4.2 VELOCITY AND ACCELERATION ANALYSIS OF MECHANISMS-4

Example:

Define the fixed link lengths:
        w₁₂:=4              
             
                     

Generate the input crank angle for every 50.





x_Bk,y_Bk are the rectangular coordinates of B with respect to C₀ with +ve x axis along QC₀.

Convert rectangular coordinates.

to polar coordinates.

              Cosine theorem. for angle BC₀C.

            Cosine theorem for the transmission
            angle BCC₀ 


                      

                             

Note that f, b and s are used as dummy variables.

                      

x numbers with A₀ as the origin.

Coordinates of F in complex numbers:

 

In figure below, The path of point F is shown ( the x sign is the position of point F when q₁₂=90⁰).

For velocity and acceleration analysis create the coefficient matrix:

The angular velocities are found from the velocity loop equations.

The angular acceleration of the links are found from the acceleration loop equations.

Velocity and acceleration of point F:

The polar plots of the velocity and acceleration vectors of point F for a complete cycle are shown below.

Velocity of point F

(Velocity of F at an instant is the vector from point 0 to a point on the curve e.g

velocity of Point F when q₁₂=90⁰ is the line drawn from 0 to the mark x)

The polar plots of the velocity and acceleration vectors of point F for a complete cycle are shown below.

Acceleration of point F

(Acceleration of F at an instant is the vector from point 0 to a point on the curve e.g

acceleration of Point F when q₁₂=90⁰ is the line drawn from 0 to the mark x)

Polar plot of the velocity and acceleration of point F. (velocity and acceleration are measured in mm/s and mm/s², respectively. The x sign (for k=18) is when q₁₂=90⁰).

In case of graphical solution for the position given, link 2 is in a fixed axis of rotation with w₁₂=4 rad/s. Therefore V_A=V_B=728 mm/s and while V_Ais to the left, V_B is to the right. Using a scale factor k_v=0.5 mm/mm/s we draw these known vectors (Fig. 4.85). Considering the loop formed by links 1,2,3 and 4 The velocity equation is:

            V_C=V_B+V_C/B

V_C/B is perpendicular to CB and V_C is perpendicular to CC₀. At the given position, since BA₀ and CC₀ are parallel, V_C and V_B must be equal  and V_C/B =0 and w₁₃=0. In such a case V_D=V_C=V_B. Next consider point E which is a permanently coincident point on links 5 and 6. For points D and E on link 5:

            V_E=V_D+V_E/D

Similarly, for points A and E on link 6:

            V_E=V_A+V_E/A

V_D and V_A are of known magnitude and direction. The relative velocities V_E/D and V_E/A must be perpendicular to ED and EA respectively. Although these two equations cannot be solved separately, when they are equated to each other, the two relative velocity magnitudes are the unknowns. Therefore simultaneous solution is made. For the velocity of point F, we can apply Mehmke's theorem or solve the two vector equations:

            V_F=V_E+V_F/E

            V_F=V_A+V_F/A ,

simultaneously. ,

Velocity Polygon

 For the acceleration analysis we use the same points that have been used for the velocity analysis. However we have to write the acceleration of these points in terms of their components. Since link 2 is rotating at a constant speed, the tangential accelerations of points A and B are zero (a_A=aⁿ_A , a_B=aⁿ_B). a_A= a_B =2912mm/s², both towards the center of rotation. Using a scale factor k_a=0.2mm/mm/s², we draw these vector (Fig.4.80). Next we write the acceleration of point C as:

            aⁿ_C + a^t_C = aⁿ_B + aⁿ_C/B + a^t_C/B

aⁿ_C is a vector along CC₀ towards C₀ and is of magnitude w²₁₃ |CC₀|=V²_C/|CC₀|= 1045mm/s². aⁿ_C/B=0 since w₁₃=0 for this position. a^t_C and a^t_C/B are perpendicular to CC₀ and CB respectively (unknown magnitudes). Note that while w₁₃=0,   a₁₃≠0 .a^t_D/B  can now be determined since a₁₃=a^t_C/B /|CB| (aⁿ_D/B =0). Hence  a_D is known. For point E we can write:

            a_E = a_D + aⁿ_E/D + a^t_E/D

and

            a_E = a_A + aⁿ_E/A + a^t_E/A

            aⁿ_E/D= w²₁₅|ED| =V²_E/D/|ED| =1366mm/s²,  along ED, towards D:

            aⁿ_E/A= w²₁₆|EA| = V²_E/A/|EA| = 1269 mm/s², along EA towards A.

Tangential acceleration components a^t_E/D and a^t_E/A are perpendicular to the lines ED and EA respectively (unknown magnitudes). The two equations can now be solved simultaneously to determine the velocity of point E. For the acceleration of point F Mehmke's theorem is applied (triangle aef on the acceleration polygon is similar to the triangle AEF of link 6). The result is shown in Fig.4.86.  Hence a_F = 1075.5/0.2 =5378 mm/s² , downwards as shown.

Acceleration Polygon

©es