COORDINATE TRANSFORMATIONS Members of a structural system are typically oriented in differing directions, e.g., Fig. 17.1. In order to perform an analysis, the ele- ment stiffness equations need to be expressed in a common coor- dinate 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 Figure 17.1 – Frame Structure 1 (Kassimali, 1999) 2 assembled.
Coordinate Transformations: Frame Elements Consider frame element m of Fig. 17.7.
Y e x y m b
X
(a) Frame
Q , u 6 6 Q , u m 4 4 Q3, u3 Q , u Q5, u5 1 1 Figure 17.7: Local – Global Q2, u2 Coordinate Relationships (b) Local Coordinate End Forces and Displacements
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1 Figure 17.7 shows that the local – QcosFsinFxXY global coordinate transformations QsinFcosFyXY (17.9) can be expressed as QFzZ x = cos X + sin Y y = -sin X + cos Y where x, X = 1 or 4; y, Y = 2 or 5; and z, Z = 3 or 6. and since z and Z are parallel, this coordinate transformation is Utilizing (17.9) for all six member expressed as force components and expressing z = Z the resulting transformations in Using the above coordinate matrix form gives transformations, the end force and QFb [t] [0] b displacement transformations can QFee[0] [t] be expressed as or
5 {Q} = [T] {F} (17.11)
T {Q} Q Q Q T where b 123 = member; {Q} = <b
e> = beginning node local coordinate element local coordinate force T T force vector; {Q}e =
2 The direction cosines used in the {u} [T]{v} (17.14) transformation matrices can easily It is also useful in matrix structural be calculated from the nodal analysis to calculate the global end geometry, i.e. displacements and forces in terms XX cos eb of the local coordinate end dis- L placements and forces as shown in YY (17.13) sin eb L Fig. D. 22 L(XX)(YY)eb eb Since the end displacements are aligned with the end forces, the local to global coordinate displacement relationships are Figure D: Global – Local Coordinate Relationships
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X = cos x - sin y Fcossin0QXx Y = sin x + cos y Fsincos0Q Yy Z = z F001Z Qz Applying the global – local or coordinate transformations to the T {F}node [t] {Q} node end node forces gives where node = b or e. FXxy cos Q sin Q Expanding the global – local FYx sin Q cos Q y(17.15) coordinate transformation to both FQZz end nodes leads to where X, x = 1 or 4; Y, y = 2 or 5; T {F}b [t] [0] {Q}b and Z, z = 3 or 6. {F}T {Q} ee[0] [t]
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3 or Continuous Beam Members T (17.17) {F} [T] {Q} When analyzing continuous beam Similarly, the global coordinate structures, the axial displacement displacement vector is related to and force degrees of freedom are typically ignored since they are zero the local coordinate displacement unless there is axial loading and the vector as continuous beam is restrained {v} [T]T {u} (17.18) 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 [tb] = [I]2x2. This leads to
{Fb} = {Qb} and {vb} = {ub} (17.19)
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Truss Members Local – global force and displacement relationships (see Fig. 17.9): (a) Local Coordinate End Forces F1 and Displacements for a Truss Member QF12cos sin 0 0 QF2300cossin F4
{Qa} = [Ta] {Fa} (b) Global Coordinate End Forces (17.21) and Displacements for a Truss Member {ua} = [Ta] {va} Similarly, Figure 17.9 Truss Member T {Faaa } [T ] {Q } T {vaaa } [T ] {u }
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4 MEMBER STIFFNESS force vectors into the global coor- RELATIONS IN GLOBAL dinate system, pre-multiply both COORDINATES sides of (17.a) by [T]T To establish the global coordinate [T]TT {Q} [T] [k][T]{v} representation of the element [T]T {Q } stiffness equations, start by f (17.23) T substituting {u} = [T] {v} into Substituting {F} = [T] {Q} into (17.23): (17.4): T {F} [T] [k][T]{v} {Ff } (17.24) {Q} [k][T]{v} {Qf } (17.a) T which results in the force quan- where {Ff} = [T] {Qf}. tities being defined in the local Equation (17.24) in matrix form: coordinate system and the {F} [K]{v}{ Ff } (17.25) displacement vector expressed in T terms of the global coordinate where [K] = [T] [k] [T] = global 17 18 system. To transform the coordinate element stiffness
matrix, i.e., the element stiffness For a continuous beam member matrix coefficients aligned with the (also see discussion following global coordinate system and equation 1 and 17.19): KFij i v1j [T] = [Tb] = [I]4x4 with all other vk = 0 and k ≠ j. {v} = {vb} = {ub} All global coordinate stiffness {F} = {F } = {Q } equations are expressed by (17.24) b b and (17.25). However, for beam {Ff} = {Ffb} = {Qfb} and truss structures, the transfor- [K] = [Kbb] = [kbb] (see 17.6) mation matrix [T], displacement vector {v}, and force vectors {F} and {Ff} must be for these members. 19 20
5 For a truss member (also see Structure Stiffness Relations discussion following equations 1 and 17.21): The structure stiffness equations [T] = [Ta] can now be determined using the
{v} = {va} global coordinate member stiffness
{F} = {Fa} equations. Generation of the structure stiffness equations is {Ff} = {Ffa} T based on the three basic [K] [Kaa ] [T a ] [k aa ][T a ] relationships of structural analysis: ccsccs22 22 (1) equilibrium, (2) constitutive EA cs s cs s (17.29) relationships, and (3) compatibility. L ccsccs22 22 Specifically, the direct stiffness cs s cs s procedure involves expressing: c cos s sin 21 22
(1)Node point equilibrium of the element end forces meeting at the node with the externally applied nodal forces; (2)Substituting the global coor- dinate constitutive (matrix stiffness) equations for the forces in terms of the stiffness coefficients times the element end displacements and fixed- end force contributions; and (3)Compatibility of the element Figure 17.10 – Illustration of Direct end displacements with the Stiffness Analysis structure displacement degrees 23 24 of freedom {d}.
