01 ce225 multi-degree of freedom systems

8/12/2019 01 CE225 Multi-Degree of Freedom Systems

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Multi-Degree-of-Freedom Systems


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Multi-Degree-of-Freedom Systems

Degree-of-Freedom is two or more Degree-of-Freedom is the number of independent

displacement coordinates necessary to describe the

motion of the system.


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k

u

m

Examples of Single-Degree-of-Freedom Systems

u

k

m

u

k

m

m

k

u


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c1

u1

c2

k1 k2

m1 m2

p1(t) p2(t)

u2

Examples of Two-Degree-of-Freedom Systems

u1

k1

m1

k2

m2

u2

2-DOF

2-DOF

u1

u2

2-DOF

Rigid bar


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Example of Two-Degree-of-Freedom System

k1

k2

m1

m2

u1

u2

2-DOF

Shearbuilding


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Example of Three-Degree-of-Freedom System

c1

u1

c2 c3

k1 k2 k3

m1 m2 m3

p1(t) p2(t) p3(t)

u2 u3


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Example of Three-Degree-of-Freedom System

p1(t)

p2(t)

m1u1

u2

c1

c2

k1

k2

p3(t) u3

k3

m2

m3

c3


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c1

u1

k1

m1 m2

p1(t) p2(t)

u2

Degree of Freedom does not necessarily mean the number

of lumped masses

u1

k1

m1

m2

u2

1-DOF

1-DOF

Rigid bar

u1 =u2

Rigid bar

u1 =u2


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Equation of Motion of a Two-

Degree-of-Freedom System


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Two-Degree-of-Freedom Spring System (Classical)

c1

u1

c2

k1 k2

m1 m2

p1(t) p2(t)

u2

c1

u1

c2

k1 k2

m1 m2

p1(t) p2(t)

u2

u1

p1(t) p2(t)

u2

c1u1c2(u2-u1)

k1u1 k2(u2-u1)

c2(u2-u1)

k2(u2-u1)

m1u1 m1u1


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Spring System

u1

p1(t)

p2(t)

u2

c1u1 c2(u2-u1)

k1u1 k2(u2-u1)

c2(u2-u1)

k2(u2-u1)

m1u1

m2u2

m1u1+ + = p1(t)c1u1 k1u1c2(u2-u1)- k2(u2-u1)-

m2u2+ + = p2(t)k2(u2-u1)c2(u2-u1)


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Spring System

c1

u1

m u(t) + c u(t) + k u(t) = p(t)

c2

k1 k2

m1 m2

p1(t) p2(t)

u2

Equation of Motion

m1 0

0 m2

u1

u2

+

c1+c2 -c2

- c2 c2

u1

u2

+

k1+k2 -k2

- k2 k2

u1

u2

=

p1(t)

p2(t )


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Spring System

c1

u1

m v(t) + c v(t) + k v(t) = p(t)

c2 c3

k1 k2 k3

m1 m2 m3

p1(t) p2(t) p3(t)

u2 u3

Equation of Motion

m1 0 0

0 m2 0

0 0 m3

+

c1+c2 -c2 0

- c2 c2+c3 -c3

0 -c3 c3

+

k1+k2 -k2 0

- k2 k2+k3 -k3

0 -k3 k3

=

p1(t)

p2(t)

p3(t )

u1

u2

u3

u1

u2

u3

u1

u2

u3


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Two-Story Shear Building

(Assumptions)

Beams and floor systems are rigid (infinitely stiff)

in flexure

Axial deformations of beams and columns are

neglected Effect of axial force on stiffness of the columns

are neglected

Mass is concentrated at the floor levels Linear viscous damping is associated with

deformational motions of each story


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Two-Story Shear Building

p1(t)

p2(t)

m1

m2

u1

u2

c1

c2

k1

k2

u2

u1


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Using Newtons Second Law of Motion

p2(t)

p1(t)


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p2(t)

p1(t)


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p2(t)

p1(t)


