Moment Splice- Backup

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

ΦFt

ΦFv

Ru

Mu

Pu

Lev

Leh

nr

gauge distance (row distance)

Φbr = 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 ΦMn = 318 kip-ftCheck ΦMn> 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

Φb

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 Φ = 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 Φ = 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

ΦFv

Puf

rnsrnb1rnb2rnbmaxrnbΦrn

nmin

ΦRn

Pufp

Design Check (ΦRn>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 Φ 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 ΦRn1 = 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 (Φ*0.6*Fu= 127.24 kips

d) Block shear strength-1 ΦRn1 = 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


Tension component (Φubs*Fu*Ant)Shear Yield component (Φ*0.6*Fy*Agv)


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 (Φ*0.6*Fu= 225.08 kips

d) Block shear strength-1 ΦRn1 = 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 Φ = 0.90Ru=Pufp Ru = 161 kips

a) KL/r 6.00KL/r < 25 TRUE

Gross plate area Agp = 5.25ΦPn=ΦFcr=ΦFy ΦPn = 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 (Φubs*Fu*Ant)Shear Yield component (Φ*0.6*Fy*Agv)


Design Check (ΦPn>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

𝑀_𝑛=(𝐴_𝑓𝑛/𝐴_𝑓𝑔 ∗𝐹_𝑢 𝑆_𝑥 𝑜𝑟 _ _𝐹 𝑢 𝑆 𝑥 ) 𝑑𝑒𝑝𝑒𝑛𝑑𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑐ℎ𝑒𝑐𝑘

𝐴_𝑓𝑛=𝐴_𝑓𝑔−𝑛_𝑟 (𝑑_ℎ+1/16 𝑖𝑛 ) 𝑡_𝑓

∅𝑅_𝑛=0.9∗𝐹_𝑦 𝐴_𝑔

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

𝑎) 𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝑅𝑢𝑝𝑡𝑢𝑟𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡=𝑈_𝑏𝑠 𝐹_𝑢 𝐴_𝑛𝑡𝑏) 𝑆ℎ𝑒𝑎𝑟 𝑌𝑖𝑒𝑙𝑑 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡=0.6𝐹_𝑦 𝐴_𝑔𝑣𝑐) 𝑆ℎ𝑒𝑎𝑟 𝑅𝑢𝑝𝑡𝑢𝑟𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡=0.6𝐹_𝑢 𝐴_𝑛𝑣

𝑎) 𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝑅𝑢𝑝𝑡𝑢𝑟𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡=𝑇𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 𝑋 𝑃𝑙𝑎𝑡𝑒 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑋 𝑏𝑜𝑙𝑡 𝑟𝑜𝑤 𝑛𝑢𝑚𝑏𝑒𝑟

𝑏𝑆ℎ𝑒𝑎𝑟 𝑌𝑖𝑒𝑙𝑑 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡=𝑇𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 𝑋 𝑃𝑙𝑎𝑡𝑒 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑋 𝑏𝑜𝑙𝑡 𝑟𝑜𝑤 𝑛𝑢𝑚𝑏𝑒𝑟


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

𝑟=√(𝐼/𝐴)

𝑀_𝑛=(𝐴_𝑓𝑛/𝐴_𝑓𝑔 ∗𝐹_𝑢 𝑆_𝑥 𝑜𝑟 _ _𝐹 𝑢 𝑆 𝑥 ) 𝑑𝑒𝑝𝑒𝑛𝑑𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑐ℎ𝑒𝑐𝑘

Alternatively, For Case 1 tension rupture component can be calculated as below by using AISC Manual, Table9-3a, Pg 1108


𝑏𝑆ℎ𝑒𝑎𝑟 𝑌𝑖𝑒𝑙𝑑 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡=𝑇𝑎𝑏𝑙𝑒 𝑣𝑎𝑙𝑢𝑒 𝑋 𝑃𝑙𝑎𝑡𝑒 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑋 𝑏𝑜𝑙𝑡 𝑟𝑜𝑤 𝑛𝑢𝑚𝑏𝑒𝑟


