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IID-1
Chapter IID Miscellaneous Connections This section contains design examples on connections in the AISC Steel Construction Manual that that are not covered in other sections of the AISC Design AISC Design Examples Examples.
Design Examples V14.1
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IID-2
EXAMPLE II.D-1 PRYING ACTION IN TEES AND IN SINGLE ANGLES Given:
Design an ASTM A992 WT hanger connection between an ASTM A36 2L33c tension member and an ASTM A992 W2494 beam to support the following loads: P D = 13.5 = 13.5 kips P L = 40 = 40 kips Use w-in.-diameter -in.-diameter ASTM A325-N or F1852-N bolts and 70-ksi electrodes.
Solution:
From AISC Manual AISC Manual Table Table 2-4, the material properties are as follows: Hanger WT
ASTM A992 F y = 50 ksi F u = 65 ksi Beam W2494 ASTM A992 F y = 50 ksi F u = 65 ksi Angles 2L33c ASTM A36 F y = 36 ksi F u = 58 ksi
Design Examples V14.1
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IID-3
From AISC Manual Tables 1-1, 1-7 and 1-15, the geometric properties are as follows: Beam W2494 d = 24.3 in. t w = 0.515 in. b f = 9.07 in. t f = 0.875 in. Angles 2L33c A = 3.56 in.2 x = 0.860 in. for single angle From Chapter 2 of ASCE/SEI 7, the required strength is: LRFD P u 1.2(13.5 kips) 1.6(40 kips)
ASD P a
= 80.2 kips
13.5 kips 40 kips = 53.5 kips
Tensile Yielding of Angles Pn
Fy Ag
(Spec. Eq. D2-1)
36 ksi 3.56 in.2 = 128 kips LRFD
0.90 P n 0.90(128 kips) 115 kips 80.2 kips
ASD
1.67 P n o.k.
128 kips
1.67 76.6 kips 53.5 kips
From AISC Specification Table J2.4, the minimum size of fillet weld based on a material thickness of x in. From AISC Specification Section J2.2b, the maximum size of fillet weld is: wmax
thickness z in. c in. z in. 4 in.
Try 4-in. fillet welds.
Design Examples V14.1
o.k. c in.
is
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IID-4
From AISC Manual Part 8, Equations 8-2: LRFD l min
=
ASD
P u
l min
1.392 D 80.2 kips
P a
=
1.392(4 sixteenths)
= 14.4 in.
0.928 D 53.5 kips 0.928(4 sixteenths)
= 14.4 in.
Use four 4-in. welds (16 in. total), one at the toe and heel of each angle. Tensile Rupture Strength of Angles U 1
1
x
from AISC Specification Table D3.1 case 2 L 0.860 in.
4.00 in. = 0.785 Ae
AnU
(Spec. Eq. D3-1)
= 3.56 in.2 0.785 = 2.79 in.2 Pn
Fu Ae
(Spec. Eq. D2-2)
= 58 ksi 2.79 in.2
= 162 kips LRFD
t 0.75 t P n 0.75162kips 122 kips 80.2
ASD
t 2.00 P n kips
o.k.
t
162kips 2.00
81.0 kips 53.5 kips
o.k.
Preliminary WT Selection Using Beam Gage g = 4 in. Try four w-in.-diameter ASTM A325-N bolts. From AISC Manual Table 7-2: LRFD T
r ut
ASD
P u
T
n 80.2 kips
r at
4 20.1 kips/bolt B r n
29.8 kips 20.1 kips
P a n 53.5 kips
4 13.4 kips/bolt o.k.
B r n / 19.9 kips 13.4 kips
Design Examples V14.1
o.k.
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IID-5
Determine tributary length per pair of bolts, p, using AISC Manual Figure 9-4 and assuming a thickness. p =
4.00 in. 2 in.
2-in.
web
8.00 in. 42 in.
2 = 3.50 in. 42 in.
