Force Analysis of Mechanisms (2)

Principle of Superposition

If there was only one known external force acting on one member of mechanism and the system was brought to static equilibrium by an input or output force (or torque). The magnitude of this type of force or torque was an unknown. If there are two or more known external forces acting on one link and these forces can be combined into a single resultant. However in real machinery there are various external forces acting on different links. For example, if we do not neglect the weight of members. There will be at least one known external force on each link. In such a case if we draw free body diagrams of the links then no simplification will be possible and one has to write three equilibrium equations for each link. The resulting 3(l-1) of linear equations will include 3(l-1) unknown joint force components and the input force or torque. Mostly simultaneous solution of equilibrium equations would be required.. Another solution method the principal of superposition. This principal states that the effect of forces is the sum of the individual effects of the forces considered separately. In other words if there are two or more external forces present, one can neglect all but one of forces and determine the joint forces and the unknown reaction force that brings the system into the equilibrium for this one of the external force. If above procedure is carried out for each of the external forces at each joint there will be different joint forces corresponding to each external force. The resultant joint force is the vectorial sum of all these forces.

Analysis of a rigid eccentric cam

Analysis of a rigid eccentric cam involves the determination of contact force, the spring force and cam shaft torque for one revolution of the cam. In simplified analysis all components of cam system are assumed to be rigid and the results are applicable to low speed systems. However if speeds are high and members are elastic. An elastic body analysis must be made. The elasticity of the members may be due to extreme length of follower or due to use of elastic materials in the system. In such cases, chatter, noise, excessive wear, fatigue failure of some of parts are the usual things. A circular disc cam with the cam shaft hole drilled off centre is known as eccentric plate cam. Figure (a) shows a simplified reciprocating eccentric cam system consisting of a plate cam with a flat face follower and a retaining spring.

Fig: Analysis of a rigid eccentric cam

Let,

e = eccentricity, the distance between the centre of the disc and of the shaft

m = mass of the follower

s = stiffness of the retaining spring

ω= angular velocity of the cam rotation

P = preload including the weight of the follower or force on the cam at x = 0

x = motion of the follower (zero at the bottom of the stroke)

If the disc is rotated through an angleθ the mass m is displaced by a distance x, so that

x = e – e cosθ

\[Velocity,\;\;\;\;\;\;\;\;\;\;\mathop x\limits^. = \frac{{dx}}{{dt}} = \frac{{dx}}{{d\theta }}\frac{{d\theta }}{{dt}} = e\omega \sin \theta \]

\[Acceleration,\;\;\;\;\;\;\;\;\;\;\mathop x\limits^. = \frac{{dv}}{{dt}} = \frac{{dv}}{{d\theta }}\frac{{d\theta }}{{dt}} = e\omega \cos \theta \]

Now, as the follower is displaced in upward direction, the acceleration x of follower is in the upward direction indicating the inertia force to be in downward direction. The spring force is also exerted in the downward direction. The force exerted by cam on follower F. However it will be in the upward direction. Thus, the various forces acting on the follower mass are,

Inertia force = m[{\ddot x}]downwards)

Spring force = sx (downwards)

Preload = P (downwards)

Force exerted by cam = F (upwards)

Thus equilibrium equation becomes,

\[\begin{array}{l}
{\rm{m\ddot x + sx + P - F = 0}}\\
{\rm{or }}\;\;\;\;\;\;\;\;F = {\rm{m\ddot x + sx + P}} = me{\omega ^2}cos\omega t + se--secos\omega t + P\\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \left( {se + P} \right) + \left( {m{\omega ^2}--{\rm{ }}s} \right)ecos\omega t
\end{array}\]

Equation shows that the force F exerted by cam on the follower consists of a constant term (se+P) with a cosine wave superimposed on it, the maximum value of which occurs atθ = 0^o and minimum atθ = 180^o. Figure(b) shows the change of this force with the angular displacement of cam. As cam shaft velocity increases, the term involving square of the velocity increases at a faster rate and then becomes zero at a speed when (se + P) + (mω² – s) ecosωt = 0

Fig: Force with the angular displacement of the cam

This can happen at θ = 180^o. At that speed there will be some impact between the cam and the follower, resulting in rattling, clicking and noisy operation. This is usually known as jump. However, this can be prevented to some extent by increasing the preload exerted by the spring or the spring stiffness.

