Pin Design Help

[SVaughn]

Hi everybody,

I have been given the task of designing a pin connection for a high load application. I have attached a screen shots to describe my situation. Please note that an additional force coming out of the page of 237 kips is not shown in the screen shot. I was hoping to get a little feedback on the way I designed my pin. It is a 5.25" diameter pin with an overall load of 333 kips. I used the following method:

Per AISC 13th Ed.
Use ASTM A36 solid bar => Fu = 58 ksi, Fy = 36 ksi (Table 2-4; pg. 2-40)
Use 0.50*Fu*Pin Diameter (Table J3.2; pg. 16.1-104)

=> 0.50*(58)*(21.65)*(2/3)*(2) = 836 kips ((2/3) used to achieve a S.F. of 3 and (2) was used since the pin is in double shear)
=> 836 kips > 333 kips DESIGN OKAY

Since the center plate (2.75" pl.) fills the entire gap between the two outer plates (2.00" pl.) I ignored bending of the pin and only considered shear.

Any and all input would be greatly appreciated. Thank you.

 

[SVaughn]

I apologize, there is a little bit of information in my original post that is incorrect. The total force on the connection is 237(2) = 474 kips and 233(2) = 466 kips (coming out of the page), each side of the pin only sees 237 kips and 233 kips. Sorry for any confusion this may have caused. Thanks.

 

[ToadJones]

I would think you still need to check bearing as well as meet the provisions for hole clearances to be considered a "pin" according to AISC.

 

[rb1957]

2/3 for SF = 3 ? ... your formulae is based on SF = 2 ? (just checking)

"out-of-plane" force = transverse shear on pin ... need to vector sum for resultant (no?)

i'd check bearing (on the central plate) and i'd look at bolt bending ... it might not be much (if you assume the bolt is well supported by the hole walls) but it does reduce your allowable shear.

 

[seedyw]

why don't you try using a Nitronic 60 stainless pin. It's stronger and won't gall on you. if you plan on a forging for your present selection, you may have a wait on your hands. A 5" Nitronic 60 will meet what you need to design for ..........I believe just looking at your numbers and it will get to you alot faster. The UT is 105 ksi; YS is 60 ksi.

 

[dhengr]

Svaughn:
That’s an interesting CAD sketch which probably doesn’t represent, very well, what you’ll end up with. Your solution for the pin shear seems to be formula driven, rather than well thought through, from a knowledgeable engineering standpoint. Why don’t you do a little less CADing and a little more thinking about how this detail really works. Do a plan view, and two side views, to scale, showing the loaded members framing into the pinned connection, etc. Put the significant dimensions on the sketches. Draw some free body diagrams of how this all works. 21" c/c for the two main pin plates is a pretty big connection and it probably won’t be made of solid, shaded, blocks of steel. How does this all fit on a foundation, without tipping over? What do the various stiffeners and weld details look like? The pin shear seems the least of the problems, and you probably wouldn’t use a single 34" long by 5.25" dia. pin.

I’ll bet you have some moments in this detail that you are neglecting in your current sketch. On that large a connection I’m not sure that you should assume that the load is split evenly btwn. the two pin plates in every case, and moments could add to that problem. Then, you have to check the bearing stresses in the pins and pin plates. Then, you have to consider how you are going fabricate and machine such a large piece. You should get with your boss and get his help so you understand how to think through these types of problems. He should know what you know and what you don’t know, so as to keep you and the company out of trouble. And, he will then know better how to guide and help you become an engineer.