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Mechanics of Aircraft structures C.T. Sun
4.1 A uniform beam of a thin-walled angle section as shown in
Fig. 4.19 is
subjected to the bending (yM 0=zM ). Find the neutral axis and
bending stress distribution over the cross-section.
Figure 4.19 Thin-walled angle section
Solution: (a) For finding the location of the centroid, we
select the corner of the thin-walled
section as the origin of a Cartesian coordinate system with the
horizontal and vertical distances between the centroid and the
origin denoted by and , respectively.
cy cz
42)2/( h
hththyc ==
42)2/( h
hththzc ==
--- ANS (b) Set up a Cartesian coordinate system (y, z) in the
pane of the section with the
origin at the centroid. The moments of inertia with respect to
this coordinate system are (assume t
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Mechanics of Aircraft structures C.T. Sun
32
2 161|)
2( thytzytdyzyzdA c
c
c
c
yhyc
yh
y cA
===
(c) Using equation (4.25) in the textbook,
zIII
MIMIy
IIIMIMI
yzzy
zyzyz
yzzy
yyzzyxx 22
+=
By substituting the known values we obtain
)915(2
])8/1()24/5)(24/5[()24/5(
])8/1()24/5)(24/5[()8/1(
3
3232
yzth
M
zth
My
thM
y
yyxx
=+
=
--- ANS Maximum positive stress:
At hzhz c 43== and
4hyy c ==
23 427
)915(2 th
Myz
thM yy
xx == Maximum negative stress:
At 4hzz c == and hyhy c 4
3==
23 421
)915(2 th
Myz
thM yy
xx ==
The absolute maximum stress is 2427
thM y
xx = (d) The neutral axis is located along 0=xx
0)915(2 3
== yzth
M yxx => 0915 = yz
So the neutral plane is located at 0915 = yz in the y-z
coordinate system (the centroid is the origin of this coordinate
system).
--- ANS
4.1.2
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Mechanics of Aircraft structures C.T. Sun
4.2 Rotate the angle section of Fig. 4.19 counterclockwise for .
Find the neutral axis and the maximum bending stress. Compare the
load capacity with that of the original section given by Fig.
4.19.
o45
Figure 4.19 Thin-walled angle section
Solution:
Remove the primes in the coordinates
Set up a temporary Cartesian coordinate system with the origin
at the corner of the thin-walled section to find the centroid. The
horizontal and vertical distances from the centroid to the origin
are denoted by and , respectively. cy czBecause of the symmetry, .
Assuming 0yc = ht
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Mechanics of Aircraft structures C.T. Sun
'' 22 zIIIMIMI
yIIIMIMI
yzzy
zyzyz
yzzy
yyzzyxx
+=
and substituting the values of moments of inertia in the
equation above, we obtain
zthM
12zIM
3y
y
yxx ==
--- ANS Maximum positive stress is at
22hz = , => 2
23thM y
xx =
Maximum negative stress is at
22
hz = , => 223thM y
xx =
The absolute maximum stress is 223thM y
xx = (c) The neutral axis (plane) is located along 0=xx ,
0zthM
12 3y
xx == => 0z =So the neutral axis coincides with the
centroidal axis. Note that this section in this particular position
is symmetric with respect to the y-z coordinate system. For
symmetric sections the neutral axis always coincides with the
location of the centroid.
--- ANS (d) The load capacity with the original section
For the same maximum bending stress in both beams,
2,
2,
42723
thM
thM originyrotatey
xx ==
=> 59.1212
27MM
origin,y
rotate,y ==
The load capacity of the rotated section is 1.59 times that of
the original section. --- ANS
4.2.2
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Mechanics of Aircraft structures C.T. Sun
4.3 The stringer-web sections shown in Figs. 4.20, 4.21, and
4.22 are subjected to
the shear force , while 0zV 0=yV . Find the bending stresses in
the stringers for the same bending moment . Which section is most
effective in bending? yM
Figure 4.20 Stringer-web section
Figure 4.21 Stringer-web section
Figure 4.22 Stringer-web section
Solution:
The contribution of the thin sheets to bending is assumed to be
negligible. Thus the
neutral axis is only depends on the cross-sectional area of the
stringers. Also, assume
4.3.1
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Mechanics of Aircraft structures C.T. Sun
y and z are the horizontal axis and vertical axis, respectively.
The origin of the system
is located at the centroid.
(a) Figure 4.20.
(1) Because of symmetry, the centroid is located at the middle
of the vertical
web.
(2) Moment of inertia 222 4)2(2 AhhAzAI
iiiy ===
0)02(2 22 === AyAIi
iiz
0== i
iiiyz zyAI
(3) Bending stress. Considering 0yM and 0=zM
zAhM
zIM y
y
yxx 24
==
The stresses at the stringer are
. At , hz =AhM
zAhM yy
xx 44 2==
. At ,hz =AhM
zAhM yy
xx 44 2==
--- ANS
(b) Figure 4.21.
