caed lec5 truss sept2014
DESCRIPTION
caeTRANSCRIPT
Truss Element
Outline
Introduction to 2-D FE Analysis
Truss Bar problem
2-D Cantilever problem
Frame element
Objectives Understand what truss element is and its degree of
freedom
Understandhow to develop the transformation matrix
Know how to formulate the stiffness matrix of truss element
CAE (FEM) Syllabus Structure
1-D Problem
1-D element
-spring
-bar
2-D Problem
1-D element
-Truss
-Beam
-Frame
2-D element
-Linear
-Quadratic
Two-dimensional analysis allows us to
determine detailed information concerning
deformation, stress and strain, within a
complex shaped two-dimensional elastic body.
Why 2-D Analysis?
Introduction
In 1-D problems, we have equations in 1-D axis
Force
In x-Dir.
x
Force
In y-Dir.
y
Force1
In the bar longitudinal-Dir.
Force2
Introduction
E1 E2 100 N
Cantilever
(Variable or constant area)
force
Metal plate with cut
In all 1-D cases, we applied one equilibrium principles
at two node for each element.
And we have obtained 2 equations for each
Element.
Introduction
For a 1-D , the 2 equilibrium equations
are written in local element coordinates as:
e2
e1
e2
e1
F
F
u
u
kk
kk
ee
ee
}F{}u{]K[ eee
Element e
1e 2e
u1e
u2e
F1e F2
e
ke
As a reminder:
e2
e2
e1
e1
e2
e1
F)uu(k
F)uu(k
e
e
And transferred to Matrix, as:
IntroductionStep from
1-D to 2-D
From local coordinates to Global Coordinates
Global coordinates
Fixed in space
x`Local coordinates
Directed according
to the element direction
Y
X
Look, now for each node we will write 2 equilibrium equations
Hence, For each element, they will be 4 equations
y`
Introduction
Original before
deformation
After
deformation
In 1-D analysis, assuming small displacements and elastic material behavior we have:
a) A strain-displacement eqn.,
and,
b) A stress-strain eqn.
(1D Hooke’s Law)
dx
du
L
LL
0
0
E
Question: What is the 2-D or 3-D form of these equations?
2-D Stress Analysis
Three cases will be considered under
the 2-D simulation :
•Truss Problem
•Beam Problem
•Frame Problem
2-D FE Analysis
2-D Truss Bar Problem
Definition: Truss An engineering structure consisting of straight members
connected at their ends by means of bolts, rivets, pins or welding
Tension and compression capability only (no bending).
Degree of Freedom
Truss has tension and compression capability
No bending
Therefore, in 2-D problem, truss element has 2 dof at each node
Displacements in x and y directions
U1y
U2
x
U2y
q
U1x
ke
node 1
node 2
Typical plane 2-D trusses
A truss structure consists only of two-force members:
direct tension and compression.
Truss Bar Problem
What is the size of the element stiffness matrix for a 2-Dtruss element?
U1y
U2x
U2y
F1x
F1y
F2x
F2y
q
U1x
ke
node 1
node 2
Truss Bar Problem
The element stiffness matrix is now a 4 x 4 instead of a 2 x 2 in 1-D analysis, because each node can now move in the x- or y- direction. In this case each node has 2 DOFs.
U1y
U2x
U2y
F1x
F1y
F2x
F2y
q
U1x
ke
node 1
node 2
The four nodal equilibrium
equations for the element can be
written in matrix form as:
Where k11, …, k44 are sixteen
unknown stiffness coefficients
to be determined.
Truss Bar Problem
y2
x2
y1
x1
F
F
F
F
y2
x2
y1
x1
U
U
U
U
44434241
34333231
24232221
14131211
kkkk
kkkk
kkkk
kkkk
Xx
y
θ4(= 0)
Xθ1
X
θ2 (= 90)X
θ3
X
Y
Since there are many cases, I need to generalize
Truss Bar Problem
x
y
There are two frames of references for the truss problems:
• Global coordinate system (X,Y)
• Local coordinate system (x, y)
1. Choose the global (X,Y) system to represent the
location of each joint (or, node).
2. Track the orientation of each member (or, element)
by angle, θ.
3. Write the transformation of the displacement from
local to global
Truss Bar Problem
UiX
UiY
X
Yi
j
θ
Uiy
UiX
θ
UiX = uix cos θ - uiy sin θ
UiY = uix sin θ + uiy cos θ
UjX = ujx cos θ - ujy sin θ
UjY = ujx sin θ + ujy cos θ
UiX = uix cos θ - uiy sin θ
UiY = uix sin θ + uiy cos θ
UjX = ujx cos θ - ujy sin θ
UjY = ujx sin θ + ujy cos θ
In Matrix form: {U}=[T]{u}
Displacements of nodes
Truss Bar Problem
UjX
UjY
UiX
UiY
X
Yi
j
θ
Local displacements in Global CS
jy
jx
iy
ix
jyjx
jyjx
iyix
iyix
u
u
u
u
uu
uu
uu
uu
cossin00
sincos00
00cossin
00sincos
cossin00
sincos00
00cossin
00sincos
jX
jX
iY
iX
jX
jX
iY
iX
U
U
U
U
form;matrix in written be can equationslinear These
U
U
U
U
=> {U}=[T]{u}
[T] is called the transformation matrix
Is it only the displacement can be analyzed at the nodes??
