E2_CONNECTION:-

Design_Method = "ASD 15th Edition"

LOADS :- Vertical shear
Incoming_Beam = "HSS12X10X3/8" Rvc := 15kip (EOR TO VERIFY)
(Connection capacity)
CONCRETE

EMBED PLATE
(BY OTHERS)
10" X 1/2" min
L5X5X3/8 EMBED PLATE
(LONG LEG OSL)

1/2" THK SPLICE PLATE

1
(TYP)

2"
2"
1

HSS12X10X3/8
1"
12

5"

1/4
3"

1" 1/4

7"
14"X1/2" THK CLOSURE PLATE
3"
1"
12

1"
32 1"
12

2" MIN
TYP
1/4 2" MIN
TYP

3/4" STUD A108

8" MIN EMBEDMENT
3 SIDES
WELD LENGTH 3 1/2"
1/4 @TOP AND BOTTOM

Notes:-

All bolts are 3/4" dia, A325N

All bolt holes are SSL hole in shear plate and STD hole in splice plate

Weld electrode is E70XX.

All W steel are grade A992, HSS are A500 grade B & connection plates , angles are grade A36, studs are A108.
Incoming Beam: Incoming_Beam  "HSS12X10X3/8"

Section Size:
2
db1 = 12  in bfb1 = 10  in t fb1 = 0.375  in t fb1des = 0.349  in Agb1 = 14.6  in

Beam material strength:- Fyb1 := 46  ksi Fub1 := 58  ksi

Thickness of angle t sp := 0.375in

Thickness of splice plate t sc := 0.5in

Thickness of closure plate t c := 0.5in

Connection material strength:- Fyp := 36  ksi Fup := 58  ksi

Weld strength:- FEXX := 70ksi

Factors:
Factors for tension yielding, bending & buckling f a = 0.6

Factors for bolts, welds, bolt bearing & web crippling f c = 0.5

Factors for shear yielding & web yielding f g = 0.67

Factors for tension / shear rupture f n = 0.5

Bolt properties :-

Bolt diameter dbolt := 0.75in

Bolt input value

Bolt strength:- Fnt = 90  ksi Fnv = 54  ksi

Shear_plate holes = "Short Slot: 0.8125"x1" (Transverse to Loading)"

Diameter of hole in shear plate Dhsp = 0.8125  in

Length of hole in shear plate Dh1sp = 1  in

Splice_plate holes = "STD Hole: 0.8125" dia."

Diameter of hole in splice plate Dhsc = 0.8125  in

Length of hole in splice plate Dh1sc = 0.8125  in

2
Area of bolt Abolt = 0.442  in

Shear capacity of bolt Scap = 11.93  kip

End clearance ecl := 2in

Bolt spacing Sb1 := 3.5in

Edge distance (Vertical) evb1 = 1.5  in

Edge distance (Horizontal) ehb1 = 1.5  in

Clip corner Cc := 1in

Check for bolts :-

Number of bolt rows nrb1 := 3

Number of bolt columns ncb1 := 1

Total number of bolts nb1 := nrb1  ncb1 nb1 = 3

First bolt distance a := 3.5in a = 3.5  in

Holeconn := "SSLT" if Holesp = 2  Holesc = 2 Holeconn = "SSLT"

"STD" otherwise

( )
Ecc Holeconn , a , nr1 , nc1 , gb1 := e if a  3.5in  nc1 = 1

a
2
(
if 2  nr1  5  Holeconn = "SSLT"  Holeconn = "STD" )
a
if 6  nr1  12  Holeconn = "SSLT"
2
a if 6  nr1  12  Holeconn = "STD"

"Cannot use conventional design" otherwise

e  a if a > 3.5in  nc1 = 1

( nc1 - 1)  gb1
e a + if nc1 > 1
2
return e

θ :=
Eccentricity (
e1 := Ecc Holeconn , a , nrb1 , ncb1 , gb1 ) e1 = 1.75  in

Coefficien C = 2.481
t
Fbg := C  Scap Fbg = 29.6  kip

( )
Result kip Fbg , R vc = "Capacity (29.6 kip) > Load(15 kip)-----------> O.K"

Check for bearing/tearing capacity of shear / splice plate:- (Considering vertical shear)

Maximum thickness of shear / splice plate (

t s1 := min t sp , t sc ) t s1 = 0.375  in

  D hsp  
Bearing/tearing capacity

( ) ( )
Rs1v := min  2.4  f c  dbolt  t s1  Fup , 1.2  f c   evb1 -   t s1  Fup , Scap
2  
at extreme bolt

