huangyhg

equipment oshpd compliance
i work in a multi-discipline manufacturing company and i am evaluating cooling towers that go with our package. whole package needs to be osphd compliant. cooling tower manufacturer claims that their towers withstand seismic loads up to 0.5g horizontal acceleration. question is how to calculate lateral seismic force. here are my numbers:
seismic design parameters (cbc 2001) zone 4, ca=0.66, ip=1.5, rp=3.0, ap=2.5
lateral force on elements of structures, nonstructural components and equipment supported by structures per cbc
(1632.2) design for total lateral force.
(32-1) fp = 4.0 ca ip wp = 4.0 x .66 x 1.5 x wp = 3.96 wp
or "alternate method"
(32-2) fp = (ap ca ip)/rp[1 + 3 hx/hr] wp = (2.5)(.66)(1.5)/3.0[1 + 3 (20/20)] wp = 3.30 wp
(i used hx=hr=20 ft because cooling tower sits on steel platform 20 ft above grade. is it correct?)
or
seismic evaluation procedures for hospital buildings per the office of statewide health planning and development (oshpd) chapter 6: (2.4.6) demand on parts and portions of the building: 鈥渆quipment supported by a structure and their attachments, as identified in the building evaluation procedures, shall be evaluated to verify that they are capable of resisting the seismic forces specified below.鈥?br />
total lateral seismic force
fp = 0.67 ( av cc wc ) = (0.67)(.4)(2.4) wc = 0.64 wc
av=0.4 oshpd figure 2.1a
cc=2.4 oshpd table 2.4.6
which of these methods should be used and what compare with given horizontal acceleration of 0.5g?
thank you very much,

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