Lecture 12
FEM in 2D: Bilinear and Cubic Shape Functions
(Lecture notes taken by Sriram Pakkam and Kishore Krishna B. S.)
· Bilinear Shape Functions:
uex,y=α1+α2x+α3y+α4xy
Let the element be rectangular. e rectangular.
η 3
y 4 2 ξ
b 1 a
x
ueξ,η= ∝1+∝2ξ+∝3η+∝4ξη
For point 1, ξ=0 and η=0
For point 2, ξ=a and η=0
For point 3, ξ=a and η=b
For point 4, ξ=0 and η=b
u1eu2eu3eu4e=101a00001a10babb0∝1∝2∝3∝4
A matrix
ue=A∝
A-1=B=10-1a1a0000-1b01ab-1ab01b1ab-1ab
ue=N1ξ,ηu1e+N2ξ,ηu2e+N3ξ,ηu3e+N4ξ,ηu4e
Where for instance,
1=N10,0
0=N1a,0
0=N1a,b
0=N10,b
N1ξ,η=1+-1a ξ+-1b η+1abξ η
=1-ξa1-ηb
=L1(ξ) L1(η)
L2(η) L2(η)
Hence it is bilinear. The functions here are analogous to the 1-D shape functions.
· Element matrices:
ke=a(Nj,Ni)
4 X 4
As in the quadratic shape function case, one can represent this element matrix as
ke= BTGB.
G is derived from-uxx-uyy=f
G=000ab000ab22000ab22aba2b2a2b2a3b+ab33
The following formula is used in the computation of the above components
hm,n=ξmηn
=am+1m+1*bn+1n+1
· Cubic Shape Functions:
Let the elements be triangles and use ten terms in 10 terms
ueξ,η=∝iξmiηni
mi+ni≤3
mi=0,1,2,3
η
3 2 ξ
uc 10x1 10x10 10x1
1 centroid u1u1ξu1ηu2u2ξu2ηu3u3ξu3ηuc=ue=A∝
where for point 1 we haveu1u1ξu1η for point 2, u2u2ξu2η and point 3,u3u3ξu3η
· Element matrix ke is of dimension 10X10:
Let B=A-1 and G is obtained from the differential equation
ke=RTBTGBR
R is a result of the partial derivatives and the chain rule
R=CZZZCZZrZrCr ccc1
C=0000cosθsinθ0-sinθcosθ c=000
Z=000000000 r=000
· Gauss Quadrature Rules:
1-D case on [-1,1]:
-11fxdx≅inωif(xi)
Choose the weights ωi and the points xi so that the error is zero, that is,
-11P2n-1(x)=ωif(xi)
Example 1: Use n = 2 and require the errors to be zero for
1,x,x2,x3
-111= ω1*1+ω2*1
2= ω1+ω2
-11x= ω1x1+ω2x2
0= ω1x1+ω2x2
-11x2= ω1x12+ω2x22
23= ω1x12+ω2x22
-11x3= ω1x13+ω2x23
0=ω1x13+ω2x23
Solving the above equations gives
ω1=ω2=1 and x1=-13 , x2=13.
Example 2: Use n = 3
xi / ωi-0.77454 / 0.5555
0 / 0.88885
0.77454 / 0.5555
Gauss Quadrature of Triangle:
P=1,s,t,s2,st,t2
By requiring the errors to be zeros for the above six polynomials and solving weights and (s,t)
s1,t1=16,16
s2,t2=23,16
s3,t3=16,23
ω1=ω2=ω3=16
See chapter seven in Gockenbach’s book for additional examples of quadrature rules.