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Structural Analysis 3

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CHAPTER 18 PLANAR KINETICS OF A RIGID BODY: WORK AND ENERGY (Sections 18.1-18.4) Objectives: a) Define the various ways a force and couple do work. – PowerPoint PPT presentation

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Title: Structural Analysis 3


1
CHAPTER 18 PLANAR KINETICS OF A RIGID BODY
WORK AND ENERGY (Sections 18.1-18.4)
Objectives a) Define the various ways a force
and couple do work. b) Apply the principle of
work and energy to a rigid body.
2
APPLICATIONS
The work of the torque developed by the driving
gears on the two motors on the mixer is
transformed into the rotational kinetic energy of
the mixing drum.
The work done by the compactor's engine is
transformed into the translational kinetic energy
of the frame and the translational and rotational
kinetic energy of its roller and wheels
3
KINETIC ENERGY
The kinetic energy of a rigid body in general
plane motion is given by T 1/2 m (vG)2 1/2
IG ?2
  • Several simplifications can occur.
  • Pure Translation the rotational kinetic energy
    is zero (? 0)
  • T 0.5 m (vG)2

4
KINETIC ENERGY (continued)
2. Rotation When a rigid body is rotating
about a fixed axis passing through point O, the
body has both translational and rotational
kinetic energy T 0.5m(vG)2 0.5IGw2 Since
vG rGw, T 0.5(IG m(rG)2)w2 0.5IOw2
If the rotation occurs about the mass center
(pure rotation), G, then vG0
T 0.5 IG w2
5
WORK OF A FORCE
Work of a weight Uw -W?y.
Work of a spring force Us -0.5k(s2)2
(s1)2
6
FORCES THAT DO NO WORK
1. Reactions at fixed supports because the
displacement at their point of application is
zero.
2. Normal and frictional forces acting on bodies
as they roll without slipping over a rough
surface since there is no instantaneous
displacement of the point in contact with ground.
3. Internal forces because they always act in
equal and opposite pairs.
7
THE WORK OF A COUPLE
A body subjected to a couple produces work only
when it undergoes rotation.
constant M UM M (q2 q1)
8
PRINCIPLE OF WORK AND ENERGY
The principle states SU1-2 T12 i.e. the work
done by all external forces and couple moments
equals the change in bodys kinetic energy
(translational and rotational).
This equation is a scalar equation. It can be
applied to a system of rigid bodies by summing
contributions from all bodies.
9
EXAMPLE
GivenThe disk weighs 40 N, has a radius of
gyration (kG) 0.6 m. A 15 N.m moment is applied.
The spring constant is 10N/m.
Find Angular velocity of the disk when point G
moves 0.5 m. The disk starts from rest and rolls
without slipping. The spring is initially
unstretched.
10
EXAMPLE (solution)
Free body diagram of the disk
Only the spring force and couple moment M do
work.
Spring will stretch twice the amount of
displacement of G, or 1 m. Why?
11
EXAMPLE (continued)
Kinematic relation vG r w 0.8w
Kinetic energy T1 0 T2 0.5m (vG)2 0.5 IG
w2 T2 0.5(40/9.8)(0.8w)2 0.5(40/9.8)(0.6)2w2
T2 2.0 w2
Work and energy U1-2 T12 4.4 2.0 w2 w
1.5 rad/s
12
GROUP PROBLEM SOLVING
Given A sphere weighing 10N rolls along a
semicircular hoop. Its w equals 0 when q 0.
Find The angular velocity of the sphere when q
45 if the sphere rolls without slipping.
13
GROUP PROBLEM SOLVING (continued)
Solution Draw a FBD and calculate the vertical
distance the mass center moves.
Now calculate the _______________
14
GROUP PROBLEM SOLVING (continued)
A kinematic equation for finding the velocity of
the mass center is needed. It is
_________ energy
Now apply the principle of work and energy
equation w 13.1
rad/s
15
End of 18.1-4
Let Learning Continue
16
PLANAR KINETICS OF A RIGID BODY CONSERVATION OF
ENERGY (Section 18.5)
  • Objectives
  • Determine the potential energy of conservative
    forces.
  • Apply the principle of conservation of energy.

