# Mechanics of materials torsion doubt

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1. Jun 5, 2016

### MarcusAu314

• Moved from another forum, so homework template missing
Good evening. I've got stuck in this question from my book of 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.

2. Jun 5, 2016

### MarcusAu314

Excuse me for the failed LATEX writting but the formula goes:
tmax=TC/J
where tmax is the maximum shear stress, T is the torque, C is the radius of the rod and J is the polar moment of inertia

3. Jun 6, 2016

### billy_joule

Calculate for both then whichever is lower is the limiting torque value and cannot be exceeded.
This can be shown by doing a free body diagram of each section of the shaft - one for the steel part, one for the brass.

It's analogous to tying ropes or strings together - you can only apply as much tension as the weakest rope can take. eg a bit of cotton thread can only support a kg or two, it doesn't matter if it's tied to a bit of fishing line or some carbon nanotube, the cotton still breaks at it's normal load.