# Torsional moment

In technical mechanics , a moment is referred to as a torsional moment when a body subjected to it is twisted ( twisted ). The mechanical stresses that occur in the body (usually a beam , a rod or a shaft ) that keep the torsional moment in balance are called torsional stresses .

## Torsion of straight bars with a general but constant cross-section

If the torsional moment vector goes in the direction of the beam axis and through the center of gravity of the cross-section and no other stress is acting, pure torsion is present. The rod is twisted about its axis, the cross sections perpendicular to the axis generally not remaining flat. The distribution of the shear stresses occurring in the material can not be determined in an elementary manner. Because of the constant cross-section, it is constant over the length of the member, with the exception of the moment introduction points (usually at the member ends).

## Torsion of straight bars with a round cross-section

The shear stress distribution in the bar cross-section is rotationally symmetrical, which is why the cross-sections perpendicular to the bar axis remain flat.

### Shear stress

The shear stress  is proportional to the torsional  moment and increases linearly with the cross-sectional radius  . The proportionality factor is the reciprocal of the polar area moment of inertia of the circle or annulus (tube as rod): ${\ displaystyle \ tau}$${\ displaystyle M_ {t}}$${\ displaystyle r}$ ${\ displaystyle I_ {p}}$

${\ displaystyle \ tau = {\ frac {M_ {t}} {I_ {p}}} \ cdot r}$

At the edge (radius  ) it is greatest: ${\ displaystyle R}$

${\ displaystyle \ tau _ {max} = {\ frac {M_ {t}} {I_ {p}}} \ cdot R}$
Rod with a circular cross-section loaded with torsional moment M t over the length L: Angle of twist α, surface
line deformed to the helical line

### Twist angle

The twist angle  is proportional to the torsional  moment and increases linearly with the rod length  . Proportionality factors are the reciprocal values ​​of the shear modulus of the material and the polar area moment of inertia  : ${\ displaystyle \ alpha}$${\ displaystyle M_ {t}}$${\ displaystyle L}$ ${\ displaystyle G}$${\ displaystyle I_ {p}}$

${\ displaystyle \ alpha = {\ frac {L} {G \ cdot I_ {p}}} \ cdot M_ {t}}$

If the length, shear modulus and polar area moment of inertia are combined to form the directional moment , the following applies: ${\ displaystyle D}$

${\ displaystyle \ alpha = {\ frac {1} {D}} \ cdot M_ {t}}$