AISC iid.pdf

IID-1 Chapter IID Miscellaneous Connections This section contains design examples on connections in the AISC Steel Construction Manual. that are not covered in other sections of AISC Design Examples. IID-2 Example II.D-1 Prying Action in Tees and in Single Angles Given: Design a WT tension-hanger connection between a 2L3×3×c tension member and a W24×94 beam connection to support the following loads: PD = 13.5 kips PL = 40 kips Use w-in. diameter ASTM A325-N bolts and 70 ksi electrodes. Material Properties: Hanger ASTM A992 Fy = 50 ksi Fu = 65 ksi Beam W24×94 ASTM A992 Fy = 50 ksi Fu = 65 ksi Angles 2L3×3×c ASTM A36 Fy = 36 ksi Fu = 58 ksi Manual Table 2-3 Geometric Properties: Beam W24×94 d = 24.3 in. tw = 0.515 in. bf = 9.07 in. tf = 0.875 in. Angles 2L3×3×c A = 3.55 in.2 x= 0.860 in. Manual Tables 1-1, 1-7, and 1-15 Solution: LRFD ASD 1.2(13.5 kips)1.6(40 kips)80.2 kips u P =+= 13.5 kips40 kips53.5 kips a P =+= IID-3 Check tension yielding of angles 2 (36 ksi)(3.55 in. )128 kips= nyg RF A Eqn. D2-1 LRFD ASD 0.90(128 kips) 115 kips80.2 kips n R= = 128 kips / 1.67 76.6 kips53.5 kips n R = = Try 4-in. fillet welds min 80.2 kips 1.3921.392(4 sixteenths) 14.4 in. u P L D = = Use four 4-in. welds (16 in. total), one at each toe and heel of each angle. Try 4-in. fillet welds min 53.5 kips 0.9280.928(4 sixteenths) 14.4 in. = = a P L D Use four 4-in. welds (16 in. total), one at each toe and heel of each angle. Check tension rupture of angles Calculate the effective net area 0.860 in. 110.785 4 in. x U L = = = () 22 (3.55 in. ) 0.7852.80 in. en AAU= Table D3.1 Case 2 Eqn. D2-2 LRFD ASD 0.75 = 2 (58 ksi)(2.80 in. )163 kips nue RF A= 2.00 = 2 (58 ksi)(2.80 in. )163 kips nue RF A= 0.75(163) 122 kips80 kips tn R= =o.k. 163 / 2.00 81.5 kips53.5 kips nt R = =o.k. . Select a preliminary WT using beam gage g = 4 in. With four w-in. diameter ASTM A325-N bolts, 80 kips 20 kips/bolt 4 = u uut P Tr n 29.8 kips20 kips n Br= = o.k. Select a preliminary WT using beam gage g = 4 in. With four w-in. diameter ASTM A325-N bolts, 53.5 kips 13.4 kips/bolt 4 a aat P Tr n = /19.9 kips13.4 kips= = n Br o.k. Manual Table 7-2 IID-4 With four bolts, the maximum effective length is 2g = 8 in. Thus, there are 4 in. of tee length tributary to each pair of bolts and LRFD ASD 2 bolts(20 kips/bolt) 10.0 kips/in. 4 in. = 2 bolts(13.4 kips/bolt) 6.7 kips/in. 4 in. = The minimum depth WT that can be used is equal to the sum of the weld length plus the weld size plus the k-dimension for the selected section. From Manual Table 1-8 with an assumed b = 4 in./2 = 2 in., 1116 0 in.,t and 14 min 4 in. in.6 in.,dk=+appropriate selections include: WT6×39.5 WT8×28.5 WT7×34 WT9×30 Try WT8×28.5; bf = 7.12 in., tf = 0.715 in., tw = 0.430 in. Check prying action (4 in.0.430 in.) 22 = w gt b 14 1.79 in.1-in. entering and tightening clearance, =o.k. (7.12 in.4 in.) 1.56 in. 22 = f