2
LRFD 2 bolts(20.1 kips/bolt) 3.50 in.
11.5 kips/in.
ASD 2 bolts(13.4 kips/bolt) 3.50 in.
From AISC Manual Table 15-2b, with an assumed b = (4.00 in. – of the WT hanger should be approximately s in.
2 in.)/2
7.66 kips/in.
= 1.75 in., the flange thickness, t = t f ,
The minimum depth WT that can be used is equal to the sum of the weld length plus the weld size plus the k dimension for the selected section. From AISC Manual Table 1-8 with an assumed b = 1.75 in., t f s in., and d min
4 in. 4 in. k 6 in., appropriate selections include:
25 7 WT 26.5 WT825 WT927.5 WT6
Try a WT625. From AISC Manual Table 1-8, the geometric properties are as follows: b f = 8.08 in. t f = 0.640 in. t w = 0.370 in. Prying Action Using AISC Manual Part 9 The beam flange is thicker than the WT flange; therefore, prying in the tee flange will control over prying in the beam flange. b
g t w
2 4.00 in. 0.370 in.
2 1.82 in. 14-in. entering and tightening clearance, and the fillet toe is cleared a
=
b f
g
2 8.08 in. 4.00 in. 2
= 2.04 in. b b
d b
( Manual Eq. 9-21)
2
Design Examples V14.1
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IID-6
w in. 2
= 1.82 in. = 1.45 in.
1.25b d b 2 2 w in. w in. 2.04 in. 1.25(1.82 in.)+ 2 2 2.42 in. 2.65 in.
a a
db
( Manual Eq. 9-27)
b
( Manual Eq. 9-26)
a
1.45 in.
=
2.42 in. = 0.599 LRFD 1 B 1 T 1 29.8 kips/bolt = 1 0.599 20.1 kips/bolt
ASD ( Manual Eq. 9-25)
= 0.806
1 =1
1 B 1 T 1 19.9 kips/bolt = 1 0.599 13.4 kips/bolt
( Manual Eq. 9-25)
= 0.810
d
( Manual Eq. 9-24)
p w in.
z in.
3.50 in.
= 0.768 Since
1.0, LRFD
1 1.0 1 1 0.806 0.768 1 0.806 5.41, therefore, 1.0 0.90 4Tb t min pF u (1 )
ASD
( Manual Eq. 9-23a)
4 20.1 kips/bolt 1.45 in.
0.90 3.50 in. 65 ksi 1 0.7681.0
0.567 in. t f 0.640 in.
1 1.0 1 1 0.810 0.768 1 0.810 5.55, therefore, 1.0 1.67 4Tb t min pF u (1 )
o.k.
1.67 4 13.4 kips/bolt 1.45 in. 3.50 in. 65 ksi 1 0.7681.0
0.568 in. t f 0.640 in.
Design Examples V14.1
( Manual Eq. 9-23b)
o.k.
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IID-7
Tensile Yielding of the WT Stem on the Whitmore Section Using AISC Manual Part 9 The effective width of the WT stem (which cannot exceed the actual width of 8 in.) is: l w
3.00 in. 2(4.00 in.)(tan 30 ) 8.00 in.
= 7.62 in. The nominal strength is determined as:
Fy Ag
Rn
(Spec. Eq. J4-1)
= 50 ksi(7.62 in.)(0.370 in.) = 141 kips LRFD
ASD
0.90
1.67
Rn 0.90(141 kips) 127 kips 80.2 kips
Rn o.k.
141 kips
1.67 84.4 kips 53.5 kips
o.k.
