Transcript for:
Soil Classification & Properties

hey well welcome to another edition of fe review civil 2022 so so glad to have you here and excited for tonight where we get to talk about dirt well soil really because we don't call it dirt you know but um the cool thing is we'll try to make some fun jokes but hopefully they won't be dirty okay and the puns won't hit rock bottom but we will try to have some fun here so um as we get started what we're going to be doing is we are just going to be working through problems uh like we do each week so this week we are going to focus on geotechnical engineering and uh there are some problems up here i did make a slight update here so i i reposted and added a problem here uh today so if you have you know previously downloaded this you can definitely uh download the new questions the link is down below so updated problems are always going to be there so we can just jump in there's a lot in geotechnical engineering uh in this problem set there's actually a lot of questions uh that i put in and a number of those are conceptual questions just so that i can talk about a few different pieces here of geotechnical engineering that you know maybe might not have a huge problem but they're concepts that you need to understand and so we'll go through that on one of the things that you notice is as we look at this this uh list of what's on the fe and actually if you go back a couple years i've been in this few years in version 9 of the reference handbook we're now into version 10 of the reference handbook from nces um in version 9 there used to be phonats now there aren't so i don't put flownuts in this problem set uh and hopefully you like that i don't know maybe you really love flownuts and we're hoping that there'd be flownuts in this but we're not going to have those but index properties something that you probably are going to see on the test uh you know uniformity coefficient even that type of thing um phase relations definitely lab and field tests i think those are going to be there effective stress and then some of these other topics they become a little bit more plug and chug but we'll we'll go through some of these problems and hopefully uh you'll get some good information as well so um as we go uh you know definitely feel free to put a comment in the box but um let's just jump in here so index property soil classifications uh number one right so what we have here is anytime you see a a question where you have to use a certain classification system the first thing that you want to do is go find that classification system right i mean you have to understand the soil properties here so yeah we can understand the soil properties like we have a non-cohesive material typically sand is not cohesive it doesn't stick together right without gravel so we have a hundred percent passing number four sieve so right away we know that we have uh a sand it's primarily sand but we have some silt we have eight percent passing number two hundred seven number 200 sieve is the distinction between your your sands and your silts and your your clays your sands and your fines right your granule material and your fines and a uniformly coefficient eight a coefficient of curvature of 2.5 and we're told the unified soil classification system is to be used here so this soil can be described as what so in the fe reference handbook and that's where we're going to go to start here the epi reference handbook has this in so if this is the latest version of the manual i think but it's it's page 267 of the pdf i'm trying to think here the this is kind of far down in the geotech and actually it's kind of interesting is civil engineering starts with geotech i don't know maybe it's because it's from the ground up i don't know but uh we get to start with geotech here and uh you know a few pages in we get to the uscs soil classification system so we'll go through some of these other pieces here but this is where we're going to start and um if you remember what we were told is we have a hundred percent passing a number four sieve so what you see here is the number four sieve is that distinction that dividing line between sand and gravel and number four sieve i mean it's i think it's like 4.76 millimeters um so it's it's if you think of a number four it's kind of like a four millimeter it's a little bigger than that but it kind of gives you an an idea maybe three sixteenths of an inch type of type of thing it kind of gives you an idea bigger than that it's going to be rock or gravel um smaller than that is going to be a sand and then the number 200 sieve is where we get into the fine grain soil so that's even pretty small it's less than a tenth of a millimeter so um this tells us we're in we're in uh you know everything's passing a number four sieve so we're in a sand and then where do we go from there well we we don't have less than five percent fines we don't have more than twelve percent fines we have eight percent fine so right away this problem is sort of maybe a bad problem or what this means is actually you have to go down and look in below so below what we see here is we actually have some sands with five to percent five to twelve percent uh fines required dual symbols so the thing is we can still use this even though we don't have our five to you know we're not less than five we're not greater than 12 we're still kind of in this range though so let's keep going with sands and what we see here is we have a uniformity coefficient here of i think well let's go back and see what we had i i don't remember so this this uniformity coefficient was eight the coefficient of curvature uh was 2.5 so the uniformity coefficient is eight and the coefficient of curvature is 2.5 okay so that tells us something here so if we come back uh what we see is we're greater than six right we're greater than six for uniformity uh where uh what was we were 2.5 so we're in between this one and three we're kind of a well-graded sand right but you have to remember we didn't have less than five percent fines so what that means is instead of having less than five percent fines uh we have eight percent fines and we're going to come down here and we're gonna have a dual symbol okay so we're kind of in this well-graded condition the uniformity coefficient coefficient a higher value means it's kind of well graded there's a good distribution size on there it's not just like uh beach sand where everything's the same so what that means is we're going to come in here we're going to say this is going to be a well graded sand but it's a well graded sand with some cells so if we come back we hope that's in the in the list above and good news is it is so we have well graded sand with silt and that's our answer so you have to understand some of these pieces but you also have to understand some of the footnotes to the tables as well so hopefully if you get one of these questions they're pretty straightforward honestly hopefully it's pretty straightforward you can just push it right through and go through the kind of the algorithm of that that classification system and get your answer so hopefully that's what happens with this next one right this next one is similar it's same type of thing it's just using uscs but now we're in a fine grain soil right we have a hundred percent the passes and number two hundreds again that's the dividing line between uh the the sandy soils and the fine grain soil so even 100 passes the number 200 sieve so this is your silts and clays a liquid limit of seventy percent of seventy percent and a plastic limit forty percent so the liquid limit oops uh let's change this here so we have a liquid limit of a seventy and a plastic limit of forty using uscis the soil can best be classified as well let's go back to uscs and see what we have to say um so sammy it is up there now so i just got it up so i apologize for that a little bit late but it is there this week so um so you should be able to find it hopefully you can uh get on that now as well okay so we're back here um we're looking at fine grain soils we have a liquid limit uh i think we said that was 70 so we are um we're not less than 70 we're more than than we're more than 50 okay and what you'll see here is we uh if our liquid limits more than 50 we are looking at um a plastic limit of 40. but the problem is this doesn't this looks like a pl almost but it's actually pi if you if you look closely that's a pi so what's pi is pi plastic limit um hopefully uh you see that it's not and what we see here is the pi is actually going to be if we come down here a little bit this is the plasticity index and this is just a index of how kind of stretchy the soil is if you will right but what this takes into consideration is kind of uh the the difference between the liquid limit and the plastic limit that's our plasticity index right so the liquid limit if you remember it's that machine you kind of crank crank crank crank crank and and you get you know until the the soil comes together maybe you remember that lab um where you did it um the the plastic limit is where you roll it down and get that that eighth inch diameter three uh you get a 360 i think three sixteenths inch diameter uh roll ten eighth inch three sixteenths until it crumbles and you find out what those limits are right but the plasticity index right the plasticity index is the difference between those two we could look at that here as well so if we come back a few pages in the the reference handbook you see the plasticity index pi is the liquid limit minus the plastic limit so let's just figure that out here the plasticity index is the liquid limit minus the plastic limit so what's that 70 minus 40 and that's going to equal you know equal 30 here so we'll plasticity index and that brings us back to our uscs so let's come to our classification system and what do we get we get a plasticity index of um of greater than seven for sure but let's look at this let's look at we're not in organic materials um we don't have any any um organ i'm sorry we don't have any organic materials but let's come let's come back over here we were at a liquid limit of 50 or more so now we're looking at the a line what's the a line right um we're looking at an a line here so let's look at this well it's if we keep coming down here what you see down here is we have what's called an a line so i can take a look at this a line but let's look we have a plasticity index which we got as 30 right so 30 is this this line across here and our liquid limit was 70 so we're firm kind of on this point and if this is the a line up here we are going to be below the a line right and if we were used if we did our plasticity index wrong right kind of use that plastic limit we would be above the a line so you got to be careful with plasticity index versus plasticity's elasticity limit so we're below the a line right here we have more than 50 percent or more than 50 of a liquid limit we are above the a line so we got fat clay going on how's that sound is that right or did i do something wrong um oh no i'm sorry we're below the a-line right sorry i think i read that wrong we are we are our liquid limits 70. i it would stink to go through all the work and then get it wrong right but yeah so we're liquid limit more than 50. we're below the a-line rate our plasticity index was 30 and our liquid was 74 below the a-line if that means we're at this this elastic cell okay so we can come back here and um we can select an elastic cell so again it's just knowing some of those terms knowing what they mean knowing where to find them and hopefully uh that's gonna going to help you with those but i honestly if you get a soil classification i hope it's just kind of plug and chug you look it up and it's not it's not too crazy okay but there's there's the uscs there's also i mean if we come back and look at this there's also an ash toe classification so you could be given something similar here uh but again what we have here is is another classification so if we have more than you know more than 35 percent passing it 0.0756.075 that's our number 200 sip right that's the the small sieve that distinguishes between uh between fine uh fines silts and clays and granular materials and again what we have here is we have you know more than 35 percent uh passing a 0.7075 sieve and then the liquid limit plasticity index same types of things right we're greater than 40 and our plasticity index is greater than 11 right or greater than 40 greater than 11 we're going to be kind of in in in this range a clay um soil is what it sounds like here right so um it's you know the the different if i'm if i'm doing that right am i looking at that right uh i'm i'm getting confused so our liquid limit is greater than 40 plasticity index uh was was around 30 i think so yeah we're in this this side of things so they did they distinguish the they're they classify the soils a little bit differently uh but it depends on kind of the two systems that you're working with here so um in this one we specifically said the uscs so you know flick with the more than 50 or below the airline and that's what we're going to go with okay so as we keep going here though what we're going to do is some of these questions are going to be kind of quick some of them are going to take a little bit longer here but um this one we could take a take a quick look at or we could take a little bit longer this one i kind of threw in there just because um sometimes when you go on a job site or when you are looking for materials and when you specify any materials as an engineer what you'll get from the contractor or maybe you'll be the project engineer for a contractor what you're going to do is you're going to get a sieve analysis of that soil and you're going to give us so you're going to go to a borrow pit or a test bit and take a sample of material get a sieve analysis done and send it off to the engineer and say is this good and then the as the engineer you're gonna have to say like yes it is or no it's not you're gonna have to approve or not approve the material based on what's in your specification so a lot of times the first test that's done especially for um granular granular materials like a uniformly graded sand it is going to be a sieve analysis so the question that is asked here is we are given a whole bunch of sieve analysis results right and the question says which one's uniformly graded sand so can you look at these sieve analyses and know what's uniformly graded i mean uniform you might think well you know these two sort of look alike i don't know this one sort of this one does just doesn't look uniform this one sort of looks uniform i don't know does it this is where you have to understand what a sip analysis shows so sieve analysis it says the percent passing in a particle diameter this is a log base scale and i mean if we look