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CALCULATION OF THE UNDETERMINED STATIC REACTIONS FOR THE ARTICULATED PL

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CALCULATION OF THE UNDETERMINED STATIC REACTIONS FOR THE ARTICULATED PL Powered By Docstoc
					  International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 –
INTERNATIONAL JOURNAL OF MECHANICAL ENGINEERING
  6340(Print), ISSN 0976 – 6359(Online) Volume 4, Issue 3, May - June (2013) © IAEME
                         AND TECHNOLOGY (IJMET)

ISSN 0976 – 6340 (Print)
ISSN 0976 – 6359 (Online)                                                        IJMET
Volume 4, Issue 3, May - June (2013), pp. 400-408
© IAEME: www.iaeme.com/ijmet.asp
Journal Impact Factor (2013): 5.7731 (Calculated by GISI)                   ©IAEME
www.jifactor.com




       CALCULATION OF THE UNDETERMINED STATIC REACTIONS
      FOR THE ARTICULATED PLAN QUADRILATERAL MECHANISM

                              Jan-Cristian Grigore1, Nicolae Pandrea2
                    1
                        (University of Piteşti, str. Targul din Vale nr.1, Romania)
                    2
                        (University of Piteşti, str. Targul din Vale nr.1, Romania)


   ABSTRACT

           Spatial mechanisms of the non-zero families constitute statically undetermined
   systems, the undetermination order is given by the number representing the family of the
   mechanism. The articulated plan quadrilateral mechanism, shown in this paper, is a third
   family mechanism, an undetermined static third order mechanism. This paper uses the
   relative displacement method and it establishes the mathematical model that allows the linear
   elastic calculation in order to determine the statically undetermined reactions.

   Keywords: coordinates pl ckeriene, matrix flexibility, stiffness matrix

 I.       INTRODUCTION

            If the plane mechanisms are stressed by vector components forces perpendicular to
   the motion plane or by vector component moment placed in the plane of motion, they are
   statically undetermined systems. In these cases, in order to determine the components of the
   reaction forces perpendicular to the motion plane as well as the components of the reaction
   moments in the motion plane, the linear elastic calculation shall be used. This paper shows
   these components using the relative displacement method [3], [4] and the pl ckeriene
   coordinates.

II.       NOTATIONS, REFERENCE SYSTEMS, TRANSFORMATION RELATIONS

           Forces acting on a rigid point, the velocities of the points of a rigid, the small
   movements of the points of the rigid are systems reduced to a point O ( Fig.1)at a torsion
   vector consisting of mainly f and moment vector m .
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        If the reduction is in point O0 , then equivalent torsion has components F , M
satisfying the conditions
                               F = f ; M = m + O0 O x f                                  (1)
        Considering reference systems with origins in points O , O 0 (Fig.1) and noting with
(f   x   , f y , f z , m x , m y , m z ),   (F , F , F , M
                                              x   y   z      x   ,M y,M z   )   vector      projections       ( f , m ), (F, M )
respective on the axis of the systems Oxyz , O0 XYZ then these scalar components are
pl ckeriene coordinates [4] of the torsion with representation by matrices column so;
       { f } = [ f x f y f z m x m y m z ]T ; {F } = Fx Fy Fz M x M y M z T [         (2)                        ]
               For a system of forces, f is the resultant force vector and m is moment resulting in
O , for rigid speeds f is the angular velocity of the rigid and m is the velocity of point O
and for small displacements of rigid, f is small rotation vector, and m is small movement
of the point O .




