Download Calculus of axial force in a mechanism

Calculus of axial force in a mechanism
using Lagrange equations
Thien Van NGUYEN*,1, Roxana Alexandra PETRE1, Ion STROE1
1
*Corresponding author
“POLITEHNICA” University of Bucharest, Department of Mechanics,
313 Splaiul Independentei, Bucharest, Romania
[email protected]*, [email protected],
[email protected]
DOI: 10.13111/2066-8201.2016.8.2.8
Received: 14 April 2016 / Accepted: 23 May 2016
Copyright©2016. Published by INCAS. This is an open access article under the CC BY-NC-ND
license (http://creativecommons.org/licenses/by-nc-nd/4.0/)
4th International Workshop on Numerical Modelling in Aerospace Sciences, NMAS 2016,
11-12 May 2016, Bucharest, Romania, (held at INCAS, B-dul Iuliu Maniu 220, sector 6)
Section 3 – Modelling of structural problems in aerospace airframes
Abstract: Lagrange equations are used to study the motion of a system under the action of known
external forces. Besides, based on these equations we can determine the internal force in an arbitrary
element of the mechanism acting by active force. If an internal force has to be found, a supplementary
mobility related to it is considered in the system. The corresponding internal force for new mobility is
found for null values of mobility and of its first and second derivatives. Also the determination of the
axial force in the connecting rod of the slider-crank mechanism is presented in this paper as an
illustration of this method.
Key Words: Dynamics, axial force, slider-crank mechanism, constraints.
1. INTRODUCTION
Upon to now, several methods have been applied to study and formulate the forward and
inverse dynamics such as: Newton-Euler classical procedure, Lagrange equations and
multipliers formalism and principle of virtual work, [1], with the goal of giving the
kinematic relations and dynamic analysis of the system. Determining the internal forces
along with the constraint force analysis are important steps in dynamic analysis, which is the
base of the structure design of mechanism.
Lu [2] uses the virtual work method to determine the generalized forces of the actuator,
in spatial parallel structures and relates them to the real forces that they exert. Geike and Mc
Phee [3] proposed a general approach which could determine the inverse dynamics solutions
for a planar 3-RRR parallel manipulator and spatial 6-DOF parallel mechanism. Using
Newton Euler method and D’Alembert principle Jiang, Li and Wang [9] established the
force analysis equations, and also put forward the dynamic analysis model of a parallel
mechanism based on the deformation compatibility method. Zhi and Wang [10] improved
and applied the solution method of reciprocal screw system to the solution procedure of the
constraint forces. Stefan STAICU [5], [7], [8] uses the principle of virtual work for solving
the inverse dynamic problem in a 3-DOF parallel mechanism. By using Lagrange equations
INCAS BULLETIN, Volume 8, Issue 2/ 2016, pp. 97 – 108
ISSN 2066 – 8201
Thien Van NGUYEN, Roxana Alexandra PETRE, Ion STROE
98
and principle of virtual work, Ion STROE and Stefan STAICU [4] proposed the approach for
calculating joint forces in mechanism. The difficulties commonly encountered in dynamic
calculus are: complicated kinematical structure possessing of large number of passive
degrees of freedom, taking account of inertial forces and constraint forces, problems related
to the solution of the inverse dynamics [6]. By using a set of generalized coordinates that are
consistent with the constraint relations, we can formulate the equations of motion and
calculate the internal forces in the multi-body system without considering the constraint
forces, which is the main advantage of Lagrange equations.
In fact, the calculus of internal forces for a static system of rigid bodies is quite familiar,
but using Lagrange equations for that aim, is an interesting issue considered in recent years.
Based on this method, the internal forces in an arbitrary element of the mechanism are
simply determined. The slider-crank mechanism is one of the most commonly used machine
subsystem in mechanical system design. It is employed as the principal element of internal
combustion engines, compressors, fly-ball governors, stamping machines, and many other
machines. Therefore, the slider-crank mechanism is considered a simple model for
calculating the axial force in connecting rod of the system at an instant moment of time in
order to illustrate that method.
2. KINEMATICS OF MULTIBODY SYSTEM
Let two bodies (i) and (j) be with motions constrained by a coupling mechanism, which is
defined by points Oi, Oj (Fig. 1).
The motion of the body (i) with respect to the inertial reference frame O0x0y0z0 is
determined by the position vector of the mass center O0Ci and by matrix  Ai0  which gives
the attitude of moving reference frame Cixiyizi, attached with (i) body, with respect to inertial
reference frame O0x0y0z0. In the similar way, the position vector O0C j and matrix  A j 0  are
defined for the body (j).
Figure 1 - System of rigid bodies
Each body, (i) or (j), has 6 degrees of freedom when it has no constraint with others. The
number of degrees of freedom is reduced by the number of constraints which are imposed by
the coupling mechanism. If the general motions of bodies (i) and (j) with respect to the
inertial reference frame O0x0y0z0 are known, then the relative motion of the body (i) with
respect to the body (j) can be determined by the position vector
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Calculus of axial force in a mechanism using Lagrange equations
O j Oi  (O0Ci  Ci Oi )  (O0C j  C j O j )
(1)
and by matrix  Aij  , which gives the attitude of the body (i) with respect to the body (j)
 Aij    Ai 0   Aj 0 
T
(2)
The matrix  Ai 0  allows expressing the unit vectors of the moving reference frame
Cixiyizi with respect to unit vectors of inertial reference frame O0x0y0z0
ii 
i0 
 
