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Modelling of flexible link manipulator dynamics using rigid link theory with
- 1. International Journal of Advanced Research in Engineering and Technology (IJARET), ISSN 0976 –
6480(Print), ISSN 0976 – 6499(Online) Volume 4, Issue 5, July – August (2013), © IAEME
1
MODELLING OF FLEXIBLE LINK MANIPULATOR DYNAMICS USING
RIGID LINK THEORY WITH FINITE ELEMENT METHOD
Mahesh A. Makwana, Arvind N. Nakiya, AnuragNema
Assistant Professor, Department of Food Engineering, College of Food Processing Technology and
Bio Energy, Anand Agricultural University, Anand, India
ABSTRACT
As flexible link manipulator (FLM) more often used in applications viz.,space and terrestrial
applications, micro-surgical operation, collision control and contouring control which may be used in
grinding robots, nuclear maintenance, e.g., to perform decontamination tasks and are widely used
where volume constrain is there. Hence, more rapid and easier method to solve dynamics of FLM is
much in concern. This paper presents dynamic modeling techniques using finite element method for
flexible robots using rigid link theory. In this approach main emphasis is to discretise whole flexible
link in very small parts each considering as a rigid link. Then applying Lagrange-Euler formulationof
rigid link with finite element method dynamics of FLM can be solved.
Keywords: Discretisation, Dynamics, Finite Element Method, Flexible link manipulator (FLM),
Rigid link manipulator
1. INTRODUCTION
1.1 Hyper-Redundant Robots
These are robots in the form of serpentine or snake or rodshape. Snakes, tentacle, trunk, and
fish are examples of biological hyper-redundant bodies. The redundancy means different ways to
perform the same movement and is usually denoted in terms of degrees of freedom.
Robots with a snake-like locomotion have these advantages:
1. The redundancy allows them to still function after losingmobility in one or more sections;
2. Stability on all terrain because of low center of gravity;
3. Terrainability which is the ability to traverse rough terrain;
4. Traction is very high as the whole body is involved;
5. High efficiency in energy use as there is no need to liftthe body;
6. Small size that can penetrate small cracks;
7. They are capable of being amphibious by sealing thewhole body, the same body motion on
land is then[1].
INTERNATIONAL JOURNAL OF ADVANCED RESEARCH IN
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ISSN 0976 - 6480 (Print)
ISSN 0976 - 6499 (Online)
Volume 4, Issue 5, July – August 2013, pp. 01-09
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- 2. International Journal of Advanced Research in Engineering and Technology (IJARET), ISSN 0976 –
6480(Print), ISSN 0976 – 6499(Online) Volume 4, Issue 5, July – August (2013), © IAEME
2
In some specific conditions there is definite need of FLM as
(i) Automated inspections in submarine i.e., Submarine cable;
(ii) In collapsed building to search and rescue among wreckage;
(iii) In space exploration;
(iv) firefighting (acting like intelligent fire hose that can autonomously locate fire);
(v) Manufacturing and machine maintenance in a convoluted environment
(vi) General manufacturing, they can act as robotic arm with great dexterity;
(vii) Minimally invasive surgery, as laparoscope or endoscopethat can follow a very complex
path withoutcolliding or penetrating organs;
2. DYNAMICS OF RIGID-LINK
Manipulator dynamics is concerned with the equations of motion, the way in which the
manipulator moves in response to torques applied by the actuators, or external forces. The history
and mathematics of the dynamics of serial-link manipulators is well covered by Paul and
Hollerbach[2].
There are several methods by which dynamics of the rigid manipulator has solved as:
1. Newton-Euler formulation
2. Langrange-Euler formulation
3. Generalized d’Alembert equation of motion
There are two problems related to manipulator dynamics that are important to solve:
• Inverse dynamics in which the manipulator’s equations of motion are solved for given
motion to determine the generalized forces and
• Direct dynamics in which the equations of motion are integrated to determine the generalized
coordinate response to applied generalized forces.
