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Either we scouring or it will not cause deposition of the sediment in the bed. So, that was the
principle and by that principle we could do design of alluvial channel and that particular
velocity, we termed as critical velocity. And then, but these 2 theories or the equations
derived or equations used in these 2 theories are empirical and based on some observation
and that way it has some drawbacks, and in, again in the entire procedure there are some
drawbacks, those drawbacks also we did discuss in the last class. And then, as these are
empirical, so people were trying to develop some analytical basis for designing alluvial
channel. And then, USBR, United State Bureau of Reclamation, so they started working on
one method, called tractive force method. And we stated in the last class, what the
tractive force method is and we will be continuing with that tractive force method and say, 1st
in the last class we did define what is, say, a tractive force? So, it is the force exerted
by the flowing water on the perimeter of the channel or say, on the material of the sides
and bed. So, the, in suppose uniform flow condition if we talk about, then weight of
the fluid in the direction of the flow is actually the tractive force and that we will
try to equate with the resistance force to find, what is the critical tractive force
and all. Well, then we were talking about unit tractive
force, that means, this tractive force per unit area. Well, so that we refer as unit
tractive force, which is also called as drag force or shear force. So, these different
nomenclatures are used and then, we were discussing in the last class about distribution of tractive
force.
Means, say we could see the unit tractive force, that unit tractive force means that
was coming. In fact, as average unit tractive force because the force exerted by the fluid,
we were considering on the entire surface. Well, the surface, which is in contact with
the water during that length of the channel and then we divided it by the entire surface
length, which is in contact with the fluid motion, which is in contact with fluid, here
our fluid is water, of course, we are talking about. And so that way, the stress what we
are getting, tractive force, that per unit area is in fact, a on average basis, so average
tractive force. But in reality, the tractive force exerted
by the fluid on the side of the canal will not be uniform all over the surface. It will
be more, may be at the bottom, at certain point at the centre point, it may be more
than, if it is a trapezoidal channel, then on the side towards the top, the tractive
force will be less, but towards the bottom on the side, that will be more. So, that way,
tractive force distribution we were talking about and we did draw a diagram for trapezoidal
channel where we could just draw a trapezoidal channel and we were showing, how the tractive
force distribution can be plotted on the side and on the bed of the channel well. But that
distribution, we draw like this just to recall, because from here we will be moving ahead
again. So, that, suppose it is a trapezoidal channel
of course, we were considering a channel where our y, that is the depth, y is, say, I mean,
1-4th of the bed or we can say, the bed, which is 4y of the depth, well 4 times the depth,
in that case we got that tractive force distribution. Of course, pattern will be same, it is like
that and here it is almost uniformly distributed, it is almost uniformly distributed and here,
again, it will be like that, here it is more. So, maximum tractive force we could get at
this point and so this maximum tractive force, and if we just compare the maximum tractive
force, if we compare the maximum tractive force at the bed and at the side, you can
see, that at the bed it is more and this can be expressed in terms of w, y and S. It is
the unit, this w is unit, weight of water, so this unit weight of water, then depth,
then the bed slope. So, based on that, we could derive the expression, but there will
be a term, say 0.77, that means, it is not exactly equal to wyS, but it will be a percentage
of wyS, this will be... And maximum can be wyS, that I will show just in the next slide,
but this can be, say 0.77, 0.8, something like that, this can be, and so that factor
we can write, say x; x into wyS, where x can be, say 0.7, 0.8 like that, 0.6, anything
it can be and that way this can be, can have different value.
Well, then similarly, y also can be, I mean, similarly in the side tractive force also,
this can be expressed as w, y and s, and similarly here also, a product will be coming. So, x
into that, I am writing here and now this was given for this particular channel, where
width of the channel is 4 times the depth of the channel. And we did discuss what the
value of x there, and then this slope, ok, my diagram is not proper, but this slope is,
say if it is 1, this is 1.5; for that sort of side slope, if it is 1 this is 1.5, and
for that sort of side slope and for this shape, where B is equal to, where B is equal to 4y,
for that we gave some value of x in the last class. And that way, we were discussing, how
shear stress or other we, we call it is a how unit tractive force get distributed tractive
stress, get distributed on the side. Well, now, say in reality, we will be having
different type of channel, say B by y ratio. Here, our B by y ratio is 4B by y, if we make
it is 4, but always the B by y will not be 4, and here our side slope is equal to 1 is
to 1.1, 0.5 is to 1 rather, but this will not be always same. So, if the side slope
changes, say 1 is to 2, like that if it is changes, z value then, and the B by y ratio
changes, then how this will vary, how this maximum tractive shear stress or unit shear
force varies, so that we need to know. And USBR, united state bureau of reclamation,
again gave some curve for that, which curve can be? This curve can be used, this curves
can be used for design of tractive force, let us see how these curves are.
