Dynamics · PDF fileHibbeler, R.C., Mechanics of ... Normal and Tangential Components...
Transcript of Dynamics · PDF fileHibbeler, R.C., Mechanics of ... Normal and Tangential Components...
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Dynamics
Monday
Tuesday
Wednesday
Thursday 10:00 - 11:00 5D-43
7. Faculty Name: Dr. Yavuz YARDIM8. Office No.: 5D-439. Tele Ext. : 98410. E-mail: [email protected]. Time for office hours : Day Time
Sunday 12:00-12:502A-1
Tuesday 12:00-12:502A-1
Dynamics
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Text Book & References :Text Book References
1. Hibbeler, R.C.,Mechanics of Materials (4th Edition).
Riley, W.F., Mechanics of Materials (6th Edition), John Wiley & Sons.
1. Gere, J.M., Timoshenko, S.P. and Ismail, A.,Mechanics ofMaterials, Nelson Thornes, Cheltenham, U.K., 2003.
2. Hibbeler, R.C.,Mechanics of Materials (4th Edition), PrenticeHall, New Jersey, 2004.
3. Megson, T.H.G.,Structural and Stress Analysis,Butterworth-Heinemann, London, 2005.
4. Morrow, H.W. and Kokernak, R.P.,Statics and Strength ofMaterials (5th Edition),Prentice Hall, New Jersey, 2004.
5. Riley, W.F., Mechanics of Materials (6th Edition), JohnWiley & Sons, New York, 2006.
Learning Outcomes
1. analyze of dependent motion of two particles.2. investigate accelerated motion of a particle using the equation of
motion.3. develop and apply the principle of work and energy that involve
force, velocity, and displacement.4. analyze rigid-body motion about a fixed axis.5. study the analysis of undamped forced vibration and viscous
damped forced vibration.
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Week No. Title
Course
Outcome
Assessment
of
Outcomes
1.
Introduction to General
Principals of dynamics and
applications in Civil Engineering.
Analyze of
dependent
motion of two
particles.
Investigate
accelerated
motion of a
particle using
the equation of
motion.
Assignment
1
Quiz 1
In-Semester
Exam (I)
Final Exam
2
Rectilinear motion, Erratic
Motion, General Curvilinear
Motion , Rectangular
Components
Assignment
1
Quiz 1
In-Semester
Exam (I)
Final Exam
Weekly Teaching & Assessment Plan:
3 to 4
Motion of a Projectile Curvilinear Motion:Normal and Tangential Components CurvilinearMotion: Cylindrical Components AbsoluteDependent Motion
Assignment 1
Quiz 1
Assignment 1Quiz 1In-Semester Exam (I)Final Exam
5 to 6
Kinematics of a ParticleEquation of Motion, Concepts of position, displacement, velocity, and acceleration, particle motion along a straight line, particle motion along a curved path, Analysis of dependent motion of two particles,
In-Semester Exam (I)
Investigate accelerated motion of a particle using the equation of motion.
Develop and apply the principle of work and energy that involve force, velocity, and displacement.
Assignment 2Quiz 2In-Semester Exam (I)Final Exam
7 to 9
Kinematics of a ParticlePrinciples of relative motion of two particlesusing translating axes. State Newton’s Laws ofMotion and Gravitational attraction and to beable to define mass and weight, velocity, anddisplacement, power and efficiency, conservationof energy to solve kinetic problems.
Assignment 2( 7th week)
Assignment 2-3Quiz 2In-Semester Exam (II)Final Exam
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10 to 12
Kinematics of a Rigid BodyTypes of rigid-body planar motion, kineticequations of motion for a symmetric rigidbody, force and couple do work, force,velocity, and displacement, principle of workand energy.
Assignment 3In-Semester Exam (II)
Analyze rigid-body motion about a fixed axis.
Assignment 3In-Semester Exam (II)Final Exam
13 to 14
Vibrations: Fundamentals of vibration, equations ofmotion, simple harmonic motion, naturalfrequency and applications.