                                                          Fig. 1. System of forces

          With notations:
( X 0 ,Y0 , Z 0 ) - point coordinates O ; α i , β i , γ i , i = 1, 2, 3 , direction cosines of axes
Ox, Oy , Oz ; [G ] , [R ], [T ] translation matrices, position respectively
                        0           − Z0      Y0            α1 α 2           α3 
                                                                                                [R ]        [0]
                [G ] =  Z 0           0      − X0  ; [R ] =  β β             β 3  ; [T ] =                             (3)
                                                             1    2                        [G ]⋅ [R ]   [R ]
                                                                                                                 
                        − Y0
                                     X0        0           γ 1 γ 2
                                                                               γ3 
               Obtaining [4] transformation relations between the matrices column {F }{ f }
                                                                                     ,
                                              {F } = [T ]⋅ { f } ;{ f } = [T ]−1 ⋅ {F }                                     (4)
where

                                                  [T ]−1 =  [T ] T [0]T 
                                                                 T
                                                                           
                                                                                                                            (5)
                                                           [R ] [G ] [R ] 
                                                                T




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III.     RELATIONS BETWEEN RELATIVE MOVEMENTS AND EFFORTS AT THE
    ENDS OF A BAR

         Consider the straight bar AB (see Fig. 2), with length l , constant section, the area A ,
  modules of elasticity E , G and either Axyz local reference system, Ax , Ay the central
  principal axes of inertia of the normal section A .




                                            Fig. 2. Straight bar AB , under the influence of efforts

                Noting with           (f   A   , m A ) torsion effort from A ; with ( f B , m B ) torsion effort from B ;
  with d A , d B , m B vectors defined by the relations:

                                                                                   *
                                      d A = AA′ ; d B = BB ′ + AB x θ AB ; m B = m B + AB x f B                                               (6)
  and          noting           the          projections on the axes                          trihedral          Axyz     of vectors
  f A , m A , f B , m B ,θ A     , d A , θ B ,d B ,   respectively with                      (fAx , f Ay , f Az ) ,   (m Ax , m Ay , m Az ) ,
  (f   Bx   , f By , f Bz ) ,     (m   Bx   , m By , m Bz ),   (d   Ax   , d Ay , d Az ) ,    (θ   Ax   ,θ Ay ,θ Az ) ,   (θ   Bx   ,θ By ,θ Bz ) ,
  (d   Bx   , d By , d Bz ) , we obtain               column matrix of                 pl ckeriene           coordinates in the local
  system Axyz

                      { f A } = [ f Ax f Ay f Az m Ax m Ay m Az ]T ; { f B } = [ f Bx f By f Bz mBx mBy mBz ]T                                (7)
                       {d A } = [θ Ax θ Ay θ Az d Ax d Ay d Az ]T ; {d B } = [θ Bx θ By θ Bz d Bx d By d Bz ]T                                (8)

  and equality resulting from the equilibrium condition
                                       { f A }+ { f B } = {0}                                                                                 (9)

         Stiffness matrix [k AB ] and matrix flexibility [h AB ] = [k AB ]
                                                                                                              −1
                                                                                                                   [4] expressed in the
  reference system Axyz are given by the equalities


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                                0               0             0           Al
                                                                                2
                                                                                         0        0  
                                0               0         6I z l              0     12 I z    0 
                                                                                                    
                      E
                                0           − 6I y l          0               0       0     12 I z 
            [k AB ] = 3 ⋅  G           2
                                                                                               0  ;
                                                                                                     
                     l          I zl            0             0               0       0
                          E                          2                                              
                                0           4I z l            0               0       0    − 6I y l 
                                                                  2
                                                                                               0 
                                0               0        4I z l               0     6I z l          
                                                                          E                      
                                  0    0                   0          6
                                                                           G
                                                                                    0         0
                                                                                                  
                                                          3l                       6             
                                  0    0                                  0                   0        (10)
                                                          Iy                       Iy            
                                       3l                                                     6 
                                  0 −                      0              0        0             
                                        I                                                     Iz
                    [h AB ] = l ⋅  6 z                                                           
                             6E        0                   0              0        0          0 
                                  A                                                              
                                     2l
                                         2
                                                                                               3l 
                                  0                        0              0        0        −    
                                      Iz                                                      Iz 
                                                         2l
                                                               2
                                                                                    3l            
                                  0    0                                  0                   0 
                                  
