AN ANALYTICAL METHOD OF MODELING DEMOLITIONS
B.S., United States Air Force Academy, 1991
A thesis submitted to the
University of Colorado at Denver
in partial fulfillment
of the requirements for the degree of
Master of Science
Mechanical Engineering
1999
by
Antoine Derrick Johnson
1999 by Antoine Derrick Johnson
All rights reserved.
This thesis for the Master of Science
degree by
Antoine Derrick Johnson
has been approved
by
Date
r
Johnson, Antoine Derrick (M.S., Mechanical Engineering)
An Analytical Method of Modeling Demolitions
Thesis directed by Professor James C. Gerdeen
ABSTRACT
The author analyses the effects of an explosion on a bolt. The pressure required
to shear a bolt and the pressure required to shear a section of pipe are derived.
Finally, an amount of explosive is determined.
This abstract accurately represents the content of the candidate's thesis. I
recommend its publication.
Signed
James C. Gerdeen
iv
DEDICATION
I dedicate this first to the men and women of the 8th Maintenance Squadron
Propulsion Flight; they did the job well, and deserved more than they got.
I also dedicate this to my mother, for her support.
Finally, I'd like to dedicate this to my friend Don. He was supposed to see me
graduate. Through his life, I learned how to live.
"Stand back. Be silent. Be still. That's it. And look upon this moment. Savor
it. Rejoice with great gladness. Great gladness. Remember it always, for you
are joined by it. You are one, under the stars. Remember.... For it is the doom
of men that they forget." Merlin, in "Excalibur." Copyright MCMLXXXI, Orion
Pictures Company.
ACKNOWLEDGMENTS
I would like to thank the following people:
My sisters, Kimberly and Heather, and my father, who had to deal with the worst
of me for these three years.
Ms. Loretta Duran, for her assistance over the last two years. I could not
(literally!) have graduated without her.
Ms. Annette Dye, who provided impressive illustrations and provided significant
technical input.
Mrs. Kathleen Van Orsdel, who read this document and provided editorial
comments.
Drs. Zalabak and ShockleyZalabak for their inspiration.
Colonel (Dr.) James Head, Lieutenant Colonel Anthony Heyward, and Major Paul
Bowman, for getting me into this fine institution.
Mr. Fred Saar, for listening patiently to the inspired narration of this document.
My supervisors, Kevin and Hal, and all of the other people at AlliedSignal.
CONTENTS
Figures......................................................................ix
Chapter
1 Problem Definition........................................................ 1
2 Fun with Plasticity.......................................................4
2.1 Introduce theorems.................................................. 5
2.1.1 Generalized Stress and Strain.....................................5
2.1.2 Lower Bound, or Equilibrium Theory................................5
2.1.3 Upper Bound, or Kinematic Theory..................................5
2.1.4 Dynamic Collapse..................................................6
2.2 Beam theory...........................................................6
2.2.1 Equations.........................................................6
2.2.2 Square yield concept..............................................7
2.2.3 An AsideStrain Rate effects on yield strength....................9
2.3 Derive Velocity Equation..............................................9
2.4 Explosives vs. the circular pipe, Round 1............................10
2.4.1 Determine Initial yield moment...................................10
2.4.2 Determine Collapse moment........................................11
2.4.3 Determine Collapse Load..........................................11
2.4.4 Determine Shape factor.......................................... 12
2.4.5 Integrate spatially and solve for shear force................... 12
2.4.6 An asidethe explosive pressure representation...................12
2.4.7 Integrate via time and solve for
velocity, time and distance..........................................14
2.4.8 Integrate spatially (again!)
and solve for moment.................................................16
2.4.9 Check applicability with square yield............................16
2.4.10 Results..........................................................17
3 Simulation Physics........................................................18
3.1 Impulse definition, and why it is good...............................18
vii
4 Fun with Waves.........................................................19
4.1 Shockwave..........................................................19
4.2 Hugeniots..........................................................19
4.3 Rarefaction wave...................................................21
4.4 Reflection........................................................ 22
5 Fun with Explosives.................................................. 23
5.1 DefinitionsActivation Energy, Detonation
Velocity, and ChapmanJouguet Pressures............................23
5.2 Explosive Equations................................................24
6 Rates of Chemical Reaction, and the
Arrhenius Equation.....................................................25
6.1 Definitions........................................................25
6.2 How to get the Reaction Rate, or
Chemical Kinetics 101..............................................25
6.2.1 Determining the Order..........................................25
6.2.2 Determining the Reaction Rate Constant k.......................26
7 Resolutions?...........................................................28
7.1 Problem Definition, again..........................................28
7.2 Problem results....................................................29
8 OK, but...the obvious flaws and
weaknesses of the theory...............................................30
8.1 Accuracy of the Chemical reaction rate.............................31
9 Summary................................................................33
Appendix
A. Clohessy's Circular Beam...............................................34
B. Pipeline Design........................................................35
C. Evaluating ta and td from Cooper's charts..............................39
D. Explosives vs. the Circular Pipe, Round II.............................41
Bibliography..............................................................48
viii
FIGURES
Figure
1.1  Pipeline Layout...........................................
1.2  Side and Top Views........................................
2.1  The bolt. Distance between bolt head and line is 0.3 meters.
2.2  Square Yield Surface......................................
2.3  Uniform Velocity Profile..................................
D.l Varying Velocity Profile...............................................41
ix
VO 00 VI NJ M
1 Problem Definition
The problem is inspired by Alistair McLean's "Navarone" movie series. In the
two movies, a squad of commandos inevitably destroys a target of strategic
importance, altering the course (or so it would seem) of World War II.
In "The Guns of Navarone," the team destroys a monolithic gun emplacement
that is preventing the landing of the British fleet. The guns are placed on an
unassailable cliff, heavily surrounded by the enemy. The Allies cannot destroy
the guns the old fashioned way (aerial bombardment), so a team sneaks behind
enemy lines to dislodge the guns. The guns collapse under their weight into the
seaan innovative approach. The fleet then can pass beyond its Charybdis, and
presumably the eastern half of the Mediterranean is freed.
In "Force 10 from Navarone," the team (Force 10) sets out to destroy a bridge
that separates a German division from Greece and the remainder of the Balkans.
The survivors from the original (Major Mallory and Staff Sergeant Miller) are
attached to Force 10 to kill a double agent. Neither half trusts the other.
Ultimately, the separate halves accomplishes the other's mission. The team
succeeds again in an innovative fashionby weakening a small dam.1 The water
pressure is sufficient to weaken the dam, which releases the river behind it. The
sudden onrush of the river is enough to topple the bridge supports, and the
bridge falls down.
The problem presented is hoped to be of similar interest. Imagine a pipeline in
the cold arctic north. The allies have lost a major beachhead in the north, and
the enemy will be there all winter long if something cannot be done to force
them out. The pipeline is well guarded, and protected against the obvious aerial
assault to destroy it. Obviously the pumping stations would be a target, but
each one has massive numbers (enough to prevent the forward assault, anyway)
of hostile enemy troops ready to protect it. The entire pipeline is protected, in
fact, except a three or four mile stretch across an ocean inlet that we will call
"Ocean Bay." Here the pipeline lies submerged 50 to 70 feet in ice covered
waters.
The direct naval strike is not an option, because the enemy has a large naval
fleet there. However, a small team of "brave men with explosives" might be
able to sneak inland, follow the river to the sea, and destroy it. Our intelligence
1 "Oh! I can do a tremendous job on the dam. Yes! See, with the dam, you
have your natural elements working for you. It's like having an enormous bath.
All you have to do is pull the plug!" Staff Sergeant Miller, in "Force 10 from
Navarone", Copyright 1974, Orion Pictures Corporation.
1
sources have assured us that standard enemy operating procedure calls for flow
shutoff during air attack. At this point, since the pipe flow devices would not
notice any pressure loss, the destruction could be made. The resuming oil flow,
combined with a steady 3 or 4 mile current and a mysteriously damaged pipe
restraint, might cause the pipe to break underwater in an unreachable
environmenta total loss significant enough to encourage the enemy to spend
winter elsewhere. This thesis will look at the engineering aspects of such a
problem.
Figure 1.2 Side and Top Views
The layout is sketched above. Assuming the inlet in the chosen section is 100
feet deep, the pipe crosses this section in 70 feet of water. For civil engineering
reasons, there is a 30 degree turn in the chosen 100 meter section of pipe,
which is conveniently in the path of the current. The pipe is clamped at the end
of these sections. The pipe is made of type 304 stainless steel, with a 0.55
meter inner radius and a 0.57 meter outer radius.2 The yield strength for this
2 See Appendix: Pipeline Design.
2
steel is 1103 MPa.3 The post is bolted by eight bolts arranged radially around
the post. This design is similar to a light pole. The bolts are installed into a
large concrete weight which helps steady the pipe, and allows construction with
minimal surface preparation.
The idea is to determine how much explosive is needed to destroy the bolts
without significantly damaging the pipe or pole above. If the pipe is buckled or
damaged, the resulting leak would alert the personnel at the pumping station, so
it is something that the team must avoid. The 50 cm diameter steel bolt is also
made of type 304 stainless steel. Destroying the bolts will unbalance the column
sufficiently (conveniently) for the oil flow to topple the structure. The complete
destruction of this pipe, along with a related mission to destroy other oilrelated
assets, would force the enemy to abandon the wintry coast.
3 Data from Machine Design: An Integrated Approach, table 08, pg. 997.
Obviously, an underwater pipe constructed of steel would have corrosion
problems; steel was chosen based on availability of explosive related data.
3
2 Fun with Plasticity
Plasticity covers materials after they have reached the elastic limit. Most
mechanical engineering efforts do not focus on the plastic region of a material
primarily because permanent deformation, let alone yielding, is a bad thing.
Metal forming and designing crash structures are two significant exceptions. The
problem involves plasticity to determine the deformations and collapse loads to
destroy the bolt or damage the pipe.
The following analogue is offered to clarify the problem and the solution
proposed here. Imagine a small boy on a cold Saturday morning, with a little
red wagon and lots of snow for a snowball fight. He has to pull the wagon
across the street to his friends house, so they will be supplied with plenty of
snowballs for the morning's festivities. The force that the little boy can pull the
wagon with can be called F. The wagon has a mass m, so the acceleration of the
wagon is
a = F/m
Now, the road that the boy has to cross is clear. Therefore, while there is a
resisting frictional force, but it is not significant compared to the boy's strength.
Since this is a simple example, the resisting frictional force will be ignored. Now,
suppose that the boy were to give the wagon a push of a certain duration. The
wagon would accelerate to some maximum velocity and finally stop due to
inertia and friction. At that point, the wagon should have crossed the street,
which has a certain width of w. (That was the little boy's goal, anyway). The
distance that the wagon covers is a function of the acceleration that the boy
gives, integrated twice to find velocity and distance. Initial conditions usually
imply that the velocity is zero, and the distance is zero, but these can change.
The boy gets the wagon across the street to his friend, and the snowball fights
that day are marvelous.
Now, the comparisons can be made. The "wagon" is an engineering structure
known as a beam. The bolts that keep the pipe support upright will be modeled
as a beam. The "distance across the street" is the width or depth of the beam
that the explosive needs to cut through. Instead of the little boy supplying a
force, the explosives supply a pressure distribution for a certain time, so that
sufficient force is generated to shear completely through the beam and therefore
cause it to fail. Frictional effects (the resistance of pressure on the other face of
the beam) are neglected, just as the friction involved in dragging the wagon
across the road was neglected. Finally, the inertia of the beam is included.
Initial conditions of velocity (is the beam moving prior to start?) and distance
(almost always zero) will be assumed.
4
2.1 Introduce theorems
Plasticity covers many topics. This section will define generalized stress and
generalized strain, topics necessary to understand the two collapse theorems
used in plasticity. These are the Lower and Upper Bound theories, because they
predict the lower and upper estimates for collapse. Finally, the paper will
introduce some basic concepts of dynamic plasticity theory.
2.1.1 Generalized Stress and Strain
Generalized stresses and strains are not what they might seem. If one starts
with the concept of energy, then the rate of consumption of energy is termed
the power. The power is integrated over time to determine the energy
consumed.
Returning to elementary Strength of Materials theory, we have the concept of
strain energy, which is the time integral of the product of stress and strain. The
concept of a generalized stress and strain fit here. Instead of using the stress
and strain, a product that creates a "power" term is definedfor instance, a pair
consisting of moment and angular velocity, or a force and velocity pair. The
portion that deals with the effort is termed the generalized stress, and the
portion that deals with the motion is termed the generalized strain.
2.1.2 Lower Bound, or Equilibrium Theory
The Lower Bound theorem is best described as a theorem of moments. Assume
a bar exists with a specified load. If a moment can be placed such that the
moment and the force are in equilibrium, without the moment exceeding the
collapse moment previously predicted anywhere in the structure, then the load
will not cause collapse. This theorem is probably most useful for determining
the collapse load; with various uniform cross sections and something called a
shape factor4, it is relatively easy to determine if the load will collapse the
structural element.
2.1.3 Upper Bound, or Kinematic Theory
The Upper Bound theorem equates the rates of energy generation (motion) to
energy dissipation (response). A motion is proposed that describes the collapse
4 The shape factor for a beam is the ratio of the fully plastic (collapse) moment
to the initial yield moment. It characterizes the shape. The initial yield moment
in the beam is the moment which causes the tensile stress at the top of the
beam to equal the yield stress. The collapse moment is that moment which
causes the yield stress to exist throughout the beam.
5
mechanism, along with the forces involved, and the energy required (Fe)
represents the energy generation part. The energy dissipation portion is
represented by the strain energy involved. If the energy generated exceeds the
energy dissipated, the motion is possible.
2.1.4 Dynamic Collapse
Assuming that the above equations are correct, why would anyone want to
introduce dynamic effects? Well, the load is a dynamic one. A good example of
a static load is resting a large metallic object (for instance, a ship) on the pipe.
Assuming it is gradually applied, no significant dynamic effects will occur.
On the other hand, the times involved in the application of the explosive loads
are significantly less than 30 milliseconds. Well before one can even begin to
say "Mississippi," the explosive event is over. A good rule of thumb is that if the
load is applied for a time less than or equal to the period of a structure's lowest
fundamental frequency, then dynamic effects should be considered. The
equations in this paper assume that is the case.
Essentially, what the static approach neglects, and what is added by dynamic
plasticity, is the acceleration of the beam, plate, or other affected structure. The
result is that motion of the beam alone can cause collapsethis effect can
reduce the force required. Also, a shear effect, called a shear slide, can occur,
where the shear is discontinuous across a section. This is the effect that the
problem seeks.
For the problem, what will be needed is an explosive pressure to shear the bolt,
and an idea of the pressure needed to shear the pipe. These are dynamic
effects which will require the dynamic approaches.
In this paper, the author will derive calculations for the bolt, and reserve
calculations for the pipe for the appendix.
2.2 Beam theory
2.2.1 Equations
A beam is an object with one axis dimension significantly greater that the other
two, that is loaded perpendicularly to the long axis. The first example of the
beam that we are exposed to is probably the "teetertotter", a beam with two
(relatively) equal loads on its ends and pinned in the middle to pivot. Other
examples of beams are railroad tracks, beams in houses and construction, and
various machine elements that carry weight as beams, for instance a shaft in a
turbine.
Jones has defined the equation of motion of a beam as:
6
(2.1)
9 ?
32M/Sx2 = 5Q/5X = p + m W
where M is the bending moment in the beam, Q is the shear force acting in the
beam, x describes the length of the beam, p is the pressure on the beam, m is
the mass of the beam5 and W (W dot dot) is the second derivative of the
deflection with respect to time of the beam. After integration (twicesee
section 2.3, Derive Velocity Equation), equation 2.1 describes the beam's
deflection. As mentioned above, we will model the bolts as beams.
