Hello friends welcome back to NPTEL online certification course on Soil and Water Conservation
Engineering, I am Rajendra Singh professor in Agriculture and Food Engineering Department
of IIT Kharagpur. We are in week 4 lecture 20 and today we will solve Problems on Broad
Base Terrace. Now, just to remind you of the course content
that we covered in this week up till now in lecture 16 we started the week with introducing
terrace, in lecture 17 we covered the concepts corresponding to bench terraces.
In lecture 18 we used the concepts learnt in lecture 17 to design bench terraces
that is we solved problems dealing with the design of bench terraces. In lecture 19 we
dealt with broad based terraces and todays lecture we will solve certain problems dealing
with the design of broad based terraces. Let us start with problem number one and this
problem states that for a broad based terrace constructed on 7 percent land slope, the channel
depth is 0.3 meters and the width of terrace is 6 meters. Compute the cut and fill heights
and the slope ratios for the front slope and back slope assuming a balanced cross section.
So; that means, it is a broad based terrace, a constructed on 7 percent land slope, channel
depth and channel width of the terrace of are already given. And we have to compute
the cut and fill heights; at the slope ratios for the front slope and the back slope and
we have to assume a balanced cross section. And balanced cross section means the cut is
equal to fill that is this is the information which is already given in this particular
problem . So, once again the data given are land slope
S equals to 7 percent; channel depth, h equal to 0.3 meters; terrace width, W is 6 meters
and balanced cross section. So, we have been given h is known to us, W is known to us and
of course, the land slope which is here S that is known to us.
So, these are three things which are known to us and of course, balance cross section
so, that simply means; cut is equal to fill that is also known to us so; that means,
this cut is equal to this fill that is known to us. And for this section when it is a or
width of the when all the three widths are same that when means when when the section
is balanced or the cross section is balanced. Then we know the relationship between cut
fill, the channel depth and terrace width, which is given by equation 3 which we saw
in the previous lecture. And that reads as C plus f equal to h plus S times W. Where
C is cut f is fill C is cut f is fill h is the depth of the channel and W is the terrace
width and S is the land slope. Ah So, this is what it is and as per this problem statement
it is a balance cross section so, C is equal to f that is cut is equal to f.
So, from this relationship C equal to f is h plus SW divided by 2 because, C plus
you can also write 2 C equals to or 2 f equal to h plus SW or in this case writing C equal
to f equal to h plus SW by 2. And in this we know the value of h, we know the value
of S, we know the value of W, so; that means, by just by putting the values of h S and W,
we can get the values of C and f that is cut and fill.
And so, C equals to f which is equal to h is 0.3, S is 7 percent; that means, 0.07 W
is width of the terraces which is given a 6 meters. So, by putting the value you get
C equal to f is equal to 0.36 meters. So, that is cut fill is 0.36 meters so;
that means, from the this is the bottom of the channel, that is measured at the channel
centerline the bottom of the channel channel and this is the original land slope, the vertical
difference between two is cut. Similarly here the the terrace centerline
the top of the ridge and the original land slope at this particular vertical line, the
vertical difference is the fill. So, cut and fill value we have obtained in this problem
now we have to obtain the slopes . So, from the geometry for the front slope
the change in elevation over W is equal to h. So, this is front slope so that this is
W and elevation change from here to here this is h basically. So, front slope will be S
f is W by h and W value is known h value is known W is 6 h is 0.3 so; that means, it is
20. So, front slope ratio comes out to be 20 is
to 1, that is 20 horizontal to 1 vertical. So, we are always in this subject typically
we represent slopes as horizontal is to vertical ratio. For the back slope if you see; the
change in elevation over W has two components, one is of course, h h this h is already there,
but the same time there is a elevation difference of the original land slope basically.
So, that we have to take into account over the width which is same. So, it is h plus
SW. So, that is why back slope is W divided by h plus WS. So, 6 divided by 0.3 which is
h W is 6 S is 0.07 and by calculating this we get 8.33 S b equals to 8.33.
So, back slope ratio is 8.33 is to 1 so, front slope ratio is comes out to be 20 is to 1,
back slope ratio comes out to be 8.33 is to 1. So, that is means cut and fill is equal
to 0.36 meters, front slope ratio is 20 is to 1 and back slope ratio is 8.33 is to 1.
