Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

interior metal pedestrian doors 1

Status
Not open for further replies.

aluminumb

Mechanical
Sep 11, 2015
7
Does anyone have any input and/or recommendations as to the engineering practices for designing interior metal pedestrian doors. For example if the sheeting that is backed with structural members can be pushed all the way to yield stress. This would mean the use of no safety factor etc. The door would have horizontal structural members spaced at 24 inches on center. The sheeting would span vertically between the horizontal members and would have a sole purpose of only supporting and carrying the wind pressure or suction to the horizontal members. Any advise or a recommended source to research the topic would be appreciated. The biggest obstacle is an occasional 60 psf fan pressure that the doors may be periodically subjected to. Thanks.
 
Replies continue below

Recommended for you

You could calculate the ultimate capacity of the sheet using yield line theory, then apply a reasonable factor of safety.

If the width of door is in the same order as the spacing of the supporting members, the sheet will tend to span in both vertical and horizontal directions.

If the deflection is large compared to the thickness of the plate, you may wish to consider "Large Deflection of Plates" as outlined in "Theory of Plates and Shells" by Timoshenko and Woinowsky-Krieger.

BA
 
Thanks BA for your hints on lean plate/sheet engineering concepts.
I looked into yield line theory and it does seem to be a system that takes a little practice in order to apply. Did not seem to be much for application examples out there, but none the less, interesting. One thing that seems unknown or arbitrary is to mix ultimate capacity with factor of safety, as one seems like lrfd & the other from asd.

The two-directional spanning also seem possible in this case since the door is not too much wider than the span of the supporting members for the sheet. We do use SolidWorks simulation occasionally and this may be a more suited scenario. When we do a simple hand calculation, we at least model with fixed ends, as opposed to just simple span analysis.

This scenario would most fall into a thin plate/sheet with a large potential deflection(several times more than the thickness of the sheeting). I found some information about it on the web but struggled to find the proper procedures in order to solve with confidence.

I was hoping to be able to keep as simplified as possible so that it would be rather easy to explain to an attorney if ever questioned. Is there anything that says that one could not use the good old allowable stress system and push it to yield with the use of a factor of safety of 1.0? The consequences are low if the sheeting did deflect under load. It would not in theory go beyond yield. It would be easy to explain to someone else as long as this is not breaking any code or specification requirements.

Thanks!
 
Aluminumb:
It seem to me that there are at least a hundred other things to worry about before you worry about how to do a stress analysis on these doors or how to explain that process to an attorney, for whatever the later is high on the list. Are you going to manufacture these doors or are you helping an attorney with a legal case? Get a couple metal clad interior doors, used or new and cut them into cross sections so you can study how they are constructed. What are your doors going to be made of, all the various components and their dimensions, thicknesses, etc? Are they solid or semi-solid core or hollow core doors; are they fire rated? What type of service are they expected to see, single family and aprt. residential, or commercial like schools, warehouses, etc? Are they just a door to be hung individually or are they pre-hung in their own frame? The two styles and the top and bot. rails have to be fairly strong for mounting and trimming to size, and 24" seem like a very large spacing for a thin skin. The thin skin will span in both directions, no doubt. Hollow core doors are notoriously prone to damage and need frequent replacement in all but the least abusive environments. What is the core material, a stressed skin solution spanning btwn. styles and a few rails, and bonded to some filler mat’l., is a pretty good solution and stands up to some abuse. What tooling and manufacturing equip. is going to be needed to build these?
 
dhengr brings up some interesting points which I had not thought much about. If the core is filled with a material such as polystyrene, then pressure from one side would be resisted by door panels on both sides of the door, probably in approximately equal proportions. If the core is hollow, then only one panel resists pressure.

aluminumb said:
I was hoping to be able to keep as simplified as possible so that it would be rather easy to explain to an attorney if ever questioned. Is there anything that says that one could not use the good old allowable stress system and push it to yield with the use of a factor of safety of 1.0? The consequences are low if the sheeting did deflect under load. It would not in theory go beyond yield. It would be easy to explain to someone else as long as this is not breaking any code or specification requirements.

1) Yield Line theory is pretty simple to apply but it does have some drawbacks, particularly if you are trying to explain it to a lawyer. I have used yield line theory very often in the design of steel plates, but never for extremely thin veneer. It would not be appropriate where there are large deflections as it assumes the veneer acts only in bending.

2) The good old allowable stress system assumes elastic behavior and cannot be pushed to yield with a F.S. of one without abandoning the usual theory.

3) The primary consequence of large deflection is that load is resisted by a combination of bending and membrane stress, not bending stress alone. With large deflections, the plate behaves differently so any bending theory would be inappropriate.

If the service load is 60 psf, a load factor of 1.5 gives a factored load of 90 psf. The usual procedure is to assume a yield line pattern, then determine the factored moment per unit length required to resist the factored load and finally, determine the sheet thickness required to provide the factored moment calculated above. It may be necessary to experiment with more than one yield line pattern, but the calculations are fairly simple; simply equate internal and external energy work for the panel.




BA
 
dhengr:
This happens to be an old design(probably 20 years old or so). Interior metal pedestrian doors with a potential, but short duration 60 psf fan pressure load. There are no reported issues or concerns about the performance of these existing doors, therefore no legal case involved. Although some similar conditions were presented to design for just recently, and it was questioned why the sheeting would be thicker and the members were closer spaced than what was used in the past. So basically was being question on the engineering practices that are being used today versus 20 years ago. The doors(mild steel) are 2 inch thick and produced with a frame and are not required to have a fire-rating. No load transfer resistance would be possible between the face wythes. Full service fab shop to produce doors.
Although looks like more homework and analysis in order to be compliant with valid calculations. I appreciate the heads up on the analysis of bi-directional effects and the thin sheeting with respect to large deflection an validity.
Currently checking to see if SolidWorks simulation software incorporates enough related theory in order to produce a valid result to check alongside some of the existing theories that have been presented in this discussion thread.

Thanks!
 
For an interior panel, without considering bi-directional effects or large deflections, the sheet would fail when it reached yield at each support and at midspan. The sheet must develop a unit moment of:

m = wL2/16 in all three locations

If w = 1.5*60 = 90 psf or 0.625 psi then m = 0.625(24)2/16 = 22.5"# per inch

m = Z*Fy
If Fy = 30000 psi, Z = m/30000 = 0.00075 = t2/4 so t = 0.055" or 16 gage.

End panels would be more critical unless the sheets could be clamped at the exterior supports or unless the span could be reduced below 24".



BA
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor