3D Coordinate
Transformation
Calculations
Space Truss Member
Transformation of the element
stiffness equations for a space
frame member from the local to
the global coordinate system
can be accomplished as the
product of three separate
transformations. In these slides,
we will develop the details for
these calculations considering
both a space truss member and
a space frame member.
Fig. 1. Typical Space Truss
Member
Consider the space truss member
shown in Fig. 1 with six local coordinate degrees of freedom identified.
We desire to transform the element
stiffness equations
1
{F'}  [ k ' ]{ '}
(1)
into the global coordinate element
stiffness equations
{F}  [k ]{}
(2)
The global coordinate element
stiffness equations of (2) are related
to (1) via
{F}  []T {F'}
{ '}  []{}
member;
l x '
[  ]  l y'
l z '
mx '
m y'
mz '
nx ' 
n y' 
nz ' 
(see (5.2) in
MGZ); {F} = <Fx1 Fy1 Fz1 Fx2 Fy2
Fz2>T; {F} = <Fx1 Fy1 Fz1 Fx2 Fy2
Fz2>T; {} = <u1 v1 w1 u2 v2 w2
>T; {} = <u1 v1 w1 u2 v2 w2>T; and
the 6x6 local coordinate element
(3)
[ k ]  [ ] T [ k ' ] [ ]
[  ]

[ ]  
[  ]

2
stiffness matrix [k] contains only
four nonzero coefficients, i.e., k11=
= local-to-global coordinate
k14= k41= k44=EA/L.
transformation matrix for a truss
Only
3
lx ' , mx ' , and nx '
of [] multiply
4
1
nonzero coefficients in the element
stiffness matrix.
Consequently,
only these rotation components are
required to construct the global
coordinate stiffness matrix and these
three elements are the same as the
direction cosines for the member
itself and can be found from the
coordinates of the member ends as
lx'  C x 
x 2  x1
L
mx'  C y 
y 2  y1
L
Substituting (4) into [] and using
the third equation of (3) leads to
 C2
x

 CxCy

EA  C x C z
[k ] 
L   C 2x

  CxCy

 C x C z
Cy Cx
CzCx
 C 2x
 Cy Cx
C 2y
CzCy
 CxCy
 C 2y
Cy Cz
C 2z
 Cx Cz
 C y Cz
 Cy Cx
 CzCx
C 2x
Cy Cx
 C 2y
 CzCx
CxCy
C 2y
 C y Cz
 C 2z
Cx Cz
C y Cz
 CzCx 

 CzC y 

 C 2z 
C z C x 
CzCx 

C 2z 
(6)
z z
nx '  C z  2 1
L
(4)
The length L can also be calculated
from the member end point
coordinates as
L  ( x 2  x1 ) 2  ( y 2  y1 ) 2  (z 2  z1 ) 2
(5)
5
For geometric nonlinear analysis
and for loads directly applied to the
truss members, it is necessary to
specify the orientation of the cross
section axes y and z. Figure 2
shows the rotational geometry of a
typical space truss member. It is
desired to determine the relationship between the local coordinate
axes and the global coordinate
axes. This relationship can be
determined using two two-dimensional coordinate transformations
for space truss members. The first
two-dimensional transformation is
about the y-axis and relates the
7
global axes to the 1-axes, i.e.
Fig. 2. Rotation Axes for a Space
6
Truss Member
x1   cos  0 sin   x 
  
 
1
0   y 
 y1    0
 z   sin  0 cos   z 
 1 
 
 {x1}  [   ]{x}
(7)
The direction cosines of (7) can be
expressed in terms of Cx and Cxz as
cos  
Cx
C xz
sin  
Cz
C xz
(8)
Equation (7) expresses the rotation of the global x- and z-axes
such that global x-axis lies under
the projection of the x-axis onto
the x-z plane. To complete the
desired transformation, the x1-axis
must be rotated into the x-axis.
8
2
This second two-dimensional transformation is about the z1- or z-axis
and can be expressed as
x '  cos  sin  0
  

 y'   sin  cos  0
 z'  0
0
1
  
x1 
 
 y1 
z 
 1
 {x '}  [   ]{x1}

 Cx
x ' 
 Cx Cy
 
  y'  

