# COURSE OUTLINE - Rowan University

SESSION 4 FRAMING (SHEAR) CONNECTIONS CONTINUED 1 SEMINAR OUTLINE Session 4 Framing Connections Continued Session 5 Moment Connections Session 6 End-Plate Moment Connections/ Bracing Connections Closure 2

Bolted / Bolted Double Angles " 1 14 3" 3" " 1 14 1" 2 3"

1" 2 W14x30 A992 tw = 0.27 in. 2L 5 x 3 x 5/16 x 0'-8 1/2A36 3/4 A325-N Bolts 3 Bolted / Bolted Double Angles Eccentricity not considered in these connections For Bolt Rupture: Vn = Fv Ab x 6 No New Limit States

4 Bolted / Welded Double Angles 2L 3 x 3 x 5/16 x 0'-8 1/2" A36 W14x30 A992 tw = 0.27 in. 2" 3/4 A325-N Bolts E70XX 812"

4" 1/4" Knife Connection Return @ Top 5 Bolted / Welded Double Angles Knife Connection Beam to Column Flange Bottom Cope to Permit Erection

New Limit States: Coped Beam Web Strength at Tension Flange Weld Strength on OSLs 6 Bolted / Welded Double Angles Coped Beam Web Strength at Tension Flange W14x30 Vn = b Fy Snet / e b = 0.9 2"

Snet from Table 8-49 4" 7 Bolted / Welded Double Angles Weld Strength on OSLs e O Vn /2 L/6 L

Tension, ft CL Web 8 Bolted / Welded Double Angles fv e Vn 2

L Mo 0 1 5 2 Vn ft L L e 2 6 3 2 O Vn /2 L/6 L

Tension, ft CL Web Vn e ft 1.8 2 L 9 Bolted / Welded Double Angles e 2

t fw f f Vn 2 2L 2 v O 2 L 12.9 e

2 Vn /2 L/6 L Tension, ft CL Web 2

Vn 2 L (1.392 D) 2 L 12.9 e 2 10 Bolted / Welded Double Angles Example: Calculate Vn based on the weld rupture strength of the OSLs 2L 3 x 3 x 5/16 x 0'-8 1/2" A36

W14x30 A992 tw = 0.27 in. 2" e 812 " 1/4" e = 3 + 0.27/2 =3.135 in. L = 8.5 in. Return @ Top

3/4 A325-N Bolts E70XX 11 Bolted / Welded Double Angles Example 2 Vn 2 L (1.392 D) 2 L 12.9 e

2 2 2 (8.5) (1.392 4) 8.5 2 12.9 (3.135) 2 57.0 k Note: Weld returns at top of angles have been neglected 12

SHEAR TAB OR SINGLE PLATE Single Plate 13 Shear Tab or Single Plate Bolt Line Mb Mw Mw = Vu ew Mb = Vu eb

a eb ew 14 Shear Tab or Single Plate Rotation is obtained by bolts plowing the plate, which requires limiting the plate thickness Single Plate 15

Shear Tab or Single Plate Geometric Limitations: tp < db /2 + 1/16 Lh Lv Lh > 1 1/2 in. Lv > 1 1/2 in. 2 1/2 in. < a < 3 1/2 in. 2 < n = No. of Bolts < 9 L > T/2 Plate Material: A36 Steel

[email protected]" L a 16 Shear Tab or Single Plate Eccentricities depend on: i) Connected Element Rigid Flexible ii) Hole Type Standard Short Slots 17

Shear Tab or Single Plate 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. 18 Shear Tab or Single Plate Flexible Support / Standard Holes eb = |(n - 1) a| > a

Flexible Support / Short Slotted Holes eb = |(2n/3) a| > a Bolt Line a eb 19 Shear Tab or Single Plate Rigid Support / Standard Holes eb = |(n - 1) a|

Rigid Support / Short Slotted Holes eb = |(2n/3) a| Bolt Line a eb 20 Shear Tab or Single Plate Weld Strength E70XX Electrode Weld Size > 3/4 tp on each side

Weld size is then sufficient to develop plate in tension yielding 21 Shear Tab or Single Plate Plate Limit States: Shear Yielding Shear Rupture Block Shear Bearing / Tear Out Plate Buckling 22