2.1 Lagrange-Euler Formulation
The general motion equation of a manipulator can conveniently be expressed through the
direct application of the Lagrange-Euler formulation to non-conservative system. Many investigators
utilize the Denavit-Hartenberg matrix representation to describe the spatial displacement between the
neighboring link coordinate frame to obtain link kinematics information and they employ the L-E
equation to derive dynamic equation of manipulator.
The derivation of the dynamic equation of an n degree of freedom manipulator is based on the
understanding of:
• The 4X4 homogeneous coordinate transformation matrix, i-1
Ai, which describes the spatial
relationship between the ith
and (i-1)th
link coordinate frame. It relates the point fixed in link i
expressed in homogeneous coordinates with respect to ith
coordinate system to the (i-1)th
coordinates system[3].
• The Lagrange-Euler equation:
ௗ
ௗ௧
ቂ
డ
డపሶ
ቃ െ
డ
డ
ൌ τi=1, 2,… ,n (1)
Where
L= Lagrangian function= kinetic energy K – potential energy P
K= total kinetic energy of the robot arm
- 3. International Journal of Advanced Research in Engineering and Technology (IJARET), ISSN 0976 –
6480(Print), ISSN 0976 – 6499(Online) Volume 4, Issue 5, July – August (2013), © IAEME
3
P= total potential energy of robot arm
qi=generalized coordinates of the robot arm
ݍሶi=first time derivative of the generalized coordinates, qi
τi= generalized force or (torque) applied to the system at joint i to drive link i
From the above Lagrangian equation one is required to properly choose a set of generalized
coordinates to describe the system. Generalized coordinates are used as a convenient set of a
coordinates which completely describe the location of a system with respect to reference coordinate
frame.
3. DYNAMICS OF FLEXIBLE LINK
There are methods by which we can solve the dynamics of the flexible manipulator as[4]:
1. Newton-Euler formulation
2. Lagrange-Euler formulation
3. Generalized d’Alembert equation of motion
4. Recursive Gibbs-Appell formulation
5. Finite dimensional approximation
6. Hamilton’s principle and FE approach
7. Assume mode method and Lagrange approach
Because of below Advantages of Lagrange Approach over other methods it is more preferred.
1. The Lagrange Approach automatically yields as many equations as there are degrees of freedom.
It has the convenience of energy methods, but whereas energy conservation only yields just one
equation, which isn’t enough for a multi-degree-of freedom system, Lagrange yields as many
equations as you need.
2. The Lagrange equations naturally use the generalized coordinates of the system. By contrast,
Newton’s Equations are essentially Cartesian which ends up having to convert everything into
Cartesian components of acceleration and Cartesian components of forces to use Newton’s Equation.
Lagrange bypasses that conversion.
3. The Lagrange approach naturally eliminates non-contributing forces. In case of the direct
(Newtonian) approach ability to minimize the number of variables depends very much on skill
whereas Lagrange takes care of it because the generalized forces only include force components in
directions of admissible motion.
3.1 Dynamics using Lagrange-Euler approach
In order to obtain a set of differential equations of motion to adequately describe the
dynamics of a flexible link manipulator, the Lagrange-Euler approach can be used. A system with n
generalized coordinates qi must satisfy n differential equations of the form:
ௗ
ௗ௧
ሾ
డ
డపሶ
ሿ െ
డ
డ
డ
డపሶ
ൌ ߬i=1,2,…,n (2)
Where L is the so called Lagrangian which is given by L=K-P; K represents the kinetic
energy of the system and P the potential energy. Also, D is the Rayleigh’s dissipation function which
allows dissipative effects to be included, and τi is the generalized force acting on qi.
To solve dynamics using above approach directly for flexible link much mathematical
calculations and computational complexities comes in solving the problem. Instead of using above
approach if someone use approach presented in this paper (L-E theory of rigid link with FEM), there
is much reduction in mathematical calculations and computational difficulties.
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4
4. DYNAMICS WITH THREE EQUAL LINKS
For an example we will solve dynamics for three links only and on the basis of that, one can
expand equation up to number of links as per need.
