SINGLE DEGREE OF FREEDOM SYSTEM Equation of Motion, Problem Statement &

Solution MethodsPertemuan 19

Matakuliah : Dinamika Struktur & Teknik GempaTahun : S0774

Systems with two degree of freedomRecap

Pendulum systems (double pendulum)

Torsional systems

String systems

Linear systems

Definite and semi-definite systems

Analysis of various 2DOF systems such as:

Systems with two degree of freedom

Dynamic coupling

Static coupling

Principal co-ordinates

Co-ordinate coupling

Systems with two degree of freedomProblem-1

Obtain the equations of motion of the system shown in the figure.The vibration is restricted in plane of paper

m -mass of the systemJ -mass MI of the systemG -centre of gravity

G

K1 K2

m,J

a b

Systems with two degree of freedom

G

K1K2

m,J

a bThe system has two generalized co-ordinates, x and Cartesian (x), Polar ()

Problem-1

(x-a)(x+b)

x

G

Static equilibriumline

a b

Dr. S. K. Kudari, Professor, BVB College of Engg. & Tech., Hubli

Systems with two degree of freedom

Eqns. of motion

The Lagrange’s equation is :

iiii

Qx

U

x

T

x

T

dt

d

22 θJ2

1xm

2

1T

222

1 bθxK2

1aθ-xK

2

1U

θ

xx i

generalized co-ordinates

Problem-1

(x-a)(x+b)

x

G

Static equilibriumline

a b

Dr. S. K. Kudari, Professor, BVB College of Engg. & Tech., Hubli

Systems with two degree of freedom

The Lagrange’s equation is :

1Qx

U

x

T

x

T

dt

d

22 θJ2

1xm

2

1T

222

1 bθxK2

1aθ-xK

2

1U

θ

xx i

xmx

T

dt

d

)1(

)1(

bθxK

aθ-xKx

U

2

1

0x

T

Problem-1

Systems with two degree of freedom

equations of motion (1st)

)1()1( bθxKaθ-xKx

U21

bθKxKaθK-xKx

U2211

)bK-aθ(K)Kx(Kx

U2121

0b)θK-a(K-)xK(Kxm 2121 First Eqn. of motion

Problem-1

Systems with two degree of freedom

equations of motion (2nd)

θJθ

T

dt

d

1

0θ

T

(b)bθxKa)(aθ-xKθ

U21

θbKxbKθaKxaKθ

U 222

211

)θbKaKb)xKaKθ

U 22

2121

((

Problem-1

22 θJ2

1xm

2

1T

222

1 bθxK2

1aθ-xK

2

1U

θ

xx i

1Qθ

U

θ

T

θ

T

dt

d

Systems with two degree of freedom

equations of motion

0b)θK-a(K-)xK(Kxm 2121

Second Eqn. of motion

0)θbKaKb)xKaKθJ 22

2121 ((

0)θbKaKb)xKaKθJ 22

2121 ((

equations of motion are:

Problem-1

First

Second

Systems with two degree of freedom

equations of motion0b)θK-a(K-)xK(Kxm 2121

0)θbKaKb)xKaKθJ 22

2121 ((

Matrix form

0

0

θ

x

)bKa(Kb)Ka(K

b)Ka(K)K(K

θ

x

J0

0m2

22

121

2121

Problem-1

Systems with two degree of freedom

Matrix form

0

0

θ

x

)bKa(Kb)Ka(K

b)Ka(K)K(K

θ

x

J0

0m2

22

121

2121

Mass/inertia matrix Stiffness matrix

Stiffness matrix shows that co-ordinate x and are dependent on each other. Any change in x reflects in change in As seen from the matrix, the equations of motion are coupled with stiffness. This condition is referred as STATIC COUPLING

coupling in mass matrix is referred as DYNAMIC COUPLING

Problem-1

Systems with two degree of freedom

Matrix form

0

0

θ

x

)bKa(Kb)Ka(K

b)Ka(K)K(K

θ

x

J0

0m2

22

121

2121

Mass/inertia matrix Stiffness matrix

From the above equations, it can be seen that system do not have dynamic coupling

But, it has static coupling

Problem-1

Systems with two degree of freedom

Matrix form

0

0

θ

x

)bKa(Kb)Ka(K

b)Ka(K)K(K

θ

x

J0

0m2

22

121

2121

Mass/inertia matrix Stiffness matrix

To have static uncoupling the condition to be satisfied is: K1a=K2b

Problem-1

Systems with two degree of freedom

Matrix form

0

0

θ

x

)bKa(K0

0)K(K

θ

x

J0

0m2

22

1

21

The uncoupled Eqns. of motion are

0)xK(Kxm 21

0)θbKaKθJ 22

21 (

Contains only one coordinate, x

Contains only one coordinate,

Under such conditions, x and are referred as PRINCIPAL COORDINATES

Problem-1

Systems with two degree of freedom

Solution to uncoupled Eqns. of motion:

0)xK(Kxm 21

m

KKω 21

1

0xm

KKx 21

From Eqn.1:

0)θbKaKθJ 22

21 (

From Eqn.2:

0θJ

bKaKθ

22

21

J

bKaKω

22

21

2

Problem-1

Systems with two degree of freedom

Obtain the equations of motion of the system shown in the figure. The centre of gravity is away from geometric centre by distance e The vibration is restricted in plane of paper

m -mass of the systemJ -mass MI of the systemG -centre of gravityC -centre of geometry

C

K1 K2

m,J

a b

G

e

Problem-2

Systems with two degree of freedom

Due to some eccentricity e, the changes are:x=x+eJ=J+me2

Substitute in Eqns. of motion of earlier problem having e=0:

Problem-2

K1(x-a)K1(x+b)

x

G C

x+e

Static equilibriumline

a b

Systems with two degree of freedom

equations of motion for system having e=0

0b)θK-a(K-)xK(Kxm 2121

0)θbKaKb)xKaKθJ 22

2121 ((

Substitute x=x+e and J=J+me2=Jn in above Eqns.

0b)θK-a(K-)xK(Kθexm( 2121 )

0b)θK-a(K-)xK(Kθmexm 2121

0)θbKaKb)xKaKθJ 22

2121n ((

Problem-2

Systems with two degree of freedom

New equations of motion are

0b)θK-a(K-)xK(Kθmexm 2121

0)θbKaKb)xKaKθJ 22

2121n ((

0

0

θ

x

)bKa(Kb)Ka(K

b)Ka(K)K(K

θ

x

J0

mem2

22

121

2121

n

Matrix form

Dynamic coupling

Static coupling

Problem-2

Systems with two degree of freedom

Derive expressions for two natural frequencies for small oscillation of pendulum shown in figure in plane of the paper. Assume rods are rigid and mass less

a a

a

m

m

K

Problem-3

Systems with two degree of freedom

a a

a

m

m

K

a a

amg

mg

Ka(2-1)

1 2

22θJ

1θJ1

Equilibrium diagramProblem-3

Systems with two degree of freedom

Equilibrium diagram

a a

amg

mg

Ka(2-1)

1 2

22θJ

1θJ1

0))(acosθθKa(θ)mg(asinθθJ 112111 For first mass

as is smaller

0)θ(θKamgaθθJ 122

111

0θKaθKamgaθθ)(ma 12

22

112

0KaθKa)θ(mgθma 211

First Eqn. of motion

Problem-3

Systems with two degree of freedom

Equilibrium diagram

a a

amg

mg

Ka(2-1)

1 2

22θJ

1θJ1

For second mass

as is smaller

Second Eqn. of motion

0))(acosθθKa(θ)mg(2asinθθJ 212222

0θKaθKa2mgaθθm(2a) 12

22

212

0Ka)θ(2mgKaθθ(4ma) 211

Problem-3

Systems with two degree of freedom

0KaθKa)θ(mgθma 211 First Eqn. of motion

Second Eqn. of motion0Ka)θ(2mgKaθθ(4ma) 211

0

0

θ

θ

Ka)(2mgKa

KaKa)(mg

θ

θ

4ma0

0ma

2

1

2

1

Eqns. of motion in matrix form

Problem-3

For static coupling Ka=0, which is not possible

Systems with two degree of freedom

0KaθKa)θ(mgθma 211 First Eqn. of motion

Second Eqn. of motion0Ka)θ(2mgKaθθ(4ma) 211

Problem-3

Solution to governing eqns.:

φωtsinAx 11 φ)sin(ωAx 22 t

Assume SHM

The above equations have to satisfy the governing equations of motions

Equations of motion

Systems with two degree of freedom

0φ)sin(ωKaAφ)sin(ωAmaωKa)(mg 212

0φ)sin(ωA4maωka)(2mgφ)sin(ωKaA 22

1

0φ)sin(ω tIn above equations

String systems

0KaAAmaωKa)(mg 212

0A4maωka)(2mgKaA 22

1

Characteristic Eqns.:

Systems with two degree of freedom

0

4maωKa)(2mg

Ka........

Ka

maωKa)(mg2

2

The above equation is referred as a characteristic determinant Solving, we get :

Frequency equation

String systems

222 Ka4maωKa)(2mgmaωKa)(mg

Systems with two degree of freedom

21 ω....and...ω

Solve the frequency Eqn. for Natural frequencies of the system

String systems

As the system has two natural frequencies, under certain conditions it may vibrate with first or second frequency, which are referred as principal modes of vibration

Summary

Dynamic coupling

Static coupling

Principal co-ordinates

Co-ordinate coupling

Thank You