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November 6, 2019
Revised: Nov 8, 2019
Fundamentals of Connection Design
Session 3: Shear Connections, Part I
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
November 6, 2019
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American Institute of Steel Construction
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AISC Live Webinar Series
November 6, 2019
Revised: Nov 8, 2019
AISC Live Webinars
Fundamentals of Connection Design
Session 3: Shear Connections, Part I
AISC Live Webinars
Copyright Materials
Course Description
This presentation is protected by US and International Copyright laws. Reproduction,
distribution, display and use of the presentation without written permission of AISC is
prohibited.
Shear Connections, Part I
November 6, 2019
© The American Institute of Steel Construction 2019
The information presented herein is based on recognized engineering principles and is
for general information only. While it is believed to be accurate, this information should
not be applied to any specific application without competent professional examination
and verification by a licensed professional engineer. Anyone making use of this
information assumes all liability arising from such use.
This session will provide an overview of a variety of shear connection types, addressing
the advantages and disadvantages of each. The limit states for block shear and flexural
strength in coped beams will be presented. Shear end-plate and double-angle
connection designs will also be discussed. Design examples will be presented to
demonstrate the concepts.
AISC Live Webinars
Learning Objectives
• Identify several types of shear connections.
Fundamentals of Connection Design
• Explain where the point of rotation is modeled for various shear connections.
Session 3: Shear Connections, Part I
November 6, 2019
• List the steps in designing shear end-plate connections.
• List the steps in designing double-angle connections.
Brad Davis, PhD, SE
Associate Professor, University of Kentucky
Owner, Davis Structural Engineering
Copyright © 2019
American Institute of Steel Construction
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AISC Live Webinar Series
November 6, 2019
Revised: Nov 8, 2019
Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Schedule
•
•
•
•
October 23, 2019
October 30, 2019
November 6, 2019
November 13, 2019
Fundamental Concepts Part I
Fundamental Concepts Part II
Shear Connections Part I
Shear Connections Part II
SHEAR
CONNECTIONS
PART I
9
10
Topics
Connection Classification
• Types of Shear Connections
• Design Considerations
• Additional Limit States for Shear
Connections
• Shear End-Plate Connections
• Double-Angle Connections
qs
qs
M
MF
q, M
q, M
MF
MF
0.8qs
0.2MF
q
11
qs
Manual Figure 10-1
12
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American Institute of Steel Construction
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AISC Live Webinar Series
November 6, 2019
Revised: Nov 8, 2019
Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Types of Shear Connections
•
•
•
•
•
•
•
Design Considerations
• Shear connection design assumes the
connection is pinned.
• Where is the pin?
Shear End-Plate
Double-Angle
Single-Angle
Single-Plate or Shear Tab
Tee Shear Connections
Unstiffened Seated Connections
Stiffened Seated Connections
13
Design Considerations
14
Design Considerations
• Where is the pin?
Answer: At the most flexible side of the
connection.
• Where is the pin?
2 Angles
15
16
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Design Considerations
Design Considerations
• Ductility Considerations
• Beam Length Tolerance +/- 1/4 in.
– Angle or end plate thickness < 5/8 in.
+/- ¼ in.
– Wide gage
– Wide vertical weld spacing
– Avoid welding along the top of angles or end plate
Beam Length +/- ¼ in.
• Stability Consideration
Setbacks are usually 1/2 in.
In calcs, end edge distances are taken as 1/4 in. less
than detailed.
– Depth of Connection > T / 2
(T is clear distance between fillets – Tabulated
in Manual Table 1-1)
17
Design Considerations
18
Design Considerations
• Beam Length Tolerance + 1/4 in.
• Beam Length Tolerance + 1/4 in.
Minus 1/4” in Design Calcs
1/2" setback
19
20
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Design Considerations
Add’l Limit States for Shear Connections
• Effective Weld Length
• Block Shear in Coped Beams
When a weld terminates in the “air,” the
dimensioned weld length is reduced by the weld
size for calculations except for angles welded to a
beam web.