Well, first you can see, this is, here in this graph, on this side it is B by y ratio
is plotted on this side, say B by y ratio, they examine, say up to 10 we can have and
then on this side, that is on the y axis, on the y axis it is unit tractive force in
terms of wyS; w means again, unit weight of water.
So, suppose if I get a point here, 0.4 means, it is 0.4 wyS. Let me draw the curve and then,
at it is actually a family of curve, it is not a single curve, it is family of curve,
which were drawn for different side slope. Well, and for, even for rectangular channel,
and for rectangular channel if I draw for a side, and this unit tractive force is for
side tractive force, we are looking about that is in a channel, it is at the side. So,
for rectangular channel it is like this, that is, from 0 it is starting and it is, it is
going like this and then it is going like this, it is a smooth curve of course, it is going like this. So, this
sort of curve was obtained for rectangular channel. Well, I, I must say, that my curve
may not be exactly like the curve, that was given, but just I am trying to show what the
trend is and this is for rectangular. And then for trapezoidal, again, say it will
be starting from about little higher from 0.5 and then it is going like this, that it
will be going like this, this curve is going like this, my space is less painted, this
little larger, but still let me try. And then, so this curve is for say z equal to 1.5, and
then this is rectangular, this is rectangular and of course, if we draw for a, if we draw
for a point, say z equal to 1, it will be almost following the same line, but it will
be starting from here; it will be starting from here. So, this we can say, it is z equal
to 1 and then if z increases to 2, if z increases to 2 the curve starts little higher and then
it will be, but maximum limit is always remaining within 0.7 and 8. So, it is going like that,
this is for z equal to 2. Well, so that way a family of curve was given, curve was given
by USBR and following that we can just let us try our design, and that is quite helpful.
Say, we, initially if we take, that our B by y ratio is 3, then from, from that point
if we consider, that B by y ratio is 3, then what will be and if we consider that our,
it will be a rectangular channel, then we can see what is the side tractive stress.
That is, of course, after that we will need to check whether that channel will be sufficient
for carrying the required discharge that will be coming later. And then, for again, the
unit bed tractive stress, unit bed tractive force, say this is, we can plot that as given
by USBR. When it is bed, then for rectangular this will be starting from here and this will
be like this, this is going up to almost 1, it is going up to almost 1 like that and if
I extend it, it will be going up to 1. Then, say for trapezoidal, for trapezoidal,
it was interesting to observe because it is bed, although it is trapezoidal, but say side
slope, whether it is more or less, it does not influence that much, but still for the
z equal to 2 and 1.5, we are drawing for bed. So, it starts from, say very close to 3, but
it follows almost that one. Then from 5 onwards, it just leave this axis and then it goes like
that and for higher value, for higher value of, in fact, it will be passing more steeper
like this, and for higher value of B by y ratio it goes almost up to this point 1, almost
up to 1. So, what we can see, that for a B by y ratio
of say 8, 9 or around that, this tractive force is becoming equal to wyS, that is, which
we could get for wide rectangular channel. And just equating that, our component of gravity
force acting in the direction of flow and that when we put as our tractive force tau,
and then when we divide it by the contact area, then we get the tractive force per unit
area, that symbol here itself we call as tau. And that value we got, that it is equal to
w, r and s, where r is the hydraulic radius and for wide rectangular channel, we got it,
that r is approximately equal to y, and that is why, we can consider this as wyS. And as
our B by y ratio is increasing, means we are talking about a wide rectangular channel and
B is larger as compared to y, that is why, B by y ratio is quite larger and as it is
increasing, we are finding, that it is becoming this, this fraction or the product factor
is becoming almost equal to 1 and what we could derive theoretically is valid here.
