Direct Method Truss Analysis Matrice Di Rigidezza Composizione
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7/31/2019 Direct Method Truss Analysis Matrice Di Rigidezza Composizione
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CCOORRNNEELLLLU N I V E R S I T Y 3MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Truss analysis
Internal forces in a truss element act along
the member However, displacements at the nodes can
have both components (x- and y-directions,
in local coordinates). This is due to rotation
'( ) '( )
1 2 0e e
y yF F= =
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CCOORRNNEELLLLU N I V E R S I T Y 4MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Truss analysis
To analyze a truss
element in the globalcoordinates xand y, youneed to account for bothcomponents ofdisplacement:
Also note that the crosssection of truss elementscan vary as shown.
( ) ( ) ( ) ( )
1 1 2 2, , ,
e e e e
x y x yu u u u
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CCOORRNNEELLLLU N I V E R S I T Y 5MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Stiffness of a truss element The internal force in the
truss is given (see free body
diagram) as:
Assuming elasticdeformations:
The (small) strain is given as:
Finally:
( ) ( )
2 1
e e e e ep F F A = = =
( ) ( )
2 1
e e e e e ep F F A E = = =
2 1
e ee
e
u u
L
=
ep
( ) ( )
2 1 2 1
2 1
( )
( )
e ee e e e
e
e e e
A EF F u u
L
k u u
= = =
=
( )
1
eF ( )
2
eF
( ) ( )( ) ( )
1 1
( ) ( ) ( ) ( )
2 2
e ee e
e e e e
F uk k
F k k u
=
e e
e
e
A Ek
L=
http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803 -
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CCOORRNNEELLLLU N I V E R S I T Y 7MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Element stiffness in global coordinates
We need to be able to
transform displacementsfrom the xand yaxes todisplacements along the xand yaxes. We start withthe reverse:
Transformation matrix T(e)
( ) ( )( )
'( ) ( )
1 1
'( ) ( )
1 1
'( ) ( )
2 2
'( ) ( )2 2
{ }{ ' }
cos sin 0 0
sin cos 0 0
0 0 cos sin
0 0 sin cos
e ee
e ee ex x
e ee ey y
e e e e
x x
e ee ey y
T dd
u u
u u
u u
u u
=
The angle is measuredanti-clockwise from
xto x
e
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CCOORRNNEELLLLU N I V E R S I T Y 8MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Coordinate transformation
( ) ( )( )
'( ) ( )
1 1
'( ) ( )
1 1
'( ) ( )
2 2
'( ) ( )
2 2
[ ] { }{ ' }
cos sin 0 0
sin cos 0 0
0 0 cos sin
0 0 sin cos
e ee
e ee ex x
e ee ey y
e e e e
x x
e ee e
y y
T dd
u u
u u
u u
u u
=
{ ' } [ ]{ }
e e e
d T d=
{ } [ ] { ' }e e T ed T d=
:
[ ] [ ] [ ]e T e
Note that
T T I=
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CCOORRNNEELLLLU N I V E R S I T Y 9MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
4 4
cos sin 0 0 cos sin 0 0
sin cos 0 0 sin cos 0 0
0 0 cos sin 0 0 cos sin
0 0 sin cos 0 0 sin cos
1 0 0 0
0 1 0 00 0 1 0
0 0 0 1
T ee
e e e e
e e e e
e e e e
e e e e
TT
I
=
Coordinate transformation
Verify that:4 4
[ ] [ ] [ ]e T e
xT T I=
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CCOORRNNEELLLLU N I V E R S I T Y 10MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Stiffness of a truss element
Similarly for the forces:
( )
'( ) ( )
1 1
'( ) ( )
