# Torsion of Shafts

## The torsion of solid or hollow shafts - Polar Moment of Inertia of Area.

### Shear Stress in the Shaft

When a shaft is subjected to a torque or twisting a shearing stress is produced in the shaft. The shear stress varies from zero in the axis to a maximum at the outside surface of the shaft.

The shear stress in a solid circular shaft in a given position can be expressed as:

τ = T r / J (1)

where

τ= shear stress (Pa, lb_{f}/ft^{2}(psf))

T = twisting moment (Nm, lb_{f}ft)

r = distance from center to stressed surface in the given position (m, ft)

J = Polar Moment of Inertia of Area (m^{4}, ft^{4})

Note

- the "
*Polar Moment of Inertia of an Area*" is a measure of a shaft's ability to resist torsion. The "*Polar Moment*of Inertia" is defined with respect to an axis perpendicular to the area considered. It is analogous to the "Area Moment of Inertia" - which characterizes a beam's ability to resist bending - required to predict deflection and stress in a beam. *1 ft = 12 in**1 ft*^{4}= 20736 in^{4}*1 psf (lb*_{f}/ft^{2}) = 144 psi (lb_{f}/in^{2})

"*Polar Moment of Inertia of an Area*" is also called "*Polar Moment of Inertia*", "*Second Moment of Area*", "*Area Moment of Inertia*", "*Polar Moment of Area*" or "*Second Area Moment*".

#### Polar Moment of Inertia vs. Area Moment of Inertia

- "Polar Moment of Inertia" - a measure of a beam's ability to resist torsion - which is required to calculate the twist of a beam subjected to torque
- "Area Moment of Inertia" - a property of shape that is used to predict deflection, bending and stress in beams

### Circular Shaft and Maximum Moment or Torque

Maximum moment in a circular shaft can be expressed as:

T_{max}=τ_{max}J / R (2)

where

T_{max}= maximum twisting torque (Nm, lb_{f}ft)

τ_{max}= maximum shear stress (Pa,lb)_{f}/ft^{2}

R = radius of shaft (m, ft)

Combining (2) and (3) for a **solid shaft**

T_{max}= (π / 16)τ_{max}D^{3}(2b)

Combining (2) and (3b) for a **hollow shaft**

T_{max}= (π / 16)τ_{max}(D^{4}- d^{4}) / D (2c)

### Circular Shaft and Polar Moment of Inertia

Polar Moment of Inertia of a circular solid shaft can be expressed as

J = π R^{4}/ 2

= π (D / 2)^{4}/ 2

= π D^{4}/ 32 (3)

where

D = shaft outside diameter (m, in)

Polar Moment of Inertia of a circular hollow shaft can be expressed as

J = π (D^{4}- d^{4}) / 32 (3b)

where

d = shaft inside diameter (m, ft)

### Diameter of a Solid Shaft

Diameter of a solid shaft can calculated by the formula

D = 1.72 (T_{max}/τ_{max})^{1/3}(4)

### Torsional Deflection of Shaft

The angular deflection of a torsion shaft can be expressed as

α = L T / (J_{ }G) (5)

where

α= angular shaft deflection (radians)

L = length of shaft (m, ft)

G = Shear Modulus of Rigidity - or Modulus of Rigidity (Pa, psf)

The angular deflection of a torsion solid shaft can be expressed as

α= 32 L T / (G π D^{4}) (5a)

The angular deflection of a torsion hollow shaft can be expressed as

α= 32 L T / (G π (D^{4}- d^{4})) (5b)

The angle in degrees can be achieved by multiplying the angle *θ* in radians with *180 / π.*

*S*olid shaft (*π* replaced)

* α_{degrees} ≈ 584 L T / (G D^{4}) (6a) *

*Hollow shaft ( π replaced)*

* α_{degrees} ≈ 584 L T / (G (D^{4}- d^{4}) (6b) *

### Torsion Resisting Moments from Shafts of Various Cross Sections

Shaft Cross Section Area | Maximum Torsional Resisting Moment - T_{max} -(Nm, lb_{f} ft) | Nomenclature | |
---|---|---|---|

Solid Cylinder Shaft |
| ||

Hollow Cylinder Shaft |
| ||

Ellipse Shaft | (π / 16) τ_{max} b^{2} h | h = "height" of shaft b = "width" of shaft h > b | |

Rectangle Shaft | (2 / 9) τ_{max} b^{2} h | h > b | |

Square Shaft | (2 / 9) τ_{max} H^{3} | ||

Triangle Shaft | (1 / 20) τ_{max} b^{3} | b = length of triangle side | |

Hexagon Shaft | 0.123^{ }τ_{max }D ^{3} |

### Example - Shear Stress and Angular Deflection in a Solid Cylinder

A moment of *1000 Nm* is acting on a solid cylinder shaft with diameter *50 mm* *(0.05 m)* and length *1 m*. The shaft is made in steel with modulus of rigidity *79 GPa (79 10 ^{9} Pa)*.

Maximum shear stress can be calculated as

*τ _{max} = T r / J_{}*

* = T (D / 2) / (**π D ^{4} / 32) *

* = (1000 Nm) ((0.05 m) / 2) / (**π (0.05 m) ^{4}*

*/ 32)*

* = 40764331 Pa*

* = 40.8 MPa*

The angular deflection of the shaft can be calculated as

*θ = L T / (J G) *

* = L T / (**(**π D ^{4} / 32) G)*

* = (1 m) (1000 Nm) / (**(**π (0.05 m) ^{4}*

*/ 32) (79 10*

^{9}Pa))* = 0.021 (radians)*

* = 1.2 ^{ o}*

### Example - Shear Stress and Angular Deflection in a Hollow Cylinder

A moment of *1000 Nm* is acting on a hollow cylinder shaft with outer diameter *50 mm (0.05 m)*, inner diameter

*30 mm*

*(0.03 m)*and length

*1 m*. The shaft is made in steel with modulus of rigidity

*79 GPa (79 10*.

^{9}Pa)Maximum shear stress can be calculated as

*τ _{max}* = T r / J

_{ }

* = T (D / 2) / (**π (D ^{4} - d^{4}) / 32) *

* = (1000 Nm) ((0.05 m) / 2) / (**π ((0.05 m) ^{4}*

*-*

*(0.03 m)*

^{4}*) / 32)*

* = 46.8 MPa*

The angular deflection of the shaft can be calculated as

*θ = L T / (J _{} G) *

* = L T / ** ((**π D ^{4} / 32) G)*

* = (1 m) (1000 Nm) / (**(**π **((0.05 m) ^{4}*

*-*

*(0.03 m)*

^{4}*)*

*/ 32) (79 10*

^{9}Pa))* = 0.023 radian)*

* = 1.4 ^{ o}*

### Example - Required Shaft Diameter to Transmit Power

A *15 kW* electric motor shall be used to transmit power through a connected solid shaft. The motor and the shaft rotates with *2000 rpm*. The maximum allowable shear stress - *τ _{max}* - in the shaft is

*100 MPa*.

The connection between power and torque can be expressed

*P = 0.105 n_{rpm} T (7) *

*where *

*P = power (W)*

*n _{rpm} = speed of shaft (rpm)*

Re-arranged and with values - the torque can be calculated

*T = (15 10 ^{3} W) / (0.105 (2000 rpm))*

* = 71 Nm*

Minimum diameter of the shaft can be calculated with eq. 4

*D = 1.72 ((71 Nm) / (100 10 ^{6} Pa))^{1/3} *

* = 0.0153 m*

* = 15.3 mm *