Example 3: Rectangular Cylinder[edit | edit source]
In this case, the form of is not obvious and has to be
derived from the traction-free BCs
Suppose that and are the two sides of the rectangle, and .
Also is the side parallel to and is the side parallel to .
Then, the traction-free BCs are
A suitable must satisfy these BCs and .
We can simplify the problem by a change of variable
Then the equilibrium condition becomes
The traction-free BCs become
Let us assume that
Case 1: or [edit | edit source]
In both these cases, we get trivial values of .
Case 2: [edit | edit source]
Apply the BCs at ~~ (), to get
The RHS of both equations are odd. Therefore, is odd. Since,
is an even function, we must have .
Hence, is even. Since is an odd function, we must
Apply BCs at (), to get
The only nontrivial solution is obtained when , which means that
The BCs at are satisfied by every terms of the series
Applying the BCs at again, we get
Using the orthogonality of terms of the sine series,
The warping function is
The torsion constant and the stresses can be calculated from .