DIFFERENTIAL MODEL AND IMPACT RESPONSE OF A FLEXIBLE BEAM
ATTACHED TO A RIGID SUPPORTING STRUCTURE
A Thesis
Presented to
The Graduate Faculty of the University of Akron
In Partial Fulfillment
of the Requirements for the Degree
Master of Science
Harish Chandra
May, 2008
DIFFERENTIAL MODEL AND IMPACT RESPONSE OF A FLEXIBLE BEAM
ATTACHED TO A RIGID SUPPORTING STRUCTURE
Harish Chandra
Thesis
Approved:
Accepted:
Advisor
Dr. D. Dane Quinn
Dean of the College
Dr. George K. Haritos
Faculty Reader
Dr. T.S.Srivatsan
Dean of the Graduate School
Dr. George R. Newkome
Department Chair
Dr. Celal Batur
Date
ii
ABSTRACT
Often electronic components such as laptops and cellular phones are dropped accidentally
during usage and extensive damage is developed due to the impulsive force generated at
the contact point. While external damage is easy to detect, internal damage to the
electronic circuitry go undetected, yet may cause failure of the system. An accurate
description of the impulsive force is necessary to understand the dynamics of the system.
This research study involves development of a differential model of a flexible beam
attached to a rigid supporting structure and studying its response due to impacts. An
Euler Bernoulli beam theory is used to model the beam, and Routh’s graphical method
for two dimensional impacts is used to calculate the impulse at the contact point. The
dynamics of impact at the contact point is used to develop the boundary conditions and
Galerkin’s approach is used to find an approximate solution. An example is presented in
which the response due to drop at different angles of approach is studied. The position of
the beam on the frame, the coefficient of friction (µ ) and the coefficient of restitution (e )
are varied to see their influence on the beam response. Finally the influence of the
boundary conditions on the stresses and strains developed in the beam is discussed.
iii
ACKNOWLEDGEMENTS
First and foremost I would like to thank my advisor Dr.Dane Quinn for his invaluable
guidance, patience and support throughout the course of my study. I remain indebted to
him for awarding me a Research Assistantship during my first semester and helping me
get a Teaching assistantship thereafter till the end of my study.
I would like to thank the Department of Mechanical Engineering at the University of
Akron [Dr. Celal Batur] for awarding me a Teaching Assistantship during my graduate
study for the Master of Science degree. I would like to thank Dr. T.S.Srivatsan and
Dr.Graham Kelly for serving on my thesis committee. I appreciate all the support I got
from the staff members specially Stacy and Stephanie.
I also extend warmest thanks to my parents and friends who have been with me and
encouraged me during my study.
iv
TABLE OF CONTENTS
Page
LIST OF FIGURES …………………………………………..…………….…………viii
CHAPTER
I.
INTRODUCTION ………………………………………………………………1
1.1
Overview…………………………………………………………………..1
1.2
Motivation………………………………………………………………....2
1.3
Collisions …………………………………………………………………3
1.4
1.3.1
Rigid body collisions……………………………………………...6
1.3.2
Collision law properties…………………………………………...7
1.3.3
Rigid body assumptions…………………………………………...7
Mathematical modeling…………………………………………………...8
1.4.1
Steps in Mathemaical Modeling…………………………………..9
II.
LITERATURE REVIEW …………………………………………………… 12
III.
MATHEMATICAL MODELING……………………………………………….17
3.1
Introduction………………………………………………………………17
3.2
Problem formulation……………………………………………………..17
3.3
Boundary conditions……………………………………………………..20
3.3.1
Initial Conditions………………………………………………...20
3.3.2 Transformation of boundary conditions.........................................21
v
3.4
The dynamics of impact………………………………………………….22
3.5
Impact velocities………………………………………………………... 23
3.6
Routh’s method to find the impulse……………………………………...25
3.7
3.6.1
Procedure to calculate impulse…………………………………..26
3.6.2
The impact process diagram……………………………………..29
Galerkin Reduction……………………………………………………....33
3.7.1
IV.
V.
Galerkin method as applied to the beam model………………….33
3.8
Stresses and Strains……………………………………………………...36
3.9
Review…………………………………………………………………..36
THE DYNAMICS OF IMPACT AND IMPACT VELOCITIES……………….38
4.1
Overview………………………………………………………………… 38
4.2
An impact problem
4.3
Routh’s graphical method………………………………………………...40
4.4
Using the impulse to find change in velocities…………………………...42
4.5
Mode shapes equations…………………………………………………...43
4.6
Results…………………………………………………………………….45
…………………………………………………….38
RESULTS AND CONCLUSION………………………………………………..48
5.1
Overview…………………………………………………………………48
5.2
Angle of approach v/s impact……………………………………………48
5.3
Influence of (µ ) on the impulse at contact point………………………...51
5.4
Influence of the coefficient of restitution (e) on the impulse…………….53
5.5
Influence of the angle of impact on the beam velocity…………………..54
vi
5.6
Influence of the angle of impact on the deflection, stress and strain…….56
5.7
Influence of the location of the beam on the boundary conditions………57
5.8
Change in stress and strain with change in position of beam……………59
5.9
Conclusion……………………………………………………………….60
5.10
Underlying simplifications………………………………………………..61
5.11
Recommendations for future work ………………………………………62
REFERENCES…………………………………………………………………………..63
vii
LIST OF FIGURES
Figure
Page
1.1
Impact configuration………………………………………………………………1
1.2
Single degree of freedom system………………………………………………….5
3.1
Rigid frame with a flexible beam………………………………………………...18
3.2
Illustration of Impact……………………………………………………………..24
3.3
Position of Mass center…………………………………………………………..27
3.4
Contact velocities………………………………………………………………...27
3.5
Impact process diagram illustrating Slip-Stick…………………………………..30
3.6
Impact process diagram illustrating Stick………………………………………..31
4.1
An Impact Problem………………………………………………………………37
4.2
Impact process diagram………………………………………………………….40
4.3
Maximum Deflection…………………………………………………………….45
4.4
Maximum Stress…………………………………………………………………46
4.5
Maximum Strain…………………………………………………………………46
5.1
Angle of Impact v/s Magnitude of impact……………………………………….48
5.2
Impulse path at θ = 15 o …………………………………………………………..49
5.3
Impulse path at θ = 45o ………………………………………………………….49
5.4
Impulse path at θ = 75o …………………………………………………………..49
5.5
µ v/s Pt …………………………………………………………………………..50
5.6
Impulse path for µ = 0.2 …………………………………………………………51
viii
5.7
Impulse path for µ = 0.8 ………………………………………………………….51
5.8
Impulse path for µ = 1 ……………………………………………………………52
5.9
Impulse path for µ = 2 ……………………………………………………………52
5.10
e v/s P…………………………………………………………………………...53
5.11
Angle of impact v/s Change in magnitude of velocity at end ‘A’………………54
5.12
Angle of impact v/s Change in magnitude of velocity at end ‘B’……………….54
5.13
Angle of impact v/s Deflection………………………………………………....55
5.14
Angle of impact v/s Stress………………………………………………………55
5.15
Angle of Impact v/s Strain……………………………………………………...56
5.16
Frame with beam attached at different locations………………………………...56
5.17
Location of beam v/s change in magnitude of velocity at ‘A’ ………………….57
5.18
Location of beam v/s change in magnitude of velocity at ‘B’ ………………….57
5.19
Stress v/s Location of beam……………………………………………………...58
5.20
Strain v/s Location of beam……………………………………………………...58
ix
CHAPTER 1
INTRODUCTION
1.1 Overview
Impact can be described as the interaction between two or more bodies and plays a
vital role in many mechanical engineering applications. An accurate description of this
interaction is necessary to understand the dynamics of mechanical systems. The objective
of this thesis is to develop a differential model of the response of a flexible beam attached
to rigid supporting structure upon impact. Figure 3.1 illustrates the problem being
studied.
Q
ê2
ê1
B
R
A
G ( x, y )
P
ĵ
iˆ
S
Figure 1.1 Impact configuration
This work studies the effect of the impulse produced at the contact point on the
response of the beam. The process of impact is described by Routh’s model, a simplified
description of the post - collision velocity of the body can be predicted, given the pre1
collision velocity and geometry. The impact then specifies boundary conditions on the
flexible beam, so that the effect of the impact on the stress, strain and deflection of the
beam are presented.
The first chapter of this thesis introduces the reader to rigid body collisions and
impacts. A few basic terms as applied to rigid body dynamics are explained. The second
chapter discusses the previous work that has been done on collision models and their
results. The main work done in this thesis is explained in chapters three, four and five. In
Chapter 3 the basic equations that form the differential model of the beam are developed.
A simplified model for the approximation of impact is presented. The relationship
between impact and the post-collision velocities is established. In Chapter 4 the equations
developed in the previous chapter are used and an impact problem is explained in detail.
The stresses and strains developed in the beam due to changes in velocity upon impact
are studied. Chapter 5 concludes the thesis with the study of the influence of the angle of
impact of the block on the post-collision velocities and hence the stresses and strains
developed and their location. Finally limitations and recommendations for possible future
work are stated.
