Computer (matrix) version of the stiffness method

1. The computer version of the stiffness matrix is a generalization of the classical version leading towards the computer applications and the finite element method. The basic ideas remain the same, though some assumptions are released. The computer version is based on the following assumptions – displacements and strains remain small compared to the dimensions of the structure – the

equilibrium is related to the undeformed configuration (1st order theory) – material is linearly elastic with classical Hooke’s law – plane cross-sections remain plane during bending – from these three conditions it follows that the superposition rule is valid Compared to the classical approach the assumption of no deformability in the axial direction of the bars is released. Hence, the influence of the axial forces on the displacements is taken into account. The main concept of the method remains, i.e. the bar structure is subdivided (discretised) into a finite number of bars (elements). The ends of the elements are called nodes. e – element number i, k – node numbers The deformed state of the structure is described by the generalized nodal displacements, i.e. linear displacements and angles of cross-section rotations. Type and number of generalized displacements in the node depends on the type of structure. The nodal displacements are also called the nodal degrees of freedom. They can be assembled into a vector of nodal displacements qn, those for the element – into the vector of element displacements qe and, finally, those for the entire structure – into the global vector of displacements q.

=

i

in

v

uq

=

i

i

i

n

w

v

u

q

i Truss Frame k e

Plane truss

i

y

x

ui

vi

i

y

x

ui vi

z

wi 3D truss

=

i

i

i

n v

u

ϕ

q

=

z

y

x

i

i

i

n

w

v

u

ϕ

ϕ

ϕq

The displacements presented in the above figures are referred to the global sets of co-ordinates xy or xyz. In practice it is much easier to form the basic equations on the level of elements, hence, the local systems of co-ordinates are introduced. The values corresponding to the local co-ordinates are additionally denoted with tildes. The slope-defection formulae from the classical stiffness method are replaced by the element stiffness matrix, which relates the element displacements in local co-ordinates to the reactions at the element supports in local co-ordinates:

eee qKR ~~~= [ ]

nneije k×

=~~

K

where n is the number of element degrees of freedom. The elements of the stiffness matrix represent reactions at the element supports created by unit displacements. For instance, the

element eijk~

represents the reaction number i created by the action of the displacement 1~ =jq

First, let us consider the plane truss element. In this case the vectors of element displacements and reactions have four components

=

=

4

3

2

1

~

~

~

~

~

~

~

~

~

q

q

q

q

v

u

v

u

k

k

i

i

eq

−

−

=

=

k

k

i

i

e

T

N

T

N

R

R

R

R

4

3

2

1

~

~

~

~

~R

Plane frame

i

y

x ui

vi

ϕi

i y

x

ui vi

z

wi 3D frame ϕy

ϕz

ϕx

Interpretation of the rotation angle

ϕx

Plane yz

y~

x~

i

k

11~ ,

~qR

22~ ,

~qR

44~ ,

~qR

33~ ,

~qR

e

l

From the Hooke’s law

l

EAR

l

EAR

l

EAN −==⇒⋅−= 31

~ ,

~ 1

From the Hooke’s law

l

EAR

l

EAR

l

EAN =−=⇒⋅= 31

~ ,

~ 1

No internal forces are created. The same

situation occurs in the state 1~4 =q .

Summarising of these results yields the following element stiffness matrix

−

−

=

0000

0101

0000

0101

~

l

EAeK

From the physical interpretation of the element stiffness matrix it follows, that for instance the first column of this matrix represents the vector of reactions in the element created by the action of the

displacement 1~1 =q .

Now, let us consider the stiffness matrix for the plane bar element under bending (i.e. the plane beam element).

=

=

6

5

4

3

2

1

~

~

~

~

~

~

~

~

~

~

~

~

~

q

q

q

q

q

q

v

u

v

u

k

k

k

i

i

i

e

ϕ

ϕq

−

−

=

=

ki

ki

ki

ik

ik

ik

e

M

T

N

M

T

N

R

R

R

R

R

R

6

5

4

3

2

1

~

~

~

~

~

~

~R

x~

y~

i

k

1~ 1 =q

e

State 1~1 =q

N

N

x~

y~

i

k

1~ 3 =q e

State 1~3 =q

N

N

x~

y~

i

k

1~ 2 =q

e

State 1~2 =q

y~

x~

i

k

11~ ,

~qR

22~ ,

~qR

44~ ,

~qR

33~ ,

~qR

e

55~ ,

~qR

66~ ,

~qR

l

Reactions number 1 and 4 created by the action of displacements number 1 and 4 are obtained in the same way as in the case of the plane truss element. Note, that these displacements do not yield any other reactions. Also, these reactions are zero under the action of all the other displacements. Hence, in the classical plane beam element the action of axial forces and displacements is fully decoupled from the combined action of shear and bending. The reactions due to the transverse displacements and cross-section rotations can be found using the slope-deflection formulae from the classical stiffness method. This yields:

( ) 652322

25633

~2~6~4~6~~3~~2

232

2~q

l

EIq

l

EIq

l

EIq

l

EI

l

qqqq

l

EI

l

EIMR ikkiik +−+=

−−+=−+== ψϕϕ

( ) 652322

25366

~4~6~2~6~~3~~2

232

2~q

l

EIq

l

EIq

l

EIq

l

EI

l

qqqq

l

EI

l

EIMR ikikki +−+=

−−+=−+== ψϕϕ

( ) 62533223

2563222

~6~12~6~12~~2~~6

26~

ql

EIq

l

EIq

l

EIq

l

EI

l

qqqq

l

EI

l

EITR ikkiik +−+=

−−+=−+=−= ψϕϕ

( )

