. Test pile showing placement of strain gages. ever, even though the soil resistance had now been ex-

TABLE 4.1. SUSPENDED PILE DATA

Pile Cushion Ram

E Ap L Ac t Weight Velocity
Case Material (psi) (in.2 ) (ft) Material (in.') (in.) (lb) (ft/sec)

Class A
LT-48 Concrete 6.12x10' 254 65 Fir 62.8 9.0 4160 13.91
Class A
LT-41 Concrete 6.12x10' 254 6.5 Micarta 89.1 9.0 4160 8.03
LT-39 Steel 30x10' 21.46 85 Oak 225.0 7.5 2128 11.42
Class Y
LT-15 Concrete *3.96x10' 225 65 Oak 225.0 9.5 2128 13.98

*Esonic = 4.64x1o• psi

PAGE TWENTY-TWO
 5000r--------------------------------------- Furthermore, it is not known how much the rigidity
of the pedestal shown in Figure 4.3 affects the cushion's
behavior. Therefore, the wave equation was used to
check the results. The second method required the fol-
lowing information: l) the stresses determined experi-
mentally at the head of the pile vs time, 2) the velocity
m DYNAMIC S-S CURVE of the ram at impact, and 3) the physical properties of
4000 "' STATIC S-S CURVE
the pile system required for solution by the wave
equation.
As shown in Figure 4.5, both the cushion and ram
are omitted and the previously determined stresses meas-
ured experimentally at gage l {see Figure 4.2) are
placed on the head of the pile. The wave equation is
then used to determine the motion of the ram and the
3000 pile, from which the compression of the cushion at any
instant of time is known. By plotting the measured
(/)
cushion forces against the corresponding compressions
a.. of the cushion, the dynamic stress-strain curve may be
(/) determined. The curves obtained by this method are
(/)
w illustrated in Figures 4.6, 4.7, and 4.8. Comparing these
0:
I-
with Figure 4.4, it is noted that the curves are generally
I/)
2000'
similar in shape.

.Dynamic Coefficient of Restitution

Although the cushion is needed to limit the driving
stresses in both hammer and pile, it reduces the avail-
able hammer energy because of internal damping. The
load diagram shown in Figure 4.1 illustrates this energy
1000 loss since the energy input is given by the area ABC
5000r-----------------------------------------~

SLOPE OF CURVE AT
ONE-HALF OF THE
0.08 0.12 0.16 0.20 4000
MAXIMUM STRAIN:: EAVG
E4y 0 = 37,300 PSI
STRAIN (IN./ IN.) ; -

B
Figure 4.4. Dynamic and static stress-strain curves for
a fir cushion.

eluded, the shape of the stress wave still did not agree in 3000
with the theoretical shape, and so the device illustrated ~
Cll
in Figure 4.3 was used to see if the cushion's stress-strain Cll
1.1.1
diagram was actually a straight line. Q:

Iii
Using this method, the dynamic stresses and strains
were measured for several cushion materials. It was
later discovered that for a given material, the dynamic 2000
stress-strain curves were almost identical to the cor-
responding static curves. This is demonstrated in Figure
4.4 in which the dynamic and static curves for a fir
cushion are compared.
Since the stress-strain curves are not linear as
1000
assumed, the shape of the theoretical stress wave in the
pile is not likely to agree with the experimental shape
and so the "dynamic" curves were used.

0.16 0.20 0.24 0.26

(lN./IN.)

Figure 4.5. Idealized test pile with known forces applied Figure 4.6. Dynamic stress-strain curve for fir cushion
at head of the pile. (Case LT-48).
PAGE TWENTY-THREE
 1000,-----------------------------------------~

4000 900

800
3500

700
3000

600
~ 2500
(/)
iii
0.
l!la: lll 500
li; 2000
"'a:
1-
- - 4 - S L O P E OF CURVE AT
ONE· HALF OF THE

SLOPE OF CURVE AT
ONE- HALF OF THE
"' 400 MAXIMUM STRAIN : EAvG
EAvG= 17,300 PSI

MAXIMUM STRAIN: EAVG

1500 EAvG=212,000 PSI

300
1000

750
200

500

250 100

0.005 0.010 0.015 0.020 0.025 0.030

STRAIN (lN./IN.)
.01 .02 .03 .04 .05 .06 .07
Figure 4.7. Dynamic stress-strain curve· for a mica-rta STRAIN (lN./IN.)
cushion (Case LT-41).
Figure 4.8. Dynamic stress-strain curve for an oak
cushion (Case LT-39).
w~ile the energy_ output is given by area BCD. Usually
this energy loss IS accounted for by a coefficient of resti- experiments indicate that e is actually much lower,
tution of the cushion "e," in which probably around 0.6.
.. / Area under BCD
e- Idealized Dynamic Stress-Strain Curves
V Area under ABD
The major difficulty in using the dynamic curves
When the dynamic stress-strain curve for the cushion
derived in the previous section is that numerous points
is known, such as for the previous problem the coeffi. on the curve must be specified in the input data, unless
cient of restitution can be computed. As sho~n in Figure the curve can be input in equation form. Although the
4.6, the a_rea under the dynamic curve ABC is computed
by summmg elemental areas ijkl until point B is reached
(i.e., until the strain reaches a maximum) , then the area
under BCD is determined by summing elemental areas
mnop until point D is reached.
Table 4.2 summarizes the results found for the
curvt;s of Figures 4.6, 4.7, and 4.8. These coefficients en
en PARABOLIC CURVE DEFINED
of_ restitution ~grt;e dose~y with values recommended by w BY THE ORIGIN, PLUS
Hirsch. 55 It IS mterestmg that although e = 0.8 is a: POINTS P1 8 P2 .
t-
commonly recommended for a micarta capblock, these en
Ku=SLOPE OF THE
'#;~44--STRAIGHT LINE DEFINED BY
SMAX• E 2 , AND e.
TABLE 4.2. DYNAMIC CUSHION PROPERTIES
Commonly
Case Cushion Dynamic Recommended A
Material e e

LT-48 Fir 0.35 0.40''

LT-41 Micarta 0.60 0.80 43
Figure 4.9. Idealized dynamic stress-strain curve for
LT-39 Oak 0.47 0.4855
cushion (parabolic).
PAGE TWENTY-FOUR
 200
- - SOLUTION USING LINEAR CUSHION BLOCK
- - - . SOLUTION USING PARABOLIC INPUT
400
- SOLUTION USING KNOWN FORCE ON HEAD OF PILE
(FROM THE TRUE FORCE VS COMPRESSION CURVE 180
SHOWN IN FIGURE 4.6)

300
160

§;
~
ILl 200
0 140
0:
~
100
120

a:
~
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.00 1.10 1.20 1.30 1.40 ILl 100
0
COMPRESSION (IN.) 0:
0
LL
Figure 4.10. Dynamic force vs compression curves for
80
a fir cushion (Case LT-48).
- - SOLUTION USING LINEAR
CUSHION BLOCK
--.... SOLUTION ~SING PARABOLIC
increasing load curve for each of the curves is nearly 60 INPUT
parabolic, the unloading segment is rather complex. a.-e--G SOLUTION USING KNOWN
FORCE ON HEAO of PILE
Therefore, for convenience, the unloading segmep.t will (FROM THE TRUE FORCE VS
he approximated· by . a straight line having a slope such 40 COMPRESSION CURVE SHOWN
IN FIGURE 4.8)
that the areas under the two curves result in the use of
the correct coefficient of restitution for the cushion
material being used. 20

Thus, the curve shown in Figure 4.9 can he defined

by two different points on the loading curve (other than
0
0.0) and "e" of the material. The points on the curve 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
are used to define the equation of the loading curve, and COMPRESSION !IN.)
as long as the cushion strain increases, the increased
input energy is computed as described earlier. When Figure 4.12. Dynamic force vs compression curve for
the strain in the cushion hegins to decrease, the total an oak cushion (Case LT-39).
input energy and the coefficient of restitution are used
to determine the slope of the unloading curve in order
to- give the correct value of "e."
Az (Sm~-O) (Ez-€1)
As shown in Figure 4.9, the total input energy is
given by the area under the parabolic curve, A1 + A2 ,
while the output energy is given by the area under the ( Sn;x) (Ez-€1)
unloading curve, A2 • Since e is defined by
e2 = A2/(A1+Az), 2e 2 (Al + Az)
then Smax
A2 = 2
e (Al+Az). Since the slope of the straight line BD is given by:
But A2 is also given by
_ Smax
Ku -
400
(Ez-el)
where Ku defines the slope of the unloading curve, e
is the coefficient o-f restitution of the material, ( A1 + A2 )
is the total area under the curve ABD (calculated by the
computer), and Smax is the maximum stress in the cush-
a:: - - SOLUTION USING LINEAR
ion determined by the wave equation.
g CUSHION BLOCK
ILl 200 - - SOLUTION USING PARABOLIC
~ INPUT Figures 4.10, 4.11, and 4.12 compare experimental
~ - SOLUTION USING KNOWN force vs compression curves obtained for the first three
FORCE ON HEAD OF PILE
(FROM THE TRUE FORCE VS cases listed in Table 4.1, with those resulting from the
100 COMPRESSION CURVE SHOWN
IN FIGURE 4.7) parabolic idealization of Figure 4.9, and the straight line
shown in Figure 4.1. Note that the parabolic curves
closely represent the actual force-displacement curves
0.2 Q3 0.4
while the linear curves are not nearly so close. In each
COMPRESSION (IN.)
case the parabolic curves tend to- "over-shoot" the true
maximum force, while the linear curve does not. The
Figure 4.11. Dynamic force vs compression curve for effect this has on the stress wave in the pile will he
a mica:rta cushion (Case LT-41). discussed in Chapter V.
PAGE TWENTY-FIVE
 Chapter V
STRESS WAVES IN PILING

Comparison of Actual test piles cracked while setting up the experiment. There-
fore, any reflected tensile forces greater than the pre-
and Experimental Stress Waves stressing force opened a small gap at the crack such that
As noted in Chapter IV, the shape and magnitude the prestressing strands alone could transmit the tensile
of the stress wave in a pile is greatly dependent upon stress down the pile. This is seen by the relative agree-
the properties of the cushion used. This will become ment shown in Figures 5.1 through 5.6. Note that the
apparent by comparing the actual stress wave determined stress-waves shown for the concrete piles (Figures 5.1
experimentally with results found by using the idealized through 5.4) do not agree nearly so well as those for
cushion properties mentioned earlier. the steel pile (Figures 5.5 and 5.6) .
The solution for stresses in the pile should be more Still, the results agree closely in each case, not only
accurate if the effects of the cushion and ram can be in magnitude, but also in the over-all shape of the wave,
omitted. To accomplish this, the force measured at the thus indicating that the numerical solution to the wave
head of the pile and the stresses at other gage points were equation is quite accurate. Further, any inaccuracies are
then determined by using the wave equation. The cases likely due to faulty assumptions concerning the dynamic
solved by this method are listed in Table 4.1. Compari- behavior of other variables such as the cushion, soil, etc.
sons between the experimental results and wave equation
solutions at two points on the pile are shown in Figures As mentioned earlier, the stress-strain curve for the
5.1 through 5.6. cushion is normally assumed to be linear as in Figure
One of the major factors which influenced these 4.1. The true stress-strain curves shown in Figures 4.6
comparisons was the fact that the prestressed concrete through 4.8 indicate that the curves are not actually
1.20 ,-------------------------~ 1.20,----------------------------,

1.00 1.00

~ 0.60
0.60

~ 0.60
0.80
·.
~ 0.40 t5 0.40

~ 0.20 ~ 0.20

~ ~
0:
I&J 0.20
."
~ 0.20
:;;
~ 0.40 "'"'
~ 0.40

~
g
0.60

0.80
.. ~
..
«
0.60

..
0.80
'••·,'--------~
oeeo THEORETICAl. SOLUTION USING
1.00 1.00 - EXPERIMENTAL SOLUTION
KNOWN FORCES AT THE PILE
HEAD TO ELIMINATE ERRORS oooo THEORETICAL SOLUTION USING
1.20 CAUSED BY THE RAM AND 1.20 KNOWN FORCES AT THE PILE
CUSHION. HEAD TO ELIMINATE ERRORS
CAUSED BY THE RAM AND
1.40 1.40 CUSHION.

1'60 0~~-;;---:.-o--!:•---:•o--!:7--:8o--:-.----:"o---:-11----:12:----!::-,.~~•=--'=-1o---:!,.:--1!::-7---:!18:--1!::-9--:!zo· 1"60 ~--'--,'-..1.,-'.'--'----'----:----'.--'--.---'lo:---'--11---',o--'=1:;::3=~.=:;::~::::::i,.=l:;::7::::::i~=~L:•::'J,o

0
TIME (SEC X I0-3) TIME (SEC X 10-3)

Figure 5.1. Theoretical vs experimental solution for Figure 5.3. Theoretical vs experimental solution for
Case LT-48, Gage #3. Case LT-41, Gage #3.
1.20,-------------------------- 1.20.--------------------------

1.00 1.00

0.80 0.80

-:- 0.60 ~ 0.60

ez Q 88 oo., 88•.,
8
o<>Oo 0 .,
880
OAO ~

it
0.40 oo ..

I
0

0.20 ~~---------~~::_ _ ___:...,__ _ _ _ ~s:::"""-4 0.20 k------'-------'"""'...,.------.,,?----·-·_ ____:,"'.,),c-1

0: 0:
~ 0.20 0.20

~ 0.40
z "'
~
0
0.40

~ 0.60 =::i 0.60

g g
0.80
- EXPERIMENTAL SOLUTION
oooo THEORETICAL SOLUTION USING
080
···,--------------,
- EXPERIMENTAL SOLUTION
1.00 1.00
KNOWN FORCES AT THE PILE u e e THEORETICAL SOLUTION USING
HEAD TO ELIMINATE ERRORS
120 CAUSED BY THE RAM AND
CUSHION.
1.20
KNOWN FORCES AT THE PILE
HEAD TO ELIMINATE ERf!ORS
CAUSED BY THE RAM AND
1.40 1.40 CUSHION. .

I.SOO'.--c---,o--:---':---:..-.o--:--.o--!:.--:"o--ccll--:'12:----~,.-------:14=--~15-------:,.=--=,7-------:18:-~l.---:"20 1.60 L__,__--'--__,__,___,__--'------'--''---'-----':---'---:~:;==;;:=;:~;=;::~=;=-J

0 2 3 6 7 8 9 10 II 12 13 14 15 16 17 18 19 20

TIME (SEC X 10-3) TIME (SEC X IQ-3)

Figure 5.2. Theoretical vs experimental solution for Figure 5.4. Theoretical vs experimental solution for
C(JISe LT-48, Gage #5. Case LT-41, Gage #5.
PAGE TWENTY-SIX
 1.20,-----------------------------,
1.00

0.80 0.8

~ 0.60

~ 0.40 ~ 0.4
;::
~ 0.20 z
-' 0

~
~
~ 0.20
0
·.·..
"
~ 0.40
~
ffi
1-
0.4 ·.·
~ 0.60 !:1
<3 ~
0.80

1.00 - EXPERIMENTAL SOLUTION

..
~ 0.8
<.?
• u e THEORETICAL SOLUTION USING
1.20 KNOWN FORCES AT THE PILE L2 - SOLUTION WITH KNOWN FORCES PLACED ON HEAD Of PILE
HEAD TO ELIMINATE ERRORS ooo e SOLUTION USING A LINEAR CUSHION SPRING RATE
CAUSED BY THE RAM AND aaao SOLUTION USING A PARABOLIC CUSHION SPRING RATE
1-40 CUSHION.

1.60~~~~--'-----'--~-"-"--'----'----'---'-'=:;:::::=:::::;::~=;::::::.;:=:::::;::~
0 5 6 8 9 10 II 12 13 14 15 16 17 18 19 20 2 3 4 5 6 7 8 9 10 II 12 13 14 15 16 17 18 19 20
TIME (SEC X 10"3) TIME (SEC X I0-3)

Figure 5.5. Theoretical vs experimental solution for Figure 5.7. Theoretical vs experimental solution for Case
Case LT-39, Gage #3. LT-48, Gage #3.
1.20,------------------------,
1.00
though it agrees with the actual stress-strain curve most
0.80
of the time, it cannot follow the reversed curvature at
~ 0.60
the peak of the actual curve and thus "over-shoots" the
true peak force. Figures 4.10 through 4.12 show how
closely the parabolic curves follow the true cushion
forces, and also how far off the straight line assumption
is. The parabolic curve always peaks above the true
force vs compression curve, while the spring rate of the
straight line can be raised or lowered so that the true
nao
maximum cushion force is not exceeded.
1.00 Thus the use of the straight-line assumption seems
1.20 reasonable since it gives fairly accurate results. The
1.40
linear spring constants used for the curves shown in
Figures 5.7 through 5.12 were first varied between wide
I.Ei0};-0-------!-'!-7---.!----::--7.---:7;--:-.---:.~1~0-:':,~1~2-::-,~I.L_-=-,---:~=----=-17---:,.:---=-1.~20 limits to obtain the most accurate maximum stresses.
TIME (SEC X 10-3)
These spring rates were then used to determine what
dynamic modulus of elasticity was required to give the
Figure 5.6. Theoretical vs experimental solution for Case desired spring rate, using the equation: Edynamic =
LT-39, Gage #5. (K cushion) (Length)/(Area of cushion). As shown
in Table 5.1, these results give a lower value of E for
oak than for fir, which in this case is correct since the
linear and this assumption might therefore cause fir capblock was highly stressed ( 4,170 psi) while the
inaccuracies. oak capblock was stressed only slightly ( 765 psi).
To determine how the shape of the curve affects the Further consideration of the dynamic stress-strain
solution, the previous three _problems were run using the curves revealed that the dynamic modulus of elasticity
cushion stress-strain curves shown in Figures 4.1 of the capblock is almost exactly lO percent greater than
(straight line), 4.6 through 4.8 (true stress-strain that given by the slope of the stress-strain curve (Figures
curves), and 4.9 (parabolic curve). These solutions are 4.6 through 4.8) taken at a point halfway between zero
compared in Figures 5. 7 through 5.12. In each case, and the maximum strain. As noted by Hirsch, 02 the
it is noted that the straight line solution is more accurate static and dynamic stress-strain curves are quite similar,
than the solution using the parabolic curve. This is so that curves like those shown in Figures 4.6 through
because a simple parabolic curve was used which, even 4.8 are easily determined for any other cushion material.

TABLE 5.1 DYNAMIC PROPERTIES OF NEW CUSHION BLOCKS OF VARIOUS MATERIALS

Slope at
Linear Spring Depth of Area of Midpoint SMAX in
Cushion Rate - K Cushion Cushion Ed~·Jutmie of Curve Cushion
Case Material (lb/in.) (in.) (in.') (psi) (psi) (psi)

LT-48 Fir 295,000 9.0 62.8 42,200 37,300 4170

LT-41 Micarta 2,320,000 9.0 89.1 234,000 212,000 3850
LT-39 Oak 585,000 7.5 225.0 19,500 17,300 765

PAGE TWENTY-SEVEN
 1.2,----------------------------,

..... ,/"De
0.8
.." 0.8
a"
8
0 eoe
9

ea
'o

~ 0.4

z
0
;::
[il
~
"'
.0
ffi,__ 0.4

,."'
0

0.8 ~ 0.6

I.Z
-
------------------- ------:]
SOLUTION WITH KNOWN FORCES PLACED ON HEAD OF PILE
e oe o SOLUTION USING A LINEAR CUSHION SPRING RATE
~

I.Z
-
eoo o
SO!..UTION WITH KNOWN FORCES PLACE_O ON HEAD OF PILE
SOLUTION USING A LINEAR CUSHION SPRING RATE
oa1u1 SOLUTION USING A PARABOLIC CUSHION SPRING RATE
IHI!H> SOLUTION USING A PARABOLIC CUSHION SPRING RATE
-·-··---- --- ------------· ·- - -------··--
1"6L----'--'--,----',-.~-:-.-.':--:'-7----'8,-------:'-9~~o,-------',-11~,.,------1=-3------:':14,------1=-5------:':16:-;1';-7----=~8~1.:-----:-zo
2 3 4 5 6 7 8 9 10 II 12 13 14 15 16 17 18 19 20 0
TIME (SEC X 10-3) TIME (SEC X I0-3)

Figure 5.8. Theoretical vs experimental solution for Case Figure 5.10. Theoretical vs experimental solution for
LT-48, Gage #5. Case LT-41, Gwge #5.