The forces can be analyzed as well
Truss Bar Problem
i
jFjX
FjY
FiX
FiY
X
Y
θ FiX = fix cos θ - fiy sin θ
FiY = fix sin θ + fiy cos θ
FjX = fjx cos θ - fjy sin θ
FjY = fjx sin θ + fjy cos θ
In Matrix form: {F}=[T]{f}
Forces at nodes
Truss Bar Problem
Similar to displacement
jy
jx
iy
ix
jyjx
jyjx
iyix
iyix
f
f
f
f
ff
ff
uu
uu
cossin00
sincos00
00cossin
00sincos
cossin00
sincos00
00cossin
00sincos
jX
jX
iY
iX
jX
jX
iY
iX
F
F
F
F
form;matrix in written be can equationslinear These
F
F
U
U
=> {F}=[T]{f}
[T] is called the transformation matrix
y
x
(x2,y2)
(x1,y1)1
2
θ
φle
x2-x1
y2-y1
Direction cosines?
212
2
12 yyxxle q sincos 12
el
yym
qcosl
el
xxl 12cos
q
qsinmthen
Truss Bar Problem
Now, the Transformation matrix becomes
lm
ml
lm
ml
T
00
00
00
00
cossin00
sincos00
00cossin
00sincos
Local force-stiffness-displacement matrices
For 1-d problem:
jx
ix
jx
ix
u
u
kk
kk
f
f
For 2-d problem:
jy
jx
iy
ix
jy
jx
iy
ix
u
u
u
u
kk
kk
f
f
f
f
0000
00
0000
00
jy
jx
iy
ix
jy
jx
iy
ix
u
u
u
u
k
f
f
f
f
0000
0101
0000
0101k is defined as equivalent
stiffness
L
EAk
We want to relate the forces to the deformation by using the spring eqn.
What is the element (spring) equation?
f = k . u
If we have multi forces, multi displacements in the same element,
it should be written in matrix format
{f}= [k] {u}Remember: this is
in local coordinates
But, we have{f}=[T]¯¹{F} & {u}=[T]¯¹{U}
Truss Bar Problem
Compensate for {f} and {u}, getting:
[T]¯¹{F} = [k] [T]¯¹{U}
Multiply both sides by [T], getting:
{F} = [T][k] [T]¯¹{U}
where, [T][k] [T]¯¹= [K] , element stiffness matrix
This will result in the general 2-D element relation, as
Truss Bar Problem
{F} = [K] {U}
FiX
FiY
FjX
FjY
22
22
22
22
mlmmlm
lmllml
mlmmlm
lmllml
k
UiX
UiY
UjX
UjY
=
l
AEk ee
e
Force matrix
Displacement
matrix
Element stiffness matrix
Truss Bar Problem
After getting the forces and displacements, Further information:
• the Reaction forces
From the equilibrium equation applied on the element,
R = ku – f
Generally, for set of elements with 2-D,
{R} = [K]{U} – {F}
• the Stresses in the element
From the definition of the stress
σe = fe /Ae = ke u / Ae = ke (ui – uj)e / Ae
Truss Bar Problem
Reactions and Stresses
Reaction forces
FUKRG
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
Y
X
Y
X
Y
X
Y
X
Y
X
F
F
F
F
F
F
F
F
F
F
U
U
U
U
U
U
U
U
U
U
R
R
R
R
R
R
R
R
R
R
5
5
4
4
3
3
2
2
1
1
5
5
4
4
3
3
2
2
1
1
5
5
4
4
3
3
2
2
1
1
KG
Stress in each element
𝜎 =𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑜𝑟𝑐𝑒
𝑎𝑟𝑒𝑎=𝑓
𝐴
𝜎 =𝑓
𝐴=𝑘(𝑢𝑖𝑥 − 𝑢𝑗𝑥)
𝐴=
𝐸𝐴𝐿 (𝑢𝑖𝑥 − 𝑢𝑗𝑥)
𝐴
𝜎 =𝐸(𝑢𝑖𝑥 − 𝑢𝑗𝑥)
𝐿
𝑈 = 𝑇 𝑢𝑢 = 𝑇 −1 𝑈𝑢𝑖𝑥𝑢𝑖𝑦𝑢𝑗𝑥𝑢𝑗𝑦
=
cos 𝜃 sin 𝜃 0−sin 𝜃 cos 𝜃 0
00
00
cos 𝜃−sin 𝜃
00
sin 𝜃cos 𝜃
𝑈𝑖𝑥
𝑈𝑖𝑦
𝑈𝑗𝑥𝑈𝑗𝑦
Any questions? Transformation matrix
Directional cosines
Element stiffness matrix (for Truss)
UjX = ujx cos θ - ujy sin θ
UjY = ujx sin θ + ujy cos θ
UiX = uix cos θ - uiy sin θ
UiY = uix sin θ + uiy cos θ
UjX = l ujx – m ujy
UjY = m ujx + l ujy
UiX = l uix - m uiy
UiY = m uix + l uiy
The 4 displacement equations
can be re-written as
In Matrix form: {U}=[T]{u}
or {u}=[T]¯¹{U}
In Matrix form, still same:
{U}=[T]{u}
or {u}=[T]¯¹{U}
Truss Bar Problem
FjX = fjx cos θ - fjy sin θ
FjY = fjx sin θ + fjy cos θ
FiX = fix cos θ - fiy sin θ
FiY = fix sin θ + fiy cos θ
FjX = l fjx – m fjy
FjY = m fjx + l fjy
FiX = l fix - m fiy
FiY = m fix + l fiy
The 4 force equations
are re-written as
In Matrix form: {F}=[T]{f}
or {f}=[T]¯¹{F}
In Matrix form, still same:
{F}=[T]{f}
or {f}=[T]¯¹{F}
Truss Bar Problem