ME 5550 Intermediate Dynamics
Principle of Virtual Work Example
The figure shows a slider crank
mechanism under the action of an
external torque T acting on link AB ,
external forces P and Q acting at B and
C , and a linear spring attached at B .
Problem:
ME 555 Intermediate Dynamics
Equations of Motion of Example System II
In previous notes for Example System II, we
found R B the angular velocity of the bar and
%
[ I G ]e that the inertia matrix (associated with I G )
%
%
%
resolved in bar-fixed direction
ME 5550 Intermediate Dynamics
Partial Velocities and the Slider Crank Mechanism
System Configuration
The figure shows a simple slider
crank mechanism with no offset. Given
the physical dimensions of the links
( R, L) , the configuration of the system
at a
ME 5550 Intermediate Dynamics
Principle of Virtual Work
Principle of Virtual Work
If a mechanical system whose configuration is defined by a set of independent
generalized coordinates qk (k = 1,., n) is in static equilibrium, then the
generalized force as
ME 555 Intermediate Dynamics
Angular Momentum of a Rigid Body about an Arbitrary Point
The angular momentum of a rigid body B
about its mass center G is defined as
H G = ( rP / G R vP ) dm
%
%
%
B
In previous notes, it was shown that H G may be
%
written
ME 555 Intermediate Dynamics
Angular Momentum of a Rigid Body about its Mass Center
To calculate the angular momentum of a rigid body
about its mass center, consider the body B shown in the
figure to the right. The point P represents any point in
the body
ME 555 Intermediate Dynamics
Moments and Products of Inertia and the Inertia Matrix
Moments of Inertia
A rigid body B is shown in the diagram below. The unit vectors (e1 , e2 , e3 ) are fixed in
% % %
the body and are directed along a convenient set of ax
ME 555 Intermediate Dynamics
Kinetic Energy of a Rigid Body
The figure at the right depicts a rigid body B moving
relative to a fixed frame R. The kinetic energy of B is
defined as
K = 1 ( R vP R vP ) dm .
2
%
%
B
A more useful definition can be derived b
ME 555 Intermediate Dynamics
Newton/Euler Equations of Motion for a Rigid Body
Using the theory of systems of particles, it can be shown that the equations of motion for
rigid body motion in an inertial frame R may be written as
F = m
%
i
i
( M G )i =
i
ME 5550 Intermediate Dynamics
Angular Momentum and Kinetic Energy of a Misaligned Disk
ME 5550 Example 12: (Disk welded to the shaft)
Reference frames: (R is the fixed frame)
S : i , j, k (rotates with the shaft; aligned with the shaft)
D : i , e 2 , e 3
ME 555 Intermediate Dynamics
Orientation of a Rigid Body Using Euler Parameters
Euler's Theorem on Rotation
Consider the rigid body shown in the figure to
the right. Let R : ( N1, N 2 , N 3 ) represent the base
% % %
reference frame and B : ( e1 , e2 , e3
ME 555 Intermediate Dynamics
Partial Velocities and Partial Angular Velocities
Partial Velocities
If the velocity of some point P within a mechanical system can be written in terms of a set
of generalized coordinates qi (i = 1, n) and their time derivativ
ME 555 Intermediate Dynamics
Linearization of Differential Equations of Motion
The equations of motion (EOM) derived using Newtons laws or Lagranges
equations may be linear or nonlinear. If they are nonlinear, it may be possible to
linearize the equations
ME 555 Intermediate Dynamics
Angular Momentum and Kinetic Energy of Example System II
Example system II consists of a T-shaped frame
F and a slender bar B that rotates like a propeller
relative to F . The angular momentum of the bar
about its mass center
ME 555 Intermediate Dynamics
Introduction to Lagrangian Dynamics
Newton/Euler Equations of Motion
One approach to finding the equations of motion (EOM) of a mechanical system is
to use the Newton/Euler equations of motion
F = m
%
i
i
R
aG
%
( M ) = ( I%
ME 5550 Intermediate Dynamics
Generalized Forces
General Definition
Given a mechanical system whose configuration is defined by a set of generalized
coordinates qk (k = 1,., n) , we can define a generalized force associated with each of the
"n" generalize
ME 555 Intermediate Dynamics
Lagrange's Equations Example System II
In previous notes for Example System II, we
found R B the angular velocity of the bar and H G
the angular momentum of B resolved in bar-fixed
directions B : (e1 , n2 , e3 ) to be
B = (S
ME 555 Intermediate Dynamics
Bearing Loads on a Simple Crank Shaft
The figure to the right shows a
simple crank shaft consisting of seven
segments, each considered to be a
slender bar. Each segment of length
has mass m .
There are six
segments of length
a
ME 555 Intermediate Dynamics
Degrees of Freedom of Mechanical Systems
System Configuration and Generalized Coordinates
The configuration of a mechanical system is defined as the position of each of the
bodies within the system. In general, both translatio
ME 555 Intermediate Dynamics
Lagrange's Equations Examples
Example #1
The system at the right consists of two
bodies, a slender bar B and a disk D, moving
together in a vertical plane. As B rotates
about O, D rolls without slipping on the fixed
circular o
ME 555 Intermediate Dynamics
Lagranges Equations for Multi-Degree-of-Freedom Systems
The configuration of systems with N degrees of freedom (DOF) can be defined in
terms of N generalized coordinates, say qk (k = 1,K, N ) . The differential equations
of mo
ME 555 Intermediate Dynamics
Natural Frequencies and Mode Shapes
To calculate the natural frequencies and mode shapes for multiple degree-offreedom (DOF), rigid-body systems, we first linearize the equations of motion
(EOM) and express them in the matrix
ME 5550 Intermediate Dynamics
Angular Momentum and Kinetic Energy of a Simple Crank Shaft
ME 5550 Example #11
The figure shows a simple crank shaft consisting of seven segments, each considered to be a
slender bar. Each segment of length has mass m . Ther