Shear Rupture of the WT Stem Base Metal t min =
6.19 D
( Manual Eq. 9-3)
F u
4 sixteenths 6.19 65 ksi 0.381 in. > 0.370 in.
shear rupture strength of WT stem controls over weld rupture strength
Block Shear Rupture of the WT Stem A gv
2 shear planes 4.00 in. 0.370 in. = 2.96 in.2
Tension stress is uniform, therefore U bs = 1.0. Ant
Agt 3.00 in. 0.370 in. = 1.11 in.2
Rn = 0.60 F u Anv+U b F s u Ant 0.60 F y A gv+U b F s u Ant
(Spec. Eq. J4-5)
Because the angles are welded to the WT-hanger, shear yielding on the gross area will control (that is, the portion of the block shear rupture equation that addresses shear rupture on the net area does not control). Rn
0.60Fy Agv U bs Fu Ant
= 0.60 50 ksi 2.96 in.2
1.0 65 ksi 1.11 in. 2
= 161 kips
Design Examples V14.1
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IID-8
LRFD
ASD
0.75
2.00 Rn
Rn = 0.75(161 kips) = 121 kips > 80.2 kips
o.k.
161 kips
2.00 80.5 kips 53.5 kips
o.k.
Note: As an alternative to the preceding calculations, the designer can use a simplified procedure to select a WT hanger with a flange thick enough to reduce the effect of prying action to an insignificant amount, i.e., q 0. Assuming b ' 1.45 in. From AISC Manual Part 9: LRFD
ASD
0.90 t min
=
1.67 4Tb
pF u 4(20.1 kips/bolt)(1.45 in.) 0.90(3.50 in./bolt)(65 ksi)
= 0.755 in.
( Manual Eq. 9-20a)
t min
=
4Tb pF u
( Manual Eq. 9-20b)
1.67 4 (13.4 kips/bolt)(1.45 in.) (3.50 in./bolt)(65 ksi)
= 0.755 in.
A WT625, with t f = 0.640 in. < 0.755 in., does not have a sufficient flange thickness to reduce the effect of prying action to an insignificant amount. In this case, the simplified approach requires a WT section with a thicker flange.
Design Examples V14.1
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IID-9
EXAMPLE II.D-2 BEAM BEARING PLATE Given:
An ASTM A992 W1850 beam with a dead load end reaction of 15 kips and a live load end reaction of 45 kips is supported by a 10-in.-thick concrete wall. Assuming the concrete has f c = 3 ksi, and the bearing plate is ASTM A36 material determine the following: a. If a bearing plate is required if the beam is supported by the full wall thickness b. The bearing plate required if l b = 10 in. (the full wall thickness) c. The bearing plate required if l b = 62 in. and the bearing plate is centered on the thickness of the wall
Solution:
From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam W1850 ASTM A992 F y = 50 ksi F u = 65 ksi Bearing Plate (if required) ASTM A36 F y = 36 ksi F u = 58 ksi Concrete Wall f c = 3 ksi From AISC Manual Table 1-1, the geometric properties are as follows: Beam W1850 d = 18.0 in. t w = 0.355 in. b f = 7.50 in. t f = 0.570 in. k des = 0.972 in. k 1 = m in.
Design Examples V14.1
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IID-10
Concrete Wall h = 10.0 in. Solution a:
LRFD Calculate required strength.
ASD Calculate required strength.
Ru = 1.2(15 kips) + 1.6(45 kips) = 90.0 kips
Ra = 15 kips + 45 kips = 60.0 kips
Check web local yielding using AISC Manual Table 9-4 and Manual Equation 9-45a.
Check web local yielding using AISC Manual Table 9-4 and Manual Equation 9-45b.
R1 k des R2 90.0 kips 43.1 kips
Ru
l b req = =
17.8 kips/in.
l b req =
0.972 in.
R1 / k des R2 / 60.0 kips 28.8 kips
Ra
=
= 2.63 in. < 10.0 in.
11.8 kips/in.
0.972 in.
= 2.64 in. < 10.0 in.
o.k.
o.k.
Check web local crippling using AISC Manual Table 9-4.
Check web local crippling using AISC Manual Table 9-4.
l b
10.0 in.
l b
18.0 in. = 0.556
d
d
=
Since
l b d
l b req = =
> 0.2, use Manual Equation 9-48a.
