1、TECHNICAL NOTE 2612 STRESS PROBLEMS IN PRESSURIZED CABINS By W. Flugge Stanford University Washington February 1952 Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-Provided by IHSNot for ResaleNo reproduction or networking permitted without license f
2、rom IHS-,-,-NATIONAL ADVISORY COMMITTEE FOR mRONAUTICS TECHNICAL NOTE 2612 STRESS PROBUMS IN PRESSURIZED CABINS SUMMARY The report presents information on the stress problems in the analysis of pressurized cabins of high-altitude aircraft not met with in other fields of stress analysis relating to a
3、ircraft. The material may be roughly divided into shell problems and plate problems, the former being concerned with the curved walls of the cabin or pressure vessel and the latter being concerned with small rectangular panels of its walls, framed by stiffeners, but not necessarily plane. INTRODUCTI
4、ON The analysis of pressurized cabins of high-altitude aircraft pre- sents particular stress problems not usually met with in other fields of stress analysis relating to aircraft, It is the purpose of the present report to gather information on these problems and to make it easily accessible to airc
5、raft engineers. Some of the work in this field is presented in references 1to 10. This report contains a choice of subjects taken from the theory of plates and shells which is expected to be useful for the designer of pressurized airplane cabins or similar Lightweight pressure vessels. This material
6、 may be roughly divided into shell problems and plate problems, the former being concerned with the curved walls of the cabin or pressure vessel and the latter, with small rectangular panels of its walls, framed by stiffeners, but not necessarily plane. As far as shell problems are concerned, some u
7、se has been made of a manuscript for a book on “stresses in Shells,“ which the author is preparing. (See reference 3.) The prospect that this book will be available sonie time in 1952 makes it possible to discuss in the present report several problems which are too complex to explain here in all the
8、ir mathematical details. The pressurized cabin is a rather new element in the airplane structure ah,d will, in all probability,. undergo future development. Pn view of this situation, no attempt has been made to present anything like a textbook on the subject giving time-tested methods for solving P
9、rovided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACA TN 2612 various problems, but rather an attempt has been made to show the general P lines of thought which have proved to be useful and to give suggestions for their application. w This investigatio
10、n was carried out at Stanford University under the sponsorship and with the financial assistance of the National Advisory Committee for Aeronautics, SYMBOIS X9Y9z rectangular coordinates , fl, 6 angular coordinates UY VIW displacements a radius of cylinder or sphere a,b sides of rectangle; axes of e
11、llipse or ellipsoid 2 span of beam thickness of plate or shell pressure difference between interior and exterior of cabin distributed load on shells (force per unit area of middle surface), in directions , 8, and radial normal forces in shells (force per unit length of section), in direction , 8, or
12、 x bending moment in plates and shells (moment per unit length of section) twisting moment in plates (moment per unit length of section) normal stress shear stress elastic modulus Poisson s ratio Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACA T
13、N 2612 - SHELL PR0BL;EMS Cylindrical Shell Circular cylinder.- The fuselage of a high-altitude passenger plane is usually of circular cross section and is, for most of its length, almost cylindrical. Some useful information regarding its strength may be found, therefore, when a circular cylinder clo
14、sed at both ends by some kind of bulkhead which permits the air pressure inside tc be greater than that outside (fig. 1) is considered. The pressure difference will be called p. For a homogeneous shell of thickness t the stresses produced by this pressure are given by the well-known boiler formulas
15、for hoop stress a and axial stress a,: The shell of a pressure cabin is reinforced by rings and stringers, which participate in carrying the load. The stringers will always be spaced closely enough to make the distribution of the longitudinal stress on the skin between them practically uniform. With
16、 the rings this may be different. The limiting case, that is, that they too are closely ,spaced, will be considered here. In finding the stresses, start fromthe internal forces per unit length of section acting in the shell, When a slice of length Ax = 1. is cut out of the shell (fig. 2), the hoop f
