Deflection in plate with load in middle of plate (5'x5')

[SilverBeam75]

Hi,

I do not have the Roark's book...and I need to evaluate the thickness required for a 5'x5' steel plate A36 simply supported on all edges. My load in the middle is 33 kips. The maximum deflection permitted is 0.25 in. How can I evaluate the thickness required ?

What is the formula ?

Thank you for the help

 

[desertfox]

Hi silverbeam

Have a look at this site:-

http://www.xcalcs.com/cgi-bin/tutti/x2calcs.cgi?d=i_4_0_2_0_5&l=en

desertfox

 

[DHKpeWI]

See attachment from Roark's. I think 1C will apply in your case.

These formulas are only valid when the deflection is less than 1/2 of the plate thickness.

 

[SilverBeam75]

Hi,

Thank you very much Desertfox, I'm not sure if I understand how to use is it.

Steel plate: E = 2.9E+7 psi (29000 ksi) and the thickness is 1 inch.

My W = 22000 lb applied on a surface 12in x 12in in the center of the plate which is 60in x 60in.

My q = 22000/(12x12) = 152 psi

The b = 1.28 (from table 1c)

my Beta = 0.16 and alpha = 0.022

Therefore: Ymax = -(alpha)*q*b^4/E*t^3

Ymax = -(0.022)*152*1.28^4/(2.9E+7)*(1)^3 = -309E-9 in.

I don't think its a correct value. Do you see my error ?

Thank you for your help,

 

[BAretired]

If the plate is simply supported, you will develop a large downward reaction at each corner. If the corners are free to lift off the supports, the central deflection will be affected.

BA

 

[JedClampett]

You're using the wrong formula out of Roark. The factors you're using are for 1d and for a increasing load along the length. Deflection isn't calculated for 1c. The closest case with deflection is 1b or for the load over a round area in the middle of the plate.
I get about .35 inches.

 

[SilverBeam75]

Thank you JedClampett

Can you provide the calc ?

 

[JedClampett]

Alpha = .1267 from Roark
W = 22,000 lb
b = 60 inches
t = 1 inch
E = 29,000,000 psi

Deflection = -(.1267 * 22,000 * 60^2)/(29,000,000 * 1^3)

 

[desertfox]

Hi SilverBeam75

Have a look at this file I have uploaded.
You need to transpose the formula if you want thickness to be your unknown and put all the other parameters in.
When its comes to "e" in the formula ie radius for the load I would just use 6".

desertfox

 

[BAretired]

According to "Theory of Plates and Shells" by Timoshenko & Woinowsky-Krieger, the expression for deflection of a simply supported rectangular plate is:

[ω] = [α]Pa²/D

[ω] = 0.01160 for b/a = 1 (length to width)
P = 33,000#
D = Eh³/12
a = 60"

D = [α]Pa^2/0.25 = 0.01160*33,000*60^2/0.25 =5,512,000

from which h = 1.32" (thickness required to limit deflection to 1/4").

The four corners must be held down against uplift.




BA

 

[BAretired]

In the above, D = Eh³/12 should read:

D = Eh³/12*(1-[ν]²)

i.e. h (thickness) = [12D(1-[ν]²)/E]^1/3

h = 1.28"

You need a thickness of 1.28" to hold deflection to 1/4".

BA

 

[SilverBeam75]

Very clear now

Thank you all for the help.

Its really appreciated.

 

[SilverBeam75]

Roark's theory is O.K., I would like to understand what BAretired suggested.

I have found the eBook "Theory of Plates and Shells" by Timoshenko & Woinowsky-Krieger SECOND EDITION

Where exactly is your formula coming from ?

 

[BAretired]

Article 34 - p. 141 including Table 23 - p. 143.

BA

 

[SilverBeam75]

How do you go from 4Pb^3/pi^4aD (page 141 art. 34)

to ? = ?Pa2/D ?