ME 478 Introduction to Finite Element...
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Polynomial Form • The polynomial representation of the field variable must contain the same number of terms as the number of nodal degrees of freedom. • The polynomial representation for an M‐degree of freedom element should contain all powers of the independent variable up to and including M − 1. If this is the case the polynomial is complete. • In 2D and 3D, polynomial representations of the field variable satisfy the compatibility and completeness requirements if the polynomial exhibits the property known as geometric isotropy. A mathematical function satisfies geometric isotropy if the functional form does not change under a translation or rotation of coordinates.
Transcript of ME 478 Introduction to Finite Element...
Polynomial Form• The polynomial representation of the field variable must contain the same number of terms as the number of nodal degrees of freedom.
• The polynomial representation for an M‐degree of freedom element should contain all powers of the independent variable up to and including M − 1. If this is the case the polynomial is complete.
• In 2D and 3D, polynomial representations of the field variable satisfy the compatibility and completeness requirements if the polynomial exhibits the property known as geometric isotropy. A mathematical function satisfies geometric isotropy if
the functional form does not change under a translation or rotation of coordinates.
An M order 2D polynomial is complete in the following form.
Where (2)N [(M 1)(M 2)] / 2t
Pascal triangle
A complete polynomial of order M must contain all terms shown above the horizontal line.
This is represented by the dashed lines in the figure below.
One can use complete polynomials, or incomplete polynomials if the incomplete polynomial is symmetric. So the polynomial must contain all the terms above the straight lines or above the dashed lines.
In 3D Pascal triangle is the Pascal Pyramid.
In 3D Pascal triangle is the Pascal Pyramid.
Triangular Elements
Inverting the matrix and taking the dot product with the vector gives.
The shape function are.
where A is the area of the triangular element.
The area can be restated as.
Given a coordinate system that is aligned with the triangle can simplify the shape function.
The shape functions become.
While simpler, all of these must be transformedinto an aligned global coordinate system. This is many coordinate rotations.
Complete interpolation function with 1 term
(x , y) = a0 + a1x + a2 y
The field variable can take on a constant value, as required for completeness, and that the first partial derivatives with respect to the independent variables x and y are constants.
The three‐node triangular element is commonly known as a constant strain triangle.
The area coordinates of an arbitrary point P are:
Given:
So a field variable may be expressed as:
Plane Stress
• The body is small in one coordinate direction in comparison to the other dimensions; the dimension in the z direction (the thickness) is either uniform or symmetric about the xy plane;
• Thickness t, if in general, is less than one‐tenth of the smallest dimension in the xy plane, would qualify for “small.”
• The body is subjected to loading only in the xyplane.
• The material of the body is linearly elastic, isotropic, and homogeneous.
Plane Stress cont.
are non‐zero.
are zero.
The equilibrium equations are.
The stress/strain relations are.
or
Isoparametric Representation of the CST
u1
u3
u5
u2u4
u6
1
2
3
1 1 2 3 3 5
1 2 2 4 3 6
1
x
y
NNNu N u N u N uu N u N u N u
Isoparametric Representation of the CST
1
2
3
1 5 3 5 5
2 5 4 6 6
1 2 3
1 2 3
1( ) ( )( ) ( )
or
where0 0 0
N=0 0 0
x
y
NNNU u u u u u uU v u u u u u
U Nu
N N NN N N
1 1 2 2 3 3
1 1 2 2 3 3
1 3 2 3 3
13 23 3
1 3 2 3 3
13 23 3
( ) ( )
( ) ( )
x N x N x N xy N y N y N xx x x x x xx x x xy y y y y yy y y y
Determine the shape functions at point P.
Solve for and .
1 2 3
1 2 3
13 23 3
13 23 3
3.85 1.5 7 44.8 2 3.5 7
3.85 2.5 3 7
4.8 5 3.5 7
N N NN N N
x x x x
y y y y
2.5 3 0.155 3.5 2.2
0.30.2
0.30.20.5
Strains
1 5 3 5 5( ) ( )xU u u u u u u
u u x u yx y
u u x u yx y
u x y uxuu x yy
x y
x y
J
J is the Jacobian.
13 13
23 23
x yx y
J
1
23 131
23 13
Where
1det
u uxuyy
uuxu yy
y yx x
J
J
JJ13 23 23 13det x y x y J
1 det2
A J If points 1, 2 and 3 are ordered in a counter clockwise manner, thendet J is positive.
