Slider Crank Mechanism Simulation

For the simulation, the following parameters were defined:

Gravity: -9.81m/s^2 (Y- direction)

Units:

Length : Meters

Force: Newtons

Simulation Time= 5 seconds

BASIC SETUP:

3 points were defined:

P0(0,0,0)

P1(0,-10,0)

P2(-20,0,0)

COM1(0,-5,0)

COM2(-10,-5,0)

2 Bodies were defined:

CYL1 (Center of mass coordinates at COM1)

CYL2 (Center of mass coordinates at COM2)

The graphic of cylinder (Diameter=2m) was associated with both the bodies and rest of the body properties were derived from the graphic itself.

Joints:

2 revolute Joints were defined:

Rev_J1: Revolute joint between CYL1 and Ground Body with its allignment along global Z-axis. 

Rev_J2: Revolute joint between CYL1 and CYL2 with its allignment along global Z-axis.

NOTE: IN ORDER TO AVOID ANY CONFUSION REGARDING THE PLOTS, HERE IS THE SERIES IN WHICH THE PLOTS ARE SHOWN IN THIS ANSWER:

Plot1: Velocity of P2 wrt Ground vs time

Plot2: Reaction torque at P0 without BISTOP torque

Plot3: Range of motion incase of 1000kN load 

Plot4: Range of motion incase of initial angular velocity

Plot 5:Reaction torque at P0 without BISTOP torque

Plot 3 and 4 are just for explanation purposes and weren\'t asked in the question.

Q1:

To simulate and produce sliding motion of P2 in the slider crank mechanism, an inplane joint was used.

Here is the simulatation for the following:

https://drive.google.com/open?id=1xVOdWRt0Bfo1t0YPQgzZ3EXfvrfbQyed

The details of the inplane joint used are given below:

The inplane joint is defined between the CYL2 and the ground body. The allignment of the joint is along global Z axis and the Inplane defined in the corresponding XY Plane is the Global X axis. This means that the joint has to move only along the global X axis in the XY plane defined by the allignment of the joint. This provides the required sliding motion of the joint and hence thew point P2 without introducing any redundancies.

Below is the DOF calculations to prove that this model is not over constrained:

DOF due to 2 rigid bodies (CYL1 and CYL2)= 6+6= 12

Constraints introduced due to 2 revolute joints= 5+5=10

Constraints introduced due to one inplane joint= 1

Total DOF of the system= 12-10-1=1

Thus there is no redundancy.

To verify the sliding motion we can introduce a motion (2*TIME) on Rev_J1 without introducing redundancy (the above simulation in the video link was done by introducing this motion only).

NOTE: The above motion element is not used in the rest of the simulation results.

Q2:

A rotational velocity of -0.5 rad/s (Clockwise direction along the negative Z axis) was defined as an initial condition on Rev_J1 and the simulation was run. To obtain the velocity of P2 wrt ground, an output was defined between 2 points, the first one being P2 and the second one being the global origin. The following plot was obtained.

This plot is of the velocity magnitude of P2.

As seen in the above plot, the maximum velocity of P2 is coming out to be 5m/s.

In this particular simulation, the DOF of the model is 1 only.

Q3:

The initial condition on Rev_J1 is removed and an action reaction force element is defined on Rev_J2  (Point P1), whose values are set as the following:

Fx= -1000000 N

Fy= 0

Fz= 0

For obtaining the reaction torque on the joint Rev_J1, an output is defined by the expression JOINT(301001,0,5,0).

In the above expression 301001 represents the id of Rev_J1 and 5 return the magnitude of the torque due to the joint.

The plot obtained was:

For comparing the range of motions, a displacement output was defined between P2 and the ground. Through the range covered by the sliding motion in both the cases were compared.

Case 1 (with 1000 kN load):

Case 2 (With initial velocity):

It can be clearly seen that the range of motion is greater in case 2 i.e. the one with the initial angular velocity.

To prove this analytically the following calculations were done,

Since h2>h1 thus the range of motion in case 2 is greater than the one in case 1.

Q4:

Now a BISTOP torque is inroduced in the model which is applied on Rev_J1 and is defined by the expression, `BISTOP(AZ({j_0.i.idstring},{j_0.j.idstring}),WZ({j_0.i.idstring},{j_0.j.idstring}),-10*PI/180,0,10^4,2,100,.1)`. The expression means that the BISTOP torque will start acting beyond the range -10 degrees to 0 degrees of the crank rotation motion.

If we compare both the plots we can see that the overall magnitude of the torque in case of BISTOP logically increases as the BISTOP torque acts to stop the motion of the crank beyond its limits whereas the load is trying to rotate the crank. Thus beyond the limits of the crank motion, 2 torques are being apllied, one due to the load and another one due to the BISTOP torque.