So, these are the curves that were provided by USBR. Now, that another point is more important
here, that say, we are using this curves and now one curve is giving us the tractive force
for the side, and this curves are giving for tractive force for the side and these curves
are giving tractive force for the bed, that is, a tractive force exerted, we can say like
this, tractive force exerted by the fluid on the channel, this is what we are talking
about. And for design purpose, we should know again, how much is the resistance of this
tractive force? And then, there is a question, that can we have some relationship between
tractive force at side and tractive force at bed, can we have some relation between
these 2, so that when we will try to find a permissible tractive stress, when we will
try to find a permissible tractive stress, then for some of our analysis, we may get
there tractive force at bed is equal to this much permissible tractive force, permissible
to be more precise, we can say permissible unit tractive force at the bed is this much.
Now, if we design it from that consideration that may not be safe because on the side it
may not be safe. So, we need to know, what is the permissible
tractive shear stress, unit tractive shear force at the side? So, now, that if we could
derive by having a relationship between tractive force at bed and tractive force at side, then
it would have been much better. So, and that can be derived very easily, very easily from,
from our conceptual understanding of the geometry of the channel and just they equating the
forces, we can do that, and let us see.
So, to just see that, let us take a channel. Well, let us consider 1 channel, say this
is our channel and the water is flowing in this channel, water is flowing in this channel
and then, when water will be flowing, let me just show, that suppose this is the water
level and it is flowing, I am not drawing it because something else I will have to draw,
but well, for just to appreciate, that say water level is there, you just imagine that,
we have water level at this level, we can draw a line to show just very small, draw
some dot, say this is a what the water level, oh I can draw it here, I can draw it here,
say here the water level and then the water is flowing, so there, that will not create
any problem. So, that way, water is flowing with a particular depth and then, in this
channel, here is our, say, side of the channel, we are digging it. And then, this way after
digging it, it is making, we are making the channel.
Now, let us see that, say, side slope of this channel is suppose theta; it is suppose theta.
Well, just to analyze, let us consider, say we have a small area, let us consider a small
area, say a; let us consider a small area a. And then, when water is flowing in this
direction, when water is flowing in this direction, then the tractive force exerted on this area,
say area is a and let the tractive force exerted on this area is say tau s, that is, tractive
force means, we are always, we are meaning, by unit tractive force means tractive stress.
So, tractive stress exerted by the fluid on this particle is say tau s, then if we say,
that this area is a, this area is a, this small area, then the tractive force on these
area will be on this particular area, will be tau s into a; this area will be tau s into
a, fine. Now, let us see, whether this area or that
area, if we talk about a very small portion, then it may ultimately come down to a sediment
particle, whether this is getting subjected to some other forces also. Well, the weight
of this particle is acting vertically downward; weight of this particle will be acting vertically
downward. If I draw the same area here, then it is convenient to show, that it will be
acting vertically downward. Say, this is also area a and this weight of this will be acting
vertically downward, and let me write that weight as, say w dash. This is I am not writing
unit weight, I am writing the complete weight of the sediment of that particular portion
a, and that is why I am writing w dash because this is under water. We are considering the
force exerted, tractive force exerted means, this is under water, it will be submerged
by the water. So, weight of the particle will be basically submerged weight and that is
why we are writing this as w dash, well. So, this w dash is the weight and then this particle
or particles, or the area itself will be having a tendency to slide down along the slope,
or rather we can say, that there is another force, which is acting in this direction and
because of which our interest is to know, that what are the forces acting to move this
particle from its location, move this particle from its location rather. So, one is the tractive
stress and now we are talking is a tractive force, which is tau s into a; another is the
force, that is trying to move it along the slope in the downward direction because of
the gravity, and that gravity is basically this weight of the particle.
And then, what that angle is? We know that this angle is theta. So, between this line
and that line, so if I draw this line and if I draw a normal to this particular slope,
then this angle will be theta. So, the angle just acting in this direction will be, say
force acting in this direction will be w sine theta, theta is away from this part. So, this
is theta. So, and theta is the angle between this line and that line, and this line is
perpendicular to this plane means this line, and then it is perpendicular to this. So,
this is theta and this is our w sine theta, w sine theta, that particular force we can
write w sine theta, and here I am writing this as w sine theta.
These 2 forces are, forces are acting, w dash, well we are writing w dash. So, what are the
net forces that are acting on this particular area to disturb this sediment? Well, there
are some other forces, which will try to resist this motion, but now we are talking about
or we are looking into the forces, which are trying to disturb this, I mean particles.
So, these 2 forces, that is, w dash sine theta and this tau s a are the forces, that is trying
to disturb the particle from its position. And then, the, what is the resultant of that?
