IIB-1
Chapter IIB
Fully-Restrained (FR) Moment Connections
The design of fully restrained (FR) moment connections is covered in Part 11 of the
AISC Steel Construction Manual.
IIB-2
Example II.B-1
Bolted Flange-Plate FR Moment Connection
(beam-to-column flange)
Given:
Design a bolted flange-plated FR moment connection between a W18×50 beam and a W14×99
column flange to transfer the following forces:
RD = 7 kips
RL = 21 kips
MD = 42 kip-ft
ML = 126 kip-ft
Use d-in. diameter ASTM A325-N bolts in standard holes and E70 electrodes.
Material Properties:
Beam W18×50
Column W14×99
Plate
ASTM A992
ASTM A992
ASTM A36
Fy = 50 ksi
Fy = 50 ksi
Fy = 36 ksi
Fu = 65 ksi
Fu = 65 ksi
Fu = 58 ksi
Manual
Table 2-2
Table 2-3
Geometric Properties:
Beam W18×50 d = 18.0 in.
in.3
Column W14×99 d = 14.2 in.
bf = 7.50 in.
tf = 0.570 in.
bf = 14.6 in.
tf = 0.780 in.
tw = 0.355 in.
Sx = 88.9
Manual
Table 1-1
IIB-3
Solution:
LRFD
ASD
Ru = 1.2 ( 7 kips) + 1.6 ( 21 kips)
Ra = 7 kips + 21 kips
= 28 kips
= 42 kips
Mu = 1.2 ( 42 kip-ft ) + 1.6 (126 kip-ft )
M a = 42 kip-ft + 126 kip-ft
= 168 kip-ft
= 252 kip-ft
Check the beam available flexural strength
Assume two rows of bolts in standard holes.
Afg = b f t f = ( 7.50 in.)( 0.570 in.) = 4.28 in.2
Section
F13.1
Afn = Afg − 2 ( db + 1 8 in.) t f = 4.28 in.2 − 2 ( 7 8 in. + 1 8 in.)( 0.570 in.) = 3.14 in.2
Fy
Fu
50 ksi
= 0.769 < 0.8 , therefore Yt = 1.0.
65 ksi
=
(
)
Fu Afn = ( 65 ksi) 3.14 in.2 = 204 kips
(
)
Yt Fy Afg = (1.0)( 50 ksi ) 4.28 in.2 = 214 kips > 204 kips
Therefore the nominal flexural strength, Mn, at the location of the holes in the tension flange is
as follows:
Mn =
Fu Afn S x
Afg
=
( 65 ksi ) ( 3.14 in.2 )(88.9 in.3 )
4.28 in.2
LRFD
ASD
Ω = 1.67
φ = 0.90
M n / Ωb =
φb M n = 0.90 ( 353 kip-ft ) = 318 kip-ft
318 kip-ft > 252 kip-ft
Specification
Eqn. F13-1
= 4240 kip-in. or 353 kip-ft
o.k.
353 kip-ft
= 211 kip-ft
1.67
211 kip-ft > 168 kip-ft
Design single-plate web connection
Try a PL a×4×0’-9, with three d-in. diameter ASTM A325-N bolts and 4-in. fillet welds.
o.k.
IIB-4
LRFD
ASD
Design shear strength of the bolts
Allowable shear strength of bolts
Single shear;
Single shear;
φrn = 21.6 kips/bolt
rn/Ω = 14.4 kips/bolt
= 42 kips/(21.6 kips/bolt) = 1.94 bolts
Manual
Table 7-1
= 28 kips/(14.4 kips/bolt) = 1.94 bolts
Manual
Table 7-5
Bearing strength of bolts
Bearing strength of bolts
Bolt spacing = 3 in.
Bolt spacing = 3 in.
φrn = (91.4 kips/in./bolt)(a in.)
rn/Ω = (60.9 kips/in./bolt)(a in.)
= 34.3 kips/bolt
= 22.8 kips/bolt
= 42 kips/(34.3 kips/bolt) = 1.23 bolts
= 28 kips/(22.8 kips/bolt) = 1.23 bolts
Plate shear yielding
Plate shear yielding
φ = 1.00
Ω = 1.50
φRn = 0.60 φFy Ag
Eqn J4-3
rn/Ω = 0.60 Fy Ag /Ω =
= 0.60(1.00)(36 ksi)(9 in.)(a in.)
= 0.60(36 ksi)(9 in.)(a in.)/(1.50)
= 72.9 kips > 42 kips
= 48.6 kips > 28 kips
o.k.
Plate shear rupture
o.k.
