Design of Moment SpliceRef: AISC Design Guide, Example-II B-1, pg-556; LRFD, Pg, 1789
1) Inputs
A) Beam Column and Plate properties
Beam (same section connected) Value UnitSection = W18x50 0.00Depth (d) = 17.99 in
Width = 7.50 in
Web thickness = 0.36 in
Flange thickness = 0.57 in
Yield strength = 50.00 ksi
Ultimate strength = 65.00 ksi
Section PropertiesGross Area A = 14.70
Moment of Inertia = 800.00
Elastic Section Modulus = 88.90
Plastic Modulus = 101.00
Plate
length = 7.00 in
thickness = 0.75 in
Fy = 36.00 ksi
Fu = 58.00 ksiB) Bolt and Weld
Bolt Type = ASTM-A328-N
Bolt dia = 0.88 in Area A = 0.60
Electrode Strength = 70.00 ksi
Tensile strength = 67.50 ksi Ref: Table 8-15
Shear Strength = 36.00 ksi Ref: Table 8-11
C) Forces
Ru = 42.00 Kips
Mu = 252.00 Kip-ft
Pu = 0.00 0.00
d) Miscellaneous
Bolt Spacing s = 3.00 in
Vertical Edge distance = 1.50 in
Horizontal edge distance = 1.50 inΦb 0.90number of bolts in a row n = 4.00 nos
Number of rows = 2.00 nosTotal number of bolt N = 8.00 nos
g = 4.00 in
bf
tw
tf
Fy
Fu
in2
IXX in4
SX in3
ZXX in3
Lp
tp
Fy
Fu
db
in2
FEXX
Ξ¦Ft
Ξ¦Fv
Ru
Mu
Pu
Lev
Leh
nr
gauge distance (row distance)
Ξ¦br = 0.75
2) Design Checks
A) Flexural Strength check of Beam (AISC Specifications F13.1, Pg 120) PASSED
Inputs Value Unit
Beam Fu 65.00 ksi
Beam Fy 50.00 ksi
Sx 88.90 in3
7.50 in
tf 0.57 in
Hole diameter = 0.94 inch
Flexure reduction factor = 0.90
Moment demand = 252.00 Kip-ft
assuming two rows of 0.875 inch diameter bolts in standard holes
a) Gross flange area = 4.27
b) Net Flange area = 3.13
Fy/Fu 0.77Yt = 1
= 204 kips
= 214 kips
TRUE
c) Nominal flexural strength Mn = 4237 Kip-in= 353 kip-ft
Design Flexural Strength Ξ¦Mn = 318 kip-ftCheck Ξ¦Mn> Mu 1
B) Tension Flange plate Connection Check
Bolt dia db = 0.88 in Area A = 0.60 in2Hole dia dh = 0.94 inbolt spacing s = 3.00 invertical edge distance Lev = 1.50 inHorizontal edge distance Leh = 1.50 inClear edge length lc1 = 1.03 inClear Spacing lc2 = 2.06 in
Fu
Fy
SX
bf bf
tf
dh
Ξ¦b
Mu
Afg in2
Afn in2
Note: To calculate net flange area for tension and shear, bolt dia shall be taken as 1/16 in greater than the
nominal hole dia. 3- Pg-74
Note: Strength should be reduced for members with holes in tension flange.
