Buckling Under External Pressure of Cylindrical Shells Evenly Stiffened by Rings Only

Author:

BIJLAARD P. P.1

Affiliation:

1. Bell Aircraft Corporation,

Publisher

American Institute of Aeronautics and Astronautics (AIAA)

Reference42 articles.

1. T h e only complete test d a t a available to t h e writer are those of reference 19. T h e total length of the shell is Lt= 33.75 in., the ring spacing L = 5.27 in., a = 13.375 in., and t = 0.062 in. F o r t h e solid rings b = 0.656 in. and t' = 0.1875 in. T h e eccentricity e of the shell, determined according to reference 21, was 0.374 L T h e yield stress of the material was 54,400 psi. Panel buckling occurred at p = 80 psi. Panel Buckling (Mode A)

2. in Eq. (38) depends on \/b = 2.47/0.656 = 3.76 and k = b2t'crr/(ir2Nr). Assuming o> = 10,000 psi, one ends up with pcr= 80 psi. Using nowp = 80 psi, one obtains from page404of reference 14, with v = 0.3, 2a= [3(1 - v2)]mL/{at)l/2= 7.45 and /3 - 2a/L = 7.45/5.27 = 1.413. T h e n from Eq. (247) of reference 14, since all values % are practically equal to one, h = t = 0.062 in., and A = AT= btf= 0.123 in., one obtains P = Pr= 0.825 p = 66 Ib./in., so t h a t o> = Pra/Ar= 7,200 psi. Further, Nr= Et*/[12(1 - v2)} = 18,100 Ib.in., so t h a t k = 0.00325. With these values of \/b and k, from reference 12, SIU= 0.2203 Nr/b = 6,080 Ib.in./in. Measuring b up to the middle plane of the shell, in Eq. (38), so t h a t b = 0.656 + 0.031 = 0.687 in., it gives (3r= 23,100 lb.in./in. F r o m Eqs. (39), (40), and (41), & = - 3 8 0 , Pv= 23, and 13R= 22,743 Ib.in./in. Hence, in E q . (7), |8 = &R/2 = 11,371 Ib.in./in., c = L / 2 - 2.635 in., and IV = £*3/[12(l - ^2)] = 654 Ib.in., so that, by trial and error, I = 3.57 in. and l/a = 0.267. F r o m Eq. (29) again, n = 17. F r o m Eq. (27) now (p„)8h= 181 psi, so t h a t for steel shells E q . (18) yields (l/a)c= 2.23. T h e n E q . (31) gives per= 0.78 (181) = 141 psi, so t h a t (TyCT= (a/t)pcr= 30,400 psi. This is in the elastic range, so t h a t Et/E = 1. T h e slenderness of the equivalent column is s5= ir \/E/o-ycr- 98.7; rs= 0.289 X t = 0.018 in.; e = 0.374/ = 0.0232 in. From Eq. (95) ee= [98.7(0.018)/2.47]2e - 0M9e - 0.0120 in. F r o m E q . (27) of reference 20, with

3. = 1 + (21,000 - 0.235

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