The drilling shaft maintains a torque or moment of 15 kipft but the amount
of the torque in each of the two sections is not known. Summing the moments around
the pipe axis gives
T_{A} + T_{S} = T_{total} =
15 kipft
Thus, there are two unknowns but only one equation from static equilibrium.
A second equation can be developed from the angle of twist compatibility relationship.
Each section must twist the same, or
θ_{steel} = θ_{alum}
Recall, the angle of twist for any circular shaft is
Using this in the compatibility equation gives
Rearranging and substituting values gives
T_{S} = 4.586 T_{A}
This indicates the steel section takes almost 80% of the torque. Thus, the aluminum
will not help much. This result can be substituted back into
the moment equilibrium equation, giving
T_{A} + 4.586 T_{A} = 15,000
lbft
T_{A} = 2,685 lbft
T_{S} = 12,310 lbft