 
 ji    Ai 0   j0 
 
 
ki 
k0 
(3)
For the unit vectors of the moving reference frame Cjxjyjzj, the relation can be written as
following:
i j 
i0 
 
 
 
 j j    A j0   j0 
 
 
k j 
k0 
(4)
The attitude of the body (i) with respect to the body (j) is given by matrix  Aij  using
the relation bellow:
i j 
ii 
 
 
 


j

A
 i   ij   j j 
 
 
k j 
ki 
(5)
In contrast, the attitude of the body (j) with respect to the body (i) is given by matrix
 A ji  with the following relation:
i j 
ii 
 
 
 
 j j    A ji   ji 
 
 
ki 
k j 
(6)
By comparing relations (5) and (6), it follows:
 Aij    A ji 
 Ai 0 T for
T
(7)
T
relation (3), and multiplying left with  A j 0  for
relation (4), then adding them, we obtain:
Multiplying left with
INCAS BULLETIN, Volume 8, Issue 2/ 2016
Thien Van NGUYEN, Roxana Alexandra PETRE, Ion STROE
 Ai 0 T
ii 
 
T
 ji    A j 0 
 
ki 
i j 
 
 jj 
 
k j 
100
(8)
By multiplying left with  Ai 0  for relation (8), the below relation will be obtained:
i j 
 
(9)
 jj 
 
k j 
From relations (5) and (9), relation (2) is obtained, which is used to compute the
matrix  Aij  , if the matrices  Ai 0  and  A j 0  are known. The terms of matrices  Ai 0  and
 A j 0  depend on the attitude angles of the bodies (i) and (j) with respect to the inertial
ii 
 
T
 ji    Ai 0   A j 0 
 
ki 
reference frame. If the orientation of the body (i) with respect to the inertial reference frame
is defined by 1i ,  2i , 3i angles, which correspond to the sequence of rotations 1-2-3 with
respect to a reference frame parallels with the inertial reference frame O0x0y0z0, then the
matrix below
 c2i c3i s1i s2i c3i  s3i c1i c1i s2i c3i  s1i s3i 
 Ai 0   c2i s3i s1i s2i s3i  c1i c3i c1i s2i s3i  s1i c3i 
 s1i c2i
c1i c2i
 s2i

(10)
and the angular velocity
 c2i c3i s3i 0  1i 
ωi 0    c2i s3i c3i 0 2i 

 s2i
0 1  
3i 
are obtained. In the above relations, notations of the following form were used:
(11)
s1i  sin 1i , c1i  cos 1i
s2i  sin 2i , c2i  cos 2i
(12)
s3i  sin 3i , c3i  cos 3i
Coupling mechanism between the bodies (i) and (j) imposes restrictions on the relative
motion of (i) with respect to (j).
3. LAGRANGE EQUATIONS
3.1. Equations of Motion of the System
A significant advantage of the Lagrange approach for developing equations of motion for
complex systems comes as we leave the Cartesian xi coordinate system and move into a
general coordinate system. We introduce a general notation for the relationship between h
Cartesian variables of position xi and their description in generalized coordinates (for some
systems, the number of generalized coordinates is lager then the number of degrees of
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freedom and this is accounted for by introducing constraints on the system). In general case,
each xi could be dependent upon every qk.
xi  xi  q1 , q2 ...qk ..., qh 
(13)
When constraints are expressed by functions of coordinates, the motion of the systems
can be studied with Lagrange equations for holonomic systems with dependent variables,
whereas other conditions of constraint are expressed by velocities, the motion is described
with Lagrange equations for non-holonomic systems. For a non-holonomic system, the
Lagrange equations corresponding to a system of h generalized coordinates
d  E