17
APPLICATIONS
Torsional spring at the top of the door winds up
as the door is lowered. When the door is raised,
the spring potential energy is transferred into
gravitational potential energy of the doors
weight, thus making it easy to open.
18
CONSERVATION OF ENERGY
The conservation of energy theorem is a simpler
energy method for solving problems. Once again,
the problem parameter of distance is a key
indicator of when conservation of energy is a
good method for solving the problem.
If it is appropriate, conservation of energy is
easier to use than the principle of work and
energy.
19
CONSERVATIVE FORCES
A force F is conservative if the work done by the
force is independent of the path.
Typical conservative forces encountered in
dynamics are gravitational forces (i.e., weight)
and elastic forces (i.e., springs).
What is a common force that is not conservative?
20
CONSERVATION OF ENERGY
When a rigid body is acted upon by a system of
conservative forces, the work done by these
forces is conserved. Thus, the sum of kinetic
energy and potential energy remains constant T1
V1 T2 V2 Constant
i.e. as a rigid body moves from one position to
another when acted upon by only conservative
forces, kinetic energy is converted to potential
energy and vice versa.
21
GRAVITATIONAL POTENTIAL ENERGY
The gravitational potential energy is a function
of the height of the bodys center above or
below a datum.
The gravitational potential energy is found by
Vg W yG
Gravitational potential energy is ve when yG is
ve, since the weight has the ability to do ve
work when the body is moved back to the datum.
22
ELASTIC POTENTIAL ENERGY
Spring forces are also conservative forces.
The potential energy of a spring force (F ks)
is found by the equation Ve ½ ks2
Notice that the elastic potential energy is
always ve
23
EXAMPLE 1
Given The rod AB has a mass of 10 kg. Piston B
is attached to a spring of constant k 800 N/m.
The spring is un-stretched when ? 0. Neglect
the mass of the pistons.
Find The angular velocity of rod AB at ? 0 if
the rod is released from rest when ? 30.
24
EXAMPLE 1 (solution)
Initial Position
Final Position
Potential Energy put datum in line with rod when
?0. Gravitational and elastic energy0 at 2.
gt V2 0 Gravitational energy at 1 - (10)(
9.81) ½ (0.4 sin 30) Elastic energy at 1 ½
(800) (0.4 sin 30)2 So V1 -
9.81J 16.0 J 6.19 J
25
EXAMPLE 1 (continued)
Initial Position
Final Position
Kinetic Energy
The rod released from rest (vG10, ?10). Thus,
T1 0. At 2, the angular velocity is ?2 and the
velocity is vG2 .
26
EXAMPLE 1 (continued)
Thus, T2 ½ (10)(vG2)2 ½ (1/12)(10)(0.42)(?2)
2 At 2, vA0. Hence, vG2 r ? 0.2 ?2
. Then, T2 0.2 ?22 0.067 ?22 0.267 ?22
conservation of energy T1 V1 T2 V2
0 6.19 0.267?22 0 gt ?2 4.82
rad/s
27
EXAMPLE 2
Given The weight of the disk is 30 N and its kG
equals 0.6 m. The spring has a stiffness of 2 N/m
and an unstretched length of 1m.
Find The angular velocity at the instant G moves
3 m to the left. The disk is released from rest
in the position shown and rolls without slipping.
28
EXAMPLE 2 (solution)
Potential Energy No changes in the gravitational
potential energy
The elastic potential energy at 1 V1 0.5 k
(s1)2 where s1 4 m. Thus, V1 ½ 2 (4)2
16 N.m.
The elastic potential energy at 2 is V2 ½ 2
(3)2 9 N.m.
29
EXAMPLE 2 (continued)
Kinetic Energy
Released from rest vG1?10, T10 At 2, ?2 and
vG2
T2 ½ m (vG2)2 ½ IG (?2) 2 ½ (30/9.8)
(vG2)2 ½ (30/9.8) 0.62 (?2) 2
Disk is rolling without slipping vG2 (0.75 ?2)
T2 ½(30/9.8)(0.75 ?2)2 ½(30/9.8) 0.62 (?2)2
1.41 (?2)2
30
EXAMPLE 2 (continued)
conservation of energy
T1 V1 T2 V2 0 16.0 1.41 ?22 9
Solving , ?2 2.23 rad/s
31
GROUP PROBLEM SOLVING
Given A 50 N bar is rotating downward at 2
rad/s. The spring has an
unstretched length of 2m and a spring constant of
12 N/m
Find The angle (measured down from the
horizontal) to which the bar rotates before it
stops its initial downward movement.
32
GROUP PROBLEM SOLVING (solution)
Potential Energy Put the datum in line with the
rod when ? 0.
Gravitational potential energy at 2
Elastic potential energy at 2 So, V2
33
GROUP PROBLEM SOLVING (continued)
Kinetic Energy
At 1 (when ? 0)
At 2
34
GROUP PROBLEM SOLVING (continued)
conservation of energy
Thus, ? 49.9 deg.
35
End of 18.5
Let Learning Continue
36
CHAPTER 19 PLANAR KINETICS IMPULSE AND MOMENTUM
(Sections 19.1-19.2)
Todays Objectives a) Develop formulations for
the linear and angular momentum of a
body. b) Apply the principle of linear and
angular impulse and momentum.
37
APPLICATIONS
As the pendulum swings downward, its angular
momentum and linear momentum both increase. By
calculating its momenta in the vertical position,
we can calculate the impulse the pendulum exerts
when it hits the test specimen.
The space shuttle has several engines that exert
thrust on the shuttle when they are fired. By
firing different engines, the pilot can control
the motion and direction of the shuttle.
38
LINEAR AND ANGULAR MOMENTUM
The linear momentum of a rigid body is L m vG

