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Module2 stiffness- rajesh sir

This document discusses the stiffness method for structural analysis. It begins by introducing the stiffness method and its key concepts, such as using joint displacements as the primary unknowns and establishing a relationship between member forces and displacements through stiffness matrices. It then provides examples of calculating stiffness matrices for different structural elements, including beams, trusses, frames, grids, and space frames. The document also explains how to develop the total stiffness matrix of a structure by assembling the member stiffness matrices using a displacement transformation matrix. Overall, the document provides an overview of the fundamental concepts and procedures for applying the stiffness method to analyze structures.

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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
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
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
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
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
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
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= + +
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
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 Bθ
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 Bθ The actual rotation at B is Bθ 2 12 B wL M = − (Note the sign convention: anticlockwise positive) •Apply unit rotation corresponding to Bθ Let the moment required for this unit rotation be Bm 4 B EI m L = anticlockwise
Dept. of CE, GCE Kannur Dr.RajeshKN • Moment required to induce a rotation of Bθ B Bm θis 2 4 0 12 B wL EI L θ− + = 3 48 B B B wL EI M m θ∴ = − = Bm (Moment required for unit rotation) is the stiffness coefficient here. 0B B BM m θ+ = (Joint equilibrium equation)
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 − =
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
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
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
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 ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎣ ⎦
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 ⎡ ⎤ −⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎣ ⎦
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 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎣ ⎦
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
Dept. of CE, GCE Kannur Dr.RajeshKN 20 { } [ ]{ }M M MA S D= [ ]MS is the unassembled stiffness matrix of the entire structure
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
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
Dept. of CE, GCE Kannur Dr.RajeshKN 23 { }MDδ { } [ ]{ } [ ] [ ] { } { } F M MJ J MF MR R D D C D C C D δ δ δ δ ⎧ ⎫ = = ⎡ ⎤ ⎨ ⎬⎣ ⎦ ⎩ ⎭ • 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 δ δ δ δ δ ⎧ ⎫⎡ ⎤= = ⎨ ⎬⎣ ⎦ ⎩ ⎭ { }JDδ• External virtual work produced by the virtual displacements { }JAand real loads is
Dept. of CE, GCE Kannur Dr.RajeshKN 24 { } { } T M MU A Dδ δ= • Internal virtual work produced by the virtual (relative) end displacements { }MDδ { }MAand actual member end actions is { } { } { } { } T T J J M MA D A Dδ δ= • Equating the above two (principle of virtual work), { } [ ]{ }M MJ JD C D= { } [ ]{ }M M MA S D=But and { } [ ]{ }M MJ JD C Dδ δ=Also, { } { } { } [ ] [ ] [ ]{ } TT T T J J J MJ M MJ JA D D C S C Dδ δ=Hence, { } [ ]{ }J J JA S D=
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, 
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
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
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= + +
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 indeterminacy = 2
Dept. of CE, GCE Kannur Dr.RajeshKN 30 Fixed end actions Equivalent joint loads
Dept. of CE, GCE Kannur Dr.RajeshKN Joint displacements Free (unknown) joint displacements { }FD { }RDRestraint displacements
Dept. of CE, GCE Kannur Dr.RajeshKN Joint displacements Free (unknown) joint displacements { }FD { }RDRestraint displacements •Each column in the submatrix consists of member displacements caused by a unit value of an unknown displacement applied to the restrained structure. [ ]MFC •Each column in the submatrix consists of member displacements caused by a unit value of a support displacement applied to the restrained structure. [ ]MRC
Dept. of CE, GCE Kannur Dr.RajeshKN 0 1 1 0 0 0 0 1 [ ] 0 0 1 0 1 0 0 1 M FC ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ DF1 DF2 =1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN 1 L 1 L 0 0 1 0 0 0 1 L− 1 L− 1 L 1 L 0 0 1 L− 1 L−
Dept. of CE, GCE Kannur Dr.RajeshKN 35 [ ] [ ] [ ]MJ MF MRC C C= ⎡ ⎤⎣ ⎦ 0 0 1 1 0 0 1 0 1 01 0 0 0 1 1 0 0 0 1 1 L L LL L −⎡ ⎤ ⎢ ⎥ −⎢ ⎥= ⎢ ⎥− ⎢ ⎥ −⎣ ⎦ [ ] 1 1 1 0 1 0 1 0 0 0 1 1 0 0 1 1 MR L L L L C L L L L −⎡ ⎤ ⎢ ⎥− ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ −⎣ ⎦ DR1 DR2 DR3 DR4 =1 =1 =1 =1 DR1 DR2 DR3 DR4 =1 =1 =1 =1 DF1 DF2 =1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN 36 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ { } [ ] { } 1 F FF FD S A − ∴ = Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 4 2 0 0 0 0 0 1 1 0 2 4 0 0 1 02 0 0 0 1 0 0 2 1 1 0 0 0 1 2 0 1 EI L ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 6 12 1 2 EI L ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ is a null matrix, since there are no support displacements { }RD
Dept. of CE, GCE Kannur Dr.RajeshKN { } 1 6 1 12 1 2 29 F EI PL D L − ⎛ ⎞⎡ ⎤ ⎧ ⎫ = ⎨ ⎬⎜ ⎟⎢ ⎥ ⎣ ⎦ ⎩ ⎭⎝ ⎠ Free (unknown) joint displacements 2 2 01 . 11 818 11 0 11 PL E EII PL⎧ ⎫ ⎧ = = ⎫ ⎨⎨ ⎬ ⎭ ⎩⎩ ⎬ ⎭ 2 3 6 2 3 32 0 0 3 3 L L L LEI L L L ⎡ ⎤− − = ⎢ ⎥ −⎣ ⎦ [ ] [ ] [ ][ ]T FR MF M MRS C S C= 4 2 0 0 1 1 0 0 1 1 0 2 4 0 0 1 0 1 02 1 0 0 0 1 0 0 2 1 0 0 1 1 0 0 1 2 0 0 1 1 L EI L L −⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥−⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥ −⎣ ⎦ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN { } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + + is a null matrix.{ }RD Support reactions [ ] [ ] [ ][ ]T RF MR M MFS C S C= 2 6 0 2 02 3 3 3 3 LEI L ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ − −⎣ ⎦ 1 1 0 4 2 0 0 0 0 1 0 1 0 2 4 0 0 1 01 2 0 0 1 1 0 0 2 1 1 0 0 0 1 1 0 0 1 2 0 1 T L EI L L −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ { } { } [ ]{ }R RC RF FA A S D∴ = − +
Dept. of CE, GCE Kannur Dr.RajeshKN [ ] [ ] [ ][ ]T RR MR M MRS C S C= 1 1 0 4 2 0 0 1 1 0 1 0 1 0 2 4 0 0 1 0 1 01 2 1 0 0 1 1 0 0 2 1 0 0 1 1 0 0 1 1 0 0 1 2 0 0 1 1 T L L EI L L L − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 3 12 6 12 0 6 4 6 02 12 6 18 6 0 0 6 6 L L L LEI LL −⎡ ⎤ ⎢ ⎥− ⎢ ⎥= − − −⎢ ⎥ ⎢ ⎥ −⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN 2 2 2 6 0 2 0 02 3 3 3 118 3 3 3 P PL LEI PL L EI P P −⎧ ⎫ ⎡ ⎤⎪ ⎪− ⎢ ⎥⎪ ⎪ ⎧ ⎫⎢ ⎥= − +⎨ ⎬ ⎨ ⎬ −⎢ ⎥ ⎩ ⎭⎪ ⎪− ⎢ ⎥⎪ ⎪ − −⎣ ⎦−⎩ ⎭ 2 2 0 2 2 0 0 02 3 3 318 33 3 3 3 P P PL PL PL EI P EI L P P P P P ⎧ ⎫ −⎧ ⎫ ⎧ ⎫ ⎪ ⎪⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ = − + = +⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎩ ⎭− −⎩ ⎭ ⎩ ⎭ ⎪ ⎪ ⎩ ⎭ { } { } [ ]{ }R RC RF FA A S D∴ = − + 2 3 10 3 2 3 P PL P P ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ =
Dept. of CE, GCE Kannur Dr.RajeshKN 41 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } 2 3 4 2 0 0 0 0 3 2 4 0 0 1 0 02 2 0 0 2 1 1 0 19 18 2 0 0 1 2 0 1 M PL EI PL A L EI ⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥− ⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥= +⎨ ⎬ ⎨ ⎬ ⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪−⎩ ⎭ ⎣ ⎦ ⎣ ⎦ 2 3 4 2 0 0 0 3 2 4 0 0 02 2 0 0 2 1 09 18 2 0 0 1 2 1 PL EI PL L EI ⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥= +⎨ ⎬ ⎨ ⎬ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭⎣ ⎦ 3 0 3 0 2 19 9 2 2 PL PL ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪ = +⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭ 1 1 13 0 PL ⎧ ⎫ ⎪ ⎪−⎪ ⎪ ⎨ ⎬ ⎪ ⎪ = ⎪ ⎪⎩ ⎭ is a null matrix{ }RD Member end actions { } { } [ ][ ]{ }M ML M MF FA A S C D∴ = +
Dept. of CE, GCE Kannur Dr.RajeshKN 42 3 0 0 0 0 0 2 0 6 4 6 0 1 1 0 0 4 0 6 2 6 02 0 0 0 2 0 0 3 3 1 1 1 1 2 0 0 3 3 0 0 1 1 L L L L L L LEI L L LL L L ⎡ ⎤ ⎢ ⎥ −⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥= ⎢ ⎥ −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− − −⎣ ⎦ ⎢ ⎥ − −⎣ ⎦ [ ] [ ] [ ] [ ] 2 2 2 2 2 3 2 6 6 2 3 3 2 0 0 3 3 6 0 12 6 12 02 2 0 6 4 6 0 3 3 12 6 18 6 3 3 0 0 6 6 FF FR RF RR L L L L L L L L L L S SL LEI S SL L L L L L L L L L ⎡ ⎤− − ⎢ ⎥ −⎢ ⎥ ⎢ ⎥ ⎡ ⎤− = =⎢ ⎥ ⎢ ⎥ − ⎣ ⎦⎢ ⎥ ⎢ ⎥− − − − ⎢ ⎥ − − −⎣ ⎦ [ ] [ ] [ ][ ]T J MJ M MJS C S C= 0 0 0 0 0 4 2 0 0 0 0 1 1 0 1 1 0 0 2 4 0 0 0 1 0 1 01 2 1 0 0 0 0 0 2 1 0 0 0 1 1 1 1 1 1 0 0 1 2 0 0 0 1 1 0 0 1 1 L L L L LEI L LL L L L ⎡ ⎤ ⎢ ⎥ −⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎢ ⎥ − −⎣ ⎦ Alternatively, if the entire [SJ] matrix is assembled at a time,
Dept. of CE, GCE Kannur Dr.RajeshKN 43 •Problem 2: A B C D 40 kN10 kN/m 2 m 4 m 4 m 2 m 100 kN Kinematic indeterminacy = 3 (Not considering joint D in the overhanging portion) Degrees of freedom A B C D DF2DF1 DF3
Dept. of CE, GCE Kannur Dr.RajeshKN 44 [ ] 2 1 0 0 1 2 0 02 0 0 2 14 0 0 1 2 M EI S ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ Unassembled stiffness matrix [ ] 2 12 1 24 Mi EI S ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ Member stiffness matrix of beam member
Dept. of CE, GCE Kannur Dr.RajeshKN 45 Fixed end actions Equivalent joint loads + actual joint loads A B 13.33 kNm 13.33 20 kN 20 kN B C 20 20 20 kN 20 kN A B 13.33 kNm 6.67 20 kN 20 +20 = 40 kN 20 +100 = 120 kN 2x100 – 20 = 180 kNm
Dept. of CE, GCE Kannur Dr.RajeshKN 1 0 0 0 0 1 1 0 DF1 =1 DF2 =1 L
Dept. of CE, GCE Kannur Dr.RajeshKN 47 [ ] 1 0 0 0 1 0 0 1 0 0 0 1 M FC ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ DF1 DF2 DF3 =1 =1 =1 0 0 0 1 DF3 =1
Dept. of CE, GCE Kannur Dr.RajeshKN 48 { } [ ] { } 1 F FF FD S A − ∴ = Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 1 0 0 2 1 0 0 1 0 0 0 1 0 1 2 0 0 0 1 02 0 1 0 0 0 2 1 0 1 04 0 0 1 0 0 1 2 0 0 1 T EI ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 1 0.5 0 0.5 2 0.5 0 0.5 1 EI ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ is a null matrix.{ }RD∴
Dept. of CE, GCE Kannur Dr.RajeshKN { } 1 1 0.5 0 13.33 1 0.5 2 0.5 6.67 0 0.5 1 180 FD EI − −⎛ ⎞⎡ ⎤ ⎧ ⎫ ⎪ ⎪⎜ ⎟⎢ ⎥= −⎨ ⎬⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟ −⎢ ⎥⎣ ⎦ ⎩ ⎭⎝ ⎠ Joint displacements 43.435 60.33 210.6 1 EI −⎧ ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ ⎪−⎩ ⎭ =
Dept. of CE, GCE Kannur Dr.RajeshKN 50 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } 43.435 60.33 210.6 13.33 2 1 0 0 1 0 0 13.33 1 2 0 0 0 1 02 1 20 0 0 2 1 0 1 04 20 0 0 1 2 0 0 1 M EI A EI ⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥−⎪ ⎪ ⎢ ⎥ ⎢ ⎥= +⎨ ⎬ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪− −⎧ ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ ⎪− ⎣ ⎩ ⎭ ⎦ ⎣ ⎦ ⎭ ⎩ 0 25 25 200 kNm ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ −⎪ ⎪ ⎪ ⎪−⎩ ⎭ = is a null matrix{ }RD Member end actions { } { } [ ][ ]{ }M ML M MF FA A S C D∴ = +
Dept. of CE, GCE Kannur Dr.RajeshKN •Problem 3 Analyse the beam. Support B has a downward settlement of 30mm. EI=5.6×103 kNm2 [ ] 4 2 0 0 0 0 2 4 0 0 0 0 0 0 2 1 0 02 0 0 1 2 0 06 0 0 0 0 4 2 0 0 0 0 2 4 M EI S ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ Unassembled stiffness matrix [ ] 2 12 1 2Mi EI S L ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ Member stiffness matrix of beam member
Dept. of CE, GCE Kannur Dr.RajeshKN Fixed end moments 2525 Equivalent joint loads 2525 Support settlements { }RD(Restraint displacements) A B C D30mm 1RD Free (unknown) joint displacements { }FD 1FD 2FD 3FD
Dept. of CE, GCE Kannur Dr.RajeshKN 53 BA C D 1 1FD = 0 1 1 0 0 0 and consist of member displacements due to unit displacements on the restrained structure. [ ]MFC [ ]MRC
Dept. of CE, GCE Kannur Dr.RajeshKN 54 [ ] [ ] [ ]MJ MF MRC C C= ⎡ ⎤⎣ ⎦ 0 0 0 1 3 1 0 0 1 3 1 0 0 1 6 0 1 0 1 6 0 1 0 0 0 0 1 0 −⎡ ⎤ ⎢ ⎥ −⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ 31 2 1 1 1 11 FF F RDD D D = = == A B C D1 1RD = 1 3− 1 3− 1 6 1 6 0 0
Dept. of CE, GCE Kannur Dr.RajeshKN 55 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 0 0 0 4 2 0 0 0 0 0 0 0 1 0 0 2 4 0 0 0 0 1 0 0 1 0 0 0 0 2 1 0 0 1 0 0 0 1 0 0 0 1 2 0 0 0 1 03 0 1 0 0 0 0 0 4 2 0 1 0 0 0 1 0 0 0 0 2 4 0 0 1 T EI ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 6 1 0 1 6 2 3 0 2 4 EI ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN [ ] [ ] [ ][ ]T FR MF M MRS C S C= 0 0 0 4 2 0 0 0 0 1 3 1 0 0 2 4 0 0 0 0 1 3 1 0 0 0 0 2 1 0 0 1 6 0 1 0 0 0 1 2 0 0 1 63 0 1 0 0 0 0 0 4 2 0 0 0 1 0 0 0 0 2 4 0 T EI −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 3 2 1 2 3 0 EI −⎧ ⎫ ⎪ ⎪ = ⎨ ⎬ ⎪ ⎪ ⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN { } 1 6 1 0 0 3 2 1 6 2 25 1 2 0.03 3 3 0 2 4 25 0 EI EI − −⎛ ⎞ ⎡ ⎤⎧ ⎫⎡ ⎤ ⎡ ⎤ ⎪ ⎪⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎢ ⎥= − − −⎨ ⎬⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎪ ⎪⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎩ ⎭⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎣ ⎦ 20 4 2 0 0.045 3 5600 4 24 12 25 0.015 116 3 2 12 35 25 0 EI − ⎡ ⎤⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎢ ⎥= − − − − −⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥−⎢ ⎥ ⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦ 3 1642 3 108 116 5.6 10 6 0.007583 0.0005 0.007 311 −⎧ ⎫ ⎪ ⎪ = =⎨ ⎬ × × ⎪ ⎪ −⎧ ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎩ ⎭⎭ 20 4 2 0 84 3 4 24 12 25 28 116 2 12 35 25 0 EI − ⎡ ⎤⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎢ ⎥= − − − − −⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥−⎢ ⎥ ⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦ { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN 58 { } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + + Support reactions [ ] 4 2 0 0 0 0 0 0 0 2 4 0 0 0 0 1 0 0 0 0 2 1 0 0 1 0 0 1 3 1 3 1 6 1 6 0 0 0 0 1 2 0 0 0 1 03 0 0 0 0 4 2 0 1 0 0 0 0 0 2 4 0 0 1 EI ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = − − ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ [ ] [ ] [ ][ ]T RF MR M MFS C S C= [ ]3 2 1 2 0 3 EI = −
