Informazioni sul documento
- Università
- Politecnico di Milano
- Corso di laurea
- Mechanical Engineering
- Materia
- Control and Actuating Devices for Mechanical Systems
- Classificazione
- Altro materiale
- Formato originale
- Testo
- Testo ricercabile
Altro di Control and Actuating Devices for Mechanical Systems per il corso di Mechanical Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.
Altro di Control and Actuating Devices for Mechanical Systems per il corso di Mechanical Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.
Qualità dell’importazione: il testo è stato estratto direttamente dal documento originale.
Passaggi rappresentativi riconosciuti nelle diverse parti del materiale. Il testo completo resta presente nella pagina per la ricerca, mentre l’anteprima compatta rende più semplice la lettura.
1 PART A: • Write the equation of motion of the system First of all we need to choose the degree of freedom according to which we are going to express all the energy contribution. We can choose the degree of freedom on which we must add the controller to simplify everything. Once we have decide which is the degree of freedom we must decide its direction This is an extremely important topic to highlight. The direction must choose so that: − In case of an actuator with two chambers , the piston moves according to the direction of the two Q: we’ll have one Q that enters and one that goes out and the piston must move from the side where Q enters to the chamber where Q goes out. Thanks to this choice we’ll have the hydraulic actuator force coherent with the motion of the piston and so we don’t need to add a minus (-) to this expression − In case of an actuator with only one chamber, the piston must move far from the entrance of Q − In case of a motor the we must have the rotation of the motor coherent (equal) to the rotation of 𝐶𝑚 In this part I have to write the equation of all th e energy contribution and then I have to express the kinematic relationship between the values of the equation and the degree of freedom (𝑞) that I have chosen. After doing this I have to substitute this relationship into the energy equation and then I ha ve to derive them in order to obtain the Lagrange equation. The equation that I will have at the end of this procedure is the equation of motion not linearized: 𝑚𝑞̈+𝑟𝑞̇+𝑘𝑞=𝐹 Kinetic energy: Concentrate mass 𝑚𝑐 𝐸𝑘,𝑚𝑐= 1 2𝑚𝐶𝑣2 Concentrate inertia 𝐽𝑐 𝐸𝑘,𝐽𝑐 = 1 2𝐽𝐶𝜗̇2 Lagrange contribution 𝑑 𝑑𝑡( 𝜕𝐸𝑘 𝜕𝑞̇) Elastic potential energy: Spring linear 𝑉𝑠𝑝𝑟𝑖𝑛𝑔= 1 2𝑘∆𝑙2 Spring torsional 𝑉𝑠𝑝𝑟𝑖𝑛𝑔,𝑡 = 1 2𝑘∆𝜗2 Gravity 𝑉𝑔𝑟𝑎𝑣=𝑚𝑔ℎ Lagrange contribution: 𝜕𝑉 𝜕𝑞 Notes:…
Prima pagina del documento.