UNIT I_5.pdf

Axi-symmetricAnalysis
Cylindrical coordinates:
x  rcos; y  r sin; z  z
r, , z
• quantities depend on rand zonly
• 3-D problem 2-D problem
135

Axi-symmetric Analysis – Single-VariableProblem
00
22
11 


 
 z
u(r, z)   a u  f (r, z)  0
z

r
u(r, z)  
 a

1   ra
r r

Weakform:
e
wqn ds
e


  
 z

z



 r
u 
wa
 r

wa u  a wu  wf (r,z)rdrdz
0  00
22
11
where nz
137
z
r
u(r, z)
u(r, z)
nr  a22
qn  a11

FiniteElement Model – Single-VariableProblem
u  u jj
j
Ritzmethod:
e
n
e e
K u  f Qe
 ij j i i
j1
where j (r, z) j (x, y)
w i
Weakform
22
138
e
j j

a
i i
K e
  a  a 

rdrdz
00 i j 
r r z z
  
ij   11
where
i i
e
f e
  frdrdz

i i n
e
Qe
  q ds


Single-Variable Problem – HeatTransfer
Weakform
Heat Transfer:

1  rk
T(r, z)  
 k
T(r, z)   f (r, z)  0
 
z  z
r r  r
   
e


0 
wk
T  
wk
T   wf (r, z) rdrdz
   
 
 r  r  z  z 
wqn ds
e
nz
139
T(r, z) T(r, z)
qn  k nr  k
r z
where

3-Node Axi-symmetricElement
T (r, z)  T11 T22 T33
1
2
3

140

1 2 3
e
1 r z
2A
r2z3  r3z2
 
 
  z  z
 
 
r r
 3 2 
  3 1 1 3
2 3 1
e
1 r z
2A
r z  r z
 
 z 
   z
 
 
r  r
 1 3 
  1 2 2 1
3 1 2
e
1 r z
2A
r z  r z
 
 
  z z
 
 
r r
 2 1 

4-Node Axi-symmetricElement
T (r, z)  T11 T22  T33  T44
1 2
3
4
a
b


r
141
z
2
3
a b
  1
 1
   
 1

1     
 a  b  a  b 
 
 
  1
 
4  a  b
 

Single-Variable Problem –Example
Step1: Discretization
Step2: Element equation
e
142
j
i
ij 

 
K e
  i
j 
rdrdz
r r z z 
  

i i
e
f e
  frdrdz
 i i n
e
Qe
  q ds


Reviewof CSTElement
• Constant Strain Triangle (CST)- easiest and simplest
finite element
– Displacement field in terms ofgeneralized coordinates
– Resulting strain field is
– Strains do not vary within the element. Hence, the name
constant strain triangle(CST)
• Other elements are not solucky.
• Canalso be called linear triangle becausedisplacement field is
linear in xand y - sidesremainstraight.
143

Constant Strain Triangle
• Thestrain field from the shapefunctions lookslike:
– Where, xi and yi are nodal coordinates (i=1, 2,3)
– xij =xi - xj and yij=yi - yj
– 2Ais twice the area of the triangle, 2A=x21y31-x31y21
• Node numbering is arbitrary except that the sequence123
must go clockwise around the element if Ais tobe positive.
144

145
Constant Strain Triangle
• Stiffness matrix for element k =BTEBtA
• TheCSTgivesgood results in regions of theFE
model where there is little straingradient
– Otherwise it does notwork well.

Linear Strain Triangle
• Changesthe shape functions and results in
quadratic displacement distributions and
linear strain distributions within theelement.
146

Linear Strain Triangle
• Will this element work better for the problem?
147

ExampleProblem
• Consider the problem we were lookingat:
5in.
1in.
0.1in.
I  0.113
/12  0.008333 in4
 
M  c

1 0.5
I 0.008333
 60 ksi
E
 
  0.00207
2
 
ML

25
2EI 2  29000  0.008333
 0.0517in.
148
1k
1k

Bilinear Quadratic
• TheQ4element is aquadrilateral element
that hasfour nodes. In terms ofgeneralized
coordinates, its displacement field is:
149

Bilinear Quadratic
• Shapefunctions and strain-displacement
matrix
150

ExampleProblem
• Consider the problem we were lookingat:
5in.
1in.
0.1in.
E
151
I 0.008333
 0.0345in.
3EI 3290000.008333
0.2125

 60ksi
10.5
I  0.1 13
/ 12 0.008333in4
 
PL3
 
 0.00207
 
M  c

0.1k
0.1k

Quadratic Quadrilateral Element
• The8 noded quadratic quadrilateral element
usesquadratic functions for thedisplacements
152

• Shapefunction examples:
• Strain distribution within theelement
153

154
• Should we try to use this element to solve our
problem?
• Or try fixing the Q4element for our purposes.
– Hmm…toughchoice.

