Milan S. Cajić
PhD Student
University of Belgrade
Faculty of Mechanical Engineering
Research Assistant
Mathematical Institute of the SASA (Serbian
academy of Sciences and Arts)
Mihailo P. Lazarević
Full Professor
University of Belgrade
Faculty of Mechanical Engineering
Determination of Joint Reactions in a
Rigid Multibody System, Two Different
Approaches
In this paper two different methods for determination of frictionless joint
reaction forces and moments are presented. The considered multibody
system has an open kinematic chain structure. The first method refers to the
determination of resultant joint reaction forces and moments based on the
Rodrigues approach suitable for computation in a symbolic form. The
second method presented is the method based on the so-called vectors of
the body mass moments and vector rotators coupled for a pole and
oriented axes. Both approaches are presented and discussed on the threelike rigid multibody system.
Keywords: multibody dynamics, resultant joint reaction forces and
moments, vectors of body mass moments, vector rotators
1.
INTRODUCTION
Determination of joint reactions is of great importance
in dynamics of multibody systems, machines and
mechanisms. In this paper, the Coulomb friction forces
are not considered but it is necessary if we want to get a
more realistic picture of the dynamics of a multibody
system (see e.g. [1-4]).
In the recently published paper [5] the broad
overview of research papers that deals with the
determination of joint reactions have been presented.
The papers based on Newton-Euler formalism with
recursive and non-recursive kinematic and dynamic
relations are cited there. The main part of that paper is
determination of joint reactions based on fictitious
bodies method and general method based on Lagrange’s
equations. Therefore, for better understanding of this
problem the reader is referred to the [5] and [6].
In [8-12] a different approach to the problem of
determination of joint reactions and kinetic pressures is
used. By introducing the vector method based on the
vectors of the body mass moments [8, 9] and vector
rotators coupled for pole and oriented axes it is possible
to obtain vector expressions for determination of joint
reactions and kinetic pressures for a rigid body dynamics.
In this paper, both methods are illustrated on the threelike rigid multibody system. Expressions for the linear and
angular momentum as well as expressions for their
derivatives are derived. The procedure for determination of
the resultant joint reaction forces and moments is presented
and some comments on both methods are given.
2.
determination of the resultant joint reaction forces and
moments is given. If the robotic system has the form of
a multibody system with open kinematic chain structure
, then we can denote the first and the fixed base body
with (V1 ) and the following bodies in the system we
denote with subscripts incremented for one with regard
to the subscript of its preceding adjacent body
(V2 ,....Vk ,...Vi ,Vi +1 , Vi + 2 ,...,Vn ) . Also, the geometry of
the system has been defined by unit vectors
ei , i = 1, 2,..., j ,.., n where unit vectors ei describe the
axis of rotation (translation) of the i -th segment with
respect to the previous segment and as well as vectors
ρi and ρii , (see [6, 7]). In a general case, the
configuration of the multibody system can be defined
by the vector of joint (internal) generalized coordinates
(
)
q of the dimension n, ( q ) = q1 , q 2 ,..., q n where q are
the relative angles of rotation (in the case of revolute
joints) and the relative displacements (in the case of
prismatic joints). If all the variables qi, i = 1, 2, ..., n are
zero, it is said that the robotic system is in the reference
configuration (position).
DETERMINATION OF JOINT REACTIONS USING
RODRIGUES APPROACH
In this section, formulation of a general method for
Received: June 2014, Accepted: May 2015
Correspondence to: Mihailo P. Lazarević
Faculty of Mechanical Engineering,
Kraljice Marije 16, 11120 Belgrade 35, Serbia
E-mail: [email protected]
doi:10.5937/fmet1602165C
© Faculty of Mechanical Engineering, Belgrade. All rights reserved
Figure 1. Rigid multibody system with the open kinematic
V j  )
 
To determine the resultant joint reaction force and
the moment in some k-th joint let us first consider the
equation for linear momentum of all segments of a rigid
multibody system (see Figure 1) behind the k-th
segment as:
chain structure ( Vk  →
FME Transactions (2016) 44, 165-173 165
K=
∑
i:k ( ≤i )
Ki = ∑ mi vCi
(1)
i
where vCi and mi denote the velocity of the mass
center and the mass of the body ( Vi ). After
diferentiation of (1) we obtain:
dK
(2)
= ∑ mi aCi = FR( rv ) + ∑ mi g + ξ k Fuk
dt
i
i
From (2) we can obtain a formula for determining of
the resultant joint reaction force FR( rv ) in the k-th joint as:
FR( rv ) = ∑ mi ( aCi − g ) − ξ k Fuk ek
(3)
d( j)
as well as square matrix  e j  represents dual object


of vector:
 e ( j )  = e , e , e
 j 
ξj ηj ζ


(

vCi =
i
aCi =
i
∑ Tα (i ) qα +
α =1
α
0
aCi
β
where q , q
i
∑
i
∑ Γαβ (i ) qα q β
Resultant joint reaction force FR( rv ) with respect to
inertial fixed reference frame Oxyz (null base) can be
calculated as:
{ (( ) )}
0
FR rv
( 0)
= ∑ mi  aCi −  A0, i  { g} 


i
−ξ k Fuk  A0, k  {ek } ,
(6)
where the appropriate Rodrigues matrices of trans–
formation are:
2
(
( ))
 A j −1, j  = [ I ] +  edj ( j )  1 − cos q j