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





Tensile strength = 67.50 ksi

Shear Strength = 36.00 ksi

C) Forces

Ru = 60.00 Kips

Mu = 0.00

Pu = 0.00

d) Miscellaneous



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

ΦFt

ΦFv

Ru

Mu

Pu

Lev

Leh


Φbr = 0.75Φb 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


c) Strength of bolt group = 61.9 kips

Force Demand = 60.0 Kipsd) D/C ratio (ΦRn>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 (ΦMn>Mu) = 0.34 TRUE

C Flexural Rupture of Plate

Net Plastic section modulus of the plate Znet in3

Fu

ΦFv

Φrn

rns

rnb

rnb1

rnb2

rnbmax

ΦRn

Ru

Ru

Fy

Lp

tp

ZX in3

Φb

Mu

ΦMn

Plate, Fu Fu = 58.00 ksi

Net plastic section modulus = 9.00Reduction factor Φ = 0.75

a) Flexural rupture strength = 391.50 kip-inb) Required flexural strength Mu = 150.00 kip-inc) D/C ratio (ΦMn>Mu) = 0.38 TRUE

d) Shear Yielding of Plate

Plate thickness = 0.38 in

Plate length 12.00

Plate Fy = 36.00 ksireduction factor Φ 1

Gross shear area = 4.50 in2

a) Shear Yield strength = 97.20 Kipsb) Required strength Ru = 60.00 Kipsc) D/C ratio (ΦRn>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 Φ = 0.75

Effective length Le = 8.00


a) Shear rupture strength = 78.30 Kipsb) Required strength Ru = 60.00 Kipsc) D/C ratio (ΦRn>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

ΦMn

tp

Lp

Fy

Agv

ΦRn

tp

Lp

Fu

Anv

ΦRn

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 ΦRn = 79.93 kipse) Demand force Ru = 60 Kips

= 0.75 TRUEReferences:


2) LRFD Vol-I



Design Check (ΦRn>Ru)


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

𝑀_𝑢=(𝑅_𝑢 𝑔)/2

∅𝑅_𝑛=𝐶∗∅𝑟_𝑛

∅𝑟_𝑛𝑠=𝐴∗∅𝐹_𝑣

∅𝑀_𝑛=0.9∗𝐹_𝑦 𝑍_𝑥𝑍_𝑥=(𝑡_𝑝 〖𝐿 _𝑝〗 ^2)/4

PASSED

PASSED

PASSED

𝑆ℎ𝑒𝑎𝑟 𝑌𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ (∅𝑅_𝑛)=∅∗0.6𝐹_𝑦 𝐴_𝑔𝑣

𝑆ℎ𝑒𝑎𝑟 𝑅𝑢𝑝𝑡𝑢𝑟𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡( _ )=∅𝑅 𝑛 ∅∗0.6𝐹_𝑢 𝐴_𝑛𝑣

∅𝑀_𝑛=0.75∗𝐹_𝑢 𝑍_𝑛𝑒𝑡

𝐴_𝑔𝑣=𝑡_𝑝 𝐿_𝑝

𝐴_𝑛𝑣=𝑡_𝑝 𝐿_𝑒𝐿_𝑒=𝐿_𝑝−𝑁(𝑑_ℎ+1/16)𝑑_ℎ=𝑑_𝑏+1/16

Check shear rupture May be wrong in calculating Anv

𝑎) 𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝑅𝑢𝑝𝑡𝑢𝑟𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡=∅𝑈_𝑏𝑠 𝐹_𝑢 𝐴_𝑛𝑡𝑏) 𝑆ℎ𝑒𝑎𝑟 𝑌𝑖𝑒𝑙𝑑 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡=∅∗0.6𝐹_𝑦 𝐴_𝑔𝑣𝑐) 𝑆ℎ𝑒𝑎𝑟 𝑅𝑢𝑝𝑡𝑢𝑟𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡=∅∗0.6𝐹_𝑢 𝐴_𝑛𝑣

