CHAPTER 3

Stiffness Matrix Method

3-1- DEFINITION

The stiffness method is a method of analysis, where the main unknowns

are the displacements of joints. These unknowns are determined from

equilibrium. The method can be used for determination of displacements

and internal forces due to

external loads,

environmental changes (temperature and shrinkage), and

support movement.

The stiffness method is applicable to skeletal structures (beams, plane and

space frames and trusses, and grids), and continuum structures (plates,

shells, and three dimensional solids).

(Question: what is the difference between skeletal and non-skeletal

structures (continuum structures)?)

The steps of this method are:

1- Modeling

Model the structure with a number of members and

joints,

Define types of members and connections,

Identify the relevant unknown components of joint displacements.

2- Load vector

For each member determine the member fixed end

forces (which are the reactions due to loads if all the

relevant joint displacements were prevented).

3- Stiffness matrix

For each member determine the member end forces

due to joint displacements (i.e. member stiffness

equations).

4- Equilibrium equations

Write the equations of equilibrium of all joints (under

the forces in steps 2 and 3, and joint loads).

5- Solving equilibrium equation

Solve equilibrium eq. to determine joints

displacements.

6- Internal forces

Use the resulting joint displacements to determine the

total member end forces.

The above steps can be illustrated by the following example:

EXAMPLE

Determine the displacements

and member end forces in the

shown frame due to the

given loads.

STEP 1: Modeling.

3 t2 t / m

2.00 6.00

3.00

1

2 3

1

2D1

D2

D3 D4

STEP 2: Load vector

Fixed-end forces {F}

STEP 3: Stiffness matrix

Stiffness relations [K]

STEP 4: Equilibrium {F} = [K] {D}

STEP 5: Equations solution Get {D}

STEP 6: Internal forces Q, N, and M

3-2- STEP 1: MODELING

a- Numbering

Organization of the solution requires numbering the members, joints, and

forces and displacements components. Several numbering schemes are

possible. Choice of the most appropriate scheme depends on several

considerations including the method of assembling and solving equations

of equilibrium (banded matrix methods, frontal method, etc.). When

banded matrix methods are used for solving equilibrium equations,

members may be numbered in any convenient order, but

joints should be numbered in such an order that

the maximum difference between the numbers of the two joints of each

member is as small as possible. For example, the numbering in Fig. Is

more appropriate than that in Fig.

The unknown displacement components are usually numbered in a

certain sequence (for example in plane frame; x, y, and rotation) starting

at joint 1 and proceeding in ascending order through the joints. These

displacement components are called degrees of freedom.

2 t / m

6

6

0

6

6

0

b- Degrees of Freedoms

Several models of skeletal structures may be used depending on the

nature of the structure and the loads. Common skeletal models include

space frames, space trusses, plane frames, plane trusses, beams, and grids.

* In space frames, each joint has six degrees of freedom: three

translations (in X, Y, and Z directions) and three rotations (about X, Y,

and Z axes).

* In space trusses, all member ends are assumed to have hinged

connections, i.e. three degrees of freedom at each joint (rotation about X,

Y, and Z axes).

* In plane frames, a joint has three degrees of freedom since it has two

translations and one rotation.

* In plane trusses, a joint has two degrees of freedom (the translation

only) since rotation are not considered.

* In beams, axial and transverse forces and displacements are uncoupled,

and separate analyses may be carried out for axial effects and transverse

effects.

* In grids, in-plane displacements are not considered; therefore, also

rotations about an axis normal to the plane of the grid are not considered.

Thus, a grid joint has only three degrees of freedom (an out-of-plane

translation, and two rotations about in-plane axes).

c- Local and Global Axes

The axes which are convenient in dealing with members individually are

called local (member) axes, but the axes which are convenient in dealing

with the structure as a whole are called global (structure) axes.

Displacement and force components may be expressed using one of the

previous two systems.

The relation between the components in the two systems of axes is

expressed in matrix form which called transformation matrix [T].

