8/8/2019 Sm Lecture 22

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8/8/2019 Sm Lecture 22

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28/09/2006 2 SM lecture 22

Rotating Discs

A rotating disc is subjected to stress induced by centripetal acceleration.

Assume a thin disc. The radial and hoop stress are constant through the thickness and the outof plane stress z is zero.

Problem is axi-symmetric and the equilibrium of a small segment of the disc can be

considered. (Approach as for thick cylinders).

i.e.

R = centripetal force = = mr r r t r 2 2

= 2 2r t r

For equilibrium in radial direction:

( )

rr

r

d

drr r r t t r t r t r +

+ + =2

20

2 2

In the limit becomes:rd

drr

rr

+ + =

2 2 0 (1)

(When = 0, same as thick cylinder equation (1))

From thick cylinder equations (4) to (6), setting z=

0 gives:

( ) r rdu

dr E= =

1(u is rdirection displacement) (2)

( ) = = u

r Er

1(3)

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28/09/2006 3 SM lecture 22

( )

z rE

= + (4)

Eq (2) & (3) then gives:

r

du

dr

u

r

E= +

1 2(5)

= +

du

dr

u

r

E

1 2(6)

Substituting Eq (5) & (6) into Eq (1) gives:

( )d udr r

du

dr

u

r Er

2

2 2

2

21 10+ +

=

Solution of this differential equation takes the form:

( ) r

AB

rr= +

2

2 23

8(7)

and( )

= + +

AB

rr

2

2 21 3

8(8)

where A and B are constants found from boundary conditions.

Solid uniform thickness disc - no external pressure (0 rro)

B = 0 because if B 0, r and are infinite at r= 0 (See Eq 7 & 8).

At outside of disc, r = r0, r = 0

Gives from Eq 7, A =( )3

8

2

0

2+

r

from (7) : ( )

r

r r=+

3

8

2

0

2 2 (9)

and from (8) : ( ) ( )[ ]

= + +

2

0

2 2

8 3 1 3r r (10)

Max. stress at centre where r = 0

and( )

r r= =+3

8

2

0

2(11)

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Uniform thickness disc with central hole - no internal or external pressure (rirr0)

Boundary conditions r = 0 at inner and outer edges i.e. at r= ri and at r= r0.

Substituting into Eq 7 gives A=( )

( )3

8

2 2

0

2+

+

r ri

and B =( )3

8

2

0

2 2+

r ri

Substituting A and B back into Eq 7 and 8 gives:

( )

r i

ir rr r

rr=

++

3

8

2 2

0

22

0

2

2

2(12)

and

( ) ( )

=

+

+ +

+

+

3

8

1 3

3

2 2

0

22

0

2

2

2

r r

r r

r rii

(13)

Max. r occurs at r r ri= 0

when( )

( )

r i

r rmax =+

3

8

2

0

2(14)

Max. occurs at r= ri

when

( ) ( )

( )

max =

+

+

+

3

4

1

3

2

0

2 2

r ri (15)