dt   qk
 E
 Qk 

  qk
p
 a
i ik
,
 k  1, 2,..., h 
(14)
i 1
are completed with the constraints
h
a
ik qk
 i  1, 2,..., p  ,
 bi  0,
(15)
k 1
where E is the kinetic energy expressed with respect to an inertial reference frame,
conventionally considered as fixed, and
 Wk
Qk 
(16)
 qk
are the generalized forces, while  Wk is the virtual work produced by the forces acting upon
the system, corresponding to the virtual displacement  qk . By solving system (14), of h
equation, and (15), of p equations, coordinates qk and Lagrange multipliers i will be found.
From (14), the equations for the holonomic system can be obtained by replacing
functions aik . In the case of a holonomic system, constraints are of the form:
i  q1 ,..., qh , t   0,  i  1,2,..., p 
(17)
From the above formula, the differential form is obtained:
h
i
 q
k 1
qk  bi  0,  i  1, 2,..., p 
(18)
k
By comparing relations (18) and (15), it follows:
i
aik 
 qk
(19)
Then equations (14) become:
d  E  E

 Qk 


dt  qk  qk
qk
Let define the analytical function
U 
p
   ,  k  1, 2,..., h 
i
i
(20)
i 1
p
 
i
i
(21)
i 1
then equations (20) can be written in the form:
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Thien Van NGUYEN, Roxana Alexandra PETRE, Ion STROE
d  E

dt   qk
102
 E
 U
 Qk 
,  k  1, 2,..., h 

 qk
  qk
(22)
Starting from these h differential equations and using p relations of constraint, we
determine just the generalized coordinates qk and the Lagrange multipliers.
3.2. Calculus of Internal Forces
For a mechanical system with h degrees of freedom represented by independent generalized
coordinates qk (k=1,2…,h), the Lagrange equations are expressed as following:
d  E

dt  qk
 E U

 Qk* , (k  1,..., h)

 qk qk
(23)
An internal force Qh 1 , as new generalized force, can be found if a new fictitious mobility
according to internal force is considered. Then the mechanical system is considered as the
one with h+1 degrees of freedom. Equation for the new mobility is
d  E  E
U

 Qh 1 .


dt  qh 1  qh 1 qh 1
(24)
Considering again the mechanism, the internal force h1 is easily obtained from (24)
in the following form
 d  E  E
U 
h 1   



 dt  qh 1  qh 1 qh 1  qqhh 11 00
(25)
qh 1  0
4. EXAMPLE: SLIDER-CRANK MECHANISM
As an example, the one degree of freedom system of slider-crank mechanism is considered
(Fig.2). The slider-crank mechanism consisting of a crank (1) characterized of the length
OA= r, mass m1; connecting rod (2) characterized by the length AB= l, mass m2 and a slider
3 characterized by mass m3. The crank OA rotates with the motion law   o t 
o
2
t 2 by
the active torque Mo.
A
1
2
C2
C1
Mo
B
3
o
Figure 2 - Slider-crank mechanism
The Lagrange equations of motion of a system are written in the form:
d  E  E
U
 Qk 
, k  1, 2...h
(26)


d t  qk  qk
qk
We first apply Lagrange equations to derive the equations of motion of the mechanism.
This is one degree of freedom system, so we choose the generalized coordinates as q  ( ) .
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It is convenient to describe the positions of the mechanism mass centers with Cartesian
coordinates, and then to express the kinetic energy E and the force function U using the polar
angle. Moments of inertia of the crank 1 and the connecting rod 2 are written:
Jo 
m1.r 2
,
3
and J C 2 
(27)
m2 .l 2
.
12
(28)
respectively. Thus, the kinetic energy is
1
1
1
1
E  1T .J o .1  rCT2 .m2 .r .C 2  2T .J C 2 .2  rCT3 .m3 .r .C 3
2
2
2
2
 E
 sin 2 
m1.r 2 . 2
l 2 .cos 2 
r.sin 2  .cos 
 m2 .
 2

 2
6
6(l  r 2 .sin 2  ) 2. l 2  r 2 .sin 2 

1
r 2 .cos 2 
r.cos
 m3 .  2

 2 2(l  r 2 .sin 2  )
2
l  r 2 .sin 2 

 2 2
 .r . 