at this what you see is uh with this one you have a chunk of material that passes right here a chunk of material that passes right here and and this kind of gives you an idea that this is what we would call or geotechnical engineers would call like a gap graded soil right there's there's kind of a couple different pieces that are that are put in there a couple different particle sizes that get that get in there and this is a gap-graded soil okay so this one it's not a uniformly graded sand okay so we can get rid of that one and then we get to uh bc and d right and the question is what is what do these mean right what this this curve means here is that as you go down and get progressively smaller with those sieves you remember the zips are you get a bunch of pans put some soil on the top then you shake them right in the particles kind of sieve out and you kind of figure out what sizes each of the or you know what sizes of particles you have in the soil right so with what we're doing here is is is this shows you know this curve shows that you have a pretty good distribution this is actually not uniform this is actually what we what's considered like a well graded soil right but then we get down to these two we see uniform soils these are both uniform soils so these are both uniform soils these this is a uniform soil and this is a uniform soil sometimes we'll see that called a poorly grated soil actually if you come back to uh the reference handbook for a second what you see here is you have poorly graded sand it's not called uniformly graded and it's called poorly graded um it might also be called uh uniformly graded in the sense that this is your beach sand this is the the sand that's all kind of uniform and it's you know you can use it for uh you know for leveling at times you can use it under uh certain uh certain pads or equipment uh but this is this is a uniform a uniform soil okay so it what it means is you you kind of get to a certain sieve and all of a sudden all it all goes through but then it stops on the next sip if that makes sense it's like it all passes through one sieve and then stops on the next step it's kind of the idea here and this is this is going to be uniform soil now which one is going to be sand and this is again where if you if you know a couple of a couple of sieve sizes um one sieve size is going to be that number four sieve it's remember mercedes is about five millimeters about four millimeters five millimeters um if we look down here we got one two three four if i'm counting right five so our number four sieve is somewhere around here and that number 200 if you remember that was like 0.075 that's going to be somewhere down around here so this is firmly in the middle of the sand it's firmly looks pretty good and this is going to be all right our uniformly graded sand so this is like a number four sieve um this is like a number 200 sieve right and how is this different than the this this one right here it's different because again if we look at that number 2007 our number 200 sieve is going to be somewhere around here probably not getting it perfect but what you see here is if this is our number 200 sieve most of most of the material on this one right here is going to be is going to be in that fines area this is going to be more of a uniform silt or clay okay so um plot c on this one is going to be our uniform our uniform sand okay we could also take a look at this because what we see here is we could go and we could actually start to calculate a few things right if we wanted a more numerical approach what we could do here is we could say well let's take a more numerical approach to this because if we come back to our classification system right what we see here cu is less than four for poorly graded sand um cc could be less than one or greater than three for poorly graded sand again that's a uniformly uh graded sand it's very uniform in size the particle size don't have a lot of variation they're not well graded in in terms of they're all it's kind of like those glass beads that are all the same size they're marbles right they're all fitting together they're all kind of perfectly um the same size right that's that's a uniformity coefficient of of like one right if they're all exactly the same size so what we're going to do here is we're going to say well what do we look how do we find that and this is cool too because i i think this is going to show up on the test i think somewhere cu cc they're going to show up on the test and they're just basic formulas this is d60 over d10 and that's uniformity coefficient and the coefficient curvatures d30 squared over d10 over d60 so if we look at the this this plot here what we can do is we can actually come back and we can say well d60 what's that d30 what's that right and lastly d10 uh what's that right so we can actually go and calculate these values right so if we look at these values right our d60 is right about maybe 0.9 that's our d60 our our d30 um is this like the kind of particle size at 30 percent right our d30 is gonna be oh man where is this this is one two three four five six seven maybe 0.7 if i'm getting that right that's our d30 and then oh where's this this is maybe um am i looking at this right uh let me just double check my numbers here i'll make sure i'm not way off but our d10 is going to be like uh i'm i'm messing something up here oh sorry i i'm i'm counting wrong um so so our d10 is going to be down at this point here and what do we get like one two three four so so i'm sorry so this is point one this is point two this is point three point four point five point six maybe that's point six okay so sorry so zero point six that's going to be like our d10 right so these are those those points where we can go and now we can calculate we can say um we can go and we can actually calculate some of these values so we can calculate the cc which was uh what cc if we come back to our our um table here for a second um what we see is that cc is uh d60 over d10 or i'm sorry cu is d60 over d10 right so c let me start with cu is d60 over d10 so what's that it's like 0.9 over uh 0.6 so that's about 1.5 okay that makes sense i think it does uh so that's about 1.5 and if we keep going here uh we can say like cc which was this is equal to d uh d 30 squared over d10 times the 60. so these are these are things that i think are probably going to show up somewhere okay and can you uh can you solve for them okay so let's take a look i mean let's look at this d30 is going to be about 0.7 squared over 0.6 times 0.9 and if we do that out let's see 0.7 squared over 0.6 over 0.9 i get about 0.9 okay so again that puts us firmly in that uscs uh poorly graded sand uh uniform sand range right so that puts us firmly kind of in this cu less than four it's a 1.5 that's a pretty uniform soil um cc less than one is also again an indicator of a pretty uniform soil poorly graded sand um or i i i'm up here with gravel i'm sorry i should probably be down here with sand less than six less than one same same type of thing though poorly graded and uh and that kind of gets us in those ranges so again these uniformity coefficients coefficient curvature i think they will be there and uh potentially show up on the test so good things to know um but i i like this problem because it kind of gets in there a few of the different pieces right it talks about those those um the sieves it talks about the how to interpret these these plots a little bit and i i i put this one together um to try to get those different pieces in all right i mean man we're fine today we've got three questions and we're we're going okay so let's keep going three major phases of soil okay so three major phases of soil this is um supposed to hopefully not be that difficult uh but this is just a concept question right i mean you're gonna get some concept questions hopefully these are the questions you can answer in 30 seconds i put this one together just because i wanted to get you thinking about some different things right clay silt sand gravel these are different you know fine versus granular materials right fines versus granular tails that's not a phase of a material what's a phase of water well you have ice liquid and steam right or in other words a solid uh liquid and gas and that's the same thing that we have with soil right with soil we have solids water and air right so earth fire water i mean that was just that for fun you know but uh what else cohesive non-cohesive organic these are just categories of soils but they're not phases phases kind of get into the kind of the components but this is solids water and air and this is where we get into phase diagram so now we're into phase relations part b and when we start taking a look at this this actually starts us off where honestly the geotech reference starts so there's a phase diagram here i made another one that i stuck in uh the problem sets but i just took this this thing you know this diagram i made one kind of similar for the next few problems so yeah the answer here is solids water and air but this also comes to light when you see kind of this question this next question i mean this next question hopefully is just a warm up question it's not really that super super challenging yet but the idea here is we have different phases right we have these these solids the water the air they all make up a composite solid right a composite soil i should say that you have to understand the different components with and probably one of the most basic tests that you did in your soils lab is you took a piece the can of soil you stuck you know you weighed it you know you found either the mass or the weight of it but before with water in it and then you stuck it in an oven and then you waited again and a day later why'd you do that because you wanted to evaporate all the water so if you evaporate all the water you can figure out how much water was in it and you can figure out the water content right so this is kind of one of those basic things it just kind of get you warming up is to talk a little bit about the different phases of soil right but this is where we get into phase relations so this question starts off with a soil sample obtained in the fields a weight of 135 pounds right so what is that that's 135 pounds and honestly if you're doing one of these problems i would suggest sketching a little diagram like this so you sketch a little diagram on on your little notepad that they give you and you you start to just fill in the pieces that you have right uh and and this is this is one of those things so we know we have 135 pounds of field weight that includes both water and soil right so it's a volume of one cubic foot right so 135 pounds one cubic foot and what's next so what's next is uh we oven dry the sample and we get a weight of 122 pounds what's left if we when we oven dry the sample it pushes all the moisture out it evaporates at all and what word drives it all out what we get is 122 pounds of solid so what does that mean that means if if you're doing this you can start to figure out the volume or the weight of the voids here which is the water and the air the way the voids is going to be 135 minus 122. if i can do that in my head and still i know it's late but if i can do that in my head 12 13 plus 122 i think it gives me 135 all right what does that mean that means that we have the water volume or the water weight here because air weighs nothing okay in in the grand scheme of what we're doing here um we're saying air uh doesn't weigh anything and that's okay so air does air is not gonna weigh anything um that means this whole uh this whole void weight is gonna be our water weight right 13 pounds over here and then the question really is just asking for what's the water content so again this question isn't meant to hopefully stump you hopefully this is an easy one but the question is do we divide 13 into 135 or 13 into 122 and if you don't remember uh that's okay these these relationships show up right in uh in the reference handbook so the water content right so if we look at the water content over here the water content is going to be the weight of the water divided by the weight of the solid times 100 percent right the weight of water by the way weighs solid times 100 so what does that mean that means this water content is the weight of the water divided by the weight of the solid times 100 percent so we get the weight of the water right what's the weight of the water the weight of the water is 13 pounds uh divided by the weight of the solids which is 122 pounds uh in a perfect world all our units are gonna work out wonderfully and never have any problems okay so the good news is we we worked our units out here 13 divided by 122. oh it is about um well times 113 divided by 122 times 100 i got about 10.6 okay so 10.65 so uh maybe i could have written this a little bit better this is close enough to 11 then i'm going to use 11 percent here so i probably could have made my number a little bit better giving you decimals in the answers but it still rounds to 11. i mean if you if you did it if you did the wrong way right if you did the wrong way 13 over uh 135 you'd get closer to 10 right but that way we said is wrong because what you really want to do is you want to go get these formulas right you want to get the the correct formulas and use them okay so water content one of those real basic ones um hopefully if you get something as easy as this it's going to work but it will get a little bit more complicated as we go through some of these phase relation problems because some of the questions and some of the equations actually get a little bit more uh complicated right and sometimes you need to know some conversions and they're not quite as straightforward as this one uh was so let's go to let's go to the next one so where do we what do we have here we have a soil sample with a volume of 150 cubic centimeters so again what i like to do is is i have my phase diagram here i just again i copied this in uh i made my own but this is one of those things where i i made my own here and i'm going to go in and sort of plug these in so it's just like you have free body diagrams and statics and structures you know um what we're going to do here is we're going to make this phase diagram anytime we get a problem like this because it's going to help you organize what you have what you don't have what you need to have and what do we have we have a soil sample it's 150 cubic centimeters okay 150 cubic centimeters and a mass of 240 grams 240 grams and you might be saying but wait this says w it does and that's okay you're only given one one you know phase diagram this says weights or masses so if it's a mass you just substitute w for m okay not a big deal uh just something you have to deal with and what a weight or a force equals a mass times acceleration so what's