# Exercicio de metodos dos elementos finitos

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Exercícios
Determinar a função de forma para os elementos
1)
1 2 3 4
2)
1 2 3 4 5
a) Pela condição
Nixi=1Nixj=1
b) Usando o polinômio de Lagrange
Resolução:
1.a)

ξ1=-1 ξ2=-13 ξ3=13 ξ4=1
Ni=ai+bi∙ξ+ci∙ξ2+di∙ξ3
Para o nó 1:
N1ξ1=a1+b1∙ξ1+c1∙ξ12+d1∙ξ13=1N1ξ2=a1+b1∙ξ2+c1∙ξ22+d1∙ξ23=0
N1ξ3=a1+b1∙ξ3+c1∙ξ32+d1∙ξ33=0
N1ξ4=a1+b1∙ξ4+c1∙ξ42+d1∙ξ43=0

Para o nó 2:
N2ξ1=a2+b2∙ξ1+c2∙ξ12+d2∙ξ13=0
N2ξ2=a2+b1∙ξ2+c2∙ξ22+d2∙ξ23=1
N2ξ3=a2+b2∙ξ3+c2∙ξ32+d2∙ξ33=0
N2ξ4=a2+b2∙ξ4+c2∙ξ42+d2∙ξ43=0

Para o nó 3:
N3ξ1=a3+b3∙ξ1+c3∙ξ12+d3∙ξ13=0
N3ξ2=a3+b3∙ξ2+c3∙ξ22+d3∙ξ23=0
N3ξ3=a3+b3∙ξ3+c3∙ξ32+d3∙ξ33=1
N3ξ4=a3+b3∙ξ4+c3∙ξ42+d3∙ξ43=0Para o nó 4:
N4ξ1=a4+b4∙ξ1+c4∙ξ12+d4∙ξ13=0
N4ξ2=a4+b4∙ξ2+c4∙ξ22+d4∙ξ23=0
N4ξ3=a4+b4∙ξ3+c4∙ξ32+d4∙ξ33=0
N4ξ4=a4+b4∙ξ4+c4∙ξ42+d4∙ξ43=1

N1=-116+116∙ξ-916∙ξ2+916∙ξ3=116∙-1+ξ+9∙ξ2-9∙ξ3
N2=916-2716∙ξ-916∙ξ2+2716∙ξ3=916∙1-3∙ξ-ξ2+3∙ξ3
N3=916+2716∙ξ-916∙ξ2-2716∙ξ3=916∙1+3∙ξ-ξ2-3∙ξ3
N4=-116-116∙ξ+916∙ξ2+916∙ξ3=116∙-1-ξ+9∙ξ2+9∙ξ3

1.b)
Niξe=j=i,j≠1Nξ-ξjξi-ξj
N1ξ1=ξ-ξ2ξ1-ξ2∙ξ-ξ3ξ1-ξ3∙ξ-ξ4ξ1-ξ4
N1ξ1=ξ+13-1+13∙ξ-13-1-13∙ξ-1-1-1=116∙-1+ξ+9∙ξ2-9∙ξ3
N2ξ2=ξ-ξ1ξ2-ξ1∙ξ-ξ3ξ2-ξ3∙ξ-ξ4ξ2-ξ4
N2ξ2=ξ+1-13+1∙ξ-13-13-13∙ξ-1-13-1=916∙1-3∙ξ-ξ2+3∙ξ3
N3ξ3=ξ-ξ1ξ3-ξ1∙ξ-ξ2ξ3-ξ2∙ξ-ξ4ξ3-ξ4
N3ξ3=ξ+113+1∙ξ+1313+13∙ξ-113-1=916∙1+3∙ξ-ξ2-3∙ξ3
N4ξ4=ξ-ξ1ξ4-ξ1∙ξ-ξ2ξ4-ξ2∙ξ-ξ3ξ4-ξ3
N4ξ4=ξ+11+1∙ξ+131+13∙ξ-131-13=116∙-1-ξ+9∙ξ2+9∙ξ3

2.a)

