DYNAMIC OF STRUCTURES

CHAPTER 10GENERAL APPROACH FOR LINIER

SYSTEMS

Department of civil engineering, University of North Sumatera

Ir. DANIEL RUMBI TERUNA, MT;IP-U

STATIC CONDENSATION

The equation of motion base on shear building modelling is not suitable for complex structures.

q Discretization

A frame structures can be idealized as an assemblage of element- beams, columns, walls interconnected at nodal points or nodes

The displacement of nodes are the degree of freedom.In, general, a node in a two dimensional frame has three DOFs- two translations and one rotation.A node in a three dimensional frame has six DOFs- three translations ( the x, y, z components) and three rotations (about the x, y, z axes).In, general, axial deformation of beams can be neglected in analyzing most buildings, and axial deformation of columns need not be considered for low rise buildings

STATIC CONDENSATION

For example, a two story, two bay planar frame has six nodes and 18 DOFs (Fig. 1.a). With assumption of axial deformations is neglected, two bay frame has eight DOFs (Fig. 1.b)

(a) (b)

6u

3u

2u

1u4u

5u

7u 8u

Fig.1 Degrees of freedom: (a) axial deformation included, 18 DOFs; (b) axialdeformation negleted, 8 DOFs

STATIC CONDENSATION

In Fig. 2 , shows the external dynamic forces are applied at the nodes.

The moments to are zero in most practical cases.

Fig.2 External dynamic forces

)(1 tp

)(2 tp

)(5 tp)(4 tp

)(7 tp )(8 tp)(6 tp

)(3 tp

)(3 tp )(8 tp

)(tp

we will relate the external forces on the stiffness component of thestrutures to the resulting displacement (Fig. 3.a).

The stiffness influence coeficient is the forces requred along DOFs due to unit displacement at DOF

STATIC CONDENSATION

q Elastic Forces

For linear systems this relationship can be obtained by the method ofsuperpotition and the concept of stiffness influence coeficients (directequilibrium method)

We apply a unit displacement along DOF , holding all otherdisplacements to zero as shown; to maintain these displacement forcesmust be applied along all DOFs.

sjf

ju

j

ijk ij

For example, the forces shown in Fig. 3.b arerequired to maintain the deflected shape associated with and allother

)8.......,,2,1(1 =iki

11 =u0=ju

STATIC CONDENSATION

11 =u

1sf

2sf

3sf 4sf 5sf

6sf 7sf 8sf

11 =u

11k

21k81k71k61k

51k41k

31k

(a)

(b)

Fig.3 (a) stiffness component of frame; (b) stiffness influence coeffcient for

The force at DOF associated with displacement (Fig.3.a) is obtained by superpotition:

STATIC CONDENSATION

Ntojuj 1, =

N

=

NNNNjNN

Nj

Nj

sN

s

s

u

u

u

kkkk

kkkk

kkkk

f

f

f

M

KK

MMMM

KK

KK

M2

1

21

222221

111211

2

1

kuf s =

i

NiNjijiisi ukukukukf +++++= ............2211

sif

(1)

One such equation exists for each . The set of equation can be written in matrix form:

Ntoi 1=

or

(2)

(3)

we will relate the external forces acting on the mass component ofthe sructures to the acceleration (Fig. 4.a).

The mass influence coeficient is the external forces in DOF due to unit acceleration along DOF .

STATIC CONDENSATION

q Inertia Forces

We apply a unit acceleration along DOF , while acceleration in all otherDOFs are kept zero. According to D’Alembert’ principle, the fictitiousinertia force oppose these acceleration; therefore external forces will benecessary to equlilibrate these inertia forces.

Ijf

ju&&

j

ijm ij

For example, the forces shown in Fig. 4.b arerequired in the various DOF to equilibrate the inertia forces with andall other

)8.......,,2,1(1 =imi

11 =u&&0=ju&&

STATIC CONDENSATION

11 =u&&

1If

2If

3If 4If 5If

6If 7If 8If

(a)

11 =u&&11m

21m81m71m61m

51m41m

31m

(b)

Fig.4 (a) mass component of frame; (b) mass influence coeffcient for

Lump mass at node

The force at DOF associated with acceleration (Fig.4.a) is obtained by superpotition:

STATIC CONDENSATION

Ntojuj 1, =&&

N

=

NNNNjNN

Nj

Nj

IN

I

I

u

u

u

mmmm

mmmm

mmmm

f

f

f

&&

M

&&

&&

KK

MMMM

KK

KK

M2

1

21

222221

111211

2

1

umf I &&=

i

NiNjijiiIi umumumumf &&&&&&&& +++++= ............2211

Iif

(4)