 z'   C xz
 
  Cz
 C xz
Cy
C xz
0

Cz 
 x 
 C y Cz   
 y
C xz   
C x   z 
C xz 
(11)
(9)
The direction cosines of (9) can be
expressed in terms of Cy and Cxz as
cos   C xz
(10)
sin   C y
Combining (7) – (10) leads to
Fig. 3. Rotation Axes for a Vertical
Truss Member
{x '}  [  ]{x}  [   ]{x1}  [   ][   ]{x}
9
The transformation matrix given in
(11) is valid for all space truss
member orientations with the
exception of a vertical truss
member as shown in Fig. 3. For
the vertical truss member, Cx = Cz =
Cxz = 0 and (11) is not numerically
defined. Inspection of Fig. 3 shows
that
 0
[  vert ]   C y
 0
Cy
0
0
0
0
1
(12)
11
10
Space Frame Element
A space frame member has twelve
degrees of freedom as discussed in
MGZ (Sections 4.5 and 5.1). Our
purpose is to extend the space truss
transformations (11) and (12) to
include the principal cross section
axes of the member, which is not
necessary for the space truss
member.
The form of the rotation matrix []
depends upon the particular
orientation of the member axes. In
many instances a space frame
12
3
member will be oriented so that the
principal axes of the cross section
lie in horizontal and vertical planes
(e.g., I-beam with its web in a
vertical plane). Under these
scenarios, the y- and z-axes can
be selected in the exact same
fashion as the space truss member.
again be selected to be the same as
the space truss member. This
choice is possible because all axes
in the cross section are principal
axes.
However, a space frame member
may have its principal y- and zaxes in general directions. Two
There are other instances in which a
space frame member has two axes
of symmetry in the cross section and
the same moment of inertia about
each axis (e.g., circular or square
member either tubular or solid). In
these cases, the y- and z-axes can
Fig. 4. Rotation
About the x-Axis
13
methods are typically used to
describe this more general case.
The first choice is to specify the 
angle about the x-axis. This choice
is normally reserved for cases in
which  can easily be visualized,
e.g., horizontal or vertical members.
Another choice is to specify the
coordinates of a point p as shown in
Fig. 4 and calculate the  angle.
In either case, the transformation
process for a space frame member
can be expressed in terms of three
two-dimensional transformations as
shown schematically in the Example
5.6 figure. This three-step process
15
can be expressed as
14
{x }  [   ]{x 2 }  [   ][   ]{x1}
 [   ][   ][  ]{x}  [  ]{x}
(13)
where
0
0  x 2 
x ' 1
  
 
 y'  0 cos  sin    y 2 
 z'  0  sin  cos   z 
  
 2
 {x '}  [   ]{x 2 }


Cx
Cy

  C x C y cos   C z sin 
[]  
C xz cos 
C xz

 C x C y sin   C z cos 
 C xz sin 

C xz

(14)


Cz

 C y C z cos   C x sin  

C xz

C y C z sin   C x cos  

C xz

(15)
16
4
Obviously, if the angle  is input
directly, then (15) is used directly in
computing the local to global
coordinate transformation of the
three-dimensional frame element
stiffness equations. However, if the
angle  is not readily available,
then the coordinates of a reference
point p must be input in order to
calculate  (this procedure is
typically easier than trying to
visualize the angle ). In order to
calculate  from the coordinates of
point p of Fig. 4, you first compute
x pS  x p  x1
y pS  y p  y1
z pS  z p  z1
(16)
in which xp, yp, zp = global
coordinates of point p; and x1, y1, z1
= global coordinates of node 1.
Next, calculate coordinates of p in
terms of {x2} coordinate system as
x pS 
x 2p 




 y 2p   [   ] [   ]  y pS 
z 
z 
 pS 
 2p 
x 2p  C x x pS  C y y pS  C z z pS
y 2p  
z 2p  
CxCy
C xz
x pS  C xz y pS 
C y Cz
C xz
z pS (17)
C
Cz
x pS  x z pS
C xz
C xz
Equations (17) provide the
coordinates of point p with respect
17
18
to the 2-axes. These coordinates
are then used to calculate the
direction cosines for the angle 
of Fig. 4 as
sin  
z 2p
y 22p  z 22p
cos  
y 2p
y 22p  z 22p
(18)
Equations (18) can be substituted
into (15) to produce the desired
transformation for a space frame
member.
As in the space truss member, (15)
will not work for a vertical space
frame element since Cx = Cz =
Cxz = 0. In such cases, the rotation
matrix for a vertical space frame 19
Fig. 5. Rotation of Axes for a
Vertical Space Frame Member
element can be obtained
inspection from Fig. 5 to give

