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# RE: Seismic loads on mechanical and electrical components

• To: <seaint(--nospam--at)seaint.org>
• Subject: RE: Seismic loads on mechanical and electrical components
• From: "Gunnar Hafsteinn Isleifsson" <gunnarhi(--nospam--at)post4.tele.dk>
• Date: Thu, 6 Dec 2001 02:29:28 +0100

```The design criteria is meant to be all-encompassing. I have though excluded chimneys and tanks. Worst case scenario for mechanical or electrical components, acc. to ASCE 7-95, table 9.3.3.2,  is thus when ap/Rp equals 2.5/3.0 (stacks, panelboards, etc). In my case Ca = 0.3 and Ip = 1.0, so the design equation reduces to

Fp = ap/Rp*Ca*Ip*[1 + 3*x/h]*Wp = 0.25*[1 + 3*x/h]*Wp
or
Fp = 0.25*Wp	at grade level (using x = level of attachment)
Fp = 1.0*Wp		at roof level (x = h)

This reminds me of the reason I wrote in the first place  8-)

Do any of you guys have the 1994 NEHRP Provisions (FEMA 222) at hand to check if it is an error in ASCE 7-95 when it says that Fp shall be applied both vertically and laterally. As mentioned before; UBC94, UBC97, 1997 NEHRP (FEMA 302) and 2000 NEHRP (FEMA 368) all say that Fp shall only be applied horizontally.

Regards,
Gunnar Hafsteinn Isleifsson
Denmark

-----Original Message-----
Sent: Thursday, December 06, 2001 01:36
To: 'seaint(--nospam--at)seaint.org'
Subject: RE: Seismic loads on mechanical and electrical components

The 2.5 ap is for flexible equipment. Do you really have that? If you are on
tall and slender and able to wobble? I would venture to guess that most
equipment is rigid if it comes pre-assembled from a manufacturer and
contained within sheet metal walls on a base frame.

Otherwise, ap =1.0 and Fp=1/1.5*Ca*Ip*Wp=0.67CaIpWp

-gerard
SJ, CA

-----Original Message-----
From: Gunnar Hafsteinn Isleifsson [mailto:gunnarhi(--nospam--at)post4.tele.dk]
Sent: Wednesday, December 05, 2001 4:05 PM
To: seaint(--nospam--at)seaint.org
Subject: RE: Seismic loads on mechanical and electrical components

I don't have the UBC97 at hand at the moment, but I seem to recall that the
worst case ratio of ap to Rp equals 2.5/1.5 which results, at base level, in
Fp = 1.67*Ca*Ip*Wp  or far higher than the minimum base shear. Whether to
use height of CG or height of attachment point would also have a big impact
on the seismic force, especially for a tall component.
The UBC97 says that this force shall be used in LRFD or ASD load
combinations with "rho" equal to one. I understand this so that the seismic
effects, E = rho*Eh + Ev, shall be calculated as

Horizontal effects:	Eh = Fp
Vertical effects:		Ev = 0.5*Ca*Ip*Wp (for LRFD, can be
taken as zero in ASD)

Regards,
Gunnar Hafsteinn Isleifsson
Denmark

[snip]

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