**TORSION**

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**Basic
kinematics of torsion:**

It
is assumed that when twisted each cross-section in a circular bar rotates as a
rigid body. If the amount of rotation of each cross-section is given by the
function _{}, the total twist _{} in the bar of length *l*
shown is given

_{}

During
the torsion of a bar, axial lines drawn on the surface of a bar become helical
and circumferential lines remain circumferential. Therefore, as can be seen in
the figure, material elements are sheared in the process. This shear strain
must be accompanied by a shear stress (from Hook’s law we know_{}).

The
shear strain can be related to the rotation of the cross sections. If over an
increment of length _{}, the cross section rotates an amount _{}, then as can be seen in the figure the shear strain is given
by

_{}

Therefore, at the limit one can write

_{}

As a result, the shear strain starts at zero at the center and reaches a maximum at the outer radius.

**The shear stress:**

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The shear
stress, through Hook’s law, is related to the shear strain. Therefore, the
shear stress is given by

_{}

Again, like the
shear strain, the shear stress is zero at the core of the bar and increases
linearly to a maximum at the outer surface.

**Calculating the
torque from the shear stress:**

The torque transmitted through the cross-section of a bar is the resultant of the moments created by the shear stresses on the cross-section.

The differential shear load *dV*
that results from the shear stress applied over the area *dA* of the cross
section is _{}. The moment created by such a shear load is *dM = rdV*.
The equivalent torque T on the cross section is the resultant of all the
moments crated by the shear stresses so that

_{}

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Using the expression for shear stress in
terms of twist in the bar we get

_{}

where *I _{p} *is the polar
area moment of inertia of the cross section. Using this relation, the
expression for shear stress in terms of the torque follows from

_{}

**Total twist in
a bar:**

The total twist in a bar is given by

_{}

The twist per unit length in terms of
the torque is given by

_{}

Therefore, the total twist can be
calculated from

_{}

In the case of a uniform bar for which the _{}is constant one can move the argument out and integrate to
get

_{}

ã Mehrdad Negahban and the University of Nebraska, 1996-2000.

All rights reserved

Copy and distribute freely for personal use only

Department of Engineering Mechanics, University of Nebraska, Lincoln, NE 68588-0526

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