COORDINATE
TRANSFORMATIONS
Members of a structural system
are typically oriented in differing
directions, e.g., Fig. 17.1. In order
to perform an analysis, the element stiffness equations need to
be expressed in a common coordinate system – typically the global
coordinate system. Once the
element equations are expressed
in a common coordinate system,
the equations for each element
comprising the structure can be
1
assembled.
Figure 17.1 – Frame Structure
(Kassimali, 1999)
2
Coordinate Transformations:
Frame Elements
Consider frame element m of Fig.
17.7.
Y
y
e
m
x

b
X
(a) Frame
Q6, u6
Q3, u3
m
Q1, u1

Q4, u4
Q5, u5
Figure 17.7: Local – Global
Coordinate Relationships
Q2, u2
(b) Local Coordinate End Forces
and Displacements
3
4
1
Figure 17.7 shows that the local –
global coordinate transformations
can be expressed as
Q x  cos  FX  sin  FY
Q y   sin  FX  cos  FY
(17.9)
Q z  FZ
x = cos X + sin Y
y = -sin X + cos Y
and since z and Z are parallel, this
coordinate transformation is
expressed as
z=Z
Using the above coordinate
transformations, the end force and
displacement transformations can
be expressed as
5
T
where {Q}b   Q1 Q2 Q3  =
beginning node local coordinate
force vector; {Q}e = <Q4 Q5 Q6>T =
end node local coordinate force
vector; {F}b = <F1 F2 F3>T =
beginning node global coordinate
force vector; {F}e = <F4 F5 F6>T =
end node global coordinate force
 cos  sin  0 


vector; [t] =   sin  cos  0  =
0
1 
 0
local to global coordinate transformation matrix which is the same at
each end node for a straight
7
where x, X = 1 or 4; y, Y = 2 or 5;
and z, Z = 3 or 6.
Utilizing (17.9) for all six member
force components and expressing
the resulting transformations in
matrix form gives
Qb 


Qe 
 [t] [0] Fb 
[0] [t]  F 

  e 
or
{Q} = [T] {F}
(17.11)
member; {Q} = <<Q>b <Q>e>T =
element local coordinate force
vector; {F} = <<F>b <F>e>T =
element global coordinate force
 [t] [0]
vector; and [T] = 
 =
[0] [t] 
element local to global coordinate
transformation matrix.
Utilizing (17.9) for all six member
force components and expressing
the resulting transformations in
matrix form gives
8
2
The direction cosines used in the
transformation matrices can easily
be calculated from the nodal
geometry, i.e.
Xe  X b
L
Ye  Yb
sin  
L
cos  
(17.13)
(17.14)
{u}  [T]{v}
It is also useful in matrix structural
analysis to calculate the global end
displacements and forces in terms
of the local coordinate end displacements and forces as shown in
Fig. D.
L  (X e  X b ) 2  (Ye  Yb ) 2
Since the end displacements are
aligned with the end forces, the
local to global coordinate
displacement relationships are
Figure D: Global – Local
Coordinate Relationships
9
X = cos x - sin y
 FX 
 
 FY  
 
 FZ 
Y = sin x + cos y
Z = z
Applying the global – local
coordinate transformations to the
end node forces gives
cos   sin  0  Q x 
 sin  cos  0  Q 

  y
0
1   Qz 
 0
or
{F}node  [t]T {Q}node
where node = b or e.
FX  cos  Q x  sin  Q y
FY  sin  Q x  cos  Q y
10
(17.15)
FZ  Q z
where X, x = 1 or 4; Y, y = 2 or 5;
and Z, z = 3 or 6.
13
Expanding the global – local
coordinate transformation to both
end nodes leads to
{F}b 


{F}
 e
[t]T [0] 


T
 [0] [t] 
{Q}b 


{Q}e 
14
3
or
T
{F}  [T] {Q}
Continuous Beam Members
(17.17)
Similarly, the global coordinate
displacement vector is related to
the local coordinate displacement
vector as
{v}  [T]T {u}
(17.18)
When analyzing continuous beam
structures, the axial displacement
and force degrees of freedom are
typically ignored since they are zero
unless there is axial loading and the
continuous beam is restrained
against longitudinal motion, i.e., is
not free to expand. Regardless, in
continuous beam structures the
local and global coordinate systems
typically coincide resulting in [tb] =
[I]2x2. This leads to
{Fb} = {Qb} and {vb} = {ub} (17.19)
15
14
Truss Members
Local – global force and
displacement relationships (see
Fig. 17.9):
0
0 
 Q1  cos  sin 
 
0
cos  sin 
Q 2   0
{Qa} = [Ta] {Fa}
{ua} = [Ta] {va}
(a) Local Coordinate End Forces
 F1 
F 
 2
 
