R n = 2.4dtFu
φ= 0.75
with;
(EQ. J3-1a AISC LRFD)
φ n = 0.75(2.4(3 / 8" )3 / 16" (58ksi)) = 7.35k > 0.83k = Vu (32.7KN > 3.69KN)
O.K.
R
Check combined tension and shear;
The tension limit stress is;
59 - 1.9f v ≤45 ksi (310.3MPa)
(per AISC LRFD Table J3.5)
Therefore,
0.83k
Pu
fv =
=
= 7.55ksi (52.1MPa)
A b 0.11 - in2
59ksi - 1.9(7.55ksi) = 44.7ksi < 45ksi Therefore, Ft = 44.7ksi (308.2MPa)
φ n = φ t A b = 0.75(44.7ksi)0.11 - in2 = 3.69 k > 0.83k = Tu (16.4KN > 3.69KN)
O.K.
R
F
Use a single 3/8" φbolt for connections to pipe
Design connections to concrete;
Note: For anchors in concrete without special inspection, Section 9.2.1 in FEMA 302 requires an
additional load factor of 2.0. Therefore;
Pu = Vu = 2(0.83k) = 1.66k (7.38KN)
Try a single 3/8" (9.53mm) φA307 bolt in a bearing type connection;
Check capacity in shear;
Steel;
Vs = (0.75A b Fu )n
(EQ. 9.2.4.2-1 FEMA 302)
For 3/8" (9.53mm) φbolt (A 307)
where; Ab = 0.11-in2 (71.0mm2)
Fu = 60ksi (413.7MPa)
n = 1-bolt
Vs = 0.75(0.11 - in2 )60ksi(1.0 - bolt) = 4.95k > 1.66k = Vu (22.0KN > 7.38KN)
O.K.
Concrete;
φ c = (φ 00A b λ fc' )n
(EQ. 9.2.4.2-2 FEMA 302)
V
8
where; φ= 0.65
λ= 1.0
(normal weight concrete)
fc'= 4,000psi (27.6MPa)
n = 1-bolt
∴ φ c = 0.65(800)0.11 - in2 (1.0) 4,000psi = 0.65(5.57 k ) = 3.62 k (16.1KN)
V
φ c = 3.62 k > 1.66k = Vu (16.1KN > 7.38KN)
O.K.
V
Check capacity in tension;
Steel;
Ps = (0.9A b Fu )n
(EQ. 9.2.4.1-1 FEMA 302)
2
2
where; Ab = 0.11-in (71.0mm )
Fu = 60ksi (413.7MPa)
n = 1-bolt
∴ Ps = 0.9(0.11 - in )60ksi(1 - bolt) = 5.94 k > 1.66k = Pu (26.4KN > 7.38KN)
2
O.K.
Concrete;
φ c = φ f c' (2.8A s ) n
λ
(EQ. 9.2.4.1-2 FEMA 302)
P
where; φ=0.65
λ= 1.0
(normal weight concrete)
fc'= 4,000psi (27.6MPa)
As = π 3")3" = 28.3-in2 (18.25X103 mm2) (for a 3-in. (76.2mm)
(
embedment)
J3-6
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