Single Angle Connections

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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 1/2”A36
3/4” A325-N Bolts
3
Bolted / Bolted Double Angles
Eccentricity not considered in these
connections
For Bolt Rupture: fVn = f 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"
1/4"
Return
@ Top
4"
“Knife” Connection
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 OSL’s
6
Bolted / Welded Double Angles
Coped Beam Web Strength at
Tension Flange
f Vn = fb Fy Snet / e
fb = 0.9
Snet from Table 8-49
W14x30
2"
4"
7
Bolted / Welded Double Angles
Weld Strength on OSL’s
e
O
Vn /2
L/6
L
Tension, f t
CL
Web
8
Bolted / Welded Double Angles
fv

Vn 2

L
 Mo  0
1  5  2  Vn
f t  L  L  
e
2  6  3  2
e
O
Vn /2
L/6
L
Tension, ft
CL
Web
Vn e
ft  1.8 2
L
9
Bolted / Welded Double Angles
e
fw  f  f
2
t
Vn

2
2L
2
v
O
L  12.9e
2
2
L/6
L
Tension, ft
f Vn 
Vn /2
CL
Web
2
2 L (1.392D)
L  12.9e
2
2
10
Bolted / Welded Double Angles
Example: Calculate f Vn based on the weld
rupture strength of the OSL’s
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
f Vn 

2
2 L (1.392D)
L  12.9e
2
2 (8.5) (1.392 4)
2
2
8.5 2  12.9 (3.135)2
 57.0k
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 > 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
Lh
Lv
[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
fVn = (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
Plate Geometry
1
12"
[email protected]" 9" > T/2
Check Plate
Buckling
3"
L / 64 = 9 / 64
= 0.14 < 1/4 OK
112"
112"
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
fVn = C x frv
= 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:
f 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:
fVn = 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"
112"
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
Bolts
ea
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
fVn = 0.9 Fy Sg / ea
L
Sg = tp L2/6
38
Single Angle Connections
ea
• Flexural Rupture
fVn = 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)
fVn
fVn
Table 8-18
Table 8-19
fVn = C (frv)
41
Single Angle Connections
• Bearing and Tear Out
fVn = C (frvb)
where frvb is the bearing / tear out
strength at the outermost bolt
42
Single Angle Connections
db (in.)
3/4
7/8
1
Min. t (in.)
3/8
3/8
1/2
43
Single Angle Connections
Additional Limit States for Welded
Connections
• Eccentric Shear Strength of Weld
2 tweld
Return
fV n
fV 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
f = 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 @ <d/2
N d  0.2
1.5




N
t


2
R n  68 t w 1  3   w  
 d  t f  

N d  0.2
Fy t f
tw
1.5
 4N



t

2
R n  68 t w 1  
- 0.2 w  
 t f  
  d
Fy t f
tw
51
Design Model for Angle Flexure
Supported Beam
The N-distance is
determined from the
limit states of web
yielding and web
crippling, but not less
than kdetailing.
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
Nmin
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(68tw )
1.5

 tf 
tw
 1 
Fy t f
 t w 
when N/d > 0.2
N min
Ru
d

 
4  0.75(68t2w )

 t f 
tw
 1 
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
Ru  fRn  0.9(0.6Fy )Latangle
Therefore
t req
Ru

0.9(0.6Fy )La
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
ts
36
36
tw
50
36
1.4 tw
50
50
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
112"
3/4 in.
A325N Bolts
3"
114"
134"
3"
2"
3"
OSL
71
Single Angle Connection Example
Angle Flexural Yield
Supporting Side OSL Controls
fVn = 0.9 Fy Sg / ea
Sg = (3/8) (15.0)2 / 6 = 14.06 in3
ea = 0.38 / 2 + 3 = 3.19 in.
a
Vu
fVn = 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
fVn = 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
frv = 0.75 x 48 x 0.4418
= 15.9 k
fVn = C (frv) = 6.29 x 15.9
= 100 k > 90 k OK
78
Single Angle Connection Example
Bearing / Tear Out
Critical Bolts
Vu
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
frbv = 0.75 (28.5)
= 21.4 k > frv = 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.
al = 0.38/2 + 6 – 0.855
al
15"
Vu = 90 k
= 5.34 in.
a = 5.34 / 15 = 0.356
0.19"
82
83
Single Angle Connection Example
C = 1.64 by interpolation
Dreq’d = 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
Nmin
Ru

 2.5 k design
1.0 Fy t w
35

 2.5x1.01
1.0x 50 x 0.360
 0.58  kdetailing
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 )
1.5

 tf 
tw
 1 
Fy t f
 t w 
18.06 
35


3  0.75 ( 68x0.3602 )
1.5

0.360
 0.605 
 1

50x0.605
 0.360 
 5.5 in.
88
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 La
4x35x0.375
0.9x36x8.0
= 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 )La
35

0.9( 0.6x 36)8.0
 0.225 i n.  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 = ¼ in.
Use ¼ 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
L
W
97
Stiffened Seat Example
W14x90 Satisfies column web requirement.
W24x68
A992
t wb = 0.415 in.
W14x90
A992
t wc = 0.440 in.
ts
L
W
98
99
Stiffened Seat Example
Web Crippling
Using the Factored Uniform Load Table in
Part-4 assuming N/d < 0.2
Wmin
R u  fR 3

 se tback
fR 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  fR 1

 se tback
fR 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.
f Rn = 82.3 > Ru = 80 k
10"
4"
102
Stiffened Seat Example
103
Stiffened Seat Example
Column Web Strength at Weld:
fRn = 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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