MULTI-DEGREE of FREEDOM SYSTEMS (MDOF)

In many cases, it is necessary to use many coordinates to adequetly

describe the configuration of the structure and the deformed shape.

Such systems are said to be multi-DOF discrete systems.

Fig. shows a beam whose movement is defined by the displacements

u1(t), u2(t),……. ui(t),……, un(t) of n points. These points can be placed

arbitrarily along the beam and necessarily at places where there are

changes in the geometry of the beam or the distribution of the loading.

The set of displacements associated with the n points constitutes the

number of DOFs of the system.

Simplified Model of a Building

Consider the three-story building shown in

Fig. Assumptions:

The mass of the columns are negligible

when compared to the total mass of the

building, which is concentrated at the floor

levels.

The floor and the beams are rigid in

longitudinal direction and in flexure.

The columns are infinitely stiff axially but

flexible transversally.

From these assumptions it can be

concluded that the building has only a shear

deformation mode in the transverse direction.

This simplified model of the building is called

a shear building idealization.

Simplified Model of a Building

Consider the free-body diagram shown in Fig.

Assuming a linear system, the elastic forces fs can be related to the floor

displacements u.

For this purpose we introduce the lateral stiffness kj of the jth story; it relates

the story shear Vj to the story deformation or drift, j=uj-uj-1, by

)(

)(

)(

3333

2222

1111

tpfff

tpfff

tpfff

SDI

SDI

SDI

(1)

jjj kV

The story stiffness is the sum of the lateral stiffnessess of all columns in the

story. For a story height of h and a column with modulus E and second

moment of area Ic, the lateral stiffness of a column with clamped ends,

implied by the shear-building idealization, is 12EIc/h3. Thus the story

stiffness is

columns

cj

h

EIk

3

12

The equations of equilibrium of the three masses are,

)(

)()(

)(

23333

323122222

21211111

uukff

uukuukfff

uukukfff

i

SS

s

S

i

SS

s

S

i

SS

(2)

Similarly, the damping forces fDi can be written with respect to the relative

velocities of the DOFs u

)(

)()(

)(

23333

323122222

21211111

uucff

uucuucfff

uucucfff

i

DD

s

D

i

DD

s

D

i

DD

(3)

We can relate the elastic forces fsi to the relative floor displacements ui.

The force fsi at the first floor is made up of two contributions: fsis above

and fsii below the first story. Thus,

s

S

i

SS fff 111

Considering Vj=jkj, and noting that 1=u1 and 2=u2-u1

)( 212111 uukukfS

Similarly, the forces fs2 and fs3 will be determined as,

The inertia forces associated with the masses m1, m2 and m3 undergoing

accelerations ü1, ü1 and ü3, respectively, can be written

333

222

111

umf

umf

umf

I

I

I

(4)

After substitution of Equations (2), (3) and (4) into Eq. (1), we get:

)()()(

)()()()()(

)()()(

323323333

232312232312222

1212112121111

tpuukuucum

tpuukuukuucuucum

tpuukukuucucum

(5)

which can be written in the matrix equation as,

)(tpfff SDI

in which fI is the vector of inertia forces

3

2

1

3

2

1

00

00

00

u

u

u

m

m

m

uMf I

(6)

in which fD is the vector of damping forces

(7)

and fS is the vector of elastic forces

3

2

1

33

3322

221

0

0

u

u

u

cc

cccc

ccc

uCfD

3

2

1

33

3322

221

0

0

u

u

u

kk

kkkk

kkk

KufS (8)

Eq. (5) becomes*,

)()()()( tptKutuCtuM (9)

Equation of Dynamic Equilibrium

It can be concluded that the dynamic equation of equilibrium of a discrete

multi-DOF system written in matrix form is entirely equivalent to the

SDOF system one, with each term replaced by a vector or a matrix of

order equal to the number of DOFs n as follows:

)(tpfff SDI (10)

in which,

)(

)(

)(

tKuf

tuCf

tuMf

S

D

I

(11)

Hence,

)()()()( tptKutuCtuM (12)

Eq. (12) expressess the dynamic equation of equilibrium of a discrete

multi-DOF system. fI, fD and fS represent the vectors of inertia forces,

damping forces and elastic forces, respectively. Each system of forces is a

summation of a set of influence coefficients mij, cij and kij whose definitions

are:

The stiffness influence coefficient kij is the force in the direction of DOF i,

caused by a unit displacement imposed in the direction of DOF j, while

displacements in the direction of all other DOFs are kept equal to zero. The

matrix of all stiffness influence coefficients K is called the stiffness matrix.

The stiffness influence coefficient cij is the force in the direction of DOF i,

caused by a unit velocity imparted in the direction of DOF j, while velocities

in the direction of all other DOFs are kept equal to zero. The matrix of all

damping influence coefficients C is called the damping matrix.