 F3 
F4 
(17.21)
and Displacements for a Truss Member
(b) Global Coordinate End Forces
and Displacements for a Truss Member
Figure 17.9 Truss Member
Similarly,
{Fa }  [Ta ]T {Qa }
{va }  [Ta ]T {u a }
15
16
4
MEMBER STIFFNESS
RELATIONS IN GLOBAL
COORDINATES
To establish the global coordinate
representation of the element
stiffness equations, start by
substituting {u} = [T] {v} into
(17.4):
{Q}  [k][T]{v}  {Qf } (17.a)
which results in the force quantities being defined in the local
coordinate system and the
displacement vector expressed in
terms of the global coordinate
17
system. To transform the
matrix, i.e., the element stiffness
matrix coefficients aligned with the
global coordinate system and
K ij  Fi v 1
j
force vectors into the global coordinate system, pre-multiply both
sides of (17.a) by [T]T
[T]T {Q}  [T]T [k][T]{v}
 [T]T {Qf }
(17.23)
Substituting {F} = [T]T {Q} into
(17.23):
{F}  [T]T [k][T]{v}  {Ff } (17.24)
where {Ff} = [T]T {Qf}.
Equation (17.24) in matrix form:
{F}  [K]{v}  {Ff }
(17.25)
where [K] = [T]T [k] [T] = global
coordinate element stiffness
18
For a continuous beam member
(also see discussion following
equation 1 and 17.19):
[T] = [Tb] = [I]4x4
with all other vk = 0 and k ≠ j.
All global coordinate stiffness
equations are expressed by (17.24)
and (17.25). However, for beam
and truss structures, the transformation matrix [T], displacement
vector {v}, and force vectors {F}
and {Ff} must be for these
members.
19
{v} = {vb} = {ub}
{F} = {Fb} = {Qb}
{Ff} = {Ffb} = {Qfb}
[K] = [Kbb] = [kbb]
(see 17.6)
20
5
For a truss member (also see
discussion following equations 1
and 17.21):
Structure Stiffness Relations
[T] = [Ta]
{v} = {va}
{F} = {Fa}
{Ff} = {Ffa}
[K]  [K aa ]  [Ta ]T [k aa ][Ta ]
 c2

EA  cs


L  c 2

 cs
c  cos
s  sin
cs
c 2
s2
cs
cs
c2
s 2
cs
cs 

s 2 
 (17.29)
cs 

s 2 
The structure stiffness equations
can now be determined using the
global coordinate member stiffness
equations. Generation of the
structure stiffness equations is
based on the three basic
relationships of structural analysis:
(1) equilibrium, (2) constitutive
relationships, and (3) compatibility.
Specifically, the direct stiffness
procedure involves expressing:
21
22
(1)Node point equilibrium of the
element end forces meeting at
the node with the externally
applied nodal forces;
(2)Substituting the global coordinate constitutive (matrix
stiffness) equations for the
forces in terms of the stiffness
coefficients times the element
end displacements and fixedend force contributions; and
(3)Compatibility of the element
end displacements with the
structure displacement degrees
23
of freedom {d}.
Figure 17.10 – Illustration of Direct
Stiffness Analysis
24
6
Equilibrium Equations
Member Stiffness Relations
(Constitutive Equations)
 FX  P1  F4(1)  F1(2)
2
 P1  F4(1)  F1(2)
(17.30a)
(m)
 FY  P2  F5(1)  F2(2)
2
 P2  F5(1)  F2(2)
(17.30b)
 M Z  P3  F6(1)  F3(2)
2
 P3  F6(1)  F3(2)
m
Since the end forces Fi in (17.30)
are unknown, the member
stiffness relations
 F1 
F 
 2
 F3 
 
F4 
 F5 
 
 F6 
 K11
K
 21
K
  31
 K 41
 K 51

 K 61
K12
K13
K14
K15
K 22
K 32
K 23
K 33
K 24
K 34
K 25
K 35
K 42
K 43
K 44
K 45
K 52
K 62
K 53
K 63
K 54
K 64
K 55
K 65
(m)
 v1 
v 
 2
v 
  3
v4 
 v5 
 
 v6 
(17.30c)
where superscripts (1), (2) designate
25
element (member) 1, 2; respectively.
are substituted into (17.30) to give
(m)
(m)
 Ff1 
F 
 f2
F 
  f3
 Ff 4 
 Ff 5 
 