The angular momentum of a rigid body is HG
IG w
39
LINEAR AND ANGULAR MOMENTUM (continued)
Translation.
When a rigid body undergoes rectilinear or
curvilinear translation, its angular momentum is
zero because ? 0.
Therefore L m vG HG 0
40
LINEAR AND ANGULAR MOMENTUM (continued)
Rotation about a fixed axis.
The bodys linear momentum and angular momentum
are L m vG HG IGw
HO ( rG x mvG) IGw IO w
41
LINEAR AND ANGULAR MOMENTUM (continued)
General plane motion.
The linear and angular momentum computed about G
are required. L m vG HG IG?
The angular momentum about point A is HA
IG? (d)mvG
42
PRINCIPLE OF IMPULSE AND MOMENTUM
The principle is developed by combining the
equation of motion with kinematics. The
resulting equations allow a direct solution to
problems involving force, velocity, and time.
43
PRINCIPLE OF IMPULSE AND MOMENTUM (continued)
The previous relations can be represented
graphically by drawing the impulse-momentum
diagram.
44
EXAMPLE
Given A disk weighing 50 N has a rope wrapped
around it. The rope is pulled with a force P
equaling 2 N.
Find The angular velocity of the disk after 4
seconds if it starts from rest and rolls without
slipping.
45
EXAMPLE (solution)
Impulse-momentum diagram
Kinematics (vG)2 r w2
46
GROUP PROBLEM SOLVING
Given A gear set with WA 15 N WB 10 N kA
0.5 m kB 0.35 m M 2(1 e-0.5t) N.m
Find The angular velocity of gear A after 5
seconds if the gears start turning from rest.
Plan Time is a parameter, thus
47
GROUP PROBLEM SOLVING (continued)
Solution
Impulse-momentum diagrams
Gear A
Gear B
48
GROUP PROBLEM SOLVING (continued)
Kinematics
Angular impulse momentum relation for gear A
about point A yields For gear B