- Bolted at Web
- Welded at Web
• Coped Beam Flexural Strength
Lw
Leff = Lw – 2 w
Shear End-Plate
Double-Angle
21
Block Shear in Coped Beams
22
Block Shear in Coped Beams
Specification Section J4.3
ϕ = 0.75
Rn = 0.6FuAnv + UbsFuAnt ≤ 0.6FyAgv +UbsFuAnt
Equivalent to:
Rn  min
 min
23
Shear Rupture
Shear Yielding
0.6 Fu Anv
0.6 Fy Agv
 Tensile Rupture
 U bs Fu Ant
Ubs = 1.0 when tensile stress is uniform
= 0.5 otherwise
24
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Block Shear in Coped Beams
Ubs = 1.0
Ubs = 1.0
Coped Beam Flexural Strength
Ubs = 0.5
Single Cope
More Examples in Commentary Figure C-J4.2
Evaluation Criterion
Limit States
Mu = Ru e < ϕbMn
ϕb = 0.9
• Flexural Yielding (C or T)
• Web Flexural Local
Buckling (Single Cope)
• Web LTB (Double Cope)
25
Coped Beam Flexural Strength
Double Cope
26
Single Coped Beam Flexural Strength
Single Coped Beam Flexural Strength
Single Coped Beam Flexural Strength
Manual Pages 9-6 through 9-9
λ ≤ λp
Mn = Mp = FyZnet
λp < λ ≤ 2λp
Mn = Mp - (Mp-My)(λ/λp-1)
λ > 2λp
Mn = FcrSnet
λ = web slenderness = ho/tw
(Manual 9-6)
(Manual 9-7)
(Manual 9-8)
 p  0.475 k1E / Fy
k1  max
fk
1.61
1.65
h 
k  2.2  o 
c
h
k  2.2 o
c
where
Fcr  0.903Ek1 / 2
Snet = net elastic section modulus at the cope, in.3
Znet = net plastic section modulus at the cope, in.3
27
Single Cope
c
 1.0
ho
c
if
 1.0
ho
if
c
 1.0
d
f 
1 c / d
c
min
if  1.0
3.0
d
2
c
d
if
28
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Double Coped Beam Flexural Strength
Double Coped Beam Flexural Strength
Web Lateral-Torsional Buckling
Manual Page 9-9 and Specification Section F11
Web Lateral-Torsional
Buckling, Manual Page 9-9
λ ≤ λp
Mn = Mp = FyZ ≤ 1.6FyS
λp < λ ≤ λr
Mn = Cb[1.52-0.274λ(Fy/E)]My ≤ Mp
λ > λr
Mn = (1.9ECb/ λ)Sx ≤ Mp
where
Z  twho2 / 4
  Lb ho / tw2
If cb ≥ ct
 p  0.08E / Fy
(Spec. F11-1)
(Spec. F11-2)
(Spec. F11-3)
Note: First printing
of the Manual Page
9-9 shows < . Should
be > as shown here.
29
Single Cope Flexural Strength Example
W14x30
1/2"
8"
3"
d = 13.8 in. tw = 0.270 in.
bf = 6.73 in. tf = 0.385 in.
Vu = 40 kips
W14x30
ho = 13.8 – 3.0 = 10.8 in.
Snet = 8.37 in.3 from Manual Table 9-2
Znet = 15.1 in.3 from Companion to the AISC Steel
Construction Manual, Volume 2, Table 9-A.
1/2"
8"
30
Coped Beam Flexural Strength Example
Example: Evaluate coped beam flexural strength.
Vu = 40 kips
L  d 

Cb   3  ln b 1  ct  1.84
d  d 

c c
Lb  t b
Otherwise
2
cb 
Lb  dct 
Cb   3  ln 1   1.84
ct 
d  d 
Note: When cb > ct flexural tensile
yielding must be checked at the
bottom cope.
S  twho2 / 6
r  1.9E / Fy
Lb  ct
3"
W14x30 A992 Steel
Note: The distance ho above ≠ ho that is tabulated in
Manual Table 1-1 W-Shapes Dimensions.
31
32
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Coped Beam Flexural Strength Example
Coped Beam Flexural Strength Example
1/2"
8"
Vu = 40 kips
1/2"
8"
3"
Vu = 40 kips
W14x30
Snet = 8.37 in.3
3"
W14x30
Znet = 15.1 in.3
33
Coped Beam Flexural Strength Example
1/2"
Local Web Flexural Strength
8"
ho 10.8 in.