R s1v = 11.93  kip

Bearing/tearing capacity Rs2v := ( )
min  2.4  f c  dbolt  t s1  Fup , Scap if nrb1 = 1
at intermediate bolt
min ( 2.4  f c)  dbolt  t s1  Fup , ( 1.2  f c)  ( Sb1 - D hsp)  t s1  Fup , Scap otherwise
 

R s2v = 11.93  kip

Bearing/tearing capacity (
Rsv := min R s1v , R s2v  C ) R sv = 29.6  kip

( )
Result kip Rsv , R vc = "Capacity (29.6 kip) > Load(15 kip)-----------> O.K"

Check for shear capacity of the shear / splice plate :-

Gross shear area

 (  )
Ag :=  nrb1 - 1  Sb1 + 2  evb1  t s1 Ag = 3.75  in
2

An := Ag - nrb1 Dhsp +
1  2
Net shear area in  t An = 2.77  in
 16  s1

Shear capacity ( ) ( )
Fcap := min  0.6  f g  Ag  Fyp , 0.6  f n An  Fup Fcap = 48.1  kip

( )
Result kip Fcap , Rvc = "Capacity (48.12 kip) > Load(15 kip)-----------> O.K"

Check for shear capacity of HSS wall :-

Gross shear area (

Agh := 2  db1 - 2  2.25  t fb1  t fb1des ) Agh = 7.198  in
2

Shear capacity (
Fcaph := 0.6  f g  Agh  Fyb1 ) Fcaph = 132.45  kip

( )
Result kip Fcaph , Rvc = "Capacity (132.45 kip) > Load(15 kip)-----------> O.K"

Check for block shear capacity of shear / splice plate (considering vertical shear) :-
Gross shear area (
Agv :=  nrb1 - 1  Sb1 + evb1  t s1 ) Agv = 3.19  in
2

Anv := Agv - nrb1 - 0.5  Dhsp +

1 
Net shear area

( in  t
16  s1
) Anv = 2.37  in
2

Ant := ehb1 - ncb1 - 0.5  D h1sp +

1 
Net tension area
 
( 16 
)
in  t s1 Ant = 0.36  in
2

U bs = 1 (Reduction coefficient) (Refer J.4.3 AISC 15th Ed)

Block shear capacity ( ) ( )

Bscap := min  0.6  f n  Agv  Fyp , 0.6  f n Anv  Fup + f n  U bs  Ant  Fup ( ) Bscap = 44.96  kip

( )
Result kip Bscap , Rvc = "Capacity (44.96 kip) > Load(15 kip)-----------> O.K"

Check for combined flexural and shear stress of shear / splice plate ( Gross ) :-

Eccentricity between first column of e2 := a e2 = 3.5  in

bolt and connection interface
Depth of shear/splice plate (
Dp := nrb1 - 1  Sb1 + 2evb1 ) D p = 10  in

2
Gross shear area Agv := Dp  t s1 Agv = 3.75  in

Induced moment on plate Ms := Rvc  e2 Ms = 4.38  kip  ft

due to vertical shear
2
Dp  t s1 3
Section modulus of plate Zmodg := Zmodg = 9.375  in
4

Moment capacity of plate ( )

Mc := f a  Zmodg  Fyp Mc = 16.84  kip  ft

( )
Shear capacity of plate ( )
Vgc := 0.6  f g  Agv  Fyp Vgc = 54  kip

2 2
 Rvc   Ms 
Ratio of combined stress on plate Rx :=   +  R x = 0.145
 Vgc   Mc 
( )
Result ratio 1 , Rx = "(1) > (0.145)-----------> O.K"

Check for combined flexural and shear stress of shear / splice plate ( Net ) :-

Anv :=  nrb1 - 1  Sb1 + 2evb1 - nrb1 D hsp +

1 
Net shear area (

) 
in  t
16  s1
Anv = 2.766  in
2

3
Section modulus of plate Zmodn = 7.006  in

Moment capacity of plate ( )

Mcap := f n  Zmodn  Fup Mcap = 16.93  kip  ft

Induced moment on plate Ms1 := Rvc  e2 Ms1 = 4.38  kip  ft

due to vertical shear
Shear capacity of plate ( )
Vnc := 0.6  f n  Anv  Fup Vnc = 48.122  kip

2 2
 Rvc   Ms1 
Ratio of combined stress on plate Ry :=   +  R y = 0.164
 Vnc   Mcap 
( )
Result ratio 1 , Ry = "(1) > (0.164)-----------> O.K"

