Torsion or twisting is a common concept in mechanical engineering systems. This section looks at the basic theory associated with torsion and examines some typical examples by calculating the main parameters. Further examples include determination of the torque and power requirements of torsional sy...
Torsion or twisting is a common concept in mechanical engineering systems. This section looks at the basic theory associated with torsion and examines some typical examples by calculating the main parameters. Further examples include determination of the torque and power requirements of torsional systems.
Introduction to Torsion
Torsion
Torsion is the term used for the twisting of a structural member when it is acted upon by
TORQUE so that rotation is produced about the longitudinal axis at on end of the member with
respect to the other.
Torque - Fr - Twisting
moment
Fixed end
The amount of twist ( torsion) that the shaft experiences will increase as we move away from
the fixed end of the shaft.
Assumptions;
·The shaft has a uniform cross section
·The shaft material is uniform throughout and the shear stress is proportional to the shear
strain (Elastic region)
·The shaft is straight and initially unstressed
·The axis of twisting moment is the axis of the shaft
·Plain transverse sections remain the same after twisting
General equation for torsion of cross sectioned circular shafts;
T = τ = Gθ Where T = torque (Nm)
J r L J = polar 2
nd
moment of area
τ = Max. shear stress (MPa)
r = radius of shaft (m)
G = Modulus of Rigidity (GPa)
θ = Angle of twist (radians)
L = length of shaft (m)
Torsional stiffness - T/ θ (Applied torque per radian)
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Introduction to Torsion
Transmission of power
P = 2πnT (n = revs/sec, T = applied torque
Example 1
Calculate the torsional stiffness of a 0.5m long shaft, 15mm diameter. G = 90GPa.
Using;
T = Gθso that T/θ = GJ / L
J L
And J = πd
4
/ 32 for a shaft
J = (π x 0.015
4
)/ 32 = 4.97 x 10
-9
T/θ = (90 x 10
9
x 4.97 x 10
-9
) / 0.5
= 894.62Nm / rad
Example 2
A solid steel shaft is 2.5m long and 40mm in diameter. The maximum stress in the shaft must
not exceed 60 MPa. Determine the maximum torque that can be applied and the angle of twist
at this torque (in degrees)
Assume G = 80 GPa
J = (π x 0.04
4
)/ 32 = 2.514 x 10
-7
Using T = τ
J r
Therefore T = J τ = (2.514 x 10
-7
x 60 x 10
6
) / 0.02
r
= 753.98 Nm
Using;
τ = Gθ(NOTE we can neglect one term)
r L
θ = τ L = (60 x 10
6
x 2.5) / (0.02 x 80 x 10
9
)
r G
=0.09375 rads
Change radians to degrees ;
0.09375 x 57.3 = 5.372 degrees
Note – Conversion Rads to degrees - multiply by 57.3
(360/2π ) = 57.3
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Introduction to Torsion
Credits
This resource was created Leicester College and released as an open educational resource
through the Open Engineering Resources project of the Higher Education Academy
Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE
and part of the JISC/HE Academy UKOER programme.
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