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COORDINATE TRANSFORMATIONS Members of a Structural System Are Typically Oriented in Differing Directions, E.G., Fig

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COORDINATE TRANSFORMATIONS Members of a Structural System Are Typically Oriented in Differing Directions, E.G., Fig 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 3 4 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} = <<Q>b <Q>e> = beginning node local coordinate element local coordinate force T T force vector; {Q}e = <Q4 Q5 Q6> = vector; {F} = <<F>b <F>e> = end node local coordinate force element global coordinate force T vector; {F}b = <F1 F2 F3> = [t] [0] beginning node global coordinate vector; and [T] = = T [0] [t] force vector; {F}e = <F4 F5 F6> = end node global coordinate force element local to global coordinate cos sin 0 transformation matrix. vector; [t] = sin cos 0 = Utilizing (17.9) for all six member 001 force components and expressing local to global coordinate transfor- the resulting transformations in mation matrix which is the same at matrix form gives each end node for a straight 7 8 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 9 10 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] 13 14 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) 15 14 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 } 15 16 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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