Solution CH3 (1)
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3.1 Show that there is no warping in a bar of circular cross-section.
Solution:
(a)Saint-Venant assumed that as the shaft twists the plane cross-sections are warped
but the projections on thex-yplane rotate as a rigid body, then,
zyu =
zxv = (3.1.1)
),( yxw =
where ),( yx is some function of x and y, called warping function, and is
the angle of twist per unit length of the shaft and is assumed to be very small.
(b)From the displacement field above, it is easy to obtain that
0==== xyzzyyxx
So from the stress-strain relationship, we have
0==== xyzzyyxx
Therefore the equilibrium equations reduce to
0=
+
yx
yzxz
This equation is identically satisfied if the stresses are derived from a stress
function ),( yx , so that
yxz
=
,x
yz
=
(3.1.2)
(c)From the displacement field and stress-strain relationship, we can obtain
yx
w
z
u
x
wxz
=
+
= (3.1.3)
xy
w
z
v
y
wyz
+
=
+
= (3.1.4)
So it forms the compatibility equation
2=
yx
xzyz
,
or in terms of Prandtl stress function
Gyx
22
2
2
2
=
+
(3.1.5)
(d)Boundary conditions,
0=ds
d, or .const= But for a solid sections with a single contour boundary,
this constant can be chosen to be zero. Then we have the boundary condition
0= on the lateral surface of the bar.
(e)
For a bar with circular cross-section, assume the Prandtl stress function as
3.1.1
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)1(2
2
2
2
+=a
y
a
xC which satisfies the boundary conditions stated above.
Substitute into (3.1.5), we obtain GaC 2
2
1=
Then )(2
222ayx
G+=
Using (3.1.2), we have
yyG
xz
=
=
1, and x
xGyz
=
=
1
Comparing with (3.1.3) and (3.1.4), we have
yyx
wxz =
= => 0=
x
w. Thus, )(yfw =
xxyw
yz =+
= => 0=
yw , Thus, )(xgw =
Hence we conclude . This means that the cross-section remains plane
after torsion. In other words, there is no warping.
constw =
Therefore can be verified, and it successfully expresses the
statement.
0),( =yxw
--- ANS
3.1.2
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3.2 Show that the Prandtl stress function for bars of circular solid sections is also
valid for bars of hollow circular sections as shown in Fig. 3.34. Find the torsion
constant in terms of the inner radius and outer radius , and compare
with the torsion constant obtained using (3.59) for thin-walled sections. What is
the condition on the wall thickness for the approximate to be within 1
percent of the exact ?
J ia 0a
J
J
ia
0a
Figure 3.34 Bar of a hollow circular section
Solution:
Recall:
(a)Saint-Venant assumed that as the shaft twists the plane cross-sections are warped
but the projections on the x-y plane rotate as a rigid body, then,zyu =
zxv = (3.2.1)
),( yxw =
where ),( yx is a function of x and y, called warping function, and is the
angle of twist per unit length of the shaft and is assumed to be very small.
(b)From the displacement field above, it is easy to obtain that
0==== xyzzyyxx
So from the stress-strain relationship, we have
0==== xyzzyyxx
Therefore the equilibrium equations reduce to
0=
+
yx
yzxz
This equation is identically satisfied if the stresses are derived from a stress
function ),( yx , so that
3.2.1
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yxz
=
,x
yz
=
(3.2.2)
(c)From the displacement field and stress-strain relationship, we can obtain
yx
w
z
u
x
wxz
=
+
= (3.2.3)
xy
w
z
v
y
wyz
+
=
+
= (3.2.4)
So it forms the compatibility equation
2=
yx
xzyz ,
or in terms of Prandtl stress function
Gyx
22
2
2
2
=
+
(3.2.5)
(d)Boundary conditions,
0=dsd , or .const=
---
1.
To show that the Prandtl stress function for bars of circular solid sections is also
valid for bars of hollow circular sections, we have to show that the Prandtl stress
function for hollow circular sections satifies equilibrium equations, compatibility
equations as well as traction boundary conditions.(1)Equilibrium equations
Prandtl stress functions by their definition must satify equilibrium
equations..
(2)Compatibility equations
Use the Prandtl stress function as it stated for bars of circular solid sections
)1(2
0
2
2
0
2
+=a
y
a
xC (here we use . Assuming0a )1( 2
2
2
2
+=ii a
y
a
xC
would be fine too).
Then substitute into (3.2.5), we have GaC2
02
1= . Thus we have
)1(2 2
0
2
2
0
22
0 +=a
y
a
xaG . (3.2.6)
Therefore we have a stress function for bars of hollow circular sections
satisfying the compatibility equation
(3)
Traction boundary conditions
3.2.2
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Substituting and into the above error equation, we haveappJ J
01.0)(2
)(
)(2
)(2
)()((4
220
20
44
0
44
003
0
+
=
+=
i
i
i
iiiapp
aa
aa
aa
aaaaaa
J
JJ
Because and are positive real number, we haveia 0a
01.0)(2
)(22
0
20
+
i
i
aa
aa => 01)(040816.2)(
0
2
0
+a
a
a
a ii
We can obtain the solution of the above equation as
2235.18174.00
a
ai
Since we have the solutioniaa >0 8174.00
a
ai
Therefore the condition on the wall thickness t is
0000 1826.08174.0 aaaaat i ==
(OR iiii aaaaat 2235.08174.0
10 == )
--- ANS
3.2.4
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3.3 Consider the straight bar of a uniform elliptical cross-section. The semimajor
and semiminor axes are aand b, respectively. Show that the stress function of
the form
)1( 2
2
2
2
+= b
y
a
x
C
provides the solution for torsion of the bar.