∑
eξ j
α =1
i
=
α =1
( 0)
Tα i qα ,
0
Tα i qα +
{
i
i
∑∑
{
α =1 β =1
(0)
Γαβ
(i )
}
(9)
α β
q q
()
}
(10)
()
(b )
can be obtained in the
where quasibase vectors T
b (i )
following form
(b )
 i
Tb i = ξb ebd   ∑  Ab,k  { ρ kk } + ξ k q k {ek }
()
 
 k =b
+  Ab,i  { ρi } + ξb {eb } ,
(
(11)
)}
α =1 β =1
quasibase vectors (for details see [6]) and vectors
Γαβ ( i ) are in the following form:
Γαβ ( i ) = ξα eα × Tβ ( i ) , ∀α ≤ β ,
(5)
Γαβ ( i ) = ξα eβ × Tα ( i ) , ∀α > β ,
Γαβ ( i ) = 0,
∀α > i
0
(8)
a = α, β , b = α, β,
(0)
(b )
Γαβ i = ξ a ea × Tb i
(4)
are generalized velocities, Ti (α ) are
eη j 

−eξ j 

0 
j
and vectors Γαβ ( i ) with respect to the null base are
∑ Tα (i ) qα ,
α =1
i
{ } { ( )}
{ ( ) } ∑ { ( ()) }
( 0)
vCi =
where aCi and g are acceleration of mass center and
case of a prismatic joint, where ξ k = 1 − ξ k is also
introduced. The velocity vCi and acceleration aCi of
mass center of i-th segment can be evaluated from the
following expressions:
−eζ
In a similar maner, the velocity and acceleration of
the center of mass with respect to the null base are:
i
acceleration of gravity, respectively. Also, in (3) Fuk is
an actuator force, ek denotes unit vector describing the
axis of rotation (translation) of the k-th segment with
respect to the previous segment, whereas parameter ξ k
deter–mines the type of the k-th joint. The parameter
ξ k = 0 if the considered joint is revolute and ξk = 1 in
 0

 eζ j
 
 −eη j
ed ( j )  =
 j 
),
j
Further, angular momentum of i-th segment with
respect to the pole in k-th joint of a multibody system is
given in the form:
LOK ( i ) = LCi + Ok' Ci × mi vCi
(12)
Total angular momentum of all segments behind the
k-th joint is:
LOK = ∑  LCi + Ok' Ci × mi vCi 
(13)


i
After differentiation of the last expression, we obtain
dLOK
dLCi dOk' Ci
=∑
+
× mi vCi
dt
dt
dt
(14)
i
+Ok' Ci × mi aCi ,
and
dOk' Ci = vCi − vOk , Ri( k ) = Ok' Ci
dt
(15)
( )
where LCi is angular momentum with respect to the
mass centre of the i-th body, we can present (14) in the
following form
166 ▪ VOL. 44, No 2, 2016
FME Transactions
+  e j

d( j) 
j
 sin q ,
(7)
 dLCi dLOK

= ∑
− vOk × mi vCi + Ri( k ) × mi aCi  (16)


dt

i  dt
Resultant reaction moment M OkR of all forces
acting on segments behind k-th joint is:
dLOk M OkR =
(17)
+ vOk × ∑ mi vCi
dt
i
Equations (14) and (15) yields :
 dLCi 
M Ok ( rv ) = ∑ 
+ Ri( k ) × mi ( aCi − g ) 
 dt


i 
−ξ k M uk ek .
(18)
When we take into account the actuator torque M uk
and torque due to the gravitational forces mi g we have
the final form of the resultant joint reaction moment in
the k-th joint as
 dLCi 
M Ok ( rv ) = ∑ 
+ Ri( k ) × mi ( aCi − g ) 
 dt

(19)

i 
−ξ k M uk ek .
On the other side, first derivative dLCi dt can be
obtained in the final form , [6] as:
dLCi  d 
= ωi  J Ci {ωi } +  J Ci {ε i }
(20)
 
dt
where ωi , ε i and J Ci are:
1) angular velocity of the i-th body in vector form
ωi =
i
∑ ξα eα qα ,
(21)
α≤i
α =1
τ ξ 
 
{τ } = τη  ,
 
τ ζ 
 0

τ  =  τ
   ζ
 −τη
d
−τ ζ
0
τξ
τη 

−τ ξ 

0 
(25)
For more details, the reader is reffered to [6] and [7].
2.1 Determination of joint reaction forces and
moments for the three-like rigid multibody system
When we consider the simple three rigid body system,
(Fig. 2) where the linear momentum of all bodies in the
system is:
3 3
K = ∑ Ki = ∑ mi vCi
i =1
(26)
i =1
where velocities of body mass centers with respect to
the local inertial coordinate systems are
vC1 = T(1)1q1; vC 2 = T(1)2 q1 + T( 2 )2 q 2 ;
(27)
vC 3 = T(1)3 q1 + T( 2 )3 q 2 + T( 3)3 q 3
and quaibase vectors
T(1)1 = ξ1e1 × [ ρ11 + ρ1 ] + ξ1e1 ,
T(1)2 = ξ1e1 × [ ρ11 + ρ22 + ρ2 ] + ξ1e1 ,
T(1)3 = ξ1e1 ×  ρ11 + ρ22 + ρ33 + ρ3  + ξ1e1 ,
T( 2 )2 = ξ 2 e2 × [ ρ22 + ρ2 ] + ξ 2 e2 ,
T( 2 )3 = ξ 2 e2 ×  ρ22 + ρ33 + ρ3  + ξ 2 e2 ,
T( 3)3 = ξ3e3 ×  ρ33 + ρ3  + ξ3e3 ,
(28)
as well as in a matrix form
ωi ξ 