<Â§>
n~Ti
Distance =
0.3 meters
Figure 2.1 The bolt. Distance between bolt head and line is 0.3 meters.
2.2.2 Square yield concept
Of interest is the method of determining when the material has yielded. The
reader should be familiar with the concept of a yield stress, which is the stress
beyond which permanent deformation takes place. It is used as a factor of merit
in design because in most cases, permanent deformation due to applied loads is
not a good thing. A normal6 material has a yield stress, and then hardens, so
that further on, another (ultimate) stress defines when the material ruptures.
A perfectly plastic material is an engineering fictionevery material will
eventually strain harden before it fails.7 A perfectly plastic material reaches a
yield stress and from then on behaves sort of like silly puttyit does not get any
stronger or weaker, it just deforms.
The "square yield" is so termed because it includes the yield due to shear and to
plastic collapse in a squarelike so (see below).
5 The mass must be independent of x, i.e. constant cross section, for this
equation to be valid.
6 The author has ignored the "brittle" materials. Traditionally, if a material has
less than 2% deformation at rupture, it is brittle and therefore is not really a
candidate for yield stress calculations. Brittle materials have yield stresses
they are just far more likely to fail because of large normal stresses due to crack
propagation and other effects well before a yield theory could be applied.
7 In deference to my thesis advisor, there are actually materials that have nearly
constant strain for a given stress over a certain temperature range.
7
Q/Qo
m/Mo
Figure 2.2 Square Yield Surface
Note that Q0 is the shear force per unit length causing yield and M0 is the
moment causing plastic collapse. The yield surface follows Drucker's postulate
that a yield surface cannot be concave. A concave yield surface would ultimately
mean that energy is gained through deformation, i.e. the material is self
deforming. It is convenient to represent the surfaces as Q/Q0 and M/M0, where
both quantities are functions divided by the materiallength. A material cannot
exceed the Q/Q0 or M/M0 boundary, because at that point it has yielded or
collapsed. This particular characteristic is important for determining whether or
not the motion proceeds as assumed.
Significant among this square yield8 concept is the following ratio, v, where
v=Q0L72Mo
(2.2)
L is the length of the beam. This gwes a method to determine which effect
(shearing or plastic collapse) is stronger. A "v" value less than 1 indicates that
the beam is more likely to shear rather than collapse (bend). Similarly, a value
of v greater than one indicates the beam is more likely to collapse (bend) than
yield due to shear.
8 Note that the square yield is an upper limit approximation where failure is
assumed by shear alone or moment alone. Yielding by a combination of shear
and moment is represented by an elliptical yield surface.
8
2.2.3 An AsideStrain Rate effects on yield strength
Both Q0 and M0 ultimately depend on the value of a0l which is the tensile yield
strength of the material. However, this tensile yield strength is measured using
normal conventional methods, which do not need to take strain rate into
account. It is commonly known that high strain rate deformation can
significantly9 increase the yield stress from the statically determined value. In
order to simplify the problem solution, the author assumes that the value of a0
used has already been adjusted for this effect. If not, multiply a given cr0 by a
factor of three as a first order estimate.
2.3 Derive Velocity Equation
The following velocity profile will be used on the beam.
W
rniftjuuiwm
Figure 2.3 Uniform Velocity Profile
The velocity equation will be used to determine how much time the shock affects
the item we are attempting to demolish. Starting with the beam equation,
i? f
52M/3x2 = dQJdx = p + m W
the second derivative of the moment, M, with respect to x, is equal to the
inverse of the pressure force and the product of the mass of the beam times an
acceleration, w dot dot. The pressure force is always a function of time, if not
constant, i.e. p = p(t). m is always uniform and unchanging. W dot dot is
constant in this first example, but could also be a function of the distance x
along the beam. It is assumed that the functions of x and t for the beam are
separable, i.e. function M(x,t) = function Q(x,t) = function X(x)T(t).
Replace the second derivative of the moment with the first derivative of the
shear force, Q and integrate this equation with respect to distance
Q= px + mWx +Ci
(2.3)
9 Usually by a factor of 2 or 3.
9
to find an equation for Q. For a fully clamped beam (our bolt, at first attempt),
we have a shear value of Q0 at the left end and Q0 at the right end. This gives
an equation for w dot dot and allows a solution for the constant from the
integration.
The next step is to find an equation for pressure. The author has created a
curve fitted equation from Cooper10 to represent the pressure. This is inserted
into the equation for w dot dot, and integrated with respect to time.
The result is a velocity equation for the beam (w dot). When the velocity equals
zero, it means that the pressure and shear forces have slowed the beam down
it is the end of the accelerative effect of the beam. This time is important
because it is the maximum amount of time the explosive will have effect on the
material. Note that this time is not the same amount of time that the explosive
event lastedthe additional time is due to the acceleration of the beam.
Integrating again, an equation exists for the distance or deformation. This
distance can be interpreted as the distance that the explosive will cut through a
material. The shear equation is integrated again (spatially) to find the moment
in the beam. Finally, the results are compared to the square yield to verify that
the plasticity laws are satisfied.
2.4 Explosives vs. the circular pipe, Round I...
The following process should be able to be used in most types of problem with
suitable modification. The process is broken into eight steps that covers the
material above.
The pipe has a 0.55 meter inner diameter and a 0.57 meter outer diameter, and
was manufactured from cold rolled type 304 steel. According to Norton, this
steel has a tensile yield strength of 1103 MPa.
2.4.1 Determine Initial yield moment
The initial yield moment defines the minimal moment which first induces plastic
yield in the material. Since this is a beam in bending, the tensile stress is equal
to a=My/I; therefore M is equal to al/y. The appropriate stress is the yield stress
(oy); I is the moment of inertia, and y is the distance from the neutral axis.
Completing this calculation, the value determined for the yield moment is:
10 Cooper, chapter 28.
10
M
(2.4)
where Ro and Rf are the outer and inner radii, respectively. Given values for Ro,
Ri and c0 above, the initial yield moment is 21.36 MNm.
2.4.2 Determine Collapse moment
The collapse moment, M0, determines when the section has achieved the same
values of yield stress completely across the section. Perhaps the easiest method
to find it is to integrate the quantity a0 over half of the cross section, and
multiply the result by two. The resultant for a pipe is the quantity:
The resulting collapse moment for our pipe is 27.68 MNm.
2.4.3 Determine Collapse Load
The collapse load is the load that will create moments in the beam equivalent to
the collapse moment. This load is derived by application of the plasticity laws.
The collapse load for a fully clamped (both the pipe and the bolt can be
considered to be fully clamped) beam is
where M0 is defined above, and L is the half the length of the beam in question.
Note that this is not quite the same as shearing completely through two sides of
the beam. This load is
where Q0 is equal to cr0/2*A.
Tresca condition, the shear stress for plastic yielding is a0/2. Although this is a
failure mechanism that is of interest, it does not describe plastic collapse, which
is necessary for the development of the theorem.
Collapse moment:
3
(2.5)
P 0 :=
(2.6a)
(2.6b)
11
Thus, for our 10 meter section of the pipe and the 0.33 meter section of the
bolt, the lengths are 5 meter and 0.17 meter. The collapse load is equal to
36.78 MN/m, which is the load per unit length. This suggests for the pipe that
the maximum pipe pressure is approximately 72.07 MN/m2.
2.4.4 Determine Shape factor
The shape factor is the ratio between the collapse moment and the initial yield
moment. It is readily available for most cross sections, which prevents the
effort of having to derive it for an individual cross section. For the pipe cross
section, the shape factor is represented by the following formula:
Note that for a solid circular cross section, the shape factor simplifies to 16/371.
2.4.5 Integrate spatially and solve for shear force
Starting with the beam equation, the shear equation is found by integrating with
respect to x. The equation has one constant, Ci, which is found by noting that
there is no shear in the middle of the beam. Thus, the value of Q. is zero.
However, the requirement exists for the beam to shear at the ends (this is why
the explosive is being used in the first place!) Thus the value of the shear force
per unit length, Q0U, is applied to this formula. Now, by solving the shear
equation, an equation for w dot dot is determined.
In this equation, the pressure is only a function of time. The w dot dot equation
can be integrated to determine the shear of the bodythis will be covered in a
later section.
2.4.6 An asidethe explosive pressure representation
One of the more significant problems in formulating this theory was determining
a suitable value for the pressure. 11
11 Q0 is the force per unit length needed to shear the pipe. We want to shear an
area of 7i(Ro2 R2) on two ends, i.e. double shear. The shear stress according to
the Tresca condition is equal to half of the tensile yield stress.
Shape factor: (M0/Mj) Shape_Factor
(2.7)
(2.8)
m ml
12
All of the examples in the book by Jones reference a constant pressure. A mean
pressure could be determined, but this does not allow use of (more) realistic
explosive pressure data.
The equations from Cooper, chapter 28, create a method to compare any given
explosive to an equivalent amount of TNT12. Cooper provides data charts that
sketch the pressure decay at any given distance from the explosive fireball. The
author modeled this chart into a pressure vs. time relationship, assuming that a
parabolic decay exists for the pressure over the duration of the explosive event,
and then returns to atmospheric pressure exponentially, so that the pressure
has returned to normal in about 3 times the explosive event duration.
Mathematically:
a := P 0 t d
b :=
pa p0
1 d
(2.9 ac)
P0 is the initial maximum explosive pressure. td is the duration of the explosive,
which can be related to the explosive weight. Pa is the atmospheric pressure, or
in this case the pressure underwater at the location of the bolts. ta is the time of
arrival of the explosive blastthis allows the bolt explosives to propagate to the
pipe. Note that p is zero at times before ta. At t = 3 XJ2, the pressure has
reached its approximate negative minimum, and the pressure equation becomes
+ p a C2:=(PaP0)
(2.10 ab)
2
where t = 3td/5. This ensures that the pressure returns to normal after about
three times the value of t^
3t,
t t.+
p(t) := c y
12 TNT (trinitrotoluene) is probably the best researched explosive. It is the
standard to which other explosives are compared.
13
2.4.7 Integrate via time and solve for
velocity, time and distance
The two separate pressure functions introduce a large amount of algebra to be
conquered. Users of various mathematical integration and computational
algebra systems will benefit enormously from the capabilities of these machines.
The first effort is to integrate the w dot dot equation. When w dot reaches zero,
the beam stops moving. It is at that point that the effective time of the
explosive can be determined. Note that this is not the explosive duration (ta)
mentioned above. In effect, we are using the explosive to accelerate the beam
and shear it; when the acceleration stops, or even becomes negative (such as a
negative pressure) the beam can still move. We are concerned about a negative
velocity, because that is when the beam starts doing illogical things (such as
decreasing the shearing distance).
The author recommends integrating for velocity with final pressure function first,
because the quantity of interest is the time that the velocity of the beam reaches
zero. The distance equation uses this velocity to determine how much shear
occurs. It is more likely that the zero velocity value is in the second equation
because it does not have the explosive driving the beam. Note that the zero
velocity value could be in the previous equation, in which case the second
equation for pressure is unnecessary. Both equations must have the same
values for velocity at the switchover point (i.e. at t = ta + 3W2) for the pressure
function.
For the velocity equation, we have the following equations:
W 3 :=
~ C 2't 6 ^
3 t,
t t+
t P a'^ Q ot
where K :=
m
3 ^ a \
24a +
2 t d/
mL
+ K
+ 27td c2,tpjili+t
24 m
m
(2.11 ab)
We also have the equation for the velocity using the predecay pressure
relationship: , ,
1 b,')
 Q o't (2.12)
w
at +
l \ 3
2 =
m
ITT L
14
It might be necessary (and probably is the easiest method) to find the zeros by
plotting W dot against time. This tells the maximum time with the given amount
of explosives.
Once again, we can integrate these equations, the condition being that the value
of W2 is zero at time equals tg, and W3 matches W2 at time (ta + 3td/2). After
integration, we find that:
eL
3t,
t t+.
J +
p at
W
3 ,=
m
Qpt
2 mL
+ K1 + K 2 (213a)
where
and
C 2t2 +
3fcd
+ t
Ki
rn
+ K(U^ + ta k3
Q ot a2 a t2
2mL 2m1 j
(2.13 bc)
ar
W j :=
td
b (t t
~Y2~
m
Qo't
2 m L
+ K
(2.13d)
These equations determine how far the explosive shear progresses in the
material. In general, W3 will be the applicable equation, again because the
second pressure function will come into play.
15
2.4.8 Integrate spatially (again!)
and solve for moment
After the algebraic manipulations of the previous equations, this equation is
almost trivial. The next step is to integrate the shear equation spatially. This
time, the boundary condition on the end is the fully plastic moment. Thus, our
shear equation becomes, after integration:
M= M0v[l(x/L)2] M0
(2.14)
2.4.9 Check applicability with square yield
The last significant step is to check the equations for applicability. The whole
purpose behind the square yield surface is to ensure that the total value of the
shear force or the bending moment does not exceed the maximums predicted by
static theory. Thus, the way to do this is to solve the shear equation for the
quantity Q/Q0 (this paper refers to this as the shear ratio) and the moment
equation M/M0 (the moment ratio). Both ratios have to be less than or equal to
one for all time. The shear equation seems to make this (its maximum value is 
X/L) but the moment equation does not. The set of equations derived are valid
only for the beams with the quantity
(2.15) 13
13 A tacit assuption has been made at this point. The assumption is that the
bending moment on the end of the beam is equal to the fully plastic moment at
the end of the beam. For a relatively long beam, this is true. For a short beam,
this moment is significantly less than M0, i.e. it fails completely by shear. If the
moment at the end is equal to MSh0rt at the end, where Msh0rt < Myieid < M0, then
equation (2.14) becomes
M = M0v(l x2/L2) MShoit
and equation (2.15) becomes M/M0 = v(l x2/L2) Msh0rt/Mo, or
V < 1 + Mshort/M0.
Note that this becomes v s 1 for Msh0rt 0.
16
Fortunately, the bolt fits in this category, but the pipe does not.
2.4.10 Results
The entire effort in this chapter is focused to find an equation that determines
the cutting distance for the explosive, using the theorems of plasticity. Equation
(2.13a) was derived by applying various boundary conditions for shear slides
and bending moments. A pressure equation was stipulated and applied.
Because of this form of the pressure equation, continuity conditions at
changeover were required in the equations derived. After integration, the final
equation was determined. Applicability was checked by using the values of the
shear ratio (Q/Q0) and the moment ratio (M/M0).
The time that it takes to determine the explosive distance was not defined;
because of the exponential nature of the pressure and other factors, an explicit
analytical solution is not available. The best method to get the time is to graph
the velocity behavior over distance, and note the time when the velocity is equal
to zero.
17
3 Simulation Physics
3.1 Impulse definition, and why it is good
In physics, impulse is the term used to describe a change in momentum. Note
that it is not the rate of change of momentumthat quantity is clearly defined
as a force. Impulse is the change in momentum during an instantaneous
amount in time.14
In this paper, the author uses the impulse concept to accelerate the beam. This
acceleration is a direct result of the pressure on the beam. If all of the
momentum of the shock wave goes into the beam, then the new initial velocity
of the beam is equal to the momentum of the shock wave divided by the mass of
the beam. This is particularly important for the appendix example of the pipe,
because the pipe needs an initial velocity, and this is the method to determine it.
14 See Marion, Classical Dynamics of Particles and Systems.
18
4 Fun with Waves
4.1 Shockwave
The concept of a shock must be understood. A shock is a discontinuity that
propagates through a medium. In our case, the shock is a pressure
discontinuity. If a material exists with a disturbance traveling at under the
material speed of sound, then a shock does not form because the material has
time to react to the change. A shock results when the material does not have
that time to react, i.e. the disturbance moves faster than its speed of sound.
Most material properties (pressure, temperature, etc.) differ discontinuously
over the shock interface.
