AISC chapk.pdf
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1、K-1 CHAPTER K Design of HSS and Box Member Connections Examples K.1 through K.6 illustrate common beam to column shear connections that have been adapted for use with HSS columns. Example K.7 illustrates a through plate shear connection, which is unique to HSS columns. Calculations for transverse an
2、d longitudinal forces applied to HSS are illustrated in Examples K.8 and K.9. An example of an HSS truss connection is given in Example K.10. Examples on HSS cap plate and base plate connections are given in Examples K.11 through K.13. K-2 Example K.1 Welded/bolted Wide Tee Connection to an HSS Colu
3、mn Given: Design a connection between a W1650 beam and a HSS884 column using a WT524.5. Use w-in. diameter ASTM A325-N bolts in standard holes with a bolt spacing, s, of 3 in., vertical edge distance Lev of 14 in. and 3 in. from the weld line to the bolt line. Design as a flexible connection. PD = 6
4、.2 kips PL = 18.5 kips Note: A tee with a flange width wider than 8 in. was selected to provide sufficient surface for flare bevel groove welds on both sides of the column, since the tee will be slightly offset from the column centerline. Solution: Material Properties: Beam ASTM A992 Fy = 50 ksi Fu
5、= 65 ksi Tee ASTM A992 Fy = 50 ksi Fu = 65 ksi Column ASTM A500 Gr. B Fy = 46 ksi Fu = 58 ksi Manual Table 2-3 Geometric Properties: Beam W1650 tw = 0.380 in. d = 16.3 in. tf = 0.630 in. T = 13 s in. Tee WT524.5 ts = tw = 0.340 in. d = 4.99 in. tf = 0.560 in. bf = 10.0 in. k1 = m in. Column HSS884 t
6、 = 0.233 in. B = 8.00 in. Manual Tables 1-1, 1-8, and 1-12 Calculate the required strength LRFD ASD Pu = 1.2(6.20 kips) + 1.6(18.5 kips) = 37.0 kips Pa = 6.20 kips + 18.5 kips = 24.7 kips K-3 Calculate the available strength assuming the connection is flexible LRFD ASD Determine the number of bolts
7、Determine the single bolt shear strength rn = 15.9 kips Determine single bolt bearing strength based on edge distance Lev = 14-in. 1.25 in. o.k. rn = 49.4 kips/in.(0.340 in.) = 16.8 kips Determine single bolt bearing capacity based on spacing s = 3.00 in. 3(w) = 2.25 in. rn = 87.8 kips/in.(0.340 in.
8、) = 29.8 kips Therefore bolt shear controls, 37.0 kips 2.33 15.9 kips u min n P C r = Using e = 3 in. and s = 3 in., determine C. Try 4 bolts, C = 2.81 2.33 o.k. Determine the number of bolts Determine the single bolt shear strength rn / = 10.6 kips Determine single bolt bearing strength based on ed
9、ge distance Lev = 14-in. 1.25 in. o.k. rn / = 32.9 kips/in.(0.340 in.) = 11.2 kips Determine single bolt bearing capacity based on spacing s = 3.00 in. 3(w) = 2.25 in. rn / = 58.5 kips/in.(0.340 in.) = 19.9 kips Therefore bolt shear controls, 24.7 kips 2.33 /10.6 kips a min n P C r = Using e = 3 in.
10、 and s = 3 in., determine C. Try 4 bolts, C = 2.81 2.33 o.k. Manual Table 7-1 Table J3.4 Manual Table 7-6 Manual Table 7-5 Manual Table 7-7 Check WT stem thickness limit () 34 11 1616 max in. in.in.0.438 in.0.340 in. 22 b d t=+=+= o.k. Manual Section 9 Note: beam web thickness is greater than WT ste
11、m thickness. If the beam web were thinner than the WT stem, this check could be satisfied by checking the thickness of the beam web Determine WT length required A W1650 has a T-dimension of 13s in. Lmin = T/2 = (13s in.)/2 = 6.81 in. Determine WT length required for bolt spacing and edge distances L
12、 = 3(3.00 in.) + 2 (1.25 in.) = 11.5 in. 37.0 kips o.k. 78.0 kips 24.7 kips o.k. Calculate the stem shear rupture strength ( )()(1/16)0.6 nhu RLn dtF=+ = 11.5 in. 4(0.875 in.)(0.340 in.)(0.6)(65 ksi) = 106 kips Section J4.2 LRFD ASD Rn = 0.75(106 kips) = 79.6 kips 106 kips / 2.00 n R = 53.0 kips 79.
