now, today i will discuss about the designof sheet pile, or basically in this lecture, i will explain how to determine the depthof the sheet pile, because this sheet pile is very important structure in geotechnicalengineering. this is another type of retaining structure. now, this structure is very flexible,and the difference between the retaining wall and the sheet pile is, said that the depthof the retaining wall, or the foundation depth
Offshore Work Great Yarmouth, of retaining wall, is very less compared tothe depth of the sheet pile, because the resistance of this sheet pile is getting from the soilpressure itself, that is why the required depth of the sheet pile is more, as comparedto the retaining wall. now for example, if i first, explain thatwhat is sheet pile. now the basic difference,
if this is our retaining structure, so thisis one cantilever retaining wall. and this is say the dredge level, or this is the existingground level, and this is your fill material. now, here if we consider this is the depthof this retaining wall, and similar type. so, here this portion is void, and this isthe soil, exist fill soil. and similar type of retaining wall for the sheet pile if iconsider. so, this is the sheet pile, and here the ground surface. here we can say thisis the dredge level. and here if i consider this is the d is the depth of the retainingsheet pile. so, this is the retaining wall, and this is sheet pile. now here we can seethat, this type of structure is very flexible; that is why it is used for this excavationpurpose, it is used.
this type of sheet pile can be used in excavationpurpose. so, then after the excavation one place is over, then this can be used or thiscan be taken out from the soil, and this type of sheet pile, can be used for any other locationfor the excavation purpose, even in for the waters, in offshore structure, or in for retainingthe soil near the sea course, that sometimes use this sheet pile. now, here this is alsothe soil, and this portion it can be the, this can be the void or this can be the water.so, now, here we can see, that here also the lateral pressure is acting on the retainingwall. here also lateral pressure is also acting on the retaining wall, and here. so that means,when we are designing the retaining wall tradition and retaining wall, then we consider thatthis side is active, pressure is acting, and
this side this is the passive resistance isacting. so, this side we can say, this is the passive resistance p p, and here, thisis the active resistance p a. now this active pressure is acting from this side, and passivepressure is acting from this side. so, as during the design also, as the depthof this retaining structure, foundation is very small compared to the height of the retainingwall, then sometimes we neglect this passive resistance, and we design by considering thisactive resistance only. but here; that means here, this resistance is, this passive force,which is coming from this side, is active force; that is registered by this retainingwall itself; that means, this retaining wall is, if it is a gravity retaining wall, thenthe size of retaining wall is huge. if it
is a cantilever retaining wall, then to resistthis moment, this reinforcement is used. so; that means, that that is why the requireddepth of the retaining wall, is less compared to the sheet pile. whereas, in sheet pilehere also, this lateral force is acting this side, and this side also soil pressure isacting. but here, the force which is coming from this side.suppose this is the active pressure is acting this side in here also. the passive pressureis acting this side; that means, the load which is acting from this side; that is resistedby the load, which is or the stress which is active pressure acting from the bottomof the wall. so; that means, the. so, the most of the resistance, this sheet piles isgetting from the soil itself. so, that is
why, the required depth of the sheet pileis more, as compared to the retaining wall. as the resistance from the soil below thisdredge level, the soil resistance is giving the stability of this retaining of this sheetpile wall. so, that is why here, we cannot neglect this passive resistance, because thatresistance is not small, as the depth of the sheet pile is more as compared to the retainingwall. so, that passive resistance, we have to considerin the design. so, now so, first if i consider that, how this sheet piles are joined. supposethis is our one sheet pile segment, which is joined with another sheet pile segment,similar sheet pile segment, by this ball and socket joint. so, this is your first pile,this is second pile, which is joined in this
portion by ball and socket, and then nextone is placed here. so; that means, the shape of the sheet pile structure may be in thisform. so, this is the plan view of the sheet pile structure. so, that is why, and thenthis side is the elevation or cross sectional view, and this is the plan view of the sheetpile structure, and then if i take the. so, in the design purpose, you consider this isvertical and which is very flexible. so, next one we can consider the differenttypes of sheet pile. so, that the sheet pile can be; one is cantilever sheet pile, anotherone is anchored sheet pile. so, this may be the one type of cantilever sheet pile is.suppose this is the sheet pile structure. this is dredge level, and this is the foundation,or this is the existing ground level. so,