1.2 Motivation
Often electronic components are dropped accidentally during usage and can cause
extensive damage. The damage could be exterior in nature such as dents or cracks on the
surface or may lead to failure of the electronic circuits that play a vital role in its
functioning. While external damage may do little to affect the performance of the object,
internal damage may cause the object to cease proper orientation. Therefore the proper
2
functioning of portable electronic products such as mobile phones, laptops, and digital
cameras depends largely on the integrity of the internal components and the electronic
circuitry. Due to the rapid progress made in the technological field in the last few years
electronic devices have become smaller and lighter and the internal components have
become so complex that they are more susceptible to damage due to accidental drop.
When an object is accidentally dropped an impulsive force is created when it makes
contact with the ground. This is a potential cause of failure. Under these circumstances
their reliability due to shock and impact becomes a critical issue affecting performance.
They are also exposed to shock during various stages of their production and operation.
When the product is accidentally dropped on the ground, impact forces are transmitted
from the product case to the printed circuit board (PCB) and other components within the
case. This can cause fracture problems. Also sometimes the product may fail due to
motion of the internal components.
With escalating production and manufacturing costs it would help if the damage
caused to these components can be identified or predicted before it is delivered to
customers. This would help in the production of components that are more reliable under
shock and more resistant to damage due to drops. One of the methods used to analyze the
dynamics of the problem is mathematical modeling. It is a powerful and useful approach
to identify the potential weakness of products in the initial stages of fabrication.
1.3 Collisions
Collisions can be defined as the action of two or more bodies coming together or
striking one another for a small amount of time. A collision results in an impulse. In the
3
event of a collision the mass and the change in velocity is easily measured but the impact
force is not. Once the time of collision is known the average impact force can be
calculated. In other words one of the first steps in order to find the impact force is
collision detection. Collisions are classified as single point and multipoint depending on
the way they come into contact. If the bodies come into contact at a single point it’s
called a single point collision. Otherwise it’s called a multipoint collision. During a
collision between two bodies there is large force acting for a very brief period of time.
Integration of that force over that brief time gives the impulse. Collisions can be either
elastic or inelastic. A perfectly elastic collision is defined as one in which there is no loss
of kinetic energy in the collision. Elastic collisions occur only if there is no conversion of
kinetic energy to other forms. Collisions between atoms are perfect examples of elastic
collisions. In an elastic collision the total kinetic energy is the same before and after
collision, hence:
m1u12 m2 u 22 m1v12 m2 v 22
+
=
+
2
2
2
2
Momentum balance is maintained in all collisions. Therefore total momentum remains
constant after collision:
m1u1 + m2 u 2 = m1v1 + m2 v 2
An inelastic collision is one in which there is some form of energy conversion during the
collision. During an inelastic collision part of the kinetic energy is converted to internal
energy. Finally, in a plastic collision the objects stick together after collision.
Impact is characterized by a sudden high force or shock applied over a short
interval. Across a collision only the impulse is important. Impact has been studied in
4
detail in a lot of previous works since it forms the subject of a wide variety of
engineering applications. The topic of interest for most design and structural engineers is
the reaction forces that develop during collision and the response of structures to these
forces. There are various methods to determine the response of a system to a shock load.
The two most popular ones are the frequency domain approach or the shock response
spectrum and the time domain approach or the time history of the system. In the
frequency domain approach the steady-state maximum response of the system to a given
shock pulse is determined. This is accomplished by considering the response of a single
degree of freedom model composed of a spring mass and damper. In the time domain
approach the equations of motion governing the system are first written, they are then
integrated with respect to time to determine the response. Mechanical components can
often be modeled with partial differential equations; these equations are discretized to
yield a system of ordinary differential equations in time which can then be integrated.
Figure 1.2: Single degree of freedom system
A shock is defined as a transient physical excitation that causes a sudden jump in
velocity. It can be caused by a drop, collision with another object, sudden disturbance
like an
earthquake or explosion. The magnitude of shock is measured using an
5
accelerometer. A shock response spectrum is used for evaluating the response to a
mechanical shock. It is a graph depicting the response of a single degree of freedom
system such as a spring-mass-damper system to an arbitrary transient acceleration as
input. The horizontal coordinate represents the natural frequency of any given single
degree of freedom system and the vertical coordinate represents the acceleration
response. Shocks of high magnitude have the potential of damaging an entire structure.
Also the damage caused depends upon whether the material is brittle or ductile. A brittle
material under shock fractures whereas a ductile material bends. An example of the
damage caused to a brittle material under shock is when a crystal glass is dropped to the
floor it shatters. Whereas when a copper pitcher, a ductile item is dropped on the floor it
bends. Some materials are not damaged by a single shock but experience fatigue failure
under numerous low level shocks.
1.3.1 Rigid Body Collisions
A rigid body is a solid object that has a definite shape and cannot be deformed in
any way. In other words its shape does not change during collision. In order to find the
post collision velocities of two colliding rigid bodies certain laws are used. These laws
are referred to as collision laws. Given the velocities of the centers of mass and the
angular velocities of the bodies at the instant before collision, a collision law is a rule
which predicts the corresponding velocities after the collision. A collision law requires
physical details of the colliding bodies such as material properties, geometric
characteristics of the bodies, friction properties in the region of contact, etc.
6
1.3.2 Collision Law properties
1. A collision law should be consistent with the fundamental laws of mechanics and
dynamics like conservation of mass and momentum, linear and angular momentum
balance.
2. It should consider bodies with arbitrary shape, mass distributions, material and
surface properties, orientations and velocities.
3. It should be able to give results that are in agreement with observed behavior for
simple models.
4. It should be consistent with other simpler laws like the laws of friction and other
phenomena which are modeled using lesser known laws.
5. It should be dependent on reasonably small number of input parameters and
should involve simple calculations.
6. It should be able to capture a wide variety of observed behavior for given input
parameters.
7. The input parameters should have simple physical interpretations.
1.3.3 Rigid Body assumptions
The colliding bodies are treated as rigid before and after collision. This means that
the colliding bodies move almost like rigid bodies before and after collision with
deformation being neglected in the calculation of linear and angular momentum. Any non
rigid behavior taking place during collision causes small deformations for majority of the
body with larger deformations occurring at the contact area. Collision occurs for a very
short duration with displacements and rotations being negligible, accelerations being
large with definite changes in velocities. Impulses other than those occurring at the
7
contact area are neglected with other body forces also being negligible. The mass,
moments of inertia and dimensions of at least one colliding body are finite and known.
Point contact between the bodies is assumed. Even though in reality contact occurs over a
region, this region is assumed to have smaller dimensions when compared to the length
of the smaller colliding body.
There is no kinetic energy created in a collision. The net kinetic energy of rigid
body motion in the bodies after collision is less or equal to the kinetic energy before
collision. The bodies do not pass through each other and there is no interpenetration as
this violates the point contact assumption. These are treated as reasonable assumptions
according to the laws of laws of mechanics.
1.4 Mathematical Modeling
Engineers often use their knowledge of science, mathematics, and appropriate
experience to find suitable solutions to a problem. Creating an appropriate mathematical
model of a problem allows them to analyze it, and to test potential solutions. Usually
multiple reasonable solutions exist, so engineers must evaluate the different design
choices on their merits and choose the solution that best meets their requirements. A
mathematical model usually describes a system by a set of variables and a set of
equations that establish relationships between the variables. The values of the variables
can be real or integer numbers, boolean values or strings. The variables represent some
properties of the system, for example, system outputs in the form of signals, timing data,
counters, event occurrences etc. Mathematical models are of great importance in physics.
It is common to use idealized models in physics to simplify things. It is a effective tool
that engineers use to anlyse, control and optimize physical systems Throughout history,
8
more and more accurate mathematical models have been developed. The laws of physics
are represented with simple equations such as Newton's laws; Maxwell's equations
etc.These laws form a basis for making mathematical models of real situations. Many real
situations are very complex and thus modeled approximately on a computer; a model that
is computationally feasible to compute is made from the basic laws or from approximate
models made from the basic laws. Mathematical modeling problems are often classified
into black box or white box models, according to how much background information is
available of the system. A black-box model is a system of which there is no background
information available. A white-box model is a system where all necessary information is
available. Practically all systems are somewhere between the black-box and white-box
models, so this concept only works as an intuitive guide for approach. Usually it is
preferable to use as much background information as possible to make the model more
accurate. Therefore the white-box models are usually considered easier, because if you
have used the information correctly, then the model will behave correctly.
1.4.1 Steps in Mathematical Modeling
1. Problem identification: The system to be modeled is isolated from its
surroundings and the effects of the surroundings are noted. Known constants and
variables are identified.
2. Assumptions: Assumptions are made to simplify the modeling. Considering all
effects of the system results in complexity and hence a mathematical solution
becomes difficult. Assumptions should only be made if they yield results that are
simpler than those got without the assumptions. Sometimes certain implicit
9
assumptions are made which are taken for granted and are seldom mentioned.
Examples include assuming physical properties as continuous functions,
considering all materials to be linear, homogeneous and isotropic, ignoring
relativistic, chemical, nuclear and other effects.
3. Basic laws of nature: A basic law is a physical law that applies to all physical
systems regardless of the material from which the system is constructed. They are
laws that can be observed but not derived from more fundamental laws. Examples
include conservation of energy, conservation of momentum; the second and a third
law of thermodynamics.Among the above only conservation of momentum plays a
significant role in problems involving vibrating systems.
4. Constitutive equations: They provide information about the materials of which a
system is made. They are used to develop force-displacement relationships for
mechanical components used in vibration problems.