It was also recommended that the dynamic modulus be (b) and if B is set equal to zero, no damping is present,
increased as the cushion consolidated. Ga as seen in Figure 5.13 (a).
The model shown in Figure 5.13 (c) has one major
Internal Damping in Piling advantage over the previous model in that it is able to
As noted earlier, differences between experimental account for damping by considering the difference be-
and theoretical results were assumed to be the result of tween the material's static modulus of elasticity E, and
inaccurate soil information. Other parameters were also its sonic modulus of elasticity E.. This is because a
varied in an attempt to obtain more accurate results, 56 slowly applied load gives the dashpot time to relax with-
one of which was the material damping or internal damp- out causing the spring K. to exert a force, thereby result-
ing capacity of the pile material. ing in a spring rate equal to K 0 • However, when the
loads are applied rapidly the dashpot has no chance to
Smith" 7 first suggested that the internal damping deform, resulting in a spring rate of Ko- K.. Thus for
in the pile might prove significant, and proposed the the model of Figure 5.13 (c), K 0 is determined from
following equation by which hysterisis in the pile could the static modulus of elasticity E, while Kn + K. would
be accounted for: use the sonic value E •.
F(m,t) = C(m,t)K(m) It is interesting to note that when K. is infinitely
BK(m) large, model (c) becomes equivalent to model (b) , and
+ 12 ~t [C(m,t) -C(m,t-1)] if K. = 0, model (c) becomes equivalent to model (a).
in which B is the internal damping constant. He also In order to derive the equation, Figure 5.14 is pro-
recommended that B be given a value of about 0.002 in vided. Figure 5.14 (a) illustrates the damping model
order to produce a narrow hysteresis loop. This equa- wherein point "m" (on the upper mass) has moved a
tion was derived from the model shown in Figure 5.13 distance x 1 , point "n" (between the dashpot and spring)

I.Z , - - - - - - - - - - - - - - - - - - - - - - - - - -

0.8 0.8

0.4

..
.
"•
...
8
6 0° 9
0
e: 0 o'" 0 11
- SOLUTION WITH KNOWN FORCES PLACED ON HEAD OF PILE
.,.,., " SOLUTION USING A LINEAR CUSHION SPRING RATE
'-' - SOLUTION WITH KNOWN FORCES PLACED ON HEAD OF PILE
eee" SOLUTION USING A LINEAR CUSHION SPRING RATE
oaoa SOLUTION USING A PARABOLIC CUSHION SPRING RATE nea SOLUTION USING A PARABOLIC CUSHION SPRING RATE

1.60 L__,____._~_~__.____.__,_~==============~
0 3 4 5 6 7 8 9 10 II 12 13 14 t5 16 17 18 19 20
1.6!--'----:-----'---'---:---:---:--'=::;:=:::;::=:;=:::=:=;::::;:::=;::::::;::=;::::;:=:;:=~~
0 34 67891011121314151617181920
TIME (SEC X 10-3) TIME (SEC X 10-3)

Figure 5.9. Theoretical vs experimental solution fo·r Case Figure 5.11. Theoretical vs experimental solution for
LT-4l,Gwge #3. Case LT-39, Gage #3.
PAGE TWENTY-EIGHT
 1.2,.-------------------------,
has moved a distance x2, and point "o" (on the lower
mass) has moved a distance of x3 • Assume that at time
t = t0 there exists a force F 0 to in the spring K0 • There
is also a force in the spring K. given by F.to, and a
force in the dashpot equal to F n to.
As shown in Figure 5.14 (b), after a single time
interval passes, point m moves an additional distance
Axh point na moves Ax 2 , and point o moves Ax 3 • At
this time, t =t1 = t 0 + At, and the forces in K0 , K.,
and B are designated F 0 t1, F.t1, and Fnt1, respectively.
At time t 0 :
F 8 to= K.(xl-x2). Eq. 5.1 1.2
,----------------
- SOLUTION WITH KNOWN FORCES PLACED ON HEAD OF PILE
eou SOLUTION USING A LINEAR CUSHION SPRING RATE

At time t1 = t 0 + At1: Gao a SOLUTION USING A PARABOLIC CUSHION SPRING RATE

F.t1 = K.[ (x 1 + Axl) - (x2 + Ax2) ]. -:':16~17:---:18';----1~9--;:20

I.S OL__._I--'-2--'3-4'----5L.._..L6--'-7--'8----'9'----10::----"-,-:':,2---:13:---:14';----=-15

F.t1 = K.[(xl-x:d + (Ax 1-Ax2)]. Eq. 5.2 TIME {SEC X 10~3)

Substituting Equation 5.1 into 5.2: Figure 5.12. Theoretical vs experimental solution for
F.t1 = F.to+K.(Axl-Ax2). Eq. 5.3 Case LT-39, Gage #5.
By definition, at all times:
F t _ B (Ax2-Axa) Eq. 5.4
D 1 - At . .
Because point n must be in equilibrium: K B

F.t1 = Fntl. Eq. 5.5

Substituting Equation 5.3 and 5.4 into 5.5:
(a) NO DAMPING PRESENT (b) INTERNAL DAMPING PROVIDED

Fnto + K.(Axl-Ax2) = B (Ax2-Axa)

BY DASHPOT

At

Solving for Ax2:

FntoAt+ K,Ax 1At+ BAxa
Ax2 = K.At+ B
Eq. 5.6

Substituting Equation 5.6 into 5.4 produces: (c) INTERNAL DAMPING PROVIDED BY AN
ELASTIC SPRING AND DASHPOT CONNECTED
F t 1 = Fnto+K (Ax 1-Ax3 ) IN SERIES
D (K.At/B) + 1 Eq. 5· 7
Figure 5.13. Various idealizations for the spring seg-
The solution begins by setting F D to equal to zero, ment of a pile.
and calculating it for the next time interval from Equa-
tion 5.7. The quantity K. is a constant and (Ax 1-Axa)
is simply the change in compression during a single time ing force given by Equation 5. 7 were checked. From
interval. Therefore, returning to the earlier terminology, Equation 5. 7,
Equation 5. 7 can be written: . K " -- F D to+ 0 : F n t1 --
( a ) Lettmg + 0 --
1 0
DF(It+ 1 ) = DF(I,t) +DK(I)[C(I,t+1)-C(I,t)]
' [DK(I)At/B]+LO Eq.5.8 This is correct since F n begins at zero and cannot in-
where DF(I,t) is the damping force in dashpot number
crease in magnitude when K. 0. =
_ Fnto+ oo
"I" during time interval "t," DK (I) is the dynamic
spring rate of damping spring "I," C (l,t) is the com-
(b) Letting K. = oo : F n t1
00
+
1
oo I oo. =
pression in spring I during time interval number t, At Since this is indeterminate,
is the time increment, and B is a damping constant. d
The static force in spring I will be computed as ciK.[F,to + K,(Axl-Axa)]
before, by d
dK. [KBAt + 1]
F(I,t+1) = K(I)[C(I,t+1)]. Eq. 5.9
· Thus by adding the Equations 5.8 and 5.9, the total force 1 im O+ (Ax 1 -Axa) _ B(Axt-Axa)
acting on each mass can be determined for the next time K.~ oo At/B + 0 At
interval.
This checks since it is the equation found when K. = oo
Since as far as is known this derivation does not and only the dashpot remains. In this case the models
appear elsewhere, the boundary conditions for the damp- of Figures 5.13 (b) and (c) would be identical
PAGE TWENTY-NINE
 Substituting this into the previous equation one finds
Fot1 [K.][(xl-x2)+(..!lxl-llx2)]
[K.] [ (x1 + ..!lx!) - (x2 + ..!lx2)]
[K.] [Total compression at time t].
This is correct since it is the equation for the spring and
when B = oo, the dashpot is "locked" and no damping
occurs.
( e ) Le ttlng
. Ll _ . F t _ Foto+ K.(..!lx1-..!lx3)
t - 0 . 0 1 - +
0 1
This result agrees because it gives the same result as
letting B = oo. (See part (d) above.)
(f) Letting Llt-3>oo :F0 t1
Frito+K.(..!lxl-llxa) = O.
oo +B -
0 This checks because the force stored in the damping
spring would be released by relaxation of the dashpot
0 if Llt = 00.
(g) Let Llx 1 =
..!lx2 and assume that the damping
spring has an initial force stored at t = t0 • Although
this fmce ehould diminish with time, it cannot go to zero
during a single time interval, unless ..!lt = oo.
Foto+ K.(O)
(a) POSITION OF MASSES
AT TIME t =to
(b) POSITION OF MASSES
AT TIME t=to•L.t _Ktt +LO

Figure 5.14. Idealized pile segment with standard linea.r

This is correct since the force in the spring is reduced,
solid damping.
but will never actually reach zero unless Llt oo. =
Figures 5.15 through 5.18 compare the effects of
damping in a pile using the damping models shown in
Figure 5.13. The results given are for test pile number
1000
LT-15 which is described in Table 4.1. This particular
800
pile was of lightweight concrete with E 3.96 X 10 6 =
600
and E. = 4.63 X 10 psi. This problem was chosen
6
400 since E. was relatively larger than E, indicating the
possibility of rather high damping.
However, one is often more interested in the maxi-
mum stresses found in the pile, which usually occurs
400 during the first or second pass of the stress wave along
600 the pile. During this time the effects of damping are
800 small and can usually be neglected.
1000 - EXPERIMENTAL SOLUTION I
"" "" THEORETICAL SOLUTION WITH DAMPING NE<?!,EC't~.!?J
0 I 2 3 4 5 6 7 8 9 10 II 12 13 14 15 16 17 18 19 20
TIME (SEC X \0-3 )

Figure 5.15. Comparison of experimental and theoreti- 1000

cal solutions for stresses at Gage #3 with damping 800

omitted (Ca:se LT-15). 600

400
F 0 to+ K.(..!lxl- .:lx3)
(c) Letting B = 0: F 0 t1 =
1 + K •..!lt
0
1
00
= 0.
600

This checks since if the dashpot has no damping ability, 800

the damping force must be zero. 1000

Foto + K.(..!lxl-..!lxs)
(d) Letting B oo : F 0 t1 = K •..!lt +
= I 2 3 4 5 6 7 8 9 10 II
TIME (SEC X 10~3 )
12. 13 14 15 16 17 18 19 20

1
00
Figure 5.16. Comparison of experimental and theo·reti-
F Dto + K.(..!lxl- Llxs) cal solutions for stresses at Gage #3 for different damp-
But Foto = Fsto :____ K. Cx1-x2r ing models (Case LT-15).
PAGE THIRTY
 - EXPERIMENTAL SOLUTION
700 ...... THEORETICAL SOLUTION WITH 700
DAMPING NEGLECTED
600

500

400

300 300

200 200

100 100

100 100

200 200

300 300

400 400
500
500 J.._£ r-~--,E"'XP""ER=IM=EN"-'-TA~LSJ,~IQN ~-
oeoe STANDARD LINEAR SOLID DAMPING WITH 8:0.8

600
~7-7-~~-7~~~~~~·=••~•~S~MIT~H~'S~O~~P~IN~GM~OO~EL~W~IT~H~B~~0~~0~05~
0 I 2 3 4 56 7 8 91011121314161617181920 0 I 2 3 4 5 6 7 8 9 10 II 12 13 14 15 16 17 18 19 20
TIME (SEC X 10-3} TIME (SEC X I0- 3 I

Figure 5.17. Comparison of experimental and theoreti- Figure 5.18. Comparison of experimental and theoreti-
cal solutions for stresses at Gage #5 with damping cal solutions for stresses at Gage #5 for different damp-
omitted (Case LT-15). ing models (Case LT-15).

This conclusion may not be accurate for timber neglected. This is illustrated in Figures 5.15 and 5.17
piles since wood has a i:nuch higher damping capacity where fairly large errors resulted when damping was
than either the steel or concrete piles for which experi- neglected. On the other hand, Figures 5.16 and 5.18
mental data were available. This higher damping ca- suggest that in certain cases damping should be account-
pacity might affect the results earlier in the solution ed for using either of the damping models of Figure 5.13.
which might in turn lower the accuracy of the results. The most surprising result of this study is not the
Nevertheless, if more testing should indicate that the accuracy of the damping models, but rather that both
damping models are accurate for timber piling too, then models give nearly identical results even though Smith's
the problem, or rather the uncertainties of damping model is extremely simple while the other is rather com-
effects will no longer be a problem. plex. Again, this may also prove incorrect for timber
piling or other piling which has a large damping ca-
In any case, if the wave is to be studied for an ex- pacity. For example, one of the above methods might
tended period of time, damping in the pile cannot be be more accurate than the other.

Chapter VI
SOIL PROPERTIES

Idealized Soil Resistance Curves the soil, including the effect of loading rate, is given by
R(m,t) = [D(m,t) - D'(m,t)] K'(m)[1
The load-defoTmation characteristics assumed for + J(m)V(m,t-1)]
the soil in Smith's numerical solution am shown in
Figure 6.1 (a). This curve excludes the damping effects where m denotes the segment number of the pile, t is the
of the soil caused by rapid loading, and illustmtes only time interval number, D(m,t) is the displacement of
the soil resistance caused ~y static loading. As shown, segment m at time interval number t, K' (m,t) is the
the two parameters required to define the load-defm:ma- plastic deformation of the soil, J( m) is the soil damp-
tion curve are the ground quake "Q(m)" and the ulti- ing constant, K' ( m) is the soil spring constant, V ( m,t)
mate static soil resistance "Ru ( m) ." is the velocity of mass number m at time interval number
t, and R(m,t) is the soil resistance acting on that ele-
When the soil is located along the side of the pile, ment at time t.
it is assumed to resist any rebound of the pile as well as
any downward motion. This is typified by the curve In cases in which more accurate soil data are avail-
OABCDEFG. However, the soil located at the tip of the able, the general soil resistance curve of Figure 6.1 (b)
pile can only exert upward forces, as represented by the may be used to advantage. This curve also uses the
curve OABCFCB. variables Q(m) and Ru(m), but the curve no- longer
must be linear. In this case, the ground quake Q(m)
The spring rate for the curve between point 0 and is divided into ten equal segments, and the static soil
A may now be determined from resistances corresponding to these ten points comprise
the input data required to- establish the curve. Also,
K '( m ) = Ru(m)
Q(m)
as shown in Figure 6.1 (b), the slope of the unloading
curve is given by K' (m). A more complete discussion
In order to include the damping effects of the soil, of the use of this method is given in the appendix.
a third variable J (m) is defined as the damping con- To- check out the programming changes involved in
stant of soil spring "m." Thus the total resistance of this method, several problems were first solved using
PAGE THIRTY-ONE
 -------------------------------------------------------------------------------------------

for example, although the soil quake was doubled and

the curve made nonlinear, the maximum change in stress
was less than 9 percent, and the permanent set increased
B
less than 8 percent. Only a drastic change in the soil
T
Ru(m)
resistance curve was found to cause an appreciable dif-
ference in the solution.

DEFORMATION
Therefore, if the soil resistance curve for the prob-
lem even slightly resembles the curve of Figure 6.2 (a) ,
the linear resistance equation will probably be satisfac-
Ru(m)
tory. Whenever it becomes necessary, the nonlinear soil
resistance can be used as explained in the appendix.
t
Significance of the Soil Quake "Q"
(a) ELASTIC-PLASTIC OR"LINEAR" SOIL RESISTANCE CURVE The properties of the soil under the action of dy-
namic loading are probably the least understood of the
many variables affecting the problem. 64 Although a
1'--------- Q ( ml --------1 number of values for the soil quake may be used, the
value Q =
0.1, recommended by Chellis65 is probably
the most widely accepted for general use, except when a
more accurate value can be determined. As might be
expected, the trouble stems mainly from the large num-
ber of variables influencing the value of Q· at any given
driving location, the most Dbvious of course being the
type of soil encountered. Much work is presently being
-j~ I DEFORMATION done to define these factors and to more accurately de-
termine the actual values for both "Q" and "J" to in-
r------- Q(m) -----j crease the solution's accuracy. 6 6,67
(b) GENERALIZED SOIL RESISTANCE CURVE
Figure 6.1. Load-deformation chamcteristics assumed
for the soil. Q..
:::! ;;2 200
o~
(f)LJJ
the regular elastic-plastic curve of Figure 6.1 (a). These (.)(.)
problems were then solved again using the generalized j::Z
soil resistance method with soil resistance values lying <!j:!
on the same curve, the two solutions then being checked ~ 100
!;; (/)
for identical results. lJJ
a:
A number of other problems were also solved to K' = 2000 kip/in.
see what changes might result when the shape of the soil
resistance curve was altered. For example, the linear OL-----L--~~L----~-----J
soil resistance curve used in a problem originally solved 0 0.1 0.2 0.3
by Smith58 is shown in Figure 6.2 (a). This problem DEFORMATION (IN.)
was then solved using the nonlinear curve of Figure (a) ELASTIC-PLASTIC SOIL RESISTANCE CURVE
6.2 (b). (AFTER REFERENCE 58)

The solutions for these two problems, shown in

Table 6.1, are typical of the results found for the other
cases studied, in that a rather large change in the soil
curve changed the results only slightly. In this case,
.... ~ 200
-~
0~
(/)
TABLE 6.1. COMPARISON OF RESULTS FOUND BY
USING ELASTIC-PLASTIC VS NONLINEAR SOIL !2~
1-Z
RESISTANCE CURVES ,_,_
<{<(

(/) ~ 100
Maximum Force (kip) (/) Ru=200 ki
Maximum lJJ
At At At Point a:
Type Head Center Point Displace-
of Soil of of of ment
Resistance Pile Pile Pile (in.)
0.1 0.2 0.3
Elastic Plastic 290 300 405 0.203 DEFORMATION (IN.)
Nonlinear 290 301 370 0.218 (b) GENERALIZED SOIL RESISTANCE CURVE
Percent Change 0.0 +0.3 -8.7 . +7.4 Figure 6.2. Soil resistance vs deformation curves.
PAGE THIRTY-TWO
 While it is beyond the scope of this paper to attempt were found for the other Michigan cases studied, except
to determine values for Q, it is interesting to see how that the tensile force often varied substantially more than
the value of Q affects the solution. After a number of indicated for the case of Table 6.2.
the Michigan research problems with varying values of
Q were studied, Case BLTP~6; 57.9 was chosen as
being fairly representative. The problems were solved Significance of the Soil Damping
with Q ranging from 0.1 to 0.5, as seen in Table 6.2.
To determine whether Q would have similar effects at Michigan Case BLTP-6;57.9 was also chosen to
all magnitudes of soil resistance, Rutotal was also varied. illustrate the damping effects of the soil. These damp-
The results of this parameter study are given in Table 6.2. ing constants were given values ranging from 0.0 to 0.5,
and as was done in the previous section, the total soil
One of the trends noted in Table 6.2 is the small resistance was varied from 50 to 400 kip to see if trends
effect Q has on the maximum compressive force found in found at low resistances would also be noted when the
the pile. The effect on tensile force is more pronounced, soil resistance was large. Since the soil damping con-
although no conclusion could he reached as to whether stants most commonly used are those recommended by
the tensile stress will increase or decrease as Q changes Smith, 68 i.e., a soil damping constant of 0.05 sec/ft
since the results did not indicate an apparent trend. along the side of the pile and 0.15 sec/ft at the point
Maximum ENTHRU values are also relatively inde- of the pile, the variation of J = 0.0 to 0.5 very likely
pendent of the soil quake, with ENTHRU tending to covers the values typical for many conditions and soils.
decrease as the soil quake increases. These results are given in Table 6.3.