Find the expression of Cand show that
22
33
ba
baJ
+=
3
2
ab
Tyzx
= ,
ba
Txzy 3
2
=
and the warping displacement
xyGba
abTw33
22 )(
=
Solution:
Recall:
1.
Saint-Venant assumed that as the shaft twists the plane cross-sections are warped
but the projections on the x-y plane rotate as a rigid body, then,
zyu =
zxv = (3.3.1)),( yxw =
where ),( yx is warping function, and is the angle of twist per unit length
of the shaft and is assumed to be very small.
2. From the displacement field above, it is easy to obtain that
0==== xyzzyyxx
From the stress-strain relationship, we have
0==== xyzzyyxx
Therefore the equilibrium equations reduce to
0=
+
yx
yzxz
which is identically satisfied if the stresses are derived from a stress function
),( yx , so that
yxz
=
,x
yz
=
(3.3.2)
3.
From the displacement field and stress-strain relationship, we can obtain
3.3.1
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yx
w
z
u
x
wxz
=
+
= (3.3.3)
xy
w
z
v
y
wyz
+
=
+
= (3.3.4)
The compatibility equation becomes
2=
yx
xzyz ,
or in terms of Prandtl stress function
Gyx
22
2
2
2
=
+
(3.3.5)
4. The boundary condition along the bounding surface is
0=ds
d, or .const=
---
(a)Let the stress function be of the form )1(2
2
2
2
+=b
y
a
xC . In order to show this
stress function provides the solution for torsion of the bar, we have to show that
this stress function satisfies the equilibrium equations, compatibility equations and
traction boundary conditions.
(1)Equilibrium equations
)2( 2byC
yxz ==
, )2( 2axC
xyz ==
Substituting the above stress expressions into the equilibrium equations, we
have
000 =+=
+
yx
yzxz
(2)Compatibility equations
Substituting )1(2
2
2
2
+=b
y
a
xC into (3.3.5) we get
22
22
ba
baGC
+= . (3.3.6)
Therefore we have a stress function satisfying compatibility equation
(3)Traction boundary conditions
To satisfy the traction boundary condition we must show 0=ds
d on the
traction free lateral surface.
Since the boundary of the cross section is given by the equation
3.3.2
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012
2
2
2
=+b
y
a
x,
it is easy to see that 0)1(2
2
2
2
=+=b
y
a
xC on the free surface and therefore
it satisfies the required condition 0=ds
d
Since equilibrium equations, compatibility equations and traction boundary
conditions are all satisfied, the stated stress function provides the solution
for torsion of the bar.
--- ANS
(b)
Torsion constant J
(1)We have the torque produced by the stresses is
=
A
dAy
yx
xT )( (3.3.7)
Substituting )1(2
2
2
2
+=b
y
a
xC into (3.3.7), then we have,
+==AA
dAb
y
a
xCdA
b
yCy
a
xCxT )())
2()
2((
2
2
2
2
22
Note that the integral part of the above equation is the area of the elliptical
cross-section. It can be easily obtained that abdA
b
y
a
x
A
=+ )( 22
2
2
So we have the torsion abCT =
By substituting C and utilizing GJT= , we have
22
3322
22
)(
ba
ba
G
abba
baG
G
abCJ
+=+
=
=
(3.3.8)
--- ANS
(2)322
2)
2()
2(
ab
Ty
b
y
ab
T
b
yC
yxz
=
==
= , (3.3.9)
andba
Tx
a
xC
xyz 32
2)
2(
==
= (3.3.10)
--- ANS
(c)The warping displacement can be derived from (3.3.3), (3.3.4), (3.3.9), (3.3.10)
From (3.3.9) and (3.3.10), we have3
2
abG
Tyxz
= and
baG
Txyz 3
2
= .
Also we need to know33
22 )(
baG
baT
GJ
T
+==
So from (3.3.3) and (3.3.4), we can rewrite in
3.3.3
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33
22
33
22
3
)()(2
baG
yabT
baG
ybaT
abG
Tyy
x
wxz
=
++=+=
(3.3.11)
33
22
33
22
3
)()(2
baG
xabT
baG
xbaT
baG
Txx
y
wyz
=
+==
(3.3.12)
From (3.3.11), we can obtain
)()(
),(33
22
yfxybaG
abTyxw +
=
(3.3.13)
Then differentiating (3.3.13) with respect to y, we have
)()(),(
33
22
yfxbaG
abT
y
yxw+
=
.
Comparing this equation with (3.3.12) we have 0)( = yf , that is .)( constyf =
For a symmetric cross-section 0)0,0( =w , that is, .0)( =yf Thus, the warping displacement is
.)(
),(33
22
xybaG
abTyxw
=
--- ANS
And it is easy to also find that the warping function
xyba
ba
baG
baT
xybaGabT
yxwyx
22
22
33
22
33
22
)(
.)(),(
),(+
=
+
==
3.3.4
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3.4 A thin aluminum sheet is to be used to form a closed thin-walled section. If the
total length of the wall contour is 100 cm, what is the shape that would achieve
the highest torsional rigidity? Consider elliptical (including circular), rectangular,
and equilateral triangular shapes.
Solution:
(a)We denote as torsional rigidity, for the same material in comparison, only
the torsion constant needs to be taken into consideration.
GJ
J
For the closed thin-walled section, the torsion constant isJ
=
tds
AJ
/
42
(3.4.1)
where A is the area enclosed by the centerline of the wall section.We now have a thin aluminum sheet with its thickness , all shapes of products
made from this aluminum sheet will have the same thickness, . Also, the total
length of the wall contour is 100cm. Then
t
t
tds / is the same for all shapes of the
cross-section. Consequently, only A needs to be taken into consideration in the
evaluation of the torsional rigidity.
