REG NACA-TN-1462-1948 BENDING OF RECTANGULAR PLATES WITH LARGE DEFLECTIONS.pdf

1、 _ _L_L _g.NATIONAL ADVISORY COMMITTEEFOR AERONAUTICSTECHNICAL NOTENo. 1462BENDING OF RECTANGULAR PLATESWITH LARGE DEFLECTIONSBy Chi-Teh WangBrown UniversityWashingtonApril 1948._. . /Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NATIONAL ADVISORY

2、 COMNUIEEE FOR AERONAUTICSTEC_ICAL NOTE NO. 1462BENDING OF RECTANGULAR PIATESWITH LARGE DEFLECTIONSBy Chi-Teh WangSUMMARYVon K_rm_ns equations for thin plates with large deflections aresolved for the special cases of rectangular plates having ratios of lengthto width of 1.5 and 2 and loaded by unifo

3、rm noznal pressure. The boundaryconditions are such as to approximate panels with riveted edges undernormal pressure greater than that of the surrounding panels. Centerdeflections, membrane stresses, and extreme-fiber bending stresses aregiven as functions of the pressure for center deflections up t

4、o twice thethickness of the plate. For small deflections the results are consistentwith those given by Timoshenko.INTRODUCTIONA general numerical method for solving Von Kirm_ns equations forthin plates with large deflections was developed in reference I. Asquare plate loaded by uniform normal pressu

5、re with simply supported edgeswas studied. The boundary conditions approximate the case when a rivetedsheet-stringer panel is under normal pressure greater than that of thesurrounding ones.It was subsequently decided to extend the investigation to rectangularplates of various ratios of length to wid

6、th. Two special cases are studiedin this report; namely, rectangular plates for which the ratio of lengthto width has the value of 1.5 or 2, (For rectangular plates having alength equal to or greater than twice their width, experimental evidence(reference 2) indicates that they can practically be re

7、garded as infinitelylong.) Center deflections, membrane stresses, and extreme-fiber bendingstresses are given as fuI_ctions of the pressure for center deflections upto twice the thickness of the plate. For small deflections, the resultsare consistent with those obtained by Timoshenko (reference 3).T

8、his work was conducted at Brown University ua%der the sponsorship_nd with the financial assistance of the National Advisory Committee forAeronautics.The author is indebted to Professor W. Prager for his kind interest.Provided by IHSNot for ResaleNo reproduction or networking permitted without licens

9、e from IHS-,-,-2 NACAIN No. 1462SYMBOLSa b length and width of plate; a, shorter side ofrectangular plateh thickness of platex, y, z coordinates of a point in plateU V horizontal displace-_nts in x- and y-directions ofpoints in middle surface; nondimensional forms areua/h 2 and va/h2 respectivelyW d

10、eflection of middle surface out of its initial plane;nondimensional form is w/hP normal load on plate per unit area; nondimensional formis pa4/Eh 4ED2 _2 _2V = -+-_.X2 _y2Youngs modulus and Poissonts ratio respectivelyflexural rigidity of plate k 12(I -_24 _4 B4 _4v = 2 a:2 * - y4_x _, _yl, TXyW mem

11、brane stresses in middle surface; nondimensionalforms are axa2_h2 , _ya2_h2 and , a2/Eh 2xyre spe ctively(TX“ “ T “ _y xy extreme-fiber bending and shearing stresses; nondimensionalforms are _x“a2#h 2, _y“a2_h 2, and _xy“a2_h 2,re spe ctively_x w, _yt, Txy_, Cy, ,6x 7xyFmembrane strains in middle su

12、rface; nondimensional formsare _ a2/h 2, _ Wa2/h2, and 7xyla2/h 2, respectivelyx Yextreme-fiber bending and shearing strains; nondimensionalforms are _x“a2/h 2, _ “a2/h 2, - “a2L27xy /n , respectivelyYstress function; nondimsnsional form is F_h 2Provided by IHSNot for ResaleNo reproduction or networ

13、king permitted without license from IHS-,-,-NACATN No. 1_62 3,A2 first-, second-, . . . and nth-order differences,respectivelyfirst-order differences in x- and y-directions,respectivelyFU_JS_E_J_ EQUATIONsThe deformation of a thin plate, the deflections of which are largein comparison with its thick

14、ness but are still small ,as compared with theother dimensions, is governed by Von K_ s equations:(1)3_w _4w 34w P h _f_ 3% 82F 8_ 2 82F _2w-_+ 2 +_= +- +whe re Eh 3D : _e12(l- _2)The median-fiber stresses are_#mGxt 8y2xy _x By(2)and the media_-fiber strains areProvided by IHSNot for ResaleNo reprod

15、uction or networking permitted without license from IHS-,-,-4 NACATN No. 1462_F _7_)E _x _y(3)The extreme-fiber bending and shearing stresses areEh _%_“ + _v “ - Eh _2wxy 2(z+ _)_x_(4)For a riveted panel under normal pressure greater than that of thesurrounding panels, the boundary conditions are fo

16、rmulated in reference 1and are as follows:alongw=0- - 0-_-=0_y2 _x2y +b=-_. and=o L-_ _2 dy = 0JC5)Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACA TNNo. 1462 5along any lineAnd,alo_ x = + a2x = Constant.w=0and_%= 08x2=0 _ - _ _x2 2dx=Oalong any

17、line y = Constant.F wThese expressions can be made nondlmensio_x_l by writing h-_ _Pa4 _(_ x (_)_-_h“EV_J _,_, and_ in_laceof F, w, p, o, x, y, _d _,respectively, w_re a is the smaller side of tlle rectangular plate.(These latter symbols are used to effect a reduction in the emount ofwriting involve

18、d in the equations.) The resulting differential equationscan then be transformed into finite-difference equations. In terms offinite differences, the differential equations (i) are replaced by thefollowing equations :_x4W : 2F+_% +Ay4w lO.8(AZ)_p+_0.8 % +_y _%-(7)Provided by IHSNot for ResaleNo repr

19、oduction or networking permitted without license from IHS-,-,-6 NACA TN No. 1462where AS = Ax = 2_v and B2 is taken to be 0.i, which value is character-istic for aluminum alloys.In terms of finite differences, the boundary conditions arealong y = + _, andWm, k = 0m,k2 MY Ui,n_. (8)where n = k denote

20、s points along the edges y = + b_ _, n = 0 denotespoints along the center lines y = O, and m = i denotes any point alongthe line x = Constant in the plate. Similarly, along x = + 2 theboundary conditions areandWk, n = 0:0- “ 0f! 1-_x _ - (z_#) = om=O m, i(9)wherepoints along the center line x = 0the

21、 line y = Constant in the plate.#m = k denotes points along the edges x = +_2 m = 0 denotesand n = i denotes any point alongProvided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACATN No. 1462 7The normal median-fiber stresses areOt = -X(Az)2z 2F1_ ! = -Y

22、 (az)2and the extreme-fiber bending stresses are, i (Z_x2w_x = - 2(1 - _2)(Az)2cr“ = _ i fA2wY 2(1 - _2)(Az)2 y, (io)(ii)RECTANGUIA2 PLATE WITH LENGTH-WIDTH RATIO OF i.5UNDER UNIFORM NORMAL PRESSUREThe plate is divided into 24 square meshes. (See fig. i.) Thepoints 2, 5, 8, 9t, I0, ii, and ii“ are f

23、ictitious points outsidethe plate in order to make possible a better approximation to the boundaryconditions. Since the plate is symmetrical with respect to the centerlines, only one-quarter of the plate needs to be considered. In order toget satisfactory results in the subsequent computations, it i

24、s convenientto use a number of figures beyond those normally considered Justifiablein view of the precision of the basic data.Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-8 NACATN No. 1462With _2 = 0.i or _ = 0.316225, the compatibility equationsb

25、ecome20F0 - 16FI + 2-F2 - 16F3 + 8F4 + 2-F6 : K0-SF0 + 21FI -8F 2 + 4F3 -16F 4 + 4F5 + 2-F7 + F2, = KI-SF0 + 4FI + 21F3 -16F 4 + 2F5 -8F 6 + 4F7 + F9 = K32F0 - 8FI + 2-F2- 8F3 + 22-F4- 8F5 + 2F6 - 8F7 + 2F8 + FI0 + F5 = 4F0 - 8F3 + 4F4 + 20F6 - 16F7 + 2F8 - 8F9 + 4FI0 + F9 = 6FI + 2_F3_ 8F4 + 2F5 _

26、8F6 + 21F7 - 8F8 + 2F9 - 8FI0 + 2FII + FS, + FIO , = rwhere 0 i 3 K4 K6 and K7 are equal to “ -21w 7 + wI + 2w3 -8w 4 + 2w5 - 8w6 -8w8 + 2w9 -8w10 + 2Wll + w8, + Wl0, = p+_o._E_ -_+ _!+ w7 -w8 -Wl0)(13)Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-

27、NACATN No. 1462 9where p! 12(1 _2)(AZ)4p 0.0421875P since AZ = _, and _ _-. _ -. and 7 are _x2F, _F, and _ at the points indicated by thesubscripts, respectively.The conditions for zero edge displacements are)+_5- _8 + FII)- _8 _ -F8+ FT) = $3 I( 14)_F I - 2FO) -_.(2F 3 - 2Fo)+ _4Fh.- 4F3)- U,_:_F0

28、 h.F3+ P_F6)+ (h.F 7 - h.F_-_(_EF 3 - 4F6+ _Fg) + (2FIo - 2F9) -_(F 6 - 2F9+ F9,) = S 4(F2- _1 + FO) -“(_4- _1) + (_- _F_+ _3) -“(2F1- 4F4+ _?)+ C2F8- 4F7+ 2F6)-_-C_B4- _.F7+ 2FIo) + (FII - 2FIo+ F9)-.(_-_o+“_o,):_wherewo) _:Cw_-_0_C_-_)_:Cw_-w_Cw_w_)_s_:L-_-w_)_C-_-_)_(w_wo)_: (_o-w_)_(._w_)_(w_w_

29、)_Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-l0 NACA TN No. 1462The boundary conditions areW2 = 0, W5 = 0, W8 = 0, W9 = 0, Wl0 = 0, Wll = 0w2f - 2w 2 + wI = 0w9 - ew5 + wk = 0w8 - 2w8 + w7 = 0w9, - ew9 + w6 = 0Wl0 T - 2wlO + w7 = 0F2! - 9_F2 + F

30、I -_(2F 5 - 2F2J = 0Fsr - 9_F5 + F4 - _(F 8 - 9_F5 + F2) = 0Fs-_8 F7-_n -_8 Fs_=oFg,-_9 +F6-_(_io-%):o_o, - =_o,_7- _(_- _o +_9)=oNow the boundary-value problem determines the values of w uniquely andthe values of F to within an unknown constant. Since the actual valueof the constant is irrelevant,

31、it may be defined by letting Fll = O.On combining the boundary conditions with equations (12), (13), and (l_),the final equations areProvided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACATN No. 146220F0 - 16F1 + 2F2 - 16F3 + 8F4 + 2F6 = K0- 8F0 + 20F1

32、 6.632456F2 + 4F3 - 16F4 + 4.632456F5 + 2F7 = K1- 8F0 + 4F1 + 21F3 - 16F4 + 2F5 - 8F6 + 4F7 + F9 = K32F0 -8F 1 + 2.316228F2 -8F 3 + 21F4 -6.632456F 5 + 2F6 -8F 7+ FI0 = K4F0 -8F 3F1 + 2F3- 2F0 - 3.3675A4F 1 - 2.432456F 2 + 2F 3FoF3+ 4F4 + 1.8F 5+ 2IF1 + 0.gF 2 -+ 21% + 0.gF 9-ii+ 2.316228F 8+ 4F4 +

33、 19F 6 - 16F 7 + 2F8 - 6.632456F 9 + 4.632456FI0 : K6-8F 4 + 2.316228F 5 -83“6 + 19F 7 - 6.632456F 8 + 2.316228F 9- 6.632456FIo = K7= S12F3 - 3.367544F 4 - 2.432456F 5 + F6 + 2F7 + 0.gF 8 = S22F6 - 3.367544F 7 -2.432456F 8 -_F9 + 2FIo + FII = S3- 2F0 + 2F1 - 4F3 + 4F4 - 3.367544F 6 + 4F7 - 2.432456F

34、 9 + 1.8FIo : S4F0 - 2F1 +F 2 + ZF 3 - 4F4 + 2F5 + 2F 6 - 3.367944F 7 + 2F8 + 0.gF 9- 2.432456FI0 = S5andProvided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-12 NACA TN No. 1462_0+_._c_,0+0,o+_,o_wo-E_+_(0,o+_,o_w_-_+_._(o,o+_o)_w_+(8+_,o)w_+_,+(_+_3)w7+w

35、9=_ x_6)_o-0+_o._,_)w_+_,_+E_+_._(o_+_+_)w_iBy following the procedures outlined in reference i, equations (15) and (16)can be solved simultaneously for w and F by the method of successiveapproximations.RECTAN_ PLATE WITH LENGTH-WIDTH RATIO OF 2UNDER UNIFORM NORMAL PRESSUREThe plato is first divided

36、 into 8 square ashes and then into 32 squaremeshes. (See figs. 2 and 3, respectively.) On referring first to figure 2jpoints 1I, 3_, 4v, 51 , and 5“ are fictitious points outside the plate inorder to giveabetter approximation to the boundary conditions. Consideron_-quarter of the plate; the compatib

37、ility equations are20F 0 -20F 2 -16FI -16F2 + 8F3 + 2F4 + 2Fi = K0 F 18FO - 16F3 - 8F4 + 4FI + 4F5 + F2 + ?-F3 + 4 = KI (1v)Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACA TN No. 1462 13where K0 _I KI are equal %orespectively.at points 0 and i,T

38、he equilibrium equations are+ ooE,20w 2 - 8w0 - 16w 3 - 8w4 + 4wI + 4w 9 + w2 + 2w3, + w4,where= P + i0.8 _2_0- 2w2- 27 2(w5 + w2 -w3 -w4_p : 12(1 -_2)(AZ)4 = 0.675p sincexyrespectively.i _ and_Z _, and _at the points indicated by the subscriptsThe conditions for zero edge displacements are(_- _o)-.

39、 _o)+(m-_ -.(z9-_(F2 - 2F4+ F4,)+ (2F5 -2F4): (v2- Wo)2+ w22Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-14 NACA TN No. 1462The boundary conditions arewI = O, w3 = O, w4 = O, w3 = 0wI + w0 : 0w3 , + w2 = 0w4, + w2 = 0FI, - 2_FI + F0 - _(o_v3 -

40、2FI) = 0F3 - 2F3 + F2 - _i - 2F3 + FS) _:0F4, - 2F4 + F2 - _-F 5 - 2F4) = 0The problem can now be solved uniquely for the values of w and the valuesof F to within an unknown constant. As the actual value of the constantis irrelevant, it may be defined by letting F5 = 0.Combined with the boundary con

41、ditions, equations (17), (18), and (19)are then18F 0 - 13.264912/_ 1 -16F 2 + 9.264912F 3 + 2F4 : K0-SF0 + 4.632456F l + 18F 2 - 13.264912F 3 - 6.632456F 4 = K1-1.367544F 0 - 2.532456F I + 2F2 + 1.8F 3 = w02F0 + 0.9F I - 1.367544F 2 - 2.532456F 3 + F4 = w2_- _“o + _i - 3-367944F2+ 4F3 - 2“532456;4:

42、w2- WO)2 + w22(20)(21)Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACA TN No. 1462 15Equations (20) and (21) can now be solved simultaneously for w aad Fby the method of successive approximations.With the values of w thus computed as a first app

43、roximation, thecase wherein the plate is divided into 32 square meshes is,now to bestudied. On referring to figure 3, points 2, 5, 8, ii, 12, 1314, and 14“ are again fictitious points outside the plate in orderto give a better approximation to the boundary conditions. Considerone-quarter of the plat

44、e; the compatibility equations are20F 0 - 16F I + 2F 2 - 16F 3 + 8F 4 + 2F 6 = K0-_b_Fo + 21F I - 8F 2 + 4F 3 - 16F 4 + 4F 5 + 2F 7 + F 2, = KI-8F0 + 4F 1 + 21F 3 - 16F 4 + 2F 5 - 8F 6 + 4F_,I+ F9 = K32F 0 - 8F I + 2F 2 - 8F 3 + 22F 4 - 8F 5 + 2F 6 - 8F 7 + 2F 8 + FI0 + F 5, = K 4F0 - 8F 3 + 4F4 + 2

45、0F6 - I6F7 + P_8 - 8F 9 + 4FIo + FI2 = K6FI + 2F3 - 8F4“+ _5 - 8F6 + 21F7 - 8F8 + 2F9 - 8FIo + 2Fil + FI3 + F8 = K7F 3 - 8F 6 + 4F7 + 20F 9 - 16Fio + 2FII - 8F12 + 4F13 + FI2, = K 9F4+ 2F6-_ + _8- a_9+ a_lO- 8Fll+ _12-_13 +2F_4+ _i+ F13: _0at the points indicated by the subscripts.where K= _AxyW) 2

46、Ax2W_Ay2W _Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-16 NACA TN No. 1462The equilibrium equations are -_20w 0 - 16w I -16w3 + 8w4 + 2w2 + 2w6 = p + lO.8_O_-w 3 - 2Wo)0-Sw 0 + 21w I 8w2 + 4w3 - 16w 4 + 4w 5 + 2w7 + w2,+ _l(W2 - 2Wl + Wo) - 27i(

47、w5 -w2 -w4 + Wl_-8Wo + 4Wl + 21w 3 -16w4 + 2w5 - 8w6 + 4w7 + w9 = P + lO.8E3(WO-2w 3 + w6)2w0 - 8wI + 2w2 - 8w 3 + 22w 4- 8w5 + 2w6 - 8w7 + 2w8 + wlO+ w5, = p + i0.8c_4_7 - 2w4 + Wl) + _4(w5 - 2w4 + w3_-274(w8 -w5 -_w7 + w4)w0 - 8w3 + 4w4 + 20w 6 - 16w 7 + 2w8 - 8w 9 + 4wlO + w12wI + 2w3 - 8w4 + 2w5- 8w6 + 21w7- 8w8 + 2w9- 8Wlo+ 2w + + = p + i0.8 E 7(w4 -2w7+ WlO)ii w13 w8+ _ _ + - _ +w3