In the last class I have discussed about the

load carrying capacity of single pile, but generally pile these are used in a group.

So, now today I will discuss about the load carrying capacity of the pile or in group

or the group action of the pile. Then I will discuss the other method by which we can determine

the load carrying capacity of the pile. Now, first I will discuss about the group action

of the pile. In the group action of the pile, now when

pile is used in a group then we can calculate the efficiency of the pile group, now that

efficiency if we write that this is the efficiency of group or the group efficiency, that we

can write in this form this Q u g divided by n into Q u, where Q u g is load carrying

capacity of pile group and Q u is the, you can say this is the ultimate load carrying

capacity of the pile group and Q u is the ultimate of single pile. And n is the number

of piles; that means, is here n is the number of piles in the group Q u is the ultimate

load carrying capacity of the single pile, and Q u g is the ultimate load carrying capacity

of the pile group. So, here to grade these group efficiency we have to calculate the

ultimate load carrying capacity of the pile in group, consider this pile as a whole and

then this Q use the single pile. Now, this now for this smaller spacing between piles generally efficiency or the

group efficiency less than u 1 or less than 100 percent. So, that means…

So, that it indicates that that; that means, here these in a group the pile that can fail

in as failure can be occurred as individual pile failure or failure can be occurred as

a group or as a block type of failure of the considering the total all the piles that as

a group and that can fail as we whole or it is a block type of failure. So, that is one

type of failure another is individual pile failure. Now if the spacing is very small

then this will occur this group type of failure generally occur and in that case the efficiency

is less than 1. Now for the larger spacing it is group efficiency is equal to 1. So,

now, for the largest type of spacing where individual pile failure will occur.

So, we will get the group failure is equal to the individual pile load carrying capacity

into the number of piles that means, Q u g will be equal to is n into u g that means,

the contribution from each pile will, summation of contribution of each pile will give the

group load carrying capacity of the pile in that case this efficiency of book will be

1 and if it is less spacing is very small then we will get the, because the over load

due to the over liking over lapping of the stress zone or the influence zone for the

single pile individual piles. Though; that means, the group failure will occur in that

will less than the load carrying capacity of the individual pile, if you sum the all

the load carrying capacity of the individual pile that will be generally more than the

pile load carrying capacity as a. So, in that case if the spacing is very small then the

group load carrying capacity is smaller or the lower than the summation of call individual

piles load carrying capacity. So, in that case this efficiency will be less

than 1. Now another thing that for the driven pile is loose to medium sand loose to medium sand this efficiency can be

greater than 1 this is because there is one pile is driven into a loose to medium say

and then sand become dense. So, in that case because of the installation of the pile the

group can capacity of the pile surrounding the soil pile soil surrounding this pile at

get dense. So, that is why the efficiency that will increase.

So, there is a three possible cases that for the smaller spacing between the pile. So,

as the summation of all individual pile load carrying capacity that is more than the group

carrying capacity group load carrying capacity of the pile. So, within that case efficiency

will be less than 1 now if the larger spacing in that case basically the individual pile

failure will occur and if the spacing is very less in the group type of the block type of

failure will occur. So, if the spacing is very large. So, individual pile will failure

will occur in that case the group load carrying capacity is equal to the summation of the

all individual pile load carrying capacity. Now, in third case in the driven pile if it

is driven in the in the loose to medium sand in that case the density the sand the sand

become dens. So, because of this nature the group carrying capacity or the efficiency

of the pile that will increase and that efficiency is sometimes even greater than 100 percent

or greater than 1. Now for in we will discuss about how to calculate the group carrying

capacity of the pile. So, in that case we can write that pile group

in clay; first we will calculate the group carrying capacity of the pile if it is in

the clay. So, as I have mention there is a two types of failure one is that block failure

and second one individual pile failure. These two types of failure will occur if the spacing

is generally more spacing between the two piles. So, more than this individual pile

failure occur, and if it is less than the block failure is occur generally, but in then

now for the another thing that is condition for the clay that generally block failure

will occur if spacing is less than 2 to 3 D, now if the spacing less than 2 to 3 D this

the diameter of the pile and if thus largest spacing individual pile failure will occur.

In that case if we calculate Q u g that is the group load carrying capacity of the pile

that is C u b at the base then N C then A b at the base plus P into L into C u s.

