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SESSION 3. FRAMING (SHEAR) CONECTIONS. Type I, FR Moment Connection. M = 0.9M. F. Typical Beam Line. Moment, M. Type III, PR Moment Connection. M = 0.5M. F. Type II, Simple Shear. Connection. M = 0.2M. F. Rotation, q. MOMENT ROTATION CURVES. Framing Connections. CONNECTION TYPES.
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SESSION 3 FRAMING (SHEAR) CONECTIONS
Type I, FR Moment Connection M = 0.9M F Typical Beam Line Moment, M Type III, PR Moment Connection M = 0.5M F Type II, Simple Shear Connection M = 0.2M F Rotation, q MOMENT ROTATION CURVES Framing Connections
CONNECTION TYPES • Shear End-Plate • Double Angles • Single Angle • Shear Tab or Single Plate • Tee • Seated – Unstiffened and Stiffened
1 1/4" Minimum Edge Distance End Plate Note : End Plate Thickness Range is 1/4" to 3/8" SHEAR END-PLATE Advantages: • Simple – Few Parts • No Holes in Beam Disadvantages: • Requires Beam to be Cut to Exact Length Comments: • Not commonly used in US; very common in Australia and Europe
Horizontal short slots may be used in angles 2 Angles 2 Angles DOUBLE ANGLES Advantages: • Beam Length can Vary • Weld or Bolt to Beam Disadvantages: • Double Sided Connections into Column Webs are an Erection Problem
2 Angles DOUBLE ANGLES
2 Angles DOUBLE ANGLES
2 Angles Return @ Top DOUBLE ANGLES Double Angle Knife Connection
Return @ Top Bolted and Welded Alternatives SINGLE ANGLE Advantages: • Eliminates Erection Problem • Fewer Parts Disadvantages: • Larger Angle Required • Larger Bolts or Weld Comment: • Not recommended for laterally unbraced beams.
Single Plate SHEAR TAB or SINGLE PLATE Advantages: • Simple – Few Parts • No Welding on Beam Disadvantages: • Stiffer than Other Types • Requires Careful Design Comments: • Two design models available, with very different results.
Min. Clearance "k" Distance + 1/4" Return @ Top Tee Concrete Wall TEE CONNECTION Advantages: • One Sided Disadvantages: • Tee can be Heavy • Stiffer than Other Types except Shear Tab Comment: • Primarily used to connect to concrete wall or existing construction
4" Stabilizer Clip 2" Alternate Clip Position Return @ Top Seat Angle UNSTIFFENED SEATED CONNECTION Advantages: • Few Parts • Few Bolts Disadvantages: • Requires Stability Angle • Limited strength Comment: • Commonly used to connect to the web of a column.
4" Stabilizer Clip 2" Alternate Clip Position Seat Plate Stiffener Optional Trim Lines STIFFENED SEATED CONNECTION Advantages: • Few Parts • Few Bolts Disadvantages: • Requires Stability Angle • Introduces a Column Web Limit State Comment: • Commonly used to connect to the web of a column.
DESIGN CONSIDERATIONS • Where is the pin?
DESIGN CONSIDERATIONS Answer: At the most flexible side of the connection.
DESIGN CONSIDERATIONS • Where is the pin?
DESIGN CONSIDERATIONS • Ductility Considerations • Angle thickness < 5/8 in. • Wide gage • Wide vertical weld spacing with minimum horizontal returns
DESIGN CONSIDERATIONS • Beam Length Tolerance +/- 1/4 in. To accommodate: Setbacks in calculations are usually 1/2 in. End edge distances are taken 1/4 in. less than detailed.
1/2" setback DESIGN CONSIDERATIONS • Beam Length Tolerance
minus 1/4" DESIGN CONSIDERATIONS • Beam Length Tolerance
DESIGN CONSIDERATIONS • Effective Weld Length When a weld terminates in the “air”, the dimensioned weld length is reduced by the weld size for calculations.