(1) Because of symmetry (when neglecting the effects of webs),
the centroid is
located at the center of the section as shown in the figure.
(2) Moment of inertia 222 4)(4 AhhAzAI
iiiy ===
222 ))2
((4 AhhAyAIi
iiz === 0==
iiiiyz zyAI
(3) Bending stress. Considering 0yM and 0=zM
zIM
zIII
MIy
IIIMI
y
y
yzzy
yz
yzzy
yyzxx =+
= 22
The stresses at the stringers are (y position is not
involved)
4.3.2
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Mechanics of Aircraft structures C.T. Sun
. At , hz =AhM
zAhM yy
xx 44 2==
. At , hz =AhM
zAhM yy
xx 44 2==
--- ANS
(c) Figure 4.22.
(1) Again, when neglecting the effects of webs, the centroid is
located at the
middle of the vertical web.
(2) Moment of inertia 222 4)(4 AhhAzAI
iiiy ===
222 2)(2 AhhAyAIi
iiz === 22))((2 AhhhAzyAI
iiiiyz ===
(3) Bending stress. Considering 0yM and 0=zM
zAhM
yAhM
zAh
My
AhM
zIII
MIy
IIIMI
yy
yy
yzzy
yz
yzzy
yyzxx
22
222222
22
])2(24[2
])2(24[2
+=+=+
=
The stresses at the stringer are
At , , hz = hy =
0)(222 222
=+=+= hhAhM
zAhM
yAhM yyy
xx . At , , hz = 0=y
AhM
hAhM
zAhM
yAhM yyyy
xx 2)0(
222 222=+=+= . At , , hz = 0=y
AhM
hAhM
zAhM
yAhM yyyy
xx 2)0(
222 222==+= . At , , hz = hy =
0)(222 222
==+= hhAhM
zAhM
yAhM yyy
xx --- ANS
4.3.3
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Mechanics of Aircraft structures C.T. Sun
(d) Comparing the above results, sections in Figure 4.20 and
Figure 4.21 are both
more effective than the section in Figure 4.22 for this
particular loading.
--- ANS
4.3.4
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Mechanics of Aircraft structures C.T. Sun
4.4 Compare the bending capabilities of the two sections of
Figs. 4.21 and 4.22 if
, . 0=yM 0zM
Figure 4.21 Stringer-web section
Figure 4.22 Stringer-web section
Solution:
The thin sheets are assumed to be negligible in bending. Thus,
the location of the
centroid of the cross-section only depends on stringers. The
coordinates (y, z) are set
up with the origin at the centroid with y and z designating the
horizontal axis and
vertical axis, respectively.
(a) Figure 4.21.
(1) The centroid is located at the center of of the space
defined by the four
stringers.
(2) Moment of inertia 222 4)(4 AhhAzAI
iiiy ===
4.4.1
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Mechanics of Aircraft structures C.T. Sun
222 ))2
((4 AhhAyAIi
iiz === 0==
iiiiyz zyAI
(3) Bending stress.
Considering and 0=yM 0zM we have
yIMz
IIIMI
yIII
MI
z
z
yzzy
zyz
yzzy
zyxx =
+= 22
The stresses in stringers 1 and 4 are:
At 2hy = , y
AhM
2z
xx = The stresses in stringers 2 and 3 are
At 2hy = , y
AhM
2z
xx =
--- ANS
(b) Figure 4.22.
(1) The centroid is located at the middle of the vertical
web.
(2) Moment of inertia 222 4)(4 AhhAzAI
iiiy ===
222 2)(2 AhhAyAIi
iiz === 22))((2 AhhhAzyAI
iiiiyz ===
(3) Bending stress.
For and 0=yM 0zM
zAhMy
AhM
zAh
MyAh
MzIIIMI
yIII
MI
zz
zz
yzzy
zyz
yzzy
zyxx
22
222222
2
])2(24[2
])2(24[4
+=+=
+=
The stress in stringer 1 is
At , hz = hy =
4.4.2
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Mechanics of Aircraft structures C.T. Sun
AhMhh
AhMz
AhMy
AhM zzzz
xx 2)2(
22 222=+=+=
Stringer 2:
At , , hz = 0=yAhMh
AhMz
AhMy
AhM zzzz
xx 2)0(
22 222=+=+=
Stringer 3:
At hz = , ,0=yAhMh
AhMz
AhMy
AhM zzzz
xx 2)0(
22 222==+=
Stringer 4:
At hz = , , hy =AhMhh
AhMz
AhMy
AhM zzzz
xx 2)2(
22 222==+=
--- ANS
(c) Comparing the above results, we see that sections in Figure
4.21 and Figure 4.22
have the same bending efficiency; they both reach the same
maximum bending
stress under the same moment.
--- ANS
4.4.3
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Mechanics of Aircraft structures C.T. Sun
4.5 Figure 4.23 shows the cross-section of a four-stringer box
beam. Assume that
the thin walls are ineffective in bending and the applied
bending moments are
cmNM y = 000,500 cmNM z = 000,200 .