Shocks also have the property that they change over interfaces. If we have an
interface of two materials, the shock will change in some fashion at that
interface. We assume for the time being that the shock will propagate
indefinitely unless it hits an interface at some time, i.e. there are no decay
effects while the shock propagates through a material.15
4.2 Hugeniots
Now that a conclusive idea of the magnitudes of the pressure needed to shear
the bolt exists, how does the explosive generate the pressure? This is the
function of a Hugeniot.
Fortunately, the conservation equations still remain intact with the shock.
Energy is still conserved, as is momentum and mass. The equations that relate
properties before the shock to those after the shock are called the Rankine
Hugeniot equations, after the men who helped develop them. They are listed
below:
Mass
p i U u 0
p 0 u u 1
(4.1)
15 This is obviously a questionable assumptionif nothing else, the fluid viscosity
of the material will kill the shock wave's "kinetic" energy. However, there is
another effect, called a rarefaction wave, that eventually overtakes and "kills"
the shocksee below.
19
Momentum
p 1 p 0 P o(u 1 u
o) (u u o)
(4.2)
Energy
pl'ul_pO'uO 1/
(4.3)
Five quantities exist in these equationspressure p, density p, shock velocity U,
internal energy e, and particle velocity u. The values with zero subscripts are
behind the shock, and therefore the "one subscripts" are in front of (have not
been affected) by the shock. The particle velocity can be thought of as the
velocity of the medium.
The equations get interesting, however, when they are combined with another
equation, specifically known as the Hugeniot, which relates the shock velocity, U,
in the material to the particle velocity u in the material. With this additional
relationship, it becomes possible to actually solve some problems. Most
Hugeniots look something like this:
where C0 is something called the "bulk velocity of sound." Its significance is that
this is the minimum shock velocity. The value of "s" is chosen to meet
experimental data; it does not have any other physical significance.
Each explosive has a specific initial pressure immediately after the shock. This
pressure is called the ChapmanJouguet pressure, or Pg. The (created) shock
velocity due to complete explosion (detonation) is called the detonation velocity,
D.
U := c Q P o'u + S L2
(4.4)
20
Particle velocity, u
This is a graph of the Uu hugeniotit relates the shock velocity to the particle
velocity. It is used to determine the shock velocity. Water was used for the
sample.
The Pu diagram (above) is used along with other Pu diagrams to find pressures
and other values at intersections of materials.
4.3 Rarefaction wave
Eventually, of course, the shock must pass, and the material must return to
normal. This effect is called a rarefaction wave, and it expands the "shocked"
material, returning it to a normal state. Rarefaction waves also have the
fortunate16 behavior that they travel faster than the shock that created them.
For the purposes of this paper, the author assumes that once the rarefaction
wave has passed, any and all shock related events are over.17
16 This is why we assume that the shock propagates without friction. The
rarefaction wave will eventually catch and destroy the shock.
17 Cooper (p. 229) suggests the imagery of a spongethe shock wave squeezes
the sponge, and the rarefaction returns the sponge to normal. However, the
21
4.4 Reflection
A shock wave can reflect much like any other wave when it meets an interface
between two materials. In fact, just like optics, a certain amount of energy is
reflected, and a certain amount is transmitted into the new material. More
importantly for us, reflected shock waves will decrease the net strength of the
incoming shock wave, which is bad because it decreases the net effect of the
force on the beam. In this analysis, we have ignored the first reflection at the
beam itself, primarily for simplicity reasons (it does not hurt to consider the
explosive is on the other side).
Far more significant is the reflection propagating through the beam, reflecting,
at the edge of the beam and traversing through the beam again and exiting at
the edge of the beam itself. Such an effect is significant because two
intersecting wavefronts could either cancel or drastically increase the stress,
causing a fracture of the beam itself. The author has ignored this effect since
the pressure relation is modeled at the beginning of the beam over time, which
should account for this type of effect.
Neither effect is terribly important to us because we want to completely destroy
the bolt, not partially damage it.
sponge can still have a velocity afterward, as will our materialthis is not
analyzed in the paper.
22
5 Fun with Explosives
5.1 DefinitionsActivation Energy,
Detonation Velocity, and ChapmanJouguet
Pressures
An explosive is characterized by a reaction that produces energy at a rate faster
than the environment can absorb it. Our reaction might be:
The energy required can be supplied in many ways. Another shock wave from a
smaller explosive, a process that releases heat energy, or a spark might do.
Energyl is the activation energywithout it, the process will not start. Energy2
is the heat of reaction of the explosiveit is manifested in the shock wave.
Energy2 is always greater than energyl, usually much greater. Energy2 is
greater because the components C and D are chemically simpler, i.e. they do not
have energy tied up in more complex chemical bonds.
When the explosive explodes, it creates a shock wave that travels at a velocity D
through the explosive. This velocity is a characteristic of the explosiveit is
called the detonation velocity. Immediately after explosion and the propagation
of the shock wave through the explosive, a pressure P exists that is also a
particular characteristic of the explosive. This is the ChapmanJouguet pressure,
or Pcj. Based on data, Cooper18 determined the following relationship between
shock pressure and shock velocity, which is valid for most explosives:
This allows us to estimate the Pc, pressure. The shock speed could be estimated
by a flyer plate experiment.19
18 Cooper, pg. 265
19 The flyer plate experiment collides two blocks of the same material together to
measure the resultant shock velocity. This is a method to generate values for
the shock properties. This experiment works well with inert materials.
Obviously, this is unsatisfactory for an explosive materialif a shock is
generated, it is likely to complete detonation.
(5.2)
23
5.2 Explosive Equations
Equations are also needed to determine the value of P0. As a first estimate, the
value of Pg could be used. Based on work from Cooper20, two groups can be
determined. Cooper equates these groups to create an equation for the
pressure ratio of P0 to Pa.
Note that Ta is the ambient temperature in Kelvin, and Pa is the ambient
pressure in bars, where 1 bar = 1 X 105 N/M2. W is the equivalent amount of
TNT. Therefore, if one kg of TNT releases 10 J of energy, and 1 kg of an
experimental explosive releases 20 J of energy, then one kg of TNT is equivalent
to V2 kg of the experimental explosive.
Based on Cooper's data, this equation can be directly related to the value of
P0/Pa, which is the second ratio. Similar equations can be used to predict the
values for ta and td. An example, using data points from the charts 281 through
285 is in the appendix. Note that the temperature and pressure have to be
corrected for reference pressure, which Cooper gives as 288K and 1.01325 bar.
20 Cooper's work created these factors based on the Buckingham pi theorem,
and data from Explosive Shocks in Air, by G.F. Kinney and KJ. Graham. This
work was published by SpringerVerlag Publishing Company, in Berlin, Germany,
Z :=
R
Z is the scaled distance, and is one of the ratios.
(5.3)
1985.
24
6 Rates of Chemical Reaction,
and the Arrhenius Equation
6.1 Definitions
A chemical reaction rate determines how long it takes for a certain stoichiometric
amount of materials to react and form the desired product. Stoichiometric is the
chemical term for a perfect recipe, i.e. the correct amount of substances.
A chemical reaction rate is usually defined in terms of a constant, k, which is a
constant of proportionality, and the amounts of distinct substances involved,
usually raised to an exponential factor. So, a reaction might be described as
, . a b
rate :=kq j q 2
(6.1)
where qx is the amount of one reacting substance, and q2 is the amount of the
other reacting substance.
A reaction rate will determine how much explosive is needed to cause a pulse of
a certain duration.
6.2 How to get the Reaction Rate, or
Chemical Kinetics 101
The easy way to get the information is to find it in books or other informational
literature. The following information below will help determine how to generate
the information if it is not available.
Most chemical reactions can be "speeded up" within limits with increased
temperature. Explosive reactions are no different. In fact, at a certain
temperature well (hopefully!) known to its manufacturer, the explosive will
literally selfinitiate.
6.2.1 Determining the Order
The first step is to determine the "order" of the reaction with the reactants. If
we have a relation with two reactants, call them A and B, the reaction rate
equation looks like equation (6.1) above. Typically, the only two components
that need to be considered are the explosive and oxygendetonation of an
explosive is usually an oxidation reaction.
25
Now, the test is to find the exponents. This is done by varying the
concentrations of the substances involved and measuring the reaction time. For
instance, if a reaction proceeds at twice the rate when one species is doubled,
then the value of the quotient is one. If the reaction were to increase to four
times as fast, then the quotient would be two.21 Typically the concentrations are
not temperature related, i.e. a reaction that proceeds twice as fast at
temperature T1 with increased concentration will take place twice as fast at
temperature T2 with increased concentration.
6.2.2 Determining the Reaction Rate Constant k
The second step is to determine the reaction rate constant, "k," which includes
the temperature dependence of the explosive.
The Arrhenius equation defines the reaction rate constant "k" of a chemical
reaction. It was named after the Swedish chemist Svante Arrhenius, who
proposed the relationship in 1899. The reaction rate constant, "k," depends on
three thingsa "preexponential" factor, an activation energy, and the
temperature of the reaction.
The preexponential factor can be thought of as the probability that two
molecules are in a good condition to combine with each other. This is where the
shape of the molecules and reacting atoms might be considered.
The activation energy is an energy barrier that must be overcome for the
chemical reaction (the explosion) to occur. This energy barrier is usually
overcome by oldfashioned heating, but scientists (physicists) have used laser
energy or shock, for instance.
6.2.2.1 The Easy wayGroup additivity.
One way that the activation energy and preexponential factors can be
determined is with is termed "group additivity." Suppose that one could
determine that the explosion were related to breaking a Carbon=Carbon bond.
Group additivity builds on this concept by grouping chemicals into groups or
families that have similar reactions. This allows us to determine the heat of
reaction, (the heat energy released by the completion of a reaction) and the
entropy of reaction (the entropy released by the completion of the reaction), as
well as the value for Cp. These values can be determined at standard values and
temperatures (298 K and 1 atmosphere), and related to the values of activation
energy and preexponential constant with data from Benson.
21 In general, this is true, but the details get more difficult. At times, reactions
do not always proceed directly as written. The method is a valid first cut,
however.
26
6.2.2.2 Bludgeoning the Uninitiated
Statistical Thermodynamics.
The more elegant and common method to determine activation energy and pre
exponential factors is the statistical mechanics method. Quantum theory says
that an atom and molecule have a certain state in which they exist, with a
reasonable probability (not exactly!) that they will reside in that state. The
function that does this is called a wave function. For instance, a water molecule
might have a certain probability of having a certain amount of energy.
The states are compared to transition states, which correspond to the highest
energies that a molecule might have. On the other side are the products
themselves.
Using the methods of statistical mechanics, the activation energy (the difference
between the transition states and the original state) and the preexponential
factor (the size and shape of the molecule) can be calculated. The temperature
here has the fundamental property that it measures the energy of the molecules
involved in the collision, i.e. it is a direct measure of the internal energy of the
molecules.
27
7 Resolutions?
7.1 Problem Definition, again
The central problem of this thesis has been to determine how to destroy a bolt
without damaging a pipe stationed above it. We are now able to answer that
question.
In both cases, each item has a distance s that the shock wave must travel
through. Since both items are circular in cross section, we will characterize that
distance as the diameters. Internal decay due to a material has been
neglectedi.e. we have tacitly assumed that a shock, once created, will
propagate "forever" as long as it remains in a material. Thus, a shock wave
sufficient to shear the material will propagate merrily without decay throughout
the material, leaving destruction in its wake.
It is also known that the rarefaction wave travels faster than the shock. Once
the rarefaction wave intercepts the shock, the shock dies. Our goal is to get the
shock to the other side of the item before the rarefaction wave kills it.
The distances traveled by both effects must be equal. Mathematically, what we
have said is
(7.1)
U r t ^D't 2
Where Ur is the rarefaction velocity and D is the detonation velocity.
Numerically, if the detonation velocity is 2000 km/s and the rarefaction velocity
is 2500 km/s, the time for the detonation to cover 1 meter is 1 microsecond
(10'6 seconds); the time for the rarefaction wave to do it is 0.4 microseconds. If
the detonation gets a 0.6 microsecond head start, both will cover a meter at the
same time. This ensures the explosive effects do not propagate past the
distance desired.
The hypothesis of this paper is that is that to destroy something, a bolt, a pipe,
whatever, with a characteristic length of s, the explosive duration must be:
(7.2)
Ur
This allows the explosive to operate as a "ideal pressure generation device" for
enough time before the effects of a rarefaction wave become apparent. The
28
amount of time necessary can be related to the chemical reaction rate, R. Thus,
the final amount of explosives needed can be expressed as:
where R is the reaction rate, D is the detonation velocity, Ur is the rarefaction
velocity, and s is the distance that is being cut through.
7.2 Problem results
So, to reinforce the method, the first effort is to postulate that a beam exists.
Measure its profile and determine the plastic moment M0. Use the plasticity laws
to determine the collapse load and pressures. Use the beam equations with
appropriate boundary conditions to determine a distance profile for the
explosive. Verify that the proposed equations do not violate the shear ratio or
moment ratio laws. Determine the time of explosion (most likely graphically)
and the distance of explosion. Determine the ChapmanJouguet pressure of
various explosives and compare the P0 needed. Propagate this pressure through
several materials if necessary using various hugeniot relations to get the
pressure at the material in question. Determine amounts of explosive using the
time of detonation and the time of arrival, as well as the velocities of the
rarefaction wave (from the hugeniot equations) and the detonation velocity.
(7.3)
29
8 OK, but...the obvious flaws and
weaknesses of the theory
The most significant weakness with the theory is the author's inability to test the
theory, due to various federal and state regulations. The equations and
concepts are sound, but without experimental data, it is difficult to determine if
the theory works. In light of the fact that persons with the proper licenses and
permits to test the theory probably also have access to more accurate theories
and data, it is unlikely that this will be resolved soon. Nonetheless, it would be
interesting to see.
To refine the theory, a decay effect needs to be generated for the shockthis is
the most glaring defect with the theory. If a simple exponential decay rate was
predicted, for instance, then the pressure at the end of the distance s would be
the pressure predicted by this theory. The pressure to generate would obviously
be
P
(8.1)
where lambda is the decay factor, and s is the distance propagated through the
material. Further investigation using the shock reflections (the shock will reflect
off the edge of the beam, then back again on the front of the beam) which
causes a net velocity. This would probably require another method and also
significantly increase the amount of algebra involved.
Drag effects have been neglected. There is no form of resistance that indicated
when the beam will stop due to the viscosity of the fluid that the beam travels
through. This would be more significant in determining how far a complete
fragment might travel, however, i.e. after separation, which was the desired
effect.
Shear deformation effects have been neglected. A short beam in bending has a
shear deformation. The author is reasonably confident that the inclusion would
not significantly improve the calculation. One way to include it is to change the
value of Q0 to include the shear stress generated by a collapse pressure. Thus, a
difference of pressures would take the place of the single shear pressure.
Strain hardening effects have been neglected. The easiest method to include
them, after experimentation, would be to adjust the ao by some constant after
30
adjustment for strain rate. The author suspects that such an adjustment might
only have a minor effect.
Shape effects have been neglected. The theory assumes that the explosive
propagates equally in all directions. This is not true, particularly in shaped
charges. The pressures and detonation velocity could differ spatially22. A more
realistic prediction must include this type of effect.
The material to be destroyed has been characterized with a single parameter, s.
Ideally, one would study a three dimensional representation of the item. This is
where the stress wave theory would be very important.23
The buckling load for the pipe was not determined. Although it seems
reasonable to show that the rarefaction wave intersects the shock wave and its
effects long before it reaches the pipe, it does not conclusively prove that the
explosive will not damage the pipe. Future studies should take note of this.
Size effects also have not been covered. The future of explosives is not in big
thingsit is in little things. An explosive that can destroy something small, say
the size of a pebble, but not significantly damage the pebble a centimeter away,
will become crucial. Already medicine is using the effects of interfering shock
waves to destroy such things as kidney stonesa well placed "explosive" in the
center of a malignant tumor would be sufficient to destroy it, but also might
force the body to "heal" the area without cancerous effects.
