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Daniel
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PROJECT : PAGE :
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Tian LiCLIENT :
DESIGN BY :
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JOB NO. : DATE :
REVIEW BY :
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Brace Connection Design Based on AISC-ASD 9th
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INPUT DATA & DESIGN SUMMARY
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BRACE SECTION (Tube or Pipe)
= >
HSS8X8X5/8
TubeArminth
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BRACE AXIAL LOAD AT SERVICE LEVEL
T =450kips16.402.980.588.00
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(For seismic T = W0 x ASD / 1.7 suggested. CBC 2213.4.2)
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ANGLE BETWEEN BRACE & COLUMN
q =500
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COLUMN INTERFACE DIMENSION
2 b =24in
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COLUMN CENTER TO INTERFACE
ec =7.33in
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BEAM CENTER TO INTERFACE
eb =14.8in
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( 5/8" Gusset Plate with 5/16" Fillet Weld, 4 leg x 25" Length at Brace,
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and 2 leg x 24" at Column Interface, 2 leg x 50" at Beam Interface. )
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THE CONNECTION DESIGN IS ADEQUATE.
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ANALYSIS
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DETERMINE BEST FILLET WELD SIZE PER BRACE THICKNESS (ASD Sec.J2.2b)
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w =5/16in
> wMIN =
 3/16in
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< wMAX =
 7/16in
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[Satisfactory]
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DETERMINE REQUIRED WELD LENGTH AT BRACE (ASD Sec.J2.4 )
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L = T / [(4) (0.3) Fu (0.707 w)]
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= 450.00 / [(4) (0.3) (70) (0.707x5/16)] =
24.25in
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( USE25in )
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CHECK SHEAR RUPTURE CAPACITY OF SLOTED BRACE (ASD Sec.J4)
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Tt,rup,brace =(0.3Fu)Anu =
1010.94kips>T
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[Satisfactory]
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Where Fu =58
ksi (LRFD Tab.1-4, Pg. 1-21)
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Anu =
4 t L = 4 x 0.581 x 25 =
58.10in2
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DETERMINE REQUIRED THICKNESS OF GUSSET PLATE PER Tt,rup,brace ABOVE (ASD Tab. J2.4)
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tg = 5/8in
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CHECK SHEAR RUPTURE CAPACITY OF GUSSET PLATE (ASD Sec.J4 & CBC 2213A.4.2)
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Tt,rup,gusset = (0.3Fu)Anv =
543.8kips>T
[Satisfactory]
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Where Fu =58
ksi (A36 Steel)
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Anv =
2 tg L = 2 x 5/8 x 25 =
31.25in2
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CHECK TENSION CAPACITY AT SLOTED BRACE (ASD Sec.D1).
CHECK GUSSET BLOCK SHEAR CAPACITY (ASD J4)
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Tt,brace = (0.5FuUAn) =
458.75kips>T
Ts,guss = [0.3FuAnv + 0.5FuAgt] =
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[Satisfactory]
=544
+1.7[0.5FuAgt]
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Where U =1
(ASD Sec.B3)
>T =450.0
[Satisfactory]
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An =A - 2 t tg =15.82in2
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DETERMINE CONNECTION INTERFACE FORCES, USING THE UNIFORM FORCE METHOD & TO REMAIN FREE OF INTERFACE MOMENTS
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b =12.00in, as given
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a = (eb + b) tanq - ec =
24.61in
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r = [(eb + b)2 + (ec + a)2] 0.5 =
41.69in
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Vc = (b / r) T =
129.5kips
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Hc = (ec / r) T =
79.1kips50
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Vb = (eb / r) T =
159.7kips
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Hb = (a / r) T =
265.6kips
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CHECK WELD CAPACITY AT INTERFACES (ASD Sec.J2.4 )
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fVc = Vc / (4 b 0.707 w) =
12.21ksi
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fHc = Hc / (4 b 0.707 w) =
7.46ksi
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fVb = VB / (4 a 0.707 w) =
7.34ksi
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fHb = HB / (4 a 0.707 w) =
12.21ksi
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fv,c = [(fvc)2 + (fHc)2] 0.5 =
14.31ksi<0.3 Fu =21.00ksi
[Satisfactory]
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fv,b = [(fvb)2 + (fHb)2] 0.5 =
14.25ksi<0.3 Fu =21.00ksi
[Satisfactory]
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LICENSE NUMBER: ?????
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