62533223

2563225

~6~12~6~12

~~2~~6

26~

ql

EIq

l

EIq

l

EIq

l

EI

l

qqqq

l

EI

l

EITR ikkiki

−+−−=

=

−−+−=−+−== ψϕϕ

Summarising these calculations yield the following stiffness matrix for the plane beam element

−

−−−

−

−

−

−

=

l

EI

l

EI

l

EI

l

EIl

EI

l

EI

l

EI

l

EIl

EA

l

EAl

EI

l

EI

l

EI

l

EIl

EI

l

EI

l

EI

l

EIl

EA

l

EA

e

460

260

6120

6120

0000

260

460

6120

6120

0000

~

22

2323

22

2323

K

It is worth to note, that the element stiffness matrices are symmetric. This is the consequence of the reactions reciprocity theorem (Rayleigh’s theorem) stating that

jiij kk =

The physical interpretation of the entries in this matrix is unchanged, i.e. for instance the second

column collects the values of reactions created by the action of the displacement 1~2 =q .

After the derivation of these two examples of the element stiffness matrices let us now deal with the loading. Generally, the loading is split into actions directly at the nodes and the actions along the elements. The former one will be considered at the later stages. The latter one must be transferred to the nodes in the form of appropriate reactions. They are assembled into a vector of reactions due to the span loading

[ ] 6,...,2,1 ~~

1600 ==×

iR ieR

This vector enters the equation of element equilibrium as an additional term

eeee 0

~~~~RqKR +=

For instance, in the case of the uniformly distributed load on the beam we have

−

−

−

=

=

12

2

012

2

0

~

~

~

~

~

~

~

2

2

06

05

04

03

02

01

0

ql

ql

ql

ql

R

R

R

R

R

R

eR

and in the case of the non-uniform heating ∆t with warmer upper side and the uniform heating t0

∆−

−

∆

=

=

h

tEI

tEAh

tEI

tEA

R

R

R

R

R

R

t

t

t

t

e

α

α

α

α

0

0

~

~

~

~

~

~

~

0

0

06

05

04

03

02

01

0R

In the latter case note that the distribution of the bending moment is constant with the fibres under tension on the colder side. Hence, the shear force is zero leading to zero reactions number 2 and 5. On the other hand, the uniform heating leads to the constant compressive axial force. To obtain the stiffness matrices for another types of beams with degenerate supports, e.g. the clamped-hinged beam, the clamped-slider beam, etc. the process of static condensation can be used. In general, the process uses the fact, that some reactions in the degenerate supports are zero. In the case of the hinge the moment is zero. From this condition the corresponding displacements can be eliminated and the dimension of the matrix is reduced from 6 by a number of zero reaction. The equilibrium conditions can be rearranged as:

eeee 0

~~~~RqKR +=

+

=

ec

e

ec

e

eccec

cee

ec

e

0

101

1

1111~

~

~

~

~~

~~

~

~

R

R

q

q

KK

KK

R

R

and the reactions in degenerate supports

0R =ec

~

y~

x~

i

k

01

~R

02

~R

04

~R

03

~R

e

05

~R

06

~R

l

y~

x~

i

k

01

~R

02

~R

04

~R

03

~R

e 05

~R

06

~R

l

∆t +

– t0

++=

++=

ecececceec

eecceeee

011

1011111~~~~~

~~~~~~

RqKqK0

RqKqKR

From the second equation vector of displacements

( )eceececcec 011

1 ~~~~~ RqKKq +−=−

can be substituted into the first equation

( ) 10011

1

11111

~~~~~~~~~eeceececcceeee RRqKKKqKR ++−=

−

to get

( ) ( )ececcceeeececcceee 0

1

11011

1

1111

~~~~~~~~~~RKKRqKKKKR

−−−+−=

where the expressions in the parentheses are the modified stiffness matrix and the modified vector of reactions due to loading, respectively.

ececcceee

ececcceee

0

1

1100

1

1

111

~~~~'