1 1
'( ) ( )
2 2'( ) ( )
2 2
[ ]{ ' } { }
cos sin 0 0
sin cos 0 0
0 0 cos sin
0 0 sin cos
ee e
e ee ex x
e ee ey y
e e e e
x xe ee e
y y
TF F
F F
F F
F F
F F
=
{ } [ ] { ' }e e T eF T F=
{ ' } [ ]{ }e e e
F T F=
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CCOORRNNEELLLLU N I V E R S I T Y 12MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Truss element stiffness( ) ( ) ( ) ( )
[ ] [ ] [ ' ][ ]e e T e e
K T K T =
( ) ( )
1 0 1 0
0 0 0 0[ ' ] ,
1 0 1 0
0 0 0 0
e eK k
=
( )
cos sin 0 0
sin cos 0 0[ ]
0 0 cos sin
0 0 sin cos
e e
e e
e
e e
e e
T
=
2 2
2 2
( ) ( )
2 2
2 2
cos sin cos cos sin cos
sin cos sin sin cos sin[ ]
cos sin cos cos sin cossin cos sin sin cos sin
e e e e e e
e e e e e e
e e
e e e e e e
e e e e e e
K k
=
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CCOORRNNEELLLLU N I V E R S I T Y 13MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Truss element stiffness
( ) ( )2 21 1
( ) 2 21 ( )
( ) 2 2
2
2 2( )
2
cos sin cos cos sin cos
sin cos sin sin cos sin
cos sin cos cos sin cos
sin cos sin sin cos sin
e ee e e e e ex x
e e e e e e ey e
e e e e e e e
x
e e e e e ee
y
F u
F ukF
F
=
( )
1
( )
2
( )
2
e
x
e
x
e
y
u
u
Note the 2x2symmetric submatrix structure This implies that you can reverse the numbering of nodes (1 and 2)
without any changes in the element stiffness.
( ) ( )2 22 2
( ) 2 22 ( )
( ) 2 2
1
2 2( )
1
cos sin cos cos sin cos
sin cos sin sin cos sin
cos sin cos cos sin cos
sin cos sin sin cos sin
e ee e e e e ex x
e e e e e e ey e
e e e e e e e
x
e e e e e ee
y
F u
F uk
F
F
=
( )2
( )
1
( )
1
ex
e
x
e
y
u
u
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CCOORRNNEELLLLU N I V E R S I T Y 14MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Assembly process
The assembly process is identical to the
one discussed for `spring structures and itwill not be repeated here in its generalform (no need to show at this point
complicated looking matrix operations). We will however provide soon a simple
example demonstrating this assemblyprocess.
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CCOORRNNEELLLLU N I V E R S I T Y 15MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Generalizing the application of essential BCs
You already have seen through an example
how essential boundary conditions areapplied to the global system of eqs:
In essence, we partition the stiffness matrix
in a way that separates known from unknowndegrees of freedom as follows:
[ ]{ } { }K d F=
EE EF E
T
EF F F F
K K fd
K K d f
=
:Ed Known displacements
:Fd Unknown displacements
:Ef Unknown reaction forcescorresponding tonodes/directionswith prescribed displacement:Ff Applied (known) forces
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CCOORRNNEELLLLU N I V E R S I T Y 16MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Generalizing the application of essential BCs
EE EF E
TEF F F F
K K fd
K K d f
=
EE EF F E
TEEF F F F
K d K d f
K d K d f
+ =
+ =
The unknown displacements are obtained from the 2nd eq. as:
1( )
T TE EEF F F F F F F EF
K d K d f d K f K d + = =
With known , we can return to the 1st eq. to compute thereaction forces:
EE EF FEf K d K d= +
Fd
Note that the matrix is symmetric and positive definite,so a solution for always exists!
