Transcript shear

SESSION 3
FRAMING (SHEAR)
CONECTIONS
1
MOMENT ROTATION CURVES
Type I, FR Moment Connection
M = 0.9M F
Moment, M
Typical Beam Line
Type III, PR Moment Connection
M = 0.5M F
M = 0.2M F
Type II, Simple Shear
Connection
Framing Connections
Rotation, q
2
CONNECTION TYPES
•
•
•
•
•
•
Shear End-Plate
Double Angles
Single Angle
Shear Tab or Single Plate
Tee
Seated – Unstiffened and Stiffened
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SHEAR END-PLATE
1 1/4" Minimum
Edge Distance
Advantages:
• Simple – Few Parts
• No Holes in Beam
End Plate
Disadvantages:
Note: End Plate Thickness Range is 1/4" to 3/8"
• Requires Beam to be Cut to Exact Length
Comments:
• Not commonly used in US; very common
in Australia and Europe
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DOUBLE ANGLES
Horizontal short slots
may be used in angles
Advantages:
• Beam Length
2 Angles
can Vary
• Weld or Bolt to Beam
Disadvantages:
• Double Sided Connections into Column
Webs are an Erection Problem
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DOUBLE ANGLES
2 Angles
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DOUBLE ANGLES
2 Angles
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DOUBLE ANGLES
2 Angles
Return
@ Top
Double Angle Knife Connection
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SINGLE ANGLE
Return
@ Top
Advantages:
• Eliminates Erection
Problem
• Fewer Parts
Disadvantages:
• Larger Angle Required Bolted and Welded Alternatives
• Larger Bolts or Weld
Comment:
• Not recommended for laterally unbraced
beams.
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SHEAR TAB or SINGLE PLATE
Advantages:
Single Plate
• Simple – Few Parts
• No Welding on Beam
Disadvantages:
• Stiffer than Other Types
• Requires Careful Design
Comments:
• Two design models available, with very
different results.
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TEE CONNECTION
Min. Clearance
"k" Distance + 1/4"
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
Return
@ Top
Tee
Concrete Wall
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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"
Return
@ Top
Alternate
Clip Position
Seat Angle
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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.
4"
Stabilizer
Clip
2"
Alternate
Clip Position
Seat Plate
Stiffener
Optional
Trim Lines
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DESIGN CONSIDERATIONS
• Where is the pin?
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DESIGN CONSIDERATIONS
Answer: At the most flexible side of the
connection.
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DESIGN CONSIDERATIONS
• Where is the pin?
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DESIGN CONSIDERATIONS
• Ductility Considerations
– Angle thickness < 5/8 in.
– Wide gage
– Wide vertical weld spacing with
minimum horizontal returns
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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.
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DESIGN CONSIDERATIONS
• Beam Length Tolerance
1/2" setback
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DESIGN CONSIDERATIONS
• Beam Length Tolerance
minus 1/4"
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DESIGN CONSIDERATIONS
• Effective Weld Length
When a weld terminates in the “air”, the
dimensioned weld length is reduced by
the weld size for calculations.
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DESIGN CONSIDERATIONS
• Effective Weld Length
Shear
End-Plate
Lw
Leff = Lw – 2 tw
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NEW LIMIT STATES
• Block Shear in Coped Beams
- Bolted at Web
- Welded at Web
• Coped Beam Flexural Strength
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Block Shear in Coped Beams
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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]
 = 0.75
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Block Shear in Coped Beams
Rn =
Ten. Rupture +
Opp. Yield
max Shear Rupture min Opp. Rupture
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Single Coped Beam
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Double Coped Beam
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Coped Beam Flexural Strength
Mu = Ru e < b Mn
Flexural Yielding
b Mn = 0.90 Fy Snet
Snet = net section modulus
Local Web Buckling
Mn =  Fbc Snet
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Coped Beam Flexural Strength
For Single Cope
Limitations: c < 2 d
dc < d / 2
Fbc = 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
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Coped Beam Flexural Strength
For Double Cope
Limitations: c < 2 d
dct < 0.2 d
dcb < 0.2 d
Fbc = 50,840 [tw2 / (c ho)] fd < 0.9 Fy
fd = 3.5 – 7.5 (dc / d)
dc = max (dct , dcb)
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Coped Beam Flexural Strength
Example: Determine if Adequate.
