10 CE225 MDOF Seismic Inertial Forces

8/11/2019 10 CE225 MDOF Seismic Inertial Forces

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MDOF Seismic Inertial Forces


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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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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

(Loading Applied on Mass)

p1(t)

p2(t)

m1

m2

u1

u2

c1

c2

k1

k2

u2

u1


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Equation of Motion of Two-Story Shear

Building

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)


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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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Equation of Motion of a Two-

Degree-of-Freedom System

(Base Excitation, Undamped)


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2-DOF System Two-storey building with

rigid girder (Influence of Support Excitation)

= + 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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11/42

fixedr

ef

erencea

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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vgdisplacement of support/ground from a fixed reference axis

v1displacement of mass1 relative to the base

v2displacement 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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If damping is considered:

+ = -

m1

0

0

m2

1

1

vg

= - 1 vg

Equation of Motion (Ground Acceleration)

m1 0

0 m2

u1

u2

c1+c2 -c2

- c2 c2

+

k1+k2 -k2

- k2 k2

u1

u2

u1

u2

m u + c u +k u m


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Equation of Motion (Loading on Mass)

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)

+ = -

m1

0

0

m2

1

1

vg

= - 1 vg


m1 0

0 m2

u1

u2

c1+c2 -c2

- c2 c2

+

k1+k2 -k2

- k2 k2

u1

u2

u1

u2

m u + c u +k u m


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Equivalence:

m1

m2

vg

m1

m2

==== -m1vg

-m2vg


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Disregard damping; solve eigenvalues and

eigenvectors. Uncouple the equation of motion:

+ = -

m1

0

0

m2

1

1

vg

= - 1 vg


m1 0

0 m2

u1

u2

c1+c2 -c2

- c2 c2

+

k1+k2 -k2

- k2 k2

u1

u2

u1

u2

m u + c u +k u m

1

T m 1 +

z1

1

T c 1

+z1 1

T k 1

z1

= 1

T m- 1 vg

2

T m 2

+z2

2

T c 2

+z2

2

T k 2

z2

= 2

T m- 1 vg


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1

T m 1

+z1

1

T c 1

+z1

1

T k 1

z1

= 1

T m- 1 vg

2

T m 2

+z2

2

T c 2

+z2

2

T k 2

z2

= 2

T m- 1 vg

The above equations can be combined as:

T

m +z T

c +z T

k z = T

m- 1 vg


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1

T m 1

+z1

1

T c 1

+z1

1

T k 1

z1

= 1

T m- 1 vg

2

T m 2

+z2

2

T c 2

+z2

2

T k 2

z2

= 2

T m- 1 vg

m1* c 1* k 1* p 1*

m1*

+z1

c 1*

z1

+ k 1* z

1= - p 1

*vg

m2* c 2*

k 2*

p 2*

m2*

+z2

c 2*

z2

+ k 2* z

2= - p 2

*vg


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m1*

+z1

c 1*

z1

+ k 1* z

1= - p 1

*vg

m2

*

+z2 c 2*

z2 + k 2* z

2 = - p 2*

vg

The general form of the above equations is:

mi*

+zi

c i*

zi

+ k i* z

i= - p i

*vg

These are equations of motion of single-

degree-of-freedom systems, vibrating under

each mode.


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Divide each term by :

mi*

+zi

c i*

zi

+ k i* z

i= - p i

*vg

mi*

+zi i

vg2 i zi + i z

2

i= -

p i*

mi*

=p

i

*

mi*i =

iT m 1

iT m i

i is called mode participation factor (for mode i)


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Since vgis a function of time, we can express it as

maxgv

+zi i

2 i zi + i z

2i

=

=

gv f(t)

gv= kcg

maxgv

t

gv = f(t)kcg

f(t)kcgi-


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The general form for the acceleration response is:

+zi i

2 i zi + i z

2i = f(t)kcgi-

+

zi

i2 i zi + i z

2

i

= f(t)kcgi- + i (t, , )


24/42

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 - k2 + k2v2

2 x 1

=

- m1 vg

- m2 vg

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

=

- m1vg

- m2vg

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

=

- m1vg

- m2vg

2 x 1

m1v1+ 0 v2( k1+k2 )v1 - k2v2+ = - m1vg

0 v1 m2+ v2 - k2 + k2v2 - m2vg=


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QuizCOMBINE THE TWO EQUATIONS INTO ONE MATRIX EQUATION

m1v1( k1+k2 )v1 - k2v2+ = - m1vg

m2v2 - k2 + k2v2 - m2vg=


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QuizCOMBINE THE TWO EQUATIONS INTO ONE MATRIX EQUATION

m1v1( k1+k2 )v1 - k2v2+ = - m1vg

m2v2 - k2 + k2v2 - m2vg=

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 kg x1

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/mm4= 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

10 CE225 MDOF Seismic Inertial Forces

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