Plate shear rupture
Where φ = 0.75
Ω = 2.00
Eqn J4-4
φRn = 0.60 φFu Anv
rn/Ω = 0.60 Fu Anv /Ω =
(3 bolts)(d in.+ z in.+ z in.) = 3 in.
(3 bolts)(d in.+ z in.+ z in.) = 3 in.
Anv = (9 in. – 3 in.)(a in.) = 2.25 in2
Anv = (9 in. – 3 in.)(a in.) = 2.25 in2
= 0.60(0.75)(58 ksi)(2.25 in2)
= 0.60(58 ksi)(2.25 in2)/(2.00)
= 58.7 kips > 42 kips
= 39.2 kips > 28 kips
o.k.
o.k.
Block shear rupture strength of the plate
Leh = 14 in.; Lev = 12 in.; Ubs = 1.0; n = 3
LRFD
(
φRn = φFu AntU bs + min φ0.6 Fy Agv , φFu Anv
Tension rupture component
φFuAnt = 32.6 kips/in(a in)
)
ASD
⎛ 0.6 Fy Agv Fu Anv ⎞
Rn Fu AntU bs
=
+ min ⎜
,
Ω
Ω
Ω
Ω ⎟⎠
⎝
Tension rupture component
FuAnt/ Ω = 21.8 kips/in(a in)
Manual
Table 9-3a
IIB-5
Manual
Table 9-3b
Shear yielding component
Shear yielding component
0.60FyAgv / Ω = 81.0 kips/in(a in)
φ0.60FyAgv = 121 kips/in(a in)
Shear rupture component
Manual
Table 9-3c
Shear rupture component
φ0.60FuAnv = 131 kips/in(a in)
0.60FuAnv / Ω = 87.0 kips/in(a in)
φRn = (121 kips/in + 32.6 kips/in)(a in)
Rn/Ω = (81.0 kips/in + 21.6 kips/in)(a in)
= 38.5 kips > 24 kips
= 57.6 kips > 42 kips
o.k.
o.k.
Weld Strength
Weld Strength
φRn = 1.392D l (2)
Rn / Ω = 0.928Dl (2)
Manual
Part 8
= 1.392(4 sixteenths)(9 in.)(2)
= 0.928(4 sixteenths)(9 in.)(2)
= 100 kips > 42 kips
= 66.8 kips > 28 kips
o.k.
o.k.
Connecting Elements Rupture Strength at Welds
Shear rupture strength of base metal
tmin
⎛ 2 ⎞⎛ D ⎞
0.6 FEXX ⎜⎜
⎟⎜ ⎟
2 ⎟⎠ ⎝ 16 ⎠ 3.09 D
⎝
=
=
0.6 Fu
Fu
Section J4.2
Column flange; tf = 0.780 in.
tmin =
3.09 D ( 3.09 )( 4 sixteenths )
=
= 0.190 in.
Fu
65 ksi
Plate; t = a in.
tmin =
3.09 ( 2 ) D
Fu
=
( 6.19 )( 4 sixteenths )
58 ksi
= 0.427 in. > a in. proration required.
LRFD
φRn =
ASD
8 in.
( 50.1 kips ) = 44 kips
0.427 in.
3
44 kips > 42 kips
Rn / Ω =
o.k.
8 in.
( 33.4 kips) = 29.3 kips
0.427 in.
3
29.3 kips > 28 kips
o.k.
IIB-6
Design tension flange plate and connection
Design of bolts
LRFD
Puf =
M u ( 252 kip-ft )(12 in./ft )
=
= 168 kips
18.0 in.
d
ASD
M a (168 kip-ft )(12 in./ft )
=
= 112 kips
18.0 in.
d
Paf =
Try a PL w× 7
Try a PL w×7
Determine critical bolt strength
Determine critical bolt strength
For shear, φrn = 21.6 kips/bolt
For shear, rn / Ω = 14.4 kips/bolt
For bearing on flange;
For bearing on flange;
φrn = (102 kips/bolt ) t f
rn / Ω = ( 68.3 kips/bolt ) t f
= (102 kips/bolt )( 0.570 in.)
= ( 68.3 kips/bolt )( 0.570 in.)
= 58.1 kips/bolt
= 38.9 kips/bolt
For bearing in plate;
For bearing in plate;
φrn = ( 91.4 kips/bolt ) t f
rn / Ω = ( 60.9 kips/bolt ) t f
= ( 91.4 kips/bolt )( 0.570 in. )
= ( 60.9 kips/bolt )( 0.570 in. )
= 52.1 kips/bolt
= 34.7 kips/bolt
Shear controls, therefore the number of bolts
required is as follows:
nmin =
Puf
φrn
=
168 kips
= 7.78 bolts
21.6 kips/bolt
Manual
Part 12
Manual
Table 7-1
Manual
Table 7-6
Shear controls, therefore the number of bolts
required is as follows:
nmin =
Paf
rn / Ω
=
112 kips
= 7.78 bolts
14.4 kips/bolt
Use eight bolts.