Shear Tab or Single Plate Plate Buckling L tp > L / 64 > 1/4 in. Compression 23 Shear Tab or Single Plate Example: Determine required plate and weld sizes. E70XX

W14x30 tw = 0.27 in T = 12 in. Vu = 40 k 3/4 in. A325-N Bolts 24 Shear Tab or Single Plate Example Try 3 3/4 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. 25 Shear Tab or Single Plate Example 112" Plate Geometry [email protected]" 9" > T/2 Check Plate Buckling L / 64 = 9 / 64 = 0.14 < 1/4 OK

112" 3" 1 12 " 26 Shear Tab or Single Plate Example Check bolts for eccentric loading: Rigid Support / Standard Holes eb = |(n - 1) a| = |(3 1) 3| = 1.0 in. Using Table 8-18 with s = 3 in. ex = 1.0 in.

n=3 By extrapolation: C = 2.71 27 28 Shear Tab or Single Plate Example Vn = C x rv = 2.71 x 15.9 = 43.1 k > Vu = 40 k OK 29 Shear Tab or Single Plate Example

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 30 Shear Tab or Single Plate Example 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 31 Shear Tab or Single Plate Example 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 32 Shear Tab or Single Plate Example 112"

PL 1/4 x 4 1/2 x 0'-9" [email protected]" 9" 1 12 " 3/16 3" 1 12 " 33 SINGLE ANGLE CONNECTIONS

Horizontal short slots may be used in angle Return @ Top One Angle Bolted and Welded Alternatives 34 Single Angle Connections Eccentricity Assumptions for OSL e a =e b

CL Web Single Row b ea Bolts Angle CL Web Double Row

35 Single Angle Connections Eccentricity Assumptions for OSL 2 t weld Return ew + c.g.weld CL Web Welded

36 Single Angle Connections 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 on the supporting member side. 37 Single Angle Connections Additional Limit States for Bolted Connections

ea Flexural Yielding Vn = 0.9 Fy Sg / ea L Sg = tp L2/6 38 Single Angle Connections ea

Flexural Rupture Vn = 0.75 Fu Snet / ea L Snet: See LRFD Manual, Table 12-1 39 40 Single Angle Connections Eccentric Shear of Bolt Group (Instantaneous Center of Rotation Method)

Vn Vn Table 8-18 Table 8-19 Vn = C (rv) 41 Single Angle Connections Bearing and Tear Out Vn = C (rvb) where rvb is the bearing / tear out strength at the outermost bolt

42 Single Angle Connections db (in.) Min. t (in.) 3/4 7/8 1 3/8 3/8 1/2

43 Single Angle Connections Additional Limit States for Welded Connections Eccentric Shear Strength of Weld 2 tweld Return V n V n Table 8-44

44 45 Welded Unstiffened Seat Connections Stabilizing Angle L4x4x1/4 Alternate Location 46 Welded Unstiffened Seat

Connections Limit States: Beam Local Web Yielding Beam Local Web Crippling Seat Angle Bending Seat Angle Shear Yielding Weld Rupture 47 Beam Local Web Yielding

N+2.5k N 48 Beam Local Web Yielding N+2.5k Section K1.3 Web Local Yielding = 1.0 Rn = (2.5kdesign + N)Fyw tw

N 49 Beam Local Web Crippling 50 Beam Local Web Crippling Section K1.4 Web Local Crippling @

R n 68 t w 1 3 d tf N d 0.2 1.5 4N tw 2

R n 68 t w 1 - 0.2 d tf Fy t f tw 1.5 Fy t f

tw 51 Design Model for Angle Flexure The N-distance is determined from the limit states of web yielding and web crippling, but not less than kdetailing. Supported Beam 3" 4

N Critical Section for Bending, Shear r a = 3/8" ta Supporting Column 52 Design Model for Angle Flexure Setback = 1/2 in. Beam Tolerance = 1/4 in.

