[ Pobierz całość w formacie PDF ]
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 ] zanotowane.pldoc.pisz.plpdf.pisz.plgrolux.keep.pl
|