{ωi } = ωiη  ,


ωi ζ 
ω d
 i
 0

=ω
  iζ
 −ωiη
−ωi ζ
0
ωi ξ
ωiη 

−ωi ξ  (22)

0 
2) angular acceleration of the i-th body is
εi =
i
i
∑ ξα eα qα + ∑
α =1
α
∑ ξα ξ β eβ × eα qα q β
α =1 β =1
(23)
α ≤ i, β ≤ α
3) mass inertia tensor
J Ci of the i-th body is
τ 2 + τ 2
η ζ
def
2
d
 −τ τ
 J Ci  = −  τ  dm =
 
 η ξ
(V )
(V )  −τ τ
 ζ ξ
∫
Figure 2. Three-like rigid multibody system with vectors
using Rodrigues approach
∫
−τ ξ τη
τ ζ2 + τ ξ2
−τ ζ τη
−τ ξ τ ζ 

−τητ ζ dm

τ ξ2 + τη2 

 Jζ
− Jξη − Jξζ 


Jη
− Jηζ 
(24)
 J Ci  =  − Jηξ


Jξ 
 − Jζξ − Jζη
where τ is the position vector with respect to the body
mass centre
FME Transactions
After derivation of a linear momentum by the
following formula:
3
3
dK
= ∑ mi aCi = FR( rv ) + ∑ mi g
(29)
dt i =1
i =1
we can easily find the resultant joint reaction force for
the first and fixed joint of the three body system as
3
FR( rv ) = ∑ mi ( aCi − g i ) − ξ k Fuk ek
(30)
i =1
in case of revolute joint we have (30) without the last
term in the form:
VOL. 44, No 2, 2016 ▪ 167
FR( rv ) = m1 ( aC1 − g1 ) +
+ m2 ( aC 2 − g 2 ) + m3 ( aC 3 − g3 )
(31)
Accelerations of body mass centers are
aC1 = T1(1) q1 + Γ11(1) q1q1
aC 2 = T2(1) q1 + T2( 2 ) q2 + Γ11( 2 ) q1q1 + Γ12( 2 ) q1q 2
+Γ 21( 2 ) q 2 q1 + Γ 22( 2 ) q 2 q 2 ,
aC 3 = T3(1) q1 + T3( 2 ) q2 + T3( 3) q3 + Γ11( 3) q1q1 + (32)
+Γ12( 3) q1q 2 + Γ13( 3) q1q 3 + Γ 21( 3) q 2 q1 +
+Γ 22( 3) q 2 q 2 + Γ 23( 3) q 2 q 3 + +Γ31( 3) q 3 q1 ,
+Γ32( 3) q 3 q 2 + Γ33( 3) q 3 q 3
and Γαβ ( i ) vectors are
Γ11(1) = ξ1e1 × T1(1) , Γ11( 2 ) = ξ1e1 × T1( 2 )
Γ12( 2 ) = ξ1e1 × T2( 2 ) = Γ 21( 2 )
Γ 21( 2 ) = ξ1e1 × T2( 2 ) , Γ 22( 2 ) = ξ 2 e2 × T2( 2 )
Γ11( 3) = ξ1e1 × T1( 3) , Γ a 2 + b 2 12( 3) =
= ξ1e1 × T2( 3) = Γ 21( 3)
Γ13( 3) = ξ1e1 × T3( 3) = Γ31( 3) , Γ 21( 3) = ξ1e1 × T2( 3)
Γ 22( 3) = ξ 2 e2 × T2( 3) , Γ 23( 3) = ξ 2 e2 × T2( 3)
Γ31( 3) = ξ1e1 × T3( 3) , Γ32( 3) = ξ 2 e2 × T3( 3) = Γ 23( 3)
Γ33( 3) = ξ3e3 × T3( 3)
(33)
Derivative of angular momentum is given by:
3  dL
dLOK

= ∑  Ci − vOk × mi vCi + Ri( k ) × mi aCi 


dt

i =1  dt
In case the first joint is revolute, then for the
resultant joint reaction moment we have
 dL

M Ok ( rv ) =  C1 + R1( k ) × m1 ( aC1 − g ) 
 dt



 dL

(37)
+  C 2 + R2( k ) × m2 ( aC 2 − g ) 
 dt



 dLC 3 
+
+ R3( k ) × m3 ( aC 3 − g )  − ξ k M uk ek ,
 dt



where dLCi dt for the three body system can be
dermined as
dLC1  d 
= ω1  J C1 {ω1} +  J C1 {ε1} ,
 
dt
dLC 2  d 
(38)
= ω2  J C 2 {ω2 } +  J C 2 {ε 2 } ,
 
dt
dLC 3  d 
= ω3  J C 3 {ω3 } +  J C 3 {ε 3 }
 
dt
where ωi and ωid for i = 1, 2, 3 are determined by using
the (21) and (22) and ε i are
(34)
(
2
( )
( ) ( q )
+Γ 22
3
2
+ 2Γ 23( 3) q 2 q 3 + Γ33( 3) q 3
( )
2

− g3  .

