Maximum shear stress

  • Thread starter aaronfue
  • Start date
  • #1
122
0

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



[itex]\tau[/itex]max = [itex]\frac{Tc}{J}[/itex]
T = internal torque acting at cross section
c = outer radius
J = polar moment of inertia → J = [itex]\frac{\pi}{2}[/itex]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 = [itex]\frac{\pi}{2}[/itex]0.054 m4 = 13.36×10-6m4

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

Attachments

  • shaft.JPG
    shaft.JPG
    11.4 KB · Views: 1,196

Answers and Replies

  • #2
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,167
507
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



[itex]\tau[/itex]max = [itex]\frac{Tc}{J}[/itex]
T = internal torque acting at cross section
c = outer radius
J = polar moment of inertia → J = [itex]\frac{\pi}{2}[/itex]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 = [itex]\frac{\pi}{2}[/itex]0.054 m4 = 13.36×10-6m4

[itex]\tau[/itex]max = [itex]\frac{(4.62546 N*m)*(0.054 m)}{13.36×10^{-6} m^4}[/itex]
Once you calculate the torsional shear stress, there is the vertical shear stress to consider from the vertical load P.

Please state the question.
 
  • #3
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,670
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:
  • #4
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,167
507
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.
 

Related Threads on Maximum shear stress

Replies
5
Views
3K
Replies
47
Views
4K
  • Last Post
2
Replies
36
Views
2K
Replies
1
Views
6K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
8
Views
963
  • Last Post
Replies
1
Views
2K
Top