Aircraft Structures 3E Episode 12 pptx

426 Stress analysis of aircraft components corresponding to a1 is given by A1 E1 = - I and = El&] (10.44) (10.45) where El is the modulus of elasticity of the lamina in the direction of the flament. Also, using the suffixes f and m to designate filament and matrix parameters, we have af = EfEl, 0, = E,E~ (10.46) Further, if A is the total area of cross-section of the lamina in Fig. 10.66, Af is the cross-sectional area of the filament and A, the cross-sectional area of the matrix then, for equilibrium in the direction of the filament o1A = UfAf + amA, or, substituting for q, af and a, from Eqs (10.45) and (10.46) E~E~A = EfEfAf + Em€lAm so that Writing Af/A = vf and A,/A = v,, Eq. (10.47) becomes El = VfEf + VmEm (10.47) (10.48) Equation (10.48) is generally referred to as the law of mixtures. A similar approach may be used to determine the modulus of elasticity in the transverse direction (Et). In Fig. 10.67 the total extension in the transverse direction is produced by q and is given by Etlt = Em& + Eflf t t fUit t t Fig. 10.67 Determination of Et. 10.6 laminated composite structures 427 or which gives Rearranging this we obtain (10.49) The major Poisson’s ratio qt may be found by referring to the stress system of Fig. 10.66 and the dimensions given in Fig. 10.67. The total displacement in the transverse direction produced by ul is given by At = YtEllt i.e. At = YtEllt = U,E~I~ + VfEllf from which Yt = vmvm + Wf (10.50) The minor Poisson’s ratio ut] is found by referring to Fig. 10.67. The strain in the longitudinal direction produced by the transverse stress at is given by at ut ut ufl- = urn- = q- Et Et Ef From the last two of Eqs (10.51) m Ef Em q=-u Substituting in Eq. (10.50) or, from Eq. (10.48) Now substituting for urn in the first two of Eqs (10.51) ut1 Yt -=- Et Et or (10.51) (10.52) 428 Stress analysis of aircraft components 80mm f * Fig. 10.68 Determination of Glt. Finally, the shear modulus qt(= Gd) is determined by assuming that the constituent materials are subjected to the same shear stress qt as shown in Fig. 10.68. The displacement A, produced by shear is in which G, and Gf are the shear moduli of the matrix and filament respectively. Thus whence (10.53) Example IO. 17 A laminated bar whose cross-section is shown in Fig. 10.69 is 500mm long and com- prises an epoxy resin matrix reinforced by a carbon filament having moduli equal to 5000 N/mm2 and 200 000 N/mm2 respectively; the corresponding values of Poisson’s ratio are 0.2 and 0.3. If the bar is subjected to an axial tensile load of 100 kN, deter- mine the lengthening of the bar and the reduction in its thickness. Calculate also the stresses in the epoxy resin and the carbon filament. Fig. 10.69 Cross-section of the bar of Example 10.1 7. 10.6 Laminated composite structures 429 From Eq. (10.48) the modulus of the bar is given by 80 x 40 80 x 50 + 5000 x - E1 = 200000 x - 80 x 10 80 x 50 i.e. El = 44000N/m111' The direct stress, q, in the longitudinal direction is given by = 25.0 N/m2 100 x io3 a1 = 80 x 50 Therefore, from Eq. (10.45), the longitudinal strain in the bar is 25.0 44 000 &I = - = 5.68 x 10-~ The lengthening, A,, of the bar is then AI = 5.68 x x 500 i.e. A1 = 0.284- The major Poisson's ratio for the bar is found from Eq. (10.50). Thus x 0.3 = 0.22 '80 x 10 x 0.2 + - vlt =m 80 x 50 80 x 40 Hence the strain in the bar across its thickness is &, = -0.22 5.68 10-~ = -1.25 10-~ The reduction in thickness, A,, of the bar is then A, = 1.25 x x 50 i.e. A, = 0.006111m The stresses in the epoxy and the carbon are found using Eqs (10.46). Thus nrn (epoxy) = 5000 x 5.68 x of (carbon) = 200000 x 5.68 x = 2.84N/mm2 = 113.6N/mm2 10.6.2 Composite plates In Chapter 5 we considered thin plates subjected to a variety of loading conditions. We shall now extend the analysis to a lamina comprising a filament and matrix of the type shown in Fig. 10.66. 