6 Equilibrium Equations Member Stiffness Relations (Constitutive Equations) FPFF(1) (2) X141 Since the end forces F m in (17.30) 2 i (1) (2) are unknown, the member PF1 41 F (17.30a) stiffness relations (m) (m) (1) (2) FKKKKKK1111213141516 FPFFY2 5 2 FKKKKKK2212223242526 2 FKKKKKK3313233343536 (1) (2) FKKKKKK PF2 5 F2 (17.30b) 4414243444546 FKKKKKK5 515253545556 FKKKKKK (1) (2) 6 616263646566 MPFFZ3 63 vF(m) (m) 2 1f1 vF (1) (2) 2f2 PF3 63 F (17.30c) vF3f3 (17.31) where superscripts (1), (2) designate vF4f4 vF5f5 element (member) 1, 2; respectively.25 26 vF6f6
are substituted into (17.30) to give Compatibility
(1) (1) (1) (1) (1) (1) PKvKvKv1 41 1 42 2 43 3 Imposing the compatibility (1) (1) (1) (1) (1) (1) (1) Kv44 4 Kv 455 Kv 46 6 Ff4 (continuity) conditions Kv(2) (2) Kv (2) (2) Kv (2) (2) 11 1 12 2 13 3 vvv0(1) (1) (1) (2) (2) (2) (2) (2) (2) (2) 123 (17.35) Kv14 4 Kv 155 Kv 16 6 Ff1 (1) (1) (1) vd;vd;vd46 1235 (1) (1) (1) (1) (1) (1) PKvKvKv2 51 1 52 2 53 3 (2) (2) (2) vd;vd;vd123 123 Kv(1) (1) Kv (1) (1) Kv (1) (1) F (1) (17.36) 54 455 5 56 6 f5 vvv0(2) (2) (2) (2) (2) (2) (2) (2) (2) 465 Kv21 1 Kv 22 2 Kv 23 3 (2) (2) (2) (2) (2) (2) (2) on the constitutive equation version Kv24 4 Kv 255 Kv 26 6 Ff2 of the equilibrium equations leads to (1) (1) (1) (1) (1) (1) PKvKvKv3 61 1 62 2 63 3 (1)(2) (1)(2) P(KK)d(KK)d112 (1) (1) (1) (1) (1) (1) (1) 44 11 45 12 Kv Kv Kv F (17.39a) 64 4 655 66 6 f6 (K(1) K (2) )d (F (1) F (2) ) (2) (2) (2) (2) (2) (2) 46 13 3 f4 f1 Kv31 1 Kv 32 2 Kv 33 3 (2) (2) (2) (2) (2) (2) (2) 27 28 Kv34 4 Kv 355 Kv 36 6 Ff3
7 (1) (2) (1) (2) FF(1) (2) P(KK)d(KK)d21254 21 55 22 f4 f1 (1) (2) (1) (2) (1) (2) (17.39b) (K K )d (F F ) {Pf } Ff5 F f2 56 23 3 f5 f2 FF(1) (2) (1) (2) (1) (2) f6 f3 P(KK)d(KK)d31264 31 65 32 (1) (2) (1) (2) (17.39c) (K66 K 33 )d3 (Ff6 F f3 ) PP1f1 Or collectively as {P} P2f2 P PP3f3 {P} [S]{d} {Pf } (17.41) [S]{d} ({P} {Pf }) {P} The structure stiffness coefficients
Sij are defined in the usual manner, KK(1)(2)(1)(2)(1)(2) KK KK 44 11 45 12 46 13 i.e., SP force at dof i due to ij i d1 [S]KKKKKK(1)(2)(1)(2)(1)(2) j 54 2155 22 56 23 a unit displacement at j with all KK(1)(2)(1)(2)(1)(2) KK KK 64 31 65 32 66 33 other displacements dk = 0 and k≠j.
