Solving the torsion problem for isotropic matrial with a rectangular cross section using the FEM and FVM methods using triangular elements
Nasser Abbasi
Introduction
We consider bar made of isotropic martial with rectangular cross section
subjected to twisting torque
.
The following diagram illustrate the basic geometry.

Experiments show that rectangular cross sections do wrap and that cross sections do not remain plane as shown in this diagram (in the case of a circular cross section, cross section do NOT wrap).

This is another diagram showing a bar under torsion

The twist rate (called
in this problem) and defined as
where
is the twist angle is assumed to be constant.
Cross section can wrap also in the
direction (i.e. the cross section does not have to remain in the
plane) but if this happens, all cross sections will wrap in the
section by the same amount.
Material is isotropic
The input to the problem are the following (these are the known or given):
The width
and height
of the cross section.
Material Modulus of rigidity or sheer modulus
which is the ratio of the shearing stress
to the shearing strain
The applied torque
the torsion constant for the a rectangular cross section. For a rectangular
section of dimensions
it is given by

Hence the torsional rigidity
is known since
is given (material) and
is from above (geometry).
The stress distribution in the cross section (stress tensor field). Once this is found then using the material constitutive relation we can the strain tensor field.
The angle of twist
as a function of
(the length of the beam).
First we solve for the Prandtl stress function
by solving the Poisson equation
Where
is the sheer modulus and
is the twist rate (which was assumed to be constant).
The boundary conditions
(
at any point on the edge of the cross section and at the ends of the beam) is
an arbitrary constant. We take this constant to be zero. Hence at the cross
section boundary we
have
The analytical solution to the above equation is from book Theory of elasticity by S. P. Timoshenko and J. N. Goodier
where the linear twist
Hence (2) becomes
Where
is given by (1)
Hence
and
Timoshenko gives the maximum sheer stress, which is
as
Given that
is Young's modulus for the material,
is Poisson's ratio for the material, and
we can now obtain the strain components from the constitutive equations
(stress-strain equations) since we have determined the stress components from
the above solution.
Hence only
and
are non-zero.

If we look at a cross section of the bar at some distance
from the end of the bar, the angle that this specific cross section has
twisted due to the torque is
.

This angle is given by the solution to the equation
But
is the linear twist and is given by
hence the above equation becomes
Hence
Where
is the constant of integration. Assuming
at
we obtain that
and using the expression
given in equation (1) above we can determine
for each

we see that
Hence
Where
TODO
Mathematica Structural Mechanics help page
MIT course 16.20 lecture notes. MIT open course website.
Theory of elasticity by S. P. Timoshenko and J. N. Goodier. chapter 10