beam column connection

7/30/2019 beam column connection

1/106

1

SESSION 4

FRAMING (SHEAR)

CONNECTIONS

CONTINUED


2/106

2

SEMINAR OUTLINE

Session 4 Framing Connections Continued

Session 5 Moment ConnectionsSession 6 End-Plate Moment Connections/

Bracing Connections

Closure


3/106

3

Bolted / Bolted Double Angles

114"

3"3"

114

"

W14x30 A992

tw = 0.27 in.12"

12"

3"

2L 5 x 3 x 5/16 x 0'-8 1/2A36

3/4 A325-N Bolts


4/106

4

Bolted / Bolted Double Angles

Eccentricity not considered in these

connections

For Bolt Rupture: Vn = Fv Ab x 6

No New Limit States


5/106

5

Bolted / Welded Double Angles

2L 3 x 3 x 5/16 x 0'-8 1/2"

W14x30A992tw = 0.27 in.

A36

812"

2"

4"

Return

@ Top1/4"3/4 A325-N Bolts

E70XX

Knife Connection


6/106

6


Knife Connection

Beam to Column Flange

Bottom Cope to Permit Erection

New Limit States:

Coped Beam Web Strength atTension Flange

Weld Strength on OSLs


7/106

7


Coped Beam Web Strength at

Tension Flange

Vn = b Fy Snet / e

b = 0.9

Snet from Table 8-49

4"

2"

W14x30


8/106

8

Vn/2

L/6

e

L

CL

Web

O

Tension, ft


Weld Strength on OSLs


9/106

9


2

nt

nt

o

nv

L

eV1.8f

e

2

VL

3

2L

6

5f

2

1

0M

L

2Vf

Vn/2

L/6

e

L

CLWeb

O

Tension, ft


10/106

10


22

2

n

22

2

n

2

v

2

tw

e12.9L

D)(1.392L2V

e12.9LL2V

fff

Vn/2

L/6

e

L

CLWeb

O

Tension, ft


11/106

11


Example: Calculate Vn based on the weldrupture strength of the OSLs

e = 3 + 0.27/2 =3.135 in.

L = 8.5 in.

3/4 A325-N Bolts

E70XX

2L 3 x 3 x 5/16 x 0'-8 1/2"

W14x30A992tw = 0.27 in.

A36

812"

2"

e

Return@ Top

1/4"


12/106

12

Bolted / Welded Double Angles Example

Note: Weld returns at top of angles have

been neglected

k57.0

(3.135)12.95.8

4)(1.392(8.5)2

e12.9L

D)(1.392L2V

22

2

22

2

n


13/106

13

Single Plate

SHEAR TAB OR SINGLE

PLATE


14/106

14

Mw

= Vu

ew

Mb = Vu eb

Shear Tab or Single Plate

Mw M b

Bolt

Line

e b

ew

a


15/106

15


Rotation is obtained

by bolts plowingthe plate, which

requires limiting the

plate thickness

Single Plate


16/106

16


Geometric Limitations:

tp < db /2 + 1/16

Lh

> 1 1/2 in.

Lv > 1 1/2 in.

2 1/2 in. < a < 3 1/2 in.

2 < n = No. of Bolts < 9L > T/2

Plate Material: A36 Steel

a

L

Lv

2@3"

Lh


17/106

17


Eccentricities depend on:

i) Connected Element

Rigid

Flexible

ii) Hole Type

Standard

Short Slots


18/106

18


Rigid Elements:

Column Flange

Girder w/ Plates on Both Sides Concrete Wall

Flexible Elements

Girder Web on One Side

Note: Not recommended for column webs.


19/106

19


Flexible Support /

Standard Holes

eb = |(n - 1)a| > a

Flexible Support /

Short Slotted Holes

eb = |(2n/3)a| > a

a eb

Bolt

Line


20/106

20


Rigid Support /

Standard Holes

eb = |(n - 1)a|

Rigid Support /

Short Slotted Holeseb = |(2n/3)a|

a eb

Bolt

Line


21/106

21


Weld Strength

E70XX Electrode

Weld Size > 3/4 tp on each side

Weld size is then sufficient to develop plate

in tension yielding


22/106

22


Plate Limit States:

Shear Yielding

Shear Rupture Block Shear

Bearing / Tear Out

Plate Buckling


23/106

23


Plate Buckling

tp > L / 64 > 1/4 in.