So, in that case here the similar to the individual this expression. So, this is the from the

tip resistance and this is the friction resistance. So, this is the tip resistance and this is

frictional resistance. So, now we can write that C u b is the undrained strength of the

soil at the base of the pile of the clay of the clay at the base of the

pile. So, this is tip resistance is coming from the base. So, this is at the base of

the pile So, similarly Q u s this is the undrained

strength of the clay along the surface or length of the pile group or block and the

here also we can write this is pile group. Similarly in C values we can write is equal

to 9 like the single pile then we can write A b is equal to cross section area of the

block this is the cross section a of the block L is the embedded length of the pile pile and P b or P; that is the perimeter of

the block. So, here this u g we are calculating by considering

the block failure to the, if we consider this is the block failure, and then we calculate

this this is the tip resistance and this is the friction resistance. So, C u b at the

undrained resistance at the base of the pile n C is 9 A b is the cross section area of

the block P is the perimeter L is the embedded length, and Q is the undrained resistance

along the surface for the pile group or block, so same as the single block.

Now, if we can similarly we can write we can determine the or we can write the expression

for the pile in the sand the same while using the same type of expression for the sand because

this is similar to the single pile and already thing is the here the cross section area the

expression of same in the cross section area in case of single pile we are consider there

is a cross section area of the single pile, here the cross section area we have to consider

the block when you calculate the tip resistance. Similarly in the calculation of the single

pile friction resistance we use the area is the surface area of the single pile here we

have to consider the surface area of the block; that means, the perimeter and the length of

the block. So, that is the only difference when you consider

the group and the individual failure individual pile load carrying capacity. So, this is the

load carrying capacity for the clay similarly while using the same expression like the single

pile in the in the sand we can determine the load carrying capacity of the pile in the

sand. Generally and sometime as I have mentioned that in driven pile is efficiencies greater

than 1. But one we design this things we can consider that efficiency equal to 1 and you

can design. So, now so that means, we are now giving the expression for the clay and

similarly the same expression as a given for the single file that we can used by slide

modification for the cross section area for the base and the cross section area of the

surface area of, to consider the cross section area of the block and the when we calculate

the steep resistance. And when the cross section area of the surface area of the block when

we calculate the friction resistance that is the difference between the single and group

others the expression are almost same. So, now we will solve one example which we

can determine that how we can calculate the load carrying capacity of the pile, and the

single pile and the group pile and then we will determine then the other factors also

and the spacing also. Suppose problem is that that we have 1 block

this is the element of the pile in that thing this is the total 16 piles of their so 4 columns

and 4 rows, total 16 piles. Now, it is a in the clay soil. The diameter of the pile or

D of the pile is 300 millimeter this is the diameter of the pile, now the thing is that

this is the total number of 16 piles with diameter

is 300 millimeter, length of the pile or embedded

pile is 10 meter now undrained Q u undrained strength of the soil is 50 kilo newton is

unconfined compressive strength of the soil this Q so; that means, C we can calculate

C u and then C is 10 of the soil that will be Q u divided by 2. So, that is 25 kilonewton

per meter square; so C u, so we are considering the same C u for the surface as well as at

the base. So, that. So, that mean C u b and C u S similarly C u b is equal to C u S equal

to C u equal to 25 kilonewton per meter square. So, Q u equal to 50 kilonewton per meter square

C u will be Q u by 225 kilonewton per meter square.

Now, we have to determine the spacing, what will the spacing such that, that grouped efficiency

will be exactly 1, why we can check any other value also so they can take the 0.9 0.8. So,

here we will design these things such that the group efficiency will be exactly 1 what

will be the spacing. So, as I have mentioned that here we have to consider the single and

as well as the block failure. This is the block basically for the group.

So, suppose we consider this is the spacing between the space this S and similarly this

is one is also S, this is S, this is S, this is S. Now distance from this because this

is the center of the each pile. So, distance from this center to this h is 0.15 meter similarly

this one is also 0.15 meter as similarly this one is also 0.15 meter and this one is also

0.15 meter, because the total diameter is 0.3 meter. So, half of this is 0.15 meters.

So, as calculate the group efficiency Q u g is n into Q u. Now here Q u g is 1 n is

equal to 16 in is the number of pile. So, now, if Q u g is equal to 1 so that means,

Q u g if efficiency is equal to 1 so that means, Q u g will be n into Q u or Q u g equal

to 16 into Q u. So, now, first we will calculate the Q u g and here another thing the condition

is that we can neglect the tip resistance of this pile that mean the resistance we are

getting. So, we can neglect it, neglect neglect the bearing at the tip of the piles, because

these are the in the clay. So, that in the these are the friction pile. So, the resistance

coming from the friction will be more as compared to the resistance coming from the tip.

