Mechanics of Aircraft Structures solution manual C.T. Sun 2nd ed

Solution Manual for
Mechanics of Aircraft Structures
2nd
edition, 3rd
printing, 2007
C.T. Sun
School of Aeronautics and Astronautics
Purdue University
W. Lafayette, Indiana U.S.A.
Prepared with the assistance of Mr. Hsin-Haou Huang, graduate
student in School of Aeronautics and Astronautics,
Purdue University
June, 2007
Name: Mohamed Naleer Abdul Gaffor Email: muhammed_naleer@yahoo.com IP: 184.162.144.24
Address: 1650, BLVD DE MAISONNEUVE Apt. 904, Montreal, QC H3H2P3, CAN

Mechanics of Aircraft structures
C.T. Sun
1.1 The beam of a rectangular thin-walled section (i.e., t is very small) is designed
to carry both bending moment M and torque T. If the total wall contour length
(see Fig. 1.16) is fixed, find the optimum b/a ratio to achieve the
most efficient section if
)(2 baL +=
TM = and allowableallowable τσ 2= . Note that for closed
thin-walled sections such as the one in Fig.1.16, the shear stress due to torsion is
abt
T
2
=τ
Figure 1.16 Closed thin-walled section
Solution:
(1) The bending stress of beams is
I
My
=σ , where y is the distance from the neutral
axis. The moment of inertia I of the cross-section can be calculated by considering
the four segments of thin walls and using the formula for a rectangular section
with height h and width w. )Adwh
12
1
(I 23
∑ += in which A is the
cross-sectional area of the segment and d is the distance of the centroid of the
segment to the neutral axis. Note that the Parallel Axis Theorem is applied. The
result is )3(
6
])
2
()(
12
1
[2
12
1
2
2
233
ba
tbb
atattbI +≈⋅+⋅⋅+⋅= , assuming that t is
very small.
(2) The shear stress due to torsion for a closed thin-walled section shown above is
abt
T
2
=τ .
1.1.1

C.T. Sun
(3) Two approaches are employed to find the solution.
(i) Assume that the bending stress reaches the allowable allowableσ first and find
the corresponding bending maximum bending moment. Then apply the stated
loading condition of MT = to check whether the corresponding maxτ has
exceeded the allowable shear stress allowableτ . If this condition is violated, then
the optimized b/a ratio is not valid.
(a)
)3(
3
)3(
6
2| 2
2
batb
M
ba
tb
b
M
I
My
b
y +
=
+
⋅
==
=
σ
When given )(2 baL += as a constant, a can be expressed in terms of b
and L as b
L
a −=
2
. Then we can minimize
6
)43(
3
)3( bLtbbatb
S
−
=
+
= in order to maximize σ , i.e.,
8
3
0)83(
6
0
L
bbL
t
b
S
=⇒=−⇒=
∂
∂
, so
82
L
b
L
a =−=
where the optimum ratio is 3=
a
b
Thus, 2max
3
32
)8/38/3()8/3(
3
)3(
3
tL
M
LLLt
M
batb
M
=
+⋅⋅⋅
=
+
=σ
(b) Check maxτ with MT = and b/a = 3 and check whether maxτ is within
the allowable shear stress allowableτ .
2
3
32
)8/3()8/(22
max2max
allowable
allowable
allowable
tL
M
tLL
M
abt
T
σ
τ
σστ
=>
===
⋅⋅⋅
==
The result above means that under this assumption, shear stress τ would
reach the allowable stress allowableτ before σ reaches allowableσ . Consequently,
the optimal ratio obtained is not valid and different assumption needs to be
made.
(ii) Assume now that failure is controlled by shear stress. We assume that
allowableττ =max is reached first and then find the corresponding bending stress
according to the loading condition TM = .
(a)
abt
T
2
=τ
Again we minimize btbLabtS )2(2 −== in order to maximize τ , i.e.,
1.1.2

C.T. Sun
4
0)4(0
L
bbL
b
S
=⇒=−⇒=
∂
∂
, so
42
L
b
L
a =−=
and the optimum ratio is 1=
a
b
and 2max
8
)4/()4/(22 tL
T
tLL
T
abt
T
=
⋅⋅⋅
==τ
(b) Then corresponding maxσ under the optimum condition stated above can
be obtained using TM = . We have
allowableallowable
allowable
tL
T
LLLt
T
batb
M
τσ
ττσ
2
2
3
2
312
)4/4/3()4/(
3
)3(
3
max2max
=<
===
+⋅⋅⋅
=
+
=
This means that when the structure fails in shear, the bending stress is
still within the allowable stress level. Thus the optimum ratio 1=
a
b
is
valid.
(4) In conclusion, 1=
a
b
achieves the most efficient section for the stated conditions.
--- ANS
1.1.3

C.T. Sun
1.2 Do problem 1.1 with TM α= where ∞= to0α .
Figure 1.16 Closed thin-walled section
Solution:
(1) The bending stress of beams is
I
My
=σ , where y is the distance from the neutral
axis. The moment of inertia I of the cross-section can be calculated by considering
the four segments of thin walls and using the formula for a rectangular section
with height h and width w. )Adwh
12
1
(I 23
∑ += in which A is the
cross-sectional area of the segment and d is the distance of the centroid of the
segment to the neutral axis. Note that the Parallel Axis Theorem is applied. The
result is )3(
6
])
2
()(
12
1
[2
12
1
2
2
233
ba
tbb
atattbI +≈⋅+⋅⋅+⋅= , assuming that t is
very small.
(2) The shear stress due to torsion for a closed thin-walled section shown above is
abt
T
2
=τ .
(3) Two approaches are employed to find the solution.
(i) Assume that the bending stress reaches the allowable allowableσ first and find
the corresponding bending maximum bending moment. Then apply the stated
loading condition of TM α= to check whether the corresponding maxτ has
exceeded the allowable shear stress allowableτ . If this condition is violated, then
the optimized b/a ratio is not valid.
(a)
)3(
3
)3(
6
2| 2
2
batb
M
ba
tb
b
M
I
My
b
y +
=
+
⋅
==
=
σ
When given )(2 baL += as a constant, a can be expressed in terms of b
1.2.1

C.T. Sun
and L as b
L
a −=
2
. Then we can minimize
6
)43(
3
)3( bLtbbatb
S
−
=
+
= in order to maximize σ , i.e.,
8
3
0)83(
6
0
L
bbL
t
b
S
=⇒=−⇒=
∂
∂
, so
82
L
b
L
a =−=
where the optimum ratio is 3=
a
b
Thus, 2max
3
32
)8/38/3()8/3(
3
)3(
3
tL
M
LLLt
M
batb
M
=
+⋅⋅⋅
=
+
=σ
(b) Check maxτ with TM α= and b/a = 3 and check whether maxτ is
within the allowable shear stress allowableτ .
allowableallowable
tL
M
tLL
M
abt
T
τ
α
σ
α
σ
αα
α
τ
21
1
3
32
)8/3()8/(2
/
2
max2max
==
==
⋅⋅⋅
==
We have allowableallowableallowable ττ
α
ττ ≤⇒≤
2
max
2≥α (since 0>allowableτ is always satisfied)
(ii) Assume now that failure is controlled by shear stress. We assume that
allowableττ =max is reached first and then find the corresponding bending stress
according to the loading condition TM α= .
(a)
abt
T
2
max =τ
Again we minimize btbLabtS )2(2 −== in order to maximize τ , i.e.,
4
0)4(0
L
bbL
b
S
=⇒=−⇒=
∂
∂
, so
42
L
b
L
a =−=
and the optimum ratio is 1=
a
b
and 2max
8
)4/()4/(22 tL
T
tLL
T
abt
T
=
⋅⋅⋅
==τ
(b) Then corresponding maxσ under the optimum condition stated above can
be obtained using TM α= . We have
allowable
allowable
allowable
tL
T
LLLt
T
batb
M
ασ
σ
αατ
ατ
αα
σ
4
3
)
2
(
2
3
2
3
2
312
)4/4/3()4/(
3
)3(
3
max2max
=⋅==
==
+⋅⋅⋅
=
+
=
Since allowableallowableallowable σασσσ ≤⇒≤
4
3
max
3
4
≤α (since 0>allowableσ is always satisfied)
1.2.2

C.T. Sun
(4) From the above two approaches, we have the conclusions.
(i) For
3
4
0 ≤< α , the failure is controlled by shear and the optimum ratio
of 1=
a
b
achieves the most efficient section..
(ii) For 2≥α , the failure is controlled by bending and the optimum ratio of
3=
a
b
achieves the most efficient section.
(iii) For 2
3
4
<< α , the optimal ratio lies between 1 and 3. The most
straightforward way in finding the best ratio for a given α in this range
is to calculate the maximum bending moments and torques for different
values of b/a ratios between 1 and 3 and pick the ratio that produces the
greatest minimum failure load, either T or M.
--- ANS
1.2.3

C.T. Sun
1.3 The dimensions of a steel (300M) I-beam are b = 50 mm, t = 5 mm, and h = 200
mm (Fig. 1.17). Assume that t and h are to be fixed for an aluminum(7075-T6)
I-beam. Find the width b for the aluminum beam so that its bending stiffness EI
is equal to that of the steel beam. Compare the weights-per-unit length of these
two beams. Which is more efficient weightwise?
Figure 1.17 Dimensions of the cross-section of an I-beam
Solution:
(1) The expression of area moment of inertia I for an I-beam is:
2])
2
)((
12
[)(
12
233
×++−=
h
btt
b
th
t
I , by applying Parallel Axis Theorem.
(2) First obtaining the area moment of inertia of the steel (300M) I-beam with given b,
t, and h.
4233
80905732])
2
200
)(550(5
12
50
[)5200(
12
5
mmISteel =×⋅+⋅+−=
(3) For the given condition SteelumAlu EIEI )()( min =
we have 4
St
Al
St
Al mm227900008090573
71
200
I
E
E
I =×==
which allows to calculate the width b for the aluminum beam with the following
result:
22790000b8.1000203.3089531
2])
2
200
)(5b(5
12
b
[)5200(
12
5
I 233
Al
=+=
×⋅+⋅+−=
and mm197b =
---- ANS
(4) Then we compare the weights-per-unit length of these two beams.
1.3.1

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The weights-per-unit length is defined as
Aw ⋅= ρ , where ρ = density , and A = cross-sectional area
(i) For the Steel beam
)/(108.7)/(8.7 333
mmgcmgSt
−
×==ρ
)(147555025)5200( 2
mmASt =××+×−=
)mm/g(5.111475108.7Aw 3
StStSt =××=⋅= −
ρ
(ii) For the Aluminum beam
)/(1078.2)/(78.2 333
mmgcmgAl
−
×==ρ
)mm(2945597.19625)5200(A 2
Al =××+×−=
)mm/g(2.87.29441078.2Aw 3
AlAlAl =××=⋅= −
ρ
For a unit length of both materials, the aluminum beam is much lighter than the
steel beam. It means that the ALUMINUM BEAM IS MORE EFFICIENT!
--- ANS
1.3.2

