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Moved from another forum, so homework template missing

Good evening. I've got stuck in this question from my book of

This is the diagram of the problem

https://scontent-lax3-1.xx.fbcdn.net/v/t1.0-9/13327492_1734753976809139_591495701010647559_n.jpg?oh=c90b0b5caea0dff2cfa2524a864a764f&oe=57D2EC5B

And according to the section: answers to selected problems of the same book the answer is: T=9.16 kip*in.

So this is my attempt for the solution yet I know I'm not correct since the answer is different to the indicated in the book.

First of all what I did is that I know the formula for maximum shear stress in torsion is:

[itex]\tau_{max}=\frac{TC}{J}[\itex]

But I realized that I'm working with two different materials and not a single one so when I performed the operation for the Steel I got like almost 10 kip*in so I need to know how to calculate a torque when I have a rod of two distinct materials.

*Mechanics of Materials*by Beer and Johnston about torsion.**The allowable shearing stress is 15 ksi in the 1.5-in.-diameter steel rod AB and 8 ksi in the 1.8-in.-diameter brass rod BC. Neglecting the effect of stress concentrations, determine the largest torque that can be applied at A.**This is the diagram of the problem

https://scontent-lax3-1.xx.fbcdn.net/v/t1.0-9/13327492_1734753976809139_591495701010647559_n.jpg?oh=c90b0b5caea0dff2cfa2524a864a764f&oe=57D2EC5B

And according to the section: answers to selected problems of the same book the answer is: T=9.16 kip*in.

So this is my attempt for the solution yet I know I'm not correct since the answer is different to the indicated in the book.

First of all what I did is that I know the formula for maximum shear stress in torsion is:

[itex]\tau_{max}=\frac{TC}{J}[\itex]

But I realized that I'm working with two different materials and not a single one so when I performed the operation for the Steel I got like almost 10 kip*in so I need to know how to calculate a torque when I have a rod of two distinct materials.