L

Compression


24/106

24


Example: Determine required plate and

weld sizes.

W14x30

tw = 0.27 in

T = 12 in.

Vu = 40 k

E70XX

3/4 in. A325-N Bolts


25/106

25

Shear Tab or Single Plate Example

Try 33/4 in. A325-N Bolts

Vn = (0.75 x 48 x 0.4418) (3)

= 15.9 x 3

= 47.7 k > Vu = 40 k

Try 1/4 in. plate

Maximum Plate Thickness

= d/2 + 1/16 = 3/8 + 1/16 = 0.4375 in.


26/106

26

Plate Geometry

Check Plate

BucklingL / 64 = 9 / 64

= 0.14 < 1/4 OK


2@3"9" > T/2

112"

112"

3"112"


27/106

27

Check bolts for eccentric loading:

Rigid Support / Standard Holes

eb

= |(n - 1)a| = |(31)3| = 1.0 in.

Using Table 8-18 with

s = 3 in. ex = 1.0 in. n = 3By extrapolation: C = 2.71



28/106

28


29/106

29

Vn = C x rv

= 2.71 x 15.9= 43.1 k > Vu = 40 k OK



30/106

30

Plate Limit States

t = 1/4 in.

Shear Yielding:

Vn = 0.9 (0.6 Fy) Ag

= 0.9 (0.6 x 36) (0.25 x 9)

= 43.7 k > 40 k OK



31/106

31

Shear Rupture:

Vn = 0.75 (0.6 Fu) An = 41.6 k OK

Block Shear: 44.8 k OK

Bearing/Tear Out: 53.4 k OK

Use PL 1/4 x 4 1/2 x 0-9 A36

with (3) 3/4 in. A325-N Bolts



32/106

32

Required Weld Size

tweld = 3/4 (1/4) = 3/16 in.

Notes: - Min. weld requirements may

control

- Beam web (tw

= 0.27 in.) will not

control bearing and tear out



33/106

33


112"

2@3"

1

1

2"

112"

9"

3"

3/16

PL 1/4 x 4 1/2 x 0'-9"


34/106

34

SINGLE ANGLE

CONNECTIONS

Bolted and Welded Alternatives

One Angle

Horizontal short slots

may be used in angle

@ TopReturn


35/106

35

Eccentricity Assumptions for OSL

Single Row Double Row

Single Angle Connections

ea=ebb

e a

CL Web CL Web

Bolts

Angle


36/106

36

Eccentricity Assumptions for OSL

Welded


e w2 tweldReturn

CL Web

+

c.g.weld


37/106

37


Notes:

Eccentricity is ignored on the beam side

when the connection is a single row. Standard holes or short slots can be used

on the beam side.

Only standard holes should be used onthe supporting member side.


38/106

38


Additional Limit States for BoltedConnections

Flexural Yielding

Vn = 0.9 Fy Sg / ea

Sg = tp L2/6

L

ea


39/106

39


Flexural Rupture

Vn = 0.75 Fu Snet / ea

Snet:

See LRFD Manual, Table 12-1

L

ea


40/106

40


41/106

41

Eccentric Shear of Bolt Group

(Instantaneous Center of Rotation Method)

Table 8-18 Table 8-19

Vn = C ( rv)

fVnfVn



42/106

42


Bearing and Tear Out

Vn = C ( rvb)

where rvb is the bearing / tear out

strength at the outermost bolt


43/106

43


db (in.) Min. t (in.)