1 2 3 4 5 6 7 8 9 10

20 19 18 17 16 15 14 13 12 11

1

2

3

4

5

67

1 12 13 24

2 2311 14

3 2210 15

4 219 16

5 208 17

6 197 18

4

1

5

1

32

4 5

2 3 6

21

17

13

9

5

1

7

6

8 9 12

4

8

2

6

3

7

12

16

20

24

10

14

18

15

11

19

13

22 23

16 17

7 10 11 14 15 18

20

19

Global and local force and displacement components

d1

d1

End 1

Plane or grid member

f1g

gf1

Y

c) Plane truss member

l

b) Plane frame member

l

f3

Globalg

f4

g gf2 d2

d4gg

d1f1l

d3g g

f2l

Global

d6

f4

g

d3f3g g

f5

d2f2gg f6

d5

g

g g

lf1 d1

g

gd4

g

lf2

f3

Local

f4

d2l

d4l l

d3l l

f4

d6

Local

llf3 d3

f5

ld2

ll

l

d5

f6l

lld4

a) Geometry

( X2,Y2)

Length =L

( X1,Y1)

End 2

X

Z

( X2,Y2,Z2)

Space member

End 1

( X1,Y1,Z1)Z

X

Y

X

Y End 2

In the case of plane frame, if the components of member

end forces in global direction are:

}{}{ 654321

ggggggg fffffff

and the corresponding components in local directions are:

}{}{ 654321

lllllll fffffff

as shown in Fig. , then;

}{][}{ lg fTf

where;

100000

0000

0000

000100

0000

0000

][

cs

sc

cs

sc

T

in which; c = cos θ = (X2 – X1) / L , and

s = sin θ = (Y2 – Y1) / L

where (X1 , Y1) and (X2 , Y2) are the coordinates of the joints at the start

and the end of member respectively with respect to global axes, and L is

the length of the member.

Similarly, the relation between the displacement components in global

directions {dg} and the displacement components in local directions {d

l}

for the joints of plane frame members is:

}{][}{ lg dTd

where [T] as defined above.

In the case of plane trusses, each of {fg}, {f

l}, {d

g}, and

{dl} has only four elements and the transformation matrix

reduces to:

cs

sc

cs

sc

T

00

00

00

00

][

In the case of grids, the transformation matrix is:

sc

cs

sc

cs

T

0000

0000

001000

0000

0000

000001

][

In the case of space frames, each joint has six degrees

of freedom then the transformation matrix is:

*

*

*

*

][

t

t

t

t

T

where

[Ø] = [0]3*3 , and

zyx

zyx

zyx

nnn

mmm

lll

t ][ *

where (lx, mx, nx), (ly, my, ny), and (lz, mz, nz) are the direction cosines of

the local axes x, y, and z with respect to the global axes X, Y, and Z.

In the case of space trusses, all member ends are

hinged. Therefore the transformation matrix is:

**000

**000

**000

000**

000**

000**

][

x

x

x

x

x

x

n

m

l

n

m

l

T

where the asterisks refer to unneeded elements, and (lx, mx, nx) are the

direction cosines which given by:

lx = (X2 – X1) / L, mx = (Y2 – Y1) / L, and nx = (Z2 – Z1) / L

3-2- STEP 2: LOAD VECTOR (MEMBER FIXED-END FORCES)

Member fixed-end forces means the reactions at the ends of members due

to loads, environmental changes, or support movement, when all the

unknown displacements at member end joints are prevented. These

reactions can be determined by classical methods such as column analogy

or consistent deformations.

Components of fixed-end forces in local directions may be arranged in a

vector }{ l

mf , which is called member load vector in local directions. The

corresponding components in global directions may be arranged in a

vector }{ g

mf which is called member load vector in global directions, and

can be determined from the transformation relation

}{ g

mf = [T] }{ l

mf

Fig. Shows some cases of member fixed-end forces.

WL/2

WL /12W t / m

L

WL/2

WL /122 2

PL/8

P/2 P/2

L

PPL/8

WL /30

L

W t / m2

WL /202

WL2

20

WL

30

2

( )

LL

2

20

WL2

WL

30)(

0.333 WL0.167 WL

WL/4

5WL /96

L

W t / m2

5WL /962

WL/4

2PL/9

L

P

P

P

2PL/9P

P.a.b /L

bPb/L

P

Pa/La

2 2P.b.a /L

2 2

P.a.b(

2

L

L)2

2P.b.a

L2

P.b.a2

P.a.b(

L

2L L2

2

)

L

M/4 M/4

1.5 M / L

M

1.5 M / L

W

W

P3

W

(5)

