This project consisted of using both MATLAB and Siemens NX to carry out simulations and analysis of mechanism systems.
A constrained rigid body system, in which one of the bodies is in the frame is referred to as a mechanism. When considering a constrained rigid body system, the degrees of freedom of that mechanism is important.
For this project the simulation and analysis of a hypathtical two-bar linked planar manipulator which welds a structure in a production line was carried out. Also a four-bar link planar mechanism thats would be driven by an electric motor was also simulated analyed.
MATLAB software will be used to create a simulation of both these link bar mechanisms, with defining variables, such as the velocity, length, mass and inertia. For theoretical calculations, MATLAB is a popular software when it comes to the analysis of experimental data. The mechanisms will also be modelled in Siemens NX, which a motion simulation will then be carried out to produce graphical results which will be compared against results produced in MATLAB.
The two-bar link planar manipulator shown in Figure 1, consists of link OA, which is fixated to the point of and to the link AB at point A. To ensure that point B travels in a straight-line path in the XY coordinates, from (0.3m,0.0m) to (0.0m,0.3m) with a constant velocity of 2 m/s, the angles 𝜃1 and 𝜃2 will be controlled.
Figure 1: Two bar link manipulator
Before carrying out the simulation, a mathematical model of the system needs to be developed. The parameters given of the two-bar link system are the following:
r1 = 0.3m
r2 = 0.2m
Velocity at point B = 2 m/s
Where:
r1= Link OA length
r2= Link AB length
A mathematical analysis will be performed to obtain equations that describe the system.
For the closed loop OABC:
Equation 1
Equation 2
Solving for both 𝑟1𝑐𝑜𝑠𝜃1 and 𝑟1𝑠𝑖𝑛𝜃1 gives:
Equation 3
Equation 4
By squaring and adding up both sides of the equations the mechanisms position can be found:
Equation 5
Where:
Now 𝜃2 can be calculated:
Equation 6
Where:
Equation 7
The angular velocity can now be calculated using both Equation 1 and Equation 2 from above differentiating with respect to time:
r1=0.3;
r2=0.2;
YB=sqrt(2);
i=0;
XB=-YB;
for X=0.3:-0.003:0;
Y= -X+0.3;
A=2*r2*X;
B=2*r2*Y;
C=((r2)^2)+(X^2)+(Y^2)-((r1)^2);
R=sqrt((A^2)+(B^2));
Z=C/R;
zz=sqrt(1-(Z^2));
i=i+1;
TH2=atan2(B,A)+atan2(zz,Z);
if TH2<0; TH2=TH2+2*pi;end
TH2b=atan2(B,A)-atan2(zz,Z);
if TH2b<0; TH2b=TH2b+2*pi;end
TH1=atan2((Y-r2*sin (TH2)), (X-r2*cos (TH2)));
T1(i)=TH1; T2(i)=TH2; T2b(i)=TH2b;
w1=((XB*cos (TH2)) + (YB*sin (TH2)))/(r1*sin (TH2-TH1));
w2=((XB*cos (TH1)) + (YB*sin (TH1)))/(r2*sin (TH1-TH2));
W1 (i)=w1;
W2 (i)=w2;
The two-link bar manipulator was modelled on Siemens NX software in order to carry out a motion simulation of the arm. To create the two-bar link manipulator, assembly components with the appropriate parameters were modelled.
Figure 2
Figure 3
Figure 4
Figure 5
Figure 6
The four-bar link mechanism shown Figure 7 in consists of three links, link AB, BC, CD and DA which represents the ground. Link AB is driven by a motor with the torque force Tm starting the movement in the mechanism, CD then carries the load torque Tl. A counter-clockwise direction is taken in a horizontal axis to calculate the angles.
Figure 7 :Four bar linked mechanism
Before carrying out the simulation, a mathematical model of the system needs to be developed.
The parameters given of the four-bar link system are the following:
| LINK | MASS (kg) | LENGTH (mm) | INERTIA(Kg.m2) | rG (mm) |
|---|---|---|---|---|
| AB | 0.2 | 10 | 0.00001 | 5 |
| BC | 0.4 | 35 | 0.004 | 17.5 |
| CD | 0.4 | 20 | 0.00008 | 15 |
| DA | - | 30 | - | - |
Table 1: Tables of parameters for the four-bar linked mechanism
A mathematical analysis will be performed to obtain equations that describe the system.