Use eight bolts.
Check flange plate tension yielding
Pn = Fy Ag = ( 36 ksi )( 7 in. )( 3 4 in. ) = 189 kips
LRFD
Mu
( 252 kip-ft )(12 in./ft )
=
Puf =
d + tp
(18.0 in. + 3 4 in.)
= 161 kips
φ = 0.90
Eqn. D2-1
ASD
Mu
(168 kip-ft )(12 in./ft )
=
Puf =
d + tp
(18.0 in. + 3 4 in.)
= 108 kips
Ω = 1.67
IIB-7
φPn = 0.90 (189 kips ) = 170 kips
Pn / Ω =
170 kips > 161 kips
189 kips
= 113 kips
1.67
113 kips > 108 kips
o.k.
o.k.
Check flange plate tension rupture
An M 0.85 Ag = 0.85(7 in.)(0.75 in.) = 4.46 in2
An = ⎡⎣ B − 2 ( db +
Ae = 3.75 in.2
1
8
in.) ⎤⎦ t p = ⎡⎣( 7 in.) − 2 (
(
7
8
in. +
1
8
in.) ⎤⎦ (
3
Eqn. J4-1
in.) = 3.75 in.
2
4
)
Pn = Fu Ae = ( 58 ksi ) 3.75 in.2 = 217 kips
Eqn. D2-2
LRFD
ASD
φ = 0.75
Ω = 2.00
φPn = 0.75 ( 217 kips ) = 163 kips
Pn / Ω =
163 kips > 161 kips
217 kips
= 109 kips
2.0
o.k.
109 kips > 108 kips
o.k.
Check flange plate block shear rupture
There are two cases for which block shear rupture must be checked. The first case involves the
tearout of the two blocks outside the two rows of bolt holes in the flange plate; for this case Leh
= 12 in. and Lev = 12 in. The second case involves the tearout of the block between the two
rows of the holes in the flange plate. Manual Tables 9-3a, 9-3b, and 9-3c may be adapted for
this calculation by considering the 4 in. width to be comprised of two, 2-in. wide blocks where
Leh = 2 in. and Lev = 12 in. Thus, the former case is the more critical.
LRFD
ASD
(
φRn = φFu AntU bs + min φ0.6 Fy Agv , φFu Anv
Tension component
φU bs Fu Ant = 43.5 kips
(
Shear yielding component
φ0.6 Fy Agv = 170 kips
(
Shear rupture component
φ0.6 Fu Anv = 183 kips
(
3
3
4
3
4
4
)
in.)( 2)
Rn Fu AntU bs
=
Ω
Ω
⎛ 0.6 Fy Agv Fu Anv ⎞
+ min ⎜
,
Ω
Ω ⎟⎠
⎝
Tension component
Fu Ant / Ω = 29.0 kips
( 34
Shear yielding component
0.6 Fy Agv / Ω = 113 kips ( 3 4 in.)( 2)
in.)( 2)
Shear rupture component
0.6 Fu Anv / Ω = 122 kips
in.)( 2)
Shear yielding controls, thus
Shear yielding controls, thus
φRn = (170 kips + 43.5 kips)( 3 4 in.)( 2)
Rn / Ω = (113 kips + 29.0 kips)( 3 4 in.)( 2)
= 320 kips > 161 kips
o.k.
= 213 kips > 108 kips
Manual
Table 9-3b
in.)( 2)
in.)( 2)
( 34
Manual
Table 9-3a
o.k.
Manual
Table 9-3c
IIB-8
Determine the required size of the fillet weld to supporting column flange
Eqns. J2-4,
and J2-5
The applied tension load is perpendicular to the weld,
therefore θ = 90° and 1.0 + 0.5sin1.5 θ = 1.5.
LRFD
Dmin =
=
ASD
Puf
Dmin =
2 (1.5 )(1.392 ) l
161 kips
2 (1.5 )(1.392 )( 7 in. )
=
= 5.51 sixteenths
Use a-in. fillet welds, 6 > 5.51
Paf
2 (1.5 )( 0.928 ) l
108 kips
2 (1.5 )( 0.928 )( 7 in. )
= 5.54 sixteenths
o.k.
Use a-in. fillet welds, 6 > 5.54
o.k.