Total angular momentum of a three body sytem with
a pole in the first joint is:
3  L + O' C × m v  = L + R × m v
LOK =
 Ci
k i
i Ci 
C1
1 C1
1( k )


i =1
+ LC 2 + R2( k ) × m2vC 2 + LC 3 + R3( k ) × m3vC 3 .
∑
168 ▪ VOL. 44, No 2, 2016
Even if expressions for determination of joint
reactions are given here explicitly, it is possible to use
some of the software for symbolic computation (e.g.
Wolfram Mathematica), or to make computation
algorithm using some of the programming languages (e.g.
see [6]), and to determine in an easy way joint reactions
for arbitrary number of rigid bodies in a system.
3.
+ 2Γ12( 3) q1q 2 + 2Γ13( 3) q1q 3
2
+ξ 2ξ1e1 × e2 q 2 q1 + ξ 2ξ 2 e2 × e2 q 2 q 2 ,
ε 3 = ξ1e1q1 + ξ 2 e2 q2 + ξ3e3 q3 + ξ1ξ1e1 × e1q1q1 (39)
+ξ 2ξ1e1 × e2 q 2 q1 + ξ 2ξ 2 e2 × e2 q 2 q 2
+ξ3ξ1e1 × e3 q 3 q1 + ξ3ξ 2 e2 × e3 q 3 q 2
+ξ3ξ3e3 × e3 q 3 q 3 ,
( )
+Γ11( 3) q1
ε 2 = ξ1e1q1 + ξ 2 e2 q2 + ξ1ξ1e1 × e1q1q1
( )
( )
ε1 = ξ1e1q1 + ξ1ξ1e1 × e1q1q1 = ξ1e1q1 ,
Finaly, after using the (32) and (33) we obtain:
2 

FR( rv ) = m1  T1(1) q1 + Γ11(1) q1 − g1 


2

+ m2  T2(1) q1 + T2( 2 ) q2 + Γ11( 2 ) q1

2 
+2Γ12( 2 ) q1q 2 + Γ 22( 2 ) q 2 − g 2 

1
2
3
+ m3 T3(1) q + T3( 2 ) q + T3( 3) q +
(36)
(35)
DETERMINATION OF JOINT REACTIONS BY
APPLYING THE METHOD BASED ON VECTORS
OF BODY MASS MOMENTS AND VECTOR
ROTATORS
This method was first introduced in [8] where vectors of
body mass moments were introduced by definitions. In
monograph [9], the more extensive study of this method
and nonlinear dynamics analysis of the heavy rotors and
gyro-rotors have been carried out. This method was
applied in [10-12] for vector analysis properties of
vector rotators of rigid body dynamics with coupled
rotations around axes without intersection as well as
vector analysis of nonlinear oscillations of gyro-rotors.
FME Transactions
3.1
Vectors of body mass moments
Here, we define three kinetic vectors fixed to a certain
point and axis passing through the given rigid body
point (see Figure 3). The rotation axis of a body is
passing through a revolute joint that connects two
bodies in the system and it is oriented by the unit vector
ni where subscript i is the same as the subscript of the
corresponding body.
the axial moment of the body mass inertia and the other
two components represent the deviational moments:
(O )
O O (O ) ℑn i = J n i n + D(nui ) u + D(nv i ) v
(43)
i
i
(O )
where J n i is:
i
( Oi ) =
J n
i
2
 ( Oi ) , n 

i

∫∫∫ [n, ρ] dmn =  ℑn
V
(44)
The rigid body deviation moment vector at the point
for the axis oriented by the unit vector has the following
form:
(O )
Dn i =
i
  ( Oi )  
, n   (45)
i