We can draw the resultant of that just by completing this and then joining this. So,
this is what the resultant and that resultant if I write as f1, say disturbing force, so
that how we can write? This will be d square, I mean tau s into a, tau s square a square
plus w dash square sine square theta and then taking root of that, we will be getting this
force. So, we can write the force disturbing or we can write this disturbing
force, the disturbing force. Well, the disturbing force f1 is equal to root over, say, tau s
square a square plus w dash square sine square theta, well. So, that is what our disturbing
force, so I am writing this as f1. Well, then when there is a disturbing force, then if
it is not moving means, there is a resisting force also. And at the time of just incipient
motion condition, these 2 forces, what we mean by incipient motion condition, that particle
are just in the verge of motion; it is just in the verge of motion.
So, what is that? The disturbing force is just equal to the resisting force and our
interest is to find out permissible stress, means, up to that point when our force is,
when our particles are not moving, then we are fine, our channel design is fine. So,
that way, we are talking about that situation and of course, our permissible force may be
little less than this particular force, but when it is just in the verge of motion, then
we can call that as a critical condition. And the permissible condition should not be
the critical condition, it should be little less than that, but anyway let us see, what
is the expression for this critical condition. And so, considering incipient motion condition,
what we can have, say considering incipient motion condition, we did discuss about incipient
motion condition earlier also, that we will equate, but the resistance force, the 1st
we have find out the disturbing force, then resisting force, the resisting force because
why we are writing this, that considering incipient motion condition. Now, resisting
force, as we know, that it is coming from the frictional resistance here and the cohesion
part, we are not counting in this point. So, for resisting because we are talking about
alluvial channel, because we are talking about alluvial channel, and this resisting force
will be proportional to suppose, your, your, force exerted is increasing proportionally,
there frictional resistance is increasing, and of course, beyond a particular limit it
cannot increase and that is the limiting condition, well.
So, at incipient motion condition, this will be at its maximum and what we can see, that
suppose this is the bed, this is the not bed, this is the side, then this will be opposing
this motion and frictional motion, opposing this particular force, that will be the normal
weight or normal reaction. So, this will be having a normal force in this direction, let
me draw on this phase because we are seeing it from the backside, that is fine, we can
see. So, in this direction, in the normal direction,
what will be the component? This is w cos theta and as we know, the frictional resistance,
frictional resistance will be in any direction. We can say the frictional resistance is equal
to mu into the normal force; mu into the normal force. So, the, not that normal force is cos
theta, weight is this one, so reaction will be like that and then it is mu into cos theta,
so this mu cos theta is basically opposing this forces. We are, I am not drawing exactly
the direction here, but we can say, that this is, say mu cos theta.
Now, what is mu? That is again we need to refer to the frictional property of the soil,
so, but let me write here as mu cos theta, here itself I am showing on this phase, it
is easy to show, and here just I am drawing it because that is, we cannot see the backside
from this side, that is why I am drawing it here, well. So, this is equal to mu cos theta
resisting force, say f2, if I write, this is equal to mu cos theta.
Now, if angle of internal friction of the soil is say phi, then what the mu is? That
we all know that mu is nothing but the tan phi. If the angle of internal friction is
phi, then it is tan phi, so we can write that. Well, this, sorry, this is not cos theta,
this is w cos theta, this I did a mistake in writing here, this is not cos theta, this
is w dash cos theta. Again, dash is always coming, w dash cos theta, this effective weight
of the, I mean, soil will be significant there. So, this will be, this I should write as mu
w dash cos theta; this will be w dash cos theta, well. And this mu, we can write, as
I have explained, that this is equal to, tan, tan of phi into w dash cos theta, where phi
is, phi is nothing but angle of internal friction; angle of internal friction.
Well, so now, for incipient motion condition, we can equate these 2; we can equate these
2. So, equating this 2, what we can write? That w dash cos theta and tan of phi, this
is what the resisting force, and the other one is, say tau s a plus w dash sine theta
with a root sine, then squaring these 2, what we can write, that w dash square cos square
theta tan square phi is equal to tau s a. Just to our target is to separate this tau
s and then, this is tau s square, these are also square because after taking square only
we were taking the roots. So, it will be tau square a square w dash square sine square
theta. Well, now this expression, where we got is this one, tau square a square dash
square w dash square sine square theta. We are writing the same expression here, then
from here what we can write, that tau s, tau s is equal to, that is the side tractive stress,
we can write from this part as w dash cos theta tan phi by a. First we are writing tau
s is equal to this minus that and then we are taking w dash cos theta tan phi common
and then of course, we are taking root and then, if we take the other part here, then
it become 1 minus tan square theta by tan square phi.