the difference weight or mass times gravity is going to give us our weight so here you know we especially when we get to metric sometimes we'll we'll use masses okay so what does that mean that means what we're going to do here is we're going to come in and we are going to go uh next we're going to keep reading the problem so when aven dried the sample as a mass of 140 grams so we had a hundred grams of water right so when oven dried the sample is a mass of 140 grams when it's dry no water it's 140 grams so 140 grams down here that means this weight of the water is 100 grams the weight of the voids um was 100 grams right and but what we also see here is this was completely saturated at that at that point right so this is completely saturated what does that mean about the volume of the air completely saturated means there was no air right okay so it was full of water and no air so that means uh that that's gonna tell us something here too right um what we're trying to find though is the specific gravity so we're going to do is we're going to come back and we're going to scroll down and look for specific gravity right so so hopefully there's just something here that we can plug in plug and chug and solve right but so the gs specific gravity here is going to be the weight of the solids divided by the volume of the solids divided by gamma water and i i specifically chose this one because i don't know about you but for me sometimes i have to think about my units and my conversions because i don't always use and think in terms of 9.81 meters per second squared i don't always think in terms of milliliters and centimeters cubed and all those things and i think it's probably good to review that a little bit so the specific gravity equation gives us the weight of the solids um divided by the volume of the solids all of that's divided by gamma w so the question is now this is you know we're given this phase diagram over here and it says okay weights or masses but what we have is a mass so do we convert everything to weight and then go there do we convert all of this to masses and we have to be consistent okay so the way that i think this is is is i think of i think a bit like this so i think of this is this the equivalent here is going to be well what what we said a force equals a mass times an acceleration so a weight equals a mass times gravity right does that make sense and similarly um if we're thinking about the unit weight here right the unit weight uh we're going to have kind of like the density is equal to rho g okay or the i'm sorry the unit weight is going to equal the density times rho g so what we see here all of a sudden is is we get the mass of the solids oh sorry for the wrong thing mass of the solids times gravity over the volume of the solids right divide all that by um a rho of water times g okay and the cool thing is here the g's go away and we just get an equation where we get the mass over uh this density rather than unit weight what's the density of water okay that's one of those things where hopefully you don't have to look it up okay so hopefully you don't have to look it up if you do need to look it up um the good news is we're told that the unit weight here of water is 9.81 but also hopefully one of those one of those one of those uh numbers that you know is is what the row of water hopefully you know is one gram per watt per milliliter or one gram per cubic centimeter okay so if you know that that's going to help you as well right and what this is going to tell us now is is a couple things we still have the mass of the solids so we know we know the mass of the solids but we don't know the volume of the solids yeah so how do we get that this is this is where um we do know the volume or the weight of the water we can find the volume of the water right because we know that one gram of water is equal to what one cubic centimeter right so if we have one so we can find the volume of the water equal to what is is equal to essentially we know we have 100 grams of water divided by one gram per cubic centimeter which is going to equal 100 cubic centimeters of water right i mean is does that make any sense i mean the numbers here are fairly fairly simple in terms of we don't need a calculator to do that but it hopefully the units make sense so what that means is we have a hundred cubic centimeters of water okay in that also means that if we if we plug in uh and use our phase relationship here we know that remember what i said earlier this is completely saturated right so if it's completely saturated that means the volume of the air is zero that means the volume of the voids is a hundred cubic centimeters which means the volume of our solids is a hundred or fifty cubic centimeters let's follow that right once we found this this volume of the water we're now able to kind of work backwards to get the volume of the solids and once we have that um the rest becomes kind of plug and chug here so i skipped down a little bit here but let's let's go through and finish our sorry wrong question um let's finish our plug-in chug here and what do we get so if we finish this plug-in chug we get the mass of the solids or uh 140 grams divided by the volume of the solids which is 50 cubic centimeters right and if we divide that by a one what divided by rho of water row waters one gram uh per cubic centimeter right so the cool thing is what we see here is units start falling out right our grams go away our cubic centimeters go away and uh when we solve this we get a value of 140 divided by 50 essentially so 140 uh divided by 50 and when i plug that in i get like 2.8 okay so 2.8 is kind of the upper end of a specific gravity of a of a soil but it should it should work okay to make a little bit of sense but again these phase these phase diagrams are going to be helpful in terms of trying to figure out what you know what you don't know and what you need to figure out okay so hopefully that one uh makes a little bit of sense and let's go to the next one all right we'll keep going here phase phase relations right so phase relations uh oh man one day he'll have in-law relations and i love my in-laws um so where are we going to go so we're going to go and we're going to say a soil sample obtained in the field has a mass of 20 kilograms so again a field think it has water in it 20 kilograms this is going to be our total 20 kilograms okay a total of volume of we just shape this put in a 0.01 cubic meter box and we got 0.01 cubic meters okay we know that the specific gravity of this soil is equal to 2.7 so we already have that so when oven dried the sample has a mass of 17.5 kilograms so we're given the 17.5 kilograms over here so 17.5 kilograms okay so if you've been tracking with me right away what am i doing i'm saying okay i know 20 minus 17.5 before i even read the rest of the question i'm thinking like what's that weight of water 2.5 kilograms right so so again just trying to figure out what i know what i don't know what i need to know but the degree of saturations this this question after degree of saturation of the soil is most nearly so how do we get the degree of saturation of the soil um so what we get there is is let's go back and take a look for that equation so degree of saturation is what we're looking for um we come back here and you know we have some uh dry unit weights unit weight of solids uh what else do we get we have we have avoid ratios porosity and what else so what else do we have um if we go to the next page this is where we get the degree of saturation is either the volume of the water or the volume of the voids or the water content times specific gravity over the void ratio and there's a lot of things we don't know there so what i'm going to do is i'm just going to come back and kind of write those equations in and try to see what we have what we don't have and or what we can get here so so the two equations that we had degree of saturation is equal to either the volume of the water divided by the volume of the voids or the water content times the specific gravity over e okay so this question is definitely more complicated and right away you can see that wait there's there's some bigger pieces here that i have to figure out because i don't know the void ratio i don't necessarily know the water content yet um because what's the water content or do i know the water content can i find the water content maybe i can let's go look what's the water content right the water content is is what the weight of the water divided by the weight of the solid so i can get the water content that's pretty cool so let's go look so i don't know this is where you kind of have to piece this out and see what do i know what can i figure out and uh how can i make this this work here so okay uh let's let's maybe we can go down this route if we go down this route though let's look for a second what's e um e is the volume of the voids divided by the volume of the solids okay so let's let's just keep going here so the water content we said is going to be uh the sorry um the water content is going to be the the weight or the mass of the water right the mass of the water so we just did this one earlier mass of the water divided by mass of the solids um the mass of the water divided by mass of the solids so it's at 2.5 over 17.5 times a hundred percent and honestly i think a lot of times i'll just leave the 100 off but um we get what we get about we get oh 2.5 over 17.5 and that's actually one seventh right so if i if i don't multiply you know times 100 percent i get 1 7. that's okay um it's like what 14.3 okay that's good um and so what else do we need well we could go and we could find the volume of the water here right or we could find but ultimately what we need is we need to find the volume of this this the voids and the volume of the solids so one of the things that we just learned or maybe remembered from long ago is that specific gravity formula also comes into play here right so if we know the specific gravity sometimes you're just looking and seeing okay what do i have what pieces can plug in where i mean i know you know i could go from a gamma i could go here and i could find my gamma my solids to find my you know volume of my solids or i could just plug into this equation because i know my specific gravity and what pieces can i use here uh to to make this work and honestly i think this equation is probably going to work out uh for us but but let's take a look right so um so if we take a look at that uh let's let's let's try it okay so what do we get uh we have the specific gravity uh equals the weight of the solids um divided by the volume of the solids divided by gamma w and what we said the last time this is going to be mass of the solids divided divided by the volume of the solids divided by rho water so the mass of the solids was 17.5 kilograms divided by the volume of the solids which is something we don't know yet okay we don't know that value yet um divided by row of water okay so row of water what we just had we just said rho water was equal to one gram per cubic centimeter but the problem or maybe a problem is we have kilograms and grams so what do we have to do we have to convert it and maybe know this conversion already maybe you don't maybe this is one of those ones that you want to write down somewhere in your the recesses of your memory so that you remember it but what do we get i mean if we multiply we can do this conversion here but essentially what we're going to get is we're going to multiply by 10 cubed and divide by a thousand why is that because we have a thousand grams per kilogram and 10 cubed centimeters cubed per uh per watt per meter cubed so that's going to get us our conversion which is going to mean we're going to a thousand kilograms per cubic meter okay so we have that 1 000 kilograms per cubic meter and that's going to equal 2.7 so we have to come back and solve for this vs value right the volume of the solids and if we can solve for the volume of the solids we've got this problem close we're we're in its grasp okay so the volume of the solids is going to equal we got to rearrange a few things we're going to multiply both sides by what we're going to we're going to multiply both sides by bs divide by 2.7 so we're essentially getting 17.5 divided by 2.7 divided by a thousand is going to give me 0.00648 cubic meters so i like to once i get something i can use i like to put that up here 0.0065 okay around that but that means now i can find my volume of my voids right because what's my volume of my voice my volume my voids right is just going to be the volume of the voids plus the volume of the solids equals the volume total or we could say the volume of the voids equals the volume minus the volume of the solids right so that volume of the voids is going to equal 0.01 minus 0.0065 cubic meters and i'll keep my units in there because they're so important and we get the volume of the voids equal to what's at .01 minus .0065 i like to put these into my calculator just because otherwise i tend to screw them up with zero zero three five cubic meters okay so now we have that value and we can come here 0.0035 and keeping this this in in in in focus here what equation are we going to use well we could either come back and solve for our volume of water we could solve for our our um our void ratio honestly i think the volume of the water is going to be the easiest one to solve for here and the volume of the water is going to equal what the volume of the water is just going to equal our 2.5 kilograms divided by the density of the water which is our thousand kilograms per cubic meter that's the 2.5 divided by a thousand i shouldn't need to put that in my calculator but it's late and i'm you know i mean you're going to be going a mile a minute on the test here but 0.0025 uh 0.0025 0.0025 so what does that mean it means the volume of the air is 0.001 okay because the volume of the air plus the volume of the water has to equal that three 0.0035 okay and now again what we could go and solve for our our void ratio but this equation is going to be the easier one to use and we can say the degree of saturation equals uh you know the degree of saturation s equals the volume of the water which is 0.0025 divided by the volume of the voids the volume of the voids is 0.0035 and let's see what we get well i got about 0.71 so i got 0.71 which i i'm gonna go with so i'm gonna i'm gonna take that as our answer uh and um we should be good we could solve it the other way and just for completeness let's let's do that but let's let's go and look just for completeness i mean not that you're gonna have time to double check things on the exam but what's this this is the void ratio e is the volume of the voids divided by the volume of the solids right so e we could have just we could have stopped here we could have found e right