Module2 stiffness- rajesh sir

download Module2 stiffness- rajesh sir

of 91

Embed Size (px)

description

GCE Kannur

Transcript of Module2 stiffness- rajesh sir

  • 1. Dept. of CE, GCE Kannur Dr.RajeshKN 1 Structural Analysis - III Dr. Rajesh K. N. Asst. Professor in Civil Engineering Govt. College of Engineering, Kannur Stiffness Method

2. Dept. of CE, GCE Kannur Dr.RajeshKN 2 Development of stiffness matrices by physical approach stiffness matrices for truss,beam and frame elements displacement transformation matrix development of total stiffness matrix - analysis of simple structures plane truss beam and plane frame- nodal loads and element loads lack of fit and temperature effects. Stiffness method Module II 3. Dept. of CE, GCE Kannur Dr.RajeshKN 3 Displacement components are the primary unknowns Number of unknowns is equal to the kinematic indeterminacy Redundants are the joint displacements, which are automatically specified Choice of redundants is unique Conducive to computer programming FUNDAMENTALS OF STIFFNESS METHOD Introduction 4. Dept. of CE, GCE Kannur Dr.RajeshKN 4 Stiffness method (displacements of the joints are the primary unknowns): kinematic indeterminacy kinematic indeterminacy joints: a) where members meet, b) supports, c) free ends joints undergo translations or rotations in some cases joint displacements will be known, from the restraint conditions the unknown joint displacements are the kinematically indeterminate quantities o degree of kinematic indeterminacy: number of degrees of freedom 5. Dept. of CE, GCE Kannur Dr.RajeshKN 5 in a truss, the joint rotation is not regarded as a degree of freedom. joint rotations do not have any physical significance as they have no effects in the members of the truss in a frame, degrees of freedom due to axial deformations can be neglected 6. Dept. of CE, GCE Kannur Dr.RajeshKN 6 Example 1: Stiffness and flexibility coefficients of a beam Stiffness coefficients Unit displacement Unit action Forces due to unit dispts stiffness coefficients 11 21 12 22, , ,S S S S Dispts due to unit forces flexibility coefficients 11 21 12 22, , ,F F F F 7. Dept. of CE, GCE Kannur Dr.RajeshKN 7 Example 2: Action-displacement equations for a beam subjected to several loads 1 11 12 13A A A A= + + 1 11 1 12 2 13 3A S D S D S D= + + 2 21 1 22 2 23 3A S D S D S D= + + 3 31 1 32 2 33 3A S D S D S D= + + 8. Dept. of CE, GCE Kannur Dr.RajeshKN 8 1 11 1 12 2 13 3 1 2 21 1 22 2 23 3 2 1 1 2 2 3 3 ... ... ............................................................ ... n n n n n n n n nn n A S D S D S D S D A S D S D S D S D A S D S D S D S D = + + + + = + + + + = + + + + 1 11 12 1 1 2 21 22 2 2 1 2 11 ... ... ... ... ... ... ...... ... n n n n nn nn n n nn A S S S D A S S S D S S S DA = A = SD Action matrix, Stiffness matrix, Displacement matrix Stiffness coefficient ijS { } [ ] { } [ ] [ ] 1 1 A F D S F = = oIn matrix form, { } [ ]{ }A S D= [ ] [ ] 1 F S = Stiffness matrix 9. Dept. of CE, GCE Kannur Dr.RajeshKN 9 Example: Propped cantilever (Kinematically indeterminate to first degree) Stiffness method (Direct approach: Explanation using principle of superposition) degrees of freedom: one Kinematically determinate structure is obtained by restraining all displacements (all displacement components made zero - restrained structure) Required to get B 10. Dept. of CE, GCE Kannur Dr.RajeshKN 10 Restraint at B causes a reaction of MB as shown. Hence it is required to induce a rotation of B The actual rotation at B is B 2 12 B wL M = (Note the sign convention: anticlockwise positive) Apply unit rotation corresponding to B Let the moment required for this unit rotation be Bm 4 B EI m L = anticlockwise 11. Dept. of CE, GCE Kannur Dr.RajeshKN Moment required to induce a rotation of B B Bm is 2 4 0 12 B wL EI L + = 3 48 B B B wL EI M m = = Bm (Moment required for unit rotation) is the stiffness coefficient here. 0B B BM m + = (Joint equilibrium equation) 12. Dept. of CE, GCE Kannur Dr.RajeshKN 12 Stiffnesses of prismatic members Stiffness coefficients of a structure are calculated from the contributions of individual members Hence it is worthwhile to construct member stiffness matrices [ ] [ ] 1 Mi MiS F = 13. Dept. of CE, GCE Kannur Dr.RajeshKN 13 Member stiffness matrix for prismatic beam member with rotations at the ends as degrees of freedom [ ] 2 12 1 2 Mi EI S L = [ ] [ ] 1 1 1 2 13 6 1 26 6 3 Mi Mi L L LEI EI S F L L EI EI EI = = = 2 1 2 16 2 1 2 1 2(3) EI EI L L = = Verification: B A 1 14. Dept. of CE, GCE Kannur Dr.RajeshKN 14 [ ] 11 12 21 22 3 2 2 12 6 6 4 M M Mi M M S S S S S EI EI L L EI EI L L = = Member stiffness matrix for prismatic beam member with deflection and rotation at one end as degrees of freedom 15. Dept. of CE, GCE Kannur Dr.RajeshKN [ ] [ ] 13 2 1 2 1 2 2 33 2 6 3 6 2 Mi Mi L L L LLEI EI S F EI LL L EI EI = = = ( ) 