
Let us consider a cylindrical wire of length ‘L’ and radius 'r'. The wire is fixed at its upper end and twisted through an angle ‘θ’ by applying a torque at the lower end.
COUPLE
PER UNIT TWIST
Let
us consider a cylindrical wire of length ‘L’
and radius 'r'. The wire is fixed at its upper end and twisted through an angle
‘θ’ by applying a torque at the lower end. The wire can be assumed to be made
up of a number of hollow cylindrical tubes (co‒axial) whose radii varies from 0
to r.
Let
us consider one such cylinder of radius x
and thickness dx as shown in Fig 4.6.

Due
to the twisting torque the line AB which is initially parallel to the axis OO‘
of the cylinder is displaced to a position AB' through an angle ϕ as shown in
Fig.4.6.
The
result of twisting the cylinder is a shearing strain.
The
angle of shear = ∠BAB'
= ϕ
Here
BB' = хθ = Lϕ
(or)
ϕ
= хθ / L ……….(1)
Rigidity
modulus n = Tangential stress / Shearing
strain
(or)
Rigidity
modulus n = Shearing stress / Angle
of shear (ϕ)
Shearing
stress = nϕ ..(2)
Substituting
for ϕ from eqn.(1) in eqn.(2), we have
Shearing
stress = nxθ / L ………..(3)
We
know shearing stress = Shearing Force / Area
Shearing
Force = Shearing Stress × Area on which the shearing force is acting
(i.e)
F = nxθ/L . 2πx.dx ………(4)
Where
2πxdx is the area over which the shearing force acts, as shown in
Fig.4.7.
Moment
of the force about the OO' axis of the cylinder = Shearing force × Distance
= nxθ/L
. 2πx.dx.x
= 2πnθ.x3.dx / L ………(5)
Twisting
couple of the whole wire can be derived by integrating eqn.(5) within the
limits 0 to r (since the radii varies from 0 to r)
Twisting
couple on the wire = C

If
twist θ is unity i.e. θ = 1 radian.
Then,
we can write
The
torque per unit twist C = nπr4
/ 2L ……..(6)
Note:
If the material is in the form of hollow cylinder of inner radius "r1"
and outer radius “r2" then Torque per unit twist C = nπ[r24‒r14] / 2L
Applied Physics I: Chapter 4: Oscillations and Waves : Tag: Applied Physics : - Couple per unit twist
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