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Week 8
§ 3-9 Strain Energy in Torsion and Pure Shear
Fig. 3-34 Prismatic bar in pure torsion.
PGI
TL=φ
Fig. 3-35 Torque-rotation diagram for a bar in pure torsion (linearly elastic material).
PGITL
=φ
L
GIGI
LTU P
P 22
22 φ==
(單位 Joule=N.m 或 ft-lb)
1
φW= 2
TU =
(類似軸向受力元件之
2δTWU == )
Week 8
Non-uniform Torsion
( )∑∑==
==n
i iPi
iin
ii IG
LTUU
1
2
1 2
( )[ ]( )xGI
xTdUP2
2
=
( )[ ]( )∫=
L
P xGIdxxTU
0
2
2
重要假設:linearly elastic material
Strain Energy Density in Pure Shear
Fig. 3-36 Element in pure shear.
Shear Force
V=τht
©
2
0
0
1
2
Week 8
δ=γh
the Strain energy
2δVWU ==
2
2thU τγ=
volume of the element=h2t
Strain-energy density
2τγ
=U
代入τ=Gγ
G2
2
u τ= 或
2
2γGu =
(單位 J/m3 或 psi) (pa)
例 3-10
Fig. 3-37 Example 3-10. Strain energy produced by two loads.
©
2
0
0
1
3
Week 8
TLI
(
(
(
(
(a)Ta (b) Tb (c)Ta & Tb (d)Ta=100 N.m,
b=150N.m =1.6m,G=80Gpa
P=79.52×103 mm4
a)P
aa GI
LT2
2
=U
b)P
b
P
b
b GILT
GI
LT
422 2
2
=
=U
c)
( )
P
b
P
ba
P
a
pP
b
iP
iin
ic
GILT
GILTT
GILT
GI
LTbTa
GI
LT
IGLT
U
422
22
22
)(222
222
1
++=
+
+
== ∑=
d)將數值代入
Ua=1.26 J Ub=1.41 J
Uc=4.56 J
4
Week 8
例 3-11
Fig. 3-38 Example 3-11. Strain energy produced by
a distributed torque.
(a) T(x)=tx
( )[ ]( ) ( )∫ ∫ ===
L
P
L
PP GILtdxtx
GIxGIdxxTU
0
32
0
22
621
2
(b) 數值代入
U=580 in-lb
5
Week 8
例 3-12
Fig. 3-39 Example 3-12. Tapered bar in torsion.
(work by the applied torque)=(Strain energy of the bar)
W=U
( ) ( )[ ]
( )
( )
( )[ ]( )
( )
−
−=
−
+
==
⋅
++=⇒
⋅+
+=
=
∫∫
33
2
0 4
2
0
2
4
4
113
16
)(
162
32
32
BAAB
L
ABA
L
P
ABAP
ABA
P
ddddGLT
AppendixCL
ddd
dxGT
xGIdxxTU
XL
dddXI
XL
dddxd
xdxI
π
π
π
π
積分查表
6
Week 8
§ 3-10 Thin-Walled Tubes
Shear Stress & Shear Flow
Fig. 3-40 Thin-walled tube of arbitrary cross-sectional shape.
Fb=τbtbdx Fc=τctcdx Fb=Fc
τbtb=τctc Shear Flow f=τt=const
7
Week 8
Torsion Formula for Thin-Walled Tubes Dash line median line
Fig. 3-41 Cross section of thin-walled tube.
Total shear force on the element f ds The moment
dT=rfds
the total torque
∫=Lm
rdsfT0
其中 Lm為中線總長度
中線(虛線)所夾面積 Am
fAmT
AmrdsLm
2
20
=⇒
=∫
Shear flow
tAmT
tAmTf
2
2
=⇒
==
τ
τ
8
Week 8
Thin-walled circular tube
Am=πγ2
tT
22πγτ =
Fig. 3-42 Thin-walled circular tube.
©2001 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark
Thin-walled rectangular tube
Am=bh
bht
Tvert
12=τ
bht
Thoriz
22=τ
Fig. 3-43 Thin-walled rectangular tube.
9
Week 8
Strain Energy & Torsion Constant
Fig. 3-41 Cross section of thin-walled tube.
Fig. 3-40
Thin-walled tube of arbitrary cross-sectional shape.
10
Week 8
The volume of the element abcd(Fig 3-40)
tdsdx
the strain-energy density (Eq.3-55a)
G2
2τ
the total strain energy of the element
dxtds
Gf
dxtds
Gt
tdsdxG
dU
2
2
2
2
22
2
=
=
=
τ
τ
the total strain energy
∫
∫
∫∫∫
=
=
=
==
Lm
m
Lm
LLm
tds
GALT
AmTf
tds
Gt
dxtds
GdUU
02
2
0
22
00
2
8
2
2
2τ
τ
定義 torsion constant
∫=
Lm
tds
J
0
4
GJ
LTU2
2
=
11
Week 8
Special case t=const
LmtA
J m24
=
Thin-walled circular tube Lm=2πr Am=πr2
J=2πr3t Rectangular tube
Am=bh
21
2122
210
20
10
2
222
htbttthbJ
tb
th
tds
tds
tds bhLm
+=
+=+= ∫∫∫
Angle of Twist
Fig. 3-44 Angle of twist φ for a thin-walled tube.