you can call this is our cantilever sheetpile. then another one, this is ground level, where one anchored is attached with the sheetpile, this is anchored sheet pile. now use of these anchor as for the cantilever sheetpile, the required depth of this is this is the depth of the sheet pile, required depthis more. so, that is why, to sometimes reduce these required depth, this anchors are usedin the sheet pile. so, this type of sheet pile is called anchored sheet pile. sometimesthis retaining wall, there is a free cantilever sheet pile also. so, this is the cantileversheet pile, and sometimes this is free cantilever sheet pile. so, that this portion top portionis free. so, that basically these are the two types of sheet pile; one is cantileversheet pile, one is anchored sheet pile. in
the first, i will discuss about the cantileversheet pile, then how to determine the depth of this sheet pile, all this things. so, first problem, or first discussion withcantilever sheet pile, piles wall with is in granular soil. so, that means, this sheetpile wall can be in the granular soil, it can be in the cohesive soil. so, first wewill discuss about the sheet pile is in the granular soil. so, suppose if this is thesheet pile wall, and this is ground surface, and this line is a dredge line. so, this portionis may be in the void, or this is water this side, and soil this side. so, this one isthe height of the sheet pile, above the dredge line, and this one is the depth of the sheetpile, which is up to the base of the sheet
pile. now, as mentioned that here this depthof the sheet pile is not negligible one. so, when this, due to application of this lateralforce, it will deflect. so, that means, the deflection pattern of this sheet pile, willbe something like that. so, that means, rotate it within point below the ground level, so;that means, this is the deflection pattern of the sheet pile.so, now from this deflection pattern, if we can say this is the a, this distance is afrom the dredge level. so, you can see you can say, that up to this point, from the topof the sheet pile, there we have passive, there we have active pressure that will act,and in the this side. so, if i consider the two sides of the retaining wall; one is thisside, or this is the right side, this is left
side. so, that means, right side, from thetop to this point o say. so, this portion active resistance will act, and from thiso to the base of this retaining wall, can say if this is a this is b. so, from a too in the right hand side, active resistance will act, because the active resistance, thisis deflection is towards the direction of the force. and from o to b in the right handside, this passive resistance will act. so, that from a to o from the right hand sideactive resistance, and from o to b the passive resistance will act. similarly, from thisleft hand side, this side, from d to o this passive resistance will act, and from o tob active resistance will act. so, now if i draw the active pressure diagram. this isthe cantilever sheet pile, and if i draw the
active pressure diagram, so this is the retainingwall. so, as i have mentioned, this is the sheet pile wall. as i have mentioned thatfrom a to o. so, this is o, this point is d o a, this is base b. so, from a to o theactive resistance, active pressure will act. so, this is the active pressure that willact from a to o. so, this value we can calculate, this form that, at this level the active pressurewill be k a into gamma into h, if gamma is the unit weight of the soil. similarly, atthis point, because this distance is a, this will be k gamma h plus a. similarly in thisportion, the right hand side here, if i extend this pressure diagram.so, here also the active pressure will act in the left hand side from o to b. so, thiswill give us the active pressure. so, the
value of this active pressure, if i writethis value here, this is k a gamma into h plus d, so this value. but at this point,this is gamma k gamma h plus a. similarly here this side, this value is k a gamma intod. now if i draw the passive resistance of this same of passive earth pressure. thisis a point, this is same as d point, this o point, this is b point. so, if at this side,this is active resistance. then, similarly this side, this area that will be passiveresistance. so, this is the passive pressure diagram, when this side is active definitelythe opposite side will be passive. so, this is the passive resistance. so, this valueis k p into gamma into a, because this value is a, where k a is the coefficient of activeearth pressure, k p is the coefficient of
passive earth pressure.and similarly if we extend this value, had extend this line, this will be in this form.similarly, when this side is active; that means, definitely this portion will be passive.so, now, if we extend, so this portion will be passive. so, this value is k p gamma intoh plus d. so, now this is the net diagram, is that, this is active from a to o, the righthand side. similarly in the opposite direction, because this portion is void or so that soilis not present here. so, the passive resistance will act from this point to d to o, so thed to o is the passive, opposite to this portion. here, this is active from o to b, and definitelypassive will be in other hand side, this is passive. now this is the passive and activepressure diagram for this cantilever retaining
wall based on from this deflection and alsowe can see. now, if we draw the net pressure diagram,considering the active and the passive. so, this is retaining wall. so, this is the dredgeline o, this is a, sorry this is d a and b. so, we can draw, this will be the net pressurediagram, because we can see from this figure, that here this is passive, and this side isactive, as the net pressure; that means, the passive minus active, as passive pressureis more compare to the active pressure. so, this side, the pressure, net pressure diagram,this side the pressure will exist. and similarly here; that means, up to this point, here activepressure is increasing, and then up to this point there will be active pressure. so, thisis the line straight forward we can draw.
after that the net pressure diagrams, meansthis active and the passive both will act. so, that means, when this will start goingfrom right to left, and that means, when this pressure, active pressure and then the passivepressure, if i take the name. so, at that point where this line will cross this sheetpile; that means, from here it will start, and then it will go this side and cross thisdiagram, and then again it will cross, so two portion it will cross.so; that means, the final diagram if i draw, so this is up to this point is active, thenthe net pressure diagram, this passive will start acting. then it will go this side, thenagain the passive force is more in this portion, then it will go this side, and then it willfollow this pattern. so, this will be the
net pressure diagram. so, here we can seethis is active pressure zone, this portion is passive pressure zone. again this portionis passive pressure zone. so, from this figure we can see, so this is active pressure zone,then again this portion is more this is passive force, passive pressure zone. and then itwill go in this side, so this passive force will act, so this is also passive pressurezone. so, there is a three parts; one is active pressure zone, then passive pressure zone,then again passive pressure zone. so, that means, this forces that will act p a p p 1,and then another force that will act p p 2. now during the analysis. so, this is the actualnet pressure diagram in cantilever sheet pile wall. now, during the analysis, if we considerthis pressure, this type of pressure diagram,
then it is slightly complicated. now to simplifythis things, we can consider one passive pressure, one net pressure diagram, or this net pressurediagram, where same sheet pile if i consider a d and this is b. then this portion as itis active pressure zone, then instead of taking this circle or non-linear, this non-linearvariation, or the. then we consider, we take one linear type of variation. so, the samenet pressure diagram is representing in this form, where which is in same two points iscrossing, but it is in linear form. so, that means, here also this is, active pressurezone, this is passive pressure zone, and this is also passive pressure zone. so, these arethe net pressure diagram. so, this is the converted net pressure diagram. now basedon this net pressure diagram, we will do the
analysis. now, if we start this, draw the net pressurediagram again. this is dredge line. so, this is the net pressure diagram. so, we can seethat this force is acting this side. so, this p a is active force will act in this side.now here, so further simplify what we will do. we will extend this line, and you considerthat this is p p 1 and this is p p 1, this is 2, or we can say suffix b, because thisis acting at b. now, this distance is a from the base line, where this is passing firstcrossing this sheet pile, this diagram. so, this is a, now we consider this distance isy, capital y. and this one is this is y 1, y and this is a z, so this one is y minusz. so, this distance, where this is point
one point. so, this is a this point is, thisis b, this is c 1, and this is the c 2. so, d 2 c 1 is a b 2 b 2, this point where theedge of this triangle is z and then this one is y minus z, so total from c 1 to b is. so,we can write the d to c 1 is equal to a, then b to c 1 is equal to y, and this h point fromthe b point is z. so, now if i write this expression this isp a, p a will be k a, because this distance say height is h, and total is the depth isthe, d is the depth of the sheet pile below the dredge line. so, this p a is k a intogamma into h. so, if this gamma is the unit weight of the soil, where gamma is the soil,and k is the coefficient of active earth pressure. now, this point at this c 1 point, point thatnet pressure is 0; that means, the active
pressure is equal to the passive pressure.now if i write in this form; that at c 1 point what will be the active pressure. at c 1 pointthe active pressure will be k a gamma in to h plus a, because at c 1 point is active pressureis equal to passive pressure, because at c 1 point is, because this is net pressure diagram.so, that means, at this point active pressure will be h plus a into gamma into k a; thatis equal to the passive pressure, which is k p gamma into small a, because this is thepassive and active pressure. so, now, if i draw the previous figure; that means, at thispoint, that this point, any point, the active pressure is equal to passive pressure. so,passive pressure at this point is k p gamma into a, and active pressure will be k p gammak a gamma into h plus a. so, if we equate
this two, because at this point this passivepressure and active pressure, this is equal to, this is equal; that means, net pressureis 0. so, if i simplify this thing, a value will be k a gamma h divided by k p minus ka into gamma. so, k a h k p k a gamma is that is equation to p a divided by k p minus ka into gamma. so, we will get first, now we will get the a value. now, next one we willdetermine this p b 1 p p 1 b. now this p p 1 b this value, that will be the net pressurethat this p p 1 b; that is k p gamma h plus d minus k a gamma d. so, we can see from thisfigure; that at this point the passive force is k p gamma h plus d, and here the activeforce is k p gamma into d. so, if we take the net, then this will bek p gamma h plus d minus k a gamma d. so,
this is k p gamma h plus d minus k a gammad. so, this is the net pressure diagram for this p p 1 b. similarly p p 2 b, that is equalto. because that a this p a into k p minus k a into gamma. so, we can write this value,that p p b, that will be k p minus k a into gamma into y. so, this is y the net pressureis k p minus k into gamma into y. so, this is the net pressure for this p p b 2. now,we will start to calculate the forces f h. what is f h, because what are the forces thoseare acting here. so, now if i consider this things in three parts, that this is the pa, is the net pressure, this is the total pressure for this portion; that means, froma d c 1, this total pressure is acting as a p a. and then if i consider this triangle,this triangle this is say 1 2 3. so, if i
consider 1 2 3 triangle, and subtract fromthis triangle c 1 2 and b. then ultimately; that means, you have firstconsidering this total triangle, then we are subtracting this triangle c 1 2 b, then whatwill happen this portion will be cancelled out, so this thing minus this thing; thatmeans, this are acting opposite direction, because this force is acting this side, andthis force is acting this side. so, these are the three forces for one for this portionlower portion, one for this triangle middle triangle, one for lower triangle which isacting this side, one for the middle triangle, which is acting opposite side, and one forthe upper portion. now, what we are doing that, we are taking this triangle 1 2 3, andthen subtracting from this 1 2 3 triangle,
forces with c 1 2 and b. so, ultimately thisportion is cancelled out, and this is acting opposite direction. so, now if i do in thisform, then f h will be r a, then this triangle plus half into p p 1 b plus p p 2 b into thisz. z is the height of this triangle 1 2 3 into z, then subtracting this other triangle,so that height is y half into base is p p 2 b into y. so, that forces will be 0, forthe equilibrium these forces will be 0. so, now from this forces. so, if i write thatagain that is r a, or we can write this force, this is p a, p a, because here this is p a.so, you can write this is p a plus p p 2 b plus this triangle. this p a is the totalforce for this upper portion. now, we can write, if i write this expression again pa plus half p p 1 b plus p p 2 b into z minus
half p p 2 b into y, this is equal to 0. so,now from this expression we can write, z will be p p 2 b into y, p p 2 b into y minus 2 p adivided by p p 1 b plus p p 2 b. so, from this expression we can also calculate thez. now we will take the moment, at base capital b, that is also equal to 0. now, if i takethe moment, and we can say that from this center point, from this point c 1, the distanceof these forces, that is acting at the top portion is y bar. so, that distance is, fromthe c 1, where the net pressure is 0. so, that from the c 1 point, this force p a isacting at a distance y bar, so that we can easily determine, and this is the capitaly. so, from the base b, the distance of p a inb capital y plus small y bar. so, and so you
can write this p a, if you take the momentinto capital y plus small y bar, plus for the first angle; that is half into p p 1 bplus p p 2 b into z; that is the area, into that will act as the distance z by 3 fromthe base. then minus, the area for the mixed angle is half into p p to b into y, and thatwill act y by 3 from the base. so, that is equal to 0. so, ultimately after simplifying,so after simplification we can write 6 p a into capital y plus y bar plus z square pp 2 b plus p p 1 b minus p p 2 b y square, that is equal to 0. so, in this expressionwhat are the unknowns, that unknowns is z, that will get from this expression, and thenp p 2 b p p 1 b this will calculate, and p will calculate, and then y that will alsocalculate, and ultimately we will determine
the value of y. so, from this expression,we will determine capital y, then if we add a, then d will be capital y plus a.so, we have to calculate capital y from this final expression, last expression. then ifwe add the a, the a calculation that we will do from this expression number one. so, now,if i give the m, this is number one. from here we can determine a, then this is equationnumber two. from here you will determine z, this is equation number three. from here wewill determine the capital y. so; that means, and this z we will put here, so d we willget capital y plus a. now once we get this things, then we have to increase day d by20 to 40 percent for give a additional factor of safety. for this factor of safety purpose,you have to increase this d by 20 to 40 percent
that is a factor of safety. now, this fashionwill get the depth of sheet pile, based on this calculation, from this three expression,and this is for the sandy soil. now for the clay soil, what we will do, thatif the same thing we calculate for the cohesive soil, that where phi u equal to 0, and samesheet pile. this is ground level. this dredge level. so, now for the cohesive soil, so thiswill be the pressure diagram for the cohesive soil. so, this is minus, this is plus, andfrom here we will get this type of pressure diagram. so now, why you will get this typeof pressure diagram. first we will explain this is h, this one is d, d this is the depthof the sheet pile. now, here first expression of p a; that is equal to q bar into k a minus2 cu root k a. so, q into k a minus 2 cu into
