6.9 Rotating Discs of Uniform Thickness
181
C =
3 + ν
8
ρw
2 b
2
(6.46)
Hence, Eqs. (6.44) and (6.45) become,
σ r =
3 + ν
8
ρw
2
(b
2
− r
2
)
(6.47)
σ θ =
3 + ν
8
ρw
2 b
2
−
1 + 3ν
8
ρw
2 r
2
(6.48)
The stresses attain their maximum values at the centre of the disc, i.e., at r = 0.
Therefore,
σ r = σ θ =
3 + ν
8
ρw
2 b
2
(6.49)
6.9.2 Circular Disc with a Hole
Let a = Radius of the hole.
If there are no forces applied at the boundaries a and b, then.
(σ r ) r=a = 0, (σ r ) r=b = 0.
from which we find that
C =
3 + ν
8
ρw
2
(b
2
+ a
2
)
and
C 1 = −
3 + ν
8
ρw
2 a
2 b
2
Substituting the above in Eqs. (6.44) and (6.45), we obtain
σ r =
3 + ν
8
ρw
2
b
2
+ a
2
−
a
2 b
2
r 2
− r
2
(6.50)
σ θ =
3 + ν
8
ρw
2
b
2
+ a
2
+
a
2 b
2
r 2
−
1 + 3ν
3 + ν
r
2
(6.51)
The radial stress σ r reaches its maximum at r =
√
ab, where