The most pronounced and consistent trend is the As was previously determined for Q, the soil damp-
marked increase in maximum point displacement cor- ing constants also have little effect on the maximum
responding to increasing values of Q. It is also noted ENTHRU values. The maximum compressive forces do
that the percent increa·se in maximum point displace- ·
ment is relatively small for a small soil resistance, hut
greatly increases as the total soil resistance becomes TABLE 6.3. INFLUENCE OF SOIL DAMPING ON
DIFFERENT SOIL RESISTANCES FOR CASE BLTP-6;
large. This is also shown in Figure 6.3. Similar results 57.9 (Q = 0.1 FOR ALL CASES)

Maxi- Maxi-
TABLE 6.2. INFLUENCE OF SOIL QUAKE AT DIF- mum mum Maxi-
FERENT SOIL RESISTANCES FOR CASE BLTP-6; 57.9 Point Com- mum
WITH NO SOIL DAMPING Total Soil Displace-Maximum pressive Tensile
Resistance J ment ENTHRU Force Force
(kip) (sec/ft) (in.) (kip ft) (kip) (kip)
Maxi- Maxi-
mum mum Maxi-
Point Com- mum 50 0.0 1.49 6.80 225 109
Total Soil Displace- Maximum pressive Tensile 0.1 1.11 6.89 221 68
Resistance Q ment ENTHRU Force Force 0.2 0.85 7.03 221 41
(kip) (in.) (in.) (kip ft) (kip) (kip) 0.3 0.72 7.21 221 18
0.4 0.63 7.23 222 6
0.5 0.56 7.25 222 5
50 0.1 1.49 6.80 225 109
0.2 1.51 6.80 222 109 100 0.0 0.84 6.96 230 68
0.3 1.51 6.73 221 114 0.1 0.58 7.12 222 31
0.4 1.54 6.71 221 119 0.2 0.49 7.20 223 11
0.5 1.58 6.69 221 124 0.3 0.43 7.25 223 14
0.4 0.38 7.27 224 12
100 0.1 0.84 6.96 230 68 0.5 0.34 7.28 225 17
0.2 0.88 6.88 224 85
0.3 0.90 6.86 223 97 150 0.0 0.56 7.10 235 91
0.4 0.93 6.84 222 98 0.1 0.42 7.23 223 23
0.5 0.97 6.83 222 97 0.2 0.34 7.26 224 21
0.3 0.28 7.28 225 26
150 0.1 0.56 7.10 235 91 0.4 0.24 7.27 239 24
0.2 0.57 7.05 227 90 0.5 0.21 7.26 251 22
0.3 0.61 6.93 225 128
0.4 0.64 6.88 223 163 200 0.0 0.41 7.21 223 79
0.5 0.69 6.85 223 188 i).1 0.28 7.28 225 35
0.2 0.22 7.28 239 37
200 0.1 0.41 7.21 240 79 0.3 0.18 7.25 255 31
0.2 0.44 7.13 230 67 0.4 0.15 7.22 267 27
0.3 0.48 7.06 226 77 0.5 0.13 7.20 274 26
0.4 0.52 6.99 224 107
0.5 0.56 6.90 224 118 300 0.0 0.22 7.28 250 82
0.1 0.12 7.23 272 53
300 0.1_.,;< 0.22 7.28 250 82 0.2 0.09 7.18 286 41
8:i.
o..f
0.30
0.36
0.42
7.24
7.16
7.10
234
229
225
108
111
59
0.3
0.4
0.08
0.07
7.14
7.11
293
298
33
31
0.5 0.07 7.07 302 30
0.5 0.47 7.05 224 73
400 0.0 0.11 7.20 260 127
400 0.1 0.11 7.30 260 127 0.1 0.07 7.13 308 61
0.2 0.21 7.28 239 114 0.2 0.06 7.07 313 41
0.3 0.29 7.24 233 158 0.3 0.05 7.02 314 35
0.4 0.36 7.18 228 158 0.4 0.05 6.96 314 33
0.5 0.41 7.12 226 102 0.5 0.05 6.90 314 33

PAGE THIRTY-THREE
 1.8

1.6
RUT =50K

1.4

1.2
~
1-
z
11.1
:::E
11.1 1.0
RUT •IOOKIP
~
..J
a..
<I)
i5
1- 0.8
z
~ RUT= 150KIP
:::E
:::>
:::E 0.6 RUT= 200KIP RUT •eoK 1P
x
<{
:::E RUT= 300KIP

0.4 RUT • 400KIP 0.4

IWT •IOOICIP

0.2 0.2 IWT•IIIOICIP

-----RUT •200KI'
---=::::::==:::::::=======RUT RUT o!OQKIP
o4QOKIP
0 0 o~----O~.~-----O..J.2 0L~-----O~A-----...lQ~l5----------_..J
______
0 0.1 0.2 0.3 0.4 0.5
SOIL QUAKE-(IN.) SOIL DAMPING CONSTANT - J (SEC I FT)

Figure 6.3. Maximum point: displacement vs quake (Case Figure 6.4. Maximum point displacement vs so·il damp-
BLTP-6; 57.9). ing constant (Case BLTP-6; 57.9).

have a tendency to increase as J increases, especially The maximum point displacements again show the
when the soil resistance is large. While the tensile forces most consistent trend as J is varied, as shown in Figure
still do not follow any definite pattern, they are some- 6.4. The other cases studied showed this same trend,
what more regular than those determined by varying i.e., as J increases, the maximum displacement decreases
".Q." rapidly.

Chapter VII
CONCLUSIONS

The correlation between the numerical solution and tion such as the ram velocity was not reported. Still, it
the experimental data presented in Chapter V indicates was possible to study the behavior of the pile-driving
the potential accuracy of Smith's method, but the prob- hammers discussed. For example, the efficiency of the
lem involves so many important parameters that it is cushion assembly was remarkably consistent, in that they
extremely important to know as much as possible about were nearly independent o·f the type of pile, pile length,
their actual behavior. and soil resistance. The correlation between the wave
equation and the field data shown in Chapter III further
As shown in Chapter III, it is possible to determine illustrates that Smith's method is accurate, especially
valuable info.rmation from the wave equation even when the required data are known and need not be
though exact values for some of these parameters are assumed.
unknown. For example, several problems can be solved
in which the unknown parameter varies between upper Much of the value of this method of analysis is its
and lower limits as was done to determine the effect of flexibility. As illustrated in Chapter III, the wave equa-
the ram's elasticity. This study shows that only for steel tion can be used for any number of studies which other-
on steel impact does the elasticity of the ram affect the wise would not be possible.
solution.
It was shown that the stress-strain curve for a cush-
In order to study the Michigan dai~ over 5,000 ion is not a straight line. Instead, it follows a curve
problems had to be solved because certain key informa- which is closely parabolic. However, a straight line
PAGE THIRTY-FOUR
which has a slope equal to that of the true stress-strain and diesel) can he determined by the simple equation:
curve taken at a point halfway between zero and the
maximum strain gives accurate results. The cushion's
E = WR x h x e
where E = energy output in ft-lh
dynamic coefficient of restitution was found to agree W R = ram weight in lh,
with commonly recommended values. h ram stroke or equivalent stroke in ft,
The effect of internal damping in the concrete and and
steel piles was shown to he negligible in these cases, e - hammer efficiency (found to he 60 ';{
although it can he accurately accounted for by the wave for the Vul. No. l, 87% for the Vul.
equation if desired. SOC and SOC, and 100% for the diesel
hammers investigated by the Michigan
The data from the Michigan Study of Pile Driving Study).
Hammers were extrapolated to evaluate the true energy This is believed to be a most significant finding in
output of different pile driving hammers. It was found view of the existing controversy over the manufacturers'
that the energy output for all types of hammers (steam rated energies for diesel hammers.

Recommendations
RECOMMENDATIONS FOR FURTHER RESEARCH

The following areas are recommended for further cien<_::y of the pile-driving hammers pre.sently in use. This
research: type of research should be most interesting to the ham-
l. A complete evaluation of the data collected by mer manufacturers since present equipment could he
the Michigan State Highway Commission, including cor- optimized to drive piling faster and/ or reduce the driv-
relation of hammer energy, permanent set of pile per ing stresses during driving. The possibility that today's
blow, etc. This would require a major research effort pile-driving hammers are as efficient as possible through
because of the quantity of data reported. Also, because trial and error is remote.
certain variables were not determined, several theoreti- 3. Further research is needed to insure that the
cal solutions must be solved for each attempt correlation damping models proposed in Chapter IV are also ac-
until the unknown parameter can he "pinned down" with curate for timber piling, and to determine what damping
reasonable accuracy. For example, the solutions for constants should be used.
over 5,000 problems were required to complete the 28-
case study made in Chapter III. 4. Major research efforts are needed to investigate
every aspect of the soil resistance acting on the pile
2. A study to determine how to improve the effi- during driving.

PAGE THIRTY-FIVE
 References
1. Smith, E. A. L., "Pile Driving Analysis by the Wave 22. Smith, E. A. L., "Pile Driving Analysis by the Wave
Eq~ation," Proc. ASCE, Aug., 1960, p. 35. Equation," Proc. ASCE, Aug., 1960.
2. Federal Construction Council, "Foundation Piling," 23. Smith, E. A. L., "Pile Driving Analysis by the
Building Research Advisory Board, National Acade- Wave Equation," Transactions, ASCE, Vol. 127,
my of Sciences-National Research Council, Publi- 1962, Part I, p. 1171.
cation 987, 1962, p. 28.
24. Samson, Charles H., Jr., "Pile briving Analysis by
3. Ibid, p. 8. the Wave Equation (Computer Procedure)," Report
4. Cummings, A. E., "Dynamic Pile Driving Formulas," of the Texas Transportation Institute, Texas A&M
Journal of Boston Soc. of Civil Engrs., Jan., 1940. University, May, 1962.
5. Forehand, P. W. and J. L. Reese, "Pile Driving 25. Samson, Charles H., Jr., "Investigation of Behavior
Analysis Using the Wave Equation," Master of of Piles During Drivinl!;, Lavaca Bay Causeway,"
Science in Engineering Thesis, Princeton University, Unpublished Report of TTl to Texas Highway Dept.,
1963, p. 5. 1962.
6. Isaacs, D. V., "Reinforced Concrete Pile Formula," 26. Samson, Charles H., Jr., "Pile Stress Analysis-
lnst. Aust. Eng. J., Vol. 12, 1931, p. 415. Harbor Island Bay Bridge," Unpublished Report of
7. Fox, E. N., "Stress Phenomena Occurring in Pile the Texas Transportation Institute, July, 1962.
Driving," Engineering (London) Vol. 134, 1932,
p. 312. 27.· Hirsch, T. J., "Stresses in Long Prestressed Con-
crete Piles During Driving," Report of the Texas
8. Smith, E. A. L., "Pile Driving Analysis by the Wave Transportation Institute, Texas A&M University,
Equation," Proc. ASCE, Aug., 1960. September, 1962.
9. Chellis, R. D., "Pile Foundations," McGraw-Hill 28. Forehand, P. W. and J. L. Reese, "Pile Driving
Book Co., New York, 1951, p. 20. Analysis Using the Wave Equation," Master of
10. Fox, E. N., "Stress Phenomena Occurring in Pile Science in Engineering Thesis, Princeton Univer-
Driving," Engineering (London) Vol. 134, 1932. sity, 1963.
11. Taylor, D. W., "Fundamentals of Soil Mechanics," 29. Samson, C. H., Jr., T. J. Hirsch, and L. L. Lowery,
John Wiley & Sons, New York, 1956, p. 770. "Computer Study of Dynamic Behavior of Piling,"
12. Isaacs, D. V., "Reinforced Concrete Pile Formula," Journal of the Structural Division, ASCE, Vol. 89,
lnst. Aust. Eng. J., Vol. 12, 1931, p. 312. No. ST4, Proc. Paper 3608, August, 1963.
13. Fox, E. N., "Stress Phenomena Occurring in Pile 30. Hirsch, T. J., C. H. Samson, Jr., and L. L. Lowery,
Driving," Engineering (London) Vol. 134, 1932, "Driving Stresses in Prestressed Concrete Piles,"
p. 263. Structural Eng. Con£. of ASCE, San Francisco, Oct.,
14. Glanville, W. H., G. Grime, E. N. Fox, and W. W. 1963.
Davies, "An Investigation of the Stresses in Rein- 31. Hirsch, T. J., "Field Tests of Prestressed Concrete
forced Concrete Piles during Driving," British Bid~!;. Piles During Driving," Report of the Texas Trans-
Research Bd. Tech. Paper No. 20, D. S. I. R., 1938. portation Institute, Texas A&M University, August,
15. Cummings, A. E., "Dynamic Pile Driving Formula," 1963.
Journal of Boston Soc. of Civil Engrs., Jan., 1940, 32. Hirsch T. J. "Computer Study of Variables which
p. 6. Affect 'the B~havior of Concrete Piles During Driv-
16. Smith, E. A. L., "Pile Driving Impact," Proceed- ing," Report of the Texas Transportation Institute,
ings, Industrial Computation Seminar, September, Texas A&M University, August, 1963.
1950, International Business Machines Corp., New
York, N. Y., 1951, p. 44. 33. Samson, C. H., Jr., F. C. Bundy, and T. J. Hirsch,
"Practical Applications of Stress-Wave Theory in
17. Smith, E. A. L., "Impact and Longitudinal Wave Piling Design," Presented to the annual Texas Sec-
Transmission," Transactions, ASME, August, 1955, tion ASCE meeting; San Antonio, Texas, October,
p. 963. 1963. .
18. Smith, E. A. L., "What Happens When Hammer 34. Hirsch, T. J., and C. H. Samson, Jr., "Driving Prac-
Hits Pile," Engineering News Record, McGraw-Hill tices for Prestressed Concrete Piles," Report of the
Publishing Co., Inc., New York, N. Y., September 5, Texas Transportation Institute, Texas A&M Univer-
1957, p. 46. sity, April, 1965.
19. Smith, E. A. L., "Tension in Concrete Piles During 35. Hirsch, T. J., "Fundamental Design and Driving
Driving," Journal, Prestressed Concrete Institute, Considerations for Concrete Piles," 45th Annual
Vol. 5, 1960, pp. 35·40. Meeting of Highway Research Board, Washington,
20. Smith, E. A. L., "The Wave Equation Applied to D. C., January, 1966.
Pile Driving," Raymond Concrete Pile Co., 1957. 36. Hirsch, T. J., and Thomas C. Edwards, "Impact
21. Smith, E. A. L., "Pile Calculations by the Wave Load-Deformation Properties of Pile Cushioning
Equation," Concrete and Constructional Engr., Lon- Materials," Report of the Texas Transportation In-
don, June, 1958. stitute, Texas A&M University, July 1965.
PAGE THIRTY-SIX
 37. Smith, E. A. L., "Pile Driving Impact," Proceed- Materials," Report of the Texas Transportation In-
ings, Industrial Computation Seminar, September, stitute, Texas A&M UniveJ:sity, July, 1965.
1950, International Business Machines Corp., New
York, N. Y., 1951. 54. Hirsch, T. J., and Thomas C. Edwards, "Impact
Load-Deformation Properties of Pile Cushioning
38. Smith, E. A. L., "Impact and Longitudinal Wave Materials," Report of the Texas Transportation In-
Transmission," Transactions, ASME, August, 1955. stitute, Texas A&M University, July, 1965.
39. Smith, E. A. L., "Pile Driving Analysis by the Wave 55. Ibid. p. 2Y.
Equation," Proc. ASCE, Aug., 1960.
56. Samson, C. H., Jr., T. J. Hirsch, and L. L. Lowery,
40. Forehand, P. W., and J. L. Reese, "Pile Driving "Computer Study of Dynamic Behavior of Piling,"
Analysis Using the Wave Equation," Master of Journal of the Structural Division, ASCE, Vol. 89,
Science in Engineering Thesis, Princeton Univer- No. ST4, Proc. Paper 3608, August, 1963.
sity, 1963.
57. Jacobsen, L. S., "Steady Forced Vibrations as In-
41. Samson, C. H., Jr., T. J. Hirsch, and L. L. Lowery, fluenced by Damping," ASME Transactions, Vol.
"Computer Study of Dynamic Behavior of Piling," 52, 1930, p. 168.
Journal of the Structural Division, ASCE, Vol. 89,
No. ST4, Proc. Paper 3608, August, 1963. 58. Smith, E. A. L., "Pile Driving Analysis by the Wave
Equation," Transactions, ASCE, Vol. 127, 1962,
42. Smith, E. A. L., "Pile Driving Analysis by the Wave Part I, p. 1167.
Equation," Transactions, ASCE, Vol. 127, 1962,
Part I, p. 1152. 59. Samson, C. H., Jr., T. J. Hirsch, and L. L. Lowery,
"Computer Study of Dynamic Behavior of Piling,"
43. Chellis, R. D., "Pile Foundations," McGraw-Hill Journal of the Structural Division, ASCE, Vol. 89,
Book Co., New York, 1951, p. 29. No. ST4, Proc. Paper 3608, August, 1963, p. 417.
44. Michigan State Highway Commission, "A Perform- 60. Ibid, p. 427.
ance Investigation of Pile Driving Hammers and
Piles," Office of Testing and Research, Lansing, 61. Ibid, p. 429.
March, 1965. 62. Hirsch, T. J., and Thomas C. Edwards, "Impact
45. Housel, W. S., "Pile Load Capacity: Estimates and Load-Deformation Properties of Pile Cushioning
Test Results," Journal of the Soil Mechanics and Materials," Report of the Texas Transportation
Foundations Division, ASCE, Proc. Paper 4483, Institute, Texas A&M University, July, 1965, p. 21.
September, 1965. 63. Ibid, p. 20.
46. Michigan State Highway Commission, "A Perform- 64. Samson, C. H., Jr., T. J. Hirsch, and L. L. Lowery,
ance Investigation of Pile Driving Hammers and "Computer Study of Dynamic Behavior of Piling,"
Piles," Office of Testing and Research, Lansing, Journal of the· Structural Division, ASCE, Vol. 89,
March, 1965, p. 330. No. ST4, Proc. Paper 3608, August, "1963, p. 418.
47. Smith, E. A. L., "Pile Driving Analysis by the Wave 65. ~Chellis,
R. D., "Pile Foundations," McGraw-Hjll
Equation," Transactions, ASCE, Vol. 127, 1962, Book Co., New York, 1951, p. 23.
Part I.
66. Chan, P. A., "A Laboratory" Study of Dynamic
48. Fo·rehand, P. W. and J. L. Reese, "Pile Driviiig Load-Deformation and Damping Properties of Sands
Analysis Using the Wave Equation," Master of Concerned with a Pile-Soil System," a dissertation,
Science in Engineering Thesis, Princeton Univer- Texas A&M University, unpublished, January, 1967.
sity, 1963.
67. Hirsch, T. J., and Thomas C. Edwards, "Use of the
49. Michigan State Highway Commission, "A Perform- Wave Equation to Predict Pile Load Bearing Ca-
ance Investigation of Pile Driving Hammers and pacity," Report of the Texas Transportation Insti-
Piles," Office of Testing and Research, Lansing, tute, Texas A&M University, unpublished, August,
March, 1965, p. 246. 1966.
50. Hirsch, T. J., and Thomas C. Edwards, "Impact 68. Samson, C. H., Jr., F. C. Bundy, and T. J. Hirsch,
Load-Deformation Properties of Pile Cushioning "Practical Applications of Stress-Wave Theory in
Materials," Report o·f the Texas Transportation Piling Design," Presented at the annual Texas Sec-
Institute, Texas A&M University, July, 1965, p. 1. tion ASCE meeting, San Antonio, Texas, October,
1963, p. 1162.
51. Smith, E. A. L., "Pile Driving Analysis by the Wave 69. Rand, Mogens, "Explosion Adds Driving Force to
Equation," Transactions, ASCE, Vol. 127, 1962, Diesel Hammer," Engineering Contract Record,
Part I, p. 1162.
December, 1960.
52. Samson, C. H., Jr., T. J. Hirsch, and L. L. Lowery, 70. Rand, Mogens, "Test Performed with the Delmag
"Computer Study of Dynamic Behavior of Piling," Diesel Hammer, Type D-12," December 13, 1955,
Journal of the Structural Division, ASCE, Vol. 89, Publication of the Department of Mechanical Engi-
No. ST4, Proc. Paper 3608, August, 1963. neering, University of Toronto.
53. Hirsch, T. J., and Thomas C. Edwards, "Impact 71. Rand, Mogens, "How the Diesel Pile Hammer
Load-Deformation Properties of Pile Cushioning Works," Roads & Streets, May, 1961.
PAGE THIRTY-SEVEN
 ---------- --~----------------------------------------------.

