- Torsion (mechanics)
In

solid mechanics ,**torsion**is the twisting of an object due to an appliedtorque . In circular sections, the resultant shearing stress is perpendicular to the radius.For solid or hollow shafts of uniform circular cross-section and constant wall thickness, the torsion relations are::$frac\{T\}\{J\}\; =\; frac\{\; au\}\{R\}\; =\; frac\{Gphi\}\{l\}$where:

*R is the outer radius of the shaft.

*$au$ is the maximum shear stress at the outer surface.

*"Φ" is the angle of twist inradian s.

*"T" is the torque (N·m or ft·lbf).

*"l" is the length of the object the torque is being applied to or over.

*"G" is the shear modulus or more commonly themodulus of rigidity and is usually given ingigapascal s (GPa), lbf/in^{2}(psi), or lbf/ft^{2}.

*"J" is thetorsion constant for the section . It is identical to thepolar moment of inertia for a round shaft or concentric tube only. For other shapes J must be determined by other means. For solid shafts the membrane analogy is useful, and for thin walled tubes of arbitrary shape the shear flow approximation is fairly good, if the section is not re-entrant. For thick walled tubes of arbitrary shape there is no simple solution, and FEA may be the best method.

*the product "GJ" is called thetorsional rigidity .The shear stress at a point within a shaft is::$au\_\{phi\_\{z\; =\; \{T\; r\; over\; J\}$where:

*"r" is the distance from the center of rotation

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

stress concentrations such as rough spots. Thus, shafts for use in high torsion are polished to a fine surface finish to reduce the maximum stress in the shaft and increase its service life.The angle of twist can be found by using::$phi\_\{\}\; =\; \{T\; l\; over\; JG\}$

**Polar moment of inertia**The polar moment of inertia for a solid shaft is::$J\; =\; \{pi\; over\; 2\}\; r^4$

where "r" is the radius of the object.

The polar moment of inertia for a pipe is::$J\; =\; \{pi\; over\; 2\}\; (r\_\{o\}^4\; -\; r\_\{i\}^4)$

where the "o" and "i" subscripts stand for the outer and inner

radius of the pipe.For a thin cylinder:"J" = 2"π" "R"

^{3}"t"where "R" is the average of the outer and inner radiusand "t" is the wall thickness.**Failure mode**The shear stress in the shaft may be resolved into

principal stresses viaMohr's circle . If the shaft is loaded only in torsion then one of the principal stresses will be in tension and the other in compression. These stresses are oriented at a 45 degree helical angle around the shaft. If the shaft is made ofbrittle material then the shaft will fail by a crack initiating at the surface and propagating through to the core of the shaft fracturing in a 45 degree angle helical shape. This is often demonstrated by twisting a piece of blackboard chalk between one's fingers.**ee also***

torsion spring or -bar

*torsional vibration

*torque

* membrane analogy

*Saint-Venant's theorem

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