                                                          Iy                       Iy            
                                                                                                  
       where I y , I z are the principal central moments of inertia of normal areas on axis
Ax and I x is defined by equality

                                                 Ix = I y + Iz                                           (11)
       With these notations [4] to obtain equalities

                          { f A } = [k AB ] ⋅ {d AB } ; {d AB } = [k AB ] ⋅ { f A }                      (12)
where {d AB }   is the relative displacement

                                            {d AB }= {d A }− {d B }                                      (13)

        Switching to the reference OXYZ is done using relations (3), (4), (5) and these
equalities are obtained
                  {D A }= [TAB ]{d A } ; {DB }= [TAB ]{d B } ; {D AB }= {D A }− {DB }
                  {FA }= [TAB ]{ f A } ; {FB }= [TAB ]{ f B }                         (14)
                  {K AB }= [TAB ]{k AB }[TAB ] ; [H AB ] = [TAB ]{k AB }[TAB ]
                                              −1                              −1


                             {FA }= [k AB ]{D AB } ; {D AB }= [H AB ]{FA }            (15)



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  International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 –
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IV.       CALCULATION OF THE REACTIONS

         Considering articulated plan mechanism ABCD from Fig. 3 acted by external forces
  and moments [4] which are distributed in points B, C , D and are marked by column matrix
  of pl ckeriene coordinate. Noting generally with {E } efforts at the ends of the bars, and
  with {R} reactions by isolating bars and nodes B, C , D obtain formal scheme from fig. 4, for
  which the following equations of equilibrium can be written




                                       Fig. 3. Articulated quadrilateral plan mechanism

                  {E A }+ {E B } = {0} ; {E B }+ {E B } = {PB } ; {E B }+ {EC } = {0} ;
                               1                 1               2                 2       2
                                                                                                     (16)
                   {EC }+ {EC }= {PC } ; {EC }+ {E D }= {0}
                       2           3                     3           3

  from which resulting

                             {E B 2 }= {E A }+ {PB }; {EC 3 }= {E A }+ {PB }+ {PC }                  (17)
             {RB }= {PB }+ {E A }; {RC }= {PB }+ {PC }+ {E A }; {RD }= {PB }+ {PC }+ {PD }+ {E A }   (18)
  and using relations (15) the following expressions are derived:

                    {E A }= [K AB ]{{D A }− {DB }}; {E A }+ {PB } = [K BC ]{{DB }− {DC }}
                                                     1                                 2       2
                                                                                                     (19)
                    {E A }+ {PB }+ {PC } = [K CD ]{{DC }− {DD }}
                                                             3             3

  or

                   {D A }− {DB }= [H AB ]{E A } ; {D B }− {D C }= [H BC ]{{E A }+ {PB }}
                                   1                             2             2
                                                                                                     (20)
                    {D C }− {D D }= [H CD ]{{E A }+ {PB }+ {PC }}
                           3           3

         { } { }{ }
  where D Bi , D Ci , D Di , i = 1, 2 , are the movements at the ends of indexed bars (Fig. 3)
  with indices 1, 2, 3 .




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           Fig. 4. Formal mechanism quadrilateral scheme, the representations efforts
                                     and reaction forces

         Consider the insertion of A , ({D A } = {0}) in the linear elastic calculation and in this
way the movements of the other sections relate to the section of A . Noting with
{U B },{U C }, {U D } column matrices attached to kinematic couplings of rotation [6] to obtain
expressions

                                  {U B } = [0      0 1 YB − X B                     0] ,
                                                                                        T


                                  {U C } = [0      0 1 YC − X C                     0] ,                  (21)
                                  {U D } = [0 0 1 0 − X D 0]
                                                                                    T


and noting with ξ B , ξ C , ξ D    rotations of the joints, the following equalities are derived:

                                        {D }= {D }+ ξ {U } ;
                                             B2             B1       B       B

                                        {D }= {D }+ ξ {U } ;
                                             C3             C2       C       C                            (22)
                                        {0} = {D }+ ξ {U }
                                                       D3        D       D