Quiz 3Study the analysis of undamped forced vibration and viscous damped forced vibration
Quiz 3Final Exam
15Vibrations: undamped forced vibration and viscousdamped forced vibration.
Final Exam
Definition
Mechanics:
Branch of physical sciences
concerned with the state of rest or
motion of bodies subjected to
forces.
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Assessment Percentage % DueAssignments 10% Week 3, 6, 10
Quizzes 15% Week 4, 8 & 12In-Semester Exam I 17.5% Week 6
In-Semester Exam II 17.5% Week 12
Final Exam 40% Week 16 &17
Rigid Body Mechanics
• Statics – Bodies at rest
• Dynamics – Accelerated motion of
bodies
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Statics Dynamics
Rigid Bodies Deformable Bodies
Solid Mechanics Fluid Mechanics
Engineering Mechanics
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Basic Quantities
Length
– meter
– foot
Time
– second
Mass
– kilogram
– slug
Force
– newton
– pound
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mechanical quantities
• All mechanical quantities can be defined in
terms of mass, length, and time.
• The weight of an object is the force
of gravity on the object and may be
defined as the mass times the acceleration
of gravity, w = mg. Since the weight is
a force, its SI unit is the newton.
MassIn physics, mass is a property of a physicalbody which determines the strength of itsmutual gravitational attraction to otherbodies, its resistance to being acceleratedby a force, and in the theory of relativitygives the mass–energy content of asystem. The SI unit of mass is the kilogram(kg).
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Mass
Length
Needed to locate the position of a point in space and describe the size of a physical system.
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Time
Time is a measure in which events can be ordered from the past through the present into the future, and also the measure of durations of events and the intervals between them. Time is often referred to as the fourth dimension, along with the three spatial dimensions.
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Force
Generally considered as a push of a pull exerted by one body on another. Interaction occurs when there is direct contact between the bodies. Gravitational, electrical and magnetic forces do not require direct contact. Force is characterized by magnitude, direction and point of application.
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Idealizations
1. Particle - an object having mass but the size
is neglected.
2. Rigid Body - a combination of a large
number of particles which remain in a fixed
position relative to each other, both before
and after the application of a force.
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1. Modern version of metric system.
2. Base units are length, time and mass, meter
(m), second (s), and kilogram (kg)
3. Acceleration of gravity:
SI Units
2
mg 9.81
s=
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4. Force is derived quantity measured in unit
called a newton
SI Units
2s
mkg1N1
⋅=
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Systems of Units
Name Length Time Mass Force
SImeter
(m)
second
(s)
kilogram
(kg)
newton
(N)
Kgm/s^2
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Prefixes for SI units
Exponential form Prefix SI symbol
Multiple 1,000,000,000 109 giga G 1,000,000 106 mega M 1,000 103 kilo k Submultiple 0.001 10-3 milli m 0.000001 10-6 micro µ 0.00000001 10-9 nano n
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Each of the terms of an equation must be
expressed in the same units.
s = v t + 1/2 a t 2
s is position in meters
v is velocity in m/s
a is acceleration in m/s2
t is time in seconds
[m] = [m/s] • [s] + [m/s2] • [s2] = [m]
Dimensional Homogeneity
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Convert 2 km/h to m/s. How many ft/s is this?SOLUTION:Since 1 km = 1000 m and 1 h = 3600 s, the conversion factors are arranged so that a cancellation of units can be applied.
sm556.05555.0
sm
36002000
hkm2
s3600h1
kmm1000
hkm
2hkm2
===
=
Recall that 1 ft = 0.3038 m
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Procedure for Analysis
1. Read the problem carefully and correlate
the actual physical situation with the
theory studied.
2. Draw necessary diagrams and tables.
3. Apply relevant principles, generally in
mathematical form.
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Procedure for Analysis
4. Solve the equations algebraically (without
numbers) as far as possible, then obtain a
numerical answer.