Dept. of CE, GCE Kannur Dr.RajeshKN { } { } [ ] { } 0.007583 0 3 2 1 2 0 0.0005 0.03 3 2 0.0031 R EI EI A −⎧ ⎫ ⎪ ⎪ ⎡ ⎤ = − + − + −⎨ ⎬ ⎢ ⎥⎣ ⎦⎪ ⎪ ⎩ ⎭ (Support reaction corresponding to DR . i.e., reaction at B) ( )0.0116 0.045 0.0334 62 5 3 3 .3 EI EI = − = − −= [ ] [ ] [ ][ ]T RR MR M MRS C S C= [ ] 4 2 0 0 0 0 1 3 2 4 0 0 0 0 1 3 0 0 2 1 0 0 1 6 1 3 1 3 1 6 1 6 0 0 0 0 1 2 0 0 1 63 0 0 0 0 4 2 0 0 0 0 0 2 4 0 EI −⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = − − ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 2 EI⎡ ⎤ = ⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN 60 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } { } 0 4 2 0 0 0 0 0 0 0 1 3 0 2 4 0 0 0 0 1 0 0 1 3 0.007583 0 0 0 2 1 0 0 1 0 0 1 62 0.0005 0.03 0 0 0 1 2 0 0 0 1 0 1 66 0.0031 25 0 0 0 0 4 2 0 1 0 0 25 0 0 0 0 2 4 0 0 1 0 M EI A −⎛⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎧ ⎫ ⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ = + + −⎜⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎜⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ −⎩ ⎭ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎜ ⎟⎜ ⎟ ⎠ 83.7 55.4 55.4 40.3 40.3 0 ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ −⎪ ⎪ ⎨ ⎬ −⎪ ⎪ ⎪ ⎩ = ⎪ ⎪ ⎪ ⎭ Member end actions
Dept. of CE, GCE Kannur Dr.RajeshKN 61 [ ] [ ] [ ] [ ] FF FR RF RR S S S S ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ 6 1 0 3 2 1 6 2 1 2 0 2 4 03 3 2 1 2 0 3 2 E I −⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ −⎣ ⎦ 2 0 0 2 0 1 1 0 0 0 4 0 0 2 0 0 0 1 1 0 2 1 0 1 2 0 0 0 0 0 1 1 2 0 1 23 1 3 1 3 1 6 1 6 0 0 0 4 2 0 0 2 4 0 EI −⎡ ⎤ ⎢ ⎥−⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎢ ⎥ ⎣ ⎦ [ ] [ ] [ ][ ]T J MJ M MJS C S C= 4 2 0 0 0 0 0 0 0 1 3 0 1 1 0 0 0 2 4 0 0 0 0 1 0 0 1 3 0 0 0 1 1 0 0 0 2 1 0 0 1 0 0 1 6 0 0 0 0 0 1 0 0 1 2 0 0 0 1 0 1 63 1 3 1 3 1 6 1 6 0 0 0 0 0 0 4 2 0 1 0 0 0 0 0 0 2 4 0 0 1 0 EI −⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥ −⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ Alternatively, if the entire [SJ] matrix is assembled at a time,
Dept. of CE, GCE Kannur Dr.RajeshKN 62 [ ] [ ] [ ]MJ MF MRC C C= ⎡ ⎤⎣ ⎦ 0 0 0 1 3 1 1 3 0 0 1 0 0 1 3 0 1 3 0 0 1 0 0 0 0 1 6 1 6 0 0 1 0 0 0 1 6 1 6 0 0 1 0 0 0 0 1 3 1 3 0 0 1 0 0 0 1 3 1 3 −⎡ ⎤ ⎢ ⎥ −⎢ ⎥ ⎢ ⎥− = ⎢ ⎥ −⎢ ⎥ ⎢ ⎥− ⎢ ⎥ −⎢ ⎥⎣ ⎦ [ ] [ ] [ ][ ]T J MJ M MJS C S C= 0 0 01 3 1 1 3 0 0 4 2 0 0 0 0 0 0 01 3 1 1 3 0 0 1 0 01 3 0 1 3 0 0 2 4 0 0 0 0 1 0 01 3 0 1 3 0 0 1 0 0 0 0 1 6 1 6 0 0 0 2 1 0 0 1 0 0 0 0 1 6 1 6 0 0 1 0 0 0 1 6 1 6 0 0 0 1 2 0 0 0 1 0 0 0 1 6 1 6 03 0 1 0 0 0 0 1 3 1 3 0 0 0 0 4 2 0 1 0 0 0 0 1 3 0 0 1 0 0 0 1 3 1 3 0 0 0 0 2 4 0 0 1 T EI − −⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ ⎢ ⎥ −⎢ ⎥ ⎣ ⎦⎣ ⎦ 1 3 0 0 0 1 3 1 3 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ −⎢ ⎥⎣ ⎦ Alternatively, if ALL possible support settlements are accounted for,
Dept. of CE, GCE Kannur Dr.RajeshKN 63 [ ] [ ] [ ] [ ] FF FR RF RR S S S S ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ 0 1 1 0 0 0 0 0 0 1 1 0 2 0 0 2 4 2 0 0 0 0 0 0 0 1 4 0 0 2 2 2 0 0 1 3 1 3 0 0 0 0 2 1 0 0 0 1 2 1 2 0 1 0 0 0 0 0 1 2 0 0 0 1 2 1 2 03 1 3 1 3 1 6 1 6 0 0 0 4 2 0 0 0 2 2 0 0 1 6 1 6 1 3 1 3 0 2 4 0 0 0 2 2 0 0 0 0 1 3 1 3 EI ⎡ ⎤ ⎢ ⎥ −⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥⎢ ⎥− − −⎣ ⎦ ⎢ ⎥ − −⎣ ⎦ 6 1 0 2 2 3 2 1 2 0 1 6 2 0 0 1 2 3 2 2 0 2 4 0 0 0 2 2 2 0 0 4 3 2 4 3 0 0 2 0 0 2 4 2 0 03 3 2 1 2 0 4 3 2 3 2 1 6 0 1 2 3 2 2 0 0 1 6 3 2 4 3 0 2 2 0 0 0 4 3 4 3 EI − −⎡ ⎤ ⎢ ⎥− ⎢ ⎥ −⎢ ⎥ ⎢ ⎥ −⎢ ⎥= ⎢ ⎥− ⎢ ⎥ − − − −⎢ ⎥ ⎢ ⎥− − − ⎢ ⎥ − − −⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN 64 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ 1 0 6 1 0 0 2 2 3 2 1 2 0 0 1 6 2 25 0 0 1 2 3 2 2 0.03 3 3 0 2 4 25 0 0 0 2 2 0 0 EI EI − ⎡ ⎤⎧ ⎫ ⎢ ⎥⎪ ⎪− −⎛ ⎞ ⎧ ⎫⎡ ⎤ ⎡ ⎤⎢ ⎥⎪ ⎪⎪ ⎪ ⎪ ⎪⎜ ⎟⎢ ⎥ ⎢ ⎥⎢ ⎥= − − − −⎨ ⎬ ⎨ ⎬⎜ ⎟⎢ ⎥ ⎢ ⎥⎢ ⎥⎪ ⎪ ⎪ ⎪⎜ ⎟ −⎢ ⎥ ⎢ ⎥⎩ ⎭⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎢ ⎥⎪ ⎪ ⎢ ⎥⎪ ⎪⎩ ⎭⎣ ⎦ 20 4 2 0 0.045 3 5600 4 24 12 25 0.015 116 3 2 12 35 25 0 20 4 2 84 3 4 24 12 3 116 2 12 35 25 EI EI − ⎡ ⎤⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎢ ⎥= − − − − −⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥−⎢ ⎥ ⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦ − −⎧ ⎫⎡ ⎤ ⎪ ⎪⎢ ⎥= − − ⎨ ⎬⎢ ⎥ ⎪ ⎪−⎢ ⎥ ⎩ ⎭⎣ ⎦ 1642 3 108 116 671 EI −⎧ ⎫ ⎪ ⎪ = ⎨ ⎬ ⎪ ⎪ ⎩ ⎭ 0.007583 0.0005 0.0031 −⎧ ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ ⎩ = ⎪ ⎭ Joint displacements
Dept. of CE, GCE Kannur Dr.RajeshKN 65 { } 0 2 0 0 4 3 2 4 3 0 0 0 0.0075830 2 0 0 2 4 2 0 0 0 0.00050 3 2 1 2 0 4 3 2 3 2 1 6 0 0.03 3 3 0.003150 1 2 3 2 2 0 0 1 6 3 2 4 3 0 50 0 2 2 0 0 0 4 3 4 3 0 R EI EI A −⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎡ ⎤ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− −⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = − + +− − − − −⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪− − − −⎩ ⎭⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥− − − −⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦ { } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + + { } 0 28.373 74.667 0 28.373 112 0 21.653 84 50 19.973 9.333 46.294 83.627 62.347 79.31 36.5650 13.44 0 RA −⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = − + + =−⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪⎪ ⎪ −⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭⎭ ⎪ Support reactions
Dept. of CE, GCE Kannur Dr.RajeshKN 66 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } 0 4 2 0 0 0 0 0 0 0 1 3 1 1 3 0 0 0 2 4 0 0 0 0 1 0 0 1 3 0 1 3 0 0 0.007583 0 0 0 2 1 0 0 1 0 0 0 0 1 6 1 6 02 0.0005 0 0 0 1 2 0 0 0 1 0 0 0 1 6 1 6 06 0.0031 25 0 0 0 0 4 2 0 1 0 0 0 0 1 3 1 3 25 0 0 0 0 2 4 0 0 1 0 0 0 M EI A −⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ − ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎧ ⎫ −⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ = + +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ −⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎪ ⎪ ⎢ ⎥ ⎢ ⎥ − ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎩ ⎭ ⎣ ⎦ ⎣ ⎦ 0 0 0.03 0 0 1 3 1 3 ⎛ ⎞⎡ ⎤ ⎧ ⎫⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟⎢ ⎥ ⎪ ⎪ −⎜ ⎟⎨ ⎬⎢ ⎥ ⎜ ⎟⎪ ⎪⎢ ⎥ ⎜ ⎟⎪ ⎪⎢ ⎥ ⎜ ⎟⎪ ⎪⎢ ⎥ ⎩ ⎭⎜ ⎟−⎣ ⎦⎝ ⎠ { } 83.7 55.4 55.4 40.3 40.3 0 MA ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ −⎪ ⎪ ⎨ ⎬ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ = Member end actions
Dept. of CE, GCE Kannur Dr.RajeshKN [ ] 2 4 1 4 0 0 0 0 1 4 2 4 0 0 0 0 0 0 6 4.8 3 4.8 0 0 2 0 0 3 4.8 6 4.8 0 0 0 0 0 0 4 8 2 8 0 0 0 0 2 8 4 8 MS EI ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ •Problem 4 Unassembled stiffness matrix Equivalent joint loads 85.33 42.67 1 2 3
Dept. of CE, GCE Kannur Dr.RajeshKN 68 • consists of member displacements due to unit displacements on the restrained structure. [ ]MFC 1 1FD = 2 1FD = [ ] 1 0 0 1 0 0 0 1 0 1 0 0 MFC ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ DF1 DF2 =1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN 69 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ { } [ ] { } 1 F FF FD S A − = since there are no support displacements. Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 2 4 1 4 0 0 0 0 1 0 1 4 2 4 0 0 0 0 0 1 1 0 0 0 0 0 0 0 6 4.8 3 4.8 0 0 0 0 2 0 1 0 1 1 0 0 0 3 4.8 6 4.8 0 0 0 1 0 0 0 0 4 8 2 8 0 1 0 0 0 0 2 8 4 8 0 0 EI ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ = ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN 70 8 4 4 8 1 0 0 0 0 0 0 20 0 1 0 1 1 0 0 108 0 8 0 4 EI ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ = ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 2 1 1 92 EI ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ { } [ ] { } 1 1 2 1 0 1 9 85.3332 F FF F EI D S A − − ⎛ ⎞⎡ ⎤ ⎧ ⎫ ∴ = = = ⎨ ⎬⎜ ⎟⎢ ⎥ ⎣ ⎦ ⎩ ⎭⎝ ⎠ 36 4 0 341.3328 1 4 8 85.333 682.66 10.03 4272 91 203 . 784 0EI EIEI − −⎧ ⎫ ⎧ ⎫⎡ ⎤ = = =⎨ ⎬ ⎨ ⎬⎢ ⎥− ⎩ ⎭ ⎩ ⎭⎣ − ⎦ ⎧ ⎫ ⎨ ⎬ ⎩ ⎭
