COORDINAE RANSFORMAIONS Members of a structural system are typically oriented in differing directions, e.g., Fig. 17.1. In order to perform an analysis, the element stiffness equations need to be expressed in a common coordinate system typically the global coordinate system. Once the element equations are expressed in a common coordinate system, the equations for each element comprising the structure can be 1 assembled. Figure 17.1 Frame Structure (Kassimali, 1999) 2 Coordinate ransformations: Frame Elements Consider frame element m of Fig. 17.7. y Y m e x b X (a) Frame Q1, u 1 Q 3, u 3 Q 2, u 2 Q 6, u 6 Q 4, u m 4 Q 5, u 5 (b) Local Coordinate End Forces and Displacements Figure 17.7: Local Global Coordinate Relationships 3 4 1
Figure 17.7 shows that the local global coordinate transformations can be expressed as x = cos X + sin Y y = -sin X + cos Y and since z and Z are parallel, this coordinate transformation is expressed as z = Z Using the above coordinate transformations, the end force and displacement transformations can be expressed as 5 Q cosf sinf x X Y Q sinf cosf Q y X Y z F Z (17.9) where x, X = 1 or 4; y, Y = 2 or 5; and z, Z = 3 or 6. Utilizing (17.9) for all six member force components and expressing the resulting transformations in matrix form gives Qb [t] [0] Fb Q [0] [t] e F e or {Q} = [] {F} (17.11) where {Q} Q Q Q = b 1 2 3 beginning node local coordinate force vector; {Q} e = <Q 4 Q 5 Q 6 > = end node local coordinate force vector; {F} b = <F 1 F 2 F 3 > = beginning node global coordinate force vector; {F} e = <F 4 F 5 F 6 > = end node global coordinate force cos sin 0 vector; [t] = sin cos 0 = 0 0 1 local to global coordinate transformation matrix which is the same at each end node for a straight 7 member; {Q} = <<Q> b <Q> e > = element local coordinate force vector; {F} = <<F> b <F> e > = element global coordinate force [t] [0] vector; and [] = = [0] [t] element local to global coordinate transformation matrix. Utilizing (17.9) for all six member force components and expressing the resulting transformations in matrix form gives 8 2
he direction cosines used in the transformation matrices can easily be calculated from the nodal geometry, i.e. XeX cos b L Ye Y sin b L 2 2 L (XeX b) (YeY b) (17.13) {u} []{v} (17.14) It is also useful in matrix structural analysis to calculate the global end displacements and forces in terms of the local coordinate end displacements and forces as shown in Fig. D. Since the end displacements are aligned with the end forces, the local to global coordinate displacement relationships are 9 Figure D: Global Local Coordinate Relationships 10 X = cos x - sin y Y = sin x + cos y Z = z Applying the global local coordinate transformations to the end node forces gives FX cosqx sin Qy FY sin Qx cosqy FZ Qz (17.15) where X, x = 1 or 4; Y, y = 2 or 5; and Z, z = 3 or 6. F cos sin 0 Q X x F Y sin cos 0 Qy F 0 0 1 Q Z z or {F} node [t] {Q} node where node = b or e. Expanding the global local coordinate transformation to both end nodes leads to {F} b [t] [0] {Q} b {F} e {Q} e [0] [t] 13 14 3
