R u = f r = fs2 + f b = (2.83k/in ) 2 + (2.58k/in ) 2 = 3.83k /in < 4.19 k /in

2

(0.67KN/mm < 0.73KN/mm)

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 =

=

F

(314.6X103 mm3 < 435.9X103 mm3)

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.

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.

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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