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p2(t)

p1(t)

m2u2

k2(u2-u1)

k1u1 c1u1

m2u2+ += p2(t)

m1u1+ + = p1(t)

m1u1

c1u1 k1u1

c2(u2-u1)

k2(u2-u1) c2(u2-u1)

k2(u2-u1)c2(u2-u1)

c2(u2-u1)- k2(u2-u1)-


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m2u2+ + =p2

(t)

m1u1+ + = p1(t)c1u1 k1u1

k2(u2-u1)c2(u2-u1)

c2(u2-u1)- k2(u2-u1)-

m2u2- + =p2(t)k2u2c2u1 c2u2+ k2u1-

m1u1+ - = p1(t)k2u2(c1+c2)u1 c2u2- + (k1+k2)u1


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m2u2- + =p2(t)k2u2c2u1 c2u2+ k2u1-

m1u1+ - = p1(t)k2u2(c1+c2)u1 c2u2- + (k1+k2)u1

m1u1+ 0 u2 + (k1+k2)u1k2u2-(c1+c2)u1 c2u2-

= p1(t)

0 u1+ m2u2 - c2u1 + c2u2

+

k2u1- + k2u2= p2(t)

m1 0

0 m2

u1

u2

+c1+c2 -c2

- c2 c2

u1

u2

+

k1+k

2-k2

- k2 k2

u1

u2

=

p1(t)

p2(t )


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m1 0

0 m2

u1

u2

+

c1+c2 -c2

- c2 c2

u1

u2

+

k1+k2 -k2

- k2 k2

u1

u2

=

p1(t)

p2(t )

m u+

c u+

k u = p(t)

mass

matr ix

acc.

vector

damping

matr ix

vel.

vector

st i f fness

matr ix

disp.

vector

loading

vector


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m1 0

0 m2

u1

u2

+

c1+c2 -c2

- c2 c2

u1

u2

+

k1+k2 -k2

- k2 k2

u1

u2

=

p1(t)

p2(t )

m u+

c u+

k u = p(t)

This is the equation of motion of the two-story shear

building

The matrix equation represents two ordinary differential

equations

Each equation contains both u1 and u2, and, therefore,

coupled.


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General Form of the Equation of Motion

m11 m12

m21 m22

u1

u2

+

u1

u2

+

u1

u2

=

p1(t)

p2(t )

Physical meaning of each element of the matrices mij, cij, kijis the force at the i

thmass due to a unit

acceleration, velocity or displacement at the jthmass,

respectively, with all other accelerations, velocities

and displacements equal to zero.

c11 c12

c21 c22

k11 k12

k21 k22


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p2(t)

p1(t)

m2u2

k2(u2-u1)

k1u1 c1u1

m2u2+ += p2(t)

m1u1+ + = p1(t)

m1u1

c1u1 k1u1

c2(u2-u1)

k2(u2-u1) c2(u2-u1)

k2(u2-u1)c2(u2-u1)

c2(u2-u1)- k2(u2-u1)-


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Equation of Motion of a Three-



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Equation of Motion of a 3-DOF System

m1 0 0

0 m2 0

0 0 m3

u1

u2

u3

+

c1+c2 -c2 0

- c2 c2+c3 -c3

0 -c3 c3

u1

u2

u3

+

k1+k2 -k2 0

- k2 k2+k3 -k3

0 -k3 k3

u1

u2

u3

=

p1(t)

p2(t)

p3(t )

p1(t)

p2(t)

m1u1

u2

c1

c2

k1

k2

p3(t) u3

k3

m2

m3

c3


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Problem: Formulate the equation of motion of the given system.

Level Weight (ton) Stiffness

(ton/cm)

5 100 53

4 100 61.5

3 100 96

2 100 115

1 100 114

5

4

3

2

1

m = W/g = W ton/980 cm/sec2

100 0 0 0 0 y1 -115 -115 0 0 0 y1 0

0 100 0 0 0 y2 -115 211 -96 0 0 y2 0

0 0 100 0 0 y3 + 0 -96 157.5 -61.5 0 y3 = 0

0 0 0 100 0 y40 0 -61.5 114.5 -53

y4 00 0 0 0 100 y5 0 0 -53 53 y5 0

1/980

ton-sec2/cm cm ton/cmcm ton


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Equation of Motion of a Two-


(Base Excitation, Undamped)


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SDOF SystemOne-storey building with rigid girder

(Influence of Support Excitation)

fixedr

eferen

cea

xis

vg(t)

vT(t)

v(t)

fs(t)

2

fs(t)

2

fD(t)

fI(t)

m

k

2

k

2fD(t)= c v(t)

fs(t)= k v(t)

fI(t)= m vT(t)