So, these are the answers which we have obtained for this particular problem .
Now, we go to second problem; which says that design a 300 meter long variable graded terrace
for a land having average slope of 3 percent. The maximum permeable velocity of water in
the terrace channel is 60 centimeter per second. Intensity of 1 hour rainfall exopected during
10 year recurrent interval is 10 centimeter per hour. The region is less humid and soils
are normal. So; that means, we have to design a 300 meter
long variable graded terrace for a landslope of 3 percent. The maximum permeabile velocity
is given here 60 centimeter per second. Ah Intensity of 1 hour rainfall for 10 year
reccurrence interval is given as 10 centimeter per hour and the region identified as less
humid and soils ae normal. So, thats means remember we had a condition
for vertical interval. So, that simply shows that we can straightaway use the one of the
equations recommended for hm graded terraces .
So, the data once again with the the symbol length of variable graded terrace l small
l is 300 meters, land slope S is 3 percent, maximum permissible velocity of water which
is nothing, but the non erosive velocity we call is 60 centimeter per second and intensity
of 1 hour 10 year rainfall is 10 centimeter per second, the region is less humid and with
normal soils . So, first thing we need to estimate is vertical
intervals and vertical intervals and for this purpose we may use equation 2 previous
lecture, which there was equation given for vertical interval for less humid regions and
normal soil . So, as per the relationship vertical interval equals to S plus 6 by 10,
where S is the land slope, the land slope which is 3 percent in this case .
So, putting the value of S we can calculate the vertical interval as 3 plus 6 9 by 10
that is 0.9 meters. And the relationship between HI and VI we already know because, we know
that HI and VI they are related with reference to S percent land slope . So, this
is VI this is HI so VI by HI equal to S by 100.
So, this we have seen many times so, from this relationship HI is equals to 100 VI by
S. So, same here 100 VI by S and VI value we already know S value we already know so,
putting these values we get horizontal interval equals to 30 meter. So, for this particular
given conditions vertical interval is 0.9 meters and horizontal interval is 30 meters
. Now, step two is let the grades of variable
graded terrace be we can use the standard table we already saw table 2, in lecture 19
0.5 0.4 0.2 and 0.2 for each quarter length from outlet respectively.
And this we can see we can also see the table basically by looking into the table
because here the channel grades are recommended for for variable graded terraces, based on
us NRCS recommendations. So, our terrace length is 300 meters; that means, we fall here in
this 260-360 range and then the recommended grades are in percent 0.50 for lower quarter
0.40 for second quarter 0.30 for third quarter and 0.20 for the upper quarter.
So, right from upstream to downstream the slope is increasing point 0.20 0.30 0.40
0.50 slowly . So, this is these are the recommended grades which we can accept basically.
And so; that means, these are the grades which we have chosen. Next step three we need
to determine the peak of runoff and obviously, for this we use rational method we already
know that whenever we need peak. Peak rate of runoff we use rational method which is
Q equals to CIA by 36. Where C is the runoff coefficient, Q is is
in cubic meter per second that is peak runoff rate in cubic meter per second, C is the runoff
coefficient, I is the rainfall intensity equal to time concentration unit is centimeter per
hour. And A is the drainage area and the unit used is hectares and using this this is the
rational formula using this we can use the cultivate the peak rate of runoff .
