Mechanisms and robots analysis with matlab pdf download

For the closed kinematic chain, a path is chosen from link 0 to link n. A systematic procedure, using the contour method, is presented below. The equa- tions for velocities and accelerations are written for any closed contour of the mech- anism.

However, it is best to write the contour equations only for the independent loops of the diagram representing the mechanism. Determine the position analysis of the mechanism. Draw the contour diagram representing the mechanism and select the indepen- dent contours. For the contour diagram the numbered links are the nodes of the diagram and are represented by circles, and the joints are represented by lines that connect the nodes.

Determine a path for each contour. For each closed loop write the contour velocity relations, Eq. For a closed kinematic chain in planar motion Eq. Project on a Cartesian reference system the velocity and acceleration equations. Solve the algebraic system of equations and determine the unknown kinematic parameters.

Compute the velocities and accelerations of the characteristic points and joints. The mechanism has two independent contours. The contour diagram of the mechanism is represented in Fig. Clockwise paths are chosen for each closed contours I and II.

Contour I: Figure 3. For the velocity analysis, using Eq. The sign of the relative angular velocities is selected arbitrarily as positive. Equation 3. The numer- ical computation will then give the correct orientation of the unknown vectors. Chapter 4 Dynamic Force Analysis 4. Figure 4. The time derivative of Eq. Any particle of the system is acted on by two types of forces: internal forces exerted by other particles that are also part of the system and external forces exerted by a particle or object not included in the system.

Let fi j be the internal force exerted on the jth particle by the ith particle. Equation 4. The sum of any such pair must be zero. The origin of the Cartesian reference frame is O. The mass center C of the rigid body is located in the plane of the motion. Let Cz be the parallel axis through the mass center C. The mass mo- ment of inertia ICz is a constant property of the body and is a measure of the rota- tional inertia or resistance to change in angular velocity due to the radial distribution of the rigid body mass around the z-axis through C.

Substituting this expression into Eq. For this case the Eq. The forces and moments are known and the equations are solved for the motion of the rigid body direct dynamics.

The motion of the rigid body is known and the equations are solved for the forces and moments inverse dynamics. The dynamic force analysis in this chapter is based on the known motion of the mechanism. The equations of motion for a rigid body are analogous to the equations for static equilibrium: The sum of the forces equals zero and the sum of the moments about any point equals zero when the inertial forces and moments are taken into account.

The dynamic analysis problem is reduced to a static force and moment balance problem where the inertia forces and moments are treated in the same way as exter- nal forces and moments. In Fig. The forces and moments acting on the system include an external driven moment M, and the forces transmitted from the frame at joint A, F01 , and at joint C, F The force analysis can be accomplished by examining individual links or a sub- system of links.

In this way the joint forces between links as well as the required input force or moment for a given output load are computed. The equivalence between the kinetic diagram and the free-body diagram will be employed in the solution of dynamical problems. The method is based on the decoupling of a closed kinematic chain and writing the dynamic equilibrium equations.

The kinematic links are loaded with external forces and inertia forces and moments. The closed kinematic chain has been transformed into two open kinematic chains, and two paths I and II are associated. The two paths start from Ai.

A sim- ilar analysis is performed for the path II of the open kinematic chain. The number of equilibrium equations written is equal to the number of unknown scalars intro- duced by joint Ai joint forces at this joint. For a joint, the number of equilibrium conditions is equal to the number of relative mobilities of the joint. The point A is selected as the origin of the xyz reference frame. The center of mass of link 1 is C1 , the center of mass of link 2 is C2 , and the center of mass of slider 3 is C.

If x is greater than zero sign x returns 1, if x is zero sign x returns zero, and if if x is less than zero sign x returns —1. It results that the reaction force F03 acts at C. For the slider 3 the vector sum of the net forces external forces Fext , gravitational force G3 , joint forces F23 , F03 is equal to m3 aC3 Fig. The reaction force F01 does not appear in this moment equation. The force analysis starts with the link 3 because the external driven force Fext is given. For the slider 3 the vector sum of all the forces external forces Fext , gravita- tional force G3 , inertia forces Fin 3 , joint forces F23 , F03 is zero Fig.

The dynamic force analysis can start with any joint. For the path I, shown in Fig. If the path I is followed Fig. Equations 4. The diagrams of the slider 4 are shown in Fig. There are eight equations Eqs. For the link 3 the vector sum of the net forces, gravitational force G3 , joint forces F43 , F03 , F23 , is equal to m3 aC3 Fig. The diagrams of the slider 2 are shown in Fig. The force analysis ends with the driver link 1.