One such equation exists for each . The set of equation can be written in matrix form:

Ntoi 1=

or

(5)

(6)

STATIC CONDENSATION

The mass is distributed throughout an actual structures, but it can be idealized as lumped mass or concentrated at the nodes of the discretized structures; usually, such a lumped mass idealization is satisfactory

fed

a b c

(a)

fmem

dm

am

bm

cm

(c)

am

dm em

bm

fm

cm

Fig. 5 Lumping of mass at structural nodes

(b)

Each structural element is replaced by point masses at its two nodes,with the distribution of the two masses being determined by statics.

STATIC CONDENSATION

11 =u&&

The lump mass at a node of the structures is the sum of the masscontribution of all the structural element connected to the node.

once the lumped masses at the node have been calculated, the massmatrix for the structures can readly br formulated.

The external forces associated with acceleration (Fig.4.b) are ,where (Fig.5.c), and

111 mm =cba mmmm ++=1

8......,4,3,201 == iformi

The coefficient is equal to the rotationalinertia of the mass lumped has negligible influence on the dynamic ofpractical structures; thus is taken zero

887766554433 mmmmmm =====

In general for a lumped mass idealization, the mass matrix is diagonal:

00 ormmjiform jjjij ===

Usually the mass of the structure is idealized as lumped masses at nodes,and the mass matrix contain zero diagonal element in the rotational DOFs.These are the DOFs that can be eliminated from the dynamic analysis ofthe structures provided that the dynamic excitation does not include anyexternal forces in the rotational DOFs, as in the case of erthquakeexcitation.

The vertical DOFs of the building can also be eliminated from dynamicanalysis because the inertia effects associated with the vertical DOFs ofbuilding frames are usually small.

The equation of motion for a system excluding damping can be rewritten as

( )tpkuum =+&& (1)

Or in partitioned form:

STATIC CONDENSATION

Where denotes the DOFs with zero mass and the remaining DOFs. The two partitioned equations are:

( )

=

+

000

0

0000

0

0

tp

u

u

kk

kk

u

um tt

t

tttttt

&&

&&(2)

0u tu

( )tpukukum tttttttt =++ 00&&

00000 =+ ukuk tt (4)

(3)

From equation (4) gives relationship between and :0u tu

ttukku 01

000−−= (5)

STATIC CONDENSATION

Where is the condensed stiffness matrix given by:

(6)

ttk̂

( )tpukum ttttttt =+ ˆ&&

(7)

Solution of equation (6) provided the displacement in the dynamic DOFs, and the displacment in the condensed DOFs are determined from Eq. (5)

)(0 tu( )tut

tTttttt kkkkk 0

1000

ˆ −−=

Substituting equation (5) in Eq. (3) gives

STATIC CONDENSATION

EXAMPLE 1

q Element Stiffness Matrix

( )tp1

elemen (1)

EI1u

EI

4/mL2/mL

2u3u4u

elemen (2)

11 =u31k

11k

41k

21k

The stiffness coefficient is the force required along DOF due to unit displacement at DOF

ijk ij

3/96 LEI

3/96 LEI

2/24 LEI

2/24 LEI

13k

13 =u

23k

31k

2/24 LEI

2/24 LEILEI /4

LEI /8

EI EI

( )tp2

4/mL2/mL2/L 2/L

43k

EXAMPLE 1

2/24 LEI

12 =u

32k

12k42k

22k

3/96 LEI

3/96 LEI

2/24 LEI

2/24 LEI

2/24 LEI 2/24 LEI

3/96 LEI

14 =u

34k

14k24k

44k

2/24 LEI

2/24 LEILEI /4

LEI /8

2/24 LEI

LEI /4

LEI /8

EXAMPLE 1.

−−−

−−−

=

=

22

222

44434241

34333231

24232221

14131211

22/03

2/33

032412

331212

8

LLL

LLLL

L

LL

L

EI

kkkk

kkkk

kkkk

kkkk

k

q Determined the stiffness matrix

With all the stiffness influence coefficient determined, the stiffness matrix is

q Determined the mass matrix

With the DOFs defined at the location of the lumped masses, the diagonal mass matrix is give by Eq. (9)

(8)

EXAMPLE 1.

=

=

0000

0000

002/0

0004/

44434241

34333231

24232221

14131211

mL

mL

mmmm

mmmm

mmmm

mmmm

m

q Determined the equation of motion

(9)

The governing equation are

( )tpkuum =+&& (10)

Where , and are given by Eqs. (8)

and (9), and

{ }Tuuuuu 4321= mk( ) { }Ttptptp 00)()( 21=

EXAMPLE 1.

