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Torsion of solid and hollow shafts - Polar Moment of Inertia of an 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 inthe 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 (MPa, psi) 

T = twisting moment (Nmm, in lb) 

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

J = Polar Moment of Inertia of an Area (mm4, in4 ) 

Note

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  the "Polar Moment of Inertia of an Area" is a measure of a beam's ability to resist torsion. The "Polar Moment  of Inertia" isdefined 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.

"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 ".

Circular Shaft and Maximum Moment

Maximum moment in a circular shaft can be expressed as:

T max  = τ max  J / R (2) 

where 

T max  = maximum twisting moment (Nmm, in lb) 

τ max  = maximum shear stress (MPa, psi) 

R = radius of shaft (mm, in) 

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

T max  = (π / 16) τ max  D3  (2b) 

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

T max  = (π / 16) τ max  (D4 - 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  

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  = π (D / 2)4 / 2  

= π D4 / 32 (3) 

where 

D = shaft outside diameter (mm, in) 

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

J = π (D4 - d 4 ) / 32 (3b) 

where 

d = shaft inside diameter (mm, in) 

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

θ = L T / (J  G) (5) 

where 

θ = angular shaft deflection ( radians) 

L = length of shaft (mm, in) 

G = modulus of rigidity (Mpa, psi) 

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

θ = 32 L T / (G π D4 ) (5a)

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

θ = 32 L T / (G π (D4- d 4 )) (5b) 

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

Solid shaft (π  replaced)

θ degrees ≈  584 L T / (G D4 ) (6a)

Hollow shaft (π  replaced) 

θ degrees ≈  584 L T / (G (D4- d 4 ) (6b)

Torsion Resisting Moments of Shafts of Various Cross Sections

Shaft Cross Section Area

Maximum TorsionalResisting Moment

- T max  - 

(Nm, in lb) 

Nomenclature

Solid Cylinder Shaft ( π / 16) σ max  D3 

Hollow Cylinder Shaft ( π / 16) σ max  (D4 - d 4 ) / D 

Ellipse Shaft ( π / 16) σ max  b2  h 

h = "height" of shaftb = "width" of shaft

h > b

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Shaft Cross Section Area

Maximum TorsionalResisting Moment

- T max  - (Nm, in lb) 

Nomenclature

Rectangle Shaft (2 / 9) σ max  b2  h  h > b

Square Shaft (2 / 9) σ max  b3 

Triangle Shaft (1 / 20) σ max  b3  b = length of triangle side

Hexagon Shaft 1.09 σ max  b

3

  b = length of hexagon sideExample - 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 and length 1 m. The shaft is made in steelwith 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) 

= 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) 

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  = (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, inner diameter 30 mm and length 1 m. Theshaft 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 (D4 - 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 D4 / 32) G) 

= (1 m) (1000 Nm) / ((π  ((0.05 m)4 - (0.03 m)4 ) / 32) (79 10 9 Pa)) 

= 0.023 (radians) 

= 1.4 o