The Reduction is done as per F13.1 3
FuAfn
YtFyAfg
Plate thickness tp = 0.75 inBeam flance thickness tf = 0.57 in Beam Fu 65.00 ksiplate Fu 58.00 ksiBolt shear strength = 36.00 ksiΦbearing = 0.75
B.1 Minimum number of bolts required PASSED
a) Calculate the flange force = 168.09 kipsb) Critical Bolt strength
i) shear strength of one bolt = 21.65 kips/bolt
ii) Bolt bearing = 26.05 kips/bolt
iii) Bolt bearing 68.77 kips/bolt
iii) Bolt bearing 58.35 kips/bolt
Bolt bearing strength = 26.05 kips/boltCritical bolt strength 21.65 kips/bolt
c) Minimum number of bolts = 7.77 nos
Therefore, required minumum bolt number 8
Provided bolt numbers= 8
B.2 Flange Plate Tensile Yielding PASSED
Plate, Fy Fy = 36.00 ksiPlate length Lp = 7.00 inPlate thickness tp = 0.75 inGross area of plate Ag = 5.25 in2Reduction factor Ξ¦ = 0.90Moment demand Mu = 252.00 Kip-ftBeam depth d = 17.99 in
a) Design tensile strength = 170.10 kips
b) Tensile force in plate = 161.37 kips= TRUE
B.3 Flange Plate Tensile Rupture PASSED
hole diameter dh = 0.94 inPlate, Fu Fu = 58.00 ksiReduction factor Ξ¦ = 0.75
a) Effective net area AeNet plate area An = 3.75 in2Gross plate area Ag = 5.25 in2Effective net area Ae = 3.75
b) Design tensile rupture strenΦRn = 163.13 Kips= TRUE
Ξ¦Fv
Puf
rnsrnb1rnb2rnbmaxrnbΦrn
nmin
Ξ¦Rn
Pufp
Design Check (Ξ¦Rn>Pufp)
Design Check (Ξ¦Rn>Pufp)
B.4 Flange Plate Block Shear Rupture PASSED
Plate thickness tp = 0.75 inbolt spacing s = 3.00 inLeh Leh = 1.50 inLev Lev = 1.50 inPlate Fu Fu = 58.00 ksiPlate Fy Fy = 36.00 ksiUbs = 1reduction factor Ξ¦ 0.75number of bolts in a row n = 4.00 nosnumber of bolt column nr 2.00Ru=Pufp Ru = 161 kips
g = 4 inCase-1
Net tension area = 0.75 in2
Net shear area = 4.88 in2
Gross shear area = 7.88 in2a) Tension component = 65.25 Kipsb) Shear Yield component = 255.15 kipsc) Shear rupture component = 254.47 kips
d) Block shear strength-1 Ξ¦Rn1 = 319.72 kips= TRUE
Case-3Net tension area Ant = 3.0Net shear area Anv = 4.88Gross shear area Agv = 7.88
a) = 130.50 Kipsb) = 127.57 kipsc) Shear rupture component (Ξ¦*0.6*Fu= 127.24 kips
d) Block shear strength-1 Ξ¦Rn1 = 257.74 kips
= TRUE PASSED
C) Beam Flange Block Shear Rupture Check PASSED
In the beam block shear rupture, the case-1 as illustrated above governshole diameter dh = 0.94 inFlange Thickness tf = 0.57 in bolt spacing s = 3.00 inLeh Leh = 1.50 inLev Lev = 1.50 inBeam Fu Fu = 65.00 ksiBeam Fy Fy = 50.00 ksi
gauge (distance between two rows)
Ant
Anv
Agv
Design Check (Ξ¦Rn>Pufp)
Tension component (Ξ¦ubs*Fu*Ant)Shear Yield component (Ξ¦*0.6*Fy*Agv)
Design Check (Ξ¦Rn>Pufp)
Ubs = 1reduction factor( shear, rupture) 0.75number of bolts in a row nr = 4.00 nosRu=Pufp Ru = 168 kipsCase-1Net tension area Ant = 0.57 in2Net shear area Anv = 3.85 in2Gross shear area Ang = 5.72 in2