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1.4 Use AS4/3501-6 carbon/epoxy composite to make the I-beam as stated in
Problem 1.3. Compare its weight with that of the aluminum beam.
Figure 1.17 Dimensions of the cross-section of an I-beam
Solution:
Proceed in the same manner as that of problem 1.3.
(1) The expression of area moment of inertia I for a I-beam is:
2])
2
)((
12
[)(
12
233
×++−=
h
btt
b
th
t
I
(2) First, obtain the area moment of inertia of the steel (300M) I-beam with given b, t,
and h. We have
4233
80905732])
2
200
)(550(5
12
50
[)5200(
12
5
mmISteel =×⋅+⋅+−=
(3) For the condition SteelComposite EIEI )()( =
we have 4
St
Com
St
Com mm115580008090573
140
200
I
E
E
I =×==
The moment of inertia of the composite beam is given by
11558000b8.1000203.3089531
2])
2
200
)(5b(5
12
b
[)5200(
12
5
I 233
Com
=+=
×⋅+⋅+−=
Thus the width of the cross-section is obtained as mm7.84b =
---- ANS
(4) Then, we compare the weights-per-unit length of these two beams.
The weights-per-unit length is defined as
Aw ⋅= ρ , where ρ = density , and A = cross-sectional area
1.4.1

C.T. Sun
(i) For the composite beam
)/(1055.1)/(55.1 333
mmgcmgCom
−
×==ρ
)mm(1822567.8425)5200(A 2
Com =××+×−=
)mm/g(8.27.18211055.1Aw 3
ComComCom =××=⋅= −
ρ
(ii) Compare the weights per unit length with that of the aluminum beam
)mm/g(2.8w)mm/g(8.2w AlCom =<=
This indicates that the AS4/3501-6 CARBON/EPOXY COMPOSITE BEAM IS
MORE EFFICIENT than the aluminum beam!
--- ANS
1.4.2

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1.5 Derive the relations given by (1.4) and (1.5).
Remark:
btV
atV
y
x
⋅⋅=
⋅⋅=
τ
τ
:)5.1(
:)4.1(
Solution:
(1) Consider a very small section within the curved panel with thickness t and length
Lδ . τ is the constant shear stress, so we have the shear force )tL(V ⋅⋅= δτΔ
acting on the cross section.
(2) It is possible to take apart the shear force into x and y direction shown in the
figure, where
xt
LttLVVx
Δ⋅⋅=
⋅⋅⋅=⋅⋅⋅=⋅Δ=Δ
τ
θδτθδτθ )cos(coscos
similarly, ytVy δτ ⋅⋅=Δ
(3) Now consider the length to be extremely small, therefore xx dVV →Δ as well as
. The horizontal component and the vertical component of the shear
force , can be verified as f
dVV = ∫
yy dVV →Δ
xV yV ollowing:
a
xx ⋅⋅=⋅⋅= ∫ ττ
0
atdxt
btdytdVV
b
yy ⋅⋅=⋅⋅== ∫∫ ττ
0
1.5.1

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1.6 The sign convention (positive direction of resultants) used in the beam theory
depends on the coordinate system chosen. Consider the moment-curvature
relation
2
2
dx
wd
EIM −=
in reference to the coordinate system shown in Fig. 1.18. If w is regarded as a
positive displacement (or deflection) in the positive y-direction, find the positive
direction of the bending moment. State the reason.
Figure 1.18 Coordinate system for a beam
Solution:
(1) The moment-curvature relation 2
2
dx
wd
EIM −= gives that 2
2
dx
wd
is always
opposite in sign to M. (It is quite obvious that both E and I are always positive.).
(2) We can assume a moment M applying to the beam as shown below, which makes
the beam concave upwards. It is not difficult to observe that the slope
dx
dw
increases with increasing x and thus a positive 2
2
dx
wd
.
(3) By applying the statement (1), it is concluded that the deformation described in (2)
is produced by a negative moment while a positive moment makes the beam
concave downward as shown below..
1.6.1

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1.7 Compare the load-carrying capabilities of two beams having the respective
cross-sections shown in Fig. 1.19. Use bending rigidity as the criterion for
comparison. It is given that a = 4 cm, t = 0.2 cm, and the two cross-sections
have the same area.
Figure 1.19 Cross-sections of two beams
Solution:
When using the bending rigidity ( EI ) as a criterion for comparison, Young’s modulus
E and the area moment of inertia I should be estimated.
(1) Young’s modulus E :
Assume the Young’s modulus of the beam having the left-hand-side
cross-section and the right-hand-side cross-section are and ,
respectively.
lE rE
(2) Moment of inertia I :
(i) Left cross-section:
444
33.214
12
1
12
1
cmaIl =×==
(ii) Right cross-section:
33
12
)2(
12
a
tb
ba
b
Ir
−
−+= --- (a)
or { 2])
22
(
12
1
[
12
2243
×+⋅++=
ba
bba
t
Ir }
where b remains unknown. There is another condition, two cross-section
have the same area, which will help to solve b.
222
164 cmaAl === , 2.0422 22
⋅+⋅=⋅+⋅= btabAr
1.7.1

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let Al=Ar => , then we havecmb 7568.2=
( ) ( ) 433
r cm1844
12
2.07568.2
7568.224
12
7568.2
I =⋅
−
−×+=
or { 42243
r cm1842])
2
7568.2
2
4
(7568.27568.2
12
1
[4
12
2.0
I =×+⋅++= }
(3) Performance:
The ratio of the moments of inertia of the two cross-sections can be expressed as
r
l
r
l
r
l
rr
ll
r
l
E
E
12.0
E635.8
E
E18.184
E33.21
IE
IE
)EI(
)EI(
====
The cross-section to the right is much better if the same material is used for
both beams.
(i) If lr E12.0E <
The left cross-section outperforms the right one.
(ii) If lr E12.0E =
They are equivalent.
(iii) If lr E12.0E >
The right cross-section outperforms the left one.
--- ANS
1.7.2

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2.1 Consider a unit cube of a solid occupying the region
10 ≤≤ x , 10 ≤≤ y , 10 ≤≤ z
After loads are applied, the displacements are given by
xu α=
yv β=
0=w
(a) Sketch the deformed shape for 03.0=α , 01.0−=β .
(b) Calculate the six strain components.
(c) Find the volume change VΔ [ VΔ = V (the volume after deformation) – V0
(the original volume) ] for this unit cube. Show that Vzzyyxx Δ≈++ εεε .
Solution:
(a) Since , there is no deformation in the z-direction and the deformation can
be represented in the x-y plane.
0=w
The new position of point B after deformation is given by
03.1103.01|1' 0,1 =⋅+=+= == yxux
00)01.0(0|0' 0,1 =⋅−+=+= == yxvy
New coordinates of B’ = ( 1.03 , 0 )
Similarly, new positions of A, C, D can be obtained as follows:
A’ = ( 0 , 0 )
C’ = ( 1.03, 0.99 )
D’ = ( 0 , 0.99 )
--- ANS
A, A’ B
1
B’
1.03
D, 1
D’
0.99
C
C’
x
y
A=(0,0) A’=(0,0)
B=(1,0) B’=(1.03,0)
C=(1,1) C’=(1.03,0.99)
D=(0,1) D’=(0,0.99)
2.1.1

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(b) Strain components are:
nents,Normal strain compo
03.0==
∂
= αε
∂x
u
xx
01.0−==
∂
∂
= βε
y
v
yy
0=
∂
∂
=
z
w
zzε
Shear strain components,
0=
∂
+
∂∂
∂
==
vu
γγ
xy
yxxy
0=
∂
∂
+
∂
∂
==
x
w
z
u
zxxz γγ
0=
∂
∂
+
∂
∂
==
y
w
z
v
zyyz γγ
--- ANS
(c) The volume change is defined by
03.11'z'y'xVVV 0 0197.0111199.0 ××=−⋅⋅=−=Δ − × × =
Also, 0197.002.00)01.0(03.0 =Δ≈=+−+=++ Vzzyyxx εεε
--- ANS
It can also be verified by:
zzyyxxzzyyzzxxyyxxzzyyxx0
zzyyxxzzyyzzxxyyxxzzyyxx0
zzyy
εεεεεεεεεεεε
εεεεεεεεεεεε
εε
++++++=
−+++++++=
⋅⋅−⋅+⋅⋅+⋅
Since the deformation is very small, we have
)(V
)11(V
1111)1(1)1(1)1(VVV xx0 εΔ ⋅+=−=
1<<ε => higher order terms can
be dropped. Therefore we have
VV zzyyxxzzyyxx εεεε ε ++=++≈Δ )( ε0 , since V0 = 1
2.1.2

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2.2 Consider the following displacement field:
yu α=
xv α−=
0=w
Sketch the displaced configuration of a unit cube with the faces originally
perpendicular to the axes, respectively. This displacement field does not yield
any strains; it only produces a rigid body rotation. Show that the angle of
rotation is
α−=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
∂
−
∂
∂
y
u
x
v
2
1
Solution:
(a) Consider a unit cube, the coordinates before deformation corresponding to each
corner are:
A(0,0,0), B(1,0,0), C(1,1,0), D(0,1,0),
E(0,0,1), F(1,0,1), G(1,1,1), H(0,1,1)
x
y
z
A B
CD
E F
GH
After deformation, we have the coordinates for each point as follows:
For point A’ (‘ denotes the point after deformation) :
000|' 0,0,0 =⋅+=+= === αzyxuxx
0)0(0|' 0,0,0 =⋅−+=+= === αzyxvyy
000|' 0,0,0 =+=+= === zyxwzz , Thus, we have A’(0,0,0)
Another example for G’, we have
2.2.1

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αα +=⋅+=+= === 111|' 1,1,1 zyxuxx
αα −=⋅−+=+= === 1)1(1|' 1,1,1 zyxvyy
101|' 1,1,1 =+=+= === zyxwzz , Thus, G’(1+α ,1-α ,1)
Similarly, we can verify other points
B’(1,-α ,0), C’(1+α ,1-α ,0), D’(α ,1,0),
E’(0,0,1), F’(1,-α ,1), H’(α ,1,1)
(1) In x-y plane
A(0,0,0), B(1,0,0), C(1,1,0), D(0,1,0)
A’(0,0,0), B’(1,-α ,0), C’(1+α ,1-α ,0), D’(α ,1,0)
y
xA,A’ B
B’
C
C’
D D’
α
α
(2) In y-z plane
A(0,0,0), D(0,1,0), H(0,1,1), E(0,0,1)
A’(0,0,0), D’(α ,1,0), H’(α ,1,1), E’(0,0,1)
HE E’
z
yA,A’ D D’
H’
2.2.2

