1Lecture 4-Equivalent M & KMEEM 3700 1 of 18

MEEM 3700MEEM 3700Mechanical VibrationsMechanical Vibrations

Mohan D. Rao Chuck Van Karsen

Mechanical Engineering-Engineering MechanicsMichigan Technological University

Copyright 2003

Lecture 4-Equivalent M & KMEEM 3700 2 of 18

Spring Force is Linear and Proportional to Displacement.

F = KX

KFF

XX

fx

F

Sections 1.7 & 1.8 in the text

2Lecture 4-Equivalent M & KMEEM 3700 3 of 18

LF

AEK= L

A= Area of Cross-SectionE= Modulus of ElasticityL= Length

Stiffness Elements: Bar in Tension/Compression

Lecture 4-Equivalent M & KMEEM 3700 4 of 18

E = modulus of elasticity. I = area moment of inertia relative to bending axis. Kb = spring rate of beam.

F

L F = 3EI _____

L3

Kb = 3EI_____L3

3F= for a cantilever beam

3EIL

Similarly obtain Kb for other boundary conditionsfrom deflection equations

3Lecture 4-Equivalent M & KMEEM 3700 5 of 18

max

FL

max

LF

F = Kmax F = Kmax

Kb = 48EI______

L3Kb =

192EI_______L3

Fixed-Fixed Simply Supported

3

maxFL

= 48EI

3

maxFL

= 192EI

Lecture 4-Equivalent M & KMEEM 3700 6 of 18

LM

M

( )t L R tM = K - = K t= K

4t

GJK = J= d / 32 L

for a round shaft of diameter d

(J = Polar area moment)

tK

R

L

M=Moment

Nmunits=rad

4Lecture 4-Equivalent M & KMEEM 3700 7 of 18

Keq = K1 + K2

K1 K2

F

Equivalent System

Keq

F

How to combine Stiffness Elements?

Lecture 4-Equivalent M & KMEEM 3700 8 of 18

Equivalent System

Keq

F XXF

K1

K2

X1K1

F

F

X2K2

F

F

( )

1 2

1 1 2 2

1 2

1 2

1 2

1 2

1 1 1

eq

eq eq

eq

eq

X X XF K X K X K X

K X K XX

K K

K K K

K KKK K

= += = =

= +

= +

= +

5Lecture 4-Equivalent M & KMEEM 3700 9 of 18

FL

K X

Keq = K +3EI_____L3

Keq

F

Equivalent System

X

Lecture 4-Equivalent M & KMEEM 3700 10 of 18

F2

K1

K3

K2

F1

Case 1Case 11 eqF =K X

2 3eq 1

2 3

K KK =K +K +K

Case 2Case 22 eqF =K X

2 1eq 3

2 1

K KK =K +K +K

6Lecture 4-Equivalent M & KMEEM 3700 11 of 18

U = Potential Energy

21U = Kx2

2t

1or U = K 2

for a torsional spring

Spring Force

x

Kx

dx

2

0

x

1x x2

x

U K dx

U K dx K

== =

Lecture 4-Equivalent M & KMEEM 3700 12 of 18

Y

X

2eqK = Kcos

d

X

x cosd =Equivalent potential energy

( )2 21 1x cos x2 2 eq

U k k= =

eqK

in the x-direction

7Lecture 4-Equivalent M & KMEEM 3700 13 of 18

mg

bF mg gAx= +

x

Static equilibrium

Archimedes principle: The buoyant force acting on a floating body is equal to the weight of the liquid displaced by the body.

Spring Force (F)= Weight of the fluid displaced = mass*g =density*volume*g. Volume displaced= cross-sectional area*displacement x.

Hence, F= gAx= Keq x, x is measured from the static equilibrium.

Keq

F

Keq=gA

Lecture 4-Equivalent M & KMEEM 3700 14 of 18

M

Mass

X

Rigid Body Behavior

Gain / Lose Kinetic Energy

RotatingInertia

J

J = mass moment of inertia ( )2kgm

8Lecture 4-Equivalent M & KMEEM 3700 15 of 18

Kinetic Energy of translation

MX 21K.E. = mx

2&

Kinetic Energy of rotation

M,J 21K.E. = J2

&

Lecture 4-Equivalent M & KMEEM 3700 16 of 18

Kinetic Energy of rolling motion

rX

2 21 1K.E. = mx + J2 2

&&

22

2

2 2eq eq

1 1 xNOTE: x = r therefore K.E. = mx + J2 2 r

1K.E. m x where m = m+ J/r2

=

&&

&

translation rotation

9Lecture 4-Equivalent M & KMEEM 3700 17 of 18

d

CG

M , J

20

1K.E. = J 2

&

20 cgJ =J +Md

Parallel Axis Theorem (PAT)CG M , J

d

CG0

0

J0 is the mass Moment of Inertia of the body about the pivot 0. Jcg is the same about the center of mass

Lecture 4-Equivalent M & KMEEM 3700 18 of 18

Linear TorsionalViscous Damper

Ct

Linear Viscous Damper

C

Force (or Torque)

Velocity(Translation or Angular)

C or Ct

Viscous damping

Linear

tM = C &

F Cx= &