Giving the equations:
By squaring and adding up both sides of the equations, the following equations are obtained:
Therefore:
Where:
With the solution to equation being:
Where:
For MATLAB to be able to produce results the XY position of each link needs to be calculated. Link AB:
Link BC:
Link CD:
To analyse the angular velocity to the derivative of Equation 3 and Equation 4 are carried out, giving the following:
To calculate the angular acceleration, the derivative of the equations above in the velocity analysis can be taken to obtain:
The force conditions on points A, B, C and D need to be calculated to obtain the centre of mass acceleration:
The forces, as well as the torque acting in both the XY direction in links AB, BC, and CD need to be calculated individually to carry out a dynamic analysis.
Figure 8: Link AB
Figure 9: Link BC
Figure 10: Link CD
r1=0.01;
r2=0.035;
r3=0.02;
r4=0.03;
rg1=r1/2;
rg2=r2/2;
rg3=r3/2;
ig1=0.00001;
ig2=0.0004;
ig3=0.00008;
w1=100;
m1=0.2;
m2=0.6;
m3=0.4;
a1=0;
i=0;
for TH1=0:0.001:2*pi
A=cos(TH1)-(r4/r1);
B=sin(TH1);
C=(r4/r3)*cos(TH1)-((r1^2-r2^2+r3^2+r4^2)/(2*r1*r3));
R=sqrt(A^2+B^2);
Z=C/R;
zz=sqrt(1-Z^2);
i=i+1;
TH3=atan2(B,A)+atan2(zz,Z);
if TH3<0; TH3=TH3+2*pi;end
TH3b=atan2(B,A)-atan2(zz,Z);
if TH3b<0; TH3b=TH3b+2*pi;end
TH2=atan2((-r3*sin(TH3)-r1*sin(TH1)),(r4-r3*cos(TH3)-r1*cos(TH1)));
T1(i)=TH1; T2(i)=TH2; T3(i)=TH3; T3b(i)=TH3b;
w3=w1*(r1*sin(TH1-TH2))/(r3*sin(TH2-TH3));
w2=w1*(r1*sin(TH3-TH1))/(r2*sin(TH2-TH3));
W3(i)=w3; W2(i)=w2;
a3=(w1^2*r1/r3)*((1-(w2/w1))*cos(TH1-TH2)*sin(TH2-TH3)-((w2/w1)-(w3/w1))*cos(TH2-TH3)*sin(TH1-TH2))/(sin(TH2-TH3))^2;
a2=-(w1^2*r1/r2)*((1-(w3/w1))*cos(TH1-TH3)*sin(TH2-TH3)-((w2/w1)-(w3/w1))*cos(TH2-TH3)*sin(TH1-TH3))/(sin(TH2-TH3))^2;
A3(i)=a3; A2(i)=a2;
xg1= -rg1* (a1*sin(TH1) +w1^2*cos (TH1));
yg1= rg1* (a1*cos (TH1) -w1^2*sin (TH1));
xg2= -r1* (a1*sin (TH1) +w1^2*cos (TH1))- rg2* (a2*sin (TH2) +w2^2*cos(TH2));
yg2= r1* (a1*cos (TH1) -w1^2*sin (TH1))+ rg2* (a2*cos (TH2)-w2^2*sin (TH2));
xg3= (r3-rg3)*(a3*sin(TH3)+w3^2*cos(TH3));
yg3= -(r3-rg3)*(a3*cos(TH3)-w3^2*sin(TH3));
Matrices can now be used to solve the dynamic equations used above. The matrix will be written in the form:
[𝐴] 𝑋 [𝐵] = [X]
Where:
A= is the multiplying variable of the instantaneous geometry of the mechanism
B= is the multiplying variable defines the velocity’s and acceleration links
X= The sum of the multiplying variables A and B and where the plot for the MATLAB results will be obtained from.
