Off-line compensation of the tool path deviations on robotic machining: Application to incremental sheet forming
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Citations
Robots in machining
Single point incremental forming: state-of-the-art and prospects
Impact & improvement of tool deviation in friction stir welding
Design, programming and orchestration of heterogeneous manufacturing systems through VR-powered remote collaboration
Elasto-geometrical modeling and calibration of robot manipulators: Application to machining and forming applications
References
A Kinematic Notation for Lower-Pair Mechanisms Based on Matrices
Theory of Robot Control
Asymmetric single point incremental forming of sheet metal
Modeling, Identification and Control of Robots
An integral manifold approach to the feedback control of flexible joint robots
Related Papers (5)
Enhanced stiffness modeling, identification and characterization for robot manipulators
Frequently Asked Questions (15)
Q2. How is the size of the stiffness matrix reduced?
The size of stiffness matrix is then reduced by deleting rows andcolumns corresponding to the nodes for which displacements are null.
Q3. What are the methods proposed in the literature?
The methodologies proposed in the literature are based either on Lumped-parameter [14], [15], [16] or more realistic Finite Element models [17], [18], [19].
Q4. What is the vector of nodal wrench that is applied at node v?
The vector of nodal wrench that is applied at node v is:Fv = [ fv mv ]T(4)with:fv = [ fv,x fv,y fv,z ]T the force applied at node v.mv = [ mv,x mv,y mv,z ]T the torque.
Q5. What is the solution vector for the node displacements?
The node displacements are given by the relation:0∆XG = ( 0KG )−1 0FG. (15)The solution vector 0∆XG contains the values of all node displacements and Lagrange multipliers.
Q6. What is the reason for the deviations in the robot structure?
These deviations are mainly due to the elastic deformations of the robot structure which lack of stiffness in comparison to dedicated machines [4], [5].
Q7. What is the simplest way to calculate the node displacements?
For a given load on the structure (link own weight, payload, external forces, etc.), it is possible to calculate the elastic displacements 0∆XE at the node corresponding to the TCP.
Q8. What is the relation of the node displacements?
Finally rigid link motions can be expressed by the following relation:0A∗c 0∆X∗c = 0M,1. (12)It is possible to calculate the node displacements 0∆X∗c , that verify the relation 0F∗c = 0K∗c 0∆X∗c under the constrains given by the relation (12).
Q9. What is the main advantage of this approach?
The main advantage of this approach is that a realistic and complete 3D elastic model can be derived automatically for any industrial robot manipulators including openand closed-loop structures.
Q10. What is the optimum design of the robot?
The identi ed stiffness values obtained with the optimum design are used to evaluate the elastic displacements of the robot for the veri cation load con guration.
Q11. What is the relation of the forward geometric model?
The Forward Geometrical Model is de ned by the following relation that links the controlled pose 0XE to the joint con guration q k and the vector of the geometrical parameters ξ:0XE =[0PE 0ΦE]= f ( qk, ξ )(20)where 0PE and 0ΦE give respectively the cartesian position and the orientation of the end-effector relative to R0.
Q12. What is the axial and radial stiffness of a beam?
The calculation of the stiffness matrix components inR0 is done by using the transformation matrix 0Gj according to:0Kj = ( 0Gj ) Kj ( 0Gj )−1 . (7)0Gj is given by:0Gj =0 Aj 03,3 03,3 03,3 03,3 0 Aj 03,3 03,3 03,3 03,3 0 Aj 03,3 03,3 03,3 03,3 0 Aj (12×12). (8)For a beam, the stiffness matrix depends on its geometrical and mechanical parameters (TABLE 2).
Q13. What are the axial and radial stiffness of a link?
Before establishing relation between forces/displacements, it is necessary to include rigid link motion when a link is considered as rigid.
Q14. What is the relation for the forward elastic model?
The Forward Elastic Model whose elastic parameters have to be identi ed can be described by the relation (21) for a joint con guration qk:0∆XE =[0∆PE 0∆ΦE]= h ( qk, ξ, 0Feq,Λ,Γ ) = ( 0Keq )−1 0Feq (21)where:ξ: the geometrical parameters.
Q15. What is the advantage of the non-symetrical geometry of the FE analysis?
The non-symetrical geometry of this last one allows to validate the robustness of both the FE analysis and the elasto geometricalmodeling.