TORSION OF SANDWICH PANELS OF TRAPEZOIDAL, TPIANGUIAP, AND RECTANGULAR CROSS SECTIONS

Transcription

1 !CULTURE ROOM TORSION OF SANDWICH PANELS OF TRAPEZOIDAL, TPIANGUIAP, AND RECTANGULAR CROSS SECTIONS June 1960 No This Report Is One of a Series Issued in Cooperation with the ANC-23 PANEL ON COMPOSITE CONSTRUCTION FOR FLIGHT VEHICLES of the Departments of the AIR FORCE, NAVY, AND COMMERCE A FOREST PRODUCTS LABORATORY MADISON 5, WISCONSIN UNITED STATES DEPARTMENT OF AGRICULTURE FOREST SERVICE ID Cooperation the University of Wisconsin

2 TORSION OF SANDWICH PANELS OF TRAPEZOIDAL, TRIANGULAR, AND RECTANGULAR CROSS SECTIONS! By SHUN CHENG, Engineer Forest Products Laboratory, 2 Forest Service U. S. Department of Agriculture Introduction A theoretical analysis is made of the torsion of a sandwich panel with a trapezoidal cross section. The facings are of uniform and equal thickness. The core is tapered so that its cross section taken perpendicular to the x axis is a trapezoid as shown in figure 1. Its cross section taken perpendicular to the y axis is rectangular. Such sandwich panels are employed in aircraft for control and stabilization of flight. McComb (2)3 considered the torsion of shells having reinforcing cores )3 of rectangular, triangular, or diamond shape cross section. The shells were like facings of a sandwich panel except that they were wrapped around and completely enclosed the cores. Both the shells and the cores were taken to be isotropic. McComb used the Saint Venant theory of torsion, which assumes that the distribution of the shear stress is the same on all cross sections perpendicular to the axis of twist. -This progress report is one of a series (ANC-23, Item 57-4) prepared and distributed by the Forest Products Laboratory under U. S. Navy, Bureau of Aeronautics Order No. NAer and U. S. Air Force Contract No. DO 33(616)58-1. Results reported here are preliminary and may be revised as additional data become available.?maintained at Madison, Wis., in cooperation with the University of Wisconsin. 3 Underlined numbers in parentheses refer to Literature Cited at the end of the text. Report No

3 Seide (4) considered the torsion of sandwich panels having isotropic facings of equal thickness, orthotropic cores, and rectangular cross sections. He also used the Saint Venant theory. The author (1) also considered the torsion of sandwich panels having isotropic facings of equal thickness, orthotropic cores, and rectangular cross sections. The torque was applied by forces concentrated at the corners of the panel and a rigorous solution obtained. Numerical results of this solution were compared with those of a similar Saint Venant solution. The present report is concerned with the torsion of sandwich panels with trapezoidal cross sections. The facings are taken to be isotropic membranes. The core is taken to be a honeycomb structure with axes of the cells placed substantially perpendicular to the facings. Because of this orientation, the stresses in the core associated with strains in the plane of the panel are so small that they may be neglected. The derivation of a system of suitable differential stress strain relations is carried out by means of the variational theorem of complementary energy in conjunction with Lagrangian multipliers (3) (6). These equations apply to any type of loading. Saint Venant torsion is assumed in their solution for sandwich panels of trapezoidal, triangular, and rectangular cross sections. The formula for the torsional stiffness of a sandwich panel of rectangular cross section so obtained is similar to Bredt' s formula for thin tubular sections (5) and agrees with the infinite series solution given in the previous work (1). Notation x, y, z rectangular coordinates for core (fig. 1). x l, y i, z 1rectangular coordinates for facings (fig. 1). b width of sandwich h half thickness of the smaller side of core (fig. 1). t thickness of facings. Report No

4 a slope of facing (fig. 1). E Young' s modulus of elasticity of the facings. v Poisson' s ratio of the facings., shear modulus of the facings. 2(1 + v ) Gxz Gyz shear moduli of the core. ox, Ty, T stresses in facings. Txz Tyz stresses in core. p load per unit area. U strain energy. v1, v 2, v 3,..v6 generalized boundary displacements. w, (3, 'I Lagrangian multipliers. C1, C2, C3,..C6 constants. A1, B1, Dlconstants. 10, Ko, K1 modified Bessel functions of first and second kind of order zero and one. xz Gt cos a tan 2 a T applied torque angle of twist. length of sandwich D 2 t, 2 2Eth (1+-2T.I ) 4Gth (1+2n. (1- v2 ) v ) Report No