root k a, k a is the coefficient of activeearth pressure. now where q bar is equal to effective vertical stress at any depth, andhere phi u is 0, so your k a that is equal to k p, that will be equal to 1, as phi uequal to 0 for this expression, so k a equal to k p that will be 1.now, we can write that p a; that is equal to q bar minus 2 cu, where cu is the undrainedcohesion of the soil. and p p that will be q bar plus 2 cu, so p a will be q bar minus2 cu p p will be q bar plus 2 cu. so, now at e point, or at say d point this is a dand this is b. so, at d your passive, the net force, net pressure, the net pressurewill be that p p minus p a at d; that will be p p, because at this. here the p p thatpassive resistance that will act in this side,
and active will act in this side. so, butat this point, at this d point, if i consider the passive, that q is 0, because in thisside, this is, it is no soil, so q u we can consider this is 0. so, at this side, thisis 0 plus 2 cu minus q bar. this is the q bar will be, this q bar is gamma h. the gammais the unit weight of the soil, q bar minus 2 cu, if it is a same soil, cu is here also,and cu is also here. so, that net force that is 4 cu minus q bar.so, at this point, this value is 4 cu minus q bar. similarly, at b point the net pressureis p p minus p a at b, that will be equal to. so, at this point b, here this side, atright hand side there will be passive pressure, and the left hand side there will be activepressure, because at this point this right
hand side is active pressure, and left handside is passive pressure, but at the base right hand side is passive pressure, and lefthand side is active pressure. so, passive pressure if you write that is q bar, plusthis load; that is gamma into d plus 2 cu, this is minus, then this q bar will be gammainto d into minus 2 cu. so, now net pressure will be 4 cu plus q bar. so, this value is4 cu plus q bar. so, again we have to take the forces so; thatmeans, if i take this f h, summation of the force, and we can write that this force isp a at total active force, and which is acting at a distance y bar from the d point. similarly,we can write that this distance is z; that means, where this is taking a, this distanceis this. so, what we are doing by taking this
force, this force is acting, so there is anotherforce is acting this side, and another force is acting this side. so, we are taking addingthis force, and then adding the force of this triangle, then you are subtracting this force,which is acting opposite direction. so, ultimately this portion will be cancelled out. so, inthis way, so if h we can write; this is f h is 0 this is p a total force of the upperportion plus half into z into half into z; that means, this triangle half into z into4 cu minus q plus 4 c u plus q bar. so, we can write this is 4 cu minus q barplus 4 cu plus q bar, then minus d is the depth into 4 cu minus q bar is equal to 0,and 4 cu means this d, then for this rectangle this portion is 4 cu minus q bar into d; thatis the area. so, from this upper portion force,
plus this triangular force, when minus thisrectangular force. then ultimately we can write this is p a plus half into 8 cu minusd into 4 cu minus q bar that is equal to 0. so, z will be d into 4 cu minus q bar minusp a that is divided by 4 cu. so, this is one expression one, from which we can determinethe z value. now, next one, we will take the moment with respect to this b; that is also0. so, moment that will be p a into capital y plus y bar, then minus that d is half thistriangular 1 plus; that is z half into z, and this one will be here also, this is zwe missed that z 1, this is half 8 cu into z z with z then we can determine the z fromthis expression. so, half into z into again 4 cu minus q bar plus 4 cu plus q bar andthat will act at a distance z by 3 from the
base b. then minus this d into 4 cu minusq bar, and that will act distance d by 2 from the base, because this is the rectangle, thatis equal to 0. so, after simplifying this we will get d square4 cu minus q bar minus 2 d p a; that is minus p a 12 cu y bar plus p a into 2 cu plus qbar that is equal to 0. so, after putting the value of z here, so this is expressiontwo. so, this is the final expression, so where this z value is replaced by this expressionone. so, we put the z value in this expression, and then after simplifying this, we will getthis final expression. so, from, again this q bar is equal to gamma into h. now, againfrom this final expression, expression two we will determine the value of d, and thenincrease, we determine the value of d, and
then increase it by 20 percent to 40 percentfor the factor of safety. so, these are the two, this is the determination of the depthof the sheet pile for two different cases, for cantilever sheet pile; one is for thegranular soil, one is for the clay soil cohesive soil. so, next class i will solve few example,and then we will find how to determine the depth of this sheet pile wall for differentsoil condition, need for the cantilever sheet pile. then later on, i will discuss abouthow to determine the depth of the anchored sheet pile also.thank you.
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