Appendix A
PROGRAM INPUT DATA

CARD 101 (Required)

IDl and ID2 All "ID" values are for identifica- NOP(I) VALUE FUNCTION
tion only and can be either alpha- NOP(3) Used to specify the input method
betic or numeric. for the internal spring stiffness.
1/.lt - Time interval. If left blank, .ltcr/2 (XKAM(I).
will be used. (1/sec)
= 1 Read one stiffness for each inter-
MP - Total number of segments in the nal spring from card series 300.
system to be analyzed.
VELMI -'- Initial velocity of the ram. (ft!sec) = 2 Read the stiffness values for only
the first five and last five internal
MH Element number of the first pile springs on a single_ card 300, and
segment. assign the fifth value to all re-
NR maining internal springs.
Number of divisions of the ram.
EEM(NR) Coefficient of Restitution of spring (NOP(3) = 2 is used under the
number NR, directly under ram. same conditions as NOP(2) = 2.
------------------
EEM(NR+l) Coefficient of Restitution of spring NOP(4) Used to specify what soil resist-
number NR+l. ance distribution act along the pile.
GAMMA(NR) The minimum force in the spring = 1 Read RUM(I) for each element
beneath the ram once that force from card series 400, and set the
has reached a maximum. · (kip) point bearing soil resistance RUM
For example, if the diesel hammer (MP+l) equal to RUP.
explosive pressure causes 158.7 kip
minimum force in this spring, set = 2 Set all side resistances equal to
GAMMA(NR) = 158.7 kip. If the zero, and set RUM(MP+l)
minimum force the spring can RUP.
transmit is zero (for example,
when no tensile force can exist be- = 3 Distribute RUT-RUP uniformly
tween the ram and anvil) set the along the side of the pile from
corresponding GAMMA(!) = 0.0. segment MO thru MP, and- set
If the spring represents a continu- RUM(MP+L) = RUP.
ous body such as the spring be-
tween any two pile segments, it can = 4 Distribute RUT-RUP triangularly
transmit tensile forces between the along the pile between segments
elements. This is signified by set- MO and MP, and set RUM(MP+l)
ting GAMMA(!) equal to any = RUP.
negative value, Usually -1.0 kip. = 5 Read one 450 series card for each
GAMMA mass upon which a nonlinear re-
(NR+l) Same as above, but for spring num- sistance vs displacement curve
ber NR+l. acts. If a linear curve also hap-
NSTOP Total number of time intervals the pens to be acting on an element,
program is to run. it must also be input on a 450
series card.
NOP(I) VALUE FUNCTION
NOP(l) Used to read cards 103-106 and NOP(5) Used to specify the input method
print out the data for problem for GAMMA(I). Note: The sig-
identification. nificance of GAMMA(!) is dis-
cussed in the "500 card series."
=1 No identification card is to be used.
= 1,2 Read GAMMAl and GAMMA2
=2 Read and print a single ID card. from card 101 and assign GAM-
(card 103) MAl to internal spring number
=3 Read and print two ID cards. NR, and assign GAMMA2 to spring
(cards 103 and 104) number NR+l. Then set GAM-
MA(!) of the remaining springs
=4 Read and print ID cards 103, 104, to -1.0.
and 105.
Read and print ID cards 103, 104, ==3 Same as for NOP(5) =2, except
=5 that GAMMA(NR+2) is also set
105, and. 106. equal to 0.0.
NOP(2) Used to specify the input method = 4 Same as for NOP(_5) =2, except
for the segment weights W AM (I). GAMMA(NR+2) =0.0 and GAM-
= 1 Read one weight for each segment MA(NR+3) =0.0. This option is
(card series 200) . used when a large number of ele-
ments such as an anvil, follower,
== 2 R_ead the segment weights for only load cell and pile cap are encoun-
the first five and last five seg- tered, since these elements cannot
ments of the pile system from a transmit a tensile force to the next
single card· (card 200), and equate element. This option can be used
all remaining segment weights to to set up to eight consecutive val-
the sixth weight in the system. ues of GAMMA(!) =0.0 by setting
(NOP(2) = 2 is used when a large NOP(5) =8.
number of equal weights are pres-
ent except for the first or last few = 9 Read GAMMA(!) for each spring
weights.) from card series 500.
PAGE THIRTY-EIGHT
 NOP(I) VALUE FUNCTION NOP(I) VALUE FUNCTION
NOP(6) Used to specify the input method = 2 In this case, the force at the head
for EEM(I). of the pile at all times is known,
probably by experimental methods,
= 1 Read EEM1 and EEM2 from card and this force curve is to be ap-
101 set EEM(NR) =EEM1, and piied at the head of the pile. The
EEM(NR+1) =EEM(2). Then set force at each time interval FOR-
EEM(I) for all other springs equal GIN (t) is read from card series
to 1.0 (perfectly elastic). 1300 (kip).
= 2 Read EEM(I) for each spring = 3 Same as when NOP(14) =2, except
from card series 600. that galvanometer readings rather
NOP(7) than forces at each time interval
Used to specify the input method are input and the cushion forces
for BEEM(I). are determined by the computer.
= 1 Set all BEEM(I) =0.0. In this case, the information on the
1400 header card is needed, fol-
= 2 Read BEEM(I) for each spring lowed by the galvanometer deflec-
from card series 700. tion at each time interval from
NOP(8) Used to specify the input method card series 1400.
for VEL(I). NOP(15) Used to specify how gravity is to
= 1 Read VELMI from card 101 and be accounted for in the solution.
set VEL(I,t=O) for all segments = 1 The effect of gravity is to be neg-
of the ram (usually one segment) lected.
equal to VELMI. Set all other
VEL (I) = 0.0. = 2 Gravity is to be considered, with
the initial displacement of each
= 2 Read VEL(!) for each segment segment, D(I,O), and the initial
f:rom card series 800. soil resistances RAM(I,O) assumed
NOP(9) Used to specify input method for to be zero.
Q(I). = 3 Gravity is to be considered, and
D(I,O) and RAM(I,O) are to be
= 1 Read QSIDE and QPOINT from approximated by Smith's suggest-
card 102 and set all Q(I) along ed method.'"
side of the pile equal to QSIDE. = 4 Gravity is to be considered, and the
Set Q(MP+1) under pile tip equal values for D(I,O) and RAM(I,O)
to QPOINT. are computed by Samson's suggest-
= 2 Read Q(I) for each element includ- ed method.61
ing Q(MP+1) from card series
900. NOP(16) Used to specify the number of
problems to be solved using the
NOP(10) Used to specify input method for basic data given on cards 101
SJ(I). through the 1700 card series.
= 1 Only one problem is to be solved
= 1 Read SIDEJ and POINTJ from using this set of data.
card 102. Set all SJ (I) along side
of pile equal to SIDEJ and SJ(MP = 2 Run more than one problem with
+1) under pile tip equal to changes in these data as specified
POINTJ. on card 1600.
= 2 Read SJ ( (I) for each element in- NOP(17) Used to specify whether the ulti-
cluding SJ(MP+1) from card se- mate pile capacities predicted by
ries 1000. various . pile driving equations are
desired.
NOP(ll) Used to specify the input method
for DYNAMK(I). = 1 No capacities are to be computed.
= 1 Set all DYNAMK(I) =0.0. = 2 Using the information from card
1700 an!f the information provided
= 2 Read DYNAMK(I) for each spring by the wave equation solution,
from card series 1100. solve for the ultimate resistance
NOP(12) Used to specify input method for to failure as predicted by several
A(I). popular pile driving equations.
= 1 Read AREA from card 102 and set CARD 102 (Required)
all A(I) equal to AREA. ID3 Identification.
= 2 Read A(I) for each internal spring ID4 identification.
from card series 1200.
RUT The total static soil resistance act-
NOP(13) Used to specify which method of ing on the pile (kip).
internal damping is to be used in
the pile. RUP The total static' soil resistance act-
ing beneath .the point (kip).
= 1 Use Smith's method (refer to Fig- MO - Number of first element upon
ure 5.13b). which soil resistance acts.
= 2 Use standard linear solid method QSIDE
(refer to Figure 5.13c). Soil quake along side of pile, if a
single value exists. If not, set
NOP(14) Used to specify how the ·force in QSIDE=O.O (in.).
the cushion after impact is to be QPOINT Soil quake beneath pile point (in.).
determined.
SIDEJ Soil damping factor in shear along
= 1 Calculate cushion forces from the the side of the pile if a single value
wave equation applied to the mov- exists. If not, set SIDEJ =0.0
ing ram after impact. (sec/ft).
PAGE THIRTY-NINE
 POINTJ - Soil damping factor in compres- b) If NOP(4) =2, set all side fric-
sion beneath· the pile point (sec/ tion=O.O and set RUM(MP+l) =
ft). RUP.
NUMR - Number of elements for which the c) If NOP(4) =3, distribute (RUT-
soil spring does not have a linear RUP) uniformly along the pile
stress-strain curve. starting from segment number MO
!PRINT - Print frequency. For example, if to number MP, and set RUM(MP
the solution at every 5th time in- +l)=RUP.
terval is wanted, set IPRINT=5. d) If NOP(4) =4, distribute (RUT-
AREA - A constant used to convert the RUP) triangularly between MO
forces into stresses or other more and MP set RUM(MP+1) =RUP.
convenient values (such as chang- e) If NOP(4) =5, read NUMR
ing lb. to kip by setting AREA= cards, each of which can define a
1000.0). linear or nonlinear force-displace-
NS1-NS6 The element numbers for which ment curve for the soil (see card
solutions vs time interval will be series 450).
printed. Maximum values and 450 CARD SERIES (Required if NOP(4) =5)
other information are always
printed for each element after When NOP(4) =5, the soil resistance vs displacement
NSTOP time intervals have curve is nonlinear. This requires ten soil resistances to be
elapsed. read for each soil spring, one for each displacement cor-
responding to a multiple of Q/10. As shown on data card
CARDS 103-106 (Required only if NOP(l) =2,3,4,5) 451, I is the number of the element upon which the non-
If NOP(l) =1, no identification card will be read. If linear resistance is. acting, XKIM(I) is the unloading
NOP(l) =2, read card 103 containing 72 columns of alpha- spring rate (kip/in.), and R(I,J) are the soil resistances
betic or numeric identification and print this information (kip) at each of the displacements Q/10, 2Q/10, . . . ,
above the problem. If NOP(l) =3, read and print two 9Q/10, Q. Whenever NOP(4) =.5, one 450 series card is
identification cards, up to a maximum of four cards required for each element upon which soil resistance acts.
(NOP0.)=5). 500 CARD ·SERIES (Required when NOP(5) =2)
200 CARD SERIES (Required) IDG1, IDG2 - Identification.
IDWl, IDW2 - Throughout this Input, variables GAMMA(!) The minimum force possible in
beginning with the letters "ID" are spring I after a peak compressive
for identification, in this case to force has passed, except that any
help identify what s e g m e n t negative GAMMA(!) is construed
weights are being used. to mean that that spring can trans-
WAM(I) - The weight of element number I mit a tensile force of any magni-
(kip). a) If NOP(2) =1, the com- tude (kip).
puter will read MP s e g m e n t 600 CARD SERIES (Required when NOP(6) =2)
weights, ten segment weights to a
card from cards 201-230, up to a IDE1, IDE2 Identification.
maximum of 300 segments. For EEM(I) The coefficient of restitution for
example, if the system is divided MP-1 internal springs. This deter-
into 37 segments, four 200 series mines the slope of the unloading
cards must be included in the data: curve (dimensionless).
201 through 204. b) If NOP(2) =2,
in this case the pile must have a 700 CARD SERIES (Required when NOP(7) =2)
constant weight per foot along its IDB1, IDB2 Identification.
length. Since the pile is usually BEEM(l) - The damping coefficient of the
divided into equal segment lengths, MP-1 internal springs (in. sec/ft).
only a few of the element weights
are different. Therefore, only the 800 CARD SERIES (Required when NOP(8) =2)
top five weights (the ram, anvil, IDYl, IDV2 - Identification .
. . . ) and the bottom five weights VEL(I) - The initial velocities of each of the
( . . . , pile segment, pile point) MP weights (ft/sec).
must be read from the card 200.
The computer then sets all other 900 CARD SERIES (Required when NOP(9) =2)
element weights equal to the sixth IDQ1, IDQ2 Identification.
value punched in the card.
Q(I) The soil "quake" for MP+1 soil
300 CARD SERIES (Required) springs (in.) .
IDKl, IDK2 - Identification. 1000 CARD SERIES (Required when NOP(lO) =2)
XKAM(I) - The internal spring rate of spring IDJ1, IDJ2 Identification.
I (kip/in.). SJ(I) The soil damping factor for MP+l
a) If NOP(3) = 1, the computer soil spring (sec/ft).
reads MP-1 spring rates from
cards 301-330. 1100 CARD SERIES (Required when NOP(11) =2)
b) If NOP(3) =2, the first and last IDDK1, IDDK2 - Identification.
five XKAM(I) are read from card DYNAMK(I) - The dynamic spring rate of MP-1
300, and the remaining XKAM(I) internal springs (kip/in.).
are set equal to the sixth
XKAM(I) value, i.e., XKAM 1200 CARD SERIES (Required when NOP(12) =2)
(MP-4). IDA1, IDA2 - Identification.
A(I) - The cross-sectional area of each of
400 CARD SERIES (Required if NOP(4) =1) the MP-1 internal springs (in.').
IDRLl, IDRL2 - Identification.
RUM(I) - The ultimate static resistance of 1300 CARD SERIES (Required when NOP(13) =2)
the soil acting on pile segment I FORCIN (INTV) - The force acting on the head of the
(kip). a) If NOP(4) =1, read MP pile (kip) at time interval INTV,
ultimate soil resistances, from for NSTOP intervals with a maxi-
cards 401-430, and set RUM(MP mum NSTOP equal to 100 time
+1) equal to RUP. intervals.

PAGE FORTY
1400 CARD SERIES (Required when NOP(14) =2) that the effects of ram velocities
CARD 1400 Header Card. of 10, 12, 14, 16, 18, and 20 ft/sec
APILE - The area of the head of the pile are being studied. The value of
(in.'). DVl would be
EMODUL - The modulus of elasticity of the (12 ft/sec- 10 ft/sec)
pile (kip/in.'). 10 ft/sec
RGAGE - The strain gage resistance (ohm). or DV1=0.20. In this case, NOPP
RCAL - Calibration resistance (ohm). (1) would equal 6 since 6 separate
problems are to be run.
ACTIVG - Number of active gages. The variables controlled by DVl
GFACTR - Gage factor for the gages used. to DK1 are also listed in Table A.l.
D1 - Displacement of the galvanometer
trace when RCAL is thrown into TABLE A.l. LIST OF PARAMETER VARIATIONS
the bridge at the head of the pile AND THEIR CONTROLLING OPTIONS
(in.). Controlling Per Cent Increase Parameter
D2 Through Option in Original Value Controlled
D5 Galvo displacements corresponding
to RCAL at any other four strain NOPP(l) DVl VELMI (Initial ram
gage points (in.). velocity)
CARDS 1401 UP TO 1410 NOPP(2) DWl W(1)
NOPP(3) DW2 W(2)
DGALVI(INTV) - The galvanometer deflection for NOPP(4) DWl W(3) through W(MP)
the gage at the head of the pile, at NOPP(5) DKl XKAM(1)
interval number INTV (in.). NOPP(6) DK2 XKAM(2)
CARD 1500 (Required when NOP(15) =4) NOPP(7) DKI XKAM(3) through
XKAM(MP-1)
Fl and F2 - Forces known to lie on the true NOPP(8) DQI O.SIDE
dynamic force vs compression NOPP(9) DQP QPOINT
curve .of the cushion (kip). NOPP(lO) DJI SIDEJ
Cl and C2 The cushion compressions corre- NOPP(ll) DJP POINTJ
sponding to Fl and F2, respective- NOPP(l2) DRI RUT
ly (in.). NOPP(13) DRP RUP
NOPP(14) DRI RUT & RUP
CARD 1600 (Required when NOP(16) =2) NOPP(15) DEl EEM(l)
NOPP(I) When a number of cases are to be NOPP(l6) DE2 EEM(2)
solved for which only a few pa-
rameters will change, NOPP(I) CARD 1700 (Required when NOP(17) =2)
designates which parameter to AREAP Cross-sectional area of pile (in.').
vary and how many different val-
ues it should be assigned. For ex- XLONG Length of pile (ft).
ample: NOPP(l) =5 indicates that ELAST Modulus of elasticity of pile
five problems are to be solved, for (kip/in.').
which only the ram's initial veloci- CENR Value for use in ENR pile driving
ty will vary. Each NOPP(I) con- formula.
trols a single variable as shown in QAVG Average ground "Quake" (in.).
Table A.l.
DVl Through WRAM Ram weight (kip).
DKl These parameters control the per- WPILE Pile weight (kip).
cent change in the variables men- ENERGY Actual energy output of the ram
tioned above. For example, assume (ft lb).

