# Maximum shear stress

1. Jun 18, 2013

### aaronfue

1. The problem statement, all variables and given/known data

A locked door handle is composed of a solid circular shaft AB with a diameter of b = 108mm and a flat plate BC with a force P = 69N applied at point C as shown. Let c = 532mm , d = 126mm , and e = 153mm .

2. Relevant equations

$\tau$max = $\frac{Tc}{J}$
T = internal torque acting at cross section
J = polar moment of inertia → J = $\frac{\pi}{2}$c4

3. The attempt at a solution

I've converted all mm to m.
I just wanted to see if someone could verify that I am on the right track.

1. The torque caused by the P force:
69 N × 0.532 m = 36.71 N*m → acting along the length of the solid shaft.

2. Internal torque, T:
T = 36.71 N*m *0. 126 m = 4.62546 N*m → acting at section m-m

3. J = $\frac{\pi}{2}$0.054 m4 = 13.36×10-6m4

$\tau$max = $\frac{(4.62546 N*m)*(0.054 m)}{13.36×10^{-6} m^4}$

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2. Jun 18, 2013

### PhanthomJay

Is the problem asking for max shear stress at the section m-m? You left out the actual question.
Yes, but I wouldn't say it acts 'along' the length: it is an internal torque that acts about the axis AB
Why are you multiplying a torque by a distance? Your result is in N*m^2 and incorrect.
Once you calculate the torsional shear stress, there is the vertical shear stress to consider from the vertical load P.

3. Jun 19, 2013

### SteamKing

Staff Emeritus
Jay:

The shear stress due to an applied torque on a circular shaft is T * r / J. You overlooked the units of the polar moment of inertia J, which is m^4.

So,

tau = T*r/J which has units of N-m * m / m^4 = N/m^2 = Pa, a unit of stress, I believe.

tau = Tr/J is somewhat analogous to finding the bending stress in a beam, where

sigma = M * y / I

Last edited: Jun 19, 2013
4. Jun 19, 2013

### PhanthomJay

The OP incorrectly identified the torque in item 2 of Post 1. The vertical shear in the rod was also neglected, if that is what the problem is asking.