Aircraft Structures for Engineering Students, 5th Edition Solution Manual

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Aircraft Structures
for Engineering Students
Fifth Edition
Solutions Manual
T. H. G. Megson

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 = 80N/mm2, σy = 0 (or vice versa) and τxy = 45N/mm2. Thus, from Eq. (1.11)
2 2
I
80 1 80 4 45
2 2
σ = + + ×
i.e.
σI = 100.2 N/mm
From Eq. (1.12)
2
2 2
II
80 1 80 4 45
2 2
σ = − + ×
i.e.
σII = – 20.2 N/mm
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
2 45
tan 2 1.125
80 0
θ ×
= =
−
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)
.
2
max
100.2 ( 20.2) 60.2N / mm
2
τ − −
= =
and will act on planes at 45° to the principal planes.

4 Solutions Manual
S.1.2
The principal stresses are given directly by Eqs (1.11) and (1.12) in which σx =
50N/mm2, σy = –35 N/mm2 and τxy = 40 N/mm2. Thus, from Eq. (1.11)
2 2
I
50 35 1 (50 35) 4 40
2 2
σ −
= + + + ×
i.e.
σI = 65.9 N/mm
and from Eq. (1.12)
2
2 2
II
50 35 1 (50 35) 4 40
2 2
σ −
= − + + ×
i.e.
σII = –50.9 N/mm
From Fig. 1.8(b) and Eq. (1.10)
2
2 40
tan 2 0.941
50 35
θ ×
= =
+
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 35BCsin 40ABsin +40BC cos
α
α
α
α= − + (i)
Fig. S.1.2

Solutions to Chapter 1 Problems 5
Dividing through Eq. (i) by AB
0 50cos 35 tan sin 40sin 40 tan cos
α
α α
α
α α= − + +
which, dividing through by cos
α, simplifies to
0 = 50–35 tan2
from which
α + 80 tan
α
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)

6 Solutions Manual
Fig. S.1.3(c)
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 σx = 0, σy = 10 N/mm2, τxy = 0 (Fig. S.1.4(a)). The two remaining cases
are transformed by considering the equilibrium of the triangular 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

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Solutions to Chapter 1 Problems 11
Fig. S.1.5
In Eq. (i) OB = σI, BC is the radius of the circle which is equal to τmax
2 2 2 2
1 1 1 1 maxCP CQ Q P .xy
τ τ= − = −
and
Hence
2 2
1 max maxx xy
σ σ τ τ τ= − + −
Similarly
2 2 2 1OP OB BC CP in which CP CPy
σ = = − − =
Thus
2 2
I max maxy xy
σ σ τ τ τ= − − −
S.1.6
From bending theory the direct stress due to bending on the upper surface of the shaft
at a point in the vertical plane of symmetry is given by
6 2
4
25 10 75 75 N / mm
150 / 64
x
My
I
σ π
× ×
= = =
×
From the theory of the torsion of circular section shafts the shear stress at the same
point is
6 2
4
50 10 75 75N / mm
150 / 32
xy
Tr
J
τ π
× ×
= = =
×

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12 Solutions Manual
Substituting these values in Eqs (1.11) and (1.12) in turn and noting that σy = 0
2 2
I
75 1 75 4 75
2 2
σ = + + ×
i.e.
2
I 121.4N / mm
σ =
2 2
II
75 1 75 4 75
2 2
σ = − + ×
i.e.
2
II 46.4N / mm
σ = −
The corresponding directions as defined by θ in Fig. 1.8(b) are given by Eq. (1.10)
i.e.
2 75
tan 2 2
75 0
θ ×
= =
−
Hence
I31 43 ( )
θ
σ′= °
and
II121 43 ( )
θ
σ′= °
S.1.7
The direct strains are expressed in terms of the stresses using Eqs (1.42), i.e.
1 [ ( )]x x y zv
E
ε σ σ σ= − + (i)
1 [ ( )]y y x zv
E
ε σ σ σ= − + (ii)
1 [ ( )]z z x yv
E
ε σ σ σ= − + (iii)
Then
1 [ 2 ( )]x y z x y z x y ze v
E
ε ε ε σ σ σ σ σ σ= + + = + + − + +
i.e.
(1 2 ) ( )x y z
v
e E
σ σ σ
−
= + +
whence
(1 2 )
y z x
Ee
v
σ σ
σ+ = −
−

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Solutions to Chapter 1 Problems 13
Substituting in Eq. (i)
1
1 2
x x x
Ee
v
E v
ε σ
σ
  
= − −  −  
so that
(1 ) 1 2
x x
vEe
E v v
ε σ= + − −
Thus
(1 2 )(1 ) (1 )
x x
vEe E
v v v
σ
ε= +
− + +
or, since G = E/2(1 + ν) (see Section 1.15)
2x xe G
σ λ ε= +
Similarly
2y ye G
σ λ ε= +
and
2z ze G
σ λ ε= +
S.1.8
The implication in this problem is that the condition of plane strain also describes
the condition of plane stress. Hence, from Eqs (1.52)
1 ( )x x yv
E
ε σ σ= − (i)
1 ( )y y xv
E
ε σ σ= − (ii)
2(1 ) (seeSection 1.15)
xy
xy xy
v
G E
τ
γ
τ
+
= = (iii)
The compatibility condition for plane strain is
2 2 2
2 2 (see Section 1.11)
xy y x
x y x y
γ ε ε∂ ∂ ∂
= +
∂ ∂ ∂ ∂ (iv)
Substituting in Eq. (iv) for εx, εy and γxy from Eqs (i)–(iii), respectively, gives
2 2 2
2 2
2 (1 ) ( ) ( )
xy
y x x yv v v
x y x y
τ σ σ
σ σ
∂ ∂ ∂
+ = − + −
∂ ∂ ∂ ∂ (v)

