584 ___· �I �__I�Y�Rsse� GENERAL FORMULATION-LINEAR SYSTEM CHAP. 17 in (a) and (17-131): IIp, v + (T 1 -P)(BU + H3) -V = (¥-/o)Tk'(Y -) (17-132) = A1BU + A1H3 + A2U 2 The variation of HIp considering U as the independent variable is drfp = AUT[BT(P, -P) + (BTATk'AIB)U + BTAk'(A 1H3 + A2l 2 -'')] (g) = AUT[(BTKlB)U -BTH 4] Requiring [,I to be stationary for arbitrary AU results in (17-126). Note that we could have used the reduced form for V. i.e., equation (d). Also, we still have to determine the constraint forces. REFERENCES 1. FENVES, S. J., and F. H. BRANIN, JR., "Network-Topological Formulation of Struc-tural Analysis," J. StructuralDiv., A.S.C.E., Vol. 89, No. ST4, August, 1963, pp. 483-514. 2. DIMAGrIO, F. L., and W. R. SPILLARS. "Network Analysis of Structures," J. Eng.Mech. Div., A.S.C.E., Vol 91, No. EM3, June, 1965, pp. 169-188. 3. ARGYRIS, J. H., "The Matrix Analysis of Structures with Cut-Outs and Modifica-tions," Proc. Ninth InternationalCongress Appl. Mech., Vol. 6, 1957, pp. 131--142. 18 Analysis of Geometrically Nonlinear Systems 18-1. INTRODUCTION In this chapter, we extend the displacement formulation to include geometric nonlinearity. The derivation is restricted to small rotation, i.e., where squares of rotations are negligible with respect to unity. We also consider the material to be linearly elastic and the member to be prismatic. The first phase involves developing appropriate member force-displacement relations by integrating the governing equations derived in Sec. 13-9. We treat first planar deformation, since the equations for this case are easily integrated and it reveals the essential nonlinear effects. The three-dimensional problem is more formidable and one has to introduce numerous approximations in order to generate an explicit solution. We will briefly sketch out the solution strategy and then present a linearized solution applicable for doubly symmetric cross-sections. The direct stiffness method is employed to assemble the system equations. This phase is essentially the same as for the linear case. However, the governing equations are now nonlinear. Next, we described two iterative procedures for solving a set of nonlinear algebraic equations, successive substitution and Newton-Raphson iteration. These methods are applied to the system equations and the appropriate re-:.,rrence relations are developed. Finally, we utilize the classical stabilitycriterion to investigate the stability of an equilibrium position. 18-2. MEMBER EQUATIONS-PLANAR DEFORMATION Figure 18-1 shows the initial and deformed positions of the member. The centroidal axis initially coincides with the X direction and X2 is an axis of symmetry for the cross section. We work with displacements (ul, u2, c3), 585587 SEC. 18-2. MEMBER EQUATIONS-PLANAR DEFORMATION FR-, ANALYSIS OF GEOMETRICALLY NONLINEAR SYSTEMS CHAP. 18 Boundary Conditions For x = 0:distributed external force (b2), and end forces (F1, F2. M713) referred to the initial (X 1-X2-X 3) member frame. The rotation of the chord is denoted by U = Al or IF,| = -FA, p3 and is related to the end displacements by U2 = UA2 or IF2 + Fu 2.,0o= -FA2 (c) tB 2 UA2 (18-1)-P3= L = 0)A 3 Or IMIo -MA3 For x = L: The governing equations follow