49
GROUP PROBLEM SOLVING (continued)
and wA 14.4 rad/s
50
End of 19.1-2
Let Learning Continue
51
CONSERVATION OF MOMENTUM (Section 19.3)
Objectives a) Understand the conditions for
conservation of linear and angular
momentum b) Use the condition of conservation of
linear/ angular momentum
52
APPLICATIONS
A skater spends a lot of time either spinning on
the ice or rotating through the air. To spin
fast, or for a long time, the skater must develop
a large amount of angular momentum.
If the skaters angular momentum is constant, can
the skater vary his rotational speed? How?
The skater spins faster when the arms are drawn
in and slower when the arms are extended. Why?
53
CONSERVATION OF LINEAR MOMENTUM
This equation is referred to as the conservation
of linear momentum. The conservation of linear
momentum equation can be used if the linear
impulses are small or non-impulsive.
54
CONSERVATION OF ANGULAR MOMENTUM
If the initial condition of the rigid body (or
system) is known, conservation of momentum is
often used to determine the final linear or
angular velocity of a body just after an event
occurs.
55
EXAMPLE
Given A 10 kg wheel (IG 0.156 kgm2) rolls
without slipping and does not bounce at A.
Find The minimum velocity, vG, of the wheel to
just roll over the obstruction at A.
Note Since no slipping or bouncing
occurs, the wheel pivots about point A. The force
at A is much greater than the weight, and since
the time of impact is very short, the weight can
be considered non-impulsive. The reaction force
at A, since we dont know either its direction or
magnitude, can be eliminated by applying the
conservation of angular momentum equation about A.
56
EXAMPLE (solution)
y
Impulse-momentum diagram
x


Conservation of angular momentum (HA)1
(HA)2 r ' (mvG)1 IG w1 r (mvG)2 IG w2 (0.2
- 0.03)(10)(vG)1 0.156 w1 0.2(10)(vG)2
0.156 w2
Kinematics no slip, w vG / r 5 vG
.Substituting and solving
(vG)2 0.892 (vG)1
57
EXAMPLE (continued)
To complete the solution, conservation of energy
can be used. Since it cannot be used for the
impact (why?), it is applied just after the
impact. In order to roll over
the bump, the wheel must go to position 3 from 2.
When (vG)2 is a minimum, (vG)3 is zero. Why?
T2 V2 T3 V3
½(10)(vG)22 ½(0.156)w22 0 0 98.1
(0.03)
Substituting w2 5 (vG)2 and (vG)2 0.892
(vG)1 and solving yields (vG)1 0.729 m/s,
(vG)2 0.65 m/s
58
GROUP PROBLEM SOLVING
Given A slender rod (Wr 5 N) has a wood block
(Ww 10 N) attached. A bullet (Wb 0.2 N) is
fired into the center of the block at 1000 m/s.
Assume the pendulum is initially at rest and the
bullet embeds itself into the block.
Find The angular velocity of the pendulum just
after impact.
59
GROUP PROBLEM SOLVING (continued)
Solution
To use conservation of angular momentum,
First draw a FBD.
IA ? IA 7.2kg.m2
60
GROUP PROBLEM SOLVING (continued)
Apply the conservation of angular momentum
equation
  • Solving yields
  • w2 7.0 rad/s

61
End of 19.3
Let Learning Continue
62
CHAPTER 22 VIBRATIONS (section 22.1)
Objectives Discuss Undamped single degree of
freedom (SDOF) vibration of a rigid body
63
Vibration definition
  • Vibration is the periodic motion of a body or
    system of connected bodies displaced from a
    position of equilibrium
  • Types of vibrations
  • Free vibrations motion is maintained by
    conservative forces (like gravitational or
    elastic restoring forces).
  • Forced vibrations motion is caused by external
    periodic or intermittent forces.