Connecting Elements Rupture Strength at Welds
Tension rupture strength of base metal
⎛ 2 ⎞⎛ D ⎞
0.6 FEXX ⎜⎜
⎟⎟ ⎜ ⎟
⎝ 2 ⎠ ⎝ 16 ⎠ = 1.86 D
tmin =
Fu
Fu
Column flange; tf = 0.780 in.
tmin =
1.86 D (1.86 )( 6 sixteenths )
=
= 0.171 in.
Fu
65 ksi
Flange plate; tf = 0.75 in.
tmin =
3.71D (1.86)( 2)( 6 sixteenths)
=
= 0.384 in.
Fu
58 ksi
Design compression flange plate and connection
Try PL w×7
Assume K = 0.65 and l = 2.0 in. (12 in. edge distance and 2 in. setback).
Kl 0.65 ( 2.0 in.)
=
= 6.00 < 25
r
⎛ 3 4 in. ⎞
⎜⎝
⎟
12 ⎠
Therefore, Fcr = Fy
A = ( 7 in. )( 3 4 in. ) = 5.25 in.2
Section J4.3
IIB-9
LRFD
ASD
Ω = 1.67
φ = 0.90
(
φPn = φFy Ag = 0.90 ( 36 ksi ) 5.25 in.
= 170 kips > 161 kips
2
)
Pn / Ω =
o.k.
Fy Ag
Ω
=
( 36 ksi ) ( 5.25 in.2 )
1.67
= 113 kips > 108 kips
Eqn. J4-6
o.k.
The compression flange plate will be identical to the tension flange plate; a w in.×7 in. plate
with eight bolts in two rows of four bolts on a 4 in. gage and a in. fillet welds to the
supporting column flange.
Note: Tension due to load reversal must also be considered in the design of
the fillet weld to the supporting column flange.
The column must be checked for stiffening requirements. For further information, see AISC
Design Guide No. 13 Wide-Flange Column Stiffening at Moment Connections – Wind and
Seismic Applications (Carter, 1999).
IIB-20
Plate tp = a in.
tmin =
3.09 ( 2 ) D
Fu
=
( 6.19 )( 4 sixteenths )
58 ksi
= 0.427 in.
LRFD
φRn =
ASD
3 in.
Rn
= 8
( 33.4 kips) = 29.3 kips
Ω 0.427 in.
8 in.
( 50.1 kips ) = 44 kips
0.427 in.
3
44 kips > 42 kips
o.k.
29.3 kips > 28 kips
o.k.
A complete-joint penetration groove weld will transfer the entire flange force in tension and
compression.
Note: The column must be checked for stiffening requirements. For further information, see
AISC Design Guide No. 13 Wide-Flange Column Stiffening at Moment Connections – Wind
and Seismic Applications. (Carter, 1999).
Table J2.5
IIB-21
Example II.B-4
Four-Bolt Unstiffened Extended End-Plate FR Moment
Connection (beam-to-column flange).
Given:
Design a four-bolt unstiffened extended end-plate FR moment connection between a W18×50
beam and a W14×99 column-flange to transfer the following forces:
RD = 7 kips
RL = 21 kips
MD = 42 kip-ft
ML = 126 kip-ft
Use ASTM A325-N snug-tight bolts in standard holes and E70 electrodes.
a.
b.
Use design procedure 1 (thick end-plate and smaller diameter bolts) from AISC Steel
Design Guide 16 Flush and Extended Multiple-Row Moment End-Plate Connections.
Use design procedure 2 (thin end-plate and larger diameter bolts) from AISC Steel
Design Guide 16 Flush and Extended Multiple-Row Moment End-Plate Connections.
Material Properties:
Beam W18×50
Column W14×99
Plate
ASTM A992
ASTM A992
ASTM A36
Fy = 50 ksi
Fy = 50 ksi
Fy = 36 ksi
Fu = 65 ksi
Fu = 65 ksi
Fu = 58 ksi
Manual
Table 2-3
Table 2-4
Geometric Properties:
Beam W18×50 d = 18.0 in.
Column W14×99 d = 14.2 in.
bf = 7.50 in.
bf = 14.6 in.
tf = 0.570 in.
tf = 0.780 in.
tw = 0.355 in. Sx = 88.9 in.3
Manual
Table 1-1
IIB-22
Solution A:
LRFD
ASD
Ru = 1.2(7 kips) + 1.6(21 kips) = 42 kips
Ra = 7 kips + 21 kips = 28 kips
Mu = 1.2(42 kip-ft) + 1.6(126 kip-ft)
Ma = 42 kip-ft + 126 kip-ft
= 252 kip-ft
= 168 kip-ft
Check beam flexural strength
LRFD
ASD
For a W18×50
For a W18×50
φbMn = 379 kip-ft > 252 kip-ft
o.k.