P2

P2

(3)

W

(1)

P1

P1

P1

P2

(4)

(2)

3-3- STEP 3: STIFFNESS MATRIX (MEMBER END FORCES DUE TO

JOINT DISPLACEMENTS)

Displacements of joints cause member end displacements and hence

member deformations and internal forces between joints and member

ends. These member end forces depend on the type of connection

between joints and member end.

a- Force-Displacement Relation in Local Directions

In the case of plane truss member shown in Fig. , let

the joints undergo displacements whose components in

local directions are ld1 , ld2 , ld 3 , and ld4 . The lateral

displacements can occur without deformation (small

displacements and hinged connections). On the other hand,

the axial displacements ld1 and ld 3 result in an elongation

( ld 3 - ld1 ) and hence tension N = ( ld 3 - ld1 ) EA / L.

The corresponding forces from joints to member ends are: l

df 1 = -N = ( ld 3 - ld1 ) EA / L l

df 2 = l

df 4 = 0

l

df 3 = N = ( ld 3 - ld1 ) EA / L

or in matrix form;

}{ l

df = }{][ ll dk

where

0000

0101

0000

0101

][

lk

The matrix ][ lk is known as the member stiffness matrix in local

directions. It can be noticed that the elements of the ith

column of ][ lk

are the forces }{ l

df when l

id = 1 while all other components of }{ ld are

zeros. This observation is usually used as a convenient basis for deriving

the matrix ][ lk for members of different types.

Consider a plane frame member with fixed ends as shown in Fig. . To

derive elements of the first column of ][ lk , let ld1 = 1 while all other

components of }{ ld are zeros. The corresponding end forces are the

elements of the first column of ][ lk , and these elements are:

(EA/L) { 1 0 0 -1 0 0 }T

Elements of the second column of ][ lk are the forces correspond to ld2 =

1, while all other components of }{ ld are zeros. Then the elements of the

second column of ][ lk are:

(6EI/L2) { 0 2/L 1 0 -2/L 1 }

T

Note that these forces can be determined using the method of consistent

deformations. Similarly the components of the third, fourth, fifth, and

sixth columns can be determined and the complete stiffness matrix of the

member in local axes directions is:

LEILEILEILEI

LEILEILEILEI

LEALEA

LEILEILEILEI

LEILEILEILEI

LEALEA

k l

/4/60/2/60

/6/120/6/120

00/00/

/2/60/4/60

/6/120/6/120

00/00/

][

22

2323

22

2323

It may be observed that each column of ][ lk represent a set of forces in

equilibrium. It may also be observed that the matrix ][ lk is symmetric,

i.e. l

ji

l

ij kk . These two observations may be used for deriving some

elements of ][ lk .

Members, joints and degreesof freedom (D.O.F.)

