Post on 09-Sep-2014
DYNAMICSDYNAMICSBDA 2013BDA 2013
Dr. Waluyo Adi Siswanto
Lecture 1 – Introduction, Rectilinear MotionLecture 1 – Introduction, Rectilinear Motion
Lecture Module 1 Dr. Waluyo A. Siswanto 2
Course ObjectivesCourse Objectives
To understand basic concepts in dynamics (kinematics and kinetics) of particles and rigid bodies.
To predict the effects of force and motion while carrying out the creative design functions of engineering.
To apply problem-solving procedures that will be applicable in succeeding courses and throughout the engineering careers.
Lecture Module 1 Dr. Waluyo A. Siswanto 3
Learning OutcomesLearning Outcomes
At the end of this course, students will be able to: Determine the position, velocity and acceleration of
particles and rigid bodies using kinematics approach. Use law of motion to relate forces and acceleration of
particles and rigid bodies. Apply the principle of work and energy to solve
kinetics problem on particles or rigid bodies. Solve the kinetics problems using the principles of
impulse and momentum. Apply all methods that have studied in real situation. Communicate and work effectively in a group through
given assignments and presentations.
Lecture Module 1 Dr. Waluyo A. Siswanto 4
What you can do…What you can do…
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ReferencesReferences Siswanto W.A., 2008, Principles of Engineering
Dynamics, Penerbit UTHM Meriam J.L. and Kraige L. G., 2007. Engineering
mechanics – Dynamics, 6th Edition, John Wiley & Sons, Inc.
Bear F.P. and Johnson E. R., 2007. Vector Mechanics for Engineers – Dynamics, 8th S.I. Edition, Mc Graw Hill
Hibbeler R.C., 2004. Engineering Mechanics –Dynamics, 3rd S.I. Edition, Prentice Hall.
Bedford A. & Wallace T. Fowler, 2005. Engineering Mechanics-Dynamics, S.I. Edition, Prentice Hall.
Mohd Imran Ghazali, Izzuddin Zaman, et.all., 2007, Koleksi Soalan Dinamik, Penerbit UTHM.
Saifulnizam Jamian; Izzuddin Zaman, et.all., 2007, Dynamics, Penerbit UTHM.
Lecture Module 1 Dr. Waluyo A. Siswanto 6
Subject Delivery (S4 and S6) Subject Delivery (S4 and S6)
LecturingExplaining theories, methods and understanding related to the topics
TutoringDiscussing sample problems related to the topics
All course materials, announcements, and students marks will be published in E-Learning
S4: Tuesday (BS A6, 20.00 – 21.50) Friday (BS A6, 08.00 – 09.50)
S6: Monday (BS C2, 08.00 – 08.50) Wednesday (DK D, 10.00 – 12.50)
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Assessment SchemeAssessment Scheme
Students attendance * Individual assignment (3) (10%) Group Assignment (10%) Test 1 (15%) Test 2 (15%) Final Examination (50%)
* Attendance must not less than 80%
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Time Tables Time Tables BDA 2013 Dynamics, S4BDA 2013 Dynamics, S4
Part 1Ass 1
07/07 10/07 14/07 17/07 21/07 24/07 28/07 31/07
1 2 3 4
Ass 1: First assignmentAss 2: Second assignmentAss 3: Third assignment
Ass 1 Due: First assignment deadline submissionAss 2 Due: Second assignment deadline submissionAss 3 Due: Third assignment deadline submission
Proj: Project commencing Test 1: First Test Project Presentation: Presentation of the project Test 2: Second Test
Part 2Ass 2 Ass 2 Due
Test 2: 17/9 or 18/9
25/08 28/08 01/09 04/09 08/09 11/09 15/09 18/09
7 8 9 10
Part 3Ass 3 Ass 3 Due
29/09 02/10 06/10 09/10 13/10 16/10 20/10 23/10
11 12 13 14
04/08 07/08 11/08 14/08
5 6 Ass 1 Due
Test 1: 13/8 or 14/8
ProjProject Presentation
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Time Tables Time Tables BDA 2013 Dynamics, S6BDA 2013 Dynamics, S6
Part 1Ass 1
06/07 08/07 13/07 15/07 20/07 22/07 27/07 29/07
1 2 3 4
Ass 1: First assignmentAss 2: Second assignmentAss 3: Third assignment
Ass 1 Due: First assignment deadline submissionAss 2 Due: Second assignment deadline submissionAss 3 Due: Third assignment deadline submission
Proj: Project commencing Test 1: First Test Project Presentation: Presentation of the project Test 2: Second Test
Part 2Ass 2
24/08 26/08 31/08 02/09 07/09 09/09 14/09 16/09
7 8 9 10
Part 3Ass 3 Ass 3 Due
28/09 30/09 05/10 07/10 12/10 14/10 19/10 21/10
11 12 13 14
03/08 05/08 10/08 12/08
5 6
ProjProject Presentation
Ass 1 DueTest 1: 13/8 or 14/8
Ass 2 DueTest 2: 17/9 or 18/9
