(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

Question

(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?

Accepted Answer

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

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

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

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

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

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

[latex]\sigma _{H}=A+\frac{B}{r^{2}}+\frac{\alpha E Kr}{3}-\alpha EKr[/latex] (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 [latex]σ_{r} = 0[/latex] at r = R.
Therefore substituting in (3),

[latex]0=A-\frac{\alpha EKR}{3} ext{ and} A=\frac{\alpha EKR}{3}[/latex]

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

[latex]\sigma _{r}=\frac{\alpha E K}{3} (R-r)[/latex]
[latex]\sigma _{H}=\frac{\alpha E K}{3} (R-2r)[/latex]

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

[latex]\sigma _{r_{max}}=\sigma _{H_{max}}=\frac{\alpha E KR}{3}[/latex]
[latex]=\frac{12 imes 10^{-6} imes 206.8 imes 10^{9} imes K imes 0.075}{3}[/latex]

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

∴ [latex]150=K imes 0.075[/latex]
∴ [latex]K=2000^{\circ} /m[/latex]

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

(b) With the modified form of temperature gradient

[latex]\int{Tr dr} =\int{(a+br)rdr} =\int{(ar+br^{2})dr}[/latex]
[latex]=\frac{ar^{2}}{2}+\frac{br^{3}}{3}[/latex]

Substituting in (1) and (2),

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

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

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

[latex]30=a+b(0)[/latex] (7)
and a=30

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

[latex]180=a+b(0.075)[/latex] (8)
(8) - (7) [latex]150=0.075b[/latex] ∴ [latex]b=2000[/latex]

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

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

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

Now for finite stresses at the centre,

B=0

Also, at r = 0.075, [latex]\sigma _{r}=0 [/latex] and [latex]T=180 °C[/latex]

Therefore substituting in (9),

[latex]0=A-12 imes 10^{-6} imes 206.8 imes 10^{9}(15+667 imes 0.075)[/latex]
[latex]0=A-12 imes 206.8 imes 10^{3} imes 65[/latex]
∴ [latex]A=161.5 imes 10^{6}[/latex]

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

i.e. [latex]\sigma _{r_{max}}=\sigma _{H_{max}}=A-\alpha ET=124 MN/m^{2}[/latex] , 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).

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

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