∫∫∫ n,  ρ, [n, ρ] , n dm = n, ℑn
V
3.2 Model of a three-like rigid body system
Figure 3. Vectors of body mass moments for the single
body rotation around the fixed axes
(O )
We give a definition of the vector M n i of
i
the
body mass at the point Oi for the axis oriented by the
unit vector ni [8-9]:
(O )
M n i := ∫∫∫ ndm = Mn , dm = σ dV
(40)
i
V
(O )
Then we define vector Sn i of the body mass static
i
(linear) moment at the point Oi for the axis oriented by
the unit vector ni in the form [8-9]:
(O )
Sn i := ∫∫∫ [ n , ρ ]dm = n , ρc M, dm = σ dV
In this section we observe the same system as in the
previous one but with different notation for the body
vectors (see Figure 4). Allow us first to consider rotation
of the rigid body (V1 ) around fixed axis oriented with the
unit vector n1 with angular velocity ω1 = ω1n1 . In a
similar way angular velocities around moving axes
oriented with unit vectors n2 and n3 are ω2 = ω2 n2
and ω3 = ω3 n3 , respectively. Joints, whose positions with
respect to the first joint O1 are determined with r12 and
r23 vectors, are denoted with O2 and O3 and they
coincide with points which are the shortest orthogonal
distance between axes oriented with unit vectors n1 , n2
and n3 . Axes of rotation are without intersections.
(41)
i
V
(O )
The third vector ℑn i of the body mass inertia at
i
the point Oi for the axis oriented by the unit vector ni is
defined as [8-9]:
(O )
(42)
ℑn i := ∫∫∫  ρ, [ n , ρ ]dmn, dm = σ dV
i
V
where ρ is the position vector of the elementary body
mass and ρc is the position vector of the centre of body
mass with respect to the pole Oi . The body mass inertia
(O )
moment vector ℑn i can be decomposed into three
i
components: the component
( Oi ) which is co-linear
J n
i
with the axis and two other components D(nui ) and
O
D(nv i ) in the directions u and v which are normal to the
orientation axis n . The co-linear component represents
O
FME Transactions
Figure 4. Three-like rigid multibody system with vectors
used in a vector method
Here ρic and ρi , i = 1, 2, 3 are position vectors of the
body mass centre and elementary body mass with
respect to the joint of i -th body, respectively. Position
vectors of elementary mass with origin in joint O1 are
introduced with vectors ri , i = 1, 2, 3 in the form:
(46)
r1 = ρ1 , r2 = r12 + ρ2 , r2 = r12 + r23 + ρ3
and velocities of mass particles vi , i = 1, 2, 3 in the form
v1 = [ω1 , ρ1 ] , v2 = [ω1 , r12 ] + [ω1 + ω2 , ρ 2 ] ,
v3 = [ω1 , r12 ] + ω1 + ω2 , r23  + ω1 + ω2 + ω3 , ρ3 
(47)
VOL. 44, No 2, 2016 ▪ 169
3.3 Vector equations of linear momentum and
angular momentum
+ ω1  ρ3C , n1 , r23  M 3 + ω2  ρ3C , n2 , r23  M 3
( O )( 3)
( O )( 3)
( O )( 3)
,
+ω1ℑ n
+ ω2 ℑ n
+ ω3 ℑ n
3
Using the basic definition of linear momentum and angular
momentum and basic equations for velocity of mass
particles, the following vector equations can be written:
1) For linear momentum in the form
K = ∫∫∫ v1dm1 + ∫∫∫ v2 dm2 + ∫∫∫ v3 dm3
V
V
1
V
2
3
( O )(1)
( O )( 2 )
= ω1S n
+ ω1 [ n1 , r12 ] ( M 2 + M 3 ) + ω1S n
1
2
( )
( O )( 2 )
( )
1
+ ω2 S
2
n
( )
1
+ ω1 n1 , r23  M 3 + ω2 n2 , r23  M 3
(48)
( )
1
V
3
V2
3
( O )( 2 )
ℑ n 2
= ∫∫∫ ρ 2 , [ n2 , ρ 2 ] dm2 ,
( 2)
( )
( )
( )
( Oi )
where Sn , i = 1,2,3 are body linear mass moments for
2
2
V2
( O )( 3)
ℑ n 3
= ∫∫∫ ρ3 , n1 , ρ3  dm3 ,
( 1)
i
corresponding points and axes which are written as:
( O )(1)
= ∫∫∫ [ n1 ,ρ1 ] dm1 = n1 ,ρ1C  M1 ,
S n
1
V3
V
V3
( O )( 2 )
= ∫∫∫ [ n1 , ρ 2 ] dm2 ,
S n
( O )( 3)
ℑ n 3
= ∫∫∫ ρ3 , n3 , ρ3  dm3
( 3)
2
( )
1
(51)
( O )( 3)
ℑ n 3
= ∫∫∫ ρ3 , n2 , ρ3  dm3 ,
( 2)
1
( )
3
( O )( 2 )
ℑ n 2
= ∫∫∫ ρ 2 , [ n1 , ρ 2 ] dm2 ,
( 1)
2
3
3
( )
2
( O j )( j )
, i , j = 1, 2, 3 are corresponding rigid body
where ℑ n
( i)
mass inertia moment vectors for axes oriented by direc–
tions of component rotations through the joints Oi , i = 1,
2, 3 on self-rotating axes written in the following form:
( O )(1)
ℑ n 1
= ∫∫∫ ρ1 , [ n1 , ρ1 ] dm1 ,
( 1)
( O )( 3)
( O )( 3)
( O )( 3)
+ ω1S n
+ ω1S n
+ ω1S n
,
1
3
( )
V
V3
( O )( 2 )
= ∫∫∫ [ n2 , ρ 2 ] dm2 ,
S n
2
( )
1
V
(49)
( O )( 3)
= ∫∫∫ n1 , ρ3  dm3 ,
S n
( )
3
1
V
( O )( 3)
= ∫∫∫ n2 , ρ3  dm3 ,
S n
( )
3
2
V
( O )( 3)
= ∫∫∫ n3 , ρ3  dm3
S n
( )
3
3
V
2) For angular momentum the expression is in the form
L = ∫∫∫ [ r1 , v1 ] dm1 + ∫∫∫ [ r2 , v2 ] dm2 + ∫∫∫ r3 , v3  dm3
V
V
V
( O )(1)
( O )( 2 ) 
= ω1ℑ n
+ ω1  r12 , [ n1 , r12 ] M 2 + ω1  r12 , S n
( )
( ) 

( O )( 2 ) 
+ω2  r12 , S n
+ ω1  ρ 2C , [ n1 , r12 ] M 2

(
)


( O )( 2 )
( O )( 2 )
+ω1ℑ n
+ ω2 ℑ n
1
2
1
1
2
2
2
2
( )
( )
1
2
+ω1  r12 , [ n1 , r12 ] M 3 + ω1  r12 , n1 , r23  M 3
( O )( 3) 
+ω2  r12 , n2 , r23  M 3 + ω1  r12 , S n
( ) 

