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MODELING and SIMULATION OF HYDRAULIC POWER STEERING SYSTEM WITH MATLAB. MOHAMMAD AKTERUZZAMAN Advisor: DR. SHUVRA DAS. MODELING and SIMULATION OF HYDRAULIC POWER STEERING SYSTEM. MODELING OF the MECHANICAL and HYDRAULIC COMPONENTS of a POWER STEERING SYSTEM.
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MODELING and SIMULATION OF HYDRAULIC POWER STEERING SYSTEMWITH MATLAB MOHAMMADAKTERUZZAMAN Advisor:DR. SHUVRA DAS
MODELING and SIMULATION OF HYDRAULIC POWER STEERING SYSTEM • MODELING OF the MECHANICAL and HYDRAULIC COMPONENTS of a POWER STEERING SYSTEM. • SIMULATION OF THE MODEL BY MATLAB. • Model REPRESENTS THE DYNAMIC RESPONSES OF THE power Steering System AND is CAPABLE OF ESTIMATING the effect of parameters on system response. • Model is used to study the effect of various system parameters on system response.
PRIOR WORK Discussion of Reference Model : • Ali Keyhani : He presents the identification of the dynamic model for a power steering system constructed using a rotary valve based on Mathematical (ODE). • Jose J. Granda : Analyze a multi energy non linear system using a bond graph model. • Joel E. Birching : He describes a method of applying the orifice equation to a steering valve along with the procedure for experimentally determining the flow Co-efficient for this equation.
Prior Work • AMESim (Object oriented software) :This case study gives us a good understanding of how AMESim can be used to construct parameterize and analyze complex hydro-mechanical dynamic model like power steering system. • N.Riva, E.Suraci (ADAMS based work) : A methology has developed to simulate the vehicle dynamics through Adams Car and Matlab co-simulation.
Prior Work • We took Ali Keyhani’s dynamic power steering model consisting of ordinary differential equations for Mechanical and Hydraulic system. • Some of the design Parameters are difficult to obtain. Ali Keyhani used experimental data & least square approach to determine these parameters.
ALI KEYHANI MODEL :Mechanical subsystem • The equations for the steering column, pinion and rack can be written : Equation 1 : Equation 2 :
ALI KEYHANI MODEL :Mechanical subsystem Where Td=Torque generated by the driver, Theta1=rotational displacement for the steering column, K2=tire stiffness B2=Viscous damping coefficient B1=friction constant of the upper-steering column X=displacement of the rack m= mass of pinion Ap= Piston area K1=torsion bar stiffness J1=Inertia constant of the upper steering column
ALI KEYHANI MODEL :Mechanical subsystem The following assumptions were made : -the pressure forces on the spool are neglected. -the stiffness of the steering column is infinite. -the inertia of the lower steering column (valve spool and pinion) is lumped into the rack mass.
ALI KEYHANI MODEL hydraulic subsystem By applying the orifice equations to the rotary valve metering orifices and mass conservation equations to the entire hydraulic subsystem the following equation are obtained : Equation 1 : Equation 2 : Equation 3 :
ALI KEYHANI MODEL hydraulic subsystem Where Ps and Po =supply and return pressure of the pump. Pl and Pr = cylinder pressure on the left and right side. Q = supply flow rate of the pump A1 and A2 are the metering orifice area Rho = density of the fluid Beta=bulk modulus of fluid L=length of the cylinder Cd= discharge co-efficient
ALI KEYHANI MODEL hydraulic subsystem The following assumption were made : -there is no pressure drop on the fluid transmission lines between the pump and the valve and the cylinder. -the wave dynamics on the fluid transmission lines are neglected -the bulk modulus of the fluid is considered constant -the inertance of the fluid is neglected -there is no leakage at the piston-cylinder interface -the return pressure dynamics are negligible
Information lacking in ALI KEYHANI’s Work -Missing relationship for variation of A(theta), Torque and Flow rate Q. -His established parameters do not say from which type of vehicle they were obtained.
Value of Q • Q=1.5 GPM (gallon per minute) for reasonable minimum with the quicker steering ratios for pavement cars. • Q=2.5 GPM for dirt . (reference : power steering Tech, www.woodwardsteering.com) □ Q=.0002 m3/s (reference : H.Chai. Electromechanical Motion Devices, Upper addle River, NJ:Prentice Hall PTR,1998)
Value of Torque • Td=0-8 N-m is not enough to excite the lower steering column modes. (reference : Ali Keyhani) Td=0-2 N-m is required at the handwheel during normal driving ranges. Td=15 N-m in extreme cases. (reference : H.Chai. Electromechanical Motion Devices, Upper addle River, NJ:Prentice Hall PTR,1998)
Model • Using the equations and input data a MATLAB based program was written • Model parameters were adjusted to obtain the results reported by Ali Keyhani
Results from different study • Include the other results that I had suggested. (driver torque Vs. assist torques, also in previous pages copy fig from reference)
Response graph :Effect on theta(radians) Torque Td=2, 9, 15 N-m
Response graph :Effect on cylinder pressure Torque Td=2, 9, 15 N-m
Response graph :Effect on rack Assist pressure vs. rotation angle (theta)Torque Td=2, 9, 15 N-m
Response graph :Effect on pump pressureTorque Td=2, 9, 15 N-m
Response graph :Effect on assist pressure w.r.t rotation on degreePump flow rate, Q=0.00014,0.00016,0.00024 m3/S.
Response graph :Effect on Pump pressurePump flow rate, Q=0.00014,0.00016,0.00024 m3/S.
Response graph :Effect on displacement (X)Pump flow rate, Q=0.00014,0.00016,0.00024 m3/S.
Response graph :Effect on cylinder pressurePump flow rate, Q=0.00014,0.00016,0.00024 m3/S.
Response graph :Effect on thetaPump flow rate, Q=0.00014,0.00016,0.00024 m3/S.
Response graph :Effect on Assist pressure on rotation angle (degree)J1=.0000322, .0000598 N-m-s2/rad
Response graph :Effect on pump pressure J1=.0000322, .0000598 N-m-s2/rad
Response graph :Effect on rack displacement (X) in meterJ1=.0000322, .0000598 N-m-s2/rad
Response graph :Effect on right cylinder pressure( N/m2)J1=.0000322, .0000598 N-m-s2/rad
Response graph :Effect on Rotation( radians)J1=.0000322, .0000598 N-m-s2/rad
Response graph : Effect assist pressure( N/m2) on rotation angle ( degree)When m=4.76, 8.84 Kg
Response graph : Effect pump pressure( N/m2) When m=4.76, 8.84 Kg
Response graph : Effect displacement(X) When m=4.76, 8.84 Kg
Response graph : Effect on cylinder pressure When m=4.76, 8.84 Kg
Response graph : Effect on assist pressure with rotation when K1=27.651,31.33N-m/rad
Response graph : Effect on pump pressure When K1=27.651,31.33N-m/rad
Response graph : Effect on rack displacement When K1=27.651,31.33N-m/rad