8.1 Accuracy of the Chemical reaction rate
The reaction rate relationship tends to assume that the reaction occurs as the
stoichiometric equation is written. For instance, if we have a reaction
2H2 + 02 > 2H20
(8.2)
one might assume that the reaction functions strictly on H2 and 02 atoms. The
reverse reaction has been shown to depend on the amount of H+ in the system.
There are side reactions that make up the reactive chains. Similar reactions
might also affect the forward reactions that create the explosive detonation.
22 For instance, in a shaped charge, an explosive "jet" is formed by the collapse
of the liner. This jet propagates more or less in a straight direction, without
significant expansion.
23 However, it also introduces effects not covered in this thesis. Interactions
between the shocks and the rarefactions would have to be modeled and
individually managed. Including this effect (and 22, above) would force the
problem into that of computer simulation.
31
Thermal effects of the shock have been neglected. For instance, in a typical
shock condition, there is a significant area on the Pv diagram between the jump
condition (the straight line) and the Hugeniot. This energy has to go
somewherethe answer is that it is usually converted to heat. A massive
temperature rise usually takes place after a shock, and can be significant enough
to actually melt, or even vaporize, the material in question. If thermal effects
can be added to the theory such that temperature profiles and pressure profiles
are precisely predicted, the theory might be able to quantitatively explain how
explosive welding works.
32
9 Summary
The author analyses the effects of an explosion on a bolt. The pressure required
to shear a bolt is derived. The pressure required to shear a section of pipe is
derived. Finally, an amount of explosive is derived.
The paper has discussed the two principal effects of the explosive on the bolt,
namely the shear slides and bending hinges. The pressure to shear a bolt is
modeled in equations 2.9 and 2.10, and used to determine distances in
equations 2.13. The appendix lists the distances in equation 10.13 for the pipe.
The amount of explosive is derived in equation 7.3.
Both the reaction rates and the shell failure criteria need further research. The
reaction rates will require significant amounts of effort and are probably not
available outside of government research. This is a flaw of the proposed
method. Also, the buckling failure mode of the pipe was not studied. It is quite
probable that the pipe would not fail as a beam, but as an individual shell. This
was not modeled or addressed in the paper.
The analysis is very dependent on the pressure profile described in equations
(2.9) and (2.10). These equations drove the possibilities of solution. A different
profile will change the solution significantly. Although the profile was derived
from experimental data (Cooper), it may or may not reflect the performance of
an actual explosive. Effects of the yield stress o0 due to strain rate were
estimated. A more accurate method might include determining yield stress as a
function of the velocity w, which is ultimately a function of the strain rate.
Effects due to drag and fluid viscosity were not included. More accuracy, at the
cost of computational modeling, could be gained by explicitly including pressure
and viscosity effects.
Jones uses a similar method for modeling dynamic plasticity effects on these
structural elements, as well as modeling dropped masses (vs. pressure
impulses). The method could be extended to other items, such as plates and
shells, and has potential for improving the design of explosive items.
Appendix
A. Clohessy's Circular Beam
Upon reflection, it is difficult to imagine a uniform stress along the beam. One
way that this could be done is to stage a line charge at some distance away from
the beam's surface. Another method that could be used is to use an annulus
with a small slot on top.
The explosive is centered in the plane of the material, By doing so, a uniform
pressure is generated over the material, and since the explosive would be
generated radially, no untoward bending effects due to curvature should occur.
This would be a sound method to verify the theory.
34
B. Pipeline Design
Assume maximum flow velocity of 2.25 m/s, which is about 5 mi/hr
vel :=2.25
Flow area is equal to 81% of wetted area using Chezy formula:
chezy,
The flow rate is therefore A*vel, or A flow'vel = 1862
This equates to approximately 25,000 gallons per minute. To put this in
perspective, a small swimming pool (301 X 60') is approximately 10,000 cubic
feet, which is about 78,000 gallons. Our flow could fill this swimming pool in
about 3 minutes!
Now, our goal is to determine the external radius of the pipe. Obviously, within
some factor of safety, the pipe has to hold a certain pressure to keep the flow
moving at the rate mentioned above. If we assume that the pump station is
approximately 12 km away, and have a rough idea of the grade and friction loss,
we can determine the input pressure. This input pressure will be used, along
with the worst conditions (underwater at 70 ft) to determine the size of the pipe
wall. This gives us sound conditions to know how to clamp the pipe also.
We will find the Reynolds's number to characterize the flow. Then, we will solve
for the pressure drops back to the other station, which will give the worst
pressure that the pipe will have to accept. Then, this pressure can determine
the thickness of the pipe.
The Reynolds number is equal to the equivalent diameter*the mass flow rate per
unit area, divided by the absolute, or dynamic viscosity.
The equivalent diameter, De, is equal to four times the hydraulic radius.
The hydraulic radius is equal to the area of flow divided by the wetted perimeter:
For SAE 20W oil at zero degrees Fahrenheit (well below freezing), the viscosity
in centistokes is 2590 cs. The oil in question is slightly less viscous at 0
degrees, but it is in this range.
** wet 2r chezy ^ wet
A flow
p
r wet
35
Conversions:
mics:=2590 rhosmall:=
640
1000
mu_ = 1.65^10^
HP
mucp r^ small'nu cs l1 ,mucp 001
p = 1.658 poy:=640
G is the mass flow rate per unit area, or rho velocity. G:=P oil'vel G= 1
.44* 103
_De'G A
So the Reynolds number is: Re: Re= 116>10^ which suggests laminar
flow.
Using the Darcy friction formula, f := f= 055 This is the friction factor we
Xv6
will use.
g:=9.87 (gravity)
The next step is to back solve for the pressure losses. Pzero is the pressure
leaving the Alpha station. PI is the pressure before entering the water. P2 is
the pressure at the 15 m north shore. P3 is the pressure at the 20 m south
shore. P4 is the pressure exiting the water on the south shore. P5 is the
pressure at Bravo station. The assumption is that there are no shock losses due
to sudden changes in the pipe diameter.
4f'L45 vel2 5
L45l=2000 p4:='P oil P4= 5.32610
De 2
4'f'L 34 vel2
L34:=5000 P3 :=P4+ 20p oilgH P oii p3= 1.9^10
De 2
L23 vel2
L23:=4000 P2:=P35P0il'g + 4f,_T"poil P2 = 302^10
De 2
L 12 Vel2
L12:=500 p j :=p215p oilg + 4fp oij p j = 3.06?10
De 2
4*f*L ni Vp1^ f.
L01 := 1000 p 0 :=p 1oil p 0 = 3.329*10
De 2
36
Stresses in the Pipeline.
Using the equations of elasticity and neglecting the weight of the oil, the
following equations can be derived.
rout
.57
m
,55
P atm'
1.013105
P water ,P atm
orlrpipe
+ 209.81000 rpipe :=rin''^"*rm" rout
2 _ 2 \
1 P0' 1 rout P water' x2 out
2 2
1 rpipe \ rpiPe
K'in)3'329,16
00 f'pipe)
Ap 1
* _ 2 \
rout
P0' 1+ 2
\ rpipe
~ P water'
v :=
.28
cz=
rout
2
rpipe /.
0 P water'
37
107
So, the three principal stresses are 18 MPa for the hoop stress in tension, 4.1
MPa for the axial stress, and 3.3 MPa in compression radially. Because these
are principal stresses, the yield condition (using VonMises law) is
Y= 7.82>107
The yield strength for this particular steel in tension is 1103 MPa, which, even
allowing for a 20% factor of safety, is significantly less (78 MPa vs. 910 MPa)
than the yield strength. The axial stress must be relieved with bearings and
other pipe attachments; this is not covered in this paper.
38
C. Evaluating ta and td from Cooper's charts
T 10
LThis is the length of the beam
Interpolation Functions
These functions are interpolated from charts in Explosives Engineering, by Paul
W. Cooper. The interpolations represent figures 28.2, 28.3, and 28.5. The
charts are based on data from Explosive Shocks in Air, by G.F. Kinney and K.J.
Graham.
zx:=
MO
r 2
410'2
MO'1
1.51CT1
410"1
1
1.75
10
10.25
zy:=
4102
M02 310"2 1 1.8510"2
6.7510"2 110'1 61 O'2
2.5101
7.71 O'2 8.410"2 MO1
MO1
1 tdx := 910"2 tdy := MO"1 tax:= 2.310"1
1.7101 110'1 2.21 O'1 910"1
110'1 2.710'1 1 1
1.2510'2 1 10 3 4 7 10
MO'2
110'2
MO'1
2.510'1
1
10
14
100
200
39
These functions are used to find the peak pressure (P0), time of arrival (ta) and
time of detonation (td). For instance, following the examples in the book for a
charge of Cyclotol 10m from the target,
T_ := 288 R:=10 Tn:=297 W:=1.4220 P:=0.834 Z:=
U a a j
/WTa\3
Z = 0.462
P0 = 0.762
P0 :=linterp(zx,zy)Z) Pa
This suggests that we are in the ballpark, which is sufficient for our needs.
Scaled time of arrival
t :=linterp(tax,tay,Z) t = 4.122
*a :=ta'
W
^p \ 3 I\ 6
ta= 12.141
\ o/
tds :scspline(tdx,tdy)
t d := interp(tds, tdx, tdy, Z)
:=t,
W
/p \3 /x \'
t d = 3.479
/
o /
40
D. Explosives vs. the Circular Pipe, Round II
Unlike the original effort, a variable velocity distribution exists in x for the
proposed solution of the pipe. Thus, instead of
IttTUlHiUmtl
Figure 2>3, Repeated
we have
The minor difference of the velocity distribution increases the math significantly,
therefore only the significant steps will be shown. Note that is the location of
the original plastic hinge.
Starting with the bending moment equation (2.1), we perform the first spatial
integration. This gives us the following equation:
Q := p'X+ m W j_x (10.1)
and, since there is no shear at x=0, Q1 is zero for the entire section to the left of
This makes equal to
W x
Â£
m
(10.2)
41
Now, we solve for the section beyond Â£0 Note that both shear and moment
must be continuous with the previous section.
First, we have to introduce the velocity profile in this region. The velocity profile
is:
tf
w:=Ws+ WrWs
(L x)
?0
(10.3)
which means that the acceleration is simply:
w:=Ws+ WrW v v 1
L$0
(10.3)
Solving again for shear, we get the following equations:
& I
0
W 1 W g
Q2 := "P'x *" rrWgx + ^Lx + +
Cl
(10.4a)
Cl := p0'0 nr
i * i
Ws5 Q +
v t i
W 1 w
______Â£. l^Â£'
L5 0 \ 2/
(10.4b)
This is the shear in the section to the right of with a matching condition of
Qs=0 at Â§o.
i
This gives us the following equation for Ws:
* C
Ws:=
P(L \ 0 ~ 2Q
m(L \ 0
(10.5)
which is the acceleration of interest. However, conditions still need to be applied
for the moments on the beam. After integration, we find that:
42
+ C2
(10.6a)
ptmWs , ,e . , 6 2 x3
M :=.I \ 0 x 35 0 Lx + 3Â£, 0 x
2'(LU) \ 2
where
C2 :=
2'M0'(L5 o) + (P + mW sj(4 oJ'Lf O3
(10.6b)
2 L0
These are the moment equations.1 The astute reader will note that Ws is defined
as:
_P(L!o)2'Q0
m(L 5 o)
from the shear equation and
(10.5)
2Mo(LS0)
7 L3 24' \ o L2 + 27S 02L 10? Q3
(10.7)
from the moment equation. Irregardless of which definition is chosen, equations
10.5 and 10.7 are equal. Equating the two equations, the following equation is
generated for So:
10v
10v
10v
(10.8)
Finally, the shear ratio and the moment ratio need to be checked:
Q (* 5 o)'[3'(L~ 4 o) *] (10.9)
5o
which after simplification requires that So > L/2, and
1 But see footnote 13 in the main document.
43
(10.10)
which again, after simplification, requires Â£0 > 9L/10. The most stringent of the
two equations is the second equation, thus we need a hinge to occur somewhere
after > 9L/10.
The next effort is to find values for the shear. Replacing Q0 with v, equation
10.5 becomes:
Next, this equation is integrated, assuming an initial condition of WsX at time t
equals zero:
and finally integrated again, assuming an initial distance of zero at time t equals
zero.
This is the equivalent of equations 2.13 for the pipe. Note that an effort must be
made to solve for the time that makes equation 10.12 equal to zero. This t
determines the distance that the shear will progress. Finally, a method must be
available to determine the ratio of Wx to Wsi. The value of Wx can be
determined (assumed) to be equal to ux, where ux is the initial particle velocity
after a shock of speed D propagates into the pipe material.
(10.11)
(10.12)
(10.13)
44
Bibliography
1. Benham, P.P., R. 3. Crawford, and C.G. Armstrong. Mechanics of Engineering
Materials. 2nd Edition. Essex, England: Longman Group Limited, 1996.
2. Blake, Alexander. Design of Mechanical Joints. Copyright 1985, Marcel
Dekker, Inc. Published New York, NY: Marcel Dekker, 1985.
3. Benson, Sidney W. Thermochemical Kinetics. 2nd Edition. Copyright 1976,
John Wiley and Sons. Published New York, NY: Wiley Interscience, 1976.
4. Cooper, Paul W. Explosives Engineering. New York, NY: VCH Publishers,
1996.
5. Clohessy, William(?) Private conversation, July 1999.
6. Drumheller, D.S. Introduction to Wave Propagation in Nonlinear Fluids and
Solids. Copyright 1998, D.S. Drumheller. Published Cambridge, United
Kingdom: Cambridge University Press, 1998.
7. Farlow, Stanley J. Partial Differential Eouations for Scientists and Engineers.
Copyright 1993, Stanley J. Farlow. Published New York, NY: Dover Publications,
Inc., 1993.
8. Feynman, Richard P., Leighton, Robert B., Sands, Matthew. The Fevnman
Lectures on Physics. Volume 2Mainly Electromagnetism and Matter. Copyright
1964, California Institute of Technology. Published Reading, Massachusetts:
AddisonWesley Publishing Company, 1977.
9. Jeppersen, Neil. Chemistry. Copyright 1997, Barron's Educational Series,
Inc. Hauppauge, New York: Barron's Educational Series, 1977.
10. Jones, Norman. Structural Impact. Cambridge, United Kingdom:
Cambridge University Press, 1997.
11. Mader, Charles L. Numerical Modeling of Explosives and Propellants. 2^
Edition. Copyright 1998, CRC Press LLC. Boca Raton, Florida: CRC Press LLC,
1998.
12. Marion, Jerry B., Thorton, Stephen T. Classical Dynamics in Particles and
Systems. Copyright 1988, 1970, Harcourt Brace Jovanovich, Inc. Orlando,
Florida: Harcourt Brace Jovanovich, 1988.
45
13. Masterton, William L., and Emil J. Slowinski. Chemical Principles. 4a
Edition. Copyright 1977, W.B. Saunders Company. Published Philadelphia,
Pennsylvania: W.B. Saunders Co, 1977.
14. Menkes, S. B. and H. J. Opat. Broken Beams. "Experimental Mechanics."
volume 13.
15. Neal, B.G. The Plastic Methods of Structural Analysis (Third S.I. Edition!.
Copyright 1977, B.G. Neal. Published London, England: Chapman and Hall
Limited, 1977.
16. Norton, Robert L. Machine Design: An Integrated Approach. Copyright
1996, Prentice Hall, Inc. Upper Saddle River, NJ: Prentice Hall, 1996.
17. Ragab, AbdelRahman A. F., and Salah Eldin A. Bayoumi. Engineering Solid
Mechanics: Fundamentals and Applications. Boca Raton, Florida: CRC Press
LLC, 1999.
