Strajnikki vestnik - Journal ofMechanlcul Engineering64(2018)6, 412-421 © 2018 Jaurual of MechanicalEnEinearing. AllriehVsneserved. D0l:10.5545/sv -jme.2017.5023 Original Scientific Paper Recelved for review: 2017-10-27 Received reoisedform: 2017-01-10 Acceptef fer publication:2017-01-21 A Trajectory Compensation Model for Roll Hemming Applications Eduardo Esquivel González1* - José Pablo Rodríguez Arzate2 - Giuseppe Carbone3 - Marco Ceccarelli3 - Juan Carlos Jáuregui1 1 Autonomous University of Querétaro, Faculty of Engineering, México 2 Technological Institute of Celaya, Department of Mechatronics, México 3University of Cassino, Laboratory of Robotics and Mechatronics, Italy This paper presents a trajectory compensation model to correct the deviation in the roll hemming applications. First, the main defects and problems of roll hemming technology are established. A trajectory compensation proposal is analyzed as well as the kinematic and stiffness model of the robot and the material deformation model. The implementation of the model on an industrial robot is tested and simulated. Consequently, the viability of the model is discussed and compared with other works. Highlights • An offllno componsation stratogy is implomontod for roll hommlng. • Tho componsation stratogy rollos on tho dovlatlon duo to tho robot. • Tho componsation stratogy rollos on tho dovlatlon duo to tho panol. • Tho rosults shoo tho noo trajoctory oithin tho main paramotors of tho procoss. Keywords: roll hemming, wrinkling, tool center point, stiffness, trajectory 0 INTRODUCTION Robots in automotive applications offer low-cost solutions for most of the manufacturing processes, opening new possibilities ranging from simple tasks, like pick and place, painting and sealing, to more complex tasks, like milling or welding [1]. This allows replacing the computer-navigated control (CNC) machines and stamping machines by using robotic manipulators in metal forming processes. They are also applicable for new processes like roll hemming, commonly used for doors, hoods and deck lids of the automotive industry. In the process, a serial robot moves a roller through the pre-hemming steps over the contour in order to attach the exterior panel to the interior panel of a door [2]. The roll hemming offers flexibility but different defects may appear in the final panel's shape. A common visible defect is the formation of waves on the flange, called wrinkles, related to the velocity and force of the roller. The origin of such defects relies on the capacity of the roller to deform the panel depending on the robot's pose. For example, an extended configuration pose demands higher torque and applies lower force than a constricted one that requires less torque and applies higher force. The stiffness of a robot (1 N/um) is lower compared to a CNC machine (50 N/um) [3], decreasing its capacity to follow a designated trajectory under external forces. As consequence, the final quality of the panel may vary along the trajectory. Most of the works of roll hemming process [4] are related to the finite element analysis and the prediction of the deformation patterns of the panel and few works [5] are focused on the dynamic performance of the robot. Accordingly, this paper proposes a compensation strategy for the roll hemming process based on the variable stiffness of the robot to minimize the tool deviation along the trajectory. This paper proposes an offline compensation model for roll hemming with emphasis on determining the stiffness of the robot. A simulation of the process has been developed and the experimental tests show the error's results. If this compensation strategy is implemented, then a more accurate solution is achieved and a better quality of the product can be guaranteed. This paper is structured in three main sections. The methods section establishes the trajectory compensation proposal and the analysis of each element concluding with the integration of the trajectory compensation model. The experimental section describes how the stiffness values were determined and how the forces and speed of the process were related to the wrinkling defect of the panel. Finally, the results section shows the values obtained from the computation of the algorithm and the deviation due to the error of the trajectory. 