 Next part of your application of soil mechanics is your underground conduits, shafts and tunnels. Basically how this stress variations in conduits, shafts and tunnels we are going to discuss. So then first part we will start with this conduits. Now conduits it may be classified as number one ditch conduits, number two projecting conduits that means positive projecting conduits, then negative projecting conduits, third one is your special one it is called imperfect ditch conduits imperfect ditch conduits. Basically where we are using this conduits, what is the conduits, why we say that it is conduits, conduits basically used for you can say that sewer pipes, drains, culverts conduits basically used in case of kind of sewer pipelines, sewer pipes or sewer lines or drains, culverts So it has been classified into three parts, if I classify what are the different types of conduits, first one is your ditch conduits, second one is your projecting conduits and this projecting conduits has two parts, one is your positive projecting conduits, second is your negative projecting conduits, third one is your imperfect ditch conduits. Now if I draw the line, draw the show this what is ditch conduits and the projecting conduits and imperfect ditch conduits, graphically or figure wise I am just showing it this is your ground level ground level and this called ditch conduits, second one if I draw it this kind of figure, if this is my ground level, so this is your positive projecting conduits, this has been particularly used in highways or roads, then third one is your if I draw negative projecting conduits, it has been used particularly in case of railway culverts, this is my ground level and this is called this type of things is called negative projecting conduits, this is for railway culverts. Now if I say ditch conduits, this is example is your ditch conduits is your sewers, water main pipes, water mains in this case example is your highways and this case railway culverts, a ditch conduit is installed, it is particularly it is installed in a narrow ditch, in a narrow ditch excavated in a undisturbed soil below the ground level, if you look at here in a narrow ditch excavated in a undisturbed soil sample soil below the ground level, which is then covered with earth backfill, in case of positive projecting conduits, install in installed in a shallow bedding with its projecting above the natural ground level, top projecting above the natural ground level, similarly in case of negative projecting conduits, conduits installed in a narrow shallow ditch with its top below the ground level, with its top below ground level, which is then covered with an embankment. Look at here, the classification of this conduits I am taking into three parts, basically this classification of this conduits is three parts. So if I classify these conduits into three parts, basically conduits has been used sewer lines, drainage and culverts. So classification is your ditch conduits and projecting conduits, these two conduits has been widely used, ditch conduits and projecting conduits, the projecting conduits has two parts, two again, one is your positive projecting conduits, second one is your negative projecting conduits, third one which is a special conduits it is called imperfect ditch conduits, we will discuss later on. Now we are concentrating with this first two ditch conduits as well as projecting conduits, if you look at this figure, if you look at the figure what you mean by ditch conduits, if a narrow ditch below the ground surface, excavation has been made and narrow ditch above the narrow ditch, below the below the narrow ditch a conduits has been installed. So this is called ditch conduits. So installed in a narrow ditch and best example is your sewer lines or water main pipelines it go far below the ground surface, below the ground surface, that case it is these are called ditch conduits, then positive projecting conduits, if you look at here the ground surface g l is your ground level, ground level is at somewhere else above your ditch conduits. In case of projecting conduits, if you look at positive projecting conduits the ground level is there above ground level, if this is my ground surface above ground surface the conduit will be distinctly, distinctly visible that means above the ground surface the conduit will be some part will be there. So it generally provided where it is generally provided in case of highways or road, if this is my ground surface if a road has to be constructed above this ground surface generally above the ground level this conduit has been installed and above this this construction road construction has been made. So installing a shallow bedding this is a shallow bedding with its top projecting above the natural ground level with its top this part is your top projecting above the natural ground level this g l is nothing but is your natural, natural ground level, natural ground level natural ground level. Now come back to negative projecting conduits this has been highly used particularly in case of railway culverts, railway culverts because the railway line has to pass railway culvert what will happen? Install in a narrow shallow ditch if you look at these two if I take these two this is a long ditch and this is a very small or narrow ditch in this case this part has been installed so that ground level is slightly higher slightly higher so in this case it is called it is called negative projecting conduits that means whole conduits it has been divided into two parts one is your ditch conduits other is your projecting conduits and projecting conduits has two parts that means projecting conduits has two parts one is your positive projecting conduits second one is your