18. Ugural, A.C., and S.K. Fenster. Advanced Strength and Applied Elasticity:
The SI version. Copyright 1981, Elsevier Science Publishing Company. NY, NY:
SpringerVerlag New York, 1998.
19. Zukas, Jonas. Impact Dynamics.
20. Zukas, Jonas A. (ed.) and William P. Walters. Explosive Effects and
Applications. Copyright 1998, SpringerVerlag New York, Inc.
46

PAGE 1
AN ANALYTICAL METHOD OF MODEUNG DEMOLITIONS by Antoine Derrick Johnson B.S., United States Air Force Academy, 1991 A thesis submitted to the University of Colorado at Denver in partial fulfillment of the requirements for the degree of Master of Science Mechanical Engineering 1999
PAGE 2
by Antoine Derrick Johnson All rights reserved.
PAGE 3
This thesis for the Master of Science degree by Antoine Derrick Johnson has been approved by W1lham H. Clohessy Date
PAGE 4
Johnson, Antoine Derrick (M.S., Mechanical Engineering) An Analytical Method of Modeling Demolitions Thesis directed by Professor James C. Gerdeen ABSTRACT The author analyses the effects of an explosion on a bolt. The pressure required to shear a bolt and the pressure required to shear a section of pipe are derived. Finally, an amount of explosive is determined. This abstract accurately represents the content of the candidate's thesis. I recommend its publication. Signed iv
PAGE 5
DEDICATION I dedicate this first to the men and women of the ath Maintenance Squadron Propulsion Flight; they did the job well, and deserved more than they got. I also dedicate this to my mother, for her support. Finally, I'd like to dedicate this to my friend Don. He was supposed to see me graduate. Through his life, I learned how to live. "Stand back. Be silent. Be still. That's it. And look upon this moment. Savor it. Rejoice with great gladness. Great gladness. Remember it always, for you are joined by it. You are one, under the stars. Remember.... For it is the doom of men that they forget." Merlin, in "Excalibur." Copyright MCMLXXXI, Orion Pictures Company.
PAGE 6
ACKNOWLEDGMENTS I would like to thank the following people: My sisters, Kimberly and Heather, and my father, who had to deal with the worst of me for these three years. Ms. Loretta Duran, for her assistance over the last two years. I could not (literally!) have graduated without her. Ms. Annette Dye, who provided impressive illustrations and provided significant technical input. Mrs. Kathleen Van Orsdel, who read this document and provided editorial comments. Drs. Zalabak and ShockleyZalabak for their inspiration. Colonel (Dr.) James Head, Lieutenant Colonel Anthony Heyward, and Major Paul Bowman, for getting me into this fine institution. Mr. Fred Saar, for listening patiently to the inspired narration of this document. My supervisors, Kevin and Hal, and all of the other people at AlliedSignal.
PAGE 7
CONTENTS Figures ................................................................................................... ix Chapter 1 Problem Definition ................................................................................ 1 2 Fun with Plasticity ................................................................................ 4 2.1 Introduce theorems ................................... ..................................... 5 2.1.1 2.1.2 2.1.3 2.1.4 2.2 2.2.1 2.2.2 2.2.3 2.3 2.4 2.4.1 2.4.2 2.4.3 2.4.4 2.4.5 2.4.6 2.4.7 2.4.8 2.4.9 2.4.10 Generalized Stress and Strain ..................................................... 5 Lower Bound, or Equilibrium Theory ............................................ 5 Upper Bound, or Kinematic Theory .............................................. 5 Dynamic Collapse ..................................................................... 6 Beam theory .................................................................................. 6 Equations ................................................................................. 6 Square yield concept ................................................................. 7 An AsideStrain Rate effects on yield strength ............................. 9 Derive Velocity Equation .................................................................. 9 Explosives vs. the circular pipe, Round I .......................................... 10 Determine Initial yield moment.. ............................................... 10 Determine Collapse moment .................................................... 11 Determine Collapse Load ......................................................... 11 Determine Shape factor ........................................................... 12 Integrate spatially and solve for shear force ............................... 12 An asidethe explosive pressure representation ......................... 12 Integrate via time and solve for velocity, time and distance ........................................................... 14 Integrate spatially (again!) and solve for moment ................................................................... 16 Check applicability with square yield ......................................... 16 Results .................................................................................. 17 3 Simulation Physics ............................................................................. 18 3.1 Impulse definition, and why it is good ............................................. 18 vii
PAGE 8
4 Fun with Waves ................................................................................. 19 4.1 Shock wave ................................................................................. 19 4.2 Hugeniots .................................................................................... 19 4.3 Rarefaction wave .......................................................................... 21 4.4 Reflection ............................................................................. ....... 22 5 Fun with Explosives ............................................................................ 23 5.1 DefinitionsActivation Energy, Detonation Velocity, and ChapmanJouguet Pressures ....................................... 23 5.2 Explosive Equations ...................................................................... 24 6 Rates of Chemical Reaction, and the Arrhenius Equation ............................................................................ 25 6.1 Definitions ................................................................................... 25 6.2 How to get the Reaction Rate, or Chemical Kinetics 101 .................................................................. 25 6.2.1 Determining the Order ............................................................. 25 6.2.2 Determining the Reaction Rate Constant k ................................. 26 7 Resolutions? ...................................................................................... 28 7.1 Problem Definition, again .............................................................. 28 7.2 Problem results ............................................................................ 29 8 OK, but. .. the obvious flaws and weaknesses of the theory .................................................................... 30 8.1 Accuracy of the Chemical reaction rate ............................................ 31 9 Summary .......................................................................................... 33 Appendix A. Clohessy's Circular Beam .................................................................... 34 B. Pipeline Design .................................................................................. 35 C. Evaluating ta and td from Cooper's charts .............................................. 39 D. Explosives vs. the Circular Pipe, Round II .............................................. 41 Bibliography ............................................................................................ 48 viii
PAGE 9
FIGURES Figure 1.1 Pipeline Layout .................................................................................. 2 1.2Side and Top Views ............................................................................ 2 2.1 The bolt. Distance between bolt head and line is 0.3 meters .................. 7 2.2 Square Yield Surface .......................................................................... 8 2.3 Uniform Velocity Profile ...................................................................... 9 D.1 Varying Velocity Profile .................................................................... 41 ix
PAGE 10
1 Problem Definition The problem is inspired by Alistair Mclean's "Navarone" movie series. In the two movies, a squad of commandos inevitably destroys a target of strategic importance, altering the course (or so it would seem) of World War II. In "The Guns of Navarone," the team destroys a monolithic gun emplacement that is preventing the landing of the British fleet. The guns are placed on an unassailable cliff, heavily surrounded by the enemy. The Allies cannot destroy the guns the old fashioned way (aerial bombardment), so a team sneaks behind enemy lines to dislodge the guns. The guns collapse under their weight into the seaan innovative approach. The fleet then can pass beyond its Charybdis, and presumably the eastern half of the Mediterranean is freed. In "Force 10 from Navarone," the team (Force 10) sets out to destroy a bridge that separates a German division from Greece and the remainder of the Balkans. The survivors from the original (Major Mallory and Staff Sergeant Miller) are attached to Force 10 to kill a double agent. Neither half trusts the other. Ultimately, the separate halves accomplishes the other's mission. The team succeeds again in an innovative fashionby weakening a small dam.1 The water pressure is sufficient to weaken the dam, which releases the river behind it. The sudden onrush of the river is enough to topple the bridge supports, and the bridge falls down. The problem presented is hoped to be of similar interest. Imagine a pipeline in the cold arctic north. The allies have lost a major beachhead in the north, and the enemy will be there all winter long if something cannot be done to force them out. The pipeline is well guarded, and protected against the obvious aerial assault to destroy it. Obviously the pumping stations would be a target, but each one has massive numbers (enough to prevent the forward assault, anyway) of hostile enemy troops ready to protect it. The entire pipeline is protected, in fact, except a three or four mile stretch across an ocean inlet that we will call "Ocean Bay." Here the pipeline lies submerged 50 to 70 feet in ice covered waters. The direct naval strike is not an option, because the enemy has a large naval fleet there. However, a small team of "brave men with explosives" might be able to sneak inland, follow the river to the sea, and destroy it. Our intelligence 1 "Oh! I can do a tremendous job on the dam. Yes! See, with the dam, you have your natural elements working for you. It's like having an enormous bath. All you have to do is pull the plug!" Staff Sergeant Miller, in "Force 10 from Navarone", Copyright 1974, Orion Pictures Corporation. 1
PAGE 11
sources have assured us that standard enemy operating procedure calls for flow shutoff during air attack. At this point, since the pipe flow devices would not notice any pressure loss, the destruction could be made. The resuming oil flow, Figure 1.1Pipeline Layout combined with a steady 3 or 4 mile current and a mysteriously damaged pipe restraint, might cause the pipe to break underwater in an unreachable environmenta total loss significant enough to encourage the enemy to spend winter elsewhere. This thesis will look at the engineering aspects of such a problem. Figure 1.2Side and Top Views The layout is sketched above. Assuming the inlet in the chosen section is 100 feet deep, the pipe crosses this section in 70 feet of water. For civil engineering reasons, there is a 30 degree turn in the chosen 100 meter section of pipe, which is conveniently in the path of the current. The pipe is clamped at the end of these sections. The pipe is made of type 304 stainless steel, with a 0.55 meter inner radius and a 0.57 meter outer radius. 2 The yield strength for this 2 See Appendix: Pipeline Design. 2
PAGE 12
steel is 1103 MPa.3 The post is bolted by eight bolts arranged radially around the post. This design is similar to a light pole. The bolts are installed into a large concrete weight which helps steady the pipe, and allows construction with minimal surface preparation. The idea is to determine how much explosive is needed to destroy the bolts without significantly damaging the pipe or pole above. If the pipe is buckled or damaged, the resulting leak would alert the personnel at the pumping station, so it is something that the team must avoid. The 50 em diameter steel bolt is also made of type 304 stainless steel. Destroying the bolts will unbalance the column sufficiently (conveniently) for the oil flow to topple the structure. The complete destruction of this pipe, along with a related mission to destroy other oilrelated assets, would force the enemy to abandon the wintry coast. 3 Data from Machine Design: An Integrated Approach, table C8, pg. 997. Obviously, an underwater pipe constructed of steel would have corrosion problems; steel was chosen based on availability of explosive related data. 3
PAGE 13
2 Fun with Plasticity Plasticity covers materials after they have reached the elastic limit. Most mechanical engineering efforts do not focus on the plastic region of a material primarily because permanent deformation, let alone yielding, Is a bad thing. Metal forming and designing crash structures are two significant exceptions. The problem involves plasticity to determine the deformations and collapse loads to destroy the bolt or damage the pipe. The following analogue is offered to clarify the problem and the solution proposed here. Imagine a small boy on a cold Saturday morning, with a little red wagon and lots of snow for a snowball fight. He has to pull the wagon across the street to his friends house, so they will be supplied with plenty of snowballs for the morning's festivities. The force that the little boy can pull the wagon with can be called F. The wagon has a mass m, so the acceleration of the wagon is a= F/m Now, the road that the boy has to cross is clear. Therefore, while there is a resisting frictional force, but it is not significant compared to the boy's strength. Since this is a simple example, the resisting frictional force will be ignored. Now, suppose that the boy were to give the wagon a push of a certain duration. The wagon would accelerate to some maximum velocity and finally stop due to inertia and friction. At that point, the wagon should have crossed the street, which has a certain width of w. (That was the little boy's goal, anyway). The distance that the wagon covers is a function of the acceleration that the boy gives, integrated twice to find velocity and distance. Initial conditions usually Imply that the velocity is zero, and the distance is zero, but these can change. The boy gets the wagon across the street to his friend, and the snowball fights that day are marvelous. Now, the comparisons can be made. The "wagon" is an engineering structure known as a beam. The bolts that keep the pipe support upright will be modeled as a beam. The "distance across the street" is the width or depth of the beam that the explosive needs to cut through. Instead of the little boy supplying a force, the explosives supply a pressure distribution for a certain time, so that sufficient force is generated to shear completely through the beam and therefore cause it to fall. Frictional effects (the resistance of pressure on the other face of the beam) are neglected, just as the friction involved in dragging the wagon across the road was neglected. Finally, the Inertia of the beam is included. Initial conditions of velocity (is the beam moving prior to start?) and distance (almost always zero) will be assumed. 4
PAGE 14
2.1 Introduce theorems Plasticity covers many topics. This section will define generalized stress and generalized strain, topics necessary to understand the two collapse theorems used in plasticity. These are the Lower and Upper Bound theories, because they predict the lower and upper estimates for collapse. Finally, the paper will introduce some basic concepts of dynamic plasticity theory. 2.1.1 Generalized Stress and Strain Generalized stresses and strains are not what they might seem. If one starts with the concept of energy, then the rate of consumption of energy is termed the power. The power is integrated over time to determine the energy consumed. Returning to elementary Strength of Materials theory, we have the concept of strain energy, which is the time integral of the product of stress and strain. The concept of a generalized stress and strain fit here. Instead of using the stress and strain, a product that creates a "power" term is definedfor instance, a pair consisting of moment and angular velocity, or a force and velocity pair. The portion that deals with the effort is termed the generalized stress, and the portion that deals with the motion is termed the generalized strain. 2.1.2 Lower Bound, or Equilibrium Theory The Lower Bound theorem is best described as a theorem of moments. Assume a bar exists with a specified load. If a moment can be placed such that the moment and the force are in equilibrium, without the moment exceeding the collapse moment previously predicted anywhere in the structure, then the load will not cause collapse. This theorem is probably most useful for determining the collapse load; with various uniform cross sections and something called a shape factor4 it is relatively easy to determine if the load will collapse the structural element. 2.1.3 Upper Bound, or Kinematic Theory The Upper Bound theorem equates the rates of energy generation (motion) to energy dissipation (response). A motion is proposed that describes the collapse 4 The shape factor for a beam is the ratio of the fully plastic (collapse) moment to the initial yield moment. It characterizes the shape. The initial yield moment in the beam is the moment which causes the tensile stress at the top of the beam to equal the yield stress. The collapse moment is that moment which causes the yield stress to exist throughout the beam. 5
PAGE 15
