Page 127 - Structural Steel Designers Handbook AISC, AASHTO, AISI, ASTM, and ASCE-07 Design Standards

Brockenbrough_Ch03.qxd 9/29/05 5:05 PM Page 3.59

CONNECTIONS

CONNECTIONS 3.59

2
Gross Tension. The gross area A gt is 1.94 × 2 = 3.88 in . The gross tension design strength is
φR gt =φF y A gt = 0.90 × 36 × 3.88 = 126 kips
2
Net Tension. The net tension area is A nt = 3.88 − 0.25 × 1.0 × 2 = 3.38 in . The effective net ten-
sion area A e is less than the net area because of shear lag, because only one of the two angle legs is
connected. From the AISC Specification, Sec. D3.3,

U = max(0.60, 1 − 1.09/3) = 0.637
A e = U × A nt = 0.637 × 3.38 = 2.15 in 2
The net tension design strength is

φR nt =φF t A e = 0.75 × 58 × 2.15 = 93.5 kips
Block Shear Rupture. This failure mode involves the tearing out of the shaded block at the
end of the angles in Fig. 3.36. Ultimate strength is characterized by yielding on the longitudinal
line through the bolts (line ab) and a simultaneous fracture on the perpendicular line from the bolts
longitudinal line to the angle toe (line bc). Because yielding on the longitudinal section may some-
times exceed fracture on this section, AISC Specification J4.3 limits the strength to the lesser of the
two. Thus, the block shear limit state is
φR bs = 0.75[U bs F u A nt + min(0.6F y A gv , 0.6F u A nv )] (3.57)
For line ab, the gross shear area is A gv = 5 × 0.25 × 2 = 2.5 in and the net shear area is A nv = 2.5 −
2
2
(1.5 × 0.25 × 1.0)2 = 1.75 in . For line bc, the gross tension area is A gt = 1.5 × 0.25 × 2 = 0.75 in 2
2
and the net tension area is A nt = 0.75 − 0.5 × 1.0 × 0.25 × 2 = 0.5 in . The term U bs accounts for the
fact that for highly eccentric connections, the tension force distribution on section bc will not be uni-
form. In this case, U bs is taken as 0.5. In the present case, the force distribution is essentially uniform
because the angle gage line and the angle gravity axis are close to each other. Thus U bs = 1.0, and
the block shear strength is
φR bs = 0.75[1.0 × 58 × 0.5 + min(0.6 × 36 × 2.5, 0.6 × 58 × 1.75)] = 62.2 kips

Shear/Bearing/Tear-out on Bolts and Parts. As pointed out in Art. 3.3.4, bearing, tear-out, and
bolt shear are tied together for each bolt. Therefore, it is not possible to check bolt shear for the bolt
group as a whole, and bearing/tear-out for each part separately, and then take the minimum of these
limit states as the controlling limit state. The procedure is as follows for each bolt.
ANGLES TO PIECE W16. For the upper bolt, the design strengths for the various limit states are as
follows:
2
Bolt shear: φR v = 0.75 × 48 × (π/4) × 0.875 × 2 = 43.3 kips
Bearing on angles: φR p = 0.75 × 2.4 × 0.875 × 2 × 0.25 × 58 = 45.7 kips
Bearing on W16 × 57: φR p = 0.75 × 2.4 × 0.875 × 0.43 × 65 = 44.0 kips
Tear-out on angles: φR to = 0.75 × 1.2(2 − 0.5 × 0.9375) × 2 × 0.25 × 58 = 40.0 kips
Tear-out on W16 × 57: φR to = 0.75 × 1.2(3 − 0.9375) × 0.430 × 65 = 51.9 kips

The design strength for shear/bearing/tear-out of the upper bolt is the lowest value, 40.0 kips.
For the lower bolt, the design strengths for the various limit states are as follows:
Bolt shear: φR v = 43.4 kips
Bearing on the angles: φR p = 45.7 kips

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