 40
tw 0.270 in.
Vu = 40 kips
3"
Buckling Adjustment Factor, f:
c
8 in.

 0.580  1.0, so
d 13.8 in.
c
f  2  2(0.580)  1.16
d
W14x30
Limiting Slenderness for Yielding:
 p  0.475 k1E / Fy Need k1 = f k ≥ 1.61
Plate Buckling Coefficient, k:
1.65
k1  max
1.65
 10.8 in. 
 2.2 

 8 in. 
1/2"
8"
Vu = 40 kips
3"
W14x30
Modified Plate Bending Coefficient, k1:
c
8 in.

 1.0, so
ho 10.8 in.
h 
k  2.2  o 
c
Coped Beam Flexural Strength Example
Local Web Flexural Strength
Slenderness:

34
fk  (1.16)(3.61)  4.19
 4.19
1.61
 3.61
35
36
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Coped Beam Flexural Strength Example
Local Web Flexural Strength
Coped Beam Flexural Strength Example
Single Coped Beam Flexural Strength
1/2"
 p  (0.475) k1E / Fy
8"
1/2"
3"
8"
 (0.475) (4.19)(29,000 / 50)
3"
 23.4
 p  (  40)  2 p  I.B.
Vu = 40 kips
W14x30
Vu = 40 kips
W14x30
M p  Fy Znet  (50)(15.1)  755 kip-in.
M y  Fy Snet  (50)(8.37)  419 kip-in.
Mn = Mp - (Mp-My)(λ/λp-1)
= 755 – (755-419)(40/23.4 -1)
= 517 kip-in.
M n  (0.9)(517 kip-in.)  465 kip-in.
(Manual Eq. 9-7)
M u  (40 kips)(8.5 in.)  340 kip-in.  M n , OK
37
Shear End-Plate Connections
38
Shear End-Plate Connections
Advantages
• Simple – Few Parts
• No Holes in Beam
Disadvantage
• Requires Beam to be
Cut to Exact Length
39
Manual Figure 10-6
40
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Shear End-Plate Connection Limit States
Beam
Shear End-Plate Connection Limit States
Plate
1. Shear Yielding
2. Coped Beam Flexural Strength
3. Web Base Metal Strength at Weld
Weld
4. Weld Rupture Strength
3, 4 1
76 55 67
5. Shear Yielding
6. Shear Rupture
7. Block Shear
2
8
8. Shear Transfer
Between Plate and
Support
Vu
7
7
6 55 6
41
Shear End-Plate Connection Example
Shear End-Plate Connection Example
Example: Determine the design strength, ϕVn.
1. Shear Yielding at Cope
8"
1
14"
2@3"
1
14"
3"
Spec. J4-3:
W14x30
A992
Vu
PL 1/4 x 6 x 0'-8 1/2" A36
Girder: A992 steel
tw = 0.5 in.
8"
d = 13.8 in.
dc = 3.0 in.
3/4 in. Gr. A325-N Bolts, E70XX Electrode
1
3 2"
42
3/16
W14x30 Fy = 50 ksi Fu = 65 ksi
d = 13.8 in. tw = 0.27 in.
43
Vu
3"
W14x30
A992
ϕVn = (0.6 Fy)(d-dc)tw
= (1.0)(0.6)(50)(13.8-3.0)(0.27)
= 87.5 kips
44
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Shear End-Plate Connection Example
Shear End-Plate Connection Example
2. Coped Beam Flexural Strength
3. Web Shear Rupture Strength
at Weld
From previous example
ϕMn = 465 kip-in.
Plate L = 8.5 in.
tweld = 3/16 in.
Beam Web tw = 0.270 in.
With e = cope length + plate thickness
= 8.0 + 0.25 = 8.25 in.
8"
3"
W14x30
A992
Vu
Spec. Table J2.5 and Eq. J4-4:
3/16
ϕVn = (0.6Fu)(L - 2tweld)tw
= 0.75(0.6)(65)[8.5 – (2)(3/16)] (0.270)
= 64.2 kips
ϕVn = 465 / 8.25
= 56.4 kips
45
Shear End-Plate Connection Example
46
Shear End-Plate Connection Example
4. Weld Rupture Strength
Plate Limit States
8"
From Manual Table J2.4, the
minimum weld size is 1/8 in.
