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The relationship between torsional forces, shear strain and polar moments of inertia.

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Introduction

Torsion occurs when an object, such as a bar with a cylindrical or square cross section (as shown in the figure), is twisted. The twisting force acting on the object is known as torque, and the resulting stress is known as Shear stress. If the object undergoes deformation as a result of and in the direction of the application of the force, the resulting deflection is known as Strain. Twisting a simple piece of blackboard chalk between ones fingers until it snaps is an example of a torsional force in action. A common example of torsion in engineering is when a transmission drive shaft (such as in an automobile) receives a turning force from its power source (the engine).

Torsion acting on a bar. Image courtesy Wikipedia - \url http://en.wikipedia.org/ — Torsion acting on a bar. Image courtesy Wikipedia - http://en.wikipedia.org/

Circular Shafts

When a pure torque acts upon a shaft, shear stresses will be set up which act in directions perpendicular to the radius at all points.

The complimentary shear stresses on longitudinal planes will cause the distortion of filaments which were originally in the longitudinal direction. Assume that the points lying on a radius before twisting will remain on that radius, and that the angle of twist is $\theta$ uniformly over the length of the shaft.

The diagram shows the shear strain $\phi$ of elements which are at a distance r from the axis ( When T is constant $\phi$ is constant). The original line OA has twisted to OB, and the angle ACB is now $\theta$ , the relative angle of twist for cross sections a distance l apart .

$Arc\;\;AB = r\theta \, \;\approx \;l\phi$

(1)

But,

$\phi=\frac{s}{C}\;\;\;\;\;\;\;(\,C \;is\: the \:modulus\: of \;rigidity\,)$

(2)

From the above two equations

$\frac{s}{r} = \frac{C\,\theta }{l}$

(3)

The torque T can be equated to the sum of the moments of the tangential stresses on the element $2\, \pi \, r\, \delta r$

$\therefore\;\;\;\;\;\;T = \int s\, (2\pi rdr)\, r$

(4)

substituting from equation (3):

$T = \int \frac{C\, \theta }{l}(2\pi r^3)\, dr$

(5)

$= \frac{C\, \theta }{l}\;\frac{\pi r^4}{2}$

(6)

But the polar moment of inertia, J ,is

$J= \frac{\pi r^4}{2}$

(7)

Combing equations (3) and (6):

$\frac{T}{J} = \frac{s}{r} = \frac{C\, \theta }{l}$

(8)

This equation shows that the shear stress is proportional to the radius (the equation is analogous to that proven in the section on linear bending stress).

For a solid shaft $J = \frac{\pi \, D^4}{32}$ and for a hollow shaft

$J = \frac{\pi}{32}(D^4 - d^4)$

(9)

The maximum stress for a solid shaft $s_{max} = \frac{16\, T}{\pi \, D^3}\;\;at\;r = \frac{D}{2}$ and for a hollow shaft $s_{max} = \frac{16\,D\,T}{\pi \, (D^4 - d^4)}\;\;\;\;at\;\;\frac{d}{2}$

Torsional stiffness $\mu$ is defined as torque per radian twist. i.e.

$\mu =\frac{T}{\theta } = \frac{C\,J}{l}$

(10)

Example 1

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Shafts of varying diameter

Angle of twist in larger shaft = $\frac{T}{J_1}\;X\;\frac{l_1}{C_1}$ and for the thinner =

$\frac{T}{J_2}\;X\;\frac{l_2}{C_2}$

(16)

$\therefore\;\;\;\;Total\;twist\;\theta = T\left(\frac{l_1}{C_1J_1} + \frac{l_2}{C_2J_2} \right)$

(17)

If the shafts are made of the same material $C_1 = C_2$

Composite shafts

The angle of twist per unit length is the same for both materials.

$\therefore\;\;\;Total\,\;T = T_s + T_b$

(18)

$but\;\;\;T_s = \frac{\theta \,C_s\,J_s}{l}\;\;\;\;and\;\;\;T_b = \frac{\theta \,C_b\.J_b}{l}$

(19)

$\therefore\;\;\;T = \frac{\theta }{l}\left(C_s\,J_s + C_b\,J_b \right)$

(20)

$\frac{\theta }{l} = \frac{T}{(C_s\,J_s + C_b\,J_b)}$

(21)

From which,

$t_s = \frac{C_s\,J_s}{C_s\,J_s + C_b\,J_b}\;T$

(22)

$t_b = \frac{C_b\,J_b}{C_s\,J_s + C_b\,J_b}\;.T$

(23)

$q_s = \frac{T_s}{J_s}\;X\;Radius_s$

(24)

$q_b = \frac{T_b}{J_b}\;X\;Radius_b$

(25)

Combined Bending and Twisting.

This occurs frequently in practice when a shaft is subjected to both bending and inertia forces. Stresses are set up due to gravity, torque, and shear forces, although the latter is usually unimportant since its maximum value occurs at the neutral axis where the bending stress is zero.

For the purposes of design it is necessary to calculate all the principle stresses( maximum shear stress, shear strain energy etc.) so that these can be used to compare them with the criterion of failure.

If $f_b$ is the greatest bending stress and s is the greatest shear stress due to bending, then:

$f_b = \frac{32M}{\pi D^3}$

(26)

$s = \frac{16T}{\pi D^3}\;for\;a\;solid\;shaft$

(27)

For loading in the vertical plane, these stresses will occur together at the end of a vertical diameter, since there is no normal stress on longitudinal planes of the shaft, they are maximum principle stress.

$f_{max} = \frac{1}{2}\, f_b + \frac{1}{2}\sqrt{(f_b^2 + 4s^2)}$

(28)

$= \frac{16M}{\pi D^3} + \frac{1}{2}\sqrt{\left[\left( \frac{32M}{\pi D^3} \right)^2 + 4\left( \frac{16T}{\pi D^3} \right)^2 \right]}$

(29)

$\therefore\;\;\;\;f_{max} = \left(\frac{16}{\pi D^3} \right)\sqrt{(M^2 + T^2)}$

(30)

Note: that $\frac{1}{2}[M + \sqrt{(M^2 + T^2)}]$ is the equivalent bending moment that would produce the same maximum stress.

The maximum shear stress q is given by;-

$q = \frac{1}{2}\sqrt{f_b^2\;+\a;4s^2}$

(31)

$\therefore\;\;\;\;q = \left(\frac{16}{\pi D^3} \right)\sqrt{(M^2 + T^2)}$

(32)

and this is the same shear stress as would be produced by a pure torque of magnitude $\sqrt{(M^2 + T^2)}$

Example 2

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Strain Energy in Torsion.

When a shaft of length l is gradually twisted through an angle $\theta$ under the influence of a Torque T. The work done (equals the total strain energy U) is given by:-