        By adding relations (20), taking into account the equations (22) and using notations

                  [H AD ] = [H AB ] + [H BC ] + [H CD ] ;
                                                                                                 ξ B 
                   [K AD ] = [H AD ] ; [U ] = [{U B } {U C }
                                        −1
                                                                                 {U D }]; [ξ ] = ξ C 
                                                                                                        (23)
                                                                                                 ξ D 
                                                                                                  
                   ~
                  {∆}= [H ]{P }+ [H ]{{P }+ {P }}
                            BC      B             CD        B        C
obtain the equation
                                     [E A ] = [K AD ][U ]{ξ } − [K AD ]{∆}
                                                                        ~
                                                                                                          (24)
        Using the notations


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                                 ~
                                {U } = [Y − X 0 0 0 0];
                                  B
                                      T
                                               B                  B
                                 ~
                                {U } = [Y − X 0 0 0 1];
                                 C
                                      T
                                               C                  C
                                                                                                               (25)
                                 ~
                                {U } = [Y − X 0 0 0 1];
                                  D
                                      T
                                               D                  D

                                    {U }= [{U } {U } {U }] ;
                                     ~      ~    ~    ~
                                                        B               C          D
                                                                                       T
                                                                                                               (26)
knowing [6] that reactions satisfy the relations

                      ~               ~               ~
                     {U } {R } = 0 ; {U } {R } = 0 ; {U } {R } = 0
                        B
                            T
                                 B                     C
                                                            T
                                                                    C                  D
                                                                                           T
                                                                                               D
                                                                                                               (27)
taking into account the equality relations (18) and the notation

                                                   ~ T
                                          
                                          
                                                   { }
                                                  U B {PB }       
                                                                  
                                                                                                               (28)
                                              ~
                                                 { }
                                                  T
                                {PT } = −  U C {{PB }+ {PC }} 
                                           U T {{P }+ {P }+ {P } 
                                            ~
                                               { }
                                           D                  D }
                                                   B    C
                                                                  
obtain the expression

                                                [U ]{E
                                                 ~
                                             A } = {P }
                                                     T                                                         (29)
with that of (24) the matrix of rotations in the joints is deduced

                                 {ξ } = [U ⋅ K AD ⋅ U ]−1 {{PT }+ [U ][K AD ]⋅ {∆}}
                                         ~                         ~            ~
                                                                                                               (30)
and then from (24) the reaction is deduced form A , {R A } = {E A }
       Relations (30), (29), (24) can be expressed in a simpler form if the following
notations are made
                            {E A } = [E Ax     E Ay          E Az       M Ax    M Ay       M Az   ]T
                                                                                                               (31)
                            {∆} = [θ x
                             ~     ~      ~
                                          θy
                                                   ~
                                                   θz
                                                             ~
                                                             ∆x
                                                                      ~
                                                                      ∆y       ]
                                                                              ~ T
                                                                              ∆z

             0      0  K13 K14 K15     0 
             0      0  K 23 K 24 K 25  0 
                                                K13 K14 K15             K 31 K 32 K 36 

             K 41
                   K
                     42
                         0    0    0
                                       K 46 
                                              1)
                                                 
                                                            [ ]
  [K AD ] = K 31 K 32 0 0 0 K 36  ; K (AD =  K 23 K 24 K 25  ; K (AD = K 41 K 42 K 46 
                                                                  
                                                                       2)
                                                                                           
                                                                                                       [ ]     (32)
             K 51 K 52 0     0    0   K 56      K 63 K 64 K 65          K 51 K 52 K 56 
             0      0  K 63 K 64 K 65  0 
                                           
