1

Structural Steel ConnectionsStructural Steel Connections

Overview of Connections

Beam End ConnectionsBeam-to-GirderBeam-to-Column

Other Connection Types

Connection Basics: Bolt and weld strength, and other design issues

Design Example

Beam End ConnectionsBeam End Connections

End Fixity

Simple (Shear) Connections

Moment Connections- Fully Restrained (FR)- Partially Restrained (PR)

Primary Applications

Beam-to-Girder Connections- usually Simple connections

Beam-to-Column Connections- Simple or Moment (FR)- Moment (PR) rarely used

BeamBeam--toto--Girder ConnectionsGirder Connections

Double Angle: Bolted- Bolted

2

Double Angle: Welded - Bolted

Shear End Plate

3

Single Plate (Shear Tab)

4

BeamBeam--toto--Column ConnectionsColumn Connections

Simple ("Pinned") FramingSimple ("Pinned") Framing

5

2 Angles

Double Angle

2 Angles

Double Angle

6

Flange Seat Flange Seat

Single Plate

Single Plate (Shear Tab)

7

BeamBeam--toto--Column ConnectionsColumn Connections

Moment ConnectionsMoment Connections

8

Bolted Flange Plate

V M

Welded Flange

9

Moment End Plate

10

Other ConnectionsOther Connections

11

12

13

Specification References for Connection Design

2005 AISC Specification:Chapter J - Design of Connections

Additional References for Connections in Seismic-Resistant Steel Structures:

2005 AISC Seismic Provisions for Structural Steel Buildings.

Prequalified Connections for Special and Intermediate Steel Moment Frames for Seismic Applications

2005 AISC Specification

Chapter J - Design of Connections

Connection Basics: Bolt and Weld Strength and other Design Issues

14

Bolt Strength

Three cases to consider:

1. Bolts loaded in tension

2. Bolts loaded in shear

3. Bolts loaded in shear + tension

For any loading case:

Limit State = Bolt Fracture

Examples:

Bolts loaded in tension

Examples: Bolts loaded in shear Examples: Bolts loaded in shear and tension

15

Bolt Strength Bolt Strength -- TensionTension

Tension Strength Requirement:

LRFD: Ru Rn = 0.75

ASD: Ra Rn / = 2.00

Nominal Strength in Tension:

Rn = Fnt AbAb = nominal area of bolt (area of unthreaded portion)

Fnt = nominal tensile stress = 0.75 Fu

Bolt Strength Bolt Strength -- TensionTension

Rn = Fnt AbFnt = nominal tensile stress = 0.75 Fu

A325 Bolt: Fu = 120 ksi Fnt = 0.75 x 120 ksi = 90 ksi

A490 Bolt: Fu = 150 ksi Fnt = 0.75 x 150 ksi = 113 ksi

Example:

Compute the Available Tension Strength of a 7/8" A325 Bolt

7/8 Bolt: Ab = (.875"/2)2 = 0.601 in2

Rn = Fnt Ab

A325: Fnt = 90 ksi

Rnt = 90 ksi x 0.601 in2 = 54.1 kips

LRFD: Rn = (0.75) (54.1 kips) = 40.6 kips

ASD: Rn / = 54.1 kips / 2.00 = 27.1 kips

16

Bolt Strength Bolt Strength -- ShearShear

Shear Strength Requirement:

LRFD: Ru Rn = 0.75

ASD: Ra Rn / = 2.00

Nominal Strength in Shear (per shear plane):

Rn = Fnv AbAb = nominal area of bolt (area of unthreaded portion)

Fnv = nominal shear stress

Nominal Strength in Shear (per shear plane):

Rn = Fnv AbAb = nominal area of bolt (area of unthreaded portion)

Fnv = nominal shear stress

Fnv = 0.40 Fu threads included in shear plane

Fnv = 0.50 Fu threads excluded from shear plane

A325-N Fnv = 0.40 x 120 ksi = 48 ksi

A325-X Fnv = 0.60 x 120 ksi = 60 ksi

A490-N Fnv = 0.40 x 150 ksi = 60 ksi

A490-X Fnv = 0.50 x 150 ksi = 75 ksi

17

Example:

Compute the Available Shear Strength of a 7/8" A325-N Bolt in Double Shear:

7/8 Bolt: Ab = (.875"/2)2 = 0.601 in2

Rn = Fnv Ab

A325-N: Fnv = 48 ksi

Rnt = 48 ksi x 0.601 in2 x 2 shear planes = 57.7 kips

LRFD: Rn = (0.75) (57.7 kips) = 43.3 kips

ASD: Rn / = 57.7 kips / 2.00 = 28.9 kips

Bolt Strength Bolt Strength -- Shear + TensionShear + Tension

Tension Strength Requirement:

LRFD: Ru Rn = 0.75

ASD: Ra Rn / = 2.00

Nominal Strength in Tension:

