Transcript for:
Columns and Foundation Engineering

hi guys a pleasant day um today we will be continuing our discussion in rcd so today we will tackle columns topic then afterwards it will be followed by Foundation Engineering okay all right so actually loaded columns so we have here a column here are its reinforcements so this column is subjected to an actual own po so since there is APO this will be resisted by the column strength Sorry by the concrete Sprint PC and is still strength PS so this should be an equilibrium condition therefore if we take summation of forces vertical that summation should be equal to zero so the downward Force po should be equal to the upward forces PC and PS and we all know from our strength of materials that forces can be expressed as stress times area all right so we are going to express this pcnps as the product of stress and area so stress concrete area concrete for PS stress still area still now we all know that the stress of concrete is 0.85 F Prime C and the area well gross area of column minus area of Steel in this way you will be getting the pure concrete area now for the second expression the stress is FY that is the area is a s now this is our po so we are going to discuss two kinds of columns we have the tide column this one and we have the spiral column this one so the difference between the two is the transverse reinforcement that we have this one it has ties this one it has spiral reinforcement Okay so this is our po let's talk about the two two types of column we have the type column which is when you get DPN it is just 0.8 of Po uh-huh this PO is replaced by its equivalent expression all right and this is the nominal P when you when you want to have the Pu you are just going to multiply this nominal by the reduction Factor okay so for the spiral column it has a difference multiplier to PO so for spiral column the multiplier is higher than what the type column has okay so re-expressing the PO substituting its equivalent expression we have this that we have this expression now to have the Pu multiply the p n by the reduction Factor so I hope this is clear to us now let's talk about the reduction Factor for for the tide column the reduction factor is 0.7 from old code then 0.65 if you are referring to the 2010 and 2015 nscp now yeah we have also the the so-called reinforcement ratio the the ratio of reinforcement to the gross area of the volume and take note that this ratio is accepted if its value will range from one percent to eight percent or from 0.01 to 0.08 all right now for the column spiral the reduction factor is 0.75 then also for spiral column it has its reinforcement ratio similar to The Tide column however when it comes to a spiral column it has another reinforcement ratio the so-called spiral reinforcement ratio so if I were you I would memorize this formula because well actually during the time when I took my board exam The Examiner did not provide the section of nscp um we we had a column problem there good thing I memorized the formulas so that is why I was able to answer the problem correctly all right so now let's talk about this the spacing of the reinforcements maximum spacing of ties so from the provision we have six times the diameter of the main bars 48 times the diameter of the ties and list volume Dimension and among the three values you are going to choose the smallest value the least among them and for this spiral column the spacing of spirals can be obtained by using this expression spiral reinforcement ratio equals volume of spiral over volume of concrete core and take note um this is spacing also has its minimum and maximum limits here are those the clear spacing of spirals is limited to 25 mm to 75 mm meaning the value of the clear spacing of spirals should only run between 25 and 75 mm all right now I hope this is clear to us let's go to the next slide okay for the column dimension let's let's use the Pu expression okay so this is valid for the tide and spiral column so pu is equal to the reduction Factor so of course the reduction Factor should be present here because we are talking about the ultimate one okay blank why is that so why why is this blank guys by the way this expression is the PO all right well I make this one a blank one because for Titan spiral if you are going to use the tide this one is point eight if you are going to use the spiral you this should be 0.85 all right so the value here will depend on the column that you are solving okay so now from this expression these two this expression is valid for both Titan spiral okay I will just cross multiply this AJ to this row so that we will be having a s equals rho a g so as you can see we can re-express a s as raw a G let us replace this a s guys base expression so now right away so the here's the reminder for you for time 0.8 is the multiplier for spiral 0.85 it's the multiplier so 0.8.85 will be the factor to be put here so that will depend on the column that you are solving now let us replace this as by this expression guys so yeah we have now this expression so as you can see we replace as by row Ag and you will notice that we can factor out a g so let's factor out a g okay so we have now this one so if we are after with the dimension of the column we have to concentrate on solving the AG so to have this a g I will just divide both sides of the equation by this expression therefore we have a g equals pu over this expression all right so this expression will be used later when we are going to solve the column dimension all right so let's move to the next slide okay so situation one a square tide column carries an actual dead load of 560 and an actual live load of 750. assume F Prime C 28 FY 275 use 20 mm diameter bars all right so here are the tasks number one determine the factor loaded the column will carry all right that's very easy number two determine the minimum required dimension of the column rounded up to the nearest 10 mm use reinforcement ratio of two percent and lastly determine the required number of 20 mm diameter bars okay we are now crying because the question the task is very easy all right so number one determine the factored load that the column will carry now we will we will be using the the load combination factors here so we have seven load combination factors but in this case since the loads are only dead and live we are just going to use the one uh the the one that is frequently used in solving the problems so we have This 1.2 dead load and 1.6 live load but guys do not worry because later I think in situation 3 I will be showing to you all the load combination factors and we are going to use all of them so this time we just start with the with with the with an easy problem okay so yeah let's use this expression now substitute the values 560 for the dead load correct 750 for the live load correct okay we have now the answer 1872. now for number two we are going to compute the dimension of the column so from our previous slide we have this expression all right now since this is a tied column we are going to use 0.8 Factor so yeah we have now the 0.8 here and let us substitute all the values that we have since this is a column we are going to use b as its Dimension so what is the gross area of the column this time B squared yeah B squared guys then for pu it is 1872. I multiply it by 1000 to make it in Newtons okay the reduction Factor 0.65 um well it is in the discussion a while ago in our introduction guys so F Prime C 28 the raw is two percent or point zero two all right now we have the B 124 895.9208 this is this is still B squared so take the square root of both sides we have 353.406 mm now we are going to round this up therefore we are going to to declare our column Dimension as 360 mm so I hope this is clear to us guys um this cannot be rounded down because you will make your column less uh or you will decrease the capacity of your column if you do that thing all right so we tend to round up the dimensions of our members all right now number three I hope this is clear to us guys this is very easy column problems are easy problems okay number three determine the required number of 20 mm diameter bars hmm who are we going to compute the number of 20 mm diameter bars guys huh do you still remember what you are what you were doing during your college days okay I'll show you two solutions okay solution one is the use of reinforcement ratio so we have very enforcement ratio of two percent then I over for the diameter squared then times n number of bars guys so this represents the total area of reinforcements then the gross area of column we have just obtained the 360 so the gross area is 360 squared so now we have n equals 8.251 okay my question to you is are we going to round this up or round this down hmm what do you think well if I will declare my number of bars as eight bars this this column will will have a a lacking of two or of 0.251 number of bar so that is not correct so this end should be rounded up all right so you will decrease so you will declare your end as nine bars okay let's go to the solution too for the solution 2 we will be using the the Pu expression so as you can see since this is a tied column this is 0.8 and the reduction factor is 0.65 Okay so 0.65 for the reduction factor F Prime C 28 correct we are just checking our inputs guys okay AG 360 okay good I over 4 20 squared and all right because we don't have yet the number of bars so we we express this as n okay two five two seven