Dept. of CE, GCE Kannur Dr.RajeshKN 71 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + 0 2 4 1 4 0 0 0 0 1 0 0 1 4 2 4 0 0 0 0 0 1 42.667 0 0 6 4.8 3 4.8 0 0 0 1 10.0391 2 85.333 0 0 3 4.8 6 4.8 0 0 0 0 20.078 0 0 0 0 0 4 8 2 8 0 1 0 0 0 0 0 2 8 4 8 0 0 EI EI ⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎧ ⎫ = +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ − ⎩ ⎭⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎩ ⎭ ⎣ ⎦ ⎣ ⎦ is a null matrix{ }RD 0 0 0 120.47 42.667 200.781 85.333 401.568 0 160.624 0 80.312 ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = +⎨ ⎬ ⎨ ⎬ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ 0 15.059 67.765 35.138 20.078 10.039 ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨= ⎬ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ Member end actions
Dept. of CE, GCE Kannur Dr.RajeshKN 72 •Homework 2: 32 1 80kNm 3m3m 4m 6m 20kN m C A B D
Dept. of CE, GCE Kannur Dr.RajeshKN •Problem 5 [ ] 2 1 0 0 0 0 1 2 0 0 0 0 0 0 2 1 0 02 0 0 1 2 0 0 0 0 0 0 2 1 0 0 0 0 1 2 M EI S L ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ Unassembled stiffness matrix
Dept. of CE, GCE Kannur Dr.RajeshKN 74 [ ]MFC = 0 0 1 1 0 1 1 0 0 0 1 0 0 1 1 0 0 1 L L L L ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 1 2 3 DF1 DF2 DF3 =1 =1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN 75 2 1 0 0 0 0 0 0 1 1 2 0 0 0 0 1 0 1 0 1 1 0 0 0 0 0 2 1 0 0 1 0 02 0 0 0 1 1 0 0 0 1 2 0 0 0 1 0 1 1 0 0 1 1 0 0 0 0 2 1 0 1 1 0 0 0 0 1 2 0 0 1 L L EI L L L L L L L ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ [ ] [ ] [ ][ ]T FF MF M MFS C S C= 1 0 3 2 0 3 0 1 1 0 0 0 2 1 02 0 0 0 1 1 0 1 2 0 1 1 0 0 1 1 0 2 3 0 1 3 L L EI L L L L L L L ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 2 4 1 3 2 1 4 3 3 3 12 L EI L L L L L ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN 76 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ { } 1 2 4 1 3 0 2 1 4 3 0 3 3 12 F L EI D L L L L L P − ⎛ ⎞ ⎧ ⎫⎡ ⎤ ⎪ ⎪⎜ ⎟⎢ ⎥= ⎨ ⎬⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟⎢ ⎥ ⎩ ⎭⎣ ⎦⎝ ⎠ { } [ ] { } 1 F FF FD S A − = , since there are no support displacements. 2 13 3 3 0 3 13 3 0 84 3 3 5 L L L EI L L L P − − ⎧ ⎫⎡ ⎤ ⎪ ⎪⎢ ⎥= − − ⎨ ⎬⎢ ⎥ ⎪ ⎪− −⎢ ⎥ ⎩ ⎭⎣ ⎦ 2 3 3 84 5 PL EI L −⎧ ⎫ ⎪ ⎪ −⎨ ⎬ ⎪ ⎩ = ⎪ ⎭ Joint displacements
Dept. of CE, GCE Kannur Dr.RajeshKN 77 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } 2 2 1 0 0 0 0 0 0 1 1 2 0 0 0 0 1 0 1 3 0 0 2 1 0 0 1 0 02 3 0 0 1 2 0 0 0 1 0 84 5 0 0 0 0 2 1 0 1 1 0 0 0 0 1 2 0 0 1 M L L EI PL A L EI L L L ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ −⎧ ⎫ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ = −⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 2 2 1 0 0 0 0 5 1 2 0 0 0 0 2 0 0 2 1 0 0 32 0 0 1 2 0 0 384 0 0 0 0 2 1 2 0 0 0 0 1 2 5 EI PL L EI ⎧ ⎫⎡ ⎤ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ −⎪ ⎪⎢ ⎥ = ⎨ ⎬⎢ ⎥ −⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎩ ⎭⎣ ⎦ { } [ ][ ]{ }M M MF FA S C D= 2 12 9 92 984 9 12 EI PL L EI ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ −⎪ ⎪ = ⎨ ⎬ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 4 3 3 314 3 4 PL ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ − = ⎪ ⎪ ⎨ ⎬ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ Member end actions
Dept. of CE, GCE Kannur Dr.RajeshKN 78 [ ] 0.2 0.0 0.0 0.0 0.2 0.0 0.0 0.0 0.2 MS ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ Unassembled stiffness matrix •Problem 6: 1 2 3 50 kN 80 kN 5 m 5 m 5 m 4 m 4 m 3 m 3 m
Dept. of CE, GCE Kannur Dr.RajeshKN [ ]MFC = 0.8 -0.6 -0.8 0.6 0.8 0.6 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ 1 cos 36.87=0.8 36.87 1 2 3 0.8 10.6 53.13 cos 53.13=0.6 DF1 DF2 =1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN 80 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ { } 2.930 1.302 -50 1.302 5.208 -80 FD ⎧ ⎫⎡ ⎤ = ⎨ ⎬⎢ ⎥ ⎩ ⎭⎣ ⎦ { } [ ] { } 1 F FF FD S A − = , since there are no support displacements. -250.651 -481.771 ⎧ = ⎫ ⎨ ⎬ ⎩ ⎭ Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 0.384 -0.096 -0.096 0.216 ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN 0.2 0.0 0.0 0.8 -0.6 250.651 0.0 0.2 0.0 -0.8 0.6 481.771 0.0 0.0 0.2 0.8 0.6 ⎡ ⎤⎡ ⎤ −⎧ ⎫⎢ ⎥⎢ ⎥= ⎨ ⎬⎢ ⎥⎢ ⎥ −⎩ ⎭ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + [ ][ ]{ }M MF FS C D= 17.708 17.708 97.917 ⎧ ⎫ ⎪ ⎪ ⎨− ⎪ ⎩ = ⎬ − ⎪ ⎭ Member Forces:
Dept. of CE, GCE Kannur Dr.RajeshKN 82 •Problem 7: [ ] 0.2 0.0 0.0 0.0 0.5 0.0 0.0 0.0 0.2 MS ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ Unassembled stiffness matrix 1 2 3
Dept. of CE, GCE Kannur Dr.RajeshKN [ ]MFC = 0.600 0.800 0.000 -1.000 -0.600 0.800 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ 1 53.13 5 m cos 53.13=0.6 0.6 1 0 53.13 5 m 0 36.87 cos 36.87=0.8 cos 36.87=0.8 0.8 DF1 DF2 =1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN 84 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ { } 6.944 0.000 15.000 0.000 1.323 -10.000 FD ⎡ ⎤ ⎧ ⎫ = ⎨ ⎬⎢ ⎥ ⎣ ⎦ ⎩ ⎭ { } [ ] { } 1 F FF FD S A − = , since there are no support displacements. 104.167 -13.228 ⎧ = ⎫ ⎨ ⎬ ⎩ ⎭ Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 0.144 0.000 0.000 0.756 ⎡ ⎤ =⎢ ⎥ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN { } 0.2 0.0 0.0 0.600 0.800 104.167 0.0 0.5 0.0 0.000 -1.000 -13.228 0.0 0.0 0.2 -0.600 0.800 MA ⎡ ⎤⎡ ⎤ ⎧ ⎫⎢ ⎥⎢ ⎥= ⎨ ⎬⎢ ⎥⎢ ⎥⎩ ⎭ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } [ ][ ]{ }M M MF FA S C D= 10.4 6.6 -14.6 ⎧ ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ = ⎪ ⎩ ⎭ Member Forces:
Dept. of CE, GCE Kannur Dr.RajeshKN [ ] 0.866 0.000 0.000 0.000 1.000 0.000 0.000 0.000 0.500 MS ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ Unassembled stiffness matrix •Problem 8 (Homework 3):
Dept. of CE, GCE Kannur Dr.RajeshKN 1 0.866 0.5 1 0.866 0.5 [ ]MFC = 0.866 0.500 1.000 0.000 0.500 -0.866 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ DF1 DF2 =1 =1
Dept. of CE, GCE Kannur Dr.RajeshKN 88 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ { } 0.577 -0.155 5 -0.155 1.732 0 FD ⎡ ⎤ ⎧ ⎫ = ⎨ ⎬⎢ ⎥ ⎣ ⎦ ⎩ ⎭ { } [ ] { } 1 F FF FD S A − = , since there are no support displacements. 2.887 -0.773 ⎧ ⎫ ⎨ ⎩ = ⎬ ⎭ Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 1.774 0.158 0.158 0.591 ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦
Dept. of CE, GCE Kannur Dr.RajeshKN { } 0.866 0.000 0.000 0.866 0.500 2.887 0.000 1.000 0.000 1.000 0.000 -0.773 0.000 0.000 0.500 0.500 -0.866 MA ⎡ ⎤⎡ ⎤ ⎧ ⎫⎢ ⎥⎢ ⎥= ⎨ ⎬⎢ ⎥⎢ ⎥⎩ ⎭ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } [ ][ ]{ }M M MF FA S C D= 1.830 2.887 1.057 ⎧ ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩ ⎭ = Member Forces:
Dept. of CE, GCE Kannur Dr.RajeshKN 90 •Homework 4: AE is constant. 20 kN 1 2 3 600 450 A B C D 1m
Dept. of CE, GCE Kannur Dr.RajeshKN 91 • 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 Summary