Rn = F'nt AbAb = nominal area of bolt (area of unthreaded portion)

F'nt = nominal tensile stress modified to include effects of shear

18

Bearing Strength at Bolt Holes

Limit States:

- Excessive elongation at bolts holes

- Tear-out at bolt holes

Bearing Strength at Bolt Holes

Excessive hole elongation

Tear-out

Bearing Strength at Bolt Holes

LcLc

Standard holes: Rn = lesser of2.4 d t Fu1.2 Lc t Fu

19

Welds

Fillet Weld

w = leg size

w

Welds

Groove weld: Complete joint penetration (CJP)

Groove weld: Partial joint penetration (PJP)

20

Welds

Plug Weld

Welds

Limit State for welds: Fracture

Welds

Available Strength:

LRFD: RnASD: Rn /

Rn = lesser ofFBM ABM Base metal

Fw Aw Weld metal

21

22

Other Connection Design IssuesOther Connection Design Issues

Connection Elements in Tension Connection Elements in Shear Block Shear Rupture

Connection Elements in Tension

P

Gusset Plate in Tension

Check yield of Gusset PlateRn = Ag Fy = 0.9 = 1.67

Pu Rn (LRFD)

Pa Rn / (ASD)

Connection Elements in Tension

P

Gusset Plate in Tension

Check fracture of Gusset PlateRn = Ae Fu = 0.75 = 2.00

Pu Rn (LRFD)

Pa Rn / (ASD)

Connection Elements in Tension

Whitmore Section

23

Vu

Connection Elements in Shear

Check shear yield of connection plateRn = Ag (0.6 Fy) = 0.9 = 1.67

Vu Rn (LRFD)

Va Rn / (ASD)

Vu

Connection Elements in Shear

Check shear fracture of connection plateRn = Ae (0.6 Fu) = 0.75 = 2.00

Vu Rn (LRFD)

Va Rn / (ASD)

Block Shear Rupture

P

Ubs Ant Fu + lesser ofRn = Anv (0.6 Fu)

Agv (0.6 Fy)

LRFD: = 0.75 Pu RnASD: = 2.00 Pa Rn /

Block Shear Rupture

P

Ubs Ant Fu + lesser ofRn = Anv (0.6 Fu)

Agv (0.6 Fy)

Ant = net area of tension surface of block

Agt = gross area of shear surface of block

Ant = net area of shear surface of block

tension surface of block

shear surface of block

24

Block Shear Rupture

P

Ubs Ant Fu + lesser ofRn = Anv (0.6 Fu)

Agv (0.6 Fy)

tension surface of block

shear surface of block

Ubs = 1 for uniform stress on tension surface

Ubs = 1 for non-uniform stress on tension surface

Example:Example: Beam to Column Shear Connection Beam to Column Shear Connection Single Plate ConnectionSingle Plate Connection

Ref: AISC Manual - Section 10

Two options:

Conventional Configuration

Extended Configuration

Single Plate Connection: Conventional ConfigurationSingle Plate Connection: Conventional ConfigurationLimitations

L T / 2

Single row of bolts.

2 n 12

a 3.5"

Standard or Short-Slotted Holes Permitted.

Leh 2 db for plate and for beam web.

Lev must satisfy Spec. Table J3.4

Plate or beam web must satisfy: t db/2 + 1/16".

25

L T / 2

Single Plate Connection: Conventional ConfigurationSingle Plate Connection: Conventional ConfigurationDesign

Check bolt shear and bearing.

Check plate for shear yielding, shear fracture, and block shear rupture.

For standard holes, ignore eccentricity on bolts forn 9

Size fillet weld with leg size of 5/8 tp (develops strenghtof either A36 or Gr. 50 plate)

Example: LRFDa=3"

Leh=1.75"

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

1/41/4

L = 1

5"

Bolts: 7/8" A325-N

Welds: E70

W24x76 (A992)

W10x112 (A992)

Vu

Compute Design Strength of this Single Plate Connection

26

Check Limitationsa=3"

Leh=1.75"

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

1/41/4

L = 1

5"

Bolts: 7/8" A325-N

Standard Holes

Welds: E70

W24x76 (A992)

W10x112 (A992)

Vu

L T / 2W24x76: T = 20"L = 15"

OK

Single Row of Bolts OK

2 n 12 n=5 OK

Standard Holes OK

n 9 - ignore eccentricities OK

Check Limitationsa=3"

Leh=1.75"

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

1/41/4

L = 1

5"

Bolts: 7/8" A325-N

Standard Holes

Welds: E70

W24x76 (A992)

W10x112 (A992)

Vu

Leh 2 dbLeh = 1.75"2 db = 1.75" OK

Lev must satisfy Spec. Table J3.4

Lev = 1.5"Per Table J3.4:Lev 1.25" OK

Check Limitationsa=3"

Leh=1.75"

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

1/41/4

L = 1

5"