So, the one condition that we can we are neglecting that friction components so that means, Q

u g will be equal to that I have mention that is the 16 Q u. So, the single pile capacities

16 into alpha into C u into A S area at the surface. So, now, here this is 16 into alpha

into C u is 25, area is pi into 0.3 into 10 is the length.

So, area is pile D L. So, pi into 0.3 into point into L is the 10 meter and this alpha

we have to calculate. So, in the last class I have given one chart. So, calculate the

how to calculate the at a will how to calculate this alpha value and here based on this C

u value I am giving another figure by which we can determine the value of alpha. And that figure is proposed or this is proposed

by the Tomlinson 1979 and this is for the driven pile. So, in this figure. So, this

is. So, this point we can say this is alpha this is 1.5 this one is 1 this is 0.5 and

this is 0; and this side this is 0.5 sorry this is 25 50 75 100 125 this is the value

of C u it is in kilonewton per meter square. So, the point that we are get in. So, these

are the points. So, we can if you join this points this is the chart. So, this is from

here we can determine what will be the value of alpha corresponding to 25 because our C

u value is 25 kilonewton kilonewton per meter square. So, now, what will the value of alpha

as corresponding to 25 kilonewton per meter square. So, with this value you can calculate

this alpha is coming 0.95. So, we can calculate the value of Q u g in

terms of 16 into Q u and this is 16 into alpha is 0.95 into 25 is C u into pile into 0.3

into 10. So, this is this part is 16 and that this basically this part is 0.95 alpha into

25 into pi into 0.3 into 10 this is this total thing is equal to the individual pile load

carrying capacity ultimate pile load carrying capacity of the individual pile. So, the total

load carrying capacity is 3581.42 kilonewton; that is the load carrying capacity of the

pile in group. So, here all the calculation we are doing

for the compressive load carrying capacity of the pile. So, next term we will we can

calculate this Q u g in terms of blocks failure this is this is in terms of individual failure

now we other time in terms of block failure we can write this is P perimeter into L into

C u. When we are calculating this block failure here we consider that our alpha is basically

1 the reason is that here is the alpha is the addition. So, that mean this here the

addition your this is the addition factor and this addition your considering when it

is a single pile; that means, there the addition is between the pile surface and the soil.

So, that is why have to consider alpha value because your two different materials we are

considering, it is the pile different materials and the soil is the another different materials,

but when you are considering the block this a block failure. So, there is blocks this

said this is the interaction between the soil and soil. So, both are same material that

is why we are considering here we consider this block failure we consider alpha equal

to 1, but when we consider the individual pile failure that is the interaction between

the pile surface of pile material and the soil to the two different material that is

why we consider different alpha value, but here interaction between the two same soil

the here we will consider the alpha equal to 1, because the interaction between the

soil and soil when this is a block. So, now here we can write this P is the perimeter

is 4 into 3 S because if I look this block total. So, when we are considering the block

failure this most of the surface here interaction is soil to soil, individual piles interaction

in the soil and pile. So, that is we are considering alpha in the block, but the side one is 3

S plus 0.15. plus 0.15 So, 3 S plus 0.3 is one side another is also 3 S plus 0.3, because

here this is the square type of arrangement so; that means, we can write the perimeter

is 3 S plus 0.3 into 4 because this is one side another 4 side into L into 25. So, this

is 3000 S plus 300 now for efficiency is equal to 1 we can write that 3000 S plus 300 that

is equal to 3581.42. So, now from here we can calculate that S

is coming 1094 millimeter or S is equal to 3.65 D where D is the diameter of the pile,

this is the diameter of the pile. Now then another thing that we have to check with that

IS code recommends that 2 IS 2911 this is part I 1979 that for the minimum spacing or

S minimum that is equal to 2.5 D for point bearing pile piles and equal to 3 D for friction

pile friction pile. So, here it is minimum spacing here you designing

this for the friction pile the minimum requirement is 3 D, but here our calculation is coming

3.65 D; that means, is o k. So, to get efficiency equal to exactly 1 we have to provide a spacing

3.65 D. So, that spacing we can provide to get spacing exactly one exactly efficiency

exactly 1. So, if we want to design it for the different efficiency then we have to put

that value and then corresponding spacing we have to calculate.