Well, so when we took tan phi common, then here we are having already, say tan square
phi is coming here and then it becomes and we are taking cos theta common. So, it is
becoming at, in fact, we took cos square theta. So, that way, it is becoming, cos, sine square
theta by cos square theta. So, this tan square theta is coming and then, later when we took
root over this, cos square theta become cos theta, tan square phi become tan phi and all
are just getting this thing. This one more step we can, we could write in between, well.
So, that way we are getting this tau s, the side tractive force, we are getting an expression.
Let us give a number one, today’s equation. Well, now, if we talk about the shear stress
at the bed, so shear stress or rather, let me use the word tractive, unit tractive stress,
unit tractive stress or tractive force, unit tractive force because already we are using
the term unit, so tractive force at bed. Now, what is the difference between this particular
expression in bed and side? Well, if we talk about bed, then our theta for the bed is becoming
0, if we just flatten it this become bed. So, in bed other things will be remaining
same, this, this force will be there, the reaction force will be there, but the, here
if we take a bed, same area, then this force will be there, and there is this tau b into
a, then weight component will be directly vertical. So, there will be no component in
that direction and then the resistance force will be coming here, it is directly equal
to w, means basically, if we make theta is equal to 0, then we get the expression for
bed. So, what we can write, unit tractive force at bed tau b is equal to, so if we make
theta equal to 0, our expression will become, say we can write it as w dash by a, we can
write w dash by a and then, on the w dash by a and then it is tan of phi. This part
means, w dash by a cos theta is becoming 0; theta is becoming 0 means, cos theta equal
to 1 and then it become tan phi, and what about this part? When theta becoming 0, this
part is becoming 0 and that is why it is 1 minus, this means only root over 1. So, finally,
this is becoming 1 and we get this expression, so this is what, the expression 2.
Now, our target was to find an expression or to find a relation between this shear stress
or unit tractive stress, tractive force at side and unit tractive force at bed. So, we
can now do 1 by 2. So, that will give us, say tau s by tau b; tau s by tau b is equal
to this and that part will get cancelled and we will be getting only cos theta. Then, root
over 1 minus tan square theta by tan square phi, well. So, this is what, very, very important
expression, that we use for design purpose; thus, we use for design purpose.
And then, as you can see that cos theta root over this can further be simplified, but of
course, we are leaving it in this level, that is, tau s and this ratio. This ratio has a
particular name, we call this as k; we call this ratio as k. And then, what we can see?
That k is equal to, say tau s by tau b, this is equal to cos theta root over 1 minus tan
square theta by tan square phi. That means, name we are giving as k and then one point
we should note here, that theta will always be having some value, and of course, for trapezoidal
channel and then, this value will be always a fractional less than 1. So, tau s is always
equal to some fraction multiplied by tau b; well what it means?
So, this ratio we name as k and so you can see, that k is tau s by tau b and cos theta
root over 1 plus tan square theta by phi. And more importantly, that, when we see, that
cos theta, theta value will be always, say, a fraction. So, it will be, it is between
1 and 0, so it will be a fraction and I mean cos theta value will be a fraction. So, the
highest being 1 now, I mean, so if this is a fraction, then what we can derive from that,
that tau s by tau b is always a fraction because tan theta by tan square phi, this will be
of course, a positive term and less than 1. So, 1 minus something less than 1, then root
over terms. So, this product will be what? Whatever it is, this thing, this product will
be a fraction and that way we will be finding, that this tau s is equal to some fraction
multiplied by tau b and that way we are getting, the tractive force at side is less than the
tractive force at bottom; tractive force at side is less than the tractive force at bottom
and this is very important, which help us in proceeding for design.
Well, now, let us just see how we can have, now let us talk about design of tractive force
method or rather design by tractive force method. And till now, we have discussed about
tractive force that is how we can have theoretically a relationship between the tractive force
at the bed and tractive force at the side. And that way, once we can have some idea about
the permissible tractive force of the bed, then we can know, what will be the permissible
tractive force, unit tractive force at the side, permissible unit tractive force of the
side. So, this is we have done and we can get it, then how much is the tractive force
exerted by the fluid on the side and bottom bed of the channel, that we can get from the
curves given by USBR. Well, now, let us see how this information helps us. This information
helps us in designing the channel by tractive force method, well.
