Shafts Torsion
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}) = 1/144 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 / (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
Shaft Cross Section Area | Maximum Torsional Resisting Moment - T _{ max } - (Nm, lb _{ f } ft) | Nomenclature | |
---|---|---|---|
Solid Cylinder Shaft | (π / 16) τ _{ max } (2 r) ^{ 3 } = (π / 16) τ _{ max } D ^{ 3 } | ||
Hollow Cylinder Shaft | (π / 16) τ _{ max } ((2 R) ^{ 4 } - (2 r) ^{ 4 } ) / (2 R) = (π / 16) τ _{ max } (D ^{ 4 } - d ^{ 4 } ) / D | ||
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 | 0.123τ _{ 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
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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