Aircraft Structures 1 2011 Part 6 ppt

186 Structural instability t C 2.5mrn I - X 2.5mm - 75 rnm - 0 - PCR(rx) - Pxs - PCR(~~) 0 PYS PYS -Pxs Io (P - PcR(e) )/A Fig. 6.22 Column seclion of Example 6.1. =o (6.91) 6.1 2 Flexural-torsional buckling of thin-walled columns 187 If the column has, say, Cx as an axis of symmetry, then the shear centre lies on this axis and ys = 0. Equation (6.91) thereby reduces to (6.92) The roots of the quadratic equation formed by expanding Eqs (6.92) are the values of axial load which will produce flexural-torsional buckling about the longitudinal and x axes. If PCR(,,,,) is less than the smallest of these roots the column will buckle in pure bending about the y axis. Example 6.2 A column of length lm has the cross-section shown in Fig. 6.23. If the ends of the column are pinned and free to warp, calculate its buckling load; E = 70 OOON/mm2, G = 30 000 N/mm2. Fig. 6.23 Column section of Example 6.2. In this case the shear centre S is positioned on the Cx axis so that ys = 0 and Eq. (6.92) applies. The distance X of the centroid of area C from the web of the section is found by taking first moments of area about the web. Thus 2( 100 + 100 + 1OO)X = 2 x 2 x 100 x 50 which gives i = 33.3mm The position of the shear centre S is found using the method of Example 9.5; this gives xs = -76.2mm. The remaining section properties are found by the methods specified in Example 6.1 and are listed below A = 600mm2 Zxx = 1.17 x 106mm4 J = 800mm4 = 0.67 x 106mm4 I? = 2488 x 106mm6 Zo = 5.32 x 106mm4 188 Structural instability From Eqs (6.90) P~~(~~) = 4.63 x io5 N, P~~(~~.~) = 8.08 x io5 N, P~~(~) = 1.97 x io5 N Expanding Eq. (6.92) (P - PCR(.~.~))(P - PCR(8))zO/A - p2xg = 0 (i) Rearranging Eq. (i) P2(1 - Axt/zO) - P(pCR(.~.~) + PCR(B)) + PCR(s.~)pCR(8) = (ii) Substituting the values of the constant terms in Eq. (ii) we obtain P2 - 29.13 x 105P + 46.14 x 10" = 0 (iii) The roots of Eq. (iii) give two values of critical load, the lowest of which is P = 1.68 x 10'N It can be seen that this value of flexural-torsional buckling load is lower than any of the uncoupled buckling loads PCR(xx), PCR(yy) or PcR(e). The reduction is due to the interaction of the bending and torsional buckling modes and illustrates the cautionary remarks made in the introduction to Section 6.10. The spans of aircraft wings usually comprise an upper and a lower flange connected by thin stiffened webs. These webs are often of such a thickness that they buckle under shear stresses at a fraction of their ultimate load. The form of the buckle is shown in Fig. 6.24(a), where the web of the beam buckles under the action of internal diagonal compressive stresses produced by shear, leaving a wrinkled web capable of supporting diagonal tension only in a direction perpendicular to that of the buckle; the beam is then said to be a complete tensionJield beam. W ut A Qc ff D ut (a) (W 1 Fig. 6.24 Diagonal tension field beam 6.1 3 Tension field beams 189 Ylll 6.1 3.1 Complete diagonal tension is _.__*___- The theory presented here is due to H. Wagner'"4. The beam shown in Fig. 6.24(a) has concentrated flange areas having a depth d between their centroids and vertical stiffeners which are spaced uniformly along the length of the beam. It is assumed that the flanges resist the internal bending moment at any section of the beam while the web, of thickness t, resists the vertical shear force. The effect of this assumption is to produce a uniform shear stress distribution through the depth of the web (see Section 9.7) at any section. Therefore, at a section of the beam where the shear force is S, the shear stress r is given by S td r=- (6.93) Consider now an element ABCD of the web in a panel of the beam, as