ANALYSIS OF SPATIAL MECHANISM
IN DYNAMIC EQUILIBRIUM
CONDITION USING MATLAB
KAUSHIK V. PATEL1
Lecturer in Mechanical Engineering Department, B.S.Patel Polytechnic , Ganpat Vidhyanagar, Mehsana, Gujarat, India
DR ANURAG VERMA2
Principal, G H Patel College of Engineering & Technology, Vallabh Vidhyanagar, Gujarat, India-388120
Abstract
Over the last two decades, spatial mechanisms evolved from rather marginal machines to widely used mechanical architectures. The use of computer in kinematic and dynamic simulation has a powerful tool for the analysis and design of multi body systems in fields such as automobile industry, aerospace, robotics, machinery, biomechanics, and others. The purpose of this work is to describe use of kinematic and dynamic simulation, and advanced topics. For a kinematic chain it is important to know how forces and moments are transmitted from the input to the output, so that the links can be properly designated. Idea present here has been implemented with MatLab Program and it is used for analysis of mechanism in dynamic equilibrium condition.
Keyword: Spatial, Contour, Dynamic, Reaction, Equilibrium
Introduction:
The links are represented with bars or triangles. The one degree of freedom joints are represented with a cross circle. General analytical model and inertial properties of each moving link is represent in fig 3.The link 1 is connected to ground 0 at A and to link 2 at B, Fig. 1. The link 2 is connected to link 1 at B and to link 3 at B. Next, link 3 is connected to link 2 at B, link 0 at C, and link 4 at D. Link 3 is a ternary link because it is connected to three links. At B there is a joint between link 1 and link 2 and a joint between link 2 and link 3. Link 4 is connected to link 3 at D and to link 5 at D. The last link, 5, is connected to link 4 at D and to 0 at A. In this way the table in Fig. 1.12a is obtained. At A there is a multiple joint, two rotational joints, one joint between link 1 and link 0, and one joint between link 5 and link 0. The number of DOF for the mechanism is M = 3(5)
−2(7)= 1. If M = 1, there is just one driver link. One can choose link 1 as the driver link of the mechanism. Once the driver link is taken away from the mechanism the remaining kinematic chain (links 2, 3, 4, 5) has the mobility equal to zero. The dyad is the simplest system group and has two links and three joints.
Theory:
The position analysis of a kinematic chain requires the determination of the joint positions, the position of the centers of gravity, and the angles of the links with the horizontal axis.
Let (xA, yA) be the coordinates of the joint A with respect to the reference frame xOy, and (xB, yB) be
the coordinates of the joint B with the same reference frame. Using Pythagoras the following relation can be written
(xB−xA)2+(yB−yA)2= AB2= L2AB (A)
Where LAB is the length of the link AB. Let φ be the angle of the link AB with the horizontal axis Ox. Then, the
slope m of the link AB is defined as
m = tan φ = yB –yA/xB−xA (B)
Let n be the intercept of AB with the vertical axis Oy. Using the slope m and the intercept n, the equation of the
y = m(X)+n (C )
Where x and y are the coordinates of any point on this link.
The classical method for obtaining the velocities and accelerations involves the computation of the derivative with respect to time of the position vectors. The method of contour equations avoids this task and uses only algebraic equations.
Fig: 1.a First Contour method for velocity
Fig: 1.b First Contour method for acceleration
Fig: 2.b Second Contour method for acceleration
The velocity equation for simple closed loop kinematic chain is
∑ , 0 (D)
And
∑ , ∑ , 1 0 (E)
Where
ωi,i−1 is the relative angular velocity of link (i) with respect to link (i−1);
rAi = is the position vector of the joint Ai;
VA i, i−1 = V rel Ai,I Ai,i−1 is the relative velocity of Ai,i on link (i) with respect to Ai,i−1 on link (i−1).