or {F} [] {Q} (17.17) Similarly, the global coordinate displacement vector is related to the local coordinate displacement vector as {v} [] {u} (17.18) Continuous Beam Members When analyzing continuous beam structures, the axial displacement and force degrees of freedom are typically ignored since they are zero unless there is axial loading and the continuous beam is restrained against longitudinal motion, i.e., is not free to expand. Regardless, in continuous beam structures the local and global coordinate systems typically coincide resulting in [t b ] = [I] 2x2. his leads to {F b } = {Q b } and {v b } = {u b } (17.19) 15 14 russ Members Local global force and displacement relationships (see Fig. 17.9): F1 Q1 cos sin 0 0 F 2 Q 2 0 0 cos sin F3 F4 {Q a } = [ a ] {F a } (17.21) {u a } = [ a ] {v a } Similarly, {F a} [ a] {Q a} {v a} [ a] {u a} (a) Local Coordinate End Forces and Displacements for a russ Member (b) Global Coordinate End Forces and Displacements for a russ Member Figure 17.9 russ Member 15 16 4
MEMBER SIFFNESS RELAIONS IN GLOBAL COORDINAES o establish the global coordinate representation of the element stiffness equations, start by substituting {u} = [] {v} into (17.4): {Q} [k][]{v} {Q } (17.a) f which results in the force quantities being defined in the local coordinate system and the displacement vector expressed in terms of the global coordinate 17 system. o transform the force vectors into the global coordinate system, pre-multiply both sides of (17.a) by [] [] {Q} [] [k][]{v} [] {Q f } (17.23) Substituting {F} = [] {Q} into (17.23): {F} [] [k][]{v} {F f } (17.24) where {F f } = [] {Q f }. Equation (17.24) in matrix form: {F} [K]{v} {F } (17.25) where [K] = [] [k] [] = global coordinate element stiffness f 18 matrix, i.e., the element stiffness matrix coefficients aligned with the global coordinate system and K F ij i v 1 j with all other v k = 0 and k j. All global coordinate stiffness equations are expressed by (17.24) and (17.25). However, for beam and truss structures, the transformation matrix [], displacement vector {v}, and force vectors {F} and {F f } must be for these members. 19 For a continuous beam member (also see discussion following equation 1 and 17.19): [] = [ b ] = [I] 4x4 {v} = {v b } = {u b } {F} = {F b } = {Q b } {F f } = {F fb } = {Q fb } [K] = [K bb ] = [k bb ] (see 17.6) 20 5
For a truss member (also see discussion following equations 1 and 17.21): [] = [ a ] {v} = {v a } {F} = {F a } {F f } = {F fa } [K] [K aa ] [ a ] [k aa][ a ] 2 2 c cs c cs 2 2 EA cs s cs s L 2 2 (17.29) c cs c cs 2 2 cs s cs s c cos s sin 21 Structure Stiffness Relations he structure stiffness equations can now be determined using the global coordinate member stiffness equations. Generation of the structure stiffness equations is based on the three basic relationships of structural analysis: (1) equilibrium, (2) constitutive relationships, and (3) compatibility. Specifically, the direct stiffness procedure involves expressing: 22 (1)Node point equilibrium of the element end forces meeting at the node with the externally applied nodal forces; (2)Substituting the global coordinate constitutive (matrix stiffness) equations for the forces in terms of the stiffness coefficients times the element end displacements and fixedend force contributions; and (3)Compatibility of the element end displacements with the structure displacement degrees of freedom {d}. 23 Figure 17.10 Illustration of Direct Stiffness Analysis 24 6