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(3) In z-x plane
A(0,0,0), E(0,0,1), B(1,0,0), F(1,0,1)
A’(0,0,0), E’(0,0,1,) B’(1,-α ,0), F’(1,-α ,1)
FB B’
x
zA,A’ E E’
F’
--- ANS
(b) Then we verify the strains are zero under this circumstance:
0=
∂
=
∂x
u
ε ,xx 0=
∂
∂
=
v
ε , 0=
∂
∂
=ε
z
w
zz
y
yy
0=−=
∂
∂
+
∂
∂
= ααγ
x
v
y
u
xy , 0=
∂
∂
+
∂
∂
=
y
w
z
v
yzγ , 0=
∂
∂
+
∂
∂
=
x
w
z
u
xzγ
--- ANS
(c) If we denote the counterclockwise rotation
to be positive, we have the angle of
rotation equals to ( )1
βββ −=
If the deformation is small enough, we
21
2
avg .
have αβ −=
∂
=
∂x
v
, and1 αβ =
∂
∂
=
u
y
2
Therefore the angle of rotation is
( )
( ) ααα −=−−=
⎟
⎠
⎜
⎝ ∂∂
2
1
22
21
yx
avg
--- ANS
y
xβ1
β2
βββ ⎟
⎞
⎜
⎛ ∂
−
∂
=−=
11 uv
2.2.3

C.T. Sun
2.3 Consider the displacement field in a body:
u = 0.02x + 0.02y – 0.01z cm
v = + 0.01y – 0.02z cm
w = -0.01x + 0.01z cm
Find the locations of the two points (0,0,0) and (5,0,0) after deformation. What
is the change of distance between these two points after deformation? Calculate
the strain components corresponding to the given displacement field. Use the
definition of xxε to estimate the change of distance between the two points.
Compare the two results.
Solution:
(a) Consider the point (0,0,0), after deformation :
0|' 0,0,0 =+= === zyxuxx
0|' 0,0,0 =+= === zyxvyy
0|' 0,0,0 =+= === zyxwzz
So the corresponding location after deformation is (0,0,0)
(b) Consider the point (5,0,0), after deformation :
1.5502.05|' 0,0,5 =×+=+= === zyxuxx
0|' 0,0,5 =+= === zyxvyy
05.0501.00|' 0,0,5 −=×−=+= === zyxwzz
(c) The change of distance between these two points after deformation.
(1) before deformation:
distance between (0,0,0) and (5,0,0), 5=beforeD
(2) after deformation:
distance between (0,0,0) and ( 5.1 , 0 , -0.05 ),
( ) ( ) ( ) 100245.5005.00001.5
222
=−−+−+−=afterD
(3) change of distance Dδ :
100245.05100245.5 =−=−= beforeafter DDDδ
(d) Calculate the strain components
2.3.1

C.T. Sun
02.0=
∂
∂
=
x
u
xxε , 02.0002.0 =+=
∂
∂
+
∂
∂
=
x
v
y
u
xyγ ,
01.0=
∂
∂
=
y
v
yyε , 02.0002.0 −=+−=
∂
∂
+
∂
∂
=
y
w
z
v
yzγ ,
01.0=
∂
∂
=
z
w
zzε 02.001.001.0 −=−−=
∂
∂
+
∂
∂
=
x
w
z
u
xzγ
(e) The normal strain in the x-direction is 02,0
x
u
xx =
∂
∂
=ε . The change of distance
between the two points can be estimated by
=> ( ) 1.0)05(02.012 =−⋅=−⋅=Δ xxD xxx ε
(f) Compare the two results
From the displacement field calculation, we have 100245.0=Dδ ,
And directly from the strain calculation, we have 1.0=ΔD
They are basically the same when the strain components are small.
2.3.2

C.T. Sun
2.4 Consider the problem of simple shear in Example 2.1 and Fig. 2.5. From the
deformed shape, find the normal strain for material along the line CB by
comparing the deformed length ''BC and undeformed length CB.
Set up new coordinates (x’,y’) so that the x’-axis coincides with CB, and y’ is
perpendicular to the x’-axis. The relation between (x,y) and (x’,y’) is given by
θθ sincos' yxx +=
θθ cossin' yxy +−=
where is the angle between x’ and the x-axis.o
45=θ
Write the displacements u’ and v’ in the x’ and y’ directions, respectively, in
terms of the new coordinates x’ and y’. The relation between (u’,v’) and (u,v) is
the same as between (x’,y’) and (x,y). Then calculate the strains using u’ and v’,
i.e.,
'
'
'
x
u
xx
∂
∂
=ε
'
'
'
y
v
yy
∂
∂
=ε
'
'
'
'
'
x
v
y
u
xy
∂
∂
+
∂
∂
=γ
xx'ε with the normal strain (alongCB) obtained earlier.Compare
Solution:
(a) The result of example 2.1 gives the new positions of A, B, C, D, which are
A’(0.01,0), B’(1.01,1.015), C’(0,0), D’(1,0.015).
From which we obtain
( ) ( ) 414214.120101
22
==−+−=CB
( ) ( ) 431896.10015.1001.1''
22
=−+−=BC
and the normal strain along the line CB is
0125.0
414214.1
414214.1431896.1''
=
−
=
−
=
CB
CBBC
ε
--- ANS
(b) The relation between (x,y) and (x’,y’) is given by
θθ sincos' yxx +=
θθ cossin' yxy +−=
which can be written in matrix form as
2.4.1

C.T. Sun
[ ]
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
y
x
y
x
β
'
'
--- (1)
where the transformation matrix is , and[ ] ⎥
⎦
⎤
⎢
⎣
⎡
−
=
θθ
θθ
β
cossin
sincos o
45=θ
From this equation we can get the equivalent form:
[ ] [ ]
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧ −
'
'
'
'1
y
x
y
x
y
x T
ββ --- (2)
Here it is easy to prove that [ ] [ ] ⎥
⎦
⎤
⎢
⎣
⎡ −
==
−
θθ
θθ
ββ
cossin
sincos1 T
Since displacements transform like coordinates, we can write
[ ]
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
v
u
v
u
β
'
'
--- (3)
From example 2.1 we have the displacement field : yu 01.0= , ,
where we can also write in matrix form,
xv 015.0=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
=
⎭
⎬
⎫
⎩
⎨
⎧
y
x
v
u
0015.0
01.00
--- (4)
So the displacements u’ and v’ can be derived, from equation (3) and (4), and by
applying (2), set , as follows:o
45=θ
[ ] [ ] [ ] [ ]
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−−
−
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
=
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
'
'
0125.00025.0
0025.00125.0
'
'
cossin025.0sin01.0cos015.0
sin015.0cos01.0cossin025.0
'
'
0015.0
01.00
0015.0
01.00
'
'
22
22
y
x
y
x
y
x
y
x
v
u
v
u T
θθθθ
θθθθ
ββββ
--- ANS
(c) Strains in the new transformed coordinate and displacements are:
0125.0
'
'
' =
∂
∂
=
x
u
xxε
0125.0
'
'
' −=
∂
∂
=
y
v
yyε
00025.00025.0
'
'
'
'
' =+−=
∂
∂
+
∂
∂
=
x
v
y
u
xyγ
--- ANS
xx'ε is the same as the strain measured along CB.Note that
2.4.2

C.T. Sun
2.5 A cantilever beam of a rectangular cross-section is subjected to a shear force V
as shown in Fig. 2.17. The bending stress is given by
I
Mz
xx =σ
where )( xLVM −−= . Assume a state of plane stress parallel to the x-z plane,
i.e., 0=== yzxyyy ττσ . Find the transverse shear stress ( )zxxz ττ = by
integrating the equilibrium equations over the beam thickness and applying the
boundary conditions 0=xzτ at
2
h
z ±= .
Hint: From the equilibrium equation
0=
∂
∂
+
∂
∂
zx
xzxx τσ
we have
x
M
I
z
xz
xxxz
∂
∂
−=
∂
∂
−=
∂
∂ στ
Figure 2.17 Cantilever beam subjected to a shear force
Solution:
(a) Bending moment is )( xLVM −−= , so V
x
M
=
∂
∂
(b) From the equilibrium equation for a state of plane stress parallel to the x-z plane,
we have
z
I
V
x
M
I
z
xz
xxxz
−=
∂
∂
−=
∂
∂
−=
∂
∂ στ
Therefore, ∫ +−=⎟
⎠
⎞
⎜
⎝
⎛
−= Cz
I
V
dzz
I
V
xz
2
2
τ
2.5.1

C.T. Sun
(c) Applying the B.C.: 0=xzτ at
2
h
z ±=
we have
0
22
2
=+⎟
⎠
⎞
⎜
⎝
⎛
− C
h
I
V
=>
I
Vh
C
8
2
=
So ( 222
4
82
zh )I
V
Cz
I
V
xz −=+−=τ also
12
3
bh
I =
=> ( ) ( )22
3
22
4
2
3
4
8
zh
bh
V
zh
I
V
xz −=−=τ
--- ANS
2.5.2

C.T. Sun
2.6 The state of stress in a body is uniform and is given by
MPaxx 4=σ , MPaxy 2=τ , 0=xzτ
MPayy 3=σ , 0=yzτ , 0=zzσ
Find the three components of the stress vector t on the surface ABCD as shown
in Fig. 2.18. Find the normal component nσ of the stress vector.
Hint: From the equilibrium equation
Figure 2.18 Shape of a wedge
Solution:
(a) The stress vector t can be expressed as {} [ ]{ }nt σ= ,
in which ijji σσ =
{ }
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
z
y
x
i
t
t
t
t is the stress vector on surface ABCD,
[ ]
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
zzzyzx
yzyyyx
xzxyxx
ij
σττ
τστ
ττσ
σ are the stress components associated with the coordinate
x-y-z,
and { } is the normal vector to the surface ABCD,
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
z
y
x
j
n
n
n
n
2.6.1

C.T. Sun
(b) Calculate the normal vector to the surface ABCD, { }jn
Assume that the positions of point B, C, D are B(1,0,1), C(1,0,0), D(0,1,0)
We have ( 1,0,0BC −= ) and ( )0,1,1CD −=
thus { } ⎟
⎠
⎞
⎜
⎝
⎛
=
×
×
= 0,
2
1
,
2
1
CDBC
CDBC
n
T
(c) Now applying {} [ ]{ }nt σ= , we have
MPa
0
54.3
24.4
0
2/5
2/6
0
2/1
2/1
000
032
024
t
t
t
z
y
x
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
--- ANS
(d) The normal component {} { }nt
T
n ⋅=σ ,
MPan 5.5
2
11
0
2/1
2/1
0
2
5
2
6
==
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎭
⎬
⎫
⎩
⎨
⎧
=σ
--- ANS
2.6.2