3/16 in. weld OK so far.
Vu
tp = 1/4 in.
3"
W14x30
A992
A36 Steel:
Fy = 36 ksi Fu = 58 ksi
3 2"
1
14 "
2@3"
1
14"
6"
5. Shear Yielding
3/16
ϕVn = (1.392)(D)(L - 2tweld)
= (1.392)(3)[8.5 – (2)(3/16)](2 welds)
= 67.9 kips
ϕVn = (0.6Fy)(2 L tp) (Spec. J4-3)
= 1.0 (0.6 x 36) (2 x 8.5 x 1/4)
= 91.8 kips
47
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Shear End-Plate Connection Example
3 2"
6. Shear Rupture
d h  3 / 4  1 / 16  1 / 16  7 / 8 in.
Anv  (1/ 4) 8.5  3(7 / 8)  (2)
 2.94 in.2
Shear End-Plate Connection Example
PL 1/4 x 6 x 0’-8 1/2”
ϕ = 0.75
1
14 "
2@3"
1
14"
1
14"
7. Block Shear Strength
3"
3"
Shear Rupture
Rn  min
 Tensile Rupture
Shear Yielding
6"
Vn  0.6 Fu Anv
 (0.75)(0.6)(58)(2.94)
 min
 76.7 kips
0.6 Fu Anv
0.6 Fy Agv
 U bs Fu Ant
(Spec. J4-5)
49
Shear End-Plate Connection Example
50
Shear End-Plate Connection Example
7. Plate Block Shear
7. Plate Block Shear
- Shear Rupture
0.6FuAnv = (0.6)(58)(1/4)[7.25 – (2.5)(7/8)](2)
= 88.1 kips
1
14"
- Shear Yielding
0.6FyAgv = (0.6)(36)(1/4)(7.25)(2)
3"
= 78.3 kips
3"
Tensile Rupture
FuAnt = 58(1/4)[1.25 – (0.5)(7/8)](2)
= 23.6 kips
1
14"
Rn  min
 min
 min
Shear Rupture
Shear Yielding
0.6 Fu Anv
0.6 Fy Agv
 Tensile Rupture
1
14"
3"
3"
 U bs Fu Ant
88.1 kips
78.3 kips
1
14"
 (1.0)(23.6 kips)  102 kips
1
14"
Vn = (0.75)(102) = 76.4 kips
51
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Shear End-Plate Connection Example
Shear End-Plate Connection Example
8. Shear Transfer Between Plate and Girder
8. Shear Transfer Between Plate and Girder
A992 girder with tw = 0.5 in.
1-1/4"
A36 plate with t = 0.25 in.
3"
Therefore, the plate won’t will control. 3"
A
B
C
Vu
Bearing: rn = 26.1 kips with d = 3/4 in.
Bolt Shear Rupture
rn = 23.9 kips
A
3"
3"
1-1/4"
Bolt Rupture
rn  min Plate Bearing
B
C
Vu
Plate Tearout
23.9 kips
Tearout with dh = 13/16 in.
At A: rn = 14.7 kips
At B and C: rn = 38.1 kips
rnA  min 26.1 kips
23.9 kips 
rnB  rnC  min 26.1 kips
14.7 kips 
53
38.1 kips
Rn  (0.75)  (14.7)(2)  (23.9)(4)   93.8 kips
54
Shear End-Plate Connection Example
Double-Angle Connections
Connection Design Strength
8"
1
3 2"
3"
1
14 "
2@3"
1
14"
W14x30
A992
Vu
PL 1/4 x 6 x 0'-8 1/2"
A36
Welded/Bolted
3/16
Bolted/Bolted
Bolted/Welded
Coped Beam Flexural Strength Controls
ϕVn = 56.4 kips
55
56
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Double-Angle Connections
Solution of Erection Safety Issue
Advantages
• Beam length can vary.
• Weld or bolt to beam.
• Strong
Disadvantage
• For double-sided connections at a column or
girder web, shared bolts cause an erection
safety issue.