So, for this case, the drainage area that is the terrace area which is drainage area
is nothing, but multiplication of length of the terrace and the horizontal interval, because
the area between two terraces that is used as a drainage area for the designing the
graded channel. And that means a length is given as 300 meters HI we have calculated
as 30 meters. So, the area comes out to be to be 9000 square
meters and or because the unit required in q I equals to CIA hectares. So, we are again
converting this 9000 square meter into 0.9 hectare divided by 10000, where 10000 square
meters equal to 1 hectare. So, it comes out to be 0.9 hectare. Now we need to determine
the time of concentration because I definition says that it is peak rainfall intensity for
duration equal to time of concentration. And for time of concentration we use Kirpich
formula that is the standard formula we use and as per this formula T c is equal to 0.0195
L to the power 0.77 S to the power minus minus 0.385 where L is the total flow length and
S is the total slope over the flow length. So, the total flow length typically is length
time plus horizontal interval that every time you have seen that in between two two
graded terraces will be built here. And then the second one is let us say here so; obviously,
this is the horizontal interval this is the horizontal interval between these two. So,
this is the area that x is the terrace area. So, the flow has to traverse this distance
and along the along the channel perpendicular longitudinal distance has traverse this length
of the terrace. So, that is why that is the total flow length
is HI plus the length along the channel. So, this is 300 plus 30 or 330 meters. And the
total slope is elevation difference again the elevation difference here, plus the grade
of the channel we have to consider and the total length for calculating the slope. So,
H is the difference in elevation between most remote point and outlet. So, most remote point
could be considered the upstream of the previous terrace.
Now, coming to elevation difference between adjacent terraces over HI is land slope is
3 percent. So, 0.03 times 30 so, 0.9 meters it comes out to be straight away. And over
the channel length we have already decided the slopes that is 0.5 percent this should
this should be read as point 4 percent. So, 0.4 percent, then 0.3 percent and 0.2 percent
so, from table we read point 0.2 0.3 0.4 and 0.5 percent.
So, that is why the and and total 300 meters so, the quarter length is 75 meters. So that
means, 0.005 times 75 0.004 times 75 0.003 times 75 0.002 times 75 and then calculate
slopes of these . So, the total elevation difference is this plus sum of these four.
So, all sums comes out to be 1.95 meters. So, total elevation difference is 1.95 meters,
total length is 330 meters slope over the total flow length is 0.006 that is the slope
we calculate . So, that simply means that now putting the
because, we know L now we know S. So, putting the value in Kirpichs formula that is 0.0195
330 to power 0.77 0.006 to the power minus 0.385 we get time of concentration equals
to 12.15 meters for the given area. Now, we have been given 1 hour intensity
is 10 centimeter per hour and we have to obtain the intensity corresponding to 12.15 meters.
So, like in the previous class where we solved problem on bench terraces and we use
the standard nomograph. So, the same standard nomograph we have to use here, which relates
1 hour rainfall intensity, 1 hour rainfall intensity to rainfall intensity for different
durations. And the durations are given here 5 minutes 15 minutes 30 minutes 60 minutes
120 minutes and so on. So, for for our case it is 1 hour intensity
is 10 centimeter per hour so; that means, 1 hour intensity. So, our value will lie on
this horizontal line, then corresponding to 10 centimeter per hour and we have to look
for 12.15 minutes. So, this is 5 minute line so, this is one point of reference and this
is 15 minutes, this is one point of reference. And we have to find the this is 10 will be
somewhere here so, 12.5 will be somewhere here so more than 20 somewhere here. So, roughly
we are approximately we are saying this value is 22 centimeter per hour.
So, we are assuming rather from this nomograph we are reading the value that corresponding
to time of concentration, the intensity, the peak intensity is 22 centimeter per hour and
that is the value we will use in our CIA by 36 .
So; that means, now we can use the runoff rational formula and for the we we assume
C equals to 0.3 so, now, CIA by 36. So, C we assumed at 0.3 I we have just now approximated
at 22 the area we have calculated as 0.9 hectares. So, putting the value we get Q value equals
to 0.165 cubic meter per second, this is this should be read a cubic meter per second not
the m 3. So, it is a cubic meter per second or cumec
which is the unit of discharge, because here we are calculating the peak runoff rate. Now,
once we have calculated that discharge which our terrace would be able to handle
or grade channel should be able to handle. Then we have to come to defining the terrace
cross section . So, for terrace cross section we need to select the cut, front and back
slopes . So, for this again we can use the standard table, that is table 3 provided in
lecture number 19 that is the US SCS recommendation and from this table we we will just see the
table, the table is here. So, from this table the land slope is ours
is 3 percent so, closer to this 3 percent. And for this the channel slopes are recommended
as 6 is to 1 that is cut slope this is ridge fore slope or what we are calling as front
slope 8 is to 1 and back slope is 8 is to 1.
