Question 4.4: (a) Derive expressions for the hoop and radial stresses deve...

(a) The hoop and radial stresses are given by eqns. (4.29) and (4.30) as follows:

$\sigma_{r}=A-\frac{B}{r^{2}}-\frac{E a}{r^{2}} \int T r d r$               (4.29)

$\sigma_{H}=A+\frac{B}{r^{2}}+\frac{E a}{r^{2}} \int T r d r-E a T$                   (4.30)

$\sigma _{r}=A-\frac{B}{r^{2}} -\frac{\alpha E}{r^{2}}\int{Trdr}$                (1)
$\sigma _{H}=A+\frac{B}{r^{2}}+ \frac{\alpha E}{r^{2}}\int{Trdr}-\alpha ET$           (2)
In this case $\int{Trdr}=K\int{r^{2}dr}=\frac{Kr^{3}}{3}$

the constant of integration being incorporated into the general constant A.

∴         $\sigma _{r}=A-\frac{B}{r^{2}} -\frac{\alpha E Kr}{3}$    (3)

$\sigma _{H}=A+\frac{B}{r^{2}}+\frac{\alpha E Kr}{3}-\alpha EKr$    (4)

Now in order that the stresses at the centre of the disc, where r = 0, shall not be infinite, B must be zero and hence B/r² is zero. Also $σ_{r} = 0$ at r = R.
Therefore substituting in (3),

Substituting in (3) and (4) and rearranging,

$\sigma _{r}=\frac{\alpha E K}{3} (R-r)$
$\sigma _{H}=\frac{\alpha E K}{3} (R-2r)$

The variation of both stresses with radius is linear and they will both have maximum values at the centre where r = 0.

$\sigma _{r_{max}}=\sigma _{H_{max}}=\frac{\alpha E KR}{3}$
$=\frac{12\times 10^{-6}\times 206.8\times 10^{9}\times K\times 0.075}{3}$

Now  T = Kr and T must therefore be zero at the centre of the disc where r is zero. Thus, with a known temperature rise of 150°C, it follows that the temperature at the outside radius must be 150°C

∴       $150=K\times 0.075$
∴       $K=2000^{\circ} /m$

i.e.   $\sigma _{r_{max}}=\sigma _{H_{max}}=\frac{12\times 10^{-6}\times 206.8\times 10^{9}\times 2000\times 0.075}{3}$
$=124   MN/m^{2}$

(b) With the modified form of temperature gradient

$\int{Tr dr} =\int{(a+br)rdr} =\int{(ar+br^{2})dr}$
$=\frac{ar^{2}}{2}+\frac{br^{3}}{3}$

Substituting in (1) and (2),

$\sigma _{r}=A-\frac{B}{r^{2}}-\frac{\alpha E}{r^{2}}[\frac{ar^{2}}{2}+\frac{br^{3}}{3} ]$              (5)

$\sigma _{H}=A+\frac{B}{r^{2}}+\frac{\alpha E}{r^{2}}[\frac{ar^{2}}{2}+\frac{br^{3}}{3} ]-\alpha ET$      (6)
Now  $T=a+br$

Therefore at the inside of the disc where r = 0 and T = 30°C,

$30=a+b(0)$             (7)
and     a=30

At the outside of the disc where T = 180°C,

$180=a+b(0.075)$                      (8)
(8) – (7)           $150=0.075b$    ∴    $b=2000$

Substituting in (5) and (6) and simplifying,

$\sigma _{r}=A-\frac{B}{r^{2}}-\alpha E(15+667r)$   (9)

$\sigma _{H}=A+\frac{B}{r^{2}}+\alpha E(15+667r)-\alpha ET$      (10)

Now for finite stresses at the centre,

B=0

Also, at r = 0.075,      $\sigma _{r}=0$ and  $T=180 °C$

Therefore substituting in (9),

$0=A-12\times 10^{-6}\times 206.8\times 10^{9}(15+667\times 0.075)$
$0=A-12\times 206.8\times 10^{3}\times 65$
∴     $A=161.5\times 10^{6}$

From (9) and (10) the maximum stresses will again be at the centre where r = 0,

i.e.    $\sigma _{r_{max}}=\sigma _{H_{max}}=A-\alpha ET=124   MN/m^{2}$ , as before.

N.B. The same answers would be obtained for any linear gradient with a temperature difference of 150°C. Thus a solution could be obtained with the procedure of part (a) using the form of distribution T = Kr with the value of T at the outside taken to be 150°C (the value at r = 0 being automatically zero).