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MODULE 1. BASIC CONCEPTS

MODULE 2. SYSTEM OF FORCES

MODULE 3.

MODULE 4. FRICTION AND FRICTIONAL FORCES

MODULE 5.

MODULE 6.

MODULE 7.

MODULE 8.

## LESSON 27.

**27.1 POLAR MODULUS**

Let *T *be the torsional moment of resistance of the section of a shaft of radius *R* and *Ip* the polar moment of inertia of the shaft section.

The shear stress intensity *q* at any point on the section distant *r *from the axis of the shaft is given by

*q* = \[{T \over {{I_{p}}}}.r\]

The maximum shear stress *fs* occurs at the greatest radius *R*

Therefore *f _{s} *= \[{T \over {{I_{p}}}}\] .

*R*

*T* = *f _{s}*. \[{{{I_p}} \over R}\] or

*T*=

*f*

_{s}. ZpWhere *Z _{p}* = \[{{{I_p}} \over R}\] = \[{{Polarmomentofinertiaoftheshaftsection} \over {Maximumradius}}\]

This ratio is called the polar modulus of the shaft section. The greatest twisting moment which a given shaft section can resist

= maximum permissible shear stress × polar modulus

Hence for a shaft of a given material the magnitude of the polar modulus is a measure of its strength in resisting torsion.

Given a number of shafts of the same length and material, the shaft which can resist the greatest twisting moment is the one whose polar modulus is greater.

ü Shafts of the same material and length having the same polar modulus have the same strength.

For a solid shaft of diameter *D*,

*I _{p}* = \[{\pi{{D^4}} \over {32}}\] ,

*R*= \[{D \over 2}\]

*Z _{p} *= \[{\pi{{D^3}} \over {16}}\]

For a hollow shaft whose outer and inner diameters are *D _{1}*

_{ }and

*D*.

_{2}*I _{p}* = \[{\pi \over {32}}\left[ {{D_1}^4 - {D_2}^4} \right]\]

*R _{1}* = \[{{{D_1}} \over 2}\] , Therefore

*Z*= \[{\pi \over {16{D_1}}}\left[ {{D_1}^4 - {D_2}^4} \right]\]

_{p}Torsional moment of resistance = *f _{s }Z_{p} *=

*f*. \[{\pi \over {16{D_1}}}\left[ {{D_1}^4 - {D_2}^4} \right]\]

_{s}

Hollow shafts are more efficient than solid shafts in resisting torsional moment.

In a solid shaft the shear stresses are maximum at the outer boundary of the section and zero at the centre. Hence considerable material of the solid shaft is subjected to shear stresses much below the maximum shear stress. This means the shera resisting capacity of the material of the shaft is not utilized.

On the other hand in hollow shaft material being present more away from the centre of the section, the shear resistance of the material is higher.

Using a hollow shaft results in not only saving of material but also in reduction of weight.

**Torsional Rigidity**

Let a twisting moment *T* produce a twist of *Ө* radians in a length *l*

We know the relation, \[{T \over {{I_p}}} = {{C\theta } \over l}\]

Therefore *Ө *= \[{{Tl} \over {C{I_p}}}\]

For a given shaft the twist is therefore proportional to the twisting moment

T.In a beam a bending moment produces a bend or deflection, in the same manner, a torque produces a twist in a shaft.

The quantity

CIis called Torsional rigidity._{p}

CI_{p}_{ }stands for the torque required to produce a twist of 1 radian per unit length of the shaft.The quantity \[{{C{I_p}} \over l}\] is called torsional stiffness. It is the torque required to produce a twist of 1 radian over the length of the shaft.

Torsional flexibility is the reciprocal of the torsional stiffness and is equal to \[{l \over {C{I_p}}}\] and is the angle of rotation produced by a unit torque.

**Limitations**

The theory presented above is valid for linearly elastic bars of circular sections (solid or hollow).

The stresses determined from the formulae derived above are acceptable in regions far away from regions of stress concentrations (like holes and abrupt changes in diameter) and also far away from sections subjected to concentrated torques.

However the angle of twist is not affected by stress concentrations and the formulae can be applied for determining the twist.

It should also be noted that we derived the torsion formula assuming prismatic circular bars. The formula may however be used, when the changes in diameter are small and gradual.

**27.2 POWER TRANSMITTED BY A SHAFT**

Let a shaft turning at *N* *rpm* transmit *P* *kilo watts*. Let the mean torque to which the shaft is subjected to be *T Nm*.

Therefore, Power Transmitted = *P* = Mean torque × Angle turned per second

= *T* \[{N \over {60}}\] *2П* *Watts* =* T * \[{N \over {60}}\] . \[{\pi{2} \over {1000}}\] Kilowatts

Therefore, *P* = \[{{2{\pi}NT}\over {6000}}\] *kW*

Sometimes angular speed is expressed as the frequency f of rotation which means the number of revolutions per unit of time. The unit of frequency is *Hertz (Hz).*

1 *Hz* = 1 revolution per second

Power transmitted = *P* = \[{{2{\pi}fT}\over {1000}}\]