PAGE FORTY-ONE
 Appendix B
EXAMPLE PROBLEM

Introduction c. Soil damping factor "J" and soil quake

The following example problem is given to illustrate "Q"-not known.
the steps necessary to arrive at a solution. In the previ- 4. Miscellaneous Data
ous chapters, the functional components involved were a. Load Cell Weight = 580 lb.
discussed separately; for example, the driving hammer, b. Additional Helmet Weight 1,080 lb.
pile, soil properties, etc. However, the input data is B. Input Data Calculations
more easily handled by grouping according to similar Card 101
physical quantities rather than functional quantities. For l. IDl-ldentification Tag, use BLTP-6
example, one series of cards is used to input all segment 2. ID2-Identification Tag, use 57.9.
weights, another for the spring rates. The order in which
the input data is set up for the example problems is by 3. Segment Lengths- Although s e g m en t
no means unique, but it probably should be followed lengt~s of lO ft are usually satisfactory, a
until the programmer becomes familiar with the opera- 5 ft length will be used to increase the ac-
tions involved. curacy of the solution.
4. Time Interval-The normal time interval
It should be noted that any variable without a deci- of l/4000 to 1/5000 iterations/sec must be
mal point (such as MP, MH, NR, NSTOP, and NOP(I) halved since the normal segment length of
on card 101) is always an integer and must be entered 10 ft was reduced by half. Therefore, use
as far to the right in its field as. possible. Also, the ~t = 1/10,000 sec or l/ ~t 10,000. =
decimal point does not have to be punched for any varia-
ble which has a decimal place already shown on the 5. MP-The total number of segments as
data sheet unless it is desired to change its position. shown in Figure 3.4 is 3 above the pile
For example, if the initial ram velocity (IVEL on card plus 14 pile segments. Thus, MP 17. =
101) is 13.48 ft/sec, the numbers 1, 3, 4, and 8 should 6. Since the ram velocity at impact was not
be punched in columns 19 through 22, respectively. How- recorded, the following ram velocities will
ever, to enter a velocity of 127 ft/sec into IVEL, punch be studied: IVEL = 8, 12, 16, and 20
1, 2, and 7 in columns 19, 20, and 21, and punch a deci- ft/sec.
mal point in column 22. 7. MH-The first pile segment weight = 4.
Except for this last case, decimal points need never 8. NR-Number of divisions of the ram l. =
be punched. 9. EEMI-Coefficient of restitution of cush-
ion = 0.4, EEM2--coefficient of restitu-
Example Problem tion of load cell = 1.0.
Since case BLTP-6; 57.9 (from the Michigan Pile 10. Since springs 1, 2, and 3 cannot transmit
Study) was one of the problems most often used in this tensile forces, GAMMA ( 1), ( 2) , and ( 3)
report, the input data required for its solution will be are 0.0. The remaining GAMMA (I) are
determined first. Figures 3.3 and 3.4 show the real set equal to· - 1.0. This is done by setting
system and the idealized system. GAMMA1 = GAMMA2 = 0.0 and desig-
A. Given Information-Case BLTP-6; 57.9 nating NOP(S) = 3 so that GAMMA(3)
l. Hammer Data-Vulcan #1 will also be set 0.0. =
a. Manufacturer's Rated Energy = 15,000 11. To allow the wave time to make two com-
ft lb, normal stroke = 3 ft. plete passes up and down the pile, NSTOP
b. Ram Weight = 5,000 lb, velocity at im- is set = 173 iterations. This is found from
pact not measured. the velocity of travel of the stress wave and
c. Driving Cap Weight = 1,000 lb. the value of ~t.
d. Cushion Data = Oak block, 6-~ in.
deep by 11-~ in. in diameter, direction v 'IE/ .. /30,000,000
of grain unknown, condition of cushion
wave =v p =
V (0.283/386)
unknown (somewhere between new and = 202,000 ips or
"crushed and badly burnt"). - 202,000
2. Pile Data-CBP 124 H-section Vwave - = 16,800 ft/sec.
12
a. Area = 15.58 in. 2 • Total distance wave must travel = 4(72.5)
b. Weight = 53 lb/ft. = 290 ft.
c. Total Length = 72.5 ft.
d. Driven Length = 57.9 ft. .
T otal time . d 290 ft
reqmre = 16,800 ft/sec
.0173
e. Modulus of Elasticity = 30 x 106 •
3. Soil Data sec.
a: Ultimate Soil Resistance = 300 kip
NSTOP
Total time
(static value from load test after soil ~t
"set-up").
.0173 sec
b. From driving log, 75 percent of the soil -:-::--;::-::-::-:::-:::.,....---;-;---:--
( 1/10,000) sec/iteration
= 173 iterations.
resistance is assumed point bearing and
25 percent side resistance. Therefore, use NSTOP = 173 iterations.
PAGE FORTY-TWO
 12. Option Calculations-NOP(I) 5. MO-Sinee the length of pile in the ground
a. NOP(1)-No header cards to be read was 57.9 ft, the first segment upon which
in and printed out, so NOP(1) = L soil resistance acts is given by:
b. NOP(2)-Read segment weights from
card series 200 (long form), so NOP(2)
MO = MP +1_ ( Depth Driven )
Segment Length
=L
c. NOP(3)-Read spring constants from 17 + 1- 57.9
card series 300 (long form) , so N OP ( 3) 5.0
=L 18 11.6
d. NOP (4) -Assume triangular soil dis- 18 12
tribution along the side of the pile, so so MO 6
NOP(4) = 4. 6. QSIDE and QPOINT-Smith's recommend-
e. NOP(S)-Since GAMMA(3) is to be ed value of 0.1 in. will be used due to lack
set equal to 0.0, NOP(S) = 3. of better soils data.
f. NOP(6)-Since all the internal springs 7. SIDEJ and POINTJ-For the same reasons
are considered perfectly elastic, except above for values of Q, use SIDEJ = 0.05
for the first one or two for which values sec/ft and POINTJ = 0.15 sec/ft.
of "c" are given by EEM1 and EEM2, 8. NUMR-Since the soil springs all act as
set NOP(6) = 1 (short form, no series shown in Figure 6.1 (a) , NUMR = 0.
600 cards). 9. Set IPRINT = 5 to print out the solution
g. NOP(7)-Assume zero internal damp- at every 5th iteration.
ing in the steel pile, thus set NOP(7) 10. AREA-A single factor will be used to
= 1 and do Iiot include the 700 card change all forces from lb to kip, thus AREA
series. = 1000.0.
h. NOP(8)-0nly the ram has an initial 11. NS1 through NS6-In this case, the solu-
velocity, so NOP(8) = 1, no 800 card tions for segments 1, 2, 3, 4, 11, and 17
series. are desired and, therefore, NS1 through
i. NOP(9) and NOP(10)-Since more NS6 are given these values.
exact soils information is not available
Smith's recommended values for Q and Cards 201-202
J will be used and input on card 102 Segment Weights-As shown in Figure 3.4,
(short form). Thus, NOP(9) several weights normally present during
= NOP(10) = L driving have been added between the pile
and the driving cap to obtain experimental
j. NOP(ll)-No damping, set NOP(ll)
=L data.
a. W ( 1) = ram weight = 5.0 kip.
k. NOP(12)-Use a single factor to
change force to stress for all springs- b. W ( 2) = driving cap weight + ¥z of
NOP(12) = L the load cell weight =1.29 kip.
I. NOP(13)-Use the damping procedure c. W(3) = ¥z load cell weight + helmet
illustrated in Figure 5.13(a), so NOP = 1.37 kip.
(13) = l. d. W(4) through W(17)
weights =
= =
pile segment
(53 lb/ft) (5 ft) 0.265
m. NOP(14)-Calculate the force at the
pile head from the action of the ram kip.
so NOP(14) = L Cards 301-302
n. NOP(15)-Neglect gravity effects- Segment Stiffness
NOP(15) = L
a. Because of the lack of data concerning
o. NOP(16)-Since several parameters cushion stiffness, several values of K ( 1)
are to be varied, set NOP(16) = 2, will be run: K ( 1) - 500, 1,000, and
thus card 1600 must be included in the 1,500 kip. in.
data.
b. The helmet was found to be extremely
p. NOP(17)-Do not calculate driving re- stiff compared to the load cell, so K(2)
sistance predicted by pile driving equa- was taken as the stiffness of the load
tions. NOP(17) = L cell alone. From dimensions of the load
Card 102 cell given in the Michigan Report and
using K = AE/L, the spring rate of the
L ID3-Identification Tag, use 12H53.
load cell was found to be 86,500 kip/in.
2. ID4-Identification Tag, use L = 72.
c. The spring rate of each 5 ft pile segment
3. RUT-Since the Michigan Report noted a is found by: ·
soil "set-up" of about 2.0, the static resist-
ance actually encountered during driving K = AE (15.58) (30x10 3 )
was probably around half of the measured L 5x12
400 kip, so RUT = 200 kip. = 7,790 kip/in.
4. RUP-Assuming 75 percent of the total soil So K(3) through K(16) 7,790
resistance at the point, RUP = 150 kip. kip/in.
PAGE FORTY-THREE
 Card I600 2. Parameter Change Constants-DVI, DEI,
I. Parameter Options-NOPP (I) -Note that DE2, etc. These values specify the desired
all values of NOPP(I) are set =I except increase in a given parameter based on the
parameter's original value. They may he
when an option is used to vary its assigned
parameter, in which case NOPP(I) can calculated from the equation:
equal 2 through 9. _ Second Vaue · Initial Value
Constant - I mha
. . l Va l ue
a. Since IVEL is to he given the four
values of 8, I2, I6, and 20 ft/sec, Thus, since the initial value of IVEL is
NOPP(I) = 4. 8 ft/sec and the second value is I2 ft/sec

h. NOPP(2) through NOPP(4) = I since DVI = I2 8 8 : = 1.0

no segment weights are to he varied.
The value for DKl is therefore given by
c. NOPP(5) = 3 since three different DKI = 1000-500 = 500 = I O
cushion stiffnesses are to he used (K (I) 500 500 .
=500, 1,000, and I,500 kip/in.)
All other values such as DWl, DW2, etc.,
d. NOPP(6) through NOPP(7)-I since may he left blank or given any value for
no other parameter changes are re- later use since they are not used as long as
quired. the corresponding NOPP(I) = I.

PAGE FORTY-FOUR

L __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
 Appendix C
PROGRAM LISTING

PAGE FORTY-FIVE
 $EXECUTE IBJOB
$18JOB
SIBFTC MAIN
C - PROGRAM CONSISTS Of APPROXIMATELY 1200 LINES OUTPUT
C - LINES/PROBLEM = 50 +2•MP +NSTOP/IPRINT (UNLESS JS CHANGES)
C - RUN TIME FOR PROGRAM IS ABOUT 1 MINUTE
C - RUM TIME FOR ONE PROBLEM IS ABOUT = (MP•NSTOP)/60,000 (MINUTES)
C NOP(l) = OtltNO IDENTIFICATION CARDS (SERIES 103)
C = 2, READ IDENTIFICATION CARD 103 {72 COLS OF ALPHAMERIC POOP)
C = 3, READ 2 IDENTIFICATION CARDS
C = 4, ETC. UP TO 4 CARDS
C NOP(2) = 0
C = l,REAO NEW WAM(I),I=l,MP
C = 2t READ CARD 200 MAXIMUM DIFFERENT WAM(I} = TEN
C NOP(3J = 0
C = l,READ NEW XKAM(Il,l=l,N
C = 2, READ CARD 300 MAXIMUM DIFFERENT XKAM(I) = TEN
C ~0P(4) = O,USE OLD SOIL RESISTANCE VALUES,STANDARD OR GENERAL METHOD
C =!,READ NEW STANDARD RUM(I),I=l,MPP
C = 2,ZERO SIDE RESISTANCE, SET RUM(MPP) = RUT
C = 2,ZERO SIDE RESISTANCE, SET RUM(MPPJ = RUP
C = 3rUNIFORM SIDE RESISTANCEtRUT-RUP) WITH RUM(MPP) = RUP
C = 4rTRIANGULAR SIDE RESISTANCE(RUT-RUP) WITH RUM(MPPJ = RUP
C = 5,READ NUMR CARDS AND USE GENERAL SOIL BEHAVIOR ROUTINE
C NOP(5) = O,USE OLD GAMMA(!)
C = lr2 SET GAMMACNR)=GAMMAl AND GAMMACNR+l)=GAMMA2 (SOP}
C = 3, USE SOP ABOVE AND SET GAMMACNR+2) = 0.0
C = 4, USE SOP ABOVE AND SET GAMMAStNR+2) AND (NR+3) = 0.0
C = 4, ETC.
C = 9, USE LONG FORM INPUT
C NOTE THAT NOPf5) IS USED TO SET ADDITIONAL GAMMA(.IJS = 0.0
C NOP(6) = O, USE OLD EEMtiJ,I=l,N
, C = lrUSE SHORT FORM .INPUT
>
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111 C = 2t USE LONG FORM INPUT
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C NOPl7) = O, USE OLD BEEMII), l=ltN
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X
C = 2, USE LONG FORM INPUT
..j
C NOP(8) = O,USE OLD VEL(I), l=l,MP
C = ltUSE SHORT FORM INPUT
C = 2, USE LONG FORM INPUT
C NOP(9) = O,USE OLD Q(l), I=l,MPP
C = lrUSE SHORT FORM INPUT
C = 2, USE LONG FORM INPUT
C NOPClOl = O,USE OLD SJ(J}, I=l,MPP
C = l,USE SHORT FORM INPUT
C = 2, USE LONG FORM INPUT
C NOPCll) = O,USE OLD DYNAMK(I), I=l,N
C = ltDYNAMK=O.O
C = 2t USE LONG FORM INPUT
C NOP(l2) = O,USE OLD All), I=lrN
C = lrUSE SHORT FORM INPUT
C = 2, USE LONG FORM INPUT
C NOPC13) = O,I,US~ SMITHS EEM ROUTINE
C = 2, USE LINEAR SOLID DAMPING
C NOP(l4) = O,l,USE FOMfMIJ COMPUTED FROM RAMS BEHAVIOR
C = z, READ NSTOP VALUES OF FORCIN(INTV) (CARD SERIES 1300)
C = 3,READ HEADER CARD+ NSTOP GALVO DEFLECTIONS(IN.) CARDS 1400
C = 4,READ CARD 1500 AND USE PARABOLIC FOM(l) VS. CEEM(l)
C NOP(l5) = ltNO GRAVITY
C = 2tGRAVITY WITH DEM(J,O) = 0.0
C = 3,GRAVITY WITH DEMCI,O} BY SMITH
C = 4,GRAVITY WITH DEMti,O) BY EXACT
C = S,GRAVITY WITH DEMti,Ol AS USED FOR PREVIOUS PROBLEM
C NOPC16) = O,l,NO PARAMETER. C.HANGES
C = 2t READ CARD 1600 WITH PARAMETER CHANGES
C NOPC17l =O,l,NO PILE DRIVING FORMULA OUTPUT
C = 2t READ CARD 1700 WITH PILE DRIVING CONSTANTS
c
c
C NUMBER OF CASES= NOPP11l•NOPPC2)• ••• * NOPP(l4)
.....,..,..----- ·---·-·--·~-·---·--------

c
C NOPP ( ll l,RAM VELOCITY = VELMI
c 2,RAM VELOCITY=VELMI,tl.O+OVll*VELMI
c 3,RAM VELOCITY=VELMI,(l.O+OVll*VELMI,tl.0+2.*DVl)•VELMI
c 4,ETC.
C NOPP(2) WAMt 1) CHANGE
C NOPP(3) WAM(2) CHANGES
C NOPP(4) WAM(3,MP) CHANGES
C NOPP(5) XKAM(l) CHANGES
C NOPP(6) XK.AM(2) CHANGES
C NOPPC7J XKAM(3,N) CHANGES
C NOPP 8) QSIOE CHANGES
C NOPP(9) QPOINT CHANGES
C NOPP(lO) SIOEJ CHANGES
C NOPP( 11) POINTJ CHANGES
C NOPP(l2) RUM(l,MPl CHANGES
C NOPPC13) RUMfMPP) CHANGES
C NOPPll4) BOTH RUM(l,MP) AND RUM(MPP) CHANGE
C NOPP(l5) EEMCl) CHANGES
C NOPP(l6) EEM(2) CHANGES
c
c
c
COMMON WAMllOO), XKAM(l00), RUMtlOO), BEEMllOO}, EEMtlOO) 1
COMMON GAMMA(lOO), XKIM(lOO),CEEMAS(lOO}, NFOM{lOO), XOEMtlOO) 2
.COMMON OEMClOO), XCEEMClOO), CEEM(lOO), FOMtlOO}, XFOMtlOO) 3
COMMON VEL(lOO), DIM(100), RAM(l00), RMAXtlOO), RSTAT(l00) 4
COMMON RtlOO,lO) , ITRIG(l00) 1 Q(lQO),FORCIN{lOO), DFOM(l00) 5
COMMON FOMAX(l00),IFOMAXC100), FOMIN(l00l,IFOMIN(l00), A(lOOl 6
COMMON OEMAX(l00) 1 10EMAX(l00), SJtlOO), NOP( 22),DYNAMKClQOl 7
COMMON CEEMIN(l00) 1 HOLOEM(l00),ANSVECt 50),SE(50,51) , IROW( 51) R
COMMON RUMAtlOO), WAMCI100), XKAMCtlOO), QA(l00), SJAtlOO) 9
COMMON ICOL( 51), NOPP( 20),ENTHRUClOO),ENTMAX(l00), IDS( 50) 10
, COMMON QSIOE , QPOINT, SIOEJ , POINTJ, NQOIV , 'NORAMS, NSTOP 50
> INTV , ISECTN 1 NUMR , Fl , F2 , Cl , C2 51
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COMMON IPRINT, OELTEE, EEMl , EEM2 GAMMA!, GAMMA2, INT 52
COMMON INTT ' I ITST IX ,' NR , MO , MP 53
COMMON NPAGE , N ,' QUAKE ', RUP , RUT , VELMI 101 54
COMMON 102 f 103 ' 104 I OWl IDW2 IDKl ' IDK2 55
COMMON IORLl , IORL2 , IOGl ' IOG2 ', I DEl ' IDE2 ' IDBl 56
COMMON 1082 , IOVl f IOV2 ' IDQl IDQ2 ' IOJl ,' IDJ2 57
COMMON IOOKl , IOOK2 , IOAl ' IDA2 ', KGRAOO,' J5 TMIN 58
THAX , SHIN , SMAX ,' NOPNT,S, AREA , NSl
t
COMMON , NS2,NS6 59
COMMON NS3 , NS4 , NSS IOEEM , MH , VEll ACCELR 60
COMMON 8 ' c , AREAP '
• XLONG , ELAST , ACELMX ' 61
COMMON OVl,OEl,OE2,DRI,ORPrDQI,OQP,DJI,OJP,OWl,OW2,0WI,OKl,DK2,0KI
c
c
c
c
NPAGE =0
9 CONTINUE
NSl = 0
CALL INPUT
MP .:: MP
MO = MO
NR : NR
MH=MH
N = MP-1
MPP : MP+l
c INITIALIZE PARAMETER CONSTANTS
DELTAA = OELTEE
WAMA = WAM(l)
WAMB = WAM(2)
XKAMA = XKAM(l)
XKAM8 .:: XKAH{2)
00 1 l=lrMP
RUMAtl) = RUM(I)
WAMCCI) = WAM(I)
XKAMC(I) = XKAM(I)
 - -------------------------------------------------------------------------------------------------------------------------------------------------------------