1"
2
8"
3"
A992 Steel
Vu = 40 k
W14x30
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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
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Coped Beam Flexural Strength Example
Fbc = 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
Fbc = 23,590 (0.270 / 10.8)2 (1.16) (3.61)
= 61.7 ksi > 0.9 Fy = 0.9 (50) = 45 ksi
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Coped Beam Flexural Strength Example
1"
2
8"
Mn =  Fy Snet
= 0.9 x 50 x 8.37
= 376.6 in.-kips
Vu = 40 k
3"
W14x30
Mu = Vu e = 40.0 (8.5)
= 340 in.-kips < 376.6 in.-kips
Adequate
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SHEAR END-PLATES
1 1/4" Minimum
Edge Distance
End Plate
Note: End Plate Thickness Range is 1/4" to 3/8"
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Shear End-Plate Limit States
Beam:
Beam Gross Shear
Coped Beam Flexural Strength
Web Strength at Weld
Weld:
Weld Rupture
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Shear End-Plate Limit States
Plate:
1. Gross Shear
2. Net Shear
3. Block Shear
4. Bearing and Tear Out
Bolts:
3
5. Bolt Shear
Girder or Column:
Bearing and Tear Out
32 11 23
4, 5
3
2 11 2
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Shear End-Plate Example
Determine  Vn.
3/4” A325-N Bolts, E70XX
8"
3"
3 12 "
114"
2@3"
114"
Vn
PL 1/4 x 6 x 0'-8 1/2"
A36
W14x30
A992
3/16
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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
Vn = 0.9 (0.6 Fy) ho tw
= 0.9 (0.6 x 50) (10.84) (0.27)
= 79.0 k
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Shear End-Plate Example
Coped Beam Flexural Strength
From previous example
Mn = 376.6 in.-kips
with e = cope length + plate thickness
= 8.0 + 0.25 = 8.25 in.
 Vn = 376.6 / 8.25 = 45.6 k
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Shear End-Plate Example
Beam Web Strength at Weld
Plate L = 8.5 in.
tweld = 3/16 in.
Vn = 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.)
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Shear End-Plate Example
Weld Limit State
- 3/16 in. Fillet Weld
Minimum Weld Size 3/16 in. OK
Vn = (D x 1.392) (L - 2 tweld)
= (2 x 3 x 1.392) [8.5 – (2 x 3/16)]
= 67.9 k
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Shear End-Plate Example
Plate Limit States:
tp = 1/4 in.
A36 Steel
114"
Fy = 36 ksi
2@3"
Fu = 58 ksi
1
14"
Plate Gross Shear
Vn = 0.9 (0.6 Fy) (2 L tp)
= 0.9 (0.6 x 36) (2 x 8.5 x 1/4)
= 82.6 k
3 2"
6"
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Shear End-Plate Example
Plate Net Shear
dh´ = 3/4 + 1/16 + 1/16 = 7/8 in.
1
An = (8.5 - 3 x 7/8) (1/4)(2)
14"
= 2.94 in.2
2@3"
Vn = 0.75 (0.6 Fu) (An)
= 0.75 (0.6 x 58) (2.94)
= 76.7 k
3 12 "
114 "
6"
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Shear End-Plate Example
Plate Block Shear
PL 1/4 x 6 x 0’-8 1/2”
114 "
3"
3"
1
14"
Rn =
Ten. Rupture
Opp. Yield
+
max Shear Rupture min Opp. Rupture
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Shear End-Plate Example
114 "
Plate Block Shear
- Tension Rupture
3"
Fu Ant = 58(1.25 - 0.5 x 7/8)(2 x 1/4) 3"
= 23.6 k
- Shear Rupture
114 "
0.6FuAnv = (0.6 x 58) (7.25 – 2.5 x 7/8) (2 x 1/4)
= 88.1 k
Shear Rupture Controls
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Shear End-Plate Example
- Opposite Rupture
Fu Ant = 23.6 k
- Opposite Yield
Fy Agt = 36 (1.25) (2 x 1/4)
= 22.5 kips
Opposite Yield Controls
114 "
3"
3"
114 "
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Shear End-Plate Example
Plate Block Shear
114 "
Vn = 0.75 (Shear Rupture
3"
+ Tension Yield)
3"
= 0.75 (88.1 + 22.5)
= 83.0 k
114 "
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Shear End-Plate Example
Plate Bearing / Tear Out
