R u = f r = fs2 + f b = (2.83k/in ) 2 + (2.58k/in ) 2 = 3.83k /in < 4.19 k /in
2
O.K.
(0.67KN/mm < 0.73KN/mm)
Use four 3/16-in. (4.8mm) welds to secure tank to perimeter beams
Check transverse perimeter beams;
The transverse perimeter beams support relatively little of the tanks dead weight, but resist the overturning
reaction. This reaction was previously calculated to be Pu = 20.7k (92.1KN). These beams were sized in
the gravity load design as W10x22' and will be checked here. It is conservatively assumed that the beams
s
support only the overturning seismic reaction.
The maximum moment occurs at mid span as;
Pu L 20.7 k (10' )(12"/1)
= 621in- k (70.17KN-m)
Mu =
=
4
4
Note: Beam is laterally supported throughout its span
621in- k
Mu
= 19.2 - in3 < 26.6 = ZW10x22
Z req'd =
=
φ y 0.9(36ksi)
F
(314.6X103 mm3 < 435.9X103 mm3)
O.K.
Keep W10x22 (W254mmX1.05KN/m) for transverse beams
Check longitudinal beams;
The longitudinal beams support all of the tanks dead weight (transferred to it from the interior perimeter
beams), and also resist the overturning reaction of Pu = 20.7k (92.1KN). These beams were sized in the
gravity load design as W10x22' and will be checked here.
s
Per load combination ` = 0.9D + E'the center load P1u in figure J2-4 is reduced to an uplift load of;
U
20.7 19.1 = 1.6k (7.12KN)
k
k
(uplift)
By inspection, this reduces the end reactions and the maximum moment acting within the beam. Therefore,
the W10x22 is still adequate.
Keep W10x22 for longitudinal beams
Design transverse beam to column connection;
The worst case beam reaction is Ru/2 = 20.7k/2 = 10.4k (46.3KN). By inspection, the same single plate
connection used for the other beam to beam or beam to column connections is adequate.
Use two 7/8-in. (22.2mm) φA325-N bolts with 1/4-in. (6.4mm) single plate and 3/16-in. (4.8mm)
welds similar to Figure J2-5
Check column for combined loading (see Figure J2-7);
Determine design loads;
Calculate reactions;
From symmetry;
R1H = R2H =(1/2)(45.2k + 4.15k) = 24.7k (109.9KN)
∑M
= 0;
2
(R 1V )10'- 45.2 k (15' ) - 4.15k (10' ) = 0
R1V = 72k (320.3KN)
(tension)
∑F
= 0;
y
R2V = 72k (compression)
Calculate compressive force in column;
Summation of loads at point 2;
Due to the 45 degree inclination of the brace;
(Fbrace ) horz = (Fbrace ) vert = 24.7k (109.9KN)
∑F
= 0;
y
(Fcol ) vert = R 2V - (Fbrace ) vert = 72 k - 24.7 k = 47.3k (210.4KN)
Superimposing the dead load;
From load combination ` = 1.2D + E'
U
;
J2-8
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