Check for buckling capacity of shear / splice plate :-

Fyp
Dp 
1ksi
λ1 := λ1 = 0.305
2
 Dp 
10  t s1  475 + 280  
 e2 

Q := 1 if λ1  0.7 Q=1

( 1.34 - 0.486λ1) if 0.7 < λ1  1.41

 1.30  if λ > 1.41

 2 1
 λ1 

Critical buckling stress Fcr2 := Q  Fyp Fcr2 = 36  ksi

2
Dp  t s1 3
Elastic section modulus Z1 := Z1 = 6.25  in
6

( fa)  Fcr2 Z1
Buckling capacity of shear/splice plate Pbuc := Pbuc = 38.49  kip
e2

( )
Result kip Pbuc , Rvc = "Capacity (38.49 kip) > Load(15 kip)-----------> O.K"

Check for maximum thickness of shear plate :-

Coefficient (
C1z := C1zx nrb1 , ncb1 , Sb1 , gb1 ) C1z = 6.871  in

Moment strength of bolt group Mmax1 := 1.11  Fnv  Abolt  C1z Mmax1 = 15.16  kip  ft
 6Mmax1
Maximum thickness of shear tab Tpmax := if ncb1 > 1 Tpmax = 0.438  in
2
Fyp  Dp

dbolt 1
+ in otherwise
2 16

t sa := t sp if ncb1 > 1 t sa = 0.375  in

(
min t sp , t sc ) otherwise

( )
Result in Tpmax , t sa = "t.avl (0.44 in) > t.req(0.37 in)-----------> O.K"

Check for weld b/w shear plate and beam web :-

Length of weld (
Lw := nrb1 - 1 Sb1 + 2  evb1 ) Lw = 10  in

Rvc kip
Shear per inch on weld FL := FL = 0.750 
2Lw in

5 FL 
Size of fillet weld s := max  t sp ,  16 s = 3.75  in
 8 0.707  0.6  f c F EXX  ( )
Adopt 1/4 inch FW

 FL 
 6.19  
 0.707  ( 0.6  f c) FEXX 
Min shear plate thickness t min :=    16  ksi t min = 0.086  in
Fup
 
( )
Result in t sp , t min = "t.avl (0.37 in) > t.req(0.09 in)-----------> O.K"

Check for shear capacity of shear plate @ beam web :-

2
Gross shear area Ags := Lw  t sp Ags = 3.75  in

Shear capacity (
Fscap := 0.6  f g  Ags  Fyp ) Fscap = 54  kip

( )
Result kip Fscap , Rvc = "Capacity (54 kip) > Load(15 kip)-----------> O.K"

Check for weld b/w closure plate and splice plate :-

Length of weld Lw1 := db1 + 0.5in Lw1 = 12.5  in

Length of weld with dispersion (

Lwd := t fb1des + 2.5  t c ) Lwd = 1.599  in

Induced moment due to vertical shear (

M1 := Rvc  ecl + 2ehb1 + 0.5in ) M1 = 6.875  kip  ft

M1
Couple force about weld C.G Cfs := C fs = 7.932  kip
db1 - Lwd

Rvc kip
Shear per inch on weld FL := FL = 0.600 
2Lw1 in
 Cfs kip
Tension force per inch on weld FT := FT = 4.961 
Lwd in

2 2 kip
Resultant stress per inch on weld Rs := FT + FL R s = 4.997 
in

 FT 
Resultant angle θs := atan   θs = 83.103  deg
 FL 

1 Rs 
Size of fillet weld s := max  in ,  16 s = 4.00  in
( )
 4 0.707  0.6  f  1 + 0.5  sin θ 1.5  F
 c  s (

 EXX  ( ) )
Adopt 1/4 inch FW

 Rs 
 3.09  

t min :=   (
0.707  ( 0.6  f c)  1 + 0.5  sin ( θs)
1.5 
)
 FEXX   16  ksi t
Min closure plate
Fup
 min = 0.192  in
thickness  

( )
Result in t c , t min = "t.avl (0.5 in) > t.req(0.19 in)-----------> O.K"

 Rs 
 6.19  

t min :=   (
0.707  ( 0.6  f c)  1 + 0.5  sin ( θs)
1.5 
 )
FEXX 
Min splice plate
Fup
  16  ksi t min = 0.385  in
thickness  

( )
Result in t sc , t min = "t.avl (0.5 in) > t.req(0.38 in)-----------> O.K"

