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)