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Aircraft Structures for Engineering Students (7th Edition) Solutions Manual [2025/2026 Edition] – Megson

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Ace your aerospace and mechanical engineering exams with this comprehensive Solutions Manual for Aircraft Structures for Engineering Students (7th Edition) by T.H.G. Megson. Perfect for 2025/2026 learners, this detailed guide provides step-by-step problem solutions, structural analysis examples, and real-world aircraft design applications. Ideal for university coursework, revision, and engineering project support in aerospace, civil, and mechanical structures. Master topics in stress analysis, elasticity, beam theory, and composite structures with clarity and confidence.

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AERO3001
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Uploaded on
October 29, 2025
Number of pages
447
Written in
2025/2026
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All 28 Chapters Covered




SOLUTIONS

,Table of contents
Part A: Fundamentals of Structural Analysis

1. Basic elasticity

2. Two-dimensional problems in elasticity

3. Torsion of solid sections

4. Virtual work and energy methods

5. Energy methods

6. Matrix methods

7. Bending of thin plates

8. Columns

9. Thin plates

10. Structural vibration

Part B: Analysis of Aircraft Structures

11. Materials

12. Structural components of aircraft

13. Airworthiness

14. Airframe loads

15. Fatigue

16. Bending of open and closed, thin-walled beams

,17. Shear of beams

18. Torsion of beams

19. Combined open and closed section beams

20. Structural idealization

21. Wing spars and box beams

22. Fuselages

23. Wings

24. Fuselage frames and wing ribs

25. Laminated composite structures

26. Closed section beams

27. Open section beams

28. Wing problems

, 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
2 τxy = 45 N/mm . Thus, from Eq. (1.11),


80 1 pffiffiffiffiffi2ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffi2ffi
σ I = 2 + 2 80 + 4 × 4 5
i.e.,
σI = 100.2 N/mm2

From Eq.
(1.12),
80 1 pffi ffiffi ffiffi2ffiffiffiffiffiffiffi ffiffiffiffiffiffiffi ffiffi ffiffiffi2ffi
σ II = — 80 + 4 × 45
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) 2
τmax = 2 = 60.2 N/mm

and will act on planes at 45° to the principal planes.


S.1.2
The principal stresses are given directly by Eqs (1.11) and (1.12) in which σx = 50
N/mm2,
σy =–35 N/mm2, and τxy = 40 N/mm2. Thus, from Eq. (1.11),


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