6.4. DYNAMIC FORCE ANALYSIS OF MACHINERY

In this section we shall assume that the motion of the machine parts are specified beforehand, e.g. the position velocity and acceleration of each rigid body is known or can be calculated by performing kinematic analysis. We shall also assume that the mass and the moment of inertia of each machine member is known or can be calculated from the given data. There may be external forces of known magnitude and direction or friction forces present. However, we shall assume that there is one external force (such as the input torque) of an unknown magnitude but of a known direction and a point of application. The system is in a state of dynamic equilibrium under the action of these forces. We would like to determine the joint forces, forces acting on the members and the magnitude of the unknown external force.

The above problem is commonly known as kinetostatics or Wittenbauer's second problem. Such a formulation is valid under steady state conditions and when the mechanism involved is a constrained mechanism. The input speed(s) must be almost constant for these assumptions to be valid or the changes in the velocity and acceleration of the input link is determined  by some other means.

In kinetostatics we determine the velocity and acceleration of each machine member by performing kinematic analysis. If the mass and mass distribution of the members are known, we can calculate the inertia forces and torques. Next we apply D'Alambert's principle so that we can treat the fictitious inertia forces as if they are external forces. The problem reduces to static force analysis of machinery, which we have seen in Section 5.2.

Example

Figure shows a double slide. Link 2 is moving with a constant velocity at 2 m/s in positive x direction. Link 3 is a thin rod of mass m₃ = 5 kg and length AB = 500  mm. The masses of links 2 and 4 are negligible. Determine the force F₁₂ acting on link 2 and the joint forces that occur when the mechanism is in the given state of motion and when x = 200 mm. Assume the mechanism is operating on a horizontal plane so that we can neglect the gravitational acceleration. Also neglect friction.

As a first step, kinematic analysis must be performed. Writing the loop equation:

Equating the real and imaginary parts:

Substituting x = 200 mm we obtain: q = 113.578⁰ and y = 458.258 mm.

Differentiation of the above equations give us the velocity:

= 4.364 s^-1 (CCW)

 = -872.872 mms^-1

The second differentiation yields ():

  = 8.312 s^-2   (CCW)

 = -10.390 ms^-2

The position, velocity and acceleration of point G₃ is:

substituting the known values:

a_G3= 5.194e^i(-p/2)= 5.194 ms^-2 Ð270⁰

In the second step, we must determine the inertia forces and torques. From Table 1.4 for a thin rod . Hence I_{G 3}=0.10417 kgm² and:

= 25.97 N  Ð 90⁰

865 N-mm (CW)

These forces are as shown in Fig.5.48a. If a graphical solution is to be performed, inertia force and torque can be replaced by a single resultant (Fig.5.48b), which is at a distance h from the center of gravity. h is given by:

 = 33.31 mm

For the force analysis we proceed as if the inertia forces were external forces and apply the methods that were described before for a force analysis.

Graphical Method.

Link 4 is a two force, link 3 and link 2  are three force members.and. For link 3, since R₃ⁱ is known, we determine the point of concurrency of the three forces and draw the force polygon for the equilibrium equation R₃ⁱ +F₂₃+F₄₃=0 as shown in Fig.5.48. For link 4,  G₁₄=-F₃₄=F₄₃. For link 2,  F₃₂+F₁₂+G₁₂=0 .  Since F₃₂ = -F₂₃ , The force polygon for the equilibrium equation is that drawn for link 3 with all the arrows reversed. Hence G₁₂= - R₃ⁱ  and F₁₂= -F₄₃ .

Analytical Method

     In the analytical method, since the moment and force equilibrium equations are going to be used separately, one need not combine the inertia force and torque into a single resultant. The free body diagrams of the moving links are as shown in Fig.5.50.

