Solutions Manual for Aircraft Structures for Engineering Students | 7th Edition (2026) | Megson | Covers All 28 Chapters

Aircraft Structures
for engineering students
Solutions Manual




T. H. G. Megson




A member of the Hodder Headline Group
LONDON  SYDNEY  AUCKLAND
Copublished in North, Central and South America by
John Wiley & Sons Inc., New York  Toronto

, Contents

Part I Elasticity
Solutions to Chapter 1 Problems ± Basic elasticity 1
Solutions to Chapter 2 Problems ± Two-dimensional problems in elasticity 12
Solutions to Chapter 3 Problems ± Torsion of solid sections 19
Solutions to Chapter 4 Problems ± Energy methods of structural analysis 27
Solutions to Chapter 5 Problems ± Bending of thin plates 68
Solutions to Chapter 6 Problems ± Structural instability 76


Part II Aircraft Structures
Solutions to Chapter 8 Problems ± Airworthiness and airframe loads 107
Solutions to Chapter 9 Problems ± Bending, shear and torsion of open and
closed, thin-walled beams 121
Solutions to Chapter 10 Problems ± Stress analysis of aircraft components 170
Solutions to Chapter 11 Problems ± Structural constraint 208
Solutions to Chapter 12 Problems ± Matrix methods of structural analysis 240
Solutions to Chapter 13 Problems ± Elementary aeroelasticity 262




Aircraft Structures for Engineering Students, 3rd Edition: Solutions Manual by T.H.G. Megson.
Published in 1999 by Arnold, 338 Euston Road, London NW1 3BH, UK.
# T.H.G. Megson. All rights reserved.

, Solutions to Chapter 1
Problems
S.1.1

The principal stresses are given directly by Eqs (1.11) and (1.12) in which x ˆ 80
N/mm2 , y ˆ 0 (or vice versa) and
xy ˆ 45 N/mm2 . Thus, from Eq. (1.11)
80 1 p
I ˆ ‡ 802 ‡ 4  452
2 2
i.e.
I ˆ 100:2 N=mm2
From Eq. (1.12)
80 1 p
II ˆ ÿ 802 ‡ 4  452
2 2
i.e.
II ˆ ÿ20:2 N=mm2
The directions of the principal stresses are de®ned by the angle  in Fig. 1.8(b) in
which  is given by Eq. (1.10). Hence
2  45
tan 2 ˆ ˆ 1:125
80 ÿ 0
which gives
 ˆ 248 110 and  ˆ 1148 110
It is clear from the derivation of Eqs (1.11) and (1.12) that the ®rst value of 
corresponds to I while the second value corresponds to II .
Finally, the maximum shear stress is obtained from either of Eqs (1.14) or (1.15).
Hence from Eq. (1.15)
100:2 ÿ ÿ20:2†

max ˆ ˆ 60:2 N=mm2
2
and will act on planes at 458 to the principal planes.

Aircraft Structures for Engineering Students, 3rd Edition: Solutions Manual by T.H.G. Megson.
Published in 1999 by Arnold, 338 Euston Road, London NW1 3BH, UK.
# T.H.G. Megson. All rights reserved.

, 2 Solutions to Chapter 1 Problems

S.1.2

The principal stresses are given directly by Eqs (1.11) and (1.12) in which x ˆ
50 N=mm2 , y ˆ ÿ35 N=mm2 and
xy ˆ 40 N=mm2 . Thus, from Eq. (1.11)
q
50 ÿ 35 1
I ˆ ‡ 50 ‡ 35†2 ‡ 4  402
2 2
i.e.
I ˆ 65:9 N=mm2
and from Eq. (1.12)
q
50 ÿ 35 1
II ˆ ÿ 50 ‡ 35†2 ‡ 4  402
2 2
i.e.
II ˆ ÿ50:9 N=mm2
From Fig. 1.8(b) and Eq. (1.10)
2  40
tan 2 ˆ ˆ 0:941
50 ‡ 35
which gives
 ˆ 218 380 I † and  ˆ 1118 380 II †
The planes on which there is no direct stress may be found by 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 ˆ 50AB cos ÿ 35BC sin ‡ 40AB sin ‡ 40BC cos i†
Dividing through Eq. (i) by AB
0 ˆ 50 cos ÿ 35 tan sin ‡ 40 sin ‡ 40 tan cos
which, dividing through by cos , simpli®es to
0 ˆ 50 ÿ 35 tan2 ‡ 80 tan

A
α



50 N/mm2

τ

B
C
40 N/mm2

35 N/mm2
Fig. S.1.2

Aircraft Structures for Engineering Students, 3rd Edition: Solutions Manual by T.H.G. Megson.
Published in 1999 by Arnold, 338 Euston Road, London NW1 3BH, UK.
# T.H.G. Megson. All rights reserved.