Page 138 - Structural Steel Designers Handbook AISC, AASHTO, AISI, ASTM, and ASCE-07 Design Standards

Brockenbrough_Ch03.qxd 9/29/05 5:05 PM Page 3.70

CONNECTIONS

3.70 CHAPTER THREE

The flange force for design of the splice is thus

F = F + max( F , F ) = 98 .2 + max(110 55 . ) = 208 kips
0
,
f
f t
f x
f y
Suppose that M f > / 8 φM py . Let M f = 1500 in⋅kips, γ= 1 + 7.5/14.6 = 1.51, and check M f =
3
2
2
3
3
1500 in⋅kips < / 8γ φM py = / 8(1.51) × 3760 = 3200 in⋅kips. Proceeding,

ε= 1 ×14 6 . ×1 51 1 − 1 − 1500   = 149. in

.
4  3200 
T = 0 75. × 1 8 50( . × ) × 0 780 2. × ×1 49 157. = kips
F = 2T = 314 kips
fy
This is still less than the maximum possible value of F fy = (1500/7.5) × 2 = 400 kips.
Returning to the splice design example, the splice will be designed for a factored load of 208 kips.
Since the columns are of different depths, fill plates will be needed. The theoretical fill thickness is
1
11
3
1
1
(15 / 2 − 14 / 8)/2 = / 16 in, but for ease of erection AISC suggests subtracting either / 8 in or / 16 in,
3
1
1
11
whichever results in / 8-in multiples of fill thickness. Thus, use actual fills / 16 − / 16 = / 2 in thick.
Since this splice is a bearing splice, the fills either must be developed, or the shear strength of the
bolts must be reduced. It is usually more economical to do the latter in accordance with AISC
3
Specification Sec. J6 when the total filler thickness is not more than / 4 in. With the reduced bolt shear
design strength, φr v = 44.2[1 − 0.4(0.5 − 0.25)] = 39.8 kips. The number of bolts required is 208/39.8 =
5.23 or 6 bolts. By contrast, if the fillers were developed, the number of bolts required would be
208[1 + 0.5/(0.5 + 0.780)]/44.2 = 6.54 or 8 bolts. By reducing the bolt shear strength instead of
developing the fills, [(8 − 6)/8]100 = 25% fewer bolts are required for this splice.
Next, the splice plates are designed. These plates will be approximately as wide as the narrower
5
1
column flange. Since the W14 × 99 has a flange width of 14 / 8 in, use a plate 14 / 2 in wide. The fol-
lowing limit states are checked.
Gross Area. The required plate thickness based on gross area is t p = 208/(0.9 × 36 × 14.5) = 0.44 in.
1
Try a / 2 in plate.
2
Net Area. The net area is A n = (14.5 − 2 × 1.125) × 0.5 = 6.125 in , but this cannot exceed 0.85
2
2
2
of the gross area or 0.85 × 14.5 × 0.5 = 6.16 in . Since 6.16 in > 6.125 in , the effective net area is
2
A e = A n = 6.125 in . The design strength in net tension is
φR n = 0.75 × 58 × 6.125 = 266 kips > 208 kips OK
Block Shear Rupture. Since b − g < g, failure will occur as shown in Fig. 3.42 on the outer parts
of the splice plate.
A gv = 8 × 0.5 × 2 = 8.0 in 2
A gt = (14.5 − 7.5) × 0.5 = 3.5 in 2
A nv = 8.0 − 2.5 × 1.125 × 0.5 × 2 = 5.1875 in 2
A nt = 3.5 − 1 × 1.125 × 0.5 = 2.9375 in 2
F u A nt = 58 × 2.9375 = 170 kips
0.6F y A gv = 0.6 × 36 × 8.0 = 173 kips
0.6F u A nv = 0.6 × 58 × 5.1875 = 181 kips
U bs = 1.0 (uniform tension)
φR bs = 0.75[1.0 × 170 + min(173, 181)] = 257 kips > 208 kips OK
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