5 Mathematical Analysis The trapezoidal sandwich panel is shown in figure 1. Let the xy plane be the middle plane of the core and xiyi plane be the plane of the facing plate with zi axis along the normal to this plane, as shown in figure 1. The equation of the xiyi plane is given by z = h + y tan a (1) For points in the plane zi = 0 = x, = cos a ' (2) Since xj. = x, only x will be used hereafter. It is assumed that:4 (1) The core stiffnesses associated with plane stress components are negligibly small.. (2) The facings are treated as isotropic solid membranes. Under these assumptions the strain energy of the sandwich plate is given by U +1-Iff r 2 r 2 ( xz yz GXz Gyz ) dx dy dz (3) where the subscripts f and c refer to the integrations throughout the face layers and the core. As Tx, try, and Txy of the core are assumed to be negligibly small, it follows from the differential equations of equilibrium of the core that 4 These assumptions have been used in many previous analyses and are known to represent usual sandwich construction having honeycomb cores. Report No

6 the transverse shear stresses do not vary across the thickness of the core: at XZ 8z 0T = o, Yz - 0 az Equation 3 becomes, with (2) and (4), 1 b U =if 0 0 t [1 (6.3C2 + Cr - 2v0-- o- ) + 1COS a x Y1 G ztxz Tyz ).1 dx dy G xz G yz (4) (5) From summation of moments and forces of a differential element of the sandwich plate, as shown in figure 2, the following equations of equilibrium are found COS a ao-x at _ A (GZ t COS a) xz t (2Z t COS a) - 0 (6) 0Y t (2z cos a + t) acryii - 2ZTyz t (2z + t ) DT = 0 cos a T3Z (7) 8y atxz atv z at so- 2z + 2T yztan a + 2z + 2t tan a + 2t sin Y1 a 8x ay 3x ay + p = 0 (8) Equations (6), (7), and (8) are three equations for five unknowns-- o- cr, T, Txz, and T. To obtain further equations, use is made x of the stress-strain relations. This is done here through the use of the variational theorem of complementary energy (3) (6), the Euler equations of which are equivalent to the required stress-strain relations. The expression to be varied takes the following form: Report No

7 I =fit [1 (o- 2 + o- 2 v o- ) + 1 T2 ] T 2 + cos a E x Y1 x Y1 G Gxz xz 1 2%, at at vz T yz ) w k2z xz + 2 -ryz tan a + 2z + Gyz Ox ay acry 2t at tan a + 2t 1 sin a + ex ay 81:rx [ t (2z + t cos a ) ax + - COS a - 2Z T xz t (2Z + t COS cr) + y [t (2z cos a + aryl 8y - 2z-r+ t (Zz t ) at)3 dx dy yz cos a ax b (v i Txz v 2 T + v 3 crx)x=o dy - x=.9 o I (v4 Tyz + v 5 T + v 6 o- ) yi y =o dx y=b where w, and y are Lagrangian multipliers and are functions of x and y and w is subsequently found to be the deflection in the z direction. Values v1, v 2,... v 6 are the generalized boundary displacements of the problem. After carrying out the variation 61 = 0, integrating by parts and transforming the appropriate surface integrals to line integrals by means of Green' s theorem, the variational equations (Euler' a equations) are: (9) 2 (crx l- (2z + t cos a) = 0 8 (1 0) v Tx) E cos a sin a ay - y sin a - (z cos a + av = 0 (11) ay ay 1 p tan a (z + t cos a 813 T tan a aw G cos a 8x 2 ay (z + ) 8= 0 2 cos a (12) Report No

8 1 aw T xz - 13 = 0 Gxz ax (13) z a(wz) T y W tan a - zy = 0 yz ay (14) These equations must be satisfied by values of (Tx, Cry T Txz T yz 1 w, /3, and y, which render the complementary energy a minimum. The boundary conditions (three at each edge of the plate) arising from the independent vanishing of each term of the two line integrals given in (9) are at x = 0 and x = 1 v 1 = 2wz v2 = t (2w tan a + 2z N + t y) Cos a 2t zp + t 2 g v3 - cos a (17) at y = 0 and y = b v4 = 2wz v5 = 2tzi3 + t 2 f3 cos a v6 = 2w sin a + 2tzy cos a + t 2 y Equations (10) to (14), together with equations of equilibrium (6), (7), and (8), represent a complete system of equations for the eight functions CrX Cr y1 T Txz Tyz w, fi, and Solutions 1. Trapezoidal Cross Sections For a sufficiently long plate under torsion, the stresses and strains may be considered not to vary along the longitudinal axis and therefore are independent of x. Then crx = 0 and equations of equilibrium (6), (7), and (8) become Report No

9 T XZ t t cos a ) dt (21) dc- Y1 ZzT yz dy t (2z cos a + t) d irm T yz tana+z +t dy (22) dcr ' 1 sin a = 0 (23) dy The solution of equations (22) and (23) for Tyz and aryl, and the use of the boundary conditions that both stresses are zero when y = 0 and y = b leads to 1 =T yz= 0 = (24) Equations (10), (11), and (14) thus become 813 =0, or f3 = f (y) (25) 8x 8w. sm a + y sin a + (z cos a + = 0 ay 8y (26) w tan a - a(wz) ay - zy = 0 (27) Equation (27) may be reduced to Y = Ow (27a) Substituting y = - aw into equation (26) we obtain By ay = o By or y = F(x) only and hence w = - yf(x) + F 1 (x) (27b) because of equation (27a). Report No

10 From equations (13) and (25) and the assumption that the stresses are independent of x, it follows that W = xf2(y) F3(Y) (27c) In view of the above results of (27a), (27b), and (27c) we obtain w = - (Cix + C 5) y + C 4x + C6 (28) = Cix + C 5 (29) where C 1, C 4, C 5, and C 6 are constants of integration to be determined. From equations (12), (13), (21), (28), and (29), it follows that 4G d2t (2z tan a - t sin a) dt xz zt cry2z (2z + t cos a) dy tg cos a (2z + t cos a)2-2g xy C iz(4z + t cos a + ) COS a t (2z + t cos a)2 (30) This equation, with equation (2), can be written as (1- t d 2 T + 2z dz 2(1+ 2z cos a) dt GXZ T cos a) ) dz tg cos a tan 2 a ( cos a) 2-2C 1 Gxz z + t z cos a + t ) 4z cos a t tan g a (1 + t cos a)2 2z Since facings are treated as membranes, is relatively small as 2z compared with 1. With this consideration, the above equation may be reduced to z d2 T dt G T xz -2C1Gxz z dz 2 dz tg cos a tan 2 a t tan 2 a (31) Report No

11 The complete solution for T with three arbitrary constants C 1, C2, and C3 is T = ZCIG cos a (z ) + C 21,0 (2 ) + C 3K0 (2 j) (32) The last two arbitrary constants C2 and C3, which are determined from the boundary conditions that T = o at y = o and y = b, can be expressed in terms of Ci as + K0 (2 ITITI) C2 C I -2 G cos a + b tan a +4-1 Ko Ifr(h+ b tan a)) Io (2 ) K0 (2 Fir) 10 (2 t/r(h+ b tan a) K0 (2 Lir (h b tan cx)) Io (2 I xr17. ) (h +4.) Io (2 Vr(h+ b tan a)) (h 4- b tan a 4) C3 = 2 G cos a (33) Cl I0 (2 %5F). ) K0 (2 irr17) I0 Vr(h + b tan a)) Ko Lir(h + b tan a)) It can be shown that the resultant of the forces distributed over the ends of the plate is zero, and these forces represent a couple, the magnitude of which is b b T = 4t T zdy + t 2 (1 + cost T cos a dy If the displacement v1 and v 2 are set to be zero at x = 0, then w and y from equations (28) and (29) become (34) w = - Cixy + C 4x (35) clx ( 3 6) Report No

12 The torsional stiffness T may be obtained from b T8 =(Vi f T xz + V2 T ) dy (37) 0 x =1 Substituting for vi and v2 from equations (15), (16), (35), (36), and for Txz from equation (21), then integrating by parts and making use of equation (34), we obtain = 0 ( 3 8) When the value of T from equation (32) is substituted into equation (34) and the integration carried out, we obtain T 2tG cos a z + cos 2 z2 z 0 [4 (z3 + 2 zr + tan a cos a (2 + 7)] 4t C2 r 2 tan a{. C1 (rz + 1) I 1 (2 rz ) - (rz) Io (2 V1Z-)] _ C3 [ (rz + 1) Ki (2 rz ) + (rz) Ko (2 FZ 11} C1 t 2 FT (1 + cos t a) [ C r tan a cos a CT (2 ) - C3 K 1 (2 t RT )] Cl h- b tan a h ( 3 9 ) where h and h + b tan a are lower and upper limits of the integration of equation (34) with respect to z. Report No

13 For rz > 25 equation (39) can be expressed approximately as t2tg cos a z 3 z 2 t(1 + cos 2 a) (Z 2 Z )1 [4( + 0 ta.n a 3 2r ) cos a 2 r 5 + 4t(rz)4 - C2 2 FT [ e (1 - + r 2 tan a Cl 512 rz 161:7 _ L rz rz 8192 (rz)2a- ) C 3-2 Irz e ( Cl 16 ii. z 512 rz 8192 rz, IFT t2(rz)7r (1+ cos 2 a) + )] 4' 8192 (rz) (rz) 2./72- r tan a cos a C z [ Ci e (1 - ) 16,,/ 1 -"-512rz 8192 rz. rz C 3-2,/ / e ( Cl 161/ 512 rz 8192 rz + b tan a 105 )1 (40) Report No

14 2. Triangular Cross Sections For h = 0 and z = y tan a, equation (30) may be written as y (y t cos a)2 d2 T + (y2 t2 cos 2 a) dt _ Gxz Y21-2 tan a dy 24 tan 2 a dy tg sin a t cos a + t - 2 Gxz CI y 2 (y + 4 tan a 4 sin a) t oo The above differential equation can be solved for T by letting T = Anyn. n = o For a very small t, as was assumed, the above equation may be reduced to d 2 T dt GxzT Y + - dy2 dy tg sin a - 2G 1 Gxz y t (41) Because equation (41) can be obtained from (31) by placing h equal to zero, equations (32) to (40) are applicable to triangular cross sections by setting h = Rectangular Cross Sections By setting a = 0, equation (30) is reduced to d2t 4h Gxz T 4Ci h Gxz dy2 tg (2h + t) 2t (2h + t) (42) where Ci = according to equation (38). It can be verified by direct substitution that the general solution of equation (42) is given by T = AI sinh y 1 y + B 1 cosh Ni y + Dl(43) where yi - 4hGxz, D i = G (2h + t) Gt (2h + t)2 (44) Report No

15 The first two arbitrary constants Al, and Bi, which are determined from the boundary conditions that T = 0 at y = 0 and y = b, can be expressed in terms of D 1 as A l - (cosh yl b - 1) D1, B 1 = - D1 sin h yi b (45) thus T = D [ (co. sh y b 1) sinh y y - cosh y y + 1] Y 1 J (46) All equations of equilibrium are satisfied identically except equation (21), which becomes T t (211 + t) d-r (47) 2h dy Introducing (46) into (47), results in tgxz [ (cosh y b - 1) T xz = cosh y 1 y - sinh y] D 1 (48) hg sinh b The resultant torque due to stresses is equal to the applied torque T b b T = f t (2h + t) -rdy - f 2h-rxydy (49) Substituting T from (46) and -r xz from (48) into equation (49) and then integrating, gives D 1 = 2t (2h + t) [b - T 2 (cosh y i b - 1) Y 1 sinh y l b ] (50) Report No

16 The torsional stiffness is obtained from equations (50) and (44) 2 (cosh yi b - 1) T = 2t (2h + t) 2Gb [1-0 yib sinhyib (51) It is evident that equation (51), similar to equation (142) of Forest Products Laboratory Report No (1), is independent of plate length, Young' s modulus Ec, and shear modulus Gyz of the core. The results of numerical computations based on equation (51) are given in table 1. The results based on the infinite series solution (1) are also listed in table 1 for comparison with the results obtained by the present method. The numerical computations were for sandwich plate of various widths and the following properties: t = inch h = 0.25 inch Gxz = 25,000 pounds per square inch G = 4 x 10 6 pounds per square inch The results show a maximum difference in torsional stiffness between the two methods of 3 percent. Their differences can be attributed to small errors or lack of using sufficient terms in the infinite series as given by equation (142) of Forest Products Laboratory Report No It may be concluded that the present result, which is in a form similar to R. Bredt' s formula for thin tubular sections (5), is in excellent agreement with the infinite series solution. Report No

17 Literature Cited (1) Cheng, S Torsion of Rectangular Sandwich Plates. Forest Products Laboratory Report No (2) McComb, H. G., Jr Torsional Stiffness of Thin-Walled Shells Having Reinforcing Cores and Rectangular, Triangular, or Diamond Cross Section: U. S. National Advisory Committee for Aeronautics, Tech. Note (3) Reissner, E On Bending of Elastic Plates, Vol. V, No. 1, April, Quarterly of Applied Mathematics. (4) Seide, P On the Torsion of Rectangular Sandwich Plates. Journal of Applied Mechanics, Vol. 23, No. 2. (5) Timoshenko, S. and Goodier, J. N Theory of Elasticity. New York, McGraw-Hill. (6) Wang, C. T Applied Elasticity. New York, McGraw-Hill. Report No

18 Table 1.--Torsional rigidity of rectangular sandwich plate as computed by two methods b by present method : 0 by infinite series solutionl 0 7F In.. Lb.-in.2 : Lb.-in.2 1 : 3,500 3,360 2 : 19,000 : 19, ,920 41, ,280 : 66,800 5 : 93,200 : 92,300 6 : 119,500 : 118, , , : 224, , ,200 : 476,000 -Obtained from table 2 of Forest Products Laboratory Report No Report No. 1874

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