5. Be sure to use a consistent set of units.
6. Report the answer with no more significant
figures than the accuracy of the given data.
7. Decide if answer seems reasonable.
8. Think about what the problem taught you!
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An Overview of Mechanics
Statics:The study of bodies in equilibrium.
Dynamics:1. Kinematics – concerned with the geometric aspects of motion2. Kinetics - concerned with the forces causing the motion
Mechanics:The study of how bodies react to forces acting on them.
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• Dynamics includes:
Kinematics: study of the geometry of motion. Relates displacement, velocity, acceleration, and time without reference to the cause of motion.
Fthrust
Flift
Fdrag
Kinetics: study of the relations existing between the forces acting on a body, the mass of the body, and the motion of the body. Kinetics is used to predict the motion caused by given forces or to determine the forces required to produce a given motion.
Kinematic relationships are used to help us determine the trajectory of a golf ball, the orbital speed of a satellite, and the accelerations during acrobatic flying.
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• Particle kinetics includes:
• Rectilinear motion: position, velocity, and acceleration of a particle as it moves along a straight line.
• Curvilinear motion: position, velocity, and acceleration of a particle as it moves along a curved line in two or three dimensions.
Rectilinear Motion: Position, Velocity & Acceleration
• Rectilinear motion: particle moving along a straight line
• Position coordinate: defined by positive or negative distance from a fixed origin on the line.
• The motion of a particle is known if the position coordinate for particle is known for every value of time t.
• May be expressed in the form of a function, e.g.,
or in the form of a graph x vs. t.
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• Instantaneous velocity may be positive or negative. Magnitude of velocity is referred to as particle speed.
• Consider particle which occupies position Pat time t and P’ at t+∆t,
t
xv
t
x
t ∆∆==
∆∆=
→∆ 0lim
Average velocity
Instantaneous velocity
• From the definition of a derivative,
dt
dx
t
xv
t=
∆∆=
→∆ 0lim
e.g.,
2
32
312
6
ttdt
dxv
ttx
−==
−=
Rectilinear Motion: Position, Velocity & Acceleration
Rectilinear Motion: Position, Velocity & Acceleration
• Consider particle with velocity v at time t and v’ at t+∆t,
Instantaneous accelerationt
va
t ∆∆==
→∆ 0lim
tdt
dva
ttv
dt
xd
dt
dv
t
va
t
612
312e.g.
lim
2
2
2
0
−==
−=
==∆∆=
→∆
• From the definition of a derivative,
• Instantaneous acceleration may be:
- positive: increasing positive velocity
or decreasing negative velocity
- negative: decreasing positive velocity
or increasing negative velocity.
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Rectilinear Motion: Position, Velocity & Acceleration
• From our example,326 ttx −=
2312 ttdt
dxv −==
tdt
xd
dt
dva 612
2
2−===
- at t = 2 s, x = 16 m, v = vmax= 12 m/s, a = 0
- at t = 4 s, x = xmax= 32 m, v = 0, a = -12 m/s2
• What are x, v, and a at t = 2 s ?
• Note that vmaxoccurs when a=0, and that the slope of the velocity curve is zero at this point.
• What are x, v, and a at t = 4 s ?
Concept Quiz
2 - 36
What is true about the kinematics of a particle?
a) The velocity of a particle is always positiveb) The velocity of a particle is equal to the slope of
the position-time graphc) If the position of a particle is zero, then the
velocity must zerod) If the velocity of a particle is zero, then its
acceleration must be zero
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EXAMPLE
Plan: Establish the positive coordinate, s, in the direction the
particle is traveling. Since the velocity is given as a
function of time, take a derivative of it to calculate the
acceleration. Conversely, integrate the velocity
function to calculate the position.
Given: A particle travels along a straight line to the rightwith a velocity of v = ( 4 t – 3 t2 ) m/s where t is in seconds. Also, s = 0 when t = 0.
Find: The position and acceleration of the particle when t = 4 s.
