User:Eas4200c.f08.wiki.b/HW4

=Mtg. 20 - Friday 10/10=

The engineering, or ad-hoc, derivation of the rate of twist of the following equation

 $$\displaystyle \theta = {1 \over 2G \bar A} \oint {q \over t} ds $$

applies to a uniform bar with a non-circular cross section.



The displacement of segment $$\displaystyle PP' $$ due to an angle $$\displaystyle \alpha $$ is defined as

 $$\displaystyle {PP' \over OP} = \tan \alpha $$

But for a small angle $$\displaystyle \alpha $$, the following approximation is appropriate:

 $$\displaystyle \tan \alpha \approx \alpha $$

Hence,

 $$\displaystyle {PP' \over OP} \approx \alpha $$

The projected displacement of $$\displaystyle PP' $$ in the direction perpendicular to $$\displaystyle OP' $$ can be written

 $$\displaystyle PP'' = PP' \cos \alpha $$

By substitution

 $$\displaystyle PP'' = (OP \tan \alpha) \cos \alpha $$

By the associative property of multiplication

 $$\displaystyle PP'' = (OP \cos \alpha) \tan \alpha $$

Recall that $$\displaystyle OP = r $$ and $$\displaystyle OP'' = \rho $$, thus we can express it as

 $$\displaystyle PP'' = (r \cos \alpha) \tan \alpha $$

Strain, $$\displaystyle \gamma $$ can be written as

 $$\displaystyle \gamma = {PP' \over dx} = {\rho \alpha \over dx} = \rho \theta $$

where, as previously stated, $$\displaystyle \theta $$ is the rate of twist.