Page 142 - Structural Steel Designers Handbook AISC, AASHTO, AISI, ASTM, and ASCE-07 Design Standards

Brockenbrough_Ch03.qxd 9/29/05 5:05 PM Page 3.74

CONNECTIONS

3.74 CHAPTER THREE

Bearing on outer plate: φr p = 0.75 × 2.4 × 0.875 × 0.50 × 58 = 45.7 kips
Bearing on inner plate: φr p = 0.75 × 2.4 × 0.875 × 0.75 × 58 = 68.5 kips
Tear-out on W14 × 99 flange: L c = 1.75 − 0.50 × 0.9375 = 1.281 in

φr to = 0.75 × 1.2 × 1.281 × 0.710 × 65 = 53.2 kips

Tear-out on outer plate: L c = 1.5 − 0.5 × 0.9375 = 1.031 in
φr to = 0.75 × 1.2 × 1.031 × 0.50 × 58 = 26.9 kips

Tear-out on inner plate: L c = 1.031 in
φr to = 0.75 × 1.2 × 1.031 × 0.75 × 58 = 40.4 kips

Tear-out between bolts will not control in this case, since 3 − 0.9375 = 2.0625 > 1.281 or 1.031 in.
From the above, the shear/bearing/tear-out strength of the flange connection is

φR vpt = 2 × 54.2 + 2 × 26.9 + 2 × 27.1 + 2 × 54.2 = 325 kips > 311 kips OK
In the expression for φR vpt , the first term is for the two bolts in the center, which are controlled by
shear; the second and third terms are for the outer two bolts controlled by outer-plate edge distance
and bolt shear; and the fourth term is for the two inner bolts, again controlled by bolt shear.
This completes the calculation for the flange portion of the splice. The bolts, outer plate, and
inner plates, as chosen above, are OK.
WEB SPLICE DESIGN
Member Limit States. The calculations for the web connection involve the same limit states as the
flange connection, except for tension rupture of the chord net section, which involves flanges and web.
Bolt pattern must be established to check the web. The minimum number of bolts in double
shear required is (178/2)/27.1 = 3.28. Try four bolts.
Bearing/tear-out will be checked after the web splice plates are designed.
Block shear rupture will be checked as follows. As a first trial, assume the bolts have a 3-in
pitch longitudinally.

A nv = (4.75 − 1.5 × 1) × 0.440 × 2 = 2.86 in 2
A nt = (3 − 1 × 1) × 0.440 = 0.88 in 2
A gv = 4.75 × 0.440 × 2 = 4.18 in 2
A gt = 3 × 0.440 = 1.32 in 2
F u A nt = 65 × 0.88 = 57.2 kips
0.6F u A nv = 0.60 × 65 × 2.86 = 112 kips
0.6F y A gy = 0.60 × 50 × 4.18 = 125 kips
U bs = 1.0
φR bs = 0.75[57.2 + min(115, 125)] = 127 kips < 178 kips No good

Since the block shear limit state fails, the bolts can be spaced out to increase the capacity. Increase
the bolt pitch, from the 3 in assumed above to 6 in, and repeat the calculations.

A nv = (7.75 − 1.5 × 1) × 0.440 × 2 = 5.50 in 2
A nt = 0.88 in 2

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