## Pxb

Now for finite stresses at the centre,

Therefore substituting in (9),

0 = A - 12 x 10~6 x 206.8 x 109(15 + 667 x 0.075) 0 = A - 12 x 206.8 x 103 x 65 A = 161.5 x 106

From (9) and (10) the maximum stresses will again be at the centre where r = 0, i.e. or = oHmm = A — aET = 124 MN/m2, 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).

### Example 4 J

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, 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

B aET

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, v = 0.3 and a = 11 x 10~6 per °C.

Solution

T — a + b log, r 200 = a + b log, 0.2 = a + ¿>(0.6931 - 2.3026) 200 = a - 1.6095 b (1)

also 100 = a + b log, 0.3 = a + b(\ .0986 - 2.3026)

(2) - (1), 100 =-0.4055 b b = -246.5 = -247 Ea 207 x 109 x 11 x 10"6

Also

Therefore substituting in the given expression for radial stress,

B t, ar = A--- — 1.6 x 106r rAt r = 0.3, or = 0 and T = 100

0 0