For definiteness we will again consider a simple case where l1, l2, and l3 are all equal to a
length l. The more general case involves a lot more arithmetic and the form of the final result is the
same, only numerical constants will be changed. Further, we will ignore gravity for now, and assume
the links to be uniform sticks of mass m and inertia (l/12)ml2
about their center of mass. Once again
we start by finding the rotational and translational velocities of each of the links. Evidently the
angular velocities of the three links are, ߠሶଵ,൫ߠሶଵ ߠሶଶ൯,൫ߠሶଵ ߠሶଶ ߠሶଷ൯ [4].
Fig.1: Representation of 3-link
The square of the magnitude of the instantaneous linear velocity of the center of mass of link 1 is
[(1/2)r1]:
หሺ1/2ሻݎଵ ߠሶଵห
ଶ
=݈ଶ
(1/4)ߠሶଵ
ଶ
(3)
The square of the magnitude of the velocity of the center of link 2 is:
หݎଵߠሶଵ ሺ1/2ሻݎଶ ൫ߠሶଵ ߠሶଶ൯ห
ଶ
=݈ଶ
ቂߠሶଵ
ଶ
ܿݏሺߠଶሻߠሶଵ൫ߠሶଵ ߠሶଶ൯ ሺ1/4ሻ൫ߠሶଵ ߠሶଶ൯
ଶ
ቃ
=݈ଶ
ቂߠሶଵ
ଶ
ቀ
ହ
ସ
ቁ ܿݏሺߠଶሻߠሶଵߠሶଶ ൬ቀ
ଵ
ଶ
ቁ cos൫ߠሶଶ൯൰ ሺ1/4ሻ൫ߠሶଶ൯
ଶ
ቃ (4)
The square of the magnitude of the velocity of the center of link 3 is:
หݎଵߠሶଵ ݎଶ ൫ߠሶଵ ߠሶଶ൯ ሺ1/2ሻݎଷ ൫ߠሶଵ ߠሶଶ ߠሶଷ൯ห
ଶ
=݈ଶ
ቂߠሶଵ
ଶ
2ܿݏሺߠଶሻߠሶଵ൫ߠሶଵ ߠሶଶ൯ ൫ߠሶଵ ߠሶଶ൯
ଶ
ܿݏሺߠଷሻ൫ߠሶଵ ߠሶଶ൯൫ߠሶଵ ߠሶଶ ߠሶଷ൯ ቀ
ଵ
ସ
ቁ ൫ߠሶଵ ߠሶଶ
ߠሶଷሻଶ
cos ሺߠଶ ߠଷሻߠሶଵ൫ߠሶଵ ߠሶଶ ߠሶଷ൯ቃ (5)
- 5. International Journal of Advanced Research in Engineering and Technology (IJARET), ISSN 0976 –
6480(Print), ISSN 0976 – 6499(Online) Volume 4, Issue 5, July – August (2013), © IAEME
5
ൌ ݈ଶ
ߠሶଵ
ଶ
൬
9
4
൰ 2ܿݏሺߠଶሻ ܿݏሺߠଷሻ cosሺߠଶ ߠଷሻሻ ߠሶଵߠሶଶ ൬
11
2
൰ 2ܿݏሺߠଶሻ 2ܿݏሺߠଷሻ
cosሺߠଶ ߠଷሻሻ ߠሶଶ
ଶ
൬
5
4
൰ ܿݏሺߠଷሻሻ ൬ሺ 1/4ሻߠሶଷ
ଶ
ሻ൰ ߠሶଷߠሶଵሺ1/2ሻܿݏሺߠଷሻ
ܿݏሺߠଶ ߠଷሻሻ൨ ሺ6ሻ
Now add up the kinetic energy due to rotation and that due to linear translation of the center of mass
for all three links to obtain the Lagrangian:
ܮ ൌ 1/2݈݉ଶ
ቂߠሶଵ
ଶ
൫4 3ܿݏሺߠଶሻ൯ ܿݏሺߠଶ ߠଷሻ cosሺߠଷሻ ߠሶଵߠሶଶ ቀ
ଵଽ
ଷ
ቁ 3ܿݏሺߠଶሻ
cosሺߠଶ ߠଷሻ 2ܿݏሺߠଷሻሻ ߠሶଶ
ଶ
ቀ
ହ
ଷ
ቁ ܿݏሺߠଷሻሻ ߠሶଶߠሶଷ ቀ
ଶ
ଷ
ቁ ܿݏሺߠଷሻሻ ൬ሺ 1/3ሻߠሶଷ
ଶ
ሻ൰
ߠሶଷߠሶଵሺ2/3ሻܿݏሺߠଶ ߠଷሻ ܿݏሺߠଷሻሻቃ (7)
So this is the Lagrangian for this system and from it we will be able to calculate the relation
between joint torques and joint accelerations. Let us use the shorthand notation for trigonometric
terms introduced in the discussion of the statics of this device, e.gݏଶଷ ൌ sinሺߠଶ ߠଷሻ.
