Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Introduction
The principles for the direct stiffness method are now in place. In this section of notes we will derive the stiffness matrix, both local and global, for a truss element using the direct stiffness method. Here a local coordinate system will be utilized initially anddirect stiffness method. Here a local coordinate system will be utilized initially and the element stiffness matrix will be transformed into a global coordinate system that is convenient for the overall structure.
Inclined or skewed supports will be discussed.pp
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The truss element shown below is assumed to have a constant cross sectional area (A), a modulus of elasticity (E), and an initial length (L). The nodal degrees of freedom are the axial displacements directed along the length of the truss element. In addition the followingaxial displacements directed along the length of the truss element. In addition the following assumptions are made:
1. The truss element cannot sustain shear forces
2 A ff t f t di l t i d
xd2ˆ
d̂
2. Any effect of transverse displacements are ignored
3. Hooke’s law applies
Deformed shape
x2xd1
f̂f̂Node(a truss hinge)
x
Element
xf2xf1
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
With truss elements we again assume just as we did with spring and axial force elements that
ˆˆ
This is an approximation function. We are assuming through the use of this function that displacements are distributed in an approximately linear fashion along the element
xaau ˆˆ 21 +=
that displacements are distributed in an approximately linear fashion along the element.
( )xu ˆˆ
x̂As noted earlier the total number of unknown coefficients must equal the degrees of freedom associated with the element.
i h i l f l d i l l l di ffi iWith axial force elements and spring elements local coordinate axes were sufficient to formulate most problems. This does not work for truss elements where careful attention must be paid to local and global coordinate orientations.
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Transformation of Vectors
In nearly all finite element analyses it is a necessity to introduce both local (to the element) and global (to the component) coordinate axes.
P
Y
Gl b l
XZ
Global axes
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Looking at forces, the transformation from local to global coordinates is as follows:
fGlobal Local
fY
−=
y
x
Y
X
ff
ff
θθθθ
cossinsincos
Wh li d t t l t
fX
θ
ff ˆ00i θθ
GlobalLocal
When applied to a truss element:
fff
fff
x
y
x
X
Y
X
ˆˆˆ
sincos0000cossin00sincos
2
1
1
2
1
1
−
−
=
θθθθθθ
f2Y
{ } [ ] { }ff
ff yY
ˆ*T
cossin00 22
=
θθ
f1Y f2X
ˆ
yf 2̂xf 2̂
2 { } [ ] { }ff
f1Xθ
xf1̂yf1
1Note: Local x-axis is always along the element and θ is always measured counterclockwise from global x-axis to local x-axis.
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
From Statics we learned that a truss component is a two force member, i.e., this is an element that will not sustain shear forces: f2Y
f1Y f2X
02̂ =yfxf 2̂
2
f1Xθ
xf1̂01̂ =yf
1
2
The forces at either end must be directed along the truss component, and
ˆ00i ff θθ
Global Local
1
−
−
=
ˆ0
i00sincos0000cossin00sincos
2
1
2
1
1
x
x
X
Y
X
f
f
ffff
θθθθ
θθθθ
0cossin002Yf θθ
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
This leads to the following system of equations
000ˆcos 11 ++−=X ff θ
0ˆcos00
000ˆsin
000cos
22
11
11
−++=
+++=
++
xX
xY
xX
ff
ff
ff
θ
θ
θ
and the solution leads to the following matrix formulation
0ˆsin00 22 +++= xY ff θ
g
Y
X
x ff
f 1
1
1 00sincosˆ θθ
=
Y
Xx
fff
2
22sincos00ˆ θθ
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
A similar relationship holds for nodal displacements between local and global coordinate systems.
d2Y
d̂d1Y d2X
xd2ˆyd2
ˆd1̂
2
d1X
dd
dd xX
ˆˆ
00cossin00sincos 11
−
θθθθ
θxd1
yd1 Global Local
1
ddd
ddd
y
x
y
Y
X
Y
ˆˆ
cossin00sincos0000cossin
2
2
1
2
2
1
−=
θθθθ
θθ
{ } [ ] { }dd ˆ*T=
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Inverting this last matrix expression yields:
ˆ
−
=
X
Y
X
x
y
x
ddd
ddd
2
1
1
2
1
1
sincos0000cossin00sincos
ˆˆ
θθθθθθ
or
−
Y
X
y
x
dd 2
2
2
2
cossin00ˆ θθ
It becomes obvious that:
{ } [ ] { }dd Tˆ =
Similarly
[ ] [ ] 1*TT −=
S a y
{ } [ ] { }ff Tˆ =
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The element stiffness matrix can now be formulated in terms of the global coordinate system as follows.