shown in Fig. 6.24(a). The element is subjected to tensile stresses, at, produced by the diagonal tension on the planes AB and CD; the angle of the diagonal tension is a. On a vertical plane FD in the element the shear stress is r and the direct stress a,. Now considering the equilibrium of the element FCD (Fig. 6.24(b)) and resolving forces vertically, we have (see Section 1.6) a,CDt sin a = TFDt which gives 27 sin 2a - 7 a, = - sin a cos a (6.94) or, substituting for r from Eq. (6.93) and noting that in this case S = W at all sections of the beam 2w td sin 2a a, = Further, resolving forces horizontally for the element azFDt = atCDt cos a whence 7 a, = a, cos- a or, substituting for at from Eq. (6.94) r a, = - tan a or, for this particular beam, from Eq. (6.93) W a, = ~ td tan a FCD (6.95) (6.96) (6.97) Since T and at are constant through the depth of the beam it follows that 0; is constant through the depth of the beam. The direct loads in the flanges are found by considering a length z of the beam as shown in Fig. 6.25. On the plane mm there are direct and shear stresses az and r acting 190 Structural instability Fig. 6.25 Determination of flange forces. in the web, together with direct loads FT and FB in the top and bottom flanges respectively. FT and FB are produced by a combination of the bending moment Wz at the section plus the compressive action (a,) of the diagonal tension. Taking moments about the bottom flange aztd2 2 WZ = FTd - - Hence, substituting for a- from Eq. (6.97) and rearranging wz w F==-+- d 2tana Now resolving forces horizontally FB - FT + aztd = O which gives, on substituting for nz and FT from Eqs (6.97) and (6.98) wz w d 2tana FB= (6.98) (6.99) The diagonal tension stress a, induces a direct stress a,, on horizontal planes at any point in the web. Thus, on a horizontal plane HC in the element ABCD of Fig. 6.24 there is a direct stress a,, and a complementary shear stress 7, as shown in Fig. 6.26. B Fig. 6.26 Stress system on a horizontal plane in the beam web. 6.13 Tension field beams 191 From a consideration of the vertical equilibrium of the element HDC we have ayHCt = a,CDt sin a which gives 2 au = a, sin a Substituting for at from Eq. (6.94) aJ = Ttana! (6.100) or, from Eq. (6.93) in which S = W W a,, = -tan a . td (6.101) The tensile stresses a,, on horizontal planes in the web of the beam cause compression in the vertical stiffeners. Each stiffener may be assumed to support half of each adjacent panel in the beam so that the compressive load P in a stiffener is given by P = a,tb which becomes, from Eq. (6.101) Wb P = ana d (6.102) If the load P is sufficiently high the stiffeners will buckle. Tests indicate that they buckle as columns of equivalent length or I, = d/dm I, = d forb < 1.5d for b > 1.5d (6.103) In addition to causing compression in the stiffeners the direct stress a,, produces bending of the beam flanges between the stiffeners as shown in Fig. 6.27. Each flange acts as a continuous beam carrying a uniformly distributed load of intensity aut. The maximum bending moment in a continuous beam with ends fixed against rotation occurs at a support and is wL2/12 in which w is the load intensity and L the beam span. In this case, therefore, the maximum bending moment M,,, occurs Fig. 6.27 Bending of flanges due to web stress. 