 Ff 6 
(17.31)
26
Compatibility
Imposing the compatibility
(continuity) conditions
(1)
(1) (1)
(1) (1)
P1  K (1)
41 v1  K 42 v 2  K 43 v 3
(1)
(1) (1)
(1) (1)
(1)
 K (1)
44 v 4  K 45 v 5  K 46 v 6  Ff 4
(2) (2)
(2) (2)
(2) (2)
 K11
v1  K12
v 2  K13
v3
(1)
v1(1)  v (1)
2  v3  0
(2) (2)
(2) (2)
(2) (2)
(2)
 K14
v 4  K15
v 5  K16
v 6  Ff1
v (1)
4
(1) (1)
(1) (1)
(1) (1)
P2  K51
v1  K 52
v 2  K53
v3
 d1 ;
v 5(1)
 d2 ;
(17.35)
v (1)
6
 d3
(2)
v1(2)  d1 ; v (2)
2  d 2 ; v 3  d3
(1) (1)
(1) (1)
(1) (1)
 K54
v 4  K 55
v 5  K56
v 6  Ff(1)
5
(2) (2)
(2) (2)
(2) (2)
 K 21 v1  K 22 v 2  K 23 v 3
(2)
(2) (2)
(2) (2)
(2)
 K (2)
24 v 4  K 25 v 5  K 26 v 6  Ff 2
(2)
(2)
v (2)
4  v5  v6  0
(17.36)
on the constitutive equation version
of the equilibrium equations leads to
(1)
(1) (1)
(1) (1)
P3  K (1)
61 v1  K 62 v 2  K 63 v 3
(1)
(1) (1)
(1) (1)
(1)
 K (1)
64 v 4  K 65 v 5  K 66 v 6  Ff 6
(2)
(2) (2)
(2) (2)
 K (2)
31 v1  K 32 v 2  K 33 v 3
(2)
(2) (2)
(2) (2)
(2)
 K (2)
34 v 4  K 35 v 5  K 36 v 6  Ff 3
K16 
K 26 

K 36 

K 46 
K 56 

K 66 
(2)
(1)
(2)
P1  (K (1)
44  K11 ) d1  (K 45  K12 ) d 2
(2)
(1)
(2)
 (K (1)
46  K13 ) d 3  (Ff 4  Ff1 )
27
(17.39a)
28
7
(1)
(1)
(2)
P2  (K54
 K (2)
21 ) d1  (K 55  K 22 ) d 2
(1)
(1)
(2)
 (K 56
 K (2)
23 )d 3  (Ff 5  Ff 2 )
(2)
(1)
(2)
P3  (K (1)
64  K 31 ) d1  (K 65  K 32 ) d 2
(2)
(1)
(2)
 (K (1)
66  K 33 ) d 3  (Ff 6  Ff 3 )
(17.39b)
(17.39c)
Or collectively as
{P}  [S]{d}  {Pf }
 [S]{d}  ({P}  {Pf })  {P} (17.41)
 K (1)  K (2)
11
 44
(1)
[S]   K 54  K (2)
21
 (1)
(2)
 K 64  K 31

(2)
K (1)
45  K12
(1)
K 55
 K (2)
22
(2)
K (1)
65  K 32
(2) 
K (1)
46  K13 
(1)

K 56
 K (2)
23

(2)
K (1)
66  K 33 
29
Assembly of [S] and {Pf} Using
Member Code Numbers
The explicit details given for the direct
stiff-ness procedure highlights how the
basic equations of structural analysis are
utilized in matrix structural analysis. A
disadvantage of the procedure is that it
is tedious and not amenable to
 F(1)  F(2) 
f1 
 f4

(2) 
{Pf }   Ff(1)
F

5
f2 
 (1) (2) 
F  Ff 3 
 f 6

 P1  Pf1 


{P}   P2  Pf 2 


 P3  Pf 3 
The structure stiffness coefficients
Sij are defined in the usual manner,
i.e., Sij  Pi d 1 force at dof i due to
j
a unit displacement at j with all
other displacements dk = 0 and k≠j.
30
for frame members; and NNP 
Number of Node Points, i.e., number
of nodes used in the 2D structure
discretization). The ID( , ) array
identifies the nodal equation
numbers in sequence:
ID(1,node) = X-axis nodal displ. number
ID(2,node) = Y-axis nodal displ. number
ID(3,node) = Nodal rotation displ. number
computer implementation.
Computer implementation is based
on using a “destination array”
ID(NNDF,NNP) (NNDF  Number of
Nodal Degrees of Freedom, three
31
For example, consider the gable
frame structure shown on the next
slide. The destination array is also
shown on this slide.
32
8
The ID( , ) array along with the
element node number array
IEN(NEN, NEL) (NEN  Number of
Element Nodes, which is two for
discrete structural elements; and
NEL  Number of Elements used to
discretize the structure) is used to
construct the element location
matrix array LM(NEDF, NEL)
(NEDF  Number of Element
Degrees of Freedom, which is six
for the 2D frame elements), i.e.,
Gable Frame Structure
1
ID
2
1
0
0
0
2
1
2
3
3
4
5
6
4
7
8
9
5
0
0
10
Node
3
34
33
LM(i, iel) = ID(i, IEN(1,iel)); 1 = b node
LM(j, iel) = ID(i, IEN(2,iel)); 2 = e node
i = 1, 2, 3 ; j = 4, 5, 6
iel = element number
The IEN( , ) and LM( , ) arrays for the
gable frame example are
iel
1
2
3
4
IEN
1
2
1
2
3
1
2
3
5
0
1
4
0
0
2
5
0
0
3
6
10
2
3
4
4
LM
4
1
4
7
7
5
6
2
5
8
8
3
6
9
9
The LM( , ) array is used to assemble
the global element stiffness matrix and
fixed-end force vector as illustrated on
the next two slides.
35
LM: (1,m) (2,m) (3,m) (4,m) (5,m) (6,m)
LM(1, m)
LM(2, m)
LM(3, m)
LM(4, m)
LM(5, m)
LM(6, m)
 K11
K
 21
 K 31
K
 41
 K 51
K
 61
K12
K 22
K 32
K13
K 23
K 33
K14
K 24
K 34
K15
K 25
K 35
K 42
K52
K 62
K 43
K 53
K 63
K 44
K 54
K 64
K 45
K 55
K 65
 Ff1 
F 
 f2
 Ff 3 
 
