Solutions manual for aircraft
structures for engineering students |
7th edition (2026) | megson | covers
all 28 chapters | pdf download
Contents
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, UH.
Ⓒ T.H.G. Megson. All rights reserved.
,Part i elasticity
Solutions to chapter 1 problems − basic elasticity 1
Solutions to chapter 2 problems − t wo-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 б8
Solutions to chapter б problems − structural instability 7б
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 2б2
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, UH.
Ⓒ 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 σx = 80
n/mm 2 , σy = 0 (or vice versa) and τxy = 45 n/mm 2 . Thus, from eq. (1.11)
80 1 √ 2
σ 80 + 4 × 45 2
I = +
2 2
I.e.
σi = 100.2 n/mm 2
From eq. (1.12)
80 1√
σ 80 2 + 4 × 45 2
ii = —
2 2
I.e.
Σii = —20.2 n/mm 2
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) 2
τmax = = б0.2 n/mm
2
And will act on planes at 45° 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, UH.
Ⓒ T.H.G. Megson. All rights reserved.
, 2 Solutions to Chapter 1 Problems
s.1.2
The principa l stresses are given directly by eqs (1.11) and (1.12) in which σx =
50 n/mm 2 , σy = —35 n/mm 2 and τxy = 40 2
q n/mm . Thus, from eq. (1.11)
50 — 35 1
σ = + (50 + 35)2 + 4 × 40 2
I
2 2
I.e.
Σi = б5.9 n/mm 2
And from eq. (1.12) q
50 — 35 1
Σ = — (50 + 35)2 + 4 × 40 2
ii
2 2
I.e.
Σii = —50.9 n/mm 2
From fig. 1.8(b) and eq. (1.10)
2 × 40
tan 2ϴ = = 0.941
50 + 35
Which gives
ϴ = 21° 38 ' (σi ) and ϴ = 111° 38 ' (σii )
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 = 50ab cos 2 — 35bc sin 2 + 40ab sin 2 + 40bc cos 2 (i)
Dividing through eq. (i) by ab
0 = 50 cos 2 — 35 tan 2 sin 2 + 40 sin 2 + 40 tan 2 cos 2
Which, dividing through by cos 2, simplifies to
0 = 50 — 35 tan 2 2 + 80 tan 2
a
50 n/mm2
B
C
2
40 N/mm
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, UH.
Ⓒ T.H.G. Megson. All rights reserved.
structures for engineering students |
7th edition (2026) | megson | covers
all 28 chapters | pdf download
Contents
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, UH.
Ⓒ T.H.G. Megson. All rights reserved.
,Part i elasticity
Solutions to chapter 1 problems − basic elasticity 1
Solutions to chapter 2 problems − t wo-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 б8
Solutions to chapter б problems − structural instability 7б
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 2б2
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, UH.
Ⓒ 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 σx = 80
n/mm 2 , σy = 0 (or vice versa) and τxy = 45 n/mm 2 . Thus, from eq. (1.11)
80 1 √ 2
σ 80 + 4 × 45 2
I = +
2 2
I.e.
σi = 100.2 n/mm 2
From eq. (1.12)
80 1√
σ 80 2 + 4 × 45 2
ii = —
2 2
I.e.
Σii = —20.2 n/mm 2
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) 2
τmax = = б0.2 n/mm
2
And will act on planes at 45° 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, UH.
Ⓒ T.H.G. Megson. All rights reserved.
, 2 Solutions to Chapter 1 Problems
s.1.2
The principa l stresses are given directly by eqs (1.11) and (1.12) in which σx =
50 n/mm 2 , σy = —35 n/mm 2 and τxy = 40 2
q n/mm . Thus, from eq. (1.11)
50 — 35 1
σ = + (50 + 35)2 + 4 × 40 2
I
2 2
I.e.
Σi = б5.9 n/mm 2
And from eq. (1.12) q
50 — 35 1
Σ = — (50 + 35)2 + 4 × 40 2
ii
2 2
I.e.
Σii = —50.9 n/mm 2
From fig. 1.8(b) and eq. (1.10)
2 × 40
tan 2ϴ = = 0.941
50 + 35
Which gives
ϴ = 21° 38 ' (σi ) and ϴ = 111° 38 ' (σii )
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 = 50ab cos 2 — 35bc sin 2 + 40ab sin 2 + 40bc cos 2 (i)
Dividing through eq. (i) by ab
0 = 50 cos 2 — 35 tan 2 sin 2 + 40 sin 2 + 40 tan 2 cos 2
Which, dividing through by cos 2, simplifies to
0 = 50 — 35 tan 2 2 + 80 tan 2
a
50 n/mm2
B
C
2
40 N/mm
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, UH.
Ⓒ T.H.G. Megson. All rights reserved.