QA{Il =
Q(l}
SJA(I} = SJ{I)
1 CONTINUE
NOPA = NOPP ( 1)
NOPB = NOPP( 2)
NOPC = NOPP(3 l
NOPD = NOPP( 4)
NOPE = NOPPf5 )
NOPF = NOPP( 6)
NOPG = NOPP( 7)
NOPH = NOPP( 8)
NOPI = NOPP( 9)
NOPJ = NOPPtlO)
NOPK = NOPP ( 11}
NOPL = NOPP(l2)
NOPM = NOPP{l3)
NOPN = NOPP(l4)
NOPO = NOPP(l5)
NOPQ = NOPP(l6)
c BEGIN PARAMETER VARIATIONS
DO 98 IQ = l,NOPQ
DO 98 10 = l,NOPO
11 DO 98 IN = l,NOPN
IM = IN
Il = IN
DO 98 IK = l,NOPK
DO 98 IJ = I,NOPJ
00 98 II = ltNOPI
DO 98 IH = l,NOPH
DO 98 IG = l,NOPG
00 98 IF = l,NOPF
DO 98 IE = I,NOPE
DO 98 ID = l,NOPO
: DO 98 IC = l,NOPC
~
m DO 98 IS = l,NOPB
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00 98 lA = lrNOPA
OELTEE = OELTAA
00 4 l=l,MP
VEL(l) = 0.0
WAMCI) = WAMC(l) •(1.0 + FLOATCI0-1) *OWl)
XKAH(I) = XKAMC(I) *(1.0 + FLOATliG-1) * DKI)
Q(l) = QA(I) *(1.0 + FLOAT(IH-1) * DQI)
SJ(I) = SJA(I) •(1.0 + FLOATtiJ-1) * DJI)
RUM(I) = RUMA(I) *(1.0 + FLOAT(IL-1) * ORI)
4 CONTINUE
00 3 I=l,NR
VEL(l) = VELMI •(1.0 + FLOATCIA-1) * OVl)
3 CONTINUE
VEll = VEL ( 1)
WAH(l) = WAMA *(1.0 + FLOATIIB-1) * OWlt
WAM(2) = WAMB •(1.0 + FLOATtiC-11 * 0W2)
XKAM(l) = XKAMA •(1.0 + FLOAT{IE-1) * OKl)
XKAH(2) = XKAMB *(1.0 + FLOATCIF-1) * DK2)
Q(MPP) = QPOINT •(1.0 + FLOATCII-1) * DQP)
SJtMPP) = POINTJ •(1.0 + FLOATliK-1) * DJP)
RUHtMPPl = RUP •(1.0 + FLOATCIM-1) * DRP)
EEM(NR)= EEM1 •11.0 + FlOAT(I0-1) *DEll
EEMCNR+l)= EEM2 •Cl.O + FlOAT(IQ-1) * DE2)
IFINOP(4)-5)13rl6rl3
13 00 15 I=l,MPP
15 XKIMCil = RUM(I)/Q(I)
16 CONTINUE
C IF OELTEE IS LEFT BLANK, 1/2 THE CRITICAL TIME INTERVAL WILL BE USED
IFCDElTEEl32,32r31
32 00 33 1=1,N
33 OELTEE = AMAX1(0ElTEE,39.296•SQRTlXKAM(I)/WAM(I)),
1 39.296•SQRT(XKAM(J)/WAM(I+l)))
31 CONTINUE
C END PARAMETER VARIATIONS
ClPC2 = 0.0
 ACELMX = 0.0
CALL PRINT l
CAll REP 1
J5 = IPRINT
KXT=l
INTV = 0
INTT =1
MP = MP
N = MP-1
MPP = MP+l
NOP15P = NOP(l5)+1
GO T0(50,50.49,48,47,43,50,50,50) 1 NOPl5P
43 DO 42 I = l,MP
42 DEM{I) = HOLDEMfl)
44 RAMlHP) = DEM(MP)•XKIH(MP)
RAM(MP+l) = DEM(MPl•XKIM(MP+l)
HOLDEH(MP) = DEMlMP)
HOLDEM(l) = OEM(l}
CEeMfl) = OEM(l) - OEM(2)
FOM(l) = CEEM(l)•XKAM(l)
00 45 1 = 2,N
'HOLOEMtl) = DEMII)
CEEMtll = OEM(I)-OEM(I+l)
FOM(IJ = CEEM(I)•XKAM{l)
45 RAM(I) = FOM(l-1)-FOM(I)+WAM(I)
GO TO 49
47 CAll EXACTG
GO TO 49
48 CAll SMlTK
49 CON-TINUE
WRITEt6,8002){0EM(IJ,I=l,MPJ
WRITE(6,800l)l01M(I),I=l,MP)
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WRITE(6,8006) CXKIH(l) ,I=l 1 HPP)
50 CONTINUE
NSM = MP-1
NSM=MINOtNS6,NSH)
WRITEC6tll04)NSl,NS2,NS3,NS4,NS5,NSM,NSl,NS2,NS3 1 N$4 1 NS5,NS6,MPP
C BEGIN ITERATION LOOP
12 CAll REP N
INTT=INTT
GO T0(22,9 ),INTT
22 CONTINUE
CMAX = 0.0
DO 24 I=NR,N
24 CMAX = CHAX+CEEM(l)
ClPC2 = AHAXllClPC2,CMAX)
IFliNTV-999)25,23,25
23 J5 = 25
25 CONTINUE
lf(((INTV/J5)*J5)-INTV)94,26,94
26 CONTINUE
27 FOHA = FOMCNSl)/A(NSl)
FOMB = FOM(NS2)/AfNS2)
FOMC = FOM(NS3)/ACNS3)
FOMO = FOM(NS4)/A(NS4)
FOME = FOMCNS5)/A{NS5)
FOMF = FOMlNSH)/ACNSM)
RAMP = RAH(MP)/1000.0
C WRITEI6,99)1NTV ,FOMA,FOMB,FOMC,FOMO,FOME, CEEM(llrOEM(NS3),
C 1 OEMCNS4l,OEMCNS5),0EMlNS5P),tENTHRU(I),I=2,4),ENTHRU(N),ACCELR
WRITE(6 1 99)fNTV,FOMA 1 FOM8 1 FOMC,FOM0 1 FOME,FOMF,OEMtNS1) 1 0EM(NS2),
lOEMCNS3~tDEMtNS4l,OEMfNS5),0EMCNS6),RAMP
94 CONTINUE
lf(INTV-NSTOP )12,14,14
14 WRITEC6,105)
MP = MP
N = MP-1
 MH = MH
D020I=l,N
FOMAX(I) = FOMAX(l)/A(l)
FOMIN(I) = FOMIN(I)/A(I)
WRITE(6,106)1,1FOMAX(I),FOMAXti),IFOMIN(I) 1 FOMIN{I),
1 ENTHRUtiJ,ENTMAX(I)
20 CONTINUE
C BLOWS= 1.0/0IM(MP) OLD STATEMENT.
C WRITE(6,2107JQIM(MP},BLOWS OLD STATEMENT
WRITE ( 6,2108 )OEM.AX ( MH-1), DE MAX ( MP)
SMIN = SKIN/12.0
SMAX = SMAX/12.0
ERESl = SQRT(SMIN/SMAX)
WRITE(6,109)SMIN,SMAX,ERESl
EINPUT = CWAMllJ•VEL1••2)/64.4
WRITE(6,ll0)EINPUT
WRITE(6,lll)ACELMX
C BEGIN ULTIMATE LOAD FORMULAS
IF(NOPtl7)-1}98,98,5
5 CONTINUE
C4 = 0.1
AEL = AREAP•ELAST/XLONG
NRP = NR+l
C3 = QAVG
S = OlMlMPP)
W =WRAM
U = ENERGY
P = WPILE
RWAVE =- 0.0
00: 6. t=·l t MPP
RWAVc = RWA\/E+RUMflt/tOOO.O
6 CONTINUE
SEGL = XLONG/(flOATIMP-MH+l))
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= DO 10 I=MH,MP
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10 SUMR : SUMR+RUMtl )~SEGL•CFLOATli -MH)+0.5)
SUMR : SUMR+RUMCMPP)•XLONG
HILEYL = SUMR/RWAVE
RENEWS = U/(S+CENR)
REYTEL = U/fS+(C4•P/W))
RTERZG: AEL•f-S+SQRTfS••2+(2.0•U•(W+P•EEM1••2)/(AEL•(W+P)))))
REDTEN: AEL•(-S+SQRT(S••2+(2.0•U•W/(AEL•(W+P)))))
RHILYD =AEL•t-(S+C3)+SQRT((S+C3)••2+(2.0•U•(W+P*EEM1••2)/
1 (AEL•CW+P)))))
RHILYC=U•lW+P•EEM1••2)/((S+0.5•(ClPC2+C3))•(W+P))
RCOAST =tAEl/2.)•(-S+SQRT(S••2+(4.•U•(W+P•EEM1~•2)/(AEL•fW+P)))))
WRITE(6,107) .
WRITE(6,108)RENEWS,REYTEL,RTERZG,R£DTEN,RHILYO,RHILYC,RCOAST,RWAVE
C END ULTIMATE LOAD FORMULAS
98 CONTINUE
GO TO 9
99 FORMATflX,I3,6F9.2,6F9.3,F9.1)
C 99 FORMATllX, 13, 5Fl0.2, 5Fll.7,F9.1)
105 FORMATtlHQ,//, 18X, 63HMAXIMUM COMPRESSIVE AND TENSILE STRESSES
lPSI) IN THE SEGMENTS ,//,19X, 7HSEGMENT , lX, 5H TIME ,
2 3X, 6HSTRESS , SX, 4HTIME,3X,6HSTRESS,7X,6HENTHRU,7X,
3 lOHHAX ENTHRU , //)
106 FORMATl20X,I4,I8,F9.l,I9 1 F9.1,2F13.1)
107 FORMAT( 16X,30H ULTIMATE PILE LOADS (KIPS)
108 FORMAT( 21X,25H BY ENG NEWS FORMULA = , Fl5.3,/ ,
1 22X,25H BY EYTELWEIN = , Fl5.3 1 / ,
2 22X,25H BY TERZAGHI = , Fl5.3,/ ,
3 22X,2SH BY REOTENBACHER = , Fl5.3,/ ,
4 22X,25H BY HILEY (DUNHAM) = , Fl5.3,/ ,
5 22X,25H BY HILEY CCH~LLIS) = , Fl5.3,/ ,
6 22X,25H BY PACIFIC COAST = , Fl5.3,/ ,
7 22X,25H BY THE WAVE EQUATION = , F15.3)
109 FORMAT(l7X,7HSMIN = FlO.l, 7HSHAX: FlO.lt lOHERES(l) = F10.7)
110 FORHAT(l6X,l8H EINPUT 1FT LBS) = F9.1)
111 FORHAT(l6X 1 24H MAX ACCELERATION (GS) = F9.1)
 ,.. ............. ..__.... _____ - -- ---· --~--- ~-- ~ ~· .. _ -----

1104 FORMATC3H T,6{6XtlHFtl2)t1X, 6(6X 1 1HO,I2) 1 6X,lHR,I2,//)

Cll04 FORMATlll5H TIME F(l) f(2) F(3) Fl4) F(5} 0(2) 0(3) 0(
C 14) 0(5) OlP) ENT(2) ENT(3) ENT(4) ENTIN) ACCCMH-1) )
Cll04 FORMATt5H TIME,5l2X,4HFOM( 13, lHl } ,5(3X,4HOEM( 13 1 lH) ) ,
C 1 3X, 12HENTHRU (1) /1)
2107 FORMATtlH I tl7X,24HPERMANENT SET OF PILE = Fl3.8,8H INCHES/
1 ,17X,27HNUMBER OF BLOWS PER INCH = Fl3.8)
2108 FORMATllH I tl7X,24HLIMSET FOR lMH-1) = Fl3.8,8H INCHES/
1 t17X,27HMAX DISPLACEMENT OF POINT= Fl3.8)
8001 FORMAT(33HOINITIAL VALUES FOR DIM(I),I=l,MP /(6El9.8))
8002 FORMATC33HOINITIAL "VALUES FOR DEMCI),.=l,MP /(6El9.8~)
8003 FORMAT(33HOINITIAL VALUES FOR FOMli),I=l,MP ll6El9.8))
8004 FORMATt33HOINITIAl VALUES FOR CEEMtl), I=l,N /(6El9.8J)
8005 FORMATC35HOINITIAL VALUES FOR RAM(J),I=l,MP+l 1(6El9.8))
8006 FORMAT(38HOCONSTANT VA-t:UES FOR XKIM(I),I=l,MP+l /(6El9.8)}
END
$lBFTC INPUTT
SUBROUTINE INPUT
COMMON WAMllOOl, XKAMllOO}, RUM(l00), BEEMllOO), EEM(l00) 1
COMMON GAMMAllOO), XKIMllOO);CEEMAS(lOO), NFOMllOO), XOEMllOO) 2
COMMON OEMtlOO), XCEEM(lOO), CEEMllOO), FOMllOO), XFOM(l00) 3
COMMON VELllOO), OIMllOO), RAMllOO), RMAXClOO), RSTAT(l00) 4
COMMON R(l00,10) , ITRIG(l00), Q(l00),FORCIN(l00), OFOM(l00) 5
COMMON FOMAXllOO),IFOMAXtlOO), FOMIN(lOO),IFOMINllOO), AllOO) 6
COMMON OEMAX{lOO),IOEMAXClOO)t SJ(l00), NOPl 22),0YNAMK(l00) 1
COMMON CEEMINllOO),HOLDEMClOO),ANSVEC( 50),SE(50,51) , IROW( 51) 8
COMMON RUMAilOO), WAMC(l00}, XKAMCClOO), QA(l00), SJA(l00) 9
COMMON ICOL{ 51), NOPP( 20),ENTHRUllOOJ,ENTMAX(l00}, IDS( 50) 10
COMMON QSIDE , QPOINT, SIOEJ ~ POINTJ, NQOIV , NORAMS~ NSTOP 50
COMMON INTV , ISECTN, NUMR , Fl , F2 , Cl , C2 51
COMMON IPRINT, OELTEE, EEMl , EEM2 , GAMMAl, GAMMA2~ INT 52
COMMON INTT , I , ITST , IX , NR , MO , MP 53
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COMMON NPAGE , N , QUAKE , RUP , RUT , VELMI , 101 54
Gl
111
COMMON 102 , 103 , 104 , IOWl , IOW2 , IOKl , IOK2 55
:!! COMMON lORLl , IORL2 , IOGl , IOG2 , JOEl , IOE2 , IDBl 56
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COMMON 1082 , IOVl , IOV2 , IDQl , IOQ2 , IDJl , IDJ2 57
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COMMON IDOKl , IODK2 , IOAl , IOA2 , KGRADO, J5 , TMIN 58
COMMON TMAX , SMIN , SMAX , NOPNTS, AREA , NSl , NS2,NS6 59
COMMON NS3 , NS4 , NS5 , IOEEM , MH , VEll , ACCELR 60
COMMON ~ , C , AREAP , XLONG , ELAST ,· ACELMX 61
COMMO~ DVl,DE!,OE2,DRI,ORP,DQI,OQP,OJI,OJP,DWl,OW2,0WI,OKl,DK2,0KI
c
READC5,100)101,102,0ELTEErMPrVELMI,MH,NR,EEMl,EEM2,GAMMAl,
1 GAMMA2,NSTOP,(N0P(l)r1=1,20)
REA0(5,101)103r104,RUT,RUP,MO,QSIOE,OPOINT,SIDEJ,POINTJ 1 NUMR,
1 IPRINT,AREA,NSl,NS2,NS3,NS4,NS5,NS6
RUT = RUT•lOOO.O
RUP = RUP•lOOO.O
NR = MAXO ( NR ,.l)
N = MP-1
MPP = MP+l
WAM(MPP) = -0.0
XKAM(MP) = -0.0
XKAM(MPP) = -0.0
IF(NOP(l)-2)9,7,7
7 NOIOS = 12•(N0P(l}-l)
REA0(5,103)(10S(I),I=l,NOIOS)
9 CONTINUE
IF(N0Pl2)-l) 1,1,14
1 REA0(5,102)10Wl,IOW2,(WAM(l),I=l,MP)
GO TO 2
14 NRPl = NR+l
NRP5 = NR+5
NRP6 = NR+6
MPM3 = MP-3
REAOl5,lllliOWl,IOW2,WAM{l)~(WAM(Llrl=NRPl,NRP5),
1 ( WAMt I) • .I=-MPH3,.MP·)
111 FORMAT( A5,A4 ,-3Pl0F6.4)
00 76 l=l,NR
76 WAMCI) = WAM(l)
 ''• ":.':::?'~' ~' ... • .•,,._, -~ .....,... .•"="'

DO 11 I=NRP6,MPM3
11 WAM(I) = WAM(NRP5)
2 CONTINUE
IF(N0P(3)-l) 3,3,15
3 REA0{5,L04)IDKltiDK2,tXKAMtl),l=l,N)
GO TO 4
15 NRMl = NR-1
NRP5· = NR+5
NRP6 = NR+6
MPM3 = MP-3
REA0(5,112)10Kl,IDK2,XKAM(l),(XKAM(l),I=NR,NRP5lt
1 (XKAHtil,I=MPM3,N)
112 FORMA Tl AS·, A4,-3Pl0F6.0)
DO 78 I=l,NRML
18 XKAM{I) = XKAM(l)
DO 79 I=NRP6,MPM3
79 XKAM{l) = XKAM(MPM3)
4 CONTINUE
1FlNOP{4)-1)22,5,5
5 NOP4 = NOP(4}
DO 6 I=l,MP
6 RUM(I) = 0.0
RUM(MPP) = RUP
GO TOil0,22,lltl3tl7,22t22,22,22),NOP4
10 REA0(5,106)10RlltlDRL2,(RUM(I)ri=l,MPP)
C INPUT RUM({) IN UNITS OF KIPS - THE COMPUTER WILL CONVERT TO LBS.
GO TO 22
l l RCONST = (RUT-RUP)/FLOATlMPP-MO)
DO 12 I=MO,MP
12 RUM(l) = RCONST
GO TO 22
13 DO 16 I=MO,MP
16 RUM( I ) = C2.0•lRUT-RUP)•(FLOATII-M0)+0.5) )/tFLOATCMPP-MO) >••2
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1'1 C GENERAL R(I,J) INPUT
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17 DO 20 l=l,MPP
20 XKIM(I} = 0.0
DO 21 K=l,NUMR
21 READ(5,115)1, XKIM(I),(RCI,J),J=l,lO)
22 CONTINUE
C THE RCI,J) INPUT CARDS CAN BE IN RANDOM ORDER
C THE RCI,J) ARRAY NEED NOT BE ZEROED SINCE IF XKIM(I)=O THE GENERAL
C SOIL RESISTANCE ROUTINE FOR SEGMENT(!) IS NOT CONSIDERED
C NUMR =TOTAL NUMBER OF SEGMENTS WIGEN. R COONT FORGET TO ADO MPP)
CC I =THE SEGMENT NUMBER FOR WHICH R(I,J) VALUES ARE BEING INPUT
C RtL,Jl = STATIC RESISTANCE ON SEGMENT I AT EACH OF TEN POINTS J
IFINOP(S)-1)29~27,26
26 IF(NOP(5)-9)24t25,24
25 READ(5,106)1DGl,IDGZ~lGAMMAfi),I=1,N)
GO TO 29
24 !GAMMA = NOP(51+NR-l
DO 23 I=l ,N .
23 GAMMA(!) = -1000.0
DO 19 I=NR,IGAMMA
19 GAMMA(I) = 0.0
GAMMA(NR) = GAMMAl
GAMMACNR+l) = GAMMA2
GO TO 29
21 DO 28 1=1,N
28 GAMMA{I) = -1000.0
GAMMA(NR) = GAMMAl
GAMMAfNR+l) = GAMMA2
29 GAMMA{MP) = -0.0
GAHMA{MPPJ = -0.0
IF(NOPI6)-1J33,31,30
30 REA0(5,107JIDEl,IDE2,(EEM(I),I•1,N)
GO TO 33
31 00 32 l=l ,N
32 EEM(I) = 1.0
EEMINR) = EEMl
 ;;;:;;;:;:;;_:;:::::;.:;::=;::=;;...:_;;..,,.-"';'
;;;;;;;;;;;;;;;;;:;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;:;;;:=... v ···--·-~----- . . .:.,.;,:;.,...;o:..:.. ._.._;;.~~._:::.~..,.......--"'" ··:~·--·- ~- --·---~ ..... _.... -··--~

EEMCNR+U = EEM2
33 EEMCMP) = -0.0
EEMlMPPl = -0.0
JF(NOP(7}-1)37,35,34
34 READC5,107)1DB1,I082,(BEEHti),I=l,Nl
GO TO 37
3 5 DO 36 I = 1 , N
36 SEEM( I) = 0.0
37 BEEMlMP) = -0.0
SEEMfHPP) = -0.0
C DO NOT TRY TO USE LAST PROBLEMS VALUES OF VEL(I)
1FlNOP(8)-1)39,39,38
38 READC5,108)10Vl,IDV2,( VELti),l=lrMP)
GO. TO. 71
39 DO 40 I=NR,MPP
40 VEL(!) = 0.0
DO 41 1=1 ,NR
41 VEL{!) = VELMI
71 VEL(MPP) = -0.0
IF(N0P(9)-1)45r43,42
42 REAOC5,107)IDQl,IOQ2,(Q(I),J=l,MPP)
GO TO 45
43 PO 44 1=1,MPP
Q(l) = QSIOE
44 CONTINUE
Q(MPP) = QPOINT
45 IFtNOP(l0)-1)49,47,46
46 REA0(5,107)10Jl,IOJ2,(SJ(I),I=l,MPP)
~0 TO 49
47 DO 48 I=1,MP
48 !SJ(l) = SIDEJ
SJCMPP) = POINTJ
49 iJFCNOP(ll)-1)53,51,50
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~
0 GO TO 53
51 DO 52 I = 1 , N
52 DYNAMK(I) = 0.0
C STATEMENT 52 SETS DYNAMK(I) = 0.0 SO SMITHS ROUTINE WILL BE USED
53 DYNAMKIMP) = -0.0
DYNAMK(MPP) = -0.0
IFINOP(l2)-1157,55,54
54 REA0(5,109)1DAltiDA2,(A(I),I=l,NJ
GO TO 57
55 DO 56 I=l,N
56 A(ll =AREA
57 A(MP) = -0.0
A(MPP) = -0.0
IF(N0Pl4)-1)61,58,58
58 IFlNOP(4)-5)59,61,61
59 DO 60 l=l,MPP
60 XKIM(I) = RUM(I)/Q(l)
6l CONTINUE
NOP14 = NOPtl4)+1
GO T0(65,65,62,63,65),NOP14
C READ NSTOP VALUES OF FOM(l,T) - MAXIMUM NSTOP = 300
62 READ(5,120)(f0RCIN(I),I=l,NSTOP)
GO TO 65
63 READ(5,122)AREAPrEMODUL,RGAGE,RCAL,ACTIVG,GFACTR,Ol,02,03,04,05
REA0(5,12l)(FORCIN(I),I=l,NSTOP)
CE = (AREAP•EMOOUL•RGAGE•lOOO.O)/(ACTIVG•GFACTR•RCAL)
AtNSl) = CE/01
A(NS2) = CE/02
A(NS3) =- CE/03
AlNS4) = CE/04
A(NS5) = CE/D5
DO 64 I=lrNSTOP
64 FORCIN(I) = FORCiN(I)•ACll
65 CONTINUE
 >-'' ,