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Introduction to the Stiffness Method in Structural Analysis by Dr. Rajesh K. N.

Development of stiffness matrices for truss, beam, and frame elements; total stiffness matrix and effects.

Primary unknowns are joint displacements; key terms include kinematic indeterminacy, degrees of freedom.

Discussion on degrees of freedom in trusses and frame structures, addressing rotations and significance.

Stiffness and flexibility coefficients detailed with equations for beam analysis under unit loads.

Example analysis of a propped cantilever using the stiffness method, focusing on induced rotations.

Calculation of member stiffness matrices for prismatic beam members, addressing variable degrees of freedom.

Construction of stiffness matrices for various frame components (plane, grid, space frames).

Formalization of stiffness method using virtual work principles; relationships between matrices.

Discussion on matrices relating joint displacements, support displacements, and compatibility conditions.

Examples illustrating calculations for joint displacements and reactions using stiffness properties.

Beginning problem-solving scenarios: joint displacement calculations and the implications of support conditions.

Examples of structural problems, detailing joint load conditions and stiffness matrix calculations.

Derivation of member end actions and support reactions through displacement and stiffness parameters.

Various structural analysis problems, applying stiffness method principles, culminating in a summary of method.

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 Bθ
  • 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 Bθ The actual rotation at B is Bθ 2 12 B wL M = − (Note the sign convention: anticlockwise positive) •Apply unit rotation corresponding to Bθ 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 Bθ B Bm θis 2 4 0 12 B wL EI L θ− + = 3 48 B B B wL EI M m θ∴ = − = Bm (Moment required for unit rotation) is the stiffness coefficient here. 0B B BM m θ+ = (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 { }MDδ { } [ ]{ } [ ] [ ] { } { } F M MJ J MF MR R D D C D C C D δ δ δ δ ⎧ ⎫ = = ⎡ ⎤ ⎨ ⎬⎣ ⎦ ⎩ ⎭ • 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 δ δ δ δ δ ⎧ ⎫⎡ ⎤= = ⎨ ⎬⎣ ⎦ ⎩ ⎭ { }JDδ• 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 Dδ δ= • Internal virtual work produced by the virtual (relative) end displacements { }MDδ { }MAand actual member end actions is { } { } { } { } T T J J M MA D A Dδ δ= • Equating the above two (principle of virtual work), { } [ ]{ }M MJ JD C D= { } [ ]{ }M M MA S D=But and { } [ ]{ }M MJ JD C Dδ δ=Also, { } { } { } [ ] [ ] [ ]{ } TT T T J J J MJ M MJ JA D D C S C Dδ δ=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 indeterminacy = 2
  • 30.
    Dept. of CE,GCE Kannur Dr.RajeshKN 30 Fixed end actions Equivalent joint loads
  • 31.
    Dept. of CE,GCE Kannur Dr.RajeshKN Joint displacements Free (unknown) joint displacements { }FD { }RDRestraint displacements
  • 32.
    Dept. of CE,GCE Kannur Dr.RajeshKN Joint displacements Free (unknown) joint displacements { }FD { }RDRestraint displacements •Each column in the submatrix consists of member displacements caused by a unit value of an unknown displacement applied to the restrained structure. [ ]MFC •Each column in the submatrix consists of member displacements caused by a unit value of a support displacement applied to the restrained structure. [ ]MRC
  • 33.
    Dept. of CE,GCE Kannur Dr.RajeshKN 0 1 1 0 0 0 0 1 [ ] 0 0 1 0 1 0 0 1 M FC ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ DF1 DF2 =1 =1
  • 34.
    Dept. of CE,GCE Kannur Dr.RajeshKN 1 L 1 L 0 0 1 0 0 0 1 L− 1 L− 1 L 1 L 0 0 1 L− 1 L−
  • 35.
    Dept. of CE,GCE Kannur Dr.RajeshKN 35 [ ] [ ] [ ]MJ MF MRC C C= ⎡ ⎤⎣ ⎦ 0 0 1 1 0 0 1 0 1 01 0 0 0 1 1 0 0 0 1 1 L L LL L −⎡ ⎤ ⎢ ⎥ −⎢ ⎥= ⎢ ⎥− ⎢ ⎥ −⎣ ⎦ [ ] 1 1 1 0 1 0 1 0 0 0 1 1 0 0 1 1 MR L L L L C L L L L −⎡ ⎤ ⎢ ⎥− ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ −⎣ ⎦ DR1 DR2 DR3 DR4 =1 =1 =1 =1 DR1 DR2 DR3 DR4 =1 =1 =1 =1 DF1 DF2 =1 =1
  • 36.
    Dept. of CE,GCE Kannur Dr.RajeshKN 36 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ { } [ ] { } 1 F FF FD S A − ∴ = Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 4 2 0 0 0 0 0 1 1 0 2 4 0 0 1 02 0 0 0 1 0 0 2 1 1 0 0 0 1 2 0 1 EI L ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 6 12 1 2 EI L ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ is a null matrix, since there are no support displacements { }RD
  • 37.
    Dept. of CE,GCE Kannur Dr.RajeshKN { } 1 6 1 12 1 2 29 F EI PL D L − ⎛ ⎞⎡ ⎤ ⎧ ⎫ = ⎨ ⎬⎜ ⎟⎢ ⎥ ⎣ ⎦ ⎩ ⎭⎝ ⎠ Free (unknown) joint displacements 2 2 01 . 11 818 11 0 11 PL E EII PL⎧ ⎫ ⎧ = = ⎫ ⎨⎨ ⎬ ⎭ ⎩⎩ ⎬ ⎭ 2 3 6 2 3 32 0 0 3 3 L L L LEI L L L ⎡ ⎤− − = ⎢ ⎥ −⎣ ⎦ [ ] [ ] [ ][ ]T FR MF M MRS C S C= 4 2 0 0 1 1 0 0 1 1 0 2 4 0 0 1 0 1 02 1 0 0 0 1 0 0 2 1 0 0 1 1 0 0 1 2 0 0 1 1 L EI L L −⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥−⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥ −⎣ ⎦ ⎣ ⎦
  • 38.
    Dept. of CE,GCE Kannur Dr.RajeshKN { } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + + is a null matrix.{ }RD Support reactions [ ] [ ] [ ][ ]T RF MR M MFS C S C= 2 6 0 2 02 3 3 3 3 LEI L ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ − −⎣ ⎦ 1 1 0 4 2 0 0 0 0 1 0 1 0 2 4 0 0 1 01 2 0 0 1 1 0 0 2 1 1 0 0 0 1 1 0 0 1 2 0 1 T L EI L L −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ { } { } [ ]{ }R RC RF FA A S D∴ = − +
  • 39.
    Dept. of CE,GCE Kannur Dr.RajeshKN [ ] [ ] [ ][ ]T RR MR M MRS C S C= 1 1 0 4 2 0 0 1 1 0 1 0 1 0 2 4 0 0 1 0 1 01 2 1 0 0 1 1 0 0 2 1 0 0 1 1 0 0 1 1 0 0 1 2 0 0 1 1 T L L EI L L L − −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− − ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ − −⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 3 12 6 12 0 6 4 6 02 12 6 18 6 0 0 6 6 L L L LEI LL −⎡ ⎤ ⎢ ⎥− ⎢ ⎥= − − −⎢ ⎥ ⎢ ⎥ −⎣ ⎦
  • 40.
    Dept. of CE,GCE Kannur Dr.RajeshKN 2 2 2 6 0 2 0 02 3 3 3 118 3 3 3 P PL LEI PL L EI P P −⎧ ⎫ ⎡ ⎤⎪ ⎪− ⎢ ⎥⎪ ⎪ ⎧ ⎫⎢ ⎥= − +⎨ ⎬ ⎨ ⎬ −⎢ ⎥ ⎩ ⎭⎪ ⎪− ⎢ ⎥⎪ ⎪ − −⎣ ⎦−⎩ ⎭ 2 2 0 2 2 0 0 02 3 3 318 33 3 3 3 P P PL PL PL EI P EI L P P P P P ⎧ ⎫ −⎧ ⎫ ⎧ ⎫ ⎪ ⎪⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪ ⎪⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ = − + = +⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎩ ⎭− −⎩ ⎭ ⎩ ⎭ ⎪ ⎪ ⎩ ⎭ { } { } [ ]{ }R RC RF FA A S D∴ = − + 2 3 10 3 2 3 P PL P P ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ =
  • 41.
    Dept. of CE,GCE Kannur Dr.RajeshKN 41 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } 2 3 4 2 0 0 0 0 3 2 4 0 0 1 0 02 2 0 0 2 1 1 0 19 18 2 0 0 1 2 0 1 M PL EI PL A L EI ⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥− ⎧ ⎫⎪ ⎪ ⎢ ⎥ ⎢ ⎥= +⎨ ⎬ ⎨ ⎬ ⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪−⎩ ⎭ ⎣ ⎦ ⎣ ⎦ 2 3 4 2 0 0 0 3 2 4 0 0 02 2 0 0 2 1 09 18 2 0 0 1 2 1 PL EI PL L EI ⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎪ ⎪ ⎪ ⎪⎢ ⎥−⎪ ⎪ ⎪ ⎪⎢ ⎥= +⎨ ⎬ ⎨ ⎬ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭⎣ ⎦ 3 0 3 0 2 19 9 2 2 PL PL ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪ = +⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪−⎩ ⎭ ⎩ ⎭ 1 1 13 0 PL ⎧ ⎫ ⎪ ⎪−⎪ ⎪ ⎨ ⎬ ⎪ ⎪ = ⎪ ⎪⎩ ⎭ is a null matrix{ }RD Member end actions { } { } [ ][ ]{ }M ML M MF FA A S C D∴ = +
