Solids of Revolution (Axisymmetric Solids) · Solids of Revolution (Axisymmetric Solids) كيناكم هدكشناد -ناهفصا يتعنص هاگشناد 2 دودحم يازجا

Solids of Revolution

(Axisymmetric Solids)

دانشكده مكانيك -دانشگاه صنعتي اصفهان روش اجزاي محدود 2

المبن تقبرن محوري

:تعريف

A structure of revolution is generated by rotating a generating cross section

about an axis of revolution.



:Rotating Flywheel :مثبل و كبربرد

Body forces: 2

rf r

zf g

(equivalent radial centrifugal

or inertial force )

(gravitational force )

Cylinder Subject to Internal Pressure:



:Press Fit :مثبل و كبربرد

Belleville (Conical) Spring:


المبن تقبرن محوري مثلثي

Axisymmetric solid “ring” element





(a) An axisymmetric solid, (b) FE mesh, (c) applied loads, (d) stress distribution.

(a)


است الزم مسئله، مبهيت به توجه بب

سود اياستوانه مختصبت دستگبه از

.برد

استخراج مبتريس سختي المبن تقبرن محوري

:تغييرمكبن ميذان( , )

( , )

u u r z

w w r z

v 0


:تغييرمكبن ميذان


:كرنش

:تنش

( )( )

r r

z z

rz rz

1 v v 0 v

v 1 v 0 vE

1 2v1 v 1 2v 0 0 0

2

v v 0 1 v


محوري تقبرن المبن


Axisymmetric Elements:

:نوشت توانمي سختي مبتريس براي ايصفحه تنش المبن مشببه


استخراج مبتريس سختي المبن تقبرن محوري مثلثي

The elements are defined in r-z plane. The definition of connectivity

of elements and the nodal coordinates are the same as CST element.

Note that r and z are replaced respectively by x and y. Using 3 shape

Function N1, N2 and N3.

3

(r1,z1)

r,u

z,w

u1

w1 u2

u3

w3

(r2,z2)

(r3,z3)

1

2 w2



( )

( )

1 2 3

1 2 3

u u u 1 u

w w w 1 w

Using iso-parametric representation:

321

321

)1(

)1(

zzzz

rrrr

The chain rule of differentiation gives:

u u

r

u u

z

J

w w

r

w w

z

J,

ij i j13 13

ij i j23 23

r r rr zwhere

z z zr z

J

&{ }T

1 1 2 2 3 3

1 2 3

1 2 3

u w u w u w

N 0 N 0 N 0

0 N 0 N 0 N

u = Nd

d

N

&



det 13 23 13 23 er z z r 2A J

1 1

u wu w

r rand

u u w w

z z

J J det

23 131

23 13

z z1

r r

JJ

where

Thus strain vector would be in the form of:

( ) ( )

det

( ) ( )

det

( ) ( ) ( ) ( )

det

23 1 3 13 2 3

23 1 3 13 2 3

23 1 3 13 2 3 23 1 3 13 2 3

1 1 2 2 2 2

z u u z u uu

r

r w w r w ww

z

w u r u u r u u z w w z w w

r z

u N u N u N u

r r

J

J

J



or Bd

Thus

det det det

det det det

det det det det det det

23 31 12

32 13 21

32 23 13 31 21 12

31 2

z z z0 0 0

r r r0 0 0

r z r z r z

NN N0 0 0

r r r

J J J

J J JB

J J J J J J



e

T

A

2 rdA k B E B

1,

31 2 3 N N N

The integral is logarithmic, numerical integration is preferred.

On the other hand, as a simple approximation, B and r can be evaluated at the

centroid of the triangle and used as representative values for the triangle. At the

centroid of the triangle,.

3

1 2 3r r rr

e

T T

e

A

2 r dA 2 rA k B EB B EB det( )eA

2

J

where is the radius of the centroid. Denoting as the element strain-

displacement matrix B evaluated at the centroid, we get

Br



Body Force Term

T T e

A

2 rdA d fu f

( )

( ) ( )

T

r z

A A

1 1 2 2 3 3 r 1 1 2 2 3 3 z

A

2 rdA 2 uf wf rdA

2 N u N u N u f N w N w N w f rdA

u f



Surface Traction

T T e

e

2 rdl u d TT

For a uniformly distributed load on

edge 1-2, we have:

,

1 r

1 ze

1 2

2 r

2 z

u aT

w aT2 l

u bT

w bT

d T

6

2,

6

2 2121 rrb

rra

u1

w1

u2

w2

r

1

2

21l

3

2

12

2

1221 )()( zzrrl

1 1 2 2r N r N r



Problem 1: Calculate the equivalent body force applied on

A triangular element under gravity force in z direction and

An inertia force due to rotation around z axis.

Problem 2: An axisymmetric body with a linearly distributed

Load on the conical surface is shown in Fig. 1. Determine the

Equivalent point load at nodes 2,4 and 6.

2

4

6 0.4Mpa

0.2Mpa z

r

r2=20mm, z2=70mm

r4=40mm, z4=55mm

r6=60mm, z6=40mm



Example: The cylinder is then subjected to an internal pressure of

2 Mpa. Using two elements on the 10 mm length shown, find the

displacements at the inner radius. E = 200 GPa, ν = 0.3.



E = 200000 MPa,

ν = 0.3.

Element connectivity: nodal coordinates:

. . .

. . .

.

. . .

5

2 69 1 15 0 1 15

1 15 2 69 0 1 1510

0 0 0 77 0

1 15 1 15 0 2 69

E( )( )

1 v v 0 v

v 1 v 0 vE

1 2v1 v 1 2v 0 0 0

2

v v 0 1 v

E



For element 1:

For element 2:

. .

. .

. . . .

. . .

1

0 05 0 0 0 0 05 0

0 0 1 0 0 1 0 0

0 1 0 05 0 1 0 0 0 05

0 0071 0 0 0071 0 0 0071 0

B

. .

. .

. . . .

. . .

1

0 05 0 0 05 0 0 0

0 0 0 0 1 0 0 1

0 0 05 0 1 0 05 0 1 0

0 00625 0 0 00625 0 0 00625 0

B



k =

1 2 3 4 7 8 Global dof



k =

3 4 5 6 7 8 Global dof



Using the elimination approach, on assembling the matrices with

reference to the degrees of freedom 1 and 3, we get

. .

. .

17

3

u4 03 2 34 251410

u2 34 4 35 2514

.

.

2

1

23

u 0 014 10mm

u 0 0133 10

The element stresses can be calculated from

(1)

.

.

2

2

0 014 10

0

0 0133 10

0

0

0

d(2)

.

20 0133 10

0

0

0

0

0

d

(1)

(2)

.

.

.

.

.

.

2

2

166

58 210 MPa

5 4

28 4

169 3

66 910 MPa

0

54 1

ΕΒd

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Solids of Revolution (Axisymmetric Solids) · Solids of Revolution (Axisymmetric Solids) كيناكم هدكشناد -ناهفصا يتعنص هاگشناد 2 دودحم يازجا