five all right so okay now as you can see on the right side of the expression the only unknown is n um yeah so let's let us just use our calculator so store in your calculator this one so I will multiply these two factors and I will store their product to a okay now this tool I will multiply these two factors and I will store there for product to be all right then for this 360 squared I will store this to see all right then lastly this pi over 420 squared will be stored at d now we are going to to input this expression in our calculator okay so the the reason why we use the letters in our calculator is to avoid inputting a length expression in our calculator you will see here okay so as you can see we we just input this expression but in this case in terms of letters like this one this one is being represented by a this one is being represented by B this one being represented by C this one being represented by D and this n is represented by X all right can you follow okay 275 this D then this end is represented by X all right now as you can see if you input values for X well what is this x x again this x is the n the number of bars guys you will get a force pu that is in Newtons so I will divide this by 1000 so that the value that we will be obtaining is in terms of kilo newtons now I will press Cal and your calculator will ask for the value of a I will just Express equals then for the value of B I will just press the equal sign for the value of C I will just press equals then for the value of D I will just press equals then for the value of x I will put the number of bars from the choices okay um in this in well I have I know you have your module there in our module the choices for this problem the choices are six seven eight and nine all right so six bars seven bars eight bars nine bars so press six for X okay the corresponding or the value that you will get is eighteen fifty then seven the corresponding pu in terms of kilo Newton guys because we have here the 1000 as divisor okay when you input 8 all right this one is this one is the value that you will be getting then lastly number nine this is the value of the PO now that your column is carrying this pu equals 1872. as you can see when you use six bars the Pu is 1850 so the Pu is not enough to resist this load 1872 so you will not take this six okay when you see this seven this seven gives you seven bars will give you 1891.19 kilo newtons and as you can see having seven bars will give you a column that has an enough strength to carry the load 1872. therefore the answer for this problem is seven seven bars all right the problem is solution one and solution to give different answers so which of the two is correct well the correct one is this second solution because this second solution gives you the corresponding strength of the column to the to the number of bars that you are using and besides guys besides I will go to the previous slide besides guys you cannot use this solution one because in number three there is no condition of raw equals to percent so now I will show you how we are going to use our calculator so first I am going to get the product of these two values so 0.65 times 0.8 then I will store their product to a all right now for this two values so 0.85 times 28 so I will store their product to be then 360 360 squared store to C then lastly this one 0.25 Pi or pi over 4 times 20 squared so I will store the result 2D all right now we are going to input this expression but this time we are going to use the letters of the calculator to avoid inputting lengthy expression all right so a to represent this then parenthesis B then parenthesis C so we are now here guys then the next one is this one all right minus D but this is only the area of one bar we need to multiply the N the number of bars so this n will be represented by X all right then plus we are now here two seven five then this area guys D times n which is now the X now having this expression our pu is in Newtons I want this p u B in kilonewtons so we are going to divide the expression by 1000 all right now we need to press the calc then our calculator is asking for the value of a just press equals for the B just press equals similar thing will be done for C and D all right now the calculator asks for the value of x well this x represents n so we are going to set x equals to the list value among the choices which is six six bars guys okay so equals now you have now the Pu corresponding to six parts or when our column has six bars the Pu is 1850 which is not enough to support this 1872 therefore press calc again equals equals equals equals then replace X by seven bars okay now you see that the new pu is 1851 which sorry 91 which is enough to support the 1872 kilonewtons that is acting on our column so therefore our answer in this problem is seven bars well I will not look at the results of 8 bars and nine bars because we saw that seven bars is already enough all right oh well we are after with the economical design here so seven bars is the correct answer for this problem okay you can also use the table function of this calculator so just press mode then press number seven for you to have the table mode well if you are using different calculator do not worry because your calculator has this functions also now for function for the table so I will input the expression all right all right then plus 275 as which is represented by this DX then divisor of 1000 then equals okay we will be starting with the smallest value among the choices then end at nine the largest value okay then step or the interval one okay now this table shows the pu corresponding to the number of bars that we will be using for the columns when six parts this is the capacity for seven bars this is the capacity I think there's something wrong with our input yeah there's lacking here the X okay all right so again so we start with 6 and with nine steps one all right so for six bars the capacity is eighteen fifty like what we have obtained just a while ago then for seven bars this is the capacity for eight bars and for nine bars here are the capacities all right so I hope you get the way of utilizing our calculator for this kind of problem all right so we are now going back to our slide okay now so here are the results guys so this will be the display if you are going to use the table function all right okay we are now in our situation two okay so we have we call we have the column Dimensions 407 and 600. main bars 10 pieces of 25 mm all right nice reinforcement 12 mm spaced at 100 mm on centers so this time the spacing of the ties is already provided concrete strength 28 FY 415 so we only have one FY here so we are going to use this for both longitudinal and dice reinforcement okay clear concrete cover 40 and this time we have again the permissible or allowable concrete sheer stress but we are not going to use the 0.17 square root of f sorry 0.17 Lambda squared of f Prime C the shear strength based on code the nscp 2015 guys all right so here's the figure okay so number four calculate the actual load strength of the column so nominal all right this is easy the factor is 0.8 since this is a tight column there is no reduction Factor since we are just after with a nominal value okay number five nominal share capacity of the section V and X meaning when the shear is parallel to x-axis then six calculate the nominal Shear capacity of the section v and y okay y meaning when the shear is parallel to y-axis by the way this is an actual board problem all right so let's start calculating by the way since we are going to calculate the shear capacity um so simplify detailed since it is not again specified in this problem what formula to use therefore we are going to use the simplified then is this column actually loaded huh as you can see there is no pu that is acting okay so this is not actually loaded so you will be using for number five and number six VC formula under simplified calculation without actual all right so situation two okay calculate the nominal actually load strand okay so let us compute the gross area so 400 by 600 okay now the area of the the area of the reinforcements 10 pieces 25 mm diameter all right now use this one again this is 0.8 since this is 18 column if this is spiral we are going to use Point 85 all right so substitute F Prime C is 28 correct the gross area the a s all right the FY 415 okay then as all right so this is our answer guys for number four all right so this is very easy hmm now calculate the nominal Shear capacity of the section VN X so this time this nominal Shear is parallel to x-axis let us compute the effective depth the effective depth is always parallel to the shear Force guys okay then BW is always perpendicular to the shear Force so this time this is the B sub W the width of the web this is the top view of the column by the way okay so we can obtain d by subtracting or by taking L subtracted by concrete cover by diameter of the ties and lastly by half diameter of main bar all right let's substitute the values that we have l600 concrete cover 40 12 for the ties diameter then 25 for domain by main bar diameter so we have the effect of that so let us now compute the VC this is equal to so since this is a force we can express this as product of stress and area this stress can be 0.17 Lambda squared of f Prime C then the area is BW times D however there is a a provided stress which is 0.88 the permissible Shear strength of concrete guys okay so we're going to