Since:

F₄₃=F₄₃ Ð180⁰

F₃ⁱ= 25.97 N  Ð90⁰

For link 3 the equilibrium equations are:

F₄₃  - F_23x  = 0

F₃ⁱ   - F_23y  = 0

and the moment equilibrium about point A is:                            

     500F₄₃sin(180-113.578) + 250(25.97)sin(90-113.578)-865 = 0

or

    458.258 F₄₃ = 3461.981

F₄₃ = 7.555 N  Ð180⁰ F₃₂=  -F₂₃ = 27.04 N  Ð 106.22⁰

F_23x = 7.555 N  Ð0⁰ F₁₂ = 7.555 N  Ð0⁰

F_23y = 25.97 N  Ð270⁰ G₁₂ = 25.97 N  Ð 270⁰

F₂₃  = 27.04 N  Ð-73.78⁰ G₁₄ = F₄₃

6.5. DYNAMIC FORCE ANALYSIS OF A FOUR-BAR MECHANISM

  In order to design links and joints one must determine the worst loading conditions of each link and joint. In order to select the driving motor characteristics, input torque for the whole cycle is required. In such cases analytical methods suitable for numerical computation is utilized. In this part a general dynamic analysis of a four-bar mechanism will be explained and the results will be applied to a particular four-bar for a complete force analysis.

  Referring to the figure, the equations for the position, velocity and acceleration analysis of the four-bar mechanism are:

Position Analysis

Note that rcosf and rsinf terms must be solved for r and f simultaneously and the correct quadrant must be ensured. The term s = ±1 depending on whether the mechanism is of open or cross configuration.

Velocity Analysis:

Acceleration  Analysis

The above equations can be written in different forms.

Acceleration of the centers of gravity

Using the above equations, one can determine the angular acceleration of the links and the linear accelerations of the centers of gravity for any input condition (input position, velocity and acceleration).

  For dynamic force analysis in addition of the inertia forces, we shall assume known external forces F₁₃ and F₁₄ acting on links 3 and 4 and an unknown torque, T₁₂ acting on link 2 as shown in Fig. 5.52.  The system is in dynamic equilibrium under theaction of these forces. We would like to determine the input Torque and the joint forces.

  Note that:

F₂ⁱ=-m₂a_G2 ,   F₃ⁱ=-m₃a_G3,  F₄ⁱ=-m₄a_G4  and

T₂ⁱ=-I_G2a₁₂ , T₃ⁱ=-I_G3a_G3 , T₄ⁱ=-I_G4a₁₄ ,

  The free body diagrams of each moving link can be drawn and the equilibrium equations can be written:

For link 4

    .

F_34x +G_14x+F₁₄cosf₄-m₄a_G4x=0                                                           (1)

F_34y +G_14y+F₁₄sinf₄-m₄a_G4y=0                                                 (2)

F_34ya₄sin(p/2-q₁₄) + F_34xa₄sin(-q₁₄) + F₁₄f₄sin(f₄-q₁₄-g₄) - I₄a₁₄

              - m₄a_G4xg₄sin(-q₁₄-b₄)-m₄a_G4yg₄sin(p/2-q₁₄-b₄) = 0     (3)

For link 3

F_23x -F_34x+F₁₃cosf₃-m₃a_G3x=0                                                  (4)

F_23y -F_34y+F₁₃sinf₃-m₃a_G3y=0                                                    (5)

F_34xa₃sin(p-q₁₃) + F_34ya₃sin(-p/2-q₁₃) + F₁₃f₃sin(f₃-q₁₃-g₃) - I₃a₁₃

              - m₃a_G3xg₃sin(-q₁₃-b₃)-m₃a_G3yg₃sin(p/2-q₁₃-b₃) = 0     (6)

For link 2

-F_23x +G_12x - m₂a_G2x=0                                                              (7)

  -F_23y +G_12y  - m₂a_G2y=0                                                           (8)

F_23xa₂sin(p-q₁₂) + F_23ya₂sin(-p/2-q₁₂) + T₁₂  - I₂a₁₂

              - m₂a_G2xg₂sin(-q₁₂-b₂)-m₂a_G2yg₂sin(p/2-q₁₂-b₂) = 0     (9)

Hence, we obtain nine linear equations  in nine unknowns (G_l4x,  G_l4y, F_34x, F_34y, F_23x, F_23y, G_l2x, G_l2y and T₁₂). If a computer subroutine for the matrix solution is available, these equations can be solved directly for the unknowns. However, it is much simpler to solve equations (3) and (6) simultaneously  for F_34y and F_34x and then solve for each unknown from the remaining equations.