0
Cy

[  vert ]   C y cos  0
 C y sin 
0

0 

sin  
cos 
by
(19)
Equation (19) is valid for both
orientations shown in Fig. 5 and 20
5
if  = 0, then (19) is the same as
(12) (vertical space truss
member).
In those cases where it is desired
to begin with the coordinates of a
point p that is known to lie in a
principal plane, it is possible to
calculate sin  and cos  in (19)
directly from the coordinates of
the point. Figure 6 shows the two
cases of Fig. 5 with a point p
specified. The sine and cosine of
 for both cases can be calculated
from
21
Unsymmetric Cross
Section Analysis
Fig. 6. Use of Point p for a Vertical
Space Frame Member
sin  
z pS
x 2pS

z 2pS
cos  
 x pS
x 2pS
 z 2pS
Cy
(20)
where Cy = 1 for Fig. 6(a) and Cy
= -1 for Fig. 6(b).
22
Element Transformations
and Assembly
Linear stiffness equations for
element e (e.g. 4.34) can be
expressed in terms of the principal
cross section axes as
{F e }  [ k e ]{u e }  {F eF }
(1)
where [k e ] = principal coordinate
element stiffness matrix given in
e
(4.34) of MGZ; {F } = principal
coordinate element force vector;
eF
and {F } = principal coordinate
element fixed-end force vector.
Fig. 1. Unsymmetric Angle Section
and Cross Section Coordinates
23
Equations (1) are expressed in
terms of the uncoupled principal
cross section modes of
deformation. That is, (1) axial, 24
6
(2) torsion, (3) bending about the
strong axis (defined as the z-axis),
and (4) bending about the weak
axis (defined as the y-axis). In
order to assemble various element
equations, the equations must be
expressed in terms of a common
set of coordinate axes. For
members with doubly symmetric
cross sections, the centroid and
shear centers coincide and no
additional transformations are
required. However, when the
member cross section is singly
symmetric or unsymmetric, e.g., the
unequal leg angle shown in Fig. 1,
a transformation of displacement
and force degrees of freedom is
required (e.g., Chen and Blandford
1989). Such transformations are
necessary since (ux, Fx), (y, My),
and (z, Mz) are evaluated in terms
of the cross section centroid
whereas displacement and force
pairs (uy, Fy), (uz, Fz), and (x, Mx)
are evaluated in terms of the cross
section shear center. Due to the
different evaluation points for
these displacement and force
pairs, subscripts C and S will be
introduced to refer to the centroid
and shear center, respectively.
25
Transforming the principal C and S
coordinate stiffness equations is
normally performed such that the
final equations are expressed in
terms of the principal centroidal
axes shown in Fig. 1. Expressing
the principal coordinate element
displacement vector as {u e } =
ue 1  ue 2T where {u e }i = u Cx uSy uSz
T
Sx Cy Cz = element nodal princii
pal coordinate displacement vector
with i = 1, 2 = element node points;
then the transformation from the
principal coordinates to the
centroidal principal coordinates can
be expressed as
27
26
{u e }i  [Tyz ]{u e }i
[I] [ t yz ]
[Tyz ]  

[0] [I] 
(2)
= principal coordinate
to centroid transformation matrix;
[tyz] =
 0
 z
 S
 y S
0 0
0 0 ;
0 0
[I] = 3x3 identity
matrix; {ue} = <uCx uCy uCz Cx Cy
Cz >Ti = element nodal centroidal
displacement vector; and yS, zS =
coordinate distances from the cross
section centroid to the shear center
as shown in Fig. 1. Using the transformation of (2), the element stiffness equations can be expressed in
terms of the element centroidal 28
7
dof as
[k e ]{u e }  {F eF }  {F e }
(3)
element centroid; and {Fe }  [T]T {Fe }
= element force vector measured at
the element centroid.
e
[ke] = [T]T [k ] [T] = element
stiffness matrix; {u e }= [T] {ue};
[Tyz ] [0] 
[T]  

 [0] [Tyz ]
= element principal
coordinate to centroid transformation matrix; {ue} = element displacement vector at the element
cross section centroid (i.e.,
element centroidal dof);
Transformation of the element
stiffness equations of (3) into the
global coordinate system and then
assembly into the structure stiffness
matrix follows the usual procedures.
{FeF }  [T]T {FeF } = element fixed-end
force vector measured at the
29
30
8