~

~~~~'

~

RKKRR

KKKKK

−

−

−=

−=

So we have

'~~'

~~011 eeee RqKR +=

The static condensation is very simple if just one displacement at a time is eliminated using the

single condition of the zero reaction. For instance if the displacement rq~ is eliminated, then

===

≠≠−=

rnrlk

rnrlkk

kkk

ln

rn

rr

lrlnln

e

or for 0'~

and for ~

~1~~

'~

:'~K

==

≠−=

rlR

rlRk

kRR

l

r

rr

lrll

e

for 0'~

for ~

~1~~

'~

:'~

0

000

0R

So schematically we have In the case of static reduction of several displacements this can be carried out step by step, one displacement after another. For the clamped-hinged elements we have:

l

n r

r

l

n r

r

0

0 0 0 0 0 0

0

0

0

lnk~

lrk~

rnk~

rrk~

'~

lnk lR0

~

rR0

~

'~

0lR

−−

−

−

−

−

=

000000

03

063

0

0000

03

033

0

03

033

0

0000

'~

323

22

323

l

EI

l

EI

l

EIl

EA

l

EAl

EI

l

EI

l

EIl

EI

l

EI

l

EIl

EA

l

EA

eK

−

−−

−

−

−

=

l

EI

l

EI

l

EIl

EI

l

EI

l

EIl

EA

l

EA

l

EI

l

EI

l

EIl

EA

l

EA

e

3300

30

3300

30

0000

000000

3300

30

0000

'~

22

233

233

K

Example

Elimination of 3~q in the case of the hinged-clamped beam.

Let us check the value k52' (l = 5, n = 2, r = 3)

322332

33

535252

36

4

1612~~1~~

'~

l

EI

l

EI

l

EIl

EI

l

EIk

kkkk −=⋅⋅

−−−=−=

Let us also calculate '~

06R for the case of point load at the beam centre (l = 6, r = 3)

16

3

84

12

8

~~1~~

'~

03

33

360606

PlPl

l

EIl

EIPlR

kkRR =

−⋅⋅−=−=

The 2D beam can be generalised to the 3D case. To this end we must include bending and shear in two planes as well as torsion.

y~

x~

i

k

11~ ,

~qR

22~ ,

~qR

44~ ,

~qR

33~ ,

~qR

e

55~ ,

~qR

0~

6 =R

y~

x~

i

k

11~ ,

~qR

22~ ,

~qR

44~ ,

~qR

66~ ,

~qR

e

55~ ,

~qR

0~

3 =R

'~

06R 0'~

03 =R 8

~03

PlR −=

8

~06

PlR =

77~ ,

~qR

99~ ,

~qR

x~

i

k

11~ ,

~qR

22~ ,

~qR

44~ ,

~qR

33~ ,

~qR

e

55~ ,

~qR

z~

y~ 66

~ ,~

qR

88~ ,

~qR

1010~ ,

~qR

1111~ ,

~qR

1212~ ,

~qR

The complete stiffness matrix for the 3D beam has the following form

−

−

−

−

−

−

−

−

=

l

EIl

EIl

GIyrtemmys

l

EI

l

EIl

EI

l

EIl

EAl

EI

l

EI

l

EIl

EI

l

EI

l

EIl

GI

l

GIl

EI

l

EI

l

EI

l

EIl

EI

l

EI

l

EI

l

EIl

EA

l

EA

z

y

s

yy

zz

zzz

yyy

ss

yyyy

zzzz

e

4

04

00

06

012

6000

12

00000

2000

60

4

02

06

0004

0000000

06

012

0006

012

6000

120

6000

12

0000000000

0'~

23

23

2

2

2323

2323

K

The unmodified stiffness matrix of any element is singular. This matrix is obtained for a structure without any supports for which the rigid body movements are not eliminated. For instance the plane beam element has three rigid body movements (degrees of freedom) The structures are usually assembled of many elements for which the local co-ordinate systems are oriented in various ways. To analyse efficiently the entire structure all the variables must be expressed in one selected global system of co-ordinates. Hence, the local variables for each element must be transformed to the global co-ordinates. Let us consider a 3D element with its own local co-ordinates

The nodal displacements for the node i can be related to the local co-ordinates zyx ~~~ :

=

i

i

i

en

w

v

u

~

~

~

~,q

or to the global co-ordinates xyz:

=

i

i

i

en

w

v

u

,q

u v ϕ

x~

i

k

e z~

y~ x

y

z

ui vi

wi

iu~

For instance the local displacement iu~ can be expressed in terms of global displacements

( ) ( ) ( )zxwyxvxxuu iiii ,~cos,~cos,~cos~ ++=

where the notation

( ) ( )[ ]baba ,cos,cos ∠=

was used to denote the cosine of the angle between two axes. These cosines are called the direction cosines and can be assembled into one matrix C

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

=

zzxzxz

zyyyxy

zxyxxx

,~cos,~cos,~cos

,~cos,~cos,~cos

,~cos,~cos,~cos

C

which relates the vectors of displacements in the local and global co-ordinates

enen ,,~ Cqq =

The analogue relation can be found for the vectors of nodal rotations

=

zi

yi

xi

zi

yi

xi

ϕ

ϕ

ϕ

ϕ

ϕ

ϕ

C~

~

~

For the entire vector of element displacements for the 3D beam we get

eee qTq =~

where the transformation matrix is

=

C000

0C00

00C0

000C

Te

The transformation matrix Te is quasi-diagonal and orthogonal (i.e. TeT = Te

–1), hence the inverse relation involves just the transpose of Te

e

T

ee qTq ~=

The same matrix is used to transform the vectors of forces

e

T

ee

eee

RTR

RTR~

~

=

=

The transformation of the element stiffness matrix requires a more complicated procedure. Let us start with the element equilibrium in local co-ordinates

eeee 0

~~~~RqKR +=

and let us substitute the expressions for the local vectors of forces and displacements to get

eeeeeee 0

~RTqTKRT +=

This equation can be multiplied from the left-hand side by TeT

ee

T

eeee

T

eee

T

e 0

~RTTqTKTRTT +=

Since

ITT =e

T

e

we get

e

T

eeee

T

ee 0

~RTqTKTR +=

or

eeee 0RqKR +=

with the element stiffness matrix in the global co-ordinates

ee

T

ee TKTK~

=

and the vector of nodal reactions due to the span loading in the global co-ordinates

e

T

ee 00

~RTR =

Re is the corresponding vector of reactions expressed in the global co-ordinates. In the case of the plane beam element the transformation matrix has the form