For the link 1 the vector sum of the net forces, gravitational force G1 , joint forces F01 , F21 , is equal to m1 aC1 Fig. It has two contours and F05x , eval solF One can write Eq. F45x , eval solF F34x , eval solF F03x , eval solF F23x , eval solF F12x , eval solF Following the path I Fig. F01x , eval solF The forces and moments are known and the differential equations are solved for the motion of the rigid body direct dynamics. The pendulum is connnected to the ground by a pin joint and is free to swing in a vertical plane.

The local acceleration of gravity is g. Find and solve the Newton—Euler equations of motion. Solution The system of interest is the link during the interval of its motion.

The link in rota- tional motion is constrained to move in a vertical plane. First, a reference frame will be introduced. The plane of motion will be designated the x, y plane. The y-axis is vertical, with the positive sense directed vertically upward. The x-axis is horizontal and is contained in the plane of motion.

The z-axis is also horizontal and is perpen- dicular to the plane of motion. The link is moving and hence the angle is chang- ing with time at the instant of interest. The system has one degree of freedom. The system has a single moving body. The body is connected to the ground with the rotating pin joint R at O. The mass center of the link is at the point C. As the link is uniform, its mass center is coincident with its geometric center. The motion of the link is planar, consisting of pure rotation about the pivot point.

The directions of the angular velocity and angular acceleration vectors will be perpendicular to this plane, in the z-direction. This problem involves only a single moving rigid body and the angular velocity vector refers to that body. The positive sense is clockwise. The n-axis is along the length of the link, the positive direction running from the origin O toward the mass center C. The unit vector of the n-axis is n. The t-axis will be perpendicular to the link and be contained in the plane of motion as shown in Fig.

The force driving the motion of the link is gravity. The weight of the link is acting through its mass center and will cause a moment about the pivot point. This moment will give the link a tendency to rotate about the pivot point.

The only contributor to the moment is the weight of the link. Thus we should be able to directly determine the angular acceleration from the moment equation.

The sum of the forces acting on the link should be equal to the product of the link mass and the acceleration of its mass center. This should be useful in determining the forces exerted by the pin on the link. The free-body diagram shows the link at the instant of interest, Fig. The link is acted upon by its weight acting vertically downward through the mass center of the link.

The link is acted upon by the pin force at its pivot point. The motion diagram shows the link at the instant of interest, Fig. The motion diagram shows the relevant acceleration information. The force exerted by the pin on the link is obtained from Eq. The initial angular acceleration can be determined from Eq. The initial reaction force components can be evaluated from Eq. The statement sprintf writes formatted data to string.

One can obtain a vector t and a matrix xs with the coordinates of these points using ode45 command. In general, the error tends to grow as one goes further from the initial conditions. The system is free to move in a vertical plane. Find and solve the equations of motion. The origin of the reference frame is at A.

The angles q1 t and q2 t are selected as the general- ized coordinates as shown in Fig. There are two rigid bodies in the system and the Newton—Euler equations are written for each link using the free-body diagrams shown in Fig. Equation 5. The initial conditions Cauchy problem are necessary to solve the equations.

First the reaction joint force F21 is calculated from Eq. In the case of the robot arm the set of contact forces transmitted from 0 to 1 in order to drive the link 1 can be replaced with a couple of torque T The generalized coordinates are q1 t and q2 t as shown in Fig.

To analyze the motion of the system in many cases, it is more convenient to use a set of variables different from the physical coordinates. Let us consider a set of variables q1 , q2 ,. The generalized coordinates, q1 , q2 ,. The number p is called the number of degrees of freedom of the system. The state space is the 2n-dimensional space spanned by the generalized coordi- nates and generalized velocities.

The constraints are generally dominant as a result of contact between bodies, and they limit the motion of the bodies upon which they act. A constraint equation and a constraint force are related with a constraint. The constraint force is the joint reaction force and the constraint equation represents the kinematics of the contact.

A particle P is sub- jected to a constraint of moving on the smooth surface described by Eq. The motion of the particle over the surface can be viewed as the motion of an oth- erwise free particle subjected to the constraint of moving on that particular surface. The constraint equations in velocity form or velocity constraints or motion con- straints are obtained by dividing Eqs. When the constraint is non-holonomic, it can only be expressed in the form Eqs.