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Topics (Part 1)Topics (Part 1)
KINEMATICS OF PARTICLES Rectilinear motion Curvilinear motion Rectangular components of velocity and
acceleration Tangential and normal components Radial and transverse components Relative motion of moving particle
KINETICS OF PARTICLES Force Mass and Acceleration Work and Energy
TEST 1TEST 1
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Topics (Part 2)Topics (Part 2)
KINETICS OF PARTICLES Impulse and Momentum Direct Impact Oblique Impacts
KINEMATICS OF RIGID BODIES Translation, rotation and general plane motion Absolute motion Relative velocity and acceleration Zero velocity instantaneous center
TEST 2TEST 2
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Topics (Part 3)Topics (Part 3)
KINEMATICS OF RIGID BODIES Relative motion of rotating frame
KINETICS OF RIGID BODIES Mass Moment of Inertia Moment, Inertia and Acceleration Translation and Rotation and General
Plane Motion Work and Energy in Rigid Bodies
FINAL EXAMFINAL EXAM
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INTRODUCTIONINTRODUCTION
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DynamicsDynamics
Dynamics consists two distinct parts: kinematics and kinetics.
Kinematics deals with the study of motion without reference to the force which cause motions
Kinetics relates the action of forces on bodies to their resulting motions
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Kinematics: Kinematics: Motion onlyMotion only
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Kinetics: Kinetics: Interaction force - motionInteraction force - motion
Force
Path of motion because of the force
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Vector Analysis Review: Vector Analysis Review: NotationNotation
A
length of arrow denotes magnitude of F
extension of arrow denotes line of action
arrow head denotesthe direction of F
321
F
3,2,1F
3 2 3F i j k
Vectors are defined as mathematical expression possessing magnitude and direction, which add according to the parallelogram law.
(note: scalar has only magnitude)
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Vector Analysis Review Vector Analysis Review MagnitudeMagnitude
2 2 22 1 2 1 2 1magnitude A x x y y z z = A
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Analysis Vector Review Analysis Vector Review Parallelogram Law in Two-Dimensional VectorsParallelogram Law in Two-Dimensional Vectors
A
B
A+B
P
cossin
x
y
A
A AA A
cossin
x
y
B
B BB B
cos cossin sin
A + BA + B
A Bβ
α
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Analysis Vector Review Analysis Vector Review VectorVector DirectionDirection
z
Ax
Ay
Az
ij
k
A=Ayi+ Ayj + Ayk
αβ
γ α : angle between x coord and A
β : angle between y coord and A
γ : angle between z coord and A
y
x
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Analysis Vector Review Analysis Vector Review VectorVector Direction – Unit VectorDirection – Unit Vector
cos xAA
cos yAA
cos zAA
yx zA
AA AA A A
e i j k
(cos ) (cos ) (cos )A e i j k
AAA e
2 2 2cos cos cos 1
β
yAy
A
γ
Az
A
α
Ax
A
z
x
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Particle and rigid bodyParticle and rigid body
A particle is a point mass. This means the mass is concentrated at a single point and the particle has neither dimensions (height, width, etc) nor orientation (angular position)
Under certain conditions a physical body can be modeled as a particle; for example, when considering translation of a body, or when all forces acting on a body pass through the centre of mass, or when the dimensions of a body are very much smaller than those of its path of motion
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When a body can be replaced by a When a body can be replaced by a particleparticle
Replace by a particle (no rotation involved)
The body motion cannot be replaced by a particle(rotation involved)
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Translation - RotationTranslation - Rotation
Translation Combination Translation and Rotation
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Acceleration due to gravityAcceleration due to gravity
2
2( )og ge
e
RR h
2 49,780327 1 0,005279sin 0,000023sin og
In general applications, the standard values for g of 9,81 m/s2 or 32,2 ft/s2
will be sufficiently accurate.