3/4 3/87/8 3/8

1 1/2


44/106

44

Additional Limit States for WeldedConnections

Eccentric Shear Strength of Weld

Table 8-44

2 tweldReturn

fVn fVn



45/106

45


46/106

46

Welded Unstiffened Seat

Connections

Stabilizing

Angle

L4x4x1/4

Alternate

Location


47/106

47

Welded Unstiffened Seat

ConnectionsLimit States:

Beam Local WebYielding

Beam Local WebCrippling

Seat Angle Bending

Seat Angle ShearYielding

Weld Rupture


48/106

48

Beam Local Web Yielding

N+2.5k

N


49/106

49

Beam Local Web Yielding

Section K1.3

Web Local Yielding

= 1.0Rn = (2.5kdesign + N)Fyw tw

N+2.5k

N


50/106

50

Beam Local Web Crippling


51/106

51

Beam Local Web Crippling

Section K1.4 Web Local Crippling @


52/106

52

Design Model for Angle Flexure

The N-distance is

determined from the

limit states of webyielding and web

crippling, but not less

than kdetailing.

34

"N

Critical Section

for Bending, Shear

Supported Beam

Supporting Column

ra= 3/8"

ta


53/106

53


Setback = 1/2 in.

Beam Tolerance = 1/4 in.

Use 3/4 in. setback

in calculations

34" N

Critical Section

for Bending, Shear

Supported Beam

Supporting Column

ra= 3/8"

ta


54/106

54


The maximum value of

N is then used todetermine the

eccentricity:

e = N/2 + (3/43/8)ta

= N/2 + 3/8 -ta

3

4

" N

Critical Section

for Bending, Shear

Supported Beam

Supporting Column

ra= 3/8"

ta


55/106

55


For the Limit State of Web Yielding

design

wy

umin k5.2

tF0.1RN


56/106

56


For the Limit State of Web Crippling

when N/d < 0.2

when N/d > 0.2

0.2t

t1

tF

t

)(68t0.75

R

4

dN

1.5

w

f

fy

w

2

w

umin

5.1

w

f

fy

w2

w

umint

t1tF

t

)(68t0.75

R

3

dN


57/106

57


Required angle thickness

from OSL bending

with La = angle length

e = N/2 + 3/8 - ta

ay

ureq

LF9.0

eR4t

34

" N

Critical Section

for Bending, Shear

Supported Beam

Supporting Column

ra= 3/8"

ta


58/106

58

Angle Shear Yielding

Therefore

angleaynu tL)F6.0(9.0RR

ay

ureq

L)F6.0(9.0

Rt


59/106

59

Weld Rupture

The weld is subjected

to eccentric shear.

Table 8-38 of LRFD

Vol. II will be used to

determine requiredweld size.

34"

N

Critical Sectionfor Bending, Shear

Supported Beam

Supporting Column

ra= 3/8"ta

W ld R t


60/106

60

Weld Rupture


61/106

61

4"

AlternateClip Position

StabilizerClip

Seat PlateStiffenerOptionalTrim Lines

2"

Stiffened Seated Connections


62/106

62


Limit States:

Beam Web Yielding

Beam Web Crippling

Strength of Stiffener Plate

Eccentric Shear of Connecting Side Weld

or Bolts

Column Web Failure


63/106

63


Notes:

1/2 in. setback but 3/4 in. for calculations

Seat plate > 3/8 in. For unstiffened beam webs, the seat

stiffener thickness is a function of the

beam and seat stiffener yield stresses andthe weld size.


64/106

64


For Unstiffened Beams:Seat Stiffener

Beam, Fy Fy ts

36 36 tw50 36 1.4 tw

50 50 tw

For Stiffened and Unstiffened Beams

Seat Stiffener: ts (36 ksi) > 2 tweld

ts (50 ksi) > 1.5 tweld


65/106

65


Column Web Failure

Stress concentration requires stiff web.

Stress Concentration


66/106

66

Column Web Failure

Also need to prevent flange rotation.



67/106

67


LRFD Vol. II has a simplified approach

for sections heavier than:

43 lb/ft for W14 ( > W14x43)

40 lb/ft for W12 ( > W12x40)

30 lb/ft for W10 ( > W10x30)

24 lb/ft for W8 ( > W8x24)


68/106

68


Rules:

Beam must be bolted to seat with A325 or

A490 bolts within the greater of W/2 or

2 5/8 in. from the column web.