At jump speed,

(se + P) – (mω² – s) e= 0 or 2se + P – meω² = 0

and jump will not occur if preload of the spring is increased such that

P > e(mω² – 2s)

The torque applied by the shaft to the cam, T = F. e sinωt

= [(se + P) + (mω² – s)ecosωt]esinωt

= e(se + P) sinωt +e (mω² – s)sin2ωt

Figure shows the variation of torque with the cam rotation. It can be observed from plots that area of torque-displacement diagram above and below the x-axis is the same meaning that the energy required to raise follower is recovered during the return. A flywheel may be used to handle this fluctuation of energy.

Fig: variation of torque with the cam rotation

Force on Spur Gearing

Fig: Free-body diagrams of the forces and moments acting upon two type gears of a simple gear train

The free-body diagram of the pinion has been redrawn and the forces have been resolved into the tangential and the radial components. We define as the transmitted load.

W_t = F^t_{32............................} (a)

Fig: Resolution of forces

This tangential load is really the useful component, because the radial component F^r₃₂ serves as no useful purpose. It does not the transmit power. The applied torque and the transmitted load are seen to be related by equation

T = (d/2)*W_t .................................(b)

where we have used T = T_a2 and d = d₂ to obtain a general relation. And the power H transmitted through a rotating gear can be obtained from the standard relationship of the product of torque T and angular velocity .

H = Tω = (W_td/2)ω

Force on Bevel Gearing

Fig: Forces on bevel gear

In determining the shaft and bearing loads for bevel gear applications. The general practice is to use tangential or transmitted load that occur if all the forces were concentrated at the midpoint of tooth. While actual resultant occurs somewhere between midpoint and the large end of tooth, there is only a small error in making this type assumption. For the transmitted load, this gives

W_t = T/r_av

where T is the torque andr_av is the pitch radius at the midpoint of the tooth for gear under consideration. The forces acting at the center of the tooth are shown in Fig. The resultant force W has three components i.e. a tangential force W_t, a radial force W_r, and an axial force W_a. From the trigonometry of figure,

W_r = W_ttanφcosγ

W_a = W_ttan φ sinγ

The three forces W_t, W_r and W_a are at the right angles to each other and can be used to determine bearing loads by using methods of statics.

Force on Helical Gearing

Fig: Force on Helical Gearing

Figure is a three-dimensional view of the forces acting against a helical-gear tooth. The point of application of forces is in the pitch plane and in the center of the gear face. From the geometry of the figure the three components of total normal tooth force W are

W_r = Wsinφ_n

W_t = Wcosφ_ncosψ

W_a = Wcos φ_nsinψ

Where W = total force

W_r = radial component

W_t = tangential component also called transmitted load

W_a = axial component also called thrust load

Generally W_t is given and the other forces are desired. In this case, it is not difficult to discover that

W_r = W_t tanφ_t

W_a = W_t tanψ

W =W_t/(cosφ_n.cosψ)

Linkage Force by Matrix Methods

SINGLE LINK IN THE PUREROTATION

The pin joint at 02 on link 2 has a force F₁₂ due to the mating link I, the x and y components of which are F_12xand F_12y'. These subscripts are read as "force of link I on 2" in x or y direction. This subscript notation scheme will used consistently to indicate which of "action-reaction" pair of forces at each joint is being solved for.

Fig: Forces analysis of single link in pure rotation

The equation involve are

................................(i)

.......................................(ii)

........................................(iii)

..................................................(iv)

These above equation can put in the matrix form with the cofficient of unknown variables forming the matrix A ,vector matrix B i.e. unknown variables and constant term matrix C then solve for B.

.....................................................(v)

References:
1. H.H. Mabie and C. F. Reinholtz, “Mechanism and Dynamics of Machinery”, Wiley.
2. J.S. Rao & R.V. Dukkipati Mechanisms and Machine Theory, New Age International (P) Limited..
3. J.E. Shigley and J.J. Uicker, Jr., “ Theory of Machines and Mechanisms”, McGraw Hill.
4. B. Paul, “Kinematics and Dynamics of Planar Machinery”, Prentice Hall.
5. C. E. Wilson, J.P. Sadler and W.J. Michels, “Kinematics and Dynamics of Machinery”, Harper Row.