Torsion of Shafts

Torsion of solid and hollow shafts - second moment of area

Hollow shaft

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 (MPa, psi)

T = twisting moment (Nmm, in lb)

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

J = second moment of area (mm4, in4)

Note

  • the "Second Moment of Area" is a measure of a beam's ability to resist torsion. The "Second Moment of Area" 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.

Circular Shaft and Maximum Moment

Maximum moment in a circular shaft can be expressed as:

Tmax = τmax J / R         (2)

where

Tmax = maximum twisting moment (Nmm, in lb)

τmax = maximum shear stress (MPa, psi)

R = radius of shaft (mm, in)

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

Tmax = (π / 16) τmax D3         (2b)

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

Tmax = (π / 16) τmax (D4 - d4) / D         (2c)

Circular Shaft and Second Moment of Area

Second Moment of Area of a circular solid shaft can be expressed as

J = π R4 / 2

   = π (D / 2)4 / 2

   = π D4 / 32         (3)

where

D = shaft outside diameter (mm, in)

Second Moment of Area of a circular hollow shaft can be expressed as

J = π (D4 - d4) / 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 (Tmax / τmax)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)

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- d4))         (5b)

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

Solid shaft (π replaced)

   θdegrees ≈ 584 L T / (G D4)         (6a) 

Hollow shaft (π replaced)

   θdegrees ≈ 584 L T / (G (D4- d4)         (6b) 

Torsion Resisting Moments of Shafts of Various Cross Sections

Shaft Cross Section Area Maximum Torsional
Resisting Moment
- Tmax -
(Nm, in lb)
Nomenclature
Solid Cylinder Shaft ( π / 16) σmax D3
Hollow Cylinder Shaft ( π / 16) σmax (D4 - d4) / D
Ellipse Shaft ( π / 16) σmax b2 h h = "height" of shaft
b = "width" of shaft
h > b
Rectangle Shaft (2 / 9) σmax b2 h h > b
Square Shaft (2 / 9) σmax b3
Triangle Shaft (1 / 20) σmax b3 b = length of triangle side
Hexagon Shaft 1.09 σmax b3 b = length of hexagon side

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 and length 1 m. The shaft is made in steel with modulus of rigidity 79 GPa (79 109 Pa)

Maximum shear stress can be calculated as

τmax = T r / J

  = T (D / 2) / (π D4 / 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 D4 / 32) G)

  = (1 m) (1000 Nm) / ((π (0.05 m)4 / 32) (79 109 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. The shaft is made in steel with modulus of rigidity 79 GPa (79 109 Pa)

Maximum shear stress can be calculated as

τmax = T r / J

  = T (D / 2) / (π (D4 - d4) / 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) (79 109 Pa))

  = 0.023 (radians)

  = 1.4 o

Related Topics

  • Mechanics - Kinematics, forces, vectors, motion, momentum, energy and the dynamics of objects
  • Statics - Loads - force and torque

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