MENG 4019 ‐ Practical 3 – 2022
Task: design and simulate the operation of a hydraulic curcuit. A hydraulic cylinder, stroke 2000 mm, D piston 100 mm, d rod 50 mm angled 75 degrees from horizontal is lifting a mass load of 500 kg. Pulll external force 8000 N, push external force 10000 N. The cylinder needs to be able to move at controlled speeds and stop at intermediate positions.
Firrst, we build a conceptual circuit:
1. Open Automation studio, select and insert the following components from the Hydraulic set of components. The Proportional hydraulic valve is available in the Proportina Hydraulic category
2. Connect circuit as shown below:
3. Select the Simulation tab and run a Normal Simulation. Then, Stop Simulation and run a Slow‐Motion Simulation. Click on the valve to change position.
4. Note: in the neutral position, with no load on the cylinder, the rod is slowly extending due to the difference in force originating from the difference in area between the two chambers
5. Double click on the hydraulic cylinder. Open the Data tab. Ensure the star at the top of the Component Properties is unselected. Modify the characteristics of the cylinder as in the task definition and close the Component Properties:
6. We want to control the direction and speed of the cylinder. Because we have a proportional valve, the opening of the valve is proportional to the current applied to the proportionalsolenoid that controlsthe valve. We need a control device for the current – a joystick. Insert it from the Environment & Control category:
7. We need to link an axis of the joystick to the proportional valve. Link Input Signal – Proportional to the X‐axis of the joystick ‐JY‐X. The two signals are Real, i.e. continuous, proportional with the input. Double click and link them:
8. Experiment with positioning the joystick in various positions between ‐10 and 10 and examine the flow and the speed of the piston.
9. We can insert the dynamic measurement instrument and determine the proportionality between the joystick position on X axis and the speed of the piston. The maximum speed of the piston depends on the pump output, the opening of the proportional valve and the dimensions of the cylinder (piston and rod diameters). We can see in the figure below that the pressures before and after the proportional valve differ. Thisis consistent with the role and behaviour of this category of valves
Note1: only the X‐axis of the joystick controls the speed of the piston.
Note 2: there seems to be proportionality only for a limited opening of the proportional valve. This is true and is equivalent to magnetic saturation in electrical machines.
In Hydraulics, proportionality is a property that is valid only for a small opening of the control orifices or valves.
10. We want to control the position of the piston so that it is proportional to the position (displacement) of the joystick.
We need to determine the difference between the position of the joystick and the position of the piston and set it as an error that the circuit will attempt to minimise.
Insert a Control Device from the Environment & Control category:
11. Break the association between the proportional valve and the joystick. Right click on the Association and click on delete link
12. We need to link the proportional valve with the joystick by using the Control Device. Double click on the Control Device and open the Output Variable option from the left:
13. In the Control Device Output tab from the left, link the output from the joystick and the position of the piston. We input the equation (Output X signal from Joystick – Output Signal position of the piston) as real values. Select the appropriate variables, determine the code assigned by AS to each device.
14. Simulate the circuit. Modify the position of the joystick as control input:
The circuit works, but exactly not as expected. We need to do de‐bugging and tuning.
First issue: when the joystick input is at 0 (i.e. in the middle), the piston position is fully retracted instead on being at the middle of the stroke.
We need to correct this, and we modify the Control Device Output as follows:
This means that when the cylinder is at the middle, (i.e. at output signal 5 on a zero to 10 scale) and the joystick is at zero, the error shall be zero.
15. There is, however, still, a relatively substantial error. In this case, the joystick is at +1 and we would expect that the position of the piston to be at 6 (middle, 5, +1)
16. We need to fine tune the system. This can be done in a number of steps. First, increase the quality of the joystick – normally a 5% hysteresis, reduce it to 0.5%
17. Increase the Proportional gain from 1 to 10
The error is reduced to less than 0.6 of the stroke or similar.
18. Adjust the gains so that you consistently minimise the error over the whole joystick input range while avoiding instability.
There is a structural issue with the circuit that leads to an inherent error. This comes from the properties of the proportional valve and the open‐loop control of the circuit. This can be solved by modifying the construction of the proportional valve to achieve correct proportionality between spool position and/or with a suite of sensors and fine tuning on the circuit:
19. If you have one, you can attach a joystick to the circuit and control the valve with it
Once an external joystick is attached, it will need a bit of fine tuning to align specific inputs and required outputs