GJTL
GJLTUWT
=⇒
===
φ
φ22
2
(GJ:torsion rigidity)
12
Week 8
例 3-13
Fig. 3-45 Example 3-13. Comparison of approximate and exact theories of torsion.
©
2
0
0
1
B
r
o
Thin-Wall approximation
2321 22 βππγτ
tT
tT
==
tr
=β
Torsion formula
PI
trT
+
= 22τ
( )
( )( )
( )( )
( )12214
1412
42
42222
2
2
1
23222
2244
++
=
++
=++
=⇒
+=
−−
+=
βββ
ττ
ββπβ
πτ
ππ
tT
trrttrT
trrttrtrI P
13
Week 8
例 3-14
Fig. 3-46 Example 3-14. Comparison of circular and square tubes.
Same material Same length Same wall thickness Same cross-sectional area Circular tube Am1=πr2
J1=2πr3t , A1=2πrt
Square tube
4
222
2rbAm π
==
62.0162
8
79.04
4
4,8
2
3
33
1
2
2
1
2
22
1
2
2
1
2
333
2
====
====⇒
===
ππ
πφφ
ππ
π
ττ
π
trtr
JJ
r
r
AmAm
btAtrTbJ
14
Week 8
§ 4-1 Introduction
©
2
0
0
1
B
r
o
Fig. 4-1 Examples of beams subjected to lateral loads.
Beams
15
Week 8
§ 4-2 Types of Beams,Loads,and Reaction
Simply supported beam Cantilever beam Beam with a overhang
Fig. 4-2 Types of beams: (a) simple beam, (b) cantilever beam, and (c) beam with an overhang.
©
20
01
Br
oo
ks
/C
ol
e,
Types of Loads Concentrated load Distributed load Linearly varying load Simple beam Fig 4-2a
ΣFhoriz=0 HA-P1cosα=0
HA=P1cosα
ΣMB=0 , -RAL+(P1sinα)(L-a)+P2(L-P)+qc2 / 2=0
ΣMA=0 , RBL-(P1sinα)(a)-P2b+qc(L-c / 2)=0
( )( )L
cLqc
LbP
LaLPRB
−
++−
= 2sin 21 α
( )( )L
qcL
bLPL
aLPRA 2)(sin 21 +
−+
−=
α
16
Week 8
Cantilever beam Fig 4-2b
ΣFhoriz=0
135 3P
H A =
ΣFvert=0
bqqPRA
+
+=213
12 213
ΣMA=0
−+
−+=⇒
=
−−
−−
−
3232
21312
0323
2213
12
213
213
bLbqbLbqaPM
bLbqbL
bqa
PM
A
A
Beam with an overhang Fig 4-2c
ΣMB=0 , -RAL+P4(L-a)+M1=0
ΣMA=0 , -R4a+PBL+M1=0
( )L
ML
aLPRA14 +
−=
LM
LaPRB
14 −=
17
Week 8
§ 4-3 Shear Forces &Bending Moments
Fig. 4-4 Shear force V and bending moment M in a beam.
©
2
0
0
1
B
ΣFvert=0,P-V=0 或 P=V
ΣM=0,M-PX=0 或 M=PX
Sign Convention Fig. 4-5 Sign
conventions for shear force V and bending moment M.
+ in the positive direction
Fig. 4-6 Deformations (highly exaggerated) of a beam element caused by (a) shear forces, and (b) bending moments.
©
2
0
0
1
B
r
18
Week 8
例 4-1
LMPRA
0
43
+=
LMPRB
0
4+=
Fig. 4-7 Example 4-1.
Shear forces and bending moments in a simple beam.
©
2
0
0
1
B
r
(a) To the left of the midpoint (Fig.4-7b)
ΣFvert=0,RA-P-V=0
V=RA-P= L
MP 0
4−−
ΣM=0, 0)4
()2
( =++ MLPLRA−
28
)4
()2
( 0MPLLPLRM A −=−=
(b)To the right of the midpoint (Fig.4-7c)
LMPV 0
4−−=
280MPLM +=
19
Week 8
例 4-2
Fig. 4-8 Example 4-2.
Shear force and bending moment in a cantilever beam.
©
2
0
0
1
B
Shear force
Lxq
q 0=
Form Fig.4-8b
( ) VLxq
xLxq
−==
221 2
00
when X=0,V=0
X=L,20
maxLq
−=V
Bending moment
ΣM=0, ( ) 0)3
(2
0 =
+
XXLxqLM
LXq
M6
30−
=
when X=0,M=0
X=L,LLq
M6
20
max−
=
20
Week 8
例 4-3
©
2
0
0
1
B
r
o
Fig. 4-9 Example 4-3. Shear force and bending moment in a beam with an overhang.
q=20 lb/ft
P=14K
RA=11K,RB=9K,
Form Fig.4-9b
ΣFvert=0,11K-14K-(0.2K/ft)(15ft)-V=0
V=-6K
ΣMD=0,-(11K)(15ft)+(14K)(6ft)+(0.2K/ft)(15ft)(7.5ft)+M=0
M=58.5 K-ft
21
Week 8
22
Homework
3.9-2 3.10-2 4.3-2
3.9-4 3.10-4 4.3-4
3.9-6 3.10-6 4.3-6
3.9-8 3.10-10 4.3-8
4.3-10
4.3-12