use that all right now substitute the values the B sub W this one is 400 the effective depth all right so we have now the VC so to have the VM we also need the vs well let's use the spacing formula this one since this spacing is provided and it is 100 mm as stated in the problem we are going to use to compute the AV so pi over 4 the diameter of ties then in this orientation there are three legs parallel to the shear Force so three legs since our FY is only one therefore we are going to take that 415 as FY also of the price then the D then vs all right so divide by 1000 okay vs now is in kilo newtons then we are going to use this one but this time without reduction Factor since we are just after with the nominal value okay so substituting the values this one for the VC this one for the vs all right now we have the value of the V and X 942.513 kilo newtons guys our answer for this problem all right number six okay similar question but different orientation of of the shear Force so since the VN is now parallel to Y this day changes its location so this is always parallel to Y as we just said a while ago okay so BW always perpendicular to VN okay I mean this one is parallel to v n no okay so substitute I mean compute effective depth okay so w minus concrete cover minus twice diameter minus half diameter of the main bars substitute the values we have now the effective depth compute the VC we are using the allowable first Trend provided by The Examiner okay so BW this time is 600 the longer Dimension the effective depth is this one all right so spacing this time AV has n equals 4 because we have now four legs parallel to the shear Force all right okay okay so divide by 1000 all right now B nominal sum of the two V Series S guys okay so VC check vs check okay so this is our answer for number six so I hope this is this is clear to us guys situation three a short circular column spiral reinforced okay so we have now a spiral column to be solved is to support the following loads below use f Prime c27 fy345 concrete cover is 40 mm use 10 mm diameter spiral um guys just a reminder this 40 mm is the minimum concrete cover from chapter one of NCP 2015 minimum concrete cover of beams and columns provided that they are not exposed to Earth or soil all right because when they are when the member is exposed to Earth that is the time that the concrete cover minimum will be 75 mm now we have here five loads present in this problem so number seven solve the diameter of the column rounded up to the nearest 10 mm assume reinforcement ratio of 0.028 number solve the required number of 28 mm bars and then last solve the required pitch of the Spiral pitch or spacing no before we can solve the diameter or dimension of the column we should have first the Pu so we are going to use the load combinations factors of our code using this 5 or utilizing these five loads all right so referring to ACI or nscp um the two these two codes are quite similar the only difference is that ACI involves snow load all right now we have seven load combinations here and we are going to Define each factor that is involved in the formulas all right so let us start with the B this d stands for dead load stands for weight of ice e as earthquake load f as load due to fluids and with well-defined pressures and maximum Heights F sub a for flood load H load due to lateral Earth pressure groundwater pressure or pressure of bulk materials L for live load LR we have this LR here and also here yeah for roof live load R stands for rain load s stands for snow load p as self-straining Force W as wind load and blast W sub I wind on Ice determined in accordance with chapter 10. all right so now let us use this seven load combination factors let us start with this one 1.4 dead load plus fluid pressure load okay so we do not have here fluid pressure load therefore we will just make this equal to zero and this one 900 well this is provided okay so we have the first PU now let us use the second one yeah so again guys the the value that we will be choosing among the seven values that we will be obtaining is the largest value among them so yeah if we are after with the largest value since we have this expression among these three we are going to choose the largest value so LR is not present so between FPS and PR so between the snow load and the rain load we are going to consider this 700 we are going to consider this one because this is larger one and this one will give a larger PU then if we are going to use this snow load all right so since PR is greater than PS therefore PR will be considered okay let's substitute the values guys dead load present f t not present L present live load yeah yeah 500 H or lateral Earth guys the present then among these three we are considering the r they're in load all right so we have now two to Thirty so between the two we are going to consider this one but we have five five more to go okay so the third one okay this one uh huh in the second expression we already knew that we need to use this PR but we but between these two guys between these two loads PL 500 how about the wind load we unload 800 times 0.8 that will give you 640. so this one is 640 while this one is only 500 so we are going to consider this 0.8 of PW so since live load is less than or smaller than 0.8 of PW which is 640 therefore 0.8 W will be considered so substitute the values 1.8 of that load this the among these three we are considering the rain load okay then this point a okay 640 therefore we have 28.40 so so far this one is the largest value now we have four mode four more to go okay so the fourth one all right okay substitute the values for the last term we are going to use the rain load dead load wind load 800 correct live load 500 correct PR 700 correct all right so 3 2 10. so so far this is the largest one because in the previous slide 1000 plus two thousand plus two thousand plus yeah so the fourth one is the largest pu so far okay the fifth one check one point to that load earthquake live load snow load all right so that load okay we do not have earthquake load here live load all right okay so one seven uh the sixth one so yeah 0.9 that load okay all right so wind load expression lateral Earth not present okay okay so then last guys Point time that load check one earthquake 1.6 lateral Earth all right so earthquake the present lateral Earth pressure in the present all right so we have now the seven values of pu and among them we are going to consider in our design this one 3210 kilonewtons so I hope now you know how to utilize or use this seven load combination factors all right so let's proceed number seven compute the column diameter so we are going to use this one this is uh part of our introduction this can be found in our introduction guys so since this is a short story this is a spiral column the factor that we will be using here is point 85 the larger Factor because for time that is only 0.8 all right okay so substitute the values so diameter circular area so for the gross area this is pi over 4 diameter of the column squared all right so this is the Pu that we have just obtained a while ago times 1000 to make it in Newtons yeah the reduction factor is 0.75 for spiral columns all right 27 for f Prime C just check it yeah okay three four five for the FY okay correct raw 0.028 all right correct so now we have the D which is 404 47.831 mm rounding this app we will be having 450 mm as our diameter of column all right so this is easy guys number eight solve the required number of 28 mm bars all right so Solution One using the reinforcement ratio 0.02.028 then pi over 4 diameter of bars which is 28 guys in this case 10 times number of bars then since the column is circle pi over 4 450 squared okay this is for its gross area now we have the N this is 7.232 if you if you declare your final answer is seven bars you will be lacking of 0.232 therefore uh rounding this down is not correct therefore we will be declaring eight bars we tend to round up the number of bars guys all right so solution two we are going to use the expression of pu this is 0.85 since this is a spiral column all right okay uh-huh for the gross area for the as all right and all right FY three four five okay so we are good to go so I want to input this in our calculator but this is quite lengthy therefore we are going to use the letters of our calculator so again we are going to store the values in the letters of our calculator so store in your calculator this one guys all right so store at a another one this one store at B all right next one this one guys store to see then lastly you store this area today but this time I mean actually this one is area of one bar only all right now okay so I just I I copied the the details from the previous slide so input in our calculator this expression so this one is being represented by a okay good this one being represented by B this one represented by C this one represented by D then this number of bars is represented by X all right then copy three four five then this one is D then X then again I want my pu to be in kilo newtons therefore we put a 1000 as divisor here all right now press call then for a just press equals 4B press equals 4G and the press press also equals so X number of bars from the choices so the list number of bars in our choices is six so let us start with six so when you press six the Pu that you will get is three thousand eighty Five Point forty one kilo newtons this value is not enough to resist this 