192 Structural instability at a stiffener and is given by uytb 2 MmaX =- 12 or, substituting for gy from Eq. (6.101) wb2 tan a 12d Mmax = (6.104) Midway between the stiffeners this bending moment reduces to Wb2 tan a/24d. The angle a adjusts itself such that the total strain energy of the beam is a minimum. If it is assumed that the flanges and stiffeners are rigid then the strain energy comprises the shear strain energy of the web only and a = 45". In practice, both flanges and stiffeners deform so that a is somewhat less than 45", usually of the order of 40" and, in the type of beam common to aircraft structures, rarely below 38". For beams having all components made of the same material the condition of minimum strain energy leads to various equivalent expressions for Q, one of which is (6.105) tan a=- ut + % in which uF and as are the uniform direct compressive stresses induced by the diagonal tension in the flanges and stiffeners respectively. Thus, from the second term on the right-hand side of either of Eqs (6.98) or (6.99) 2 Ot +'F W 2AF tan a CF = in which AF is the cross-sectional area of each flange. Also, from Eq. (6.102) wb us = -tana ASd (6.106) (6.107) where As is the cross-sectional area of a stiffener. Substitution of at from Eq. (6.95) and oF and crs from Eqs (6.106) and (6.107) into Eq. (6.105), produces an equation which may be solved for a. An alternative expression for a, again derived from a consideration of the total strain energy of the beam, is (6.108) Example 6.3 The beam shown in Fig. 6.28 is assumed to have a complete tension field web. If the cross-sectional areas of the flanges and stiffeners are, respectively, 350mm2 and 300mm2 and the elastic section modulus of each flange is 750mm3, determine the maximum stress in a flange and also whether or not the stiffeners will buckle. The thickness of the web is 2mm and the second moment of area of a stiffener about an axis in the plane of the web is 2000 mm4; E = 70 000 N/mm2. From Eq. (6.108) = 0.7143 4 1 +2 x 400/(2 x 350) 1 + 2 x 300/300 tan a= 6.1 3 Tension field beams 193 400mm 1200 mm -I Fig. 6.28 Beam of Example 6.3. so that Q! = 42.6" The maximum flange stress will occur in the top flange at the built-in end where the bending moment on the beam is greatest and the stresses due to bending and diagonal tension are additive. Thus, from Eq. (6.98) 5 x 1200 5 400 -k 2 tan 42.6" FT = i.e. FT = 17.7 kN Hence the direct stress in the top flange produced by the externally applied bending moment and the diagonal tension is 17.7 x 103/350 = 50.7N/mm2. In addition to this uniform compressive stress, local bending of the type shown in Fig. 6.27 occurs. The local bending moment in the top flange at the built-in end is found using Eq. (6.104), i.e. 5 x lo3 x 3002 tan42.6" 12 x 400 = 8.6 x 104Nmm Mnax = The maximum compressive stress corresponding to this bending moment occurs at the lower extremity of the flange and is 8.6 x 104/750 = 114.9N/mm2. Thus the maximum stress in a flange occurs on the inside of the top flange at the built-in end of the beam, is compressive and equal to 114.9 + 50.7 = 165.6N/mm2. The compressive load in a stiffener is obtained using Eq. (6.102), i.e. 5 x 300 tan 42.6" 400 = 3.4 kN P= Since, in this case, b < 1.5d, the equivalent length of a stiffener as a column is given by the first of Eqs (6.103). Thus 1, = 400/d4 - 2 x 300/400 = 253 mm 194 Structural instability From Eqs (6.7) the buckling load of a stiffener is then = 22.0 kN 7? x 70000 x 2000 2532 PCR = Clearly the stiffener will not buckle. In Eqs (6.107) and (6.108) it is implicitly assumed that a stiffener is fdy effective in resisting axial load. This will be the case if the centroid of area of the stiffener lies in the plane of the beam web. Such a situation arises when the stiffener consists of two members symmetrically arranged on opposite sides of the web. In the case where the web is stiffened by a single member attached to one side, the compressive load P is offset from the stiffener axis thereby producing bending in addition to axial load. For a stiffener having its centroid a distance e from the centre of the web the