Special rules for W14x43

Seat plate is not welded to the beam

Weld size is limited to shear yield

strength of the column web

Stiff d S t d C ti


69/106

69



70/106

70

END OF SESSION 4

Design Examples

(For Home Study)

Si l A l C i


71/106

71

Single Angle ConnectionsExample: Is the connection adequate?

W21x50A992

tw = 0.38 in.

Vu = 90 k

Vu = 90 k

112"

2"

112"

4@3"

3"

3"

3"

134

"114

"

OSL

L8 x 6 x 3/8 x 1-3 A36

3/4 in.

A325N Bolts


72/106

72

Single Angle Connection Example

Angle Flexural Yield

Supporting Side OSL Controls

Vn = 0.9 Fy Sg / ea

Sg = (3/8) (15.0)2

/ 6 = 14.06 in3

ea = 0.38 / 2 + 3 = 3.19 in.

Vn = 0.9 Fy Sg / ea

= 0.9 (36) (14.06) / 3.19

= 142.9 k > 90 k OK

Vu

a

Si l A l C ti E l


73/106

73

Angle Flexural Rupture

From Table 12-1

Snet = 10.1 in3



74/106

74

Si l A l C ti E l


75/106

75

Vn = 0.75 Fu Snet / ea

= 0.75 (58) (10.1) / 3.19

= 137.7 k > 90 k OK


Si l A l C ti E l


76/106

76

Bolt Ruptureeb = 0.38 / 2 + 3 + 1.5 = 4.69 in.

Table 8-19

n = 5 b = 3 in. s = 3 in.

By interpolation: C = 6.29


Vu

eb

3"

3"


77/106

77

Single Angle Connection E ample


78/106

78

3/4 in. A325-N

rv = 0.75 x 48 x 0.4418

= 15.9 k

Vn = C ( rv) = 6.29 x 15.9

= 100 k > 90 k OK


Si l A l C ti E l


79/106

79

Bearing / Tear OutCritical Bolts

Brg. = 2.4 (58) (3/4 x 3/8)

= 39.2 k

T.O. = 1.2 (58) (1.513/32) (3/8)

= 28.5 k

rbv = 0.75 (28.5)

= 21.4 k > rv = 15.9 k OK

Vu




80/106

80

Beam Web

tw = 0.38 in > 3/8 in.

will not control

Connection is adequate if supporting

element thickness is> (15.9 / 39.2) (3/8) = 0.15 in.


Single Angle Design Example


81/106

81

Single Angle Design Example

Example: Outstanding leg of previousconnection welded. Determine required

weld size. Column, tf= 0.710 in.2 t

weldReturn

Vu = 90 k

E70XX

L6 x 6 x 3/8 x 1'-3"


82/106

82

Table 8-44 with l = 15 in.

kl = 6 in.

k = 6 / 15 = 0.4 x = 0.057

xl = 0.057 x 15 = 0.855 in.

al = 0.38/2 + 60.855

= 5.34 in.

a = 5.34 / 15 = 0.356


15"

0.19"

al

xl

Vu = 90 k


83/106

83


84/106

84

C = 1.64 by interpolationDreqd = Vu / (C C1 l)

= 90 / (1.64 x 1.0 x 15)

= 3.65 4/16 = 1/4 in.

Min. weld = 1/4 in. (Column tf= 0.710 in.)

Max weld = 5/16 in.

Use 1/4 in. Fillet weld


U tiff d S t C ti E


85/106

85

Unstiffened Seat Connection Ex.

W18x46 A992

W14x90 Column

L4 x 4 x 5/8 x 0'-8"

?

0.440"

Vu = 35 k

Web

CL

W18x46

bf= 6.06 in.tf= 0.605 in.d = 18.06 in.

tw= 0.360 in.kdetailing = 1 1/4 in.kdesign = 1.01in.

Example:


86/106

86


Example: Determine

(1) if seat angle is adequate, and

(2) required weld size.

