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Torsion of Shafts
Torsion of solid and hollow shafts
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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:
σ = T r / Ip (1)
where
σ = shear stress (MPa, psi)
T = twisting moment (Nmm, in lb)
r = distance from center to stressed surface in the given position (mm, in)
Ip = "polar moment of inertia" of cross section (mm4, in4)
The "polar moment of inertia" is a measure of an object's ability to resist torsion.
Circular Shaft and Maximum Moment
Maximum moment in a circular shaft can be expressed as:
Tmax = σmax Ip / R (2)
where
Tmax = maximum twisting moment (Nmm, in lb)
σmax = maximum shear stress (MPa, psi)
R = radius of shaft (mm, in)
Combining (2) and (3) for a solid shaft
Tmax = (π/16) σmax D3 (2b)
Combining (2) and (3b) for a hollow shaft
Tmax = (π/16) σmax (D4 - d4) / D (2c)
Circular Shaft and Polar Moment of Inertia
Polar moment of inertia of a circular solid shaft can be expressed as
Ip = π R4/2 = π D4/32 (3)
where
D = shaft outside diameter (mm, in)
Polar moment of inertia of a circular hollow shaft can be expressed as
Ip = 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/σmax)1/3 (4)
Torsional Deflection of Shaft
The angular deflection of a torsion shaft can be expressed as
θ = L T / Ip 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 | (1/1.09) 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.
Maximum shear stress can be calculated as
σ = T r / Ip
= 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 / Ip G
= L T / (π D4/32) G
= (1 m) (1000 Nm) / (π (0.05 m)4/32) (79 GPa)
= 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.
Maximum shear stress can be calculated as
σ = T r / Ip
= 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 / Ip G
= L T / (π D4/32) G
= (1 m) (1000 Nm) / (π ((0.05 m)4 - (0.03 m)4)/32) (79 GPa)
= 0.023 radians
= 1.4 o
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