5. Geometric constraints: They are essential in completing the mathematical model
of a system. They can be in the form of kinematic relationships between velocity,
acceleration and dispalcement.Geometric constraints are used to formulate the
boundary conditions and initial conditions once the differential equations are
developed.
6. Mathematical solution: Once the mathematical problem is obtained, the modeling
is not complete till the appropriate mathematics is used to obtain a solution. Exact
analytical solutions, if they exist are preferable to numerical or approximate
10
solutions. Usually exact solutions exist for only for linear problems but for very few
nonlinear problems.
7. Physical interpretation of results: This is the next step after the modeling of the
desired system is complete and the mathematical solution is obtained. In certain
instances it may involve drawing general conclusions from the mathematical
solution or development of design curves or just require simple arithmetic to arrive
at a conclusion.
11
CHAPTER II
LITERATURE REVIEW.
Recent developments in the electronics industry over the last few years have a brought
about a huge increase in the usage of portable electronic products like cellular phones,
personal digital assistants and digital cameras. As these products get smaller for ease of
handling, they become more susceptible to drop impacts. Impact loads are the main cause
of mechanical failures in products [1-18]. There can be significant damage caused to the
part as a result of the impact forces developed during drop. Most of the common
electronic devices have a similar structure consisting of an upper structure and lower
structure connected by a hinge device. The upper structure includes the LCD module,
covers, metal frames and the lower structure includes a keyboard, housings, battery,
motherboard and chipsets [3]. There can be serious damage caused to electronic products
when dropped. The impact force produced can cause the internal components to fail or
cracking of the outer cover. Some of the damage can also include housing fracture, joint
breaking, connector disconnection or complete component failure [2]. Of these the one
that is of great concern is the functional failure of the part. The cause of such failure is
normally due to the loss of electrical contacts arising from the breakage of
interconnections between components within the product [1]. The resulting stresses and
strains can cause the housings to deform, cause assemblies to come apart and cause the
12
liquid crystal display (LCD) to crack [2]. As a result, studying the response of these
electronic components subjected to drop impacts is vital in predicting the life of the
component. Over the years various researchers have used different methods to study the
effect of drop on the performance of the system. The traditional method of analysis is
carrying out physical tests using a prototype. But in using such a method it is difficult to
capture the impact event as it happens over such a short span of time and involves a lot of
trial and error [5]. Also it has proven to be expensive, time consuming and requires a lot
of effort [2]. Another method that is very popular and widely used is simulation of the
impact event using a software package such as FEA and comparing it experimental or
analytical results [1, 2]. Tan et al [1] used Finite element (FE) simulation to model a ball
grid array (BGA) package that consists of a integrated circuit(IC) connected to a flexible
printed circuit board (PCB) subjected to drop impact. He used three different kinds of
mesh to model the BGA package and studied the deviation in results with the change in
meshing. The parameters that were compared were the deflection of the PCB and the
stresses that develop within the solder balls. He concluded that the type of mesh used to
model a component has an influence on the stresses and strains developed. Lim et al [2]
studied the feasibility of using FE simulation for the drop impact response of a pager by
performing a numerical simulation and using ABAQUS and verified numerical results
with experiments. He concluded through his experimental and simulation results that the
impact orientation is largely responsible for variation of strains and impact force during
collision. He attributed the variation of strain with time to the difference in local
deformation occurring at different parts of the pager. Low et al [4] studied the impact
effect on mini Hi-Fi audio products He used Pro-E to create his model and used PAM13
CRASH to study the effect of material properties on the results. However, using FEA as a
method to simulate the impact event has a few disadvantages. A major difficulty is in
meshing some of the smaller components. Some of these substructures in electronic
components need extremely fine meshes and this causes too much time for the analysis
and requires high end software that could be very expensive [4 - 6].Another problem is
setting up the simulation model accurately including the geometry of the component
being studied, selection of the right material for different components, boundary
conditions etc [5]. Some of the limitations of using software to simulate the impact event
can be overcome by studying the analytical or mathematical models of impact. Goyal et
al [16] used a linear spring-mass system to analytically model the drop impact of a
cellular phone. He concluded that the geometry of the component is an important factor
in analyzing the failure of the component. He suggested that improving the rigidity of the
case may prevent slipping of the cellular phone from its cover upon drop. Goyal [15]
also studied the clattering effects that occur when a component hits the ground at an
angle. He studied the jumps in velocity of the ends of a bar for successive impacts. He
found that when a two dimensional bar was dropped on the floor the second impact could
be twice as large as the first. He concluded that the number of times that the bar would
impact the ground depends on the co efficient of restitution and the mass distribution.
Shan et al [7] also did a comprehensive study on the effect of clattering on three
dimensional bars and found that his results were comparable with Goyal. Xiang [17] did
a study on the effect of a continual shock loads on packaged products during
transportation. Continuous shock loads cause undesirable stresses on packaged material
during transport. Xiang studied the effect of acceleration amplitude on the number of
14
shocks that lead to failure by performing a series of tests in the laboratory. He established
a relation between fragility of the product and the number of continuous shock loads to
failure. Wang et al [9] in his study of two dimensional rigid body collisions found some
methods of rigid body impact could violate the energy conservation principles. He
classified the different modes of impact using Routh’s graphical method to determine the
impulse. He blamed the violation of energy conservation to the use of Newton’s
hypothesis for the coefficient of restitution. Wang proved that using Poisson’s hypothesis
over Newton’s for the co efficient of restitution yielded better results He also stated that
identifying the correct mode of contact is vital in order to satisfy the law of conservation
of energy in an impact process. However Brach [18] blamed the violation of the principle
of energy conservation to the definition of friction. He adopted lower values of
coefficient of friction to satisfy the law of conservation of energy. Keller [8] extended the
Routh’s method to solve three dimensional problems. Barbulescu et al [11] used Kane’s
method to study spatial impact of a slender beam. In his studies he dealt with both slip at
the contact point and no slip. In the former case he assumed tangential velocity at
separation to be zero. He deduced a relationship between the loss of kinetic energy and
the angle of impact. He concluded that for conditions of no slip the energy loss increased
with increase in the angle of impact. Younis et al [13] used Galerkin’s approach to study
the dynamic response of beams to mechanical shock. He investigated the nonlinearity of
the response and attributed it to the effect of mid-plane stretching. He suggested
improving the thickness of the beam to improve shock resistance. He stated that
mechanical shock when combined with electrostatic force could cause a dynamic pull-in
in MEMS devices causing instability. He also studied the effect of packaging on MEMS
15
devices. One of his important findings was that neglecting the effect of packaging could
lead to failure of the device.
16
CHAPTER III
MATHEMATICAL MODELING
3.1 Introduction
In this Chapter we establish an impact model that calculates the impulse and hence the
change in velocities at the ends of the beam upon impact. A rigid frame modeled as the
housing of an electronic component with a flexible beam attached to it modeled as an
internal component is subjected to drop. The impact produces a change in velocity of the
housing that changes the boundary conditions of the beam. The response of the beam is
solved for the resulting boundary conditions. An Euler-Bernoulli beam is used for
analysis. Initially the problem is homogeneous with non homogeneous boundary
conditions. Using a linear function in x we transfer it into a non homogeneous problem
that calculates the change in velocity upon impact. The impulse produced upon impact
produces boundary conditions on the beam. The relation between the impulse and
boundary conditions is shown. Finally the response of the beam with the maximum
amplitude, stress and strain is presented.
3.2 Problem Formulation
The initial configuration of the system is shown in Figure 3.1, illustrating the
collision between a rigid frame with an attached flexible beam and the ground.
17
Q
ê2
ê1
B
R
A
G ( x, y )
P
ĵ
S
iˆ
Figure 3.1: Rigid frame with flexible beam
Let PQRS represent the rigid frame with G being the mass center. ‘AB’ represents a
flexible beam attached to the frame with the help of two supports as shown in the figure.
The rigid frame represents the housing of an electronic device while the flexible beam
models an internal component such as a printed circuit board. Let the point of contact S
be the origin. ‘ iˆ ’ and ‘ ĵ ’ are unit vectors in the global coordinate system fixed to the
ground, ‘ ê1 ’ and ‘ ê2 ’ represent unit vectors in the local coordinate system of the beam
relative to the frame. The housing makes direct contact with the ground producing an
impulsive load at the contact point. The impulse produces a change in velocity of the
housing. This change in velocity impulsively loads the beam.
Using Hamilton’s principle, the transverse displacement of the beam w( x, t ) with
respect to the housing is derived as
ρA
∂2w
∂4w
∂2w
+
EI
−
P
=0
∂t 2
∂x 4
∂x 2
18
(3.1)
where E is the elastic modulus of the beam, ρ is its mass density, A(x) is its cross
sectional area, I(x) is its cross section moment of inertia and P is the component of the
axial load in the transverse direction. Equation (3.1) represents the model of an EulerBernoulli beam. It is convenient to nondimensionalize the partial differential equations
governing the vibrations of continuous systems through the introduction of
nondimensional dependent and independent variables. Nondimensional variables are
introduced such as
w* =
w
L
(3.2)
x* =
x
L
(3.3)
t* =
t
T
(3.4)
Where the variable with an * is a nondimensional variable, L is a characteristic length in
the system such as the length of a beam, and T is a characteristic time scale. Equation 3.1
is nondimensionalized using the nondimensional variables of equation 3.2 - 3.4 .The
beam is considered to be of uniform cross section and therefore spatial dependence is not
considered.
