Aircraft Structures for Engineering Students 4th Edition Solutions Manual PDF

,Solution-1-H6739.tex 24/1/2007 9: 28 Page 1




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
σx = 80 N/mm2 , σy = 0 (or vice versa) and τxy = 45 N/mm2 . Thus, from Eq. (1.11)

80 1
2
σI = + 80 + 4 × 452
2 2
i.e.
σI = 100.2 N/mm2
From Eq. (1.12)
80 1
2
σII = − 80 + 4 × 452
2 2
i.e.
σII = −20.2 N/mm2
The directions of the principal stresses are defined by the angle θ in Fig. 1.8(b) in
which θ is given by Eq. (1.10). Hence
2 × 45
tan 2θ = = 1.125
80 − 0
which gives
θ = 24◦ 11 and θ = 114◦ 11
It is clear from the derivation of Eqs (1.11) and (1.12) that the first value of θ
corresponds to σI while the second value corresponds to σII .
Finally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15).
Hence from Eq. (1.15)

100.2 − (−20.2)
τmax = = 60.2 N/mm2
2
and will act on planes at 45◦ to the principal planes.