How to calculate a shear flange.

[goodguy1405]

Hello, I am trying to figure out the calculations for a shear flange. I have tried Tau=v/a. There is a thickness "t" that needs to be calculated to determine the width of the flange to shear. The material is 4140 hfs Q&T. The flange needs to shear off at >40,000 pounds of force. My "t" =.37 I have attached a drawing for a better view of what I am trying to do. Thanks for any help.

 

[desertfox]

Hi

According to my calculations the thickness "t" needs to be 0.039" thick however there are differences in your post and attachment ie "The flange needs to shear off at >40,000 pounds of force", in your attachment it states 40000psi which of course makes a great difference.
I assumed the 40000 pounds force was correct, also the yield of the material specified on the pdf is 107psi should this be 107ksi?
Now even if we get to a figure I would recommend some shear tests on the component because materials do all kinds of stuff before actual failing like work hardening etc.

 

[dvd]

What is the ultimate shear strength of the material, not the tensile yield?

Failure at the sharp notch is susceptible to surface finish issues and accurate control of notch radius, not to mention fatigue cycling. Maybe it is routine in your particular industry, but it looks impractical to me. A more controllable shearing interface might be a ring soldered in place.

Calculation is only going to get you started, if this is important to fail as-planned, test it also.

 

[goodguy1405]

Thank you for the responses. I am curious about your calculation desert. DVD yes it needs to fail at a particular force. I am planning on having a backup in place and your idea of soldering on a piece may be the way to go. If the numbers look good then that is 3/4 of the battle.

 

[desertfox]

hi

My calculation was based on half the yield stress being the shear stress limit, the shear area as I see it is pi*D*t where t is the thickness your after therefore:-

t= 40000/(pi*D* 0.5*yield stress)

If you use D as 5.35 inches then the t = 0.044 inches, I got the wrong D in my earlier calc,
However as stated earlier and by others it needs to be tested this will only get you in the ball park.

 

[goodguy1405]

Thank you. Yeah this is a good guestimate on a starting point. Vibration is going to play a factor in the equation as well and probably a few other unforeseen equations. Just have to try something and go from there. Thanks for all your help.

 

[desertfox]

hi

You need to read my first post because there are differences in the pdf file you uploaded and your written post.

 

[goodguy1405]

Could you explain the difference? Pounds of force versus Pounds per square inch? I was under the impression they are the same. Thanks for the information.

 

[desertfox]

hi

pounds per square inch is a pressure, pounds force is just a force,see link below:-

http://lurnq.com/lesson/Newtons-Law...on/The-Difference-between-Force-and-Pressure/

 

[goodguy1405]

We have decided to go with a shear pin design as the results we are wanting to achieve do not seem attainable through this design concept. I am pretty sure the calculations are the same to find the right diameter for the shear pin. Thanks for all your help. On to the next project.

 

[desertfox]

Hi

Yes the calculations are very similar but you will still face the same issues in that work hardening of the pin material etc. will still mean you need to do practical tests.

 

[dvd]

Make sure to understand the difference between single-shear and double-shear.

 

[goodguy1405]

Great Idea....Looking it up now.

 

[goodguy1405]

Company changed its mind. Too many variables could go wrong with this design idea 10,000 feet down hole. NEXT
Thanks for all the useful information.