We will next derive the partial derivates of the Lagrangian with respect toߠଵ, ߠଶ,ߠଷ, ߠሶଵ, ߠሶଶ, ߠሶଷ. Let
ܮ ൌ
ଵ
ଶ
m݈ଶ
ܮ′
as before.
െ
డ′
డఏమ
=0
െ
డ′
డఏమ
=ߠଵ
ሶ
ଵ
ଶ
ሺ3ݏଶ ݏଶଷሻ+ߠሶଵߠሶଶሺ3ݏଶ ݏଶଷሻ+ߠሶଷߠሶଵሺݏଶଷሻ (8)
െ
డ′
డఏమ
ൌ ߠሶଵ ቀ
ଶ
ଷ
ܿଶଷ ܿଷቁ ߠሶଶ ቀ
ଶ
ଷ
ܿଷቁ 2ߠሶଷ ቀ
ଵ
ଷ
ቁ (9)
The partial derivates of the Lagrangian with respect to angular velocity are:
డ′
డఏሶభ
ൌ 2ߠሶଵሺ4 3ܿଶ ܿଶଷ ܿଷሻ ߠሶଶ ቀ
ଵଽ
ଷ
3ܿଶ ܿଶଷ ܿଷቁ 2ߠሶଷ ቀ
ଶ
ଷ
ܿଶଷ ܿଷቁ(10)
డ′
డఏሶమ
ൌ ߠሶଵ ቀ
ଵଽ
ଷ
3ܿଶ ܿଶଷ 2ܿଷቁ 2ߠሶଶ ቀ
ହ
ଷ
ܿଷቁ ߠሶଷ ቀ
ଶ
ଷ
ܿଷቁ (11)
Next we will need the time rate-of-change of the last three quantities above:
ௗ
ௗ௧
ቀ
డ′
డఏሶభ
ቁ ൌ 2ߠሷଵሺ4 3ܿଶ ܿଶଷ ܿଷሻ ߠሷଶ ቀ
ଵଽ
ଷ
3ܿଶ ܿଶଷ ܿଷቁ ߠሷଷ ቀ
ଶ
ଷ
ܿଶଷ ܿଷቁ െ
2ߠሶଵ൫3ݏଶߠሶଶ െ ݏଶଷ൫ߠሶଶ ߠሶଷ൯ െ ݏଶߠሶଷ൯ ߠሶଶ൫3ݏଶߠሶଶ ݏଶଷ൫ߠሶଶ ߠሶଷ൯ ݏଷߠሶଷ൯ െ ߠሶଷ൫ݏଶଷ൫ߠሶଶ ߠሶଷ൯
ݏଷߠሶଷ൯ (12)
ௗ
ௗ௧
ቀ
డ′
డఏሶమ
ቁ ൌ ߠሷଵ ቀ
ଵଽ
ଷ
3ܿଶ ܿଶଷ ܿଷቁ 2ߠሷଶ ቀ
ହ
ଷ
ܿଷቁ ߠሷଷ ቀ
ଶ
ଷ
ܿଷቁ െ ߠሶଵ൫3ݏଶߠሶଶ െ ݏଶଷ൫ߠሶଶ ߠሶଷ൯ െ
ݏଶߠሶଷ൯ െ 2ߠሶଶ൫ݏଷߠሶଷ൯ െ ߠሶଷ൫ݏଷߠሶଷ൯ (13)
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6
ௗ
ௗ௧
ቀ
డ′
డఏሶయ
ቁ ൌ ߠሷଵ ቀ
ଶ
ଷ
ܿଶଷ ܿଷቁ ߠሷଶ ቀ
ଶ
ଷ
ܿଷቁ 2ߠሷଷ ቀ
ଵ
ଷ
ቁ െ ߠሶଵ൫ݏଶଷ൫ߠሶଶ ߠሶଷ൯ െ ݏଷߠሶଷ൯ െ ߠሶଶ൫ݏଷߠሶଷ൯
(14)
Finally, inserting these terms into Lagrange’s equation gives:
߬ଵ
′
ൌ
ௗ
ௗ௧
ቀ
డ′
డఏሶభ
ቁ െ ቀ
డ′
డఏሶభ
ቁ
= 2ߠሷଵሺ4 3ܿଶ ܿଶଷ ܿଷሻ ߠሷଶ ቀ