The stiffness influence coefficient mij is the force in the direction of DOF i,

caused by a unit acceleration imparted in the direction of DOF j, while

accelerations in the direction of all other DOFs are kept equal to zero. The

matrix of all mass influence coefficients M is called the mass matrix.

By definition, the stiffness influence coefficient kij is the force in along

DOF i due to a unit displacement imposed along DOF j, while keeping all

other displacements equal to zero.

The stiffness influence coefficients are numerically equal to the applied

forces required to maintain the specified displacement condition. They are

positive when the sense of the applied force corresponds to a positive

displacement and negative otherwise.

If we know the fixed-end reactions caused by unit displacements

imposed at any one support in a given direction, we can easily determine

the coefficients kij.

Fig. summarizes the stiffness influence coefficients for a flexural

structural element**.

Stiffness Influence Coefficients

Example: Calculate the stiffness matrix of the building shown in Fig. using

a shear building idealization. Write the equation of motion assuming that

the system has no damping.

The mass matrix is diagonal and is written as,

m

mM

0

0

By successively imposing a unit displacement to DOFs 1 and 2 while

keeping the displacement along the other DOF equal to zero, we obtain

3

2

3

111

2424

h

EI

h

EIk

3

221

24

h

EIk

3

212

24

h

EIk

3

222

24

h

EIk

Hence, the stiffness matrix of the structure is

2221

1211

kk

kkK

22

221

3

224

EIEI

EIEIEI

h

EIK

The stiffness matrix could have been obtained from the free-body diagrams

of the massess similar to the determination of Eq. (8).

3

11

122

h

EIk

In this case, the stiffness matrix is 2nd order and is written as

33

3322

221

0

0

kk

kkkk

kkk

K

22

221

kk

kkkK

İn which the lateral stiffness of stories 1 and 2 are k1 and k2 and are given

by

3

22

122

h

EIk

The equations of motion of the building in matrix form are

The stiffness of a building idealized as a shear building is always larger

than the actual stiffness because additional constraints on the rotations of

the node are introduced. This leads to an overestimation of the frequencies

which are proportional to the stiffness.

)(

)(24

0

0

2

1

2

1

22

221

3

2

1

tp

tp

u

u

EIEI

EIEIEI

hu

u

m

m

The axial stiffness of the columns and the beams can be assumed to be

infinitely stiff. The frame has six DOFs numbered as shown in Fig. DOFs u1

and u2 are are translational while DOFs u3 to u6 are rotational. The

equation of motion can be written as

Example: Write the equation of motion of the two-story building shown in

Fig. considering the flexibility of the beams and rotations of the nodes.

)(tpKuuM

where the nodal displacement and acceleration vectors are

)(

)(

)(

)(

)(

)(

6

5

4

3

2

1

tu

tu

tu

tu

tu

tu

u

)(

)(

)(

)(

)(

)(

6

5

4

3

2

1

tu

tu

tu

tu

tu

tu

u

Considering that the mass moment of inertia are negligible, the mass

matrix is diagonal and contains only two non-zero terms on the diagonal

corresponding to DOFs u1 and u2. The mass matrix is

0

0

0

0

m

m

M

and the dynamic load vector is

)(

)(

)(

)(

)(

)(

)(

6

5

4

3

2

1

tp

tp

tp

tp

tp

tp

tp

Determination of the influence coefficient kij will be done by imposing

successively a unit displacement to DOFs 1 to 6 while keeping

displacements of all other DOFs equal to zero.

1) Impose a unit displacement along DOF u1 as shown in Fig.

A unit displacement imposed along DOF u1 while keeping the

displacements along all the other DOFs equal to zero gives rise to flexural

deformation to the columns only. The forces at the end of the columns

caused by this unit displacement are:

3

2

3

111

122

122

h

EI

h

EIk

3

221

122

h

EIk 2

2

2

131

66

h

EI

h

EIk

2

2

2

141

66

h

EI

h

EIk 2

251

6

h

EIk 2

261

6

h

EIk

2) Impose a unit displacement along DOF u2 as shown in Fig.

A unit displacement imposed along DOF u2 while keeping the

displacements along all the other DOFs equal to zero gives rise to flexural

deformation to the columns of the second story only. The forces at the end

of the columns caused by this unit displacement are:

3

212

122

h

EIk

3

222

122

h

EIk 2

232

6

h

EIk

2

242

6

h

EIk 2

252

6

h

EIk

2

262

6

h

EIk

3) Impose a unit displacement along DOF u3 as shown in Fig.