what's e um we we could say i'm just curious let's let's do this e equals the volume of the voids divided by the volume of the solids so the volume of the void is 0.0035 the volume of the solids was 0.0065 so 0.035 divided by 0.065 is like 0.5 or roughly 0.538 okay and what else i mean did we really need to solve for this water content uh not really unless we're doing it this this way right but what do we get we get the uh so so we could solve um we could solve we could solve this right so we could say the the 1 7 um let me write it out so we can say this degree of saturation also alternatively could equal the 1 7 for our water content times our 2.7 divided by 0.538 and let's see if we get the same answer and we get like again we get about 0.7 uh this is like 717. it's close okay so if i if i kept more significant digits i think i'd be at that point seven one so either way you know whether you use whether which you know whichever formula you use whether it's the the first one or the second one you get the same place you might be saying but that second one looks so much easier and the reason it was easier is because we had already solved for the volume of the voids we had already solved for that volume of the solids and we've gone through this right so that's that's um that's where it's at okay so these can get complicated they can get messy but the what i would suggest is if you get a phase relations problem take the take the you know the little bit of time write down your uh your for your not your formula phase diagram here and um and go from there write down what you know what you don't know and then start looking at the equations and seeing which ones will work to get you what you want but again using these um if you're given mass you kind of want to know some of these mass units in nsi because again as we come back to is as we come back to to what we have here we're only told this 9.81 kilonewtons per cubic meter we're not told kind of the density we're told the unit weight of water but a lot of times in metric you're using grams and grams per cubic centimeter or kilograms per cubic meter so just something to think about um something to keep in the back here your mind and uh one of those conversions that hopefully will get you a right answer on the test because that's that's where you want to go i mean you want to be at the point where you're getting these these things and get them right okay so that was a long one we'll follow with a short one okay and you'll notice a theme in my my answers to these concept questions here but uh a test typically used to determine the maximum dry density of soil is what so i just threw out a whole bunch of of of tests that you might see in your in your lab right i taught a soils lab at one point and it's like we did all these tests you know the sand cone the nuclear well actually we didn't do nuclear density gauge uh but that's in that's typically what's used in the field i saw that in the field quite a bit where we'd get we get reports from the field of nuclear density gauge readings you know of people doing compaction and inevitably i remember one project where uh we had a four foot lift that was going underneath uh uh a building four feet of soil was going underneath the building and what we said was uh we had to be you know we had to have one foot maximum lift to make sure that each lift was compacted so four foot total you know total fill but we could only do it in one foot lift to make sure that each each one was was compacted and of course the contractor or weekend decided to put all four feet in and didn't get it tested so uh this was i think this is one of the first jobs i had when i was a young engineer right out of college my boss said okay go to the site and sit there while they dig it all up and put it all back in one foot lifts i said you're serious i'm like yes i'm absolutely serious as well my boss you know he's like and and we're going to charge them extra for this because they did it wrong and we're not messing around with the building that was put in in a big four foot chunk and didn't get compacted properly so nuclear density gauge okay so sorry that's the story but what nuclear density gauge what is that it's a way to check the relative density of a soil not the maximum dry density so that checks the relative compaction density just like a sand cone can kind of do the same thing but this one you might have done in your in your lab where you know you take some out you fill it with an unknown uh a known volume of sand the sieve analysis we kind of talked about the sieve analysis and in the sieve analysis does what it it shakes the soil to get you a particle size distribution okay so what's left here is a standard proctor do you remember what a proctor is uh and maybe you didn't do a proctor in your lab but what's a proctor a proctor is where you have a you know a can you got a rod you put soil in it and you you poke it until it fills up and then you measure some uh this is a really crude drawing but you measure essentially compaction effort and eventually you get a curve right you get this curve and maybe you remember this this curve the dry density versus the water content right and eventually you get some curve that looks like this you know and it's in what you're looking for is that maximum dry density based on water content because if you get too much water it just becomes soup right it just becomes a mud and a mucky mess and you can't you can't get it dense no matter how how much you try if it's too dry you don't have enough kind of lubrication there you can't get the particles closer together but if you get the optimum moisture content the optimum water content you can get the right amount of compaction so that's what we're looking for here dry density um sometimes you can use like a this might be a standard proctor curve you can use more compaction effort with a bigger hammer for a modified proctor curve and you might have a modified proctor curve that looks like this and what this does actually is it approaches this idea here of a zero araboids curve and this is kind of a zero air voids it's a hundred percent it's kind of like a hundred percent saturation zero air voids right but what you're trying to figure out is what is that maximum dry density because when you're putting those lifts right whether it's under a roadway or it's under a building or wherever it is when you're putting in that soil you want to make sure it gets put in compact it you want it you want it solid you don't want to have settlement that you could take care of just by some compaction effort so that's what we're gonna that's what we're gonna look for here but but basically uh this answer goes to that standard proctor hopefully that's one of the tests you did in your soils lab when you took a soils class and hopefully you did some of these other ones like a sieve analysis or sand cone probably didn't do a nuclear density gauge but if you get into the field and do any soils testing in the field that's one of the ones you'll probably see out there as well sometimes the this will be calibrated in the field with a sand cone but it just depends it depends on the job depends on who's doing the testing and what's what's required so again laboratory and field test what's the point the point is we want to make sure the soil gets put in appropriately right we want it put in appropriately whether it's in lifts or we want to put in the right soil so that it does what we want it to like in a dam typically you want impervious layers and then you want pervious layers you want certain layers that drain and certain layers that don't drain okay uh in a in a roadway a lot of times you'll want strong materials like a whole bunch of rocks or uh whatever to create a nice base okay but what we have here is we have soil testing and construction so it indicates that soil is an in place dry unit weight of of 115 so this is our field weight our field weight that was measured uh whether it was a sand cone or nuclear density gauge 115 pounds per cubic feet per cubic foot okay we have a maximum minimum dry weight so gamma max is going to be 122 pounds per cubic foot and gamma min is going to be a pounds per cubic foot so these are you know these values here are from the lab um this value here is from the field and basically what you need to do is when a soil is going in and you want to make sure that it might be specified like in the spec is like 95 standard proctor compaction what what that means is it needs to get installed to a certain density in order to be acceptable okay and that limits it it kind of it limits the consolidation of that material because you're putting it in at a pretty consolidated state um it it it basically creates a known kind of condition if that makes sense so what we're trying to say is what's the relative density of the soil so some of you are probably going to be tempted because it's the fe you just want to get this over with you're just going to take the 115 divided by 122. i mean that's that's an answer here um but that's actually this is the relative that's the relative uh um the the relative not density but let's go to the manual here and look for a second and see what that is so if we if we come back to the manual here for a second uh actually let me get to the manual here so if we come back to the manual here for a second uh what do we have well let's let's come back here we have relative density versus relative compaction okay so be careful between those two terms the relative compaction is the one um where you have your field density divided by your maximum that's what you're looking that's typically what's specified that's typically like a it's like a 90 you know this might be the 95 percent um you know standard compaction or relative uh relative um compaction so that might be that might be specified in in a spec right and normally that has to be within a certain water content if it's too dry you'll have to actually add water in order to get up to that much um and if it's too you know too wet you'll have to dry it out or wait for it to be dried out okay so this is that's the the relative compaction that's that's not what we're doing here but the relative density right we have another formula here the relative density is going to be we could either go with with void ratios or we go with what's measured in the field and what's measured in the field here is going to be this gamma minus gamma gamma minus gamma gamma over gamma right there's a lot of gammas going on here uh and you just want to make sure that you get them right so i'll try to write this down hopefully i get it right here but if i don't just shout at me shout me down and uh we'll fix this but dr is what it's the gamma uh the field uh minus the gamma min and i'm dropping the d's here but um gamma max minus gamma min okay and all of that times what gamma max divided by gamma field so that's kind of just a long plugging chug equation but the reason i threw this one in there is because i i wanted to just just to step back and think especially with relative compaction uh versus relative density okay so these are two different numbers and uh they're you know they're both useful numbers they're just different numbers okay so what we're going to get here is if we plug this in gamma field is 115 and i'm going to leave the pcf's off here because everything's in pcf's pounds per cubic foot so i'm not going to like get super crazy about this right now but um what we're going to do here is we're going to 115 minus what the minimum is going to be 100 divided by our gamma max is what 122 over 100 or minus 100 and then times i i think the first time i put this in i screwed it up but i put gamma field over gamma max and uh it's easy to make those mistakes because it's like you're trying to copy from one to the next but gamma max upper gamma field is going to be what gamma max is 122 over uh 100 if you can put that into your calculator right i think you get about 72 percent so uh i did it earlier if i didn't do it right somebody shot me down but michelle thanks i i'm i mean i'm glad to do this i'm hope it helps somebody you know uh and um hopefully it'll go beyond just today as well so um if you do like it you know give me a like whatever uh but not seriously if it's if it's helpful for you i'm i'm i'm grateful for that so i think this is where it goes and we get to that uh 72 okay so relative compaction versus relative density i'm typically this is what's specified which again when we're looking at this i put this with that proctor test because the proctor test is going to give us some of these you know this this maximum value down here um from the lab and that's going to kind of be our starting point and sometimes i've had this happen in the field where like no matter what the contractor does they cannot hit their 95 relative compaction and typically either they don't have enough water or um they're using a different soil or the soil change like they took one side of the sample the borrow pit and over on here this soil's a little bit different they need to go get a new proctor to recalibrate uh to make sure that they're actually getting the correct compaction okay so again lab field test this is one of the ones that i ran into all the time uh when i was you know not just teaching but doing a lot more real engineering so not that this isn't real this is teaching this isn't real engineering so to speak but it is it's just getting you ready to on the fundamentals all right question 10. here we go um we kind of did this one earlier but i think it's this is specifically just looking at that uniformity coefficient and coefficient of curvature right i think this is going to be on there and if you can remember the uniformity coefficient is d60 over d10 um if you don't remember that uh these actually show up as well uh in in in their own spot so not just in uh the uscs classification system they show up here as well i i think they're probably gonna be on the test at some point and i don't know but um this is where the i made this problem a little bit more tricky i don't know harder maybe a little bit but if we if we look here actually let me try to draw that a little bit better here um if we look here we're trying to figure out where these points of intersection are for thirty percent passing sixty percent passing and ten percent passing so some of these are easier okay so some of these are easier like for example if again if we have these lines here so ah did i how to get that one i i did okay um this one's a little bit harder though and i just want to call this one out a little bit because if we zoom in on this quite a bit i don't know how how the the lines are coming if you can see the gray here this is a log scale right so as we go kind of from one line here to the next right as we go from one line to the next it looks like