Aircraft Structures 3E Episode 14 docx

506 Matrix methods of structural analysis - - u1 v1 u2 v2 u3 v3 1 0 -1 0 0 0' 01 0 0 0 -1 1 1 1 1 -1 0 I+ - 1 1 1 1 1 1 1 1 1 1 1 1 2Jz 24 2Jz 2Jz 2.\/2 2Jz 2Jz 2J2 24 2J2 2Jz 2Jz 2a 2Jz 24 2Jz. - - 0 0 0 0 0 -1 - - 1 +- - 1 -1- 00 01 {:I:} - 0 -1 - 2.41 =o 211 = 0 u2 v2 u3 = 0 v3 = 0 If we now delete rows and columns in the stiffness matrix corresponding to zero displacements, we obtain the unknown nodal displacements u2 and v2 in terms of the applied loads Fx,2(= 0) and Fv,2(=- W). Thus 1 1 1+- 24 2Jz 2Jz 2J2 - Inverting Eq. (v) gives from which (vi) (vii) (viii) The reactions at nodes 1 and 3 are now obtained by substituting for u2 and v2 from Eq. (vi) into Eq. (iv). Thus 0 Ol 12.6 Matrix analysis of space frames 507 giving Fx,l = - Fx,2 - FY,2 = W Fy,l = 0 Fy,3 1 Fy>2 = - W Fy,3 = w Finally, the forces in the members are found from Eqs (12.32), (vii) and (viii) = - w (compression) AE s12 = t[l 01 = 0 (as expected) AE s13 = l1 L __ - ." - 1 1 I-T .r mdic%sn-tz statically indeterminate frameworks The matrix method of solution described in the previous sections for spring and pin- jointed framework assemblies is completely general and is therefore applicable to any structural problem. We observe that at no stage in Example 12.1 did the question of the degree of indeterminacy of the framework arise. It follows that problems involving statically indeterminate frameworks (and other structures) are solved in an identical manner to that presented in Example 12.1, the stiffness matrices for the redundant members being included in the complete stiffness matrix as before. -" %". ", Matrix anal) The procedure for the matrix analysis of space frames is similar to that for plane pin- jointed frameworks. The main difference lies in the transformation of the member stiffness matrices from local to global coordinates since, as we see from Fig. 12.5, axial nodal forces and have each now three global components cy.;, Fv,i, Fz,i and Fx,j, Fy,j, Fz,j respectively. The member stiffness matrix referred to global coordinates is therefore of the order 6 x 6 so that [ICii] of Eq. (12.22) must be expanded to the same order to allow for this. Hence - AE [K ] = ~ "L ui v; w; uj vj wj 10 0 -1 0 0 000 000 000 000 -100 100 000 000 000 000 (12.33) 508 Matrix methods of structural analysis ‘A, pf u, 0 0 0- A, p, u, 0 0 0 A, p2 u, 0 0 0 0 0 0 A, pj vj 0 0 0 A, p, v, 0 0 0 A, pz u,- Fig. 12.5 Local and global coordinate systems for a member in a pin-jointed space frame. In Fig. 12.5 the member ij is of length L, cross-sectional area A and modulus of elasticity E. Global and local coordinate systems are designated as for the two- dimensional case. Further, we suppose that e,, = angle between x and 2 8, = angle between x and jj e,, = angle between z and jj Therefore, nodal forces referred to the two systems of axes are related as follows (12.34) - F, = F, COS exj + F,, COS e,, + F, COS e,? F~ = F~ cos e,, + cos e,, + F, COS e,,: F~ = F, cos ezj + F,, cos e,, + F, COS ezi - - Writing A, = cos e,,, A, = COS e,,, A, = COS e, pj = cos e,,,, p- - COS e - pF = COS e,, uj = cos eaf, v, = COS e,,, uz = COS e=? Y - YY’ we may express Eq. (12.34) for nodes i andj in matrix form as (12.35) (12.36) 12.7 Stiffness matrix for a uniform beam 509 or in abbreviated form {PI = [TI{F) The derivation of [ICii] for a member of a space frame proceeds on identical lines to that for the plane frame member. Thus, as before T- [Kijl = [TI [KjI[Tl Substituting for [TI and [q] from Eqs (12.36) and (12.33) gives AE [K ] = - VL (12.37) All the suffixes in Eq. (12.37) are X so that we may rewrite the equation in simpler form, namely AE [K ] = - VL x2 SYM 2 AP P2 -A2 -xp -xu ; x2 -xp -p -pu ; xp p2 xu pu u 2 2 ; xu pu u 2 -xu -pu -u (12.38) where A, p and u are the direction cosines between the x, y, z and X axes respectively. The complete stiffness matrix for a space frame is assembled from the member stiffness matrices in a similar manner to that for the plane frame and the solution completed as before. !ss matrix for a uniform beam Our discussion so far has been restricted to structures comprising members capable of resisting axial loads