df ˆ0101ˆ
In local coordinates Element nodal forces and displacements in local coordinates
−
−
=
x
y
x
x
y
x
ddd
LEA
fff
2
1
1
2
1
1
ˆˆ
010100000101
ˆˆ { } [ ] { }df ˆk̂ˆ =→
y
x
y
x
df 2
2
2
2ˆ0000ˆ Element stiffness matrix
in local coordinates
With
{ } [ ] { }dd Tˆ = { } [ ] { }ff Tˆ =
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
then { } [ ] { }[ ] { } [ ] [ ] { }df
df
=
=
Tk̂T
ˆk̂ˆ
[ ] { } [ ] [ ] { }
{ } [ ] [ ] [ ] { }
{ } [ ] { }dkf
df
df
=
=
− Tk̂T
TkT1
Thus the element stiffness matrix in terms of global coordinates is
{ } [ ] { }dkf =
[ ] [ ] [ ][ ]ˆ1
and in a matrix format the element stiffness matrix is
[ ] [ ] [ ][ ]TT 1 kk −=
θθθθθθ 22 sincoscossincoscos
[ ] [ ] [ ][ ]
−−−−
−−
== −
θθθθθθθθθθθθθθθθθθ
22
221
sincoscossincoscossinsincossinsincos
sincoscossincoscos
TˆTL
EAkk
Here again θ is the angle between the local and global x-coordinate axes, measured counterclockwise.
−− θθθθθθ 22 sinsincossinsincos
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
− 0101In order to formulate the component stiffness
In Class Example
Node number[ ]
−=
000001010000
1 LEAk
matrix from the individual element stiffness matrices consider the following simple truss
23
2
0000
−45°
−− 4
242
42
42
P (kN)Y
global
1
23
3
90°
[ ]
−−−−−−
=
42
42
42
42
42
42
42
42
42
42
42
42
2 LEAk
(kN)X
global
Elementnumber
1 3
[ ]
−10100000
EAk
L (m) 4444
number [ ]
−
=
101000003 L
EAk
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Begin creating component stiffness matrix
from element stiffness matrices using element #1from element stiffness matrices using element #1
−
XX
dd
EAff 11
00000101
1,31,1
−=
Y
X
Y
Y
X
Y
ddd
LEA
fff
3
3
1
3
3
1
000001010000
3,1 3,3
Global force-displacement relationship still has the form
−
Y
X
Y
X
dd
FF
1
1
1
1
00000101
and for the moment we focus on
=
Y
X
Y
X
ddd
LEA
FFF
2
2
2
2
0101
{ } [ ] { }dF K=
and for the moment we focus on the contribution of the element nodal forces (f) to the global nodal forces (F).
−
Y
X
Y
X
dd
FF
3
3
3
3
00000101
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
df 2222
−−−−
−−
=
X
Y
X
X
Y
X
ddd
LEA
fff
3
2
2
42
42
42
42
42
42
42
42
42
42
42
42
3
2
2 2,32,2
−−
Y
X
Y
X
dff
3
3
42
42
42
42
4444
3
33,2 3,3
−
Y
X
Y
X
dd
FF
1
1
1
1
00000101
−−−−
=
Y
X
Y
Y
X
Y
ddd
LEA
FFF
2
2
1
222242
42
42
42
42
42
42
42
2
2
1
101
−−−+−−
Y
X
Y
X
dd
FF
3
3
42
42
42
42
42
42
42
42
3
3
00101
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
df 0000
−=
X
Y
X
X
Y
X
ddd
LEA
fff
2
1
1
2
1
1
00001010
00002,11,1
−
Y
X
Y
X
dff
2
2
2
2
1010 2,12,2
−−
Y
X
Y
X
dd
FF
1
1
1
1
001010010001
−+−−−+−−
−−=
Y
X
Y
X
ddd
LEA
FFF
2
2
222242
42
42
42
42
42
42
42
2
2
101110
00
−−−+−−
Y
X
Y
X
dd
FF
3
3
42
42
42
42
4444
3
3
00101
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
− dF 010001
−−−
−
=
X
Y
X
X
Y
x
ddd
EAFFF
2
1
1
42
42
42
42
2
1
1
00001010010001
−+−−−+−−
=
X
Y
X
Y
dddL
FFF
3
2
222242
42
42
42
42
42
42
42
3
2
00101
110
−− YY dF 344443 00
vector of nodal displacementsglobal coordinate system
vector of nodal forcesl b l di t t
Component stiffness matrix
global coordinate systemglobal coordinate system { } [ ]{ }dKF =
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
• How to deal with the unknown support reactions ?• These are nodes where the displacements are known, i.e., zero in the perfectly rigid
support case. Supports can have proscribed displacements.
