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The solid spindle $AB$ has a diameter $d_S=1\mathrm{.}8in.$ and is made of a steel with an allowable shearing stress of $18$ksi, while sleeve $CD$ is made of a brass with an allowable shearing stress of $7$ksi, outer diameter $a=2\mathrm{.}5in.$ and thickness $t=0\mathrm{.}19in.$ Determine the allowable torque $T$ that can be applied at $A.$

Using FBD$/$equilibrium equations for appropraite cut sections between A&B and between C&D$,$ the magnitude of the internal reaction $T_{AB}=20\mathrm{.}612**T$ and the magnitude of the internal reaction $T_{CD}=$