Fall 2002PURPOSE:THEORY:APPARATUS:PROCEDURE:RESULTS:ANALYSIS:CONCLUSION:Grand Valley State UniversityThe Padnos School of EngineeringMECHANICAL COMPONENTSEGR 345 Dynamic Systems Modeling and ControlBrad Vander Veen September 26, 2002Lab PartnersDale SlotmanRyan JeffriesLeanne NewmanFall 2002PURPOSE:The purpose of this lab is to explore simple translational systems, which consist of simplemechanical devices such as masses, springs, and dampers. Observations will be taken,and conclusions will be made as to how these mechanical devices affect the behavior of the system.THEORY:Suppose we have a mass-spring system as seen below in Figure 1. Spring (KS) M X Figure 1 – Mass-Spring systemThis system can be modeled by the differential equations: ''MxMgxKS (1)Solving this differential equation gives us a displacement function x(t): SSSKMgtMKKMgtx )cos()( (2) By assuming values for M and KS, this function can be plotted as seen in Figure 2 below.0 2 40.50.1.5mass_spring t( )50 t Figure 2 – Plot of Mass-Spring system (displacement vs. time)As seen in Figure 2, a mass spring system will simply oscillate with a certain amplitude and frequency.Suppose we have a mass-damper system as seen below in Figure 3. Figure 3 – Mass-Damper systemThis system can be modeled by the differential equations: ''' MxMgxKd (3)Solving this differential equation gives us a displacement function x(t): dtMKdKMgeKMgtxd)( (4) By assuming values for M and Kd, this function can be plotted as seen in Figure 4 below. 0 0.5 100.511.10mass_damper t( )10 t Figure 4 – Plot of Mass-Damper system (displacement vs. time)As seen in Figure 4, a mass-damper system will logarithmically approach the steady-statedisplacement, and will not exceed it.Suppose we have a mass-spring-damper system as seen below in Figure 5.Spring KS Damper Kd M x Figure 5 – Mass-Spring-Damper systemThis system can be modeled by the differential equations: ''' MxMgxKxKdS (5)Solving this differential equation gives us two different displacement functions. One function of for an over-damped system (real roots), and the other is for an under-damped system (complex roots).Over-damped system: StRtRKMgeCeCtx 2121)( (6) Under-damped system: StRKMgCtReCtx )sin()(4433 (7)By assuming values for M, KS, and Kd,, the constants can be solved for, and the functions can be plotted as seen in Figures 6 and 7 below.0 5 100102019.356490real_mass_spring_damper t( )100 tFigure 6 – Plot of an Over-damped Mass-Spring-Damper system (displacement vs. time)0 10 20020270imaginary_mass_spring_damper t( )200 tFigure 7 – Plot of an Under-damped Mass-Spring-Damper system(displacement vs. time)As seen in Figures 6, an over-damped mass-spring-damper system will approach the steady-state displacement, but will not exceed steady-state displacement.As seen in Figures 7, an under-damped mass-spring-damper system will reach stead-state displacement and oscillate around steady-state displacement, and then settle to steady-state displacement.APPARATUS:ITEM MANUFACTURER MODEL SERIAL NUMBER GVSU IDDigital Multi-meter Fluke 45 N/A I-96 Force Gage FDV N/A 1028 N/A Ultrasonic Sensor Allen Bradley 873C 873C-DDAV1000EZ N/A Breadboard N/A N/A N/A N/A Computer Dell N/A BP5R001 N/A Damper Closer VH440 N/A N/A Assorted Masses Assorted Wires Stepladder Spring PROCEDURE:1). Find the spring constant.F125.31 N F244.83 NApplied Forcesx1.450 m x2.526 mResulting Displacementsx x2x1 x 0.076 mChange in DisplacementF F2F1F 19.52 NChange in ForceKsFxKs256.842kgs2The Spring constant is 257 N/m.2). Find the undeformed length of the spring. y = mx + bSince the slope and one data point are know, these values can be substituted:(25.31) = (257)(.450) + bSolve for b:b = -90.3The force-displacement equation is:y = 257x - 90.3Setting this equation equal to zero, the undeformed length can be found:0 = 257(x) - 90.3 x = .351 metersNOTE: At rest, the spring was .420 meters long 3). Find the spring constant for the spring inside the damper.F183.97 N F2128.8 NApplied Forcesx1.09 m x2.170 mResulting Displacementsx x2x1 x 0.08 mChange in DisplacementF F2F1F 44.83 NChange in ForceKsFxKs560.375kgs2The Spring constant is 560 N/m.4). Find the undeformed length of the spring inside the damper.The standard equation for a line is:y = mx + bSince the slope and one data point are know, these values can be substituted:(128.8) = (560)(.170) + bSolve for b:b = 33.6The force-displacement equation is:y = 560x + 33.6Setting this equation equal to zero, the undeformed length can be found:0 = 560(x) + 33.6x = -.06 meters5). Setup of the Displacement Measuring ApparatusA program was created in Labview that could take readings from the ultrasonic range sensors and store them in a spreadsheet. See Figure 8 below for the Labview setup. Notethat ACHO pin on the DAQ card is the analog signal input.Figure 8 – Labview ProgramThe ultrasonic range sensor was setup according to the schematic seen below in Figure 9. Note that the output of the ultrasonic range sensor is hooked to pin ACHO on the DAQ card, and the AIGND pin on the DAQ card is used to establish the common ground. Figure 9 – Wiring Schematic for Ultrasonic Range Sensor6). Relate Output Voltage from Sensor with the DisplacementThe sensor was placed at various displacements from the floor between 0 and 1 meter.The ultrasonic range sensor output a certain voltage at each
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