a) = 55.57 Kipsb) = 257.30 kipsc) Shear rupture component (Ξ¦*0.6*Fu= 225.08 kips
d) Block shear strength-1 Ξ¦Rn1 = 280.65 kips= TRUE
D) Compression Flange Plate connection Check PASSED
Leh Leh = 1.50 inCompression length L = 2 in L=Leh+setback
Effective Length factor K = 0.65 Ref: AISC Specification, Table C-A-7.1
Plate length Lp = 7.00 inPlate thickness tp = 0.75 inPlate moment of inertia I = 0.25 in4Plate section area Ap = 5.25 in2radius of Gyration r = 0.22 inPlate Fy Fy = 36.00 ksireduction factor Ξ¦ = 0.90Ru=Pufp Ru = 161 kips
a) KL/r 6.00KL/r < 25 TRUE
Gross plate area Agp = 5.25Ξ¦Pn=Ξ¦Fcr=Ξ¦Fy Ξ¦Pn = 170.1 Kips Spec. Sec. J4.4
= TRUE
References:
1) AISC Design Guide-14
2) LRFD Vol-I
3) AISC Specifications 360-10, 14th Edition
4) AISC Steel Construction Manual-13th Edition
Tension component (Ξ¦ubs*Fu*Ant)Shear Yield component (Ξ¦*0.6*Fy*Agv)
Design Check (Ξ¦Rn>Pufp)
Design Check (Ξ¦Pn>Pufp)
Ref: AISC Design Guide, Example-II B-1, pg-556; LRFD, Pg, 1789
To calculate net flange area for tension and shear, bolt dia shall be taken as 1/16 in greater than the
π_π=(π΄_ππ/π΄_ππ βπΉ_π’ π_π₯ ππ _ _πΉ π’ π π₯ ) πππππππππ ππ π‘βπ ππππ£π πβπππ
π΄_ππ=π΄_ππβπ_π (π_β+1/16 ππ ) π‘_π
β π _π=0.9βπΉ_π¦ π΄_π
Alternatively, For Case 1 tension rupture component can be calculated as below by using AISC Manual, Table9-3a, Pg 1108
a) Tension rupture component, 9-3a, Pg 1108From Table,= 43.5Tension rupture= 65.25 Kips
b) Shear Yield component, Table 9-3b, 1109From Table,= 170Shear Yield 255 Kips
c) Shear rupture component, 9-3c, Pg 1111From Table,= 183Shear Rupture 274.5 Kips
π) ππππ πππ π π’ππ‘π’ππ ππππππππππ‘=π_ππ πΉ_π’ π΄_ππ‘π) πβπππ πππππ ππππππππππ‘=0.6πΉ_π¦ π΄_ππ£π) πβπππ π π’ππ‘π’ππ ππππππππππ‘=0.6πΉ_π’ π΄_ππ£
π) ππππ πππ π π’ππ‘π’ππ ππππππππππ‘=πππππ π£πππ’π π ππππ‘π πβππππππ π π ππππ‘ πππ€ ππ’ππππ
ππβπππ πππππ ππππππππππ‘=πππππ π£πππ’π π ππππ‘π πβππππππ π π ππππ‘ πππ€ ππ’ππππ
π) ππππ πππ π π’ππ‘π’ππ ππππππππππ‘=πππππ π£πππ’π π ππππ‘π πβππππππ π π ππππ‘ πππ€ ππ’ππππ
Table value can be used for shear component in all cases but may not be possible to use in all casesto calculate tension rupture component
Ref: AISC Specification, Table C-A-7.1
Spec. Sec. J4.4
π=β(πΌ/π΄)
π_π=(π΄_ππ/π΄_ππ βπΉ_π’ π_π₯ ππ _ _πΉ π’ π π₯ ) πππππππππ ππ π‘βπ ππππ£π πβπππ
Alternatively, For Case 1 tension rupture component can be calculated as below by using AISC Manual, Table9-3a, Pg 1108
π) ππππ πππ π π’ππ‘π’ππ ππππππππππ‘=πππππ π£πππ’π π ππππ‘π πβππππππ π π ππππ‘ πππ€ ππ’ππππ
ππβπππ πππππ ππππππππππ‘=πππππ π£πππ’π π ππππ‘π πβππππππ π π ππππ‘ πππ€ ππ’ππππ
π) ππππ πππ π π’ππ‘π’ππ ππππππππππ‘=πππππ π£πππ’π π ππππ‘π πβππππππ π π ππππ‘ πππ€ ππ’ππππ
Table value can be used for shear component in all cases but may not be possible to use in all cases