(29)
 2 2
2
 .r . .sin 

The fore function U has the expression
(m1  m2 ) g.r.sin 
.
2
The external generalized force acting on the mechanism is written
 L M .
Q   
M
U 


(30)
(31)
Applying (26), the differential equation governing the motion is obtained
 m1  m2 .l 2

cos 2 
 m3 .r 2 .sin 2   2
  m2  m3  .sin 2 
 
2
2
3
3
l

r
.sin





2
  (m2  2m3 ).r.sin  .cos 

l 2  r 2 .sin 2 



 2
r  



  3m2

r.sin 3 

 m2


m
.sin

.cos



m
.



3
3


 2
 l 2  r 2 .sin 2 
 4



2
2
2
2
   m2  2m3  . r.sin  .cos    m2  m3  r .sin  .cos  .(cos   sin  )  


l 2  r 2 .sin 2 

l 2  r 2 .sin 2   4
 r 2 2

m2 .l 2 .sin  .cos 
r 3 .sin 3  .cos 2 

  m2



2
2
2
   2  m3  . 2

l  r 2 .sin 2  l 2  r 2 .sin 2  12(l  r .sin  )




2 2
3
  m2 .r .l .sin  .cos 

 12(l 2  r 2 .sin 2  ) 2



 m  m2  .g.r.cos
M  1
2

(32)

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- The axial force in the connecting rod AB at an instant moment of time can be
determined based on the Lagrange equations of second kind. At that time, the generalized
coordinates of the slider-crank mechanism are represented by: q1   ; q2  u. And u is
considered as a supplemental mobility.
A
1
C1
C2'
2'
2''
N
C2''
Mo
B
3
o
Figure 3 - Slider-crank mechanism with virtual supplemental displacement on the connecting rod
 d  E  E U 
N  h 1   


 u 0
 d t  u  u u  u 0
(33)
u 0
Suppose that the connecting rod AB is divided into two parts (it has the mass: m2’ and m2’’,
respectively) at the position  (0≤ ≤ l). By geometrical relation shown in Fig. 3, we have:
r.sin   (l  u)sin 
(34)
 sin  
r.sin 
l u
(l  u )2  r 2 .sin 2 
l u
By carrying out the total differential equation (36), the following is obtained
 cos  

r. .cos  u.sin  r.(l  u ). .cos  r.u.sin 

(l  u ).cos 
(l  u ). (l  u ) 2  r 2 .sin 2 
(35)
(36)
(37)
The kinetic energy will be expressed as following:
1
1
1
1
E  1T .J o .1  rCT2' .m2' .rC 2'  2'T .J C 2' .2'  rCT2'' .m2'' .rC 2'' 
2
2
2
2
(38)
1 T
1 T
 2'' .J C 2'' .2''  rC 3 .m3 .rC 3
2
2
where 2' , 2'' , which are the angular velocity vectors of the two parts m2’ and m2’’,
respectively, are expressed as




0



,
ω2'  ω2''  α 
0
(39)


 r.(l  u ). .cos  r.u.sin  


 (l  u ). (l  u ) 2  r 2 .sin 2  
and JC2’, JC2’’, which are the tensor of inertia of the crank 1, the two parts m2’, m2’’ with
respect to C2’, C2’’, are written in the forms:
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Calculus of axial force in a mechanism using Lagrange equations
m2 . 3
,
12.l
(40)
m2 .(l   )3
,
12.l
(41)
J C 2' 
J C 2'' 
and rC 2' , rC 2'' , rC 3 , which are the position vectors of mass centers of the two parts m2’, m2’’
and slider m3, respectively, are expressed as following:


2
2
2 
 r.cos  2(l  u ) . (l  u )  r .sin  


r.


  r.sin  
.sin 
,
2(l  u )




0




(42)
(l  2u   )

2
2
2 
 r.cos  2(l  u ) . (l  u )  r .sin  


r.(l  2u   )


  r.sin  
.sin 
,
2(l  u )




0




(43)
 r.cos  (l  u ) 2  r 2 .sin 2  



0
.