So, 6 is to 1 8 is to 1 8 is to 1 are our recommended slopes from the table. So that
means, we can come back here and we we see that using a standard table we get S c is
equals to 6 is to 6 S f equals to 8 is to 1 and S b equals to 8 is to 1. So, slopes
we have take taken a decision regarding . So, now we need to choose the channel shape
and determine it is dimensions. So, let us say that we consider a triangular channel.
So, triangular channel means; we are saying that the depth of channel is h that is the
depth of M bounding or depth of cut channel is h. Here in this case so; obviously, the
total cross sectional area that is the area that will be utilized for water flow .
So, here; that means, the half is to 6 is to 1 times h because, this is the channel
here this is h, this slope is 6 is to 1, this slope is 8 is to 1. So, that simply means
that this width is 8 h and this width is 6 h and so, two triangles. So, this triangle
is half into 6 h into h and the other triangle is half into 8 h into h and this is how we
can calculate the total cross sectional area which comes out to be 7 h square.
Now, we can assume we assume a simple value of h equals to 0.3 that is h value we are
assuming. So, the cross sectional area A is one 1.27 square meters, we also require the
wetted perimeter because we have to calculate the hydraulic radius.
So that means, length of this and length of this sum of these two because, we know these
two values we can always calculate using Pythagoras theorem. So, here S 1 plus S 2 that is
h square plus 6 h square h square plus 8 h square, here it will be 8 h square and if
we solve this it will come out to be 14.1 h for 14.15 h or the value of h we are assuming
0.3 so, wetted perimeter is 4.24 meters. So, hydraulic radius will be ratio of A and P
A value we calculated as 1.27 P value we calculated as 4.24. So, value of hydraulic radius is
0.3 meters . So, from there Manning's equation can be
used for calculating the velocity which is n by n R R to the power 2 by 3 S to the power
half. So, n value we are assuming as 0.04, R we have already calculated and the grade
is already given so, 0.0025. Ah So, using this the flow velocity we get is 0.56 meter
per second which is non erosive velocity by as per recommendations. And Q of the channel
is A times V A we are know V we know so, it comes out to be 0.71 cubic meter per second.
So, design channel has sufficient capacity because, the capacity required were 0.165
cubic meters per second wherein we are calculating 0.71 cubic meter per second. So, now, we can
say that the channel has sufficient capacity so, assuming 20 percent freeboard that is
0.2 h ridge height will be 1.2 h or 0.36 meters. So, this way we have designed the various
cross sections various parameters of the channel.
So, the answers are vertical interval is 0.9 meters, horizontal interval is 30, terrace
grades of 0.4 0.3 0.4 0.3 and 0.2 for each quarter length, terrace flow depth is 0.3,
slope ratios are 6 is to 1 8 is to 1 8 is to 1, terrace ridge height is 0.36 and channel
flow velocity is 0.71, terrace channel flow capacity is 0.71 cubic meter per second.
So, these are the various dimensions we have already designed .
Now, we go to the third channel which is also a design channel, design a 800 meter long
level broad terrace, for a land having average slope of 4 percent, maximum expected rainfall
having 10 year recurrence interval is 18 centimeter. Infilteration capacity of the soil of the
area is such that 40 percent of the rainfall is absorbed in the field and the region is
sub humid and has clay loam. So, the data given are length, length of level
broad terrace is l is 800 meters, land slope is 4 percent, maximum expected rainfall having
10 years recurrence interval is 18 centimeters, infilteration capacity is 40 percent of rainfall
and it is a sub humid region with clay loam soil.
So, we need to estimate vertical interval for the purpose we have to use the same equation
because, it is a less humid region and normal soil. So, VI equals S plus 6 by 10 so, VI
comes out to be 1 meter because S is 4percent and HI once VI is known HI comes out to be
25 meters . So, now we need to determine the maximum runoff
volume and so maximum expected rainfall having 10 years recurrence interval is 18 centimeters.
And infilteration capacity 40 percent of rainfall that simply means the runoff potential is
60 percent of rainfall because, 40 percent is absorbed in infilteration .