1 0 0 0 1 0 unknown support reactions known support displacements
1 31 1 1 1
1 31 1 1 1
2 2 2 22 3
1 0 0 0 1 00 1 0 1 0 0
0 0
X X X X
Y Y Y Y
F R f fF R f f
F R f f EA
− = = + −= = + − − +
000
4 4 4 42 2 2 22 2 2 22 3
4 4 4 42 2 2 21 2 2 2 2 2
3 3 3 4 4 4 4
0 0
0 1 10 1 0 1
X X X X
Y Y Y Y
X X X
F R f f EALF R f f
F f f
= = + = − − + −= = + = = + − − + − 3
00
Xd
4 4 4 41 2 2 2 2 23 3 3 4 4 4 40 0Y Y YF P f f
= − = + − − 3Yd
k d lknown applied nodal loads unknown nodal displacements
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
With the stiffness matrix partitioned as follows
XR1 0
=
X
Y
X
KKEARRR
12112
1
1
000
−X
Y
dd
KKL
P
R
3
22212 00
where YdP 3
01100001
−
0001
−
−
=42
4200
0110
11K
−=42
42
00
12K
+−
421
4210
−
42
42
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
−−2201
−+221
and
−
=
42
4200
4401
21K JSK =
−
+=
42
42
441
22
then determine nodal displacements first by solving the following subsystem of equations:
[ ]
−
=
−
− 0
1
122
3
3
PK
dd
Y
X
−
−
−−+
=
1
01
42
42
42
42
PL
PEAL
−−=
221AEPL
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
At this point the unknown unknown reaction forces can be computed from
− PR X 00100011
−=
−−−
=
P
P
PLEARRR
X
Y
X
0000
00001010010001
42
42
42
42
2
1
1
−
−−=−=
−−−+−−−+−−
− P
P
dd
AEL
P
R
Y
X
Y
0221
10
00101
1100
3
3222242
42
42
42
42
42
42
42
2
or simply
PdP Y 22100 34444
PR
−=
P
P
RRR
X
Y
X
0
2
1
1
PR Y2
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
At this point the local displacements are extracted from the global displacements through
{ } [ ] { }dd Tˆ ={ } [ ] { }dd T
−
Y
Xx
dd
dd
1
1
1
1
00cossin00sincos
ˆˆ
θθθθ
−
=
Y
X
Y
y
x
y
dd
ddd
2
2
1
2
2
1
cossin00sincos00
coss
ˆˆ
θθθθ
θθ
1 1ˆ 01 0 0 0x Xd d =
The local nodal displacements for element #1 are
PL
ˆ11
33
3
ˆ 00 1 0 0ˆ 10 0 1 0
0 0 0 1 1 2 2ˆ
Yy
Xx
Y
dd PLdAEd
dd
= = = − = − −
( )
−−
−=
AEPL
AEd
d
y
x
221ˆ
ˆ
3
3
33 Yyd
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
{ }{ }dB ˆ=ε
The axial strain in element #1 is
{ }{ }
( )PLAEPL
LL
dB
22111
−
−=
ε
( )
PAE
PLLL
22
221
−=
−−
EA=
The negative sign indicates compression. Calculate the stress and the force in element #1
AP
E
22−=
= εσ1 1
1 3
2 2
x xf fA
P
σ= −=
1 11 3
0y yf f= −
=A 2 2P= −
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The local nodal displacements for element #3 are
=
===
−
=
000
000
100000010010
ˆˆˆ
1
1
1
1
Y
X
y
x
ddd
AEPL
ddd
=
=
−
00
00
01001000
ˆ2
2
2
2
Y
X
y
x
ddAE
dd
{ }{ }ˆB dε =
The axial strain in element #3 zero
{ }{ }01 10L L = −
0=
And element #3 is a zero force member.
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The local nodal displacements for element #2 are
=
−−
0ˆ
242
42
42
42
2 Xx dd
−−=−=
=
−−−
−=
2211
0
ˆˆˆ
3
3
2
42
42
42
42
42
42
42
42
42
42
42
42
3
3
2
Y
X
Y
x
y
ddd
AEPL
ddd
( )( ) ( )( )( ) ( )
( ) ( )
−−=−−+−−=−−+−−
=
1221112211
22
42
42
42
42
344443 Yyd
( )( ) ( )( )( ) ( )
−=−−+−−=−−−−
=
1221112211
42
42
42
42
The axial strain in element #2 is
{ }{ }d
dB
=
0
ˆε
AEP
dd
AEPL
LL X
X −=
−==
−=
1011
3
2
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
The axial stress in element #2 is
E= εσ
Fi ll th d l f i t d ith l t #2
AP
−=
Finally, the nodal forces associated with element #2 are
Aff xx −= 2
32
2
0
23
22 −= yy ff
PA
−== σ 0=
Section 6: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
In class example(s)