we're kind of in the middle maybe you know but we're actually if you think about this we're closer to this 0.2 so we have to kind of think about this a little bit right so as we zoom in here right i mean you have to know how to read this log scale so point one up to one and we're back to you know 0.9 here so this might be 0.9 not too crazy um this one has a pretty good point of intersection 0.1 this one in the middle right i mean you might be tempted to say i don't know let's can i move this over just a hair maybe i can move it over here i don't know like maybe i've moved it too much it's it i can't get it perfect maybe i had it okay um where is this right this is going to be kind of this point i mean you might be tempted to look but this is going to be a little bit closer to 0.2 actually because of the way the log scale goes this is gonna be about zero point let's say zero point two three and it doesn't look like it's you know three tenths of the way and that's because it's not but it's actually closer to zero point two three so when you're interpolating on log scales just be a little bit careful uh with linear interpolation because linear interpolation sometimes doesn't quite cut it um but just just be careful so so you know and even if you're close a lot of times there's going to be a range here but we can find our uniformity coefficient what's this you know 0.9 divided by uh what's this is d60 and d10 uh 0.1 okay so uh this is going to be about nine right and then cc um that you know the the coefficient curvature is going to be what d 30 squared divided by d10 times d60 so again this is i i maybe i hit this this topic too hard but i think it's gonna be on there so i think it's probably worth hitting a little bit and talking about so what do we get we got like maybe 0.23 squared divided by 0.1 times 0.9 and i think that gets us pretty close to 0.6 so let me just double check that 0 23 squared divided by 0.1 divided by 0.9 yeah i get about 0.59 0.59 okay so that means we do have some values here 9.6 i mean if if and the thing is if i use like 0.25 let's say like i interpolated and use 0.25 right i'm at like point seven so it's not crazy so like right so if i if i used you know 0.25 instead i i would get like a cc of maybe like 0.7 i'm not crazy far off but if there was an answer like zero nine or zero point seven yeah then then it's it's it's asking can you interpolate on a on a uh on a log scale and that gets a little bit trickier okay but the the thing that we see here too is look what this is kind of cool this is like look at this remember we talked about that uniformly graded versus um and poorly graded versus well graded we have this uniformity coefficient here of nine right uniformity coefficient of nine let's just go back to what we learned earlier or we were thinking about earlier right if we look at that uniformity coefficient of nine and come down here cu greater than four greater than six that's going to put us into that well graded right it has it's like when i think of well-graded i think of like what what do i think of well-graded it is like this where you have you know various sized particles that all kind of like fill in the voids right so like you get these little particles the big particles it's well graded and they all kind of fill in versus the poorly graded right or the the uniformly graded where basically you get a whole bunch of particles that are all what all the same size right so this is you know the cu of greater than let's say four to six and here is the cu like less than four right so this is that kind of gives you an idea as well where it's like this is like you're you know you might get your beach sand versus um your your crusher run right so just a couple things like that so let's keep going all right so we got question 10 and let's keep going okay so question 11. uniform soil is at a unit weight of 110 pounds per cubic foot the water table six feet below or six and a half feet below the surface so we know that there's a surface here there's a water table uh that's that's what that's six and a half feet down 6.5 feet okay at a depth of 11 feet so down here maybe at 11 feet right so we're down at 11 feet what's the effective vertical stress okay the effective vertical stress in the soil well this is where i think this this type of question could be on there too so so there's topic effective stress but what's the idea that's going on here let's come back here for a second to our reference handbook and what we see is we have vertical stress profiles we also have horizontal stress profiles where we deal with you know different layers of soil and water and we have different effective stresses versus you know vertical pore pore pressure and horizontal forces here we're looking at vertical stress profiles and if we look at this if we think of the total vertical stress right the total vertical stress it has to basically accounts for everything of the soil plus the water weight okay the soil plus the water weight so so in basically what's kind of happening is the soil takes part of the load the water takes part of the load and that's the total vertical stress because you have both going on but what we're looking for here is the effective vertical stress right what we're looking for kind of is going to subtract out the the component of the water so the effect of vertical stress is sigma prime is going to take out that component of the water so this is one equation i i kind of think of it similarly but basically the equation that we're looking for here is we're gonna say sigma prime this is our effective uh vertical stress is gonna equal uh let's just let's go look at it make sure we're doing it right but what's this what do we have we have we have um we have we have the the gamma of the soil times the height of the soil above the water table right plus the gamma of the soil and here we have a one soil so it's not like gamma one and gamma two we have one gamma for the whole thing it's you know here we could have gamma one and gamma two for us we only have gamma one okay uh but what we're gonna have to do is we're gonna have to subtract out the the water uh times that height so as you can see here there's an h2 and an h2 i'm typically what i'll do here is i'll say this is kind of like gamma times h1 you know plus gamma minus gamma water times h2 okay so that's going to be our effective stress sigma prime is going to be our 110 uh pcf times what our six and a half feet and if we again this is sort of sort of like those phase relations right if we know the total height's 11 6 and a half plus what gives us 11 well i don't trust myself tonight 11 minus four point they're on 6.5 i'm thinking it's 4.5 yeah so it's 4.5 okay so we got 4.5 here uh so that's our h2 all right so what are we going to do then we're going to then we're going to add in this this 110 but we're going to subtract off the weight of water and the weight of water is 62.4 pounds per cubic foot and everything's in feet times h2 which is going to be 4.5 feet and if you don't know 62.4 uh the good news is that one does show up it doesn't show up right here but if you remember um and you come back a little bit right remember when we were looking at our phase diagrams that unit weight of water was was right here so again getting familiar with the the reference handbook is going to help you remembering some of those those components is going to help you as well uh but yeah this is where i i just kind of factored this this h2 out to make the the computation a little bit easier but what do we get 110 sometimes 6.5 uh plus in parentheses 110 minus the 62.4 uh times 4.5 and i got like 9 30. so sigma prime equals 9 30 pounds per cubic foot okay close enough to call it 900 uh in a perfect world i probably would have made that 930 but i didn't actually i think i had different numbers i think i might had when i first did this i had 10 in my spreadsheet i can't remember what i had i i was doing this on a spreadsheet and i had another on number so it was closer to 900 but 930 works okay i'm good with it i hope so okay so let's keep going we got a few more of it a few more questions to answer here but hopefully they all make sense um civility of retaining structures so in the structural design piece of this i kind of did a stability analysis so i put a link to that up here and that actually takes you right to that problem so if you want to see that problem uh you can go to a little bit more of a stability analysis um in this one what i did is i just said the kind of difference with soil what's the difference with soil in the last one i did a dam section just use water here we have a soil and i took the water out of it so it's you know maybe we could have something in between and another problem where you have a you know partially drained a partially drained you know wall or something but here we have a retaining structure it's a big you know big hunk of concrete right eight feet tall and behind the concrete we put a drain in why do we put a drain in we put a drain in because we don't like that water behind the wall we want to drain it to get the water away uh so that it the wall doesn't have to hold up all that water right so a lot of times you'll put in a porous layer of fill like a stone with a drain at the bottom to try to get the water away quickly okay so what we have is retaining structure and the structure must resist soil and surcharge loads indicated so we have a surcharge and a soil here and um uh oh i think i made a mistake and i'll probably need to put an update here i don't know if any of you guys did this ahead of time but maybe you can call me out and tell me um if i did something wrong here but i think i i think i forgot the surcharge uh when i did my handcuffs we'll find out here i i think i did though so um so we'll add that in but what we have here is a soil and a surcharge and we have to come up with uh the total horizontal force exerted on one foot wall based on the horizontal active lateral earth pressure only so not the greatest question because i'm telling you to just do the just the lateral active earth pressure but then i gave you a surcharge so honestly ah i made a mistake here the the the answer does not include this surcharge so maybe what i'll do is i'll just cross this surcharge out because the answer does not include that surcharge and when i ask you this question i'm just asking for the active ladder earth pressure on the wall okay i'll show you what to do if you have a surcharge in there but let's just start with this for now okay so let's modify the question and keep going okay so what we have here is 115 pounds per cubic foot so this soil has a gamma of 115 pcf okay and what else do we know uh we know it has a friction angle this here is going to be friction angle of 32 degrees so typically in soil friction angle is known as fee so 32 degrees okay what else a drain a one foot you know so so we drain at the bottom of the structure what's the horizontal force well if we think of this again what we're going to think of is we have the structure here and this soil is going to act just like a liquid sort of not exactly but what we're going to do is we're going to use a coefficient to make it act like a liquid so we're going to get a pressure uh we're going to get a pressure this p you know active one and whatever you know this is going to be a pressure it's going to be a horizontal earth pressure and if this was just water we do what we take one half gamma h squared right do you remember that formula one half zero weight of water times the height squared to get us essentially the the area of this triangle but it's not it's not water what is it it's a soil so what do we do well because it's a soil we have to come back here and we say okay well what do we do with the soil we want a horizontal stress and so if we had a water table it would get more complicated but all that we have here is we don't have a water table we have a gamma and an h right we have a we have a pressure which turns into a horizontal force okay that horizontal force is going to take a look like this so pa1 the active pressure is going to be one half sigma sigma prime ka h1 one half uh in in other words what we're gonna get here is one half gamma one ka h1 squared so let's go there and what's the difference between this one and in the just the water version we get this extra ka right what's ka it's a it's an active earth pressure coefficient so this basically says okay well if it was water ka would be one everything would be everything would be pushing up against the wall but it's not water it's a soil so we're gonna reduce that that pressure and we're going to say well a certain portion of it's going to act right that's going to be what the active earth pressure coefficient does it says okay well how much of this of that acts horizontally and that's what we're going to get here so ka is going to equal tan squared you know 45 minus you know 10 squared of 45 minus v over 2. um so that's going to give us our ka but let me just write this down first so the the the pa 1 is equal to one half ka times gamma h1 squared but before we can solve that we have to solve for ka and ka i i it's funny i don't like this formula tan squared uh 45 minus v over 2. i when i was doing real engineering uh before teaching engineering we would always use ka equals 1 minus sine phi over one plus sine phi maybe the next math review session i do i'll i'll we'll have to say see if those are equivalent they should be okay but the the i i don't know maybe in your soils class this is what you learned as opposed to this either way you plug in and you should get a value down here if ka equals y i'll do this the first one because that's the one in the reference manual but 45 minus 32 over two and i think that works out to about 0.307 uh but yeah do you know how to enter the tan squared you just do the tangent and then square it okay so tan and 45 minus 32 over two okay and then i hit the square button and i get 0.307 and if i do if i do the 1 minus sign fee 1 minus sine of 32 over 1 plus the sign of 32 um you get the same exact value so that both of them work it's just i i put i put them up there because this is the one i used to always use in practice i don't know maybe maybe this is the one you were taught in your geotech class okay but once we have that value right now we can come down and we can say this active earth pressure is going to be one half times 0.307 times our gamma which is 115 pounds per cubic foot uh times our height which is eight feet squared and i think we're going to get that 11 30 value does that sound right 0.5 times 0.307 times 115 times 8 squared yeah i got about 11 31 okay pounds okay 1131 pounds okay so let's let's go with that for now um 11 30. and somebody's they saying but what about that 250 pounds per square foot surcharge because i know when i get