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

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

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

(OP)
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
 

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

(OP)
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,


 

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

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

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

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.

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

(OP)
Thank you JedClampett

Can you provide the calc ?

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

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)
 

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

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

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

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

ω = αPa2/D

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

D = αPa^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

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

In the above, D = Eh3/12 should read:

D = Eh3/12*(1-ν2)

i.e. h (thickness) = [12D(1-ν2)/E]1/3

h = 1.28"

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

BA

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

(OP)
Very clear now

Thank you all for the help.

Its really appreciated.

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

(OP)
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 ?

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

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

BA

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

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

to ω = αPa2/D ?

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