C.T. Sun
2.7 Find the principal stresses and corresponding principal directions for the stresses
given in Problem 2.6. Check the result with other methods such as Mohr’s
circle.
Solution:
(a) The stress given in problem 2.6 is
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
000
032
024
][ ijσ ,
To find the principal stresses, we require that [ ] [ ] 0=− Iij σσ , or
0
00
032
024
=
−
−
−
σ
σ
σ
Expanding the determinant yields ( ) 0872
=+−− σσσ , the solutions of σ are
0=σ , or
2
177 ±
=σ , (which are 1.43845 and 5.56155)
--- ANS
(i) When 01 =σ
We have the equations
⎪
⎩
⎪
⎨
⎧
=⋅
=+
=+
00
032
024
z
yx
yx
n
nn
nn
, and also we have ( ) ( ) ( ) 1
222
=++ zyx nnn
So the solutions can be obtained uniquely as
⎪
⎩
⎪
⎨
⎧
=
=
=
1
0
0
z
y
x
n
n
n
, and is the corresponding principal direction
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
1
0
0
)1(
z
y
x
n
n
n
--- ANS
(ii) When 43845.12 =σ
⎪
⎩
⎪
⎨
⎧
=−
=+
=+
043845.1
056155.12
0256155.2
z
yx
yx
n
nn
nn
222
=++ zyx nnn
2.7.1

C.T. Sun
Therefore we have the corresponding principal direction
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
−=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
0
78821.0
61541.0
)2(
z
y
x
n
n
n
--- ANS
(iii) When 56155.53 =σ
⎪
⎩
⎪
⎨
⎧
=−
=−
=+−
056155.5
056155.22
0256155.1
z
yx
yx
n
nn
nn
222
=++ zyx nnn
Therefore we have the corresponding principal direction
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
0
61541.0
78821.0
)3(
z
y
x
n
n
n
--- ANS
(b) Comparing with Mohr’s circle
Since the stresses associated with z are all zero, we know one principal stress is 0,
and its corresponding principal direction is . So here we can use the
2D analysis on the x-y plane just for other principal values.
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
1
0
0
)1(
z
y
x
n
n
n
(4,2)
(3,-2)
maxσminσ
σ
τ
According to the Mohr’s circle, we have the radius of the circle
2.7.2

C.T. Sun
( ) ( )( ) 06155.22234
2
1 22
=++−=r ,
The central coordinate of the circle is ( ) ( )0,5.3
2
22
,
2
34
, =⎟
⎠
⎞
⎜
⎝
⎛ −+
=cc yx
Therefore we have the maximum and minimum stresses, respectively,
56155.506155.25.3max =+=+= rxcσ
43845.106155.25.3min =−=−= rxcσ
These are the same as we obtained above.
--- ANS
2.7.3

C.T. Sun
2.8 A state of hydrostatic stress is given by
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
0
0
0
00
00
00
][
σ
σ
σ
σij
Show that on any surface the force (or stress vector) is always perpendicular to
the surface and that the magnitude of the stress vector is equal to 0σ .
Solution:
(a) Assume any arbitrary plane surface with its normal unit vector { } { }zyx
T
nnnn ,,= .
The stress vector acting on this surface, from equation (2.29) in the textbook,
.{} [ ]{ }
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
==
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
z
y
x
0
z
y
x
n
n
n
n
t
t
t
t σσ
Since 0σ is a scalar, the stress vector on this arbitrary surface is always parallel
to the normal vector of this surface. This leads to the conclusion that the stress
vector is always perpendicular to the surface.
--- ANS
(b) The magnitude of this stress vector t is
{} ( ) ( ) ( ) { } 00
222
σσ ==++= ntttt zyx .
--- ANS
2.8.1

C.T. Sun
2.9 An isotropic solid with Young’s modulus E and Poisson’s ratio υ is under a
state of hydrostatic stress as given in Problem 2.8. Find the corresponding strain
components.
Recall:
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
=
0
0
0
00
00
00
][
σ
σ
σ
σij
Solution:
(a) Three dimensional stress-strain relations can be expressed as:
[ ]
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
×
xy
xz
yz
zz
yy
xx
ij
xy
xz
yz
zz
yy
xx
a
τ
τ
τ
σ
σ
σ
γ
γ
γ
ε
ε
ε
66
, where are elastic compliances.ija
(b) When the material is isotropic, can be obtained individually as:ija
E
aaa
1
332211 === ,
E
aaaaaa
υ
−====== 323121231312 ,
G
aaa
1
665544 === , and others are zero.
(c) For a state of hydrostatic stress, we can obtain strain components with matrix
multiplication:
⎪
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎨
⎧
−
−
−
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−−
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
0
0
0
)21(
)21(
)21(
0
0
0
1
0
1
00
1
000
1
000
1
000
1
0
0
0
0
0
0
υ
σ
υ
σ
υ
σ
σ
σ
συ
υυ
γ
γ
γ
ε
ε
ε
E
E
E
G
G
symm
G
E
EE
EEE
xy
xz
yz
zz
yy
xx
--- ANS
2.9.1

C.T. Sun
2.10 For small strains, the volume change
V
VΔ is identified to be equal to
zzYYxx εεε ++ . The bulk modulus K of an isotropic solid is defined as the ratio
of the average stress and the volume change, i.e.,
( )
V
V
Kzzyyxx
Δ
=++ σσσ
3
1
Derive K in terms of E andυ .
Solution:
(a) Three dimensional stress-strain relations can be expressed as:
[ ]
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
×
xy
xz
yz
zz
yy
xx
ij
xy
xz
yz
zz
yy
xx
a
τ
τ
τ
σ
σ
σ
γ
γ
γ
ε
ε
ε
66
(b) If the material is isotropic, are given asija
E
aaa
1
332211 === ,
E
aaaaaa
υ
−====== 323121231312 ,
G
aaa
1
665544 === , and the rest are zero.
(c) For arbitrary stresses, we can obtain strain components with matrix multiplication:
⎪
⎪
⎪
⎪
⎪
⎪
⎭
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎧
+−−
−+−
−−
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−−
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
G
G
G
E
E
E
G
G
symm
G
E
EE
EEE
xy
xz
yz
zzyyxx
zzyyxx
zzyyxx
xy
xz
yz
zz
yy
xx
xy
xz
yz
zz
yy
xx
τ
τ
τ
συσυσ
υσσυσ
υσυσσ
τ
τ
τ
σ
σ
συ
υυ
γ
γ
γ
ε
ε
ε
)(
1
)(
1
)(
1
1
0
1
00
1
000
1
000
1
000
1
(d) With the definition of bulk modulus K, we have
( ) ( )zzYYxxzzyyxx K
V
V
K εεεσσσ ++=
Δ
=++
3
1
Substituting the strains in terms of stresses in the equation above, we obtain
2.10.1

C.T. Sun
( ) ( ) ( ) ( )[ ]
( )( )zzyyxx
zzyyxxzzyyxx
E
K
E
K
σσσ
υ
υσυσυσσσσ
++
−
=
−+−+−=++
21
212121
3
1
Thus,
)21(3 υ−
=
E
K
--- ANS
2.10.2

C.T. Sun
2.11 A block of elastic solid is compressed by normal stress xxσ as shown in Fig.
2.19. The containing walls are rigid and smooth (frictionless). Find the values
of k for plane strain and plane stress conditions, respectively, in the
stress-strain relation obtained from the compression test above.
xxxx kεσ =
Assume that andGPaE 70= 3.0=υ .
x
y
0σσ −=xx0σσ −=xx
Figure 2.19 Solid between two smooth rigid walls
Solution:
Recall: Three dimensional stress-strain relations can be expressed as:
[ ]
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
×
xy
xz
yz
zz
yy
xx
ij
xy
xz
yz
zz
yy
xx
c
γ
γ
γ
ε
ε
ε
τ
τ
τ
σ
σ
σ
66
, where are elastic constants.ijc
or
[ ]
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
×
xy
xz
yz
zz
yy
xx
ij
xy
xz
yz
zz
yy
xx
a
τ
τ
τ
σ
σ
σ
γ
γ
γ
ε
ε
ε
66
If the material is isotropic, and are given asijc ija
Gccc 2332211 +=== λ , λ====== 323121231312 cccccc ,
Gccc === 665544 , and the rest are zero.
where
( )( )υυ
υ
λ
211 −+
=
E
and
( )υ+
=
12
E
G
2.11.1

C.T. Sun
or,
E
aaa
1
332211 === ,
E
aaaaaa
υ
−====== 323121231312 ,
G
aaa
1
(a) Plane strain problem:
In plane strain problems, we have
0=== xzyzzz γγε
In this problem, we also have the following constraint condition
0=yyε
Therefore, the relation { } [ ] { }εσ 66×
= ijc can be used to obtain
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
=
=
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
+
+
+
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
xy
xz
yz
zz
yy
xx
xy
xz
yz
zz
yy
xx
G
Gsymm
G
G
G
G
γ
γ
γ
ε
ε
ε
λ
λλ
λλλ
τ
τ
τ
σ
σ
σ
0
0
0
0
0
00
0002
0002
0002
Expanding the matrix multiplication, we have
( )
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
=
=
=
=
+=
xyxy
xz
yz
xxzz
xxyy
xxxx
G
G
γτ
τ
τ
λεσ
λεσ
ελσ
0
0
2
Comparing to the problem statement, we have
GPa
E
Gk 23.94
)3.021)(3.01(
70)3.01(
)21)(1(
)1(
2 =
⋅−+
⋅−
=
−+
−
=+=
υυ
υ
λ
--- ANS
(b) Plane stress problem:
In plane stress problems, we have
0=== xzyzzz ττσ
In this problem, we also have
0=yyε
2.11.2

C.T. Sun
Therefore, the relation { } [ ] { }σε 66×
= ija can be used to obtain:
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
=
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−−
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
xy
xz
yz
zz
yy
xx
xy
xz
yz
zz
yy
xx
G
G
symm
G
E
EE
EEE
τ
τ
τ
σ
σ
συ
υυ
γ
γ
γ
ε
ε
ε
0
0
0
1
0
1
00
1
000
1
000
1
000
1
0
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎨
⎧
=
=
=
+−=
=+−=
−=
xyxy
xz
yz
yyxxzz
yyxxyy
yyxxxx
G
E
E
E
τγ
γ
γ
σσ
υ
ε
συσε
υσσε
1
0
0
)(
0)(
1
)(
1
Solving the first two equations leads to,
xxyyxxxx
EE
σ
υ
υσσε
2
1
)(
1 −
=−=
Thus, GPa
E
k 92.76
3.01
70
1 22
=
−
=
−
=
υ
under plane stress condition.
--- ANS
2.11.3