3
1
( O )( 3) 
 r , S ( O )( 3) 
+ω2  r12 , S n
+
ω
(50)
3
12 ( n )
( ) 



+ω1  r23 , [ n1 , r12 ] M 3 + ω1  r23 , n1 , r23  M 3
( O )( 3) 
+ω2  r23 , n2 , r23  M 3 + ω1  r23 , S n
( ) 

( O )( 3)  + ω  r , S ( O )( 3) 
+ω2  r23 , S n
3  23 ( n )
( ) 



+ω1  ρ3C , [ n1 , r12 ] M 3 + ω1  ρ3C , n1 , r23  M 3
3
2
3
3
3.4
Derivatives of linear momentum and angular
momentum
We take into account the equation for linear momentum
(48) and find derivative of parts, after rearranging the
terms we obtain the following vector expression
( O )(1)
( O )(1)
dK
= ω1S n 1
+ ω12  n1 ,S n 1  + ω1 [ n1 , r12 ] M 2
(
)
(
)
1
1
dt


( O2 )( 2 )
( O )( 2 ) 
2 
+ω1  n1 , [ n1 , r12 ] M 2 + ω1S n
+ ω12  n1 , S n 2
( 1)
( 1 ) 

( O )( 2 )
 ( O )( 2 ) 
( O )( 2 ) 
+ω 2 S n 2
+ ω22  n2 , S n 2
+ 2ω1ω2  n1 , S 2

( 2)
(
)
1


( n2 ) 

+ω1 [ n1 , r12 ] M 3 + ω12  n1 , [ n1 , r12 ] M 3 + ω1  n1 , r23  M 3
(52)
+ω12  n1 , n1 , r23  M 3 + ω 2  n2 , r23  M 3
( O )( 3) 
( O )( 3) 
+2 ω1ω3  n1 , S n 3
+ 2 ω2ω3  n2 , S n 3

(
)
( 3 ) 
3



+2ω1ω2  n1 , n2 , r23  M 3 + ω22  n2 , n2 , r23  M 3
( O )( 3) 
( O )( 3) 
+ω1 n1 , ρ3  M 3 + ω12  n1 , S n 3
+ 2ω1ω2  n1 , S n 3

(
)
( 2 ) 
1



( O3 )( 3)
( O )( 3 ) 
+ω 2 S n
+ ω22  n2 , S n 3
( 2)
( 2 ) 

( O )( 3)
( O )( 3) 
.
+ω3 S n 3
+ ω32  n3 , S n 3
( 3)
( 3 ) 

3
1
3
2
170 ▪ VOL. 44, No 2, 2016
3
3
By analysing the structure of linear momentum deri–
vative terms, one can see that it is possible to introduce
pure kinematic vectors, so called vector rotators [8-12],
depending on the component angular velocities and
component angular accelerations of the component
coupled rotations which are given as follows:
FME Transactions

[ n1 , r12 ]
ℜ30 = ω1 + ω12  n1 ,

[ n1 , r12 ]

( O )( 3) ( O )( 3)
+ℜ33 S n 3
+ ℜ34 S n 3
[ n1, r12 ]  ,
[ n1, r12 ] 
( 1)
+ℜ35
  n , r  
 n1 , r23 
1 23
2
ℜ3 = ω1 + ω1  n1 ,    ,

n
 n1 , r23 

 1 , r23  

  n , r  
 n2 , r23 
2 23
2
ℜ31 = ω 2 + ω2  n2 ,    ,

 n2 , r23 
 n2 , r23  

  n , r  
2 23
ℜ32 = 2ω1ω2  n1 ,    ,

 n2 , r23  

( O )( 3)
( O )( 3)

S n 3
S n 3

( )
( )
ℜ33 = ω1 1
M 3 + ω12  n1 , 1
O3 )( 3)
(
( O )( 3)

S n
S n 3
( 1)
( 1)

( O )( 3)
( O )( 3)

S n 3
S n 3

( )
( )
M 3 + ω12  n1 , 1
ℜ34 = ω1 1
O3 )( 3)
(
( O )( 3)

S n
S n 3
( 2)
( 2)

( O3 )( 3) 

 S( n2 )

ℜ35 = 2ω1ω2  n1 , ,
( O )( 3) 

S n 3
( 2)


( O )( 3)
( O )( 3) 

S n 3
S n 3


(
)
( )
ℜ36 = ω3 3
+ ω32  n3 , 3
,
( O )( 3)
( O )( 3) 

S n 3
S n 3
( 3)
( 3)




= 2 ω1ω3  n1 ,


( O )( 3) 
S n 3

( 3)
( O )( 3)  ,

S n 3
( 3)



ℜ38 = 2 ω2ω3  n2 ,


( O )( 3) 
S n 3

( 3)

( O )( 3)

S n 3
( 3)

ℜ37
( 2)
( O )( 3) ( O )( 3)
S n 3
+ ℜ36 S n 3
( 2)
( 3)
( O )( 3) ( O )( 3)
+ℜ37 S n 3
+ ℜ38 S n 3
( 3)
( 3)
In the same manner, we derive the vector expression
for angular momentum and due to the large equations
we express linear momentum derivatives by using the
vector rotators and obtain the form:
dL  ( O )(1)  * ( O )(1)
= n1ω1  n1 , ℑ n
+ℜ D ( )  1 ( n )
dt