Design of Shear SpliceRef: AISC Design Guide, Example-II B-1, pg-556; LRFD, Pg, 1789
1) Inputs
A) Beam Column and Plate properties Beam-1 Beam-2Beam 1 Value UnitSection = W24x55 W24x68Depth (d) = in
Width = in
Web thickness = 0.359 0.415 in
Flange thickness = in
Yield strength = 50.00 50.00 ksi
Ultimate strength = 65.00 65.00 ksi
Section PropertiesGross Area A =
Moment of Inertia =
Elastic Section Modulus =
Plastic Modulus =
Plate
Vertical length = 12.00 in
thickness = 0.38 in
Fy = 36.00 ksi
Fu = 58.00 ksiB) Bolt and Weld
Bolt Type = ASTM-A325-N
Bolt dia = 0.875 in Area A = 0.60
Electrode Strength = 70.00 ksi
Tensile strength = 67.50 ksi
Shear Strength = 36.00 ksi
C) Forces
Ru = 60.00 Kips
Mu = 0.00
Pu = 0.00
d) Miscellaneous
Bolt Spacing s = 3.00 in
Vertical Edge distance = 1.50 in
Horizontal edge distance = 1.50 inNumber of bolt column nc = 1.00 nosnumber of bolt row nr 4.00 nos
Total number of bolt N = 4.00 nosg = 5.00 in
Eccentricity e = 2.5 in
bf
tw
tf
Fy
Fu
in2
IXX in4
SX in3
ZXX in3
Lp
tp
Fy
Fu
db
in2
FEXX
Ξ¦Ft
Ξ¦Fv
Ru
Mu
Pu
Lev
Leh
gauge distance (row distance)
Ξ¦br = 0.75Ξ¦b 0.90C 3.07
9.00
2) Design Checks
A) Bolt group Capacity
Bolt dia db = 0.88 in Area A = 0.60 in2Hole dia dh = 0.94 inbolt spacing s = 3.00 invertical edge distance Lev = 1.50 inHorizontal edge distance Leh = 1.50 inClear edge length lc1 = 1.03 inClear Spacing lc2 = 2.06 inPlate thickness tp = 0.38 in
plate Fu 58.00 ksi
Bolt shear strength = 36.00 ksiΦbearing = 0.75Coeff. Of eccentricity C = 3.07
a) Critical Bolt strength = 20.19 kips/bolt
i) Shear strength of one bolt = 21.65 kips/bolt
ii) Bearing strength of one bolt = 20.19 kips/bolt
Bolt bearing = 20.2 kips/bolt taken conservatively
Bolt bearing 40.37 kips/bolt
Bolt bearing 34.26 kips/bolt
c) Strength of bolt group = 61.9 kips
Force Demand = 60.0 Kipsd) D/C ratio (Ξ¦Rn>Ru) = 0.97 TRUE
B Flexural Yielding of Plate
Shear Demand = 60.00 Kips
Plate, Fy = 36.00 ksi
Plate vertical length = 12.00 in
Plate thickness = 0.38 ingauge distance g = 5.00 in
Plastic Modulus of plate = 13.50Reduction factor = 0.90
a) Required flexural strength = 150.00 kip-in
b) Flexural yield strength = 437.40 kip-inc) D/C ratio (Ξ¦Mn>Mu) = 0.34 TRUE
C Flexural Rupture of Plate
Net Plastic section modulus of the plate Znet in3
Fu
Ξ¦Fv
Ξ¦rn
rns
rnb
rnb1
rnb2
rnbmax
Ξ¦Rn
Ru
Ru
Fy
Lp
tp
ZX in3
Ξ¦b
Mu
Ξ¦Mn
Plate, Fu Fu = 58.00 ksi
Net plastic section modulus = 9.00Reduction factor Ξ¦ = 0.75
a) Flexural rupture strength = 391.50 kip-inb) Required flexural strength Mu = 150.00 kip-inc) D/C ratio (Ξ¦Mn>Mu) = 0.38 TRUE
d) Shear Yielding of Plate
Plate thickness = 0.38 in
Plate length 12.00
Plate Fy = 36.00 ksireduction factor Ξ¦ 1
Gross shear area = 4.50 in2