only. Many structures, however, consist of beam assemblies in which the individual members resist shear and bending forces, in addition to axial loads. We shall now derive the stiffness matrix for a uniform beam and consider the solution of rigid jointed frameworks formed by an assembly of beams, or beam elements as they are sometimes called. Figure 12.6 shows a uniform beam ij of flexural rigidity EZ and length L subjected to nodal forces FV,;, F,,j and nodal moments Mi, Mi in the xy plane. The beam suffers nodal displacements and rotations vir vi and O;, 6,. We do not include axial forces here since their effects have already been determined in our investigation of pin-jointed frameworks. 510 Matrix methods of structural analysis Fig. 12.6 Forces and moments on a beam element. The stiffness matrix [Kv] may be built up by considering various deflected states for the beam and superimposing the results, as we did initially for the spring assemblies of Figs 12.1 and 12.2 or, alternatively, it may be written down directly from the well- known beam slope-deflection equations3. We shall adopt the latter procedure. From slope-deflection theory we have 6EI 4EI 6EI 2EI M v.+-ei+-v.+-ej I- L2 L L2 L and 6EI 2EI 6EI 4EI M v.+-e.+-v.+-e. j- ~21 ~1 L~J LJ Also, considering vertical equilibrium we obtain Fy,i + Fy, j = 0 and from moment equilibrium about node j we have Fy,iL + Mi + Mj = 0 Hence the solution of Eqs (12.39), (12.40), (12.41) and (12.42) gives 12EI 6EI 12EI 6EI -F .=F .= wi+-ei+-vj+-e. L3 L2 L3 L2 Y?Z YJ (12.39) (12.40) (12.41) (12.42) (12.43) Expressing Eqs (12.39), (12.40) and (12.43) in matrix form yields 121~~ -61~~ -121~~ -61~~ -121~~ -6/L2 61~~ 4/L 121~~ 6/L2 61~~ 2/L 1 { ;} (12.44) -6/L2 2/L 6/L2 4/L which is of the form {PI = [K&51 where [Kv] is the stiffness matrix for the beam. 12.7 Stiffness matrix for a uniform beam 51 1 It is possible to write Eq. (12.44) in an alternative form such that the elements of [KJ are pure numbers. Thus 12 -6 -12 -6 -64 62 -12 6 12 6 -62 64 This form of Eq. (12.44) is particularly useful in numerical calculations for an assem- blage of beams in which EI/L3 is constant. Equation (12.44) is derived for a beam whose axis is aligned with the x axis so that the stiffness matrix defined by Eq. (12.44) is actually the stiffness matrix referred to a local coordinate system. If the beam is positioned in the xy plane with its axis arbitrarily inclined to the x axis then the x and y axes form a global coordinate system and it becomes necessary to transform Eq. (12.44) to allow for this. The procedure is similar to that for the pin-jointed framework member of Section 12.4 in that [K,] must be expanded to allow for the fact that nodal displacements iij and Uj, which are irrelevant for the beam in local coordinates, have components uj, vi and uj, vj in global coordinates. Thus = EI ui vi 8i uj vj o 121~~ -61~~ o -121~~ -61~~ o -121~~ 61~~ o 121~~ 61~~ 00 000 0 0 -6/L2 4/L 0 6/L2 2/L 00 000 0 0 -6/L2 2/L 0 6/L2 4/L , (12.45) We may deduce the transformation matrix [TI from Eq. (12.24) if we remember that although u and w transform in exactly the same way as in the case of a pin- jointed member the rotations B remain the same in either local or global coordinates. Hence [TI = 'A /.Lo 0 00 -/.LAO 0 00 0 01 0 00 0 00 x po 0 00-/.LAO -0 00 0 01 where A and p have previously been defined. Thus since (12.46) (see Section 12.4) 51 2 Matrix methods of structural analysis [Kg] = EI - 12J/~~ SYM -12~~1~~ 12~~1~~ -12p2/~3 12~~1~~ -6p/~2 12p2/~3 12~~1~~ -12~~1~~ 6x1~~ -12~~1~~ 12~~1~~ 6p/L2 -6X/L2 4/L - 6p/L2 -6X/L2 2/L 6p/L2 6X/L2 4XIL (12.47) Again the stiffness matrix for the complete structure is assembled from the member stiffness matrices, the boundary conditions are applied and the resulting set of equations solved for the unknown nodal displacements and forces. The internal shear forces and bending moments in a beam may be obtained in terms of the calculated nodal displacements. Thus, for a beam joining nodes i andj we shall have obtained the unknown values of vi, Bi and vi, 0,. The nodal forces Fy,i and Mi are then obtained from Eq. (12.44) if the beam is aligned with the x axis. Hence