Find the bending stresses in all stringers.
Figure 4.23 Thin-walled section
Solution:
(a) Set up a temporary coordinate system with stringer 1 as the
origin. The location of
the centroid is
cm5.54)4124()20012002(
A
yAy
ii
iii
c =++++==
cm9.40)4124()1004501(
z
zAz
ii
iii
c =++++==
(b) The moment of inertia
4
222
i
2iiy
cm240901
)909091.40100(4)909091.4050(1)909091.40)(24(zAI
=+++==
Similarly, 4
i
2iiz cm87273yAI ==
4.5.1
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Mechanics of Aircraft structures C.T. Sun
4
iiiiyz cm14545zyAI ==
It is more convenient to put in the chart, for instance:
iA iy iz Stringer No. )(cm )( 2cm )(cm
2ii zA )( 4cm
2ii yA )( 4cm
iii zyA )( 4cm
1 4 -54.5 -40.9 6694 11901 89256
2 2 145.5 -40.9 3347 42314 -11901
3 1 145.5 9.1 82.6 21157 1322
4 4 -54.5 59.1 13967 11901 -12893
= 24091 87273 -14545
(c) Bending stress in the stringers.
By using the equation: zIIIMIMI
yIIIMIMI
yzzy
zyzyz
yzzy
yyzzyxx 22
+= , and
cmNM y = 000,500 cmNM z = 000,200 .
4909.24090 cmI y = 4727.87272 cmI z =
4455.14545 cmI yz = We obtain z54.21y298.1xx = Therefore the
bending stresses in the stringers are:
iy Stringer No. )(cm
iz )(cm
xx )/( 2cmN
1 -54.54 -40.91 951.92
2 145.45 -40.91 692.31
3 145.45 9.09 -384.62
4 -54.54 59.09 -1201.92
--- ANS
4.5.2
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Mechanics of Aircraft structures C.T. Sun
4.6 Find the neutral axis in the tin-walled section of Fig. 4.23
for the loading given
in Problem 4.5.
cmNM y = 000,500 cmNM z = 000,200 .
Find the bending stresses in all stringers.
Figure 4.23 Thin-walled section
Solution:
(a) From Problem 4.5 we get the centroid position as
follows.
cm5.54yc = , cm9.40zc =These are the horizontal and vertical
distances, respectively, from stringer 1.
(b) Set up the coordinate system (y,z) with the origin located
at the centroid. Neutral
plane is located at the position that centroid is the origin.
From the bending stress
formulas we find the neutral plane by setting the bending stress
to zero, i.e.,
0z538.21y298.1xx == On the cross-section, this equation
represents the line passing through the centroid
with and an angle z59.16y =o45.3)
59.161(tan)
yz(tan 11 ===
--- ANS
4.6.1
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Mechanics of Aircraft structures C.T. Sun
4.7 Find the bending stresses in the stringers at the fixed end
of the box beam loaded
as shown in Fig. 4.24. Assume that the thin sheets are
negligible in bending.
Find the neutral axis.
Figure 4.24 Loaded box beam
Solution:
(a) Name the stringers from top to bottom and left to right as
stringer 1, stringer 2,
and stringer 3, respectively. Relative to string 2 the centroid
position is given by
cm67.2643804
A
yAy
ii
iii
c ===
cm33.1343404
A
zAz
ii
iii
c ===
(b) The bending moments at the fixed end of the box beam
produced by the loads are
cmNPLM y === 200000)500)(200(22 ( is positive in positive y)
yMcmNPLM z === 200000)500)(200(22 ( is positive in negative z)
zM
(c) Set up the coordinate system (x,y,z) with the origin at the
centroid.
Moment of inertia (see table below for details): 422
i
2ciy cm4266)67.2633.132(4zAI =+==
422
i
2ciz cm17067)33.5367.262(4yAI =+==
4.7.1
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Mechanics of Aircraft structures C.T. Sun
4
icciyz cm4266zyAI ==
iA iy iz Stringer No. )(cm )( 2cm )(cm
2ii zA )( 4cm
2ii yA )( 4cm
iii zyA )( 4cm
1 4 -26.67 26.67 2844 2844 -2844
2 4 -26.67 -13.33 711 2844 1422
3 4 53.33 -13.33 711 11377 -2844
= 4266 17067 -4267 (d) Bending stress in the stringers.
Using the equation zIIIMIMI
yIIIMIMI
yzzy
zyzyz
yzzy
yyzzyxx 22
+= ,
we obtain zyxx 125.7825.31 = and the bending stresses in the
stringers are:
iy Stringer No. )(cm
iz )(cm
xx )/( 2cmN
1 -26.67 26.67 -1250
2 -26.67 -13.33 1875
3 53.33 -13.33 -625
--- ANS
(e) Neutral plane by angle . Neutral plane is located at the
position where bending stresses vanish under this
particular loading. We have
0125.7825.31 == zyxx It is the line passing through the centroid
with zy 5.2=
o8.21)5.21(tan)
yz(tan 11 ===
--- ANS
4.7.2
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Mechanics of Aircraft structures C.T. Sun
4.8 Find the deflection of the box beam of Fig. 4.24 using the
simple beam theory.