=

C0

0CTe

where the component matrix C is related to the transformation of the nodal displacement vector

=

i

i

i

en v

u

ϕ,q

In this situation the axis z and z~ coincide and the matrix C has the form

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

( )( )

−=

+

−

=

=

100

0cossin

0sincos

100

0cos2cos

02coscos

,~cos,~cos,~cos

,~cos,~cos,~cos

,~cos,~cos,~cos

αα

αα

ααπ

απα

zzxzxz

zyyyxy

zxyxxx

C

The transformation of the element stiffness matrix can be given in terms of submatrices. If we represent the stiffness matrix in the local co-ordinates for an element e with nodes i and j as

=

ejjeji

eijeiie

,,

,,~~

~~~

KK

KKK

then the transformation follows as

=

=

CKCCKC

CKCCKC

C0

0C

KK

KK

C0

0CK

ejjT

ejiT

eijT

eiiT

ejjeji

eijeii

T

e

,,

,,

,,

,,~~

~~

~~

~~

Similarly for the vector of reactions we can write down

zz =~

x~ y~

x

y

α α

=

ej

eie

,0

,00 ~

~~

R

RR

and the transformation takes the form

=

=

ejT

eiT

ej

ei

T

e

,0

,0

,0

,00 ~

~

~

~

RC

RC

R

R

C0

0CR

In the case of the plane truss element the difference with respect to the plane beam element is, that the former one does not involve nodal rotations. Hence, the transformation matrix is

−

−=

αα

αα

αα

αα

cossin00

sincos00

00cossin

00sincos

eT

The element stiffness matrices after transformation to the uniform global set of co-ordinates can be assembled to yield the stiffness matrix for the entire structure. We describe this assemble process for an exemplary case of plane beam elements. Let us represent the equilibrium conditions for an element e in global co-ordinates using the split of matrices and vectors into part corresponding to the nodes of the element (i and j)

+

=

ej

ei

ej

ei

ejjeji

eijeii

ej

ei

,0

,0

,

,

,,

,,

,

,

R

R

q

q

KK

KK

R

R

Now let us consider a node i in a structure, where n-elements meet. The node is loaded by the

external generalised forces Pxi, Pyi and Pϕi. Let the directions of displacements at this node be denoted by m, (m + 1) and (m + 2). So the nodal forces can also be denoted as

≡

=

+

+

2

1

m

m

m

i

yi

xi

i

P

P

P

P

P

P

ϕ

P

The reactions in the elements meeting at the node i are already transformed to the global set of co-ordinates xy and can be expressed as

eiejeijeieiiei ,0,,,,, RqKqKR ++=

The reactions from all the elements meeting at the node i acting on the node (taken with opposite magnitude) and the external forces must be in equilibrium

i

2 1

e

n Ri1,e

Ri2,e

Ri3,e

Pm

Pm+1

Pm+2

y

x

01

, =+−∑=

i

n

eei PR

or

( ) 01

,0,,,, =−++∑=

i

n

eeiejeijeieii PRqKqK

The displacements in the vector qi,e are common for all the elements at the node i, while those in the vectors qj,e correspond to the opposite ends of the elements and thus are not common. Hence, we can rewrite the equilibrium as

( )

−=+

∑∑∑

===

n

eeii

n

eejeijei

n

eeii

1,0

1,,,

1, RPqKqK

This set of equations written for the node i comprises three equations which can be represented as On the other hand we can also consider a contribution of one element e with the nodes i and j in the form of the element matrix

=

ejjeji

eijeii

e,,

,,

KK

KKK

and the element vector of reaction due to the span loading

=

ej

ei

e,0

,0

0 R

RR

to the global stiffness matrix K and the global vector of forces P, respectively. The global set of equilibrium equations can therefore be written down as

PKq =

The square matrix K has the dimension k×k and the vector P – k, where k is the number of all nodal displacements in the entire structure. Let us assume that the node i of the element coincides with the structure node with the displacements numbered as m, (m + 1) and (m + 2) and the element node j – with the structure node with the displacements numbered as n, (n + 1) and (n + 2). We have

Fragment of global stiffness matrix K

m

m+1

m+2

m m+1

m+2

qm

qm+2

qm+1 · =

Kij,e ∑=

n

eeii

1,K

Fragment of

global vector of

displacements q

∑=

−n

eeii

1,RP

Fragment of

global vector of

forces P

In the case of elastic support in the direction of the displacement (m + 1) at a node i an additional force Fm+1 due to the displacement in the support must be considered. This force has always the orientation opposite to the displacement because the spring counteracts it. Hence, the this force is with the positive sign on the left hand side of the global set of equations. This force is obtained from the formula

11 ++ = mm kqF

where k is the spring stiffness. So the small modification of the global stiffness matrix is enough. Namely, in the (m + 1)-equation a term kqm+1 must be added, what leads to a simple addition of the spring stiffness k to the element (m + 1)-(m + 1) of the global stiffness matrix K. Let us now consider an example of the assembly process of the matrix K and the vector P for a given plane frame.

m+1

m+2

m

m m+1

m+2

n+2

n+1

n

n+1 n+2 n

Kij,e Kii,e

Kji,e Kjj,e

–R0i,e

–R0j,e

i

2 1

e

n

Pm

Pm+1

Pm+2

Fm+1

3 1 2

3 4

1 2

4 5 y

x

M

P

q

The frame consists of four beam elements which are interconnected in five nodes. The element stiffness matrices and vectors of nodal reactions due to the span loading are found in the local co-ordinates for each element are transformed to the global co-ordinates and they can be represented as

=

1,221,21

1,121,11

1 KK

KKK

=

2,222,21

2,122,11

2 KK

KKK

=

3,223,21

3,123,11

3 KK

KKK

=

3,223,21

3,123,11

1 KK

KKK

Note, that the local element numbers of element nodes can be replaced with the global numbers: Element 1: 1 – 1, 2 – 2, element 2: 1 – 2, 2 – 3, element 3: 1 – 2, 2 – 4, element 4: 1 – 3, 2 – 5. These relations enable a proper “addressing” of the subsequent submatrices of the element stiffness matrices Ke in the global stiffness matrix of the frame K. The assembly scheme is Note, that in this scheme each box corresponds to a node and it represents 3×3 elements of the global stiffness matrix corresponding to three displacements at the given node. Let us also consider the assembly of the global force vector P. Out of four beams constituting the frame only one – No. 4 has a span loading and consequently the non-zero vector of reactions due to the loading. These vectors for the other beams are zero:

0RRRRRR ====== 030201030201