Check for shear capacity of splice plate @ closure plate :-

2
Gross shear area Ags := Lw1  t sc Ags = 6.25  in

Shear capacity ( )
Fcapsc := 0.6  f g  Ags  Fyp Fcapsc = 90  kip

( )
Result kip Fcapsc , R vc = "Capacity (90 kip) > Load(15 kip)-----------> O.K"

Check for axial capacity of splice plate @ closure plate :-

2
Gross tension area Agt := Lwd  t sc Agt = 0.8  in

Axial capacity Acapsc := f a  Agt  Fyp Acapsc = 17.2  kip

( )
Result kip Acapsc , C fs = "Capacity (17.23 kip) > Load(7.93 kip)-----------> O.K"

Check for weld b/w closure plate and HSS beam wall :-(Longitudinal side)
Length of weld Lw2 := db1 - 2  2.25  t fb1 Lw2 = 10.312  in

Rvc kip
Shear per inch on weld FL := FL = 0.727 
2Lw2 in

1 FL 
Size of fillet weld s := max  in ,  16 s = 4.00  in
(
 4 0.707  0.6  f c FEXX )
Adopt 1/4 inch FW

 
  FL 
 3.09  
Min HSS wall  0.707  ( 0.6  f c) FEXX 
t min :=    16  ksi
thickness Fub1 t min = 0.042  in
 

( )
Result in t fb1des , t min = "t.avl (0.35 in) > t.req(0.04 in)-----------> O.K"

 FL 
 3.09  
 0.707  ( 0.6  f c) FEXX 
Min plate thickness t min :=    16  ksi t min = 0.042  in
Fup
 

( )
Result in t c , t min = "t.avl (0.5 in) > t.req(0.04 in)-----------> O.K"

Check for shear capacity of closure plate @ HSS beam wall :-

2
Gross shear area Ags := Lw2  t c Ags = 5.156  in

Shear capacity ( )
Fscaptc := 2  0.6  f g  Ags  Fyp Fscaptc = 148.5  kip

( )
Result kip Fscaptc , R vc = "Capacity (148.5 kip) > Load(15 kip)-----------> O.K"

Check for weld b/w closure plate and HSS beam top wall :-(Transverse side)

Length of weld (
Lw3 := min t sc + 5t c , bfb1 - 2  2.25  t fb1 ) Lw3 = 3  in

Moment induced (
Mvc := Rvc  ecl + 2ehb1 + 0.5in + t c ) Mvc = 7.5  kip  ft

Mvc
Couple force due to moment Cf := C f = 7.74  kip
db1 - t fb1

Cf kip
Tension per inch on FT := FT = 2.581 
weld Lw3 in

1 FT 
Size of fillet weld s := max  in ,  16 s = 4.00  in
( )
 4 0.707  0.6  f c  1.5FEXX 
Adopt 1/4 inch FW

 FT 
 3.09  
 0.707  ( 0.6  f c)  1.5FEXX 
t min :=
Min HSS wall
 Fub1   16  ksi t min = 0.099  in
thickness  

( )
Result in t fb1des , t min = "t.avl (0.35 in) > t.req(0.1 in)-----------> O.K"

 FT 
 3.09  
 0.707  ( 0.6  f c)  1.5FEXX 
Min closure plate thickness t min :=    16  ksi t min = 0.099  in
 Fup 
( )
Result in t c , t min = "t.avl (0.5 in) > t.req(0.1 in)-----------> O.K"

Check for weld b/w closure plate and HSS beam bottom wall :-(Transverse side)
 Cf kip
Tension per inch on FT := FT = 2.581 
weld Lw3 in

FT
Size of weld s := s = 0.12  in
(
min f a  Fyp , f c  Fup )
Adopt CJP weld

Check for axial capacity of HSS wall @ closure plate :-

Axial capacity Acapb1 := f a  Lw3  t fb1des  Fyb1 Acapb1 = 28.8  kip

( )
Result kip Acapb1 , C f = "Capacity (28.84 kip) > Load(7.74 kip)-----------> O.K"

Check for HSS wall local crippling :-

Bearing length N := t sc N = 0.5  in

Modulus of elasticity Es := 29000 ksi

  1.5 
 2 6N  t fb1des   Es  Fyb1  t c 
Crippling strength Acri := 1.6  ( f c)  t fb1des 1 +    A = 158.27  kip
  b  t fb1des  cri
  fb1  t c   

( )
Result kip Acri , 2  Cf = "Capacity (158.27 kip) > Load(15.48 kip)-----------> O.K"