A scleronomic system, f q1 , q1 ,. A rhenomic system is a system subjected to a holonomic constraint that is an explicit function of time. Equation 6. Solution The solution for the two-link robot arm will start with the dynamics when the the forces and moments are known and the equations are solved for the motion of the links.

Figure 6. T12z; The generalized coordinates, q1 and q2 , given by Eq. Multiplication of Eq. Let m1 , m2 , m3 be the masses of 1, 2, 3, respectively. The link 1 is connected to link 2 at the pin joint B.

The last link 3 is connected to 2 by means of a slider joint. The mass centers of links 1, 2, and 3 are C1 , C2 , and C3 , respectively. In the case of the robotic arm, there are two kinds of forces that contribute to the generalized forces Q1 , Q2 , and Q3 namely, contact forces applied in order to drive the links 1, 2, and 3, and gravitational forces exerted on 1, 2, and 3 by the Earth.

The set of contact forces transmitted from 0 to 1 can be replaced with a couple of torque T01 applied to 1 at A, Fig. Similarly, the set of contact forces transmitted from 1 to 2 can be replaced with a couple of torque T12 applied to 2 at B, Fig. Next, the set of contact forces exerted by link 2 on link 3 can be replaced with a force F23 applied to 3 at C3 , Fig.

The central moments of inertia of links 1 and 2 are calculated using Fig. F23z ; The generalized coordinates, q1 , q2 , and q3 given by Eq. Let vPj and aPj denote, respectively, the velocity of Pj and the acceleration of Pj in a reference frame 0.

The quantities Kin 1 , The second generalized coordinate q2 denotes the radian measure of the angle between the axes of 1 and 2. The link 2 supports link 3, and link 3 has a transla- tional motion relative to 2.

The generalized coordinate q4 is the distance between the mass centers, C2 and C3 , of 2 and 3, respectively. The generalized coordinate q3 is the radian measure of the rotation angle between 3 and 4.

Substitution into Eq. Considering, the contact forces, the set of such forces transmitted from the New- tonian frame 0 to link 1 through bearings and by means of a motor is replaced with a couple of torque T01 together with a force F01 applied to 1 at C1. Next, the set of contact forces exerted on 2 by 3 is replaced with a couple of torque T32 together with a force F32 applied to 2 at C32 , the point of 2 instanta- neously coinciding with C3.

Similarly, torque T43 and forces F43 come into play in connection with the inter- actions of 3 and 4 at C4. For the velocities of C1 , C2 , C3 , C4 , the mass centers of links 1, 2, 3, 4, it is not necessarily useful to work with RF1 in the case of velocity of C1 , with RF2 for the velocity of C2 , and so forth.

As an alternative, it is convenient to use any vector basis that permits one to write the simplest expression. The constant angular speed of the driver link 1 is n and is given in the tables. Determine the type of motion rotation, translation, and complex motion for each link, the connectivity table, the structural diagram, the contour diagram, the independent contours, the number of degrees of freedom, the dyads, and the type of the dyads.

The constant angular speed of the driver link 1 is n. The driver link 1 has a constant angular speed n. Table 7. The generalized coordinates quantities associated with the instanta- neous position of the system are q1 t , q2 t , and q3 t.

Select suitable numerical values for the input numerical data. Find the transformation matrices Ri j. Determine the position vectors, rCi , the velocities, vCi , and the accelerations, aCi of the mass centers Ci. Find the generalized active forces Qi. Find the numerical solutions for inverse dynamics and direct dynamics. Bedford and W. Buculei, D. Bagnaru, G. Nanu, D. Erdman and G. Etter and D. Kane, Analytical Elements of Mechanics, Vol. Kane and D. Kane, P. Likins, and D. Manolescu, F. Kovacs, and A.

Marghitu and M. Meriam and L. Author : Dan B. Machine Design Analysis with MATLAB is a highly practical guide to the fundamental principles of machine design which covers the static and dynamic behavior of engineering structures and components.

MATLAB has transformed the way calculations are made for engineering problems by computationally generating analytical calculations, as well as providing numerical. Authors: Dan B. Marghitu, Mihai Dupac, Nels H. Engineering mechanics involves the development of mathematical models of the physical world.

Statics addresses the forces acting on and in mechanical objects and systems.