Ru
l b d
Verify
l b
18.0 in. = 0.335 > 0.2
d o.k.
> 0.2, use Manual Equation 9-48b.
R5 / R6 / 60.0 kips 34.7 kips
Ra
4.20 kips/in.
= 6.02 in. < 10.0 in.
o.k.
> 0.2,
Check the bearing strength of concrete .
d
=
6.30 kips/in.
d
l b
l b req =
6.03 in.
18.0 in. = 0.556
90.0 kips 52.0 kips
l b
10.0 in.
Since
R5 R6
= 6.03 in. < 10.0 in. Verify
=
l b d
o.k.
> 0.2,
6.02 in.
18.0 in. = 0.334 > 0.2
o.k.
Check the bearing strength of concrete.
Note that AISC Specification Equation J8-1 is used Note that AISC Specification Equation J8-1 is used because A2 is not larger than A1 in this case. because A2 is not larger than A1 in this case. P p 0.85 f c A1
(Spec. Eq. J8-1) P p 0.85 f c A1
Design Examples V14.1
(Spec. Eq. J8-1)
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IID-11
LRFD
ASD
c 2.31
c 0.65
P p
c P p c0.85 f c A1
c
= 0.65(0.85)(3 ksi)(7.50 in.)(10.0 in.) = 124 kips > 90.0 kips
0.85 f c'A1
o.k.
c 0.85 3 ksi 7.50 in.10.0 in.
=
2.31 = 82.8 kips > 60.0 kips
o.k.
Beam Flange Thickness Check Using AISC Manual Part 14 LRFD Determine the cantilever length from Manual Equation 14-1. b f
n=
k des
2 7.50 in.
2 2.78 in.
0.972 in.
A1
Determine the minimum beam flange thickness required if no bearing plate is provided. The beam flanges along the length, n, are assumed to be fixed end cantilevers with a minimum thickness determined using the limit state of flexural yielding.
M u
Z
2
4t 2
M u
t min =
4
u
F y
t 2 F y 4
= 0.90 t min =
2 Ru n 2
A 1F y
2 Ru n 2
A 1F y
A1
Determine the minimum beam flange thickness required if no bearing plate is provided. The beam flanges along the length, n, are assumed to be fixed end cantilevers with a minimum thickness determined using the limit state of flexural yielding.
M a
Z
Fy Z
Ra
2
2 A1
0.972 in.
Determine bearing pressure. f p =
Ru n
k des 2 7.50 in.
2 2.78 in.
Ru
f p n2
b f
n=
Determine bearing pressure. f p =
ASD Determine the cantilever length from Manual Equation 14-1.
f p n 2
Ra n 2
2
2 A1
4t 2
M a
F y Z
t min =
4
F y t 2
a
F y
= t min =
Design Examples V14.1
2 Ra n2 A1 F y
4 2 Ra n2 A1 F y
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IID-12
LRFD
ASD
2(90.0 kips)(2.78 in.) 2
=
1.67 2 (60.0 kips)(2.78 in.)2
=
0.90(7.50 in.)(10.0 in.)(50 ksi)
= 0.642 in. > 0.570 in.
n.g.
A bearing plate is required. See note following.
(7.50 in.)(10.0 in.)(50 ksi)
= 0.643 in. > 0.570 in.
n.g.
A bearing plate is required. See note following.
Note: The designer may assume a bearing width narrower than the beam flange in order to justify a thinner flange. In this case, if 5.44 in. ≤ bearing width ≤ 6.56 in., a 0.570 in. flange thickness is ok and the concrete has adequate bearing strength. Solution b:
l b = 10 in. From Solution a, web local yielding and web local crippling are o.k. LRFD Calculate the required bearing-plate width using AISC Specification Equation J8-1.