Brg: 2.4 Fu db t = (2.4 x 58) (3/4 x 1/4)
= 26.1 k
Edge: 1.2 Fu Lc t = (1.2 x 58) (1.25 – 13/32) (1/4)
= 13.6 k < 26.1 k
Other: 1.2 Fu Lc t = (1.2 x 58) (3 – 13/16) (1/4)
= 38.1 k > 26.1 k
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Shear End-Plate Example
3 12 "
Plate Bearing / Tear Out
Vn =  ( 2 x Edge + 4 x Other)
= 0.75 (2 x 13.6 + 4 x 26.1)
= 98.7 k
114"
2@3"
114"
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Shear End-Plate Example
Bolt Rupture
A325-N Fv = 48 ksi
Vn = 0.75 Fv SAb
= 0.75 (48) (6 x 0.4418)
= 6 x 15.9
= 95.4 k
3 12 "
114"
2@3"
114"
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Shear End-Plate Example
8"
3"
3 12"
114"
2@3"
114"
Vn
PL 1/4 x 6 x 0'-8 1/2"
A36
W14x30
A992
3/16
Coped Beam Flexural Strength Controls
Vn = 45.6 k
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DOUBLE ANGLE
CONNECTIONS
Welded / Bolted
Horizontal short slots
may be used in angles
2 Angles
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Welded / Bolted Double Angles
Assumption:
Pin is at face of supporting element
Beam web weld is
subjected to
eccentric shear
2 Angles
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Welded / Bolted Double Angles
Possible Limit States
Beam:
Shear Yielding
Coped Beam Flexural Strength
Block Shear
Web Strength at Weld
Weld:
Weld Rupture Due to Eccentric Shear
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Welded / Bolted Double Angles
Angles:
Gross Shear
Net Shear
Block Shear
Bearing / Tear Out
Bolt:
Shear Rupture
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Welded / Bolted Double Angles
Example: Determine  Vn for the limit
states of:
1. Beam Web Block Shear
2. Weld Rupture due to Eccentric Shear
3. Beam Web Strength at Weld
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Welded / Bolted Double Angles
Example
1
14 "
3"
3"
11
4"
1"
2
2"
1"
4
W14x30 A992
tw = 0.27 in.
2L 3 x 3 x 5/16 x 0'-8 1/2” A36
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Welded / Bolted Double Angles
Example
1. Beam Web Block Shear
0.6 Fu Anv = 0.6 (65) (8.75 x 0.27)
= 92.1 k
Shear Area
Tension area
Fu Ant = (65) (3 - 1/2 - 1/4) (0.27)
= 39.5 k < 92.1 k (Shear Rupture Controls)
Fy Agt = (50) (3 - 1/2 - 1/4) (0.27)
= 30.4 k < 39.5 k (Tension Yield Controls)
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Welded / Bolted Double Angles
Example
Beam Web Block Shear Strength
Vn = 0.75 (max rupture + min opp.)
= 0.75 (Shear Rupture + Tension Yield)
= 0.75 (92.1 + 30.4) = 122.5 k
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Welded / Bolted Double Angles
Example
2. Weld Rupture Due to Eccentric Shear
e x = aL
Dimensions
from Table 8-42
V n
c.g.
L
xL
k, xL, a, etc.
 Vn = C C 1 D L
kL
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Welded / Bolted Double Angles
Example
2. Weld Rupture Due to Eccentric Shear
Vn = C C1 D L
Where: C = effective weld coefficient
C1 = Fu of weld metal / 70
D = number of 1/16 ths
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Welded / Bolted Double Angles
Example
k = 2.5 / 8.5 = 0.29  x = 0.053 e = aL
x
xL = 0.053 x 8.5 = 0.45 in.
a = (3.0 – 0.45) / 8.5 = 0.3
V n
c.g.
L
xL
kL
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Welded / Bolted Double Angles
Example
Using Table 8-42
C = 2.06
C1 = 1.0
D=3
e x = aL
V n
Vn = C C1 D L
= (2.06) (1.0) (2 x 3) (8.5)
= 105.6 k
c.g.
L
xL
kL
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Welded / Bolted Double Angles
Example
3. Beam Web Strength at Weld
 0.6 Fu t w (1.0)
Vn 
 105.6
(1.392) (6) (1.0)
0.75 (0.6  65) (0.27) (1.0)

 105.6
(1.392) (6) (1.0)
 99.8 k
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END OF SESSION 3
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