Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
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 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.
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
X
Y
Z
Global axes
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
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
Looking at forces, the transformation from local to global coordinates is as follows:
When applied to a truss element:
fX
fY
y
x
Y
X
ff
ff
cossinsincos
ff
ffff
ffff
y
x
y
x
Y
X
Y
X
ˆ*T
ˆˆˆˆ
cossin00sincos0000cossin00sincos
2
2
1
1
2
2
1
1
f1X
f1Y f2X
f2Y
xf1̂
yf1̂
yf 2̂xf 2̂
Global Local
GlobalLocal
1
2
Note: Local x-axis is always along the element and is always measured counterclockwise from global x-axis to local x-axis. 3
Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
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:
The forces at either end must be directed along the truss component, and
0
ˆ0
ˆ
cossin00sincos0000cossin00sincos
2
1
2
2
1
1
x
x
Y
X
Y
X
f
f
ffff
f1X
f1Y f2X
f2Y
xf1̂
01̂ yf
02̂ yfxf 2̂
Global Local
1
2
4
Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
This leads to the following system of equations
and the solution leads to the following matrix formulation
0ˆsin00
0ˆcos00
000ˆsin
000ˆcos
22
22
11
11
xY
xX
xY
xX
ff
ff
ff
ff
Y
X
Y
X
x
x
ffff
f
f
2
2
1
1
2
1
sincos0000sincos
ˆ
ˆ
5
Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
d1X
d1Y d2X
d2Y
dd
dddd
dddd
y
x
y
x
Y
X
Y
X
ˆ*T
ˆˆˆˆ
cossin00sincos0000cossin00sincos
2
2
1
1
2
2
1
1
xd2ˆyd2
ˆ
xd1̂yd1̂
A similar relationship holds for nodal displacements between local and global coordinate systems.
Global Local
1
2
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
Inverting this last matrix expression yields:
or
It becomes obvious that:
Similarly
Y
X
Y
X
y
x
y
x
dddd
dddd
2
2
1
1
2
2
1
1
cossin00sincos00
00cossin00sincos
ˆˆˆˆ
dd Tˆ
1*TT
ff Tˆ 7
Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
y
x
y
x
y
x
y
x
dddd
LEA
ffff
2
2
1
1
2
2
1
1
ˆˆˆˆ
0000010100000101
ˆˆˆˆ
df ˆk̂ˆ
In local coordinates Element nodal forces and displacements in local coordinates
Element stiffness matrix in local coordinates
The element stiffness matrix can now be formulated in terms of the global coordinate system as follows.
With
dd Tˆ ff Tˆ
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
then
Thus the element stiffness matrix in terms of global coordinates is
and in a matrix format the element stiffness matrix is
Here again is the angle between the local and global x-coordinate axes, measured counterclockwise.
dkf
df
df
df
Tk̂T
Tk̂T
ˆk̂ˆ
1
TˆT 1 kk
22
22
22
22
1
sinsincossinsincossincoscossincoscos
sinsincossinsincossincoscossincoscos
TˆTL
EAkk
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
Elementnumber
1
23
Node number
1
2
3
0000010100000101
1 LEAk
101000001010
0000
3 LEAk
°
°
L (m)
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
2 LEAk
P (kN)
In order to formulate the component stiffness matrix from the individual element stiffness matrices consider the following simple truss
X
In Class Example
Yglobal
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
Global force-displacement relationship still has the form
and for the moment we focus on the contribution of the element nodal forces (f) to the global nodal forces (F).