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Assembly of [S] and {Pf} Using for frame members; and NNP Member Code Numbers Number of Node Points, i.e., number
The explicit details given for the direct of nodes used in the 2D structure discretization). The ID( , ) array stiffness procedure highlights how the identifies the nodal equation basic equations of structural analysis are numbers in sequence: utilized in matrix structural analysis. A ID(1,node) = X-axis nodal displ. number disadvantage of the procedure is that it ID(2,node) = Y-axis nodal displ. number is tedious and not amenable to ID(3,node) = Nodal rotation displ. number computer implementation. For example, consider the gable Computer implementation is based frame structure shown on the next on using a “destination array” slide. The destination array is also ID(NNDF,NNP) (NNDF Number of shown on this slide. 31 32 Nodal Degrees of Freedom, three
8 The 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 Gable Frame Structure construct the element location matrix array LM(NEDF, NEL) ID Node 123 (NEDF Number of Element 1 000 Degrees of Freedom, which is six 2 123 3 456 for the 2D frame elements), i.e., 4 789 5 0010 33 34
LM(i, iel) = ID(i, IEN(1,iel)); 1 = b node
LM(j, iel) = ID(i, IEN(2,iel)); 2 = e node LM: (1,m) (2,m) (3,m) (4,m) (5,m) (6,m) (m) i = 1, 2, 3 ; j = 4, 5, 6 LM(1,m) KKKKKK11 12 13 14 15 16 LM(2,m) KKKKKK iel = element number 21 22 23 24 25 26 LM(3,m) KKKKKK31 32 33 34 35 36 The IEN( , ) and LM( , ) arrays for the LM(4,m) KKKKKK 41 42 43 44 45 46 gable frame example are LM(5,m) KKKKKK51 52 53 54 55 56 LM(6,m) KKKKKK IEN LM 61 62 63 64 65 66 iel 12123456 112000123 (m) 223123456 LM(1,m) Ff1 334456789 LM(2,m) F 4540010789 f2 LM(3,m) Ff3 The LM( , ) array is used to assemble LM(4,m) Ff4 the global element stiffness matrix and LM(5,m) Ff5 fixed-end force vector as illustrated on LM(6,m) Ff6 the next two slides. 35 36
9 For the example gable frame: SUMMARY
(1) (2) Identify structure degrees of (1) (2) PFF freedom {d} SKK11 44 11 f1 f4 f1 (1) (2) (1) (2) For each element: SKK PFFf2 21 54 21 f5 f2 Evaluate [k], {Qf}, and [T] Calculate [K] = [T]T [k] [T] T {Ff} = [T] {Qf} (3) (4) (3) (4) PFFf9 Assemble [K] into [S] SKK9,9 66 66 f6 f6 Assemble {Ff} into {Pf} (4) PF (4) SK10,9 f10 f3 36 Form nodal load vector {P} SK (4) 9,10 63 Solve: [S] {d} = {P} – {Pf} SK (4) 10,10 33 For each element: Obtain {v} from {d} (compatibility) Calculate {u} = [T] {v}
{Q} = [k] {u} + {Qf}
Determine support reactions by 37 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
Truss Frame
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10 Temperature Restraint
Example Problems
Displacements of Statically Determinate Structures 41 42
Some Features of the Stiffness Equations the degrees of freedom associated An important characteristic of both with the elements connecting to element and global (structure) that degree of freedom. Thus, linear elastic stiffness equations is nonzero coefficients in a given row that they are symmetric. Prac- of a stiffness matrix consist only of tically, this means that only the the main diagonal and coefficients main diagonal terms and corresponding to degrees of coefficients to one side of the main freedom at that node and at other diagonal need to be generated and nodes on the on the elements stored in a computer program. meeting at the node for which the Also note that the stiffness stiffness equation is being formed. (equilibrium) equation for a given degree of freedom is influenced by 43 44
11 All other coefficients in the row are Clearly, it would be advantageous zero. When there are many in the solution phase of analysis to degrees of freedom in the cluster all nonzero coefficients complete structure, the stiffness close to the main diagonal. This matrix may contain relatively few isolation of the zero terms nonzero terms, in which case it is facilitates their removal in the characterized as sparse or weakly solution process. This can be populated. 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. This is most easily Indeterminacy achieved by minimizing the nodal Cognizance of static indeterminacy difference between connected is not necessary in the direct stiff- elements. All commercial ness approach. The displacement structural and finite element codes approach uses the analogous have built in schemes to minimize concept of kinematic the interaction of the stiffness indeterminacy, refers to the equations regardless of the user- number of displacement degrees specified input. of freedom that are required to define the displacement response of the structure to any load. Stated another way, kinematic indeterminacy equals the
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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
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