FK
Fd
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CCOORRNNEELLLLU N I V E R S I T Y 17MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
A truss example
Construct the global stiffness matrix and load vector
Partition the matrices and solve for the unknowndisplacements at point B, and displacement in x
direction at point D. Find the stresses in the three bars
Find the reactions at C, D and F
E= 107Pa
Note that point D isfree to move in the x
direction
1 2 3
1,23,4 5,6
7,8
1 210A A
=
22A A=
3A A=
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CCOORRNNEELLLLU N I V E R S I T Y 18MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
A truss example: Element 1
(1)
1/ 2 1/ 2 1/ 2 1/ 2
1/ 2 1/ 2 1/ 2 1/ 2[ ]
1/ 2 1/ 2 1/ 2 1/ 22
1/ 2 1/ 2 1/ 2 1/ 2
EAK
=
7 8 1 2
7
8
1
2
(1) 0135 =
Note: Recall that you can number the corresponding global nodes in thesequence 1 2 7 8 without any changes in .(1)[ ]K
Local node 1
Local node 2
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CCOORRNNEELLLLU N I V E R S I T Y 19MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
A truss example: Element 2
(2)
0 0 0 0
0 2 2 0 2 2
[ ] 0 0 0 02
0 2 2 0 2 2
EA
K
=
7 8 3 4
7
8
3
4
090 =
22A A=
3,4
Local node 1
Local node 2
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CCOORRNNEELLLLU N I V E R S I T Y 20MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
A truss example: Element 3
(3)
1/ 2 1/ 2 1/ 2 1/ 2
1/ 2 1/ 2 1/ 2 1/ 2[ ]
1/ 2 1/ 2 1/ 2 1/ 221/ 2 1/ 2 1/ 2 1/ 2
EAK
=
7 8 5 6
7
8
5
6
045 =
3A A=
Local node 1
Local node 2
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CCOORRNNEELLLLU N I V E R S I T Y 21MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
A truss example: Assembly (element 1)
1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2
1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0[ ]
0 0 0 0 0 0 0 02
0 0 0 0 0 0 0 0
1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2
1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2
EAK
=
1
2
3
4
5
6
7
8
1 2 3 4 5 6 7 8
(1)
1/ 2 1/ 2 1/ 2 1/ 2
1/ 2 1/ 2 1/ 2 1/ 2[ ]1/ 2 1/ 2 1/ 2 1/ 22
1/ 2 1/ 2 1/ 2 1/ 2
EAK
=
7 8 1 2
7
8
1
2
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CCOORRNNEELLLLU N I V E R S I T Y 22MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
A truss example: Assembly (element 2)
1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 21/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2
0 0 0 0 0 0 0 0
0 0 0 2 2 0 0 0 2 2[ ]
0 0 0 0 0 0 0 02
0 0 0 0 0 0 0 0
1/ 2 1/ 2 0 0 0 0 1/ 2 0 1/ 2 0
1/ 2 1/ 2 0 2 2 0 0 1/ 2 1/ 2 2 2
EAK
=
+ + +
1 2 3 4 5 6 7 8
1
2
3
4
5
6
7
8
( 2)
0 0 0 0
0 2 2 0 2 2[ ]0 0 0 02
0 2 2 0 2 2
EAK
=
7 8 3 4
7
8
3
4
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CCOORRNNEELLLLU N I V E R S I T Y 23MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
A truss example: Assembly (element 3)
1/ 2 1 / 2 0 0 0 0 1/ 2 1/ 21/ 2 1/ 2 0 0 0 0 1 / 2 1/ 2
0 0 0 0 0 0 0 0
0 0 0 2 2 0 0 0 2 2[ ]
0 0 0 0 1/ 2 1/ 2 1/ 2 1/ 22
0 0 0 0 1/ 2 1/ 2 1/ 2 1/ 2
1/ 2 1/ 2 0 0 1/ 2 1/ 2 1/ 2 0 1/ 2 1/ 2 0 1/ 2
1/ 2 1 / 2 0 2 2 1 / 2 1/ 2 1/ 2 1/ 2 1/ 2 2 2 1 / 2
EAK
=
+ + + + + + +
1 2 3 4 5 6 7 81
2
3
4
5
6
7
8
(3)
1/ 2 1/ 2 1/ 2 1/ 2
1/ 2 1/ 2 1/ 2 1/ 2
[ ] 1/ 2 1 / 2 1/ 2 1/ 22
1/ 2 1 / 2 1/ 2 1/ 2
EAK
=
7 8 5 6
7
8
5
6
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CCOORRNNEELLLLU N I V E R S I T Y 24MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
A truss example: Assembly