mechanism, along with the forces involved, and the energy required (Fe) represents the energy generation part. The energy dissipation portion is represented by the strain energy involved. If the energy generated exceeds the energy dissipated, the motion is possible. 2.1.4 Dynamic Collapse Assuming that the above equations are correct, why would anyone want to introduce dynamic effects? Well, the load is a dynamic one. A good example of a static load is resting a large metallic object (for instance, a ship) on the pipe. Assuming it is gradually applied, no significant dynamic effects will occur. On the other hand, the times involved in the application of the explosive loads are significantly less than 30 milliseconds. Well before one can even begin to say "Mississippi," the explosive event is over. A good rule of thumb is that if the load is applied for a time less than or equal to the period of a structure's lowest fundamental frequency, then dynamic effects should be considered. The equations in this paper assume that is the case. Essentially, what the static approach neglects, and what is added by dynamic plasticity, is the acceleration of the beam, plate, or other affected structure. The result is that motion of the beam alone can cause collapsethis effect can reduce the force required. Also, a shear effect, called a shear slide, can occur, where the shear is discontinuous across a section. This is the effect that the problem seeks. For the problem, what will be needed is an explosive pressure to shear the bolt, and an idea of the pressure needed to shear the pipe. These are dynamic effects which will require the dynamic approaches. In this paper, the author will derive calculations for the bolt, and reserve calculations for the pipe for the appendix. 2.2 Beam theory 2.2.1 Equations A beam is an object with one axis dimension significantly greater that the other two, that is loaded perpendicularly to the long axis. The first example of the beam that we are exposed to is probably the "teetertotter", a beam with two (relatively) equal loads on its ends and pinned in the middle to pivot. Other examples of beams are railroad tracks, beams in houses and construction, and various machine elements that carry weight as beams, for instance a shaft in a turbine. Jones has defined the equation of motion of a beam as: 6
PAGE 16
fiM!ax2 = aQJax = p + m w (2.1) where M is the bending moment in the beam, Q is the shear force acting in the beam, x describes the length of the beam, p is the pressure on the beam, m is the mass of the beam5 and W (W dot dot) is the second derivative of the deflection with respect to time of the beam. After integration (twicesee section 2.3, Derive Velocity Equation), equation 2.1 describes the beam's deflection. As mentioned above, we will model the bolts as beams. Distance= 0.3 meters Figure 2.1The bolt. Distance between bolt head and line is 0.3 meters. 2.2.2 Square yield concept Of interest is the method of determining when the material has yielded. The reader should be familiar with the concept of a yield stress, which is the stress beyond which permanent deformation takes place. It is used as a factor of merit in design because in most cases, permanent deformation due to applied loads is not a good thing. A normal6 material has a yield stress, and then hardens, so that further on, another (ultimate) stress defines when the material ruptures. A perfectly plastic material is an engineering fictionevery material will eventually strain harden before it fails.7 A perfectly plastic material reaches a yield stress and from then on behaves sort of like silly puttyit does not get any stronger or weaker, it just deforms. The "square yield" is so termed because it includes the yield due to shear and to plastic collapse in a squarelike so (see below). 5 The mass must be independent of x, i.e. constant cross section, for this equation to be valid. 6 The author has ignored the "brittle" materials. Traditionally, if a material has less than 2% deformation at rupture, it is brittle and therefore is not really a candidate for yield stress calculations. Brittle materials have yield stressesthey are just far more likely to fail because of large normal stresses due to crack propagation and other effects well before a yield theory could be applied. 7 In deference to my thesis advisor, there are actually materials that have nearly constant strain for a given stress over a certain temperature range. 7
PAGE 17
Q/Qo llo ... .. M/Mo Figure 2.2Square Yield Surface Note that Q0 is the shear force per unit length causing yield and Mo is the moment causing plastic collapse. The yield surface follows Drucker's postulate that a yield surface cannot be concave. A concave yield surface would ultimately mean that energy is gained through deformation, i.e. the material is self deforming. It is convenient to represent the surfaces as Q/Q0 and M/Mo, where both quantities are functions divided by the material' length. A material cannot exceed the Q/Q0 or M/Mo boundary, because at that point it has yielded or collapsed. This particular characterist:ic is important for determining whether or not the motion proceeds as assumed. Significant among this square yield8 concept is the following ratio, v, where v=QoL/2Mo (2.2) Lis the length of the beam. This gi'Ves a method to determine which effect (shearing or plastic collapse) is stronger. A "v" value less than 1 indicates that the beam is more likely to shear rather than collapse (bend). Similarly, a value of v greater than one indicates the beam is more likely to collapse (bend) than yield due to shear. 8 Note that the square yield is an upper limit approxienation where failure is assumed by shear alone or moment alone. Yielding by a combination of shear and moment is representee! by an elliptlcal yield surface. 8
PAGE 18
2.2.3 An AsideStrain Rate effects on yield strength Both Qo and Mo ultimately depend on the value of cr0 which is the tensile yield strength of the material. However, this tensile yield strength is measured using normal conventional methods, which do not need to take strain rate into account. It is commonly known that high strain rate deformation can significantly9 increase the yield stress from the statically determined value. In order to simplify the problem solution, the author assumes that the value of cr0 used has already been adjusted for this effect. If not, multiply a given cr0 by a factor of three as a first order estimate. 2.3 Derive Velocity Equation The following veloc ity profile will be used on the beam. Figure 2.3 Uniform Velocity Profile The velocity equation will be used to determine how much time the shock affects the item we are attempting to demolish. Starting with the beam equation, f &M/ox2 = aQ!ax = p + m w the second derivative of the moment, M, with respect to x, is equal to the inverse of the pressure force and the product of the mass of the beam times an acceleration, w dot dot. The pressure force is always a function of time, if not constant, i.e. p = p(t). m is always uniform and unchanging. W dot dot is constant in this first example, but could also be a function of the distance x along the beam. It is assumed that the functions of x and t for the beam are separable, i.e. function M(x,t) = function Q(x,t) = function X(x)T(t). Replace the second derivative of the moment with the first derivative of the shear force, Q and integrate this equation with respect to distance uI Q= px + mWx +C1 9 Usually by a factor of 2 or 3. 9 (2.3)
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to find an equation for Q. For a fully clamped beam (our bolt, at first attempt), we have a shear value of Qo at the left end and Qo at the right end. This gives an equation for w dot dot and allows a solution for the constant from the integration. The next step is to find an equation for pressure. The author has created a curve fitted equation from Cooper10 to represent the pressure. This is inserted into the equation for w dot dot, and integrated with respect to time. The result is a velocity equation for the beam (w dot). When the velocity equals zero, it means that the pressure and shear forces have slowed the beam downit is the end of the accelerative effect of the beam. This time is important because it is the maximum amount of time the explosive will have effect on the material. Note that this time is not the same amount of time that the explosive event lastedthe additional time is due to the acceleration of the beam. Integrating again, an equation exists for the distance or deformation. This distance can be interpreted as the distance that the explosive will cut through a material. The shear equation is integrated again (spatially) to find the moment in the beam. Finally, the results are compared to the square yield to verify that the plasticity laws are satisfied. 2.4 Explosives vs. the circular pipe, Round I ... The following process should be able to be used in most types of problem with suitable modification. The process is broken into eight steps that covers the material above. The pipe has a 0.55 meter inner diameter and a 0.57 meter outer diameter, and was manufactured from cold rolled type 304 steel. According to Norton, this steel has a tensile yield strength of 1103 MPa. 2.4.1 Determine Initial yield moment The initial yield moment defines the minimal moment which first induces plastic yield in the material. Since this is a beam in bending, the tensile stress is equal to cr=My/I; therefore M is equal to cri/y. The appropriate stress is the yield stress (cry); I is the moment of inertia, andy is the distance from the neutral axis. Completing this calculation, the value determined for the yield moment is: 1 Cooper, chapter 28. 10
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7t cr (R 4 R .4) M. = 0 0 I 1 4R 0 (2.4) where Ro and Ri are the outer and inner radii, respectively. Given values for Ro, Ri and cr0 above, the initial yield moment is 21.36 MNm. 2.4.2 Determine Collapse moment The collapse moment, M0 determines when the section has achieved the same values of yield stress completely across the section. Perhaps the easiest method to find it is to integrate the quantity cr0 over half of the cross section, and multiply the result by two The resultant for a pipe is the quantity: Collapse moment: 4cr (R 3 R 3 ) M 0 0 I 0 ,3 (2.5) The resulting collapse moment for our pipe is 27.68 MNm. 2.4.3 Determine Collapse Load The collapse load is the load that will create moments in the beam equivalent to the collapse moment. This load is derived by application of the plasticity laws. The collapse load for a fully clamped (both the pipe and the bolt can be considered to be fully clamped) beam is 4M o Po:=L2 (2.6a) where Mo is defined above, and Lis the half the length of the beam in question. Note that this is not quite the same as shearing completely through two sides of the beam. This load is Q 0 Po:= L (2.6b) where Q0 is equal to cr0/2*A. cr0 is the yield strength in tension. Using the Tresca condition, the shear stress for plastic yielding is cro/2. Although this is a failure mechanism that is of interest, it does not describe plastic collapse, which is necessary for the development of the theorem. 11
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Thus, for our 10 meter section of the pipe and the 0.33 meter section of the bolt, the lengths are 5 meter and 0.17 meter. The collapse load is equal to 36.78 MN/m, which is the load per unit length. This suggests for the pipe that the maximum pipe pressure is approximately 72.07 MN/m2 2.4.4 Determine Shape factor The shape factor is the ratio between the collapse moment and the initial yield moment. It is readily available for most cross sections, which prevents the effort of having to derive it for an individual cross section. For the pipe cross section, the shape factor is represented by the following formula: Shape_Factor 16 R ( R 3 R 3 ) 0 0 I .3. ( R o 4R i4) (2.7) Note that for a solid circular cross section, the shape factor simplifies to 16/37t. 2.4.5 Integrate spatially and solve for shear force Starting with the beam equation, the shear equation is found by integrating with respect to x. The equation has one constant, C1 which is found by noting that there is no shear in the middle of the beam. Thus, the value of C1 is zero. However, the requirement exists for the beam to shear at the ends (this is why the explosive is being used in the first place!) Thus the value of the shear force per unit length, Q0 11 is applied to this formula. Now, by solving the shear equation, an equation for w dot dot is determined ..... p Q 0 w.m ml (2.8) In this equation, the pressure is only a function of time. The w dot dot equation can be integrated to determine the shear of the bodythis will be covered in a later section. 2.4.6 An asidethe explosive pressure representation One of the more significant problems in formulating this theory was determining a suitable value for the pressure. 11 Q0 is the force per unit length needed to shear the pipe. We want to shear an area of 1t(Ra2 R?) on two ends, i.e. double shear. The shear stress according to the Tresca condition is equal to half of the tensile yield stress. 12
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All of the examples in the book by Jones reference a constant pressure. A mean pressure could be determined, but this does not allow use of (more) realistic explosive pressure data. The equations from Cooper, chapter 28, create a method to compare any given explosive to an equivalent amount of TNT12 Cooper provides data charts that sketch the pressure decay at any given distance from the explosive fireball. The author modeled this chart into a pressure vs. time relationship, assuming that a parabolic decay exists for the pressure over the duration of the explosive event, and then returns to atmospheric pressure exponentially, so that the pressure has returned to normal in about 3 times the explosive event duration. Mathematically: a + b ( t t a )2 p(t) := t d (2.9 ac) Po is the initial maximum explosive pressure. td is the duration of the explosive, which can be related to the explosive weight. Pa is the atmospheric pressure, or in this case the pressure underwater at the location of the bolts. ta is the time of arrival of the explosive blastthis allows the bolt explosives to propagate to the pipe. Note that p is zero at times before ta. At t = 3 'crJ/2, the pressure has reached its approximate negative minimum, and the pressure equation becomes p(t) [ ( H d ) l tta+2:= C 2" e + P a 3t d 2 5 C 2 := ;( ( P a P o ) (2.10 ab) where "t = 3trj5. This ensures that the pressure returns to normal after about three times the value 12 TNT (trinitrotoluene) is probably the best researched explosive. It is the standard to which other explosives are compared. 13
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2.4.7 Integrate via time and solve for velocity, time and distance The two separate pressure functions introduce a large amount of algebra to be conquered. Users of various mathematical integration and computational algebra systems will benefit enormously from the capabilities of these machines. The first effort is to integrate thew dot dot equation. When w dot reaches zero, the beam stops moving. It is at that point that the effective time of the explosive can be determined. Note that this is not the explosive duration mentioned above. In effect, we are using the explosive to accelerate the beam and shear it; when the acceleration stops, or even becomes negative (such as a negative pressure) the beam can still move. We are concerned about a negative velocity, because that is when the beam starts doing illogical things (such as decreasing the shearing distance). The author recommends integrating for velocity with final pressure function first, because the quantity of interest is the time that the velocity of the beam reaches zero. The distance equation uses this velocity to determine how much shear occurs. It is more likely that the zero velocity value is in the second equation because it does not have the explosive driving the beam. Note that the zero velocity value could be in the previous equation, in which case the second equation for pressure is unnecessary. Both equations must have the same values for velocity at the switchover point (i.e. at t = ta + 3td2) for the pressure function. For the velocity equation, we have the following equations: where 24 a + _t _a ) + 27 t d 2 t d K:=24m (3 t d ) C 2"t + P a 2+ t a m (2.11 ab) We also have the equation for the velocity using the predecay pressure relationship: I (at+ _b(_t_3_t_a)_') w 2 := td m (2.12) 14
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It might be necessary (and probably is the easiest method) to find the zeros by plotting W dot against time. This tells the maximum time with the given amount of explosives. Once again, we can integrate these equations, the condition being that the value of W2 is zero at time equals ta, and W3 matches W 2 at time (ta + 3td2). After integration, we find that: where and Q ot2 + Kt + K 2 (2.13a) 2 ml (2.13 be) [ a t2 b(ttat] + Q t2 1 2 12 o W 2 := + K3 t d m 2ml (2.13d) These equations determine how far the explosive shear progresses in the material. In general, W3 will be the applicable equation, again because the second pressure function will come into play. 15
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2.4.8 Integrate spatially (again!) and solve for moment After the algebraic manipulations of the previous equations, this equation is almost trivial. The next step is to integrate the shear equation spatially. This time, the boundary condition on the end is the fully plastic moment. Thus, our shear equation becomes, after integration: 13 (2.14) 2.4. 9 Check applicability with square yield The last significant step is to check the equations for applicability. The whole purpose behind the square yield surface is to ensure that the total value of the shear force or the bending moment does not exceed the maximums predicted by static theory. Thus, the way to do this is to solve the shear equation for the quantity Q/Q0 (this paper refers to this as the shear ratio) and the moment equation M/Mo (the moment ratio). Both ratios have to be less than or equal to one for all time. The shear equation seems to make this (its maximum value is X/L) but the moment equation does not. The set of equations derived are valid only for the beams with the quantity or v s: 2. (2.15) 13 A tacit assuption has been made at this point. The assumption is that the bending moment on the end of the beam is equal to the fully plastic moment at the end of the beam. For a relatively long beam, this is true. For a short beam, this moment Is significantly less than M0 i.e. it fails completely by shear. If the moment at the end is equal to Mshort at the end, where Mshort < Myleld < M0 then equation (2.14) becomes M = Mov(1 x2/L2 ) Mshort and equation (2.15) becomes M/Mo = v(l x2/L2 ) Mshort/M0 or Note that this becomes v s: 1 for Mshort 0. 16
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Fortunately, the bolt fits In this category, but the pipe does not. 2.4.10 Results The entire effort in this chapter is focused to find an equation that determines the cutting distance for the explosive, using the theorems of plasticity. Equation (2.13a) was derived by applying various boundary conditions for shear slides and bending moments. A pressure equation was stipulated and applied. Because of this form ofthe pressure equation, continuity conditions at changeover were required in the equations derived. After integration, the final equation was determined. Applicability was checked by using the values of the shear ratio (Q/Qo) and the moment ratio {M/Mo) The time that it takes to determine the explosive distance was not defined; because of the exponential nature of the pressure and other factors, an explicit analytical solution is not available. The best method to get the time is to graph the velocity behavior over distance, and note the time when the velocity is equal to zero. 17
