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1、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 co
2、nnection between a 2L33c tension member and a W2494 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 W2494 ASTM A992 Fy = 50 ksi Fu = 65 ksi Angle
3、s 2L33c ASTM A36 Fy = 36 ksi Fu = 58 ksi Manual Table 2-3 Geometric Properties: Beam W2494 d = 24.3 in. tw = 0.515 in. bf = 9.07 in. tf = 0.875 in. Angles 2L33c 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 kips
4、53.5 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. wel
5、ds (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. 110.78
6、5 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 us
7、ing 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
8、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 e
9、qual 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: WT639.5 WT828.5 WT734 WT930 Try WT828.5; bf = 7.12 in., tf = 0.71
10、5 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 th
11、e 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 =
12、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 WT935.5 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 WT935.5 tf = 0.
13、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 W1850 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
14、 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 t
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