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14 Solutions Manual
Also, from Eqs (1.6) and assuming that the body forces X and Y are zero
0
zyx
x y
τ
σ ∂∂ + =
∂ ∂ (vi)
0
y xy
y x
σ τ∂ ∂
+ =
∂ ∂ (vii)
Differentiating Eq. (vi) with respect to x and Eq. (vii) with respect to y and adding gives
2 2 22
2 2 0
xy y xyx
y x x yx y
τ σ τ
σ ∂ ∂ ∂∂ + + + =
∂ ∂ ∂ ∂∂ ∂
or
2 22
2 2
2 xy yx
x y x y
τ
σ
σ ∂ ∂∂
 = − +
 ∂ ∂ ∂ ∂ 
Substituting in Eq. (v)
22 2 2
2 2 2 2
(1 ) ( ) ( )
yx y x x yv v v
x y x y
σ
σ
σ σ
σ σ
 ∂∂ ∂ ∂
 − + + = − + −
 ∂ ∂ ∂ ∂ 
so that
2 2 22 2 2
2 2 2 2 2 2
(1 ) y y yx x x
v v
x y x y x y
σ σ
σ
σ
σ
σ   ∂ ∂ ∂∂ ∂ ∂
   − + + = + − +
   ∂ ∂ ∂ ∂ ∂ ∂   
which simplifies to
2 22 2
2 2 2 2 0
y yx x
x y x y
σ
σ
σ σ∂ ∂∂ ∂
+ + + =
∂ ∂ ∂ ∂
or
2 2
2 2 ( ) 0x y
x y
σ σ
 ∂ ∂
+ + =  ∂ ∂ 
S.1.9
Suppose that the load in the steel bar is Pst and that in the aluminium bar is Pal . Then,
from equilibrium
st alP P P+ = (i)
From Eq. (1.40)
st al
st al
st st al al
P P
A E A E
ε
ε= =

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Solutions to Chapter 1 Problems 15
Since the bars contract by the same amount
st al
st st al al
P P
A E A E
= (ii)
Solving Eqs (i) and (ii)
st st al al
st
st st al al st st al al
al
A E A E
P P P P
A E A E A E A E
= =
+ +
from which the stresses are
st al
st al
st st al al st st al al
E E
P P
A E A E A E A E
σ
σ= =
+ + (iii)
The areas of cross-section are
2 2 2
2 2
st al
75 (100 75 )
4417.9 mm 3436.1mm
4 4
A A
π
π× −
= = = =
Substituting in Eq. (iii) we have
6 2
st
10 200000 172.6N/mm (compression)
(4417.9 200000 3436.1 80000)
σ ×
= =
× + ×
6 2
al
10 80000 69.1N/mm (compression)
(4417.9 200000 3436.1 80000)
σ ×
= =
× + ×
Due to the decrease in temperature in which no change in length is allowed the strain
in the steel is
αstT and that in the aluminium is
αal T. Therefore due to the decrease in
temperature
2
st st st 200000 0.000012 150 360.0 N/mm (tension)E T
σ α= = × × =
2
al al al 80000 0.000005 150 60.0 N/mm (tension)E T
σ α= = × × =
The final stresses in the steel and aluminium are then
2
st (total) 360.0 172.6 187.4N/mm (tension)
σ = − =
2
al (total) 60.0 69.1 9.1N/mm (compression)
σ = − = −
S.1.10
The principal strains are given directly by Eqs (1.69) and (1.70). Thus
2 2
I
1 1
( 0.002 0.002) ( 0.002 0.002) ( 0.002 0.002)
2 2
ε = − + + − + + + +

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16 Solutions Manual
i.e.
I 0.00283
ε = +
Similarly
II 0.00283
ε = −
The principal directions are given by Eq. (1.71), i.e.
2( 0.002) 0.002 0.002
tan 2 1
0.002 0.002
θ − + −
= = −
+
Hence
2 45 or 135
θ = − ° + °
and
22.5 or 67.5
θ = − ° + °
S.1.11
The principal strains at the point P are determined using Eqs (1.69) and (1.70). Thus
2 2 6
I
1 1
( 222 45) ( 222 213) ( 213 45) 10
2 2
ε − 
= − + + − + + − − × 
 
i.e.
6
I 94.0 10
ε −
= ×
Similarly
6
II 217.0 10
ε −
= − ×
The principal stresses follow from Eqs (1.67) and (1.68). Hence
6
I 2
31000 (94.0 0.2 271.0) 10
1 (0.2)
σ −
= − × ×
−
i.e.
2
I 1.29 N/mm
σ =
Similarly
2
II 814 N/mm
σ = −
Since P lies on the neutral axis of the beam the direct stress due to bending is zero.
Therefore, at P, σx =7 N/mm2 and σy = 0. Now subtracting Eq. (1.12) from (1.11)
2 2
I II 4x xy
σ σ σ τ− = +

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