13、6 kips 37.0 kips o.k. 53.0 kips 24.7 kips o.k. Calculate the stem available block shear rupture strength For this case Ubs = 1.0 Section J4.3 LRFD ASD 76.2 kips/in. 0.6 231 kips/in. 0.6 210 kips/in. unt ygv unv F A t F A t F A t = = = Rn = 0.6FuAnv+UbsFuAnt 0.6FyAgv+UbsFuAnt Rn=0.340 in.(210 kips/in
14、 + 76.2 kips/in) 0.340 in.(231kips/in.+76.2 kips/in.) =97.3 kips104 kips 97.3 kips 37.0 kips o.k. 50.8 kips/in. 0.6 154 kips/in. 0.6 140 kips/in. unt ygv unv F A t F A t F A t = = = 0.6 = + 0.6 + nunvbsunt ygv bsunt RF AU F A F A U F A / n R =0.340 in.(140 kips/in. +50.8 kips/in.) 0.340 in.(154 kips
15、/in.+50.8 kips/in.) = 64.9 kips69.6 kips 64.9 kips 24.7 kips o.k. Manual Table 9-3a Manual Table 9-3b Manual Table 9-3c Eqn. J4-5 Check stem bending Calculate the required flexural strength LRFD ASD Mu = Pue = 37.0 kips(3.00 in.) = 111 kip-in. Ma = Pae = 24.7 kips(3.00 in.) = 74.1 kip-in. K-5 Calcul
16、ate the stem nominal flexural yielding strength () 2 2 3 0.340in. 11.5in. 11.2in. 44 td Z = () 2 2 3 0.340in. 11.5in. 7.49in. 66 td S = 1.6 npyy MMF ZM= ()()() 33 50ksi 11.2in.1.6 50ksi7.49in.= = 560 kip-in. 111 kip-in. o.k. = 335 kip-in. 74.1 kip-in. o.k. Calculate the stem flexural rupture strengt
17、h ()() 2 2in. 1.5in.4.5in. 4 neth td Zt d=+z () ()()() 2 3 0.340in. 11.5in. 2 0.340in.in.in. 1.5in.4.5in.7.67in. 4 =+=mz () 3 65ksi 7.67in.499kip-in. nunet MF Z= Manual Part 9 LRFD ASD Mn=0.75(499 kip-in.) 499 kip-in. / 2.00 n M = = 374 kip-in. 111 kip-in. o.k. = 250 kip-in. 74.1 kip-in. o.k. Check
18、beam web bearing 0.380 in. 0.340 in. ws tt Beam web is satisfactory for bearing by comparison with WT. Calculate weld size Since the flange width of the WT is larger than the width of the HSS, a flare bevel groove weld is required. Taking the outside radius as 2(4 in.) = 2 in. and using AISC Specifi
19、cation Table J2.2, the effective throat thickness of the flare bevel weld is E = c(2 in.) = 0.156 in. Table J2.2 K-6 The equivalent fillet weld that provides the same throat dimension is () = 1 0.15616 2 0.1563.53 16 2 D D sixteenths of an inch The equivalent fillet weld size is used in the followin
20、g calculations Check weld ductility () 1 28.00 in.2 in. 3.19 in. 22 f bk b = m ()() =+ 2 2 min 2 0.0158216 0.625 yc f s F t b wt bL () ()()() () ()() =+ 22 2 50 ksi 0.560 in.3.19 in. 0.015820.625 0.340 in. 3.19 in. 11.5 in. = 37.0 kips o.k. = 75.3 kips 24.7 kips o.k. K-7 Example K.2 Welded/bolted Na
21、rrow Tee Connection to an HSS Column Given: Design a connection for a W1650 beam to an HSS884 column using a tee with fillet welds against the flat width of the HSS. Use w-in. diameter A325-N bolts in standard holes with a bolt spacing, s, of 3.00 in., vertical edge distance Lev of 14-in. and 3.00 i
22、n. from the weld line to the center of the bolt line. Design this as a connection to a flexible support. Assume that, for architectural purposes, the flanges of the WT from the previous example have been stripped down to a width of 6.5 in. PD = 6.2 kips PL = 18.5 kips Note: This is the same problem
23、as Example K.1 with the exception that a narrow tee will be selected which will permit fillet welds on the flat of the column. The beam will still be centered; therefore the tee will be slightly offset. Solution: Material Properties: Beam ASTM A992 Fy = 50 ksi Fu = 65 ksi Tee ASTM A992 Fy = 50 ksi F
24、u = 65 ksi Column ASTM A500 Gr. B Fy = 46 ksi Fu = 58 ksi Manual Table 2-3 Geometric Properties: Beam W1650 tw = 0.380 in. d = 16.3 in. tf = 0.630 in. Column HSS884 t = 0.233 in. B = 8.00 in. Tee WT524.5 ts = tw = 0.340 in. d = 4.99 in. tf = 0.560 in. k1 = m in. Manual Tables 1-1 and 1-12 Calculate
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