And for design purpose, we define few terms that are important. We call this as a critical
tractive force. Well, so what is critical tractive force? The tractive force at which
the particles are just in incipient motion condition, this is called critical tractive
force, well. Then, definitely, we do not design in engineering any design for the critical
condition, we do it for a situation that it is safer or better than the critical condition,
we are not generally trying to reach the critical condition. So, when we talk about permissible
tractive stress or say, permissible unit tractive force, then we define it in a way, that permissible
tractive force is kept less than the critical tractive stress, well.
And now, for what type of soil, what will be the permissible critical stress? This is
very, very critical, this is really critical. I mean, we have theoretically derived one
relationship between the bed tractive stress and side tractive stress, that is fine, but
what will be the critical tractive stress for the bed or for the side, what will be
the limiting value, that is a very difficult task. And because, and why it is difficult?
Because say, we are talking about alluvial channel, but there also will be some finer
fraction and those finer fraction will be changing the value. And then, we, it is of
course, generally expressed in terms of the diameter of the particle, it is generally
expressed in term of the diameter of the particle, say if diameter of the particle is this much,
then tractive stress required is this much. In fact, when we did discuss about the uniform
flow in erodible channel, then we were studying those aspect also. We, we, we were discussing
silt theory and there we were discussing those aspects, but again say particle size only
does not matter, sometimes the, orient, orientation of the particle, then what is the shape of
individual particle, how is the frictional resistance, all those factors will be influencing.
Well compactness and other things are also coming into play.
So, like that, I mean this deriving, derivation of this tractive force is difficult, but still
for our purpose, that we need to design. So, we will be, we will have to follow some methods
and the experiences shown, that well, these approach if we follow, it gives us a, say,
suitable channel, which we can use for our purpose and it is stable, so that way we go
by those method. In fact, USBR also conducted lot of experimentation and they gave, they
gave a, a, very, I mean, with quite convincingly they could give the value of permissible tractive
force for a bed based on the particle diameter and that relation, what they gave is the tau
b, and of course, they gave in pound per feet square, is equal to 0.4. Its relation is very
simple, 0.4 into diameter and that diameter is in inch. diameter is in inch.
Now, say, now, it is in India we are using, say diameter in millimetre or say centimetre,
we will be talking about and then, we are not talking about pound per feet square, we
may be talking about Newton per meter square, kilo-Newton per meter square. So, that way,
if we use this equation, then this relation, then we need to convert the unit and this
is well known, how we can convert from inch to centimetre, just to give readily, how we
can have pound per feet square to kg per meter square, that we can convert as 4.882 kg per
meter square, 1 pound feet square is equal to, that we can have. And if we want to convert
it to Newton, then we multiply it by 10, like that kilo-Newton, thousand. So, like that
we can change these things again. Well, so that way we can have the value of
say, tractive force at bed of the particle, at bed of the channel, by using the relationship
given by USBR. This is directly giving us permissible tractive stress, well.
But again, as we have just discussed, that we were discussing Shield’s theory also
and then from Shield’s theory we got a curve, and from that curve, ultimately we could derive
that tractive stress, or this we, we gave as critical tractive stress. tau c we were
writing by Shield’s equation and there with tau c, if we keep in kilo-Newton per meter
square and say, diameter of the particle in meter, diameter of the particle in meter,
say d, diameter in meter and in tau c is in kilo-Newton per meter square, then we could
see, that Shield’s equation gave us a relation like 0.06. Then unit weight of water, g, is
the specific gravity of the soil and then, minus 1 into diameter of the particle, that
simplified equation we could get and that equation, if we simplify further by taking
the standard value of w and by taking the standard value of say, soil, may be 2.65,
like that soil is specific gravity, then we can get a value tau c is equal to nearly 0.98
into d. Well then white equation is another equation
that gives us critical tractive force or critical tractive stress tau c as 0.801 into diameter
the unit is same and a Lane’s equation is another equation, which is generally applied
when the flow is a, when it is a fully developed turbulent flow. In that case, our tractive
force is, I mean, at a little lesser tractive force, the particle starts moving for a fully
developed turbulent flow and then tau c is given as 0.78 into d. So, these values can
also be used for determining the permissible tractive force or permissible unit tractive
force, other permissible tractive stress at the bed, and what is that?