17、orce is found, and when the force prra2 acting on each bulkhead is distributed over the circumference 2rra sf the cylinder, the longitudinal force transmitted by the unit length of a section right across the shell is found. Provided by IHSNot for ResaleNo reproduction or networking permitted without
18、 license from IHS-,-,-4 NACA TN 2612 If the shell has no stiffeners, the stresses u$ and ax are found by dividing N and Nx by the wall thickness t, which, of course, results in the boiler formulas (1). In the cabin shell are rings of cross section An at distance 2 from each other and stringers of cr
19、oss section AL at an angular distance 6 (see fig. 3). If these areas are distributed over the cross section of the skin, the effective thicknesses are introduced; however, the stresses u and ux are not simply the quotients $/t$ and 4/tX (see, e. g., reference 1). The reason for this is apparent when
20、 one considers the fact that the skin is in a two- dimensional state of stress and therefore for the sane strain its stress is different from that in the stiffeners. Let the stresses in the skin be u and ax as before, in the stringers, UL, and in the rings, an. Then Hookes law will yield the followi
21、ng relations for the hoop strain , and the longitudinal strain EX: E being Youngs modulus and v Poissons ratio. On the other hand, the definition of the internal forces is: Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACA TN 2612 Solving the four
22、 equations (3) and (4) for the stresses, t.$ + V (tg - t).Nx = (f - v2)tatx + v2t(tg + tX - t) D1 - v2)tx + v2tN$ - vtNx b = (1 - v2)tgtx + v2t (tg + tx - t) Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-i NACA TN 2612 When the sings are far apart,
23、 these formulas are no longer appli- cable, The problem must then be split, with the shell without rings considered first and the influence of the rings introduced afterward (see sect ion entitled “nteraction between Shell and ,Rings1). When there are no rings t = t, and the formulas are simplified
24、considerably: - pa(1-2v) vpa ax = +- 2-t x t It appears that UL is always smaller by a factor 1 - 2V than it would be if it were obtained by simply distributing Nx over the whole section. For the skin stress crx the factor depends on the ratio aGt, and if one writes the factor k will be as shown in
25、figure 4. For AL/G = 0 the boiler formulas are valid, and ax = 0.506. For AL/aGt = 1.0, the diagram shows ox = 0.4crg. The difference between these two values of uX is small, but both are much less than the hoop stress. This is very desirable since the over-all bending of %he fuselage due to air for
26、ces acting on the control surfaces produces additional stresses ax which must be superimposed on the stress ux due to cabin pressure. . Since the stringers take an important share of the axial load, it - is not good practice to interrupt them at the rings. Care should be taken to insure that the for
27、ces carried by =the stringers can go straight through from one bulkhead to the other, or to the end of the cabin shell. Double cylinders.- The circular cross section is certainly the best one, both for aerodynamic and structural reasons. However, it has some practical disadvantages when used as a pa
28、ssenger cabin, Most serious Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACA TI? 2612 is the fact that a horizontal floor must necessarily be built in, requiring additional weight; and leaving beneath it space which is not easily used. This situa
29、tion is improvea by a cross section which, with some exaggeration, is shown in figure 5. It consists of parts of two circles and a straight horizontal tie between them. Begin with a discussion of the weight of this structure. Under the action of an internal pressure p the hoop stresses in the upper
30、cylin- der and in the lower cylinder will be : The stress in the tie follows from the equilibrium at its ends (fig. 6): u3t3 = u t cos al + a t cos a2 81 1 Id2 2 = p(al con al + a2 cos a2) If tl, t2, and t3 are chosen such that the three stresses are all equal to the same value u given by the allowa
31、ble stress in the material, tg = $(al cos al + a2 COS a2 ) The material invested in the structure is given by the area $, of its metallic cross section. The two circles contribute to it Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACA TN 2612 2tl
32、al(n - al) + 2t2a2(n - a2) = $Ea12(n - al) + 2a2 (“ - %)I and the tie contributes P 2t3c = 2 (a1 cos al + a2 cos ,a2)c Now / c = a1 sin al = a2 sin a2 and hence 2t3c = P 2(al2 cos al sin al + a22 cos 9 sin 9 u ) = 2 (a12 sin 2al + a22 sin 2%) (J Summing up the three parts, the total metallic cross s
33、ection is found to be: 1 A, = 2 2E12(n u - a1 + sin 2al) + a22(n - a2 + sin 2% )1 On the other hand, the area of the hollow cross section Ah describes the useful space in the fuselage. It is 1 h = a12(. - al + $ sin 2a1) + a22(n - a;, + 2 sin 2a2 ) It is seen that the ratio of the two areas Provided