Check for HSS wall local yielding :-

Yielding strength Ayie := ( ) (

2  f g  Fyb1  t fb1des N + 5t c ) if N + 5t c < bfb1 Ayie = 64.22  kip

( fg)  Fyb1  Agb1 otherwise

( )
Result kip Ayie , 2  Cf = "Capacity (64.22 kip) > Load(15.48 kip)-----------> O.K"

Check for weld between plate and clip angle :-

Length of the weld Lw := 9in Lw = 9  in

kLw := 3.5in kLw = 3.5  in

kLw
k := k = 0.3889
Lw

2
kLw
xLw := xLw = 0.766  in
2kLw + Lw

Lai := 5in

aLw := Lai - xLw aLw = 4.234  in

aLw
a := a = 0.4705
Lw

C1 := 1 (E70xx electrode)
Size of fillet weld required per 16th of inch D := 4

Weld Coefficients Cweld = 2.547

kip
Rn := f c  C weld  C1  Lw  D R n = 45.8  kip
in

1.414  Fup  0.5in

Reduction factor for beam Web thickness C3 := C 3 = 2.343
 D  in  F 
 
 16 EXX

Ccap := 1 if ( C 3 > 1) C cap = 1

C3 otherwise

Weld design strength Fstr := Ccap  Rn Fstr = 45.85  kip

Adopt 1/4" Fillet weld

( )
Result kip Fstr , R vc = "Capacity (45.85 kip) > Load(15 kip)-----------> O.K"

Design of stud at embed plate per ACI-349 :-

Factored load:- Vf := Rvc  1.5 Vf = 22.5  kip

Stud material strength:- Fys := 50ksi

Ultimate material strength:- Fus := 60ksi

Concrete material strength:- Fconc := 4000

Dia of stud:- D stud := 0.75in

Dia of head of stud:- D hstud := 1.5in

Thick of head of stud:- Tstud := 0.6in

Factors for shear friction (Sec 11.7) f ci := 0.85 f u := 0.9 f cia := 0.65

No of stud:- n1 := 4

Thickness of embed plate: t ep := 0.5in

Check for shear area required for stud(sec B.6.5.1) :-

Vf
n1
Shear capacity Av :=
( fci  fu)  Fys
2
Av = 0.147  in

 
 A :=    D stud
π 2 2
Area of stud A = 0.442  in
4

 A Av 
Result in  ,  = "t.avl (0.44 in) > t.req(0.15 in)-----------> O.K"
 1in in 

Check for Anchor head bearing for stud(Sec B5.1.1) :-

Ahs :=    Dhstud
π 2 2
Area of stud head Ahs = 1.767  in
 4
Area of the anchor head (Ah) (including the Ahs
area of the tensile stress component) is at F := F=4
least 2.5 times the area of the tensile stress A
component

Result ratio ( F , 2.5) = "(4) > (2.5)-----------> O.K"

Check for Thickness head bearing for stud(Sec B4.5.2) :-

Thickness of the anchor heat (Th) is at least Dhstud - Dstud

1.0 times the greatest dimension from the outer Th1 := Th1 = 0.375  in
most bearing edge of the anchor head to the 2
face of the tensile stress component.

( )
Result in Tstud , Th1 = "t.avl (0.6 in) > t.req(0.37 in)-----------> O.K"

Bearing area of head is approximately evenly distributed around the perimeter of the tensile stress component.

Head and tensile stress component are concentric. Hence OK

Determine required embedment length for the stud to prevent concrete cone failure
(B4.5.2,B.5.1.1)
The design pullout strength of the concrete, Pd, must exceed the minimum specified tensile
strength (Asfut) of the tensile stress component

Tensile strength Ftcap := A  Fus

Ftcap = 26.507  kip

Design pullout strength of the concrete Pd := f cia  4  Fconc Pd = 164.438

  D
2
 D  
2
Computation of Ld π LD + hstud  -  hstud    Pd
  2   2  

Ftcap
(
LD LD + 1.5  )
Pd  π  0.001

On solving equation ( )
LD LD + 1.5  51.311

LD  6.47
Provide embedment length 8 in
Check Embed plate thickness:

Select plate thickness such that ds /t < 2.7 , “Shear Strength of Thin Flange Composite Sections,” G. G. Goble, AISC
Engineering Journal,April, 1968.
 D stud
t := t = 0.278  in
2.7

( )
Result in t ep , t = "t.avl (0.5 in) > t.req(0.28 in)-----------> O.K"