+ [ n1 , r12 ]  r12 ,ℜ20  M 2 + [ n1 , r12 ]  ρ2C , ℜ20  M 2
( O )( 2 ) 

+ S n
 r12 , ℜ2 
( )
( O )( 2 )  r12 ,ℜ22 
+ S n


( )
1


,


1
1
1
2
1


,


2
2
( O )( 2 )  r , ℜ 
+  n1 , S n
( )   12 21 

2
2
 ( O )( 2 )  * ( O )( 2 )
+ n1ω1  n1 , ℑ +ℜ D 
( n )  2 ( n )

2
2
1
1
 ( O )( 2 )  * ( O )( 2 )
+ n2 ω 2  n2 , ℑ +ℜ D +

( n )  22 ( n )

 ( O )( 2 )   ( O )( 2 ) 
O 2 
+ω1ω2   n1 , ℑ n
+  n2 , ℑ n
+ ℑ( )( ) [ n1 , n2 ] +


(
)
(
)
 



2
2
2
2
2
(53)
2
2
+ [ n1 , r12 ]  r12 , ℜ30  M 3
(55)
+ n1 , r23   r12 , ℜ3  M 3
+ n2 , r23   r12 , ℜ31  M 3 + [ n2 , r12 ]  r12 , ℜ32  M 3
( O )( 3)  r12 , ℜ33  + S ( O )( 3) 2ω1ω2  r12 , ℜ35 
+ S n




n
( )
( )
( O )( 3)  r12 , ℜ34  + S ( O )( 3)  r12 , ℜ37 
+ S n




( )
(n )
( O )( 3)  r12 , ℜ36  + S ( O )( 3)  r12 , ℜ38 
+ S n




( )
(n )
+ [ n1 , r12 ]  r23 , ℜ30  M 3
+ n1 , r23   r23 , ℜ3  M 3 + n2 , r23   r23 , ℜ32  M 3
+ n2 , r23   r23 , ℜ31  M 3
( O )( 3)  r23 , ℜ33  M 3 + S ( O )( 3)  r23 , ℜ35 
+ S n




n
( )
( )
( O )( 3)  r23 , ℜ34  M 3 + S ( O )( 3)  r23 , ℜ37 
+ S n




( )
(n )
( O )( 3)  r23 , ℜ36  M 3 + S ( O )( 3)  r23 , ℜ38 
+ S n




( )
(n )
+ [ n1 , r12 ]  ρ3C , ℜ30  M 3
+ n1 , r23   ρ3C , ℜ3  M 3
{
{
{
}
}
}
3
3
1
2
3
3
2
Then, we can express derivatives of the linear
momentum in the following form:
( O )( 2)
dK ( O1 )(1)  
= ℜ1 S n
+ ℜ20 n1, r12  M 2 + ℜ2 S n 2
( 1)
( 1)
dt
( O2 )( 2)
( O2 )( 2) +ℜ21 S n
+ ℜ22 S n
( 2)
( 2)
+ℜ30 [ n1, r12 ] M 3 + ℜ3 n1, r23  M 3
(54)
+ℜ31 n2 , r23  M 3 + ℜ32 n2 , r23  M 3
( O )( 3) ( O3 )( 3)
+ℜ33 S n 3
+ ℜ34 S n
( 1)
( 2)
( O )( 3) ( O3 )( 3)
+ℜ35 S n 3
+ ℜ36 S n
( 2)
( 3)
( O )( 3) ( O3 )( 3)
+ℜ37 S n 3
+ ℜ38 S n
( 3)
( 3)
+ℜ31 n2 , r23  M 3 + ℜ32 n2 , r23  M 3
FME Transactions
2
1
3
3
3
3
3
3
1
3
2
3
3
{
{
{
{
{
{
{
{
}
}
}
}
}
}
}
}
3
2
3
3
3
3
VOL. 44, No 2, 2016 ▪ 171
where in (56), FROk denotes the resultant joint re–action
force in the k-th joint and Gi , denotes gravitation forces.
In eq. (58) M ROk denotes the resultant joint reaction
moment in the k-th joint, rci , Gi , denotes the moments
of gravitational forces and a actuator torque M k for the
+ n2 , r23   ρ3C , ℜ31  M 3 + n2 , r23   ρ3C , ℜ32  M 3
( O )( 3)  * ( O )( 3)
+ n1ω1  n1 , ℑ n
+ ℜ30 D n
( ) 
( )

( O )( 3)  + ℜ* D ( O
)( 3)
+ n2 ω 2  n2 , ℑ n
40 ( n )
( ) 

( O )( 3)  * ( O )( 3)
+ n3ω 3  n3 , ℑ n
+ ℜ50 D n
( ) 
( )

( O )( 3) 
 3 
( O )( 3)  +  n , ℑ
+ ω1ω2   n1 , ℑ n
+ ℑ( ) [ n1 , n2 ]
2



n)
(
)
(






{
}
3
3
1
1
3
[
3
2
2
3
3
3
3
3
3
2
1
 ( O )( 3)   ( O )( 3) 
3 
+ω2ω3   n2 , ℑ n
+ n ,ℑ
+ ℑ( ) n2 , n3  + ω1
( )   3 ( n ) 


 ( O )( 3)  +  n , ℑ( O )( 3)  + ℑ( 3) n , n  ,
ω3   n1 , ℑ n
 1 3 
( )   3 ( n ) 