430 Stress analysis of aircraft components Suppose the lamina of Fig. 10.66 is subjected to stresses 01, at and qt acting simultaneously. From Eqs (1.42) and (1.46) at et = - - qt- 71t 7lt = - Et Glt Solving the first two of Eqs (10.54), we obtain Also 71t = Gltylt I Equations (10.55) may be written in matrix form as (10.54) (10.55) (10.56) in which c21 = c22 = c33 = Gt (= CI2, see Eq. (10.52)) 1 - YPtI Et 1 - YPtl Suppose now that the longitudinal and transverse directions coincide with the x and y axes respectively of the plates in Chapter 5. Equations (10.56) then become (10.57) From Eqs (1.27), (1.28) and the derivation of Eq. (5.13) we see that a'w . a2W a'W E, = -z-, EY = -z-, rxy = -2z- ax2 aY2 axay 10.6 laminated composite structures 431 so that Eqs (10.57) may be rewritten as From Section 5.3 rtl2 1-112 tl2 ayz dz, Mxy = - J ‘ T,~Z dz -t/2 M, = J ’ axzdz, M~ = I, 412 Substituting for a,, cy and rxy from Eqs (10.58) and integrating, we obtain Writing Cllt3/12 as Dll, C12t3/12 as D12 Similarly and (10.58) (10.59) (10.60) (10.61) For a lamina subjected to a distributed load of intensity q per unit area we see, by substituting for M,, My and Mxy from Eqs (10.59)-(10.61) into Eq. (5.19), that (10.62) Further, for a lamina subjected to in-plane loads in addition to q we obtain, by a comparison of Eq. (10.62) with Eq. (5.33) - a2W a2W a2W -q+Nx-+2N -+N - ax2 ’ay’ xya~ay (10.63) Problems involving laminated plates are solved in a similar manner to those included in Chapter 5 after the calculation of the modiiied flexural rigidities D1 1, Ol2, DZ2 etc. If the principal material directions 1 and t do not coincide with the x and y directions in the above equations, in-plane shear effects are introduced which modify Eqs (10.62) 432 Stress analysis of aircraft components and (10.63) (Ref. 1). The resulting equations are complex and require numerical methods of solution. Generally, composite structures consist of several laminas with the direction of the filaments arranged so that they lie in the directions of the major loads. Thus, for a loading system which comprises two mutually perpendicular loads, it is necessary to build or lay-up a laminate with sufficient plies in both directions to withstand each load. Such an arrangement is known as a cross-ply laminate. The analysis of multi-ply laminates is complex and is normally carried out using finite difference or finite element methods. 1 Calcote, L. R., The Analysis of Laminated Composite Structures, Van Nostrand Reinhold Co., New York, 1969. Datoo, M. H., Mechanics of Fibrous Composites, Elsevier Applied Science, London, 1991. P.10.1 A wing spar has the dimensions shown in Fig. P.10.1 and carries a uniformly distributed load of 15 kN/m along its complete length. Each flange has a cross-sectional area of 500mm2 with the top flange being horizontal. If the flanges are assumed to resist all direct loads while the spar web is effective only in shear, determine the flange loads and the shear flows in the web at sections 1 m and 2m from the free end. Am. 1 m from free end: Pu = 25 kN (tension), PL = 25.1 kN (compression), 2 m from free end: Pu = 75 kN (tension), PL = 75.4 kN (compression), q = 41.7 N/mm. q = 56.3 N/mm. lrn c -1 Fig. P.10.1 Problems 433 P.10.2 If the web in the wing spar of P.10.1 has a thickness of 2mm and is fully effective in resisting direct stresses, calculate the maximum value of shear flow in the web at a section 1 m from the free end of the beam. Ans. 46.8 N/mm. P.10.3 Calculate the shear flow distribution and the stringer and flange loads in the beam shown