C.T. Sun
2.12 An aluminum 2024 T3 bar of unit cross-sectional area is subjected to a tensile
force in the longitudinal direction. If the lateral surface of the bar is confined
and not allowed to contract during loading, find the force that is needed to
produce a 1 percent longitudinal strain. Compare this with the corresponding
load for the bar under simple tension.
Solution:
[ ]
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
×
xy
xz
yz
zz
yy
xx
ij
xy
xz
yz
zz
yy
xx
c
γ
γ
γ
ε
ε
ε
τ
τ
τ
σ
σ
σ
66
[ ]
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
×
xy
xz
yz
zz
yy
xx
ij
xy
xz
yz
zz
yy
xx
a
τ
τ
τ
σ
σ
σ
γ
γ
γ
ε
ε
ε
66
If the material is isotropic, and are given byijc ija
Gccc 2332211 +=== λ , λ====== 323121231312 cccccc ,
where
( )( )υυ
υ
λ
211 −+
=
E
and
( )υ+
=
12
E
G
or,
E
aaa
1
332211 === ,
E
aaaaaa
υ
−====== 323121231312 ,
G
aaa
1
(a) Aluminum alloys are usually considered isotropic, so the above three-dimensional
stress-strain relations can be utilized. Also, we have the mechanical properties for
aluminum 2024 T3: ,GPaE 72= 33.0=υ .
(b) If the lateral surface of the bar is not allowed to contract during loading, we have
the conditions: 0== yyzz εε . Also, we need to produce a 1 percent longitudinal
2.12.1

C.T. Sun
strain, which means 01.0=xxε and all shear strains vanish.
So we can use the equation{ } [ ] { }εσ 66×
= ijc to obtain stresses.
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
=
=
=
=
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
+
+
+
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
0
0
0
0
0
01.0
0
00
0002
0002
0002
xy
xz
yz
zz
yy
xx
xy
xz
yz
zz
yy
xx
G
Gsymm
G
G
G
G
γ
γ
γ
ε
ε
ε
λ
λλ
λλλ
τ
τ
τ
σ
σ
σ
Calculate λ and for aluminum 2024-T3, we have:G
GPa
E
5.52
)33.021)(33.01(
7233.0
)21)(1(
=
⋅−+
⋅
=
−+
=
υυ
υ
λ
GPa
E
G 1.27
)33.01(2
72
)1(2
=
+
=
+
=
υ
we have
⎪
⎩
⎪
⎨
⎧
=×==
=×==
=⋅×+=+=
GPa
GPa
GPaG
xxzz
xxyy
xxxx
525.001.05431.52
525.001.05431.52
07.101.0)0677.2725431.52()2(
λεσ
λεσ
ελσ
With a unit cross-sectional area, we have the axial force
NP xx
6
1010701 ×=×= σ to produce the required strain.
--- ANS
(c) Compare with the corresponding load for the bar under simple tension.
Under simple tension, we have 0== yyzz σσ and the axial stress is
GPaE xxxx 72.001.072 =×== εσ .
The required axial force is .NGPaAP xx
6
10720172.0 ×=×== σ
--- ANS
Note that for the material (Al 2024-T3) used in this illustration, the stresses produced
for a 1% strain for both cases have exceeded the yield stress of the material. However,
the results still reveal the fact that the longitudinal stiffness of a bar increases if its
lateral displacements are suppressed.
2.12.2

C.T. Sun
2.13 Compare the axial stiffnesses of aluminum alloy 2024-T3 under plane strain
and plane stress conditions, respectively.
Solution:
[ ]
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
×
xy
xz
yz
zz
yy
xx
ij
xy
xz
yz
zz
yy
xx
c
γ
γ
γ
ε
ε
ε
τ
τ
τ
σ
σ
σ
66
[ ]
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
×
xy
xz
yz
zz
yy
xx
ij
xy
xz
yz
zz
yy
xx
a
τ
τ
τ
σ
σ
σ
γ
γ
γ
ε
ε
ε
66
If the material is isotropic, and are given byijc ija
Gccc 2332211 +=== λ , λ====== 323121231312 cccccc ,
where
( )( )υυ
υ
λ
211 −+
=
E
and
( )υ+
=
12
E
G
E
aaa
1
332211 === ,
E
aaaaaa
υ
−====== 323121231312 ,
G
aaa
1
(a) For aluminum 2024 T3, ,GPaE 72= 33.0=υ
=>
( )( )
GPa
E
52
211
=
−+
=
υυ
υ
λ ,
( )
GPa
E
G 27
12
=
+
=
υ
(b) Plane strain condition
In plane strain problem, we have
0=== xzyzzz γγε
For axial loading alone, we have
0=xyγ and 0=yyσ
2.13.1

C.T. Sun
Expanding
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
=
=
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
+
+
+
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
0
0
0
0
0
00
0002
0002
0002
xy
xz
yz
zz
yy
xx
xy
xz
yz
zz
yy
xx
G
Gsymm
G
G
G
G
γ
γ
γ
ε
ε
ε
λ
λλ
λλλ
τ
τ
τ
σ
σ
σ
we have
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
=
=
+=
++=
++=
0
0
0
)(
)2(
)2(
xy
xz
yz
yyxxzz
yyxxyy
yyxxxx
G
G
τ
τ
τ
εελσ
ελλεσ
λεελσ
Applying 0=yyσ , we have
GPa
G
GG
G
G
xxxx
xx
xx
xxxx
εε
ε
λ
λ
λ
λε
λελσ
8.80
)0677.2725431.52(
)0677.275431.52)(0677.27(4
)2(
)(4
)2(
)2(
=
⋅+
+
=
+
+
=
+
−
⋅++=
(c) Plane stress condition
In plane stress problem, we have
0=== xzyzzz ττσ
For this problem we also have
0=xyγ and 0=yyσ
Therefore, the { } [ ] { }σε 66×
= ija can be used to obtain:
2.13.2

C.T. Sun
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
=
=
=
=
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−−
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
0
0
0
0
0
1
0
1
00
1
000
1
000
1
000
1
xy
xz
yz
zz
yy
xx
xy
xz
yz
zz
yy
xx
G
G
symm
G
E
EE
EEE
τ
τ
τ
σ
σ
συ
υυ
γ
γ
γ
ε
ε
ε
⎪
⎪
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎪
⎪
⎨
⎧
=
=
=
−=
−=
=
0
0
0
1
xy
xz
yz
xxzz
xxyy
xxxx
E
E
E
γ
γ
γ
σ
υ
ε
σ
υ
ε
σε
Thus, GPaE xxxxxx εεσ 72==
(d) The above analysis indicates that the axial stiffness of an aluminum bar under the
plane strain condition is (80.8-72)/72 = 12% greater than that under the plane
stress condition.
--- ANS
2.13.3

C.T. Sun
2.14 Show that the state of stress of a solid body of any shape placed in a pressured
chamber is a state of hydrostatic stress. Neglect the effect of the gravitational
force.
Solution:
Assume a solid body (without internal voids) with an arbitrary shape is placed in a
pressured chamber with a pressure 0σ as shown the figure below.
0σ
We will show that the hydrostatic stress listed below is the solution.
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
=
0
0
0
00
00
00
][
σ
σ
σ
σij --- (1)
First the constant stress field given by (1) satisfies the 3D equilibrium equations
(2.21 – 2.23). Second, this stress field satisfies the boundary condition, i.e., the
traction at any point on the surface of the body is given by
{} { } { }n
n
n
n
nt
z
y
x
ij 0
0
0
0
00
00
00
][ σ
σ
σ
σ
σ −=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎣
⎡
−
−
−
== ---(2)
where is the unit normal vector to the surface at the point of interest.
Last, the constant stress field given by (1) implies that the corresponding strain field is
also constant and the compatibility equations are also satisfied. Thus, the hydrostatic
stress field given by (1) is the solution.
{ }
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
=
z
y
x
n
n
n
n
2.14.1

C.T. Sun
2.15 Write the strain energy density expression in terms of stress components by
using (2.95) for isotropic solids and show that the Poisson’s ratio is bounded by
-1 and 0.5.
Solution:
From equation (2.95), we have the strain energy density:
}]{[}{
2
1
}]{[}{
2
1
σσεε acW TT
== (2.15.1)
For isotropic material, the stress-strain relationship can be expressed in terms of
elastic constants or elastic compliances:
[ ]
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
×
xy
xz
yz
zz
yy
xx
ij
xy
xz
yz
zz
yy
xx
c
γ
γ
γ
ε
ε
ε
τ
τ
τ
σ
σ
σ
66
Gccc 2332211 +=== λ , λ====== 323121231312 cccccc ,
where
( )( )υυ
υ
λ
211 −+
=
E
and
( )υ+
=
12
E
G
[ ]
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
×
xy
xz
yz
zz
yy
xx
ij
xy
xz
yz
zz
yy
xx
a
τ
τ
τ
σ
σ
σ
γ
γ
γ
ε
ε
ε
66
E
aaa
1
332211 === ,
E
aaaaaa
υ
−====== 323121231312 ,
G
aaa
1
The strain energy density in terms of stress components can be derived from equation
(2.15.1) as
2.15.1

C.T. Sun
{ }[ ]
)})(1(2)(2{
2
1
2
1
222222
66
yzxzxyzzyyzzxxyyxxzzyyxx
xy
xz
yz
zz
yy
xx
ijxyxzyzzzyyxx
E
aW
τττυσσσσσσυσσσ
τ
τ
τ
σ
σ
σ
τττσσσ
++++++−++=
⎪
⎪
⎪
⎪
⎭
⎪⎪
⎪
⎪
⎬
⎫
⎪
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎪
⎨
⎧
= ×
If we choose to use the principal directions as the coordianate system, then
,1xx σσ = ,2yy σσ = ,3zz σσ = and , where0yzxzxy === τττ 3,2,1 σσσ are principal
stresses. The strain energy density can be expressed in the form
)}(2{
2
1
323121
2
3
2
2
2
1 σσσσσσυσσσ ++−++=
E
W (2.15.2)
Note that the strain energy of an isotropic material compounds of two parts,
dilatational and distortional effect, i.e.,
dv WWW += (2.15.3)
where is the strain energy density associated with the volume dilatation and
is the strain energy density associated with the shape distortion.
vW
dW
We know that the dilatation is given by
00 εσ K=
where )(
3
1
3210 σσσσ ++= is the average stress,
V
VΔ
=++= 3210 εεεε is
unit volume change,
)21(3 υ−
=
E
K is the bulk modulus.
Hence,
2
000
2
1
2
1
σεσ
K
Wv == .
The expression of is obtained by subtracting from the total strain energy W
given by equation (2.15.2). We obtain
dW dW
2d J
G2
1
W =
where
])()()[(
6
1 2
13
2
32
2
212 σσσσσσ −+−+−=J
2.15.2

