Question 4.5: An initially unstressed short steel cylinder, internal radiu...
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 log_{e} r 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
\sigma _{r}=A-\frac{B}{r^{2}} -\frac{\alpha ET}{2(1-\nu )}
\sigma _{H}=A+\frac{B}{r^{2}} -\frac{\alpha ET}{2(1-\nu )}-\frac{ E\alpha b}{2(1-\nu )}
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 \alpha = 11 × 10^{-6} per °C
T=a+b\log _{e}r
200=a+b\log _{e}0.2=a+b(0.6931-2.3026)
∴ 200=a-1.6095 b (1)
also 100=a+b\log _{e}0.3=a+b(1.0986-2.3026)
100=a-1.204 b (2)
(2)-(1), 100=-0.4055 b
b=-246.5=-247
Also \frac{E\alpha }{2(1-\nu )} =\frac{207\times 10^{9}\times 11\times 10^{-6}}{2(1-0.29)}
=1.6\times 10^{6}
Therefore substituting in the given expression for radial stress
\sigma _{r}=A-\frac{B}{r^{2}} -1.6\times 10^{6}TAt r = 0.3, \sigma _{r} = 0 and T = 100
0=A-\frac{B}{0.09}-1.6 \times 10^{6} \times 100 (3)
At r = 0.2, \sigma _{r} = 0 and T = 200
0=A-\frac{B}{0.04}-1.6\times 10^{6}\times 200 (4)
(4)-(3) 0=B(11.1-25)-1.6\times 10^{8}
B=-11.5\times 10^{6}
and from (4),
A=25B+3.2\times 10^{8}
=(-2.88+3.2)10^{8}=0.32\times 10^{8}
substituting in the given expression for hoop stress,
\sigma _{H}=0.32\times 10^{8}-\frac{11.5\times 10^{6}}{r^{2}} -1.6\times 10^{6}T+1.6\times 10^{6}\times 247At r = 0.2, \sigma _{H}=(0.32-2.88-3.2+3.96)10^{8}=-180 MN/m^{2}
At r = 0.3, \sigma _{H}=(0.32-1.28-1.6+3.96)10^{8}=+140 MN/m^{2}
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