3
3
3
2
3
3
3
1
Here we also introduce some new vector rotators which
are written in the form
( O )(1)
( O )(1) 

D n 1
D n 1

*
( )
( ) 
ℜ1 = ω1 1
+ ω12  n1 , 1
,
( O )(1)
( O )(1) 

D n 1
D n 1
( 1)
( 1 ) 

( O )( 3 )
( O )( 3 ) 

D n 3
D n 3


*
( )
( )
ℜ30 = ω1 1
+ ω12  n1 , 1
,
O3 )( 3)
O3 )( 3) 
(
(

D n
D n
( 1)
( 1)



( O )( 2 ) 


D 2
*
n1 )
n1 )
(
(
 
2
ℜ2 = ω1
+ ω1  n1 ,
,
( O2 )( 2 )
( O2 )( 2 ) 


D D 
( n1 )
( n1 ) 

( O )( 3 )
( O )( 3) 

D n 3
D n 3

*
(
)
( ) 
2
ℜ40 = ω 2 + ω22  n2 , 2
,
( O )( 3 )
( O )( 3) 

D n 3
D n 3
( 2)
( 2 ) 

( O )( 2 )
( O )( 2 ) 

D n 2
D n 2


*
( )
( )
ℜ22 = ω 2 2
+ ω22  n2 , 2
,
( O )( 2 )
( O )( 2 ) 

D n 2
D n 2
( 2)
( 2)


( O )( 3 )
( O )( 3 ) 

D 3
D 3

*
(n )
(n ) 
ℜ50 = ω3 3
+ ω32  n3 , 3

( O )( 3 )
( O )( 3 ) 

D 3
D 3
( n3 )
( n3 ) 

( O )( 2 )
D 2
3.5 Determination of joint reactions by using the
method based on vectors of body mass
moments and vector rotators
By applying theorems of change of linear momentum
and angular momentum we write the following
expressions:
3 dK (57)
= FROk + ∑ Gi
dt
i=k
3 dL
(58)
= M ROk + ∑  rci , Gi  + M k
dt
i=k
172 ▪ VOL. 44, No 2, 2016
revolute joint Ok , k = 1,2,3. From previous two
equations it is easy to determine the resultant joint
reaction force and moment in the first joint as:
dK FRO1 =
− G1 − G2 − G3
(59)
dt
dL       − rc1 , G1  −  rc 2 , G2  −  rc3 , G3  − M1 (60)
M RO1 =
dt 
The similar procedure is for determination of the
resultant joint reaction forces and moments in other two
joints of the three-like rigid body system.
It is known that joint reactions can be split in to the
static parts and the kinetic parts, where negative values
of kinetic parts are called kinetic pressures. Observing
this problem could be of great importance in the
systems where the period of change of kinetic pressure
overlaps with the period of the eigenfrequency of the
joint shaft, when resonance can occur and cause damage
of the system. This problem is more frequent in the
systems with fast-rotating body parts and it is less
frequent in robotic applications. However, we can say
that using the vector method described in this section it
is easy to obtain kinetic reactions. From (57) and (58)
by taking into account (54), (55) and by neglecting the
terms with external forces we can obtain kinetic
reactions with components in the directions of vector
rotators and with magnitudes as products of deviation
moments of masses and magnitudes of vector rotators.
4.
(56)
]
CONCLUSION
In this paper, two different analytic methods for
determination of joint reaction forces and moments are
presented. The first method based on the Rodrigues
approach is applied for determination of the resultant
joint reaction forces and moments. Taking into acount
that the obtained equations are convenient for a
symbolic computation (Mathematica, MatLab, Maple
etc.), at this moment it is easy to implement this method
in the multibody dynamic problems, especially if there
are more than three bodies in the system. The second
method based on the vectors of body mass moments and
vector rotators for corresponding poles and axes is
introduced through the example of three-rigid body
system. The expressions for linear and angular
momentum are derived and the resultant reaction force
and the moment for the fixed joint in the system are
determined. In the literature it was shown that this
method is convenient to separate joint raction forces
into its components ie. to determine kinetic pressures
for the corresponding joints. Still, for this method it is
needed to develop corresponding algorithms and
computer tools so that large analytic expressions could
be more easily applied in multibody dynamics.
FME Transactions
ACKNOWLEDGMENT
This research was sponsored by the research grants
of the Serbian Ministry of Education, Science and
Techno–logical Development under the numbers OI
174001 and TR 35006.
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ОДРЕЂИВАЊЕ РЕАКЦИЈА ЗГЛОБА У
СИСТЕМУ ВИШЕ КРУТИХ ТЕЛА, ДВА
РАЗЛИЧИТА ПРИСТУПА
Милан С. Цајић, Михаило П. Лазаревић
У овом раду су приказане две различите методе за
одређивање сила и момената реакција идеалних веза
у зглобовима. Разматрани систем више тела има
структуру отвореног кинематског ланца. Прва
метода се односи на одређивање главних вектора и
момената реакција веза у симболичкој форми која се
заснива на Родригезовом приступу и погодна је за
симболичко рачунање. Друга приказана метода је
векторска метода која је базирана на векторима
момената маса и векторима ротаторима везаних за
пол и усмерену осу. Оба примера су приказана и
дискутована на систему три крута тела.
VOL. 44, No 2, 2016 ▪ 173