Re = 6,371 . 106 m h : altitude
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KINEMATICS OF PARTICLE KINEMATICS OF PARTICLE TRANSLATIONTRANSLATION
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Translation of a particle: Translation of a particle: straight, curve and circular pathsstraight, curve and circular paths
Straight path
Curve path Circular path
Straight path
Curve path Circular path
There is no orientation of the body
Can be simplified as particle
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Average and InstantanneousAverage and Instantanneous
0lim sv
t
t
dsvdt
v stime (s)
Posi
tion
(m)
s
t
Instantaneous velocity
Average Velocity avg
svt
0lim
t
t
va
Instantaneous acceleration
Average Acceleration avg
vat
2
2
va d d sdt dt
a v s
Δt
Δs
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Calculating distance and velocity, Calculating distance and velocity, graphicallygraphically
v
tt0 t
v 0 0
s t
s t
ds dt
0
v
v
v a 0
t
t
d dt
The area under v(t) is the DISTANCE TRAVEL
The area under a(t) is the NET CHANGE IN VELOCITY
a
tt0 t1
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Differential equation of motionDifferential equation of motion
v ad dt
dsd dt dt dsdt
v v v a a a v v ad ds
s ds s ds
Multiply by v
or
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Rectilinear motion Rectilinear motion at constant velocityat constant velocity
v 0 0
s t
s t
ds dt
0 0( )v s s t t
0 0( )+ v s s t t
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Velocity at constant accelerationVelocity at constant acceleration
0
v
v
v a 0
t
t
d dt 0 0v v a t t
0 0v v + a t t
0
0
v
v
v v a s
s
d ds 2 20 0
12v v a s s
2 20 0v v + a 2 s s
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Distance at constant accelerationDistance at constant acceleration
v 0 0
s t
s t
ds dt 0 0v + a t t 0 0
s t
s t
ds ( )dt
0 0
0 0 0
0 0 0
0 0
v a
v a a
t t
t t
t t t
t t t
s s dt t t dt
dt tdt t dt
2 20 0 0 0 0 0
1( )2
s s v t t a t t at t t
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Practical situationPractical situation
Usually a particle start moving when the time is set to 0 and the distance goes from 0
0 00 and 0t s at and distance t s
0 0v v + a t t
2 20 0v v + a 2 s s
2 20 0 0 0 0 0
1( )2
s s v t t a t t at t t
0 tv v + a
2 20 2 sv v + a
20
1( )2
s v t a t
0
0
0 0 0 0
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Example Problem 1-1Example Problem 1-1
The displacement of a mechanical component follows a ruled path as a function of time. The function dis(t) = 2t3 – 24t + 6 meter.
• Derive the velocity and acceleration based on the given displacement function dis(t).
• Calculate the time to reach velocity of 72 m/s. Calculate the corresponding acceleration at that time.
• Plot the displacement, velocity and acceleration versus time for the first 4 seconds every 1 second.
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Example Problem 1-2Example Problem 1-2
A particle starts from rest at x = -2 m and moves along x-axis with the velocity history shown below. •Derive the velocity equation in every stage of t (0 to 0,5), (0,5 to 1,0) and (1,0 to 2,0)•Calculate the corresponding acceleration and displacement at t= 0; 0,5; 1,0; and 2,0
v (m
/s)
0
3
0.5 1.0 1.5 2.0
-1.0
t (s)
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Example Problem 1-3Example Problem 1-3
A train is traveling at a constant speed of 108 km/h when the driver sees a stationary obstacle on the track 135m ahead. He immediately applies the brakes at full capacity. If the train’s constant deceleration on this section of track is
4,5 m/s2, determine whether the train will stop before colliding with the obstacle.
Calculate the time interval between the driver applying the brakes and the train coming to rest.
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Example Problem 1-4Example Problem 1-4
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See you again …….See you again …….