β=
α=
ρAL2
PT 2
EI
PL2
(3.5)
(3.6)
The effect of rotary inertia and shear deflection is ignored and as a result the load in the
axial direction is P = 1 , dropping the *’s and considering α and β to be constants we
get the nondimensional form of equation (3.1) as:
19
∂4w ∂2w
∂2w
α 4 − 2 +β 2 =0
∂x
∂x
∂t
(3.7)
Equation 3.7 represents a homogenous equation governing the vibrations of the beam
‘AB’. The equation is homogeneous because the forcing term is assumed to be equal to
zero.
3.3 Boundary Conditions
The boundary conditions are the displacement and velocities at the two ends of the
beam at all times. If w( x, t ) is the displacement of a particle ‘x’ at time ‘t’ and,
∂w
is its
∂t
velocity, the boundary conditions of the non dimensional beam are given by:
w(0, t ) = 0
(3.8)
w(1, t ) = 0
(3.9)
∂w
(0, t ) = v A (t )
∂t
(3.10)
∂w
(1, t ) = v B (t )
∂t
(3.11)
3.3.1 Initial Conditions
The initial conditions represent the displacement and velocities before impact. At time
t = 0 the displacement and velocity of any particle ‘x’ over the entire span of the beam
relative to the rigid frame is zero. This would mean that v A and v B are zero before impact,
so that the initial conditions are:
w( x,0) = 0
(3.12)
∂w( x, t )
=0
∂t t =0
(3.13)
20
It is seen that the differential equation governing the motion of the beam is
homogeneous whereas the boundary conditions are non-homogeneous. We find a
solution by transforming it to a problem with a non-homogeneous differential equation
having homogeneous boundary conditions.
3.3.2 Transformation of boundary conditions
We first choose a function b( x, t ) that renders the boundary conditions homogeneous.
The solution to the problem is then assumed to be the sum of a variable u ( x, t ) and the
function b( x, t )
w( x, t ) = u ( x, t ) + b( x, t )
(3.14)
Here b( x, t ) is chosen to be a function linear in x.
∂b( x, t )
= [v A (t ) + (v B (t ) − v A (t ) )x ]
∂t
(3.15)
This function satisfies the boundary conditions of equations (3.10) and (3.11). In this
manner, we transform the boundary value problem for the variable w( x, t ) into a
boundary value problem for the variable u ( x, t ) . Taking the partial derivative of
equation (3.14), using values of x = 0 and x = 1 we get:
∂w
∂u
(0, t ) =
(0, t ) + v A (t )
∂t
∂t
(3.16)
∂w
∂u
(1, t ) =
(1, t ) + v B (t )
∂t
∂t
(3.17)
Substituting equations (3.16) and (3.17) in (3.10) and (3.11) we get:
∂u
(0, t ) = 0
∂t
21
(3.18)
∂u
(1, t ) = 0
∂t
(3.19)
Therefore the non homogeneous boundary conditions are made homogeneous by the
choice of a function that satisfies the boundary conditions. The new homogeneous
boundary conditions are written as:
u (0, t ) = 0
(3.20)
u (1, t ) = 0
(3.21)
∂u
(0, t ) = v A (t )
∂t
(3.22)
∂u
(1, t ) = v B (t )
∂t
(3.23)
3.4 The dynamics of impact
When the object hits the ground there is an impulse produced at the contact point. Due
to the sudden impact upon contact with the ground the acceleration at the ends of the
beam are expressed as:
dv A
(t ) = ∆v Aδ (t )
dt
(3.24)
dv B
(t ) = ∆v Bδ (t )
dt
(3.25)
In the above two equations δ (t ) is a unit impulse that has an infinite value when x = 0
and zero at all other places. ∆v A and ∆v B are the changes in velocities at the ends of the
beam. Therefore now substituting equation (3.14) in equation (3.7) and using equation
(3.15) we get the governing equation as:
∂ 4u ∂ 2u
∂ 2u
α 4 − 2 + β 2 = −[∆v A + (∆v B − ∆v A )x ]δ (t )
∂x
∂x
∂t
22
(3.26)
Equation (3.26) represents a non homogeneous boundary value problem for the variable
u ( x, t ) with equations (3.20) - (3.23) representing the homogeneous boundary conditions.
3.5 Impact Velocities
Upon impact there is a change of velocity at the two ends of the beam. These
velocities denoted by ∆v A and ∆v B depend upon the impulse produced upon impact at
the interface between the ground and the rigid block. Using Linear momentum balance,
we get:
∑ F = ma
(3.27)
Equation (3.27) can also be written as,
dv
∑ F = m dt
(3.28)
Where,
F = force acting on the body
m = mass of the body,
a = acceleration of the mass center of the body.
v = velocity of the mass center of the body
Equation (3.28) is written as,
t
v G (t ) − v G (0) = m −1 ∫ F (τ )dτ
(3.29)
0
v G (t ) − v G (0) = m −1 P (t )
(3.30)
Equation (3.30) gives the change in linear velocity at the mass center ‘G’ upon impact.
Using angular momentum balance we get,
∑ M G = I Gα
23
(3.31)
Where,
M G = moment acting on the body.
I G = moment of inertia about the mass center
α = acceleration about the mass center of the body.
Using equation (3.31) we get,
I G [ω (t ) − ω (0)] = r BG × P(t )
(3.32)
−1
ω (t ) − ω (0) = I G [r BG × P(t )]
(3.33)
Equation (3.33) gives the change in angular velocity at the mass center upon impact. Here
rBG is the position vector of B with respect to the mass center G and P(t ) is the impulse
acting on the body at the point of contact.
Q
B
rBG
A
R
G
P
rB
rG
Py
Px
S
Figure 3.2: Illustration of Impact
Using basic vector mechanics to express the velocities at the ends of beam with respect to
their mass centers we get,
∆v A = m −1 P(t ) + r AG (t ) × ω (t ) − r AG (0) × ω (0)
(3.34)
∆v A = m −1 P(t ) + r AG (t ) × [ω (t ) − ω (0)]
(3.35)
24
Similarly we get,
∆v B = m −1 P(t ) + r BG (t ) × [ω (t ) − ω (0)]
(3.36)
3.6 Routh’s method to find the impulse
There are several methods to calculate the impulse produced upon impact of a rigid
body. In this section we briefly describe Routh’s graphical procedure for two dimensional
collisions [9]. Routh’s method is a graphical technique for analyzing planar frictional
impact using Coulomb’s law of friction. According to this method an impulse is made of
two parts, a compression impulse, and a restitution impulse. The compression impulse or
the compression phase is from the beginning of collision to the time that their relative
velocity is zero. The restitution impulse or the restitution phase is measured from the time
the relative velocity is zero to the time that objects begin to separate. Accordingly
impulse P is expressed as:
P (t ) = Pt tˆ + Pn nˆ
(3.37)
In equation (3.37) subscripts t and n denote the tangential and normal directions
respectively, which are in the iˆ and ĵ directions in the given contact plane. This model
uses the Poisson’s hypothesis to define the coefficient of restitution. Accordingly the
coefficient of restitution denoted by ‘ e ’ is defined by:
e=
In equation (3.38)
(Pn )r
(Pn )r
(Pn )c
(3.38)
is the normal impulse during restitution and (Pn )c is the normal
impulse during compression. By using the Poisson’s hypothesis to define the coefficient
of restitution this method satisfies the basic energy conservation principles as there is no
25
increase in kinetic energy. The Routh-Poisson analysis gives an impulse in accordance
with Coulomb’s law, without an increase in total energy. Also it considers a new type of
impact called the tangential impact, an impact with zero initial approach velocity. It can
be used to distinguish between several types of contact and to identify when sliding
ceases or reverses. In this thesis the Routh’s method is used to find the impulse at the
contact point of the rigid body with the ground. Once the impulse is known the velocities
at the two ends of the beam are calculated. Finally the response of the beam is presented
knowing the velocities.
3.6.1 Procedure to calculate impulse
Once the object hits the ground an impulse P(t ) is produced. This impulse can be
divided into two components, an impulse in the normal direction Pn and an impulse in the
tangential direction Pt .Let the body with mass center ( x, y ) have initial translational and
rotational velocities x& 0 , y& 0 and θ&0 .Using linear and angular impulse-momentum laws we
write the following equations [9]:
m( x& − x& 0 ) = Pt
(3.39)
m( y& − y& 0 ) = Pn
(3.40)
mρ 2 (θ& − θ&0 ) = Pt y − Pn x
(3.41)
Here m is the mass, ρ is the radius of gyration of inertia and x& , y& are the velocities upon
collision. The kinetic energy is given by:
T=
1
1
m x& 2 + y& 2 + mρ 2θ& 2
2
2
(
)
26
(3.41a)
2
2
2
m  P
 P
  mρ
T =  t + x&0  +  n + y& 0   +
2  m
2
 m
 
 Pt y − Pn x & 
 mρ 2 + θ 0 


2
(3.41b)
Expanding and rearranging we get:
 P 2 P 2 (P y − Pn x)
m 2
mρ 2 & 2 
2
θ0  + Pt x&0 + Pn y&0 + (Pt y − Pn x)θ&0 +  t + n + t
T =  x&0 + y&0 +
 (3.41c)
2
2mρ 2 
 2m 2m

2
(
)
[
]
The isoenergetic ellipse satisfies the law of conservation of energy. Hence for this
condition to be satisfied Pt and Pn assume values such that:
[P x&
t
0
 P 2 P 2 (P y − Pn x ) 
+ Pn y& 0 + (Pt y − Pn x )θ&0 +  t + n + t
 = 0.