C.T. Sun
Thus,
2
2
0
2
1
2
1
J
GK
W += σ (2.15.4)
It is evident that W is always positive and vanishes only when all stresses components
vanish. To ensure that W be always positive, it is clear that K and G must be positive.
Noting the relations
0
)21(3
E
K >
−
=
υ
and 0
)1(2
E
G >
+
=
υ
We conclude that
5.01 <<− υ
--- ANS
2.15.3

C.T. Sun
2.16 Derive the compatibility equation for plane elasticity problems in terms of
stresses, i.e.,
0)(2
=+∇ yyxx σσ
Solution:
(a) For simplicity, we will use the following notations for differentiations with respect
to x and y: 2
2
,
x
xx
xxxx
∂
∂
=
σ
σ , 2
2
,
y
xx
yyxx
∂
∂
=
σ
σ , and
yx
xy
xyxy
∂∂
∂
=
τ
τ
2
, and so on. The
same differentiation notations are applied to the strain components
(b) For 2D problems, the compatibility equation is given by
xyxyxxyyyyxx ,,, γεε =+
(c) The stress-strain relations for isotropic solids are
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
⎥
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎢
⎣
⎡
−
−
=
⎪
⎭
⎪
⎬
⎫
⎪
⎩
⎪
⎨
⎧
xy
yy
xx
xy
yy
xx
G
EE
EE
σ
σ
σ
υ
υ
γ
ε
ε
1
00
0
1
0
1
Substituting the strain components into the compatibility equation, we have
xyxyxxyyxxyyyyxx
GEE
,,, )
1
()](
1
[)](
1
[ τσυσυσσ =+−+−
=> xyxyxyxyxxyyxxxxyyyyyyxx
EGEE
,,,,,,
)1(21
)(
1
)(
1
τ
υ
τσυσυσσ
+
==+−+− --- (1)
(d) Now we consider the equilibrium equations:
⎩
⎨
⎧
=+
=+
0
0
,,
,,
yyyxxy
yxyxxx
στ
τσ
Differentiating the first equation by x and the second one by y, we have
⎩
⎨
⎧
=+
=+
0
0
,,
,,
yyyyxyxy
xyxyxxxx
στ
τσ
--- (2)
Adding the above two equations in (2) we obtain
)(
2
1
,,, yyyyxxxxxyxy σστ +−= --- (3)
(e) Now substituting equation (3) into equation (1), we have
2.16.1

C.T. Sun
)](
2
1
)[1(2)()( ,,,,,, yyyyxxxxxxyyxxxxyyyyyyxx σσυσυσυσσ +−+=+−+−
=> )( ,,,, yyyyxxxxxxyyyyxx σσσσ +−=+
=> 0,,,, =+++ yyyyxxxxxxyyyyxx σσσσ
=> 0)(2
=+∇ yyxx σσ
where 2
2
2
2
2
yx ∂
∂
+
∂
∂
=∇
--- ANS
2.16.2

C.T. Sun
2.17 Consider a thin rectangular panel loaded as shown in Fig. 2.20. Show that the
Airy stress function
2
32
2
1 ycxycxc ++=φ
solves the problem. Find the constants c1, c2, c3.
x
y
a
b
0
Figure 2.20 Thin rectangular panel subjected to uniform tension
Solution:
is a second order equation in x and y it
atically satisfies the com Airy
we can find
(b)
(a) Since 2
32
2
1 ycxycxc ++=φ
autom patibility equation, 022
=∇∇ φ . So the given
stress function can be used to solve the problem if c1, c2, c3 that
satisfies the boundary conditions.
Stress matrix:
2
2
∂
=
φ
σ ,
y
xx
∂
=> 32cxx =σ ,
2
2
x
yy
∂
∂
=
φ
σ , => 12cyy =σ ,
yx
xy
∂∂
∂
−=
φ
τ
2
=> 2cxy −=τ
So we have,
−
−
=
12
23
2
][
cc
c
σ
(c) Applying boundary conditions:
⎥
⎦
⎤
⎢
⎣
⎡ 2c
(i) at
2
a
x = , => 1=n ,x 0=yn , 0σ=xt , 0=yt
we ]{[have tn = }{}σ
σ0σ
2.17.1

C.T. Sun
=> =>
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−
−
00
1
2
2 0
12
23 σ
cc
cc
⎪⎩
⎪
⎨
⎧
=
=
0
2
2
0
3
c
c
σ
2
b
y =(ii) at , => 0=xn , 1=yn , 0=xt , 0=yt
we have }{}]{[ tn =σ
=> =>
(d) The given Airy stress function is the solution to the problem and the values of the
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−
−
0
0
1
0
2
2
12
23
cc
cc
⎩
⎨
⎧
=
=
0
0
1
2
c
c
constants are 0=c , 0=c and1 2
2
3
o
c
σ
= .
--- ANS
2.17.2

C.T. Sun
2.18 Consider the rectangular panel shown in Fig. 2.20. Find the problem
that the Airy’s stress function solves. That is, find the tractions at the
boundary of the panel.
ba×
3
xy=φ
x
y
a
b
Figure tension
Solution:
2.20 Thin rectangular panel subjected to uniform
y to verify that the stress function satisfies the compatibility
(b) s are obtai the stress function as
(a) It is eas 3
xy=φ
equation, 022
=∇∇ φ .
The stresse ned from
2
2
∂
=
φ
σ , => xy6=
y
xx
∂
xxσ ,
2
2
x
yy
∂
∂
=
φ
σ , => 0=yyσ ,
yx
xy
∂∂
∂
−=
φ
τ
2
=>
So we have,
−
−
=
03
3
][ 2
2
y
y
σ
(c) The tractions at the boundary
2
3yxy −=τ
⎥
⎦
⎤
⎢
⎣
⎡ 6xy
2
a
1=xnx(i) On the vertical face at = , => , 0=yn ,
we have }]{[}{ nt σ=
=>
⎩
⎨ 222
2
3
3
3
6
0
1
03
6
y
ay
y
xy
y
yxy
ty
x
--- ANS
⎭
⎬
⎫
⎩
⎨
⎧
−
=
⎭
⎬
⎫
⎩
⎨
⎧
−
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
⎭
⎬
⎫⎧ 3t
2.18.1

C.T. Sun
(ii) On the vertical face at
2
a
x −= , => 1−=xn , 0=yn ,
=>
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧−
=
⎭
⎬
⎫
⎩
⎨
⎧−
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
⎭
⎬
⎫
⎩
⎨
⎧
222
2
3
3
3
6
0
1
03
36
y
ay
y
xy
y
yxy
t
t
y
x
--- ANS
(iii) On the top face at
2
b
y = , => 0=xn , 1=yn ,
=>
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
−
=
⎭
⎬
⎫
⎩
⎨
⎧−
=
⎭
⎬
⎫
⎩
⎨
⎧
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
⎭
⎬
⎫
⎩
⎨
⎧
0
4
3
0
3
1
0
03
36 22
2
2
by
y
yxy
t
t
y
x
--- ANS
(iv) On the bottom face at
2
b
y −= , => 0=xn , 1−=yn ,
=>
⎪⎭
⎪
⎬
⎫
⎪⎩
⎪
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
=
⎭
⎬
⎫
⎩
⎨
⎧
−
⎥
⎦
⎤
⎢
⎣
⎡
−
−
=
⎭
⎬
⎫
⎩
⎨
⎧
0
4
3
0
3
1
0
03
36 22
2
2
by
y
yxy
t
t
y
x
--- ANS
Note: The tractions on the edges can also be found directly from the stress
components at the same locations.
2.18.2

C.T. Sun
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 the x-y plane rotate as a rigid body, then,
zyu θ−=
zxv θ= (3.1.1)
),( yxw θψ=
where ),( yxψ is some function of x and y, called warping function, and θ 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 γεεε
So from the stress-strain relationship, we have
0==== xyzzyyxx τσσσ
Therefore the equilibrium equations reduce to
0=
∂
∂
+
∂
∂
yx
yzxz
ττ
This equation is identically satisfied if the stresses are derived from a stress
function ),( yxφ , so that
y
xz
∂
∂
=
φ
τ ,
x
yz
∂
∂
−=
φ
τ (3.1.2)
(c) From the displacement field and stress-strain relationship, we can obtain
y
x
w
z
u
x
w
xz θγ −
∂
∂
=
∂
∂
+
∂
∂
= (3.1.3)
x
y
w
z
v
y
w
yz θγ +
∂
∂
=
∂
∂
+
∂
∂
= (3.1.4)
So it forms the compatibility equation θ
γγ
2=
∂
∂
−
∂
∂
yx
xzyz
,
or in terms of Prandtl stress function θ
φφ
G
yx
22
2
2
2
−=
∂
∂
+
∂
∂
(3.1.5)
(d) Boundary conditions,
0=
ds
dφ
, or .const=φ 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=φ on the lateral surface of the bar.
(e) For a bar with circular cross-section, assume the Prandtl stress function as
3.1.1

C.T. Sun
)1( 2
2
2
2
−+=
a
y
a
x
Cφ which satisfies the boundary conditions stated above.
Substitute φ into (3.1.5), we obtain θGaC 2
2
1
−=
Then )(
2
222
ayx
G
−+−=
θ
φ
Using (3.1.2), we have
y
yG
xz θ
φ
γ −=
∂
∂
=
1
, and x
xG
yz θ
φ
γ =
∂
∂
−=
1
Comparing with (3.1.3) and (3.1.4), we have
yy
x
w
xz θθγ −=−
∂
∂
= => 0=
∂
∂
x
w
. Thus, )(yfw =
xx
y
w
yz θθγ =+
∂
∂
= => 0=
∂
∂
y
w
, 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

C.T. Sun
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
zyu θ−=
zxv θ= (3.2.1)
),( yxw θψ=
where ),( yxψ is a function of x and y, called warping function, and θ 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
So from the stress-strain relationship, we have
0=
∂
∂
+
∂
∂
yx
yzxz
ττ
This equation is identically satisfied if the stresses are derived from a stress
function ),( yxφ , so that
3.2.1

