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ETABS 2016 16.2.1
License #*13GKCULR7HUKDLF
Connection Design: C1-BP Units: kip-in Story: Story2 Design Code: AISC 360-10
Column Base Plate Connection
Summary of results Design Check Type
D/C Ratio
Result
1
Concrete bearing capacity
0.073
Passed
Reference
2
Minimum base plate thickness
1.849
Failed
3
Anchor rod strength
6.552
Failed
Design Guide 1, Secion 3.2.1
4
Weld strength
0.894
Passed
J2-2, J2-3
5
Concrete pullout strength
3.538
Failed
6
Concrete breakout strength
0.492
Passed
7
Side-face blowout strength
0.253
Passed
Material Properties Column HE300A
A572Gr50
Fy = 50 ksi
Fu = 65 ksi
Base Plate
A992Fy50
Fy = 50 ksi
Fu = 65 ksi
Geometric Properties Column HE300A
tw = 0.33465 in
d = 11.42 in
tf = 0.55118 in
bf = 11.81 in
Bolts, Plate & Weld Anchor rod Base Plate
Diameter = 0.98425 in Width = 20 in
Head/Nut type = Square Height = 20 in
Material = ASTM F1554 Grade36 Thickness = 0.98425 in
Pedestal Dimensions
Width = 23.94 in
Height = 23.94 in
Design Calculations
Modelo 4 Anclajes (19-05-18).EDB
Page 1 of 5
9/09/2018
ETABS 2016 16.2.1
License #*13GKCULR7HUKDLF
Design calculations of base plate for combined moment and compression Design Provision = LRFD Load Combination = CombU1 Pu = -25.57225 kips, Mu = 4795.8 kip-in
Base plate area N = Max[Nmin , (d + 2x3 in), [d + 2(Leh + c)]] N = Max[9.84, (11.42 + 2x3 in), [11.42 + 2(2.46 + 1.48)]]
N = 20 in
B = Max[Bmin , (bf + 2x3 in), {bf + 2(Lev + c)}] B = Max[9.84, (11.81 + 2x3 in), [11.81 + 2(2.46 + 1.48)]]
B = 20 in
S1 = N-2Leh
S1 = 20-2(2.46)
S1 = 15.08 in
S2 = B-2Lev
S2 = 20-2(2.46)
S2 = 15.08 in
Base plate dimension(B inch x N inch) = 20 x 20 A1 = B*N
A1 = 20*20
A1 = 400 in2
A2 = (B + 2a)(N + 2b)
A2 = (20 + 2*1.97)(20 + 2*1.97)
A2 = 572.98 in2
e and ecrit e=
Mu Pu
e=
4795.8 25.57
e = 187.54 in
A2 ' fp (max) = ϕ0.85fc Min(2, √ ) A1 572.98 fp (max) = 0.65*0.85*4Min(2, √ ) 400 qmax = fp (max)B ecrit =
N Pu 2 2qmax
fp (max) = 2.65 ksi qmax = 2.65*20 ecrit =
20 25.57 2 2*52.9
qmax = 52.9
kips in
ecrit = 9.76 in
e is greater than ecrit, design the base plate with large moment
Concrete bearing capacity N 2Pu (e + f) [f + ]2 ≥ 2 qmax [7.54 +
20 2 2*25.57(187.54 + 7.54) ] ≥ 2 52.9
Above inequality equation is satisfied f=
N -L 2 eh
f=
20 -2.46 2
f = 7.54 in
N N 2 Pu (e + f) Y = (f + )-√[f + ]2 2 2 qmax Y = (7.54 +
20 20 2*25.57(187.54 + 7.54) )-√[7.54 + ]2 2 2 52.9
Modelo 4 Anclajes (19-05-18).EDB
Y = 6.63 in
Page 2 of 5
9/09/2018
ETABS 2016 16.2.1
License #*13GKCULR7HUKDLF
Tu = qmax Y-Pu q=
Pu Y
D/C Ratio =
Tu = 52.9*6.63-25.57 q=
q qmax
25.57 6.63
D/C Ratio =
Tu = 325.13 kips q = 3.86
3.86 52.9
kips in
D/C Ratio = 0.07292
D/C Ratio is less than 1
Minimum base plate thickness At bearing interface N-0.95d 2
m=
B-0.8bf 2
n=
m= n=
20-0.95(11.42) 2
m = 4.58 in
