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PROBLEM (11.12) Redo Prob. 11.11 for the case in which δ = 0.5 mm and L = 0.6 m.
SOLUTION
See solution of Prob. 11.11:
A = 490.9 × 10−6 m 2
I = 19.17 × 10−9 m 4
Hence,
L r = 600 6.25 = 96
and Euler’s formula applies.
(a) ΔT =
δ
0.5 × 10−3
=
= 34.7 o C
−6
2α L 2(12 ×10 )(0.6)
(b) ΔT =
δ
π 2I
+
2α L 4 Aα L2
= 34.7 +
π 2 (19.17 × 10−9 )
−6
−6
4(490.9 ×10 )(12 ×10 )(0.6)
2
r = 6.25 mm
= 57 o C
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