bg a Since a = 1.56 in. o.k. 164 /98.2 kips53.5 kips 1.67 n R = o.k. Eqn. D2-1 Manual Part 9 Check shear rupture of the base metal along the toe and heel of each weld line 6.19 = 4 6.19 65 ksi 0.381 in. 80 kips o.k. Rn / = = 187 93.5 kips53.5 kips 2.00 o.k. Note: Alternately, a WT tension hanger could be selected with a flange thickness to reduce the effect of prying action to an insignificant amount, i.e., 0. u qAssumsing '1.42 in.b = LRFD ASD min 4.44' 4.44(20 kips/bolt)(1.42 in.) (4 in./bolt)(65 ksi) 0.696 in. = = = u Tb t pF Try WT9×35.5 tf = 0.810 in. 0.696 in. o.k. tw = 0.495 in. 0.430 in. o.k. bf = 7.64 in. 7.12 in. o.k. min 6.66' 6.66(13.4 kips/bolt)(1.42 in.) (4 in./bolt)(65 ksi) 0.698 in. = = = u Tb t pF Try WT9×35.5 tf = 0.810 in. 0.698 in. o.k. tw = 0.495 in. 0.430 in. o.k. bf = 7.64 in. 7.12 in. o.k. Manual Table 1-8 IID-7 Example II.D-2 Beam Bearing Plate Given: A W18×50 beam with a dead load end reaction of 15 kips and a live load end reaction of 45 kips is supported by a 10-in. thick concrete wall. If the beam has Fy = 50 ksi, the concrete has f c = 3 ksi, and the bearing plate has Fy = 36 ksi, determine: a. if a bearing plate is required if the beam is supported by the full wall thickness, b. the bearing plate required if N = 10 in. (the full wall thickness), c. the bearing plate required if N = 6 in. and the bearing plate is centered on the thickness of the wall. Material Properties: Beam W18×50 ASTM A992 Fy = 50 ksi Fu = 65 ksi Bearing Plate (if required) ASTM A36 Fy = 36 ksi Fu = 58 ksi Concrete Wall fc = 3 ksi Manual Tables 2-3 and 2-4 Geometric Properties: Beam W18×50 d = 18.0 in. tw = 0.355 in. bf = 7.50 in. tf = 0.570 in. k = 0.972 in. k1 = 13/16 in. Concrete Wall h = 10 in. Manual Table 1-1 IID-8 Solution A: LRFD ASD Ru = 1.2(15 kips) + 1.6(45 kips) = 90 kips Check local web yielding Nreq = 1 2 u RR R k = 90 kips - 43.1 kips 17.8 kips/in. 0.972 in. = 2.63 in. 0.2, Nreq = 5 6 - u RR R = 90 kips-52.0 kips 6.30 kips/in. = 6.03 in. 0.2, N d = 0.335 0.2 o.k. Check the bearing strength of concrete 0.60 c = cPp = c(0.85fc)A1 = 0.60 (0.85)(3 ksi)(7.50 in.×10 in.) = 115 kips 90 kips o.k. Ra = 15 kips + 45 kips = 60 kips Check local web yielding Nreq = 1 2 / / a RR R k = 60 kips - 28.8 kips 11.8 kips/in. 0.972 in. = 2.64 in. 0.2, Nreq = 5 6 - / / a RR R = 60 kips - 34.7 kips 4.20 kips/in. = 6.03 in. 0.2, N d = 0.335 0.2 o.k. Check the bearing strength of concrete 2.50 c = Pp/c = ( 0.85fc) A1/c = (0.85)(3 ksi)(7.50 in.×10 in.)/2.50 = 76.5 kips 60 kips o.k. Manual Table 9-4 Manual Table 9-4 Eqn. J8-1 IID-9 Check beam flange thickness LRFD ASD Determine cantilever length n = 2 f b - k = 7.50 2 - 0.972 in. = 2.78 in., Determine bearing pressure fp = u 1 R A Determine cantilever moment 2 1 2 u u R n M A = 2 14 Zt= M FyZ Fy 2 4 t treq = = 2 1 42 uu yy MR n FA F Determine cantilever length n = 2 f b - k = 7.50 2 - 0.972 in. = 2.78 in., Determine bearing pressure fp = a 1 R A Determine cantilever moment 2 1 2 a a R n M A = 2 14 Zt= M FyZ Fy 2 4 t treq = 2 1 42 aa yy MR n FA F = Manual Part 14 = 0.90 tmin = 2 1 2.22 u y R n A F = 2 2.22(90 kips)(2.78 in.) (7.50 in. 10 in.)