a) Shear Yield strength = 97.20 Kipsb) Required strength Ru = 60.00 Kipsc) D/C ratio (Ξ¦Rn>Ru) = 0.62 TRUE
e) Shear Rupture of Plate
Plate thickness = 0.38 in3
Plate length = 12.00 in Bolt diameter db = 0.88 inHole diameter dh = 0.9375 innumber of bolts N = 4.00 nos
Plate Fu = 58.00 ksireduction factor Ξ¦ = 0.75
Effective length Le = 8.00
Gross shear area = 3.00 in2
a) Shear rupture strength = 78.30 Kipsb) Required strength Ru = 60.00 Kipsc) D/C ratio (Ξ¦Rn>Ru) = 0.77 TRUE
C) Beam Flange Block Shear Rupture Check
Bolt diameter db = 0.875 inHole diameter dh = 0.94 inPlate thickness tp = 0.38 inPlate length Lp = 12.00 inbolt spacing s = 3.00 inLeh Leh = 1.50 inLev Lev = 1.50 inPlate Fy Fy = 36.00 ksiPlate Fu Fu = 58.00 ksiUbs = 1reduction factor( shear, rupture) 0.75
Znet in3
Ξ¦Mn
tp
Lp
Fy
Agv
Ξ¦Rn
tp
Lp
Fu
Anv
Ξ¦Rn
number of rows nr = 4.00 nosdg+1/16 1.00 inLnt = 1.00 inLgv = 10.50 inLnv = 6.50 inUBS = 1.00
Net tension area Ant = 0.38 in2Net shear area Anv = 2.44 in2Gross shear area Agv = 3.94 in2
a) Tension rupture component = 16.31 Kipsb) Shear Yield component = 63.79 kipsc) Shear rupture component = 63.62 kips
d) Block Shear strength Ξ¦Rn = 79.93 kipse) Demand force Ru = 60 Kips
= 0.75 TRUEReferences:
1) AISC Design Guide-14
2) LRFD Vol-I
3) AISC Specifications 360-10, 14th Edition
4) AISC Steel Construction Manual-13th Edition
Design Check (Ξ¦Rn>Ru)
Ref: AISC Design Guide, Example-II B-1, pg-556; LRFD, Pg, 1789
Ref: Table 8-15
Ref: Table 8-11
Calculation of Coefficient of Eccentricity
Eccentricity e1 e2 ex
2 3 2.5
C 3.32 2.81 3.07
PASSED
taken conservatively
PASSED
PASSED
AISC Manual-Table15-2, Pg 1411
π_π’=(π _π’ π)/2
β π _π=πΆββ π_π
β π_ππ =π΄ββ πΉ_π£
β π_π=0.9βπΉ_π¦ π_π₯π_π₯=(π‘_π γπΏ _πγ ^2)/4
PASSED
PASSED
PASSED
πβπππ πππππ π π‘πππππ‘β (β π _π)=β β0.6πΉ_π¦ π΄_ππ£
πβπππ π π’ππ‘π’ππ ππππππππππ‘( _ )=β π π β β0.6πΉ_π’ π΄_ππ£
β π_π=0.75βπΉ_π’ π_πππ‘
π΄_ππ£=π‘_π πΏ_π
π΄_ππ£=π‘_π πΏ_ππΏ_π=πΏ_πβπ(π_β+1/16)π_β=π_π+1/16
Check shear rupture May be wrong in calculating Anv
π) ππππ πππ π π’ππ‘π’ππ ππππππππππ‘=β π_ππ πΉ_π’ π΄_ππ‘π) πβπππ πππππ ππππππππππ‘=β β0.6πΉ_π¦ π΄_ππ£π) πβπππ π π’ππ‘π’ππ ππππππππππ‘=β β0.6πΉ_π’ π΄_ππ£
Check shear rupture May be wrong in calculating Anv
SN1234567
Reference forCriteria for reduction in effective area reduction due to bolt holes in plates, web or flangesDesign Tensile strength of one boltDesign Shear Strength of one BoltBearing strength calculation flow chartsanchor rod hole diameterNominal hole dimensionsCoefficient C for eccentrically loaded bolt group
Page
12571254
from exceel sheet for Connection design16.1-415, AISC 146-82, LRFD 99AISC Manual Table 7-7
1
2
3
Design of Shear SpliceRef: AISC Design Guide, Example-II B-1, pg-556; LRFD, Pg, 1789
1) Inputs