  • 42.
    Dept. of CE,GCE Kannur Dr.RajeshKN 42 3 0 0 0 0 0 2 0 6 4 6 0 1 1 0 0 4 0 6 2 6 02 0 0 0 2 0 0 3 3 1 1 1 1 2 0 0 3 3 0 0 1 1 L L L L L L LEI L L LL L L ⎡ ⎤ ⎢ ⎥ −⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥= ⎢ ⎥ −⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− − −⎣ ⎦ ⎢ ⎥ − −⎣ ⎦ [ ] [ ] [ ] [ ] 2 2 2 2 2 3 2 6 6 2 3 3 2 0 0 3 3 6 0 12 6 12 02 2 0 6 4 6 0 3 3 12 6 18 6 3 3 0 0 6 6 FF FR RF RR L L L L L L L L L L S SL LEI S SL L L L L L L L L L ⎡ ⎤− − ⎢ ⎥ −⎢ ⎥ ⎢ ⎥ ⎡ ⎤− = =⎢ ⎥ ⎢ ⎥ − ⎣ ⎦⎢ ⎥ ⎢ ⎥− − − − ⎢ ⎥ − − −⎣ ⎦ [ ] [ ] [ ][ ]T J MJ M MJS C S C= 0 0 0 0 0 4 2 0 0 0 0 1 1 0 1 1 0 0 2 4 0 0 0 1 0 1 01 2 1 0 0 0 0 0 2 1 0 0 0 1 1 1 1 1 1 0 0 1 2 0 0 0 1 1 0 0 1 1 L L L L LEI L LL L L L ⎡ ⎤ ⎢ ⎥ −⎡ ⎤⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥−⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥− − −⎣ ⎦ ⎣ ⎦ ⎢ ⎥ − −⎣ ⎦ Alternatively, if the entire [SJ] matrix is assembled at a time,
  • 43.
    Dept. of CE,GCE Kannur Dr.RajeshKN 43 •Problem 2: A B C D 40 kN10 kN/m 2 m 4 m 4 m 2 m 100 kN Kinematic indeterminacy = 3 (Not considering joint D in the overhanging portion) Degrees of freedom A B C D DF2DF1 DF3
  • 44.
    Dept. of CE,GCE Kannur Dr.RajeshKN 44 [ ] 2 1 0 0 1 2 0 02 0 0 2 14 0 0 1 2 M EI S ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ Unassembled stiffness matrix [ ] 2 12 1 24 Mi EI S ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ Member stiffness matrix of beam member
  • 45.
    Dept. of CE,GCE Kannur Dr.RajeshKN 45 Fixed end actions Equivalent joint loads + actual joint loads A B 13.33 kNm 13.33 20 kN 20 kN B C 20 20 20 kN 20 kN A B 13.33 kNm 6.67 20 kN 20 +20 = 40 kN 20 +100 = 120 kN 2x100 – 20 = 180 kNm
  • 46.
    Dept. of CE,GCE Kannur Dr.RajeshKN 1 0 0 0 0 1 1 0 DF1 =1 DF2 =1 L
  • 47.
    Dept. of CE,GCE Kannur Dr.RajeshKN 47 [ ] 1 0 0 0 1 0 0 1 0 0 0 1 M FC ⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ DF1 DF2 DF3 =1 =1 =1 0 0 0 1 DF3 =1
  • 48.
    Dept. of CE,GCE Kannur Dr.RajeshKN 48 { } [ ] { } 1 F FF FD S A − ∴ = Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 1 0 0 2 1 0 0 1 0 0 0 1 0 1 2 0 0 0 1 02 0 1 0 0 0 2 1 0 1 04 0 0 1 0 0 1 2 0 0 1 T EI ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 1 0.5 0 0.5 2 0.5 0 0.5 1 EI ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ is a null matrix.{ }RD∴
  • 49.
    Dept. of CE,GCE Kannur Dr.RajeshKN { } 1 1 0.5 0 13.33 1 0.5 2 0.5 6.67 0 0.5 1 180 FD EI − −⎛ ⎞⎡ ⎤ ⎧ ⎫ ⎪ ⎪⎜ ⎟⎢ ⎥= −⎨ ⎬⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟ −⎢ ⎥⎣ ⎦ ⎩ ⎭⎝ ⎠ Joint displacements 43.435 60.33 210.6 1 EI −⎧ ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ ⎪−⎩ ⎭ =
  • 50.
    Dept. of CE,GCE Kannur Dr.RajeshKN 50 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } 43.435 60.33 210.6 13.33 2 1 0 0 1 0 0 13.33 1 2 0 0 0 1 02 1 20 0 0 2 1 0 1 04 20 0 0 1 2 0 0 1 M EI A EI ⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥−⎪ ⎪ ⎢ ⎥ ⎢ ⎥= +⎨ ⎬ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎢ ⎥ ⎢ ⎥⎪ ⎪− −⎧ ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ ⎪− ⎣ ⎩ ⎭ ⎦ ⎣ ⎦ ⎭ ⎩ 0 25 25 200 kNm ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎨ ⎬ −⎪ ⎪ ⎪ ⎪−⎩ ⎭ = is a null matrix{ }RD Member end actions { } { } [ ][ ]{ }M ML M MF FA A S C D∴ = +
  • 51.
    Dept. of CE,GCE Kannur Dr.RajeshKN •Problem 3 Analyse the beam. Support B has a downward settlement of 30mm. EI=5.6×103 kNm2 [ ] 4 2 0 0 0 0 2 4 0 0 0 0 0 0 2 1 0 02 0 0 1 2 0 06 0 0 0 0 4 2 0 0 0 0 2 4 M EI S ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ Unassembled stiffness matrix [ ] 2 12 1 2Mi EI S L ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ Member stiffness matrix of beam member
  • 52.
    Dept. of CE,GCE Kannur Dr.RajeshKN Fixed end moments 2525 Equivalent joint loads 2525 Support settlements { }RD(Restraint displacements) A B C D30mm 1RD Free (unknown) joint displacements { }FD 1FD 2FD 3FD
  • 53.
    Dept. of CE,GCE Kannur Dr.RajeshKN 53 BA C D 1 1FD = 0 1 1 0 0 0 and consist of member displacements due to unit displacements on the restrained structure. [ ]MFC [ ]MRC
  • 54.
    Dept. of CE,GCE Kannur Dr.RajeshKN 54 [ ] [ ] [ ]MJ MF MRC C C= ⎡ ⎤⎣ ⎦ 0 0 0 1 3 1 0 0 1 3 1 0 0 1 6 0 1 0 1 6 0 1 0 0 0 0 1 0 −⎡ ⎤ ⎢ ⎥ −⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ 31 2 1 1 1 11 FF F RDD D D = = == A B C D1 1RD = 1 3− 1 3− 1 6 1 6 0 0
  • 55.
    Dept. of CE,GCE Kannur Dr.RajeshKN 55 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 0 0 0 4 2 0 0 0 0 0 0 0 1 0 0 2 4 0 0 0 0 1 0 0 1 0 0 0 0 2 1 0 0 1 0 0 0 1 0 0 0 1 2 0 0 0 1 03 0 1 0 0 0 0 0 4 2 0 1 0 0 0 1 0 0 0 0 2 4 0 0 1 T EI ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 6 1 0 1 6 2 3 0 2 4 EI ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦
  • 56.
    Dept. of CE,GCE Kannur Dr.RajeshKN [ ] [ ] [ ][ ]T FR MF M MRS C S C= 0 0 0 4 2 0 0 0 0 1 3 1 0 0 2 4 0 0 0 0 1 3 1 0 0 0 0 2 1 0 0 1 6 0 1 0 0 0 1 2 0 0 1 63 0 1 0 0 0 0 0 4 2 0 0 0 1 0 0 0 0 2 4 0 T EI −⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 3 2 1 2 3 0 EI −⎧ ⎫ ⎪ ⎪ = ⎨ ⎬ ⎪ ⎪ ⎩ ⎭
  • 57.
    Dept. of CE,GCE Kannur Dr.RajeshKN { } 1 6 1 0 0 3 2 1 6 2 25 1 2 0.03 3 3 0 2 4 25 0 EI EI − −⎛ ⎞ ⎡ ⎤⎧ ⎫⎡ ⎤ ⎡ ⎤ ⎪ ⎪⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎢ ⎥= − − −⎨ ⎬⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎪ ⎪⎜ ⎟ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎩ ⎭⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎣ ⎦ 20 4 2 0 0.045 3 5600 4 24 12 25 0.015 116 3 2 12 35 25 0 EI − ⎡ ⎤⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎢ ⎥= − − − − −⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥−⎢ ⎥ ⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦ 3 1642 3 108 116 5.6 10 6 0.007583 0.0005 0.007 311 −⎧ ⎫ ⎪ ⎪ = =⎨ ⎬ × × ⎪ ⎪ −⎧ ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩⎩ ⎭⎭ 20 4 2 0 84 3 4 24 12 25 28 116 2 12 35 25 0 EI − ⎡ ⎤⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎢ ⎥= − − − − −⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥−⎢ ⎥ ⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦ { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦
  • 58.
    Dept. of CE,GCE Kannur Dr.RajeshKN 58 { } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + + Support reactions [ ] 4 2 0 0 0 0 0 0 0 2 4 0 0 0 0 1 0 0 0 0 2 1 0 0 1 0 0 1 3 1 3 1 6 1 6 0 0 0 0 1 2 0 0 0 1 03 0 0 0 0 4 2 0 1 0 0 0 0 0 2 4 0 0 1 EI ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = − − ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ [ ] [ ] [ ][ ]T RF MR M MFS C S C= [ ]3 2 1 2 0 3 EI = −
  • 59.
    Dept. of CE,GCE Kannur Dr.RajeshKN { } { } [ ] { } 0.007583 0 3 2 1 2 0 0.0005 0.03 3 2 0.0031 R EI EI A −⎧ ⎫ ⎪ ⎪ ⎡ ⎤ = − + − + −⎨ ⎬ ⎢ ⎥⎣ ⎦⎪ ⎪ ⎩ ⎭ (Support reaction corresponding to DR . i.e., reaction at B) ( )0.0116 0.045 0.0334 62 5 3 3 .3 EI EI = − = − −= [ ] [ ] [ ][ ]T RR MR M MRS C S C= [ ] 4 2 0 0 0 0 1 3 2 4 0 0 0 0 1 3 0 0 2 1 0 0 1 6 1 3 1 3 1 6 1 6 0 0 0 0 1 2 0 0 1 63 0 0 0 0 4 2 0 0 0 0 0 2 4 0 EI −⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = − − ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 2 EI⎡ ⎤ = ⎢ ⎥⎣ ⎦
  • 60.
