Question: COMPOSITE ( INSIDE FLOOR ) ( A ) ( 3 . 8 5 1 8 ) H 2 1 0 0 p s f =

COMPOSITE (INSIDE FLOOR)
(A)
(3.8518)H2100psf=6.980ib=6.93kps
(6)1.9(6.93)=9.702kFs
(B)
(7.71=18)ft2100PFf=13,878lb=13.678 Kips
(8)1.4(13.986)=19.429
c.ps
(D)
OEAD: [(3.3518)ft=100f]+[131403]16=7898lb=7.396 cips
LIVE I13(26.8)kips=B.933 kips
(2)4(1.38B)+1.6(8.933)=28190 hies
OEAD: [(3.3518)ft2100psf]+[131403]16=78981b=7.396 kips
(2)14(1.38B)+1.6(8.933)=51.190FiPs
bath only have
DEAD LOALOS
OEAD LOARS
-2n20 COMPOSITE
DECKING GALV. I=.400in4
LIGHT WEIGHT CONC WIMAX. UNIT
WEIGHT =110 P.C.F.+-30 P.C.F
weight of floor: 100psf
Tank Volume (84.375)(120.5)(73)H3=742,205 in =429.5H3
weight of water in tank (62.4ibft3)(429.5Ht')=26.6 kips
Density of empty tanke 59.93kft,0.035kin2
weight of tank
astumptions
Giter roth of polysibyiveDetermine if the beams have adequate moment strength to support a water tank based off of the assumptions from tributary area (attached) and information provided on the beams. Assume A992 steel and use the LRFD method.
COMPOSITE ( INSIDE FLOOR ) ( A ) ( 3 . 8 5 1 8 )

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