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to be:
Fresultant = 46.0 kips
Æbb
0.8(65.81) = 52.65
According to Article 6.13.6.1.4b, check for flexural yielding on the gross section of the web splice plates
at the strength limit state. The flexural stress is limited to Fy.
Æf
Ag = in2
2(0.375)(75.25) = 56.44
2
îø(0.375)75.253ùø
2À
îø14ëø öøùø
ïø úø
2 2 cos
ïø ìø úø
Ix cos ¸
12 360
ðø ûø ðø íø øøûø
SPL = = in3
= 686.8
c 75.25 2À
ëø öøcosîø14ëø öøùø
ìø ïø ìø úø
2 360
íø øø ðø íø øøûø
D-74
Bolted Splice Design Section 2-2 G2 Node 20.3
Splice Plates (continued)
Muv + Muw Huw
f =
+
SPL Ag
(143 + 1306)(12) 469
= ksi
Æf
+ = 33.63
686.8 56.44
Since the thickness of the two splice plates exceeds t , say the shear resistance in the splice plates
w
is adequate.
Flange Splice Plate Design
Top Flange
The width of the outside splice plate should be at least as wide as the width of the narrowest flange at
the splice. In this case, however, the width of the top flange is the same on either side of the splice.
Therefore;
Try: 16 x 0.5 in. outer plate Try: 2 - 6 x 0.625 in. inner plates
Ag = 8.0 in2 Ag = 7.50 in2
As specified in Article 6.13.6.1.4c, the effective area, Ae, of each splice plate is to be sufficient to prevent
yielding of each splice plate under its calculated portion of the minimum flange design force. For splice
plates subjected to compression, the effective area is equal to the gross area.
The effective areas of the inner and outer splice plates are computed as:
Æu Fu
ëø öø
Ae = Eq (6.13.6.1.4c-2)
An d" Ag
ìø
Æy Fyt
íø øø
Outer plate: An = in2
[ 16.0 - 4(0.875 + 0.125) ](0.5) = 6
Inner plate: An = in2
2[ 6.0 - 2(0.875 + 0.125) ] 0.625 = 5
0.8(65)
îø ùø(6.0) = 6.57
Outer plate:
ïø úø
0.95(50)
ðø ûø
0.8(65)
îø ùø(5.0) = 5.47
Inner plate:
ïø úø
0.95(50)
ðø ûø
D-75
Bolted Splice Design Section 2-2 G2 Node 20.3
Splice Plates (continued)
As specified in Article C6.13.6.1.4c, if the combined area of the inner splice plates is within 10 percent of
the area of the outside splice plate, then both the inner and outer plates may be designed for one-half the
flange design force (which is the case here). Double shear may then be assumed in designing the
connections. If the areas differ by more than 10 percent, the design force in each splice plate and its
connection at the strength limit state should be determined by multiplying the flange design force by the
ratio of the area of the splice plate under consideration to the total area of the inner and outer splice
plates. In this case, the shear resistance of the connection would be checked for the maximum
calculated splice plate force actings on a single shear plane.
For the negative live load bending case, the controlling flange is the top flange. The flange is subjected to
tension under this live load bending condition (see page D-62). Compute the minimum resistance, FcfAe,
in the top flange for this load case. The factored tensile resistance, Pr, is taken as the lesser of the
values given by Eqs (6.8.2.1-1 and 6.8.2.1-2). The factor ± in Eq (6.13.6.1.4c-1) is generally taken equal
to 1.0.
4.19
+ 1.0(1.0)(50)
1.0
Fcf = ksi Eq (6.13.6.1.4c-1)
= 27.09
2
0.75±Æf ksi (controls)
Fyf =
0.75(1.0)(1.0)(50) = 37.5
FcfAe = kips
37.5(13.14) = 493
As discussed previously, St. Venant torsional shear and lateral flange bending are not considered in the
top flange at the strength limit state. Warping torsion is also ignored. According to Article 6.13.6.1.4c,
the capacity of the splice plates to resist tension is computed using the provisions of Article 6.8.2. The
factored tensile resistance, Pr, is taken as the lesser of:
Pr = Pny = FyAg Eq (6.8.2.1-1)
Æy Æy
Pr = kips Outer plate
0.95(50)(8.0) = 380
Pr = kips Inner plates
0.95(50)(7.50) = 356
or
Pr = Pnu = FuAnU Eq (6.8.2.1-2)
Æu Æu
Pr = kips Outer plate
0.80(65)(6.0)(1.0) = 312
450
Pr = kips Inner plates (controls) > kips OK
0.80(65)(5.0)(1.0) = 260 = 225
2
D-76
Bolted Splice Design Section 2-2 G2 Node 20.3
Splice Plates (continued)
Under the positive live load bending case, the top flange is the noncontrolling flange and is subjected to
compression. The minimum design force, FncfAe, for the top flange for this load case was computed
earlier (see page D-64) to be 600 kips. The factored compressive resistance, Rr, is taken as:
Rr = FyAs (Outer and Inner plates, respectively) Eq (6.13.6.1.4c-4)
Æc
Rr = kips
0.9(50)(8.0) = 360
600
= kips > kips OK
0.9(50)(7.50) = 338 = 300
2
Bearing Resistance at Bolt Holes
Check bearing of the bolts on the connected material under the minimum design force, F Ae = 600 kips,
ncf
for the top flange. The design bearing resistance, Rn, is computed using the provisions of Article 6.13.2.9.
According to Article 6.13.2.9, the bearing resistance for the end and interior rows of bolts is computed
using Eq (6.13.2.9-1 ) or Eq (6.13.2.9-2). Calculate the clear distance between holes and the clear end
distance and compare to 2.0d to determine the equation to be used to solve for the bearing resistance.
The center-to-center distance between the bolts in the direction of the force is 3.0 in. Therefore:
Clear distance between holes = - 1.0 = 2.0
in.
3.0
For the four bolts adjacent to the end of the splice plate, the end distance is assumed to be 1.5 in.
Therefore, the clear distance between the edge of the holes and the end of the splice plate is:
1.0
Clear end distance = - = 1.0
in.
1.5
2
The value 2.0d is equal to 1.75 in. Since the clear end distance is less than 2.0d, use Eq (6.13.2.9-2).
Rn = 1.2LctFu = k/bolt Eq (6.13.2.9-2)
1.2(1.0)(1.0)(65) = 78
= 0.8
Æbb
FncfAe = 600 k
Æbb
Bottom Flange
Try: 75.5 x 0.375 in. outer plate Try: 2 - 36.75 x 0.375 in. inner plates
Ag = 28.3 in2 Ag = 27.6 in2
Note: Since the inner splice plate must be partially split to accommodate the longitudinal flange [ Pobierz całość w formacie PDF ]