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Torsion of a square section bar Example of torsion mechanics. In the field of solid mechanics, torsion is the twisting of an object due to an applied torque. [1] [2] Torsion could be defined as strain [3] [4] or angular deformation, [5] and is measured by the angle a chosen section is rotated from its equilibrium position. [6]
The formula to calculate average shear stress τ or force per unit area is: [1] =, where F is the force applied and A is the cross-sectional area.. The area involved corresponds to the material face parallel to the applied force vector, i.e., with surface normal vector perpendicular to the force.
The torsion constant or torsion coefficient is a geometrical property of a bar's cross-section. It is involved in the relationship between angle of twist and applied torque along the axis of the bar, for a homogeneous linear elastic bar. The torsion constant, together with material properties and length, describes a bar's torsional stiffness.
The polar second moment of area appears in the formulae that describe torsional stress and angular displacement. Torsional stresses: = where is the torsional shear stress, is the applied torque, is the distance from the central axis, and is the polar second moment of area.
is the shear modulus. is the second moment of area., called the Timoshenko shear coefficient, depends on the geometry. Normally, = / for a rectangular section. (,) is a distributed load (force per length).:=
In these instances, it can be useful to express internal shear stress as shear flow, which is found as the shear stress multiplied by the thickness of the section. An equivalent definition for shear flow is the shear force V per unit length of the perimeter around a thin-walled section. Shear flow has the dimensions of force per unit of length. [1]
A notable aspect of the flow is that shear stress is constant throughout the domain. In particular, the first derivative of the velocity, U / h {\displaystyle U/h} , is constant. According to Newton's Law of Viscosity ( Newtonian fluid ), the shear stress is the product of this expression and the (constant) fluid viscosity .
Assuming that the direction of the forces is known, the stress across M can be expressed simply by the single number , calculated simply with the magnitude of those forces, F and the cross sectional area, A. = Unlike normal stress, this simple shear stress is directed parallel to the cross-section considered, rather than perpendicular to it. [13]