1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 21/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2
0 0 0 0 0 0 0 0
0 0 0 2 2 0 0 0 2 2[ ]
0 0 0 0 1/ 2 1/ 2 1/ 2 1/ 22
0 0 0 0 1/ 2 1/ 2 1/ 2 1/ 2
1/ 2 1/ 2 0 0 1/ 2 1/ 2 1 0
1/ 2 1/ 2 0 2 2 1/ 2 1/ 2 0 1 2 2
EAK
=
+
1 2 3 4 5 6 7 8
1
2
3
4
5
6
7
8
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A l P i i d BC
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CCOORRNNEELLLLU N I V E R S I T Y 26MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
1
2
3
4
6
5
7
8
1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2 0
1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2 0
0 0 0 0 0 0 0 0 0
00 0 0 2 2 0 0 0 2 2
00 0 0 0 1/ 2 1/ 2 1/ 2 1/ 22
0 0 0 0 1/ 2 1/ 2 1/ 2 1/ 2
1/ 2 1/ 2 0 0 1/ 2 1/ 2 1 0
1/ 2 1/ 2 0 2 2 1/ 2 1/ 2 0 1 2 2
d
d
d
dEA
d
d
d
d
= =
=
= =
+
1
2
3
4
6
3
0
10
0
r
r
r
r
r
N
=
A truss example: Partition and BCs
[ ]E
K
[ ]F
K
[ ]EF
K
[ ]EF
TKF
f
Ef
0Ed =
Fd
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A l R i l l i
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CCOORRNNEELLLLU N I V E R S I T Y 28MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
1
2
3
4
6
5
7
8
1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2 0
1/ 2 1/ 2 0 0 0 0 1/ 2 1/ 2 0
0 0 0 0 0 0 0 0 0
00 0 0 2 2 0 0 0 2 2
00 0 0 0 1/ 2 1/ 2 1/ 2 1/ 22
0 0 0 0 1/ 2 1/ 2 1/ 2 1/ 2
1/ 2 1/ 2 0 0 1/ 2 1/ 2 1 0
1/ 2 1/ 2 0 2 2 1/ 2 1/ 2 0 1 2 2
d
d
d
dEA
d
d
d
d
= =
=
= =
+
1
2
3
4
6
3
0
10
0
r
r
r
r
r
N
=
A truss example: Reaction calculation
[ ]E
K
[ ]F
K
[ ]EF
K
[ ]EF
TKF
f
Ef
0Ed =
Fd
EE EF FEf K d K d= +
A t l R ti l l ti
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CCOORRNNEELLLLU N I V E R S I T Y 29MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
1
52
3 7
4 8
6
0 1/ 2 1/ 2
0 1/ 2 1/ 2
0 0 02
0 0 2 2
1/ 2 1/ 2 1/ 2 F
EEF
d
f K
r
drEA
r d
r d
r
=
A truss example: Reaction calculation
1
2
3
4
6
10001000
0
1000
0
Fx
Fy
Cx
Cy
Dy
Rr N
Rr N
r R
Nr R
rR
=
A t l C t th t
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CCOORRNNEELLLLU N I V E R S I T Y 30MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
cos sin cos sin { }e
e e e e e e
e
Ed
L =
A truss example: Compute the stresses
' ' ' '
2 1 2 1
e e e ee e ex x x x
e e
u u u uE
L L
= =
Combining the 2 Eqs gives:{ }
ed
'( ) ( )
1 1
'( ) ( )
1 1
'( ) ( )
2 2
'( ) ( )
2 2
cos sin 0 0
sin cos 0 0
0 0 cos sin
0 0 sin cos
e ee ex x
e ee ey y
e e e e
x x
e ee e
y y
u u
u u
u u
u u
=
However:
A t l C t th t
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CCOORRNNEELLLLU N I V E R S I T Y 31MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
cos sin cos sin { }e
e e e e e e
e
Ed
L =
A truss example: Compute the stresses
Applying this to each element, we have:
(1)
0.033284
0.0052 2 2 2
141.42102 2 2 22
0
= =
m
mE
kPa
[ ](2 )
0.033284
0.0050 1 0 1 50
0
0
= =
m
mE kPa
(3)
0.033284
0.0052 2 2 20
0.0382842 2 2 22
0
= =
m
mEkPa
m
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Th di i l ( ) t t t
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CCOORRNNEELLLLU N I V E R S I T Y 33MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Three-dimensional (space) truss structures
'( ) '( )( ) ( )
1 1
'( ) ( ) ( ) '( )
2 2
e ee e
x x
e e e e
x x
F uk kF k k u
=
A unit vector along the direction x of a 3D truss
element has the components (direction cosines ofthe axes between x and x,y,z, respectively):
where are the nodalcoordinates in the (x,y,z) system.