(b)Comparison ofA
(1)
Elliptical cross-sectionFor the elliptical cross-section, the cross-sectional area is
abAellp = , (3.4.2)
where a and b are the semimajor and semiminor axes, respectively.
Unfortunately, the length of the perimeter of an elliptical cross-section is
much more complicated to evaluate. The formula for the length of the
perimeter can be found from many math handbook It is
=2/
0
22
sin14
dkaL ,
where tyeccentricia
bak =
=
22
For the purpose to just comparing the area enclosed by the centerline of the
wall section, We approximate the perimeter with
22
22ba
L +
= (3.4.3)
By changing the form of (3.4.3) se have
3.4.1
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2222)2
(2 aCaL
b ==
, where 2)2
(2
LC= (3.4.4)
Substituting (3.4.4) into (3.4.2) we have,
22
aCaAellp =
We can find the optimum solution by 0=
a
A, by some operations leads to
02
22
22
=
=
aC
aC
a
A , therefore we have
2
2C
a= for 0, >ba
Substitute it back to (3.4.4), we have aC
b ==2
2
(3.4.5)
That means the optimum cross-section for elliptical shapes is a circle.
Then from (3.4.5) we have 2
)2
(2
2
2
22
LLCba ====
Finally, for a circle, the area enclosed by the centerline is
222cir L0796.0)
2
L(aA ===
--- ANS
(2)Rectangular section
For rectangular section, the perimeter is)(2 qpL += , (3.4.6)
where pand qare length and width, respectively.
The cross-sectional area of rectangular sections is simply,
pqArec = , (3.4.7)
Substituting (3.4.6) into (3.4.7), we have
)2
( pL
ppqArec ==
We use 0=
p
A to find the optimal solution,
022
==
p
L
p
A, we have
4
Lp= , and from (3.4.6), it is clear that
4
Lqp == , i.e., the optimal cross-section for rectangular shapes is a square.
Finally, for a square thin-walled section, the area enclosed by the centerline
is
22
squ L0625.0)4
L
(pqA ===
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--- ANS
(3)Equilateral triangular section.
For a equilateral triangle, the length of the lateral side is3
Ll= .
The area enclosed by the centerline of this triangular thin-walled section is
222tri L048.0)
3
L(
4
3l
4
3A ===
--- ANS
(c)Comparison
From the results above we can easily tell
trisqucir AAA >>
Consequently we can conclude that the shape achieving the highest torsional
rigidity is a CIRCLE.
--- ANS
NOTE: It is interesting to compare in details with variablesa
b and
p
q from 0~1.
(We here assume a>band p>q)
For ellipse, 222222
22
)1
(2
1))
2(2)(( L
b
a
a
b
L
ba
ab
ba
baababAellp
+
=+
+
+==
For rectangle, 2
2
2
22
2
)1(4
1)
2(
)()(
)(L
p
q
p
q
L
qp
pq
qp
qppqpqArec
+
=+
=+
+==
For equilateral triangle, 236
3LAtri =
We can illustrate2
L
A in terms of
a
b and
p
q, and have the plot of torsional rigidity
of different shapes vs. variable aspect ratios.
3.4.3
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3.4.4
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3.5 The two-cell section in Fig.3.35 is obtained from the single-cell section of
Fig.3.36 by adding a vertical web of the same thickness as the skin. Compare the
torsional rigidity of the structures of Figs. 3.35 and 3.36 with
and , , respectively.
cmLL 1021 ==
cmL 51= cmL 152 =
cmt 3.0=
Figure 3.35 Two-cell thin-walled section
Figure 3.36 Single-cell section
Solution:
We denote as torsional rigidity. For the same material in comparison, only the
torsion constant needs to be considered.
GJ
J
(a)Single-cell thin-walled section
The torsion constant isJ
=
tds
AJ
/
42
(3.5.1)
where A is the area enclosed by the centerline of the wall section.
We have2
321 2001020)( cmLLLA ==+= . The torsion constant can be
simply derived as
J
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42
321
2
321
2
1 8003.0)1020(2
)200(4
/)(2
])[(4
/
4cm
tLLL
LLL
tds
AJ
cell =
+=
++
+==
(b)Two-cell thin-walled section
(1)
General Form
We denote the shear flow on the left cell by , and the shear flow on the
right cell by . The shear flow in the vertical web is
1q
2q 2112 qqq =
Also, we have the torque for two-cell section
21 21 22 qAqAT += (3.5.2)
where 311 LLA = , 322 LLA =
The twist angle of the section is obtained from eirher cell. For left cell we
have
))()2((2
1
2
1321311
311
11 LqqLLq
tLGLt
qds
AG cell++== (3.5.3)
and for the right cell
))()2((2
1
2
1321322
322
22 LqqLLq
tLGLt
qds
AG cell+== (3.5.4)
Since the entire thin-wall section must rotate as a rigid body in the plane, we
require the compatibility condition == 21 (3.5.5)
From (3.5.3) to (3.5.5), we derive the relation between and ,1q 2q
1
1
3
2
3
2
3
1
3
2
)22(
)22(
q
L
L
L
L
L
L
L
L
q
++
++
= (3.5.6)
Substituting (3.5.6) into (3.5.2) and usingG
TJ= and (3.5.3), we have
)22(
)(4
))()2((2
1
)22(
323111
23213131
321311
31
2211
LqLqLq
tqLLqLLLL
LqqLLqtLGL
G
qAqAJ
+
+=
++
+=
(3.5.7)
(2)Case 1: andcmLL 1021 == cmL 103 =
From (3.5.6), , then substituting into (3.5.7) we have12 qq =
4
323111
23213131
12 800)22(
)(4cm
LqLqLq
tqLLqLLLLJ
cell =
+
+=
3.5.2
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--- ANS
(3)Case 2: ,cmL 51 = cmL 152 = and cmL 103 =
From (3.5.6), 112 25.1
)5
10
15
1022(
)15
10
5
10
22(qqq =
++
++=
Then substituting into (3.5.7) we have
4
111
1122 2857.814
)1025.110252(
3.0)25.11015105(1054cm
qqq
qqJ cell =
+
+==
--- ANS
(c)Comparison
From the results above we have
4
11222
4 8002857.814 cmJJJcm cellcellcell ==>=
Adding a vertical web does not significantly improve the torsional rigidity.