Figure 4.24 Loaded box beam
Solution:
(a) Name the stringers from top to bottom and then left to right
as stringer 1, stringer
2, and stringer 3, respectively. From the solution of problem
4.7, we have the
following moments of inertia: 4
y cm4266I = 4
z cm17066I = 4
yz cm4266I = Let the origin of the coordinate system be located
at the fixed end. )0x( =The bending moments produced by the forces
applied at the free end are
cmNxM y = )500(400 cmNxM z = )500(400
(b) The governing equations (see p. 122 in the book) for the
bidirectional bending are
)/()500(063.0 3222
cmNxIIIMIMI
dxvdE
yzzy
yyzzy == ,
)/()500(156.0 3222
cmNxIIIMIMI
dxwdE
yzzy
zyzyz ==
Integrating twice the above differential equations, we
obtain
4.8.1
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Mechanics of Aircraft structures C.T. Sun
21
32 )
6250(063.0 CxCxxEv ++=
43
32 )
6250(156.0 CxCxxEw ++=
By applying the boundary conditions, the integration constants
are solved as
0)0( ==xv , 0)0( ==xdxdv => 021 == CC
0)0( ==xw , 0)0( ==xdxdw => 043 == CC
Then the lateral (in y-direction) and vertical (in z-direction)
deflections are,
respectively,
)6
250(063.0)(3
2 xxE
xv =
)6
250(156.0)(3
2 xxE
xw = In the expressions above, distance x is measured in cm, and
the units of Youngs
modulus and deflection are and , respectively. 2/ cmN cm
--- ANS
As an example, consider Aluminum 2024-T3, .
The deflections in y and z directions at the free end are:
)/(107272 25 cmNGPaE ==
cmxv 36.0)6
500500250(1072
063.0)500(3
25 ===
cmxw 90.0)6
500500250(1072
156.0)500(3
25 ===
4.8.2
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Mechanics of Aircraft structures C.T. Sun
4.9 Find the bending stresses in the stringers of the box beam
in Fig. 4.24 for the
bending moments given in Problem 4.5.
cmNM y = 000,500 cmNM z = 000,200 .
Figure 4.24 Loaded box beam
Solution:
(a) Name the stringers from top to bottom and then left to right
as stringer 1, stringer
2, and stringer 3, respectively. The centroid position is given
by
cm67.2643804
A
yAy
ii
iii
c ===
cm33.1343404
A
zAz
ii
iii
c ===
relative stringer 2.
(b) Moment of inertia (see the table below for details) 422
i
2ciy cm4267)67.2633.132(4zAI =+==
422
i
2ciz cm17067)333333.53666667.262(4yAI =+==
4
icciyz cm4267zyAI ==
4.9.1
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Mechanics of Aircraft structures C.T. Sun
iA iy iz Stringer No. )(cm )( 2cm )(cm
2ii zA )( 4cm
2ii yA )( 4cm
iii zyA )( 4cm
1 4 -26.67 26.67 2844 2844 -2844
2 4 -26.67 -13.33 711 2844 1422
3 4 53.33 -13.33 711 11378 -2844
= 4267 17067 -4267
(c) Bending stress in the stringers.
Subsituting the moments and moments of inertia in the bending
stress formula
zIIIMIMI
yIIIMIMI
yzzy
zyzyz
yzzy
yyzzyxx 22
+= ,
we obtain z62.140y44.23xx = Therefore the bending stresses in
the stringers are:
iy Stringer No. )(cm
iz )(cm
xx )/( 2cmN
1 -26.67 26.67 -3125
2 -26.67 -13.33 2500
3 53.33 -13.33 625
--- ANS
4.9.2
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Mechanics of Aircraft structures C.T. Sun
4.10 A cantilever beam of a solid rectangular cross-section is
loaded as shown in Fig.
4.25. Assume that the material is isotropic. Find the
deflections of the beam
using the simple beam theory and Timoshenko beam theory,
respectively. Plot
the ratio of the maximum deflections of the two solutions (at
the free end)
versus L/h. Use the shear correction factor 65=k .
P
L
x
t
h
Figure 4.25 Cantilever beam subjected to a shear force P
Solution:
(a) Simple beam theory
The displacement equilibrium equations for the simple beam
theory is:
z40
4
y pdxwd
EI = (4.10.1)
In this particular problem, we have 3121 thI y = , 0pz = .
Thus,
040
4
=dxwdEI y (4.10.2)
Integrating the equation (4.10.2) and applying boundary
conditions,
PCdxwdEI y == 030
3
(shear force)
Integrating again, we obtain
120
2
CPxdxwdEI y += . (4.10.3)
At , Lx = 1202
0)( CPLLxdxwdEIM y +====
=> PLC =1From (4.10.3),
4.10.1
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Mechanics of Aircraft structures C.T. Sun
220
21 CPLxPx
dxdwEI y ++=
At , 0=x 0dxdw0 = => 02 =C
Finally, 323
0 21
61)( CPLxPxxwEI y ++= , and 03 =C because
at .