2 2 3 2 2 6 36 3 12 6 6 423 EI EI L L EI EI L LEI L LL L L = = Verification: [ ]Mi EA S L = Truss member 16. Dept. of CE, GCE Kannur Dr.RajeshKN 16 Plane frame member [ ] 11 12 13 21 22 23 31 32 33 3 2 2 0 0 12 6 0 6 4 0 M M M Mi M M M M M M S S S S S S S S S S EA L EI EI L L EI EI L L = = 17. Dept. of CE, GCE Kannur Dr.RajeshKN 17 Grid member [ ] 3 2 2 12 6 0 0 0 6 4 0 Mi EI EI L L GJ S L EI EI L L = 18. Dept. of CE, GCE Kannur Dr.RajeshKN 18 Space frame member [ ] 3 2 3 2 2 2 0 0 0 0 0 12 6 0 0 0 0 12 6 0 0 0 0 0 0 0 0 0 6 4 0 0 0 0 6 4 0 0 0 0 Z Z Y Y Mi Y Y Z Z EA L EI EI L L EI EI L LS GJ L EI EI L L EI EI L L = 19. Dept. of CE, GCE Kannur Dr.RajeshKN 19 (Explanation using principle of complimentary virtual work) Formalization of the Stiffness method { } [ ]{ }Mi Mi MiA S D= Here{ }MiD contains relative displacements of the k end with respect to j end of the i-th member If there are m members in the structure, { } { } { } { } { } [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] { } { } { } { } { } 11 1 22 2 33 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 MM M MM M MM M MiMi Mi MmMm Mm SA D SA D SA D SA D SA D = M M MA = S D 20. Dept. of CE, GCE Kannur Dr.RajeshKN 20 { } [ ]{ }M M MA S D= [ ]MS is the unassembled stiffness matrix of the entire structure 21. Dept. of CE, GCE Kannur Dr.RajeshKN 21 Relative end-displacements in will be related to a vector of joint displacements for the whole structure, { }MD { }JD If there are no support displacements specified, { }RD will be a null matrix Hence, { } [ ]{ } [ ] [ ] { } { } F M MJ J MF MR R D D C D C C D = = { }JD { }FDfree (unknown) joint displacements { }RDand restraint displacements consists of: { } [ ]{ }M MJ JD C D= displacement transformation matrix (compatibility matrix)[ ]MJC 22. Dept. of CE, GCE Kannur Dr.RajeshKN 22 Elements in displacement transformation matrix (compatibility matrix) [ ]MJC are found from compatibility conditions. Each column in the submatrix consists of member displacements caused by a unit value of a support displacement applied to the restrained structure. [ ]MRC [ ]MFC Each column in the submatrix consists of member displacements caused by a unit value of an unknown displacement applied to the restrained structure. { }MDrelate to respectively [ ]MFC [ ]MRC and { }FD { }RD and 23. Dept. of CE, GCE Kannur Dr.RajeshKN 23 { }MD { } [ ]{ } [ ] [ ] { } { } F M MJ J MF MR R D D C D C C D = = Suppose an arbitrary set of virtual displacements is applied on the structure. { } { } { } { } T T T F J J F R R D W A D A A D = = { }JD External virtual work produced by the virtual displacements { }JAand real loads is 24. Dept. of CE, GCE Kannur Dr.RajeshKN 24 { } { } T M MU A D = Internal virtual work produced by the virtual (relative) end displacements { }MD { }MAand actual member end actions is { } { } { } { } T T J J M MA D A D = Equating the above two (principle of virtual work), { } [ ]{ }M MJ JD C D= { } [ ]{ }M M MA S D=But and { } [ ]{ }M MJ JD C D =Also, { } { } { } [ ] [ ] [ ]{ } TT T T J J J MJ M MJ JA D D C S C D =Hence, { } [ ]{ }J J JA S D= 25. Dept. of CE, GCE Kannur Dr.RajeshKN 25 [ ] [ ] [ ][ ]T J MJ M MJS C S C=Where, , the assembled stiffness matrix for the entire structure. It is useful to partition into submatrices pertaining to free (unknown) joint displacements [ ]JS { }FD { }RDand restraint displacements { } [ ]{ } { } { } [ ] [ ] [ ] [ ] { } { } FF FRF F J J J RF RRR R S SA D A S D S SA D = = [ ] [ ] [ ][ ]T FF MF M MFS C S C= [ ] [ ] [ ][ ]T FR MF M MRS C S C= [ ] [ ] [ ][ ]T RF MR M MFS C S C= [ ] [ ] [ ][ ]T RR MR M MRS C S C= Where, 26. Dept. of CE, GCE Kannur Dr.RajeshKN 26 { } [ ]{ } [ ]{ }F FF F FR RA S D S D= + { } [ ]{ } [ ]{ }R RF F RR RA S D S D= + { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D = { } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= + + Support reactions { }RCA represents combined joint loads (actual and equivalent) applied directly to the supports. If actual or equivalent joint loads are applied directly to the supports, Joint displacements 27. Dept. of CE, GCE Kannur Dr.RajeshKN 27 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + Member end actions are obtained adding member end actions calculated as above and initial fixed-end actions { } { } [ ][ ]{ }M ML M MJ JA A S C D= +i.e., { }MLAwhere represents fixed end actions 28. Dept. of CE, GCE Kannur Dr.RajeshKN 28 Important formulae: Joint displacements: Member end actions: Support reactions: { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D = { } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= + + { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + 29. Dept. of CE, GCE Kannur Dr.RajeshKN 29 Problem 1 [ ] 4 2 0 0 2 4 0 02 0 0 2 1 0 0 1 2 M EI S L = Unassembled stiffness matrix [ ] 2 12 1 2 Mi EI S L = Member stiffness matrix of beam member Kinematic in