A=[1 0 1 0 0 0 0 0 0;
0 1 0 1 0 0 0 0 0;
rg1*sin(TH1) -rg1*cos(TH1) -(r1-rg1)*sin(TH1) (r1-rg1)*cos(TH1) 0 0 0 0 1;
0 0 -1 0 1 0 0 0 0;
0 0 0 -1 0 1 0 0 0;
0 0 -rg2*sin(TH2) rg2*cos(TH2) -(r2-rg2)*sin(TH2) (r2-rg2)*cos(TH2) 0 0 0;
0 0 0 0 -1 0 1 0 0;
0 0 0 0 0 -1 0 1 0;
0 0 0 0 -rg3*sin(TH3) rg3*cos(TH3) -(r3-rg3)*sin(TH3) (r3-rg3)*cos(TH3) 0];
B =[ m1*xg1;
m1*yg1;
ig1*a1;
m2*xg2;
m2*yg2;
ig2*a2;
m3*xg3;
m3*yg3;
ig3*a3];
X=inv(A)*B ;
XA (i)=X(1,1);
YA (i)=X(2,1);
XB (i)=X(3,1);
YB (i)=X(4,1);
XC (i)=X(5,1);
YC (i)=X(6,1);
XD (i)=X(7,1);
YD (i)=X(8,1);
TM (i)=X(9,1);
To create the four bar-linked mechanism for links part models shown in Figure 11-Figure 14 had to be created in NX which would then be assembled into the mechanism
Figure 11: Link AB part
Figure 12: Link AD part
Figure 13: Link BC part
Figure 14: Links CD part
Figure 15 shows the four-bar linked mechanism assembled and an emotion simulation. The joints parameters featured for each link have the appropriate features such as the revolute and the slider. Link AD simulated to be the ground and is set in an always stationary position. Link AB is the moving force and the mechanism and rotates counterclockwise at an angular velocity of 100 rads/s.
Figure 15: Completed assembly in motion simulation
Figure 16: Variation of link orientations
Figure 16 above shows variation of link orientation with the angle’s theta 1 and theta 2. It is seen that both angles are positive, as they rotate along the excesses in an anticlockwise direction. Theta one begins above -50 degrees and increases to an angle of approximately 50 degrees indicating a continuous upward direction. Theta two, on the other hand, starts in a positive position and begins to increase to points beyond 150 degrees steadily.
Figure 17: Variation of angular velocity
Figure 17 shows the variation of angular velocity for the 21. It is shown that the mega one begins a lower place than omega one at around 1 rad/sec, with omega two starting above 10 rad/sec. An increase in velocity is noticed in the second angle when in comparison to the first this is due to rapid acceleration. The first angular velocity has lower peak point in the second angular velocity, but the second angular velocity sees a sharp decrease of this peak is reached when compared to the first angular velocity.
Figure 18: Polar load diagram for joint A
Figure 19:Polar load diagram for joint B
Figure 20:Polar load diagram for joint C
Figure 21: Polar load diagram for joint D
Figure 18-Figure 21 shows the polar loads of joints A, B, C and D and is the result of the dynamic analysis of all the forces. It is apparent that Figure 18 and Figure 19 a very similar in creating perhaps most joints having the same magnitude while moving with the constant velocity. The same can be said for joints C and D.
Figure 22:Variation of motor torque
Figure 22 above shows the variation when a torque is applied to joints A and B(see Figure 8).At each cycle of the mechanism there is an increase in the torque of the motor and the angular velocity is constant. As the link moves energy is stored in the link and when the energy generated from the torque of the motor is zero, the energy stored in the link returns to the motor and vice versa. This explains the fluctuation of the peaks shown for the torque above.
Figure 23: Variation of angular velocity
Figure 23 shows the variation of the link angular velocity, and when compared to the results produced from the MATLAB plot, the results are the same except the that the angular velocity values are positive due to NX not providing negative values.
Figure 24: Polar load diagram results for joint A in NX
Figure 25:Polar load diagram results for joint B in NX
Figure 24-Figure 26 show a close resemblance to Figure 19-Figure 21, but the magnitudes shifted slightly indicating an orientation or fault during the NX modelling process.
Figure 26:Polar load diagram results for joint C in NX
Figure 27:Polar load diagram results for joint D in NX
The simpler of the two methods was the modelling method using MATLAB. After creating the parts in NX and assembling it to create both the two bar and the four-bar mechanisms, setting up the joint parameters was fairly simple, but problems occurred during the orientation of the assembly during the motion simulation process this resulted in graphs that was similar compared to the MATLAB counterpart but was not correct. To conclude both methods (MATLAB and NX) were valuable for carrying out the dynamics and kinematic analysis of a mechanism.