C.T. Sun
y
xz
∂
∂
=
φ
τ ,
x
yz
∂
∂
−=
φ
τ (3.2.2)
(c) From the displacement field and stress-strain relationship, we can obtain
y
x
w
z
u
x
w
xz θγ −
∂
∂
=
∂
∂
+
∂
∂
= (3.2.3)
x
y
w
z
v
y
w
yz θγ +
∂
∂
=
∂
∂
+
∂
∂
= (3.2.4)
So it forms the compatibility equation θ
γγ
2=
∂
∂
−
∂
∂
yx
xzyz
,
φφ
G
yx
22
2
2
2
−=
∂
∂
+
∂
∂
(3.2.5)
(d) Boundary conditions,
0=
ds
dφ
, or .const=φ
---
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
x
Cφ (here we use . Assuming0a )1( 2
2
2
2
−+=
ii a
y
a
x
Cφ
would be fine too).
Then substitute φ into (3.2.5), we have θGaC
2
0
2
1
−= . Thus we have
)1(
2 2
0
2
2
0
22
0
−+−=
a
y
a
xaGθ
φ . (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

C.T. Sun
To satisfy the traction boundary conditions we must show 0=
ds
dφ
on the
traction free surfaces.
.)1(
2
)( 2
0
22
0
const
a
aaG
ar i
i =−−==
θ
φ that is 0| == iar
ds
dφ
.0)1(
2
)( 2
0
2
0
2
0
0 =−−==
a
aaG
ar
θ
φ that is 0| 0
==ar
ds
dφ
It shows that the B.C.’s have been satisfied.
(4) Since equilibrium equations, compatibility equations and traction boundary
conditions are all satisfied, the Prandtl stress function for bars of circular
solid sections is also valid for bars of hollow circular sections.
--- ANS
2. Compare torsion constant
(1) The torque produced by the stresses is
∫∫ ∂
∂
−
∂
∂
−=
A
dA
y
y
x
xT )(
φφ
(3.2.7)
Substituting (3.2.6) into (3.2.7) and use polar coordinates to perform
integration, we have,
)](
2
[|)
4
1
)(2()()(
44
0
4222 0
i
a
a
AA
aaGrGrdrdrGdAyxGT i
−===+= ∫∫∫∫
π
θπθθθθ
Comparing with JGT θ= , we have the torsion constant )(
2
44
0 iaaJ −=
π
(2) Using (3.59) in the textbook for thin-walled sections, we have the
approximate torsion constant
∫
=
tds
A
Japp
/
4
2
where A is the area enclosed by the centerline of the wall section.
2
0
20
)(
4
)
2
( i
i
aa
aa
A +=
+
=
π
π , and iaat −= 0
Therefore )()(
4)(
)(
4
/
4
0
3
0
0
0
4
0
2
2
ii
i
i
i
app aaaa
aa
aa
aa
tds
A
J −+=
−
+
+
==
∫
π
π
π
3. In order to have the approximate to be within 1 percent of the exact , one
must have
J J
01.0≤
−
J
JJapp
3.2.3

C.T. Sun
Substituting and into the above error equation, we haveappJ J
01.0
)(2
)(
)(
2
)(
2
)()((
4
22
0
2
0
44
0
44
00
3
0
≤
+
−−
=
−
−−−+
=
−
i
i
i
iii
app
aa
aa
aa
aaaaaa
J
JJ
π
ππ
Because and are positive real number, we haveia 0a
01.0
)(2
)(
22
0
2
0
≤
+
−
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
2235.18174.0
0
≤≤
a
ai
Since we have the solutioniaa >0 8174.0
0
≥
a
ai
Therefore the condition on the wall thickness t is
0000 1826.08174.0 aaaaat i =−≤−=
(OR iiii aaaaat 2235.0
8174.0
1
0 =−≤−= )
--- ANS
3.2.4

C.T. Sun
3.3 Consider the straight bar of a uniform elliptical cross-section. The semimajor
and semiminor axes are a and b, respectively. Show that the stress function of
the form
)1( 2
2
2
2
−+=
b
y
a
x
Cφ
provides the solution for torsion of the bar.
Find the expression of C and show that
22
33
ba
ba
J
+
=
π
3
2
ab
Ty
zx
π
τ
−
= ,
ba
Tx
zy 3
2
π
τ =
and the warping displacement
xy
Gba
abT
w 33
22
)(
π
−
=
Solution:
Recall:
1. Saint-Venant assumed that as the shaft twists the plane cross-sections are warped
zyu θ−=
zxv θ= (3.3.1)
),( yxw θψ=
where ),( yxψ is warping function, and θ 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
From the stress-strain relationship, we have
0=
∂
∂
+
∂
∂
yx
yzxz
ττ
which is identically satisfied if the stresses are derived from a stress function
),( yxφ , so that
y
xz
∂
∂
=
φ
τ ,
x
yz
∂
∂
−=
φ
τ (3.3.2)
3. From the displacement field and stress-strain relationship, we can obtain
3.3.1

C.T. Sun
y
x
w
z
u
x
w
xz θγ −
∂
∂
=
∂
∂
+
∂
∂
= (3.3.3)
x
y
w
z
v
y
w
yz θγ +
∂
∂
=
∂
∂
+
∂
∂
= (3.3.4)
The compatibility equation becomes θ
γγ
2=
∂
∂
−
∂
∂
yx
xzyz
,
φφ
G
yx
22
2
2
2
−=
∂
∂
+
∂
∂
(3.3.5)
4. The boundary condition along the bounding surface is
0=
ds
dφ
, or .const=φ
---
(a) Let the stress function be of the form )1( 2
2
2
2
−+=
b
y
a
x
Cφ . 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
( 2
b
y
C
y
xz =
∂
∂
=
φ
τ , )
2
( 2
a
x
C
x
yz =
∂
∂
−=
φ
τ
Substituting the above stress expressions into the equilibrium equations, we
have
000 =+=
∂
∂
+
∂
∂
yx
yzxz
ττ
(2) Compatibility equations
Substituting )1( 2
2
2
2
−+=
b
y
a
x
Cφ into (3.3.5) we get
22
22
ba
ba
GC
+
−= θ . (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
dφ
on the
traction free lateral surface.
Since the boundary of the cross section is given by the equation
3.3.2

C.T. Sun
012
2
2
2
=−+
b
y
a
x
,
it is easy to see that 0)1( 2
2
2
2
=−+=
b
y
a
x
Cφ on the free surface and therefore
it satisfies the required condition 0=
ds
dφ
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
dA
y
y
x
xT )(
φφ
(3.3.7)
Substituting )1( 2
2
2
2
−+=
b
y
a
x
Cφ into (3.3.7), then we have,
∫∫∫∫ +−=−−=
AA
dA
b
y
a
x
CdA
b
y
Cy
a
x
CxT )())
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
π=+∫∫ )( 2
2
2
2
So we have the torsion abCT π−=
By substituting C and utilizing θGJT = , we have
22
3322
22
)(
ba
ba
G
ab
ba
ba
G
G
abC
J
+
=+
−−
=
−
=
π
θ
πθ
θ
π
(3.3.8)
--- ANS
(2) 322
2
)
2
()
2
(
ab
Ty
b
y
ab
T
b
y
C
y
xz
ππ
φ
τ −=
−
==
∂
∂
= , (3.3.9)
and
ba
Tx
a
x
C
x
yz 32
2
)
2
(
π
φ
τ =−=
∂
∂
−= (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 have 3
2
abG
Ty
xz
π
γ −= and
baG
Tx
yz 3
2
π
γ = .
Also we need to know 33
22
)(
baG
baT
GJ
T
π
θ
+
==
So from (3.3.3) and (3.3.4), we can rewrite in
3.3.3

C.T. Sun
33
22
33
22
3
)()(2
baG
yabT
baG
ybaT
abG
Ty
y
x
w
xz
πππ
θγ
−
=
+
+−=+=
∂
∂
(3.3.11)
33
22
33
22
3
)()(2
baG
xabT
baG
xbaT
baG
Tx
x
y
w
yz
πππ
θγ
−
=
+
−=−=
∂
∂
(3.3.12)
From (3.3.11), we can obtain
)(
)(
),( 33
22
yfxy
baG
abT
yxw +
−
=
π
(3.3.13)
Then differentiating (3.3.13) with respect to y, we have
)(
)(),(
33
22
yfx
baG
abT
y
yxw
′+
−
=
∂
∂
π
.
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
xy
baG
abT
yxw
π
−
=
--- ANS
And it is easy to also find that the warping function
xy
ba
ba
baG
baT
xy
baG
abT
yxw
yx 22
22
33
22
33
22
)(
.
)(
),(
),(
+
−
−=
+
−
==
π
π
θ
ψ
3.3.4

C.T. Sun
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
A
J
/
4
2
(3.4.1)
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 of A
(1) Elliptical cross-section
For the elliptical cross-section, the cross-sectional area is
abAellp π= , (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
π
φφdkaL ,
where tyeccentrici
a
ba
k =
−
=
22
For the purpose to just comparing the area enclosed by the centerline of the
wall section, We approximate the perimeter with
2
2
22
ba
L
+
= π (3.4.3)
By changing the form of (3.4.3) se have
3.4.1

C.T. Sun
2222
)
2
(2 aCa
L
b −=−=
π
, where 2
)
2
(2
π
L
C = (3.4.4)
Substituting (3.4.4) into (3.4.2) we have,
22
aCaAellp −= π
We can find the optimum solution by 0=
∂
∂
a
A
, by some operations leads to
0
2
22
22
=
−
−
=
∂
∂
aC
aC
a
A
π , therefore we have
2
2
C
a = for 0, >ba
Substitute it back to (3.4.4), we have a
C
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
ππ 2
)
2
(
2
2
2
2
2
LLC
ba ====
Finally, for a circle, the area enclosed by the centerline is
222
cir L0796.0)
2
L
(aA ===
π
ππ
--- ANS
(2) Rectangular section
For rectangular section, the perimeter is
)(2 qpL += , (3.4.6)
where p and q are 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
( p
L
ppqArec −==
We use 0=
∂
∂
p
A
to find the optimal solution,
02
2
=−=
∂
∂
p
L
p
A
, we have
4
L
p = , and from (3.4.6), it is clear that
4
L
qp == , 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 ===
3.4.2

C.T. Sun
--- ANS
(3) Equilateral triangular section.
For a equilateral triangle, the length of the lateral side is
3
L
l = .
The area enclosed by the centerline of this triangular thin-walled section is
222
tri L048.0)
3
L
(
4
3
l
4
3
A ===
--- 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 variables
a
b
and
p
q
from 0~1.
(We here assume a>b and p>q)
For ellipse, 22
2222
22
)
1
(
2
1
))
2
(2)(( L
b
a
a
b
L
ba
ab
ba
ba
ababAellp
+
=
+
≈
+
+
==
ππ
πππ
For rectangle, 2
2
2
22
2
)1(
4
1
)
2
(
)()(
)(
L
p
q
p
q
L
qp
pq
qp
qp
pqpqArec
+
=
+
=
+
+
==
For equilateral triangle, 2
36
3
LAtri =
We can illustrate 2
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