ASD Calculate the required bearing-plate width using AISC Specification Equation J8-1.
c 0.65
c 2.31
A1 req = =
Ru
A1 req =
c 0.85 f c 90.0 kips
=
0.65(0.85)(3 ksi)
A1 req
B req =
N 2
=
0.85 f c 60.0 kips(2.31) (0.85)(3 ksi)
= 54.4 in.2
= 54.3 in.2 B req =
Ra c
54.3 in.
=
A 1 req N 54.4 in.2
10.0 in. = 5.44 in.
10.0 in. = 5.43 in. Use B = 8 in. (selected as the least whole-inch dimension that exceeds b f ).
Use B = 8 in. (selected as the least whole-inch dimension that exceeds b f ).
Calculate the required bearing-plate thickness using AISC Manual Part 14.
Calculate the required bearing-plate thickness using AISC Manual Part 14.
n = =
B
k des
2 8.00 in.
2 = 3.03 in.
t min =
( Manual Eq. 14-1)
0.972 in.
2 Ru n 2
A1F y
n = =
B
k des 2 8.00 in.
2 = 3.03 in.
t min =
Design Examples V14.1
( Manual Eq. 14-1)
0.972 in.
2 Ra n 2 A1 F y
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IID-13
2(90.0 kips)(3.03 in.)
=
2
=
0.90(10.0 in.)(8.00 in.)(36 ksi)
= 0.798 in. Use PL
d
1.67 2 (60.0 kips)(3.03 in.)
2
(10.0 in.)(8.00 in.)(36 ksi)
= 0.799 in.
in.10 in.0 ft 8 in.
Use PL
d
in.10 in.0 ft 8 in.
Note: The calculations for t min are conservative. Taking the strength of the beam flange into consideration results in a thinner required bearing plate or no bearing plate at all. Solution c:
l b = N = 6.50 in. From Solution a, web local yielding and web local crippling are o.k. Try B = 8 in. A1 = BN = 8.00 in.(6.50 in.) = 52.0 in.2 To determine the dimensions of the area A2, the load is spread into the concrete until an edge or the maximum condition
A 2 / A1
2 is met. There is also a requirement that the area, A2, be geometrically similar to A1 or, in
other words, have the same aspect ratio as A1. N 1 = 6.50 in. + 2(1.75 in.) = 10.0 in. B N
8.00in.
6.50 in. = 1.23
B1 = 1.23(10.0 in.) = 12.3 in. A2 = B1 N 1 = 12.3 in. (10.0 in.) = 123 in.2
Check
P p
A2 A1
123 in.2
52.0 in.2 = 1.54 ≤ 2 o.k.
0.85 fc A1
A2 A1
1.7 fc A1
(Spec. Eq. J8-2)
0.85 3 ksi 52.0 in.2 1.54 1.7 3 ksi 52.0 in.2 = 204 kips 265 kips
Design Examples V14.1
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IID-14
LRFD
ASD
c 0.65
c 2.31 P p
c P p 0.65(204 kips)
= 133 kips o.k.
133 kips > 90.0 kips
Calculate the required bearing-plate thickness using AISC Manual Part 14. n = =
B
k
2 8.00 in.
2 = 3.03 in.
t min =
=
( Manual Eq. 14-1)
204kips
2.31 = 88.3 kips o.k.
88.3 kips > 60.0 kips
Calculate the required bearing-plate thickness using AISC Manual Part 14. n =
0.972 in. =
B 2
k
B 2
8.00 in.
2 = 3.03 in.
k
( Manual Eq. 14-1)
0.972 in.
2 Ru n 2
A1 F y 2(90.0 kips)(3.03 in.)
t min = 2
0.90 6.50 in. 8.00 in. (36 ksi)
=
= 0.990 in.
2 Ra n 2 A1 F y
1.67 2 (60.0 kips)(3.03 in.) 2
6.50 in. 8.00 in. (36 ksi)
= 0.991 in. Use PL 1 in. 62 in. 0 ft 8 in.
