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Chapter 3 Torsion
Introduction
-- Analyzing the stresses and strains in machine parts which are subjected to torque T
Circular
-- Cross-section Non-circular
Irregular shapes
-- Material (1) Elastic
(2) Elasto-plastic
-- Shaft (1) Solid
(2) Hollow
3.1 Introduction
T is a vector
Two ways of expression
-- Applications:
a. Transmission of torque in shafts, e.g. in automobiles
Assumptions in Torque Analysis:
a. Every cross section remains plane and undistorted.
b. Shearing strain varies linearly along the axis of the shaft.
3.2 Preliminary Discussion of the Stresses in a Shaft
( ) dA T
dF T
Free-body Diagram
Where = distance (torque arm)
Since dF = dA
The stress distribution is Statically Indeterminate.
3.3 Deformations in a Circular Shaft
= (T, L) -- the angle of twist (deformation)
Rectangular cross section warps under torsion
G
3.4 Stresses in the Elastic Range3.4 Stresses in the Elastic Range
Hooke’s LawHooke’s Law
max
c
max
G Gc
G max max G
max
c
Therefore, Therefore, (3.6)
Substituting Eq. (3.9) into Eq. (3.6)
JT
max TcJ
412J c
(3.10)
(3.9)
These are elastic torsion formulas.
For a solid cylinder:
For a hollow cylinder: 4 42 1
12
( ) J c c
Ductile materials fail in shear (90o fracture)
Brittle materials are weaker in tension (45o fracture)
3.5 Angle of Twist in the Elastic Range
max
cL
maxmax maxsin
Tcce
G J
TL
JG
(3.3)
max
TcJG
(3.15)
max
c TcL JG Eq. (3.3) = Eq. (3.15)
Therefore,
Hence,
Shafts with a Variable Circular Cross SectionShafts with a Variable Circular Cross Section
0
LTdxJG
Tdx
dJG
3.6 Statically Indeterminate Shafts
-- Must rely on both
(1) Torque equations and
(2) Deformation equation, i.e. TLJG
0T
Example 3.05
3.7 Design of Transmission Shafts
P power T
2P f T
fP
T2
-- Two Parameters in Transmission Shafts:
a. Power P
b. Speed of rotation
where = angular velocity (radians/s) = 2
= frequency (Hz)
[N.m/s = watts (W)] (3.21)
max TcJ
fP
T2
max
J Tc
(3.21)
(3.9)
4 31 12 2
/J c and J c c
For a Solid Circular Shaft:
Therefore,
312 max
Tc
1 32
/
max
Tc
UT
T cR
J
max TcJ
(3.9)
If we can determine experimentally an Ultimate Torque, TU,
then by means of Eq. (3.9), we have
RT = Modulus of Rupture in Torsion
3.10 Circular Shafts Made of an Elasto-Plastic Material
Y Y
JT
c
max TcJ
Case I: < Y Hooke’s Law applies, < max Case I
Case II
Case II: < Y Hooke’s Law applies, = max
TY = max elastic torque
Since
Case III: Entering Plastic Region
312 Y YT c
312
/J C c
Y Case III
0 Y:
Y
Y
Y – region within the plastic range
Y c:
(3-29)
2
02
cT d
2 2
0
3 3 3
33
3
2 2
1 2 22 3 3
2 11
3 4( )
Y
Y
cY
elastic plastic YY
Y Y Y Y Y
YY
T T T d d
c
T cc
3
3
4 11
3 4( )
Y
YT Tc
(3.26)
By evoking Eq. (3.26)
(3.31)
312 Y YT c
3.12 Torsion of Noncircular Members
0 0 yx yz
0 0 zx zy
0 0 xy xz
A rectangular shaft does not axisymmetry.
3.13 Thin-Walled Hollow Shafts
0 A BF - F = 0 xF ( ) A A AF t x
0( ) ( )
A A B B
A A B B
t x t x
t t