negative projecting conduits. Now we will start one by one how the stress variations, stress formation they are first let us start with this ditch conduits let us draw a ditch conduits and we can see how much is your stress variation let us say width b d of the trench b d of your width of the trench this if you look at here this distance to this distance is your b d width of the trench and b c is your diameter of your conduits generally width of the trench generally b d generally b d width of the trench should not not greater than should not greater than two to three times b c that means diameter of your conduits diameter of your trench should not be greater than two to three times of diameter of your conduits the loading imposed on the buried conduits can be obtained by considering equilibrium of an element slice of thickness d h element slice of an thickness d h. So what is the load come into this particularly in this conduits now if I consider a very small slice of thickness of d h at a distance h below the ground surface now what are the forces acting on it self weight of this ditch is your d w b is your total vertical force b plus d b is your after a infinite small element d h this is your other direction this is your shear force this is your sigma s lateral stress this is your shear stress or this is your sigma s lateral stress now let us say b is equal to vertical load vertical load on top surface on the top surface on the top surface of slice slice b plus d v is equal to vertical load on bottom surface of the slice d w is equal to if you look at here this is your d w this is the slice and this is your d w d w is your width means d w is your weight weight of the slice this comes out to be gamma times b into d into d h per unit length gamma is your unit weight gamma is equal to unit weight and b d b d is your width of your ditch and d h is your thickness that means this into this area into gamma and per unit length if I take it in this direction that means per unit length per meter length this is your weight of your slice so horizontal pressure sigma x sigma x is equal to k times k is your coefficient of lateral earth pressure into v by b d into d h v by b d vertical force by your b d over your d h sigma x is your k times of lateral stress is equal to k times of vertical stress vertical stress is your force per area that is your b d per unit length into d h times this is your lateral stress or horizontal pressure you can say that sigma x is equal to horizontal pressure now the shearing resistance developed along the side of the vertical direction is equal to mu times your horizontal pressure that means shearing resistance of the side s is equal to mu times your horizontal pressure sigma x mu is equal to nothing but is your coefficient of sliding friction coefficient of sliding friction between backfill material and trench wall so this comes out to be if you look at here k is equal to coefficient of lateral earth pressure mu is equal to coefficient of sliding friction between the backfill material and trench wall if you look at here coefficient of friction is this is your trench wall that means friction between this trench wall and is your backfill material this is your backfill material so this is your coefficient of friction mu is equal to coefficient of sliding friction and this comes out to be sometimes people say mu sometimes somebody say mu dash both are fine so it will be mu dash k into v y b d into d h the sigma x whatever I got it from here has been replaced to here in case of shearing resistance to find it out now you consider the equilibrium of this element equilibrium of this element equilibrium means whatever the forces force in x direction force is equal to 0 force in y direction is equal to 0 your movement is equal to 0 by taking considering the equilibrium now it will be vertical forces v plus d v plus 2 s v plus d v is your vertical load on the bottom surface of the slice v plus d v if you look at here it is acting upward shear force also acting upward I am taking in upward as a positive v plus d v plus 2 s minus v plus gamma b d into d h minus what is your minus because if I am taking this force in this direction is positive upward direction so downward direction is your v so this is negative and what is this also vertical stress vertical stress how much v by b d into d h so v by v by b d into d h so it will be gamma gamma terms of gamma times of sorry unit weight of slide unit weight of slide d w is also downward if you look at unit weight of slide is equal to slice gamma b d into d h so this is your vertical force at the top of the slice this is your unit weight so this will be a negative because this is downward and this will be vertical upward this force is vertical upward this force is vertical upward this force is vertical upward so in the equilibrium this plus this plus this is equal to this plus this now solving this d v is equal to gamma b d d h minus 2 s so which is equal to b d d h minus 2 s so which is equal to gamma b d d h minus 2 k mu prime v by b d into d h now solution of the above differential equation now if i find it out solution of the above differential equation it is a differential equation it comes out to be comes out to be v is equal to gamma b d square into 1 minus e to the power minus 2 k mu prime h by b d h by b d by 2 k mu prime at the top of the conduits at the top of the conduits at the top of the conduits you can take as h is equal to h because h you can consider sometimes universal h is equal to you can take it capital H this comes out to be v is equal to gamma b d square into 1 minus e to the power minus 2 k mu prime into h by b d by 2 k mu prime now this is your by solving this now let me summarize how i have proceed