3210 all right so six cannot be an answer cannot be the answer so for seven uh-huh E7 gives you an enough pu therefore seven bars is our answer so again Solution One and solution to give different answers and the correct one is this one our solution too again I will repeat to you the the reason why solution one here is not correct well having the solution one this is not correct because you are using this row value which is not a condition in this current item that we are solving all right so you cannot use this one if this is not part of the condition all right so let's move to the next question again you can use Also the table function of your calculator if you want to okay situation three number nine solve the required pitch or spacing of the Spiral mm-hmm so this is the drawing of our column guys so the spacing is this one s so again s is different from s n when you see and a subscript that denotes clear now clear is spacing so meaning this is a clear spacing so when we say clear spacing that is measured out to out of the reinforcement unlike s as it's being measured Center to center of the re-enforcement so we have here the so-called concrete core and we know that when we say concrete core that is measured out to out of the reinforcement by the way this is the top view of our column so we need to get this diameter of concrete core guys so this is easy this is just gross diameter minus two concrete cover all right so the gross diameter is 450 okay then the concrete cover is 40. all right okay so 370. now let us compute for the spiral reinforcement ratio so this is the formula from the code so let us solve this all right so gross area of the column so this is pi over force in circular guys pi over four gross diameter squared then for the concrete core area pi over four diameter of the concrete core squared all right so we can cancel this pi over four so we have now this expression all right so let's substitute the values 0.45 F Prime C 27 good FY three four five good okay so 450 for the gross diameter 370 for the concrete core diameter all right we have now the spiral reinforcement ratio now if you still remember our introduction you will know that or you will yeah you will recall that the spacing can be obtained by using this expression spiral ratio is equal to volume of spiral over volume of concrete pour now referring to this concrete core yeah this this one that is enclosed by the dashed lines guys from this point going to this point actually if you are looking at the looking or referring to the top view you will looking at this point guys this one yeah so we start here then down to here yeah this one guys uh so that is one revolution all right now the volume of this concrete core is very easy to obtain or to calculate because this is just an area of cylinder this is the diameter and the height is s the spacing how about the volume of spiral this volume of spiral that's that is being pertained in this expression is the volume of spiral inside this concrete core so how are you going to get the the volume of this spiral that is inside the concrete core well if you straighten this is spiral you will be obtaining this one the diameter is diameter of spiral but how about the length of it well for the length of it you may refer to the top view Okay so measuring the center to Center or the so-called Min diameter so if this is DC I will subtract half diameter spiral then another half diameter spiral so that gives us one whole diameter spiral subtraction to the DC all right now you know from your geometry that for you to have the length or this length this is just the circumference of the Spiral so circumference is 2 pi r or just simply Pi D but I am referring to the mean diameter so this is the mean diameter guys the center to Center diameter okay so Pi D so this one pi b well of course for Simplicity we disregarded the curvature all right okay so we just obtained the length of the Spiral as the circumference of it the the circumference that is being depicted in this picture all right so I hope this is clear to us so now guys how are you going to compute the Spy the volume of the Spiral well this is just a cylinder a slender cylinder right okay so okay so cross section pi over 4 diameter of spiral squared then times the length to get the volume of it okay so its length is pi times pin diameter ID circumference guys okay now for the volume of the concrete four this is the volume of the cylinder the the one that was enclosed by this sent by this dashed lines okay so pi over 4 diameter squared times the height s all right now we have all the values except for the s so by the way we manipulated this expression so we canceled out the pi over four then we we interchange the positions of s and raw s all right all right so I hope this is clear to us so substitute now the values okay so diameter is spiral squared the DC the diameter of spiral the row is the diameter of concrete core all right so now we have the S value actually um we have also a spiral problem a spiral column problem in our board exam uh way back in year 2014 and the examiner did not give the NCP provision so good thing I know the the derivation of the spacing of spirals so that gave me the advantage to answer the problem correctly all right now guys having the spacing we need to check this value or we need to compare this value to that of to that values of the maximum and minimum s all right so we need to check if the clear spacing is exceeding 75 or lesser than 25 mm so this is the SN the clear spacing guys and it is obvious that we can obtain this SN by just subtracting DS two halves of DS guys actually to the S so s minus half diameter spiral then minus another half diameter spiral so a total of one whole diameter is spiral will be subtracted to S all right so this is s the DS is 10 so yeah our SN is 38 so check this is greater than 25 mm they're for good otherwise you are going to take 25 mm as your SN all right then this one does not exceed 75 mm therefore also good therefore our final answer is 48.954 we are being asked to give the spacing not the clear spacing so do not declare 38.5 4.954 as your final answer all right so we are done with this situation I hope you you learn this calculation now okay so situation four uh-huh the spiral column actually all of our problems here are fast word problems all right the spiral column to be designed the spiral column is to be designed to carry a safe load of 2 9. concrete strength 28 main bars FY 415 spiral FY 275 concrete cover reduction capacity 0.75 well we know this even if the examiner did not provide this one because since spiral column we know that we know that the reduction factor is indeed equal to 0.75 okay then calculate the required minimum column diameter okay so again we are going to to solve for the column dimension okay still ratio 0.025 all right so let us start solving okay so we are going to use this one since this is a spiral column we are going to use the larger factor which is 0.85 all right let us substitute the values so for the Pu it is 2 9 then times one thousand to express it in Newtons so yeah uh huh point Sorry 28 for the F Prime C check 0.025 for the raw check then since this is a circular column uh this is this will this will be the expression for the gross area okay therefore we have now the D as 415.315 from the choices provided by The Examiner we are going to choose the nearest rounded up value and among them the nearest rounded up value is 450 mm so this is our answer all right so number 11 okay number eleven calculate the required minimum diameter of the main reinforcement the column diameter is 600 mm the ratio of Steel reinforcement to the gross concrete area is 0.02 all right now let's use the raw Express one so it's still reinforcement ratio is equal to a s over a g we have this row equals 0.02 how about the a s well a s is equal to pi over 4 diameter of the bar squared well this is this is the quantity that we seek in this problem guys we are required to calculate the minimum diameter of the main reinforcement for the gross area so this is a circular column its gross area is pi over four times six hundred squared all right as you can see in the expression we still do not have the N the number of bars that we used to solve from the previous problems but this time this is a different question because this time the unknown is the diameter well if the number of bars will be unknown also therefore we will be having two unknowns in this expression well this is an actual word problem and part of this problem is for you to know that the minimum number of bars for spiral column is six and for type column the minimum number of bars is four so part of the solution is the use of the minimum number of bars for spiral column so that is six bars all right without knowing that of course you will not be able to answer this problem correctly all right so we will now be obtaining the BD or the main bar diameter and it is 34.641 so from the choices you will be choosing the nearest rounded up and that is 36 mm so all right we have now the answer okay so perhaps some of you will ask why sir did not use this expression well if we are going to use this one we will not involve this value well this is part of the condition so we will be having an incorrect answer if we will push on making or on using this expression well we have the reduction Factor 0.75 we have the F Prime c28 we have the AJ this one cross the m3600 how about for the as yeah we have