20-0.8(11.81) 2
n = 5.28 in
fp = fp Y is greater than max(m, n) so fp tp1 (min) = 1.5max(m, n)√ Fy 2.65 tp1 (min) = 1.5max(4.58,5.28)√ 50
tp1 (min) = 1.82 in
At tension interface d tf x = f- + 2 2
x = 7.54-
11.42 0.55118 + 2 2
x = 2.11 in
Tu x tp2 (min) = 2.11√ BFy 325.13*2.11 tp2 (min) = 2.11√ 20*50 tp (min) = max(tp1 (min), tp2 (min)) D/C Ratio =
tp (min) t
tp2 (min) = 1.75 in tp (min) = max(1.82,1.75) D/C Ratio =
1.82 0.98425
tp (min) = 1.82 in D/C Ratio = 1.85
Increase base plate thickness
Anchor rod strength, Reference(Design Guide 1, Secion 3.2.1) Force per anchor rod Tu Pr = n ( ) 2 Ar =
πdr 2 4
Pr =
325 4 ( ) 2
Pr = 162.56
Ar =
3.14*0.984252 4
Ar = 0.76047 in2
kips bolt
kips bolt
Rn = 0.75Fu Ar
Rn = 0.75*58*0.76047
Rn = 33.08
ϕRn = ϕRn
ϕRn = 0.75*33.08
ϕRn = 24.81
D/C Ratio =
Pr ϕRn
Modelo 4 Anclajes (19-05-18).EDB
D/C Ratio =
162.56 24.81
Page 3 of 5
kips bolt
D/C Ratio = 6.55
9/09/2018
ETABS 2016 16.2.1
License #*13GKCULR7HUKDLF
D/C Ratio is greater than 1, select another anchor rod
Weld strength, Reference(J2-2, J2-3) beff = S2-tw Ruw =
beff = 15.08-0.33465
Pr beff
Ruw =
ϕRn = ϕ1.5*0.6FEXX *1*
Ruw = 11.03
kips in
ϕRn = 12.33
kips in
0.707D 16
ϕRn = 0.751.5*0.6*70*1* D/C Ratio =
162.56 14.74
beff = 14.74 in
0.707*5.91 16
Ruw ϕRn
D/C Ratio =
11.03 12.33
D/C Ratio = 0.89423
D/C Ratio is less than 1, fillet on each side of the web is ok
Concrete pullout strength ϕNp = ϕψ4 Abrg 8fc
'
ϕNp = 0.7*1.4*1.47*8*4 D/C Ratio =
Pr ϕNp
ϕNp = 45.94 kips D/C Ratio =
162.56 45.94
D/C Ratio = 3.54
Concrete pullout strength is not sufficient
Concrete breakout strength ca1 = Leh + b
ca1 = 2.46 + 1.97
ca1 = 4.43 in
ca2 = Lev + a
ca2 = 2.46 + 1.97
ca2 = 4.43 in
ANo = 9hef 2
ANo = 9*19.692
ANo = 3487.51 in2
AN = (Min(2Ca1 , 3hef ) + Min(S1 , 3hef ))*(Min(2Ca2 , 3hef ) + Min(S2 , 3hef )) AN = (Min(2*4.43, 3*19.69) + Min(15.08, 3*19.69))*(Min(2*4.43, 3*19.69) + Min(15.08, 3*19.69)) AN = 572.98 in2 hef is greater than 11 in, soϕNcbg is 5 AN ' ϕNcbg = ϕψ3 16√fc [hef ]3 ( ) ANo 5 572.98 ϕNcbg = 0.7*1.25*16*√4*[19.69]3 ( ) 3487.51
ϕNcbg = 660.2 kips DCR =
Tu ϕNcbg
DCR =
325.13 660.2
DCR = 0.49247
D/C Ratio is less than 1, Design is OK
Side-face blowout strength ca1 = min((Leh + b), (Lev + a))
ca1 = min((2.46 + 1.97), (2.46 + 1.97))
ca1 = 4.43 in
ca2 = max((Leh + b), (Lev + a))
ca2 = max((2.46 + 1.97), (2.46 + 1.97))
ca2 = 4.43 in
Modelo 4 Anclajes (19-05-18).EDB
Page 4 of 5
9/09/2018
ETABS 2016 16.2.1
License #*13GKCULR7HUKDLF
hef > 2.5Ca1 Nsb = ϕ160ca1 λa √Abrg fc
'
Nsb = 1286.62 kips
Nsb = 0.75*160*4.43*1√1.47*4 Ca2 3Ca1 Nsb =
1 ca2 (1 + )N 4 ca1 sb
D/C Ratio =
Pr Nsb
1 4.43 Nsb = (1 + )1286.62 4 4.43 162.56 643.31
D/C Ratio =
Nsbg = 643.31 kips D/C Ratio = 0.2527
D/C Ratio is less than 1, side face blowout is not possible
Modelo 4 Anclajes (19-05-18).EDB
Page 5 of 5