2 1 2 1 2 1
2 1 2 1 2 1
2 2 2
, ,
( ) ( ) ( )
s s s
e e e e e e
e e e
e e e
e e e e e e e
x x y y z zl m n
L L L
L x x y y z z
= = =
= + +
1 1 1 22 2( , , ) ( , , )
e e e e e ex y z and x y z
Three dimensional (space) tr ss str ct res
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CCOORRNNEELLLLU N I V E R S I T Y 34MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Three-dimensional (space) truss structures
'( ) '( )( ) ( )
1 1
'( ) ( ) ( ) '( )
2 2
e ee e
x x
e e e e
x x
F uk k
F k k u
=
The displacement transformation then takes the form:
Similar transformation is applied for the forces:
( )
1
( )
1
'( ) ( )
1 1
'( ) ( )
2 2
( )[ ] 2
( )
2
{ }
0 0 0[ ]{ }
0 0 0
s s s
s s s
e
e
e
x
e
y
e e ee e
x z e e
e e e e e
x x
eT y
e
z
d
uu
l m nu uT d
u l m n u
u
u
=
'( )
1
'( )
2
[ ]{ }
e
x e e
e
x
FT F
F
=
Three dimensional (space) truss structures
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CCOORRNNEELLLLU N I V E R S I T Y 35MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Three-dimensional (space) truss structures
'( ) '( )( ) ( )
1 1
'( ) ( ) ( ) '( )
2 2
'
=
e
e ee e
x x
e e e e
x x
K
F uk k
F k k u
Similarly to the derivation for 2D trusses, the stiffness
matrix in global coordinates is then:
( )
6 6 6 2 2 62 2
[ ] [ ] ' [ ] e e T e e
x x xx
K T K T
Stiffness of a space element
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CCOORRNNEELLLLU N I V E R S I T Y 36MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Stiffness of a space element
2 2
2 2
2 2
( )
2 2
2 2 2
[ ]
s s s s s s s s s s
s s s s s s s s s s
s s s s s s s s s s
s s s s s s s s s s
s s s s s s s s s s
s s s
e e e e e e e e e e
e e e e e e e e e e
e e e e e e e e e ee e
e
e e e e e e e e e e e
e e e e e e e e e e
e e e
l m l n l l m l n l
m l m m n m l m m n
n l m n n n l m n nE AK
L l m l n l l m l n l
m l m m n m l m m nn l m
=
2 2
s s s s s s s
e e e e e e en n n l m n n
2 1 2 1 2 1
2 1 2 1 2 1
2 2 2
: , ,
( ) ( ) ( )
s s s
e e e e e e
e e ee e e
e e e e e e e
x x y y z zwhere l m n
L L L
L x x y y z z
= = =
= + +
Computing the stresses in a space truss element
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CCOORRNNEELLLLU N I V E R S I T Y 37MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
{ }s s s s s s
ee e e e e e e e
e
El m n l m n d
L =
Computing the stresses in a space truss element
' '
2 1
' '' '2 1
2 1( )
e ee x x
e
e e ee e e ex x
x xe e
u u
L
u u EE u u
L L
=
= =
Combining the 2 Eqs gives:
( )
1
( )
1
'( ) ( )
1 1
'( ) ( )
2 2
( )
2( )
2
0 0 0[ ]{ }
0 0 0
s s s
s s s
e
x
e
y
e e ee e
x z e e
e e e e e
x x
e
ye
z
u
ul m nu u
T du l m n u
u
u
=
However:
Accounting for thermal effects in truss analysis
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CCOORRNNEELLLLU N I V E R S I T Y 38MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Accounting for thermal effects in truss analysis
Consider a truss structure that is heated. We need toaccount for thermal expansion effects. Note:
Hookes law is now modified as follows (using the xcoordinate system):
2 1( ) ( )
e ee e e e e e e e e
elastic thermal e
total strain
u uE E E T
L
= = =
0
( ) ( ) 2 1
2 1
2 1 0
( )
( ) ,
e
e ee e e e e e e e e
e
e ee e e e e e e
e
u uF F p A A E T
L
A Ek u u A E k
L
= = = = =
= =
( ) ( )( ) ( )
1 1
0( ) ( ) ( ) ( )
2 2
l n
1
1
e ee e
e e e
e e e e
Therma odal vector
F uk kA E
F k k u
+ =
e e e
elastic thermal
total strain
= +
Element equations with thermal effects
http://www.amazon.com/First-Course-Finite-Elements/dp/0470035803 -