So, first we assume the diameter, the diameter we are choose and based on that we can determine

what do the spacing required to get a particular amount of efficiency group efficiency. So,

in the next section I will discuss about, these are the… So, in the first class I

have mention about this these are the four different wave by which we can determine the

load carrying capacity of the pile the first one the by the static expressions or the formally

that part I have I finished. Next one that I will discuss about the pile load test so

by pile load test also we can determine the load carrying capacity of the pile. So, next

one is the pile load test that we will discuss in that section. So, now, in the pile load test, so this part

we will discuss about this load test or the on piles. So, by pile of test also we can

determine the load carrying capacity. Now, this is basically this again this pile

load test one any fail it is suitable for cohesion less soil. Now here we will do the

different types of pile load test we can perform one is compression test, one is pool out or

tension test and one is lateral load test. So, here we will discuss about the compression

test how we can do the compression test to determine the load carrying capacity of the

pile. So, here basically you are discussing the compression test

in addition to this we can do the pool out

test or the tension test or lateral test to know the lateral load carrying capacity of

the pile. Now before we go to the pile load test we have two things that we should know

that is one is initial test and next one is the routine test.

So, initial test one term it this is carry out on the test pile this is the carry out

on the test pile to estimate the allowable load carrying capacity of the pile, all to

know the settlement of the pile corresponding to the working load. So, that is the test

pile. So, this is the test pile that we carry out on the pile to estimate the allowable

load and to predict the settlement and the working load. Now routine test, can this carried

out to check the working pile load to and to know the corresponding settlement of the

pile corresponding that working load. So, that in the routine test will perform

on the working pile and initial test that we will perform on the test pile. So, now,

this thing is the more than, the condition is that more than 200 piles the minimum 2

test is equal. So, more than 200 piles minimum 2 initial test is required and for the routine

test this minimum

number of test is generally half percent of the pile used or that can vary up to 2 percent

or more that means, the initial test that is conducted on the test piles and the rooting

test that is conducting on the working piles. Now, in test piles means this piles are constructed

to for the testing purpose this is not constructed for the load carrying capacity or load carrying

purpose of the supper structure; it is not constructed for the working condition it is

just constructed for the testing purpose and then once the test then this piles are not

used. So, these are not used to carry the load which is coming from the super structure

and routine piles are routine test are conducted on the working piles. So, working piles are

the piles where the actual load of the super structure that will come it will work that

means, the routine test will conduct for the working piles on the working pile and initial

test will conduct on the test pile. Now more than 200 piles minimum 2 test piles or initial

test these are required, and for the minimum number of the routine test is given half percent

of the pile used for it up to 2 percent or nearly more.

Now, when this this is test pile, so now, thing is that test pile and working piles,

used only to load test does not carry load of the supper structure that I have already

mentioned this things. Now the minimum test or the load that these piles are taken, the

minimum test load on the test pile that is equal to 2 times

the safe load. The minimum test load on the

minimum test load on the test pile is two times the safe load. This safe load we can

determine by using the static expression that I have already explained.

So, now this load test are attained a value for this load test is generally two to the

safe load or the load

at which

the total settlement

of the pile is 10 percent of pile diameter in case of single pile and 40 millimeter in

case of group pile. So, that mean this minimum test load on the test pile in 2 times the

safe load or the load at which the total settlement of the pile 10 per attains the 10 percent

of the pile diameter for the single pile or 40 millimeter for the group pile. Similarly

for the working pile; that means, the routine test

the test load is generally up to 1.5 times

the safe load or the load at which total settlement is 12 millimeter again for

the single pile

and 40 millimeter for group pile whichever is lesser. Here also this is also whichever

is lesser so that means the these are the all the information regarding the initial

test routine test then; that means, the initial test is conductive on the test pile the test

pile. So, the pile which is constructive for the

testing purpose not for the it will not take the load which is coming from the super structure,

and the routine test is conductive on the working pile now this working piles of the

piles which will take the load; that is coming from the supper structure. Now more than 200

piles minimum 2 test piles are required initial test that is required and minimum number of

test is half person for the pile used the routine test of 2 percent, and more and the

test pile the minimum test load is 2 times the test load or the load at which is the

settlement attention that value of 10 percent of the diameter for single pile or and 40

millimeter for the group pile whichever is lesser.