--- ANS
3.5.3
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3.6 Find the torsional rigidity if the side wall of one of the two cells in Fig. 3.35
(with ) is cut open. What is the reduction of torsional rigidity
compared with the original intact structure?
cmLL 1021 ==
cmt 3.0=
Figure 3.35 Two-cell thin-walled section
Solution:
We denote torsional rigidity by as.GJ
(a)Closed sidewall
From the solution of Problem 3.5, we have the torsion constant of the
case with
12 cellJ
cmLL 1021 == 4
323111
2321313112 800
)22(
)(4cm
LqLqLq
tqLLqLLLLJ cell =
+
+=
So we have the original torsional rigidity GGJ cell 80012 = (3.6.1)
(b)With one side wall cut open
Assuming that the cell is cut open as shown in the figure, the torsional rigidity can
be derived from
cutcellcutnotcellopencut GJGJGJ += (3.6.2)
(1) + (2)
Where
3.6.1
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=tds
AJ
cutnot
cutnotcell
/
)(4 2, and 32LLA cutnot = (3.6.3)
=>
42
300)1010(2
3.0)1010(4
cmJ cutnotcell =+
=
and
=i
iicutcell tbJ3
3
1 (3.6.4)
=> 43 27.03.0)101010(3
1cmJ cutcell =++=
So, from (3.6.2) we get
GGJGJGJ cutcellcutnotcellopencut 27.300=+=
---ANS
(c)The reduction of torsional rigidity is obtained as
%5.62625.0800
27.300800
GJ
GJGJR
1cell2
opencut1cell2 ==
=
=
--- ANS
3.6.2
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3.7 Find the torque capability of the thin-walled bar with the section shown in Fig.
3.36. Assume that the shear modulus GPaG 27= and the allowable shear
stress of MPaallow 187= .
cmt 3.0= Figure 3.36 Single thin-walled section
Solution:
Since the thickness of all walls are equal to cmt 3.0= , we can obtain the allowable
shear flow from allowable shear stress, that is
m/N1061.5003.010187tq 56allowallow ===
Then we have the torque capability as
mN224401061.5)2.01.0(2qA2T 5allowallow ===
--- ANS
3.7.1
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3.8 A two-cell thin-walled member with the cross-section shown in Fig. 3.37 is
subjected to a torque T. The resulting twist angle is . Find the shear
flows of the applied torque, and the torsion constant. The material is aluminum
alloy 2024-T3.
m/3o
Figure 3.37 Two-cell section
Solution:
(a)Assume the material is linearly elastic under the twist angle . For aluminum
alloy 2024-T3, we have the shear modulus
GPa27)33.01(2
72
)1(2
EG =
+=
+=
(b)
We denote the shear flow on the left cell , and the shear flow on the right cell. The shear flow in the vertical web is
1q2q 2112 qqq = , are the positive directions
as shown in the figure above.
Also, we have the torque for two-cell sections
21 21 22 qAqAT += (3.8.1)
where 222
21 m098.08
)5.0(
8
dAA ====
,
The twist angle of the left cell is
))((2
1
2
121
12
121
1
1
11
11 qq
t
sq
t
s
AGt
qds
AG cell+== (3.8.2)
where m785.02
ds1 ==
is the length of the left side wall, and is
the length of the vertical web.
ms 5.012 =
The twist angle of the right cell is
))((2
1
2
121
12
122
2
2
22
22 qq
t
sq
t
s
AGt
qds
AG cell== (3.8.3)
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Again, we have m785.02
ds2 ==
, the length of the right side wall.
Since the entire thin-wall section must rotate as a rigid body in the plane, we
require the compatibility condition
m/rad0524.0m/321 ==== o
(3.8.4)
From (3.8.2) to (3.8.4) and noting that 21 AA = , we derive the relation between
and by substituting all the known quantities,1q 2q
122121002.0
5.0
003.0
785398.0
002.0
5.0
001.0
785398.0qqqq =+
Substituting , in the equation above, we obtain2112 qqq =
12 q687.1q = (3.8.5)
Back substituting into (3.8.2) and (3.8.4), we have
191 q
)098175.0)(100677.27(2
))169732.1(250398.785(m/rad0524.0
==
From which we obtain
m/N500,453q1=
Subsequently from (3.8.5) we obtain
m/N765000q687.1q 12 ==
--- ANS
(c)
The applied torqueFrom (3.8.1), we compute the applied torque
mN10393.2mN239300
)765000453500)(098.0(2qA2qA2T
5
2211
==
+=+=
--- ANS
(d)The torsion constant J
From the fundamental relationship of torque and twist angle, we have GJT=
So the torsion constant can be derived as
44
9 m1069.1
)0524.0)(1027(
239300
G
TJ
=
==
--- ANS
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3.9 For the bar of Fig. 3.37, find the maximum torque if the allowable shear stress is
MPaallow 187= . What is the corresponding maximum twist angle ?
Figure 3.37 Two-cell section
Solution:
(a)Assume the material is linearly elastic under the twist angle . For aluminum
alloy 2024-T3, we have the shear modulus
GPa27)33.01(2
72
)1(2
EG =
+=
+=
(b)
We denote the shear flow on the left cell as and that on the right cell as .