F = 0 inertia force

- fS(t) - fD(t) - fI(t) = 0

- k v(t) - c v(t) - m (vg(t) + v(t) ) = 0

m v(t)+ + k v(t) =c v(t) - m vg (t)


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2-DOF SystemTwo-storey building with

rigid girder (Influence of Support Excitation)

fixedr

eferencea

xis

vg

v1T

v1

m1

k

2

k

2

m2

k

2

k

2

v2T

v2

Applications:

1. Motion of building caused byearthquake

2. Motion of equipment due tomotion of building where it ishoused


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= + v1

disp of mass1 rel to

moving support/base

total disp of mass1 rel

to fixed reference axis

vgv1T

disp of frame support

rel to fixed reference

axis

fixedr

eferencea

xis

vg

v1T

v1

m1

k

2

k

2

m2

k

2

k

2

v2T

v2


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= +

disp of mass2 rel to

moving support/base

total disp of mass2 rel

to fixed reference axis

vgv2T

disp of frame support

rel to fixed reference

axis

fixedr

eferencea

xis

vg

v1T

v1

m1

k

2

k

2

m2

k

2

k

2

v2T

v2

v2


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fixedr

eferencea

xis

v

m1 v1T

k1v1

v1

v2

v1)-k2(v2

m2 v2T v1)-k2(v2+ = 0

v1)-k2(v2

m1 v1T+ k1v1 - v1)-k2(v2 = 0

m2 v2T


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vg displacement of support/ground from a fixed reference axis

v1 displacement of mass1 relative to the base

v2 displacement of mass2 relative to the base

m2v2T v1)-k2(v2+ = 0

m1v1T + k1v1 - v1)-k2 (v2 = 0

= + v1vgv1T

= +vgv2T v2

v1 = + v1vgv1T

= +vgv2T v2

v1

m2v2 k 2v2+k2v1- =

m1v1 + (k1+ k2) v1 - k2 v2 = - m1vg

- m2vg


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m2v2 k 2v2+k2v1- =

m1v1 + (k1+ k2) v1 - k2 v2 = -m1vg

- m2vg

m1

0

0

m2

+

v1

v2

k1+k2

- k2

- k2

k2

v1

v2= -

m1

0

0

m2

1

1

vg

M v + K v = - M 1 vg


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Quiz

m1

0

0

m2

+

v1

v2

k1+k2

- k2

- k2

k2

v1

v2= -

m1

0

0

m2

1

1

vg

MULTIPLY THE MATRICES

2 x 2 2 x 1 2 x 2 2 x 1 2 x 2 2 x 1


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Quiz

m1

0

0

m2

+

v1

v2

k1+k2

- k2

- k2

k2

v1

v2= -

m1

0

0

m2

1

1

vg

MULTIPLY THE MATRICES

ANSWER

m1v1

0 v1

+ 0 v2

m2+ v2

+( k1+k2 )v1 - k2v2

- k2v1 + k2v2

=

- m1 vg

- m2 vg

2 x 1 2 x 1 2 x 1


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QuizADD THE MATRICES

m1v1

0 v1

+ 0 v2

m2+ v2

+( k1+k2 )v1 - k2v2

- k2v1 + k2v2

=

- m1 vg

- m2 vg

2 x 1 2 x 1 2 x 1


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QuizADD THE MATRICES

ANSWER

m1v1

0 v1

+ 0 v2

m2+ v2

+( k1+k2 )v1 - k2v2

- k2v1 + k2v2

=

- m1 vg

- m2 vg

2 x 1 2 x 1 2 x 1

m1v1+ 0 v2( k1+k2 )v1 - k2v2+

0 v1 m2+ v2- k

2 + k2v2

2 x 1

=

- m1 vg

- m2vg

2 x 1


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QuizWHAT ARE THE RESULTING TWO EQUATIONS?

m1v1+ 0 v2( k1+k2 )v1 - k2v2+

0 v1m2+ v2 - k2 + k2v2

2 x 1

=

- m1 vg

- m2 vg

2 x 1


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QuizWHAT ARE THE RESULTING TWO EQUATIONS?