~~~

To calculate the vector for the fourth element the loading can be split into uniformly distributed load acting along the beam and transverse to the beam

1 2 1 1 1

2

1

2

1 2 3

4

x y

x y

x x

y

y

1 2 3 2 3

5

2

4

1 2 3

4 x~

y~ x~

y~

x~

y~

x~

y~

22lq

q

l q1 = qsinϕcosϕ

l =

q2 qcos2ϕ

l

+

ϕ

21lq

21lq

22lq

12

22lq

12

22lq

K =

K11,1 K12,1

K21,1 K22,1+K11,2+

+K11,3 K12,2

K21,2 K22,2+K11,4 K12,4

K12,3

K21,3

K21,4 K22,4

K22,3

Now the vector of reactions can be set up

−

−

−

−

−

=

=

12

cos2

cos2

cossin12

cos2

cos2

cossin

~

~

~

~

~

~

~

22

2

22

2

4,06

4,05

4,04

4,03

4,02

4,01

04

ϕ

ϕ

ϕϕ

ϕ

ϕ

ϕϕ

ql

ql

ql

ql

ql

ql

R

R

R

R

R

R

R

and transformed to the global set of co-ordinates

==

4,02

4,01

04404

~

R

RRTR

T

with the transformation matrix

−=

=

100

0cossin

0sincos

ϕϕ

ϕϕ

CC0

0CT

The nodal external forces P and M are directly substituted into the vector of loading P, the force P opposite to the x axis with the negative sign in the first of three places in the box corresponding to the node 3 and the clock-wise moment M with the positive sign to the third place in the box corresponding to the node 2. The global stiffness matrix K for the entire frame has a characteristic band structure. If this band is narrow a computer storage saving is possible. Hence, it is advisable to keep the band width as small as possible. The width depends on the order of node numbering. The difference between the node numbers for a single element should be kept as small as possible. Let us consider two cases of the numbering In the first case the maximum difference of node numbers for a single element equals to 5 and in the second case – to 3. The latter one is therefore more efficient because the band width in the global stiffness matrix will be smaller. The band width b can be computed as

–P 0 0

–R01,4

–R02,4

0 0 M

P =

1 2 3 4 5

6 7 8 9 10

15 14 13 12 11

1 4 7 10 13

2 5 8 11 14

15 12 9 6 3

[ ]1max +−⋅= jisb

where s is the number of degrees of freedom per node and i and j are the node numbers for a single element. The band-structured matrix can be stored in the computer memory as a single column consisting of the elements in the band read row-wise or as a rectangular matrix with the dimensions b×n, where n is the global number of unknown displacements for the structure. The assembled global stiffness matrix K is singular, because it is representing a structure without any supports, still possessing freedom of rigid body rotations. To remove the singularity the support conditions must be introduced to the system of equilibrium equations

PKq =

The k-th equation in this set corresponds to the condition, that the reaction in an additional fictitious k-th support is zero, as was the case in the classical stiffness method. On the other hand, if the k-th support does exist, then the corresponding reaction is not zero. Thus, the k-th equation is spurious. Consequently, the corresponding displacement qk is zero, hence all the elements from the k-th column in the matrix K multiply the zero value and are spurious, too. So, the direct method of inclusion of support conditions in the equilibrium equations would be to remove the rows and columns from the matrix K corresponding to the real support as well as removing the corresponding element from the vector P. This method, though correct, is not effective numerically, because it would require a complex process of renumbering of unknowns. Instead one of several indirect methods are used in computer practice. The first one introduces a elastic support with an “almost” infinite spring stiffness. This leads to a simple addition to the element (k,k) in the matrix K of a large number, several orders of magnitude larger then the maximal element of the entire matrix K. The second method modifies the k-th equation to a form