Bolts: 7/8" A325-N

Standard Holes

Welds: E70

W24x76 (A992)

W10x112 (A992)

Vu

Plate or beam web must satisfy: t db/2 + 1/16"

db/2 + 1/16" = 1/2"tp = 3/8" OK

Design Strength: Bolt Shear and Bearing

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

L = 1

5"

Bolts: 7/8" A325-N

Standard Holes

Welds: E70

W24x76 (A992)

Vu

Bolt Shear

7/8" A325-N single shear:

rn = 21.6 kips / bolt

(Table 7-1)

27

Design Strength: Bolt Shear and Bearing

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

L = 1

5"

Bolts: 7/8" A325-N

Standard Holes

Welds: E70

W24x76 (A992)

Vu

Bolt Bearing on 3/8" Plate

Top Bolt: rn = lesser of:

2.4 d t Fu =(0.75)(2.4) (7/8) (3/8) (58) =34.3 kips

OR

1.2 Lc t Fu =(0.75) (1.2) [1.5-(15/16/2)] (3/8) (58) =20. 2 kips controls

Design Strength: Bolt Shear and Bearing

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

L = 1

5"

Bolts: 7/8" A325-N

Standard Holes

Welds: E70

W24x76 (A992)

Vu

Bolt Bearing on 3/8" Plate

Lower four bolts: rn = lesser of:

2.4 d t Fu =(0.75)(2.4) (7/8) (3/8) (58) =34.3 kips controls

OR

1.2 Lc t Fu =(0.75) (1.2) [3-15/16] (3/8) (58) =40.4 kips

Design Strength: Bolt Shear and Bearing

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

L = 1

5"

Bolts: 7/8" A325-N

Standard Holes

Welds: E70

W24x76 (A992)

Vu

Bolt Bearing on W24x76 web

W24x76: tw = .44"

Bearing on beam web does not control by inspection

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Design Strength: Bolt Shear and Bearing

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

L = 1

5"

Bolts: 7/8" A325-N

Standard Holes

Welds: E70

W24x76 (A992)

Vu

Top bolt:Bearing in 3/8" Plate Controls rn = 20.2 kips

Remaining bolts:Bolt Shear Controls rn = 21.6 kips / bolt

Rn = 20.2k + 4 (21.6k) = 106.6 kips

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

L = 1

5"

Vu

Design Strength: Shear Yield of Plate

Rn = (0.9) (Agross) (0.6Fy)

Agross = 3/8" x 15" = 5.625 in2

Rn = (0.9) (5.625) (0.6x36) =

109 kips

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

L = 1

5"

Vu

Design Strength: Shear Fracture of Plate

Rn = (0.75) (Anet) (0.6Fu)

Anet = 3/8" x (15" - 5x1") = 3.75 in2

Rn = (0.75) (3.75)x (0.6x58) =

97.9 kips

P 3/8 x 4 x 15 (A36)L

13.5"

1.75"

Design Strength: Block Shear Rupture of Plate

Anv (0.6 Fu)Agv (0.6 Fy)

Ubs Ant Fu + lesser of Rn = (0.75)

Anv (0.6 Fu) = (3.375) (0.6x58) = 117.5k

Agv (0.6 Fy) = (5.063) (0.6x36) = 109.4k controls

Rn = (0.75) [ (0.469) (58) + 109.4 ] = 102.5 kips

Ant = [ 1.75" - (1") ] (3/8") = 0.469 in2

Anv = [ 13.5" - 4(1") ] (3/8") = 3.375 in2

Agv = (13.5") (3/8") = 5.063 in2

Ubs = 1

29

a=3"Leh=1.75"

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

1/41/4

L = 1

5"

Bolts: 7/8" A325-N

Welds: E70

W24x76 (A992)

W10x112 (A992)

Vu

Design Strength: Weld Fracture

For Conventional Configuration:

Double sided fillet weld is adequate (weld stronger than plate) if: leg size 5/8 tp

5/8 tp = 5/8 x 3/8 = 0.23"

Fillet weld leg size = 1/4"

Weld OK for strength

Minimum weld size (Spec. Table J2.4): 3/16" OK

Weld: OK

Design Strength

Limit State Rn

Bolt shear and bearing

Shear yield of plate

Shear fracture of plate

Block shear rupture of plate

Weld fracture

106.6k

109k

97.9k

102.5k

OK

Controls

Example: LRFDa=3"

Leh=1.75"

4 @ 3"

= 12

"1.5

"1.5

" P 3/8 x 4 x 15 (A36)L

1/41/4

L = 1

5"

Bolts: 7/8" A325-N

Welds: E70

W24x76 (A992)

W10x112 (A992)

Vu

Compute Design Strength of this Single Plate Connection

Rn = 97.9 kips

Vu 97.9 kips

ASD:

Rn / = 97.9k / 1.5 = 65.3 kips

Va 65.3 kips

Rn = 97.9 kips

Rn / = 65.3 kips