wDL = 1,173plf (17.11KN/m)
wLL = 40psf(15' = 600plf (8.75KN/m)
)
Note: Floor live load reduction, per ANSI/ASCE 7-95, is not permitted because this is a one way slab and
the floor live load = 40psf < 100psf (1.92KN/m2 < 4.79KN/m2).
wu = 1.2(1.218klf)+1.6(0.60klf) = 2.42klf (35.29KN/m)
2.42klf (30' ) 2
= 272 ft - kips = 3264in- kips (368.8KN-m)
Mu =
8
3264 in- kips (1000lb /k )
Mu
= 101 - in3 (1.66X106 mm3)
Z req'd ≥
=
φb Fy
0.9(36,000psi)
w U L 2,420plf (30' )
= 36.3k (161.5KN)
φv Vn ≥
=
2
2
Deflection requirements;
5w L L4
5(0.60klf )(30' ) 4 (12"/1' ) 3
= 251.4 - in4 (104.6X106 mm4)
I req'd ≥
=
384E∆ allow
384(29,000ksi)1.5"
Choose W21X50 (W533.4mm X 0.73KN/m), Z = 110-in3 (1.80X106 mm3), I = 984-in4 (410X106 mm4),
φ Vn = 154-kip
v
d.
Equivalent Lateral Force Procedure
B-1 Calculate Fundamental Period, T:
Ta = C t h3/4
(EQ. 5.3.3.1-1 FEMA 302)
n
Transverse direction;
Ct = 0.035
Moment frame resisting 100% of the seismic
force.
Longitudinal direction;
Ct = 0.020
Braced frame system.
hn = 22-ft (6.71m)
Height to highest level.
Ta = 0.035(22' ) 3/4 = 0.36 sec
Therefore;
transverse
Ta = 0.020(22' ) 3/4 = 0.20 sec
longitudinal
B-2 Determine Dead Load,' '
W:
Building weights were calculated on spread sheet and are shown in Figures 2 and 3. Total seismic weight
is;
W = 129-kips (574KN)
B-3 Calculate Base Shear, V:
V = CsW
(EQ. 5.3.2 FEMA 302)
S DS
Cs =
(EQ. 3-7)
where;
R
S
Cs < D1
(EQ. 3-8)
TR
Cs > 0.044S DS
(EQ. 3-9)
Longitudinal direction;
R=6
(Table 7-1)
Cs = 0.57/6 = 0.095
< 0.37/{0.2(0.6)} = 0.31
> 0.044(0.57) = 0.025
Transverse direction;
R=8
(Table 7-1)
H4-8
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