i have consider a very small element very small element below the ground surface i have consider a very small element below the ground surface which is above your conduits d h why i am consider because i want to know what are the forces or stress coming above your d h so this small element is let us say this small element is d h width of d h it is at a distance of h from the ground surface now let us say v is your let us say v is your vertical forces of the slice of the slice vertical forces at the top and v plus d v is your vertical load on the bottom surface of the slice s is your shear resistance sigma x is your horizontal pressure so now again if i am taking this slice this weight is your d w because this slice why it is d w people will ask because this slice i have taken a infinitely small d h that is why this weight coming to it is your d w now the weight of the slice is equal to gamma gamma is your unit weight unit weight of the soil into b d b d is your width of the trench if you look at this b d is your width of the trench into d h d h is your the slice that means b d into d h this is your area into per meter per meter length unit length i can consider this length this length in this direction particularly in this directions now what is the value of horizontal pressure the value of horizontal pressure sigma x is equal to k times k times i can write it in terms of k times into sigma v so k is equal to lateral earth pressure coefficient sigma v is equal to vertical stress how the vertical stress will come into picture this will come into picture v by b d v is your vertical load on the top of the surface by b d into d h because this is your small width of your slice or small thickness of your slice d h so horizontal pressure i get it then your shearing resistance s is equal to mu prime times of sigma x because mu is your coefficient of sliding friction between the backfill material and trench wall with this s is equal to mu times of mu prime times of sigma x you can take the value of sigma x from here and put it the value of sigma x here after getting all the value of d w sigma x and sigma your prime of s shearing resistance now consider the whole slice into equilibrium conditions once it is in equilibrium conditions what will happen forces in x direction forces in y direction and movement is should be zero i have consider the forces so what will happen this forces in vertical forces if you look at here the vertical up this is called upward forces as positive and the downward force i have taken negative so v plus d v plus this s minus v minus d w v plus d v it is 2 s y 2 s once you are considering the slice the s is this side this s is this side frictional resistance from both the sides so it will be v plus d v plus 2 s minus v plus gamma v d d h this is nothing but your d w unit weight small weight of your slice with taking this it will come as a differential equation and by solving this differential equation this term will come as a v is equal to gamma this is not r this is gamma v d square 1 minus e to the power minus 2 k mu prime k is your lateral r pressure mu prime is your coefficient of sliding friction between backfill material and trench wall h by b d h is your distance from ground level to your slice or you can say that h from distance from here to here now in general we take in general it has been taken as h is equal to h now the differential equation after solving v is equal to how much is your vertical load on top surface of slice it is coming it is coming gamma b d square into 1 minus e to the power minus 2 k mu prime into h this is your capital h instead of small h I put it as a capital h by b d divided by 2 k mu prime this is your complete equation of your vertical load coming to your conduit or ditch conduits now in this case what will happen in this case what will happen this slope is a vertical slope it may possible it may possible this conduit ditch conduits may be placed may be placed with a side slope if I consider this is my kind of this is the kind of b c and this is your b d if you look at here it is not always possible it is not always possible that you can go for a vertical kind of vertical kind of ditch now generally vertical kind of ditch is insta not stable so sometimes we provide the ditch conduits with a side slope this kind of consideration we give it in this case in case of very rigid conduits the side fields are relatively compressible and the conduit would carrying practically all load so if I write it if it is a rigid conduit the side fields are compressible and conduit would carry practically all the load conduit will take all the loads so in this case what will happen w c w c is equal to weight in the conduits w c is equal to weight in the conduits particularly w c is equal to gamma b d square 1 minus 2 1 minus e to the power minus 2 k mu prime into h by b d this will be multiplied with your this is your h by b d this will multiply e to the power this is coming here h by b d because it looks odd if it looks as if it is multiplication of e into h by b d no it is e to the power minus 2 k mu prime into h by b d by 2 k mu prime so now if I consider this as a if I consider this as a whole term as a c d so now weight on conduit total weight coming on the conduits is supposed to be I am taking at a c d gamma b d square now c d is known as load coefficient for these conduits this is known as load coefficient of these conduits load coefficient of these conduits so this load coefficient c d is equal to nothing but 1 minus e to the power 2 k mu prime h by b d divided by 2 k mu prime and this is called load coefficient we will discuss this graphical and other parts may be tomorrow about your load coefficients I will stop it here thanks