01 =⋅ kq

which automatically yields the zero displacement qk in the support. This is done by zeroing the k-th element of the vector P and all the elements of the k-th row of the matrix K except for the element (k,k), which is set to 1. Additionally, also all the elements of the k-th column of the matrix K are set to zero to eliminate those that are unnecessary due to their multiplication by the zero displacement qk. Existence of hinge connections in the plane frame requires special care in the analysis. For instance in the following example elements 3 and 4 meeting at the node 3 have two different angles of cross-section rotation there. So the real number of unknowns in the node 3 is four. If the computer program allows for such a situation no problem exists. But if on the other hand the number of unknowns per node is fixed than special precautions must be made. If both the elements are taken as the clamped-hinged ones with the stiffness matrices after the static condensation, than the global stiffness matrix K will be singular because there are no non-zero entries to the row and column k corresponding the rotation(s) in the node 3, including 0 on the main diagonal. This yields the singular equation

00 =⋅ kq

One way to overcome this problem is to treat one of the elements adjacent to the hinge, for instance the element 3, as the clamped-clamped one with the element matrix without static

3 1 2

3 4

1 2

4 5

condensation. Then, the matrix K will not be singular and the angle of rotation at the hinge in the element 3 will be automatically calculated as one of the unknowns. The element 4 is therefore considered as the hinged-clamped one and its stiffness matrix is obtained from the static condensation. Influence of support displacements can also be included in the analysis.

These displacements for an element can be assembled into a vector e∆q~ . Like all the nodal

displacements, they lead to the nodal reactions in elements according to the equilibrium equation for the element

( ) eeeee 0

~~~~~RqqKR ++= ∆

The influence of these displacements can be included in the vector of reactions due to the span loading

( )eeeeee 0

~~~~~~RqKqKR ++= ∆

Thus, if there is no other loading the support displacements are the only factor influencing the reactions vector

eee ∆= qKR ~~~0

Let us consider an example Here only the element 2 has “adjacent” support displacements and only its vector of reactions will be non-zero. The applied support displacements must be expressed in the local co-ordinates of the element

ϕ

ϕ

cos

sin

∆=∆

∆=∆

y

x

and the local vector of support displacements for the element 2 is

−

∆

∆=∆

γy

x

0

0

0

~2q

Note, that the displacements must be substituted with appropriate sign according to their orientation with respect to the local axes. Therefore, the anti-clockwise rotation is taken as negative. Having this vector of displacements the corresponding vector of reactions can be calculated, transformed to the global co-ordinates and used in the assembly process of the vector of nodal forces P.

γ

1

2

∆

2

ϕ

y~

x~ ∆y

∆x

Let us now set the entire algorithm of the matrix version of the stiffness method for the plane frames. 1. Assume the global co-ordinate system. Number the nodes and element. Select the local

systems of co-ordinates. Attach the internal hinged connections to the elements leaving one element at a hinge as clamped-clamped.

2. Calculate local stiffness matrices eK~

for the elements with the static condensation for the

hinged-clamped ones. Transform the local matrices to the global co-ordinates to get Ke

ee

T

ee TKTK~

=

3. Assemble the global stiffness matrix K for the entire system from all the element matrices Ke expressed in global so-ordinates.

4. Calculate the vectors of reactions e0

~R due to the span loading: forces on the beams, uniform

and non-uniform heating, support displacements. Transform these vectors to the global co-ordinates to get R0e.

e

T

ee 00

~RTR =

5. Assemble the vector of nodal forces P for the entire system from all the element vectors R0e taken with the negative sign and all the external nodal loading.

6. Include the supports in K and P. For the rigid supports set to zero all the corresponding elements in P and all the elements in the corresponding rows and columns of K except for the main diagonal term to be set to 1. For the flexible supports add the spring stiffness to the corresponding main diagonal term in K.

7. Solve the global set of equilibrium equations

PKq =

to get the nodal displacements in global co-ordinates. 8. Select the displacements for the separate elements and form the appropriate vectors qe.