A) Beam Column and Plate properties Beam-1 Beam-2Beam 1 Value UnitSection = W24x55 W24x68Depth (d) = in
Width = in
Web thickness = 0.359 0.415 in
Flange thickness = in
Yield strength = 50.00 50.00 ksi
Ultimate strength = 65.00 65.00 ksi
Section PropertiesGross Area A =
Moment of Inertia =
Elastic Section Modulus =
Plastic Modulus =
Plate
Vertical length a = 20.00 inHorizontal length b 15.25 in
thickness = 0.38 in
Fy = 36.00 ksi
Fu = 58.00 ksiB) Bolt and Weld
Bolt Type = ASTM-A325-N
Bolt dia = 0.750 in Area A = 0.44
Electrode Strength = 70.00 ksi
Tensile strength = 67.50 ksi
Shear Strength = 36.00 ksi
C) Forces
Pu = 36.00 Kips
Mu = 0.00
Vu = 0.00
d) Miscellaneous
Bolt Spacing s = 3.00 in
Vertical Edge distance = 2.50 in
Horizontal edge distance = 2.25 inNumber of bolt column nc = 2.00 nosnumber of bolt row nr 6.00 nosTotal number of bolt N = 12.00 nos
g = 5.50 in
bf
tw
tf
Fy
Fu
in2
IXX in4
SX in3
ZXX in3
tp
Fy
Fu
db
in2
FEXX
Ξ¦Ft
Ξ¦Fv
Ru
Mu
Pu
Lev
Leh
gauge distance (row distance)
Load point from bolt edge = 9.25 inEccentricity e = 12 inΦbr = 0.75Φb 0.90C 4.19
21.50
2) Design Checks
A) Bolt group Capacity
Bolt dia db = 0.75 in Area A = 0.44 in2Hole dia dh = 0.81 inbolt spacing s = 3.00 invertical edge distance Lev = 2.50 inHorizontal edge distance Leh = 2.25 inClear edge length lc1 = 2.09 inClear Spacing lc2 = 2.19 inPlate thickness tp = 0.38 in
plate Fu 58.00 ksi
Bolt shear strength = 36.00 ksiΦbearing = 0.75Coeff. Of eccentricity C = 4.19
a) Critical Bolt strength = 15.90 kips/bolt
i) Shear strength of one bolt = 15.90 kips/bolt
ii) Bearing strength of one bolt = 29.36 kips/bolt
Bolt bearing = 41.0 kips/bolt taken conservatively
Bolt bearing 42.82 kips/bolt
Bolt bearing 29.36 kips/bolt
c) Strength of bolt group = 66.6 kips
Force Demand = 36.0 Kipsd) D/C ratio (Ξ¦Rn>Ru) = 0.54 TRUE
B Flexural Yielding of Plate on line K
Shear Demand = 36.00 Kips
Plate, Fy = 36.00 ksi
Plate vertical length = 20.00 in
Plate thickness = 0.38 inEccentricity e = 12.00 inLoad to bolt distance e1 = 9.25
Plastic Modulus of plate = 37.50Reduction factor = 0.90
a) Required flexural strength = 333.00 kip-in
b) Flexural yield strength = 1215.00 kip-in
Net Plastic section modulus of the plate Znet in3
Fu
Ξ¦Fv
Ξ¦rn
rns
rnb
rnb1
rnb2
rnbmax
Ξ¦Rn
Ru
Ru
Fy
Lp
tp
ZX in3
Ξ¦b
Mu
Ξ¦Mn
c) D/C ratio (Ξ¦Mn>Mu) = 0.27 TRUE
C Flexural Rupture of Plate on line K
Plate, Fu Fu = 58.00 ksi
Net plastic section modulus = 21.50Reduction factor Ξ¦ = 0.75
a) Flexural rupture strength = 935.25 kip-inb) Required flexural strength Mu = 333.00 kip-inc) D/C ratio (Ξ¦Mn>Mu) = 0.36 TRUE
d) Shear Yielding of Plate on line J
Plate thickness = 0.38 in
Plate length 20.00
Plate Fy = 36.00 ksiVertical length a = 20 inHorizontal length b = 15.25 in