combined bending and axial compressive stress, a,, at a distance e from the stiffener centroid is P Pe2 a, = - +- As As? in which r is the radius of gyration of the stiffener cross-section about its neutral axis (note: second moment of area I = Ar2). Thus a -'[1+(;)2] - As or P a, = - Ase where (6.109) and is termed the effective stiffener area. - 6.13.2 Incomplete diagonal tension In modern aircraft structures, beams having extremely thin webs are rare. They retain, after buckling, some of their ability to support loads so that even near failure they are in a state of stress somewhere between that of pure diagonal tension and the pre-buckling stress. Such a beam is described as an incomplete diagonal tensionfield beam and may be analysed by semi-empirical theory as follows. It is assumed that the nominal web shear T(= S/td) may be divided into a 'true shear' component T~ and a diagonal tension component TDT by writing TDT = k7, T~ = (1 - k)7 (6.110) where k, the diagonal tension factor, is a measure of the degree to which the diagonal tension is developed. A completely unbuckled web has k = 0 whereas k = 1 for a web in complete diagonal tension. The value of k corresponding to a web having a critical 6.13 Tension field beams 195 Effective depth d shear stress TCR is given by the empirical expression k = tanh 0.5log- ( ;R) (6.1 11) The ratio r/rcR is known as the loading ratio or buckling stress ratio. The buckling stress TCR may be calculated from the formula (6.1 12) where k,, is the coefficient for a plate with simply supported edges and & and Rb are empirical restraint coefficients for the vertical and horizontal edges of the web panel respectively. Graphs giving k,,, Rd and Rb are reproduced in Kuhn14. The stress equations (6.106) and (6.107) are modified in the light of these assump- tions and may be rewritten in terms of the applied shear stress r as kr cot a (TF (2A,/td) + 0.5(1 - k) kr tan a (Ts = (As/tb) + 0.5( 1 - k) (6.113) (6.114) Further, the web stress ut given by Eq. (6.94) becomes two direct stresses: crl along the direction of a given by 2kr sin 2a (TI =- + r(l - k) sin2a (6.115) and CQ perpendicular to this direction given by a, = -r(1 - k) sin2a (6.116) The secondary bending moment of Eq. (6.104) is multiplied by the factor k, while the effective lengths for the calculation of stiffener buckling loads become (see Eqs (6.103)) or where d, is the actual stiffener depth, as opposed to the effective depth d of the web, taken between the web/flange connections as shown in Fig. 6.29. We observe that Eqs (6.1 13)-(6.116) are applicable to either incomplete or complete diagonal tension I, = d,/Jl + k2(3 - 2b/d,) for b < 1.5d I, = d, for b > 1% 0000000 St if fener depth _- web Fig. 6.29 Calculation of stiffener buckling load. [...]... (~T+Fg)Sin@-at(dCOSa)Sina=O (6 .11 7) For horizontal equilibrium (FT - FB) COS 3 - gttd COS' / =0 (6 .11 8) Taking moments about B w - FTd COS @ + $gttd2Cos2a = 0 z (6 .1 19) Solving Eqs (6 .11 7), (6 .11 8) and (6 .11 9) for q, FT and FB at = td 2w a (1 -$tan@) sin 2 FT=-[z+?( W d cos @ F B = -W z - F ( [ d cos @ (6 .12 0) )] )] 1 - -tan@ 1 tan@ (6 .12 1) (6 .12 2) Equation (6 .10 2) becomes (6 .12 3) Also the shear force... section of the beam is, from Fig 6. 30 s w - (FT + &j) sinp or, substituting for F and F from Eqs (6 .12 1) and (6 .12 2) T B G ) S = W 1 tan/3 (6 .12 4) Problems 19 7 1 Timoshenko, S P and Gere, J M., Theory of Elastic Stability, 2nd edition, McGraw-Hill Book Company, New York, 19 61 2 Gerard, G., Introduction to Structural Stability Theory, McGraw-Hill Book Company, New YQrk, 19 62 3 Murray, N W., Introduction... expression K = 7.70 [1 + 0.75(b/d)'] b and d having their usual significance The relationship between the diagonal tension factor and buckling stress ratio is T/TCR k 5 0.37 7 0.40 9 0.42 11 0.48 13 0. 