to my test i'm gonna get something going on here where i've got a surcharge what happens with that surcharge um if we have a surcharge let me just take this for a second and move it up and what i'm gonna do is i'm gonna actually duplic ah that's not what i meant to do um let me try it one more time here so if i take all this if we had a surcharge i'm going to duplicate it i'm going to move this down if i can can i grab it let me try to grab it here ah it's not grabbing there it goes like here i got it okay so if i move that down the surcharge basically what that's going to do here is it's going to take uh this force here and can i get it over and no it's not gonna work um it's gonna take let me see if i can let me try one more time here um it's gonna take all of this force yeah i'm not doing it um let me just delete it and start over here it's it's going to take that force and move it over okay so essentially what we're going to do is we're going to take that surcharge and let me write the surcharge in here so we're going to have a kind of a surcharge in here okay and then we're gonna have uh the same active earth pressure here so if we had the surcharge right this would be kind of this the surcharge load okay so i kind of i'll update you know the question so if you're looking at this later and thinking like but my question doesn't have a surcharge that's because i deleted it between the time you're watching the video and uh what i just did what i just went through because i uh when i saw this beforehand i you know the answers didn't have a surcharge in there so let's just go with that okay is that okay okay um but yeah i mean the surcharge would increase that load obviously and i'm trying to think if that's in the reference handbook let's come back here and look uh i don't remember seeing it in here but oh yeah here's the surcharge right see look at that the surcharge um what does that do it just moves this diagram right it moves this diagram to the right right so it adds that extra that extra surcharge pressure uh right that extra surcharge pressure that we have to deal with which is going to impact the horizontal horizontal pressure okay so so let's keep going okay let's keep going uh and and let's keep going so where do we go next oh shear strength so this is your strength and this kind of gets into that point there's a couple different ways of looking at shear strength uh when i took my geotech class a while ago actually i was looking through notes from my geotech class uh just looking to see you know do i think what's what's fundamental it do i what i think is fundamental is that the same thing that was fundamental when i took a geotech class and uh actually a lot of it is but let's come back here for a second and we have a few other things actually retaining walls we'll come back to those in a second they kind of um they kind of they kind of look here so oh good question so visual representation of passive earth pressure um underground laterally to resist overturning yeah 100 so i will actually let's come back up here for a second so if this wall were to be was to be buried let's say this wall was buried right if this wall was buried what we would have is we'd have a passive earth pressure here right a passive uh pressure that we'd have to overcome in order to push that wall over and it's it's really hard to push soil it's really hard to push soil out of the way um it compared to the you know what we have going on here because actually let's come back to those uh pressure coefficients for a second and if i can find my way there hold on we're not scrolling that wonderfully if we look at this plus d over two well i'm just gonna i i already have this in my calculator so i'm just gonna do instead of minus v over time you do plus v over two and instead of .307 guess what i get i get 1.8 i mean it's huge it's it's a huge difference right that that passive earth pressure is going to resist the wall being pushed right it's going to resist the wall being pushed and that passive earth pressure coefficient is is huge compared to the active earth pressure coefficient so a little bit of you know embedding that toe is gonna help you quite a bit um when you're trying to resist overturning uh you know it but you have to engage it okay and that's where you know there's some debate on active versus passive versus at rest and and uh it's you know this is where you smooth wall and there's there's some other things going on in here but typically what i would what i would think they're gonna ask you is probably more like an active pressure coefficient situation okay um retaining walls we're actually gonna come back to retain walls in a minute uh well maybe a few minutes but um let's keep going here and again we'll get to our poorly graded stand so this kinda i probably could have put this in the lab test i could have put this um on other places but this is there's a test for sheer strength okay and what we have is here is a poorly graded sand so what does poorly graded sand mean you're all thinking it means a uniformity coefficient of less than four yeah that's great i'm glad you're thinking that that's not what i'm thinking here though a uniformly graded sand also means it's what it means let's go back to our our uh our you know coarse grain soils thing here for a second it means it's not fine grained which also what do you know about sand you pour it in a pile i mean there's a reason we have um we have timers that have sand in them right because the the sand doesn't stick together it's non-cohesive why is that why is that important non-cohesive because when we're talking about shear strength and trying to shear a plane of soil here if it has cohesion cohesion that's going to kind of stick it together right so a poorly graded sand typically is non-cohesive doesn't have cohesion built in okay so what we're looking for here is that that's that gives us a clue about what we're doing it's subject to triaxial test so there's a couple ways of doing shear tests um one way is kind of to put you know shear and put a pressure put a soil in uh you know put a soil kind of in in in here so we've got a soil in here and then it's like a direct shear test so we just push on one side you know push on the other side this is kind of like a direct shear test and if you were to have a direct shear test um that's great what you could do is you come back here and if we come up uh we're gonna have we have bearing capacity what else do we have we have effective stress we have shear stress so the thing is this this is you know the abcd version of what ncs puts out there that's on the spec skips all over the reference handbook it's kind of all over the place um but what we see here is we come back to shear stress and if we have a direct shear stress we could just you know if we were given our direct shear stress we could come up and calculate that friction angle at an internal friction angle but also going back to that cohesion thing for a second cohesion gets added in when we're dealing with shear stress right that's there's there's the part that's the um the the soil kind of like the sheer strength and then there's the cohesion strength piece of it for the shear of failure so if you're given a direct shear stress you can kind of use this equation and plug in that's not what we have i just had to make it a little bit more complicated i don't know maybe maybe it's a little bit more complicated maybe it's not more complicated maybe it just hits another topic you know this is this is kind of a plug-in chug equation or tau f equals you know c plus uh what was it tan uh um sigma n tan phi so sigma n that's the normal stress tan fee right so that's your direct shear we don't have direct shear we have a triaxial right so we have a triaxial test what's triaxial triaxial is kind of like you get this ring okay and you push a constant pressure on the ring and then you have a disc on the top and the disc on the bottom and you push here so so we're kind of compressing this thing in multiple directions so we have a confining pressure of for what we have here right here's the soil sample inside it but what we have is a confining pressure of 10 psi and what happens is it fails with a normal stress equivalent to 34 psi and normally you do a few different tests like this and you get 10 psi and then you keep going but for what we have is we just have one data point the friction angle for this soil is most nearly well you come down here and you say but i can't use that equation and you're absolutely right you can't use that equation directly yet but we can use that equation to solve eventually for what we're looking for here and this is going to this is going to give us something so what we do know is we have a zero cohesion right but what what do we have i mean do you just take the normal stress but we don't have a shear stress right so what do we do well if you remember maybe you don't remember but maybe you do maybe you learned something called moral implant failure plane when in i don't know kind of when we did i think mechanics what did we do did we take a look at uh uh did we take a look at more circle kind of the same thing going on here what we're going to do is we're going to make more circle we've got the sigma 1 sigma 3 and then we're going to go from there but what you see here is this fee is going to be the angle of this line that's tangent to that circle so if we come up here what we could do is we could basically create this surface here right and what we're going to have is if we come out to you know 10 psi 10 psi and 34 psi maybe i can drag this line out a little bit more and i could make this like 34 psi maybe ah not what i wanted try it again um 34 psi what this is going to do is it's going to give us a circle and let's see if i can draw a circle yeah not bad um it's going to be a circle that's going to be our more coulomb failure circle and maybe i can stretch it a little okay and that's going to give us our circle but what we ultimately want to do is we want to we want to take a point here and we want to get this tangent to the curve so so if we know this point you know this point and a line that connects them guess what we have we have a right triangle which means we can come back here and find phi once we can once we can do this right so we can find fee just based on that right triangle right so with with this what we can do is we can say well okay wait wait wait i know that the sine of phi equals what it equals the opposite over the adjacent right i mean this is just trig 101 but the opposite over the adjacent so what's that going to be it's going to be the radius over the sigma average right this this is going to be our sigma average right so how do we sigma average well sigma average is just what well it's 10 plus 34 over 2 which equals what 22 psi okay so that's our sigma average so we know sine phi equals the radius over sigma average what's this well we have 22 psi which is our sigma average the radius hopefully you can see this but we have 12 psi here 12 psi here right to go from the 10 to 12 to the you know or to the 22 to the 34. okay so that's our radius that's going to be our 12 psi here as well so we can find fee right sine inverse of what 12 over 22 is going to be um did i what did i screw up did i screw something up oh i did the sign that doesn't work sine inverse of 12 over 22 is uh 33 degrees okay so a lot of c and d answers today but that's where it goes so you know it we could we could find i mean this point here is sigma n tau right this is that point um right that's that point we could go and find that point but the the the sign is a lot easier to use if we have uh if we have a circle and a right triangle okay so that's a that's a shear strength let's keep going and uh this is this is one of those ones that i think is kind of fun um we've got a couple more that are i think hopefully a little bit easier and after this but we'll see maybe they're not easier i don't know maybe they're more plug and chug this one this one's interesting this one's interesting because uh this one gets to retaining walls it gets to bearing capacity gets a whole bunch of things and you know if we come back to the manual uh there is a bearing capacity equation in here and am i missing it where's the varying capacity equation it's it's down here somewhere right there's a bearing capacity equation and if you get a bearing capacity equation and you get a whole bunch of parameters great just go plug and chug okay plug and chug make sure that your your units all work out plug and chug you should be able to evaluate that right so i'm skipping over that problem today with an idea of let's look at bearing capacity here on a bigger picture kind of this is what you're worried about when you're designing to a certain degree i mean you want to make sure that when you do your footing or whether it's a wall footing or it's a retaining wall right so i mean coming back to our retaining wall for a second i i mean i could have made this a retaining wall footing i made it kind of a column footing but the idea here is we have you know 80 kips acting down and we have uh what 20 kip feet trying to overturn the thing so the question is we're going to get some base pressure underneath this thing okay so we're gonna get some base pressure underneath this thing and is it adequate and is it entirely in compression in other words if this moment here is big enough it can start to lift up this this part of the footing and put in tension and that would be bad okay that would be something you don't want typically um could it happen sure but we don't want that because then you don't normally soil doesn't pull back on your footing to hold it down unless you put like a rock anchor in or something the question is can we do this and i thought this was a better question than just a plug and chug one so i figured let's let's look at this and this is where we get back to this retaining wall piece of it as well so we know the bearing capacity is just like the essentially the allowable um well this here's a factor safety of bearing but but basically what we want to know is what do we have so so what i'm told here is this is an allowable bearing capacity and we're not doing an ultimate here but this is just allowable so the first thing that i kind of want to check here with this one is it doesn't make sense okay um in other words the force divided by the area is that okay well it's not that simple because we have this bending going on in here so again i can take a look at like for a one foot strip here you know for one foot strip what do we get for one foot strip um you know i i'm looking at a five foot by five foot you know square column footing but for one foot we could say well 80 kips divided by five is going to be what 16 kips and you know 20 kip feet divided by five is going to be what