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CCOORRNNEELLLLU N I V E R S I T Y 39MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Element equations with thermal effects
Expanding these equations to include nodal displacementsin the y axis gives:
We need to transform this to xand ydisplacements (ourdegrees of freedom for this element)
( )( ) ( )
'( ) '( )
1 1
'( ) '( )
1 1( )
0'( ) '( )
2 2'( ) '( )
2 2
{ ' } [ ' ]{ ' } { ' }
1 1 0 1 0
0 0 0 0 0
1 1 0 1 0
0 0 0 0 0
e ethermale e
e e
x x
e e
y ye e e e
e e
x xe e
y y
F KF d
F u
F uA E k
F u
F u
+ =
{ ' } [ ]{ }e e ed T d= { ' } [ ]{ }e e eF T F=
Element equations with thermal effects
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CCOORRNNEELLLLU N I V E R S I T Y 40MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Element equations with thermal effects
We can transform these element equations as follows:
From these equations, we conclude that:
( )( ) ( )
'( ) '( )
1 1
'( ) '( )
1 1( )
0'( ) '( )
2 2
'( ) '( )
2 2
{ ' } [ ' ]{ ' } { ' }
1 1 0 1 0
0 0 0 0 0
1 1 0 1 0
0 0 0 0 0
e ethermale e
e e
x x
e e
y ye e e e
e e
x x
e e
y y
F KF d
F u
F u
A E kF u
F u
+ =
{ ' } [ ]{ }e e ed T d=
{ ' } [ ]{ }e e eF T F=
( )[ ]{ } { ' } [ ' ][ ]{ }
e e e e e e
thermalT F F K T d + =
( )
( )
[ ]{ }
{ } [ ] { ' } [ ] [ ' ][ ]{ }
eethermal
e e T e e T e e e
thermal
KF
F T F T K T d + =
Element equations with thermal effects
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CCOORRNNEELLLLU N I V E R S I T Y 41MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Element equations with thermal effects
( )
( )
[ ]{ }
{ } [ ] { ' } [ ] [ ' ][ ]{ }
ee
thermal
e e T e e T e e e
thermal
KF
F T F T K T d + =
( )
( ) 2 21
( ) 2 21 ( )
0( ) 2
2
( )
2
{ }{ }
cos cos sin cos cos sin cos
sin sin cos sin sin cos sin
cos cos sin cos
sin
eethermal
e e e e e e e ex
e e e e e e e ey
e e e ee e e e
x
ee
y
FF
F
F
A E kF
F
+ =
( ) ( )
( )
1
( )
1
2 ( )
2
2 2 ( )
2
[ ]
cos sin cos
sin cos sin sin cos sin
e e
e
x
e
y
e e e e e
x
e e e e e e e
y
K d
u
u
u
u
( )
cos sin 0 0
sin cos 0 0[ ]
0 0 cos sin
0 0 sin cos
e e
e e
e
e e
e e
T
=
Use:
Finally we obtain:
(for 2D trusses)
Element equations with thermal effects
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CCOORRNNEELLLLU N I V E R S I T Y 42MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Element equations with thermal effects
What do you need to do to account for thermal effects in truss analysis?
For each truss element that is heated, simply add to the element force, thefollowing extra term
You will need to define at which truss elements thermal effects take place andfor each of them read the values and
0 0
{ }
cos
sin,
cos
sin
ethermal
e
e
e e e e e e
e
e
F
A E where T
=
e .eT
Principle of minimum potential energy
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CCOORRNNEELLLLU N I V E R S I T Y 43MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Principle of minimum potential energy
An alternative equivalent approach to solving
many structural problems is the principle ofminimum potential energy.
From all the possiblecompatible
displacements of a structure, the one thatminimizes the total potential energyis theexact solution.
Potential energyfor given
displacements=
Strain energyfor these givendisplacements
-
Work done by externalloads on these given
displacements
Principle of minimum potential energy
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CCOORRNNEELLLLU N I V E R S I T Y 44MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Principle of minimum potential energy
Let us see this principle applied to the trussproblems discussed earlier.