Supported Beam 3" 4 Use 3/4 in. setback in calculations N Critical Section for Bending, Shear r a = 3/8" ta

Supporting Column 53 Design Model for Angle Flexure Supported Beam The maximum value of N is then used to determine the eccentricity: e = N/2 + (3/4 3/8) ta = N/2 + 3/8 -ta 3"

4 N Critical Section for Bending, Shear r a = 3/8" ta Supporting Column 54 Design Model for Angle Flexure For the Limit State of Web Yielding

N min Ru 2.5 k design 1.0 Fy t w 55 Design Model for Angle Flexure For the Limit State of Web Crippling when N/d < 0.2 N min

Ru d 2 3 0.75 (68t w ) t tw 1 f Fy t f t w

1.5 when N/d > 0.2 N min Ru d 4 0.75 (68t 2w ) t tw

1 f Fy t f t w 1.5 0.2 56 Design Model for Angle Flexure Required angle thickness

from OSL bending t req 4 Ru e 0.9 Fy La with La = angle length e = N/2 + 3/8 - ta Supported Beam 3" 4

N Critical Section for Bending, Shear r a = 3/8" ta Supporting Column 57 Angle Shear Yielding R u R n 0.9( 0.6Fy )L a t angle Therefore

t req Ru 0.9( 0.6Fy )L a 58 Weld Rupture Supported Beam The weld is subjected to eccentric shear. Table 8-38 of LRFD Vol. II will be used to

determine required weld size. 3" 4 N Critical Section for Bending, Shear r a = 3/8" ta Supporting Column

59 Weld Rupture 60 Stiffened Seated Connections 4" Stabilizer Clip 2" Alternate Clip Position

Seat Plate Stiffener Optional Trim Lines 61 Stiffened Seated Connections Limit States: Beam Web Yielding Beam Web Crippling Strength of Stiffener Plate Eccentric Shear of Connecting Side Weld or Bolts Column Web Failure 62

Stiffened Seated Connections 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 and the weld size. 63 Stiffened Seated Connections For Unstiffened Beams: Seat Stiffener Beam, Fy

Fy 36 36 50 36 50 50 ts tw 1.4 tw tw For Stiffened and Unstiffened Beams Seat Stiffener:

ts (36 ksi) > 2 tweld ts (50 ksi) > 1.5 tweld 64 Stiffened Seated Connections Column Web Failure Stress Concentration Stress concentration requires stiff web. 65 Stiffened Seated Connections Column Web Failure

Also need to prevent flange rotation. 66 Stiffened Seated Connections LRFD Vol. II has a simplified approach for sections heavier than: 43 lb/ft for W14 40 lb/ft for W12 30 lb/ft for W10 24 lb/ft for W8 ( > W14x43) ( > W12x40) ( > W10x30)

( > W8x24) 67 Stiffened Seated Connections 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 68 Stiffened Seated Connections

69 END OF SESSION 4 Design Examples (For Home Study) 70 Single Angle Connections Example: Is the connection adequate? L8 x 6 x 3/8 x 1-3 A36 112 " W21x50 A992

t w = 0.38 in. Vu = 90 k [email protected]" Vu = 90 k 3/4 in. A325N Bolts 114 " 112 " 3" 3"

134 " 2" 3" OSL 71 Single Angle Connection Example Angle Flexural Yield Supporting Side OSL Controls Vn = 0.9 Fy Sg / ea a

Sg = (3/8) (15.0)2 / 6 = 14.06 in3 ea = 0.38 / 2 + 3 = 3.19 in. Vu Vn = 0.9 Fy Sg / ea = 0.9 (36) (14.06) / 3.19 = 142.9 k > 90 k OK 72 Single Angle Connection Example Angle Flexural Rupture From Table 12-1 Snet = 10.1 in3

73 74 Single Angle Connection Example Vn = 0.75 Fu Snet / ea = 0.75 (58) (10.1) / 3.19 = 137.7 k > 90 k OK 75 Single Angle Connection Example Bolt Rupture eb = 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 eb Vu 3" 3" 76 77

Single Angle Connection Example 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 78 Single Angle Connection Example Bearing / Tear Out Critical Bolts Brg. = 2.4 (58) (3/4 x 3/8) = 39.2 k

T.O. = 1.2 (58) (1.5 13/32) (3/8) = 28.5 k rbv = 0.75 (28.5) Vu = 21.4 k > rv = 15.9 k OK 79 Single Angle Connection Example 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. 80 Single Angle Design Example Example: Outstanding leg of previous connection welded. Determine required weld size. Column, tf = 0.710 in. 2 t weld Return L6 x 6 x 3/8 x 1'-3" Vu = 90 k E70XX 81