9/09/2018
License #*13GKCULR7HUKDLF
Connection Design: C1-BP Units: kip-in Story: Story2 Design Code: AISC 360-10
Column Base Plate Connection
Summary of results Design Check Type
D/C Ratio
Result
1
Concrete bearing capacity
0.073
Passed
Reference
2
Minimum base plate thickness
1.849
Failed
3
Anchor rod strength
6.552
Failed
Design Guide 1, Secion 3.2.1
4
Weld strength
0.894
Passed
J2-2, J2-3
5
Concrete pullout strength
3.538
Failed
6
Concrete breakout strength
0.492
Passed
7
Side-face blowout strength
0.253
Passed
Material Properties Column HE300A
A572Gr50
Fy = 50 ksi
Fu = 65 ksi
Base Plate
A992Fy50
Fy = 50 ksi
Fu = 65 ksi
Geometric Properties Column HE300A
tw = 0.33465 in
d = 11.42 in
tf = 0.55118 in
bf = 11.81 in
Bolts, Plate & Weld Anchor rod Base Plate
Diameter = 0.98425 in Width = 20 in
Head/Nut type = Square Height = 20 in
Material = ASTM F1554 Grade36 Thickness = 0.98425 in
Pedestal Dimensions
Width = 23.94 in
Height = 23.94 in
Design Calculations
Modelo 4 Anclajes (19-05-18).EDB
Page 1 of 5
9/09/2018
ETABS 2016 16.2.1
License #*13GKCULR7HUKDLF
Design calculations of base plate for combined moment and compression Design Provision = LRFD Load Combination = CombU1 Pu = -25.57225 kips, Mu = 4795.8 kip-in
Base plate area N = Max[Nmin , (d + 2x3 in), [d + 2(Leh + c)]] N = Max[9.84, (11.42 + 2x3 in), [11.42 + 2(2.46 + 1.48)]]
N = 20 in
B = Max[Bmin , (bf + 2x3 in), {bf + 2(Lev + c)}] B = Max[9.84, (11.81 + 2x3 in), [11.81 + 2(2.46 + 1.48)]]
B = 20 in
S1 = N-2Leh
S1 = 20-2(2.46)
S1 = 15.08 in
S2 = B-2Lev
S2 = 20-2(2.46)
S2 = 15.08 in
Base plate dimension(B inch x N inch) = 20 x 20 A1 = B*N
A1 = 20*20
A1 = 400 in2
A2 = (B + 2a)(N + 2b)
A2 = (20 + 2*1.97)(20 + 2*1.97)
A2 = 572.98 in2
e and ecrit e=
Mu Pu
e=
4795.8 25.57
e = 187.54 in
A2 ' fp (max) = ϕ0.85fc Min(2, √ ) A1 572.98 fp (max) = 0.65*0.85*4Min(2, √ ) 400 qmax = fp (max)B ecrit =
N Pu 2 2qmax
fp (max) = 2.65 ksi qmax = 2.65*20 ecrit =
20 25.57 2 2*52.9
qmax = 52.9
kips in
ecrit = 9.76 in
e is greater than ecrit, design the base plate with large moment
Concrete bearing capacity N 2Pu (e + f) [f + ]2 ≥ 2 qmax [7.54 +
20 2 2*25.57(187.54 + 7.54) ] ≥ 2 52.9
Above inequality equation is satisfied f=
N -L 2 eh
f=
20 -2.46 2
f = 7.54 in
N N 2 Pu (e + f) Y = (f + )-√[f + ]2 2 2 qmax Y = (7.54 +
20 20 2*25.57(187.54 + 7.54) )-√[7.54 + ]2 2 2 52.9
Modelo 4 Anclajes (19-05-18).EDB
Y = 6.63 in
Page 2 of 5
9/09/2018
ETABS 2016 16.2.1
License #*13GKCULR7HUKDLF
Tu = qmax Y-Pu q=
Pu Y
D/C Ratio =
Tu = 52.9*6.63-25.57 q=
q qmax
25.57 6.63
D/C Ratio =
Tu = 325.13 kips q = 3.86
3.86 52.9
kips in
D/C Ratio = 0.07292
D/C Ratio is less than 1
Minimum base plate thickness At bearing interface N-0.95d 2
m=
B-0.8bf 2
n=
m= n=
20-0.95(11.42) 2
m = 4.58 in
20-0.8(11.81) 2
n = 5.28 in
