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.

 

[rezeew]

= 188.44 MPa

Yes, your calculations seem to be correct. The maximum shear stress in the solid shaft with the locked door handle is 188.44 MPa. This is an important consideration when designing the door handle to ensure it can withstand the applied force without failing.