ANSWER

m1v1+ 0 v2( k1+k2 )v1 - k2v2+

0 v1m2+ v2 - k2 + k2v2

2 x 1

=

- m1 vg

- m2 vg

2 x 1

m1v1+ 0 v2( k1+k2 )v1 - k2v2+ = - m1 vg

0 v1 m2+ v2 - k2 + k2v2 - m2 vg=


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QuizCOMBINE THE TWO EQUATIONS INTO ONE MATRIX EQUATION

m1v1( k1+k2 )v1 - k2v2+ = - m1 vg

m2v2 - k2 + k2v2 - m2 vg=


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QuizCOMBINE THE TWO EQUATIONS INTO ONE MATRIX EQUATION

m1v1( k1+k2 )v1 - k2v2+ = - m1 vg

m2v2 - k2 + k2v2 - m2 vg=

ANSWER

m1

0

0

m2+

v1

v2

k1+k2

- k2

- k2

k2

v1

v2= -

m1

0

0

m2

1

1vg


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k4= 1600 kN/mk4= 1600 kN/m

k3= 1200 kN/mk3= 1200 kN/m

m4= 4500 kg

k2= 800 kN/mk2= 800 kN/m

m3= 3000 kg

m2= 3000 kg

k1= 400 kN/mk1= 400 kN/m

m1= 1500 kg

x1

x2

x3

x4


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k4= 1600 kN/mk4= 1600 kN/m

k3= 1200 kN/mk3= 1200 kN/m

m4= 4500 kg

k2= 800 kN/mk2= 800 kN/m

m3= 3000 kg

m2= 3000 kg

k1= 400 kN/mk1= 400 kN/m

m1= 1500 kgx1

x2

x3

x4

x1=1

x2=0

x3=0

x4=0

800 kN

800 kN

800 kN

- 800 kN

0

0


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k4= 1600 kN/mk4= 1600 kN/m

k3= 1200 kN/mk3= 1200 kN/m

m4= 4500 kg

k2= 800 kN/mk2= 800 kN/m

m3= 3000 kg

m2= 3000 kg

k1= 400 kN/mk1= 400 kN/m

m1= 1500 kgx1

x2

x3

x4

x1=0

x2=1

x3=0

x4=0

800 kN

800 kN

1600 kN

1600 kN

- 800 kN

2400 kN

- 1600

0


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k4= 1600 kN/mk4= 1600 kN/m

k3= 1200 kN/mk3= 1200 kN/m

m4= 4500 kg

k2= 800 kN/mk2= 800 kN/m

m3= 3000 kg

m2= 3000 kg

k1= 400 kN/mk1= 400 kN/m

m1= 1500 kg

x1

x2

x3

x4

x1=0

x3=1

x2=0

x4=0

1600 kN

1600 kN

2400 kN

2400 kN

0

- 1600 kN

4000 kN

-2400 kN


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k4= 1600 kN/mk4= 1600 kN/m

k3= 1200 kN/mk3= 1200 kN/m

m4= 4500 kg

k2= 800 kN/mk2= 800 kN/m

m3= 3000 kg

m2= 3000 kg

k1= 400 kN/mk1= 400 kN/m

m1= 1500 kg

x1

x2

x3

x4

x1=0

x4=1

x2=0

x3=0

2400 kN

2400 kN

3200 kN

3200 kN

0

0

-2400 kN

5600 kN


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800

- 800

0

0

- 800

2400

- 1600

0

0

- 1600

4000

-2400

0

0

-2400

5600

F1

F2

F3

F4

=

x1

x2

x3

x4

1 2 3 4


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1500

0

0

0

0

3000

0

0

0

0

3000

0

0

0

0

4500

F1

F2

F3

F4

=

x1

x2

x3

x4

1 2 3 4


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0

0

x4

1500

0

0

0

0

3000

0

0

0

0

3000

0

0

0

0

4500

x1

x2

x3

x4

1 2 3 4

+

800

- 800

0

0

- 800

2400

- 1600

0

0

- 1600

4000

-2400

0

0

-2400

5600

x1

x2

x3

1 2 3 4

=0

0

M x + K x = 0


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M x + K x = 0

Equation of motion:

Let x = a sin ( t )

x = a sin ( t )- 2

(a)

(b)

Substitute (b) to (a) results in

K - 2 Ma = 0

=12

.

.

n

a1 =a11a21

.

.

an1

a2 =a12a22

.

.

an2

etc.

Mode1 Mode2


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END

01 ce225 multi-degree of freedom systems

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