Blow out panel design

[strainstress]

Hello,

I am working on a blow out panel (hinged on one side). The panel will rupture at 6 psi pressure. After blowout, in order to prevent the panel from damaging building roof, the panel is restrained by four chains (See attached image). I have to figure out the size of chains required.

The approach I took was to calculate the total work done on the panel (See equation 1 attached) and equate it to the potential energy of the chain (See equation 2 attached).

In order to calculate the work done, I applied the 6 psi pressure to the panel and integrated it from 0 deg to 90 deg (See equation 1 attached).

The chain force I obtain using this approach is huge and does not make sense. I understand that the 6 psi pressure will only act before blowout and after blow out, the velocity pressure of the exiting air will perform work on the panel. Is it reasonable to assume the velocity pressure acting on the panel is 6 psi ?

If this is not the correct approach, can you please suggest what would be the correct approach to tackle this design problem ?

Thanks
StrainStress

 

[WARose]

The approach I would take (first) is to multiply the 6 psi times a Dynamic Load Factor (DLF) of 2, and let that be design load (which would then have to be factored as per the load combinations). This (a factor of 2) is the maximum response for a “ramp” loading you describe.

If that gets you something outrageous, you may be stuck evaluating Duhamel’s integral or using some sort of chart for the DLF. That may not be bad though because it’s been my experience that doing so really cuts down the DLF.

By the way, are you sure the force for the blowout panels is 6psi? The reason I ask is: in my (limited) experience with blast design: the whole point of having blast panels is that they will be knocked out (and relieve peak overpressure) before seeing those types of loads.

 

[rb1957]

personally, i'd work with 6psi loads, sine i want the lower edge of the cover to open at 6psi. and i'd forget the chains and such to retain the cover; i'd calc a load on the hinge for 6psi, double that to be reasonably sure that the hinge will retain the cover.

if you don't want the cover to fly open (pivoting 180deg about the upper hinge), something you might do is add a scissor linkage so the lower edge of the cover opens in a controlled manner. possibly held closed by springs that will vent at 6psi.

if you Really wnat to know about the loads on the cover during a vent, you're into a complex decompression analysis.

looking at your calc, you can only apply 6psi to the closed cover; once open the pressure on the cover will be different (it'll be higher for an instant then drop away very quickly as the ocver opens). the load on the cover is 6*A, reacted at the top edge and lower edge (where your chains are ?) ... 6*A/2 on either edge ... 2 chains means chain load = 6*A/4.

Quando Omni Flunkus Moritati

 

[msquared48]

Why don't you integrate some springs into the system to limit the dynamic force to the chains?

Mike McCann
MMC Engineering