That is permissible tractive force, can also be taken as 0.9 of tau c. Here, we are getting
tau c and then by this method if we calculate the value, and then we take the 0.9, that
is, 90 percent of that value, then that we can take as permissible tractive stress. And
then, if you just try this calculation, then from this calculation if we do, if, if we
just try this calculation putting for standard values, then we find, that this gate, we get
almost expression of 0.76 d. This will also ultimately lead us to a value. If we change
this unit, it will lead us to a value of, say, 0.76 d around and then, here it is tau
c and then, we need to take 0.9 times of that. So, more or less, we are at almost, I mean
near value, that we are getting and so we can see, that to be on safer side for our
design purpose, we can take a tau c as say, 0.76 tau b permissible tractive stresses,
0.76 d or 0.75 d, around that we can take. Of course, if we have confidence, if we know
the soil in a much better way and flow condition, we are sure about that it will not be the
turbulent, and then we can take our tractive force little higher value. Then, our, if we
take the tractive force little higher value, that this permissible tractive force little
higher, our size of the channel will be lesser. So, I mean, if we do not want to be that much
conservative, then we can take little higher value or otherwise, we can go for lesser value.
Well, that depends on experience and knowledge of the soil formation.
Well, then, what is design principle; what is design principle? So, design principle,
let me just explain it step by step, you can refer to the slide. The design principle is
that unit tractive force exerted should be less than the permissible unit tractive force.
That we have already explained, we are repeating, but as a design principle, the main steps
we must say, that this is the 1st principle. Well, then starting with this principle, then
we could see, that tractive force ratio, we calculated one, that we name as k tractive
force ratio; k is equal to say, tau s by tau B and this ratio, this ratio indicates, that
tractive force required to move particle at side is less than that required for the bottom,
that required for the bottom. Well, and therefore, for design purpose what is done, that the
channel is designed based on the side tractive force. This is the standard procedure we follow,
for design is done based on the side tractive force because in the side, the force required
is less. And of course, there is another point, that we, we are getting, that at the side,
the force required is less. So, this is more critical, more critical if the side is safe,
generally bottom will be safe. But another point we need to note, that if we just go
back to our this curve, that we did discuss, we are discussing this curve, then if our
B by y ratio is high than the, the force exerted, the bed is higher and for otherwise also,
normally these, these are steeper in bed, the tractive force exerted is also higher
than tractive force exerted in the side. So, we know that tractive force, resisting
tractive force of the side is less, but at the same time, force applied is also less.
But of course, for high value of B by y is, this is significant, but for low value it
is not that much significant. So, generally, we design it 1st for, say, tractive force
at the base and then, sorry, tractive force of the side, side tractive force we design
it first. We see what the value is coming and then we are getting a size, and then its
safety against bottom tractive force is checked, it is the conventional practice we follow.
However, when we have a computer, then these calculations will not take much time and in
fact, if we have the curve and then we can just represent the curve also by some equation,
and then this can be done very quickly. And then, what we can see, of course one can calculate
the depth from the consideration of both side and bed tractive force, that means, depth
we can calculate from both side considerations. We can get one depth from bed consideration,
we can get one depth by using both the tractive force value, one is permissible, another is
what is exerted. So, from both these value we can get 1 depth
and then, of course, 2 depth we will be getting out of that. What is the smaller one, that
we can consider for our design purpose, but conventionally we find, that for a side, what
depth we will be getting, that is the smaller because the side is more critical. Still,
we can go by this way also, what I have mentioned in the last.
Well, then we need to follow some design step for tractive force method. First, what we
do, that we consider a B by y ratio. Well, we need to assume a by y ratio based on the
availability of space and with some of our understanding, the depth will be this much.
Suppose, we know, that this is the discharge and for that discharge, this much will be
the depth around we want to flow in the channel, then well, that should be the B.
So, we can try with B by y ratio from our experience and then we are assuming a B by
y ratio, and the side slope is assumed. So, this first B by y ratio we are getting, then
side slope is assumed based on the material property, and the value of theta is obtained.
Well, a side slope as we did discuss that for this particular material, this will be
like that, although it is alluvial, but still based on our understandings, still there will
be some variation. Alluvial means, the particular size may be still larger and smaller. Well,
that way we decide what should be the side slope and then, once we decide the side slope,
well, we will be getting what is the angle theta.
Well, then, once we get this theta, that is z value is known, z value is known, this will
give us the z value, then maximum unit tractive force tau s at the side can be obtained in
terms of wyS from the USBR curve. We have just shown that particular curve from the
side tractive force and bed tractive force, we have sets of curve for different z value,
we can find those tractive force, that is at side and bed of course, that will be in
terms of wyS, that is, some fraction multiplied by wyS.