Determine J.
13 13
23 23
2.5 5.03.0 3.5
det 2.5( 3.5) (3.0)( 5.0) 23.75
x yx y
J
J
23 13
23 13
23 13
23 13
23 131
23 13
1det
1det
Where
1det
u uu y yxu u ux xy
v vv y yxv v vx xy
y yx x
J
J
JJ
Insert v for u.
23 1 5 13 3 5
23 2 6 13 4 6
23 1 5 13 3 5 23 2 6 13 4 6
23 1 31 3 12 5
32 2
( ) ( )1 ( ) ( )
det( ) ( ) ( ) ( )
( ) ( ) ( )1 ( )
det
uu x
vxu yy u v
y x
y u u y u ux u u x u u
x u u x u u y u u y u u
y u y u y ux u
J
J 13 4 21 6
32 1 23 2 13 3 31 4 21 5 12 6
( ) ( )( ) ( ) ( ) ( ) ( ) ( )
x u x ux u y u x u y u x u y u
23 13 12
23 13 21
32 23 13 31 21 12
0 0 01 0 0 0
det
y y yx x x
x y x y x y
ε BU
BJ
Find Be for each element using the local element numbering.=
Total Potential EnergyThe potential energy of the system is given by.
Element Stiffness
12 e
e
t dA T TU B DBU
12e e
e
U t dA T TU B DB U
12e eU t A T TU B DBU
12
eeU TU k U
Global Stiffness Matrix
12
12
ee
U
T
T
U k U
U KU
The total stiffness is the summation of each elements stiffness.
Force Terms
Traction Force
1 1 2 3
1 2 2 4
x
y
U N u N uU N u N u
1 2
1 2 3 4, , , e
l
tdl u u u u
UT T
Force Example
2 1 2i i i x i yu P U P U P
The potential Energy is:
The work can be expressed as:
The force vector is:
Minimization of the potential energy.
Gives us:
Or
Where
The shape function
=y23/2A
=y12/2A
=y31/2A
=x32/2A
=x13/2A
=x21/2A
23 31 12
32 13 21
32 13 21 23 31 12
0 0 01 0 0 0
2
y y yx x x
Ax x x y y y
23 32
31 3223 31 12
12 2132 13 212
32 2332 13 21 23 31 12
13 31
21 12
00
1 0 0 0 00
1 0 0 0 004 (1 )
10 002
0
y xy x
y y yy xEtk x x x
x yAx x x y y y
x yx y
2 223 32 23 31 32 13 23 12 32 21 23 32 23 13 31 32 23 21 12 32
2 231 13 31 12 13 21 31 32 23 13 31 13 31 21 12 13
2 2212 21 12 32 23 21 12 13 31 21 1
12
12
14 (1 ) 2
y Cx y y Cx x y y Cx x y x y x Cy x y x Cy x
y Cx y y Cx x y x Cy x y x y x Cy xEtk y Cx y x Cy x y x Cy x yA
2 21
2 232 32 32 13 23 31 32 21 23 12
2 213 31 13 21 31 12
2 221 12
x
x Cy x x Cy y x x Cy ySYM x Cy x x Cy x
x Cy
where C = (1 − )/2.
Work Equivalent loads for pressure p in direction n.
with nx and ny corresponding to the components of the unit outward normal vector to edge 2‐3. Here, we use the notation p for such loadings, as the units are those of pressure.
The work being done the becomes.
Introducing the shape function definitions for u and v gives.
So the total equivalent work becomes:
with
Body Forces
In matrix format:
Gravity is in the negative y direction and is the only body force specified.
Total FE Equation
Nodal force Traction force Body force
ke =
Global Stiffness Matrix
1
2
3
UUU
1
2
4 5 6 7, , ,U U U U8U
Example
20 10 100
21
etAxA
t
Tk B DB
in2
in
Element k
Element 2
For element 1.
For element 2.
Limitations to CST
• In bending problems, the mesh of CST elements will produce a model that is stiffer than the actual problem.
• A CST modeled beam subjected to pure bending will result in afalse shear stress.
• This spurious shear strain absorbs energy so some of the energy that should go into bending is lost.
• The CST is then too stiff in bending, and the resulting deformation is smaller than actually should be.