412 *Corr. Author's Address: Autonomous University of Queretaro, Cerro de las campanas S/N, Queretaro, Mexico, je.esquivel@outlook.com Fig. 1. An offline compensation strategy based on the variable force of the process Robot Fig. 2. The model of the roll hemming process 1 METHODS 1.1 A Trajectory Compensation Proposal An offline compensation approach is defined in this paper. The Fig. 1 shows a trajectory compensation proposal related to the robot cell and the roll hemming process, where the desired position Xd and the recorded force Frec are the inputs of the model. Once computed the new trajectory Xn, the position controller commands the robot toward the hemming process. The Xout represents the location's feedback. The Fig. 2 shows the scheme of the process as the robot deforms the flange. The first three joints of the robot are modeled as torsional springs and the last three joints are used for orientation having a lower impact on the elasticity behavior. The sheet is modeled as linear spring with a damping effect. The tool center point (TCP) moves from position A to position B along its path but the material stiffness and the elasticity of the robot affect the desired trajectory Xd producing an error e in comparison with the real trajectory Xr. A representative equation of the compensation strategy is Xn =Xd + e, Xn =Xd + ^robot + ^sheet, (1) where e is the term of the error or deviation. 1.1.1 The Deviation Due to the Robot This section describes the mathematical model to determine the deflection of the robot Arobot. The deflection vector has three terms for position and three for orientation due to the configuration and the stiffness of the robot. The general deflection equations are built from the stiffness in the Cartesian and joint spaces as F = Kx8x, T = KgSg, (2) where the vector F is the 6x 1 vector of external forces and torques applied to the tool, the matrix Kx is the 6x6 Cartesian stiffness matrix, the vector 8x is the 6 x 1 vector of linear and angular displacements, the vector T is the 6x 1 vector of joint torques, the matrix Kg is the 6x6 joint stiffness matrix and the vector 8g is the 6x1 vector of joint displacements. Both displacements are related through the Jacobian relation Sx = J(@)Sg, (3) where the variable J(0) represents the 6x6 Jacobian matrix. By inserting this equation in Eq. (2) a new equation is obtained T = Kg J-1(@)Sx (4) this equation relates the two different spaces by the Jacobian matrix. Naming the principle of virtual work FT Sx = TT Sg, (5) and combining this relation with Eq. (3) another equation is formed T = JT (®)F, (6) which is similar to Eq. (3). Again, a substitution in Eq. (4) results in a new statement JT (0)F = Kg J-1(@)Sx. (7) Table 1. Denavit-Hartenberg parameters for the robot Fanuc 200IC Oi-1 a—i di 1 2 2 2 gi 2 n 2 ai 2 g2—n 8 n Ü2 2 g3 2 n 2 a3 —d4 g4 5 n 2 2 2 g5 6 n 2 2 2 g6 T 2 2 —dt 2 This last equation relates the force F and the displacement Sx of the Cartesian space with the joint stiffness matrix. Expressing in explicit form Kobot = sx = K-1J(Q)Jt (&)F W (8) Observing this equation it is noted that the tool displacement depend on the force, the jacobian matrix and the joint stiffness matrix. The force and displacement are required to be expressed in the global reference frame. A general view of the last equation shows a similarity to Eq. (2) where forces and displacements are related through the Cartesian stiffness matrix. 1.1.2 The Kinematic Model of the Robot The Fig. 3 shows the scheme of the robot with the reference frames attached to each joint and with 6 degrees of freedom. All the joints were considered as flexible and all the links as rigid bodies, the Table 1 shows the Denavit-Hartenberg parameters established for this robot, being a1 = 75 mm, a2 = 300 mm, a3 = 75 mm, d1 = 330 mm, d4 = 320 mm, and the distance to the TCP dt = 80 mm +120 mm, this parameters describe the kinematic behavior of the robot. The Jacobian matrix is proposed in order to relate the Cartesian displacement and the joint displacement as j(&) = (J(0)i J(0)2 J(©)s)T ^ d4 0 0 ¿7 (9) ¡h d. Zi Xi X4 X, X* Z4 ^ z6 Z7 X, Fig. 3. The scheme of the robot with the reference frames where each one of the terms is expressed as -sin(&2 + 63 M \ 7(0)l = ( sin(0i + 62) - cos(62 + 63)d4 + sin62a2 -a1 I . (10) cos(62 + 63)di I The second column corresponds to /(©)2 = sin03a2 + d4 0 -cosd3a2 — a3. (il) and the third column corresponds to /(0)3 = d4 0 —a3. (12) This is the jacobian matrix considered for the deviation due to the robot. 1.1.3 A Method to Compute the Stiffness/Compliance of the Robot This paper proposes to compute the joint stiffness matrix K6 as a function of the vector of external forces and torques and the vector of linear and angular displacement expressed as Kg = f (F, Sx). (13) In order to obtain the last expression, the Eq. (8) is reordered in explicit form for the compliance vector as (TKfi P6=i[JiM (14) £ P6=itj2 A P6=i[j3Át A p6=i[j4J(t ¥ jJ j(1= V*ë6 6=i[j6 j(l6= where the compliance matrix K-1 has been integrated in the resultant vector of the displacement and substituted for the vector of compliance C C iiiiii Kg i Kg 2 Kg 3 Kg 4 Kg 5 Kg 6 it is possible to present the Eq. (i4) in the form Sx = AC, (i5) (i6) where A is formed by the Jacobian and force terms. This equation is expressed as a linear matrix equation Ax = b where the unknown