An initially unstressed short steel cylinder, internal radius 0.2 m and external radius 0.3 m, is subjected to a temperature distribution of the form T =a + b loge r to ensure constant heat flow through the cylinder walls. With this form of distribution the radial and circumferential stresses at

Question

An initially unstressed short steel cylinder, internal radius 0.2 m and external radius 0.3 m, is subjected to a temperature distribution of the form [latex]T = a + b log_{e} r[/latex] to ensure constant heat flow through the cylinder walls. With this form of distribution the radial and circumferential stresses at any radius r , where the temperature is T. are given by

[latex]\sigma _{r}=A-\frac{B}{r^{2}} -\frac{\alpha ET}{2(1- u )}[/latex]

[latex]\sigma _{H}=A+\frac{B}{r^{2}} -\frac{\alpha ET}{2(1- u )}-\frac{ E\alpha b}{2(1- u )}[/latex]

If the temperatures at the inside and outside surfaces are maintained at 200°C and 100°C respectively, determine the maximum circumferential stress set up in the cylinder walls. For steel, E = 207 GN/m², ν = 0.3 and [latex]\alpha = 11 × 10^{-6} [/latex] per °C

Accepted Answer

[latex]T=a+b\log _{e}r[/latex]
[latex]200=a+b\log _{e}0.2=a+b(0.6931-2.3026)[/latex]
∴ [latex]200=a-1.6095 b [/latex] (1)
also [latex]100=a+b\log _{e}0.3=a+b(1.0986-2.3026)[/latex]
[latex]100=a-1.204 b [/latex] (2)
(2)-(1), [latex] 100=-0.4055 b[/latex]
[latex]b=-246.5=-247[/latex]
Also [latex] \frac{E\alpha }{2(1- u )} =\frac{207 imes 10^{9} imes 11 imes 10^{-6}}{2(1-0.29)}[/latex]
[latex]=1.6 imes 10^{6}[/latex]

Therefore substituting in the given expression for radial stress

[latex]\sigma _{r}=A-\frac{B}{r^{2}} -1.6 imes 10^{6}T[/latex]

At r = 0.3, [latex]\sigma _{r} = 0[/latex] and T = 100

[latex]0=A-\frac{B}{0.09}-1.6 imes 10^{6} imes 100[/latex] (3)

At r = 0.2, [latex]\sigma _{r} = 0[/latex] and T = 200

[latex]0=A-\frac{B}{0.04}-1.6 imes 10^{6} imes 200[/latex] (4)

(4)-(3) [latex] 0=B(11.1-25)-1.6 imes 10^{8}[/latex]
[latex]B=-11.5 imes 10^{6}[/latex]

and from (4),

[latex]A=25B+3.2 imes 10^{8}[/latex]
[latex]=(-2.88+3.2)10^{8}=0.32 imes 10^{8}[/latex]

substituting in the given expression for hoop stress,

[latex]\sigma _{H}=0.32 imes 10^{8}-\frac{11.5 imes 10^{6}}{r^{2}} -1.6 imes 10^{6}T+1.6 imes 10^{6} imes 247[/latex]

At r = 0.2, [latex] \sigma _{H}=(0.32-2.88-3.2+3.96)10^{8}=-180 MN/m^{2}[/latex]

At r = 0.3, [latex]\sigma _{H}=(0.32-1.28-1.6+3.96)10^{8}=+140 MN/m^{2} [/latex]

The maximum tensile circumferential stress therefore occurs at the outside radius and has a value of 140 MN/m². The maximum compressive stress is 180 MN/m² at the inside radius

Question 4.5: An initially unstressed short steel cylinder, internal radiu...

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