The shear flow in the vertical web is
1q 2q
2112 qqq = . The positive directions for the
shear flows are shown in the figure above.
The torque for two-cell section is
21 21 22 qAqAT += (3.9.1)
where 222
21 m098.08
)5.0(
8
dAA ====
,
The twist angle of the left cell is
))((2
1
2
121
12
121
1
1
11
11 qq
t
sq
t
s
AGt
qds
AG cell+== (3.9.2)
where m785.02
ds1 ==
is the length of the left side wall, and is
the length of the vertical web.
ms 5.012 =
Also we have the twist angle of the right cell as
))((2
1
2
121
122
2
2
22
22 qq
t
sq
t
s
AGt
qds
AG cell== (3.9.3)
where m785.02
ds2 ==
is the length of the right side wall.
(c)Since the entire thin-wall section must rotate as a rigid body in the plane, we
require the compatibility condition
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== 21 (3.9.4)
From (3.9.2) to (3.9.4) and note that 21 AA = , we derive the relation between
and by substituting all the known quantities,
1q
2q
122121002.0
5.0
003.0
785398.0
002.0
5.0
001.0
785398.0qqqq =+
In view of the relation 2112 qqq = we obtain
12 q687.1q = (3.9.5)
Back substituting (3.9.5) into (3.9.2) and (3.9.4), we have
19 )098175.0)(100677.27(2
))169732.1(250398.785(q
= , and then
8662000q1=
Subsequently, (3.9.6)000,600,14q687.1q 12 == (3.9.7)
000,953,5qqq 2112 == (3.9.8)
Note the units are m/radin , and are in N/m.1221 q,q,q
(d)Stress in the wall
From the above quantities of shear flow, we can then compute the shear stress in
each wall byt
q= . We have
9
1
11 1066.8
001.08662028
tq === (3.9.9)
9
2
22 1087.4
003.0
14615612
t
q=== (3.9.10)
9
12
1212 1098.2
002.0
5953584
t
q=
== (3.9.11)
(e)From the above stresses (3.9.9) to (3.9.11), because the negative value just denote
the negative direction, the maximum absolute magnitude of shear stress is6
allow9
1 101871066.8 ==
Therefore the maximum twist angle ism/24.1m/rad0216.0max
o==
--- ANS
(f)
The maximum torque can be solved by using (3.9.1), (3.9.6), (3.9.7) and the
maximum twist angle, that is
mN98700
)0216.0)(146000008662000)(098.0(2qA2qA2T 2211
=
+=+=
--- ANS
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3.10 Find the shear flow and twist angle in the two-cell three-stringer thin-walled
bar with the cross-section shown in Fig. 3.38. The material is Al2024-T3. The
applied torque is .mN 5102
Figure 3.38 Two-cell three-stringer thin-walled section
Solution:
(a)Assume the material is linearly elastic under the applied torque. For aluminum
alloy 2024-T3, we have the shear modulus
GPaE
G 27)33.01(2
72
)1(2=
+=
+=
(b)
Denote the shear flow on the left cell as , and the shear flow on the right cell as; both are considered positive if counterclockwise. The shear flow in the
vertical web is , which is positive if it is in the same direction as .
1q2q
2112 qqq = 1q
We have the torque for the two-cell section as
21 21 22 qAqAT += (3.10.1)
where 222
1 565.08
)2.1(
8m
dA ===
,
and2
2 2.12
)2.1(2
2 mbh
A ===
The twist angle of the left. cell is
))((2
1
2
121
1
12
1
1
1
11
1
1 qqt
sq
t
s
AGt
qds
AG cell+== (3.10.2)
where md
s 88.12
1 ==
is the length of the left half circular wall, and
is the length of the vertical web.ms 2.112 =
The twist angle of the right. cell is
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))((2
1
2
121
1
122
3
32
2
2
22
22 qq
t
sq
t
sq
t
s
AGt
qds
AG cell+== (3.10.3)
Again, we have , the length of the lower wall,ms 22=
and ms 33.22.1222
3 =+= , the length of the inclined wall of thickness t3 in
the right cell.
Since the entire thin-wall section must rotate as a rigid body in the plane, we
require the compatibility condition
== 21 (3.10.4)
Solving the two equations, (3.10.2) and (3.10.4), we obtain
)005.0
2.1
007.0
332.2
007.0
2(
2.1
1)
005.0
2.1
005.0
885.1(
565.0
11222121 qqqqq +=+
Eliminating from the equation above using12q 2112 qqq = we obtain
12 132.1 qq = (3.10.5)
(c)To find the shear flow , we back substitute (3.10.5) into (3.10.1) and have
121 2121 )265.22(22 qAAqAqAT +=+=
=> mNAA
Tq /51966
)2.1)(265.2()565.0)(2(
102
265.22
5
211 =
+
=
+=
--- ANS
From (3.10.5),mNqq /58844132.1 12 ==
--- ANS
(d)For the twist angle, we can utilize the shear flows and equations (3.10.2) and
(3.10.4) to get,
mmrad
qqt
sq
t
s
AG
/0336.0/1086.5
)565.0)(1027(2
51966))132.11(005.0
2.1
005.0
885.1(
))((2
1
4
921
1
121
1
1
11
o==
+=+==
--- ANS
3.10.2
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where is the length of the lower straight wall of thickness tm2s2 = 2, and
m33.22.12s 22
3 =+= is the length of the inclined wall of thickness t3.