1FlNOP(14)-4)67,66~67
66 REA0(5,123)F1,F2,C1,C2
67 CONTINUE
00 90 1=1,20
90 NOPPll) = 1
IFlNOP(16)-2)69,68,69
68 REA0(5,124)(NOPP{l),I=1,20),0Vl,OWl,OW2,DWI,OKl,OK2,0KI,DOI,
1 DQP,OJI,OJP,DRI,ORP,OEl,OE2
69 CONTINUE
DO 8 1=1 ,zo
NOPP(I) = MAXOlNOPPlll,1)
8 CONTINUE
IF(NOPC17) .... ll74,.74t.73
73 REA0(5,125)AREAP,XLONG,ELAST,CENR,QAVG,wRAM,WPILE,ENERG"Y
XLONG = XLONG•l2.0
74 CONTINUE
100 FORMAT(A5,A4;F6.0,I3,F4.2,213t2F4.3,2F6.0,I4,20il)
101 FORMATCA5,A4,2F7.2,I3,4F4.3,2I3,F6.2,613)
102 FORMAT(A5,A4,-3PlOF6.4,/(9X,-3PlOF6.4))
103 FORMA Tl 12A6)
104 FOR~ATCA5,A4,-3PlOF6.0,/(9X,-3PlOF6.0))
106 FORMATlA5,A4,-3PlOF6.1,/(9X,-3P10F6.1))
107 FORMAT(A5 1 A4, 10F6.5,/(9X, 10F6.5))
108 FORMATIA5,A4, 10f6.3,/(9X, 10F6.3))
109 FORMATCA5,A4t 10F6.2,/(9X, 10F6.2))
115 FORMATCI3,-3Pllf6.1)
120 FORMATf-3PlOF6.1)
121 FORMAT( 10F6.4)
122 FORMAT(F7.2,3Fl.'0,7F4.2)
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123 FORMATl-3P2F6.liOP2F6.5)
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111
124 FORMAT{ 2011, 17F'3.2'J
Ill 125 FORM-ATCF6.2!F5.2,F7.2)
~ RETURN
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;u
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SIBFTC PRINT
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c PRINT 1 IS A SUBROUTINE TO PRINT INPUT DATA.
c
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SUBROUTINE PRINT 1
COMMON WAM(l00), XKAMClOO), RUM(lOO), BEEM(l00), EEMClOO) 1
COMMON GAMMA(lOO), XKIMClOO),CEEMAS(lOO), NFOMC100), XOEM(lOOl 2
COMMON DEMtlOO), XCEEM(lOO), CEEM(l00), FOM(l00}, XFOM(l00} 3
COMMON VElllOO), OIM(l00) 1 RAM(l00), RMAX(l00), RSTAT(100) 4
COMMON RllOO,lO) , ITRIG(l00), Q(lOO),FORCINtlOO), OFOM(l00) 5
COMMON FOMAX(l00),IFOMAX(l00}r FOMIN(l00),IFOMIN(l00), A(l00) 6
COMMON DEMAX(lOOl,IOEMAXIlOQJ, SJ(lOO), NOP( 22),0YNAMK(l00) 7
COMMON CEEMINtlOO),HOLDEM(lOO),.ANSVEC( 50),SEC50,5U" , IROW( 51) 8
COMMON RUMAClOO), WAMCl-100)• XKAMC(l00),. QA(l00), SJA(100) q
COMMON ICOL( 51), NOPP( 20),ENTHRU(l00),ENTMA~fl00), IOSC 50) 10
COMMON QSIOE , QPOINTr SIOEJ , POINTJ, NQOIV , NORAMS, NSTOP 50
COMMON INTV , ISECTN, NUMR , Fl , F2 , Cl , C2 51
COMMON IPRINT, DEtTEE, EEMl , EEM2 , GAMMAt, GAMMA2, INT 52
COMMON INTT , 1 , ITST , IX , NR , MO , MP 53
COMMON NPAGE , N , QUAKE , RUP , RUT , VELMI , 101 54
COMMON 102 r 103 t 104 , IOWl , IDW2 ,. lOKl , IDK2 55
COMMON IORll , IDRL2 , IOGl , IOG2 , IDE! , IDE2 , 1081 56
COMMON 1082 , IOVl , IOV2 , IOQl , IOQ2 , IOJl , IOJ2 57
COMMON IODKl , IOOK2 , IOAl , IOA2 , KGRAOO, J5 , TMIN 5R
COMMON TMAX r SMIN , SMAX , NOPNTS, AREA , NSl , NS2,NS6 59
COMMON NS3 , N$4 , NS5 r IOEEM , MH , YEll , ACCELR 60
COMMON 8 , C , AREAP , XLONG , ELAST , ACELMX 61
COMMON DV1 1 DEl,OE2,DRI,ORP,OQI,OQP,OJI,OJP,OWl,DW2,0WI,DKl,OK2,0KI
c
NPAGE = NPAGE+l
WRITE 16.,1.02) N.PAGE
IF c·N{lP( 1) -213,2 t2
2 NOIOS = 12•tNOP(ll-lt
WRITE ( 6, 10 1)
WRITE(6,103 )(IOS(I),I=lrNOIOS)
WRITEC6,10ll
3 CONTINUE
 MPP=MP+l
RCT = 0.0
DO 6 I= l,MPP
RCT = RCT+RUM(IJ/1000.0
6 CONTINUE
RCP = RUM(MPP)/1000.0
WRITE(6,105)0ELTEE,NOP(l) 1 NOP(l6)
DELTEE = 1.0/DELTEE
WRITE(6,106)MP,NOP(2},NOP(l7)
WRITE(6,107)101,102,VELMI,NOP(3) 1 NOP(l8)
WRITE(6,108)!03,104,NSTOP 1 NOP(4),NOP(l9)
WRITE(6,110)10WltiDW2,RCT.NOPt5),NOP(20)
WRITEt6,lllliOKltlOK2,RCP,NOP(6)
WRITEl6tll2)10RLl,IDRL2,MO,NOP{7)
WRITEl6,113)IOGl,IDG2,QSIOE 1 NOP(8)
WRITE(6,114}10El,IOE2,QPOINT,NOP(9}
WRITE(6,115)10Bl,fOB2,SIOEJ,NOP(l0)
WRITE(6,116)10Vl,IDV2,POINTJ,NOPC11)
WRITEt6,117liDQl,IDQ2,NUMR,NOP(l2)
WRITE(6,118)10Jl,IOJ2,IPRINT,NOP(l3)
WRITE(6,119)IODKltiOOK2,AREA,NOP{l4)
WRITE(6,120)10AltiOA2,NR,NOP(l5}
WRITE(6,10l)
WRITE ( 6,121 l
MPP = MP+l
LINES = 19
00 5 I=l,MPP
WRITE(6,122)1,WAM(I},XKAMli),RUM{I),GAMMA(J),EEMCI),BEEMCIJ,
1 VEL(I),Q{I), SJ(I),OYNAMK(I),A(J)
LINES = LINES+l
IF(LINES-58)5,4,4
4 NPAGE = NPAGE
LINES = 5
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WRITEt6,102)NPAGE
111 WRITE ( 6, 101 )
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WRITE ( 6, 121)
5 CONTINUE
IFINOP(4)-5)30,7,30
7 lf(LINES-50)9,9,8
8 NPAGE = NPAGE
LINES = -1
WRITE(6,102)NPAGE
GO TO 10
9 WRITE( 6,101)
10 WRITE(6,123)(J,J=1,10)
LINES =· LINES+6
LlNAOO = NQOIV/10
IF ( NQOI V.-LINAOO• 10)13, 14·, 13
13 LINAOO = liNAOO+l
14 LINAOO = LINAOO+l
00 29 l=l,MPP
lf(XKIM(l)-0.0)29,29,20
20 LINES = LINES+LINAOO
lf(LINES-59)24,24,23
23 NPAGE = NPAGE
WRITE(6,102)NPAGE
WRITE ( 6, 12 3) ( J t J =1, 10)
LINES = 6
24 WRITE(6,124)l,(R{I,J),J=I,l0)
29 CONTINUE
WRITE (6, 101 l
LINES = LINES+2
30 WRITE(6,101)
LINES = LINES+2
LINAOO = MP/8
lf(MP-LINA00•8)40.41,40
40 LINAOO = LINA00+1
41 LINAOO = LINA00+2
fOl FORMAT(lHO)
 ------------------------------------------------------------------------------------------
._...,. ~..:;::r"-; 7,,, ....:,._~:·· ~·· ~·---~<"---...-,-..__.......~--~--... -JO~_,..~"~
<·-; ·:..:::.....__

G. .). s
.·:::~-- ...... · -~_,.... .Jt=-:.·::-"";'_. - ~--.... -. ·----~-- ••. ~.... ....:~---·--:.-_-::::::":"~-:~--:::~:--::., ...,.,t.;L..,.... ......-...-•.. :. . __ _.,.~.~----- -·
c: f)

102 FORHAT(lHl, 20H 66X 1 7HPR08LEM 14)

103 FORMATllX,l2A6l
123 FORMATC85H .R(M,N) = STATIC SOIL RESISTANCE FOR GIVEN SEGMENTS-
1 OTHERS HAVE R(I,JJ = 0.0 II 5X,l0(8X,I2l )
105 FORMATl4X,29H CARD 101 102 1/0ELTEE = FB.0 1 12H. NOP(l) =
1 12, 12H NOPC16) = 12)
106 FORMAT(28X, 5H MP = IB,l2H NOP(2) =12tl2H NOP(l7) =12)
107 FORMATCllH 101 A6,A4,12H VELMI =FB.2,12H NOP(3) =
1 12, 12H NOPtlB) = 12)
108 FORMATlllH 102 A6,A4 1 12H NSTOP = 18 ,·12H NOP(4) =
1 12, 12H NOP(l9) = 12)
110 FORMATI11H WAM A6rA4tl2H RUT =F8.1,12H NOPt5J =12,
l. 12H NOPI201 = 12)
111 FORMATCllH XKAM A6,A4,12H RUP =F8.ltl2H NOP(6) =12)
112 FORMAT(llH RUM A6,A4t12H MO =18 t12H NOP(7) =12)
113 FORMATI11H GAMMA A6,A4tl2H QSIOE :f;8.4tl2H NOP(8) =12)
114 FORMATCllH EEM A6,A4~12H QPOINT =~8.4,12H NOP(9) =12)
115 FORMAH11H BEEM A6,A4,12H SlOE-J =F8.4,l2H NOP(lO) =12)
116 FORMATI11H VEL A6,A4tl2H POINTJ =F8.4,l2H NOP(ll) =12)
117 FORMATC11H Q A6,A4tl2H NUMR =18 tl2H NOPC12) =12)
118 FORMATCllH SOILJ A6,A4,12H !PRINT =18 rl2H NOP(l3l =12)
119 FORMATC11H DYNAMK A6,A4,12H AREA =F8.2,12H NOP(l4) =12)
120 FORMATCllH A A6,A4,12H NR =IS t12H NOPfl5) =12)
121 FORMAT(ll6H M WAMCM) XKAMCM) RUM(M) GAMMA(M) EEM(M)
1 BEEMCM) VELCM) QCM) SOILJCM) OYNAMK(M) ACM) /,
2 116H CKIPS) (KIPS/IN) lKIPS) (KIPS) (NONE) (SECIN/
3FT) CFT/SEC) fiN) CSEC/FT) (KIPS/IN) fSQ IN) )
122 FORMAT(14,-3Pfl0.4,3FlO.ltOP2Fl0.6,Fl0.3,2Fl0.6,-3PF10.3,0PFl2.3)
12-4 FORMATl/4H 7 = 13,2X,lOflO.l,(/9X,lOF10.1))
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~ SUBROUTINE REPl
1- COMMON WAMClOO), XKAM(lOO), RUMflOO), BEEMClOO), EEM(lOO) 1
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~ COMMON GAMMA(lOO), XKIM(l00),CEEMAS(100), NFOMtlOO), XOEMflOO) 2
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81- COMMON DEMI100), XCEEM(l00), CEEM(l00), FOMClOO), XFOMC100) 3

J'II
Gi COMMON VEL(l00), 01Mtl00) 1 RAM(l00), RMAX(lOO), RSTATC100) 4
X COMMON R(lOO,lO) , ITRIG(l00), Q{l00),FORCINC100), OFOMC100) 5
""' COMMON FOMAXtlOO),IFOMAXtlOO), FOMIN(lOOJ,IFOMINClOO), A(l00) 6
COMMON OEMAXClOO),IOEMAX(lOO), SJ(l00), NOP( 22),0YNAMKtl00) 1
COMMON CEEMINClOO),HOLDEMClOO),ANSVECf 50J,SEC50,51) , IROWC 51) 8
COMMON RUMA(l00), WAMCtlOO), XKAHC(l00), QA(l00), SJA(l00) 9
COMMON ICOL( 51), NOPP( 20),ENTHRU(l00),ENTMAXC10Q), IDS( 50) 10
COMMON QSIDE , QPOINT, SIOEJ , POINTJ 1 NQDIV , NORAMS, NSTOP 50
COMMON INTV , ISECTN, NUHR , Fl , F2 , Cl , C2 51
COMMON !PRINT, DELTEE, EEMl , EEM2 , GAMMA!, GAMMA2t INT 52
COMMON INTT , I , ITST t IX r NR , MO , MP 53
COMMON NPAGE , N , QUAKE , RUP , RUT , VELMI , 101 54
COMMON 102 , 103 , 104 1 IOWl , IDW2 , IDKl , IDK2 55
COMMON IDRLl , IDRL2 t IOGl , IDG2 , !DEl , IDE2 , IOBl 56
COMMON 1082 t IOVl , IDV2 , IDQl , IOQ2 , IDJl , IDJ2 57
COMMON IDDKl , IDDK2 , IOAl , IOA2 , KGRAOD, J5 , TMIN 58
COMMON TMAX , SMIN , SMAX , NOPNTS, AREA , NSl , NS2,NS6 59
COMMON NS3 , NS4 t NS5 t IDEEM , MH , VEll , ACCELR 60
COMMON B , C , AREAP , XLONG , ELAST , ACELMX 61
COMMON OVI 1 0El 1 DE2,0Rl,ORP,OQI 1 0QP 1 DJI,OJP,OW1,0W2,0Wl,OKl,OK2,0KI
c
MP = MP
MPP = MP+l
SMAX = 0.0
SHIN = 0.0
DO 64 I = l,MPP
ITR IG (I l = 1
DEM(l.). = 0.0
XD.EMC.tl = 0.0
DEMAX{I). = 0.0
IDEMAX(I) = 0
CEEM (I) = O. 0
XCEEM(I) = 0.0
CEEMAStl) = 0.0
 ~-.·-.. . ---
~

FOH(I) = 0.0
XFOM( I) = 0.0
FOMAX(I) -= 0.0
FOM IN ( I ) -= 0. 0
IFOMAX(I) -= 0
IFOMINC I) = 0
NFOM (I) = 1
RAM(l) -= 0.0
RMAX(I) = 0.0
RSTATIIJ = o.O
DIM( IJ = 0.0
ENTHRUII) = 0.0
ENTMAX(I) = O. 0
64 CONTINUE
IFtNOP(l41-4)18,65,18
65 CONTINUE
C = (Fl•C2- F2•Cl)/(Cl•C2•(Cl-C2))
B = (f2•Cl••2- FI•C2••2)/(Cl•C2•(Cl-C2))
IF(8)22,22,18
22 IF(Fl-F2)24,23,23
23 C = Fl/Cl••2
GO TO 25
24 C = F2/C2••2
25 B = 0.0
WRITE{6,104)
104 FORMAT(47HOPARABOLA BASED ON F2 AND C2 ONLY MUST BE USED
18 CONTINUE
RETURN
END
$:18FTC REPREP
S.UBR'OtJTl.NE REP N
COMMON WAMClOO), XKAMllOO), RUMllOOl, BEEM(lOO), EEM(l00)
COMMON GAMMA(l00), XKIM(lOO),CEEMASflOO), NFOM(lOOJ, XDEMC100) 1
"U
)o COMMON DEMC100) 1 XCEEM(lOO), CEEM(100)~ 2
Gl FOM(l00), XFOM(l00) 3
111 COMMON VEL(lOO), DIM(l00), RAMtlOO), RMAX(l00}, RSTAlClOO)
in 4
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COMMON RllOO,lO) , ITRIG(lOQ), Q(lOO),FORCINtlOO), DFOMI100) 5
COMMON FOMAX(l00),1FOMAX(l00), FOMIN(l00),IFOMIN(l00), A(l00) 6
COMMON OEMAX(l00),10EMAXfl00), SJ(l00), NOP( 22),0YNAMK(l00) 7
COMMON CEEMINtlOO),HOLOEMflOO),ANSVECt 50l,SE(50,51) , tROW( 51) 8
COMMON RUMA(lOO), WAMC(lQO), XKAMCflOO), QA(l00), SJA(l00) 9
COMMON ICOL( 5U, NOPP( 20),ENTHRU(l00),ENTMAX(l00), IDS( 50) 10
COMMON QSIOE , QPOINT, SIOEJ , POINTJ, NQDIV , NORAMS, NSTOP 50
COMMON INTV r ISECTN, NUMR , Fl , F2 , Cl , C2 51
COMMON IPRINT, DELTEEr EEMl , EEM2 , GAMMAl, GAMMAi, INT 52
COMMON INTT , I , ITST , IX , NR , MO , MP 53
COMMON NPAGE t.N , QUAKE , RUP , RUT , VELMI, IDl 54
COMMON 102 , 103 , 104 t IDWl , IOW2 , IDKl , IDK2 55
COMMON- IDRLl , IORL2 , IOGl , IDG2 , IDEl , IDE2 , 1081 56
COMMON 1082 , IDYl , IOVZ , IOQl , IOQ2 , IOJl ~ IDJ2 57
COMMON IODKl , IDOK2 , IDAl , IOA2 , KGRAOD, J5 , TMIN 58
COMMON TMAX , SHIN , SMAX , NOPNTS, AREA , NSl , NS?,N$6 59
COMMON N$3 , NS4 , N$5 , IDEEM t MH , YEll , ACCELR 60
COMMON B , C , AREAP , XLONG , ELAST , ACELMX 61
COMMON DVl,OEl,DE2rORitDRPtDQitDQP,OJI,oJP,DWl,DW2,DWI,OKl,OK2,0KI
c
INTV = INTV+l
MP=MP
MPP = MP+l
NOP(4) = NOP(4)
NOPC13) = NOPC13J
NOPll4) = NOP(l4)
NOP(l5) = NOP(l5)
ITESTl = 1
ITESTP = 1
00 68 I = 1,. MP
l=l
lFl I-MP Jl8, :t7, 18
17 I TE.S.TP=2
18 CONTINUE
XDEMCI) = DEMCI)
 ;:·

DEM{I) = XDEMliJ +VELCI)*12.0*0ELTEE

IF(OfMAX(IJ~OEM(IJ)20,21,21
20 DEMAXCI)= DEM(I)
IDEMAX(I) = INTV
21 GO TOC34,19),JTESTP
34 XCEEMtlJ = CEEMCIJ
C STATEMENT 34 MUST USE A COMPUTED VALUE FOR THE ACTUAL OEMCI+l)
CEEMCI) = OEMCI) -OEMCI+l) -VELCI+l)*l2.0*0ELTEE
XFOM( I) = FOM( I)
IFfBEEMCl)-0.000001)36,36,30
30 IFCOYNAMK(IJ )31,31,32
C SMITHS DAMPING METHOD
31 OFOMCIJ = BEEM(I)•XKAMtll*lCEfM(I)-XCEEMCIJ)/(OELTEE•IZ.O)
GO TO 33
C STANDARD LINEAR SOLID DAMPING
32 OFOM{IJ = (OfOMCIJ+OYNAMK{I)*(CEEMCI)-XCEEM(I)))/
1 Cl.O+DYNAMKCIJ•OELTEE/flOOO.O*BEEMCIJ))
33 FOM(I) = CEEMfi)*XKAM(I) + OFOM(I)
GO TO 43
36 1Fl0.99999-EEMCI)l38,38,39
38 FOM{l) = CEEMli)*XKAMCI)
CEEMASliJ = AMAXlCCEEMAS(I),XCEEM(I))
GO TO 43
39 CEEMASCil = AMAXllCEEMAS{I),XCEEM(I))
CEEMIN(I) = AMINlCCEEMINCIJ,XCEEM(I)J
1FlCEEMCI))l3,43,5
5 IFCCEfM(IJ-CEEMASCillll,ll,38
11 FOM(I)=AMAXlCXKAM(I)*(CEEMAS(I)-CCEEMASCIJ-CEEMCIJ)/EEMCI)**2),0.)
GO TO 43
13 IF CCEEMCIJ-CEEMIN(I)J38,14,14
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14 FOM(I)=AMINl(XKAMCI)+(CfEMI~(I)-CCEEMTMtfJ-CEEMCl))/EEMCI)**2),0.)
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IF NOPC14)=2, SET FOM(l} = FORCINCINTV)
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1 NOP14 = NOPC14)+1
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2 FOM(l) = FORCINtiNTV)
IF(FOM(l)-1.0)3,3,4
3 DEM(l) = XDEM(l)
CEEMll) = XCEEM(l}
GO TO 16
C IF NOPtl4) = 4, USE PARABOLIC FOM(l) VS. CEEM(l} CURVE
C THE RAM MUST BE A SINGLE MASS IF FOM VS. OEM IS PARABOLIC
6 IF(N0P(l4)-4)4,7,4
1 IFtCEEMlll- CEEMASll))9,8,8
a FOM(l) = C•CEEM(l}••2 + B•CEEMfl)
GO TO 12
4 IFICEEM(l)-CEEMAS(l))l6,12,12
9 FOMAX(1} = AMAXllXFOM(l),FOMAX(l))
FOM(l) = FOMAX(l)-(lCEEMAStll-CEEM(l))•FOMAX(1)**2)/(2.0•SMAX•
1 EEMtU••2)
GO TO 16
12 SMAX = SMAX+((f0M(l)+XFOM(l))/2.0)•(CEEM(ll-XCEEM(l)).
16 CONTINUE
lflGAMMA(I))46,44,45
44 FOM(I} =AMAXl (.0, FOMCI))
GO TO 46
45 IF(FOMtl) - XFOM(l))48,47,47
48 NFOM( I) = 2
4 7 I X = NF OM ( I )
GO TO (46,49),1X
49 HOLOF = FOMll)
FOr•H IT = AMA Xl ( FOM( I), GAMMA (I))
COMMEN-T THE ~O.l HOLDS. MtM .. PRESSUAE A.T GAMMA( U FOR..• 0.1 SECONDS WHILE THE
COMMENT .0025 REDUCES THE PRESSURE TO ZERO l~ .002> ADDITIONAL SECONDS.
TINT = INTV
IFlTINT - .Ol/OELTEE)46,46,90
90 FOM(J) = AMAXllO.O, GAMMAll)•(l.0-(0ELTEE•TINT-.01)/.0025),HOLDF)
46 CONTINUE
ENTHRUfl) = ENTHRU(I)+tFOM(J)+XFOM(I))*(DEM(I+l)-XOEM(I+l))/24.0