Use PL 1 in. 62 in. 0 ft 8 in.
Note: The calculations for t min are conservative. Taking the strength of the beam flange into consideration results in a thinner required bearing plate or no bearing plate at all.
Design Examples V14.1
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IID-15
EXAMPLE II.D-3 SLIP-CRITICAL CONNECTION WITH OVERSIZED HOLES Given:
Design the connection of an ASTM A36 2 L33c tension member to an ASTM A36 plate welded to an ASTM A992 beam as shown in Figure II.D-3-1 for a dead load of 15 kips and a live load of 45 kips. The angles have standard holes and the plate has oversized holes per AISC Specification Table J3.3. Use w-in.-diameter ASTM A325-SC bolts with Class A surfaces. P D = 15 kips P L = 45 kips
Fig. II.D-3-1. Connection Configuration for Example II.D-3. Solution:
From AISC Manual Tables 2-4 and 2-5, the material properties are as follows: Beam W1626 ASTM A992 F y = 50 ksi F u = 65 ksi Hanger 2L33c ASTM A36 F y = 36 ksi F u = 58 ksi
Design Examples V14.1
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IID-16
Plate ASTM A36 F y = 36 ksi F u = 58 ksi From AISC Manual Tables 1-1, 1-7 and 1-15, the geometric properties are as follows: Beam W1626 t f = 0.345 in. t w = 0.250 in. k des = 0.747 in. Hanger 2L33c A = 3.56 in.2 x 0.860 in. for single angle Plate t p = 0.500 in. LRFD Calculate required strength.
ASD Calculate required strength.
Ru = (1.2)(15 kips) + (1.6)(45 kips) = 90.0 kips
Ra = 15 kips + 45 kips = 60.0 kips
Check the available slip resistance of the bolts using AISC Manual Table 7-3.
Check the available slip resistance of the bolts using AISC Manual Table 7-3.
For w-in.-diameter ASTM A325-SC bolts with Class A faying surfaces in oversized holes and double shear:
For w-in.-diameter ASTM A325-SC bolts with Class A faying surfaces in oversized holes and double shear:
r n = 16.1 kips/bolt
r n
n
Ru
r n
90.0 kips
n
16.1 kips/bolt
5.59
= 10.8 kips/bolt Ra
r n /
6 bolts
60.0 kips 10.8 kips/bolt
5.56 6 bolts
Slip-critical connections must also be designed for the limit states of bearing-type connections. Check bolt shear strength using AISC Manual Table 7-1.
rn Fv Ab = 35.8 kips/bolt
Slip-critical connections must also be designed for the limit states of bearing-type connections. Check bolt shear strength using AISC Manual Table 7-1. rn Fv Ab = 23.9 kips/bolt
Rn
Rn rn n = (35.8 kips/bolt)(6 bolts) = 215 kips > 90.0 kips
o.k.
r n
n
= (23.9 kips/bolt)(6 bolts) = 143 kips > 60.0 kips
Design Examples V14.1
o.k.
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IID-17
Tensile Yielding Strength of the Angles Pn
Fy Ag
(Spec. Eq. D2-1)
= 36ksi 3.56 in.2
= 128 kips LRFD
ASD
t 0.90 t P n 0.90128 kips 115kips > 90.0 kips
t 1.67 P n o.k.
128 kips
1.67 76.6 kips > 60.0 kips
o.k.
Tensile Rupture Strength of the Angles U 1
x
1
l
from AISC Specification Table D3.1 Case 2
0.860 in.
15.0 in. = 0.943
Ae
AnU = 3.56 in.2 2(c in.)(m in. z in.) 0.943
(Spec. Eq. D3-1)
= 2.84 in.2 Pn
Fu Ae
(Spec. Eq. D2-2) 2
= 58 ksi(2.84 in. ) = 165 kips LRFD
ASD
t 2.00
t 0.75 t P n 0.75(165 kips)
P n
= 124 kips > 90.0 kips
o.k.
t
165 kips
2.00 = 82.5 kips > 60.0 kips
o.k.