00 =w0=x
Therefore, the deflection curve is
])(6)(2[)21
61(1)( 23230 h
xhL
hx
EtPPLxPx
EIxw
y
+=+=
--- ANS
The maximum deflection occurs at Lx = : 323
max, )(4])(6)(2[
hL
EtP
hL
hL
hL
EtPw S =+=
--- ANS
(b) Timoshenko beam theory
The displacement equilibrium equations for Timoshenko beam
theory are:
0)( 022
=+ yyy dxdwkGA
dxd
EI (4.10.4)
0)( 20
2
=++ zy pdxd
dxwdkGA
(4.10.5)
and can be combined into the following equation,
2
2
40
4
dxpd
GAEI
pdxwdEI zyzy = (4.10.6)
In this particular problem, we have 3121 thI y = , and 0=zp .
Hence we have
040
4
=dxwdEI y as the governing equation.
The concentrated loading at the free end produces a constant
shear force along the
beam, so we have
Pforceshear)dxdw
(kGA y0 ==+ (4.10.7)
Substituting (4.10.7) in (4.10.4) yields
4.10.2
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Mechanics of Aircraft structures C.T. Sun
Pdxd
EI yy =22
(4.10.8)
Integrating equation (4.10.8) twice, we obtain
102
21 BxBPxEI yy ++= (4.10.9)
Using (4.10.7) and (4.10.9), we obtain
)21(1 10
20 BxBPxEIkGA
PkGAP
dxdw
yy ++==
Integrating the equation above,
212
03
0 )21
61(1)( BxBxBPx
EIx
kGAPxw
y
+++= (4.10.10)
The following boundary conditions are used to determine the
arbitrary constants in
(4.10.10):
0)()( ==== Lxdxd
EILxM yy
=> PLB =0
0)0( ==xy (no rotation of the cross-section) => 01 =B 0)0(0
==xw => 02 =B
Then the deflection equation (4.10.10) becomes
)21
61(1)( 230 PLxPxEI
xkGAPxw
y
=
With 65=k , , thA = 3
121 thI y = , and )1(2 +=
EG , we obtain
])(6)(2[)(5
)1(12)( 230 hx
hL
hx
EtP
hx
EtPxw +=
--- ANS
The maximum deflection occurs at Lx = : 3
max, )(4)(
5)1(12
hL
EtP
hL
EtPw T ++=
--- ANS
(c) The ratio of the maximum deflections of the two solutions
versus L/h
Assume the material to be Aluminum 2024-T3 with GPaE 72= , 33.0=
. For convenience, we let =
hL .
The maximum deflection according to the simple beam theory:
4.10.3
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Mechanics of Aircraft structures C.T. Sun
33max, 055556.0)(72
4 tP
tPw S ==
The maximum deflection according to the Timoshenko beam
theory:
33max, 055556.0044333.0)(72
4)()72(5
)33.01(12 tP
tP
tP
tPw T +=++=
Maximum deflections vs. L/h
0
2
4
6
8
10
12
14
0 1 2 3 4 5 6 7L/h
(w
=
P/t)
SimpleTimoshenko
Define %100(%)max,
max,max, =T
ST
www
Error
Error (%) vs. L/h
0
10
20
30
40
50
60
70
80
90
100
0 1 2 3 4 5 6L/h
Erro
r (%
)
7
4.10.4
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Mechanics of Aircraft structures C.T. Sun
4.11 A thin-walled beam of length 2 m long with one end built
into a rigid wall and
the other end is subjected to a shear force NVz 5000= . The
cross-section is given by Fig. 4.21 with mh 2.0= and the wall
thickness . The material is aluminum 2024-T3 with
m002.0=GPaE 70= , GPaG 27= , and the
cross-sectional area of each stringer is . Assume that thin
walls carry
only shear stresses. Find the deflections at the free end using
the simple beam
theory and the Timoshenko beam theory, respectively. Compare the
transverse
shear stress in the vertical web obtained from the two
theories.
225cm
Figure 4.21 Stringer-web section
Solution:
(a) Simple beam theory
(1) The displacement equilibrium equation for the simple beam
theory is:
040
4
=dx
wdEI y (4.11.1)
Integrate the equation (4.11.1) and apply shear force boundary
condition to yield,
zy VCdxwdEI == 030
3
(shear force)
Integrate again to obtain
120
2
CxVdx
wdEI zy += ,
At the free end, , Lx = 1202
0)( CLVLxdx
wdEIM zy +====
=> LVC z=14.11.1
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Mechanics of Aircraft structures C.T. Sun
Again by integration, we have
220
21 CLxVxV
dxdwEI zzy ++= .