Since the entire thin-wall section must rotate as a rigid body in the plane, we
require the compatibility condition
== 21 (3.11.4)
From (3.10.2) to (3.10.4), we can derive the relation between and by
substituting all the known quantities,
1q 2q
)q005.0
2.1q
007.0
33.2q
007.0
2(
2.1
1)q
005.0
2.1q
005.0
88.1(
566.0
11222121 +=+
Substituting into the above equation, we obtain2112 qqq =
12 q13.1q = (3.11.5)
(c)
Since we have the condition , and, thus,m/rad035.0m/2allowable == o
m/rad035.0allowable21 =
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3.12 The two shafts of thin-walled cross-sections shown in Fig. 3.39a and b,
respectively. Contain the same amount of aluminum alloy. Compare the
torsional rigidities of the two shafts without end constraints.
(a) (b)
Figure 3.39 Cross-sections of two shafts
Solution:
(a)
Fig. 3.39a is a cross-section of an open thin-wall, its torsional rigidity isa
GJ
43
i
3
iia mmG5400G)3)(200)(3
1(3tb
3
1GGJ ===
--- ANS(b)Fig. 3.39b is a cross-section of a closed thin-wall, its torsional rigidity is bGJ
=
t
ds
AGGJ
b
2
4, where
4
3 2bA= ,
46342
1064)3(4
34GmmG
tb
b
tbG
t
ds
AGGJ
b ====
--- ANS
(c)The ratio of the torsional rigidities is
1111G5400
G106
GJ
GJ 6
a
b =
=
--- ANS
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3.13 Find the distributions of the primary warping displacement on the cross-sections
shown in Fig. 3.39b. Due to symmetry, the center of twist coincides with the
centroid of the section, and warp at the midpoint of each flat sheet section is zero.
Sketch the warping displacement along the wall.
(b)
Figure 3.39 Cross-sections of two shafts
Solution:
(a)Observation.
Because of the symmetry, the center of twist coincides with the centroid of the
section, and warp at the midpoint of each flat sheet section is zero.
So from the figure above we set 0=w at the midpoint of each flat sheet. First
we assume the warp at point A is positive of z-direction. While going from A to B,
we pass the midpoint and then the warp goes from positive into negative part, then
end at point B with the maximum negative warping. Using the same concept on
sheet BC will result in a maximum positive warping at point C. Now we consider
the sheet CA by using the same conclusion, we will surprisingly find the warping
at A is negative of z-direction. Hence it contradicts our assumption of A being
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positive. By applying the assumption of A is negative direction we will conclude
in another contradiction. Therefore, we can confidently assure that there is no
warping in this equilateral triangular thin-walled cross-section. In the following
we will approve it by further derivatives.
(b)
For the closed thin-walled section, we have
+
=
+
=
s
w
z
u
s
w ssz , (3.13.1)
where is the distance from the center of twist to the tangent line of point P of
interest, wis the warping that we are seeking. Also, we have
Gt
q
G
ssz
sz ==
(3.13.2)
where is the shear flow along s-direction, tis the thickness of the wall and G
is the shear modulus.
sq
Again, recall from the relationship between applied torque and shear flow, we
haveA
Tqs
2= (3.13.3)
Combining (3.13.1) to (3.13.3) results in
GtA
T
s
w
2=+
, or =
GtA
T
s
w
2
=> ssss
Ads
GtA
Tdsds
GtA
Twsw 2
22
)0()(000
== (3.13.4)
Also the twist angle can be derived from
= tds
AG2
1 (3.13.5)
(c)Assume the applied torque is uniformly applied to the cross-section. Also, the
material is isotropic so that the shear modulus is constant.
For the equilateral triangular section, we have
4
3 2bA= (3.13.6)
And since the section is symmetric, we can just take the sheet CA into
consideration and applied to all other sheets. Assume the origin of sis on the
midpoint of sheet CA, so 0)0( =w , then we have
12
3bsAs = (3.13.7)
From (3.13.4) to (3.13.7), we obtain
))16/3(
)8/3(
3
2(
)(4
)3)(12
3(2
22
2)(
4
2
220 b
b
bGt
Ts
tAG
Tbbs
tAG
TsAds
GtA
Tsw
s
s
===
3.13.2
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=> 0)( =sw
This approves our observation in part (a).
--- ANS
3.13.3
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433 1729080)2)(2
(12
1)2)(
2(
12
1mmth
tbth
tbIx =++=
( Since the thickness of walls is relatively small, there are some approximated
solutions such as
4233
1728270]12[2)2(12 mmbtht
b
h
t
Ix =++= , or
423 1728000)(2)2(12
mmbthht
Ix =+= are all the acceptable approximations)
The eccentric distance mm49.221729080
)60(3
I
htbe
4
x
22
=== (3.14.5)
The torsional constant is
433
32160
3
3120)
3
360(2
3
1mmtbJ
i
ii =
+
==
and the twist angle per unit length can be obtained from
G
T
GJ
T
2160== (3.14.6)
(c)Break up the contour sinto two straight parts and , as shown below1s 2s
For contour , we have1s
11
2
1esAs = (3.14.7)
and for contour (the point2s 0s2 = is at upper left corner of the section) we have
22
2
1hsAs = (3.14.8)
(d)On the contour , the warping displacement wis calculated from equation
(3.14.2):
1s
G
Ts0104.0
G2160
TesA2ds)s(w 111s
s
01 ====
In which the condition 0)0(w = has been used. This is obvious since the warping
at the middle point of the vertical web is zero because of anti-symmetry. Also note
that
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G
T62.0)mm60h(w)hs(w 1 ====
--- ANS
(e)On the contour , the warping can be obtained from equation (3.14.2) by
integrating from s
2s
1=0 to any point s2. Thus,
2s12 A2)hs(w)s(w ==
=>G
Ts
G
Tsw 22 028.062.0)( += --- ANS
So the warping displacement at the left upper corner is
G
Thsw 62.0)( 1 ==
and at the right upper edge is
G
Tbsw 04.1)( 2 ==
--- ANS
(f) Similar calculations show that the warping displacement is anti-symmetric with
respect to the x-axis. From the above calculations, the maximum warp (absolute
value) is
G
Tbsww 04.1)( 22max ===
and are located at both free edges.