from (13-88). For convenience, we drop the or IF,IL =+ FB U2= UB2 or IF2 + F1U2,xL = FB2 (d)subscript on x1, and M 3, (03, i3. Also, we consider hb = m3 = 0. CO) = OB3 or IMIL = + MB3 Integrating (a) leads to x 2 FI = FB1 -P F2+ PU2. = PC2-fX b2 dx (e) M3 -Pu2 = -C3P -C2Px + fx(Sx b2dx)dx where C2, C3 are integration constants. We include the factor P so that the dimensions are consistent. Thc axial displacement ilt is determined from the first equation in (a), rv~~~~~182PL I 2u1 -UA1 = (AE (u)2,. dx (18-2) Combining the remaining two equations in (a), we obtain 2. xxM = EI1 + G-+ G-A b 2 (f) Finally, the governing equation for u2 follows from the third equation in (e), 2U2 ,xx + t2 = #u(C 2 + C3) + .Lb2 -(b 2 dX)dd 2where~ ~ ~~P A Fig. 18-1. Notation for planar bending. where (18-3) -P 2t1EquilibriumEquations EI +--F 1, = 0 GA2 (a) The solutions for u2 and M are (Ft2, + F2) + b2 =0 dx 2 =C COS sx + C5 sinx + C2x + U2b+ C 3 F2--Mx CO I (l + GA) .x C4 sin + Cs COSx) (18-4) Force-DisplacemenltRelations 2 + C2 + GA2 b2 dx +( + GA 2 U2b,xF, u1,, _) FE =u1X + 2(U2,) (b) where Ut2,denotes the particular solution due to b2. If b2 is constant,Fi2 F2 =U2.X -CO GA2 U2b {GA 2 x (18-5)12)MP GA2 2 t.c t- = O(, X589 588 ANALYSIS OF GEOMETRICALLY NONLINEAR SYSTEMS CHAP. 18 Enforcing the boundary conditions on U2, co at x = 0, L leads to four linear equations relating (C2 '" C5). When the coefficient matrix is singular, the member is said to have buckled. In what follows, we exclude member buck-ling. We also neglect transverse shear deformation since its effect is small for a homogeneous cross section. We consider the case where the end displacements are prescribed. The net displacements are lnet = u' = (I -u2b),=O, L (18-6) Onet = O' = (C -U2b,x)x=O, L Evaluating (18-4) with A2 = oo, we obtain C2 = (01. -tC 5 C3 = UA -C4 1 -cos L co -o C4 = -C5 .....sin JtL It sin tL (18-7) Cs = (u' -u oL)sin L -(CL (o) a) D = 2(1 -cos uL) -ML sin I1L Note that D 0 as FtL -2. This defines the upper limit on P, i.e., the member buckling load: 47r2EI -PInax = L2 (18-8) The end forces can be obtained with (c-e). We omit the algebraic details since they are obvious and list the final form below. MA3 = M3 + 0 2 ) --UA2) r r I 1 MB3-MB3 -+ I b4)20A3(-uA2)ILLEI3F-3 (,blB3 -L A ) =FA 2 F2 L [ B3 + O)A3 -(UB2 U2) 2 -UA2) (18-9) [CB3 + (CA3 -B2 UA2)1 + -(U2 -FB2= F2 -L22 -P = FeI T', = -P PL I2 PL (u ) 2UB II =-j xdx --eL fo' IA where D = 2(1 -cos ML) -uL sin utL Do = FtL(sin ML -bIL cos ItL) Dq 2 = L(uL -sin /L) D03 = D(4)1 + 0'2) = (,UL)2 (1 -cos /ML) SEC. 18-2. MEMBER EQUATIONS-PLANAR DEFORMATION The i functions were introduced by Livesley (Ref. 7), and are plotted in Fig.18-2. They degenerate rapidly as ylL -2. The initial end forces depend on the transverse loading, b2. If b2 is constant, A2= FB2 -bL 2 ' bL2 (18-10)B3 = (L)2 (1 -2)) MA3 = -B3 In order to evaluate the stiffness coefficients, P has to be known. If one end, say B, is unrestrainedwith respect to axial displacement, there is no difficulty since FBI is now prescribed. The relative displacement is determined from P = FI PL u1 = UA1 + A -Ler er -(U2 )2 dx = e( jL, uA2,uB, OA, COB) 2er = 5(C)2 + [4 ( -t2) + ¢i6(cJB3 -J-)A3) C5 + l)7(O.B3 -09A3)2 + -2 ) Dq04