Torsion (mechanics)

In the field of

In non-circular cross-sections, twisting is accompanied by a distortion called warping, in which transverse sections do not remain plane.[1] For shafts of uniform cross-section unrestrained against warping, the torsion is:

${\displaystyle T={\frac {J_{\text{T}}}{r}}\tau ={\frac {J_{\text{T}}}{\ell }}G\varphi }$

where:

• T is the applied torque or moment of torsion in Nm.
• ${\displaystyle \tau }$ (tau) is the maximum shear stress at the outer surface
• J_T is the torsion constant for the section. For circular rods, and tubes with constant wall thickness, it is equal to the polar moment of inertia of the section, but for other shapes, or split sections, it can be much less. For more accuracy, finite element analysis (FEA) is the best method. Other calculation methods include membrane analogy and shear flow approximation.^[2]
• r is the perpendicular distance between the rotational axis and the farthest point in the section (at the outer surface).
• ℓ is the length of the object to or over which the torque is being applied.
• φ (phi) is the angle of twist in radians.
• G is the shear modulus, also called the
  lbf/in²
  (psi), or lbf/ft² or in ISO units N/mm².
• The product J_TG is called the
  torsional rigidity
  w_T.

Properties

The shear stress at a point within a shaft is:

${\displaystyle \tau _{\varphi _{z}}={Tr \over J_{\text{T}}}}$

Note that the highest shear stress occurs on the surface of the shaft, where the radius is maximum. High stresses at the surface may be compounded by

The angle of twist can be found by using:

${\displaystyle \varphi _{}={\frac {T\ell }{GJ_{\text{T}}}}.}$

Sample calculation

Calculation of the steam turbine shaft radius for a turboset:

Assumptions:

• Power carried by the shaft is 1000
  MW; this is typical for a large nuclear power
  plant.
• Yield stress
  of the steel used to make the shaft (τ_yield) is: 250 × 10⁶ N/m².
• Electricity has a frequency of 50 Hz; this is the typical frequency in Europe. In North America, the frequency is 60 Hz.

The angular frequency can be calculated with the following formula:

${\displaystyle \omega =2\pi f}$

The torque carried by the shaft is related to the power by the following equation:

${\displaystyle P=T\omega }$

The angular frequency is therefore 314.16

The maximal torque is:

${\displaystyle T_{\max }={\frac {{\tau }_{\max }J_{\text{zz}}}{r}}}$

After substitution of the torsion constant, the following expression is obtained:

${\displaystyle D=\left({\frac {16T_{\max }}{\pi {\tau }_{\max }}}\right)^{1/3}}$

The diameter is 40 cm. If one adds a factor of safety of 5 and re-calculates the radius with the maximum stress equal to the yield stress/5, the result is a diameter of 69 cm, the approximate size of a turboset shaft in a nuclear power plant.

Failure mode

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The shear stress in the shaft may be resolved into

• List of area moments of inertia
• Saint-Venant's theorem
• Second moment of area
• Structural rigidity
• Torque tester
• Torsion siege engine
• Torsion spring or -bar
• Torsional vibration

References