    Dept. of CE,GCE Kannur Dr.RajeshKN 60 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } { } 0 4 2 0 0 0 0 0 0 0 1 3 0 2 4 0 0 0 0 1 0 0 1 3 0.007583 0 0 0 2 1 0 0 1 0 0 1 62 0.0005 0.03 0 0 0 1 2 0 0 0 1 0 1 66 0.0031 25 0 0 0 0 4 2 0 1 0 0 25 0 0 0 0 2 4 0 0 1 0 M EI A −⎛⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎧ ⎫ ⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ = + + −⎜⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎜⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎜⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ −⎩ ⎭ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎜ ⎟⎜ ⎟ ⎠ 83.7 55.4 55.4 40.3 40.3 0 ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ −⎪ ⎪ ⎨ ⎬ −⎪ ⎪ ⎪ ⎩ = ⎪ ⎪ ⎪ ⎭ Member end actions
  • 61.
    Dept. of CE,GCE Kannur Dr.RajeshKN 61 [ ] [ ] [ ] [ ] FF FR RF RR S S S S ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ 6 1 0 3 2 1 6 2 1 2 0 2 4 03 3 2 1 2 0 3 2 E I −⎡ ⎤ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ −⎣ ⎦ 2 0 0 2 0 1 1 0 0 0 4 0 0 2 0 0 0 1 1 0 2 1 0 1 2 0 0 0 0 0 1 1 2 0 1 23 1 3 1 3 1 6 1 6 0 0 0 4 2 0 0 2 4 0 EI −⎡ ⎤ ⎢ ⎥−⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎢ ⎥ ⎣ ⎦ [ ] [ ] [ ][ ]T J MJ M MJS C S C= 4 2 0 0 0 0 0 0 0 1 3 0 1 1 0 0 0 2 4 0 0 0 0 1 0 0 1 3 0 0 0 1 1 0 0 0 2 1 0 0 1 0 0 1 6 0 0 0 0 0 1 0 0 1 2 0 0 0 1 0 1 63 1 3 1 3 1 6 1 6 0 0 0 0 0 0 4 2 0 1 0 0 0 0 0 0 2 4 0 0 1 0 EI −⎡ ⎤⎡ ⎤ ⎢ ⎥⎢ ⎥ −⎡ ⎤ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥− −⎣ ⎦ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ Alternatively, if the entire [SJ] matrix is assembled at a time,
  • 62.
    Dept. of CE,GCE Kannur Dr.RajeshKN 62 [ ] [ ] [ ]MJ MF MRC C C= ⎡ ⎤⎣ ⎦ 0 0 0 1 3 1 1 3 0 0 1 0 0 1 3 0 1 3 0 0 1 0 0 0 0 1 6 1 6 0 0 1 0 0 0 1 6 1 6 0 0 1 0 0 0 0 1 3 1 3 0 0 1 0 0 0 1 3 1 3 −⎡ ⎤ ⎢ ⎥ −⎢ ⎥ ⎢ ⎥− = ⎢ ⎥ −⎢ ⎥ ⎢ ⎥− ⎢ ⎥ −⎢ ⎥⎣ ⎦ [ ] [ ] [ ][ ]T J MJ M MJS C S C= 0 0 01 3 1 1 3 0 0 4 2 0 0 0 0 0 0 01 3 1 1 3 0 0 1 0 01 3 0 1 3 0 0 2 4 0 0 0 0 1 0 01 3 0 1 3 0 0 1 0 0 0 0 1 6 1 6 0 0 0 2 1 0 0 1 0 0 0 0 1 6 1 6 0 0 1 0 0 0 1 6 1 6 0 0 0 1 2 0 0 0 1 0 0 0 1 6 1 6 03 0 1 0 0 0 0 1 3 1 3 0 0 0 0 4 2 0 1 0 0 0 0 1 3 0 0 1 0 0 0 1 3 1 3 0 0 0 0 2 4 0 0 1 T EI − −⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥ ⎢ ⎥− −⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ − −⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ ⎢ ⎥ −⎢ ⎥ ⎣ ⎦⎣ ⎦ 1 3 0 0 0 1 3 1 3 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− ⎢ ⎥ −⎢ ⎥⎣ ⎦ Alternatively, if ALL possible support settlements are accounted for,
  • 63.
    Dept. of CE,GCE Kannur Dr.RajeshKN 63 [ ] [ ] [ ] [ ] FF FR RF RR S S S S ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ 0 1 1 0 0 0 0 0 0 1 1 0 2 0 0 2 4 2 0 0 0 0 0 0 0 1 4 0 0 2 2 2 0 0 1 3 1 3 0 0 0 0 2 1 0 0 0 1 2 1 2 0 1 0 0 0 0 0 1 2 0 0 0 1 2 1 2 03 1 3 1 3 1 6 1 6 0 0 0 4 2 0 0 0 2 2 0 0 1 6 1 6 1 3 1 3 0 2 4 0 0 0 2 2 0 0 0 0 1 3 1 3 EI ⎡ ⎤ ⎢ ⎥ −⎡ ⎤⎢ ⎥ ⎢ ⎥−⎢ ⎥ ⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥ −⎢ ⎥⎢ ⎥ ⎢ ⎥− − −⎢ ⎥ ⎢ ⎥⎢ ⎥− − −⎣ ⎦ ⎢ ⎥ − −⎣ ⎦ 6 1 0 2 2 3 2 1 2 0 1 6 2 0 0 1 2 3 2 2 0 2 4 0 0 0 2 2 2 0 0 4 3 2 4 3 0 0 2 0 0 2 4 2 0 03 3 2 1 2 0 4 3 2 3 2 1 6 0 1 2 3 2 2 0 0 1 6 3 2 4 3 0 2 2 0 0 0 4 3 4 3 EI − −⎡ ⎤ ⎢ ⎥− ⎢ ⎥ −⎢ ⎥ ⎢ ⎥ −⎢ ⎥= ⎢ ⎥− ⎢ ⎥ − − − −⎢ ⎥ ⎢ ⎥− − − ⎢ ⎥ − − −⎣ ⎦
  • 64.
    Dept. of CE,GCE Kannur Dr.RajeshKN 64 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ 1 0 6 1 0 0 2 2 3 2 1 2 0 0 1 6 2 25 0 0 1 2 3 2 2 0.03 3 3 0 2 4 25 0 0 0 2 2 0 0 EI EI − ⎡ ⎤⎧ ⎫ ⎢ ⎥⎪ ⎪− −⎛ ⎞ ⎧ ⎫⎡ ⎤ ⎡ ⎤⎢ ⎥⎪ ⎪⎪ ⎪ ⎪ ⎪⎜ ⎟⎢ ⎥ ⎢ ⎥⎢ ⎥= − − − −⎨ ⎬ ⎨ ⎬⎜ ⎟⎢ ⎥ ⎢ ⎥⎢ ⎥⎪ ⎪ ⎪ ⎪⎜ ⎟ −⎢ ⎥ ⎢ ⎥⎩ ⎭⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎢ ⎥⎪ ⎪ ⎢ ⎥⎪ ⎪⎩ ⎭⎣ ⎦ 20 4 2 0 0.045 3 5600 4 24 12 25 0.015 116 3 2 12 35 25 0 20 4 2 84 3 4 24 12 3 116 2 12 35 25 EI EI − ⎡ ⎤⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎪ ⎪ ⎪ ⎪⎢ ⎥⎢ ⎥= − − − − −⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥ ⎪ ⎪ ⎪ ⎪⎢ ⎥−⎢ ⎥ ⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦ − −⎧ ⎫⎡ ⎤ ⎪ ⎪⎢ ⎥= − − ⎨ ⎬⎢ ⎥ ⎪ ⎪−⎢ ⎥ ⎩ ⎭⎣ ⎦ 1642 3 108 116 671 EI −⎧ ⎫ ⎪ ⎪ = ⎨ ⎬ ⎪ ⎪ ⎩ ⎭ 0.007583 0.0005 0.0031 −⎧ ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ ⎩ = ⎪ ⎭ Joint displacements
  • 65.
    Dept. of CE,GCE Kannur Dr.RajeshKN 65 { } 0 2 0 0 4 3 2 4 3 0 0 0 0.0075830 2 0 0 2 4 2 0 0 0 0.00050 3 2 1 2 0 4 3 2 3 2 1 6 0 0.03 3 3 0.003150 1 2 3 2 2 0 0 1 6 3 2 4 3 0 50 0 2 2 0 0 0 4 3 4 3 0 R EI EI A −⎧ ⎫ ⎧ ⎫⎡ ⎤ ⎡ ⎤ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− −⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = − + +− − − − −⎢ ⎥ ⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪− − − −⎩ ⎭⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥− − − −⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭⎣ ⎦ ⎣ ⎦ { } { } [ ]{ } [ ]{ }R RC RF F RR RA A S D S D= − + + { } 0 28.373 74.667 0 28.373 112 0 21.653 84 50 19.973 9.333 46.294 83.627 62.347 79.31 36.5650 13.44 0 RA −⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = − + + =−⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪− ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ − −⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪⎪ ⎪ −⎨ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭⎭ ⎪ Support reactions
  • 66.
    Dept. of CE,GCE Kannur Dr.RajeshKN 66 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } 0 4 2 0 0 0 0 0 0 0 1 3 1 1 3 0 0 0 2 4 0 0 0 0 1 0 0 1 3 0 1 3 0 0 0.007583 0 0 0 2 1 0 0 1 0 0 0 0 1 6 1 6 02 0.0005 0 0 0 1 2 0 0 0 1 0 0 0 1 6 1 6 06 0.0031 25 0 0 0 0 4 2 0 1 0 0 0 0 1 3 1 3 25 0 0 0 0 2 4 0 0 1 0 0 0 M EI A −⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ − ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎧ ⎫ −⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ = + +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ −⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎪ ⎪ ⎢ ⎥ ⎢ ⎥ − ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎩ ⎭ ⎣ ⎦ ⎣ ⎦ 0 0 0.03 0 0 1 3 1 3 ⎛ ⎞⎡ ⎤ ⎧ ⎫⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟⎢ ⎥ ⎪ ⎪ −⎜ ⎟⎨ ⎬⎢ ⎥ ⎜ ⎟⎪ ⎪⎢ ⎥ ⎜ ⎟⎪ ⎪⎢ ⎥ ⎜ ⎟⎪ ⎪⎢ ⎥ ⎩ ⎭⎜ ⎟−⎣ ⎦⎝ ⎠ { } 83.7 55.4 55.4 40.3 40.3 0 MA ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ −⎪ ⎪ ⎨ ⎬ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ = Member end actions
  • 67.
    Dept. of CE,GCE Kannur Dr.RajeshKN [ ] 2 4 1 4 0 0 0 0 1 4 2 4 0 0 0 0 0 0 6 4.8 3 4.8 0 0 2 0 0 3 4.8 6 4.8 0 0 0 0 0 0 4 8 2 8 0 0 0 0 2 8 4 8 MS EI ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ •Problem 4 Unassembled stiffness matrix Equivalent joint loads 85.33 42.67 1 2 3
  • 68.
    Dept. of CE,GCE Kannur Dr.RajeshKN 68 • consists of member displacements due to unit displacements on the restrained structure. [ ]MFC 1 1FD = 2 1FD = [ ] 1 0 0 1 0 0 0 1 0 1 0 0 MFC ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ DF1 DF2 =1 =1
  • 69.
    Dept. of CE,GCE Kannur Dr.RajeshKN 69 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ { } [ ] { } 1 F FF FD S A − = since there are no support displacements. Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 2 4 1 4 0 0 0 0 1 0 1 4 2 4 0 0 0 0 0 1 1 0 0 0 0 0 0 0 6 4.8 3 4.8 0 0 0 0 2 0 1 0 1 1 0 0 0 3 4.8 6 4.8 0 0 0 1 0 0 0 0 4 8 2 8 0 1 0 0 0 0 2 8 4 8 0 0 EI ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ = ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦
  • 70.
    Dept. of CE,GCE Kannur Dr.RajeshKN 70 8 4 4 8 1 0 0 0 0 0 0 20 0 1 0 1 1 0 0 108 0 8 0 4 EI ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ = ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 2 1 1 92 EI ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ { } [ ] { } 1 1 2 1 0 1 9 85.3332 F FF F EI D S A − − ⎛ ⎞⎡ ⎤ ⎧ ⎫ ∴ = = = ⎨ ⎬⎜ ⎟⎢ ⎥ ⎣ ⎦ ⎩ ⎭⎝ ⎠ 36 4 0 341.3328 1 4 8 85.333 682.66 10.03 4272 91 203 . 784 0EI EIEI − −⎧ ⎫ ⎧ ⎫⎡ ⎤ = = =⎨ ⎬ ⎨ ⎬⎢ ⎥− ⎩ ⎭ ⎩ ⎭⎣ − ⎦ ⎧ ⎫ ⎨ ⎬ ⎩ ⎭