{ }
'( ) '( ) '( ) '( )
1 1 2 2
( / )
,
1( )
2
min
e
e e e e
d e
e e e e e e e e
x x x xExternal Work
Elastic strain energy densitywork volume
PE PE U W
PE dV F u F u
=
= +
Assemblyprocess
Principle of minimum potential energy
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CCOORRNNEELLLLU N I V E R S I T Y 45MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Principle of minimum potential energy
Lets apply this principle to one truss
element. We need to minimize with respectto the nodal displacements (localcoordinates) Recall that:'( ) '( )1 2, .
e e
x xu u
'
2 1'
,
e ee e e ex x
e
u uEL
= =
2 '( ) '( ) '( ) '( )
1 1 2 2
'
2 '( ) '( ) '( ) '( )2 1
1 1 2 2
'
1
2
'1( )
2
e
ee
e e e e e e e e
x x x x
e ee e e e e ex x
x x x xe
A dx
PE E dV F u F u
u uE dV F u F u
L
= =
Principle of minimum potential energy
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CCOORRNNEELLLLU N I V E R S I T Y 46MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
Principle of minimum potential energy
' '1 2 1 2
'
2 '( ) '( ) '( ) '( )2 1
1 1 2 2
, ' , '
'1( )
2
min mine e e ex x x x
e ee e e e e e e ex x
x x x xe
u u u u
u uPE E A L F u F u
L
=
Take partial derivatives of wrt :ePE'
1 2, '
e e
x xu u
' '( )
1 2 1
1
0 ( ' ) 0
'
e e ee e e
x x xe e
x
PE E Au u F
u L
= =
' '( )
2 1 2
2
0 ( ' ) 0'
e e ee e e
x x xe e
x
PE E Au u F
u L
= =
'( ) '( )( ) ( )
1 1
'( ) ( ) ( ) '( )
2 2
e ee e
x x
e e e e
x x
F uk k
F k k u
=
These are the same Eqs as those obtained with the direct method!
Principle of minimum potential energy
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CCOORRNNEELLLLU N I V E R S I T Y 47MAE 4700 FE Analysis for Mechanical & Aerospace DesignN. Zabaras (9/1/2011)
In general (not just for mechanics
problems!), the principle of minimumpotential energy takes the following form:
or after assembly:
Note that the mimimization gives us thefamiliar solution:
Principle of minimum potential energy
( ) ( ) ( )
{ } { }
1{ } [ ]{ } { } { }
2
min mine e T e e T
d de e
PE d K d d F
=
{ } { }
1{ } [ ]{ } { } { }
2min min
T T
d d
d K d d F
PE
=
[ ]{ } { }K d F=
Principle of minimum potential energy
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CCOORRNNEELLLLU N I V E R S I T Y 48MAE 4700 FE Analysis for Mechanical & Aerospace Design
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Principle of minimum potential energy
We will not use this method to repeat the trusscalculations.
However, it will be our starting point for computingthe stiffness of beam elements (lecture 4).
The method of minimum potential energy can beapplied to many problems not related to mechanics
however there are many problems where thistechnique is not applicable.
After discussing beam bending problems, we willneed to look for more powerful (`unfortunately alsomore mathematical) methods (weak (Galerkin)formulations).
Revisiting the 2-node truss element
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CCOORRNNEELLLLU N I V E R S I T Y 49MAE 4700 FE Analysis for Mechanical & Aerospace Design
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Revisiting the 2-node truss element
Up to now we used the direct method to express the nodal loads vs.nodal displacements for the 2-node truss element.
Let us linearly interpolate the displacement of any point xin theelement in terms of the nodal displacements:
( ) ( )1 2
'( )
'( ) '( ) '( ) ( ) ( )1
1 2 '( )
2
' ' ' '(1 ) [1 , ] [ ] { }
e e
e
e e e e ex
x x xe e e e e
x basis functionsmatrixN N
Nodaldisplacementselement basis
functions
ux x x xu u u N d
L L L L u
= + = =
The strain in this 2-node element can nowbe computed as follows:
'
'
ee xdu
dx =
' ( ) ( ) ( )'( ) ( ) ( ) ( ) ( ) ( )[ ]{ }
[ ]{ } [ ]{ } [ ]{ }' ' '
e e e ee e e e e e e
xx
du d N d dN u N d d B d dx dx dx= = = =
'( ) '( ) '( )( ) ( )( ) 1 2 11 2
'( )
2
1 1 1 1[ ] [ , ] [ , ] [ , ]
' '
e e ee ee e x x x
e e e e ee
xGradient ofbasis functions
matrix
u u udN dN B
dx dx L L L L Lu
= = = =
This is exactly the sameapproximation we usedbefore (constant strain
element)
'( )e
xu
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Truss analysis with displacement constraints
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Truss analysis with displacement constraints
Up to this point, we imposed essential boundary conditionsin terms of prescribed x- or y- nodal displacements. How
about if the support is inclined as in the figure below:
Here, we dont know the displacements at node 1 but we
know the relation between u1xand u1y. In general we writethese constrains on our nodal degrees of freedom as:Cd=q.