L w DESIGN CONSIDERATIONS • Effective Weld Length Shear End-Plate Leff = Lw – 2 tw
NEW LIMIT STATES • Block Shear in Coped Beams - Bolted at Web - Welded at Web • Coped Beam Flexural Strength
Block Shear in Coped Beams Section J4.3 (1999 LRFD Specification) Block Shear Rupture Strength When Fu Ant> 0.6FuAnv: Rn = [0.6FyAgv+FuAnt] < [0.6FuAnv +FuAnt] When Fu Ant < 0.6FuAnv: Rn = [0.6FuAnv+FyAgt] < [0.6FuAnv +FuAnt] f = 0.75
Block Shear in Coped Beams Rn = Ten. Rupture Opp. Yield max Shear Rupturemin Opp. Rupture +
Coped Beam Flexural Strength Mu = Ru e <fb Mn Flexural Yielding fb Mn = 0.90 Fy Snet Snet = net section modulus Local Web Buckling fMn = Fbc Snet
Coped Beam Flexural Strength For Single Cope Limitations: c < 2 d dc< d / 2 fFbc = 23,590 (tw / ho)2 f k < 0.9 Fy f = 2 (c / d) for c / d < 1.0 f = 1 + (c / d) for c / d > 1.0 k = 2.2 (ho / c)1.65 for c / ho< 1.0 k = 2.2 (ho / c) for c / ho > 1.0
Coped Beam Flexural Strength For Double Cope Limitations: c < 2 d dct< 0.2 d dcb< 0.2 d fFbc = 50,840 [tw2 / (c ho)] fd< 0.9 Fy fd = 3.5 – 7.5 (dc / d) dc = max (dct , dcb)
1 " 2 8" 3" V = 40 k u W14x30 Coped Beam Flexural Strength Example: Determine if Adequate. A992 Steel
Coped Beam Flexural Strength Example W14x30 d = 13.8 in. tw = 0.270 in. ho = 13.8 – 3.0 = 10.8 in. Snet = 8.37 in.3 from Table 8-49
Coped Beam Flexural Strength Example fFbc = 23,590 (tw / ho)2 f k < 0.9 Fy c / d = 8.0 / 13.8 = 0.580 < 1.0 f = 2 (c / d) = 2 x 0.580 = 1.16 c / ho = 8.0 / 10.8 = 0.740 < 1.0 k = 2.2 (ho / c)1.65 = 2.2 (10.8 / 8.0)1.65 = 3.61 fFbc = 23,590 (0.270 / 10.8)2 (1.16) (3.61) = 61.7 ksi > 0.9 Fy = 0.9 (50) = 45 ksi
1 " 2 8" 3" V = 40 k u W14x30 Coped Beam Flexural Strength Example fMn = Fy Snet = 0.9 x 50 x 8.37 = 376.6 in.-kips Mu = Vu e = 40.0 (8.5) = 340 in.-kips < 376.6 in.-kips Adequate
1 1/4" Minimum Edge Distance End Plate Note :End Plate Thickness Range is 1/4" to 3/8" SHEAR END-PLATES
Shear End-Plate Limit States Beam: Beam Gross Shear Coped Beam Flexural Strength Web Strength at Weld Weld: Weld Rupture
2 2 1 1 3 3 4, 5 3 3 1 1 2 2 Shear End-Plate Limit States Plate: 1. Gross Shear 2. Net Shear 3. Block Shear 4. Bearing and Tear Out Bolts: 5. Bolt Shear Girder or Column: Bearing and Tear Out
8" 3" 1 3 " 2 1 1 " 4 2@3" W14x30 f V n 1 A992 1 " 4 PL 1/4 x 6 x 0'-8 1/2" 3/16 A36 Shear End-Plate Example Determine f Vn. 3/4” A325-N Bolts, E70XX
Shear End-Plate Example W14x30 Fy = 50 ksi Fu = 65 ksi d = 13.84 in. tw = 0.27 in. Beam Limit States Beam Shear Yielding fVn = 0.9 (0.6 Fy) ho tw = 0.9 (0.6 x 50) (10.84) (0.27) = 79.0 k
Shear End-Plate Example Coped Beam Flexural Strength From previous example fMn = 376.6 in.-kips with e = cope length + plate thickness = 8.0 + 0.25 = 8.25 in. f Vn = 376.6 / 8.25 = 45.6 k
Shear End-Plate Example Beam Web Strength at Weld Plate L = 8.5 in. tweld = 3/16 in. fVn = 0.9 (0.6 Fy) (L - 2 tweld) tw = 0.9 (0.6 x 50) [8.5 – (2 x 3/16)] (0.27) = 59.2 k (Shear rupture will not control.)