Shear Stress-working out the diameter on the pin.

[ian3mith]

Hello, guys, I am new to the website, I am stuck on this question that I need help with.

There is a pin that connects 2 rods in a framework. The shear stress of the pin is limited to 50 kPA.

I need to calculate the minimum suitable diameter for the pin in the middle. But I am not sure how to get there. could someone explain the steps for me I would appreciate it a lot.

Thanks guys.

 

[desertfox]

Hi
So can you show us any attempt you have made to calculate this?

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[ian3mith]

@desetfox
ok I can show you but do you know how to work it out? I am only saying if because if I show you and you don't understand then there is no point if I am honest.

I have the forces and stress 50 kPA.

 

[jhardy1]

Please show us your calculations for the shear stress in the pin, and we can let you know if you are on the right track.

Hint: Stress = Force / Area

(They do still teach that in Engineering Statics 101, don't they?)

http://julianh72.blogspot.com

 

[desertfox]

Well let’s see what you have, we’re not here to give you the answer but help you obtain the right answer

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[Agent666]

Maybe for axial stress this simplified relationship can be directly applied. But not for shear, shear stress is not constant across the diameter for a pin or bolt. Shear stress usually includes an allowance for some scaling shape or form factor for describing max vs average for example.

To OP, what code are you working to might be a good starting point if trying to work out a design capacity (most design codes have applicable equations for pins, often quite different than for bolts), or are you looking for a theoretical answer here.

(not sure if your discipline is structural or mechanical where different design rules might apply).

 

[jhardy1]

Yes, I'm aware of that - but go back to the OP's original question:

There is a pin that connects 2 rods in a framework. The shear stress of the pin is limited to 50 kPA. I need to calculate the minimum suitable diameter for the pin in the middle.

With that level of problem definition, and noting that this is the "Student Engineer" forum, I'm betting that this is an "Engineering Design 101" question, where the students have to size a pin for a nominal allowable shear stress of 50 kPa. The structural design codes that I've used determine the shear strength of a pin using the shank area of the pin, and a reduction factor on the shear strength of steel, to allow for the shear stress variation. E.g. in AS 4100:

V_f = 0.62 f_yp n_s A_p

(Where n_s is the number of shear planes.)

If the OP is advanced enough to be dealing with the variation of shear stress distribution across the full cross-section area of the pin, I doubt they'd be asking us how to calculate the required diameter of the pin.

(I'll also bet that if the OP re-reads the question carefully, it will say 50 MPa!)

http://julianh72.blogspot.com

 

[ian3mith]

I am so lost, I know this is wrong but anyone can point me in the correct direction?

 

[desertfox]

Hi

See the sketch below, basically because it’s double shear the shear force is halved and once you know the calculated shear stress you use that in the formulae shown in the sketch and transpose to find the pin diameter d

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

 

[fel3]

A couple other things I noticed:
- You show shear strength (not "stress")* as being both 50 kPa and 50 MPa. You multiply the first by 1000 to get the second, but this doesn't make any sense to me. 50 MPa is in the range of fiberglass and some of the softer metals, so I believe this value more than 50 kPa. Steels and most other structural materials have shear strengths that are typically in the hundreds of MPa's.
- 50N is not a very big force. In fact, it's not much more than the average weight of an adult house cat. Therefore, if you use a strong material for the pin, the pin won't need to be very big. Using 50 MPa for the shear strength, I worked out something about the size of the cotter pins I have worked with around the house and on my cars, but it's late in the afternoon so I don't guarantee the result. But, I'm also not a cat person, so pin failure doesn't bother me all that much.

Now for an observation and a suggestion:
- I prefer to have the young engineers working for me do some calculations by hand before doing them on a computer, including just typing them in like you have done here. The reason is that pencil and paper work uses and exercises different parts of the brain than doing the same thing on a computer and apparently (according to some studies I have heard about but not read) does that better than working on a computer. For simple stuff like this, you will probably learn more and learn it faster using pencil and paper. Back in the late 1970s, when I was in college, we didn't have a choice.
- That being said, in the real world I do almost everything now on a computer. I still do some hand calcs from time-to-time, but that more for the variety of it. IMHO, the best way to do calculations like this one in a professional setting is to use Mathcad or SMath Studio. Mathcad Prime 6.0 is free to try for a month, then it reverts to the far less capable but still very usable and free for life Mathcad Prime Express. SMath Studio is simply free. Both programs allow you to do math with visible equations and they both handle units, and the unit handling is one of the key capabilities for producing accurate engineering calculations. In most ways the two programs work about the same, so it's pretty easy to jump from one to the other. Mathcad produces better looking output and SMath has more capabilities than Mathcad Prime Express, but you really can't go wrong with either program.

* "strength" is a material property; "stress" is the applied force divided by the cross-sectional area.

============
"Is it the only lesson of history that mankind is unteachable?"
--Winston S. Churchill

 

[ian3mith]

@desertfox
Thanks, i will take it into consideration and apply that in my working out. hopefully get something good!

@fel3

yeah 50 N is small, do you mind showing me your working out? is it something like I got or different?

 

[fel3]

Ian...

I didn't keep the calc, but as I recall the diameter was a little bit more than one mm.

============
"Is it the only lesson of history that mankind is unteachable?"
--Winston S. Churchill

 

[jhardy1]

By my calculation, IF the permissible average shear stress is indeed 50 kPa (not 50 MPa, which might be a reasonable indicative permissible shear stress for a relatively low-grade metal pin), and IF the applied load is indeed 50 N (not 50 kN), then you should find that you need a pin more than 10 mm and less than 50 mm diameter for a pin in double-shear. (I have to leave SOME work for the OP to undertake!)

In addition, you would normally check the pin for bending, but for that calculation, you need to know the length of the pin, the thickness and clearances of the lug plates, and the allowable bending stress in the pin. That information is not provided, so I make no comment other than that sometimes shear will govern, and sometimes bending will govern the required diameter of the pin.

http://julianh72.blogspot.com