2mρ 2 
 2m 2m
]
G
rG
rG = xiˆ + yˆj
Figure 3.3: Position of Mass center
The velocity of the object when it comes in contact with the ground is called contact
velocity and is given by:
x& c = x& + θ&y
(3.42)
y& c = y& − θ&x
(3.43)
27
B
A
G ( x, y )
(x& c , y& c )
Figure 3.4: Contact velocities
Equation (3.42) denotes the tangential component of the relative velocity of the points in
contact. This velocity represented by ‘S’ is called the sliding velocity. Similarly equation
(3.43) denotes the normal component of relative velocity and is denoted by ‘C’. In other
words S = x& c and C = y& c .Substituting equations (3.39)-(3.41) into (3.42) and (3.43) we
get:
S = S 0 + B1 P t − B3 P n
(3.44)
C = C 0 − B3 P t + B2 P n
(3.45)
Equations (3.44) and (3.45) represent the line of sticking and line of maximum
compression respectively. S 0 and C 0 represent the initial values of sliding and compression
respectively. These depend on the values of the tangential and normal component of the
initial velocity and the initial rotational velocity. They are expressed as:
S 0 = x& 0 +θ& 0y
(3.46)
C 0 = y& 0 −θ& 0x
(3.47)
B1 , B2 and B3 are constants that depend on the geometry and mass properties of the
system. They are given by:
28
1
y2
B1 = +
m mρ 2
(3.48)
x2
1
+
m mρ 2
(3.49)
xy
mρ 2
(3.50)
B2 =
B3 =
The Routh method determines the two components of impulse Pt and Pn using equations
(3.44)-(3.50).This is done graphically and explained in detail in the next section.
3.6.2 The impact process diagram:
When an object deforms, it happens in two phases, compression and restitution.
When compression ends, the normal component of the relative velocity of the points in
contact is zero (C = 0).Therefore from equation (3.45) a linear relationship between the
impulse components at maximum compression is obtained as
C 0 − B3 Pt + B2 Pn = 0
(3.51)
Similarly in the sticking case, the tangential component of relative velocity becomes zero
(S = 0). Therefore a linear relationship between the impulse components at slip is
obtained as
S 0 + B1 Pt − B3 Pn = 0
(3.52)
Now, equations (3.51) and (3.52) represent the line of maximum compression and line of
no sliding respectively. These two lines represent straight lines since they are linear and
are plotted on the impulse plane with Pt representing the horizontal axis and Pn the
vertical axis. Therefore using equations (3.51) and (3.52) we get:
29
Pt =
C
B2
Pn + 0
B3
B3
(3.53)
Pt =
B3
S
Pn − 0
B1
B1
(3.54)
We also define the line of limiting friction as
Pt = − µsPn
(3.55)
In equation (3.55) µ is the coefficient of friction and is a constant and s is the sign of the
initial sliding velocity S 0 , s =
S0
if S 0 is not zero. Figures 3.5 and 3.6 give a detailed
S0
explanation of the impact process diagram.
Pn
Friction
Compression
Stick
Pn (t f
)
Pn (t c )
Pt
Figure 3.5: Impact process diagram illustrating Slip-Stick
When impact begins the point P which is the total impulse is at the origin and lies on the
line of limiting friction. Assuming initial sliding it increases along the line of limiting
friction. It proceeds along this line until it reaches line sticking or line of maximum
compression. If it reaches the line of maximum compression first, the value of normal
30
impulse Pn (t c )) at that instant is noted. The process of impact will end when the value of
Pn is 1+e times the value of ( Pn (t c )) .This value is denoted by Pn (t f
)
Pn (t f ) = (1 + e )Pn (t c )
(3.56)
This is called the termination condition .Then the point continues along the line of
limiting friction until termination is met or P reaches the line of sticking. If P reaches
the line of sticking before termination it continues along this line till termination. This
process of slip-stick is illustrated in Figure 3.5.
Alternatively if it reaches the line of stick first then slipping ends and if the limiting
friction is more than the friction necessary to prevent sliding P will continue to stick till
the process terminates as illustrated in figure 3.6.It will eventually cross the line of
maximum compression and terminate according to the termination condition as stated
earlier. On the other hand if the limiting friction is less than the friction necessary to
prevent sliding P will cross the line of sticking and will travel along the line of reversed
limiting friction and will continue till termination. In such a scenario the object will slip
through out the impact process till termination condition is reached.
31
Friction
Compression
Stick
Pn
Pt
Figure 3.6: Impact process diagram illustrating Stick
The isoenergetic ellipse depicted in the above figures should satisfy the energy
conservation principles.Acccordingly if ∆T is the change in kinetic energy then
according to the law of conservation of energy ∆T = 0 .
∆T =
1 T
P (VC + VCO )
2
[
]
(3.56a)
Where P = [Pt , Pn ]
T
VC = [S , C ] and VC 0 = [S 0 , C 0 ]
T
T
The energy change is given by:
∆T =
 B
Where B =  1
− B3
1 T
T
P BP + 2VC 0 P
2
[
− B3 
B2 
The energy ellipse requires that ∆T = 0 .
32
]
(3.56b)
3.7 Galerkin Reduction
Galerkin’s method is a method used to find an approximate solution to continuous
systems. Often exact solutions for higher order equations do not exist. Even if they do
they are cumbersome to use, requiring solutions to higher order equations. In such a
scenario Galerkin’s method is a convenient method to use although it yields only an
approximate solution. It is a means for converting a partial differential equation (PDE) to
a system of ordinary differential equations (ODEs), which become easier to handle. It
works on the principle of restricting the possible solutions to a smaller space than the
original. These smaller systems are easier to solve and less time consuming.
Galerkin’s method approximates the solution to a boundary value problem by using a
linear combination of trial functions. In order to solve our problem we choose a trial
functions that satisfy the boundary conditions.
3.7.1 Galerkin’s method as applied to the beam model.
Assume the approximate solution to the problem to be:
N
u ( x, t ) = ∑ Ai (t )φi ( x )
(3.57)
i =1
Where the trial functions φ1 ( x ), φ 2 ( x ),...φ n (x ) are the independent comparison functions
from a complete set and A1 , A2 ,... An are undetermined coefficients. Comparison functions
are trial functions which are differentiable as many times as the order of the system and
satisfy all the boundary conditions. The above solution may not satisfy the exact
differential equation defining the eigenvalue problem, so that some error is incurred. The
error is denoted by R (u (n ) , x ) , known as the residual, and because u (n ) is a linear
combination of comparison functions, the boundary conditions are satisfied exactly. To
33
we multiply the residual R (u (n ) , x ) by
A1 , A2 ,.. An
determine the coefficients
φ1 ( x ), φ 2 ( x ),..., φ n (x ) in sequence, integrate the result over the domain of the system, and
set equal to zero.
l
∫ φ (x)R(u
i
(n )
)
, x dx = 0, i = 1, 2……..n
(3.58)
0
N
Now consider equation (3.26), substituting for u ( x, t ) ≈ u ( N ) ( x, t ) = ∑ Ai (t )φi ( x) where
i =1
Ai (t ) is the ith generalized coordinate and φi ( x) is the ith linear undamped mode shape
of the straight beam we get:
∂4  N
∂2  N
 ∂2  N


α 4 ∑φi (x)Ai (t) − 2 ∑φi (x)Ai (t) + β 2 ∑φi (x)Ai (t) +[∆vA + (∆vB − ∆vA)x]δ (t) = R(u(n) , x) (3.59)
∂x  i=1
 ∂x  i=1
 ∂t  i=1

Equation (3.59) represents the residual. It has been shown in previous studies [13] that
four modes are sufficient to capture the dynamic response of a beam pinned at both ends.
Therefore in equation (3.27) substituting N = 4 we get:
u ( N ) ( x, t ) = φ1 ( x) A1 (t ) + φ 2 ( x) A2 (t ) + φ3 ( x) A3 (t ) + φ 4 ( x) A4 (t )
(3.60)
We choose the first four mode shapes to be:
φ1 ( x) = sin
πx
l
, φ 2 ( x) = sin
2πx
3πx
4πx
, φ3 ( x) = sin
, φ 4 ( x) = sin
l
l
l
Therefore equation (3.60) becomes:
u ( x, t ) ≈ u ( N ) ( x, t ) = sin
πx
l
A1 (t ) + sin
2πx
3πx
4πx
( x) A2 (t ) + sin
( x) A3 (t ) + sin
A4 (t ) (3.61)
l
l
l
34
Equation (3.61) represents the total deflection of the beam. To find the generalized
coordinates in equation (3.61) we substitute φi ( x) = sin kπx . Using values of k = 1 to 4 we
get four sets of equations from which A1 , A2 , A3 and A4 are determined. Therefore we get:
N

(
)
k
x
R
sin
π
∑ Ai (t )φi ( x), x  dx = 0
∫0
 i =1

(3.62)
β &&
 − ∆v B ∆v A 
−
δ (t )
A1 (t ) + [0.5απ 4 + 0.5π 2 ]A1 (t ) = 
2
π 
 π
(3.63)
l
For N = 4, k = 1 we get:
Similarly for k = 2, 3 and 4 we get:
∆v 
β &&
 ∆v
A2 (t ) + [8απ 4 + 2π 2 ]A2 (t ) =  B − A δ (t )
2
π 
 2π
(3.64)
β &&
 − ∆v B ∆v A 
A3 (t ) + [40.5απ 4 + 4.5π 2 ]A3 (t ) = 
δ (t )
−
2
3π 
 3π
(3.65)
∆v 
β &&
 ∆v
A4 (t ) + [128απ 4 + 8π 2 ]A4 (t ) =  B − A δ (t )
2
4π 
 4π
(3.66)
Equations (3.63)-(3.66) represent the first four mode shape equations. δ (t ) is a unit
impulse due to an instantaneous collision with the ground. For an undamped system of
the form &x& + ω 2 n x = δ (t ) , the response to a unit impulse is denoted by h (t) and is given
by:
h(t ) =
1
ωn
sin ω n t
(3.67)
In the above equations ∆v A and ∆v B are the changes in velocities of the beam upon
impact.