ଵଽ
ଷ
3ܿଶ ܿଶଷ ܿଷቁ ߠሷଷ ቀ
ଶ
ଷ
ܿଶଷ ܿଷቁ െ ߠሶଵߠሶଶሺ6ݏଶ 2ݏଶଷሻ െ
ߠሶଶ
ଶሺ3ݏଶ ݏଶଷሻ െ ߠሶଶߠሶଷሺ2ݏଶଷሻ െ ߠሶଷ
ଶሺݏଶଷ ݏଷሻ (15)
߬ଶ
′
ൌ
݀
݀ݐ
ቆ
߲ܮ′
߲ߠሶଶ
ቇ െ ቆ
߲ܮ′
߲ߠሶଶ
ቇ
= ߠሷଵ ቀ
ଵଽ
ଷ
3ܿଶ ܿଶଷ 2ܿଷቁ 2ߠሷଶ ቀ
ହ
ଷ
ܿଷቁ ߠሷଷ ቀ
ଶ
ଷ
ܿଷቁ ߠሶଶ
ଶሺ3ݏଶ ݏଶଷሻ െ ߠሶଶߠሶଷሺ2ݏଷሻ െ
ߠሶଷ
ଶሺݏଷሻ െ ߠሶଷߠሶଵሺ2ݏଶଷሻ (16)
߬ଷ
′
ൌ
݀
݀ݐ
ቆ
߲ܮ′
߲ߠሶଷ
ቇ െ ቆ
߲ܮ′
߲ߠሶଷ
ቇ
=ߠሷଵ ቀ
ଶ
ଷ
ܿଶଷܿଷቁ ߠሷଶ ቀ
ଶ
ଷ
ܿଷቁ ߠሷଷ ቀ
ଶ
ଷ
ቁ ߠሶଵ
ଶሺݏଶଷ ݏଷሻ ߠሶଵߠሶଶሺ2ݏଷሻ ߠሶଶ
ଶሺݏଷሻ
(17)
4.1 Gravity
Gravity is again very simple to take into account. If we assume that the center of mass of
each link is in its geometric center so:
ܨଷ ൌ ݉ଷ݃
ܨଶ ൌ ሺ݉ଶ ݉ଷሻ݃
ܨଵ ൌ ሺ݉ଵ ݉ଶ ݉ଷሻ݃
Fig.2: Torque due to gravity
From these, we can derive the torques induced by gravity:
߬ଷ ൌ െ݉ଷ ൬
1
2
൰ ݎଷ ൈ ݃ ൌ ݃ሾሺ1/2ሻ݉ଷ݈ଷܿଵଶଷሿ
߬ଶ ൌ ߬ଷ െ ݉ଶ ൬
1
2
൰ ݎଶ ൈ ݃ െ ݉ଷݎଶ ൈ ݃ ൌ ݃ሾሺ1/2݉ଶ ݉ଷሻ݈ଶܿଵଶ ሺ1/2ሻ݉ଷ݈ଷܿଵଶଷሿ
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߬ଵ ൌ ߬ଶ െ ݉ଶ ൬
1
2
൰ ݎଶ ൈ ݃ െ ݉ଷݎଶ ൈ ݃
ൌ ݃ ൬
1
2
݉ଵ ݉ଶ ݉ଷ൰ ݈ଶܿଵ ൬
1
2
݉ଶ ݉ଷ൰ ݈ଶܿଵଶ ൬
1
2
݉ଷ݈ଷܿଵଶଷ൰൨
Where
ݎଵ ൌ ݈ଵሾܿଵ,ݏଵሿ
ݎଶ ൌ ݈ଶሾܿଵଶ, ݏଵଶሿ
ݎଷ ൌ ݈ଷሾܿଵଶଷ, ݏଵଶଷሿ
As explain above (Dynamics of three links), one can solve dynamics for multiple links considering
each link as rigid link and apply this approach to solve flexible link dynamics.