2 Angles
Double-Sided Connection into Column Web
57
Welded/Bolted Double-Angle Connections
58
Welded/Bolted Double-Angle Connections
Horizontal short slots
may be used in angles
2 Angles
Pin is at face of supporting element
Manual Figure 10-4(b)
Beam web weld is subjected to eccentric shear.
59
60
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Welded/Bolted Double-Angle Connections
Limit States
Beam
Shear Yielding
Coped Beam Flexural Strength
Block Shear
Web Base Metal Strength at Weld
Weld
Rupture of Eccentrically Loaded Weld
Group
Welded/Bolted Double-Angle Connections
Angles
Base Metal Strength at Weld Shear
Yielding
Shear Rupture
Block Shear
Shear Transfer
Angle Bearing / Tearout
Bolt Shear Rupture
Supporting Element Bearing / Tearout
61
62
Welded/Bolted Double-Angle Example
Welded/Bolted Double-Angle Example
Example: Determine Rn for
1. Beam Web Block Shear
2. Weld Rupture due to Eccentric Shear
3. Beam Web Strength at Weld
1. Beam Web Block Shear
1/4"
Ru
3/16
W14x30 A992
tw = 0.270”
1/4"
W14x30 A992
tw = 0.270"
Shear Area
Tensile Area
Ru
2L3x3x5/16 x 0'-8 1/2"
Rn  0.6 Fy Agv  U bs Fu Ant
Setback
 (0.6)(50)(0.270)(8.5  1 / 4)  (1.0)(65)(0.270)(3  1 / 2  1 / 4)
 110 kips
Beam
E70XX
2L 3 x 3 x 5/16 x 0'-8 1/2”
0.6 Fy Agv
 U bs Fu Ant
• Anv = Agv (no holes)
• 0.6FuAnv > 0.6FyAgv
• Shear yielding controls
2"
1/2"
1-1/4"
3"
3"
1-1/4"
Rn  min
0.6 Fu Anv
1/2"
63
Rn  (0.75)(110)  82.8 kips
Length
Tolerance
64
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Welded/Bolted Double-Angle Example
Welded/Bolted Double-Angle Example
2. Weld Rupture Due to Eccentric Shear
ex = al
1/2"
Ru
Weld
Group
Ru
+
k x
x a
k&aC
l
xl
kl
2L3x3x5/16
x 0'-8 1/2"
Rn = CC1Dl
Instantaneous Center
of Rotation Method
Design Aid: Table 8-8
Rn = CC1Dl
ϕ = 0.75
C1 from Table 8-3.
C1 = 1.00 for 70 ksi
65
Welded/Bolted Double-Angle Example
66
Welded/Bolted Double-Angle Example
Determine C from Table 8-8:
ex = al
1/2"
Ru
Weld
Group
2L3x3x5/16
x 0'-8 1/2"
Ru
l = 8.5 in.
+
xl
kl = 3 – 1/2 – 1/4 = 2.25 in.
k = 2.25 / 8.5 = 0.265
67
k = 0.265  x = 0.0465
68
Copyright © 2019
American Institute of Steel Construction
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AISC Live Webinar Series
November 6, 2019
Revised: Nov 8, 2019
Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Welded/Bolted Double-Angle Example
Welded/Bolted Double-Angle Example
Determine C from Table 8-8:
ex = al
1/2"
Ru
Ru
Weld
Group
+
l = 8.5 in.
xl = (0.0465)(8.5) = 0.395 in.
2L3x3x5/16
x 0'-8 1/2"
k = 0.265
a = 0.307
k & a  C = 2.62
ex = 3 - 0.395 = 2.61 in.  a = 2.61 / 8.5 = 0.307
69
70
Welded/Bolted Double-Angle Example
Welded/Bolted Double-Angle Example
C = 2.62
3/16 in. welds  D = 3
E70XX  C1 = 1.0
3.
1/2"
Beam Web Strength at Weld
Proposed Rational Approach
8-1/2" R
u
Strength of eccentrically
loaded weld group:
Rn = CC1Dl
= (0.75)(2.62)(1.0)(3)(8.5)(2)
= 100 kips
Web Shear Rupture Strength / in.