35
3.8 Stresses and strains
If M (x) is the bending moment and y is coordinate measured from the neutral axis in the
cross section of the beam, using elementary beam theory the normal stress at a point
in the cross-section due to bending is expressed as
σ=
M ( x) y
I
M ( x) = EI
∂ 2u
∂x 2
(3.68)
(3.69)
Equations (3.68) and (3.69) are used to get the stress response of the beam.
Potential energy in the form of strain energy is stored in all deformable systems. For a
loaded member the normal stress (σ ) and normal strain (ε ) follow the Hooke’s law. The
stress -strain curve is linear. Therefore we have
σ
= E .Using this we find the maximum
ε
strain.
3.9 Review
The previous sections dealt with calculating the value of the impulse produced
upon drop. Once the impulse is known it is substituted back in equations (3.35 and (3.36)
to get the values of changes in velocities at ‘A’ and ‘B’ respectively. Before this the
geometric dimensions of the problem including the angle of impact is used to calculate
the position vectors of the mass center of the body and the ends of the beam. Once the
change in velocities are known these are substituted back in the mode shape equations
(3.63-3.66).This gives the values of the constants which are then substituted back in the
response on the beam i.e. equation (3.60) to get the deflection of the beam. Upon
36
knowing the response of the beam upon impact, the locations of maximum stress, strain,
deflection can be got.
37
CHAPTER 1V
THE DYNAMICS OF IMPACT AND IMPACT VELOCITIES
4.1: Overview
In this chapter we apply the basic equations for the dynamics of impact in Chapter
3 to a specific impact problem. We first find the impulse developed because of impact
and hence the change in velocity due to impact. A final solution is then presented using
the Galerkin’s approach. Finally the stress, strain and amplitude responses illustrating
their maximum values are presented.
4.2 An impact problem
Consider a rigid frame (PQRS) of mass m = 2 kg, width W = 0.5m, height H =
0.3m and ρ = 1 with a flexible beam ‘AB’ of length 0.25m falling from a height and
hitting the ground at θ = 45 o . Assume that the angular velocity ω of the block is zero.
Q
ê2
B
ê1
rBG
A
G
rBS
P
rAS
ĵ
iˆ
rGS
Py
S
Px
Figure 4.1: An Impact Problem
38
R
Let iˆ and ĵ be unit vectors of the coordinate system fixed to the ground. Let ê1 and
ê2 be unit vectors of the coordinate system of the motion of the frame relative to the
ground. Since there is both translation and rotation of the frame relative to the ground
their unit vectors are expressed as:
eˆ1 = cos θiˆ + sin θˆj
(4.1)
eˆ2 = − sin θiˆ + cos θˆj
(4.2)
Let r AS , r BS and r GS be the position vectors of points ‘A’, ‘B’, and ‘G’ relative to the
ground and r AG , r BG be the position vectors of ‘A’ relative to ‘G’ and ‘B’ relative to ‘G’
respectively. Using equations (4.1) and (4.2) and the dimensions of the frame and beam
we get the following equations for figure 4.1
rGS = 0.289eˆ1 + 0.0005eˆ2
(4.3)
rAS = 0eˆ1 − 0.398eˆ2
(4.4)
rBS = 0.469eˆ1 − 0.029 × 10 −3 eˆ2
(4.5)
Also using basic vector mechanics we get:
rAS = rGS + rAG
(4.6)
rBS = rGS + rBG
(4.7)
In equations (4.6) and (4.7) rAG and rBG are the position vectors of ‘A’ relative to ‘G’ and
‘B’ relative to ‘G’ respectively. From equations (4.6) and (4.7) we get:
rAG = −0.289eˆ1 − 0.398eˆ2
(4.8)
rBG = 0.180eˆ1 − 5.29 × 10 −4 eˆ2
(4.9)
39
The frame makes contact with the ground at an angle θ = 45 o .This produces an impulse
at the contact point which affects the velocities at the ends of the beam. We use linear and
angular momentum balance equations to express the relation between the change in
velocity and the impulse produced. Equations (3.35) and (3.36) give the relationship
between the change in velocities and the impulse. We now proceed to find the impulse in
order to calculate the change in velocities.
4.3. Routh’s graphical method
We use Routh’s method for two dimensional impacts to find the value of the
impact P(t ) at the point of contact. Using equation (4.3), we find the mass center(x, y) of
the frame to be (0.289, 0). Considering the geometry of our problem and using equations
(3.48) – (3.50) we get:
B1 =
1 0.0005 2
+
= 0.50
2
2(1) 2
B2 =
B3 =
1 0.289 2
+
= 0.541
2 2(1) 2
0.289 × 0.0005
= 0.07 × 10 −3
2
2(1)
(4.10)
(4.11)
(4.12)
At time t = 0 there is an initial tangential and normal component of velocity that is
assumed. These are denoted by S 0 and C 0 respectively. The angular velocity is zero,
therefore the initial values of S 0 and C 0 are given by the initial translational velocities
x& 0 and y& 0 .They are also called the initial stick and compression velocities and their
values are assumed to be 1 and -1 respectively. Using equations (3.53) and (3.54), values
of B1 , B2 , B3 and the initial stick and compression velocities we plot the line of stick and
40
line of maximum compression on an impulse plane with Pt representing the horizontal
axis and Pn the vertical axis. The coefficient of friction µ is assumed to be 0.6. Figure
4.2 illustrates the impact process.
Friction
Stick
6
5
P(n)
4
3
Impulse Path
2
1
Line of Compression
0
-5
-3
-1
1
3
5
P(t)
Figure 4.2: Impact process diagram.
When impact begins P is at the origin, as it progresses Pn increases as it begins to slip
along the line of friction , Pt also accumulates according to the relationship between the
two as expressed in equations (3.53) and (3.54).When Pn reaches the line of maximum
compression the value of the normal impulse is noted. Termination occurs when the value
of Pn reaches 1 + e times the value of Pn obtained at maximum compression. In this case
impact terminates when Pn =2.55.The object continues to slip along the line of limiting
friction till termination is reached.
41
Upon termination the values of Pt and Pn are found to be -1.8 and 2.55 respectively.
Therefore the total impact is given by:
P = −1.8iˆ + 2.55 ˆj
(4.13)
P = 0.53eˆ1 + 3.07eˆ2
(4.14)
4.4 Using the impulse to find the change in velocities
The change in angular velocity at the mass center is given by equation (3.33).
Using this we get:
ω (t ) − ω (0) = 0.96kˆ
(4.15)
Equations (3.35) and (3.36) give the change in velocities at ‘A’ and ‘B’. Using these
equations we get:
∆v A = 0.09iˆ + 2.13 ˆj
(4.16)
∆v B = 0.45iˆ + 1.11 ˆj
(4.17)
Since we consider the beam to be pinned at both the ends there is no movement in the
tangential direction and so in further calculations only the normal component of the
velocities are considered. Using equations (4.1) and (4.2) we get the tangential and
normal components of the velocities of the beam relative to the ground to be:
∆v A = 1.44eˆ1 − 1.56eˆ2
(4.18)
∆v B = 0.97eˆ1 − 0.69eˆ2
(4.19)
where ê1 and ê2 are unit vectors in the tangential and normal directions respectively.
42
4.5 Mode shape equations
Consider the first mode as given by equation (3.63).Assuming the values of α and
β to be equal to 1 we get:
&& (t ) + 107.2 A (t ) =  − ∆v B − ∆v A δ (t )
A
1
1
 π
π 

(4.20)
Now substituting the value of the normal component of the change in velocities at ‘A’
and ‘B’ we get:
&& (t ) + 107.2 A (t ) = 0.71δ (t )
A
1
1
(4.21)
Equation 4.21 represents the first mode. Similarly for the second, third and fourth modes
we get:
&& (t ) + 1598.02 A (t ) = 0.38δ (t )
A
2
2
(4.22)
&& (t ) + 7929.6 A (t ) = 0.56δ (t )
A
3
3
(4.23)
&& (t ) + 25094.6 A (t ) = 0.06δ (t )
A
4
4
(4.24)
Equation (4.21) represents a second order linear differential equation with f 0δ (t ) being
the forcing function where δ (t ) represents a unit impulse. The standard form of the above
equation for an undamped system is written as:
{&x&(t ) + ω 2 n x(t )} = P(t )
Where ω 2 n =
k
F (t )
and P(t ) =
m
m
ω n = natural frequency and P(t ) is the forcing function.
