single degree of freedom system equation of motion, problem statement & solution methods...
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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