4.2 MATLAB program for calculation of 3-Link manipulator using Lagrange Euler Approach
% input data for reference point
x= input('Enter the coordinates of reference point x= ');
y= input('y= ');
z= input('z= ');
% position vector of fixed point
r0= [x; y; z; 1];
% input data of links
t= input('time of rotation t=');
l1= input('corresponding length of Link1');
M1=input('corresponding mass of Link1');
m1= M1/1000;
Q1=input('corresponding angle of rotation of Link1');
q1= Q1*pi/180;
l2=input('corresponding length of Link2 ');
M2=input('corresponding mass of Link2 ');
m2= M2/1000;
Q2=input('corresponding angle of rotation of Link2');
q2= Q2*pi/180;
l3= input('corresponding length of Link3 ');
M3=input('corresponding mass of Link3 ');
m3= M3/1000;
Q3=input('corresponding angle of rotation of Link3 ');
q3= Q3*pi/180;
w1= q1/t;
w2= q2/t;
w3= q3/t;
% rotation vector
q= [q1; q2; q3];
% Transformation Matrices
% Transformation matrix for 1st link w.r.t. reference point
A01= [cos(q1) -sin(q1) 0 L1*cos(q1); sin(q1) cos(q1) 0
L1*sin(q1); 0 0 1 0; 0 0 0 1];
% Transformation matrix for 2nd link w.r.t. 1st link is
A12= [cos(q2) -sin(q2) 0 L2*cos(q2); sin(q2) cos(q2) 0
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L2*sin(q2); 0 0 1 0; 0 0 0 1];
% Transformation matrix for 1st link w.r.t. reference pointis
A23= [cos(q3) -sin(q3) 0 L3*cos(q3); sin(q3) cos(q3) 0
L3*sin(q3); 0 0 1 0; 0 0 0 1];
% Transformation matrix for 2nd link w.r.t. reference pointis
A02= A01*A12;
% Transformation matrix for 3rd link w.r.t. reference pointis
A03= A02*A23;
% Transformation matrix for 3rd link w.r.t. 1st link is
A13= A12*A23;
% Kinetic Energy calculation
% for revolute joint, when alpha = and theta = 90 degree,
then Arev reduces to Q where
% Q= [0 -1 0 0; 1 0 0 0; 0 0 0 0; 0 0 0 0]
Q= [0 -1 0 0; 1 0 0 0; 0 0 0 0; 0 0 0 0];
% Uij= dA0i/dqj= A(0)(j-1)*Q*A(j-1)(i) for j<=i
% = 0
U11= Q*A01; % A00= unity vector
U21= Q*A02;
U22= A01*Q*A12;
U31= Q*A03;
U32= A01*Q*A13;
U33= A02*Q*A23;
% Assuming all products of inertia are zero, pseudoinertiais calculated as Ji
J1= [1/3*m1*L1^2 0 0 -1/2*m1*L1; 0 0 0 0; 0 0 0 0; -1/2*m1*L1 0 0 m1];
J2= [1/3*m2*L2^2 0 0 -1/2*m2*L2; 0 0 0 0; 0 0 0 0; -1/2*m2*L2 0 0 m2];
