Example

Wall

1k 2k

3k11,uP 22 ,uP

Obtain expressions for the displacements (u1, u2) of the rigid blocks, in terms of spring stiffnesses k1, k2, k3 and prescribed static loads P1, P2.

Compatibility:

Equilibrium: 121 sFPP =+ 322 ss FFP +=

No rotation of blocks

Constitutive: 111 ukFs = ( )1222 uukFs −= ( )1233 uukFs −=

Example

Wall

1k 2k

3k11,uP 22 ,uP

Combining, we have:

1

211 k

PPu +=

+

++=321

21

12

11kkk

PkPu

Example

Wall

1k 2k

3k11,uP 22 ,uP

Now consider each spring as a 2-noded finite element, as follows:

On each spring, we have:

010 ,uF

1k

111,uF 121,uF

2k

222 ,uF 131,uF

3k

232 ,uF

Geometry

Example

121,uF

2k

222 ,uF

131,uF

3k

232 ,uF

010 ,uF

1k

111,uF

( )10110 uukF −=

( )01111 uukF −=

=

−

−

11

10

1

0

11

11

FF

uu

kkkk

=

−

−

22

21

2

1

22

22

FF

uu

kkkk

=

−

−

32

31

2

1

33

33

FF

uu

kkkk

stiffness matrix displacement

vector

loading/force vector

Example

121,uF

2k

222 ,uF

131,uF3k

232 ,uF

010 ,uF

1k

111,uF

1312111 PFFF =++

23222 PFF =+ 11,uP 22 ,uP010 ,uF

=

+−−−−++−

−

2

1

10

2

1

0

3232

323211

11

0

0

PPF

uuu

kkkkkkkkkk

kk

FKu =

Example

121,uF

2k

222 ,uF

131,uF3k

232 ,uF

010 ,uF

1k

111,uF

2

1

0

3232

323211

11

0

0

uuu

kkkkkkkkkk

kk

→→→

+−−−−++−

−=K

1

0

11

11

uu

kkkk

→→

−

−

2

1

22

22

uu

kkkk

→→

−

−

2

1

33

33

uu

kkkk

→→

−

−

Direct Stiffness Method

Example 11,uP

22 ,uP010 ,uF

=

+−−−−++−

−

2

1

10

2

1

0

3232

323211

11

0

0

PPF

uuu

kkkkkkkkkk

kk

FKu =

FKu 1−=K-1 does not exist as det(K)=0

1k

2k

3k1 2 3

1 2 3 1

2

3

Example

11,uP 22 ,uP010 ,uF

=

+−−−−++−

−

2

1

10

2

1

0

3232

323211

11

0

0

PPF

uuu

kkkkkkkkkk

kk

=

+−−−−++

2

1

2

1

3232

32321

PP

uu

kkkkkkkkk

K-1 exists: FKu 1−=

Example Wall

1k 2k

3k( ) 111 ,, uutP ( ) 222 ,, uutP

Now consider blocks subject to acceleration ü1, ü2, and acceleration ü0 at the wall. Mass of blocks are m1 and m2.

121,uF

2k

222 ,uF 131,uF

3k

232 ,uF010 ,uF

1k

111,uF

( ) 001110 umuukF e =−+

0u

( ) 1110111 umuukF =−+

0ume 11um 11um 22um 11um 22um

1m 2m

( ) 1112221 umuukF =−+

( ) 2221222 umuukF =−+

( ) 1112331 umuukF =−+

( ) 2221332 umuukF =−+

Example

1121 ,, uuF

2k

2222 ,, uuF

1131 ,, uuF 3k

2232 ,, uuF

0010 ,, uuF

1k

1111 ,, uuF

Assembling:

( )( )( )

=

+−−−−++−

−+

tPtPtF

uuu

kkkkkkkkkk

kk

uuu

mm

m we

2

1

2

1

0

3232

323211

11

2

1

0

2

1

0

0

000000

em

1m 2m

FKuuM =+

111 ,, uuP 222 ,, uuP 00 ,, uuFw

em 1m 2m

Example 111 ,, uuP 222 ,, uuP 00 ,, uuFw

( )( )( )

=

+−−−−++−

−+

tPtPtF

uuu

kkkkkkkkkk

kk

uuu

mm

m we

2

1

2

1

0

3232

323211

11

2

1

0

2

1

0

0

000000

em 1m 2m

lumped mass matrix

111 ,, uuP 222 ,, uuP

1m 2m

StationaryWall

What if P1 and P2 are zero, while ü1 and ü2 are non-zero?

prescribed quantities known/measured

quantity

Example

111 ,, uuP 222 ,, uuP

1m 2m

StationaryWall

=

+−−−−++

+

00

00

2

1

3232

32321

2

1

2

1

uu

kkkkkkkkk

uu

mm

0KuuM =+ Free vibration equation

( ) 0uIKM 1 =−− 2ω

If u is periodic, equation may be re-configured as:

Eigenvalue problem

Modal frequency Mode shape

Degrees of Freedom at a Node

y

z x

The freedom of movement of a nodal point, if that point is unconstrained

For a typical structural node in a 3D space, subject to mechanical forces only, We may identify six degrees of freedom (x, y, z, θx, θy, θz)

Each nodal point may have nf degrees of freedom

F(t)

x

y

z

Summary Any structure may be broken up into nodes and elements

A system of n nodes with nf degrees of freedom on each node:

generates an eigenvalue problem with a maximum of nnf modal frequencies and nnf mode shapes

Problem is solved at the nodes and approximated across elements

Essential (displacement) boundary conditions must always be satisfied

generates an nnf x 1 displacement vector generates an nnf x 1 loading/force vector

generates an nnf x nnf stiffness matrix generates an nnf x nnf mass matrix