this just input this expression here and you will be having an expression that has DB as the only unknown so you will just input the the value of the DB and you will be getting corresponding pu but again guys you are not involving the 0.02 which is a condition in this question or task all right so this time this expression gives us the correct answer well this is different from the previous calculations that we had all right now let's go to the next question calculate the design actual strength of the column given the following data okay so this is 0.85 since spiral okay so 0.75 for the reduction Factor copy Point 85 F Prime C 28 then for the gross area 600 that's the diameter of the gross then a s yeah 36 squared then 6 bars all right so do not forget this one guys okay so FY 415 do not use this one this one is for spiral all right so for the as now we have the answer divide by 1000 to make it in kilo newtons we have five eight one three point zero one two kilo newtons hmm so yeah this is the end of situation four so I hope this is clear to us now let's move to situation five again an apple board problem foreign by 600 column is subjected to an actual load and factor and moments as shown in the figure the column is reinforced with 832 diameter vertical bars yeah okay check 38 12 mm diameter ties concrete strength 21 yield strength of steel 275 for both longitudinal bars and thighs all right so here are the tasks okay but wait let's see this okay so Factor loads at the top and u350 Vu for 20 mu1 480. now at the bottom Nu is still 350. Vu is still 420. but this time the end moment is different the end moment is larger to the to that of the top five of the top support okay so the tasks are to calculate the shear strength provided by concrete using simplified calculation which is easy calculate the shear strength provided by concrete or the VC guys using the detailed calculation again an easy one calculate the spacing of Shear reinforcements using the results of the detailed calculation all right so let's have a recap so for this year for the basic calculation we have simplified and detailed calculations and under simplified calculation we have without actual and with actual well our column is subjected to an actual load so we are going to use the with actual formula for the little calculation we have two formulas for without actual and another two formulas for the width actual so since we have two formulas here we will be obtaining two values of VC and between the two values we are going to choose the smaller value alright so this is the 2015 NS CP okay so for the simplified we are going to use this one for the detailed we are going to use the with actual one so this one I mean these two formulas which whichever will give the smaller value all right so the old code all right so by the way another recap under shirt Lambda 1 for normal weight Lambda 0.85 for sun lightweight concrete 10.75 for all lightweight concrete so I hope you still remember remember this values then Lambda equals fct all over 0.56 squared of f Prime C what is this FC average is putting them so strength of concrete we already encountered this fct when we discussed the cracking moment the FR calculation the rupture strip yeah okay so this is picture of nscp this can be found on page 454 of the nscp 2015 PDF all right so here we go let's let us start calculating the shear strength provided by concrete using simplified calculation so this is the view let us compute the effective depth of the column that is a dimension that is parallel to the shear so D is equal to l minus concrete cover minus diameter of ties minus half diameter of bars main bars substitute the values all right this is 32 yeah all right so D equals five three two okay so these are the values now we are going to focus or to consider the critical section for shear so the critical section is not the section of the support it is at the distance from the support so this one so we got this one and this one is said to have a distance D from the support okay we consider the bottom support because this bottom will give you larger reactions for the critical section since since the moment here is larger than the moment at the top support all right okay so to have the reactions of the critical section let us take summation of forces horizontal so Force to the right equals force to the left so v u equals vu what is that Vu 420 summation of forces vertical equals zero so downward Force equals upward Force therefore n u equals n u n that n u is 350. now summation of moment about the critical section equals zero so equating clockwise moment to counter clockwise moment so for the clockwise moment let us start with mu3 do we have other clockwise moments huh yeah the moment produced by this view Vu times D all right then for the counter clockwise moment we have mu2 so substitute the values so this is the effective depth that we have just computed a while ago bu for 20 okay check D then mu2 yeah all right so mu3 is 976.56 kilo Newton meter so all right let us now compute the PC using simplified calculation again at the critical section there is an actual load so we are going to use the simplified calculation with actual all right so this one so the the condition of the concrete is not specified so we assume that the concrete is of normal weight therefore Lambda is one F Prime C 21 check that the B which is the dimension perpendicular to share guys for Android check the effective depth all right and U is 350 so times 1000 to make it in Newtons and for the gross area that is 400 by 600 so 400 600. so this is 20 divide 1000 all right so this is the answer for the calculation of Shear strength provided by concrete using simplified calculation okay so let's move to number 14. this time using detailed calculation so for the detailed calculation without actual we are going to use this one as you can see the only difference of this to this is this mu I mean this mm mm is the so-called modified moment all right okay so mm is defined by the code AS mu minus n a and u quantity 4 H minus D all over it okay so let's substitute all the values now so this mu is the mu3 the MU acting along the critical section okay so mu3 the Nu the total depth of the beam sorry of the column the total length of the column then the then eight all right so we have now the mm so yeah raw as over BD so for the as this is the area of tension reinforcements so we are just considering three bars guys three bars not the total of eight bars okay so three bars based on this figure uh-huh where's that figure this one guys three bars only for the AES tension bars when we talk about raw it does not consider all the bars unless this row is a s over a g all right but this time destroy is not as over a g this is a s over BD okay so I hope this is clear to us then having these two values we can now use this one all right so substitute all the values we have the Raw v u then the D then then the mm all right BD all right so we have now the VC value then another formula to be used all right so Lambda is 1 21 for the F Prime C then b d and U 350 times 1000 to make it in Newtons then gross area 400 by 600. all right so we have two values of VC and we will be declaring the smaller value as our final answer so I hope this is clear to us guys okay all right so by the way just a recap for for without actual calculation this m is Mu all right for for without actual calculation again guys this is Mu okay there is a provision or there is an statement there is a statement in the code that says for that says b u d over m u should not be greater than one all right the ratio vud over mu should not exceed one all right um but in this case when you use mm as the denominator of this ratio that provision is no longer are required to be followed so there is no limit for the ratio vud over mm so that limitation one is only applicable for vud over mu all right oh no number 15 calculate the spacing okay so for us to have the spacing of here reinforcements or the ties we should have the vs we have the vu as well as VC so we can now easily get the vs all right so reduction Factor 0.75 this is the Vu at the critical section guys okay so AV fyhd over vs so the number of legs parallel to Shear is two okay the fih 275 D then the vs times 1000 to make it in Newtons all right we have now the S then check the S Maps so for us to know the S Max expression we need to compare the value of the vs to 0.33 square root of f Prime CBD all right well again we are using this expression 0.66 squared of f Prime C we are using its value to be compared to vs just to know the adequacy of the size of the beam its Dimensions guys all right so since vs is smaller than the value of 0.33 squared of f Prime CBD therefore s Max is the over 2 or 600. s Max equals D over 2 or 600 mm all right this day is equal to 532 mm all right so having this expression let us simplify this one so we have S max value of 266 mm or 600 mm okay so we have three values of spacing and among them we will be choosing 112.064 mm as our final answer because it is the smallest value among them all right but guys if the question in this in this problem is what is the maximum spacing of reinforcements when we encounter that kind of problem since there is the word maximum therefore we will just be focusing on the X Max values and in that case the answer is 266 mm all right so subtaining Foundation Engineering no now when we say Foundation Engineering reinforced concrete design soil mechanics no