fp = fp Y is greater than max(m, n) so fp tp1 (min) = 1.5max(m, n)√ Fy 2.65 tp1 (min) = 1.5max(4.58,5.28)√ 50
tp1 (min) = 1.82 in
At tension interface d tf x = f- + 2 2
x = 7.54-
11.42 0.55118 + 2 2
x = 2.11 in
Tu x tp2 (min) = 2.11√ BFy 325.13*2.11 tp2 (min) = 2.11√ 20*50 tp (min) = max(tp1 (min), tp2 (min)) D/C Ratio =
tp (min) t
tp2 (min) = 1.75 in tp (min) = max(1.82,1.75) D/C Ratio =
1.82 0.98425
tp (min) = 1.82 in D/C Ratio = 1.85
Increase base plate thickness
Anchor rod strength, Reference(Design Guide 1, Secion 3.2.1) Force per anchor rod Tu Pr = n ( ) 2 Ar =
πdr 2 4
Pr =
325 4 ( ) 2
Pr = 162.56
Ar =
3.14*0.984252 4
Ar = 0.76047 in2
kips bolt
kips bolt
Rn = 0.75Fu Ar
Rn = 0.75*58*0.76047
Rn = 33.08
ϕRn = ϕRn
ϕRn = 0.75*33.08
ϕRn = 24.81
D/C Ratio =
Pr ϕRn
Modelo 4 Anclajes (19-05-18).EDB
D/C Ratio =
162.56 24.81
Page 3 of 5
kips bolt
D/C Ratio = 6.55
9/09/2018
ETABS 2016 16.2.1
License #*13GKCULR7HUKDLF
D/C Ratio is greater than 1, select another anchor rod
Weld strength, Reference(J2-2, J2-3) beff = S2-tw Ruw =
beff = 15.08-0.33465
Pr beff
Ruw =
ϕRn = ϕ1.5*0.6FEXX *1*
Ruw = 11.03
kips in
ϕRn = 12.33
kips in
0.707D 16
ϕRn = 0.751.5*0.6*70*1* D/C Ratio =
162.56 14.74
beff = 14.74 in
0.707*5.91 16
Ruw ϕRn
D/C Ratio =
11.03 12.33
D/C Ratio = 0.89423
D/C Ratio is less than 1, fillet on each side of the web is ok
Concrete pullout strength ϕNp = ϕψ4 Abrg 8fc
'
ϕNp = 0.7*1.4*1.47*8*4 D/C Ratio =
Pr ϕNp
ϕNp = 45.94 kips D/C Ratio =
162.56 45.94
D/C Ratio = 3.54
Concrete pullout strength is not sufficient
Concrete breakout strength ca1 = Leh + b
ca1 = 2.46 + 1.97
ca1 = 4.43 in
ca2 = Lev + a
ca2 = 2.46 + 1.97
ca2 = 4.43 in
ANo = 9hef 2
ANo = 9*19.692
ANo = 3487.51 in2
AN = (Min(2Ca1 , 3hef ) + Min(S1 , 3hef ))*(Min(2Ca2 , 3hef ) + Min(S2 , 3hef )) AN = (Min(2*4.43, 3*19.69) + Min(15.08, 3*19.69))*(Min(2*4.43, 3*19.69) + Min(15.08, 3*19.69)) AN = 572.98 in2 hef is greater than 11 in, soϕNcbg is 5 AN ' ϕNcbg = ϕψ3 16√fc [hef ]3 ( ) ANo 5 572.98 ϕNcbg = 0.7*1.25*16*√4*[19.69]3 ( ) 3487.51
ϕNcbg = 660.2 kips DCR =
Tu ϕNcbg
DCR =
325.13 660.2
DCR = 0.49247
D/C Ratio is less than 1, Design is OK
Side-face blowout strength ca1 = min((Leh + b), (Lev + a))
ca1 = min((2.46 + 1.97), (2.46 + 1.97))
ca1 = 4.43 in
ca2 = max((Leh + b), (Lev + a))
ca2 = max((2.46 + 1.97), (2.46 + 1.97))
ca2 = 4.43 in
Modelo 4 Anclajes (19-05-18).EDB
Page 4 of 5
9/09/2018
ETABS 2016 16.2.1
License #*13GKCULR7HUKDLF
hef > 2.5Ca1 Nsb = ϕ160ca1 λa √Abrg fc
'
Nsb = 1286.62 kips
Nsb = 0.75*160*4.43*1√1.47*4 Ca2 3Ca1 Nsb =
1 ca2 (1 + )N 4 ca1 sb
D/C Ratio =
Pr Nsb
1 4.43 Nsb = (1 + )1286.62 4 4.43 162.56 643.31
D/C Ratio =
Nsbg = 643.31 kips D/C Ratio = 0.2527
D/C Ratio is less than 1, side face blowout is not possible
Modelo 4 Anclajes (19-05-18).EDB
Page 5 of 5
9/09/2018
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