Mechanical Systems- Part II
Chapter 3
Objectives
In this unit, you will learn
▪ lumped inertia and stiffness of flexible/compliant mechanical systems
▪ calculation of natural frequencies for single DOF compliant mechanisms
▪ modeling of multiple DOF mechanical systems, natural frequencies and mode shapes
▪ analysis of free and forced vibration of multiple DOF mechanical systems
▪ employing MATLAB and Simulink in solving models of multiple DOF mechanical systems
Lumped Inertia and StiffnessBEAMS in bending
13eJ J
BAR in torsion
te
GIk
l
33140em m
1335em m
Fixed-free bar See Tables 3.1 and 3.2 in the text
Fixed-guided beam
3
12e
EIkl
Lumped Stiffness of Compliant Elements
3
3 3
12
312
22
ys
y yl
EIk
lEI EI
kll
,
1 2 1
e s s lk k k
, , 3
122
5y
e e p e s
EIk k k
l
Example 3.4
Problem 3.4
4
2
y zk k
yieldsw h
Microbridge (Problem 3.9)
Multiple DOF Mechanical Systems
# DOF = # apparent DOF – # constraints
# apparent DOF = 5 f,q,x,y,z# constraints = 3
# DOF = 5-3 =2
Single DOF
Two DOF
7
• In a conservative system, the total energy is constant. • The kinetic energy T is stored in the mass by virtue of its
velocity, whereas the potential energy U is stored in the form of strain energy in elastic deformation or by a spring or work done in a force field such as gravity. The total energy being constant, its rate of change is zero.
• T + U = constant Tmax = Umax
• d/dt( T + U ) = 0
Energy Method
Multiple DOF Conservative Mechanical Systems Modeling
Energy Method
Multiple DOF Conservative Mechanical Systems Modeling
Problem 3.122-DOF, 2 coupled equations
Multiple DOF Mechanical Systems: Natural frequencies and mode shapes
Problem 3.14
Let
Determinant =0 yields two natural frequencies (2-DOF)
2
2
ω 20
2 2ω 5m k k
k m k
Mode shapes
1 2
1 1 2 2ω ωsin ω ψ sin ω ψ
φ n nn n
zA V t B V t
l
Solution
1
2
21
1 21
22
2 22
2 52 82 41 3
2 52 82 41 3
n
n
n
n
n
n
m kZ krl m k k
m kZ krl m k k
1 2ω ω
8 8;41 3 41 3
1 1n n
V V
Eigen values and eigen vectors using MATLAB
θ θ 0M K M- mass matrixK- stiffness matrix
D- dynamic matrix
2
1
λ ω
D M K
θ Θ sin ωt
2ω Θ Θ 0M K λ Θ 0D I
0 2;
0 2 2 5m k k
M Km k k
>> syms k m omsq>> X=[k-m*omsq, -2*k; -2*k, 5*k-2*m*omsq]>> solve(det(X), omsq)
Let m=0.6 kg and k =110 N/m.
>> m=0.6;>> k=110;>> M=[m,0;0,2*m];>> K=[k,-2*k; -2*k, 5*k];>> d=inv(M)*K;>> [V,D]=eig(d)
d = 183.3333 -366.6667 -183.3333 458.3333
ans = (7*k + 41^(1/2)*k)/(4*m) (7*k - 41^(1/2)*k)/(4*m)
V = -0.9202 0.6480 -0.3914 -0.7616D = 27.3568 0 0 614.3099
Double Pendulum EOMs (Problem 3.17)
For small oscillations
Applying Newton’s Second law
Coupled EOMs
Multiple DOF mechanical systems: forced response by Simulink (Problem 3.20)
Review Example 3.10.
Mathematical Model
Based on values given, number crunching leads us to
Step 3.33X106
Step1 -2.52X106
Simulink (Problem 3.20 continued)