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Solutions Manual
Solutions to Chapter 28 Problems
S.28.1
The solution is obtained directly from Eq. (28.9) in which c1 / = a, = 0 and
cm,0 = CM,0 . Thus
cos (s y)
1
cos s
CM,0
+ 0
ea
=
which gives
=
CM,0
+ 0
ea
cos (s y) CM,0
0
cos s
ea
Thus
+ 0 =
C
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Solutions Manual
Solutions to Chapter 14 Problems
S.14.1
Suppose that the mass of the aircraft is m and its vertical deceleration is a. Then referring
to Fig. S.14.1(a) and resolving forces in a vertical direction
ma + 135 2 200 = 0
N
M
ma
mU.C a
2.25
Solutions to Chapter 11 Problems
Solutions to Chapter 11 Problems
S.11.2
From Eq. (1.40) Youngs modulus E is equal to the slope of the stressstrain curve.
Then, since stress = load/area and strain = extension/original length.
E = slope of the loadextensio
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Solutions Manual
Solutions to Chapter 9 Problems
S.9.1
Assuming that the elastic deection, w, of the plate is of the same form as the initial
curvature, then
y
x
sin
w = A sin
a
a
Hence, from Eq. (7.36) in which m = n = 1, a = b and Nx = t
w=
t
(42 D
Solutions to Chapter 8 Problems
S.7.12
From Eq. (7.36) the deection of the plate from its initial curved position is
w1 = B11 sin
x
y
sin
a
b
in which
B11 =
A11 Nx
2
a2
2 D
1+ 2
a2
b
Nx
The total deection, w, of the plate is given by
w = w1 + w0
i.e.
w=
80
Solutions Manual
Solutions to Chapter 6 Problems
S.6.1
Referring to Fig. P.6.1 and Fig. 6.3
Member
12
Length
L
(cos ) 1/2
(sin ) 1/ 2
23
L
1/ 2
1/ 2
34
L
1/2
1/ 2
41
L
1/
2
1/ 2
13
2L
0
1
The stiffness matrix for each member is obtained using Eq. (6.3
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Solutions Manual
Solutions to Chapter 3 Problems
S.3.1
Initially the stress function, , must be expressed in terms of Cartesian coordinates.
Thus, from the equation of a circle of radius, a, and having the origin of its axes at its
centre.
= k (x 2 +
Solutions to Chapter 2 Problems
i.e.
1.29 + 8.14 =
2
72 + 4xy
from which xy = 3.17 N/mm2 .
The shear force at P is equal to Q so that the shear stress at P is given by
xy = 3.17 =
3Q
2 150 300
from which
Q = 95 100 N = 95.1 kN.
Solutions to Chapter 2 Prob
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
x = 80 N/mm2 , y = 0 (or vice versa) and xy = 45 N/mm2 . Thus, from Eq. (1.11)
I =
80 1
802 + 4 452
+
2
2
i.e.
I = 100.2 N/m
Solutions to Chapter 5 Problems
Then substituting in Eq. (4.20)
vB =
w
8EI
L /4
L
3(Lx 2 x 3 )dx +
0
(Lx x 2 )(L x )dx
L /4
which gives
57wL 4
6144EI
For the deection at the mid-span point the bending moment at any section due to the
actual loading is ide
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Solutions Manual
from the above. Suppose that the material is subjected to an applied load P. The actual
stress is then given by = P/A while the nominal stress is given by nom = P/Ao .
Therefore, substituting in Eq. (i) for A/Ao
=
nom
1
Then
nom (1 +
Solutions to Chapter 15 Problems
The increase in wing lift
L due to the gust is given by
1
1
CL
L = V 2 S
= 1.223 2502 50 4.8 0.024
2
2
i.e.
L = 220 140 N
Hence
(220 140 + 18 162)
= 0.64
145 000
Finally the forward inertia force fW is given by
n=1
1
f W
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Solutions Manual
Solutions to Chapter 16 Problems
S.16.1
From Section 16.2.2 the components of the bending moment about the x and y axes
are, respectively
Mx = 3000 103 cos 30 = 2.6 106 N mm
My = 3000 103 sin 30 = 1.5 106 N mm
The direct stress distri
Solutions to Chapter 25 Problems
2
26.0 N/mm
h
4.0 N/mm
200 mm
6
6.0 N/mm
200 mm
3
8000 N
Fig. S.24.3(g)
i.e.
P63 = 4000 10h
(xiv)
Thus P63 varies linearly from 6000 N (compression) at 6 to 8000 N (compression) at 3.
Solutions to Chapter 25 Problems
S.25.
Solutions to Chapter 27 Problems
Solutions to Chapter 27 Problems
S.27.1
The position of the shear centre, S, is given and is also obvious by inspection (see
Fig. S.27.1(a). Initially, then, the swept area, 2AR,0 (see Section 27.2) is determined as a func
Solutions to Chapter 24 Problems
The shear ows due to the combined action of the shear and torsional loads are then
as follows:
Bay
Spar webs: q = 12.5 5 = 7.5 N/mm
Bay
Spar webs: q = 5 3.125 = 1.875 N/mm
Skin panels: q = 9.375 N/mm
The ange loads are:
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Solutions Manual
Solutions to Chapter 22 Problems
S.22.1
The direct stresses in the booms are obtained from Eq. (16.18) in which Ixy = 0 and
My = 0. Thus
z =
Mx
y
Ixx
(i)
From Fig. P.22.1 the y coordinates of the booms are
y1 = y6 = 750 mm
y2 = y10 =
Solutions to Chapter 21 Problems
which gives
sect
1359p0
q0 q1
ds =
(L z )
Gt
1000Gt
Hence
L
0
sect
1359p0
q0 q1
ds dz =
1000Gt
Gt
L
(L z ) d z =
0
1359p0 L 2
2000Gt
(iv)
Substituting in Eq. (i) from Eqs (ii)(iv) gives
=
9p0 L 4
1359p0 L 2
p0 L 2
+
+
8Gt
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Solutions Manual
Solutions to Chapter 19 Problems
S.19.1
From Example 19.1
Ixx = 14.5 106 mm4
From Eq. (16.18) in which My = 0 and Ixy = 0
z =
Mx
y
Ixx
Therefore
z =
20 106
y = 1.38y
14.5 106
(i)
The Cx axis is 75 mm (see Example 19.1) from the upper
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Solutions Manual
Therefore
max = 170 N/mm2
The maximum shear stress in the open part is, from Eqs (18.12) and (18.13)
max = 25 000 2 18.5 106 = 0.9 N/mm2
Solutions to Chapter 20 Problems
S.20.1
From either Eq. (20.1) or (20.2)
B1 = 60 10 + 40 10 +
500
Solutions to Chapter 18 Problems
Solutions to Chapter 18 Problems
S.18.1
Referring to Fig. P.18.1 the maximum torque occurs at the built-in end of the beam and
is given by
Tmax = 20 2.5 103 = 50 000 N m
From Eq. (18.1)
max =
qmax
Tmax
=
tmin
2Atmin
i.e.
m
Solutions to Chapter 17 Problems
Solutions to Chapter 17 Problems
S.17.1
In Fig. S.17.1 the x axis is an axis of symmetry (i.e. Ixy = 0) and the shear centre, S,
lies on this axis. Suppose S is a distance S from the web 24. To nd S an arbitrary
shear load
Solutions to Chapter 4 Problems
Hence
1 d 3
( y 3x 2 y ) .
2a d z
w=
S.3.5
The torsion constant, J , for the complete cross-section is found by summing the torsion
constants of the narrow rectangular strips which form the section. Then, from Eq. (3.29)
bt