Question 5.6: The composite rod with length L consists of an inner core (w......
The composite rod with length L consists of an inner core (with shear modulus G_1 and polar second moment of area J_1) and outer tube (with shear modulus G_2 and polar second moment of area J_2) that are firmly bonded to each other. Both parts are also bonded to end plates (not shown) through which a torque T is applied. Determine
a. the maximum shear stress in the core and in the tube
b. the total angle of twist of the composite rod.
Given: Dimensions and material properties of components of a composite rod.
Find: Stress in each component and total twist.
Assume: Hooke’s law applies.
a. Picking the force method (although the displacement method is equally effective for this problem), we start with an FBD and write the equation of equilibrium:
T_1+T_2-T=0The applied torque T is given, leaving one equation for two unknown torques. To gain an additional equation, we enforce geometric compatibility:
\phi_1=\phi_2 ,
where these are the twists of the two rod components. Then with our assumption of elastic (Hookean) behavior, this equation for geometric compatibility becomes
\frac{T_1 L_1}{J_1 G_1}=\frac{T_2 L_2}{J_2 G_2} .
Recognizing that both components have the same length and solving this pair of equations for the unknown torques
T_1=T \frac{J_1 G_1}{J_1 G_1+J_2 G_2} and T_2=T \frac{J_2 G_2}{J_1 G_1+J_2 G_2}
b. The total twist is \phi=\phi_1=\phi_2 , so plugging in one of the above results:
\phi=\phi_1=\frac{T_1 L}{J_1 G_1}=\frac{T L}{J_1 G_1} \frac{J_1 G_1}{J_1 G_1+J_2 G_2}=\frac{T L}{J_1 G_1+J_2 G_2} .These results make sense for torsional elastic elements in parallel.
