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- Solution first found by St. Venant.
- Tractions at the ends are statically equivalent to equal and opposite torques .
- Lateral surfaces are traction-free.
- An axis passes through the center of twist ( axis).
- Each c.s. projection on to the plane rotates,but remains undistorted.
- The rotation of each c.s. () is proportional to .
where is the twist per unit length.
- The out-of-plane distortion (warping) is the same for each c.s. and is proportional to .
- Torsional rigidity ().
- Maximum shear stress.
where is the warping function.
If (small strain),
Therefore,
Therefore,
Therefore,
- Normal to cross sections is .
- Normal traction .
- Projected shear traction is .
- Traction vector at a point in the cross section is tangent to the cross section.
- Lateral surface traction-free.
- Unit normal to lateral surface appears as an in-plane unit normal to the boundary .
We parameterize the boundary curve using
The tangent vector to is
The tractions and on the lateral surface are identically zero.
However, to satisfy the BC , we need
or,
The traction distribution is statically equivalent to the torque .
At ,
Therefore,
From equilibrium,
Hence,
If and then
with the integration direction such that is to the left.
Applying the Green-Riemann theorem to equation (17), and using
equation (16)
Similarly, we can show that . since .
The moments about the and axes are also zero.
The moment about the axis is
where is the torsion constant. Since , we have
If , then , the polar moment of inertia.
- Find a warping function that is harmonic. and satisfies the traction BCs.
- Compatibility is not an issue since we start with displacements.
- The problem is independent of applied torque and the material properties of the cylinder.
- So it is just a geometrical problem. Once is known, we can calculate
- The displacement field.
- The stress field.
- The twist per unit length.