ξ1=-1 ξ2=-12 ξ3=0ξ4=12 ξ5=1
Ni=ai+bi∙ξ+ci∙ξ2+di∙ξ3+ei∙ξ4
Para o nó 1:
N1ξ1=a1+b1∙ξ1+c1∙ξ12+d1∙ξ13+e1∙ξ14=1
N1ξ2=a1+b1∙ξ2+c1∙ξ22+d1∙ξ23+e1∙ξ24=0
N1ξ3=a1+b1∙ξ3+c1∙ξ32+d1∙ξ33+e1∙ξ34=0
N1ξ4=a1+b1∙ξ4+c1∙ξ42+d1∙ξ43+e1∙ξ44=0
N1ξ5=a1+b1∙ξ5+c1∙ξ52+d1∙ξ53+e1∙ξ54=0
1 -1 1 -1 1 1 -1214-18116 1 0 0 0 0 1 12 14 18 116 1 1 1 1 1∙a1b1c1d1e1=10000 a1b1c1d1e1=01/6-1/6-1/31/3
Para o nó 2:
N2ξ1=a2+b2∙ξ1+c2∙ξ12+d2∙ξ13+e2∙ξ14=0
N2ξ2=a2+b1∙ξ2+c2∙ξ22+d2∙ξ23+e2∙ξ24=1
N2ξ3=a2+b2∙ξ3+c2∙ξ32+d2∙ξ33+e2∙ξ34=0
N2ξ4=a2+b2∙ξ4+c2∙ξ42+d2∙ξ43+e2∙ξ44=0
N2ξ5=a2+b2∙ξ5+c2∙ξ52+d2∙ξ53+e2∙ξ54=0
1 -1 1 -1 1 1 -1214-18116 1 0 0 0 0 1 12 14 18 116 1 1 1 11 ∙a1b1c1d1e1=01000 a1b1c1d1e1=0-4/38/34/3-8/3
Para o nó 3:
N3ξ1=a3+b3∙ξ1+c3∙ξ12+d3∙ξ13+e3∙ξ14=0
N3ξ2=a3+b3∙ξ2+c3∙ξ22+d3∙ξ23+e3∙ξ24=0
N3ξ3=a3+b3∙ξ3+c3∙ξ32+d3∙ξ33+e3∙ξ34=1
N3ξ4=a3+b3∙ξ4+c3∙ξ42+d3∙ξ43+e3∙ξ44=0
N3ξ5=a3+b3∙ξ5+c3∙ξ52+d3∙ξ53+e3∙ξ54=0
1 -1 1 -1 1 1 -1214-18116 1 0 0 0 0 1 12 14 18 116 1 11 1 1 ∙a1b1c1d1e1=00100 a1b1c1d1e1=10-504
Para o nó 4:
N4ξ1=a4+b4∙ξ1+c4∙ξ12+d4∙ξ13+e5∙ξ14=0
N4ξ2=a4+b4∙ξ2+c4∙ξ22+d4∙ξ23+e5∙ξ24=0
N4ξ3=a4+b4∙ξ3+c4∙ξ32+d4∙ξ33+e5∙ξ34=0
N4ξ4=a4+b4∙ξ4+c4∙ξ42+d4∙ξ43+e5∙ξ44=1
N4ξ5=a4+b4∙ξ5+c4∙ξ52+d4∙ξ53+e5∙ξ54=0
1 -1 1 -1 1 1 -1214-18116 1 0 0 0 0 1 12 14 18 116 11 1 1 1 ∙a1b1c1d1e1=00010 a1b1c1d1e1=04/38/3-4/3-8/3
Para o nó 5:
N5ξ1=a5+b5∙ξ1+c5∙ξ12+d5∙ξ13+e5∙ξ14=0
N5ξ2=a5+b5∙ξ2+c5∙ξ22+d5∙ξ23+e5∙ξ24=0
N5ξ3=a5+b5∙ξ3+c5∙ξ32+d5∙ξ33+e5∙ξ34=0
N5ξ4=a5+b5∙ξ4+c5∙ξ42+d5∙ξ43+e5∙ξ44=0
N5ξ5=a5+b5∙ξ5+c5∙ξ52+d5∙ξ53+e5∙ξ54=1
1 -1 1 -1 1 1 -1214-18116 1 0 0 0 0 1 12 1418 116 1 1 1 1 1 ∙a1b1c1d1e1=00001 a1b1c1d1e1=0-1/6-1/62/32/3
N1=0+16∙ξ-16∙ξ2-13∙ξ3+13∙ξ4=16∙ξ-ξ2-2∙ξ3+2∙ξ4
N2=0-43∙ξ+83∙ξ2+43∙ξ3-83∙ξ4=43∙-ξ+2∙ξ2+ξ3-2∙ξ4
N3=1+0∙ξ-5∙ξ2+0∙ξ3+4∙ξ4=1-5∙ξ2+4∙ξ4
N4=0+43∙ξ+83∙ξ2-43∙ξ3-83∙ξ4=43∙ξ+2∙ξ2-ξ3-2∙ξ4
N5=0-16∙ξ-16∙ξ2+23∙ξ3+23∙ξ4=16∙-ξ-ξ2+4∙ξ3+4∙ξ4

2.b)
Niξe=j=i, j≠1Nξ-ξjξi-ξj...