This document discusses axi-symmetric analysis and finite element modeling techniques for single-variable problems. It introduces axi-symmetric coordinates where quantities depend only on radial (r) and axial (z) coordinates, reducing a 3D problem to 2D. It presents the weak form, finite element model, and element stiffness matrix formulation for a single-variable problem. It also discusses constant strain triangle (CST) and isoparametric elements, where shape functions allow modeling of arbitrary geometries.

1. 134
UNIT– 5
DynamicAnalysis

2. 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

3. Axi-symmetricAnalysis
136

4. 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

5. 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


6. 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

7. 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 

8. 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
 

9. 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


10. 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

11. 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

12. 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.

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

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

15. 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

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

17. Bilinear Quadratic
• Shapefunctions and strain-displacement
matrix
150

18. 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

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

20. Quadratic Quadrilateral Element
• Shapefunction examples:
• Strain distribution within theelement
153

21. 154
Quadratic Quadrilateral Element
• Should we try to use this element to solve our
problem?
• Or try fixing the Q4element for our purposes.
– Hmm…toughchoice.

22. 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

23. 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

24. 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

25. 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

26. • 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

27. 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

28. 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

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

30. 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 

31. Numerical Integration
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

32. Numerical Integration
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

33. Numerical Integration
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