  • 71.
    Dept. of CE,GCE Kannur Dr.RajeshKN 71 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + 0 2 4 1 4 0 0 0 0 1 0 0 1 4 2 4 0 0 0 0 0 1 42.667 0 0 6 4.8 3 4.8 0 0 0 1 10.0391 2 85.333 0 0 3 4.8 6 4.8 0 0 0 0 20.078 0 0 0 0 0 4 8 2 8 0 1 0 0 0 0 0 2 8 4 8 0 0 EI EI ⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ −⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎧ ⎫ = +⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ − ⎩ ⎭⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ ⎢ ⎥ ⎢ ⎥ ⎩ ⎭ ⎣ ⎦ ⎣ ⎦ is a null matrix{ }RD 0 0 0 120.47 42.667 200.781 85.333 401.568 0 160.624 0 80.312 ⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ = +⎨ ⎬ ⎨ ⎬ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ 0 15.059 67.765 35.138 20.078 10.039 ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨= ⎬ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ Member end actions
  • 72.
    Dept. of CE,GCE Kannur Dr.RajeshKN 72 •Homework 2: 32 1 80kNm 3m3m 4m 6m 20kN m C A B D
  • 73.
    Dept. of CE,GCE Kannur Dr.RajeshKN •Problem 5 [ ] 2 1 0 0 0 0 1 2 0 0 0 0 0 0 2 1 0 02 0 0 1 2 0 0 0 0 0 0 2 1 0 0 0 0 1 2 M EI S L ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ Unassembled stiffness matrix
  • 74.
    Dept. of CE,GCE Kannur Dr.RajeshKN 74 [ ]MFC = 0 0 1 1 0 1 1 0 0 0 1 0 0 1 1 0 0 1 L L L L ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 1 2 3 DF1 DF2 DF3 =1 =1 =1
  • 75.
    Dept. of CE,GCE Kannur Dr.RajeshKN 75 2 1 0 0 0 0 0 0 1 1 2 0 0 0 0 1 0 1 0 1 1 0 0 0 0 0 2 1 0 0 1 0 02 0 0 0 1 1 0 0 0 1 2 0 0 0 1 0 1 1 0 0 1 1 0 0 0 0 2 1 0 1 1 0 0 0 0 1 2 0 0 1 L L EI L L L L L L L ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ [ ] [ ] [ ][ ]T FF MF M MFS C S C= 1 0 3 2 0 3 0 1 1 0 0 0 2 1 02 0 0 0 1 1 0 1 2 0 1 1 0 0 1 1 0 2 3 0 1 3 L L EI L L L L L L L ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 2 4 1 3 2 1 4 3 3 3 12 L EI L L L L L ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦
  • 76.
    Dept. of CE,GCE Kannur Dr.RajeshKN 76 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ { } 1 2 4 1 3 0 2 1 4 3 0 3 3 12 F L EI D L L L L L P − ⎛ ⎞ ⎧ ⎫⎡ ⎤ ⎪ ⎪⎜ ⎟⎢ ⎥= ⎨ ⎬⎜ ⎟⎢ ⎥ ⎪ ⎪⎜ ⎟⎢ ⎥ ⎩ ⎭⎣ ⎦⎝ ⎠ { } [ ] { } 1 F FF FD S A − = , since there are no support displacements. 2 13 3 3 0 3 13 3 0 84 3 3 5 L L L EI L L L P − − ⎧ ⎫⎡ ⎤ ⎪ ⎪⎢ ⎥= − − ⎨ ⎬⎢ ⎥ ⎪ ⎪− −⎢ ⎥ ⎩ ⎭⎣ ⎦ 2 3 3 84 5 PL EI L −⎧ ⎫ ⎪ ⎪ −⎨ ⎬ ⎪ ⎩ = ⎪ ⎭ Joint displacements
  • 77.
    Dept. of CE,GCE Kannur Dr.RajeshKN 77 { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } 2 2 1 0 0 0 0 0 0 1 1 2 0 0 0 0 1 0 1 3 0 0 2 1 0 0 1 0 02 3 0 0 1 2 0 0 0 1 0 84 5 0 0 0 0 2 1 0 1 1 0 0 0 0 1 2 0 0 1 M L L EI PL A L EI L L L ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ −⎧ ⎫ ⎢ ⎥ ⎢ ⎥ ⎪ ⎪ = −⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎩ ⎭⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ 2 2 1 0 0 0 0 5 1 2 0 0 0 0 2 0 0 2 1 0 0 32 0 0 1 2 0 0 384 0 0 0 0 2 1 2 0 0 0 0 1 2 5 EI PL L EI ⎧ ⎫⎡ ⎤ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ −⎪ ⎪⎢ ⎥ = ⎨ ⎬⎢ ⎥ −⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎩ ⎭⎣ ⎦ { } [ ][ ]{ }M M MF FA S C D= 2 12 9 92 984 9 12 EI PL L EI ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ −⎪ ⎪ = ⎨ ⎬ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ 4 3 3 314 3 4 PL ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ − = ⎪ ⎪ ⎨ ⎬ −⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ⎭ Member end actions
  • 78.
    Dept. of CE,GCE Kannur Dr.RajeshKN 78 [ ] 0.2 0.0 0.0 0.0 0.2 0.0 0.0 0.0 0.2 MS ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ Unassembled stiffness matrix •Problem 6: 1 2 3 50 kN 80 kN 5 m 5 m 5 m 4 m 4 m 3 m 3 m
  • 79.
    Dept. of CE,GCE Kannur Dr.RajeshKN [ ]MFC = 0.8 -0.6 -0.8 0.6 0.8 0.6 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ 1 cos 36.87=0.8 36.87 1 2 3 0.8 10.6 53.13 cos 53.13=0.6 DF1 DF2 =1 =1
  • 80.
    Dept. of CE,GCE Kannur Dr.RajeshKN 80 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ { } 2.930 1.302 -50 1.302 5.208 -80 FD ⎧ ⎫⎡ ⎤ = ⎨ ⎬⎢ ⎥ ⎩ ⎭⎣ ⎦ { } [ ] { } 1 F FF FD S A − = , since there are no support displacements. -250.651 -481.771 ⎧ = ⎫ ⎨ ⎬ ⎩ ⎭ Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 0.384 -0.096 -0.096 0.216 ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦
  • 81.
    Dept. of CE,GCE Kannur Dr.RajeshKN 0.2 0.0 0.0 0.8 -0.6 250.651 0.0 0.2 0.0 -0.8 0.6 481.771 0.0 0.0 0.2 0.8 0.6 ⎡ ⎤⎡ ⎤ −⎧ ⎫⎢ ⎥⎢ ⎥= ⎨ ⎬⎢ ⎥⎢ ⎥ −⎩ ⎭ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + [ ][ ]{ }M MF FS C D= 17.708 17.708 97.917 ⎧ ⎫ ⎪ ⎪ ⎨− ⎪ ⎩ = ⎬ − ⎪ ⎭ Member Forces:
  • 82.
    Dept. of CE,GCE Kannur Dr.RajeshKN 82 •Problem 7: [ ] 0.2 0.0 0.0 0.0 0.5 0.0 0.0 0.0 0.2 MS ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ Unassembled stiffness matrix 1 2 3
  • 83.
    Dept. of CE,GCE Kannur Dr.RajeshKN [ ]MFC = 0.600 0.800 0.000 -1.000 -0.600 0.800 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ 1 53.13 5 m cos 53.13=0.6 0.6 1 0 53.13 5 m 0 36.87 cos 36.87=0.8 cos 36.87=0.8 0.8 DF1 DF2 =1 =1
  • 84.
    Dept. of CE,GCE Kannur Dr.RajeshKN 84 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ { } 6.944 0.000 15.000 0.000 1.323 -10.000 FD ⎡ ⎤ ⎧ ⎫ = ⎨ ⎬⎢ ⎥ ⎣ ⎦ ⎩ ⎭ { } [ ] { } 1 F FF FD S A − = , since there are no support displacements. 104.167 -13.228 ⎧ = ⎫ ⎨ ⎬ ⎩ ⎭ Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 0.144 0.000 0.000 0.756 ⎡ ⎤ =⎢ ⎥ ⎣ ⎦
  • 85.
    Dept. of CE,GCE Kannur Dr.RajeshKN { } 0.2 0.0 0.0 0.600 0.800 104.167 0.0 0.5 0.0 0.000 -1.000 -13.228 0.0 0.0 0.2 -0.600 0.800 MA ⎡ ⎤⎡ ⎤ ⎧ ⎫⎢ ⎥⎢ ⎥= ⎨ ⎬⎢ ⎥⎢ ⎥⎩ ⎭ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } [ ][ ]{ }M M MF FA S C D= 10.4 6.6 -14.6 ⎧ ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ = ⎪ ⎩ ⎭ Member Forces:
  • 86.
    Dept. of CE,GCE Kannur Dr.RajeshKN [ ] 0.866 0.000 0.000 0.000 1.000 0.000 0.000 0.000 0.500 MS ⎡ ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥⎣ ⎦ Unassembled stiffness matrix •Problem 8 (Homework 3):
  • 87.
    Dept. of CE,GCE Kannur Dr.RajeshKN 1 0.866 0.5 1 0.866 0.5 [ ]MFC = 0.866 0.500 1.000 0.000 0.500 -0.866 ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ DF1 DF2 =1 =1
  • 88.
    Dept. of CE,GCE Kannur Dr.RajeshKN 88 { } [ ] { } [ ]{ } 1 F FF F FR RD S A S D − = −⎡ ⎤⎣ ⎦ { } 0.577 -0.155 5 -0.155 1.732 0 FD ⎡ ⎤ ⎧ ⎫ = ⎨ ⎬⎢ ⎥ ⎣ ⎦ ⎩ ⎭ { } [ ] { } 1 F FF FD S A − = , since there are no support displacements. 2.887 -0.773 ⎧ ⎫ ⎨ ⎩ = ⎬ ⎭ Joint displacements [ ] [ ] [ ][ ]T FF MF M MFS C S C= 1.774 0.158 0.158 0.591 ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦
  • 89.
    Dept. of CE,GCE Kannur Dr.RajeshKN { } 0.866 0.000 0.000 0.866 0.500 2.887 0.000 1.000 0.000 1.000 0.000 -0.773 0.000 0.000 0.500 0.500 -0.866 MA ⎡ ⎤⎡ ⎤ ⎧ ⎫⎢ ⎥⎢ ⎥= ⎨ ⎬⎢ ⎥⎢ ⎥⎩ ⎭ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦ { } { } [ ] [ ]{ } [ ]{ }( )M ML M MF F MR RA A S C D C D= + + { } [ ][ ]{ }M M MF FA S C D= 1.830 2.887 1.057 ⎧ ⎫ ⎪ ⎪ ⎨ ⎬ ⎪ ⎪ ⎩ ⎭ = Member Forces:
  • 90.
    Dept. of CE,GCE Kannur Dr.RajeshKN 90 •Homework 4: AE is constant. 20 kN 1 2 3 600 450 A B C D 1m
  • 91.
    Dept. of CE,GCE Kannur Dr.RajeshKN 91 • 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 Summary