Roark Flat Plate Question

[jkate]

I have a what I feel is a stupid question.

I have a flat plate 18"x72"x0.375" thk. When using the Roark formulas, which dimension is "a" and which is "b"?

Thanks

 

[JStephen]

There should be a diagram in the tables that says.
In some cases, "a" will always be the longer dimension, and you'll have tables of a/b or b/a that don't extend beyond 1.00.
In some of the Roark load cases, there will be a section defining all the variables, etc., then tables of different load cases, and you may need to refer to that section.
Don't confuse the circular-plate tables with the rectangular plate tables.
If that doesn't help, let us know which edition of Roark's, which table, which load case, which page, and maybe somebody can look it up.

 

[jkate]

Unfortunately, I don't have my copy of Roark's with me, it's at home, so I don't know the edition. This is definitely flat plate, and I'm looking at Case 1 (4 EDGES SUPPORTED - UNIFORM LOAD)

 

[rn14]

Does this help?

 

[rn14]

 

[jkate]

Thanks.

 

[JedClampett]

I think you're going to have a problem with a plate 18 x 72. I would put Roark back on the shelf and design it one-way, spanning the short way.
If they did have a set of factors for a/b=4, it's going to approach the one-way span case.

 

[jkate]

I was planning on cutting the 18" span into thirds using two (2) L4x4x1/4" stiffeners that were 72" long. The pressure on the plate is 15 psi and the plate thickness is 0.375in.

 

[rb1957]

sorry, but why would you cut the short side ? i'd put the stiffeners on the long side, 3 18" long stiffeners. and the stiffeners are going to reactmost of the pressure load, about 24*18*15 = 6480 lbs, max moment = (15*24)*18^2/8 = 14580 in.lbs ... maybe a Zed stiffener, rather than an L angle ?

another day in paradise, or is paradise one day closer ?

 

[jkate]

rb1957, after I sent that last reply, I looked at the numbers again and decided to do what you suggested, put the stiffeners on the long side, (3) 18" long stiffeners.

 

[IRstuff]

My Mathcad version of Roark's (6th ed) describes "a" as the "length" and the example has a 10:3 aspect ratio, for Table 26 Case 1a. The table is defined for an aspect ratio up to 5. For your parameters and a modulus of elasticity of 30E6 psi, the calculated deflection is 0.139 in, and a maximum stress of 25.6 ksi

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

Of course I can. I can do anything. I can do absolutely anything. I'm an expert!

 

[rb1957]

yes, surprisingly (to me at least), this is a plate bending problem (deflection < t/2), and it weighs only 150 lbs. If you need to reduce weight, stiffeners would be very effective (at least 1/2 the weight).

look at a strip 1" wide, 18" long, 15 lb/in load, M = wL^2/8 = 608 in.lbs, stress = 6M/t^2 = 26 ksi ... sanity check.
note this is the roark solution for a infinite aspect ratio (a/b) ... 6/8*wL^2/t^2

another day in paradise, or is paradise one day closer ?

 

[jkate]

IRstuff:

Here's what I came up with:

Roark Case 1a
ymax = alpha*q*b⁴/Et³
Sb = beta*q*b²/t²

Carbon Steel Plate is 72"long x 18" wide, x 0.375" thk.
Pressure on plate, q = 14.7 psi
Max. Allow Bending Stress in plate = 18,000 psi

If I stiffen the 72" length with 3 stiffeners, the length of one "panel" is 18".
a/b = 18"/18" = 1
alpha = 0.0443
beta = 0.2874
ymax = 0.0443*14.7*18⁴/30E6*0.375³
ymax = 0.04321in
Sb = 0.2874*14.7*18²/0.375²
Sb = 9,734 psi

 

[IRstuff]

That's pretty close to what my Mathcad sheet cranks out. alpha and beta are interpolated from the table, so there's a bit of difference in the 3rd decimal, resulting in ymax = 0.0429 in, and Sb = 9720 psi

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

Of course I can. I can do anything. I can do absolutely anything. I'm an expert!

 

[rb1957]

what sort a carbon steel has an allowable of 18ksi ?

another day in paradise, or is paradise one day closer ?

 

[jkate]

My bad. Steel is SA36. I used 66% of yield = 23,760 psi

 

[jkate]

Process Eng'g just changed the design on me. The piece of equipment is essentially an oblong tank under vacuum. See attached sketch. How can I calculate if the two halves of 18" pipe are of adequate thickness for the full vacuum? I can't just do an ASME external pressure calculation since they are half pipe, correct? Need a little advice on how to determine if the 18" half pipes need stiffening.

Thanks

 

[rb1957]

it's not clear to me where the pressure will be acting ... guessing i'd say you've got a vaccuum inside the tube ? i'd certainly want some ribs/frames supporting the 6' span ...

but this looks such a change from the previous, that maybe you say "call me when you guys have a serious idea" ?

another day in paradise, or is paradise one day closer ?

 

[jkate]

I do plan to stiffen the 6' span. See the updated attached sketch.