At the fixed end, , the rotation of the cross-section vanishes,
i.e., 0=x00 ==
dxdw
y => 02 =C Thus, we have, after integration,
323
0 21
61)( CLxVxVxwEI zzy ++=
In which the integration constant C3 is determined by the
boundary condition
00 =w at => 0=x 03 =C . The deflections curve is
)21
61(1)( 230 LxVxVEI
xw zzy
+= (4.11.2)
(2) Properties of the cross-section 442422 104)2.0)(1025(44
mAhzAI
iiiy
==== GPaE 70=
mL 2= NVz 5000=
(3) Deflections
Compute deflection curve (4.11.2):
)(107857.1109762.2
))2)(5000(21)5000(
61(
)104)(1070(1)(
2435
23490
mxx
xxxw
+=+=
Deflection at the free end:
mmm
mxw
48.010762.4
)2(107857.1)2(109762.2)2(4
24350
=
+==
--- ANS
(b) Timoshenko beam theory
(1) The displacement equilibrium equations for the Timoshenko
beam theory
are:
4.11.2
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Mechanics of Aircraft structures C.T. Sun
0)( 022
=+ yyy dxdwGA
dxd
EI (4.11.3)
0)( 20
2
=++ zy pdxd
dxwdGA
(4.11.4)
which can be combined into the following equation,
2
2
40
4
dxpd
GAEI
pdx
wdEI zyzy = (4.11.5) In the equations above, the area A in the
GA term is the effective area of the
thin-walled section that carries shear stress and should not be
confused with the
stringer cross-sectional area.
Since 0=zp we have
040
4
=dx
wdEI y as the governing equation.
The concentrated shear loading at the free end produces a
constant shear force
along the beam; so we have
zy VdxdwGA =+ )( 0 (4.11.6)
Substitution of (4.11.6) in (4.11.3) yields
zy
y Vdxd
EI =22
(4.11.7)
Integrating (4.11.7), we obtain
102
21 BxBxVEI zyy ++= (4.11.8)
Using (4.11.6) and (4.11.8), we have
)21(1 10
20 BxBxVEIGA
VGAV
dxdw
zy
zy
z ++==
Integrating the above equation with the result,
212
03
0 )21
61(1)( BxBxBxV
EIx
GAVxw z
y
z +++= (4.11.9)
Applying boundary conditions to equation (4.11.8) and (4.11.9),
we have
0)()( ==== Lxdx
dEILxM yy
=> LVB z=0
0)0( ==xy => 01 =B
4.11.3
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Mechanics of Aircraft structures C.T. Sun
0)0(0 ==xw => 02 =B Then equation (4.11.9) becomes
)21
61()( 230 LxxEI
VxGAVxw
y
zz = (4.11.10)
(2) Properties of the cross-section
In the Timoshenko beam theory the area A (in the GA term) of
the
thin-walled cross-section is 24108)002.0)(2.0(22 mhtAshear
=== 442422 104)2.0)(1025(44 mAhzAI
iiiy
==== GPaE 70= , GPaG 27=
mL 2= NVz 5000=
(3) Deflection
Compute the deflection curve (4.11.10) using the above
properties:
)(109762.2107857.1103148.2
)107857.1109762.2()108)(1027(
5000)(
35244
2435490
mxxx
xxxxw
+==
Deflection at the free end is
mmm
mxw
94.010391.9
)2(109762.2)2(107857.1)2(103148.2)2(4
352440
=
+==
--- ANS
The difference between the two theories is
%3.49%100391.9
762.4391.9w
ww(%)Error
Tim,0
Sim,0Tim,0 ===
(c) Transverse Shear Stress
(1) Simple beam theory
From the derivation of the simple beam theory, we assume 0=xy as
an approximation. As a result, the transverse shear stress can not
be directly
obtained from the stress-strain relations. It is obtained
usually from the
4.11.4
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Mechanics of Aircraft structures C.T. Sun
equilibrium equation. We have
MPamNAVshear
zxz 25.6/1025.6108
5000 264 ====
(2) Timoshenko beam theory
MPaAVGshear
zxzxz 25.6===
--- ANS
4.11.5
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Mechanics of Aircraft structures C.T. Sun
4.12 A 2024-T3 aluminum box beam with a thin-walled section is
shown in Fig.
4.26. Assume that thin walls (thickness t = 0.3 cm) are
ineffective in bending.
The cross-sectional area of each stringer is 20 cm2. Find the
deflections at the
free end using the simple beam theory for shear loads and
separately. Solve the same problem using Timoshenko beam
theory. In which loading case is the simple beam theory more
accurate in
predicting the deflection? Explain.
NVz 5000=NVy 5000=
Figure 4.26 Box beam with a triangular thin-walled section
Solution:
(a) First, we need to know the centroid of this section.
Take stringer 2 as the origin of a coordinate system. Then the
centroid is located at
cmA
yAy
ii
iii
c 202036020 =
==
cmA
zAz
ii
iii
c 20203)2040(20 =
+==
The moments of inertia with respect to the coordinate system
with the origin at the
centroid are 4222 16000))20(20(20 cmzAI
iciy =+==
4222 48000))20(240(20 cmyAIi
ciz =+== 4.12.1
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Mechanics of Aircraft structures C.T. Sun
0)]20()20(20)20(040[4 =++== i
cciyz zyAI (This should be
obvious because the section is symmetric with respect to
y-axis)
For 2024-T3, , 25 /107272 cmNGPaE == 33.0= => 25 /10068.27
cmNG =
(b) Simple beam theory
The displacement equilibrium equations for the simple beam
theory are:
040
4
=dxwdEI y , for loading (4.12.1) zV
040
4
=dxvdEI z , for loading (4.12.2) yV
Integrating the above equations, we get
zy VdxwdEI =30
3
(4.12.3)
yz VdxvdEI =30
3
(4.12.4)
Thus,
)/1(103403.4160001072
5000 2853
03
cmEIV
dxwd
y
z ===
)/1(104468.1480001072
5000 2853
03
cmEIV
dxvd
z
y ===
Integrating the above equations, we have
322
139
0 2110234.7)( CxCxCxxw +++=
652
439
0 2110411.2)( CxCxCxxv +++=
The arbitrary constants are determined by the boundary
conditions,
For )(0 xw
0)0(0 ==xw , => 03 =C
0)0(0 ==xdxdw
=> 02 =C
4.12.2
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Mechanics of Aircraft structures C.T. Sun
0)()(20
2
==== LxMLxdxwdEI y
=> 0)160001072()200(5000)200( 15
20
2
=+== CcmxdxwdEI y
=> 61 10681.8=C
So, (4.12.5) 26390 10340.410234.7)( xxxw +=
For )(0 xv
0)0(0 ==xv , => 06 =C
0)0(0 ==xdxdv
=> 05 =C
0)()(20
2
==== LxMLxdxvdEI z
=> 0)480001072()200(5000)200( 45
20
2
=+== CcmxdxvdEI z
=> 64 10894.2=C
So, (4.12.6) 26390 10447.110411.2)( xxxv +=
---
Therefore deflections at the free end can be obtained from
(4.12.5) and (4.12.6) by
setting : cmx 200=
cmcmxw
116.0)200(10340.4)200(10234.7)200( 26390
=+==
cmcmxv
039.0)200(10447.1)200(10411.2)200( 26390
=+==
--- ANS
(c) Timoshenko beam theory
The displacement equilibrium equations for Timoshenko beam
theory for
loading are:
zV
0)( 022
=+ yzyy dxdwGA
dxd
EI (4.12.7)
0)( 20
2
=++ zyz pdxd
dxwd
GA
(4.12.8)
4.12.3
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Mechanics of Aircraft structures C.T. Sun
which can be combined into the following equation,
2
2
40
4
dxpd
GAEI
pdxwdEI z
z
yzy = (4.12.9)
Note that is the projection of the cross-sectional area of the
thin sheets onto
z-axis. In this case, .
zA2243.0402 cmAz ==
In this particular problem, we have 0=zp . Hence
040
4
=dxwdEI y
is the governing equation.
The concentrated shear loading at the free end produces a
constant shear force
along the beam, so we have
zyz VdxdwGA =+ )( 0 (4.12.10)
Substituting the above in equation (4.12.7) yields
zy
y Vdxd
EI =22
(4.12.11)
Integrating equation (4.12.11), we obtain
102
21 BxBxVEI zyy ++= (4.12.12)
Using equation (4.11.10) and (4.11.12), we have
)21(1 10
20 BxBxVEIGA
VGAV
dxdw
zyz
zy
z
z ++==
Integrating the above equation,
212
03
0 )21
61(1)( BxBxBxV
EIx
GAVxw z
yz
z +++= (4.12.13)
Applying boundary conditions to equation (4.12.13)
0)()( ==== Lxdxd
EILxM yy
=> LVB z=0
0)0( ==xy => 01 =B 0)0(0 ==xw => 02 =B
Then equation (4.12.13) becomes
)21
61()( 230 LxxEI
VxGAVxw
y
z
z
z = (4.12.14)
4.12.4
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Mechanics of Aircraft structures C.T. Sun
Similarly, the deflection in y-direction due to is yV
)21
61()( 230 LxxEI
Vx
GAV
xvz
y
y
y = (4.12.15)
where is the projection of the cross-sectional area of the thin
sheets onto y-axis.