155
Isoparametric Elements and Solution
• Biggestbreakthrough in the implementation of the
finite element method is the development of an
isoparametric element with capabilities to model
structure (problem) geometries of any shapeandsize.
• Thewhole idea works onmapping.
– Theelement in the real structure is mapped to an
‘imaginary’ element in an ideal coordinatesystem
– Thesolution to the stress analysis problem is easyand
known for the ‘imaginary’element
– Thesesolutions are mapped back to the element inthe
real structure.
– All the loads and boundary conditions are also mapped
from the real to the ‘imaginary’ element in this approach

Isoparametric Element
3
1
(x1,y1)
2
(x2,y2)
(x3,y3)
4
(x4,y4)
X,u
Y
,v


2
(1,-1)
1
(-1,-1)
4
(-1, 1)
3 (1, 1)
156

Isoparametric element
Y
1 2 3 4
0 0 0 N1
X N
0
  

• The mappingfunctions areq
x1u
 ite simple:
4

 x 
2
x3 
N N N 0 0 0 0  x 
N2 N3 N4 y1 
y2

 
1
4
N  1 (1)(1)
2
4
N 
1
(1 )(1)
3
4
N  1 (1 )(1 )
4
4
N 
1
(1)(1 )
y3 

y4


Basically, thex and y coordinatesof any point in the
elementare interpolations of the nodal (corner)
coordinates.
Fromthe Q4 element,thebilinear shape functions
are borrowed to be used as the interpolation
functions. They readily satisfy the boundaryvalues
too. 157

Isoparametric element
v
1 2 3 4
0 0 0 N1
u N
0
    

• Nodal shape functions for d
u
i1s
placements
4

u 
2
u3
N N N 0 0 0 0 u 
N2 N3 N4 v1 
v2

 
v3
v4


1
4
N  1 (1)(1)
2
4
N 
1
(1 )(1)
3
4
N  1 (1 )(1 )
4
4
N 
1
(1)(1 ) 158

• Thedisplacement strain relationships:
x

u

u

 
u


X  X  X
y

v

v

 
v


Y  Y  Y
x


 
 
 xy 
y

u
X
v
Y

Y X
   





 u





v 
Y Y

   0



0


0
 


 
X X
0

   
Y Y X X 



 u 
  v 
 
 u 
 
 
 

 
 v 

But,it is too difficult to obtain
 and

X X 159

Isoparametric Element
 X  Y 
u
u





  
X Y u 
X Y u 


X 
  Y 
  
It is easier to obtain
X
and
Y
 
X Y
 
J  X Y Jacobian
 
 
defines coordinate transformation
Hencewe will do itanother way
u

u

X

u

Y
 X  Y 
u

u

X

u

Y
X
 
Ni
X
  i
Y N

 

i
Yi
X N
 
i X i
 
Y N
 
i Yi
 
X
 
u
 
Y 
u 
 
1 
u
   J  
 
u


169

161
GaussQuadrature
• Themapping approach requires usto be able to
evaluate the integrations within the domain (-1…1)of
the functionsshown.
• Integration canbe done analytically by usingclosed-
form formulas from atable ofintegrals (Nah..)
– Or numerical integration can be performed
• Gaussquadrature is the more common form of
numerical integration - better suited for numerical
analysis and finite elementmethod.
• Itevaluated the integral of afunction asasum of a
becomes
n
I  Wii
i1
finite numb1er of terms
I   d
1

GaussQuadrature
• Wi is the ‘weight’ and i is the value of f(=i)
162

Numerical Integration
Calculate:
b
I   f xdx
a
• Newton – Cotesintegration
• Trapezoidalrule – 1storderNewton-Cotesintegration
• Trapezoidalrule – multipleapplication
1
b a
f (x)  f (x)  f (a) 
f (b)  f (a)
(x a)
2
f (a)  f (b)
b b
I   f (x)dx   f1(x)dx  (b  a)
a a
b
163
xn1
xn x1 x2 xn
a x0 x0 x1
I   f (x)dx   fn (x)dx   f (x)dx   f (x)dx    f
(x)dx





2 
n1
i1
i
f (x )  f (b)
f (a)  2
h 
I 

Calculate:
b
I   f xdx
a
• Newton – Cotesintegration
• Simpson1/3 rule – 2ndorderNewton-Cotesintegration
2
2
0 1
1
0 2
0 1
2 f (x )
(x2  x0 )(x2  x1)
(x  x0 )(x  x1)
f (x )  f (x )
(x0  x1)(x0  x2 ) (x  x )(x  x )
(x  x1)(x  x2 ) (x  x )(x  x )
f (x)  f (x) 
b b
a a
6
f (x )  4 f (x )  f (x )
I   f (x)dx   f2 (x)dx  (x2  x0 ) 0 1 2
164

Calculate:
b
I   f xdx
2

(b  a)
f (a) 
(b  a)
f(b)
2 2
165
I  (b  a)
f (a)  f (b)
I  c0 f (x0 )  c1 f (x1)
a
• GaussianQuadrature
Trapezoidal Rule: GaussianQuadrature:
Choose according to certaincriteria
c0,c1,x0 , x1

Calculate:
b
I   f xdx
a
• GaussianQuadrature
   

3
 1 
 1
1
1
I   f xdx  f 
3
  f 
1
• 3pt GaussianQuadrature
I   f xdx 0.55 f 0.77 0.89  f 0 0.55 f0.77
1
1
1
• 2pt GaussianQuadrature
I   f xdx  c0 f x0  c1 f x1  cn1 f xn1
b  a
~
x 1
2(x  a)
Let:
b 1

1
a
166
~ ~
1 1 1
 f (x)dx 
2
(b  a) f 2
(a  b) 
2
(b  a)xdx

UNIT I_5.pdf

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