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1. Lecture 17:Axisymmetric Problems
APL705 Finite Element Method
v18

2. Axisymmetric Solids
• The solids with an axis of symmetry or solids of revoluDon
with axisymmetric loading (and support) are modeled as
simple 2D elements. As seen from the figures below all
stresses are independent of rotaDonal angle. Eg. Flywheel,
bearing, thick-walled pressure vessels

3. CondiDon for Axisymmetric Problems
1. The problem domain must have an axis of symmetry. It is
customary to align this symmetry axis with the z-axis of the
cylindrical (r,θ,z) coordinate system.
2. The boundary condiDons are symmetric about the axis.
Therefore all BCs are independent of θ
3. All loading condiDons are symmetric about z
hence these are also independent of
circumferenDal direcDon θ
4. Also, the material properDes must be
homogeneous or symmetric. This condiDon
always saDsfied for isotropic materials
5. Lastly, the soluDons are independent of θ

4. Axisymmetric as Special 2D Problems
• Where there is symmetry in geometry and loads we can
model such problems as special two-dimensional problems.
•

5. Axisymmetric FormulaDon
• Consider the elemental volume shown here. Now we write
the potenDal energy in axisymmetric
form
• Here we note that all integrals
are independent of θ
• where
π = 1
2 σ T
ε
A
∫ rdAdθ
0
2π
∫ − uT
f
A
∫ rdAdθ
0
2π
∫ − uT
Tr
L
∫ dl dθ
0
2π
∫ − ui
T
i
∑ Pi
π = 2π 1
2 σ T
ε
A
∫ rdA − uT
f
A
∫ rdA − uT
Tr
L
∫ dl
⎛
⎝
⎜
⎞
⎠
⎟ − ui
T
i
∑ Pi
u =[u,w]T
; f =[ fr, fz ]T
; T =[Tr,Tz ]T

6. Stress Strain RelaDons
• We can write strains for this case as
• Using derivaDves of displacements this can be wrien as
• The stress vector for this case is
• Now the stress-strain relaDon is given as where D is
ε =[εr,εz,γrz,εθ ]
ε =
∂u
∂r
,
∂w
∂z
,
∂u
∂z
+
∂w
∂r
,
u
r
⎡
⎣⎢
⎤
⎦⎥
σ =[σr,σz,τrz,σθ ]
σ = Dε
D =
E(1−ν)
(1+ν)(1− 2ν)
1 k 0 k
k 1 0 k
0 0 s 0
k k 0 1
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
;k =
ν
1−ν
;s =
1− 2ν
2(1−ν)

7. Axisymmetric Galerkin FormulaDon
• The Galerkin formulaDon for an axisymmetric modeling can
be given as
• Here the strain vector is given as
2π σ T
ε(φ)
A
∫ rdA −(2π φT
f
A
∫ rdA + 2π φT
Tr
L
∫ dl + φi
T
i
∑ Pi ) = 0
φ =[φr,φz ]T
ε(φ) =
∂φr
∂r
,
∂φz
∂z
,
∂φr
∂z
+
∂φz
∂r
,
φr
r
⎡
⎣
⎢
⎤
⎦
⎥
T

8. Axisymmetric Modeling using Triangular
Elements
• The area of axisymmetry is now divided into triangular finite
elements. By revolving these elements we actually get the
discreDzed solid object. Therefore though each element is a
triangle, it actually represents a solid ring of triangular secDon
about the axis of symmetry.
• Following the discussion on CST and replacing x, y coordinates
with r,z we have a similar formulaDon now. Using 3 shape
funcDon N1, N2 and N3 we write {u}=[N]{q} where
N =
N1 0 N2 0 N3 0
0 N1 0 N2 0 N3
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
q =[q1
,q2,q3,q4,q5,q6 ]T

9. Axisymmetric - Isoparametric FormulaDon
• We can write the displacements in r and z direcDons as
• The same funcDons are used to interpolate the r, z
coordinates also in the isoparametric formulaDon
u=(q1 −q5)ξ +(q3 −q5)η+q5
v =(q2 −q6 )ξ +(q4 −q6 )η+q6
r =(r1 −r3)ξ +(r2 −r3)η+r3
z =(z1 −z3)ξ +(z2 −z3)η+z3

10. CalculaDng Strains
• The strain relaDon is converted to a matrix form
• Here the (2x2) matrix is the Jacobian of transformaDon, J
• On taking the derivaDves we have
∂u
∂ξ
∂u
∂η
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⎫
⎬
⎪
⎪
⎭
⎪
⎪
=
∂r
∂ξ
∂z
∂ξ
∂r
∂η
∂z
∂η
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
∂u
∂r
∂u
∂z
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⎫
⎬
⎪
⎪
⎭
⎪
⎪
J =
∂r
∂ξ
∂z
∂ξ
∂r
∂η
∂z
∂η
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
J =
r13 z13
r23 z23
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥

11. Element Strains
• Using the definiDon of J we have
• The inverse transformaDon of the above is
• where J-1 is the inverse of J given by
∂u
∂r
∂u
∂z
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
= J−1
∂u
∂ξ
∂u
∂η
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
∂u
∂ξ
∂u
∂η
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⎫
⎬
⎪
⎪
⎭
⎪
⎪
= J
∂u
∂r
∂u
∂z
⎧
⎨
⎪
⎪
⎩
⎪
⎪
⎫
⎬
⎪
⎪
⎭
⎪
⎪
∂w
∂r
∂w
∂z
⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
= J−1
∂w
∂ξ
∂w
∂η
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
J−1
=
1
detJ
r13 −z13
−r23 z23
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥

12. Strain Displacement RelaDon
• From strain-displacement relaDon
• This will lead to {ε}=[B]{q}
ε =
∂u
∂r
∂v
∂z
∂u
∂z
+
∂v
∂r
u
r
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
=
z23(q1 −q5)−z13(q3 −q5)( )/detJ
−r23(q2 −q6 )−r13(q4 −q6 )( )/detJ
−r23(q1 −q5)−r13(q3 −q5)+z23(q2 −q6 )−z13(q4 −q6 )( )/detJ
N1q1 + N2q3 + N1q5( )/r
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥

13. Strain Displacement RelaDon
• The strain-displacement relaDon is
• Where B is the element strain-displacement matrix given as
ε = Bq
B =
z23
detJ
0 z31
detJ
0 z12
detJ
0
0 r32
detJ
0 r13
detJ
0 r21
detJ
r32
detJ
z23
detJ
r13
detJ
z31
detJ
r21
detJ
z12
detJ
N1 /r 0 N2 /r 0 N3 /r 0
⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
⎥