Question 3.18: A steel disc with 300 mm outside diameter, 40 mm bore diamet...

The governing differential equation is:

P.I. = Cr²  (say)

∴        r^{2} \cdot 2C + 3r \cdot 2Cr = –  400E\alpha \frac{r^{2}}{150^{2}}

i.e. C = \frac{-  50 E\alpha }{150^{2}}

∴ \sigma _{r} = A  –  \frac{B}{r^{2}}  –  50 E\alpha \frac{r^{2}}{150^2}

Boundary conditions:
At r = 20 mm, \sigma _{r} = 0

∴         0 = A  –  \frac{B}{20^{2}}  –  50 E\alpha \frac{20^{2}}{150^2}    (a)                  (3.102)
At r = 150 mm, \sigma _{r} = 0
0 = A – \frac{B}{150^{2}} – 50 E\alpha \frac{150^{2}}{150^2}    (b)                   (3.103)
Subtracting (3.103) from (3.102) gives:
0 = B\left(\frac{1}{22  500}  –  \frac{1}{400} \right) + 50E\alpha \left(1  –  \frac{400}{150^{2}} \right)

i.e.               B = 20 000Eα

Substituting B in (3.103) gives:

So        \sigma _{r} + r\frac{d\sigma _{r}}{dr} = A  –  \frac{B}{r^{2}}  –  50E\alpha \frac{r^{2}}{150 ^{2}} + r \left(\frac{2B}{r^{3}}  –  \frac{100E \alpha r}{150^{2}} \right)

i.e.      \sigma _{\theta } = A + \frac{B}{r^{2}}  –  150 E\alpha \frac{r^{2}}{150^{2}} = \left(50.88 + \frac{20  000}{r^{2}}  –  \frac{r^{2}}{150} \right) E\alpha

At r = 20:

i.e.       \sigma _{\theta } = 223.9  N/mm^{2}  (at r = 20 mm)