values are those of the vector of compliance C. If AC = Sx has no solution 0i S x | Compute the values of vector C | Fig. 4. The flow chart to find the values of the compliance by finding A-1, then we can find the minimum error e = AC — Sx by the least square solution C0 = (AtA)—1At 8x , (17) where the vector of compliance C0 relates every joint deflection to a specific torque. The precision of the compliance vector is related to the measurement of the forces and displacements of the tool. The Fig. 4 illustrates the procedure to determine the values of the compliance where the first steps is to determine the location i of the work-space to test the values of displacements and forces, the second step is to select a specific location for which the robot takes the specific configuration to reach that location, once in that configuration a specific force is applied on the tool and the displacements are measured, the procedure is repeated until covering the work-space and finally the Eq. (17) computes the values in a numerical software. 1.1.4 The Deviation Due to the Panel The deviation due to the panel can be computed as Asheet Aep + Awrinklingi (18) where the total deflection of the sheet's flange Asheet is the sum of the elastic and plastic deformation Aep and the deformation due to the wrinkling defect Awrinkling produced from the waves of the flange. This model gives a contribution to the common finite element models presented in the roll hemming overview. The Fig. 5 shows how the external force F, a scalar Fig. 5. The material deformation scheme value for the panel, generates the plastic ep and elastic ee deformations as the flange inclines the angular position 9. This scheme can be modeled in the form e = ee + ep where the total deformation of the sheet e is the sum of the elastic ee and the plastic ep deformation. The elastic deformation ee may be computed by considering the flange as a beam under bending load at the end for a specific section of the material. To model this deflection two relations are considered M 1/R = —, ' EIz' My Iz ' (19) where R is the radius of curvature, M is the bending moment, E is the Young's modulus, Iz is the second moment of area, a is the stress of the beam and y is the distance from the neutral middle line on the beam towards the external fibers. The use of these two equations results in the deflection function Fx2(x — 3l) 6EIz ' (20) where F is the load at the end of the flange, x is the distance at any given section of the flange, and l is the total length of the flange. If it is considered the distance x = l for deflection, then the Eq. (20) becomes Se = - Fl3 3EL Yl2 3Eh' (21) which states the deflection for the elastic part of the material at the end of a beam, where Y is the yield stress and h is the total height of the cross section from the neutral line to the external fiber. Chakrabarty established that the longitudinal strain in a elastic beam is ex = y/R and the transverse strains are ey = ez = -vy/R, where v is the Poisson's ratio [6]. By using these relations the deflection v on e .....Plastic y Y Fig. 6. The plastic deformation on the rectangular cross section the y axis is established as x + v(y2 -z2) 2R (22) The Fig. 6 shows the rectangular cross section during plastic deformation assuming an elastic, perfectly plastic behavior, where b is the width of the section. The bending moment Me and the radius of curvature Re at the elastic limit are established in view of Eq. (19) using an area moment of inertia Iz = at the initial yielding stress when y = h Me 2bh2Y 3 ' Re = Ehh, R f- (23) where R is the radius of curvature at any stage during the elastic/plastic bending. Then according to the material strain-hardening o Ee Y ( Y ) ' (24) where 0 < n < 1, n being the strain hardening exponent, and with the condition e > Y/E. By considering that e = y/R Y(£), 0 < y < c Y(y)n- c< y < h ' (25) The bending moment at any stage of the plastic deformation is [• h M = 2b oydy. Jo (26) By substituting Eq. (25) into Eq. (26) and integrating we obtain M = [3(Re)n - (1 -n)(R)2]. (27) M 2 + ^ R "Re The above equation applies for a non hardening material when n = 0 obtaining Re i M - M < Me R 1(3- 2M)-1/2- M > Me. (28) Fig. 7. The plastic deformation model. Considering the second partial derivative of Eq. (22) respect to x we obtain 1 R d2^ d x2. (29) The Fig. 7 illustrates a frontal view of the flange as a cantiliver beam with a terminal load, where l is the total length of the flange and 9 is the inclination angle. This assumption involves the following relations Fe. F M M (30) where Fe is the load at the elastic/plastic boundary, a is the distance to the elastic/plastic boundary and Me is the bending moment at the elastic/plastic boundary. By inserting Eq. (29) and Eq. (30) into the Eq. (28) we obtain d2^ ;d x2 '(3 - 2 a )-i/2, x > a x