soil mechanics foundations reinforced concrete design okay young Foundation structure foreign Foundation Engineering we have so many factors to consider one of those is the pressure now pressure we have the QA Qs QC qsu and QE what is this QA this is the allowable soil bearing pressure before uh now guys is equal to q u divided by the factor of safety now if you are the engineer of this of the project factor of safety meeting more you can use 1.52 now okay rectangular Foundation or the strip Foundation while footing bedding pending square footing wedding around 14 depends bearing capacity or mayor hopes no okay anyway reinforced concrete designs projects practicing Structural Engineering now we have also the Q as the overburdened soil pressure gamma H pressure times the depth no okay overburden concrete pressure gamma H2 sword charge this is the load on the natural grid line semi-ground surface then the effective soil pressure okay now for the dimension of the footing so this is the bearing pressure now in strength of materials when you say bearing pressure that means contact pressure no bearing that means contact a non-bearing area or a new contact area um and structure I we need I in binubu hat no no foundation so I'm tendency I'm Foundation Foundation my contact pressure or that's the bearing pressure my contact between the soil and the foundation contact area and foundation so we can say that the pressure is equal to the force divided by the contact area the bearing area and that's the area of the footing so if your footing has dimensions of B and L and young area i b times L red now okay now this is the dead load live load uh wind load earthquake loads depends I okay now for the stresses uh for the shear stress in particular we have two types of shear stress we have the one-way share and we have also the two-way share okay now ganito for the one we share stress this is also being called as wide beam shear stress now rectangular binka begins cases first case is Ito as you can see young dash line is by the way this is the top view of the voting column okay from the face of the column distance Ayan what is this the distance this is the effective depth of the footings okay so from from the face of the column so what can only distance so the distance okay unshaded now for the peace from the face of the column distance okay iOS okay now now between the two cases bucket bucket um so can we have the shear Force here yes of course panel summation of forces vertical equals zero under equilibrium okay now aside from that stress force over area in this case what is the area the area times the depth of the footing Dimension B times the depth of the 14. as you can see for the stress this is force over the area what's the area the B times the D Hong Kong effective the plan this is not the total depth guys okay so effective depth yes ultimate is equal to production Factor times the nominal value I am nominal Shear Force so to have that divide both sides by by fee so you will have view over fee so on fidito is I share foreign IB now L times all right slide to ensure stress or the punching share stress or the diagonal tension shear stress guys examiner uncommon um um um I'm sorry I'm Callum below a one is dimensions stress uh I'm stressed I share Force then times times the effective depth so as you can see we have perimeter of the critical section times the effective depth okay so as you can see telegram share stress it all because of force okay reduction all right so for the reinforcement as you can see the critical section for bending or flexure is along the face of the column guys NASA face in some cases along the face of the column foreign um okay foreign of reinforcement anyway so foreign then f u over area as you can see X by B so xB or BX is stress is equal to force over area or if we will cross multiply area two to stress we can say that the force is equal to stress times area and we can use the words pressure and stress interchangeably so for my foreign so if we will take summation of moment about the cutting plane counterclockwise clockwise moment for the counterclockwise that's the MU and materials a few times f of x right half of X so what is this Fu again this is the resultant force of the pressure pu now for the force this is pressure times area so pressure times area x b then we just copied X over two simplify this we will have one half skill u x squared B so this is the MU now [Music] it's a previous meetings so now we have also for one-way bending members kbh for for what's what's the value of K values this is from nscp 2010 tournament from nscp 2015. I also uh indicated here the section number of the code so yeah okay so I hope this is clear to us all right nsct 2015. foreign 2015. all right Provisions all right all right nscp 2015 guys so again as you can see 2015 no so so page 408 then later um uh um regular polygon okay so as you can see here circular or regular polygon shape concrete column or pedestals shall be permitted to be treated as Square members of equivalent area polygon equivalent area okay then when locating critical section for moment for share depending on one way share or two-way share and for the development reinforcement into equivalent Square uh column okay guys okay anyway so for when we shallow Foundation everyone guys uh-huh so as you can see the design and detailing of one-way shallow foundations including strip footing combined footing graded beams shall be in accordance with the section and applicable provisions of section 407 and 409. so Bali and provision problem 407 and 409 applied 407 so one way is love is proofs concrete references uh is minimum so foreign no all right so it's all clearness so take notes area so this is okay then times the gross area BH not DB plus so for two-way bending members oh imagine this is an isolated footing so because of the pressure acting at its bottom it bends all right it bends therefore two-way directions so I hope this is clear to us now for the one-way bending member my example is wall footing and this one way is love all right so my pressure is once the Islam is loaded all right okay so I hope this is clear to us as another example of a one-way bending member retaining wall because of the lateral Earth pressure so let us start solving so let's start analyzing solution or sorry situation one this problem guys so a rectangular spread footing supports 450 Square tide column the top of the footing is to be covered with 200 mm concrete basement floor the volume is reinforced with 8 25 mm the total depth of the footing is 800. and with an effective depth of 690 the bottom of the footing is 2.3 meters below the top of the basement floor um okay so service loads meaning and factored when we say service meaning and factored means okay so 2000 dead load 1648 for the live load specific weights or the unit weights Ayan then material properties fitting in okay allowable soil pressure okay so Ayan so as you can see concrete basement floor you know 200 mm um 800 effective depth 690 so this is 800 this is 200 so total of one thousand again the bottom of the footing is 2.3 meters below the top of the basement floor so from the top of the basement floor going to the bottom that's 2.3 meters 200 mm 800 so one meter again 2.3 all right this is the 1.3 meters then 800 mm okay I hope this is clear to us all right for number one dimension of the footing the length of the footing is twice of its width all right so pressure allowable pressure gamma times H2 the depth of soil gamma concrete H Concrete then for the sword charge gamma concrete H1 it is okay so substitute values so this is given young unit weight 18 for soil 24 Newman for concrete then it is okay so the only unknown is an effective soil pressure then an effective soil pressure into I I service into a nominal pressure then divided by the area footing okay length is twice its winds to be tapos as you can see factors all right so as you can see we can now have the B if we have B times we have also the L so what's L 5.6 right times two okay so so we have the dimensions of the footing 2.8 by 5.6 I hope this is clear to us number of 28 mm bars for reinforcement all right so this is the top view of the footing that is 1.2 1.6 for for that load live load respectively foreign resultant force of the pressure to you let's get X how using geometry so l is composed of what 2x then Z 2X and z and Z young Dimensions so for the L this is 5.6 then for the column dimension for 50 mm or in meter 0.45 so X is 2.575 oh well all right no so moment about the cutting plane so uh counterclockwise equals clockwise so I'm counter ICM you unclockwise I F U times the lever arm which is half of x then f is what pressure times area so pressure then times the area where it acts so Ayan simplify all right so substitute the values that we have okay so you say nothing this is the Q U this is the X then the B all right so we have now the MU okay now for the MU is statement all right so simplify nothing but nothing is simplify uh you divide nothing both sides substitutive values so for the moment so times then reduction factor is 0.9 automatically it is all right so simplify substitute the values between 2.8 in mm okay squared tapos Ayan so use the quadratic formulas quadratic formula for so for the B this is the negative 690 for the a this is the point five that was for the constancy this is the 63 000 something again so I hope this is clear to us so we have two values of a chambray and a young depth of compression block less than 800. 