0


(44)
rC 2'
rC 2''
rC 3
By derivation rC 2' , rC 2'' , rC 3 in respect to time,
rC 2'

r 2 . . .sin  .cos 
r 2 . .u.sin 2 

 r. .sin  
2(l  u ). (l  u ) 2  r 2 .sin 2  2(l  u ) 2 . (l  u ) 2  r 2 .sin 2 


r. . .cos  r. .u.sin 
  r. .cos  


2(l  u )
2(l  u ) 2

0







,





(45)
rC 2''

r 2 .(l  2u   ). .sin  .cos  2(l  u )3 .u  r 2 .(l   ).u.sin 2  

 r. .sin  

2
2
2
2(
l

u
)
(
l

u
)

r
.sin

2(l  u ) 2 . (l  u ) 2  r 2 .sin 2  



r.(l  2u   ). .cos  r.(l   ).u.sin 
,
  r. .cos 



2(l  u )
2(l  u ) 2


0






(46)
INCAS BULLETIN, Volume 8, Issue 2/ 2016
Thien Van NGUYEN, Roxana Alexandra PETRE, Ion STROE
rC 3

r 2 . .sin  .cos 
(l  u ).u

 r. .sin  
2
2
2
(l  u )  r .sin 
(l  u )2  r 2 .sin 2 


0

0



106



.




(47)
The force function U has the expression
m .g.r.l.sin  m2 .g.(l   ).r.u.sin 
m

U    1  m2  .g.r.sin   2

2(l  u )
l.(l  u )
 2

(48)
Applying (33), we obtain the axial force N in the connecting rod AB
  6(m2  m3 ).l 5  6.m2 .l 4 .  3m2 .r 4 .(2  l ).sin 4   3(3m2  2m3 ).r 2 .l 3 .sin 2 

  12m2 .r 2 .l 2 . .sin 2   (5m2  6m3 ).r.l 3 .cos . l 2  r 2 .sin 2  

  3m .r 3 .(2  l ).sin 2  .cos . l 2  r 2 .sin 2   6m .r.l 2 . .cos  . l 2  r 2 .sin 2 
2
2
N 
2
2
2
2
2
2

6.l .(l  r .sin  ). l  r .sin 2 





  6(m2  m3 ).r.l 7 .cos   3m2 .r 7 .(2  l ).sin 6  .cos   6m2 .r.l 6 . .cos   


  3(5m2  4m3 ).r 3 .l 5 .sin 2  .cos   6(2m2  m3 ).r 5 .l 3 .sin 4  .cos  



3 2
2
2
2
2


  18m2 .r .l . .(l  r .sin  ).sin  .cos 


2
2
2
2
2
2
2
2
6l .(l  r .sin  ) . l  r .sin 
 
6
6
2 5
2
2 5
2
  3m2 .(2  l ).r .sin   3( m2  2m3 ).r .l .cos   (5m2  6m3 ).r .l .sin 

 2(4m  3m ).r 4 .l 3 .sin 4   6m .r 2 .l 2 . .(l 2  2r 2 .sin 2  ).sin 2  
2
3
2
 
   m .r 4 .l 3 .sin 2  .cos 2   3m .r 2 .l. 2 .(l 2  r 2 .sin 2  ).cos 2 
2
  2

6l 2 .(l 2  r 2 .sin 2  ) 2

(m  m ).g .r.sin  (m1.l  2m2 . ).g .r.sin 
 1 2

2l
2l 2
 




  












 2 

 







(49)
In the inverse dynamics, we suppose that the history of rotational motion of the crank is
known by the following function:


  o t  o t 2 , where o  25 (rad / s), o 
(rad / s 2 )
2
100
For the simulative purpose let’s consider the slider-crank mechanism, which has the
following characteristics: m1=0.1(kg), r=0.1 (m), m2=0.1 (kg), l=0.2 (m), m3=0.1 (kg).
Using MATLAB software, a program was developed to solve the inverse dynamics of
the slider-crank mechanism. The variations of the active torque, rotational angle versus time
are shown in Fig. 4 and Fig. 5, the variation of the axial force with respect to ratio /l is
shown in Fig.6, and the variation of the axial forces at three specified positions versus time is
shown in Fig. 7.
INCAS BULLETIN, Volume 8, Issue 2/ 2016
107
Calculus of axial force in a mechanism using Lagrange equations
Figure 4 - The active torque of the crank M.
Figure 5 - The rotational angle .
Figure 6 - The axial force with respect to “l” at the
instant time t=0.03(s).
Figure 7 - The axial forces at the three specified
positions versus time.
5. CONCLUSIONS
With the Lagrange equations, the position, velocity and acceleration of each element of
mechanism in real-time revolution can be released from the differential equations of motion
in the kinetic analysis. In addition, the active torque consistent with the given motional law
has been determined taking account the masses and forces of inertia introduced by links of
the mechanism in the model dynamics. A new method to compute axial forces is presented
in this paper, but the method can be extended to all internal forces.
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Thien Van NGUYEN, Roxana Alexandra PETRE, Ion STROE
108
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