The solution to equation (4.20) is given by:
43
(4.25)
A1 (t ) = C1 f 1 (t ) + C 2 f 2 (t )
(4.26)
f 1 (t ) and f 2 (t ) represent the solutions to a homogeneous problem, C1 and C 2 represent
constants that are determined using initial conditions. Considering sin ωt and cos ωt to be
the solutions to the homogeneous problem we get:
x(t ) = C1 sin ω n t + C 2 cos ω n t
(4.27)
In Equation (4.27), C1 and C 2 are constants that are found using the initial conditions of
displacement and velocity. Equation (4.20) is solved in a similar manner by treating it as
an initial value problem with A1 (t ) being the generalized coordinate representing
displacement. Therefore the solution to equation (4.20) is written as:
A1 (t ) = C1 sin 10.35t + C 2 cos10.35t
(4.28)
The initial displacement is taken to be zero. The principle of linear impulse and
momentum is used to determine the velocity upon impact.
Upon impact there is change in momentum of the body. This change in momentum is
the impulse that is produced upon contact with the ground. If 0 + represents the time after
impact the impulse is mathematically expressed as
A&1 (0+ ) − A&1 (0− ) = f 0
(4.29)
The change in velocity before impact is zero, therefore Equation (4.29) becomes
A&1 (0+ ) = f 0
(4.30)
Therefore using initial displacement as zero, i.e. A1 (0) = 0 in Equation (4.28) we get
C 2 = 0 and using Equation (4.30) as the initial velocity we get C1 = −0.005 .Therefore
Equation (4.28) becomes:
44
A1 (t ) = 0.06 sin 10.35t
(4.31)
A2 (t ) = 0.009 sin 39.97t
(4.32)
A3 (t ) = 0.006 sin 89.04t
(4.33)
A4 (t ) = 0.0003 sin 158.4t
(4.34)
Similarly we get,
Finally using the above calculated values and the assumed mode shapes in equation
(3.60) we get the total response of the beam to be:
u ( x, t ) = sin πx[0.06 sin 10.35t ] + sin 2πx[0.009 sin 39.97t ]
+ sin 3πx[0.006 sin 89.04t ] + sin 4πx[0.0003 sin 158.4t ]
(4.35)
The stress response is given by:
σ ( x, t ) = − sin πx[6.21 × 10 9 sin 10.35t ] − sin 2πx[3.73 × 10 9 sin 39.97t ]
[
]
[
− sin 3πx 5.59 × 10 9 sin 89.04t − sin 4πx 0.49 × 10 9 sin 158.4t
]
(4.36)
The strain response is given by:
ε ( x, t ) = − sin πx[0.02 sin 10.35t ] − sin 2πx[0.01 sin 39.97t ]
− sin 3πx[0.02 sin 89.04t ] − sin 4πx[0.002 sin 158.4t ]
(4.37)
4.6 Results
Equation (4.35) represents the displacement of the beam upon impact. A time
interval of 0 to 5 seconds is considered and the response of the beam within this interval
is studied to find the maximum amplitude of displacement and its location. A MATLAB
script is written to first find the time at which the maximum deflection occurs. The
response of the beam at this time is plotted to find the value of maximum deflection and
its location. This plot is illustrated in figure 4.3. Similarly maximum stresses and strains
are tabulated and illustrated in figures 4.4 and 4.5 respectively.
45
0.06
0.05
Deflection
0.04
0.03
0.02
0.01
0
0
0.05
0.1
0.15
Location along length of beam (x)
0.2
0.25
Figure 4.3: Maximum Deflection
In figure 4.3 it is seen that the amplitude of maximum deflection is 0.05m at x = 0.25m
at time t = 3.18 seconds.
9
12
x 10
10
Stress
8
6
4
2
0
0
0.05
0.1
0.15
Location along length of beam (x)
Figure 4.4: Maximum Stress
46
0.2
0.25
The maximum stress developed is 11927.2 Mpa at x = 0.21m at time t = 2.31 seconds.
0.04
0.035
0.03
Strain
0.025
0.02
0.015
0.01
0.005
0
0
0.05
0.1
0.15
Location along length of beam (x)
0.2
0.25
Figure 4.5: Maximum Strain
The maximum strain developed is 0.03 at x = 0.21m at time t = 2.31seconds.
47
CHAPTER V
RESULTS AND CONCLUSION
5.1 Overview
In this chapter we discuss about how the angle of approach affects the impulse
produced at the point of contact and the velocities at the end of the beam (boundary
conditions). The effect of the coefficient of friction (µ ) and the coefficient of
restitution (e ) on the normal and tangential impulses are explained. The response of the
beam fixed at different positions on the frame is studied to see the location of maximum
stress and strain developed in the beam. The influence of the boundary conditions on the
stresses and strains developed for different angles of impact is discussed.
5.2 Angle of approach v/s impact:
The impact produced upon contact of the rigid frame with the ground depends upon
the material properties of the impacting body and the angle at which it hits the ground.
Observation of the line of sticking and line of maximum compression indicates the
dependency of impact on the geometric constants B1 , B2 and B3 .
S = S 0 + B1 P x − B3 P y
(5.1)
C = C 0 − B3 P x + B2 P y
(5.2)
These values will change with change in the coordinates of the impacting body which
would in turn change with the angle at which the body makes contact with the ground.
48
In this section we see the change in impact force with the change in angle of approach
and draw suitable conclusions. The coefficient of friction µ remains constant at 0.8 for
all angles of impact with initial conditions of S 0 = 1 , C 0 = −1 and e = 0.8 . The values
of B1 , B2 and B3 are calculated for the different angles of approach with the other
geometric constants being mass m = 2 kg, width W = 0.5m, height H = 0.3m and ρ = 1
with a flexible beam ‘AB’ of length 0.25m
Magnitude of Impulse (P)
5
4.5
4
3.5
3
2.5
2
5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90
Angle of Impact
Figure 5.1: Angle of Impact v/s Magnitude of impact
Looking at the above figure we find that the impulse starts at a value of 2.95 at θ = 5 o
and increases to a maximum of 4.61 at θ = 15 o .Thereafter it varies continuously till it
reaches a value of 2.95 at θ = 90 o . As stated earlier impulse produced depends on the
material properties of the impacting body in addition to the geometric constraints. Figures
5.2, 5.3 and 5.4 illustrate the impulse paths at a few representative points. These give a
more detailed view of the change in impulse with the orientation angle.
49
Friction
P(n)
5.5
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
-5
-3
Stick
Impulse Path
Line of Compression
-1
1
3
5
P(t)
Figure 5.2: Impulse path at θ = 15 o
Friction
Stick
6
5
P(n)
4
3
Impulse Path
2
Line of Compression
1
0
-5
-3
-1
1
3
5
P(t)
Figure 5.3: Impulse path at θ = 45 o
Friction
Stick
5.5
5
4.5
Impulse Path
4
3.5
3
2.5
Line of Compression
2
1.5
1
0.5
0
-5
-3
-1
1
3
Figure 5.4: Impulse path at θ = 75 o
50
5
In the next section we see the effect of coefficient of friction (µ ) and coefficient of
restitution (e ) on the normal and tangential impulses.
5.3. Influence of (µ ) on the impulse at contact point.
Consider the collision problem of chapter 4.The coefficient of friction (µ ) causes
an impulsive force in the tangential direction at the point of contact. This section deals
with the change in tangential impulse (Pt ) with µ for the above mentioned collision
problem.
2.5
Tangential impulse
2
1.5
1
0.5
0
-0.5
0.2 0.4
0.6
0.8
1
1.2 1.4
1.6 1.8
2
2.2 2.4
-1
-1.5
-2
-2.5
Co efficient of friction
Figure 5.5: µ v/s Pt
In the above figure it seen that the value of the tangential impulse starts at -0.5
for µ = 0.2 and then increases in the negative direction as µ increases to 0.8 before being
steady for values of µ between 1 and 1.4. For lower values of µ the impact process
terminates before the object reaches the line of stick. Therefore the object slips
throughout the process of impact. For µ varying between 1 and 1.4 the line of friction
meets the line of stick before termination because of this Pt remains constant at -2.For
51
values of µ greater that 1.4 the line of friction meets the line of stick before the line of
maximum compression. This leads to reverse slip. Since the slope of the line of stick is
steeper than the line of reverse limiting friction it continues to stick till termination. As
the value of µ continues to increase the value of tangential impulse will remain constant
at 2, but the object would start to stick sooner. The below figures show the impulse paths
followed for few representative values of friction. This explains the variation in
tangential impulse more clearly.
f=0.2
Stick
6
5
P(n)
4
Impulse Path
3
2
Line of Compression
1
0
-5
-3
-1
1
3
5
P(t)
Figure 5.6: Impulse path for µ = 0.2
Friction
Stick
6
5
P(n)
4
3
Impulse Path
2
1
Line of Compression
0
-5
-3
-1
1
3
P(t)
Figure 5.7: Impulse path for µ = 0.8
52
5
f=1
Stick
6
5
P(n)
4
Impulse Path
3
2
Line of Compression
1
0
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
P(t)
Figure 5.8: Impulse path for µ = 1
f=2
Stick
6
5
P(n)
4
Impulse Path
3
Reverse Friction
2
Line of Compression
1
0
-5
-3
-1
1
3
5
P(t)
Figure 5.9: Impulse path for µ = 2
5.4. Influence of the coefficient of restitution (e) on the impulse.
The coefficient of restitution denoted by ‘e’ is the ratio of speeds of a falling
object from when it hits the surface to when it leaves the surface. It is a measure of the
elasticity of a collision. A perfectly elastic collision has e equal to 1 while a perfectly
plastic collision (object sticks immediately upon impact) has e equal to zero. In the figure
53
below we look at how the magnitude of impulse varies for values of e between 0 and
1.Collision happens at an angle of θ = 45 o .The collision problem is dealt with in detail in
Chapter IV. It is seen from the figure that the magnitude of impulse increases with
increase in e. The figure shows that perfectly elastic collisions produce the highest
impulse. A perfectly plastic collision will produce the least impulse.