(50 ksi)× = 0.643 in. 0.570 in. n.g. A bearing plate is required. = 1.67 tmin = 2 1 3.33 a y R n A F = 2 3.33(60 kips)(2.78 in.) (7.50 in. 10 in.)(50 ksi)× = 0.643 in. 0.570 in. n.g. A bearing plate is required. IID-10 Solution B: N = 10 in. From Solution A, local web yielding and web crippling are not critical. LRFD ASD Calculate the required bearing-plate width. 0.60 c = A1 req = ' (0.85) u cc R f = 90 kips 0.60(0.85 x 3 ksi) = 58.8 in2 B req = 1 req A N = 2 58.8 in. 10 in. = 5.88 in. Use B = 8 in. (least whole-inch dimension that exceeds bf) Calculate required bearing-plate thickness. n = 2 B - k = 8 in. 2 - 0.972 in. = 3.03 in. tmin = 2 1 2.22 u y R n AF = × 2 2.22(90 kips)(3.03 in.) (10 in.8 in.)(36 ksi) = 0.799 in. Use PL1×10×0-8 Calculate the required bearing-plate width. 2.50 c = A1 req = ' (0.85) ac c R f = 60 kips(2.50) (0.85 x 3 ksi) = 58.8 in2 B req = 1 req A N = 2 58.8 in. 10 in. = 5.88 in. Use B = 8 in. (least whole-inch dimension that exceeds bf) Calculate required bearing-plate thickness. n = 2 B - k = 8 in. 2 - 0.972 in. = 3.03 in. tmin = 2 1 3.33 a y R n A F = × 2 3.33(60 kips)(3.03 in.) (10 in.8 in.)(36 ksi) = 0.799 in. Use PL1×10×0-8 Eqn. J8-1 Manual Part 14 IID-11 Solution C: N = 6 in. From Solution A, local web yielding and web crippling are not critical. Try B = 8 in. A1 = B × N = (8 in.)(6 in.) = 48 in.2 To determine the dimensions of the area A2, the load is spread into the concrete at a slope of 2:1 until an edge or the maximum condition 21 /2AAis met. There is also a requirement that the area A2 be geometrically similar to A1. The 6-in. dimension spreads 2 in. to each side to meet the concrete edge. Thus, the 8-in. dimension can also be spread 2 in. to each side. Therefore, N1 = 6 in. + 2(2 in.) = 10 in. B1 = 8 in. + 2(2 in.) = 12 in. A2 = B1 × N1 = 12 in. (10 in.) = 120 in.2 Check 21 /AA = 1.58 2 o.k. LRFD ASD 0.60 c = A1 req = 2 1 A 2 ' (0.85) u cc R f = 2 1 120 in. × 2 90 kips 0.60(0.85 3 ksi) = 28.8 in.2 90 kips o.k. Try, PL a in. Check bolt bearing strength on plate rn = (1.2LctFu) M (2.4dtFu) = 0.75 Oversized holes, bolt spacing = 3 in. rn = (78.3 kips/in.)(a in.) = 29.4 kips/bolt Oversized holes, edge distance = 14 in. rn = (40.8 kips/in.)(a in.) = 15.3 kips/bolt 15.3 kips/bolt 60 kips o.k. Try, PL a in. Check bolt bearing strength on plate rn / = (1.2LctFu)/ M (2.4dtFu)/ = 2.00 Oversized holes, bolt spacing = 3 in. rn / = (52.2 kips/in.)(a in.) = 19.6 kips/bolt Oversized holes, edge distance = 14 in. rn / = (27.2 kips/in.)(a in.) = 10.2 kips/bolt 10.2 kips/bolt 80 kips o.k. Use 6 bolts / nn Rr n = = (10.2 kips/bolt)(5 bolts) = 51.0 kips 60 kips o.k. Use 6 bolts