Plate Buckling (All Sides Built-In) - LEVY (1942) Solution

[Prathik123]

Hi All,

For a rectangular plate buckling with all sides built-in, Levy has solved this problem in his 1942 paper: "BUCKLING OF RECTANGULAR PLATES WITH BUILT-IN EDGES", S. LEVY, JOURNAL OF APPLIED MECHANICS, VOL. 9, PG. 171, 1942.

In his paper, the buckling coefficient, K is determined by solving Equation 8, which is a doubly infinite set of equations. The solved values are summarized in Fig. 2. This figure is famously found in Bruhn Fig. C5.2 or C5.3.

However, I am trying to mathematically solve Equation 8. Can someone please help me figure this out or point me in the right direction? I am trying to code this in Matlab, so if someone can give me advice via code (of any language) is also fine by me.

Any help is greatly appreciated.

 

[Prathik123]

Hi guys,

Anyone have found a way forward?

Just to summarize what I have tried so far:


Attempting to solve for a/b = 1, and with t1 and k1 to k7 determinants as a 1st trial like Levy did in Table 2, the K solution should be 10.047.
These determinants means that n = 1 and m = 1, 3, 5 and 7. Levy did not take even m variables. Anyway, there should be 5 equations since there are 5 variables. Equation 8a should give 1 equation and Equation 8b should give 4.

Now, expanding the infinite series (I hope I am doing this part correctly):
Equation 8a (m = 1 to 7, n = 1):

(k1 + t1)/(1-K) + (3k3 + 9t1)/(100-9K) + (5k5 + 25t1)/(676-25K) + (7k7 + 49t1)/(2500-49K) = 0
Reordering:
(1/(4-K))*k1 + (3/(100-9K))*k3 + (5/(676-25K))*k5 + (7/(2500-49K))*k7 + ( 1/(4-K) + 9/(100-9K) + 25/(676/25K) + 49/(2500-49K) )*t1 = 0

Equation 8b (m = 1, n = 1):
(k1 + t1)/(4-K) = 0
Reordering:
(1/(4-K))*k1 + (1/(4-K))*t1 = 0

Equation 8b (m = 3, n = 1):
(k3 + 3*t1)/(100-9K) = 0
Reordering:
(1/(100-9K))*k3 + (3/(100-9K))*t1 = 0

Equation 8b (m = 5, n = 1):
(k5 + 5*t1)/(676-25K) = 0
Reordering:
(1/(676-25K))*k5 + (5/(676-25K))*t1 = 0

Equation 8b (m = 7, n = 1):
(k7 + 7*t1)/(2500-49K) = 0
Reordering:
(1/(2500-49K))*k7 + (7/(2500-49K))*t1 = 0

Now putting all the coefficients in front of the variables into a matrix, gives me the matrix below:

| 1/(4-K) 3/(100-9K) 5/(676-25K) 7/(2500-49K) very long t1 term |
| 1/(4-K) 0 0 0 1/(4-K) |
| 0 1/(100-9K) 0 0 3/(100-9K) |
| 0 0 1/(676-25K) 0 5/(676-25K) |
| 0 0 0 1/(2500-49K) 7/(2500-49K) |

The K value which sets the determinant of the above matrix to 0 is the solution.

Is this approach correct? Because I can't get logical values when I try and compute the determinant.

Regards,
Prathik