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Shafts Torsion

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

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Hollow shaft - stress

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 /ft2(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 /ft2) = 1/144 psi (lb f /in2)

" 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 /ft2)

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

Torsion deflection of shaft

The angular deflection of a torsion shaft can be expressed as

α = L T / (JG)                                  (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

Shafts - Torsional Resisting Moments
Shaft Cross Section AreaMaximum Torsional
Resisting Moment
- T max -
(Nm, lb f ft)
Nomenclature
Torsion - solid cylinder shaft Solid Cylinder Shaft

(π / 16) τ max (2 r) 3

= (π / 16) τ max D 3

Torsion - Hollow Cylinder Shaft Hollow Cylinder Shaft

(π / 16) τ max ((2 R) 4 - (2 r) 4 ) / (2 R)

= (π / 16) τ max (D 4 - d 4 ) / D

Torsion - Ellipse Shaft Ellipse Shaft (π / 16) τ max b2h h = "height" of shaft
b = "width" of shaft
h > b
Torsion - Rectangle Shaft Rectangle Shaft (2 / 9) τ max b2h h > b
Torsion - Square Shaft Square Shaft (2 / 9) τ max H 3
Torsion - Triangle Shaft Triangle Shaft (1 / 20) τ max b 3 b = length of triangle side
Torsion - Hexagon Shaft Hexagon Shaft

0.123τ max D 3

0.189 τ max b 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

Electric motor

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

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