J3= [1/3*m3*L3^2 0 0 -1/2*m3*L3; 0 0 0 0; 0 0 0 0; -1/2*m3*L3 0 0 m3];
% acceleration related Symmetric Matrix i.e.D(theta)
D11= trace(U11*J1*U11.')+ trace(U21*J2*U21)+trace(U31*J3*U31.');
D12= trace(U22*J2*U21.')+ trace(U32*J3*U31.');
D13= trace(U33*J3*U31.');
D21= D12;
D22= trace(U22*J2*U22.')+ trace(U32*J3*U32.');
D23= trace(U33*J3*U32.');
D31= D13;
D32= D23;
D33= trace(U33*J3*U33.');
D= [D11 D12 D13; D21 D22 D23; D31 D32 D33];
% Coriolis and centrifugal terms i.e. h(theta, thetadot)
h1= -(m2*L1*L2*sin(q2)+ m3*L1*(2*L2*sin(q2)+ L3*sin(q1+q2)))*w1*w2-
m3*L3*(L1*sin(q2+q3)+ L2*sin(q3))*w1*w3-1/2*m3*L3*(L1*sin(q2+q3)+ L2*sin(q3))*w2*w3-
1/2*m3*L3*(L1*sin(q2+q3)+ L2*sin(q3))*w3^2;
h2= -1/2*(m2*L1*L2*sin(q2)+ m3*L1*(2*L2*sin(q2)+
L3*sin(q1+q2)))*w1*w2- 1/2*m3*L3*(L1*sin(q2+q3)+ L2*sin(q3))*w1*w3-
m3*L2*L3*sin(q3)*w2*w3- 1/2*m3*L2*L3*sin(q3)*w3^2- 1/2*(-(m2*L1*L2*sin(q2)+
m3*L1*(2*L2*sin(q2)+ L3*sin(q2+q3)))*w1^2- (m2*L1*L2*sin(q2)+
m3*L1*(2*L2*sin(q2)+ L3*sin(q2+q3)))*w1*w2- m3*L1*L3*sin(q2+q3)*w1*w3);
h3= -1/2*m3*L1*L3*sin(q2+q3)*w1*w2-
- 9. International Journal of Advanced Research in Engineering and Technology (IJARET), ISSN 0976 –
6480(Print), ISSN 0976 – 6499(Online) Volume 4, Issue 5, July – August (2013), © IAEME
9
1/2*m3*L3*(L1*sin(q2+q3)+ L2*sin(q3))*w1*w3- 1/2*m3*L2*L3*sin(q3)*w2*w3-
1/2*(-m3*(L1*L3*sin(q2+q3)+ L2*L3*sin(q3))*w1^2- m3*(L1*L3*sin(q2+q3)+
2*L2*L3*sin(q3))*w1*w2- m3*(L1*L3*sin(q2+q3)+ L2*L3*sin(q3))*w1*w3-
m3*L2*L3*sin(q3)*w3^2- m3*L2*L3*sin(q3)*w2*w3);
h= [h1; h2; h3];
% Gravity related terms
% center of mass of links
r11= [-L1/2; 0; 0; 1];
r22= [-L2/2; 0; 0; 1];
r33= [-L3/2; 0; 0; 1];
g= [0 -9.81 0 0];
c1= -(m1*g*U11*r11+ m2*g*U21*r22+ m3*g*U31*r33);
c2= -(m2*g*U22*r22+ m3*g*U32*r33);
c3= -(m3*g*U33*r33);
% gravity matrix
c= [c1; c2; c3];
% Finally L-E Equation of motion of 3-link is
% Kinetic energy (= kinetic + coriolis & centrifugal)+ Potential energy
T= (D*q)/t^2+ h+ c
5. CONCLUSION
From above given approach of discretising the flexible link in number of rigid links one
can solve the kinematics and dynamics of the flexible link. Usually this approach is very much
applicable in hyper-redundant or serpentine type of robot.
REFERENCES
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