Aircraft Structures
for engineering students
Fourth Edition
Solutions Manual
T. H. G. Megson
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Solutions Manual
Solutions to Chapter 1 Problems
S.1.1
The principal stresses are given directly by Eqs (1.11) and (1.12) in which ¥òx = 80 N/mm2 , ¥òy = 0 (or vice versa) and ¥óxy = 45 N/mm2 . Thus, from Eq. (1.11) ¥òI = i.e. ¥òI = 100.2 N/mm2 From Eq. (1.12) ¥òII = i.e. ¥òII = ?20.2 N/mm2 The directions of the principal stresses are de?ned by the angle ¥è in Fig. 1.8(b) in which ¥è is given by Eq. (1.10). Hence tan 2¥è = which gives ¥è = 24? 11 and ¥è = 114? 11 2 ¡¿ 45 = 1.125 80 ? 0 80 1 ? 802 + 4 ¡¿ 452 2 2 80 1 802 + 4 ¡¿ 452 + 2 2
It is cle¡¦(»ý·«)
ar from the derivation of Eqs (1.11) and (1.12) that the ?rst value of ¥è corresponds to ¥òI while the second value corresponds to ¥òII . Finally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15). Hence from Eq. (1.15) ¥ómax = 100.2 ? (?20.2) = 60.2 N/mm2 2
and will act
