How to Calculate Maximum Shear Stress: Solid Shaft with Locked Door Handle

[aaronfue]

Homework Statement

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 .

Homework Equations

\tau_max = \frac{Tc}{J}
T = internal torque acting at cross section
c = outer radius
J = polar moment of inertia → J = \frac{\pi}{2}c⁴

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.
radius, c = 0.054 m

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}

 

[PhanthomJay]

Is the problem asking for max shear stress at the section m-m? You left out the actual question.

Homework Statement

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 .

Homework Equations

\tau_max = \frac{Tc}{J}
T = internal torque acting at cross section
c = outer radius
J = polar moment of inertia → J = \frac{\pi}{2}c⁴

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.
radius, c = 0.054 m

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.

Yes, but I wouldn't say it acts 'along' the length: it is an internal torque that acts about the axis AB

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

Why are you multiplying a torque by a distance? Your result is in N*m^2 and incorrect.

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}

Once you calculate the torsional shear stress, there is the vertical shear stress to consider from the vertical load P.

Please state the question.

 

[SteamKing]

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

 

[PhanthomJay]

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

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.