For this problem, the constraint is thatthere is no normal displacement
at the support 1
Truss analysis with displacement constraints
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Truss analysis with displacement constraints
Note that at node 1 we dont have essential boundaryconditions we have a displacement constraint.
To solve this problem we use the principle of minimumpotential energy with the constraint Cd=q:
Here, we enforce the constraint using Lagrange multipliers.
Find d and such that
{ }int
inint
1min { } [ ]{ } { } { } { } ([ ]{ } { })
2
TT T
d Lagrange ConstramultiplierPotential energyenforcingof unconstra edthe constrasystem
L d K d d F C d q= +
Truss analysis with displacement constraints
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Truss analysis with displacement constraints
is the Lagrange multiplier that enforces the constraint itis nothing else but the reaction force at node 1 (normal to
the support!)
Minimization is now performed with respect to both d and .
Find d and such that
0
T d FK C
qC
=
{ }
int
inint
1min { } [ ]{ } { } { } { } ([ ]{ } { })
2
TT T
d
Lagrange ConstramultiplierPotential energyenforcingof unconstra edthe constrasystem
L d K d d F C d q= +
Apply essential boundary conditionsand then solve for {dF} and
(here, reaction force at node 1)
Displacement constraints: Implementation
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Displacement constraints: Implementation
How do you implement this in the MatLab
libraries of HW1?
Introduce the constraints in the InputData.m and thenmodify the stiffness and load vectors in the NodalSoln.m.
Apply the essential boundary conditions first before youaugment the reduced global equations (Kf) with theLagrange multiplier.
% Read information for constraints
C = zeros(1,neq-length(debc)); %The dimension of C is neq minus the% prescribed DOF via essential BCs% Here there is only one constraint
C(1) = sin(pi/6); C(2) = cos(pi/6);q = 0;
InputData.m
1 1
sin 30 cos 30 0x y
u u+ =
Displacement constraints: NodalSoln.m
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MAE 4700 FE Analysis for Mechanical & Aerospace Design
Displacement constraints: NodalSoln.mfunction [d, rf, lambda] = NodalSoln(K, R, debc, ebcVals, C, q)
% K=global stiffness, R=global force, debc=degrees of freedom with specified values, ebcVals=specified displacements
dof = length(R); % Extract the total degreess-of-freedom
df = setdiff(1:dof, debc); % Sets the difference between two sets of indices, i.e. the global degrees of freedom minus the% degrees of freedom with prescribed essential boundary conditions
Kf = K(df, df); % Remove eqs. corresponding to prescribed displacements
Rf = R(df) - K(df, debc)*ebcVals; % Modify the remaining load vector to account for the essential boundary conditions
[m n] = size(C); % Extract number of constraints
Kf = [Kf C'; % Augment global equations with the Lagrange multipliersC zeros(m)];
Rf = [Rf;q]; % Augment load vector
dfVals = Kf\Rf; % Solve the linear system of equations. Here for simplicity, we use Gauss elimination.
d = zeros(dof,1); % Restore the solution vector (i.e. include back the nodes with prescribed displacements).d(debc) = ebcVals; % Use the originally established ordering of the nodes.d(df) = dfVals(1:(length(dfVals)-m));
rf = K(debc,:)*d - R(debc); % Calculate the reaction vector at nodes with prescribed displacements
lambda = dfVals((length(dfVals)-m+1):length(dfVals)); % Calculate Langrange multipliers (reactions at nodes with constraints)
0
T d FK C
qC =