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3 Simulation Physics 3.1 Impulse definition, and why it is good In physics, impulse is the term used to describe a change in momentum. Note that it is not the rate of change of momentumthat quantity is clearly defined as a force. Impulse is the change in momentum during an instantaneous amount in time.14 In this paper, the author uses the impulse concept to accelerate the beam. This acceleration is a direct result of the pressure on the beam. If all of the momentum of the shock wave goes into the beam, then the new initial velocity of the beam is equal to the momentum of the shock wave divided by the mass of the beam. This is particularly important for the appendix example of the pipe, because the pipe needs an initial velocity, and this is the method to determine it. 14 See Marion, Classical Dynamics of Particles and Systems. 18
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4 Fun with Waves 4.1 Shock wave The concept of a shock must be understood. A shock is a discontinuity that propagates through a medium. In our case, the shock is a pressure discontinuity. If a material exists with a disturbance traveling at under the material speed of sound, then a shock does not form because the material has time to react to the change. A shock results when the material does not have that time to react, i.e. the disturbance moves faster than its speed of sound. Most material properties (pressure, temperature, etc.) differ discontinuously over the shock interface. Shocks also have the property that they change over interfaces. If we have an interface of two materials, the shock will change in some fashion at that interface. We assume for the time being that the shock will propagate indefinitely unless it hits an interface at some time, i.e. there are no decay effects while the shock propagates through a material. 15 4.2 Hugeniots Now that a conclusive idea of the magnitudes of the pressure needed to shear the bolt exists, how does the explosive generate the pressure? This is the function of a Hugeniot. Fortunately, the conservation equations still remain intact with the shock. Energy is still conserved, as is momentum and mass. The equations that relate properties before the shock to those after the shock are called the Rankine Hugeniot equations, after the men who helped develop them. They are listed below: Mass P1 Uu0 PQ uu1 (4.1) 15 This is obviously a questionable assumptionif nothing else, the fluid viscosity of the material will kill the shock wave's "kinetic" energy. However, there is another effect, called a rarefaction wave, that eventually overtakes and "kills" the shocksee below. 19
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Momentum P 1Po := P o ( u 1u o ) (uu o ) (4.2) Energy P 1 u 1 Po u o 1 e 1eo : = ( u 12u o2) P 0(uu 0 ) 2 (4.3) Five quantities exist in these equationspressure p, density p, shock velocity u, internal energy e, and particle velocity u. The values with zero subscripts are behind the shock, and therefore the "one subscripts" are in front of (have not been affected) by the shock. The particle velocity can be thought of as the velocity of the medium. The equations get interesting, however, when they are combined with another equation, specifically known as the Hugeniot, which relates the shock velocity, u, in the material to the particle velocity u in the material. With this additional relationship, it becomes possible to actually solve some problems. Most Hugeniots look something like this: 2 u := c o p o u + s u (4.4) where Co is something called the "bulk velocity of sound." lts significance is that this is the minimum shock velocity. The value of "s" is chosen to meet experimental data; It does not have any other physical significance. Each explosive has a specific initial pressure immediately after the shock. This pressure is called the ChapmanJouguet pressure, or Pcj. The (created) shock velocity due to complete explosion (detonation) is called the detonation velocity, D. 20
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500 1000 1500 2000 2500 3000 3500 4000 Particle velocity, u This is a graph of the Uu hugeniotit relates the shock velocity to the particle velocity. It is used to determine the shock velocity. Water was used for the sample. 0 500 1000 1500 2000 u 2500 3000 3500 4000 The Pu diagram (above) is used along with other Pu diagrams to find pressures and other values at intersections of materials. 4.3 Rarefaction wave Eventually, of course, the shock must pass, and the material must return to normal. This effect is called a rarefaction wave, and it expands the "shocked" material, returning it to a normal state. Rarefaction waves also have the fortunate16 behavior that they travel faster than the shock that created them. For the purposes of this paper, the author assumes that once the rarefaction wave has passed, any and all shock related events are averY 16 This is why we assume that the shock propagates without friction. The rarefaction wave will eventually catch and destroy the shock. 17 Cooper (p. 229) suggests the imagery of a spongethe shock wave squeezes the sponge, and the rarefaction returns the sponge to normal. However, the 21
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4.4 Reflection A shock wave can reflect much like any other wave when it meets an interface between two materials. In fact, just like optics, a certain amount of energy is reflected, and a certain amount is transmitted into the new material. More importantly for us, reflected shock waves will decrease the net strength of the incoming shock wave, which is bad because it decreases the net effect of the force on the beam. In this analysis, we have ignored the first reflection at the beam itself, primarily for simplicity reasons (it does not hurt to consider the explosive is on the other side). Far more significant is the reflection propagating through the beam, reflecting, at the edge of the beam and traversing through the beam again and exiting at the edge of the beam itself. Such an effect is significant because two intersecting wavefronts could either cancel or drastically increase the stress, causing a fracture of the beam Itself. The author has ignored this effect since the pressure relation is modeled at the beginning of the beam over time, which should account for this type of effect. Neither effect is terribly important to us because we want to completely destroy the bolt, not partially damage it. sponge can still have a velocity afterward, as will our materialthis is not analyzed in the paper. 22
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5 Fun with Explosives 5.1 DefinitionsActivation Energy, Detonation Velocity, and ChapmanJouguet Pressures An explosive is characterized by a reaction that produces energy at a rate faster than the environment can absorb it. Our reaction might be: A + B +energy! 7 C + D + energy2 (5.1) The energy required can be supplied in many ways. Another shock wave from a smaller explosive, a process that releases heat energy, or a spark might do. Energyl is the activation energywithout it, the process will not start. Energy2 is the heat of reaction of the explosiveit is manifested in the shock wave. Energy2 is always greater than energyl, usually much greater. Energy2 is greater because the components C and D are chemically simpler, i.e. they do not have energy tied up in more complex chemical bonds. When the explosive explodes, it creates a shock wave that travels at a velocity D through the explosive This velocity is a characteristic of the explosiveit is called the detonation velocity. Immediately after explosion and the propagation of the shock wave through the explosive, a pressure P exists that is also a particular characteristic of the explosive. This is the ChapmanJouguet pressure, or Pcj. Based on data, Cooper18 determined the following relationship between shock pressure and shock velocity, which is valid for most explosives: (5.2) This allows us to estimate the Pcj pressure. The shock speed could be estimated by a flyer plate experiment. 19 18 Cooper, pg. 265 19 The flyer plate experiment collides two blocks of the same material together to measure the resultant shock velocity. This is a method to generate values for the shock properties. This experiment works well with inert materials. Obviously, this is unsatisfactory for an explosive materialif a shock is generated, it is likely to complete detonation. 23
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5.2 Explosive Equations Equations are also needed to determine the value of Po. As a first estimate, the value of Pcj could be used. Based on work from Cooper0 two groups can be determined. Cooper equates these groups to create an equation for the pressure ratio of Po to Pa. R Z := 1 Z is the scaled distance, and is one of the ratios. (5.3) Note that Ta is the ambient temperature in Kelvin, and Pais the ambient pressure in bars, where 1 bar= 1 X 105 N/M2 W is the equivalent amount of TNT. Therefore, if one kg of TNT releases 10 J of energy, and 1 kg of an experimental explosive releases 20J of energy, then one kg of TNT is equivalent to V2 kg of the experimental explosive. Based on Cooper's data, this equation can be directly related to the value of P0/Pa, which is the second ratio. Similar equations can be used to predict the values for ta and td. An example, using data points from the charts 281 through 285 is in the appendix. Note that the temperature and pressure have to be corrected for reference pressure, which Cooper gives as 288K and 1.01325 bar. 2 Cooper's work created these factors based on the Buckingham pi theorem, and data from Explosive Shocks in Air, by G.F. Kinney and K.J. Graham. This work was published by SpringerVerlag Publishing Company, in Berlin, Germany, 1985. 24
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6 Rates of Chemical Reaction, and the Arrhenius Equation 6.1 Definitions A chemical reaction rate determines how long It takes for a certain stoichiometric amount of materials to react and form the desired product. Stoichiometric is the chemical term for a perfect recipe, i.e. the correct amount of substances. A chemical reaction rate is usually defined in terms of a constant, k, which is a constant of proportionality, and the amounts of distinct substances involved, usually raised to an exponential factor. So, a reaction might be described as t k a b rae .q I q 2 (6.1) where ql is the amount of one reacting substance, and q2 Is the amount of the other reacting substance. A reaction rate will determine how much explosive is needed to cause a pulse of a certain duration. 6.2 How to get the Reaction Rate, or Chemical Kinetics 101 The easy way to get the information Is to find It in books or other informational literature. The following information below will help determine how to generate the information if it is not available. Most chemical reactions can be "speeded up" within limits with Increased temperature. Explosive reactions are no different. In fact, at a certain temperature well (hopefully!) known to its manufacturer, the explosive will literally selfinitiate. 6.2.1 Determining the Order The first step is to determine the "order'' of the reaction with the reactants. If we have a relation with two reactants, call them A and B, the reaction rate equation looks like equation (6.1) above. Typically, the only two components that need to be considered are the explosive and oxygendetonation of an explosive is usually an oxidation reaction. 25
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Now, the test is to find the exponents. This is done by varying the concentrations of the substances involved and measuring the reaction time. For instance, If a reaction proceeds at twice the rate when one species is doubled, then the value of the quotient is one. If the reaction were to increase to four times as fast, then the quotient would be two. 21 Typically the concentrations are not temperature related, i.e. a reaction that proceeds twice as fast at temperature Tl with Increased concentration will take place twice as fast at temperature T2 with increased concentration. 6.2.2 Determining the Reaction Rate Constant k The second step is to determine the reaction rate constant, "k," which includes the temperature dependence of the explosive. The Arrhenius equation defines the reaction rate constant "k" of a chemical reaction. It was named after the Swedish chemist Svante Arrhenius, who proposed the relationship in 1899. The reaction rate constant, "k," depends on three thingsa "preexponential" factor, an activation energy, and the temperature of the reaction. The preexponential factor can be thought of as the probability that two molecules are in a good condition to combine with each other. This is where the shape of the molecules and reacting atoms might be considered. The activation energy is an energy barrier that must be overcome for the chemical reaction (the explosion) to occur. This energy barrier is usually overcome by oldfashioned heating, but scientists (physicists) have used laser energy or shock, for instance. 6.2.2.1 The Easy wayGroup additivity. One way that the activation energy and preexponential factors can be determined is with is termed "group additivity." Suppose that one could determine that the explosion were related to breaking a Carbon=Carbon bond. Group additivity builds on this concept by grouping chemicals into groups or families that have similar reactions. This allows us to determine the heat of reaction, (the heat energy released by the completion of a reaction) and the entropy of reaction (the entropy released by the completion of the reaction), as well as the value for Cp. These values can be determined at standard values and temperatures (298 K and 1 atmosphere), and related to the values of activation energy and preexponential constant with data from Benson 21 In general, this is true, but the details get more difficult. At times, reactions do not always proceed directly as written. The method is a valid first cut, however. 26
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6.2.2.2 Bludgeoning the UninitiatedStatistical Thermodynamics. The more elegant and common method to determine activation energy and pre exponential factors Is the statistical mechanics method. Quantum theory says that an atom and molecule have a certain state in which they exist, with a reasonable probability (not exactly!) that they will reside in that state. The function that does this is called a wave function. For instance, a water molecule might have a certain probability of having a certain amount of energy. The states are compared to transition states, which correspond to the highest energies that a molecule might have. On the other side are the products themselves. Using the methods of statistical mechanics, the activation energy (the difference between the transition states and the original state) and the preexponential factor (the size and shape of the molecule) can be calculated. The temperature here has the fundamental property that it measures the energy of the molecules involved in the collision, i.e. it is a direct measure of the internal energy of the molecules. 27
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7 Resolutions? 7.1 Problem Definition, again The central problem of this thesis has been to determine how to destroy a bolt without damaging a pipe stationed above it. We are now able to answer that question. In both cases, each item has a distance s that the shock wave must travel through. Since both items are circular in cross section, we will characterize that distance as the diameters. Internal decay due to a material has been neglectedi.e. we have tacitly assumed that a shock, once created, will propagate "forever" as long as it remains in a material. Thus, a shock wave sufficient to shear the material will propagate merrily without decay throughout the material, leaving destruction in its wake. It Is also known that the rarefaction wave travels faster than the shock. Once the rarefaction wave intercepts the shock, the shock dies. Our goal is to get the shock to the other side of the item before the rarefaction wave kills it. The distances traveled by both effects must be equal. Mathematically, what we have said is (7.1) Where Ur is the rarefaction velocity and D is the detonation velocity. Numerically, if the detonation velocity is 2000 km/s and the rarefaction velocity is 2500 km/s, the time for the detonation to cover 1 meter is 1 microsecond (10"6 seconds); the time for the rarefaction wave to do it is 0.4 microseconds. If the detonation gets a 0.6 microsecond head start, both will cover a meter at the same time. This ensures the explosive effects do not propagate past the distance desired. The hypothesis of this paper Is that Is that to destroy something, a bolt, a pipe, whatever, with a characteristic length of s, the explosive duration must be: s s At=D U r (7.2) This allows the explosive to operate as a "ideal pressure generation device" for enough time before the effects of a rarefaction wave become apparent. The 28
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amount of time necessary can be related to the chemical reaction rate, R. Thus, the final amount of explosives needed can be expressed as: Arrount := R [ t d + t a + s ( ul r + ] (7.3) where R is the reaction rate, D is the detonation velocity, Ur is the rarefaction velocity, and sis the distance that is being cut through. 7.2 Problem results So, to reinforce the method, the first effort is to postulate that a beam exists. Measure its profile and determine the plastic moment M0 Use the plasticity laws to determine the collapse load and pressures. Use the beam equations with appropriate boundary conditions to determine a distance profile for the explosive. Verify that the proposed equations do not violate the shear ratio or moment ratio laws. Determine the time of explosion (most likely graphically) and the distance of explosion. Determine the ChapmanJouguet pressure of various explosives and compare the Po needed. Propagate this pressure through several materials if necessary using various hugeniot relations to get the pressure at the material in question. Determine amounts of explosive using the time of detonation and the time of arrival, as well as the velocities of the rarefaction wave (from the hugeniot equations) and the detonation velocity. 29
