1.7 Example 6: Simple design exercise for a hydraulic suspension
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1. User Manual 4 Set up a duty cycle to give a step increase in force output at start of stage 1 null pO output at end of stage 1 null pO UDO00 imore 5 Select Parameters gt Batch parameters 6 Drag and drop the global parameter into the Batch parameters dialog box and set the following values for a batch run Figure 1 42 Batch parameters Drag the parameters into this list to make them control parameters Submodel GLOBAL diameter of orifice mm 7 Perform a batch run for 10 s and plot the displacement of the piston Figure 1 43 Batch run results for rod displacement HJUO0 1 red displacement m 0 16 1 owe 0 14 Ta gt D o E O12 go N 10 _ 0 0 08 0 2 4 G 8 10 Time s The batch run will use orifice diameters of 1 to 6 mm in steps of 0 5mm Zooming 37 Chapter 1 Tutorial examples 38 in on the plot it becomes clear that 3 mm gives a reasonable degree of damping 8 Remove the step from UD000 so that there is a constant force of 0 N 9 Insert a linearization time at 10 s 10 Repeat the batch run and look at the damping ratio for the oscillatory frequency Looking at the eigenvalues selecting the jac0 1 to jac0 11 files we see that below 2 5 mm the system is very highly damped However the results for the 1 mm diameter give an oscillatory frequency of about 25 Hz which is curious but could be investigated with tools such as modal shapes For diameters of 2 5 mm and2. all times The plot of displacement against force Figure 1 40 shows the non linear nature of the spring It also shows that the suspension does not bottom out but it does top out 35 Chapter 1 Tutorial examples 36 Figure 1 41 Force against pressure 1 HJOO0 1 external force on rod M 160 140 120 100 ou ED 40 2 5 1 5 0 5 0 5 1 5 25 3 HAQ01 1 pressure at port 1 bar 10 Figure 1 41 shows that maximum pressure is 160 bar and the minimum is about 40 bar which occurs when the suspension tops out We could continue by doing further analytical calculations Alternatively we could do batch runs varying the accumulation pre charge pressure and accumulator volume and the interested reader could try this However we will end the exercise by considering the damping of the suspension which is mainly provided by the two orifices For simplicity we will assume they are of the same characteristics Step 2 Setup a batch run varying the diameters of the orifices with the vehicle subject to a step change in force 1 Select Parameters gt Global i Set some global parameters parameters Real Integer Iert 2 Setup the global parameter shown Value DIAM diameter of onfice mm Set the following parameters for BOTH orifices Submodel Parameter title 1 for pressure drop flow rate pair 2 for orifice diameter equivalent orifice diameter mm DIAM Hydraulic Library 4 2
3. greater there is an oscillatory frequency of about 1 23 Hz and the damping ratio is as follows Diameter of orifice mm Damping ratio 25 0 533 0 308 35 0 194 4 5 0 091 0 067 a5 0 050 0 039 es Ce We can see the evolution of these eigenvalues in a root locus plot Hydraulic Library 4 2 User Manual Figure 1 44 Root locus plot 12 5 E 6 nA U3 02 0 1 10 0 6 PS E E OF 1 Hz 5 0 8 254 09 B 4 A more refined search between 2 0 and 3 0 mm would be a good idea but 2 5 mm seems reasonable 39
4. Chapter 1 Tutorial examples 1 7 Example 6 Simple design exercise fora hydraulic suspension Objectives Do a simple initial design study for a hydraulic suspension using e Analytical analysis e AMESim standard runs e Batch runs e Linear analysis The system is shown in Figure 1 38 The hydraulic jack with the two orifices is the damper and the accumulator is the spring It is proposed to use this suspension on the cab of a truck The load on each suspension strut is 250kg Figure 1 38 A simplified hydraulic suspension TRUCK SUSPENSION Step 1 Build the system and run a simulation l 32 Build the system using Premier submodel Much sizing can be done by simple calculations but simulation can be a great help in rapidly confirming the calculations and adding dynamics to the steady state values The two ports of the jack are interconnected and in equilibrium The pressures above and below the jack piston will be the same Using a force balance in the equilibrium position in terms of the piston area 4 and rod area Arod PA Z P A i1 Aog 250xg pist pist It follows that PA rod 250g Hydraulic Library 4 2 User Manual From this if we want an operating pressure of about 70 bar the diameter of the rod must be about 22 3 mm We will use a rod diameter of 20 mm and a piston diameter of 40 mm 2 Set the parameters of the following table Submodel Parameter title angle rod makes with horizonta
5. e to the rod The difference between the minimum and maximum oil volume is A oq X Stroke which is 0 1 L The accumulator volume should be a bit bigger than this but certainly not 10 L Step 1 Investigate the spring rate 1 Set the following values Submodel Parameter title Value gas precharge pressure bar HA001 accumulator volume L 1 Do a run and verify that these values do not disturb the equilibrium The values should have changed the spring rate but not the equilibrium position We need now to investigate the spring rate Hydraulic Library 4 2 User Manual 2 Set the following values Submodel Parameter title Value output at start of stage 1 null output at end of stage 1 null duration of stage 1 s output at start of stage 2 null output at end of stage 2 null duration of stage 2 s 3 Do arun for 120 s 4 Plot graphs of e rod displacement 1 000 e pressure at port 1 1 000 against e external force on rod 1 000 Figure 1 40 Displacement against force 1 HJOO0 1 rod displacement m 0 25 N 0 2 0 15 Tie 0 1 ae 0 05 0 2 0 2 5 3 TYRID0O 1 ext inal force on ted N A The force value of 2500 N pushes down on the suspension with a value corresponding to the weight of the car The force of 2500 effectively takes the complete weight off the suspension The slow evolution of the force duty cycle ensures that the system is very close to equilibrium at
6. l degree total mass being moved 250 3 Runa dynamic simulation for 10 s rod displacement m piston diameter mm HJ000 diameter of rod mm Figure 1 39 Pressure and displacement plots 1 HJOQU0 1 rod di placement m 0 1500 0 1490 0 1480 0 1470 0 1460 0 1450 0 1440 0 1430 0 2 4 9 10 Time 3 1 HJO00 1 pressure at port 1 bar Time z Figure 1 39 shows the system pressure and the displacement Problem 1 The starting values are poor 33 Chapter 1 Tutorial examples 34 Problem 2 The accumulator spring with its precharge pressure of 100 bar is taking no part in this simulation The only spring involved at the moment is the hydraulic fluid Solution to problem 1 1 In Parameter mode select Parameters gt Set final values This will give reasonable starting values for state variables You will find that the piston has dropped slightly from the mid position 2 Reset the following parameters Submodel Parameter title Value rod displacement m HJ000 rod velocity m s a 3 Runa simulation again and check that the system is in equilibrium with the rod in mid position Solution to problem 2 The two parameters we can vary are the precharge pressure and volume of the accumulator For the accumulator to work as a spring the precharge pressure must be lower than the equilibrium pressure The volume of fluid in the jack varies according to the piston position This is du
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