Mn /Ωb = 252 kip-ft > 168 kip-ft
o.k.
Manual
Table 3-3
Extended end-plate geometric properties:
bp = 72 in.
g = 52 in.
pf,i = 12 in.
pf,o = 12 in.
pext = 3 in.
Calculate secondary dimensions
h0 = d + p f ,o = 18.0 in. + 1 1 2 in. = 19 1 2 in.
d o = h0 −
tf
2
= 19 1 2 in. −
0.570 in.
= 19.22 in.
2
h1 = d − p f ,i − t f = 18.0 in. − 1 1 2 in. − 0.570 in. = 15.9 in.
d1 = h1 −
tf
2
= 15.93 in. −
0.570 in.
= 15.6 in.
2
γ r = 1.0 for extended end-plates
Determine the required bolt diameter assuming no prying action
For ASTM A325-N bolts, Fnt = 90 ksi
Table J3.2
LRFD
dbreq =
=
2M u
πφFnt
(∑ d )
ASD
dbreq =
n
2 ( 252 kip-ft )(12 in./ft )
π ( 0.75 )( 90 ksi )(19.2 in. + 15.6 in. )
= 0.905 in.
Use 1-in. diameter ASTM A325-N snugtightened bolts.
Determine the required end-plate thickness
=
2M a Ω
πFnt
(∑ d )
n
2 (168 kip-ft )(12 in./ft )( 2.00)
π ( 90 ksi)(19.2 in. + 15.6 in.)
= 0.905 in.
Use 1-in. diameter ASTM A325-N snugtightened bolts.
IIB-23
s=
bp g
2
=
( 7 12
in.)( 5 1 2 in.)
2
= 3.21 in.
Verify interior bolt pitch, p f ,i = 1 1 2 in. ≤ s = 3.21 in.
Y=
=
o.k.
⎤ 2
⎛ 1 ⎞
bp ⎡ ⎛ 1 1 ⎞
1
+ ⎟ + h0 ⎜
⎢h1 ⎜
⎟ − 2 ⎥ + ⎣⎡ h1 p f ,i + s ⎦⎤
2 ⎢⎣ ⎝ p f ,i s ⎠
⎝ p f ,o ⎠
⎥⎦ g
(
)
7 12 in. ⎡
1 ⎞
⎛ 1
⎛ 1 ⎞ 1 ⎤
+
+ (19 1 2 in.) ⎜
− 2⎥
⎢(15.93 in.) ⎜⎝ 1
⎟
⎝ 1 12 in. ⎟⎠
2 ⎣
1 2 in. 3.21 in. ⎠
⎦
+
2
⎡(15.93 in.)(1 1 2 in. + 3.21 in.) ⎤⎦
5 1 2 in. ⎣
= 133
π (1 in.) ( 90 ksi)
πd 2 F
Pt = b nt =
= 70.7 kips
4
4
2
M n = 2 Pt ( ∑ d n ) = 2 ( 70.7 kips )(19.2 in. + 15.6 in. ) = 4930 kip-in.
LRFD
φ = 0.75
φM n = 0.75 ( 4930 kip-in. ) = 3700 kip-in.
=
Mn / Ω =
4930 kip-in.
= 2460 kip-in.
2.00
Ω = 1.67
φ = 0.90
t p req ' d =
ASD
Ω = 2.00
1.11γ r φM np
φb FpyY
1.11(1.0 )( 3700 kip-in. )
( 0.90 )( 36 ksi )(133 in.)
= 0.977 in.
Use a 1-in. thick end-plate.
t p req ' d =
=
1.11γ r ⎛ M np ⎞
⎛ Fpy ⎞ ⎝⎜ Ω ⎠⎟
⎜⎝ Ω ⎟⎠ Y
b
1.11 (1.0 )( 2460 kip-in. )(1.67 )
( 36 ksi )(133 in.)
= 0.978 in.
Use a 1-in. thick end-plate.
IIB-24
LRFD
ASD
Calculate end-plate design strength
Calculate end-plate allowable strength
From above, φM n = 3700 kip-in.
From above, M n / Ω = 2464 kip-in.
φ = 0.90
φb M pl
γr
=
Ω = 1.67
φb Fpy t p 2Y
M pl
γr
Ωb γ r
=
0.90 ( 36 ksi )(1 in.) (133 in.)
Ωb γ r
( 36 ksi )(1 in.) (133 in. )
=
1.67 (1.0 )
2
2
=
Fpy t p 2Y
1.0
= 4310 kip-in.