Equilibrium Equations 2 X 1 4 1 F P F F FY P2F5 F2 2 P2 F5 F2 (17.30b) MZ P3F6 F3 2 P3 F6 F3 (17.30c) where superscripts (1), (2) designate Member Stiffness Relations (Constitutive Equations) Since the end forces in (17.30) P1 F4 F1 (17.30a) are unknown, the member stiffness relations 5 f5 element (member) 1, 2; respectively. 25 26 m F i (m) (m) F1 K11 K12 K13 K14 K15 K16 F 2 K21 K22 K23 K24 K25 K 26 F3 K31 K32 K33 K34 K35 K36 F4 K41 K42 K43 K44 K45 K46 F5 K51 K52 K53 K54 K55 K56 F6 K61 K62 K63 K64 K65 K66 (m) (m) v1 Ff1 v 2 F f2 v3 Ff3 v4 Ff4 v F v6 Ff6 (17.31) are substituted into (17.30) to give (1) (1) (1) (1) (1) (1) P1 K41 v1 K42 v2 K43 v3 (1) (1) (1) (1) (1) (1) (1) K44 v4 K45 v5 K46 v6 Ff4 (2) (2) (2) (2) (2) (2) K11 v1 K12 v2 K13 v3 (2) (2) (2) (2) (2) (2) (2) K14 v4 K15 v5 K16 v6 Ff1 (1) (1) (1) (1) (1) (1) P2 K51 v1 K52 v2 K53 v3 (1) (1) (1) (1) (1) (1) (1) K54 v4 K55 v5 K56 v6 Ff5 (2) (2) (2) (2) (2) (2) K21 v1 K22 v2 K23 v3 (2) (2) (2) (2) (2) (2) (2) K24 v4 K25 v5 K26 v6 Ff2 (1) (1) (1) (1) (1) (1) P3 K61 v1 K62 v2 K63 v3 (1) (1) (1) (1) (1) (1) (1) K64 v4 K65 v5 K66 v6 Ff6 (2) (2) (2) (2) (2) (2) K31 v1 K32 v2 K33 v3 (2) (2) (2) (2) (2) (2) (2) K34 v4 K35 v5 K36 v6 Ff3 27 Compatibility Imposing the compatibility (continuity) conditions (1) (1) (1) v1 v2 v3 0 (17.35) (1) (1) (1) v4 d 1; v5 d 2; v6 d3 (2) (2) (2) v1 d 1; v2 d 2; v3 d3 (17.36) (2) (2) (2) v4 v5 v6 0 on the constitutive equation version of the equilibrium equations leads to P 1 (K44 K 11 )d 1(K45 K 12 )d2 (17.39a) (K46 K 13 )d 3 (Ff4 F f1 ) 28 7
P 2 (K54 K 21 )d 1(K55 K 22 )d2 (17.39b) (K56 K 23 )d 3 (Ff5 F f2 ) P 3 (K64 K 31 )d 1(K65 K 32 )d2 (17.39c) (K66 K 33 )d 3 (Ff6 F f3 ) Or collectively as {P} [S]{d} {P f } [S]{d} ({P} {P f }) {P} (17.41) K44 K11 K45 K12 K46 K 13 [S] K54 K21 K55 K22 K56 K 23 K64 K31 K65 K32 K66 K33 Ff4 F f1 {P f } Ff5 Ff2 Ff6 F f3 P1 Pf1 {P} P2Pf2 P 3 P f3 he structure stiffness coefficients S ij are defined in the usual manner, i.e., Sij Pi force at dof i due to dj 1 a unit displacement at j with all other displacements d k = 0 and k j. 29 30 Assembly of [S] and {P f } Using Member Code Numbers he explicit details given for the direct stiffness procedure highlights how the basic equations of structural analysis are utilized in matrix structural analysis. A disadvantage of the procedure is that it is tedious and not amenable to computer implementation. Computer implementation is based on using a destination array ID(NNDF,NNP) (NNDF Number of Nodal Degrees of Freedom, three 31 for frame members; and NNP Number of Node Points, i.e., number of nodes used in the 2D structure discretization). he ID(, ) array identifies the nodal equation numbers in sequence: ID(1,node) = X-axis nodal displ. number ID(2,node) = Y-axis nodal displ. number ID(3,node) = Nodal rotation displ. number For example, consider the gable frame structure shown on the next slide. he destination array is also shown on this slide. 32 8
Gable Frame Structure Node ID 1 2 3 1 0 0 0 2 1 2 3 3 4 5 6 4 7 8 9 5 0 0 10 33 he ID(, ) array along with the element node number array IEN(NEN, NEL) (NEN Number of Element Nodes, which is two for discrete structural elements; and NEL Number of Elements used to discretize the structure) is used to construct the element location matrix array LM(NEDF, NEL) (NEDF Number of Element Degrees of Freedom, which is six for the 2D frame elements), i.e., 34 LM(i, iel) = ID(i, IEN(1,iel)); 1 = b node LM(j, iel) = ID(i, IEN(2,iel)); 2 = e node i = 1, 2, 3 ; j = 4, 5, 6 iel = element number he IEN(, ) and LM(, ) arrays for the gable frame example are IEN LM iel 1 2 1 2 3 4 5 6 1 1 2 0 0 0 1 2 3 2 2 3 1 2 3 4 5 6 3 3 4 4 5 6 7 8 9 4 5 4 0 0 10 7 8 9 he LM(, ) array is used to assemble the global element stiffness matrix and fixed-end force vector as illustrated on LM: (1,m) (2,m) (3,m) (4,m) (5,m) (6,m) (m) LM(1, m) K11 K12 K13 K14 K15 K16 LM(2, m) K21 K22 K23 K24 K25 K 26 LM(3, m) K31 K32 K33 K34 K35 K36 LM(4, m) K41 K42 K43 K44 K45 K 46 LM(5, m) K51 K52 K53 K54 K55 K56 LM(6, m) K61 K62 K63 K64 K65 K 66 (m) LM(1, m) Ff1 LM(2, m) F f2 LM(3, m) Ff3 LM(4, m) Ff4 LM(5, m) Ff5 LM(6, m) Ff6 the next two slides. 35 36 9
