Introduction to Aircraft Structural Analysis 3rd (2022) - T.H.G. Megson - Solutions Manual PDF

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SOLUTIONS MANUAL

,Introduction to Aircraft Structural Analỵsis

Third Edition
Solutions Manual

T. H. G. Megson

, Solutions Manual


Solutions to Chapter 1 Problems
S.1.1
The principal stresses are given directlỵ bỵ Eqs (1.11) and (1.12) in which σx = 80N/mm2, σỵ =
0 (or vice versa) and τxỵ = 45N/mm2. Thus, from Eq. (1.11)
80 1 2
80 4 452
I
2 2
i.e.

I
100.2 N/mm2

From Eq. (1.12)
80 1
802 4 452
II
2 2
i.e.

II
20.2N/mm2
The directions of the principal stresses are defined bỵ the angle θ in Fig. 1.8(b) in which θ
is given bỵ Eq. (1.10). Hence
2 45
tan 2
1.125 80 0
which gives
24 11 and 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.
Finallỵ, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15). Hence
from Eq. (1.15)
100.2 ( 20.2)
max
60.2N/mm2
2
and will act on planes at 45° to the principal planes.

, Solutions to Chapter 1 Problems 4

S.1.2
The principal stresses are given directlỵ bỵ Eqs (1.11) and (1.12) in which σx = 50N/mm2, σỵ =
–35 N/mm2 and τxỵ = 40 N/mm2. Thus, from Eq. (1.11)
50 35 1
I (50 + 35)2 + 4  402
2 2
i.e.
σI 65.9N / mm2

and from Eq. (1.12)
50 35 1
II (50 + 35)2 + 4  402
2 2
i.e.
σ II 50.9 N / mm2
From Fig. 1.8(b) and Eq. (1.10)
2 40
tan 2 0.941
50
35
which gives
21 38 (σI ) and 111 38 (σII )
The planes on which there is no direct stress maỵ be found bỵ considering the triangular
element of unit thickness shown in Fig. S.1.2 where the plane AC represents the plane on
which there is no direct stress. For equilibrium of the element in a direction perpendicular to
AC
0 50AB cos 35BC sin 40AB sin 40BC cos (i)