AISC J3.10 Question 1

[Hokie93]

In calculating Lc for use in the equations in AISC section J3.10 (Bearing Strength at Bolt Holes), would one use the nominal bolt hole diameter or the nominal bolt hole diameter plus 1/16"? I have notes from an AISC connection seminar (from 2001) wherein it states that for bearing and tearout calculations, the actual hole diameter is to be used. The seminar speaker was Dr. Tom Murrary. I can't find clear support for this practice in the AISC Specification or its Commentary. Using the nominal bolt hole diameter plus 1/16" would be the conservative approach but I am curious how others handles this.

 

[nutte]

The specification says to use the hole diameter. This is clear in the specification. Lc is defined as the clear distance, not the clear distance plus 1/16". The tables on page 7-28 confirm this.

Personally, I'm inclined to take out the extra 1/16".

 

[ToadJones]

hmmm...
for a 3/4" bolt, std hole is 13/16". But in some calcs one would consider an extra 1/16" for damaged material...or 7/8".
Is this what you are asking about?

 

[WillisV]

Correct, nominal hole diameter is used for the bearing and tearout checks, and nominal hole diameter + 1/16" (=bolt diam. + 1/8") is used for block shear and tension and shear rupture checks.

 

[ToadJones]

I guess confusion is in wording.

does "nominal" = bolt dia. + 1/16?

I consider nominal hole diameter to be the bolt diameter.

I consider a "standard" hole to be 1/16th over

 

[WillisV]

nominal hole dia = bolt + 1/16"
nominal bolt dia for 3/4" bolt = 3/4"
hole diameter with damage for rupture checks = nominal hole + 1/16" = bolt + 1/8"

 

[Hokie93]

Thanks for the prompt responses; my question has been answered. I spot-checked a few of the values in Table 7-6 and they are definitely based on the nominal hole diameter, not the hole diameter plus 1/16" per D3.2.

 

[BadgerPE]

I would consider the diameter of the bolt (db) to be used for bearing applications. My thoughts on this are that a bolt can't be in contact with the base material over a larger area that db*t. Therefore, if (db+1/16)*t was used to determine bearing area, an unconservatvie value would be reached.