Dynamics · PDF fileHibbeler, R.C., Mechanics of ... Normal and Tangential Components...

31
2/8/2016 1 Dynamics Monday Tuesday Wednesday Thursday 10:00 - 11:00 5D-43 7. Faculty Name: Dr. Yavuz YARDIM 8. Office No.: 5D-43 9. Tele Ext. : 984 10. E-mail: [email protected] 11. Time for office hours : Day Time Sunday 12:00-12:50 2A-1 Tuesday 12:00-12:50 2A-1 Dynamics

Transcript of Dynamics · PDF fileHibbeler, R.C., Mechanics of ... Normal and Tangential Components...

Page 1: Dynamics · PDF fileHibbeler, R.C., Mechanics of ... Normal and Tangential Components Curvilinear ... and acceleration, particle motion along a straight line, particle motion

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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

Page 21: Dynamics · PDF fileHibbeler, R.C., Mechanics of ... Normal and Tangential Components Curvilinear ... and acceleration, particle motion along a straight line, particle motion

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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.

Page 25: Dynamics · PDF fileHibbeler, R.C., Mechanics of ... Normal and Tangential Components Curvilinear ... and acceleration, particle motion along a straight line, particle motion

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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

5 10

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

Page 27: Dynamics · PDF fileHibbeler, R.C., Mechanics of ... Normal and Tangential Components Curvilinear ... and acceleration, particle motion along a straight line, particle motion

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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

Page 31: Dynamics · PDF fileHibbeler, R.C., Mechanics of ... Normal and Tangential Components Curvilinear ... and acceleration, particle motion along a straight line, particle motion

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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