= 0.96Theta = 37.30 degreeb' b' = 12.12 inPu Pu = 36.00 kipsreduction factor Ξ¦ 1
b) Required strength, Vr Vu = 21.82 Kips
a) Nominal Shear Yield strength = 98.17 Kips
Design Shear yield strength 98.17c) D/C ratio (Ξ¦Rn>Ru) = 0.22 TRUE
e) Shear Yielding of Bracket Plate on line K
Plate thickness = 0.38 1.00
Plate length 20.00
Plate Fy = 36.00 kip-inGross shear area Agv = 7.50 in2reduction factor Ξ¦ 1
b) Required strength Ru = 36.00 Kips
a) Shear Yield strength = 162.00 Kipsc) D/C ratio (Ξ¦Rn>Ru) = 0.22 TRUE
f) Shear Rupture ofbracket Plate on Line K
Plate thickness = 0.38 in3
Plate length = 20.00 in Bolt diameter db = 0.75 inHole diameter dh = 0.8125 innumber of bolts N = 12.00 nos
Plate Fu = 58.00 ksireduction factor Ξ¦ = 0.75
Znet in3
Ξ¦Mn
tp
Lp
Fy
tanΞ
Vn
Ξ¦Vn
tp
Lp
Fy
Ξ¦Rn
tp
Lp
Fu
Effective length Le = 10.20
Gross shear area = 3.82 in2
a) Shear rupture strength = 99.82 Kipsb) Required strength Ru = 36.00 Kipsc) D/C ratio (Ξ¦Rn>Ru) = 0.36 TRUE
C) Beam Flange Block Shear Rupture Check
Bolt diameter db = 0.750 inHole diameter dh = 0.81 inPlate thickness tp = 0.38 inPlate length Lp = 20.00 inbolt spacing s = 3.00 inLeh Leh = 2.25 inLev Lev = 2.50 inPlate Fy Fy = 36.00 ksiPlate Fu Fu = 58.00 ksiUbs = 1reduction factor( shear, rupture) 0.75number of rows nr = 6.00 nosdg+1/16 0.88 inLnt = 1.81 inLgv = 17.50 inLnv = 12.25 inUBS = 1.00
Net tension area Ant = 0.68 in2Net shear area Anv = 4.59 in2Gross shear area Agv = 6.56 in2
a) Tension rupture component = 29.57 Kipsb) Shear Yield component = 106.31 kipsc) Shear rupture component = 119.90 kips
d) Block Shear strength Ξ¦Rn = 135.88 kipse) Demand force Ru = 36 Kips
= 0.26 TRUEReferences:
1) AISC Design Guide-14
2) LRFD Vol-I
3) AISC Specifications 360-10, 14th Edition
4) AISC Steel Construction Manual-13th Edition
Anv
Ξ¦Rn
Design Check (Ξ¦Rn>Ru)
Ref: AISC Design Guide, Example-II B-1, pg-556; LRFD, Pg, 1789
Ref: Table 8-15
Ref: Table 8-11
Calculation of Coefficient of Eccentricity
Eccentricity e1 e2 ex12 3 12
C 3.32 2.81 2.81
PASSED
taken conservatively
PASSED
AISC Manual-Table15-2, Pg 1411
π_π’=(π _π’ π)/2
β π _π=πΆββ π_π
β π_ππ =π΄ββ πΉ_π£
β π_π=0.9βπΉ_π¦ π_π₯
π_π₯=(π‘_π γπΏ _πγ ^2)/4
PASSED
PASSED
PASSED
PASSED
πβπππ πππππ π π‘πππππ‘β (β π_π)=β β0.6πΉ_π¦ π‘π^β²
β π_π=0.75βπΉ_π’ π_πππ‘
πΏ_π=πΏ_πβπ(π_β+1/16)π_β=π_π+1/16
πβπππ πππππ π π‘πππππ‘β (β π _π)=β β0.6πΉ_π¦ π΄_ππ£π΄_ππ£=π‘_π πΏ_π
PASSED
Check shear rupture May be wrong in calculating Anv
πβπππ π π’ππ‘π’ππ ππππππππππ‘(β π _π)=β β0.6πΉ_π’ π΄_ππ£π΄_ππ£=π‘_π πΏ_π
π) ππππ πππ π π’ππ‘π’ππ ππππππππππ‘=β π_ππ πΉ_π’ π΄_ππ‘π) πβπππ πππππ ππππππππππ‘=β β0.6πΉ_π¦ π΄_ππ£π) πβπππ π π’ππ‘π’ππ ππππππππππ‘=β β0.6πΉ_π’ π΄_ππ£
Check shear rupture May be wrong in calculating Anv
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