--- ANS
G
Tssw 111 0104.0)( =
G
T
G
Tssw 62.00278.0)( 222 =
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3.15 Consider the shaft of the channel section shown in Fig. 3.40. If one end of the
shaft is built in and the other end is free, find the effective torsional rigidity as a
function of the distance from the built-in end. Assume that the length Lof the
shaft is sufficiently large so that near the free end the Saint-Venant torsion
assumptions are valid. Compare the total twist angle with that for a free-free
shaft for mL 2= .
Figure 3.40 Dimensions of a channel section
Solution:
(a)
The government equation for the twist angle under an applied torque Tendconstraints is
GJ
T
dz
dk =
2
22
, (3.15.1)
whereGJ
Ek
=2 , =
1
0
24
s
ss tdsA and =
i
iitbJ3
3
1 (3.15.2)
The general solution is
)1( /2/
1kzkz
ph eCeCGJ
T ++=+= (3.15.3)
(b)Applying boundary conditions
We assume the shaft is built in at 0=z and free at Lz= where the torque Tis
applied.
(1) First, assume the length Lof the shaft is sufficiently large so that near the
free end the Saint-Venant torsion assumptions are valid, so that GJT=
(Saint-Venant torsion) when . To satisfy this condition, we require
that
Lz
01 =C , then (3.15.3) will converge to GJT= .
(2)
Second, at the built-in end ( 0=z ), warping suppressed and . From
the equation
0=w
)()(),( zswzsw s = , we conclude that 0= . Thus, we have
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12 =C .
From (1) and (2), we have the solution becoming
eff
kz
GJ
Te
GJ
T== )1( /
Then the effective torsion constant is
kzeffe
JJ
/1 = (3.15.4)
(c)
In Fig. 3.40, the channel cross-section has the properties from (3.15.2).
49433
31016.22160
3
3120)
3
360(2
3
1mmmtbJ
i
ii
==
+
==
=1
0
24
s
ss tdsA
Because of the symmetric w.r.t x-axis, it is more convenient to measure distancefrom the middle point of the vertical web.
From the solution of Problem 3.14, we have
mmmI
htbe
x
02249.049.221729080
)60(3 422
cot ====
Therefore,
})](2
1
2
1[4)
2
1(4{24 1
21cot
01
21cot
1
0
2
+
+= bh
h
hs
ss tdshshhetdssetdsA
~~~~~~A~~~~~~ ~~~~~~~~~B~~~~~~~~~
Part A:
61032
cot0
3
1
2
cot0
12
1cot 1009.13
1|)
3
1()
2
1(4 mthestetdsse h
h===
Part B:
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6103222cot
22
cot0
22
2cot
12
1cot
1031.2)3
1()(
)](2
1
2
1[4
2
mbhbhebhettdshshe
tdshshhe
bb
bh
h
+
=+==
=> 6101
0
2108.6)(24 mPartBPartAtdsA
s
ss
=+==
Taking andGPaE 70= GPaG 27= for aluminum, then
904.0=
=GJ
Ek
Thus, the effective torsional rigidity is obtained as
904.0/904.0/
99
/ 1
32.58
1
1016.21027
1 zzkzeff
eee
GJGJ
=
=
=
--- ANS(d)The total twist angle.
For the case with end constraints,
TT
ezT
dzeT
dz zz
L
fixed
0205.0)805.02(32.58
|)904.0(32.58
)1(32.58
20
904.0/2
0
904.0/
0
==
+===
For the case with free-free end,
TTGJTLLfree 0343.0
32.582 ====
The ratio of the two twist angles is
6.0=free
fixed
--- ANS
It is clear that the end constraints reduce the twist angle. In other words, end
constraints increase the torsional stiffness.
3.15.3
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TseseTsz zz
zzz 10904.0/
1cot0 0299.0|)2
1(656.2|),( == =
=
--- ANS
At ,01 =s 0)0,0( =zz
At mhs 06.01 == , )/(1079.106.00299.0 23 mNTTzz ==
On the horizontal sheet, we can derive the normal stress with (3.16.3) and
(3.16.5)
Ts
ehsheTsz zz
zzz
)0797.01079.1(
|)2
1
2
1(656.2|),(
23
0904.0/
2cot0
+=
=
=
=
--- ANS
At ,02 =s Tzz31079.1 =
At mbs 06.02 == ,)/(1099.2)06.00797.01079.1( 233 mNTTzz
=+=
The distribution of normal stress on the cross-section is anti-symmetric. The
distribution can be illustrated as the figure below.
--- ANS
(d)
The distribution of shear flow at the built-in end ( 0=z ).
From the equation (3.85) in the textbook, the shear flow at any location s at the
built-in end ( 0=z ) is
== = s
s szz tdsw
dz
dEsq
002
2
0 ||)(
(3.16.6)
where ss Asw 2)( =
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Substituting (3.16.1) in (3.16.6), we have
=
=
=
=
s
ss
s
ssz
z
z
dsAT
dsAeT
sq
0
3
00
904.0/20
1082.8
)003.02(|))904.0
1()(
32.58)(70(|)(
(3.16.7)
Here it is important to emphasize that the s-direction is measured from the point
where shear flow vanishes. Hence sbegins from the free end of the horizontal
sheet as shown in the figure below. Also, due to the symmetry w.r.t. x axis, we
only need to consider the part above x-axis. This allows us to modify equation of
sA from (3.16.4) and (3.16.5).