in Fig. P.10.3 at a section 1.5m from the built-in end. Assume that the skin and web panels are effective in resisting shear stress only; the beam tapers symmetrically in a vertical direction about its longitudinal axis. Am. 913 = q42 = 36.9 N/mm, q35 = 464 = 7.3 N/mm, qZ1 = 96.2 N/mm, 465 = 22.3 N/mm. P’ = -Pi = 133.3 kN. P4 = P6 = -P3 = -Ps = 66.7kN 250rnm Fig. P.10.3 mrn P.10.4 The doubly symmetrical fuselage section shown in Fig. P.10.4 has been idealized into an arrangement of direct stress carrying booms and shear stress carrying skin panels; the boom areas are all 150mm’. Calculate the direct stresses in the booms and the shear flows in the panels when the section is subjected to a shear load of 50 kN and a bending moment of 100 kN m. Ans. u=,~ = = 180N/mm , uz,2 = a,,lo = = -oz.7 = 144.9N/mm2, 02,3 = oz,g = -o,p = -ffzZ,8 = 60N/m2. 2 q21 = q65 = 1.9N/mm, q32 = q54 = 12.8 N/mm, q43 = 17.3 N/=, q67 = ql0 = 11.6 N/mm, q78 = q9 = 22.5 N/mm, qg9 = 27.0 N/mm 434 Stress analysis of aircraft components ~ 400 mm - - 400 rnrn . . 400 mm X m 6 Fig. P.10.4 P.10.5 Determine the shear flow distribution in the fuselage section of P.10.4 by replacing the applied load by a shear load through the shear centre together with a pure torque. P.10.6 The central cell of a wing has the idealized section shown in Fig. P.10.6. If the lift and drag loads on the wing produce bending moments of - 120 000 Nm and -30000Nm respectively at the section shown, calculate the direct stresses in the booms. Neglect axial constraint effects and assume that the lift and drag vectors are in vertical and horizontal planes Boom areas: B1 = B4 = B5 = B8 = 1OOOmm 2 B2 = B3 = B6 = B7 = 600mm2 AFZS. ~1 = -190.7N/m2, 02 = -181.7N/mm2, 63 = -172.8N/mm2, U4 = -163.8 N/lIUll2, 65 = 14ON/mI'Il2, 66 = 164.8 N/Inm2, U7 = 189.6N/m2, Us = 214.4N/m2 Fig. P.10.6 P.10.7 Figure P. 10.7 shows the cross-section of a two-cell torque box. If the shear stress in any wall must not exceed 140 N/m2, find the maximum torque which can be applied to the box. Problems 435 Fig. P.10.7 If this torque were applied at one end and resisted at the other end of such a box of span 2500mm, find the twist in degrees of one end relative to the other and the torsional rigidity of the box. The shear modulus G = 26 600 N/mm2 for all walls. Data: Shaded areas: A34 = 6450111m', A16 = 775011~~~ Wall lengths: Wall thickness: ~3~ = 250mm, $16 = 300mm tI2 = 1.63mm, t34 = 0.56mm t23 = t45 = t56 = 0.92mm 161 = 2.03- t25 = 2.54mm Ans. T = 102417Nm, f3 = 1.46", GJ = 10 x 10'2Nmm2/rad. P.10.8 Determine the torsional stiffness of the four-cell wing section shown in Fig. P.10.8. Data: Wall 12 23 34 78 67 56 45' 45' 36 27 18 Peripheral length (mm) 762 812 812 1525 356 406 356 254 Thickness (mm) 0.915 0.915 0.915 0.711 1.220 1.625 1.220 0.915 Cell areas (mm') AI = 161 500, A11 = 291 000 AI11 291 000, Alv = 226 000 Am. 522.5 x 106GNmm2/rad. Fig. P.10.8 [...]... (N/mm2) Boom Area (mrn') 34, 56 12. 23,61,18 36,81 45 380 356 306 610 0.915 0.915 1.220 1.220 20 700 24 200 24 800 24 800 1, 3, 6, 8 2,4, 5: 1 129 0 645 Nose area N,= 51 500mm' 4 153 X\WN+- I53 3 3 +_ _ _x - - ' - mm + I-mm Fig P.10.11 P.10 .12 A singly symmetric wing section consists of two closed cells and one open cell (see Fig P.10 .12) The webs 25, 34 and the walls 12, 56 are straight, while all...436 Stress analysis of aircraft components P.10.9 Determine the shear flow distribution for a torque of 56 500 N m for the three cell section shown in Fig P.10.9 The section has a constant shear modulus throughout Wall Thickness (mm) ~ 12" 12 L 14,23 34u 34L Area (mm') I I1 I11 108400 202 500 528 000 ~~ 1084 2160 127 797 797 Ans qI2u = 25.4N/mm, q43u = Cell 1.220 1.625... to 6 Calculate 438 Stress analysis of aircraft components 