Transform these vectors to the local co-ordinates to get eq~

eee qTq =~

9. Compute the reactions at the element ends from the equilibrium conditions for the separate elements

eeee 0

~~~~RqKR +=

These reactions correspond to the values of internal forces at the element ends

−

−

=

=

ki

ki

ki

ik

ik

ik

e

M

T

N

M

T

N

R

R

R

R

R

R

6

5

4

3

2

1

~

~

~

~

~

~

~R

At the end let us summarise some “tricks” related to the matrix version of the stiffness method. 1. Flexible support as an additional element

k 0 , == EIkl

EA

2. A truss can be solved using the beam elements if one element of those meeting at a node is treated as clamped. For the hinged-hinged elements the static condensation of the matrix for clamped-hinged element can be carried out to eliminate the second rotation. 3. Curved beams can be analysed in an approximate way by modelling the curve as an assembly of a sufficiently large number of straight-line beams 4. Similarly, tapered elements can be modelled as assemblies of piece-wise constant cross-section elements. 5. Calculation of displacement at an arbitrary point of an element The arbitrary displacement can be calculated using the superposition rule applied to the influences of the support (nodal) displacements and the span loading. Hence, we have

∑=

+=6

1A0AA

~

iiiqvvv

The displacements along the element due to the nodal displacements are calculated using so called shape functions. In the case of the beam element axial nodal displacements do not cause transverse displacements in the beam. So the non-zero shape functions are due to the displacements 2, 3, 5 and 6. They have the form and the exact mathematical expressions will be derived in the following section on matrix analysis of stability of frames.

A

vA

q

A

vA0 1

~ q

2~ q

3~ q

4~ q

5~ q

6~ q

A

vA2 2

~ q

A

vA3

3~ q

A

vA5 5~ q

A

vA6 6~ q

6. Calculation of envelopes of static and kinematic quantities under a set of moving loads. For instance let us consider the envelope of bending moments in the beam presented in the figure The classical approach involves calculation of influence lines. They are found for the moments at several number of points along the beam. Each influence line is loaded by the load set in the way leading to the extreme output moments (positive and negative) calculated from the superposition rule.

( )( )2211min2

2211max2

min

max

ηη

ηη

PPqSM

PPqSM

++=

++=

These maximum values are then put together to yield the envelope. 7. Solving a set of algebraic equations

BAX =

For a simple calculation of unknowns X the main matrix of coefficients A can be decomposed into a product of two triangular matrices

21AAA =

to get

BXAA =21

The product

ZXA =2

q P1 P2

0 1 2 etc.

M0(x)

M1(x)

M2(x)

etc.

q P1 P2

η1 η2

S

0 1 2 etc.

etc. M2max

M2min

= 0

0 ·

gives an auxiliary vector and the original set of equations is represented as a combination of two sets of equations

ZXA

BZA

=

=

2

1

Each of these sets is characterised be a triangular coefficient matrix and can be easily solved equation after equation by a method of back substitution. The decomposition into two triangular matrices can be done be several methods, e.g. the classical Gauss elimination method or the Choleski method. In the latter one

SAA ==T

21

so we have

BSXS =T

and the process of back substitution is even simpler because just one triangular matrix S must be stored in the computer memory. Example Let us find the distribution of internal forces in the following frame using the matrix version of the stiffness method.

Data:

E = 205 GPa k = 57195 kN/m

β = arctan2/5 = 21.801° Element 1:

A = 27.9cm2

I = 1450cm4

l = 4m Element 2:

A = 46.1 cm2

I = 4250 cm4

l = 5.3852 m

In the solution we will use the method without constant number of unknowns in the node. The hinge connecting both elements is therefore the node with four unknown displacements including two separate angles of rotation, one for each element. Both elements will be considered as clamped-hinged ones and the corresponding stiffness matrices and vectors of reactions will be found for such elements. In the following figures displacement directions for both elements are shown in local and global co-ordinates and the angle

α for the transformation is found.

α = 270°

x~

1

3

6

5

4

2 1

y~

1

3

6

4

5

1 2

x~

x α

Local co-ordinates Global co-ordinates Transformation

20kN

4

[m]

x

6kN/m

2

5

1

2

1

2

3

4 5 6

7 8

9

10

y

k β

α = 360°–β = 338.199°

Now we can associate the numbers of nodal displacements in the elements expressed in global co-ordinates (from 1 to 6 for each element) with the numbers of nodal displacements in the global numbering (from 1 to 10 in the considered example). Thus the allocation table is created

Element number

Element directions

1 2 3 4 5 6

1 1 2 3 4 5 6

2 4 5 7 8 9 10

These allocations indicate, where the subsequent values from the element matrices and vectors in global co-ordinates enter the global matrices and vectors for the entire structure. For instance: the value k52,2 from the stiffness matrix K2 is added to the element k59 in the global matrix K, the value R06,2 from the vector of reactions R02 is subtracted from the element P10 in the global vector P. Now let us find the reactions vectors due to span loading. In this case only element 2 is loaded and the loading is split into two components:

kN/m 1725.5cos6

kN/m 0689.2cossin6

2 ==

==

β

ββ

y

x

q

q

For the clamped-hinged element the vector of reactions is

−

−

=

−

−

=

=

750.18

409.17

5708.5

0

446.10

5708.5

8

8

52

08

32

~

~

~

~

~

~

~

22,06

2,05

2,04

2,03

2,02

2,01

02

lq

lq

lq

lq

lq

R

R

R

R

R

R

y

y

x

y

x

R

The element stiffness matrices in the local co-ordinates are

4

5

2 1

x~

x α

