Kinematic Analysis of a Spatial 6-Bar Linkage Mechanism

Abstract

The door operator in metallurgical equipment is a spatial 6-bar linkage mechanism. The numerical model of motion and force of the mechanism was developed and the numerical solution was applied. A computer simulation was implemented. There is a high level of consistency between the solution and simulation. The torques required on driving crank have been studied. The influences of the self-lock conditions on the crank length and con-rod length have been explored. The effects of the length of crank and con-rode on the needed maximal torques of the crank have been studied, too. And the kinematics analysis attributes the lightening design of the slider structure on the basis of considering the sliders rigidness and intensity.

Mathematical Modeling, Design Validation, and Simulation of Morphing Mechanism for Winglet Applications

“…In the above equations 1 is the position angle of link A or E, 2 is the position angle of link B or D, 1 and 2 are the lengths of links 1 and 2 respectively, is the displacement of the slider, and ℎ is the offset of the slider [13].…”

“…here and are the displacements of joint 5 in part A and B respectively, is the length of C, and H is the distance between joints 1 and 7 [13]. In this equation is known, and thus the above equation is used to simplify equation [11] into a linear one variable equations.…”

“…In this equation is known, and thus the above equation is used to simplify equation [11] into a linear one variable equations. The reference then utilizes MATLAB to complete the solution process and simulation [13].…”

“…The angle 3 is negative by convention, as the rotation is clockwise. The above relations can be solved by expanding the two equations, as shown in Equations (3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13) and (3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14). These equations can then be solved simultaneously using the symbolic computation toolbox in MATLAB.…”

“…is fixed meaning the corresponding velocity is 0. The velocity of 10 is found by differentiating Equation(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)(14)(15)(16)(17)(18)(19). The resultant velocity equation is given below.= 8̇89(4-15)The time varying component in Equation (4-15) is .…”