2 f TI 304 I mm 1- I 508 mm 762 mm Fig P.10 .12 the distance xs of the shear centre S aft of the web 34 The shear modulus G is the same for all walls Wall Length (mm) Thickness (mm) Boom Area (mm2) Cell Area (mm’) 12, 56 23,45 3: 4 38 25 510 165 1015 304 304 0.559 0.915 0.559 2.030 1.625 1,6 2, 5 3,4 645 129 0 1935 I 93 000 258000 I1 Ans 241.4mm... and v‘ = dv/dz, we obtain from Eq (1 1.1) q12 = 1.6Gv’ q23 = i o ~ ( m 0.8868’ - 0.886~’ 0.5~’) x q34 = i 2 ~ p o o 0.866e’ - ~r) x (i) (E) (iii) (4 q41 = 1.OGu’ For horizontal equilibrium 500 x 0.886q41- 500 x O.866q23 = 0 giving q41 = q23 For vertical equilibrium 375q12 - 125 q34 - 25oq23 = 22 000 For moment equilibrium about point 1 500 x 375 x 0.886q23 + 125 x 500 x 0.886q34= 22000 x 100 or 3q23... 120 0 320 320 210 1 o 1 o 0.6 2.0 2.0 1.5 1,6 2, 5 3, 4 600 800 800 I 100000 260 000 180000 I1 11 1 E E 0 E E 0 (u (u IFig P.10.13 7 0mm 9 \, 590mm A Problems 439 Ans Pi = - P 6 = 120 0N, q12 = q 5 6 = 3.74N/mm7 q43i = 12. 16N/m11; P2 = - P 5 = 2424N, P 3 = -P4 q 2 3 = q45= 3.11 N/mm, q52 = 14.58N/mm, 961 = = 2462N, q340 = 0.06N/mm, 11.22N/mm P.10.14 Solve P 10.8 using the method of successive approximations... calculate the shear flows in the panels and the direct loads in the rib flanges and stiffeners Ans ql = 4.0 N/mm, q 2 = 26.0 N/mm, q3 = 6.0 N/mm Pz in 12 = -P3 in 43 = 120 0N (tension), P5in 154 = 2000N (tension), P3 in 263 = 8000 N (compression), P5in 56 = 12 000 N (tension), P6 in 263 = 6000 N (compression) 2 1 8000 N Fig P.10.18 P.10.19 A portion of a wing box is built-in at one end and carries a shear... built-in end of the beam shown in Fig 11.2(a) when, at this section, it carries a shear load of 22 000 N acting at a distance of lOOmm from and parallel to side 12 The modulus of rigidity G is constant throughout the section Wall Length (mm) 12 375 34 125 23 500 11.2 Built-in end of a closed section beam 447 I.Ornm 2 22 000 N vL U 1.6rnrn 4 10rnrn - 1 0rnrn 0 (b) (a) Fig 11.2 (a) Beam cross-section at built-in... a shear load Sy = 12 000 N in the plane of the web 52 are applied at the larger cross-section Calculate the forces in the booms and the shear flow distribution at this cross-section The modulus G is constant throughout Section dimensions at the larger cross-section are given below Wall Length (mm) Thickness (mm) Boom Area (mm2) Cell Area (mm’) 12, 56 23,45 3: 4 34 l 25 16 600 800 120 0 320 320 210 1... the spar web 47 The shear modulus G is constant throughout and all booms have a cross-sectional area of 2000 mm2 Cell areas (mm2): I 120 000 56' 1500 2.5 Wall Lengths (mm) Thickness (mm) I1 215000 45,67 605 3.O I11 250000 43,78 603 3.O IV 215000 32,89 605 3.O V 155000 12, 910 605 2.5 All spar webs have a thickness of 3.0mm z h q650 = 9.1 N /m, q65' = 54.6N/m11, q74= 65.9 N/mm, q 2 3 = qsg q54 q43= qs7... same for all walls of the wing box ~~ Wall Length (mm) ~ Thickness (mm) ~~ 16 25 34 12, 56 23,45 254 406 202 641 1.625 2.032 1.220 0.915 0.559 175 Cell areas: AI = 232 000 mm2, ,411 = 258 000 mm2 Ans Boom q16 = 33.9N/m, q45 = q23 = 7.2N/mm, q25 = 73.4N/mm q65 = 421 = 1.1N/IIUn, q34 = 20.8 N/mm, Area (mm2) _ ~ 1 6 2, 5 334 129 0 1936 645 Problems 437 44500 N I I I 1 Fig P.lO.10 P.10.11 Figure P 10.11 shows . for a,, cy and rxy from Eqs (10.58) and integrating, we obtain Writing Cllt3 /12 as Dll, C12t3 /12 as D12 Similarly and (10.58) (10.59) (10.60) (10.61) For a lamina subjected to a. composite structures 431 so that Eqs (10.57) may be rewritten as From Section 5.3 rtl2 1- 112 tl2 ayz dz, Mxy = - J ‘ T,~Z dz -t/2 M, = J ’ axzdz, M~ = I, 412 Substituting. four-cell wing section shown in Fig. P.10.8. Data: Wall 12 23 34 78 67 56 45' 45' 36 27 18 Peripheral length (mm) 762 812 812 1525 356 406 356 254 Thickness (mm) 0.915 0.915