Title: Aircraft Structures for Engineering Students – Solutions Manual (7th Edition) by T.H.G. Megson | Complete Step-by-Step Solutions to Structural Analysis Problems in Aerospace Engineering Including Stress, Strain, Bending, Torsion, Shear, Elasticity,

All 28 Chapters Covered




SOLUTIONS

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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

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

80 1 pffiffiffiffiffi2ffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffi2ffi
σ I = 2 + 2 80 + 4 × 45
i.e.,
σI = 100.2 N/mm2

From Eq. (1.12),
80 1 pffi ffiffi ffiffi2ffiffiffi ffiffiffiffiffiffiffiffiffiffiffi ffiffi ffiffiffi2ffi
σ II = — 80 + 4 × 4 5
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),


3

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qffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
50 — 35 1
σ I= + (5 0 + 3 5) 2 + 4 × 4 0 2
2 2
i.e.,
σI = 65.9 N/mm2

and from Eq. (1.12),
qffi ffiffiffiffi ffiffiffiffiffiffiffi ffiffiffiffi ffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffi ffi
50 — 35 1
σ II= — (5 0 + 3 5) 2 + 4 × 40 2
2 2
i.e.,
σII = —50.9 N/mm2

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 = 50ABcos α — 35BCsin α + 40ABsin α + 40BCcos α (i)
Dividing through Eq. (i) by AB,
0 = 50 cos α — 35 tan α sin α + 40 sin α + 40 tan α cos α
which, dividing through by cos α, simplifies to
0 = 50 — 35 tan 2α + 80 tan α




FIG. S.1.2

, Solutions to Chapter 1 Problems 5



from which
tan α = 2.797 or — 0.511

Hence,
α = 70°21' or — 27°5'


S.1.3
The construction of Mohr’s circle for each stress combination follows the procedure described in
Section 1.8 and is shown in Figs S.1.3(a)–(d).




FIG. S.1.3(a)




FIG. S.1.3(b)




FIG. S.1.3(c)

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FIG. S.1.3(d)


S.1.4
The principal stresses at the point are determined, as indicated in the question, by transforming each
state of stress into a σx, σy, τxy stress system. Clearly, in the first case, σ=x 0, σ=
y 10 N/mm , τxy= 0
2

(Fig. S.1.4(a)). The two remaining cases are transformed by considering the equilibrium of the trian-
gular element ABC in Figs S.1.4(b), (c), (e), and (f). Thus, using the method described in Section 1.6
and the principle of superposition (see Section 5.9), the second stress system of Figs S.1.4(b) and (c)
becomes the σx, σy, τxy system shown in Fig. S.1.4(d) while the third stress system of Figs S.1.4(e) and
(f) transforms into the σx, σy, τxy system of Fig. S.1.4(g).
Finally, the states of stress shown in Figs S.1.4(a), (d), and (g) are superimposed to give the state of
stress shown in Fig. S.1.4(h) from which it can be seen that σI =σII =15 N/mm2 and that the x and
y planes are principal planes.




FIG. S.1.4(a)

, Solutions to Chapter 1 Problems 7




FIG. S.1.4(b)




FIG. S.1.4(c)




FIG. S.1.4(d)

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FIG. S.1.4(e)




FIG. S.1.4(f)




FIG. S.1.4(g)

, Solutions to Chapter 1 Problems 9




FIG. S.1.4(h)




FIG. S.1.5


S.1.5
The geometry of Mohr’s circle of stress is shown in Fig. S.1.5 in which the circle is constructed using
the method described in Section 1.8.
From Fig. S.1.5,
σx = OP1 = OB —BC + CP1 (i)
In Eq. (i) OB =σ I, BC is the radius of the circle that is equal to τ max
and CP1 =
qffi ffiffiffi ffiffiffi ffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffi ffi qffi ffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
C Q 12 — Q 1 P12 = max τ2 — τxy2 . Hence,
qffi ffi ffi ffi ffi ffi ffi ffi ffi ffi ffiffiffiffi ffiffi ffiffiffi
σx = σ1 — τmax + τ 2 max
— τxy 2

Similarly,
σy = OP2 = OB — BC — CP2 in which CP2 = CP1

Thus, qffi ffiffi ffiffi ffiffi ffi ffiffi ffiffiffiffi ffiffi ffiffiffi
σy = σI — τmax — τ 2 — τ xy2
max