torsion introduction -- analyzing the stresses and strains in machine parts which are subjected to...
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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.
-- Must rely on “deformation” to solve the problem.
Analyzing a small element:
3.3 Deformations in a Circular Shaft
= (T, L) -- the angle of twist (deformation)
Rectangular cross section warps under torsion
' 'CD C D
A circular plane remains circular plane
L (in radians)
Determination of Shear Strain
The shear strain
max
cL
max
c
max Lc
= c = radius of the shaft
L Since
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)
1
2min max
cc
max
JT
c
2maxmax
T dA dA dAc c
( ) dA T (3.1) max
c
(3.9)
But 2 dA J
Therefore, Or, max TcJ
(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
0 02 45 2max max( )cos F A A
2 oA A 3 13. ( )EqA
Since
max 0max
0
2
2
F A
A A
(3-13)
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,
i i
i i i
T J
J G
For Multiple-Section Shafts:
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