 Ff 4 
LM(5, m)  Ff 5 
 
LM(6, m)  Ff 6 
LM(1, m)
LM(2, m)
LM(3, m)
LM(4, m)
K16 
K 26 

K 36 
K 46 

K56 
K 66 
(m)
(m)
36
9
For the example gable frame:
(2)
S11  K (1)
44  K11
(2)
Pf1  Ff(1)
4  Ff1
(1)
S21  K 54
 K (2)
21

(2)
Pf 2  Ff(1)
5  Ff 2
(4)
S9,9  K (3)
66  K 66
(4)
Pf 9  Ff(3)
6  Ff 6
(4)
S10,9  K 36
Pf10 

Example Problems
Ff(4)
3
S9,10  K (4)
63
Truss
Frame
(4)
S10,10  K 33
37
38
Self-Straining: structure that is
internally strained and in a state of
stress while at rest without
sustaining any external loading
SUMMARY
Identify structure degrees of
freedom {d}
For each element:
Evaluate [k], {Qf}, and [T]
Calculate [K] = [T]T [k] [T]
{Ff} = [T]T {Qf}
Assemble [K] into [S]
Assemble {Ff} into {Pf}
Form nodal load vector {P}
Solve: [S] {d} = {P} – {Pf}
For each element:
Obtain {v} from {d} (compatibility)
Calculate {u} = [T] {v}
{Q} = [k] {u} + {Qf}
Determine support reactions by
considering support joint equilibrium
39
40
10
Temperature Restraint
Example
Problems
Displacements of Statically
Determinate Structures
41
Some Features of the
Stiffness Equations
An important characteristic of both
element and global (structure)
linear elastic stiffness equations is
that they are symmetric. Practically, this means that only the
main diagonal terms and
coefficients to one side of the main
diagonal need to be generated and
stored in a computer program.
Also note that the stiffness
(equilibrium) equation for a given
degree of freedom is influenced 43by
42
the degree of freedom is
influenced by the degrees of
freedom associated with the
elements connecting to that
degree of freedom. Thus, nonzero
coefficients in a given row of a
stiffness matrix consist only of the
main diagonal and coefficients
corresponding to degrees of
freedom at that node and at other
nodes on the on the elements
meeting at the node for which the
stiffness equation is being formed.
44
11
All other coefficients in the row are
zero. When there are many
degrees of freedom in the
complete structure, the stiffness
matrix may contain relatively few
nonzero terms, in which case it is
characterized as sparse or weakly
populated.
Banded Stiffness Equations
45
bandwidth. This is most easily
achieved by minimizing the nodal
difference between connected
elements. All commercial
structural and finite element codes
have built in schemes to minimize
the interaction of the stiffness
equations regardless of the userspecified input.
47
Clearly, it would be advantageous
in the solution phase of analysis to
cluster all nonzero coefficients
close to the main diagonal. This
isolation of the zero terms
facilitates their removal in the
solution process. This can be
done by numbering the degrees of
freedom in such a way that the
columnar distance of the term
most remote from the main
diagonal coefficient in each row is
minimized; i.e., by minimizing the
46
Indeterminacy
Cognizance of static indeterminacy
is not necessary in the direct stiffness approach. The displacement
approach uses the analogous
concept of kinematic
indeterminacy, refers to the
number of displacement degrees
of freedom that are required to
define the displacement response
of the structure to any load.
Stated another way, kinematic
indeterminacy equals the
48
12
number of displacement degrees
of freedom that must be
constrained at the nodes in
addition to the boundary
constraints (support conditions) to
reduce the system to one in which
all the nodal displacements are
zero or have predetermined
values, e.g., specified support
settlement.
It is fairly obvious that kinematic
indeterminacy is quite different
from static indeterminacy.
49
13