EXAMPLE (continued)
Solution:
1) Take a derivative of the velocity to determine the acceleration.
a = dv / dt = d(4 t – 3 t2) / dt = 4 – 6 t⇒ a = – 20 m/s2 (or in the ← direction) when t = 4 s
2) Calculate the distancetraveled in 4s by integrating the velocity using so = 0:
v = ds / dt ⇒ ds = v dt ⇒⇒ s – so = 2 t2 – t3
⇒ s – 0 = 2(4)2 – (4)3 ⇒ s = – 32 m ( or ←)
∫∫ =t
o
s
s
(4 t – 3 t2) dtds
o
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RECTILINEAR KINEMATICS: ERRATIC MOTION
Today’s Objectives:
Students will be able to:
1. Determine position, velocity, and acceleration of a
particle using graphs.
READING QUIZ
1. The slope of a v-t graph at any instant represents instantaneous
A) velocity. B) acceleration.
C) position. D) jerk.
2. Displacement of a particle in a given time interval equals the
area under the ___ graph during that time.
A) a-t B) a-s
C) v-t C) s-t
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APPLICATIONS
In many experiments, a velocity versus
position (v-s) profile is obtained.
If we have a v-s graph for the tank
truck, how can we determine its
acceleration at position s = 1500 feet?
The velocity of a car is recorded from a
experiment. The car starts from rest
and travels along a straight track.
If we know the v-t plot, how can we
determine the distance the car traveled
during the time interval 0 < t < 30 s or
15 < t < 25 s?
APPLICATIONS (continued)
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ERRATIC MOTION
(Section 12.3)
The approach builds on the facts that slope and differentiation are linked and that
integration can be thought of as finding the area under a curve.
Graphing provides a good way to handle
complex motions that would be difficult to
describe with formulas.
Graphs also provide a visual description of
motion and reinforce the calculus concepts
of differentiation and integration as used in
dynamics.
S-T GRAPH
Plots of position vs. time can be used to find
velocity vs. time curves. Finding the slope of
the line tangent to the motion curve at any
point is the velocity at that point (or v =
ds/dt).
Therefore, the v-t graph can be constructed
by finding the slope at various points along
the s-t graph.
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V-T GRAPH
Also, the distance moved (displacement) of the
particle is the area under the v-t graph during time
∆t.
Plots of velocity vs. time can be used to find
acceleration vs. time curves. Finding the slope of
the line tangent to the velocity curve at any point is
the acceleration at that point (or a = dv/dt).
Therefore, the acceleration vs. time (or a-t) graph
can be constructed by finding the slope at various
points along the v-t graph.
A-T GRAPH
Given the acceleration vs. time or a-t
curve, the change in velocity (∆v) during a
time period is the area under the a-t curve.
So we can construct a v-t graph from an a-t
graph if we know the initial velocity of the
particle.
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A-S GRAPH
a-s graph
½ (v1² – vo²) = = area under the∫
s2
s1
a ds
A more complex case is presented by the
acceleration versus position or a-s graph. The area
under the a-s curve represents the change in
velocity
(recall ∫ a ds = ∫ v dv ).
This equation can be solved for v1, allowing you
to solve for the velocity at a point. By doing this
repeatedly, you can create a plot of velocity
versus distance.
V-S GRAPH
Another complex case is presented by the
velocity vs. distance or v-s graph. By reading
the velocity v at a point on the curve and
multiplying it by the slope of the curve (dv/ds)
at this same point, we can obtain the
acceleration at that point. Recall the formula
a = v (dv/ds).
Thus, we can obtain an a-s plot from the v-s
curve.
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EXAMPLE
What is your plan of attack for the problem?
Given: The s-t graph for a sports car moving along a straight road.
Find: The v-t graph and a-t graph over the time interval shown.
