 Hello everyone, I welcome you all for this today's session on Curse and Super-Elevation. This is part of geometry design of railway track. I am Ashok Kumar, Assistant Professor, Department of Civil Engineering, Walsh and Institute Technology, Solar Port. Learning outcome of the today's session, at the end of the session students will able to analyze the super elevation on course and calculate the equilibrium super elevation for different railway gauges. Before we get into the super elevation, let us understand the some of the parameters of course. We know that curves are required to bypass the obstacles and this includes the horizontal curves and vertical curves. Whenever there is a change in the direction, it may be in horizontal curves or vertical curves. So these two curves are connected by a smooth horizontal curve. This may be a simple curve or compound curves, reverse curves, etc. So the horizontal curves includes the design of super elevation, widening, designing of transition curves, designing of horizontal curve length, so on and so on. And vertical curves, whenever there is a change in the gradient, we have to connect those the gradients by a smooth vertical curve. It may be valley curve or cement curve. So that includes the vertical curve. We need to design some gradients. We need to design the length of vertical curves. In the railways, we always define the degree of the curve or the curve by degree of the curve or by radius. In the roadways, usually we define it by radius. But in the railways, because the radius is being larger, we always define the radius the curve by degree of the curve that is D. So what is degree of the curve? It is the angle suspended at its center by a 30.5 meter or 100 feet chord. So in some textbook, we take 30 meter chord or some textbook, we take 30.5 meter. If I take 30 meter chord length, what is the degree of the curve? Let us see that. So here R is the radius of the curve in meter and D is the degree of the curve. So we always define the curves by 1 degree, 2 degree or 3 degree curves like this. So we know that the total circumference of the circle is 2 phi R and makes at 360 degree angle at the center. So this is the 30.5 meter or 30 meter chord arc and making an angle of 360 degree at the center. So now for 30 meter arc, so D by 30 equal to 360, there is a whole the 360 degree divided by the circumference of the circle is 2 phi R. So if I simplify the D value, it comes to 1718.87 by R, R you can round it up to next value 1720 by R. In some textbook, I told you we take 30.5 meter. If I take 30.5 meter chord length or arc length, so here we get D value as 1747. So we make it to 1750 divided by R. This is how we define it the curve by a degree of the curve. Now with understanding of super elevation which you have learned in the transportation one in the roads and highways, so I hope you are able to give the correct answer for this true and false questions. In super elevation, the first question is in the super elevation, the level of the outer layer is raised above the inner rail. It is true or false and for a super elevation next question, super elevation increases with wear and tear of the rails or wheels, again it is a true or false. Again on curves, centrifugal force is proportional to the square of the train velocity and inversely proportional to the radius of the curve. And last question on the curve, centrifugal force pressure will be always maximum inner rail or outer rail. So I hope you are able to give the answer for that. If not, we will see the what is the correct answer over here. So in super elevation, the level of the outer rail is raised above the inner rail. This is what we call as the canned or super elevation, it is true. Super elevation increases the wear and tear of the rails and wheels, it is false because if you design proper taking all considerations, proper super elevation designed and constructed, there is no wear and tear of the rails or wheels, that is false. So on curves, we know that centrifugal force is proportional to the square of the velocity and inversely proportional to the radius of the curve. So this is F equal to V square upon R that we what we call as centrifugal acceleration. On curves, we know that the it is not the pressure will be not the maximum inner rail because the centrifugal force acting outwardly. So due to that, the pressure will be maximum on outer rail. So let us understand the how the principle works over here for the super elevation. So super elevation is also called as canned, we know that it is a difference in height between the outer rail and inner rail. So raising of outer rail with respect to the inner rail is called as canned or super elevation. So now when a vehicle negotiating from the horizontal from the straight to the horizontal curve, it is the tendency of the vehicle to go into that path only not taking a turn like tangential to the curve. So now this introduce the centrifugal acceleration or radial acceleration what we call. So this radial acceleration results in introducing the centrifugal force that is what we call as WV square upon GR. Now