64
Undamped free response for SDOF systems
  • If you displace mass m a distance u from
    equilibrium position and release it, vibration
    occurs

65
Undamped free response for SDOF systems (cont)
  • Equilibrium eq
  • A linear, constant coefficients, homogeneous,
    second order ordinary differential equation
  • A and B are found from initial conditions

66
Tn natural period of vibration
Undamped free response for SDOF systems
  • ?n natural circular frequency of vibration

fn 1/Tn natural cyclic frequency of vibration
in hertz (cycles per second)
67
Solving initial conditions
68
Example 1
69
Example 1 (continued)
70
Example 1 (continued)
  • By analogy

By analogy
71
Example 2
72
Example 2 (continued)
From equilibrium
73
Example 2 (continued)
  • Thus

74
Group Solving problem
75
  • Draw FBD
  • Write equation of motion

76
  • Use analogy

77
End of 22.1
Let Learning Continue
78
Energy method Example 1
79
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80
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81
Energy methods Example 2
82
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83
(No Transcript)
84
Undamped forced vibrations
85
Free Body Diagram

86
Equilibrium Equation
87
Solution for constant loading
88
Solving initial conditions
89
Sine curve
90
90
91
Three cases
Case 1 ? gt 1 (overdamping) ?1
and ?2 lt 0
91
92
92
93
Case 3 ?lt 1 (underdamping)
Define damped natural frequency
Period ?d 2?/?d
93
94
It is because ? c/(2m?n), measurement of ?
determines the viscous damping constant c.
94
95
B. Viscous Damped Forced vibration
The equation of motion
95
96
96
97
Maximum M occurs at
The resonance frequency is
97
98
tan ?
Consider the following regions (1) ? is small,
tan? gt 0, ? ? 0, xp in phase with the driving
force
(2) ? is large, tan? lt 0, ? ? 0-, ? ?, xp lags
the driving force by 90o
(3) ? ? ?n-, tan? ??, ? ? ?/2(-) ? ? ?n,
tan? ?-?, ? ? ?/2()
98
99
If the driving force is not applied to the mass,
but is applied to the base of the system
If b?2 is replaced by Fo/m
This can be used as a device to detect
earthquake.
99
100
Example
m 45 kg, k 35 kN/m, c 1250 N.s/m, p 4000
sin (30 t) Pa, A 50 x 10-3 m2.

Determine (a) steady-state displacement
(b) max. force transmitted to the
base.

100
101
The amplitude of the steady-state vibration is
101
102
The force transmitted to the base is
For max Ftr
102
103
End of 22.2
Let Learning Continue
104
Newton Laws, Energy Theorems and Momentum
Principles
  • Integration Principles

105
Group Working Problems Problem 1 Chapters 17, 18
and 19
  • Given The 30-kg disk is pin connected at its
    center. If it starts from rest,
  • Find
  • the number of revolutions it must make and the
    time needed to attain an angular velocity of
    20rad/s?

106
Problem 1 Newton Laws
  • Find angular acceleration (11.7rad/s2)
  • Find angle (17.1 radians)
  • Find number of revolutions (2.73)
  • Find time (1.71 sec)

107
Problem 1 Energy Theorems
  • Find angle (17.1 radians)
  • Find number of revolutions (2.73)
  • Find angular acceleration (11.7rad/s2)
  • Find time (1.71 sec)

108
Problem 1 Momentum Principle
  • Find time (1.71 sec)
  • Find angular acceleration (11.7rad/s2)
  • Find angle (17.1 radians)
  • Find number of revolutions (2.73)

109
Problem 2 Chapter 22 in view of Chapters 17, 18
and 19
  • Given a 10-kg block suspended from a cord
    wrapped around a 5kg disk.
  • Find natural period of vibration

110
Problem 2 Newton Laws
  • Find differential equation
  • Find period (1.57s)

111
Problem 2 Energy Theorems use conservation of
energy
  • Write conservation of energy equation
  • Differentiate equation
  • Find period (1.57s)

112
Problem 2 Momentum Principle use Md(HG)/dt
  • Write Md(HG)/dt for a displacement ? from
    equilibrium position
  • Differentiate equation
  • Find period (1.57s)

113
End of Integration Principles
Let Learning Continue
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