And then, another important value we need to know is that angle of internal friction
phi. So, that angle of internal friction we need to calculate, rather calculate means
we can have our analysis in the laboratory or there are some standard curve, which provide
the value of angle of internal friction. In fact, USBR gave some curves, which gives us
the value of angle of internal friction based on the diameter of the particle. Well, those
curves can be used, we can do experimentation in the lab, we can find the value of phi.
So, that way, somehow we need to find the value of phi, then we need to calculate the
tractive force ratio k, which we know, that there is an expression we can find that.
Well, the next step will be permissible tractive force tau b is computed. So, tau c, we can
use the expression of USBR or we can use tau c for that, Shield’s equation or White equation,
Lane’s equation, then we can, I mean, from any of these we need to calculate finally,
the permissible tractive force at the bottom. So, once we get the permissible tractive force
at the bottom, we can get the permissible tractive force of the side by the equation,
that it is, tau S is nothing, but is equal to k into tau b.
Well, then, equating permissible tractive stress at the side and the tractive stress
exerted, which we have obtained from the assumed B by y ratio and value of depth B by y ratio,
so that we can equate and from that we can get the value of y, depth y and b. Why first
we will be getting the value of y and then, we know the B by y ratio. So, we can have
the value of, I mean, this B directly from them and that we are getting tau s. How we
are getting this tau s? That we can get tau s is equal to that fraction multiplied by
wyS and after that we are getting the depth and B value. So, this is what is our required
design, but of course, this is not the final design. Then, safety of this design we need
to check against the tractive force at the bed.
So, then, once we check the tractive force at the bed, suppose we have found it to be
safe, then we need to check, whether this size is enough to carry the required discharge
or carrying capacity of the channel is checked by computing the discharge by Manning’s
equation. Now, our size of the channel with us, then
slope is there. So, we can calculate how much discharge it can carry. If this is the depth
y, then if it is just sufficient from our required discharge, then this is our final
acceptable design, and if it is less than our required discharge, then we need to repeat
the entire procedure by increasing the B by y ratio. So, we need to increase the B by
ratio y ratio and then we need to carry out the entire process to get a new dimension
for our channel. So, that way we get the channel dimension, we check and if it is sufficient,
when it is just sufficient, then we stop. Suppose, in a 1st trial we have found, that
it is sufficient, but it is much more than the sufficient, then we can reduce the B by
y ratio, rather we should try with smaller B by y ratio, so that we can just get from
insufficient than we are moving into just sufficient. So, that is what we can do and
once the computer program is developed, then we can do it very easily.
Well, then, after designing the channel, we need to a, a freeboard to the channel. What
is meant by freeboard? Say, our design depth is this one, then we need to keep some extra
over this for our safety, that is called freeboard; that is called freeboard. So, that extra depth
we need to provide. And Indian standard and USBR also suggest that, saying that standard
mean, basically, Central Board of Irrigational Power has suggested some value, that if discharge
is less than 0.75 meter cube per second, the freeboard required is 0.45. Why I am giving
this value? Because freeboard is not that negligible. Suppose, sometimes we may get
our depth of y, water is 0.5, but above that we need to, it is 0.45, almost 100 percent
freeboard, I mean. Then suppose, for a depth of 0.75 to 1.5,
not depth for, a depth of 0.75 to 1.5 meter cube per second discharge of this value, we
can have freeboard of 0.6 and then, for there is a large range now, from 1.5 to 85 meter
cube per second, that range is very large and for that we need to 8.75. And then, if
it is greater than 85, then we need to 8; almost 1 meter is 0.9 meter. So, what we should
get, that it is not uniformly with the increase of discharge, it is not proportionately increasing,
but it is increasing definitely. But for a very large range of discharge we may keep
the freeboard as 0.75 meter. So, that is how we get the freeboard and finally, we get our
required design for the channel. Well, here we are concluding our discussion
on alluvial channels and we are not taking up any numerical of course, we did not take
any numericals and I hope that we have tried to explain the steps very well, so that there
will not be much problem in doing some of the numerical example, but this practice is
essential. A nd I hope that we will be able to have better understanding of all this method,
Kennedy’s, Lacey's tractive force by taking some hands on and by doing some practical
example. Thank you very much.