that's uh four right four kip feet okay so that's great this sort of brings it back into retaining wall world right and then we get into all these equations here which are all sorts of like ah now we got to find the bearing pressure at the toe and you know kind of at the heel so can we do this the keto and e what what does all this mean and let's come back and take a look at it okay so what's this um you know let's look at that let's just write it down sorry it's not g it's keto the right this is the pressure at the toe this is the pressure at the heel a lot of times with you know retaining walls they look like this right they look like a a little shoe there right um where you have a toe on one side and a heel on the other and you're trying to push it over this way so we we kind of don't have that for us but that's okay that's that's where the tone the heel comes from um and that's that works too but basically we have a q-toe and the equation was equal the sum of the vertical forces divided by b times one plus six e over b and you're thinking man i just wish you did the plug and chug formula because it's getting late and that would have been easier and it would have been but i think this is a better question okay this is actually something that i wrestled with a little bit when i was first starting out as an engineer and all of a sudden it made sense and i hope it can make sense for you as well but basically what we're trying to say is what's the eccentricity on this footing okay so so what we have so far is the sum of the vertical forces we got 16 kips down divided by the width by feet right so our width here our width is just five feet okay and if we're doing a one foot strip of wall let's take a look or one foot strip of footing whatever 16. and then we're gonna do one plus six times e we still don't have e yet so i'm just going to leave e in here as a placeholder i'll come back um divided by five feet but all we have to do is find e right what could what could be hard about that um it's i don't know but then we get another equation and it's like ah we have to do a stability analysis again so let's look let's write that equation down and let's see what this says so e equals uh b uh b over two minus the sigma m resisting minus sigma or no let me just double check this sigma um and sigma m over turn moment over turning and divided by sigma v okay can you do this i think you can okay so let's take a look at this for a second e equals b over two so five over two five feet over two minus what so so essentially we're saying e so if we have a column you know center line here that's that's i probably should have said that this is 2.5 feet somewhere in the problem statement maybe i'll update the problem statement for this one as well you know the column is at the center um right so if the column's at the center what does that mean for e well it means that that's the the five over two um but why do we care about that because the resisting moment is what the resisting moment is 80 well actually it's 16 kips times 2.5 feet that's the resisting moment right that's our resisting moment um resisting okay that's our resisting moment that's what's holding this thing down it's it's holding this thing down but the 20 kip he's trying to push it over and that's our overturning moment is going to be you know we're using instead of 20 feet we're using 4k feet because we're just taking one foot section so four kid feet that's our overturning moment okay we're going to divide all this by the vertical forces which is 16 kips and that's going to give us an eccentricity here so in other words it's taking this 80 key it's taking that 20 and it's kind of balancing and saying what is the eccentricity what does that look like and if i did this right 5 over 2 minus what in parentheses 16 times 2.5 minus 4 and then divide that by 16. i got like 0.25 feet okay so the good news is we can come back in here um i can take that 0.25 and i can i can plug it in right so i can take this and plug it in right up here and i'm going to get my q at my toe actually i'm just going to erase this because that's going to make my life easier 0.25 feet and now we can solve this so the q at the toe equals what 16 over five times in parentheses one plus six times .25 over five and i get like 4.16 ksf hips per square foot what's the problem with 4.16 ksf 4.16 ksf is greater than 4ksf which is q allowable okay qa so what that means is the bearing is no good it's inadequate okay so with 4.16 it's more than what the soil can hold up so the bearing capacity is inadequate did i i oh man i made another typo two typos sorry i tried to double check this ah okay i need to fix this one too the bearing capacity is adequate or the bearing capacity is adequate okay so we're gonna have to have an inadequate one here and uh man two mistakes i apologize um inadequate let's let's make these both inadequate here okay so it's and then are we entirely compression or are we partially in tension which one is it okay so if we come back in here well how do we get the how do we get the heel so if we get the heel uh maybe you remembered that this can be like a plus or minus you remember that the minus is the q heal so the the q heal is going to be the same equation 16 over 5 but it's going to be 1 minus 6 times 0.25 i'm gonna make more room here it's so easy for me to make room i'm sorry if you're working with a pencil maybe it's a little bit harder um but you know i can make that room here um six over two point five over five feet and this is gonna equal i think i got like 2.24 okay so 2.24 means it's still in compression and that's good so the q heal is still still in compression we didn't get a negative value so the total the total here is going to be in compression all right there's another way you can look at this honestly the way that i like to look at it is like if you remember remember that whole um mechanics of materials maybe if you haven't looked at that review you can but p over a plus m over s or um you know plus or minus m over s right the axial a plus or minus the bending remember that combined stress thing um right and this is this is another way of looking at it this is you're gonna get the same thing but what's this it's like 16 kips over five you know square feet um plus or minus the moment which is four kips divided by well what's s s is i i i looked this up i was hoping it was in the manual it's maybe we we shouldn't use s we should use m y over i the reference handbook has my over i is b h cubed over 12 okay and this is going to be the b is going to be one foot width times five foot cubed over 12 and y is going to be 2.5 if you do this out you should actually get the same values so again this is another approach to it but it's it's looking at that basic mechanics of materials axial and bending stresses which hopefully is something that you learned when you did mechanics materials as well maybe this maybe this version's a little bit easier for you i don't know um if not here are those equations from the bearing capacity retaining wall formula all right okay we're getting close okay um which type of foundation would be most appropriate for the given structure i'm not going to spend a lot of time here but let's let's take a look a retaining structure along the base of a slope um we're going to use a wall footing okay wall footing parts there you know see wall footings are ones that go along the length of a structure and that's going to be especially on bedrock i mean you're just going to put it before the footing pour the wall and be done with it a five-story residential building a relative residential buildings are relatively light structures um cohesive soils differential settlements a concern typically the bearing capacity is probably not the best sometimes you have to worry about different pieces settling a lot of times um some of those cohesive soils you'll have you'll have some fines and the bearing capacity won't be the greatest anyways you'll end up with bigger footings and sometimes what that'll do is it'll turn you into a matte footing so i would say this would probably be best suited for a map pudding uh a bridge abutment if you've ever gone by uh where there's a bridge going up near where where i live currently and you see them pounding piles all day long actually they have another set of piles that they're going to put in in the spring and they're just sitting there right why because you get this big bridge span this is on a little area and you want to distribute that load a lot of times also on near a stream you don't have the best bearing capacity so you want to get down deep to something good so we'll we'll call a deep you know piles here um we'll use piles at a bridge and what's that leave it leaves it to a storage office building non-cohesive soils bring capacity to 4 000. that's a pretty typical building honestly uh you know if you get outside the city or outside a city i mean a pretty typical building couple stories not too crazy not huge uh not huge footings but this should work for you and you should get something where you could use a spread footing here okay so kind of a concept question i wish i could spend more time on that but let's just keep going uh today because we got a few more questions and i want to not stay up all night but um it's all good you know so let's keep going here a soil plasticity index a 40 liquid limit of i'm sorry plastic limit of 30 so pi again we have a pi of 40. a lick a plastic limit of 30. what's that mean it means the liquid limit is going to be uh right the liquid limit is the sum of those two right so if we come back to our liquid limit what do we get we get um the the liquid limit is the sum of the plasticity index and the plastic limit so what does that mean this means pi plus pl that means our liquid limit's 70. okay so this is our soil um the recompression index can be estimated most nearly as well what's a recompression index and what's a compression index and what are all these pieces right so with with clay soils or silt soils or you know fine soils what we have here is a soil consolidation curve and this gets into soil consolidation so what we have is basically we have compression going on and if you think of it this way there's you know maybe you had a soil that had like glaciers on it at one point and it compressed that soil well the glaciers left and uncompressed a little bit okay that would be like an overconsolidated soil where it's it's already compressed some but it didn't uncompress all the way okay versus a virgin soil let's say you have a river that just deposits a whole bunch of clay and you get this virgin kind of you know this virgin layer that hasn't been compressed ever well when you first put load on it it's going to go down quite a bit okay so there's there's virgin oppression versus recompression so what i'm asking you for here is well what's the recompression index okay and there's a couple different ways to look at this and honestly when you look at this page at first in the reference handbook there's a lot going on in here and there really is but what we have to take a look at is how can we get our answers right i mean on the test that's what you want to know okay and honestly in the real world that's what you want to know too right what you want to know is what's this this compression index with this what's this um recompression index and to run these tests to get void ratios and you know delta log p and looking at those types of things it they'll take time it's it's really pretty quick to get a liquid limit test and if you liquid limit test here you could you could really quickly get an estimate of your compression index and recompression index right so typically your compression index is that first compression where you're going to get a lot of compression and your recompression index is that divided by six so it's again if we come back up to this curve it's like your compression index that's the steep part where you get a lot of compression right the virgin impression right away and the recompression is uh where you're getting a little it's it's already compressed some so you're not getting quite as much okay so let's look at that and uh basically with this one not not a crazy problem this one's actually um pretty straightforward but cc uh did i no did i write that down right mcc is can be estimated to 0.009 of the liquid limit by minus 10. so that's going to be this comes from trisogies one of those uh you know icons in the the soil world but what do we get 70 minus 10 so this is 0.009 times 60 anything that's 0.54 and if we take the recompression index cr this is going to be estimated as cc over 6 and that's going to be 0.09 okay so that one's pretty good just a couple more we got a couple of you still hanging on so uh we'll see i mean you can always come back to this later too but uh the good news is we're making our way we've only got a couple more problems and then we'll be done so let's look at this one we have a buried stray i'm sorry buried stratum of clay i'm two meters thick so we got this this clay layer that's two meters thick uh and it's buried okay two meters and it's you know maybe what we have is we have some good soils above it that we that are already compressed or maybe let's just fill on top of it or something okay we have a stress increase here so we have like a stress increase a delta p a change in pressure of 34 kpa and at this at the center of the layer the average mid depth pre-construction pre-construction soil over burn pressure is 50 kpa so we have got a uh we got like a 50 kpa we got an average address there and laboratory assess tests indicate this is over consolidated so it's already been consolidated but we haven't you know the the pressure is less than what the consolidation stress is so we got 74 kpa so that's our pc that's our 74 kpa avoid ratio of e is 1.5 uh d a compression index cc of 0.3 and a c our recompression index is 0.05 okay so considering these conditions the estimated decrease in the thickness of the clay layer is most nearly so this is a consolidation problem right we're in consolidations differentiation differential setup settlement so that's where we're at and this is kind of honestly it turns into plug and chug it's a little bit intimidating at first honestly there's a lot going on in here but what do all these p p knots and pcs and all these things mean essentially what they mean though is is kind of like going back to this curve where are we the 74 is kind of this layer this you know this is the preconcert the um the maximum pass consolidation stress the pc right that's that's this value we're starting at a value of what 50 but we're going up 34. so we're kind of like straddling this point if you will so if we come back here for a second what does that mean it means if we have this curve here and we kind of have one of these