For the example gable frame: SUMMARY 11 44 11 21 54 21 S K K S K K (3) (4) 9,9 66 66 S K K S S S (4) 10,9 K36 (4) 9,10 K63 (4) 10,10 K33 f1 f4 f1 f2 f5 f2 P F F P F F (3) (4) f9 f6 f6 P F F P f10 F (4) f3 Identify structure degrees of freedom {d} For each element: Evaluate [k], {Q f }, and [] Calculate [K] = [] [k] [] {F f } = [] {Q f } Assemble [K] into [S] Assemble {F f } into {P f } Form nodal load vector {P} Solve: [S] {d} = {P} {P f } For each element: Obtain {v} from {d} (compatibility) Calculate {u} = [] {v} {Q} = [k] {u} + {Q f } 37 Determine support reactions by considering support joint equilibrium 38 Self-Straining: structure that is internally strained and in a state of stress while at rest without sustaining any external loading Example Problems russ Frame 39 40 10
emperature Restraint Example Problems Displacements of Statically Determinate Structures 41 42 Some Features of the Stiffness Equations An important characteristic of both element and global (structure) linear elastic stiffness equations is that they are symmetric. Practically, this means that only the main diagonal terms and coefficients to one side of the main diagonal need to be generated and stored in a computer program. Also note that the stiffness (equilibrium) equation for a given degree of freedom is influenced by 43 the degrees of freedom associated with the elements connecting to that degree of freedom. hus, nonzero coefficients in a given row of a stiffness matrix consist only of the main diagonal and coefficients corresponding to degrees of freedom at that node and at other nodes on the on the elements meeting at the node for which the stiffness equation is being formed. 44 11
All other coefficients in the row are zero. When there are many degrees of freedom in the complete structure, the stiffness matrix may contain relatively few nonzero terms, in which case it is characterized as sparse or weakly populated. Clearly, it would be advantageous in the solution phase of analysis to cluster all nonzero coefficients close to the main diagonal. his isolation of the zero terms facilitates their removal in the solution process. his can be done by numbering the degrees of freedom in such a way that the columnar distance of the term most remote from the main diagonal coefficient in each row is minimized; i.e., by minimizing the Banded Stiffness Equations 45 46 bandwidth. his is most easily achieved by minimizing the nodal difference between connected elements. All commercial structural and finite element codes have built in schemes to minimize the interaction of the stiffness equations regardless of the userspecified input. 47 Indeterminacy Cognizance of static indeterminacy is not necessary in the direct stiffness approach. he displacement approach uses the analogous concept of kinematic indeterminacy, refers to the number of displacement degrees of freedom that are required to define the displacement response of the structure to any load. Stated another way, kinematic indeterminacy equals the 48 12
number of displacement degrees of freedom that must be constrained at the nodes in addition to the boundary constraints (support conditions) to reduce the system to one in which all the nodal displacements are zero or have predetermined values, e.g., specified support settlement. It is fairly obvious that kinematic indeterminacy is quite different from static indeterminacy. 49 13