Single Angle Connection Example Table 8-44 with l = 15 in. xl kl = 6 in. k = 6 / 15 = 0.4 x = 0.057 xl = 0.057 x 15 = 0.855 in. V = 90 k u al = 0.38/2 + 6 0.855 = 5.34 in. a = 5.34 / 15 = 0.356 al

15" 0.19" 82 83 Single Angle Connection Example C = 1.64 by interpolation Dreqd = 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

84 Unstiffened Seat Connection Ex. Example: CL Web W14x90 Column 0.440" W18x46 A992 Vu = 35 k L4 x 4 x 5/8 x 0'-8"

W18x46 bf = 6.06 in. t f = 0.605 in. d = 18.06 in. t w = 0.360 in. kdetailing = 1 1/4 in. kdesign = 1.01in. ? 85 Unstiffened Seat Connection Ex. Example: Determine (1) if seat angle is adequate, and

(2) required weld size. A36 Angle E70xx Electrode 86 Unstiffened Seat Connection Ex. For the Limit State of Web Yielding N min Ru 2.5 k design 1.0 Fy t w

35 2.5x1.01 1.0x 50x 0.360 0.58 k detailing 87 Unstiffened Seat Connection Ex. For the Limit State of Web Crippling assuming N/d < 0.2 N min d Ru

3 0.75( 68t 2w ) 18.06 3 1. 5 tf tw 1 Fy t f t w

35 2 0 . 75 ( 68 x 0 . 360 )

0.605 0.360 1 50x0.605 0.360 5.5 in. 88 1.5 Unstiffened Seat Connection Ex. 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 89 Unstiffened Seat Connection Ex. Required angle thickness from OSL bending e = N/2 + 3/8 ta = 1.25/2 + 3/8 5/8 = 0.375 in. Supported Beam

3" 4 N Critical Section for Bending, Shear r a = 3/8" ta Supporting Column 90 Unstiffened Seat Connection Ex. With La = 8 in.

t req 4R u e 0.9Fy L a 4x 35x0.375 0.9x 36x8.0 = 0.45 in. < 5/8 in. OK 91 Unstiffened Seat Connection Ex. Required angle thickness from

OSL Shear Yielding: t req Ru 0.9( 0.6Fy )L a 35 0.9( 0.6x 36)8.0 0.225 in . 5/8 in. OK 92 Unstiffened Seat Connection Ex.

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 93 Unstiffened Seat Connection Ex. 94 Unstiffened Seat Connection Ex. 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. 95 Unstiffened Seat Connection Ex. CL Web W14x90 Column 0.440" W18x46 A992 Vu = 35 k

L4 x 4 x 5/8 x 0'-8" 1/4 96 Stiffened Seat Example Determine: 1. A36 stiffener thickness 2. E70XX weld size. W24x68 A36 t wb = 0.415 in. W14x90 A992 t wc= 0.440 in.

Ru = 80 k ts W L 97 Stiffened Seat Example W14x90 Satisfies column web requirement. W24x68 A992 t wb = 0.415 in. W14x90

A992 t wc= 0.440 in. ts W L 98 99 Stiffened Seat Example Web Crippling

Using the Factored Uniform Load Table in Part-4 assuming N/d < 0.2 Wmin R u R 3 setback R 4 80 73.7 0.75 1.13 0.75 1.88 in. 5.57 Check N/d = 1.20 / 23.73 = 0.05 < 0.2 OK (Or calculate as in Session 2.)

100 Stiffened Seat Example Web Yielding Wmin R u R 1 setback R 2 80 71.3 0.75 0.418 0.75 1.17 in. 20.8

Use W = 4 in. and t = 3/8 in. Bolt location rule will be satisfied. 101 Stiffened Seat Example R u = 80 k With: W = 4 in. 1/4 in. Weld 3" 4 Using Table 9-9:

L = 10 in. Rn = 82.3 > Ru = 80 k 10" 4" 102 Stiffened Seat Example 103 Stiffened Seat Example 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 104 Stiffened Seat Example 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 105

Stiffened Seat Example < 34" PL 3/8 x 5 x 0'-8" 10" 1/4 PL 5/8 x 5 x 0'-10" 4" 106

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