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 EST (1.11) and (1.12) in which ax = 80 N/mm2, yy = 0
(or vice versa), and ax = 45 N/mm. Thus, from Eq. (1.11),
2
80 1 pffiffiffiffiffi2ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffi2ffi
σ I = 2 + 2 80 + 4 × 45
i.e.
I = 100.2 N/mm2
From Eq. (1.12),
80 1 up ffiffiffiffi2ffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffi2ffi
Σ II = — 80 + 4 × 45
2 2
i.e.
Lini = —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 1 . 1 2 5
Tan 2θ
= =
80 — 0
Which gives
θ = 24°11' and θ = 114°11'
It is clear from the derivation of EST (1.11) and (1.12) that the first value of θ corresponds to me while
the second value corresponds to lini.
Finally, the maximum shear stress is obtained from either of EST (1.14) or (1.15). Hence
from Eq. (1.15),
100.2 — (—20.2) 2
Mix = 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 EST (1.11) and (1.12) in which ax = 50 N/mm2,
Yy =–35 N/mm2, and ax = 40 N/mm2. Thus, from Eq. (1.11),
3