4

3

2

1

Equilibrium equations

33

23

13

Member load vectors

31

21

g2

g2

g2

11+k44k41k

61

51

kg1

kg1

64

54

g1

k +k

g1

k +k

g1

g1

11k

1

g1

14kg1

2

461245 k +kk +k

32

22

g2

g2

65

55

g1

k +k

g1

k +k

66

56

g1

k +kg2

g1

k +kg2

g2g1

15kg1

3

g1 g2

16kg1

4

0 - f - f

-5 - f - f

3 - f - f

0 - f

=D2

D4

D3

D1

m4 m1

m6

m5

g1

g1

m3

m2

g2

g2

g1

m1g1 g2

Fixed 2 3

Structure and loads

6

5

4

3

2

1

3

4

2

0

0

1

0 5

local row numbers

Member stiffness matrices

53

g

64

fm

1

2

3

4

0

0

2

1

Kg

11

0 5

f

0 6

mg

2

3

4

0 4

3

2

2 1

60

gK

2

numbersglobal position

01

21

20

43

43

65

W2

2

3

40

4

3

12

32

1 2

04

3 4

00

5 6

Hinged

1D

1

1

W1P2

P1D

2D

24

D3

local column numbers

Example for assembly of equilibrium equations

To find the overall stiffness matrix

0

0

/

0

0

/

LEA

LEA

2

3

2

3

/6

/12

0

/6

/12

0

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

/2

/6

0

/4

/6

0

2

2

0

0

/

0

0

/

LEA

LEA

2

3

2

3

/6

/12

0

/6

/12

0

LEI

LEI

LEI

LEI

LEI

LEI

LEI

LEI

/4

/6

0

/2

/6

0

2

2

LEILEILEILEI

LEILEILEILEI

LEALEA

LEILEILEILEI

LEILEILEILEI

LEALEA

k l

/4/60/2/60

/6/120/6/120

00/00/

/2/60/4/60

/6/120/6/120

00/00/

][

22

2323

22

2323

EA/L EA/L

= 1

= 1

6EI/L2

12EI/L12EI/L

26EI/L

3 3

= 1

2EI/L4EI/L

6EI/L2

6EI/L2

EA/L EA/L

= 1 6EI/L2

12EI/L3

12EI/L3

= 1

6EI/L2 2EI/L

26EI/L

26EI/L

= 1

4EI/L

0

0

/

0

0

/

LEA

LEA

0

/3

0

/3

/3

0

3

2

3

LEI

LEI

LEI

0

/3

0

/3

/3

0

2

2

LEI

LEI

LEI

0

0

/

0

0

/

LEA

LEA

0

/3

0

/3

/3

0

3

2

3

LEI

LEI

LEI

0

0

0

0

0

0

000000

0/30/3/30

00/00/

0/30/3/30

0/30/3/30

00/00/

][

323

22

323

LEILEILEI

LEALEA

LEILEILEI

LEILEILEI

LEALEA

k l

EA/L EA/L

= 13EI/L

2

3EI/L3

3EI/L3

= 1

23EI/L

23EI/L

= 1

3EI/L

EA/L EA/L

= 1

3EI/L3EI/L

23EI/L

= 1

3 3

= 1

0

0

0

0

0

0

0

/3

0

/3

/3

0

3

2

3

LEI

LEI

LEI

0

/3

0

/3

/3

0

2

2

LEI

LEI

LEI

0

0

/

0

0

/

LEA

LEA

0

/3

0

/3

/3

0

3

2

3

LEI

LEI

LEI

0

0

0

0

0

0

000000

0/30/3/30

00/000

0/30/3/30

0/30/3/30

00/000

][

323

22

323

LEILEILEI

LEA

LEILEILEI

LEILEILEI

LEA

k l

= 1

3EI/L3EI/L

23EI/L

= 1

3 3

= 1

3EI/L

3EI/L2

3EI/L2

EA/L

= 1

EA/L

3EI/L

3EI/L

3EI/L3

2

= 1

3

= 1

0

0

/

0

0

/

LEA

LEA

0

0

0

0

0

0

0

0

0

0

0

0

0

0

/

0

0

/

LEA

LEA

0

0

0

0

0

0

0

0

0

0

0

0

000000

000000

00/00/

000000

000000

00/00/

][LEALEA

LEALEA

k l

EA/L

= 1

EA/L= 1 = 1

EA/L

= 1

EA/L = 1 = 1

3-4- STEP 4 : The Overall Equilibrium Equation

From the steps number 3 & 4 we can construct the overall equilibrium

equation as follow:

{ F } nx1 = [ K ] nxn * { D }nx1

Where: (n) the number of degree of freedom (DOF) which defined

from step no 1( modeling ).

3-5- STEP 5 : Solve The Equilibrium Equation

By using back Guss elimination we can solve the equilibrium

equation and find the overall global displacement { D }g

We can find the local displacement from the relation

{ D }l = [ T ]

T * { D }

g

Where [ T ] the transformation matrix for member

For plane truss member is:

cs

sc

cs

sc

T

00

00

00

00

][

For plane frame member is:

100000

0000

0000

000100

0000

0000

][

cs

sc

cs

sc

T

d2

d1

d4

d3

3-6- STEP 6 : Find The Internal Forces

For plane truss member:

L

EAN sin*)(cos*)( 2413 dddd

For plane frame member:

sin*cos* 211 DDDg

sin*cos* 122 DDDg

33 DDg

sin*cos* 544 DDDg

sin*cos* 455 DDDg

66 DDg

L

EADgDgNN )( 1421

)(12

)(12

36252321 DgDgL

EIDgDg

L

EISS

)2(2

)(6

362521 DgDgL

EIDgDg

L

EIM

)2(2

)(6

635222 DgDgL

EIDgDg

L

EIM