in this case, we have to raise the outer rail over here with respect to the inner rail what we call as small e is the rate of super elevation. So why we provide the super elevation to make the effect of the centrifugal force as well as the what is the weight which is acting downwardly from the CG of the vehicle. So we have to raise the super elevation such a way that the resultant of the both centrifugal force and your weight of the vehicle acting perpendicular to the top surface of the rails. It means which is the resultant force which is acting because of these two forces that should act perpendicular to the plane of the surfaces of the both the rails. That means both are the loads are equally distributed on both the rails. That is what we have equilibrium condition and that the raising what we get this equilibrium condition is called equilibrium super elevation. Now let us try to derive the equation for this equilibrium super elevation. So there is so many terms associated with the derivation. One is weight of the vehicle which is acting through the CG of the vehicle that is w and v is speed of the vehicle in meter per second r kmph and r is radius and g is the gauge of the track in meter that is gauge distance and small g is acceleration due to gravity in meter per second square and alpha is the angle over here, alpha is the angle and one more alpha over here. And s is the length of the inclined surface in meter and centrifugal force that is f is whatever we have the f is w v square upon g r. So let us understand these terms in detail when we do the derivation for this equation. So here you can see that the g is the your gauge distance measured from inner surface to running faces of the rails that is whatever we have this distance is taken as g and the sloping surface we have a sloping surface like this. So this is called as your s. So I have taken over here is the sloping surface s and g is the gauge distance and alpha is angle between the two the gauge distance and the sloping surface and e is the rate of super elevation. Now we have tried to resolve these forces with respect to the slope surface s. Now with respect to the slope surface s what are the forces inclined to the sloping surface we got the component of the centrifugal force with respect to the sloping surface s is f cos alpha and opposite to that we have w sin alpha. Now resolve that f cos alpha equal to w sin alpha let us take it as equation 1 and now we know that the centrifugal force f equal to w v square upon g r or taking this triangle cos alpha is g upon s or sin alpha equal to e upon s. So let us try to put all these values in the equation 1 when I put the equation all those values in equation 1 this comes to w v square upon g r into g by s equal to w into e by s. So both side the s will get cancelled and also your w will get also cancelled in this section. So the equation remains e equal to v square upon g r into g where v in meter per second. So usually we express always in kmph so to convert that into kmph we know that to put the g value as 9.81 and convert this the v value in meter per second into kmph we take it as multiplying by 1000 into divided by 16 to 60 so that comes to 0.278 square. So 9.81 divided by 0.278 square so that gives me value of 127 is the value so this is how I got the value of 127. So this is here the e it is in meter and v in kmph so usually and we know that the gauge distances the various gauge we have broad gauge, meter gauge and narrow gauge. So if I put the value of broad gauge is 1.676 so the equation further simplifies that is 1.676 and here it is in meter e is in meter usually we express e in centimeter. So if I divided by this 127 by 1.27 so whatever the e value comes in meter this is going to be converted in centimeter or you can after calculating the e value in meter you can do a multiply by 100 that also you can get the value of the e value in centimeter. So to simplify we have taken 127 by 100 that comes to 1.27 so now 1.676 by 1.27 so further it simplifies to 1.315 v square by r so here please remember it is in centimeter and v in kmph. Now for the same way you can calculate for the meter gauge just replace the g value as 1 meter and for narrow gauge g value as 0.762 so the equation will for the meter gauge is 0.8 v square by r and for the narrow gauge it is 0.6 v square upon r. So whatever this speed we have considered here it is called as equilibrium speed and the super elevation which you calculated is equilibrium count. For to restrict the maximum values we have taken the maximum super elevation is taken as one tenth thumb rule one tenth to one twelfth of the gauge is taken as the maximum super elevation for further for the broad gauge for the group A and group B and C and group D and E so we take the limiting value of super elevation under normal condition is 165 mm and with respect to the permission from the chief engineer we take it as 185 mm. So for the meter gauge it is 90 mm and for the with respect with taken from chief engineer it is taken as 100 mm. These are the references I have used for preparing this presentation. Thank you.