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8 OK, but ... the obvious flaws and weaknesses of the theory The most significant weakness with the theory is the author's inability to test the theory, due to various federal and state regulations. The equations and concepts are sound, but without experimental data, it is difficult to determine if the theory works. In light of the fact that persons with the proper licenses and permits to test the theory probably also have access to more accurate theories and data, it is unlikely that this will be resolved soon. Nonetheless, it would be interesting to see. To refine the theory, a decay effect needs to be generated for the shockthis is the most glaring defect with the theory. If a simple exponential decay rate was predicted, for instance, then the pressure at the end of the distance s would be the pressure predicted by this theory. The pressure to generate would obviously be p (8.1) where lambda is the decay factor, and s is the distance propagated through the material. Further investigation using the shock reflections (the shock will reflect off the edge of the beam, then back again on the front of the beam) which causes a net velocity. This would probably require another method and also significantly increase the amount of algebra involved. Drag effects have been neglected. There is no form of resistance that indicated when the beam wilt stop due to the viscosity of the fluid that the beam travels through. This would be more significant in determining how far a complete fragment might travel, however, i.e. after separation, which was the desired effect. Shear deformation effects have been neglected. A short beam in bending has a shear deformation. The author is reasonably confident that the inclusion would not significantly Improve the calculation. One way to include it is to change the value of Qo to include the shear stress generated by a collapse pressure. Thus, a difference of pressures would take the place of the single shear pressure. Strain hardening effects have been neglected. The easiest method to include them, after experimentation, would be to adjust the cro by some constant after 30
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adjustment for strain rate. The author suspects that such an adjustment might only have a minor effect. Shape effects have been neglected. The theory assumes that the explosive propagates equally in all directions. This is not true, particularly in shaped charges. The pressures and detonation velocity could differ spatially22 A more realistic prediction must include this type of effect. The material to be destroyed has been characterized with a single parameter, s. Ideally, one would study a three dimensional representation of the item. This is where the stress wave theory would be very important. 23 The buckling load for the pipe was not determined. Although it seems reasonable to show that the rarefaction wave intersects the shock wave and its effects long before it reaches the pipe, it does not conclusively prove that the explosive will not damage the pipe. Future studies should take note of this. Size effects also have not been covered. The future of explosives is not in big thingsIt is in little things. An explosive that can destroy something small, say the size of a pebble, but not significantly damage the pebble a centimeter away, will become crucial. Already medicine is using the effects of interfering shock waves to destroy such things as kidney stonesa well placed "explosive" in the center of a malignant tumor would be sufficient to destroy it, but also might force the body to "heal" the area without cancerous effects. 8.1 Accuracy of the Chemical reaction rate The reaction rate relationship tends to assume that the reaction occurs as the stoichiometric equation is written. For instance, if we have a reaction (8.2) one might assume that the reaction functions strictly on H2 and 02 atoms. The reverse reaction has been shown to depend on the amount of H+ in the system. There are side reactions that make up the reactive chains. Similar reactions might also affect the forward reactions that create the explosive detonation. 22 For instance, in a shaped charge, an explosive "jet" Is formed by the collapse of the liner. This jet propagates more or less in a straight direction, without significant expansion. 23 However, it also introduces effects not covered in this thesis. Interactions between the shocks and the rarefactions would have to be modeled and individually managed. Including this effect (and 22, above) would force the problem into that of computer simulation. 31
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Thermal effects of the shock have been neglected. For instance, in a typical shock condition, there is a significant area on the Pv diagram between the jump condition (the straight line) and the Hugeniot. This energy has to go somewherethe answer is that it is usually converted to heat. A massive temperature rise usually takes place after a shock, and can be significant enough to actually melt, or even vaporize, the material in question. If thermal effects can be added to the theory such that temperature profiles and pressure profiles are precisely predicted, the theory might be able to quantitatively explain how explosive welding works. 32
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9 Summary The author analyses the effects of an explosion on a bolt. The pressure required to shear a bolt is derived. The pressure required to shear a section of pipe is derived. Finally, an amount of explosive is derived. The paper has discussed the two principal effects of the explosive on the bolt, namely the shear slides and bending hinges. The pressure to shear a bolt is modeled in equations 2.9 and 2.10, and used to determine distances in equations 2.13. The appendix lists the distances in equation 10.13 for the pipe. The amount of explosive is derived in equation 7.3. Both the reaction rates and the shell failure criteria need further research. The reaction rates will require significant amounts of effort and are probably not available outside of government research. This is a flaw of the proposed method. Also, the buckling failure mode of the pipe was not studied. It is quite probable that the pipe would not fail as a beam, but as an individual shell. This was not modeled or addressed in the paper. The analysis is very dependent on the pressure profile described in equations (2.9) and (2.10). These equations drove the possibilities of solution. A different profile will change the solution significantly. Although the profile was derived from experimental data (Cooper), it may or may not reflect the performance of an actual explosive. Effects of the yield stress cro due to strain rate were estimated. A more accurate method might include determining yield stress as a function of the velocity w, which is ultimately a function of the strain rate. Effects due to drag and fluid viscosity were not included. More accuracy, at the cost of computational modeling, could be gained by explicitly including pressure and viscosity effects. Jones uses a similar method for modeling dynamic plasticity effects on these structural elements, as well as modeling dropped masses (vs. pressure impulses). The method could be extended to other items, such as plates and shells, and has potential for improving the design of explosive items. 33
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Appendix A. Clohessy's Circular Beam Upon reflection, it is difficult to imagine a uniform stress along the beam. One way that this could be done is to stage a line charge at some distance away from the beam's surface. Another method that could be used is to use an annulus with a small slot on top. The explosive is centered in the plane of the material, By doing so, a uniform pressure is generated over the material, and since the explosive would be generated radially, no untoward bending effects due to curvature should occur. This would be a sound method to verify the theory. 34
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B. Pipeline Design Assume maximum flow velocity of 2.25 m/s, which is about 5 mi/hr vel :=2.25 Flow area is equal to 81% of wetted area using Chezy formula: r:=.55 8chezy:=2.247 Aflow=0.827 Aflow:=l(echezy1sin(2chezy)) The flow rate is therefore A*vel, or A flowvel = 1.862 This equates to approximately 25,000 gallons per minute. To put this In perspective, a small swimming pool (30' X 60') is approximately 10,000 cubic feet, which is about 78,000 gallons. Our flow could fill this swimming pool In about 3 minutes! Now, our goal is to determine the external radius of the pipe. Obviously, within some factor of safety, the pipe has to hold a certain pressure to keep the flow moving at the rate mentioned above. If we assume that the pump station is approximately 12 km away, and have a rough idea of the grade and friction loss, we can determine the input pressure. This input pressure will be used, along with the worst conditions (underwater at 70 ft) to determine the size of the pipe wall. This gives us sound conditions to know how to clamp the pipe also. We will find the Reynolds's number to characterize the flow. Then, we will solve for the pressure drops back to the other station, which will give the worst pressure that the pipe will have to accept. Then, this pressure can determine the thickness of the pipe. The Reynolds number is equal to the equivalent diameter*the mass flow rate per unit area, divided by the absolute, or dynamic viscosity. The equivalent diameter, De, is equal to four times the hydraulic radius. The hydraulic radius is equal to the area of flow divided by the wetted perimeter: A flow p wet :=2r chezy p wet= 2.472 rhyd :=rhyd = 0.335 De :=4rhyd De= 1.339 Pwet For SAE 20W oil at zero degrees Fahrenheit (well below freezing), the viscosity in centistokes Is 2590 cs. The oil in question is slightly less viscous at 0 degrees, but it is in this range. 35
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Conversions: 640 _1 nu cs :=2590 rho small:=mucp = 1.658'llr 1000 mucp :=rho smalrnu cs ll :=mucp .001 ll = 1.658 p oil :=640 G Is the mass flow rate per unit area, or rho* velocity. G:=p oirve1 G= 1.441cf DeG _1 So the Reynolds number Is: Re :=Re = 1.163t11J which suggests laminar ll flow. 64 Using the Darcy friction formula, r:= Re f= 0.055 This is the friction factor we will use. g :=9.87 (gravity) The next step is to back solve for the pressure losses. Pzero is the pressure leaving the Alpha station Pl Is the pressure before entering the water. P2 is the pressure at the 15 m north shore. P3 is the pressure at the 20 m south shore. P4 is the pressure exiting the water on the south shore. PS is the pressure at Bravo station. The assumption is that there are no shock losses due to sudden changes in the pipe diameter. 145 :=2000 134 :=5000 123 :=4000 112 :=500 101 := 1000 4 f 1 34 ve12 p 3 := p 4+ 20 P '1'g + p il 01 D 2 o e 123 ve12 P2 :=p35 p '1g+4fp '1 01 D 2 01 e 112 ve12 p 1'=p215p '1g+4fp '1 01 D 2 01 e 6 p 3 = 1.99"10 6 p 1 = 3.062'>10 4f1oi ve12 Po=p 1+ p 'I D 2 01 6 p 0 = 3.329'10 e 36
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Stresses in the Pipeline. Using the equations of elasticity and neglecting the weight of the oil, the following equations can be derived. rout A..rin p atm := 1.013105 rout :=.57 r in :=.55 p water :=p atm+ 20.8 r pipe :=r in .01 + r in" rout [ ( 2) ( 2)] 1 rout 2 rout crr(rpipe) :=2Po Pwater A. A. 1 r pipe r pipe 0 .55 0.555 0.56 0.565 r pipe [ ( 21 ( 2)] 1 rout 2 rout oS (r pipe) :=2 p o 1 + .2 p water' A. + .2 A. 1 r pipe r ptpe 37
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8.1 7 L..L..._.JL...'LJ 0.56 0.565 0 .545 0.55 0 .555 r pipe So, the three principal stresses are 18 MPa for the hoop stress in tension, 4.1 MPa for the axial stress, and 3.3 MPa in compression radially. Because these are principal stresses, the yield condition (using VonMises law) is cr 1 :=oe (r in) cr 2 :=crz cr 3 := crr( r in) Y= 7.822'10 7 The yield strength for this particular steel in tension is 1103 MPa, which, even allowing for a 20% factor of safety, is significantly less (78 MPa vs. 910 MPa) than the yield strength. The axial stress must be relieved with bearings and other pipe attachments; this is not covered in this paper. 38
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C. Evaluating ta and 1:c1 from Cooper's charts L := 10 This is the length of the beam 2 Interpolation Functions These functions are interpolated from charts in Explosives Engineering, by Paul W. Cooper. The Interpolations represent figures 28.2, 28.3, and 28.5. The charts are based on data from Explosive Shocks in Air, by G.F. Kinney and K.J. Graham. ZX.12 42 11 1.5"' 1 41 1 1.75 10 10.25 tay := zy.12 11 2.5101 1 10 14 100 200 4o'2 1o'2 2.5101 11 I 1.7"'1 11 1.25102 12 3"'2 6.752 7.7"'2 tdx:= 92 11 2.71 1 10 39 1 tdy := 1 2.2"'1 3 4 1.85102 62 1 o1 tax := 2.3 1 (f 1 91 1 7 10
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These functions are used to find the peak pressure (Po), time of arrival (ta) and time of detonation (tcJ). For instance, following the examples in the book for a charge of Cyclotol 10m from the target, T 0 :=288 R:=10 T a :=297 W :=1.42 P 0 :=linterp(zx,zy,Z)P a p 0 = 0.762 P a :=0.834 R z:=z = 0.462 This suggests that we are in the ballpark, which is sufficient for our needs. Scaled time of arrival t a :=linterp(tax, tay ,Z) t a= 4.122 tds :=cspline(tdx, tdy) 1 wJ t a :=t a11 t a= 12.141 1 wJ t d :=interp(tds, tdx, tdy ,Z) t d :=t d1 1 t d = 3.479 40
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. Explosives vs. the Circular Pipe, Round II Unlike the original effort, a variable velocity distribution exists in x for the proposed solution of the pipe. Thus, instead of w r 1 Figure Repeated we have Figure 0.1Varying Velocity Profile The minor difference of the velocity distribution increases the math significantly, therefore only the significant steps will be shown. Note that is the location of the original plastic hinge. Starting with the bending moment equation (2.1), we perform the first spatial integration. This gives us the following equation: Q Q 1 := p X + m W 1 X (10.1) and, since there Is no shear at x=O, Ql Is zero for the entire section to the left of 1;0 This makes'W1 equal to 'wt P 1.m (10.2) 41
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Now, we solve for the section beyond Note that both shear and moment must be continuous with the previous section. First, we have to introduce the velocity profile in this region. The velocity profile is: ' w (Lx) w := W s + ( W 1 W s ) ( L 0) (10.3) which means that the acceleration is simply: \l' ..,c cc (lX) w := W s + ( W 1 W s ) ( L 0) (10.3) Solving again for shear, we get the following equations: J q (w 1s ) ( ;)] Q 2 := p X + .. 1 W s X + ( l O ) l X + 2 + C1 (10.4a) [ . w 1 w s ( 2) l C1 := p o. 0 m W s. 0 + L 0 L 2 (10.4b) This is the shear in the section to the right of with a matching condition of Qs=O at This gives us the following equation for Ws: "t p ( L 0) 2 Q 0 w s (10.5) which is the acceleration of interest. However, conditions still need to be applied for the moments on the beam. After Integration, we find that: 42
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. p+mWs(3;L 2 x!) M := a;a Lx + a x+ C2 2 (La) 2 3 (1a.6a) where .. t (3 5 ) 2M o(L(p+ mw s) azLa3 C2 2 3 .2 ( La) (1a.6b) These are the moment equations.1 The astute reader will note that Ws is defined as: I p ( la) 2 Q 0 w .,.......s .m (La) (1a.S) from the shear equation and (la.7) from the moment equation. Irregardless of which definition is chosen, equations 10.5 and 10.7 are equal. Equating the two equations, the following equation is generated for 3 (627V) 2 2 (24V12) 3 (7V6) a + L a + L . a= L 10 v 10 v 10 v (1a.8) Finally, the shear ratio and the moment ratio need to be checked: (10.9) which after simplification requires that > L/2, and 1 But see footnote 13 in the main document. 43
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+ O >r 3 ol X + 3 O X + O l O ( 3 >lL 2 2 2 .J 3 2 5 3) M 2 3 2 3 := v (10.10) M 0 L(LoY which again, after simplification, requires > 9l/10. The most stringent of the two equations is the second equation, thus we need a hinge to occur somewhere after > 9l/10. The next effort is to find values for the shear. Replacing Qo with v, equation 10.5 becomes: c:n; pL( Lo) 4 M 0 v w s .ml (Lo) (10.11) Next, this equation is Integrated, assuming an initial condition ofW51 at timet equals zero: W s := ( ) t + W sl ml L0 (10.12) and finally integrated again, assuming an initial distance of zero at time t equals zero. pL(Lo)4M 0v 2 W 5 := t + W s1t (10.13) This is the equivalent of equations 2.13 for the pipe. Note that an effort must be made to solve for the time that makes equation 10.12 equal to zero. This t determines the distance that the shear will progress. Finally, a method must be available to determine the ratio of W1 to Wsl The value of W1 can be determined (assumed) to be equal to u1, where u1 is the initial particle velocity after a shock of speed D propagates into the pipe material. 44
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Bibliography 1. Benham, P.P., R. J. Crawford, and C.G. Armstrong. Mechanics of Engineering Materials, 2nd Edition. Essex, England: Longman Group Limited, 1996. 2. Blake, Alexander. Design of Mechanical Joints. Copyright 1985, Marcel Dekker, Inc. Published New York, NY: Marcel Dekker, 1985. 3. Benson, Sidney W. Thermochemical Kinetics. 2nd Edition. Copyright 1976, John Wiley and Sons. Published New York, NY: Wiley Interscience, 1976. 4. Cooper, Paul W. Explosives Engineering. New York, NY: VCH Publishers, 1996. 5. Clohessy, William(?) Private conversation, July 1999. 6. Drumheller, D.S. Introduction to Wave Propagation in Nonlinear Fluids and Solids. Copyright 1998, D.S. Drumheller. Published Cambridge, United Kingdom: Cambridge University Press, 1998. 7. Farlow, Stanley J. Partial Differential Equations for Scientists and Engineers. Copyright 1993, Stanley J. Farlow. Published New York, NY: Dover Publications, Inc., 1993. 8. Feynman, Richard P., Leighton, Robert B., Sands, Matthew. The Feynman Lectures on Physics. Volume 2Mainly Electromagnetism and Matter. Copyright 1964, California Institute of Technology. Published Reading, Massachusetts: AddisonWesley Publishing Company, 1977. 9. Jeppersen, Neil. Chemistry. Copyright 1997, Barron's Educational Series, Inc. Hauppauge, New York: Barron's Educational Series, 1977. 10. Jones, Norman. Structural Impact. Cambridge, United Kingdom: Cambridge University Press, 1997. 11. Mader, Charles L. Numerical Modeling of Explosives and Propellants, 2ru1 Edition. Copyright 1998, CRC Press LLC. Boca Raton, Florida: CRC Press LLC, 1998. 12. Marion, Jerry B., Thorton, Stephen T. Classical Dynamics in Particles and Systems. Copyright 1988, 1970, Harcourt Brace Jovanovich, Inc. Orlando, Florida: Harcourt Brace Jovanovich, 1988. 45
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13. Masterton, William L., and Emil J. Slowinski. Chemical Principles. 4th Edition. Copyright 1977, W.B. Saunders Company. Published Philadelphia, Pennsylvania: W.B. Saunders Co, 1977. 14. Menkes, S. B. and H. J. Opat. Broken Beams. "Experimental Mechanics." volume 13. 15. Neal, B.G. The Plastic Methods of Structural Analysis (Third S.I. Edition). Copyright 1977, B.G. Neal. Published London, England: Chapman and Hall Limited, 1977. 16. Norton, Robert L. Machine Design: An Integrated Approach. Copyright 1996, Prentice Hall, Inc. Upper Saddle River, NJ: Prentice Hall, 1996. 17. Ragab, AbdeiRahman A. F., and Salah Eldin A. Bayoumi. Engineering Solid Mechanics: Fundamentals and Applications. Boca Raton, Florida: CRC Press LLC, 1999. 18. Ugural, A.C., and S.K. Fenster. Advanced Strength and Applied Elasticity: The SI version. Copyright 1981, Elsevier Science Publishing Company. NY, NY: SpringerVerlag New York, 1998. 19. Zukas, Jonas. Impact Dynamics. 20. Zukas, Jonas A. (ed.) and William P. Walters. Explosive Effects and Applications. Copyright 1998, SpringerVerlag New York, Inc. 46
