All-bolted single-angle connections

[PowersPE80]

For all-bolted single-angle connections can you confirm that I have calculated the connection capacity correctly?

Per AISC Table 10-10.

For 2 (3/4") bolts (LRFD design):
phi*(Rn) = C * phi(rn)
phi*Rn = 1.03 * 15.9 kips
phi*Rn = 16.4 kips

Is the 16.4 kips the total connection capacity? Or do I multiple 16.4 kips x (2 bolts), so that 32.8 is my connection capacity?

Thanks.

 

[structSU10]

That capacity is for the bolts connected to the supporting member, and are thus ONE of the capacities of the connection. The angle must be checked for the applicable limit state (shear rupture, shear yielding, block shear, etc.). Table 10-11 gives you the capacities of the other parts of the connection, and thus you use the lowest value from all that apply.

In your case, the 16.4 K seems to control.

 

[nutte]

C is the number of effective bolts you have. There is eccentricity on the outstanding leg bolts, so you don't have the full capacity of 2 bolts; you have less. So 16.4 kips would be the capacity of the OSL bolts. Of course there are other limit states to check.