We have .
yA
2363.0602 cmAy ==
The deflection due to is zV
)10340.410234.7(10697.7
))200(21
61(
)16000)(1072(5000
)24)(10068.27(5000)(
26395
23550
xxx
xxxxw
==
And for : yV
)10447.110411.2(10131.5
))200(21
61(
)48000)(1072(5000
)36)(10068.27(5000)(
26395
23550
xxx
xxxxv
===
---
At the free end the respective deflection can be obtained from
(4.12.5) and
(4.12.6) by substituting in cmx 200=
cmcmxw
131.0])200(10340.4)200(10234.7[)200(10697.7)200( 263950
===
cmcmxv
049.0])200(10447.1)200(10411.2[)200(10131.5)200( 263950
===
--- ANS
(d) Summary
(1) Deflections at the free end
Simple Beam Theory Timoshenko
Beam Theory Error (%)
(1) NVz 5000= 0.116 cm 0.131 cm 11.5 (2) NVy 5000= 0.039 cm
0.049 cm 20.4
Timoshenko
SimpleTimoshenko
ddd
Error=(%) , where 00 vorwd =
4.12.5
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Mechanics of Aircraft structures C.T. Sun
(2) Case 1 ( ) of the above results is more accurate. It is
mainly
because of the reason that is smaller than , and, as a result,
the bending
behavior for z-direction is more likely to resemble a slender
beam than that
for y-direction.
NVz 5000=yI zI
--- ANS
4.12.6
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Mechanics of Aircraft structures C.T. Sun
4.13 Consider the structure with a cutout as shown in Fig. 4.17.
Find the axial force
distribution in stringers 3-4 and 5-6. Assume that both
stringers and webs have
the same material properties of GPaE 70= and GPaG 27= . Also
assume that , the thickness of the web mmb 200= mmt 2= , and the
cross-sectional area of the stringer . Hint: The zero-stress
condition in the web at
the cutout cannot be enforced because of the simplified
assumption that shear
stress and strain are uniform across the width of the web. Use
the known
condition that the force in the side stringers is at the
cutout.
264mmA =
P5.1
1 2
3 4 5 6
P
P
P
L1 L2
Figure 4.17 Cutout in a stringer sheet panel
Solution:
(a) First, we consider the part left hand side of the
cutout.
3 4
1.5P
1.5P
L1
F1
F2
F1
b
x
The balance of forces in the x-direction yields
PFF 32 21 =+ (4.13.1) Also we have the differential equation
4.13.1
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Mechanics of Aircraft structures C.T. Sun
dxdF
t11= (4.13.2)
)()2/( 2
2
1
1
AF
AF
bEG
dxd = (4.13.3)
Combining equations (4.13.2) and (4.13.3), we have
)(6)()2/(
11
2
2
1
12
12
PFEAbG
AF
AF
bEG
dxFd
t==
=> )(6 121
2
PFEAbGt
dxFd = , let
EAbGt62 =
=> PFdxFd 2
12
21
2
= (4.13.4) The general solution of this second-order
differential equation is
PxCxCxF ++= sinhcosh)( 211
(where 2
coshxx eex
+= and 2
sinhxx eex
= ) Applying the boundary conditions,
At (fixed end) => 0=x 0= , that is 0)0(1 ==xdxdF
=> 0)0( 21 === CxdxdF => 02 =C
At => 1Lx = PLxF 5.1)( 11 == => PPLCLxF 5.1cosh)( 1111
=+== =>
11 cosh2 L
PC =
Therefore the solution of the differential equation is
)1cosh2cosh()(
11 += L
xPxF (4.13.5)
--- The axial force distribution in stringers 3-4 can be
obtained from (4.13.1) and (4.13.5),
that is
)coshcosh1()(23)(
112 L
xPxFPxF ==
with 016.19)2.0)(1064)(1070(
)102)(1027(6669
39
===
EAbGt )1( m
4.13.2
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Mechanics of Aircraft structures C.T. Sun
--- ANS
(b) Next, we consider the part to the right of the cutout:
5
F1
F2
F1x
1.5P
1.5P
From the balance of forces in the x-direction we have
PFF 32 21 =+ (4.13.6) Also we have the differential equation
dxdF
t11= (4.13.7)
)()2/( 2
2
1
1
AF
AF
bEG
dxd = (4.13.8)
Combining equations (4.13.7) and (4.13.8), we have
)(6)()2/(
11
2
2
1
12
12
PFEAbG
AF
AF
bEG
dxFd
t==
=> )(6 121
2
PFEAbGt
dxFd = , let
EAbGt62 =
=> PFdxFd 2
12
21
2
= (4.13.9) The general solution of this second-order
differential equation is
PxCxCxF ++= sinhcosh)( 211 Applying the boundary conditions,
at => 0=x PxF 5.1)0(1 == => PPCxF 5.1)0( 11 =+== => PC
5.01 =
at => 2Lx = PLxF == )( 21 => PPLCLPxF =++= 2221
sinhcosh5.0)(
4.13.3
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Mechanics of Aircraft structures C.T. Sun
=> 2
2 tanh2 LPC =
The solution of the differential equation is
)2tanhsinh(cosh
2)(
21 += L
xxPxF (4.13.10)
--- The axial force distribution in stringer 5-6 can be obtained
from (4.13.6) and (4.13.10),
that is
)tanhsinhcosh1()(23)(
212 L
xxPxFPxF +==
where 016.19)2.0)(1064)(1070(
)102)(1027(6669
39
===
EAbGt )1( m
--- ANS
4.13.4