Local co-ordinates Global co-ordinates Transformation

2 3

6

y~

x~

β

2

2 3

4

5

6

1

qx

qy 4

5

2 1

3

6

4

5

2 1

3

6

−−

−

−

−

−

=

−−

−

−

−

−

=

000000

03.13903.5573.1390

001430000014300

03.557022293.5570

03.13903.5573.1390

0014300000143000

000000

03

063

0

0000

03

033

0

03

033

0

0000

~

323

22

323

1

l

EI

l

EI

l

EIl

EA

l

EAl

EI

l

EI

l

EIl

EI

l

EI

l

EIl

EA

l

EA

K

−

−−

−

−

−

=

−

−−

−

−

−

=

48543.901003.9010

3.9014.167004.1670

0017550000175500

000000

3.9014.167004.1670

0017550000175500

3300

30

3300

30

0000

000000

3300

30

0000

~

22

233

233

2

l

EI

l

EI

l

EIl

EI

l

EI

l

EIl

EA

l

EA

l

EI

l

EI

l

EIl

EA

l

EA

K

The transformation matrices are

=

C0

0CTe

−=

100

0cossin

0sincos

αα

αα

C

−

−

=

100000

001000

010000

000100

000001

000010

1T

−

−

=

100000

09285.03714.0000

03714.09285.0000

000100

00009285.03714.0

00003714.09285.0

2T

Now the transformation of local matrices and vectors to the global co-ordinates can be done.

−

−

−

==

750.18

233.18

2929.1

0

768.11

2929.1

~02202 RTR

T

−−

−−

==

000000

0143000001430000

003.139003.139

000222903.557

0143000001430000

003.1393.55703.139

~1111

TKTKT

−−−−−−−−−

−−−−−

==

48548.8367.33403.9017.334

8.836243506046002435060460

7.33460460151300060460151300

000000

3.901243506046002435060460

7.33460460151300060460151300

~2222

TKTKT

Following the relations in the allocation table the following scheme of assembly of the global stiffness matrix and the global force vector is created.

k11,1 k12,1 k13,1 k14,1 k15,1 k16,1 –R01,1

k22,1 k23,1 k24,1 k25,1 k26,1 –R02,1

k33,1 k34,1 k35,1 k36,1 –R03,1

k44,1 k45,1 k46,1 –R04,1

k55,1 k56,1 –R05,1

k66,1 –R06,1

s y m m e t r y

k11,2 k12,2 k13,2 k14,2 k15,2 k16,2 –R01,2

k22,2 k23,2 k24,2 k25,2 k26,2 –R02,2

k33,2 k34,2 k35,2 k36,2 –R03,2

k44,2 k45,2 k46,2 –R04,2

s y m m e t r y k55,2 k56,2 –R05,2

k66,2 –R06,2

Both the elements contribute simultaneously to the values k44, k45, k54 and k55 in the stiffness matrix K. The appropriate values from the element stiffness matrices are added in these windows. Due to the elastic support in the direction of the displacement 9 requires an addition of the spring stiffness k in the value k99 of the global stiffness matrix K. The nodal force 20kN acting along the direction 4 in its positive orientation must be added to the element P4 of the vector P. Thus the global matrix K and the vector P are formed.

139.3 0 557.3 –139.3 0 0 0 0 0 0 0

143000 0 0 –143000 0 0 0 0 0 0

2229 –557.3 0 0 0 0 0 0 0

151439 –60460 0 0 –151300 60460 334.7 18.707

167350 0 0 60460 –24350 901.3 11.768

0 0 0 0 0 0

0 0 0 0 0

81545 –60460 –334.7 1.2931

s y m m e t r y 24350 –836.8 18.233

4854 –18.750

Now the support conditions must be included. In the analyzed frame the following support displacements are zero:

0108321 ===== qqqqq

Thus, the 1st, 2nd, 3rd, 8th and 10th elements in the vector P and the elements in the 1st, 2nd, 3rd, 8th and 10th rows and columns of K must be set to zero except for the elements at the diagonal of K, which are set to 1. The rotations at the hinge q6 and q7 are not zero but were eliminated on the element level by the static condensation. Thus, they also must be eliminated from the global set of equations. This can be done by setting the elements k66 and k77 at the diagonal of K to 1. Thus the rotations will be kept in the vector of unknowns and their values will be calculated as 0. Despite the fact, that these values are not true, they will not be used in further calculations, because they were indeed eliminated. They will be denoted with asterisks. Thus we get the final form of the global set of equilibrium equations

1 0 0 0 0 0 0 0 0 0 0

1 0 0 0 0 0 0 0 0 0

1 0 0 0 0 0 0 0 0

151439 –60460 0 0 0 60460 0 18.707

167350 0 0 0 –24350 0 11.768

1 0 0 0 0 0

1 0 0 0 0

1 0 0 0

s y m m e t r y 24350 0 18.233

1 0

Solving of the equations

PKq =

Yields the following vector of global displacements

=

0

0001880.0

0

*0

*0

0001346.0

0001022.0

0

0

0

q

Following the relations in the allocation table the appropriate displacements can be attributed to the elements and the element vectors of displacements in global co-ordinates can be formed.

=

=

*0

0001346.0

0001022.0

0

0

0

6

5

4

3

2

1

1

q

q

q

q

q

q

q

=

=

*0

0001880.0

0

*0

0001346.0

0001022.0

6

5

4

3

2

1

2

q

q

q

q

q

q

q

These vectors can be transformed to local co-ordinates

111~ qTq = 212

~ qTq =

to get

−=

*0

0001022.0

0001346.0

0

0

0

~1q

−=

0

0001746.0

0000698.0

*0

0001629.0

0000449.0

~2q

With these vectors the final values of nodal reactions can be found

01111

~~~~RqKR += 02222

~~~~RqKR +=

−

−

−

=

0

014.0

248.19

057.0

014.0

248.19

~1R

−

−

−

=

740.18

407.17

559.14

0

448.10

701.25

~2R

Note, that the values of rotations at the hinge denoted by 0* are indeed not necessary, because they are multiplied by zeros in 6th or 3rd column in the stiffness matrices for the element 1 and 2, respectively. These reactions can be finally used to find the distributions of the internal forces The correctness of these calculations can be checked by the usual kinematic and static check, as it was done in the case of classical flexibility and force methods. The only thing, that differs is the fact, that the present calculations in the matrix version of the stiffness method were carried out with

–

– –19.2

–25.7

–14.6

+ 0.014

+ –

10.4

–17.4

0.057

18.74

N [kN]

T [kN]

M [kNm]

the released assumption of axial non-deformability of the beams. Hence, the displacement in the kinematic check must be calculated from the enhanced version of the principle of virtual work (also including the term due to the elastic support)

k

RRdx

EA

NNdx

EI

MM

xx

)0(

99)0()0(

1 ++=⋅ ∑∫∑ ∫δ