A unit displacement imposed along DOF u3 while keeping the

displacements along all the other DOFs equal to zero gives rise to flexural

deformation to the two left-hand side columns and to the beams of the first

floor only. The forces at the end of the columns and beams caused by this

unit displacement are:

l

EI

h

EI

h

EIk 321

33

444

l

EIk 3

43

2

h

EIk 2

53

2 063 k

2

223

6

h

EIk

2

2

2

113

66

h

EI

h

EIk

4) Impose a unit displacement along DOF u4 as shown in Fig.

A unit displacement imposed along DOF u4 while keeping the

displacements along all the other DOFs equal to zero gives rise to flexural

deformation to the two right-hand side columns and to the beams of the

first floor only. The forces at the end of the columns and beams caused by

this unit displacement are:

l

EI

h

EI

h

EIk 321

44

444

l

EIk 3

34

2

h

EIk 2

64

2054 k

2

224

6

h

EIk 2

2

2

114

66

h

EI

h

EIk

5) Impose a unit displacement along DOF u5 as shown in Fig.

A unit displacement imposed along DOF u5 while keeping the

displacements along all the other DOFs equal to zero gives rise to flexural

deformation to the two left-hand side column in the second story and to the

beam at the roof level only. The forces at the end of this column and this

beam caused by the imposed unit displacement are:

l

EI

h

EIk 42

55

44

h

EIk 2

35

2

l

EIk 4

65

2045 k

2

225

6

h

EIk

2

215

6

h

EIk

6) Impose a unit displacement along DOF u6 as shown in Fig.

A unit displacement imposed along DOF u6 while keeping the

displacements along all the other DOFs equal to zero gives rise to flexural

deformation to the right-hand side column in the second story and to the

beam at the roof level only. The forces at the end of this column and this

beam caused by the imposed unit displacement are:

l

EIk 4

56

2

036 k

l

EI

h

EIk 42

66

44

2

226

6

h

EIk

2

216

6

h

EIk

h

EIk 2

46

2

The total stiffness matrix K is given below. It is symmetric. It is obvious that

many terms could have been deduced by inspection when considering

symmetry of the total stiffness matrix.

Lumped-mass discretization in which rotating inertia effects are negligible

leads to a diagonal mass matrix with zeros on the diagonal. For a

conservative system, if no external forces are applied to the rotational

DOFs, Eq. (9) can be written as

Static Condensation

0

)(

00

0 tp

u

u

KK

KK

u

uM tt

t

tttttt

in which subscript indicates rotational DOFs without mass and subscript t

indicates translational DOFs with mass which are called dynamic DOFs.

Note that KtT=Kt because of the symmetry of the total stiffness matrix.

This equation can be partitioned into the following two matrix equations:

)(tpuKuKuM tttttttt (14)

(13)

0 uKuK tt(15)

Because there are no inertia forces in Eq. (15), we can deduce the

following static relation between u and ut:

Substituting Eq. (16) into Eq. (14), we obtain

)(tpuKuM ttttt

(18)

(16) ttuKKu

(17)

in which the double subscript of Mtt has been replaced by the subscript t

since there is no ambiguity and where

ttttt KKKKK 1

Matrix Kt is called reduced or condensed stiffness matrix and corresponds

to the translational stiffness matrix.

The solution of Eq. (17) gives the displacements of the dynamic DOFs ut(t).

The displacements of the condensed DOFs, u(t), can be deduced from

Eq. (16) knowing ut(t).

Consider: l=5m, h=3m

The columns are identical and their flexural stiffnesses are: EI1=EI2=20.25x106 Nm2.

The beams are identical and their flexural stiffnesses are: EI3=EI4=40.50x106 Nm2.

The mass concentrated at ecah floor is m=20 000kg and the dynamic loads applied to

DOFs u1 and u2 are p1(t) and p2(t).

Example: Write the equation of motion of the two-story building shown in

Fig. after performing a static condensation of the stiffness matrix to

eliminate the rotational DOFs.

The stiffness matrix corresponding to the six DOFs can be determined as,

We will make use of the stiffness matrix presented in the previous example:

4.592.165.1305.135.13

2.164.5905.135.135.13

5.1304.862.165.130

05.132.164.865.130

5.135.135.135.130.180.18

5.135.13000.180.36

106

KK

KKK

t

ttt

Let us partition this matrix into four sub-matrices: Ktt, Kt, Kt and K:

We mass matrix and the dynamic load vector are,

NKK T

tt

5.135.135.135.13

5.135.1300106

)(

)()(

2

1

tp

tptpt

10

011020 3xM t

mNKtt /1818

1836106

NmK

4.592.165.130

2.164.5905.13

5.1304.862.16

05.132.164.86

106

The equation of motion is,

Hence, the condensed stiffness matrix is,

mNKKKKK t

T

tttt /72377.1071222.13

71222.1306256.311061

)(

)(

)(

)(

72377.1071222.13

71222.1306256.3110

)(

)(

10

011020

2

1

2

16

2

13

tp

tp

tu

tu

tu

tux