EXAMPLE (continued)
Solution: The v-t graph can be constructed by finding the slope of the s-t graph
at key points. What are those?
when 0 < t < 5 s; v0-5 = ds/dt = d(3t2)/dt = 6 t m/s
when 5 < t < 10 s; v5-10 = ds/dt = d(30t−75)/dt = 30 m/s
v-t graph
v(m/s)
t(s)
30
5 10
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EXAMPLE (continued)
Similarly, the a-t graph can be constructed by finding the slope at various points
along the v-t graph. Using the results of the first part where the velocity was found:
when 0 < t < 5 s; a0-5 = dv/dt = d(6t)/dt = 6 m/s2
when 5 < t < 10 s; a5-10 = dv/dt = d(30)/dt = 0 m/s2
a-t graph
a(m/s2)
t(s)
6
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CONCEPT QUIZ
1. If a particle starts from rest and
accelerates according to the graph
shown, the particle’s velocity at
t = 20 s is
A) 200 m/s B) 100 m/s
C) 0 D) 20 m/s
2. The particle in Problem 1 stops moving at t = _______.
A) 10 s B) 20 s
C) 30 s D) 40 s
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GROUP PROBLEM SOLVING I
Given: The v-t graph shown.
Find: The a-t graph, average speed,
and distance traveled
for the 0 - 50 s interval.
Plan: What is your plan?
GROUP PROBLEM SOLVING I
Plan: Find slopes of the v-t curve and draw the a-t graph. Find the area under the curve. It is the distance traveled.Finally, calculate average speed (using basic definitions!).
Given: The v-t graph shown.
Find: The a-t graph, average speed,
and distance traveled
for the 0 - 50 s interval.
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GROUP PROBLEM SOLVING I (continued)
Solution:
Find the a–t graph:
For 0 ≤ t ≤ 30 a = dv/dt = 0.4 m/s²
For 30 ≤ t ≤ 50 a = dv/dt = 0 m/s²
a-t graph
0.4
a(m/s²)
30 50t(s)0
GROUP PROBLEM SOLVING I (continued)
Now find the distance traveled:
∆s0-30 = ∫ v dt = ∫ 0.4 t dt = 0.4 (1/2) (30)2 = 180 m
∆s30-50 = ∫ v dt
= ∫ 12 dt = 12 (50 – 30)
= 240 m
s0-90 = 180 + 240 = 420 m
vavg(0-90) = total distance / time
= 420 / 50
= 8.4 m/s
v = 12
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GROUP PROBLEM SOLVING II
Given: The v-t graph shown.
Find: The a-t graph, average speed,
and distance traveled
for the 0 - 48 s interval.
Plan:
GROUP PROBLEM SOLVING II
Plan: Find slopes of the v-t curve and draw the a-t graph.
Find the area under the curve. It is the distance traveled. Finally, calculate
average speed (using basic definitions!).
Given: The v-t graph shown.
Find: The a-t graph, average speed,
and distance traveled
for the 0 - 48 s interval.
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GROUP PROBLEM SOLVING II (continued)
Solution:
Find the a–t graph:
For 0 ≤ t ≤ 30 a = dv/dt = 0.2 m/s²
For 30 ≤ t ≤ 48 a = dv/dt = -0.333 m/s²
a-t graph
-0.33
0.2
a(m/s²)
30 48 t(s)
GROUP PROBLEM SOLVING II (continued)
Now find the distance traveled:
∆s0-30 = ∫ v dt = (1/5)(1/2) (30)2 = 90 m
∆s30-48 = ∫ v dt = [(-1/3) (1/2) (t – 48)2 ]
= (-1/3) (1/2)(48 – 48)2 – (-1/3) (1/2)(30 – 48)2
= 54 m
s0-48 = 90 + 54 = 144 m
vavg(0-90) = total distance / time
= 144 / 48
= 3 m/s
48
30
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ATTENTION QUIZ
1. If a car has the velocity curve shown, determine the time t necessary for the car
to travel 100 meters.
A) 8 s B) 4 s
C) 10 s D) 6 s
t
v
6 s
75
t
v
2. Select the correct a-t graph for the velocity curve shown.
A) B)
C) D)
a
t
a
t
a
t
a
t