= 2860 kip-in.
φb M pl ⎞
⎛
φM n = min ⎜ φM np ,
γ r ⎟⎠
⎝
M pl ⎞
⎛
M n / Ωb = min ⎜ M np / Ωb ,
⎟
Ω
⎝
bγ r ⎠
= 2460 kip-in. or 205 kip-ft
= 3700 kip-in. or 308 kip-ft
308 kip-ft > 252 kip-ft
205 kip-ft > 168 kip-ft
o.k.
o.k.
Check bolt shear
Try the minimum of four bolts at tension flange and two bolts at compression flange.
LRFD
ASD
φRn = nφrn = ( 2 bolts )( 28.3 kips/bolt )
= 56.6 kips > 42 kips
Rn / Ω = nrn / Ω = ( 2 bolts )(18.8 kips/bolt )
Determine the required size of the beam
web-to end-plate fillet weld
Dmin =
φFy tw
2 (1.39 )
=
= 37.6 kips > 28 kips
o.k.
o.k.
Determine the required size of the beam webto end-plate fillet weld
0.90 ( 50 ksi )( 0.355 in. )
2 (1.39 )
= 5.75 sixteenths
Dmin =
( 50 ksi )( 0.355 in.)
2Ω ( 0.928 )
2 (1.67 )( 0.928 )
Fy tw
Manual
Table 7-1
=
= 5.73 sixteenths
Use a in. fillet welds on both sides of the beam web from the inside face of the beam flange to
the centerline of the inside bolt holes plus two bolt diameters.
Determine weld size required for the end reaction
The end reaction , Ru or Ra, is resisted by weld between the mid-depth of the beam and the
inside face of the compression flange or between the inner row of tension bolts plus two bolt
diameters, which ever is smaller. By inspection the former governs for this example.
Manual
Part 8
IIB-25
l=
d
18.0 in.
−tf =
− 0.570 in. = 8.43 in.
2
2
LRFD
Dmin =
Ru
42 kips
=
2 (1.39) l 2 (1.39)( 8.43 in.)
ASD
Dmin =
= 1.79 → 3 sixteenths (minimum size)
Ra
28 kips
=
2 ( 0.928) l 2 ( 0.928)( 8.43 in.)
= 1.79 → 3 sixteenths (minimum size)
Table J2.4
Use x-in. fillet weld on both sides of the
beam web below the tension-bolt region.
Use x-in. fillet weld on both sides of the
beam web below the tension-bolt region.
Connecting Elements Rupture Strength at Welds
Shear rupture strength of base metal
tmin
⎛ 2 ⎞⎛ D ⎞
0.6 Fexx ⎜⎜
⎟⎟ ⎜ ⎟
⎝ 2 ⎠ ⎝ 16 ⎠ = 3.09 D
=
0.6 Fu
Fu
Beam web tw = 0.355 in.
tmin =
3.09 ( 2) D
Fu
=
( 6.19)( 3 sixteenths) = 0.285 in.
65 ksi
End plate tp = 1 in.
tmin =
3.09 D ( 3.09)( 3 sixteenths)
=
= 0.160 in.
Fu
58 ksi
LRFD
ASD
⎛ 0.355 in. ⎞
Ru = 1.392 Dl ⎜
⎝ 0.285 in. ⎟⎠
⎛ 0.355 in. ⎞
Ra = 0.928Dl ⎜
⎝ 0.476 in. ⎟⎠
⎛ 0.355 in. ⎞
= 1.392 ( 3 sixteenths)( 2)( 8.43 in.) ⎜
⎝ 0.285 in. ⎟⎠
⎛ 0.355 in. ⎞
= 0.928 ( 3 sixteenths)( 2)( 8.43 in.) ⎜
⎝ 0.285 in. ⎟⎠
= 87.7 kips > 42 kips
= 58.5 kips > 28 kips
Determine required fillet weld size for the beam flange to end-plate connection
(
)
l = 2 b f + t f − tw = 2 ( 7.50 in. + 0.570 in.) − 0.355 in. = 15.8 in.
IIB-26
LRFD
Puf =
ASD
2 M u 2 ( 252 kip-ft )(12 in./ft )
=
19.2 in. + 15.6 in.
∑ dn
Paf =
= 116 kips
= 173 kips
Dmin =
2 M a 2 (168 kip-ft )(12 in./ft )
=
19.2 in. + 15.6 in.
∑ dn
Puf
1.5 (1.39 ) l
=
173 kips
1.5 (1.39 )(15.8 in. )
Dmin =
= 5.27 → 6 sixteenths (minimum size)
Paf
1.5 ( 0.928) l
=
116 kips
1.5 ( 0.928)(15.8 in.)
