Flat Plate design

[stan45]

I am currently working on a analysis of an aluminum structure that consistes of verical and horizontal aluminum tubes. & 1/8" thk. aluminum plates with 3/4" sq. holes at 1" c/c that are fastened to these tubes to create a screen wall. The structure is specified to resist a 100 lb point load distributed over a 1 sq. ft area per the elevator code. This is a elevator screen wall. The structure must meet a maximum deflection of 1" under this loading condition. I do not have access to finite elemnent analysis software. Is there any way to analize this otherwise? I was think of using Roark's equations but one of the stipulations is that the plate must be of uniform thickness. Since the plate has holes in it I am not sure how or if the equations can be modified.

 

[rb1957]

you could use a nominal thickness based on the remaining material (in a 1" square, you're removing 0.44in2; so the nominal thickness is 1/2 the plate thickness.

i don't think you'll see a deflection of 1" (for a 100 lb load) ... build a test. where are the reaction points ?

 

[stan45]

The aluminum panel is 1/8" thick and covers a open area approx. 50" x 50" (of which 44% is solid area). It is anchored to the face of the aluminum tubes with stainless steel fasteners on all four sides.

 

[rb1957]

is it (not that it matters much) 44% remaining or 44% removed ??

roark says about 14" (obviously unrealistic) for 0.06" thick.

i'd suggest building it, and planning on reinforcing it (with similar Al tubes as you're using on the edges), quartering the original panel. maybe leave out a row of holes (so it might look as though you knew what you were doing ??)

 

[BAretired]

3/4" holes at 1" centers leaves bands 1/4" wide in both directions. Take 100#, divide it by 2, i.e. 50# each way. This is carried by 12 bands, so figure a 4.17# load spread over 12" at midspan. Deflections will be a little conservative but not much.

BA

 

[stan45]

44 % remaining

 

[rb1957]

i know i'm being a bit of a dyck about this, but pi(.75)^2/4 = 0.44 ... 0.44in2 area removed, 0.56in2 area (remains on a 1in pattern)

 

[stan45]

Its a square pattern 1" c/c 3/4" square holes

 

[miecz]

Are you using Roark's article for large deflections? I don't see any formulas for concentrated loads.

 

[stan45]

In roark's Table 11.4 (1c) is uniform distibuted load over a small central rectangle. Thiss is a 100 psf load over a 1' x 1' area

 

[rb1957]

that's what you get for not reading closely !

i used roark's solution of a concentrated load (a load distributed over a small area).

what if the point load is applied to the centre of one of the holes ?? ?)

 

[BAretired]

A 1/8" thick plate seems much too thin to span 50", with or without holes.

BA

 

[stan45]

The 100 psf load over one sq. ft. is meant for a safegaurd.

 

[miecz]

I guess you're not considering large deflection theory. In this case, I get a deflection of 1.6" without any holes. The holes will increase this by a factor of 2, based simply on the amount of material removed.

 

[rb1957]

looks like we concur that the deflections might (will probably) be larger than your limit. consider 1/4ering the panel (into 25" squares). what sort of vibrations are around ??