51 15 0.53 Note that a is the angle of diagonal tension measured from the spanwise axis of the beam, as in the usual notation Am 1. 2mm, 13 0As/(l +0. 011 3As), 238 910 Nmm Part II Aircraft Structures Principles... 3784, 19 57 11 Gerard, G., Handbook of Structural Stability, Pt V, Compressive Strength of Flat Stiffened Panels, NACA Tech Note 3785, 19 57 12 Gerard, G and Becker, H., Handbook of Structural Stability, Pt VU, Strength of Thin Wing Construction, NACA Tech Note D - 16 2, 19 59 13 Gerard, G., The crippling strength of compression elements, J Aeron Sci 25 (1) , 37-52 Jan 19 58 14 Kuhn, P., Stresses in Aircraft. .. sinsin I Fig P 6 1 5 204 Structural instability For the particular case I = 2b, find the number of half waves m corresponding to the lowest critical stress, expressing the result to the nearest integer Determine also the lowest critical stress [6E/( 1 - G)] ( t/b)2 P .6 . 16 A panel, comprising flat sheet and uniformly spaced Z-section stringers, a part of whose cross-section is shown in Fig P .6 . 16 , is to be... shown in Fig P .6. 6 The eccentricity e can be varied and is to be adjusted to the value which, for given values of P and w, will result in the least maximum bending moment on the column Show that e = ( w / P p 2 )tan2 p1/4 where p2 = PIE1 Deduce the end moment which will give the optimum condition when P tends to zero Ans w12 / 16 w/unit length 12 i l l l i i l l 1 1 - P e e P Z Fig P .6. 6 200 Structural.. .19 6 Structural instability Fig 6. 30 Effect of taper on diagonal tension field beam calculations field beams since, for the latter case, k = 1 giving the results of Eqs (6 .10 6) , (6 .10 7) and (6. 94) In some cases beams taper along their lengths, in which case the flange loads are no longer horizontal but have vertical components which reduce the shear load carried by the web Thus, in Fig 6. 30 where... constants and assume that after initial buckling the stress in the plate increases parabolically from its critical value in the centre of sections Am 61 3 .8 N/mm, 844.7N/mm 3mm I Fig P .6 .17 1 , 4 I 3.5mm Problems 205 ‘i l T Fig P .6 .18 P .6 .18 Figure P .6 .18 shows the doubly symmetrical cross-section of a thin-walled column with rigidly fixed ends Find an expression, in terms of the section dimensions and... parabola w = kz (1- z ) and taking the more accurate of the two expressions for the bending moment 202 Structural instability In the case where I2= 1. 61 1 and a = 0. 21 find the percentage increase in strength due to the reinforcement, and compare it with the percentage increase in weight on the basis that the radius of gyration of the section is not altered Ans PCR = 14 .96EII/l2, 52%, 36% P 6 1 A tubular... Air Ministry Specification DTD 18 issued in 19 24, while artificially aged duralumin came under Specification DTD 11 1 in 19 29 Typical properties of the two types have been quoted above although DTD 11 1 provided for slight reductions in 0 .1 per cent proof stress and tensile strength The second group of aluminium alloys differs from duralumin chiefly by the introduction of 1 to 2 per cent of nickel, a high . Example 6 .1 and are listed below A = 60 0mm2 Zxx = 1. 17 x 10 6mm4 J = 800mm4 = 0 .67 x 10 6mm4 I? = 2488 x 10 6mm6 Zo = 5.32 x 10 6mm4 18 8 Structural instability From Eqs (6. 90). 0 (6 .11 8) Taking moments about B wz - FTd COS @ + $gttd2 Cos2 a = 0 (6 .1 19) Solving Eqs (6 .11 7), (6 .11 8) and (6 .11 9) for q, FT and FB at = td 2w sin 2a (1 -$tan@). 6. 30 s w - (FT + &j) sinp or, substituting for FT and FB from Eqs (6 .12 1) and (6 .12 2) S= W 1 tan/3 G) (6 .12 4) Problems 19 7 1 2 3 4 5 6 7 8 9 10 11 12