things going on right what what it means is we're at 74 that's our pc right we're starting at 50 and we're going to go up well what's 50 plus 34 50 plus 34 is going to get us to 84. so so we're going to have some settlement that's in the in the recompression and then we're going to have some settlement that's in the compression so this is recompression and this is the compression okay so what that means is it tells us which formula we have to use and that basically brings us down to here right so what all this means is essentially we have to deal with the recompression and the compression okay we're not just in the recompression phase we're not just to the left of it we're not just to the right of it with the the virgin impression we're in both of so we have to use the big long formula here but the good news is even though it's a big long formula it's just a plug and chug formula so uh let me try to write it down here delta h equals um equals what let me actually this is a long formula uh here i think i might be able to copy this in just because it's so long let me try that here hold on a second um i tried it last time it didn't work i think i've got a better way of doing it here so let me let me look and maybe i can just do this i don't know will that work um maybe the snipping tool will come to my rescue here and i can i mean wouldn't it be nice oh look at that it's perfect so it wouldn't be nice if you just do that on the test too um that would be so awesome installer but you know sometimes it's not that easy so let's uh let's let's take a look here let's do our plug-in chug and make it work okay so what do we get we get our our layer of 2 meters over 1 plus 1.5 that's our void ratio cr was 0.3 the log of pc so pc 74 over p naught that's 50 plus cc which was 0.05 that's the converging impression of of what of uh p naught plus delta p so that's our 84 over over 74. and i i gotta write my login here to to make it right okay so that's my log and we plug all this in you have to be careful though because what do we ask for the the units up here in meters we've got uh i'm sorry so you know what's up up top or centimeters units over here are meters so i need to multiply by 100 centimeters per meter right the meters cancel out and i'm left with centimeters so if we do all that i believe the value that we get here is going to be this 2.2 okay so again consolidation you gotta understand that consolidation curve a little bit but i thought this i liked this problem i kind of tailored it so that you can kind of be on both sides that that of that maximum consolidation stress so you could kind of get an idea of the recompression versus compression uh side of the curve right so you know we've got what's this this is the um this is the recompression and this is the compression right so they're both sides or did i flip that i think i flipped it but i'm sorry i i flipped it that's okay so the the the oh no i just typed my numbers and i put my numbers in wrong so somebody's gonna yell at me here if anybody's even listening i don't know maybe i'm just talking to myself right i plugged in wrong here cr is is 0.05 right yeah that's the 0.05 here okay um so let me fix this here because i because somebody's gonna yell at me 0.05 that's the that's the now i got to plug this in to make sure i did it right so let's just do it two divided by two point five times a hundred times what point zero five times the log um the log to get the log on t thirty tf36x you have to hit it twice um 74 over 50. that makes that makes sense because the 74 is the maximum reconsolidation in the 50s where we started so that's the recompression range uh okay that's good and then we're going to add in the 0.3 times the log of 84 over 74. and definitely don't screw up with that that conversion uh let me just make sure i did it right here okay um so there there's man this is almost this isn't quite as bad as my original statics one but um i did make a mistake um this actually gets closer to about 2.2 or i'm sorry 2.0 and what i originally had in in my notes was 72 kpa so um 2 or 2.0 it's still going to be the closest one uh but this actually gets to about 2.0 centimeters all right two more questions and then we're done one of them is a conceptual question but let's take a look at this one this one is a is kind of a plug and chug question but i like this question because it brings us back to statics and so much of this this this test is if you can understand statics but what we're given is a cohesive soil with what do we have it's on a slope um we have a 30 degree angle here and we got you know this this um this soil mass that's that's essentially 30 tons right 30 tons up here and we have a lengthy slope of the 75 feet is this thing stable or is it not what's the factor of safety here okay um so what we have is is is we can go through and we can look at this but here we do have some cohesion so that cohesion is going to add into our factor safety but if we come back to our our reference handbook here and i can't remember which way it is but uh let's see if i can find it here uh so so we're looking at slope stability right slope stability here we've got this mass kind of pushing down a pushing down and um what we have to do is we have to take that mass and we have to look at well what are the resisting forces what are the mobilizing forces and then come up with a factor see so our factor safety is going to be uh it's it's going to be it's going to look like this right our factor safety is what we're trying to solve for but our factor safety is going to be uh what are our forces divided by our mobilizing forces so if we can get each of those we should be good so what do we have let's start with the um the the resisting forces the friction forces and the the friction force is here um if we come back again this is where it becomes a little bit of plug and check but i i did want to look at this because i it's on here but what do we have with the the weight times the sine of alpha okay so we have the the weight of the soil mass times divided by times the sine of alpha and if you think about this for a second what do we have right if you think about this we have this is going to have some component that's normal and some component that's perpendicular to the slope or parallel slope normal and parallel to this the slope okay and if this angle here is 30 degrees right so this angle here is 30 degrees this angle here is 30 degrees what angle or i'm sorry um uh what's what's what's what are we doing here so um this is what this is what we're looking at am i getting that right this is the mobilizing force i'm sorry this is our mobilizing force man two hours is too much for me i think so we get our mobilizing force i was just thinking like that's our mobilizing force because why this is the sine of 30 right the sine of 30 is the one pushing down it's the opposite of this angle that's not going to be our sine of 30 i was thinking like what i scrub that's our mobilizing force right that's the one pushing this thing down the slope right that's our mobilizing force and basically what all this is is just that 30 tons and i like to convert this right away to hips because i don't like tons um so 30 tons times what two kips per ton equals uh you know times the sign of of this alpha or sine of 30 degrees so 30 times 2 times the sine of 30 which is actually one half we get 30 hips of a mobilizing force and what's resisting that the the friction force essentially is resisting that uh in this i i like this equation here because it again it makes us think of that friction problem so if you if you know that friction but cls um wm cosine alpha and then this tan phi is kind of like your friction factor right um so what's this this is um cls that's the length of the slope plus the wm cosine alpha s times tan phi okay so if we think of friction for a second right we just said that we have this is kind of our mobilizing force of um sine sine alpha well well we also have you know this force here which is kind of like our normal force right so if you remember friction what's new friction friction's in normal force times mu right so that's kind of what we're getting in here but then we get this which is kind of like the glue or the cohesion that holds it together okay so that's one way to think of uh slope stability so what's this our cohesion of um 1.5 pounds per square inch yikes we don't like pounds per square inches times 75 feet so what's this well what i'm going to do is i'm going to multiply by what 1000 pounds underneath one kip and i'm going to multiply by 144 square inches per square foot okay so that's my unit conversions and then plus the the weight of this thing that's going to be um 60 kips uh times the cosine of 30 times the tan of the friction angle the friction angle is 21. 21 21 degrees and good news is i think that actually is an answer that we can calculate the answer that i got was about 36.1 kips so if we come back and take both of these values our factor safety is going to be 36.1 over 30 which is going to equal i think that works out to about 1.2 yeah so i think we get about 1.2 there and we have a good answer okay so but again i like this problem because it gets back to that whole friction idea of how do you do a block on a plane right i mean we did this problem i think in statics or strength materials i can't remember which one but it's a i mean we didn't do this problem but a block on a plane right it's it's it's it's so similar to slope stability if you can kind of think of that those angles those force components as vectors uh it's going to help you out there in solving this type of problem as well so last question here it's a concept question and then we're done for the night and that's good because um i'm just losing ground here all right there's another bad pun okay i'm digging myself into a hole um sorry i i'm trying i'm trying i'm really trying here but um what do we have question 19 soil stabilization last thing here uh you know there could be some i don't know maybe there's uh going to be a numerical question but i think this is probably more of a concept question with with when you get into erosion sediment control it becomes huge on job sites i mean i've seen more uh more silt fence put in wrong probably than right in my life still offense is one of those things that never gets put in right but erosion sediment control maybe that's going to be on there i mean just hay on a you know mulch and probably not hay but straw on a you know straw mulch on an empty site cuts so much down on erosion right so here the question is how do we stabilize soil what are some ways we can stabilize soil typically it's associated with all the following except okay mechanical compaction if you compact it make it denser it's going to be less like less susceptible to wash out that's okay that that does stabilize the soil so uh it's it's we're looking for an except here so we're looking for a knot chemical add mixtures well if i add some glue to the soil not really glue but let's say like a cement for example if you put a binder in the soil you can hold it together okay that's gonna provide soil stabilization geotextiles a lot of times geotext cells are put there just to distribute loads and stabilize soils so again not there clearing grabbing an excavation that's typically the first thing that's done on a site after erosion sediment control you know are put in appropriately to protect all those waters that you want to protect in the wetlands and all that stuff but the clearing grabbing and excavation is what rips up the soil it takes away the vegetation it takes away the trees and that opens it up and exposes it uh you know so if you get rain it gives it the potential to wash out so it can be associated uh soil stabilization can be associated with all them except for really clearing grappling excavation there are smart ways of doing excavation but typically um you know this is what's going to cause your soil to become unstable is when you do and grabbing excavation so man that's it that's it that's it we got no more questions we're done and man this was probably the longest one yet i i thought it was going to be a little bit quicker with some of those those uh you know those those those questions uh the concept questions but we made it okay and hopefully on the test you're able to make it as well um uh oh man now you're going back to question 13. uh oh let's go back to question 13 what i do sign is oh good grief you are so right that's totally what i did is i did opposite over hypotenuse um and i failed trig 101. so thank you so much i i so appreciate that yeah opposite over man i i don't this is the hypotenuse all right so the if we look at this if we look at this right triangle this right triangle um the it's opposite over hypotenuse man i'm so thankful for that because that's going to come up in a comment below somebody's gonna be like but you don't even know trig i mean like you're right i screwed it up um but you know i'm trying here it's it's uh it's it's rare for these to go without at least some stupid mistake like that and honestly on the test you're probably gonna have some silly mistake where it's like uh oh i made a mistake you know but hopefully they don't dominate and you guys are gonna do a great job and maybe even you write down the wrong thing but you do it right anyway because i still did opposite over adjacent here all right good grief i said it wrong again opposite over hypotenuse i still did opposite over hypotenuse even though i wrote it down wrong so you know maybe maybe it comes from um when i was younger i had a condition and it was hard for me to think sometimes so my older brother he just said you should just eat dirt three times a day i i'm really glad he told me about it sorry that was supposed to be a joke i think i'm really lucky that my older brother told me that no don't go eat dirt before your test that's not going to make you any smarter um you know it's it's uh it's and when you ask for fresh ground coffee um make sure they don't put dirt in it you know there's no soil in fresh ground you don't want the ground you know them to grind sorry the puns are just bad but on what gets bigger the more you take out of it um the hole i'm digging myself okay so there you go so we'll end on those bad jokes um sorry they're they're probably not as funny today as maybe some of the other ones but um if you have questions definitely shoot me you know reach out shoot a comment below if you liked this say hey this was the best thing since uh i don't know i'm trying to look for a pun here but again i'm just i'm running out of ground so hey um thanks for thanks for uh staying to the end and uh i look forward to the next week we'll we'll jump into some water and uh we'll go from there so hey thanks i'll see you next week so until then keep working hard moving onward