As shown below, a circular rod has two sections with different cross sections. A torque toward the rear is applied at the interface, and both ends are fixed. You will find the reactions and calculate the rod's stresses.

Here, the rod diameter $(d_{AB})$ from point $A$ to point $B$ is $19$ centimeters, and the diameter $)$ from point $B$ to point $C$ is $7$ centimeters.

The length $(L_{AB})$ from point $A$ to point $B$ is $1\mathrm{.}4$ meters, and the length from point

$B$ to point $C$ is $2\mathrm{.}2$ meters. The applied torque $T$ is $2kN*m$ and is applied at point $B.$

The material is a grade of steel with $G=75$GPa and ${}_y=500$MPa.

Process$-$Oriented Problem $03\mathrm{.}003.$e $-$ Compatibility for TA

Substitute the necessary quantities into the compatibility equation ${}_{AB}+{}_{BC}=0.$ You will want to use both expressions for internal forces from analyzing the left sides of the FBDs; this will simplify your algebra.

What is your result for the reaction $T_A$ in $kN-m$ with the proper sign? $($You must provide an answer before moving to the next part.$)$

The reaction $T_A$ is $kN*m.$