Conexiones Tubulares

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 and longitudinal forces applied to HSS are illustrated in Examples K.8 and K.9. An example of anHSS 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.

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Example K.1

Welded/bolted Wide Tee Connection to an HSS Column

Given: Design a connection between a W16×50 beam and a HSS8×8×4 column using a WT5×24.5. 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 = 6.2 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: BeamASTM A992 Tee ASTM A992 Column ASTM A500 Gr. B Geometric Properties: Beam W16×50 Tee Column WT5×24.5 HSS8×8×4

Fy = 50 ksi Fy = 50 ksi Fy = 46 ksi

Fu = 65 ksi Fu = 65 ksi Fu = 58 ksi

Manual Table 2-3

tw = 0.380 in. T = 13 s in. ts = tw = 0.340 in. bf = 10.0 in. t = 0.233 in.

d = 16.3 in. d = 4.99 in. k1 = m in. B = 8.00 in.

tf = 0.630 in. tf = 0.560 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

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Calculate the available strength assuming the connection is flexible LRFD Determine the number of bolts Determine the single bolt shear strength φrn = 15.9 kips Determine single bolt bearing strength based on edge distance Lev = 14-in. ≥1.25 in. o.k. φrn = 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. φrn = 87.8 kips/in.(0.340 in.) = 29.8 kips Therefore bolt shear controls, P 37.0 kips Cmin = u = = 2.33 φrn 15.9 kips Using e = 3 in. and s = 3 in., determine C. Try 4 bolts, C = 2.81 > 2.33 ASD Determine the number of bolts Determine the single bolt shearstrength rn /Ω = 10.6 kips Determine single bolt bearing strength based on edge distance Lev = 14-in. ≥ 1.25 in. o.k. rn /Ω = 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 /Ω = 58.5 kips/in.(0.340 in.) = 19.9 kips Therefore bolt shear controls, P 24.7 kips Cmin = a = = 2.33 rn / Ω 10.6 kips Using e = 3 in. and s = 3in., determine C. Try 4 bolts, C = 2.81 > 2.33 Manual Table 7-7 Manual Table 7-5 Table J3.4 Manual Table 7-6 Manual Table 7-1

o.k.

o.k.

Check WT stem thickness limit d ( 3 4 in.) + 1 in. = 0.438 in. > 0.340 in. o.k. tmax = b + 116 in. = 16 2 2 Note: beam web thickness is greater than WT stem thickness. If the beam web were thinner than the WT stem, this check could be satisfied by checkingthe thickness of the beam web Determine WT length required A W16×50 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 = 3(3.00 in.) + 2 (1.25 in.) = 11.5 in. < T = 13s in. o.k. Try L = 11.5 in.

Manual Section 9

Manual Section 10

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Calculate the stem shear yielding strength

Rn = 0.6 Fy Ag = 0.6(50ksi)(11.5 in.)(0.340 in.) = 117 kips
LRFD ASD

Eqn. J4-3

φRn = 1.00(117 kips) = 117 kips 117 kips > 37.0 kips
o.k.

117 kips = 78.0 kips 1.5 78.0 kips > 24.7 kips o.k. Rn / Ω =

Calculate the stem shear rupture strength
Rn = [ L − n(d h + 1/16)] ( t )( 0.6 Fu )

Section J4.2

= [11.5 in. – 4(0.875 in.)](0.340 in.)(0.6)(65 ksi) = 106 kips
LRFD ASD

φRn = 0.75(106 kips) = 79.6 kips...