Block Shear Rupture Strength of the Angles Use a single vertical row of bolts. U bs Rn
1,
n 6, Lev
12
in., and Leh
14 in.
0.60 Fu Anv U bs Fu Ant 0.60Fy Agv U bs Fu Ant
Shear Yielding Component A gv
5 3.00 in. +1.50 in. c in. = 5.16 in.2 per angle
Design Examples V14.1
(Spec. Eq. J4-5)
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IID-18
0.60 F y Agv
0.60 36 ksi 5.16 in.2 = 111 kips per angle
Shear Rupture Component Anv
5.16 in.2 5.5 m in. z in. c in. = 3.66 in.2 per angle
0.60 Fu Anv
0.60 58 ksi 3.66 in.2
= 127 kips per angle Shear yielding controls over shear rupture. Tension Rupture Component Ant 1.25 in. 0.5 m in. z in. c in. = 0.254 in.2 per angle U bs Fu Ant
1.0 58 ksi 0.254 in.2
= 14.7 kips per angle LRFD
0.75 Rn 0.75 2 111 kips + 14.7 kips 189 kips > 90.0 kips
ASD
o.k.
2.00 Rn 2 111kips + 14.7 kips 2.00 126 kips > 60.0 kips
o.k.
Bearing / Tear Out Strength of the Angles Holes are standard m-in. diameter. Check strength for edge bolt. l c
1.50 in.
w
in. z in. 2
= 1.09 in. rn
1.2lc tFu 2.4 dtFu 1.2(1.09 in.)(c in.)(2)(58 ksi) 2.4(w in.)(c in.)(2)(58 ksi) = 47.4 kips 65.3 kips
(Spec. Eq. J3-6a)
Check strength for interior bolts. l c
3.00 in. w in. z in. = 2.19 in.
rn
1.2lc tFu 2.4 dtFu 1.2(2.19 in.)(c in.)(2)(58 ksi) 2.4(w in.)(c in.)(2)(58 ksi) Design Examples V14.1
(Spec. Eq. J3-6a)
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IID-19
= 95.3 kips 65.3kips Total strength for all bolts. r n = 1(47.4 kips) + 5(65.3 kips) = 374 kips LRFD
ASD
2.00
0.75 r n 0.75(374 kips)
r n
= 281 kips > 90.0 kips
o.k.
374 kips
2.00 = 187 kips > 60.0 kips
o.k.
Tensile Yielding Strength of the 2 -in. Plate By inspection, the Whitmore section includes the entire width of the 2-in. plate. Rn
Fy Ag
(Spec. Eq. J4-1)
= 36 ksi(2 in.)(6.00 in.) = 108 kips LRFD
t 0.90 Rn 0.90(108 kips) 97.2 kips 90.0 kips
Tensile Rupture Strength of the
ASD
t 1.67 Rn o.k.
t
108 kips
1.67 64.7 kips > 60.0 kips
o.k.
2 -in. Plate
Holes are oversized ,-in. diameter. Calculate the effective net area. Ae
An 0.85 Ag from AISC Specification Section J4.1 0.85 3.00 in.2 ≤ 2.55
in.2
An
3.00 in.2 (2 in.)(, 2.50 in.2 2.55 in.2
Ae
AnU
in. + z in.)
(Spec. Eq. D3-1)
= 2.50 in.2 1.0 = 2.50 in.2 Rn
Fu Ae
(Spec. Eq. J4-2)
= 58 ksi(2.50 in.2 ) = 145 kips
Design Examples V14.1
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IID-20
LRFD
ASD
0.75 Rn 0.75(145 kips) 109 kips > 90.0 kips
2.00 Rn o.k.
145 kips
2.00 72.5 kips > 60.0 kips
o.k.
