elemt fininte
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Collect HW HW: 5.1,9 by hand and 2 meshes w/NENastran, # 15 (use 2 elements by hand, 4, &
8 elements w/CST.xls) Due April 14, 2005.
What if we have a 2-d problem with surface tractions, point loads, acting in the plane of a relatively thin structure?
With
We can take the preceding problem and Discretize it:
How can we model the problem better?
Constant Strain Triangle Element 2 dof per node (x,y), 3 nodes per element (6 x 6 stiffness matrix)
Shape Functions
Shape functions are just ratios of areas:
Representing Displacements:
or
Finding the Stiffness Matrix, and assembling the Force Vector:
the element stiffness matrix:
The point loads are given by:
The surface Tractions are given by:
Kreduce
2.435 106
×
0
1.258− 106
×
7.102− 105
×
0
7.162 106
×
6.684− 105
×
3.7− 106
×
1.258− 106
×
6.684− 105
×
1.259 106
×
7.123 105
×
7.102− 105
×
3.7− 106
×
7.123 105
×
2.119 106
×
=Kreduce
K2 2,K3 2,K4 2,K5 2,
K2 3,K3 3,K4 3,K5 3,
K2 4,K3 4,K4 4,K5 4,
K2 5,K3 5,K4 5,K5 5,
:=
Looking at the figure, we see that node 1, in element 1, and node 1 and 4, in element 2 are restrained. If we use the elimination method, the reduced stiffness matrix for element 1 is:
Note that the initial row and column start at 0 rather than 1, so K00 is the first stiffness value, rather than K11
K
1.178 106
×
6.664− 105
×
1.177− 106
×
6.684 105
×
1.174− 103
×
2.089− 103
×
6.664− 105
×
1.88 106
×
7.102 105
×
3.462− 106
×
4.387− 104
×
1.581 106
×
1.177− 106
×
7.102 105
×
2.435 106
×
0
1.258− 106
×
7.102− 105
×
6.684 105
×
3.462− 106
×
0
7.162 106
×
6.684− 105
×
3.7− 106
×
1.174− 103
×
4.387− 104
×
1.258− 106
×
6.684− 105
×
1.259 106
×
7.123 105
×
2.089− 103
×
1.581 106
×
7.102− 105
×
3.7− 106
×
7.123 105
×
2.119 106
×
=
KE t⋅
4 A1⋅ 1 v2
−
⋅
y23
0
y31
0
y12
0
0
x32
0
x13
0
x21
x32
y23
x13
y31
x21
y12
⋅
1v
0
v
1
0
001 v−
2
y23
0x32
0x32
y23
y31
0x13
0x13
y31
y12
0x21
0x21
y12
⋅
⋅:=
y21 y12−:=y12 1.75:=y32 y23−:=y23 1.75−:=y31 y13−:=y13 0:=
x31 x13−:=x21 x12−:=x12 3.1−:=x13 6−:=x32 x23−:=x23 2.9−:=A1 5.25:=v .32:=t .125:=E 30000000:=
Example: 2 Constant strain triangular elements, Steel plate
K2
1.215 106
×
6.874− 105
×
1.206− 106
×
7.102 105
×
2.532− 104
×
1.292− 104
×
6.874− 105
×
2.008 106
×
6.684 105
×
4.101− 105
×
2.828 104
×
1.604− 106
×
1.206− 106
×
6.684 105
×
2.419 106
×
0
1.179− 106
×
6.684− 105
×
7.102 105
×
4.101− 105
×
0
8.224 105
×
7.102− 105
×
4.009− 105
×
2.532− 104
×
2.828 104
×
1.179− 106
×
7.102− 105
×
1.188 106
×
6.72 105
×
1.292− 104
×
1.604− 106
×
6.684− 105
×
4.009− 105
×
6.72 105
×
1.999 106
×
=
Combining Elements 1 and 2, and reducing:Defining the Force Vector:
Kall Kreduce
K22 2,
K23 2,00
K22 3,
K23 3,00
0
0
00
0
0
00
+:= Kall
4.854 106
×
0
1.258− 106
×
7.102− 105
×
0
7.984 106
×
6.684− 105
×
3.7− 106
×
1.258− 106
×
6.684− 105
×
1.259 106
×
7.123 105
×
7.102− 105
×
3.7− 106
×
7.123 105
×
2.119 106
×
= Uall
u2
v2
u3
v3
:=
u2
Fall
00025−
:= Uall Kall1−Fall⋅:= Uall
8.319− 106−
×
4.502− 105−
×
2.533 105−
×
1.017− 104−
×
=
What if we assume that there is only a body force from the weight of each element, 0.5 lbs per cubic inch? What if we assume that a body force from the weight of each
element, 0.5 lbs per cubic inch, is added to the 25 lbs?
Fall
0000
0
0.5 A1 t⋅
3A2 t⋅
3+
⋅
0
0.5 A1 t⋅
3
⋅
+:= Uall Kall1−Fall⋅:= Uall
7.515 108−
×
4.066 107−
×
1.904− 107−
×
8.508 107−
×
=Fall
00025−
0
0.5 A1 t⋅
3A2 t⋅
3+
⋅
0
0.5 A1 t⋅
3
⋅
+:= Uall Kall1−Fall⋅:= Uall
8.244− 106−
×
4.462− 105−
×
2.514 105−
×
1.009− 104−
×
=
Looking at Element 2 now:
E 30000000:= t .125:= v .32:= A2 5.565:= x23 3.1:= x32 x23−:= x13 0:= x12 3.1−:= x21 x12−:= x31 x13−:=
y13 3.59:= y31 y13−:= y23 1.79:= y32 y23−:= y12 1.75:= y21 y12−:=
K2E t⋅
4 A2⋅ 1 v2
−
⋅
y23
0
y31
0
y12
0
0
x32
0
x13
0
x21
x32
y23
x13
y31
x21
y12
⋅
1v
0
v
1
0
001 v−
2
y23
0x32
0x32
y23
y31
0x13
0x13
y31
y12
0x21
0x21
y12
⋅
⋅:=
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