Begin creating component stiffness matrix from element stiffness matrices using element #1
Y
X
Y
X
Y
X
Y
X
dddd
LEA
ffff
3
3
1
1
3
3
1
1
0000010100000101
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
dddddd
LEA
FFFFFF
3
3
2
2
1
1
3
3
2
2
1
1
00000101
00000101
1,31,1
3,1 3,3
dF K
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
dddddd
LEA
FFFFFF
3
3
2
2
1
1
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
3
3
2
2
1
1
00101
00000101
Y
X
Y
X
Y
X
Y
X
dddd
LEA
ffff
3
3
2
2
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
3
3
2
2 2,32,2
3,2 3,3
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
Y
X
Y
X
Y
X
Y
X
Y
X
Y
X
dddddd
LEA
FFFFFF
3
3
2
2
1
1
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
3
3
2
2
1
1
00101
11000
001010010001
Y
X
Y
X
Y
X
Y
X
dddd
LEA
ffff
2
2
1
1
2
2
1
1
101000001010
00002,11,1
2,1 2,2
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
Y
X
Y
X
Y
X
Y
X
Y
X
Y
x
dddddd
LEA
FFFFFF
3
3
2
2
1
1
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
3
3
2
2
1
1
00101
11000
001010010001
Component stiffness matrix
vector of nodal displacementsglobal coordinate system
vector of nodal forcesglobal coordinate system dKF
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
• 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 31 1 1 1
1 31 1 1 1
2 2 2 22 34 4 4 42 2 2 2
2 2 2 22 34 4 4 42 2 2 2
1 2 2 2 2 23 3 3 4 4 4 4
1 2 2 2 2 23 3 3 4 4 4 4
1 0 0 0 1 00 1 0 1 0 0
0 0
0 1 10 1 0 1
0 0
X X X X
Y Y Y Y
X X X X
Y Y Y Y
X X X
Y Y Y
F R f fF R f f
F R f f EALF R f f
F f fF P f f
3
3
0000
X
Y
dd
known applied nodal loads unknown nodal displacements
unknown support reactions known support displacements
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
With the stiffness matrix partitioned as follows
where
Y
X
Y
X
Y
X
dd
KK
KK
LEA
P
RRRR
3
3
2221
1211
2
2
1
1
0000
0
421
4210
42
4200
01100001
11K
42
42
42
42
0001
12K
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
42
4200
42
4201
21K JSK
42
42
42
421
22
221
1
01
0
1
42
42
42
42
122
3
3
AEPL
PEAL
PK
dd
Y
X
and
then determine nodal displacements first by solving the following subsystem of equations:
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
At this point the unknown unknown reaction forces can be computed from
or simply
P
PP
P
dd
AEPL
LEA
P
RRRR
Y
X
Y
X
Y
X
0
0
2211
0000
00101
11000
001010010001
0
3
3
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
2
2
1
1
PP
P
RRRR
Y
X
Y
X
0
2
2
1
1
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
1 1
11
33
33
ˆ 01 0 0 0ˆ 00 1 0 0ˆ 10 0 1 0
0 0 0 1 1 2 2ˆ
x X
Yy
Xx
Yy
d ddd PL
dAEddd
The local nodal displacements for element #1 are
AEPL
AEPL
d
d
y
x
221ˆ
ˆ
3
3
At this point the local displacements are extracted from the global displacements through
dd Tˆ
Y
X
Y
X
y
x
y
x
dddd
dddd
2
2
1
1
2
2
1
1
cossin00sincos00
00cossin00sincos
ˆˆˆˆ
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
EAP
AEPL
AEPL
LL
dB
22
22111
ˆ
The negative sign indicates compression. Calculate the stress and the force in element #1
The axial strain in element #1 is
AP
E
22
1 1
1 3
2 2
x xf fA
P
1 11 3
0y yf f
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
The local nodal displacements for element #3 are
ˆ
01 10
0
B d
L L
The axial strain in element #3 zero
And element #3 is a zero force member.
0000
0000
0100100000010010
ˆˆˆˆ
2
2
1
1
2
2
1
1
Y
X
Y
X
y
x
y
x
dddd
AEPL
dddd
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
The axial strain in element #2 is
The local nodal displacements for element #2 are
1221112211
1221112211
2211
00
ˆˆˆˆ
42
42
42
42
22
42
42
42
42
3
3
2
2
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
42
3
3
2
2
Y
X
Y
X
y
x
y
x
dddd
AEPL
dddd
AEP
dd
AEPL
LL
dB
X
X
1011
ˆ
3
2
22
Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
The axial stress in element #2 is
Finally, the nodal forces associated with element #2 are
AP
E
PAff xx
23
22
0
23
22
yy ff
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Section 5: TRUSS ELEMENTS, LOCAL & GLOBAL COORDINATES
Washkewicz College of Engineering
In class example(s)
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