Block Shear Rupture Strength of the 2 -in. Plate Use a single vertical row of bolts. U bs
1.0,
n 6, Lev
12 in., and Leh 3 in.
0.60 Fu Anv U bs Fu Ant 0.60 Fy Agv Ubs Fu Ant
Rn
(Spec. Eq. J4-5)
Shear Yielding Component A gv
5 3.00 in. 1.50 in. 2 in. = 8.25 in.2
0.60 F y Agv
0.60 36 ksi 8.25 in.2
= 178 kips Shear Rupture Component Anv
8.25 in.2 5.5 ,
in. z in.2 in.
= 5.50 in.2 0.60 Fu Anv
0.60 58 ksi 5.50 in.2
= 191 kips Shear yielding controls over shear rupture. Tension Rupture Component Ant 3.00 in. 0.5 , in. z in. 2 in.
= 1.25 in.2 U bs Fu Ant
1.0 58 ksi 1.25 in.2
= 72.5 kips LRFD
0.75 Rn 0.75 178 kips + 72.5 kips 188 kips > 90.0 kips
ASD
o.k.
2.00 Rn 178kips + 72.5 kips 2.00 125 kips > 60.0 kips
Design Examples V14.1
o.k.
Return to Table of Contents
IID-21
Bearing/Tear Out Strength of the 2 -in. Plate Holes are oversized ,-in. diameter. Check strength for edge bolt. l c
1.50 in.
,
in.
2
= 1.03 in. rn
1.2lc tFu 2.4 dtFu 1.2(1.03 in.)(2 in.)(58 ksi) 2.4(w )(2 35.8kips 52.2 kips
(Spec. Eq. J3-6a) in.)(58 ksi)
Check strength for interior bolts. l c
3.00 in.
,
in.
= 2.06 in. rn
1.2lc tFu 2.4 dtFu 1.2(2.06 in.)(2 in.)(58 ksi) 2.4(w 71.7kips 52.2 kips
(Spec. Eq. J3-6a) in.)(2 in.)(58 ksi)
Total strength for all bolts. r n = 1(35.8 kips) + 5(52.2 kips) = 297 kips LRFD
ASD
0.75 r n 0.75(297 kips)
2.00 r n
= 223 kips > 90.0 kips
o.k.
297 kips
2.00 = 149 kips > 60.0 kips
o.k.
Fillet Weld Required for the 2 -in. Plate to the W-Shape Beam Because the angle of the force relative to the axis of the weld is 90 , the strength of the weld can be increased by the following factor from AISC Specification Section J2.4. 1.5
(1.0 0.50 sin
) (1.0 0.50 sin1.5 90 ) = 1.50
From AISC Manual Equations 8-2, LRFD Dreq
Ru
1.50(1.392l ) 90.0 kips 1.50(1.392)(2)(6.00 in.)
3.59 sixteenths
ASD Dreq
P a
1.50(0.928l ) 60.0 kips 1.50(0.928)(2)(6.00 in.)
3.59 sixteenths Design Examples V14.1
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IID-22
From AISC Manual Table J2.4, the minimum fillet weld size is
x in.
Use a 4-in. fillet weld on both sides of the plate. Beam Flange Base Metal Check t min
3.09 D
3.09(3.59 sixteenths)
( Manual Eq. 9-2)
F u
65 ksi 0.171in. 0.345 in.
o.k.
Concentrated Forces Check for W16x26 Beam Check web local yielding. (Assume the connection is at a distance from the member end greater than the depth of the member, d .) Rn
Fywtw (5kdes l b ) = 50 ksi 4 in. 5 0.747 in. + 6.00 in.
(Spec. Eq. J10-2)
= 122 kips LRFD
ASD
1.00 Rn 1.00(122 kips) = 122 kips > 90.0 kips
1.50 Rn o.k.
122 kips
1.50 = 81.3 kips > 60.0 kips
Design Examples V14.1
o.k.
