Force to be resisted by a stiffener is;
F = Pbf - (2.5k + N)Fyw t w = 114 k - (2.5(1.0" ) + 0.420" )50ksi(0.285" ) = 72.4 k
with N = tbf
(F = 322.0KN)
Length available for welding stiffeners to the column web is;
L = 12" (assuming a 1/2" chamfer) x 2 sides x 2 stiffeners = 48" (1.22m)
The required weld size is;
For E70XX electrodes; φ w = 0.75[0.60(70ksi)] = 31.5ksi (265.0MPa)
Note;
F
72.4 k
3"
Ru
s=
=
= 0.068" <
(1.72mm < 4.76mm)
(0.707)L(φ w ) 0.707(48" )31.5ksi
16
F
Therefore, the minimum weld size governs
Check shear strength of the base metal;
φ n = φ(0.6t st Fst ) x2 = 0.9(0.6)0.375" (36ksi)2 = 14.6 - kips / in (2.56KN/mm)
R
72.4 k
F
φ n = 14.6 - kips / in > 3.02 - kips / in =
=
O.K.
(2.56KN/m > 0.53KN/m)
R
48"/2 L / 2
Stiffener to column flange;
Use full penetration groove welds
Choose two 3/8" (9.53mm) thick by 3.0" (76.2mm) wide stiffeners with a 3/16" (4.76mm) weld at
the column web and a full pen-etration groove weld at the column flange
Design the single-plate web connection;
Note:
The governing load combination (1.2D+0.5L+1.6Lr) is based solely on gravity loads.
wu = 1.2D+0.5L+1.6Lr = 1.2(343plf)+0.5(0)+1.6(240plf) = 796plf (11.6KN/m)
w u (L - d c ) 796plf (30'- 13.98" (1'/12" ))
= 11.5k (51.2KN)
Vu =
=
2
2
Try a 3/8" (9.53mm) plate;
Determine the number of 3/4" (19.05mm) diameter A325-N bolts required for shear;
From AISC LRFD 2nd ed. table 8-11;
11.5k
Ru
=
=
= 0.72 bolts, Say 2 bolts
n min
φ n 15.9 - kips / bolt
r
Determine the number of 3/4" (19.05mm) diameter A325-N bolts required for bearing, assuming Le = 1-
1/2" (38.1mm), and s = 3" (76.2mm). The .255" (6.48mm) beam web is more critical then the 3/8"
(9.53mm).
From AISC LRFD 2nd ed. Table 8-13;
H4-53
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