huangyhg

re: combined tension and shear in bolts
i'm looking at the interaction equation aisc (j3-3b), page 16.1-109, aisc steel manual, 13th ed. solving for f'nt ≤ fnt. for a325 bolts, fnt = 90 ksi tension and fnv = 48 ksi shear from aisc table j3.2 and ω = 2.00 in asd. this means that all the terms in equation j3-3b are constant except the shear stress fv.
if i have a connection that has both moment and shear and m = 118.5 ft-kips and v = 9.8 kips on a w14x38 (14.125" deep) then:
c = t = m/d = (118.5 x 12)/14.125 = 100.67 kips tension
this implies:
no. of bolts = t/fnt = 100.67 kips/19.9 kips/bolt = 5.06 bolts
for shear:
no. of bolts = v/fnv = 9.8 kips / 10.6 kips/bolt = .92 bolts
choose five (5) bolts and check the interaction equation:
f'nt = 1.3*fnt – (ω*fnt/fnv)*fv ≤ fnt = 1.3*90 – (2*90/48)*fv ≤ 90
for five (5) ?" bolts fv = v/ab = 9.8/(5*.442) = 4.43 ksi
plugging this into the interaction equation:
1.3*90 – (2*90/48)*(4.43) ≤ 90 gives 100.4 ≠ 90
for seven (7) bolts 105.1 ≠ 90
for nine (9) bolts 107.8 ≠ 90
and for two (2) bolts 75.4 ≠ 90
how can fewer bolts be better???
now there is a user note saying that if the tension or shear stress is less than 20% of the allowable we can ignore the combined effect. this only confuses me more, since i would think the formula should work in either case. can someone point out what i'm missing?


fewer bolts is not better. the f'nt you are calculating is the available tension stress based on the actual shear stress that you have in the bolts. that being said, you can't have f'nt > fnt.
your calcs bear that out. for 7 bolts, f'nt = 105.1ksi, for 9 bolts f'nt = 107.8ksi.
i would also point out that your fv would be changing with the increased number of bolts (which you don't account for).
i think you are confusing what you should be comparing.
once you solve for f'nt, you should be comparing this to the actual tension in the bolt (service or factored), not the maximum allowable f'nt.
daveatkins
all the numbers were on a per bolt basis and yes, i realize that as the number of bolts increases, the stress per bolt is going down. i guess i was expecting an interaction expression that resulted in a specific number.
so, if i'm putting a spreadsheet together for this calculation, i should have an if...then statement that "if f'nt is > 90 ksi, then use 90 ksi, if f'nt < 90 ksi use that smaller number.

jheidt2543:
you could use the if ... then statement in your spreadsheet, but i prefer to use the min function. it accomplishes the same thing, but i find it is easier when you need to check your work.
joel berg

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