= 5.27 → 6 sixteenths (minimum size)
Note that the 1.5 factor is from Specification J2.4.
Use a-in. fillet welds at beam tension flange. Welds at compression flange may be 4-in. fillet
welds (minimum size per Specification Table J2.4).
Connecting Elements Rupture Strength at Welds
Tension rupture strength of base metal
Rn = Fu Ae
tmin
Section J4.1
⎛ 2 ⎞⎛ D ⎞
0.6 Fexx ⎜⎜
⎟⎜ ⎟
2 ⎟⎠ ⎝ 16 ⎠ 1.86 D
⎝
=
=
Fu
Fu
Beam flange; tf = 0.570in.
tmin =
1.86 ( 2 ) D
Fu
=
( 3.71)( 6 sixteenths )
65 ksi
= 0.343 in.
End plate; tp = 1.00 in.
tmin =
1.86 D (1.86 )( 6 sixteenths )
=
= 0.192 in.
Fu
58 ksi
IIB-27
Solution B:
Only those portions of the design that vary from the solution “A” calculations are presented
here.
LRFD
ASD
Determine required end-plate thickness
Determine required end-plate thickness
φ = 0.90
t preq =
Ω = 1.67
γ r Mu
φb FyY
t preq =
1.0 ( 252 kip-ft )(12 in./ft )
=
=
0.90 ( 36 ksi )(133 in.)
γ r M a Ωb
FyY
1.0 (168 kip-ft )(12 in./ft )(1.67)
= 0.84 in.
= 0.84 in.
Use tp = d in.
Use tp = d in.
( 36 ksi)(133 in.)
Select a trial bolt diameter and calculate the maximum prying forces
Try 1-in. diameter bolts.
w' =
bp
2
− ( db + 116 in.) =
7.50 in.
− (1 116 in.) = 2.69 in.
2
3
⎛ tp ⎞
⎛ 7 in. ⎞
ai = 3.682 ⎜ ⎟ − 0.085 = 3.682 ⎜ 8
− 0.085 = 2.38
⎝ 1 in. ⎟⎠
⎝ db ⎠
3
⎡
⎤ πd 3 F
⎛ bp ⎞
t p 2 Fyp ⎢ 0.85 ⎜ ⎟ + 0.80 w '⎥ + b nt
8
⎝ 2⎠
⎣
⎦
Fi ' =
4 p f ,i
⎡
=
⎤
7 12 ⎞
+ 0.80 ( 2.69) ⎥
2 ⎟⎠
⎦
4 (1 1 2 )
( 7 8 ) 2 ( 36) ⎢0.85 ⎛⎜⎝
⎣
π (1) ( 90)
8
+
4 (1 1 2 )
3
= 30.4 kips
IIB-28
w ' tp2
Qmax i =
=
4ai
Fyp
( 2.69 )( 7 8 )
4 ( 2.38 )
2
⎛ F'
− 3⎜ i
⎜ w't
p
⎝
⎞
⎟⎟
⎠
2
⎛ 30.4 ⎞
362 − 3 ⎜
⎜ 2.69 ( 7 8 ) ⎟⎟
⎝
⎠
2
2
= 6.10 kips
ao = min ⎡⎣ ai , pext − p f ,o ⎤⎦ = min [ 2.38 in., 1 1 2 in.] = 1 1 2 in.
⎛ pf i ⎞
⎛ 1 1 2 in. ⎞
Fo ' = Fi ' ⎜
⎟ = ( 30.4 kips) ⎜⎝ 1
1 2 in. ⎟⎠
⎝ p fo ⎠
= 30.4 kips
Qmax o =
=
w ' tp2
4ao
Fyp
( 2.69 )( 7 8 )
4 (1 1 2 )
2
2
⎛ F '
− 3⎜ o
⎜ w't
p
⎝
⎞
⎟⎟
⎠
2
⎛ 30.4 ⎞
362 − 3 ⎜
⎜ 2.69 ( 7 8 ) ⎟⎟
⎝
⎠
2
= 9.68 kips
Calculate the connection available strength for the limit state of bolt rupture with prying
action
πdb 2 Fnt π (1 in.) ( 90 ksi)
=
= 70.7 kips
4
4
Unmodified Bolt Pretension, Tb 0 = 51 kips
2
Pt =
Modify bolt pretension for the snug-tight condition.