A36 Angle E70xx Electrode


87/106

87


For the Limit State of Web Yielding

design

wy

umin k5.2

tF0.1

RN

01.1x5.2360.0x50x0.1

35

detailingk58.0

iff S C i


88/106

88


For the Limit State of Web Cripplingassuming N/d < 0.2

5.1

w

f

fy

w

2

w

umin

t

t1

tF

t

)t68(75.0

R

3

dN

5.1

2 360.0

605.01

605.0x50

360.0

)360.0x68(75.0

35

3

06.18

in.5.5

Unstiffened Seat Connection E


89/106

89


For the Limit State of Web Crippling,

N = -5.5 in. < kdetailing

Therefore

N = kdetailing = 1.25 in.

Check

N/d = 1.25/18.06 = 0.069 < 0.2 OK

Unstiffened Seat Connection Ex


90/106

90


Required angle thickness from OSL bending

e = N/2 + 3/8 ta

= 1.25/2 + 3/85/8

= 0.375 in. 34" N

Critical Sectionfor Bending, Shear

Supported Beam

Supporting Column

ra= 3/8"

ta



91/106

91


With La = 8 in.

= 0.45 in. < 5/8 in. OK

ay

ureq

LF9.0

eR4t

0.8x36x9.0

375.0x35x4

U tiff d S t C ti E


92/106

92


Required angle thickness fromOSL Shear Yielding:

ay

u

req L)F6.0(9.0

R

t

.in225.00.8)36x6.0(9.0

35

5/8 in. OK

U tiff d S t C ti E


93/106

93


Determine required weld size.

Using Table 8-38 with

angle = 0 degrees

k = 0.0

e = 3/4 + N/2 = 3/4 + 1.25/2 = 1.375 in.

a = e/L = 1.375/4.0 = 0.344

Find C = 2.18



94/106

94


U tiff d S t C ti E


95/106

95


Dmin = Ru/CC1L = 35/(2.18x1.0x4.0)

= 4.0 1/16ths

Min. Weld = in.

Use in. fillet weld both sides of angle legs.

Returns at top.



96/106

96

Unstiffened Seat Connection Ex.WebCL

W18x46 A992

L4 x 4 x 5/8 x 0'-8"

1/4

W14x90 Column0.440"

Vu= 35 k

Stiffened Seat Example


97/106

97


Determine: 1. A36 stiffener thickness

2. E70XX weld size.

W24x68A36

t wb = 0.415 in.

twc

W14x90A992

= 0.440 in.

W

Lt s

Ru = 80 k



98/106

98


W14x90 Satisfies column web requirement.

W24x68A992

t wb = 0.415 in.

twc

W14x90A992

= 0.440 in.

W

Lt s


99/106

99



100/106

100


Web Crippling

Using the Factored Uniform Load Table inPart-4 assuming N/d < 0.2

Check N/d = 1.20 / 23.73 = 0.05 < 0.2 OK(Or calculate as in Session 2.)

in.88.10.7513.10.7557.5

7.7380

setbackR

RR

W4

3u

min



101/106

101

Web Yielding

Use W = 4 in. and t = 3/8 in.

Bolt location rule will be satisfied.

in.17.10.75418.00.758.203.7180

setbackR

RRW

2

1umin




102/106

102

With: W = 4 in.

1/4 in. Weld

Using Table 9-9:

L = 10 in.

Rn = 82.3 > Ru = 80 k


34"

Ru = 80 k

10"

4"



103/106

103




104/106

104

Column Web Strength at Weld:

Rn = 0.9 (0.6 Fy) L twc x 2= 0.9 (0.6 x 50) (10) (0.440) (2)

= 237.6 k > 80 k OK




105/106

105

Stiffener Plate (W24x68 twb = 0.415 in.)

ts (36 ksi) > 1.4 twb (50 ksi)

> 1.4 x 0.415 = 0.581 in.

ts > 2 tweld = 2 x 1/4 = 1/2 in.

Use 5/8 in. plate




106/106


< 34"

10"

4"

PL 5/8 x 5 x 0'-10"

PL 3/8 x 5 x 0'-8"

1/4

beam column connection

Documents