The acceleration equations for a simple closed kinematic chain are
∑ , 1 ∑ , 1 0 (F)
∑ , 1 , ∑ , ∑ , ∑ , 0 (G)
Where
αi,i−1 is the relative angular acceleration of link (i) with respect to link (i−1);
ωi is the absolute angular velocity of the link (i), or the angular velocity of link (i) with respect to the “fixed”
reference frame Oxyz, ωi = ωi,0;
, , , , , link i-1 (H)
, = 2 , (I)
(J)
Acceleration equation for planar kinematic chain
∑ , 0 (K)
and
∑ , ∑ , ∑ , 0 (L)
In our case for counter-I ( 0-1-2-3-0) AR, BR, BT, CR are rotational joint between link 0 and 1, rotational joint
The angular velocity of based on eq. (D) and (E)
/
For velocity analysis 0
0
With the help of rB, rC and Ф the equation becomes
0
0
0 0 0 0 0 0
And for acceleration analysis based on equation (F) and (G)
0
0
In contour II (0-3-4-5-0) rotational joint CR, DR, AR between link 0 and 3, link 3 and 4 , link 5 and 0
respectively. Also translation joint DT between link 4 and 5.
Velocity and acceleration analysis is similar to contour I.
0
0
0
0
0 0 0 0 0 0
0
0
Dynamic analysis:
Inertia forces and moments can be calculated by following equation for link 1
Mass m1= ρA B h d kg (M)
Inertia force Fin = mass m1 x aC1 N (N)
Moment of inertia of link 1 with respect to C1 is
IC1= m1(AB2+h2)/12 kgm2 (O)
Moment of inertiadriver link is
0 (P)
The planar mechanism is shown in Fig 3. The following numerical data are given: AB = 0.15 m, AC = 0.10 m, CD = 0.15 m, DF = 0.40 m, and AG =0.30 m. the height of the link 1,3,and 5 is 0.010m. width of link 2 and 4 (slider) is 0.050 m. height of link 2 and 4 (slider) is 0.020 m. the angular velocity of driver link is n=50 rpm. Assume steel material. (ρ = 8000 kg/m3). Find the velocities and accelerations of the mechanism when the angle of the driver link 1 with the horizontal axis is φ = φ1 = 30◦. Find motor moment for dynamic equilibrium and joint reaction forces.
Result:
By Using theory and mention equation MatLab program give following results.
Tabel:1 Result by Matlab Program Velocity analysis
Linear velocity (m/s) Angular velocity (red/sec)
VB=VB1=VB2 [-0.366519, 0.63483, 0] ω1 [0, 0, 5.23599]
VB32 0.313096 ω2 = ω3 [0, 0, 5.44826]
VB3B2 [0.312037, 0.0257363, 0] ω3z 5.44826
VD3=VD4 [0.0671766, -0.814473, 0] ω4= ω5 [0, 0, 0.917134]
VD54 0.757991 ω5z 0.917134
VD5D4 [-0.34018, 0.677368, 0]
Acceleration analysis
Linear Acceleration (m/s2) Angular Acceleration ( red/sec2)
aB1= aB2 [-3.32396, -1.91909, 0] α1 [0, 0, 0]
aB32cor [-0.280436, 3.40011, 0] α2= α3 [0, 0, 14.5681]
aB32 -0.140694 α32 14.5681
aB3B2 [-0.140218, -0.011565, 0] α4= α5 [0, 0, -5.77155]
aB3= aB4 [4.61708, -1.81183, 0] α52 -5.77155
aD54cor [-1.24248, -0.623982, 0]
aD54 3.41104
Dynamic Force analysis:
Link 1 Link 2
m1 0.0112 (kg) m2 0.008 (kg)
m1 aC1 [ -0.0186142, -0.0107469, 0] (N) m2aC2 [ -0.0265917, -0.0153527, 0] (N)
Fin1 [ 0.0186142, 0.0107469, 0] (N) Fin2 [ 0.0265917, 0.0153527, 0] (N)
G1 [ 0, -0.109838, 0] (N) G2 [ 0, -0.078456, 0] (N)
IC1 1.83867e-005 (kg m^2) IC2 1.93333e-006 (kg m^2)
