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, 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
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)
Rearranged 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
Related Topics
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