         The solution is:

A= + I₄a₁₄ +m₄ g₄[a_G4y cos(q₁₄+b₄) - a_G4xsin(q₁₄+b₄) ]- F₁₄f₄sin(f₄-q₁₄-g₄)

  B= + I₃a₁₃ +m₃ g₃[a_G3y cos(q₁₃+b₃) - a_G3xsin(q₁₃+b₃) ]- F₁₃f₃sin(f₃-q₁₃-g₃)

  F_34x= [Aa₃cosq₁₃+Ba₄cosq₁₄] /[a₃a₄sin(q₁₃-q₁₄)]

  F_34y= [Aa₃sinq₁₃+Ba₄sinq₁₄] /[a₃a₄sin(q₁₃-q₁₄)]

  G_14x=-F_34x+m₄a_G4x-F₁₄cosf₄

  G_14y=-F_34y+m₄a_G4y-F₁₄sinf₄

  F_23x= F_34x+m₃a_G3x-F₁₃cosf₃

  F_23y= F_34y+m₃a_G3y-F₁₃sinf₃

  G_12x= F_23x+m₂a_G2x

  G_12y= F_23y+m₂a_G2y

T₁₂ =  F_23ya₂cos(q₁₂) -  F_23xa₂sin(q₁₂) +  I₂a₁₂

+ m₂ g₂[ a_G2ysin(q₁₂+b₂) -a_G2xsin(q₁₂+b₂)]

One can as well use the principle of superposition and consider forces acting on each link at one time and later add all the force components to find the resultant joint forces. The equations thus found will be similiar to the ones obtained above (written in a different form). This is left as an exercise.

Example

The four-bar mechanism shown in is used to cut running strip of material. The input link is rotating at a constant speed of 900 rpm CCW. The link dimensions are: A₀A=85mm, AB=235mm, B₀B=550mm, BC=238mm, AC= 467mm B₀C=487 ÐBAC₃=7.6⁰ , ÐBB₀C₄=25.6⁰. Using AutoCad solid modeller, the mass properties for links 3 and 4 are found as: m₃= 2 kg, k₃= 268mm, AG₃=231mm,  ÐBAG₃=1⁰ (CW), m₄=5.1 kg, k₄=424mm , B₀G₄= 305mm , ÐBB₀G₄=7⁰ (CW). For link 2 m₂= 3 kg,  k₂= 100mm and its center of gravity is coincident with A₀. We are to determine the input torque required and the joint forces. We can neglect friction. A  cutting force of 2000 N acts vertically on both links 3 and 4 at point C when the crank angle q₁₂  is 124⁰<q₁₂<156⁰.

MathCad solution for this example is as follows:

Link dimensions:

_{"Constant speed is assumed"}

_{"Increment crank angle for every 2 degrees."}

    _{"m is crank angle in degrees"}

"Solve for position variables"

_{"The angular velocity (s-1) and acceleration (s-2) of the links are as follows:”}

_{"Acceleration of the centres of gravity of the links (in mm/s2):"}

"Mass properties (mass in kg and the radii of gyration, k_i, in mm)"

             _{This constant C is to be used to convert mm to m. (Note that 1kg m/s2= 1 N)"}

_{The external forces:}

_{This external force will act vertically on both links 3 and 4 in opposite direction}

_{Determine the joint forces and the input torque. Note that the joint forces are in Newton and the torque is in Newton-meter}

_{Input torque as a function of crank rotation:}

_{Convert the forces to polar form}

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