Connection Design: B159-CI

Units: kip-inStory: N+3.10Design Code: AISC 360-10

Beam-Column Moment Major Axis Connection

Summary of results

Design Check Type D/C Ratio Result Reference

1 Beam design flexural strength 0.523 Passed Spec. Eq F13-1

2 Strength of bolt group 0.305 Passed Pg 7-18 AISC manual

3 Shear yielding of web plate 0.165 Passed J4-3

4 Shear rupture of web plate 0.317 Passed J4-4

5 Block shear rupture strength of web plate 0.286 Passed J4-5

6 Design strength of weld 0.223 Passed J2-3

7 Web plate rupture strength at weld 0.315 Passed Manual Eq 9-2

8 Shear yielding of beam web 0.257 Passed J4-3

9 Shear rupture of beam web 0.398 Passed J4-4

10 Block shear rupture strength of beam web 0.453 Passed J4-5

11 Panel zone shear strength 1.127 Failed AISC 13-Section 2.2.1

12 Local flange bending 2.102 Failed AISC 13-Section 2.2.2

13 Local web yielding 1.38 Failed AISC 13-Section 2.2.3

14 Web crippling 1.66 Failed AISC 13-Section 2.2.3

Material PropertiesBeam W12X26 STEEL Fy = 50 ksi Fu = 65 ksi

Column W14X48 STEEL Fy = 50 ksi Fu = 65 ksi

Web Plate STEEL Fy = 50 ksi Fu = 65 ksi

Geometric PropertiesBeam W12X26 tw = 0.23 in d = 12.2 in tf = 0.38 in bf = 6.49 in

Column W14X48 tw = 0.34 in d = 13.8 in tf = 0.595 in bf = 8.03 in

Preferences s = 2.95 in Lev = 1.48 in Leh = 1.48 in

Bolts, Plate & WeldWeld Size, D(1/16) = 3.94 in

Web Plate Thickness, t = 0.49213 in

Bolt Type = A325-N diameter, db = 1.26 in

Hole Type = STD diameter, dh = 1.31 in

Design Calculations

Shear Demand

Ru=√Pu2+V u2 Ru=√0.220462+21.632 Ru=21.63kips

1−Beamdesign flexural strength ,Reference (Spec .Eq F13−1)b=b f−tw b=6.49−0.23 b=6.26∈¿h=d−2 t f h=12.2−2∗0.38 h=11.44∈¿

Sxx=bf d

2

6−bh

3

6dSxx=

6.49∗12.22

6−6.26∗11.44

3

6∗12.2Sxx=32.96¿

3

ϕ M n=ϕ FuS x ϕ M n=0.9∗65∗32.96 ϕ M n=1927.96kip−¿

D /C Ratio=M u

ϕ M n

D /C Ratio=1009.181927.96

D /C Ratio=0.52344

D /C Ratio isless than 1, DesignisOK

2−Strength of bolt group ,Reference (Pg7−18 AISC manual)Compute bearing strength per bolt

ru=√Pu2+V u2

nru=

√0.220462+21.6323

ru=7.21kips

lc 1=Lev−dh2

lc 1=1.48−1.312

lc 1=0.82013∈¿

lc=s−dh lc=2.95−1.31 lc=1.64∈¿ϕ rn1=min (ϕ1.2 lc 1t Fu , ϕ1.2 lc t Fu)ϕ rn1=min (0.75∗1.2∗0.82013∗0.49213∗65,0.75∗1.2∗1.64∗0.49213∗65)ϕ rn1=23.61kipsϕ rn1=ϕ 2.4dt Fuϕ rn1=0.75∗2.4∗1.26∗0.49213∗65 ϕ Rn=72.54 kipsmin(ϕ1.2 lc1 t Fu , ϕ1.2l c t Fu)≤ϕ2.4 dt Fusoϕr n1=23.61kipsCompute shear strength per bolt

Ab=π d2

4Ab=

3.14∗1.262

4Ab=1.25¿

2

ϕ rn2=ϕ Fnv Ab ϕ rn2=0.75∗54∗1.25 ϕ rn2=50.46kipsϕ rn1is less thanϕ Rn2

Bearingcontrolsshear

D /C Ratio=ru

min(ϕr n1, ϕrn2)

D /C Ratio= 7.21min(23.61,50.46)

D /C Ratio=0.30539

D /C Ratio isless than 1, DesignisOK

3−Shear yielding of web plate ,Reference (J 4−3)Agv=L∗t Agv=8.86∗0.49213 Agv=4.36¿

2

ϕ Rn=ϕ0.6 F y Ag ϕ Rn=1∗0.6∗50∗4.36 ϕ Rn=130.78kips

D /C Ratio=Ruϕ Rn

D /C Ratio= 21.63130.78

D /C Ratio=0.16541

D /C Ratio isless than 1, DesignisOK

4−Shear rupture of web plate ,Reference (J 4−4)

Anv=[L−n(dh+116

)]t Anv=[8.86−3 (1.31+ 116

)]0.49213Anv=2.33¿2

ϕ Rn=ϕ0.6 Fu Anv ϕ Rn=0.75∗0.6∗65∗2.33 ϕ Rn=68.13 kips

D /C Ratio=Ruϕ Rn

D /C Ratio=21.6368.13

D /C Ratio=0.31749

D /C Ratio isless than 1, DesignisOK

5−Block shear rupture strengthof web plate ,Reference (J 4−5)