34、 by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-2C NACA TN 2612 - - does not depend upon the particular choice of the dimensions of the cross section. For a simple circular cylinder of radius a there is obtained by similar reasoning: Ah = xa 2 and hence as b
35、efore. This indicates that for the same inside space the same structural weight is required and one is freed from weight considera- tions when choosing that combination of al, a2, al, and a2 which 4 seems best for other reasons The validity of this result is restricted to cross sections where the ti
36、e acts in tension, and this is exactly the configuration which is most interesting in aircraft construction. In practical applications, of course, additional stresses will change the picture to some extent, and the weight of different shapes will not be equal, but the important fact remains that the
37、se is no first-order loss or gain in choosing one or another of the sections compared. Interaction between Shell and Rings Bending of a cylindrical shell.- If the rings are not spaced closely enough to be considered as part of an anisotropic shell, the problem illustrated by figure 7 must be treated
38、. Cut the shell in the plane of the ring and at its connection with the ring. The pressure p applied to the shell will lead to a hoop strain E# which may be found from Hookers law (3) and formulas (6a) to (6b) and consequently will lead to a radial displacement w = Provided by IHSNot for ResaleNo re
39、production or networking permitted without license from IHS-,-,-0 I NACA TN 2612 The ring receives no load and, therefore, has no deformation. In order - to close the gap between the ring and the deformed shell, add shearing forces T per unit length of the edges. In the ring of cross section AR thes
40、e shearing forces produce the stress and hence the radial displacement 2 ore exactly, aal should be written instead of a , where al is the radius of the center line of the ring.) For the shell, the force T is a transverse shear Q, which pro- duces bending stresses. In order to find them, some detail
41、s of the theory of bending of an anisotropic cylinder must be developed. It is necessary to consider only the internal forces and moments shown on the shell element in figure 8: The hoop force Nfl, the bending moment s, and the transverse shear Qx. They are all functions of x (fig. 7), as is the rad
42、ial deflection w. The forces and moments must satisfy the conditions of equilibrium of the shell element. They yield two equations: which, after elimination of %, give the relation: The hoop force Nfl produces a hoop strain G#_ which ay be obtained from equa*ions (3a) and (5a) to (5c) with Nx = 0. T
43、his strain leads to a - radial displacement , Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACA TN 2612 w = acg The constant 9 = Et t x 2 , which has the dimension of a force tx - v (tx - t) per unit length, is the extensional rigidity of the shel
44、l in the direc- tion of the hoop forces. Figure 9 shows that, in the range of practical interest, D is only slightly greater than Et, and it is safe to say The bending moment Mx produces a curvature d2w/dx2 of the generators, If I is the moment of inertia of the cross section of a stringer and the a
45、ttached skin of width a6 divided by the distance a6 of the stringers, Here too the coefficient EI is slightly influenced by the fact that the skin has a two-dhensional stress system. This refinement of the theory will not be discussed here. There is another circumstance, perhaps even more serious, w
46、hich will also be neglected here: The centroid of the section to which I is referred is not exactly at the distance a from the axis of the cylinder but at a somewhat shorter distance. This influence may be studied with a more general set of equations, but since the difference of the two radii is not
47、 great, it will probably not be of first-order importance: however, it may be responsible for some second-order effects which otherwise might not be explained. By introducing the expressions for N and % into equation (9), the differential equation of the problem is obtained: Provided by IHSNot for R
48、esaleNo reproduction or networking permitted without license from IHS-,-,-12 NACA TN 2612 The general solution of this equation consists of four terms. Only F. those which are symmetrical with respect to the plane x = 0 are needed. They are: a. KX KX KX w = C1 cosh cos - + C2 sinh - sin - a a a a with The boundary conditions at x = 212 are that the slope dw/dx must be zero and that the deflection must assume a certain value wl. h his will be discussed later. ) Introducing the solution here,