Similar expressions are obtained for the forces at nodej. From Fig. 12.6 we see that the shear force S, and bending moment M in the beam are given by (12.49) Substituting Eqs (12.48) into Eqs (12.49) and expressing in matrix form yields 6 (12.50) 12 6 L2 L3 {2}=E'[12z -x - - 6 6 4 12 x+- 6 62 The matrix analysis of the beam in Fig. 12.6 is based on the condition that no external forces are applied between the nodes. Obviously in a practical case a beam supports a variety of loads along its length and therefore such beams must be idealized into a number of beam-elements for which the above condition holds. The idealization is accomplished by merely specifying nodes at points along the beam such that any element lying between adjacent nodes cames, at the most, a uniform shear and a linearly varying bending moment. For example, the beam of Fig. 12.7 would be idealized into beam-elements 1-2, 2-3 and 3-4 for which the unknown nodal displacements are v2, S2, 03, v4 and 0, (q = Beams supporting distributed loads require special treatment in that the distributed load is replaced by a series of statically equivalent point loads at a selected number of nodes. Clearly the greater the number of nodes chosen, the more accurate but more 12 L3 L2 -zx+z L3 L2 L2X+Z = v3 = 0). 12.7 Stiffness matrix for a uniform beam 513 L 4 - dunit length - wL - wL - wL 4 4 4 w W 2 3 4 - 128 I28 - wL - wL - wL wL 4 4 4 8 w W 2 3 4 Fig. 12.8 Idealization of a beam supporting a uniformly distributed load. complicated and therefore time consuming will be the analysis. Figure 12.8 shows a typical idealization of a beam supporting a uniformly distributed load. Details of the analysis of such beams may be found in Martin4. Many simple beam problems may be idealized into a combination of two beam- elements and three nodes. A few examples of such beams are shown in Fig. 12.9. If we therefore assemble a stiffness matrix for the general case of a two beam-element system we may use it to solve a variety of problems simply by inserting the appro- priate loading and support conditions. Consider the assemblage of two beam- elements shown in Fig. 12.10. The stiffness matrices for the beam-elements 1-2 and 2-3 are obtained from Eq. (12.44); thus Fig. 12.9 Idealization of beams into beam-elements. - LO Lb i Fig. 12.10 Assemblage of two beam-elements. The complete stiffness matrix is formed by superimposing [K12] and [K23] as described in Example 12.1. Hence [K] = E 0 0 0 0 (12.53) Example 12.2 Determine the unknown nodal displacements and forces in the beam shown in Fig. 12.1 1. The beam is of uniform section throughout. IW Fig. 12.11 Beam of Example 12.2. 12.7 Stiffness matrix for a uniform beam 515 The beam may be idealized into two beam-elements, 1-2 and 2-3. From Fig. 12.11 we see that v1 = v3 = 0, FJ,2 = - W, M2 = +M. Therefore, eliminating rows and columns corresponding to zero displacements from Eq. (12.53), we obtain Fy,2 = - W 27/2L3 9/2L2 6/L' -3/2L2 9/2L2 6/L 2/L 6/L2 2/L 4/L 0 -3/2L2 1/L 0 2/L Equation (i) may be written such that the elements of [Kl are pure numbers Fy,z = - W 27 9 12 -3 212 9 12 4 ( M3/L = 0 -3 12 4 2 0 8 !][ iz} Expanding Eq. (iij by matrix multiplication we have and Equation (iv) gives Substituting Eq. (v) in Eq. (iii) we obtain L3 -4 -2 { cL}=z[-2 3]{ M;} from which the unknown displacements at node 2 are 4 WL3 2ML' v2= 9 EI 9 EI 2WL2 1ML + "=GT 3 EI In addition, from Eq. (v) we find that 5WL2 1ML 9 EI 6 EI 0, = + 4WL2 1ML o3 = - - 9 EI 3 EI It should be noted that the solution has been obtained by inverting two 2 x 2 matrices rather than the 4 x 4 matrix of Eq. (ii). This simplification has been brought about by the fact that MI = M3 = 0. [...]