Ant=[Leh−12(dh+

116

)] t Ant=[1.48−12(1.31+ 1

16)]0.49213Ant=0.38823¿

2

Anv=[{(n−1)s+Lev}−{2n−12

(dh+116

)}] t

Anv=[[(3−1)2.95+1.48]−[ 2∗3−12

(1.31+ 116

)]]0.49213 Anv=1.94 ¿2

Agv=[(n−1)s+Lev ]t Agv=[(3−1)2.95+1.48 ]0.49213Agv=3.63¿2

ϕ Rn=ϕ [Fu Ant+min(0.6F y A gv ,0.6Fu Anv)]ϕ Rn=0.75 [65∗0.38823+min(0.6∗50∗3.63,0 .6∗65∗1.94)]ϕ Rn=75.7 kips

D /C Ratio=Ruϕ Rn

D /C Ratio=21.6375.7

D /C Ratio=0.28574

D /C Ratio isless than 1, DesignisOK

6−Designstrengthof weld , Reference (J 2−3)

ϕ Rn=ϕ 0.6 FexxD∗2L

22.627

ϕ Rn=0.75∗0.6∗70∗3.94∗2∗8.86

22.627ϕ Rn=97.1kips

D /C Ratio=Ruϕ Rn

D /C Ratio=21.6397.1

D /C Ratio=0.22278

D /C Ratio isless than 1, DesignisOK

7−Web plate rupture strengthat weld , Reference(ManualEq9−2)

tmin=FEXXD

22.62 Futmin=

70∗3.9422.62∗65

tmin=0.18744

D /C Ratio=tmint f

D /C Ratio=0.187440.595

D /C Ratio=0.31502

D /C Ratio isless than 1, DesignisOK

8−Shear yielding of beamweb ,Reference (J 4−3)Agv=L∗t Agv=12.2∗0.23 Agv=2.81¿

2

ϕ Rn=ϕ0.6 F y Agv ϕ Rn=1∗0.6∗50∗2.81 ϕ Rn=84.18 kips

D /C Ratio=Ruϕ Rn

D /C Ratio=21.6384.18

D /C Ratio=0.25697

D /C Ratio isless than 1, DesignisOK

9−Shear rupture of beamweb ,Reference(J 4−4)

Anv=[L−n(dh+116

)]t Anv=[12.2−3(1.31+ 116

)]0.23 Anv=1.86¿2

ϕ Rn=ϕ0.6 Fu Anv ϕ Rn=0.75∗0.6∗65∗1.86 ϕ Rn=54.32kips

D /C Ratio=Ruϕ Rn

D /C Ratio=21.6354.32

D /C Ratio=0.3982

D /C Ratio isless than 1, DesignisOK

10−Block shear rupture strengthof beamweb ,Reference (J 4−5)Leh=a−g Leh=2.46−0 Leh=2.46∈¿

Lev=d−l2

Lev=12.2−8.86

2Lev=1.67∈¿

Ant=[Leh−12(dh+

116

)] t Ant=[2.46−12(1.31+ 1

16)]0.23 Ant=0.40782¿

2

Anv=[{(n−1)s+Lev}−{2n−12

(dh+116

)}] t

Anv=[[(3−1)2.95+1.67 ]−[ 2∗3−12

(1.31+ 116

)]]0.23 Anv=0.95194 ¿2

Agv=[(n−1)s+Lev ]t Agv=[(3−1)2.95+1.67 ]0.23 Agv=1.74¿2

ϕ Rn=ϕ [Fu Ant+min(0.6F y A gv ,0.6Fu Anv)]ϕ Rn=0.75 [65∗0.40782+min(0.6∗50∗1.74,0.6∗65∗0.95194)]ϕ Rn=47.73kips

D /C Ratio=Ruϕ Rn

D /C Ratio=21.6347.73

D /C Ratio=0.45326

D /C Ratio isless than 1, DesignisOK

11−Panel zone shear strength, Reference (AISC13−Section2.2.1)

Ru=M u

d−t fRu=

1009.1812.2−0.38

Ru=85.38kips

P y=F y A P y=50∗301.18 P y=15059.06 kipsPu≤0.4 Pyϕ Rn=ϕ0.6 F y d tw ϕ Rn=0.9∗0.6∗50∗12.2∗0.23 ϕ Rn=75.76 kips

D /C Ratio=Ruϕ Rn

D /C Ratio=85.3875.76

D /C Ratio=1.13

Stiffnersare required ¿ resist the panel zoneweb shear

12−Local flange bending, Reference (AISC13−Section2.2 .2)ϕ Rn=ϕ6.25 t f

2F yC tϕ Rn=0.9∗6.25∗0.38

2∗50∗1 ϕ Rn=40.61kips

D /C Ratio=Ruϕ Rn

D /C Ratio=85.3840.61

D /C Ratio=2.1

Stiffnersare required∈the flangeof column¿ resist tensile flange force

13−Localweb yielding , Reference(AISC13−Section2.2.3)N=t f

Ru=M u

d−t fRu=

1009.1812.2−0.38

Ru=85.38kips

ϕ Rn=ϕ(5K+N )F y twϕ Rn=1(5∗1+0.38)50∗0.23 ϕ Rn=61.87 kips

D /C Ratio=Ruϕ Rn

D /C Ratio=85.3861.87

D /C Ratio=1.38

Stiffnersare required∈the coulumweb ¿ resist tensile flange force

14−Webcrippling ,Reference (AISC13−Section2.2 .3)N=t f

Ru=M u

d−t fRu=

1009.1812.2−0.38

Ru=85.38kips

ϕ Rn=ϕ135 tw2¿

ϕ Rn=0.75∗135∗0.232 ¿

ϕ Rn=51.43kips

D /C Ratio=Ruϕ Rn

D /C Ratio=85.3851.43

D /C Ratio=1.66

Stiffnersare required∈the coulumweb ¿ resist compressive flange force