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

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





Tensile strength = 67.50 ksi

Shear Strength = 36.00 ksi

C) Forces

Pu = 36.00 Kips

Mu = 0.00

Vu = 0.00

d) Miscellaneous



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

ΦFt

ΦFv

Ru

Mu

Pu

Lev

Leh


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



c) Strength of bolt group = 66.6 kips

Force Demand = 36.0 Kipsd) D/C ratio (ΦRn>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

ΦFv

Φrn

rns

rnb

rnb1

rnb2

rnbmax

ΦRn

Ru

Ru

Fy

Lp

tp

ZX in3

Φb

Mu

ΦMn

c) D/C ratio (ΦMn>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 Φ = 0.75

a) Flexural rupture strength = 935.25 kip-inb) Required flexural strength Mu = 333.00 kip-inc) D/C ratio (ΦMn>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 Φ 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 (ΦRn>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 Φ 1

b) Required strength Ru = 36.00 Kips

a) Shear Yield strength = 162.00 Kipsc) D/C ratio (ΦRn>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 Φ = 0.75

Znet in3

ΦMn

tp

Lp

Fy

tanΘ

Vn

ΦVn

tp

Lp

Fy

ΦRn

tp

Lp

Fu

Effective length Le = 10.20


a) Shear rupture strength = 99.82 Kipsb) Required strength Ru = 36.00 Kipsc) D/C ratio (ΦRn>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 ΦRn = 135.88 kipse) Demand force Ru = 36 Kips

= 0.26 TRUEReferences:


2) LRFD Vol-I



Anv

ΦRn

Design Check (ΦRn>Ru)


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

𝑀_𝑢=(𝑅_𝑢 𝑔)/2

∅𝑅_𝑛=𝐶∗∅𝑟_𝑛

∅𝑟_𝑛𝑠=𝐴∗∅𝐹_𝑣

∅𝑀_𝑛=0.9∗𝐹_𝑦 𝑍_𝑥

𝑍_𝑥=(𝑡_𝑝 〖𝐿 _𝑝〗 ^2)/4

PASSED

PASSED

PASSED

PASSED

𝑆ℎ𝑒𝑎𝑟 𝑌𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ (∅𝑉_𝑛)=∅∗0.6𝐹_𝑦 𝑡𝑏^′

∅𝑀_𝑛=0.75∗𝐹_𝑢 𝑍_𝑛𝑒𝑡

𝐿_𝑒=𝐿_𝑝−𝑁(𝑑_ℎ+1/16)𝑑_ℎ=𝑑_𝑏+1/16

𝑆ℎ𝑒𝑎𝑟 𝑌𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ (∅𝑅_𝑛)=∅∗0.6𝐹_𝑦 𝐴_𝑔𝑣𝐴_𝑔𝑣=𝑡_𝑝 𝐿_𝑝

PASSED


𝑆ℎ𝑒𝑎𝑟 𝑅𝑢𝑝𝑡𝑢𝑟𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡(∅𝑅_𝑛)=∅∗0.6𝐹_𝑢 𝐴_𝑛𝑣𝐴_𝑛𝑣=𝑡_𝑝 𝐿_𝑒

𝑎) 𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝑅𝑢𝑝𝑡𝑢𝑟𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡=∅𝑈_𝑏𝑠 𝐹_𝑢 𝐴_𝑛𝑡𝑏) 𝑆ℎ𝑒𝑎𝑟 𝑌𝑖𝑒𝑙𝑑 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡=∅∗0.6𝐹_𝑦 𝐴_𝑔𝑣𝑐) 𝑆ℎ𝑒𝑎𝑟 𝑅𝑢𝑝𝑡𝑢𝑟𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑒𝑡=∅∗0.6𝐹_𝑢 𝐴_𝑛𝑣

Moment Splice- Backup

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