Tb =
Tb 0 51 kips
=
= 12.8 kips
4
4
Table J3.1
AISC Design
Guide 16
Table 2-1,
Table 4-2
IIB-29
LRFD
ASD
Mq
φM q =
Ω
=
⎧φ ⎡⎣ 2 ( Pt − Qmax o ) d0 + 2 ( Pt − Qmax i ) d1 ⎤⎦ ⎫
⎪
⎪
⎪⎪φ ⎡⎣ 2 ( Pt − Qmax o ) d0 + 2 ( Tb ) d1 ⎦⎤
⎪⎪
max ⎨
⎬
⎪φ ⎡⎣ 2 ( Pt − Qmax i ) d1 + 2 ( Tb ) d0 ⎤⎦
⎪
⎪
⎪
⎪⎩φ ⎡⎣ 2 ( Tb )( d0 + d1 ) ⎤⎦
⎪⎭
⎧1
⎫
⎪ Ω ⎡⎣( Pt − Qmax o ) d0 + 2 ( Pt − Qmax i ) d1 ⎤⎦ ⎪
⎪
⎪
⎪ 1 ⎡2 ( P − Q ) d + 2 ( T ) d ⎤
⎪
b
max o
0
1⎦
⎪Ω ⎣ t
⎪
max ⎨
⎬
⎪ 1 ⎡2 ( P − Q ) d + 2 ( T ) d ⎤
⎪
b
max i
1
0⎦
⎪Ω ⎣ t
⎪
⎪1
⎪
⎪ ⎡⎣ 2 ( Tb )( d0 + d1 ) ⎤⎦
⎪
⎩Ω
⎭
⎧
⎡ 2 ( 70.7 − 9.68 )(19.2 ) ⎤ ⎫
⎪0.75 ⎢
⎥⎪
⎪
⎣⎢ +2 ( 70.7 − 6.10 )(15.6 ) ⎦⎥ ⎪
⎪
⎪
⎡ 2 ( 70.7 − 9.68 )(19.2 ) ⎤ ⎪
⎪
⎥ ⎪
⎪0.75 ⎢
= max ⎨
⎣⎢ +2 (12.8 )(15.6 )
⎦⎥ ⎬
⎪
⎪
⎡2 ( 70.7 − 6.10 )(15.6 ) ⎤ ⎪
⎪
0.75
⎢
⎥ ⎪
⎪
⎣⎢ +2 (12.8 )(19.2 )
⎦⎥ ⎪
⎪
⎪0.75 ⎡2 12.8 19.215 + 15.6 ⎤ ⎪
)(
)⎦ ⎭
⎣ (
⎩
⎧ 1 ⎡ 2 ( 70.7 − 9.68)(19.2) ⎤ ⎫
⎪
⎢
⎥⎪
⎪ 2.00 ⎣⎢ +2 ( 70.7 − 6.10)(15.6) ⎦⎥ ⎪
⎪
⎪
⎪ 1 ⎡2 ( 70.7 − 9.68)(19.2) ⎤ ⎪
⎥ ⎪
⎪⎪ 2.00 ⎢ +2 12.8 15.6
)( )
⎢⎣ (
⎥⎦ ⎪
= max ⎨
⎬
⎪ 1 ⎡2 ( 70.7 − 6.10)(15.6) ⎤ ⎪
⎪
⎢
⎥ ⎪
⎪ 2.00 ⎢⎣ +2 (12.8)(19.2)
⎥⎦ ⎪
⎪
⎪
⎪ 1 ⎡2 (12.8)(19.215 + 15.6) ⎤ ⎪
⎦ ⎪⎭
⎩⎪ 2.00 ⎣
⎧3275 kip-in. ⎫
⎪2059 kip-in.⎪
⎪
⎪
= max ⎨
⎬ = 3275 kip-in.
1885
kip-in.
⎪
⎪
⎪⎩669 kip-in. ⎪⎭
⎧2183 kip-in.⎫
⎪1373 kip-in. ⎪
⎪
⎪
= max ⎨
⎬ = 2138 kip-in.
⎪1257 kip-in. ⎪
⎪⎩446 kip-in. ⎪⎭
= 273 kip-ft > 252 kip-ft
3275
= 2183 kip-in.
1.5
= 182 kip-ft > 168 kip-ft
Mq / Ω =
φM q = 3275 kip-in.
o.k.
o.k.
For Example IIB-4, design procedure 1 produced a design with a 1-in. thick end-plate and 1in. diameter bolts. Design procedure 2 produced a design with a d-in. thick end-plate and 1-in.
diameter bolts. Either design is acceptable. Design procedure 1 did not produce a smaller bolt
diameter for this example, although in general it should result in a thicker plate and smaller
diameter bolt than design procedure 2. It will be noted that the bolt stress is lower in design
procedure 1 than in design procedure 2.