C.T. Sun
3.4.4

C.T. Sun
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
A
J
/
4
2
(3.5.1)
We have 2
321 2001020)( cmLLLA =×=+= . The torsion constant can be
simply derived as
J
3.5.1

C.T. Sun
4
2
321
2
321
2
1 800
3.0)1020(2
)200(4
/)(2
])[(4
/
4
cm
tLLL
LLL
tds
A
J 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
1
321311
31
1
1
1 LqqLLq
tLGLt
qds
AG cell
−++== ∫θ (3.5.3)
and for the right cell
))()2((
2
1
2
1
321322
32
2
2
2 LqqLLq
tLGLt
qds
AG cell
−−+== ∫θ (3.5.4)
Since the entire thin-wall section must rotate as a rigid body in the plane, we
require the compatibility condition
θθθ == 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 using
θG
T
J = and (3.5.3), we have
)22(
)(4
))()2((
2
1
)22(
323111
23213131
321311
31
2211
LqLqLq
tqLLqLLLL
LqqLLq
tLGL
G
qAqA
J
−+
+
=
−++⋅
+
=
(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(
)(4
cm
LqLqLq
tqLLqLLLL
J cell =
−+
+
=−
3.5.2

C.T. Sun
--- ANS
(3) Case 2: ,cmL 51 = cmL 152 = and cmL 103 =
From (3.5.6), 112 25.1
)
5
10
15
10
22(
)
15
10
5
10
22(
qqq =
++
++
=
Then substituting into (3.5.7) we have
4
111
11
22 2857.814
)1025.110252(
3.0)25.11015105(1054
cm
qqq
qq
J 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

C.T. Sun
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
23213131
12 800
)22(
)(4
cm
LqLqLq
tqLLqLLLL
J 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

C.T. Sun
∫
−
−− =
tds
A
J
cutnot
cutnotcell
/
)(4 2
, and 32 LLA cutnot =− (3.6.3)
=> 4
2
300
)1010(2
3.0)1010(4
cmJ cutnotcell =
+
⋅×
=−−
and
∑=−
i
iicutcell tbJ
3
3
1
(3.6.4)
=> 43
27.03.0)101010(
3
1
cmJ cutcell =⋅++=−
So, from (3.6.2) we get
GGJGJGJ cutcellcutnotcellopencut 27.300=+= −−−−
---ANS
(c) The reduction of torsional rigidity is obtained as
%5.62625.0
800
27.300800
GJ
GJGJ
R
1cell2
opencut1cell2
==
−
=
−
=
−
−−
--- ANS
3.6.2

C.T. Sun
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=τ .
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 56
allowallow ×=××==τ
Then we have the torque capability as
mN224401061.5)2.01.0(2qA2T 5
allowallow −=×××==
--- ANS
3.7.1

C.T. Sun
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 θ 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 θ . For aluminum
alloy 2024-T3, we have the shear modulus
GPa27
)33.01(2
72
)1(2
E
G =
+
=
+
=
υ
(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
1q
2q 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 2
22
21 m098.0
8
)5.0(
8
d
AA ====
ππ
,
The twist angle of the left cell is
))((
2
1
2
1
21
12
12
1
1
1
1
1
1
1 qq
t
s
q
t
s
AGt
qds
AG cell
−+== ∫θ (3.8.2)
where m785.0
2
d
s1 ==
π
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
1
21
12
12
2
2
2
2
2
2
2 qq
t
s
q
t
s
AGt
qds
AG cell
−−== ∫θ (3.8.3)
3.8.1

C.T. Sun
Again, we have m785.0
2
d
s2 ==
π
, the length of the right side wall.
m/rad0524.0m/321 ==== o
θθθ (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
122121
002.0
5.0
003.0
785398.0
002.0
5.0
001.0
785398.0
qqqq −=+
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
×
−−
==θ
From which we obtain
m/N500,453q1 =
Subsequently from (3.8.5) we obtain
m/N765000q687.1q 12 ==
--- ANS
(c) The applied torque
From (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 θGJT =
So the torsion constant can be derived as
44
9
m1069.1
)0524.0)(1027(
239300
G
T
J −
×=
×
==
θ
--- ANS
3.8.2

C.T. Sun
3.9 For the bar of Fig. 3.37, find the maximum torque if the allowable shear stress is
MPaallow 187=τ . What is the corresponding maximum twist angle θ ?
Figure 3.37 Two-cell section
Solution:
(a) Assume the material is linearly elastic under the twist angle θ . For aluminum
GPa27
)33.01(2
72
)1(2
E
G =
+
=
+
=
υ
(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 2
22
21 m098.0
8
)5.0(
8
d
AA ====
ππ
,
))((
2
1
2
1
21
12
12
1
1
1
1
1
1
1 qq
t
s
q
t
s
AGt
qds
AG cell
−+== ∫θ (3.9.2)
where m785.0
2
d
s1 ==
π
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
1
21
12
2
2
2
2
2
2
2 qq
t
s
q
t
s
AGt
qds
AG cell
−−== ∫θ (3.9.3)
where m785.0
2
d
s2 ==
π
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
3.9.1

C.T. Sun
θθθ == 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
122121
002.0
5.0
003.0
785398.0
002.0
5.0
001.0
785398.0
qqqq −=+
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
×
−−
=θ , and then
θ8662000q1 =
Subsequently, (3.9.6)
θ000,600,14q687.1q 12 == (3.9.7)
θ000,953,5qqq 2112 −=−= (3.9.8)
Note the units are m/radinθ , 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 by
t
q
=τ . We have
θ
θ
τ 9
1
1
1 1066.8
001.0
8662028
t
q
×=== (3.9.9)
θ
θ
τ 9
2
2
2 1087.4
003.0
14615612
t
q
×=== (3.9.10)
θ
θ
τ 9
12
12
12 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 is
6
allow
9
1 101871066.8 ×=≤×= τθτ
Therefore the maximum twist angle is
m/24.1m/rad0216.0max
o
==θ
--- 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
3.9.2

C.T. Sun
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 ⋅× 5
102
Figure 3.38 Two-cell three-stringer thin-walled section
Solution:
(a) Assume the material is linearly elastic under the applied torque. For aluminum
GPa
E
G 27
)33.01(2
72
)1(2
=
+
=
+
=
υ
(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 .
1q
2q
2112 qqq −= 1q
We have the torque for the two-cell section as
21 21 22 qAqAT += (3.10.1)
where 2
22
1 565.0
8
)2.1(
8
m
d
A ===
ππ
,
and 2
2 2.1
2
)2.1(2
2
m
bh
A ===
The twist angle of the left. cell is
))((
2
1
2
1
21
1
12
1
1
1
1
1
1
1 qq
t
s
q
t
s
AGt
qds
AG cell
−+== ∫θ (3.10.2)
where m
d
s 88.1
2
1 ==
π
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
3.10.1

C.T. Sun
))((
2
1
2
1
21
1
12
2
3
3
2
2
2
2
2
2
2 qq
t
s
q
t
s
q
t
s
AGt
qds
AG cell
−−+== ∫θ (3.10.3)
Again, we have , the length of the lower wall,ms 22 =
and ms 33.22.12 22
3 =+= , the length of the inclined wall of thickness t3 in
the right cell.
θθθ == 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
1
1222121 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 +=+=
=> mN
AA
T
q /51966
)2.1)(265.2()565.0)(2(
102
265.22
5
21
1 =
+
×
=
+
=
--- 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
qq
t
s
q
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
12
1
1
1
1
1
o
=×=
×
×−+
=−+==
−
θθ --- ANS
3.10.2

C.T. Sun
3.11 What is the maximum torque for the structure of Fig. 3.38 if the allowable twist
angle θ is ?m/2o
Figure 3.38 Two-cell three-stringer thin-walled section
Solution:
(a) Assume the material used is still Aluminum alloy 2024-T3. For aluminum alloy
2024-T3, we have the shear modulus
GPa27
)33.01(2
72
)1(2
E
G =
+
=
+
=
υ
(b) Denote the shear flow on the left cell as , and the shear flow on the right cell as
. Both are assumed positive in the counterclockwise direction. The shear flow
in the vertical web is
1q
2q
2112 qqq −= , from bottom to top.
The torque for two-cell section is
21 21 22 qAqAT += (3.11.1)
where 2
22
1 m56.0
8
)2.1(
8
d
A ===
ππ
,
and 2
2 2.1
2
)2.1(2
2
m
bh
A ===
))((
2
1
2
1
21
1
12
1
1
1
1
1
1
1 qq
t
s
q
t
s
AGt
qds
AG cell
−+== ∫θ (3.11.2)
where m88.1
2
d
s1 ==
π
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
))((
2
1
2
1
21
1
12
2
3
3
2
2
2
2
2
2
2 qq
t
s
q
t
s
q
t
s
AGt
qds
AG cell
−−+== ∫θ (3.11.3)
3.11.1

C.T. Sun
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.
θθθ == 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
)q
005.0
2.1
q
007.0
33.2
q
007.0
2
(
2.1
1
)q
005.0
2.1
q
005.0
88.1
(
566.0
1
1222121 −+=+
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
θ
m/rad035.0allowable21 =<== θθθθ . By equations (3.11.2),
)m/rad(035.0)m/rad(q1013.1
)565.0)(1027(2
q))13.11(
005.0
2.1
005.0
88.1
(
))qq(
t
s
q
t
s
(
AG2
1
allowable1
8
9
1
21
1
12
1
1
1
1
1
=<×=
×
×−+
=−+==
−
θ
θθ
We then have m/N000,095,3q1 ≤
(d) To find the maximum torque, we can use equation (3.11.1)
mN1019.1000,095,3)]2.1)(26.2()56.0)(2[(
q)A264708.2A2(qA2qA2T
7
1212211
−×=×+<
+=+=
Therefore the maximum torque is
mN1019.1T 7
max −×=
--- ANS
It should be noted that under this torque the shear stress has already exceeded the
yield condition of Al 2024-T3. Consequently, this solution may not be of practical
significance if allowable stress condition is to be satisfied too.
3.11.2

C.T. Sun
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 is aGJ
43
i
3
iia mmG5400G)3)(200)(
3
1
(3tb
3
1
GGJ === ∑
--- ANS
(b) Fig. 3.39b is a cross-section of a closed thin-wall, its torsional rigidity is bGJ
∫
=
t
ds
A
GGJb
2
4
, where
4
3 2
b
A = ,
46
342
106
4)3(4
34
GmmG
tb
b
tb
G
t
ds
A
GGJb ×====
∫
--- ANS
(c) The ratio of the torsional rigidities is
1111
G5400
G106
GJ
GJ 6
a
b
=
×
=
--- ANS
3.12.1

C.T. Sun
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
3.13.1

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