Variation of Tractive Force | Tractive Effort | Effect of Partial Balancing of Locomotives

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We have studied in earlier video that balancing is done on a locomotive. But because of balancing there are 3 effects that occurs on the locomotive, which are:

  1. Variation of tractive force
  2. Swaying couple
  3. Hammer blow

In this notes we will study the effect of variation of tractive force and also derive the equation to calculate the tractive force.

The first thing to remember here is that the Locomotive engines are reciprocating engines and we know that we cannot completely balance a reciprocating engine, we partially balance it. And partially balancing generates horizontal force and vertical force.

unbalanced horizontal force Image from video Balancing of Reciprocating Masses [Part-2] | Partial balancing | (Dynamics of Machinery)

In the video of partial balancing of reciprocating of engines ( Video Part-1 & Video Part-2 ) we have studied that the unbalanced force acting in horizontal and vertical direction is given by:

FH=(1C)mrω2cosθ  (1)FV=Cmrω2sinθ\begin{aligned} F_H &= (1-C) \sdot mr \omega^2 \sdot \cos \theta \ \ --- (1) \\ F_V &= C \sdot mr \omega^2 \sdot \sin \theta \end{aligned}

Now, let us imagine the horizontal force FHF_H in three dimensions by looking at this diagram

fig_logarithmic-decrement Unbalanced horizontal force acting on cylinders of locomotive

In the above diagram, let's represent the horizontal unbalanced force acting on cylinder 1 as FHU1F_{HU1} and similarly the horizontal unbalanced force acting on cylinder 2 as FHU2F_{HU2}.

From this, it becomes easier to understand what is traction force. It is the resultant of the two horizontal unbalanced force. Hence we get the following equation:

FT=FHU1+FHU2  (2)F_T = F_{HU1} + F_{HU2} \ \ --- (2)

Traction force definition

The resultant unbalanced horizontal force due to the two cylinder along the line of stroke of cylinder is called traction force.


Try to write the definition by explaining the things that we did in the calculation. This way it becomes easier for you to understand it.

Now for cylinder 1, and using equation (1) from above, we can say that:

FHU1=(1C)mrω2cosθ  (3)F_{HU1} = (1-C) \sdot mr\omega^2 \sdot \cos \theta \ \ --- (3)

Similarly for cylinder 2:

FHU2=(1C)mrω2cosθ  (4)F_{HU2} = (1-C) \sdot mr\omega^2 \sdot \cos \theta \ \ --- (4)

As we have studied in the last video, if cylinder 1 angle is at θ\theta then the cylinder 2 angle will be at θ+90°\theta + 90\degree. Therefore from equation (4) we can say that:

FHU2=(1C)mrω2cos(θ+90°)  (5)F_{HU2} = (1-C) \sdot mr\omega^2 \sdot \cos (\theta + 90\degree) \ \ --- (5)

From equation (2)

TF=FHU1+FHU2=(1C)mrω2cosθ+(1C)mrω2cos(θ+90°)=(1C)mrω2[cosθ+cos(θ+90°)]=(1C)mrω2[cosθsinθ]  (6)where, m=mass of the reciprocating partr=radius of the crankω=angular velocity of the crankC=portion of the reciprocating part which is to be balanced\begin{aligned} T_F &= F_{HU1} + F_{HU2} \\ &= (1-C) \sdot mr\omega^2 \sdot \cos \theta + (1-C) \sdot mr\omega^2 \sdot \cos (\theta + 90\degree) \\ &= (1-C) \sdot mr\omega^2 \sdot \big[\cos \theta + \cos (\theta + 90\degree) \big] \\ &= (1-C) \sdot mr\omega^2 \big[\cos \theta - \sin \theta \big] \ \ --- (6) \\ \\ where, \ &m = \text{mass of the reciprocating part} \\ &r = \text{radius of the crank} \\ &\omega = \text{angular velocity of the crank} \\ &C = \text{portion of the reciprocating part which is to be balanced} \\ \end{aligned}

Now we need to find the maximum and minimum of this tractive force. To do this we need to differentiate equation (6) with respect to the variables. Here our variable is θ\theta. So we need to differentiate FTF_T with respect to θ\theta.

Doing this we get:

dFTdθ=ddθ[(1C)mrω2(cosθsinθ)]=0 (1C)mrω2ddθ(cosθsinθ)=0 ddθ(cosθsinθ)=0 sinθcosθ=0 sinθ+cosθ=0 sinθ=cosθ tanθ=1\begin{aligned} {dF_T \over d\theta} &= {d \over d\theta} \big[(1-C) \sdot mr\omega^2 \sdot (\cos \theta - \sin \theta) \big] = 0 \\ \therefore \ &(1-C) \sdot mr\omega^2 \sdot {d \over d\theta} (\cos \theta - \sin \theta) = 0 \\ \therefore \ &{d \over d\theta} (\cos \theta - \sin \theta) = 0 \\ \therefore \ &- \sin \theta - \cos \theta = 0 \\ \therefore \ &\sin \theta + \cos \theta = 0 \\ \therefore \ &\sin \theta = - \cos \theta \\ \therefore \ &\tan \theta = -1 \end{aligned}

Therefore, we get θ=135°\theta = 135\degree or 315°315\degree.

Now we will put the value of both these θ\theta in equation (6) to find the variation of tractive force.

For θ=135°\theta = 135\degree we get,

FT=(1C)mrω2(cos135°sin135°)FT(min)=2(1C)mrω2\begin{aligned} F_T &= (1-C) \sdot mr\omega^2 \sdot (\cos 135\degree - \sin 135\degree) \\ & \boxed {\therefore F_{T(min)} = -\sqrt2 \sdot (1-C) \sdot mr\omega^2} \end{aligned}

Similarly for θ=315°\theta = 315\degree we get,

FT=(1C)mrω2(cos315°sin315°)FT(max)=2(1C)mrω2\begin{aligned} F_T &= (1-C) \sdot mr\omega^2 \sdot (\cos 315\degree - \sin 315\degree) \\ & \boxed {\therefore F_{T(max)} = \sqrt2 \sdot (1-C) \sdot mr\omega^2} \end{aligned}

Since we get the maximum and minimum value of tractive force, we can say that the value of TFT_F varies between 2(1C)mrω2\sqrt2 \sdot (1-C) \sdot mr\omega^2 to 2(1C)mrω2-\sqrt2 \sdot (1-C) \sdot mr\omega^2.

Which we can represent as:

FT=±2(1C)mrω2\boxed{F_T = \pm \sqrt2 \sdot (1-C) \sdot mr\omega^2}


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