This document provides an introduction to the concept of torsion in mechanical engineering systems. It defines torsion as the twisting of a structural member due to an applied torque, and discusses how the amount of twist increases as you move along the length of the member away from the fixed end. It presents the general equation for calculating torque in circular shafts based on shear stress and provides two examples of using the equation to calculate torsional stiffness and maximum torque before failure.
2. 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 2nd 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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3. 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 = Gθ so that T/θ = GJ / L
J L
And J = πd4 / 32 for a shaft
J = (π x 0.0154)/ 32 = 4.97 x 10-9
T/θ = (90 x 109 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 0.044)/ 32 = 2.514 x 10-7
Using T = τ
J r
Therefore T = J τ = (2.514 x 10-7 x 60 x 106) / 0.02
r
= 753.98 Nm
Using;
τ = Gθ (NOTE we can neglect one term)
r L
θ= τL = (60 x 106 x 2.5) / (0.02 x 80 x 109)
rG
=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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