Magnitude of Impulse(P)
4
3.5
3
2.5
2
1.5
1
0.5
0
0
0.2
0.4
0.6
0.8
1
1.2
Coefficient of restitution(e)
Figure 5.10: e v/s P
5.5 Influence of the angle of impact on the beam velocity.
The velocity at the ends of the beam (boundary conditions) will change with the
change in impulse and also with the change in position of the object which in turn will
change with the angle of approach. Recalling equations (3.42) and (3.43) we have:
v A (t ) − v A (0) = m −1 P(t ) + r AG (t ) × [ω (t ) − ω (0)]
(5.3)
v B (t ) − v B (0) = m −1 P(t ) + r BG (t ) × [ω (t ) − ω (0)]
(5.4)
54
C hange in m agnitude of velocity
at 'A'
4
3.5
3
2.5
2
1.5
1
0.5
0
15
30
45
60
75
90
Angle of Impact in degrees
Change in magnitude of velocity at
'B'
Figure 5.11: Angle of impact v/s Change in magnitude of velocity at end ‘A’
3.5
3
2.5
2
1.5
1
0.5
0
15
30
45
60
75
90
Angle of Impact in degrees
Figure 5.12: Angle of impact v/s Change in magnitude of velocity at end ‘B’
Figures 5.5 and 5.6 show the change in magnitude of velocity at the ends of the beam
with the angle of impact. While there is little variation of magnitude of change in velocity
at ‘A’ till θ = 60 o , there is a sudden jump at θ = 60 o .Similarly at end ‘B’ there is a
sudden drop in velocity at θ = 45 o jump at θ = 75 o
could have an
These sudden changes in velocity
influence on the stress, strain and deflection of the beam. This is
explained in the following section.
55
5.6 Influence of the angle of impact on the deflection, stress and strain
Figures 5.13-5.15 illustrate the continuous variation of the deflection, stress and
strain with the angle of impact. Also the influence of the sudden changes in the
magnitude of velocities at the ends of the beam is investigated. It is seen from the figures
that while there is continuous variation in the deflection, stress, strain with the angle of
impact, there are sudden significant jumps. These could be attributed to the sudden
changes in velocities at the ends of the beam.
0.12
0.1
0.08
0.06
0.04
0.02
0
15
30
45
60
75
90
Angle of Impact in degrees
Figure 5.13: Angle of impact v/s Deflection
20000
Stress in Mpa
Deflection in meters
0.14
15000
10000
5000
0
15
30
45
60
Angle of impact in degrees
Figure 5.14: Angle of impact v/s Stress
56
75
90
0.08
0.07
Strain
0.06
0.05
0.04
0.03
0.02
0.01
0
15
30
45
60
75
90
Angle of Impact in degrees
Figure 5.15: Angle of Impact v/s Strain
5.7 Influence of location of the beam on the boundary conditions
In this section we look at the change in velocity at the ends of the beam with the
change in location of the beam on the rigid frame. The following figures show the
different positions at which the beam is attached to the frame.
B
Q
B
R
A
G
P
B
Q
Q
A
A
R
R
G
P
S
S
(i)
(ii)
G
P
S
(iii)
Figure 5.16: Frame with beam attached at different locations
Let the frame ‘PQRS’ impact the ground at θ = 45 o . ‘S’ is the point of contact. The
following graphs show the change in velocity at ‘A’ and ‘B’ as their positions change
from (i) to (iii).
57
Change in magnitude of velocity
at 'A'
2.5
2
1.5
1
0.5
0
(i)
(ii)
(iii)
Location of beam
Magnitude of change in velocity at
'B'
Figure 5.17: Location of beam v/s Change in magnitude of velocity at ‘A’
2.5
2
1.5
1
0.5
0
(i)
(ii)
(iii)
Location of beam
Figure 5.18: Location of beam v/s Change in magnitude of velocity at ‘B’
It is seen from the above figures that the velocities decrease as the location of the
beam changes from figure (i) to (iii) with the impact configuration shown in (i) having
the highest velocity and, (iii) the lowest velocity. In (i) the points ‘A’ and ‘B’ are closest
to the impact point ‘S’. As a result the velocities are the highest for this configuration and
as they move away from the point of contact their velocities decrease. Thus it can be
concluded that the velocities upon impact depend upon the proximity of the point under
consideration to the point of impact.
58
5.8. Change in stress and strain with change in position of beam
It is seen from preceding sections that a sudden change in velocity has an influence
on the response of the beam. In figures 5.11 and 5.12 there is a sudden drop in velocity
from (i) to (ii).As a result of this there is a sudden change or drop in the values of stress
and strain from (i) to (ii).This is shown in figures 5.13 and 5.14.
Stress in Mpa
12000
10000
8000
6000
4000
2000
0
(i)
(ii)
(iii)
Location of beam
Figure 5.19: Stress v/s Location of beam
0.05
Strain
0.04
0.03
0.02
0.01
0
(i)
(ii)
Location of beam
Figure 5.20: Strain v/s Location of beam
59
(iii)
5.9 Conclusion
A simplified differential model of a flexible beam attached to a rigid supporting
block has been developed and its response to an impulse studied. We have used Routh’s
graphical method for two dimensional collisions for an accurate description of the
impulse developed upon impact. The impulse developed for different angles of approach
is calculated. It can be concluded that the impulse produced at the contact point is
independent of the angle at which it strikes the surface. It could depend more on the
material properties of the impacting material such as friction and co- efficient of
restitution (e). For lower values of µ the impact process terminates before the object
reaches the line of stick. For µ varying between 0.8 and 1.4 the line of friction meets the
line of stick before termination, because of this Pt remains constant. For values of µ
greater that 1.4 the line of friction meets the line of stick before the line of maximum
compression. And µ increases the object will begin to stick immediately after impact
and will continue to stick till termination is reached. We also find upon varying e that a
perfectly elastic collision will produce the highest impulse and a perfectly plastic
collision produces the least impulse. We have studied the change in boundary conditions
with change in the angle of impact. It can be concluded that the deflection, stress and
strain developed in the beam depends on the boundary conditions. There is a sudden
jump or drop in these values with corresponding jumps or drops in the velocities. We
have tried to find the influence of the location of the beam on the frame on the velocities.
There is a definite increase in the change in magnitude of velocities for the beam located
at the ends of the block. The effect of the impact force on the beam is more when it is
closest to the impact point. As the beam moves further away from the point of impact the
60
velocities also decrease. The maximum stresses and strains are developed with the beam
located at the corners of the rigid block. In our case considering the frame to be the
housing of an electronic component and the beam to be an internal component it can be
said that the internal components are under maximum stress when placed at the corners of
the housing. This could serve as a pointer in placing the smaller more flexible
components in rigid housings of cellular phones or laptops. Strategic placing of these
components could increase their time to failure and improve performance. It could serve
as a guide for production units before the actual product is manufactured eliminating the
need for a trial and error method in determining failure thereby saving production costs.
In general it can be concluded that there can be several geometric as well as material
parameters or variables that can affect the damage caused to the component upon impact
and knowing some of the critical factors could lead to a better design.
5.10 Underlying simplifications
Impact is an important event in a lot of dynamic mechanical applications and can be
described in variety of ways. It could be treated as a continuum that would make the
process too complex, time consuming and difficult to analyze. However the current topic
simplifies the impact event by a simple differential collision model with the help of
ordinary differential equations. This model predicts the post collision change in velocity
given the pre collision velocity and impact. Using a simple but accurate description of the
impact force helps us understand the influence of the geometric and material parameters
that could affect the post collision state of the component under consideration.
61
5.11 Recommendations for future work.
The results of the current model could be compared against the results got by a simulated
finite element model or against actual physical tests in the laboratory. A different
approach to finding the solution to the beam equation could be attempted and the results
compared. Also a different model of the impact force could give a more accurate
description of the impact event. The current study focuses in single impacts and does not
consider clattering. A new model considering the effects of clattering could be attempted.
Finally the model described could be applied to real world applications and the results
could be verified.
62
REFERENCES
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modeling of electronic packages subjected to drop impact”, IEEE transactions on
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response of a portable electronic product”, IEEE transactions on Components and
Packaging Technologies, Vol.25, No.3, pp.478-485, September 2002.
(3) Raymon Ju, Brain Hsiao, “Drop simulation for portable electronic products”, 8th
International LS-DYNA Users Conference, May 2-4, 2004
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(5) Y.Y.Wang, C.Lu, J.Li, X.M.Tan, Y.C.Tse, “Simulation of drop/impact reliability for
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