For the horizontal sheet, we have
12
1hsAs = , (3.16.8)bs 0:1
Moving along the vertical web, we have
2cot2
1seAs = (3.16.9)hs 0:2
On the horizontal sheet, we can derive the shear flow from (3.16.8) and
(3.16.7)
Tsx
T
hxdxTsq
s
s
z
2
14
0
24
0
301
10323.1|)2
(10646.2
2
11082.8|)(
1
1
=
==
=
--- ANS
At ,01 =s 0)0( =q
At mbs 06.01 == ,
mNTTq /10762.4)06.0(10323.1)06.0( 7241 ==
On the vertical web, we can derive the shear flow with (3.16.9) and (3.16.7).
Since the shear flow is continuous, we have
Tsx
T
dxxeTbsqsq
s
s
zz
2
252
0
25
2
0cot
3011022
10957.4|2
10914.9
)2
1(10818341.8|)(|)(
==
==
==
3.16.3
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Mechanics of Aircraft structuresC.T. Sun
=> Tssq z )10762.410957.4(|)(72
25
022
= =
--- ANS
At ,02 =s Tq7
2 10762.4)0( =
At mhs 06.02 == ,
T
Tsq
7
72522
10977.2
)10762.4)06.0(10957.4()06.0(
=
==
The distribution of the shear flow at the fixed end is sketched in the figure below.
--- ANS
3.16.4
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Mechanics of Aircraft structuresC.T. Sun
3.17 Assume that the shaft of the channel section of Fig. 3.40 is built in at both ends.
Find the torque that is necessary to produce a relative twist angle
between two ends. Assume that
o5=
mL 1= , Youngs modulus , and
shear modulus . Compare this with the free-free case.
GPaE 70=
GPaG 27=
Figure 3.40 Dimensions of a channel section
Solution:
(a)
Since both end of the channel are built-in, it allows us to set 0=z at the middle
of the channel as shown above.
(b)The governing equation for the twist angle under an applied torque Tend
constraints is
GJ
T
dz
dk =
2
22
, (3.17.1)
whereGJ
Ek
=2 , (3.17.2)
=1
0
24
s
ss tdsA and =
i
iitbJ3
3
1 for open thin-walled section. (3.17.3)
The general solution of differential equation (3.17.1) is
3.17.1
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Mechanics of Aircraft structuresC.T. Sun
)sinhcosh1( 21k
zC
k
zC
GJ
Tph ++=+= (3.17.4)
(c)Compute =i
iitbJ
3
3
1 and =
1
0
24
s
ss tdsA .
49433
31016.22160
3
3120)
3
360(2
3
1mmmtbJ
i
ii
==+==
=1
0
24
s
ss tdsA
Because of symmetry w.r.t x axis, it is more convenient to set up the s contour
with the origin at the middle point of the vertical web as shown in the figure
below.
From the solution of Problem 3.14, we have
mmmI
htbe
x
0225.0485946.221729000
)60(3 422
cot ====
Therefore,
})](2
1
2
1[4)
2
1(4{24 1
21cot
01
21cot
1
0
2
+
+= bh
h
hs
ss tdshshhetdssetdsA
~~~~~~A~~~~~~ ~~~~~~~~~B~~~~~~~~~
Part A:
61032
cot0
3
1
2
cot0
12
1cot 10092.13
1|)
3
1()
2
1(4 mthestetdsse h
h===
Part B:6103222
cot22
cot0 22
2cot
12
1cot
1031.2)3
1()(
)](2
1
2
1[4
2
mbhbhebhettdshshe
tdshshhe
bb
bh
h
+
=+==
=>
6101
0
2
10804.6)(24 mPartBPartAtdsA
s
s s
=+==
3.17.2
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Mechanics of Aircraft structuresC.T. Sun
Taking andGPaE 70= GPaG 27= for aluminum, then
904.0=
=GJ
Ek
(d)Applying boundary conditions
(1)
First, because of symmetry of with respect to , the odd function
should be dropped. This is accomplished by setting .
z
)/sinh( kz 02 =C
(2) Second, at the built-in end ( 2/Lz= ), warping is suppressed and .
Since
0=w
)()(),( zswzsw s = , we conclude that 0= . Thus, we have
0)2
cosh1( 1 =+k
LC
GJ
T
Then
k
LC
2
cosh
11 =
Since and , we havemL 1= mk 904.0=
864.0)
)904.0(2
1cosh(
11 =
=C
Thus the solution for the rate of twist angle is
)904.0
cosh864.01(32.58
)cosh864.01( zT
k
z
GJ
T==
The twist angle
related to distance z from the middle of the channel then is
)904.0
sinh78.0(32.58
)904.0
cosh864.01(32.5800
zz
T
duuT
dzzz
=
== (3.17.5)
This is the twist angle measured from the middle of the channel bar to the built-in
end.
(e) If we produce a relative twist angle , then the twist angle from the middle
of the channel to the built-in end (
o5=
mLz 5.02/ == ) is
rad0436.05.22
5=== o
o
.
From equation (3.17.5) we can determine the required torque to produce such an
angle.
TT 410795.7)
904.0
5.0sinh78.05.0(
32.580436.0 ==
Then mNT = 97.55
--- ANS
(f)
For the free-free end case
3.17.3
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Mechanics of Aircraft structuresC.T. Sun
mNz
GJGJT endfree ==== 089.55.0
0436.032.58
.
The ratio of built-in ends case and free-ends case is
110894.5
974.55==
endfree
inbuilt
T
T
It is likely that the rigidity of the built-in ends case is enhanced eleven times more
than the free-ends case.
--- ANS