And working pile the test load up to which 1.5 times the safe load or the load at which

is total settlement is 12 millimeter for the single pile or 40 millimeter for the group

pile whichever is lesser. So, these are the condition for the different test and the different

type of piles which are used for the pile loop test. Now next we will discuss about the how to

will calculate the load carrying capacity of the pile using the pile load test is 2911

this is part IV 1979. So, this IS code according to this IS code now load is applied on the

R C C cap for the pile and the applied increment of 20 percent of the safe load and corresponding

settlement of the pile is recorded by using the at least three dial gauge attach in the

pile cap. So, pile cap is used and where the load is applied with an increment of 20 percent

of the safe load and as I mentioned the safe load is calculated based on the static expressions

and the settlement corresponding to the each incremental load is recorded which is measured

by using at least 3 dial gauges. Now, the allowable load how will get the allowable

load of the pile. So, how we will can determine the allowable

load on a single pile. So, one first condition is the two third the final load at which total

settlement attains a value of 12 millimeter. So, if that means, the two third of the final

load at which the total settlement attains a value of 12 millimeter. That means, if nothing

is specified mean we can consider the permissible settlement of the single pile is 12 millimeter

and the load at which this 12 millimeters attain we have to consider two third of that

final load. Now, if any permissible settlement is mentioned

other than this 12 millimeter then we have to calculate the load two third of that final

load at which the total settlement attain that permissible limit which is specified.

Now in or second case the 50 percent of the final load at which the total settlement is

10 percent of the pile diameter, this is for uniform diameter pile and 7.5 percent of bulb diameter this for

under reamed piles. So that means, this is the conditions is the one is first one is

second one is the 50 percent of the final load at which the total settlement is 10 percent

of the pile diameter for the uniform diameter pile and 7.5 percent of the pile diameter

for under reamed pile. Now another condition is that now this we can used third one of

the third condition this is the single pile. Now, for the cyclic test this is the for the

static test complicit now if we can do the cyclic test if we want to know the tip resistance

as well as the friction resistance of the pile separately. If I go the static test for

the single pile that will give us the total resistance of the pile and if you know want

to know the tip resistance, and the friction resistance separately then we have to go for

the cyclic test. Now for the cyclic test pile load test that means, the two third of the

load

final load at which the total

settlement is equal to 6 millimeter. So, there we can replace this one, this is the similar

to the number one condition, but here this is two third the final load at which total

settlement is equal to 6 mm. So, the minimum of all this condition that we will consider

as our allowable load on a single pile that is determined from the based on pile load

test. Now, for the group pile or the group pile

same thing we can use from condition. So, that first one is similar that the final load

at which settlement is equal to 25 millimeter. So,

now, this load test we can perform on the single pile as well as on the group pile.

Generally in the group piles is the permis is nothing is specified then the permissible

limit is, settlement is 25 millimeter. Now the final load at which this settlement is

attained that will one condition. So, now if any other permissible settlement

is mentioned then we have to calculate that final load corresponding to that specified

permissible settlement, if nothing is specified then will consider for the group permissible

settlement of the pile is 25 millimeter. Now b is that is two third of final load at which

the total settlement attains a value of 40 millimeter. So, if this is the second condition

this two third of the final at which total settlement attains a value 40 millimeter.

So, the minimum of these two we will consider the allowable load carrying capacity of that

group. So, these are the condition that we have to

satisfy when you calculate, determine the load carrying capacity of the pile, single

pile as well as the pile group. So, these are the condition. So, we have discussed about

the initial test the routine test and test pile and working pile and what are the condition

by which we candidate by using those condition we have to determine the load carrying capacity

of the single pile as well as the group pile. Now this up to this we have discussed about

the load carrying capacity of the pile. So, this will give us the total load carrying

capacity of the pile, allowable load carrying capacity of the pile.

Now, if we want to determine or going to know the resistance that we are getting in individually

from the keep as well as from the friction, then we have to go for the cyclic pile load

test. In the next next class I will discuss about the cyclic pile pile load test by which

we can determine what is the contribution from the friction part, and what is the contribution

from the tip resistance of the pile individually. Thank you.

Superb sir keep upload more vedeos

what is the derence between IS code BS code

why always indians??sick of them

Very poor communication skills….not able to clear things properly

Super class sir

audio is too low, please increase