−−−

−−−

=

=

22

22200

22/03

2/33

032412

331212

8

LLL

LLLL

L

LL

L

EI

kk

kkk

ot

tott

q Determined the condensed stiffness matrix

The stiffness matrix Eq. (8) is partitioned

From Eq. (7), the condensed matrix is given by

(11)

tTttttt kkkkk 0

1000

ˆ −−=

ttk̂

−

−=

165

52

7

48ˆ3L

EIk tt

(12)

EXAMPLE 1

tuL

u

−=

857.0857.0

43.357.210

( )tpukum ttttttt =+ ˆ&& (14)

(13)

Substituting the condensed stiffness matrix in Eq. (5) gives the relation between the condensed DOF and the dynamic DOF :0u tu

(15)

Where as dynamic DOF can be derived from equation of motiontu

=

−

−+

)(

)(

165

52

7

48

2/0

04/

2

1

2

1

32

1

tp

tp

u

u

L

EI

u

u

mL

mL

&&

&&

q Two-Story of Frame

EXAMPLE 2.

6u

3u

2u

1u4u

5u

mm 21 =

mm =2

EI EI

EI2EI2

( )tp1

hL 2=

( )tp2

h

h

11k

11 =u

41k

61k

31k

51k21k

−−

−

=

=⇒=

2

2

2

2

3

3

61

51

41

31

21

11

11

/6

/6

/6

/6

/24

/72

1

hEI

hEI

hEI

hEI

hEI

hEI

k

k

k

k

k

k

kutodue i

EXAMPLE 2.

With all the stiffness influence coefficient determined, the stiffness matrix is

−−

−−−−

==

222

222

222

222

3

62066

60266

2016266

0221666

66662424

66662472

hhhhh

hhhhh

hhhhh

hhhhh

hhhh

hhhh

h

EIk

To determined the condensed stiffness matrix , the stiffness matrix in Eq. (16) is partitioned

(16)

ttk̂

EXAMPLE 2.

=

=

00kk

kkk

ot

tott

−−

−−−−

222

222

222

222

62066

60266

2016266

0221666

66662424

66662472

hhhhh

hhhhh

hhhhh

hhhhh

hhhh

hhhh

From Eq. (7), the condensed matrix is given by tTttttt kkkkk 0

1000

ˆ −−=

−

−=

61.1151.17

51.1788.254ˆ3h

EIk tt (17)

EXAMPLE 2

tuh

u

−−−−−−

=

7869.09836.0

7869.09836.0

2459.04426.0

2459.04426.0

10

( )tpukum ttttttt =+ ˆ&& (19)

(18)

Substituting the condensed stiffness matrix in Eq. (17) gives the relation between the condensed DOF and the dynamic DOF :0u tu

(20)

Where as dynamic DOF can be derived from equation of motiontu

=

−

−+

)(

)(

61.1151.17

51.1788.54

0

02

2

1

2

1

32

1

tp

tp

u

u

h

EIu

u

m

m

&&

&&

q Calculate lateral stiffness frame with static condensation

EXAMPLE 3.

3u2u 1u

bEI

cEI

hL 2=

h cEI

311

)12(2

h

EIk c=

11 =u

221 /6 hEIk c=

231 /6 hEIk c=

12 =u

221 /6 hEIk c=LEIhEIk bc /4/422 +=

LEIk b /232 =

For a frame subjected to lateral forces , the equilibrium equations can be written as

EXAMPLE 3.

With all the stiffness influence coefficient determined and with , the stiffness matrix is

=

22

22

3

66

66

6624

hhh

hhh

hh

h

EIk c

cb II =

=

0

0

66

66

6624

3

2

1

22

22

3

s

c

f

u

u

u

hhh

hhh

hh

h

EI

sf

pku =or

(21)

(22)

From equation (5)

11

1

22

22

3

2

1

1

7

66

6

6

6u

hu

h

h

hh

hh

u

u

−=

−=

−

ttukku 01

000−−=

EXAMPLE 3.

Thus lateral stiffness of the frame is

(23)

the first of three equation in Eq. (22) can be written as

( ) 1313 7

96

7

66624 u

h

EIu

hhh

h

EIf cc

s =

+−=

37

96

h

EI c

tTttttt kkkkk 0

1000

ˆ −−=

Lateral stiffness of the frame can be derived from Eq. (7)