Question 4.4: (a) Derive expressions for the hoop and radial stresses deve...
(a) Derive expressions for the hoop and radial stresses developed in a solid disc of radius R when subjected to a thermal gradient of the form T = Kr. Hence determine the position and magnitude of the maximum stresses set up in a steel disc of 150 mm diameter when the temperature rise is 150°C. For steel, α = 12 × 10^{-6} per °C and E = 206.8 GN/m².
(b) How would the values be changed if the temperature at the centre of the disc was increased to 30°C, the temperature rise across the disc maintained at 150°C and the thermal gradient now taking the form T = a + br?
(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).