~o.;·
 --~----·--·-~-------.·--~ .--..,_._;;;.::--_.::..::.::;;;:.::::::::::.::::::::.:~:;:>·· :---:~---:1·~~-------
j
-;rr:~·::;t<·~·.~~'l'~r-, .......,~.,~ ·~~J~,/"·~· 1 • ~
''"':""""'·-:;·...-~~-

r... -~·

ENTMAX(I) = AMAXllENTMAX(I),ENTHRU(I))
GO T0(22,19),1TEST1
22 IF(CEEM(1) - CEEMAS(l))l5,19,19
15 SMIN = SMIN-((F0M(l)+XFOM(l))/2.0)•(CEEM(l)-XCEEM(l))
19 CONTINUE
1F(NOP(4)-5)29 1 28,29
c GENERALIZED SOIL RESISTANCE
28 CALL GENRAM
.GO TO 55
29 CONTINUE
c SMITHS SOIL RESISTANCE
IF(XKIM(I)}50,155,50
155 GO T0(55,156),ITESTP
156 IF(XKIM(MPP ))50 1 55,50
50 IF(OIM(l) -OEM(I) +Q(l) )51,52,52
51 OIM(I) = DEMtl) -Q(I}
52 CONTINUE
70 IF(OIM(I) -DEM(I) -Q(I) }53,53,54
54 DIM(I) = DEM(I) +Q(I)
53 CONTINUE
DIM(MPP } =AMAXl CDIM(MP),OIM(MPP ))
ITST = ITRIG(I)
GO TO(l0,57),ITST
10 IF{OEMtll -DIM(I} -Q(I) )56,57 1 57
56 RAM(I} = (OEM{I)-OIM(I))•XKIM(I}•(l.O+(SJ(I) •VELII)))
GO T0(55,171),1TESTP
171 RAM(MP) = RAM(MP)+(OEM(MPJ~OIM(MPP ))•XKIMlMPP )•
, 1 (l.O+(SJ(MPP)•VEL(MP)))
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1'1
C RAM(MP+ll MAY BE TENSILE
GO ro· S5
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57 RAM(IJ ::: IOEMfll-DIMCil+ S..J(I) •Q4l) •VElliJl•XKIMCI}.
1--1 ITRIGU) = 2
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Ill 172 RAMCMPl=RAM(MP)+(OEM(I)-DIM(MPP)+SJ(MPP)•Q(MPP)•VEL(MP))*XKIM(MPP)
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c
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GO T0{58,72),1TEST1
58 VEL(l) = VEll1l-(FOMll) +RAM(l))•32.17•DELTEE/WAM{l)
ITESTl = 2
GO TO 59
72 VELlll = VElli)+(FOMll-1) -FOM(I) -RAMll)}•32.17•DELTEE/WAM(I)
59 CONTINUE
IFlNOPll5l-1)85,85,83
83 VELlll = VElfl) + 32.1T•DELTEE
85 CONTINUE
65 IFlFOMAXlll-FOMll))67i67,66
67 FOMAX(I) = FOM(I)
IFOMAX(I) = INTV
66 IFlFOMINll)-F0M(I))68,69,69
69 FOMINtll = FOM(l)
IFOMINll) = INTV
68 CONTINUE
IFlVELl2l/VEll -2.1)61,60, 60
60 WRITE(6,105)
INTT = 2
RETURN
105 FORMAT(76HO THE RATIO OF THE VElOCITY OF W(2) TO THE VELOCITY OF
lTHE RAM EXCEEDS 2.1. )
61 IF(VEL(MP)/VEll -2. 1) 63 t 62 '62
62 WRITE ( 6 tl 06)
106 FORMATI76HO THE RATIO OF THE VELOCITY OF ~{P) TO THE VELOCITY OF
lTHE RAM EXCEEDS c.l. )
INTT ::z 2
R:ETUR~
63 CONTINUE
LDCEtt = MH-1
ACCELR = lFOM(LDCELL-1)-FOM(LOCELL))/WAM(LOCELL)
71 ACELMX=AMAXl(ACELMX,ACCELR)
73 CONTINUE
RETURN
 ,_,

END
$1BFTC RAMGEN
SUBROUTINE GENRAM
c
C NQDIV =NO. OF EQUAL SEGMENTS INTO WHICH Q(IJ IS DIVIDED= 10
C RSTATCI) = STATIC SOIL RESISTANCE NEGLECTING THE SOIL DAMPING EFFECTS
C RMAXCil =A TEMPORARY MAXIMUM STATIC SOIL RESISTANCE
C PERCQ =DISTANCE FROM ZERO DISPLACEMENT TO OEM(!) IN UNITS (1.732, •• )
c
COMMON WAM(lOQ), XKAMClOOl, RUM(l00), BEEMClOO), EEMC100) 1
COMMON GAMMA(l00), XKIMC100),CEEMAS(l00), NFOMC!OO), XOEMClOO) 2
COMMON DEMllOO), XCEEMC100), CEEMCtOO), FOM(lOQ}, XFOMC100} 3
COMMON VELC_lOOlt OIM(lQQ), RAM( 100), RMA.X( 1001, RSTATC 100) 4
COMMON RClOO,lO) , ITRIGC100), Q(lOOJ,FORCIN(lOO), OFOM{l00) 5
COMMON FOMAX(l00l,IFOMAX(l00), FOMIN(l00),IFOMINC100), A(l00) 6
COMMON DEMAXtlOO),IOEMAX(!OO), SJCtOO), NOP( 22),0YNAMK(l00) 7
COMMON CEEMIN(l00),HOLOEM(l00),ANSVECC 50),SEC50,51) , IROW( 51) 8
COMMON RUMA(lOQ), WAMC(lOO), XKAMC(l00), QAClOO), SJA(l00) 9
COMMON ICOL( 51), NOPP( 20),ENTHRU(l00),ENTMAXC100), IDS( 50) 10
COMMON QSIDE , QPOINT, SIDEJ , POINTJ, NQDIV , NORAMS, NSTOP 50
COMMON INTV , ISECTN, NUMR , Fl , F2 , Cl , C2 51
COMMON !PRINT, OELTEE, EEMl , EEM2 , GAMMA!, GAMMA2, INT 52
COMMON INTT , I , ITST , IX , NR , MO , MP 53
COMMON NPAGE , N , QUAKE , RUP , RUT , VELMI , IDl 54
COMMON ID2 , 103 , 104 , IDWl , IOW2 i IDKl , IDK2 55
COMMON IORLl , IDRL2 , IOGl , IOG2 , IDEl , IDE2 , 1081 56
COMMON IDBZ t IDVl , IDV2 , IOQl , IOQ2 , IDJl , IDJ2 57
COMMON IOOKl , IODK2 , IOAl , IDA2 , KGRAOD, J~ , TMIN 5B
COMMON TMAX , SMIN , SMAX 1 NOPNTSt AREA , NSl , NS2,NS6 59
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COMMON NS3 , NS4 , NS5 , ID£EM , MH , VEll , ACCELR 60
I'll COM~ON 8 r C ,. AREAP -. XLONG- • EtAST , ACELMX 61
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~ MP = MP
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-< IF(XKIM(K)-0.1) 1,2,2
til
>< 1 RAM(K) = 0.0
GO TO 70
2 IFCDEM(Kl-DIM(K})32,3 1 3
3 DIM(K) = DEM(K)
IF(DEM(Kl-Q(K))7,6,6
6 RSTAT(K) = RlK,NQDIV)
GO TO 50
1 PERCQ = DEM(K)/(Q(K}/QDIY)
IPERCQ = PERCQ
XPERCQ = IPERCQ
IFliPERCQ)8,8,9
8 RSTAT(K) = PERCQ•R(K,l)
GO TO 50
9 RSTAT(K) = R(K,IPERCQ)+(PERCQ-XPERCQ)•(R(K,IPERCQ+l)-R(K,IPERCQ))
GO TO 50
32 RMAX(K) =AMAXl(RMAXtKltRSTAT(K))
RSTAT(K) = RMAX(K)-(DIM(K)-DEM(K})•XKIM(K)
C THE STATIC FORCE SHOULD REALLY LEAVE THE XKIM(I) SLOPE AND REMAIN
C CONSTANT IF RMAX(I)+RSTATtl) EVER EXCEEDS 0.0
IF{RMA·XlK)+RSTAT(K) )39,50,50
39 WRITE(6,200)RMAX(K),RSTAT(K),K
200 FORMATl11HORMAX(I) = Fl0.2, 6X, 12H RSTAT(I) = Fl0.2,6X,4H I =16)
C STATEMENTS 50 THRU 70 INCLUDE THE SOIL DAMPING EFFECT
50 ITST = ITRIG(K)
GO T0(51,57),ITST
51 1FlDEMlK)-Q(K))56t57,57
56 RAMfKJ = RSTATlKJ+RS.TATfK.l•SJ{KJ•VEL.tiO
GO TO 70
57 RAM(K) = RSTAT(K)+R(K,NQOIVl•SJ(KJ•VEL(K)
ITRIG(K) = 2
GO TO 70
10 IF(K-MP)80,71,73

·-:;,
 71 CONTINUE
K = MP+l
lF(XKIM(K)-0.01)80,80,72
72 OEM(K) = DEM(MP)
VEL(K) = VEL{MP)
GO TO 2
73 CONTINUE
C 73 IF(RAM(K))74,75,75 (OLD STATEMENT)
C 74 RAM(K) = 0.0 (OLD STATEMENT) 1
74 CONTINUE
75 RAM(MP) = RAMCMP)+RAM(MP+l)
c RAM(MP+l) CAN GO INTO TENSION
80 RETURN
END
$1BFTC EXCTG
SUBROUTINE EXACTG
COMMON WAMClOO), XKAM(l00), RUM(l00), BEEMClOO}, EEMC100) 1
COMMON GAMMA(lOQ), XKIM(l00),CEEMAS(l00), NFUM(lQO), XDEM(l00) 2
COMMON OEMC100), XCEEMflOO), CEEM(l00), F0M(l00), XFOMC100) 3
COMMON VEL(100), OIM(l00), RAMClOO), RMAXClQQ), RSTAT(l00) 4
COMMON R(lOO,lO) , ITRIG(l00), QC100),FORCIN(l00), DFOM(lOO) 5
COMMON FOMAX(lOO),IFOMAXClOO), FOMIN(lOO),IFOMIN(lOO), A(l00) 6
COMMON OEMAX(lOOJ,IDEMAX(lOO), SJ(l00), NOP( 22),DYNAMK(l00) 1
COMMON CEEMIN(l00),H0LOEM(l00),ANSVEC( 50),SE(50,51) , IROW( 51) 8
COMMON RUMA(l00), WAMC(lOO), XKAMC(l00), QA(lQO), SJA(l00) 9
COMMON ICOL( 51), NOPP( 20),ENTHRU(l00),ENTMAXC100), IDS( 50) 10
COMMON QSIDE , QPOINT, SIDEJ , POINTJ, NQOIV , NORAMS, NSTOP 50
COMMON INTV , ISECTN, NUMR , Fl , F2 , Cl , C2 51
COMMON !PRINT, DELTEE, EEMl , EEM2 , GAMMAl, GAMMA2, INT 52
COMMON INTT , I , ITST , IX , NR , MO , MP 53
'U COMMON NPAGE , N , QUAKE , RUP , RUT , VELMI , 101 54
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COMMON ID2 , 103 , 104 , IDWl , IDW2 , IDKl , IOK2 55
en COMMON IDRLl , IDRL2 , IOGl , IOG2 , IOEl , IOE2 , IOAl 56
~ COMMON 1082 , IOVl , IOV2 , IOQl , IDQ2 , IOJl , IOJ2
Ill
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1- COMMON NS3 , NS4 , NS5 , IOEEM , MH , VEll , ACCELR
1'11 60
Gi COMMON 8 , C , AREAP , XLONG , ELAST , ACELMX
X 61
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COMMON 0VftDEl,OE2rDRirORP,OQI,OQP,OJI,OJP,OWl,OW2,DWI,DKl,DK2,0KI
c
MP = MP .
MO = MO
MMD = MO-l
MMOO = MO - 2
MAO = MP - MO
NSDD = MP-MO+l
00 6 NSEW = ltNSDD
DO 6 NSE = 1rNSOD
6 SElNSEW,NSE) = o.O
SE(1,1) = XKAM(MO) + XKIM(MQ)
SE(2,1) = -XKAMfMO)
DO 13 K = 2,MAO
NN = K + MMOO
NNN = K + MMO
SE(K-l,K) = SE(K,K-1)
SE(K,K) = XKAM(NN) + XKAM(NNN) + XKIM(NNN)
SE(K+1,K) = -XKAM(NNN)
13 CONTINUE
SECMAO,NSOO) = SECNSODrMAO)
SECNSDOtNSOO) = XKAM(MP-l)+XKIM(MP) + XKIM(MP+l)
DET = TAMINV(SE,ICOL,NS00,50,0.00001)
IFC0.00001- ABS(OET))l4,12,12
12 WRITEt6,100)0ET
100 FORMATf33HOTHE VALUE OF THE DETERMINANT = Fl0.1)
INTT = 2
RETURN
14 CONTINUE
WAMTL = 0.0
DO 15 NSEW = 2,MO
15 WAMTL = WAMTL + WAM(NSEW)
 -----------------------~-----------------------------------------------------------------------------------------------------------------

SE(l,NSDD+l) = WAMTL
DO 16 NSEW = 2,NSDD
NUTZ = MMO+NSEW
16 SElNSEWrNSDD+l) = WAMINUTZ)
DO 17 IANS = lrNSDD
17 ANSVEC(IANS) = Q.O
DO 23 IMl = l,NSDD
DO 23 1M2 = l,NSDO
23 ANSVEC(IMl) = ANSVECIIMl)+SEliMl,IM2l•SE(IH2,NSDD+l)
NAT = 0
DO 26 NST = MO,MP
NAT = NAT+l
OEM{NST) = ANSVEC(NATl
HOLDEMCNST) = DEM(NST)
26 CONTINUE
wos = o.o
DO 27 NSt = 2,MMO
WOS = WOS+WAMlNSTl
CEEM (NST) = WOS/XKAMlNST)
FOM(NSTl = WOS
21 CONTINUE
00 28 NST = l,MMO
NEL = MO-NST
DEMlNEL) = OEM(NEL+l) + CEEM(NEL)
HOLDEM(NEl) = DEM(Nfl)
28 CONTINUE
MAM = MP-1
00 29 NST = MO,MAM
CEEMtNST) = DEM!NST) - DEMfNST+l)
FOM(NST) = CEEMtNSTl•XKAMlNST)
R4PHN.ST} = DEM'fNST)•XKIMfNS.Tl
29 CONTINUE
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SIBFTC SMTH
SUBROUTINE SMITH
COMMON WAMI100), XKAM(l00), RUMllOO), 8EEMflOQ), EEM(l00) 1
COMMON GAMMA(lOO), XKIM(lOO),CEEMA$(100), NFOM(l00), XOEM(l00) 2
COMMON DEM(lQO), XCEEM(l00), CEEM(l00), FOM{l00), XFOM(l00} 3
COMMON VEL(l00) 1 OIM(l00), RAM(l00}, RMAXClOO), RSTAT(l00) 4
COMMON RllOO,lO) , ITRIGtlOO), QClOO),FORCIN(lQO), OFOM(l00) 5
COMMON FOMAXllOO),IFOMAXllOO), FOMIN(l00),1FOMIN(l00), A(l00) 6
COMMON DEMAX(l0Q),l0EMAX(100), SJ(l00), NOP( 22),0YNAMK(100} 7
COMMON CEEMINtlOO),HOLOEM(lOO),ANSVECI 50),SEI50,5ll , IROW( 51) 8
COMMON RUMA(l00), WAMC(lOO), XKAMCtlOO), QA(lOQ), SJA£100) 9
COMMON ICOL( 51), NOPP( 20l 1 ENTHRUtlOO),ENTMAX(l00), IDS( 50) 10
COMMON QSIOE ,. QPOINT, SIDEJ , POINTJr. NOOIV , NORAMS, NSTOP 5.0
COMMON INTV , ISECTN, NUMR , Fl , F2 , Cl , C2 51
COMMON !PRINT, DELTEE, EEMl , EEM2 t GAMMAl, GAMMA2, INT 52
COMMON INTT , I , ITST , IX , NR , MO , MP 53
COMMON NPAGE , N , QUAKE , RUP , RUT t VELMI , 101 54
COMMON 102 , 103 , 104 , IDWl , IDW2 , IDKl t IOK2 55
COMMON IORLl , IORL2 , IOGl , IDG2 , lOEl , IOE2 , IDBl 56
COMMON 1082 , IDVl , IOV2 , IOQl , IOQ2 , IOJl , IOJ2 57
COMMON IOOKl , IODK2 , IDAl , IOA2 , KGRAOD, J5 , TMIN 58
COMMON TMAX , SMIN , SMAX , NOPNTS, AREA , N~l , NS2,NS6 59
COMMON NS3 , N$4 , N$5 , !DEEM , MH , VELl , ACCELR 60
COMMON B , C , AREAP , XLONG , ELAST , ACELMX 61
COMMON DVl,DEl,OE2 1 DRI,ORP,DQI,OQP,DJI,DJP,DWl,DW2,DWI,OKl,OK2,0KI
c
MP = MP
N = MP-1
WAMTL = 0.0
RAMTL = 0.0
00 5 JT = 2,MP
WAMTL = WAMTL + WAMtJT)
5 RAMTL = RAMTL + RUM(JT)
RAMTL = RAMTL + RUM(MP+l)
 ;'·

00 8 JT = 2tN
RAM(JT) = (RUM(JT)*WAMTL)/RAMTL
8 FOM(JT) = FOM(JT-l)+WAMlJT)-RAM(JT)
RAM(l) = RUM(l)*WAMTL/RAMTL
RAM(MP) = RUMCMP)*WAMTL/RAMTL
RAM(MP+l) = RUM(MP+l)*WAMTL/RAMTL
OEM(MP) = (RAM(MP)+RAMlMP+l))/(XKIM{MP)+XKIM(MP+l))
HOLDEM(MP) = DEM(MP)
DO 11 JT = l,N
JTM = MP-JT
CEEM(JTM) = FOM{JTM)/XKAM(JTM)
OEM(JTM) = DEM(JTM+l) + CEEMCJTM)
HOLOEM(JTM) = DEMCJTM)
DlM(JTM)=OEMlJTM)-WAMTL*Q(JTM)/RAMTL
11 CONTINUE
RETURN
END
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