                        YB        − X B 1           1                                     1           1 
                  [A] = YC
                        
                                   − XC 1  ; [B ] =  Y
                                                     B
                                                                                            YC          YD 
                                                                                                              (33)
                        YD
                                  − X D 1
                                                    − X B
                                                                                          − XC        − XD
                                                                                                           
and then it follows
                                                                                        ~
                                                                                      θz 
                                                                                   −1  ~ 
                        {ξ } = [B]−1 ⋅ [K AD ]
                                          (1)               −1
                                                               ⋅ [A] ⋅ {PT } + [B ]  ∆ x 
                                                                    −1
                                                                                                               (34)
                                                                                        ~
                                                                                      ∆ y 
                                                                                       




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                                    E AX 
                                    E  = [A]−1 ⋅ {P } ;
                                    AY              T

                                    M AZ 
                                         
                                                         ~                                      (35)
                                    E AZ              θ x 
                                    M  = − K (2 ) ⋅ θ 
                                                    [    ~
                                                             ]
                                       AX       AD     y
                                                         ~
                                                        ∆ z 
                                    M AY 
                                                       
         Calculation of the other reactions {R B }, {RC }, {RD } are made using the relations (18).

V.       CONCLUSION

          The matrix {PT } defined by the relation (28) depends only on the components of forces
  {PB }, {PC }, {PD } compatible with the movement of the mechanism, respectively on the
  components PBX , PBY , PBZ and the analogues ones. The statically determined components of
  the reaction {R A } = {E A } are given by the first relation (35) and as expected they depend on
  the components compatible with external forces movement and they do not depend on the
  stiffness of the elements of the mechanism.
                        ~
                    {}
          The matrix ∆ is the result matrix partitions

                          {∆ }= [θ~
                           ~
                             1    z
                                      ~
                                      ∆x
                                           ~
                                           ∆y   ] ; {∆ }= [θ~
                                                T    ~
                                                         2       x
                                                                     ~
                                                                     θy
                                                                          ~
                                                                          ∆z   ]
                                                                               T
                                                                                               (36)

                          ~
                       { }
           The matrix ∆ 1 with components in the plane of motion and depending on the
                                                                                      ~
  components compatible with movement PBX , PBY , M BZ and the analogues, and ∆ 2 with   { }
  incompatible components with moving parts and is not compatible with motion-dependent
   PBZ , M BX , M BY and analogues.
           It follows from this and from the second relation (35) that statically indeterminate
  reactions depend exclusively on the components of the external forces incompatible with
  moving parts.
           Concerning the matrix {ξ }, movements of kinematic couplings resulting from the set
  and relation (35) it depends on the stiffness of elements as well as on the components of the
  external forces compatible with movement.
           Based on the relations established in this paper we can develop an algorithm and a
  program for numerical calculation of statically indeterminate components, an objective that
  will result in a subsequent paper.

VI.      ACKNOWLEDGEMENTS

         This paper is a continuation of research conducted under the grant "PD -683 / 2010",
  and we want to thanked the Romanian Government (UEFISCDI), which certainly those
  funding research.



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International Journal of Mechanical Engineering and Technology (IJMET), ISSN 0976 –
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REFERENCES

[1] Bulac, I, Grigore, J.-C., (2012) Mathematical model for calculation of liniar elastic
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[2] Buculei, M., Băgnaru, D., Nanu, Gh., Marghitu, D.,(1994) Computational methods to
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[3] Bulac, I, Grigore, J.-C., (2012) Mathematical model for calculation of liniar elastic
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[4] Pandrea, N.,(2000) Elements of solid mechanics coordinate pl ckeriene, Romanian
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[5] Popa, D.,(1997) Contributions to the study of dynamic elastic-bar mechanisms. Thesis,
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[6] Popa, D.,(1997) Contributions to the study of dynamic elastic-bar mechanisms. Thesis,
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[7] Voinea, R., Pandrea, N.,(1973) Contributions to a general mathematical theory of
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[8] Chikesh Ranjan and Dr R. P. Sharma, “Modeling, Simulation & Dynamic Analysis of
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    ISSN Online: 0976 – 6359.




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