Weld Rupture Strength / in.
0.6Futw (1.0 in.)
=  Rn Weld
(1.392)(D)(1.0 in.)(2 welds)
(0.75)(0.6)(65)(0.270)(1.0)
 (100)
(1.392)(3)(1.0)(2)
 94.6 kips
Rn   Rn Weld
3/16
2L3x3x5/16
x 0'-8 1/2”
71
72
Copyright © 2019
American Institute of Steel Construction
18
AISC Live Webinar Series
November 6, 2019
Revised: Nov 8, 2019
Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Bolted/Bolted Double-Angle Connections
Bolted/Bolted Double-Angle Connections
Bolt eccentricity ignored
in bolted/bolted doubleangle connections.
Shear transfer between angles and beam web and
angles and supporting element as previous:
Min of Bearing, Tearout & Bolt Shear Rupture
at each hole/bolt.
Beam to Girder
Beam to Column Flange
No Additional Limit States
73
Bolted/Welded Double-Angle Connections
74
Double-Angle Knife Connections
2 Angles
Return
@ Top
Weld returns per Specification Section J2.2b User Note.
Bolted/Welded to Column Flange
Bolted/Welded Double-Angle Knife Connection
75
76
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American Institute of Steel Construction
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November 6, 2019
Revised: Nov 8, 2019
Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Bolted/Welded Double-Angle
Bolted/Welded Double-Angle
Bolted to Beam / Welded to Column Flange
Coped Beam Web Strength at Tension Flange
• Referred to as a “Knife” Connection
• Bottom Cope to Permit Erection
b = 0.9
Vn = Mn / e
where
Additional Limit States
M n  min
Vu
M p  Fy Znet
e
1.6M y  1.6Fy Snet
Snet = elastic section modulus from Manual Table 9-2
Znet = plastic section modulus from AISC Companion
to the AISC Steel Construction Manual
Table 9-A.
• Coped Beam Web Strength at
Tension Flange
• Weld Strength on Outstanding Legs (OSLs) –
angle-to-column flange connection.
77
78
Bolted/Welded Double-Angle
Weld Strength on OSLs
Bolted/Welded Double-Angle
e = Leg Width
Vu / 2
Vertical Weld Force
fv 
l
Return at Top
Web
contact
pressure
Vu / 2
l
l fv
Vu / 2
Max Horizontal Weld Force
e
fv and fh
in
kips/in.
Vu / 2
Elastic Vector Method
• Uniform vertical weld force.
• Linearly varying horizontal
weld force.
• Neglect returns.
 M   0  f h  1.8
l
79
Vu e
l2
(5/6)l
fh
80
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American Institute of Steel Construction
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AISC Live Webinar Series
November 6, 2019
Revised: Nov 8, 2019
Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Bolted/Welded Double-Angle
Maximum Weld Force as a function of Vu
V
fu  f h2  f v2  u2 l 2  12.96e 2
2l
Design Weld Force (kips/in.) with q = 0○
Bolted/Welded Double-Angle Example
(kips/in.)
Example: Calculate the weld design rupture
strength at OSLs, ϕVn.
2L3x3x5/16 x 0'-8 1/2"
f n  1.392 D
W14x30
8-1/2"
Weld Group Design Shear Strength, Vn
Vn 2
l  12.96e 2
2
2l
2(1.392 Dl )
Vn 
1  12.96e 2 / l 2
f n 
4"
(Manual 10-1a)
E70XX
e = 3 in. and l = 8.5 in.
81
Bolted/Welded Double-Angle Example
1/4
82
End of Session 3
Weld design rupture strength at OSLs
Vn 

2(1.392 Dl )
Thank You for
Attending
1  12.96e 2 / l 2
(2)(1.392)(4)(8.5)
1  12.96(32 ) / (8.52 )
Next Up
 58.5 kips
83
84
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American Institute of Steel Construction
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
Next Session
• November 13, 2019 Shear Connections Part II
TOPICS
•
•
•
•
Single-Angle Connections
Single-Plate (Shear Tab) Connections
Unstiffened Seated Connections
Stiffened Seated Connections
85
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
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Fundamentals of Connection Design
Session 3: Shear Connections, Part I
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