So, the level terrace system needs to be designed to handle, the maximum runoff potential which
is 60 percent of the rainfall. So, it is 60 percent of the rainfall; that means, 0.6 or
0.18 or 0.108 meters that is the maximum runoff to be stored .
Now, the storage area which is the maximum runoff depth times horizontal interval so,
putting these values we get 2.7 square meters. And here also we need to choose a channel
shape and determine it is dimension . So, let us consider a trapezoidal section with
cut slope, front slope, back slope with 6 is to 1 8 is to 1 and 8 is to 1.
So, these ratio we are assuming here and a trapezoidal section we are considering so;
that means, the when we calculate the flow area that we have to consider three sections
a 1 a 2 and a 3 because slope here is 6 is to 1 here it 8 is to 1 .
So, the cross sectional area so, three a 1 a 2 a 3 the first one is triangle, half 6
6 h into h, the second one is a rectangle b times h and third one is rectangle half
8 h times h so, it is bh plus 7 h square. Since storage area required is 2.7 square
meters, the cross section area of the channel should be equal to the required storage area.
So, assuming b equals to 0.4 meters, let us say that bottom width is 40 centimeters or
0.4 meters. And if we put in the in this bh plus 7 h square is 2.7, if you put in this
equation b equals to 0.4 we get 7 h square plus 0.4 h equal to 2.7 or we can write 7
h square plus 0.4 h minus 2.7 equals to 0. So, quadratic equation we can use solve
using a straight formula which we know minus b plus minus square root of b square minus
4 ac by 2 a. So, by putting the values of b b this is b this is a and this is c. So,
putting these values here we get the value of h as 0.65 or 0.59. So, we select eiher
of the two so, we select h equals to 0.65 meters .
So, assuming 20 percent freeboard that is 0.2 h, ridge height comes out to be 0.78 and
terrace width 8.78 6.24 meters. So, terrace the answers are vertical interval is 1 meter,
horizontal interval is 25 meters, flow depth is 0.56 meters, channel bottom width 0.4 meters,
slopes ratios are 6 is to 1 8 is to 1 and 8 is to o1, channel ridge height is 0.78 meters
and channel width is 6.24 meters. So, these are this is all various parameters of this
level broad base terrace . So, now, we take the last problem and that
says the compute the depth of cut and fill and side slope of a broad base terrace for
30 centimeter depth of flow on 8 percent land slope. Consider that the tillage operations
are to be performed using machines and the width of machine is 4.27 meters. So, we have
been given the depth of flow, land slope and the width of the machine .
So, that simply means the data given are land slope S equals to 8 percent, channel depth
h equal to 0.3 meters that is here h is given here, this is h, this is h, this is the this
is h it is given here and machinerty width is given as 4.27 meters. So, because machinery
width width of terrace is typically kept W equal to machinery width so, W comes out to
be 4.27 meters. So, from first problem we saw this equation
C plus f equals h plus S times W, now now putting the known value of h S and W because,
h S and W is known, C plus f we come comes out to be 0.68 meters. So, this is what we
get C plus f equal to 0.18 0.64 meters . And since we normally use a balanced cross
section, where cut is equal to fill so; that means, C equal to f will be 0.32 meters. And
then front slope because front slope we know the width is width and
the fall is h which is 0.3 so, it is 4.27 by 0.3that is 14.2. So, front slope ratio
comes out to be 14.2 is to 1, wherein for the back slope when we calculate back slope
the elavation difference is h plus WS. So, here
the slope ratio will be W divided by h plus WS, W is 4.27, h is 0.3, W is 4.27, S is 0.08.
So that means, it comes out to be 6.67 so back slope ratio comes out to be 6.67 is to
1. So that means, the cut and fill comes out to be 0.32 meters, the front slope ratio comes
out to be 14.2 is to 1 and backslope ratio comes out to be 6.67 is to 1 .
So, with this we have seen that we can design broad base
terraces, be it graded terraces or be it be it a level terrace. And also we can find
what should be the cut and fill for a given condition and what should be the ratios
the slope ratios. So, these are the problems, but as I have said earlier also that there
can be other forms of the problem. So, if you solve some more problems for practice
probably things will be even clearer, but I think that with this much practice itself
you
will be able to design broad base terraces. Thank you very much.