... Finite element method for continuum structures In the previous sections we have discussed the matrix method of solution of structures composed of elements connected only at nodal points For skeletal structures consisting of arrangements of beams these nodal points fall naturally at joints and at positions of concentrated loading Continuum structures, such as flat plates, aircraft skins, shells etc, do not... ranging from two-dimensionalflat plate structures to three-dimensional folded plates and shells For three-dimensional applications the element stiffness matrix [Ke]is transformed from an in-plane xy coordinate system to a three-dimensional system of global coordinates by the use of a transformation matrix similar to those developed for the matrix analysis of skeletal structures In addition to the above,... in combination with triangular elements to build up particular geometrical shapes Figure 12 .14 shows a quadrilateral element referred to axes Oxy and having corner nodes, i, j , k and I; the nodal forces and displacements are also shown and the YA Fy,Il F., i*ui v, t i > 12.8 Finite element method for continuum structures 529 displacement and force vectors are , (12.95) As in the case of the triangular... If Young’s modulus E = 200000N/mm2 and Poisson’s ratio v = 0.3, calculate the stresses at the centre of the element Ans ux = 51.65N/mm2, uy = 55.49N/mm2, rxJ 13.46N/mm2 = Elementary aeroelasticity Aircraft structures, being extremely flexible, are prone to distortion under load When these loads are caused by aerodynamic forces, which themselves depend on the geometry of the structure and the orientation... Dynamic response 13.1 load distribution and divergence 541 In this chapter we shall concentrate on the purely structural aspects of aeroelasticity; its effect on aircraft static and dynamic stability is treated in books devoted primarily to aircraft stability and control''2 r i o a d distribution and divergence Redistribution of aerodynamic loads and divergence are closely related aeroelastic phenomena;... i.e motion without strain, while the linear terms enable states of constant strain to be specified; Eqs (12.82) ensure compatibility of displacement along the 12.8 Finite element method for continuum structures 523 edges of adjacent elements Writing Eqs (12.82) in matrix form gives {%;} = [o (12.83) 0 0 1 x y Comparing Eq (12.83) with Eq (12.55) we see that it is of the form (12.84) Substituting values... (12.90) 7"y and ux va, E, = - - - E E -ay -vox (see Chapter 1) E y - 7 7 - E Thus, in matrix form, {E} = { :} [ = ; YXY v : v 0 1 0 0 2(1+v) I(z } Tx, (12.91) 12.8 Finite element method for continuum structures 525 It may be shown that which has the form of Eq (12.68), i.e (4 [Dl{&> = Substituting for {E} in terms of the nodal displacements {$} we obtain {o} = [D][B]{a"} (see Eq (12.69)) In the case... method a displacement pattern is chosen for each element which may satisfy some, if not all, of the compatibility requirements along the sides of adjacent elements 12.8 Finite element method for continuum structures 51 7 Fig 12.12 Finite element idealization of a flat plate with a central hole Since we are employing matrix methods of solution we are concerned initially with the determination of nodal forces... 0 3 2 = 0 and D33 = cy derive the stiffness matrix for the element From Eqs (12.82) i.e i.e (ii) i.e u3 = al From Eq (i) and from Eqs (ii) and (iv) + 2a2 + 2a3 12.8 Finite element method for continuum structures 527 Then, from Eqs (iii), (iv) and (v) a = 3 2u3 - u1 - u2 4 Substituting for a l ,a2 and a3in the first of Eqs (12.82) gives 2u3 - 4 I - 2.42 U u = u l + ( 7 u2 - u1 )x+( )Y or u= (1-,-++ X... the load In the case of distributed loads, equivalent nodal concentrated loads must be calculated4 The solution procedure is identical in outline to that described in the previous sections for skeletal structures; the differences lie in the idealization of the structure into finite elements and the calculation of the stiffness matrix for each element The latter procedure, which in general terms is applicable . for continuum structures In the previous sections we have discussed the matrix method of solution of structures composed of elements connected only at nodal points. For skeletal structures consist-. uniform beam Our discussion so far has been restricted to structures comprising members capable of resisting axial loads only. Many structures, however, consist of beam assemblies in which. points fall naturally at joints and at positions of concentrated loading. Continuum structures, such as flat plates, aircraft skins, shells etc, do not possess such natural subdivisions and must