B.3 Calculate base shear, V
V = CsW
FEMA 302 Eq. 5.3.2
Cs = SDS/R,
Eq. 3-7
but need not exceed Cs = SD1/TR,
Eq. 3-8
but shall not be less than Cs = 0.044SDS
Eq. 3-9
Transverse Direction:
Cs = (0.6)/5 = 0.12
Cs = (0.43)/(0.19)(5) =0.45 > 0.12
Cs = 0.044(0.43) = 0.02 < 0.12
V= CsW = (0.12)(388 kips) = 47 kips (209 KN)
Longitudinal Direction:
Cs = (0.6)/5 = 0.12
Cs = (0.43)/(0.19)(5) =0.45 > 0.12
Cs = 0.044(0.43) = 0.02 < 0.12
V= CsW = (0.12)(388 kips) = 47 kips (209 KN)
B.4 Calculate the vertical distribution of seismic forces
This building is a combination of one & two story area. The main building is one story while the
mezzanines act as two-story areas. It is assumed that the mezzanine diaphragms will not act to drive the
overall building response. Therefore, the building is analyzed as a one-story structure.
The upper roof metal decking acts as a flexible diaphragm distributing the shears to the vertical resisting
elements according to tributary areas. In the transverse direction, the end CMU walls resist of the
lateral force while of the force is resisted by the firewall. In the longitudinal direction, it is assumed that
each of the braced bays will resist of the shear force. The shear forces developed at the mezzanine level
will be distributed to the vertical resisting elements in relation to their rigidities due to the rigid diaphragm
action of the concrete topping. In addition, torsional forces developed must be considered.
Shear Forces to Roof & Mezzanine Diaphragms:
The seismic coefficient, Cs, is the same in both directions for this structure.
Roof level:
Fr = Cs x weight tributary to roof level
Fr = (0.12)(187 kips) = 22.4 kips (99.6 KN)
Mezzanine level:
Fm = Cs x weight tributary to the mezzanines
Fm = (0.12)(201 kips) = 24.1 kips (107.2 KN)
Due to symmetry, the mezzanine level forces are assumed to act evenly between each mezzanine.
Fm = (24.1 kips) = 12.1 kips / mezzanine (53.8 KN)
H1-12
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