IC1α1 [ 0, 0, 0] (N m) IC2α2 [ 0, 0, 2.8165e-005] (N m)
Min 1 [ 0, 0, 0] (N m) Min2 [ 0, 0, -2.8165e-005] (N m)
Link 3 Link 4
m3 0.032 (kg) m4 0.008 (kg)
m3 aC3 [ -0.0492489, 0.0193262, 0] (N) m4 aC4 [ 0.0369367, -0.0144946, 0] (N)
Fin3 [ 0.0492489, -0.0193262, 0] (N) Fin4 [ -0.0369367, 0.0144946, 0] (N)
G3 [ 0, -0.313824, 0] (N) G4 [ 0, -0.078456, 0] (N)
IC3 0.000426933 (kg m^2) IC4 1.93333e-006 (kg m^2)
IC3α 3 [ 0, 0, 0.00621962] (N m) IC4α4 [ 0, 0, -1.11583e-005] (N m)
Min3 [ 0, 0, -0.00621962] (N m) Min4 [ 0, 0, 1.11583e-005] (N m)
Link 5
m5 0.04 (kg)
m5 aC5 [ 0.0553516, 0.0183855, 0] (N)
Fin5 [ -0.0553516, -0.0183855, 0] (N)
G5 [ 0, -0.39228, 0] (N)
IC5 0.000833667 (kg m^2)
IC5α 5 [ 0, 0, -0.00481155] (N m)
Min5 [ 0, 0, 0.00481155] (N m)
Result of Joint reactions and equilibrium moment
Link 5
-.149492*yP-.373731e-1-.297670*xP= 0 (1)
-.149492*F45x+.297670*F45y = 0 (2)
F05x+F45x-.553516e-1 = 0 (3)
F05y+F45y-.410666 = 0 (4)
.112198*F05y+.223409*F05x+(xP+.112198)*F45y-1.*(yP+.265909e-1)*F45x-99.9952 = 0 (5)
Link 4
F34x-1.*F45x-.369367e-1 = 0 (6)
F34y-1.*F45y-.639614e-1 = 0 (7)
-1.*(xP+.149492)*F45y+(yP-.476701e-1)*F45x+.111583e-4 = 0 (8)
By solving equation 1 to 8
We obtain F05x, F05y, F45x, F45y, F34x, F34y, xP, yP
F05 = [ 268.165, 135.057, 0] (N)
F45 = [ -268.109, -134.647, 0] (N)
F34 = [ -268.072, -134.583, 0] (N)
rP = [ -0.149492, 0.0476701, 0] (m) Link 3
.121244*yQ-.727461e-2-.100000e-1*xQ= 0 (9)
.121244*F23x+.100000e-1*F23y= 0 (10)
268.122+F03x+F23x= 0 (11)
134.250+F03y+F23y= 0 (12)
(13)
Link 2
-1.*F23x+F12x+.265917e-1 = 0 (14)
-1.*F23y+F12y-.631033e-1 = 0 (15)
-1.*(xQ-.121244)*F23y+(yQ-.700000e-1)*F23x-.281650e-4 = 0 (16)
By solving equation 9 to 16, we obtain following result F03x, F03y, F23x, F23y, F12x, F12y, xQ, yQ
F03 = [ -256.745, -272.179, 0] (N)
F23 = [ -11.3762, 137.93, 0] (N)
F12 = [ -11.4028, 137.993, 0] (N)
rP = [ 0.121243, 0.07, 0] (m) Link 1
F01 = [ -11.4214, 138.092, 0] (N)
Mm = [ 0, 0, 17.5356] (N m)
Conclusion and Future Work:
The work will assist all those interested in the design of mechanisms, manipulators, building machines, textile machines, vehicles, aircraft, satellites, ships, biomechanical systems (vehicle simulators, barrier tests, human motion studies, etc.), controlled mechanical systems, mechatronical devices and many others. Analysis of can be solved by Euclidian Distance Function and result can compare with Constraint Condition methods. Mechanisms and robots have been and continue to be essential components of mechanical systems. Mechanisms and robots are used to transmit forces and moment and to manipulate objects. A now ledge of the kinematics and dynamics of these kinematic chains is most important for their design and control. MATLAB is a modern tool that has transformed the mathematical calculations methods because MATLAB not only provides numerical calculations but also facilitates analytical calculations using the computer.
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