800 total thickness come on so I hope this is fair to us on calculator anyway checking so our beta one is 0.85 since F Prime C does nothing since 28.85 now so C is this one then for the fs 600 B minus C over C so as you can see it is greater than F wider for yield greater than one thousand so tension control okay so note FY is less than 1000 megapascal always Force footings slabs and retaining walls foreign so summation of forces horizontal equals zero imagine four C at Force t opposite in direction simplify or re-express so this is for the C this is for the FY sorry for the p f White so check inputs 22 for the F Prime C this is the a Ayan okay the B the f y so my a is kinda okay long okay Lang yes now for the as minimum temperature is number of bars member okay imagine directions foreign young largest among them so it only is nothing so since we have now the final as this is the total as if I pi over for diameter of the bar squared times number of bars 28 mm diameter so n is what uh 20.231 so choices 19 20 24 19 20 24 25. uh okay so Round Up Round Up from the choices all right so bear in this case all right very easy right okay required spacing of longitudinal bars okay for spacing of longitudinal bars so it will win so you know and the spacing is equal to wind divided by n take note that this n should be the exact value not the rounded up you know it's a reviewer [Music] okay so the sweet is the B this is 2.8 meters or in mm to 8 then divided by the exact value of n not the rounded up huh okay so rounded up so this is the S now is 3H or 450 three times the thickness of the footing or 450 mm of course safe and one total thickness by 800. okay so as you can see we have three values of spacing we we are going to use the smallest value so we have the final answer 300 138.348 now from the choices beforehand foreign cage and temperature bars temperature bars or shrinkage bars take notes no shrinkage and temperature bars are not really required since the footing is to be enforced on both ways so by perpendicular bars so shrinkage and temperature bars area okay no actually my statement so this is the as minimum note the as of temperature and shrinkage bars I shrinkage and temperature bars okay this is equal to the as minimum of one-way bending member so one-way bending member okay two-way bending member man so one-way bending member or two-way bending member as temperature bars unit equals a is minimum one-way bending member two-way bending member take note temperature bars all right foreign actually I see 420. send a CP so 2015. now for the FY depended before 275 megapascal 400 megapascal all right 280 280 c400 Asian c415 7cp 2010 all right now 2015.80 less than 420 megapascal at greater than or equal to 420 megapascal for 15. so theoretically speaking 415 at 400 NATO I see 420 corresponding before so 400. all right okay I hope this is clear to us foreign okay so sige so guys okay yeah or main bars okay okay anyway so solve the nothing Ayan okay so gross area no so L times H by the way so 2.8 times temperature bars foreign I hope this is clear to us guys okay uh-huh okay so five six computational gross area foreign [Music] so five six times another spacing my checking pattern s Max is 2.8 meters or two eight mm 2.8 foreign foreign all right okay so values so 77.6 mm guys spacing all right okay so yeah so I hope this is clear to us guys all right so number five what is the wide Beam share is stress so first first problem dimensions of footing second problem is what my second problem uh number of main bars third num spacing of main bars fourth spacing of temperature bars okay fifth wide beam shear stress so two cases so this L is composed of 2m 2D then 1z so 5.6 and the eating effective depth 690 mm or 0.69 in meters all right I am okay reactions okay now to have the shear Force the wide beam Shear Force summation to force is vertical so downward equals upward so opening upward and being upward Force area where it is acting so q u times B times M this is the B the 2.8 then the m is the plan is okay we multiply the the force by 1000 to make it in Newtons dimensions in mm in mega Pascal all right so this is our answer actually okay now um so Tigger check everyone uh since [Music] one assumption normal weight concrete okay now so this is the allowable white beam now as you can see the actual white beam shear stress exceeded the allowable wide beam shear stress therefore the footing will have a wide beam Shear failure so as a designer when he is all right for number six let's solve the punching shear stress so Mama punch okay the situation we're not okay okay so Ayana all right now for the Z plus d Dimension this is the Z this is the D all right now let us get the punching Shear Force summation of forces vertical so upward equals downward so for the downward Force this is the view for the upward Force that's the resultant force of the pressure q u so on respawn Force i q u Times Square yeah as you can see for the area this is the area of the footing minus area of the hole okay Z Plus D Squared something values okay B the L the Z plus d okay so we have now the punching share for smithsonite area so anyone area times so perimeter times the effective depth all right as we can see in the figure so unforce in Newtons so my times one thousand so we have the stress in mega Pascal all right okay now for the calculation of two ways here here [Music] as you can see um the ratio of the long side to short side of the column long side to short side a in our case the column is square so beta is alongside fourth side I equal no okay anyway concentrated load or reaction area and Alpha so the value of alpha sub s Ito is 44 interior column interior column moment for Edge column again then for Corner columns corner foundations orientation Foundation column columns okay anyway so this time interior so Alpha is 40. okay so okay uh again 1.9 Okay so in our case beta is yeah now is square longer side of the column so Ayan list okay so Alpha is 40 as you can see then the effective depth between perimeter non-critical section definition this is the perimeter of the critical section for punching so since four times the Z plus d so as you can see we have three values we will be choosing the list among this so the actual punching stress is exceeding the allowable therefore the footing will have a punching Shear failure meaning okay so redesign 1.9 something I hope this is clear to us guys [Music] all right so again let's move to our next problem so a rectangular footing supports a circular column concentrically dimension of the footing 2.5 by 4 by 0.8 so this is the total thickness 0.8 column diameter 0.5 meter concrete 28 FY 415 concrete cover to centered of the reinforcement 100 mm reinforcements outer face unit weight of concrete unit weight of soil allowable stresses at ultimate loads and so for wide beam action 0.89 for two-way action 1.8 so as you can see young allowable stresses I I given a coincidence that the footing can carry based on beam action concentrated load the footing can carry based on two-way action so it's a one-way little two-way punching no so reverse calculations action okay equivalent square column okay so area of a square equals area of the circular column okay for instance area of circle is z squared whereas area of sorry yeah Square then for the circle Pi over four diameter squared a long diameter and a 0.5 meter or in mm-500mm so Z is what uh take the square root again for the effective depth the total depth is 800 mm minus the steel cover 100 ion so my 700 and it all four four three okay so when we actually one way action okay so concentrated loaded the footing can carry based on white beam action on um longer Dimension so for meters he thought 2.5 so 2.5 meters 700 mm so 4 2 m alone D then the Z so m is this one so Ayana all right now for the allowable stress given it all so this is the reduction Factor this is the 2.5 because it 2.5 by 4 2.5 or in mm25 then the effective depth so oh how much I hope this is clear to us guys okay now having the Vu we can now take summation of forces vertical so downward equals upward so pressure times the area where it acts so B times M substitutive values the M the the Vu so we have q u equals 43.263 kPa now if we will refer to the entirety of the footing we can say that the pressure is equal to pu over area footing footing [Music] or this is the force being carried by the column then pass to the footing foreign for the area of the footing this is four times two point five so four times two point five is ten so Momo moved around decimal point so we have four three three two point six three four kilo newtons This Is The Answer class all right very easy right okay so for the ZD uh sorry okay anyway so this is the Z the equivalent square column Dimension then the D okay okay so stress is what uh force over the area so re Express so substitute the values guys foreign summation of forces vertical so the downward then this is the upward substitute the values so Ayan I hope this is clear to us so this is talk to you so um so we have p u equals one q u times area 14. so yeah move the decimal point we have four nine seventy point four six four all right I hope this is clear to us foreign okay column action load and factored that the footing can carry if the depth of Earth field is 2 meters above the footing so take note of the tattoo meters above the footing the allowable soil pressure at service load is 200 so [Music] um 800. effective so soil times H1 and soil for concrete uh concrete unit weight times H2 then foreign foreign so 14 52. all right I hope this is clear to us guys Foundation Hindi concentric and location next problem nothing a concrete pad is placed horizontal it is subjected to a downward Force Q equals 300 kilo newtons acting centrically with a equals 0.7 meter and b equals 0.4.4 meter the pad is four by three being the longer side as parallel to what sorry as far A little one two as the parallel one to x-axis okay so this is the four meters guys okay okay solve the maximum pressure acting on the pad solve the minimum pressure acting on the pad solve the pressure acting on the center of the pad solve the pressure acting at the location of Force Q okay x-axis okay now as you can see since the center now young effects aside from that my momentum x-axis with respect to x-axis IQ times the lever arm a so Q times a yeah x axis now moment about x-axis so the Q is 300 the a is 0.7 Ayan so are you now so why my moment train you know so I don't know QB come on Q times B itty okay so my computer moment of inertia guys so for the moment of inertia moment of inertia about x-axis perpendicular so CW so Naka l w Cube over 12 okay so annoying L4 W Trina okay for the my pointing y-axis perpendicular so CL so WL Cube over 12 numbers okay CJ so um guys dito foreign all right now for the maximum pressure acting on the pad or actually when we say pressure we can say that that's the stress also pressure is now as you can see we have q over a plus minus m c over I then plus minus m c over I okay see you again by the way three four meters between three meters okay so again taking compression as positive negative okay compression since okay so Ayana positive compression maximum pressure is compression so lagging positive okay so this is the queue then the area of the pad so three by four by four then for the MX a momento [Music] from axis from x-axis going to farthest fibers that's 1.5 okay then for the M1 guys so for the my that's 1 120 then for the C going y-axis going to farthest fiber so that's two meters okay I hope this is clear to us guys so simplify we have 75 okay for the minimum guys for the minimum panel okay for the minimum intentional [Music] okay separately so 300 over 12 so this is 25 guys 25. 210 times 1.5 over 9 so we have 35 okay 35 now Cesar 120 times 2 over 16 this is 15. okay detention compressions 35. it on positive 15. 35 negative negative 35. maximum okay uh um 75. okay uh positive five comma positive five positive 5. so my negative 10 my negative 25 . compression five megapascal into since negative attention thank you um side view by compressions calculations I hope this is clear to us foreign okay pressure acting on the center of the band all right so for the pad for that for the if the for the center of the pad sorry um distance from the axis zero uh foreign uh tension MX tensions MX is [Music] foreign I'm not why uh intentions anyway compare compression same X body compressions MX foreign compression okay I hope this is clear to us okay location of the force queue so originally foreign from x-axis and distance MC over IIA is again so MX X A and C and you see young distance all right okay uh by the way okay okay so I hope this is clear to us guys so this is the answer maximum pressure compression cmx compressions all right problem okay so it's a full word problem my moment provided an examiner guys all right Hindi provided okay so a rectangular footing is subjected to a factored actual load three thousand kilo newtons Factor the factored moment 600 kilonewtons okay the effective depth of the footing is 650. column diameter is 3680 footing is three by five so this is the three young shorter this is the five young longer total depth 650 750 effective okay so as you can see thingama my M Yuka so minimum um dimension of perpendicular axis BH Cube over 12. yeah okay into equivalent Square so compute the critical nominal one-way shear stress okay so shear stress okay stress okay so equivalent square column okay column diameter moment foreign I'm up C CMU compression negative foreign from axis of moment going to the farthest fiber that's 1.5 and half of B so three by five half of B so 1.5 M over BD squared six M over b d square is the strength MC over I para General okay so first term 200 second term 80. okay so negative negative 280 kPa this should be kPa anyway so for the Q minimum so on pu negative compression purpose yeah so negative 11 120 kilopascal NATO foreign so what will happen uh positive positive 600 kPa so positive compression is so yeah so since compression and so compression tension so um open column s and zero taposi so from from the axis of the moment or a new distance so this is 1.5 meters okay are you right so see ing triangular portion compression um all right so number 10 what is the critical nominal white beam shear stress so in this case actually um this time as you can see dimensions [Music] nothing so L is composed of 2m 2D then Z as you can see now for the D given Ito 650 equivalent square column Dimension young L5 Casino okay so we have now the m for this uh B is composed of 2N and 2D then Z so substitutive values we have the N okay okay guys as you can see this is the B Ayan n it is [Music] Q over a anyway portion right now I hope this is clear to us now guys can you talk foreign going to this point it I as you can see what half of the plus d so half of Z plus d okay okay substitute values this is two hundred and a half of Z plus d then the I so we have this using similar triangles convenient only right so summation of forces vertical equals zero so downward equals upward the downward for the downward force is v u the upward force is the result of measure now actually for the resultant of the pressure so that's the four that's a force right so pressure times area but this time Independence [Music] pressure diagram okay so on downwards upward Force I equals a volume now pressure diagram volume solid geometry foreign foreign [Music] so we have one house base q u Max then base again times n so area times L times Max plus 2 Q U1 over 4. okay so I hope this is clear to us [Music] okay so q u Max a previous slide it is okay so this is the this is the punching sorry not punching wide beam Shear Force seconds two so for the stress okay so this is the force stress young area a young area so L by effective depth okay so times one thousand para Newtons 0.75 for share okay L is one thousand is five thousand in mm then D 615 mm so we have this okay for the case one is okay okay [Music] [Music] okay so summation of force is vertical the downward is Vu the upward is the volume of the pressure diagram again trapezoid times the wind M okay ICB so one half bases are Q Max Q Min tapos times b t with M okay compute okay so disregarded and science huh okay um by the way okay so this is the the video now for the area [Music] okay so it uh now as you can see it 0.653 for case one four case to 0.299.299 again guys in this case a moment um Direction a momentum Direction yeah moment maximum purely maximum Union critical section for white bib but in this case orientation moment all right anyway so okay so are you now consider the largest so most critical there for the white beam sure stress is 635 megapascal all right okay critical nominal punching shear stress um cases what is considering the the force load only um because of the moment right Square then they of D plus Z plus d okay anyway so by the way from face of the column B over to distance I call them Dimension plus a total of d all right so for the Q U2 my Q U2 my Q u3 for the Q U2 I know distance near from the axis of the moment as you can see half of Z plus half of d so half of Z plus half of D this is still 200. m c over I no okay uh by the way to compressions okay so negative bone compression Plus numerically okay okay now depend this uh um free body diagram better this time and portions portions okay anyway s summation of forces vertical so upward equals downward for the upward civil union pressure diagram pressure diagram okay now I'm downward detail you know downwards you know CPU as you can see by the way are some volume one half the places are due to Q3 then Z plus d as you can see huh foreign as you can see guys as you can see we we get the sum of Q2 and Q3 it's called Q2 and Q3 or actually after we sum them up average Young 200. MC over i m c over I well I can see security uh Q2 Q U2 plus q u 3 divided by 2 so we are taking the average of them 200 kilo pascalion or actually you Max at Q you mean average Nila average [Music] 280 plus 120 that's 400 divided by 2 200 okay anyway so um um foreign all right foreign foreign [Music] okay apart the Z plus d then 650 we have now the answer so I hope this is clear to us guys okay all right so critical moment now critical moment now for case one uh as you can see case one is [Music] whereas Magneto is uh actually okay moment o you know nothing consideration okay now as you can see uh surface distance nothing so for the L guys for the L uh let us solve the X so L is composed of two x and z okay so five to X then the Z okay guys f u one times its lever arm half of x you know so negative volume and pressure diagram volume so qmax times okay so yeah volume down pressure diagram so one half q Max Q Min times B tapos times x foreign okay anyway so my w okay my w okay so negative p over a then since it is F YouTube the height is W you know one half base times height is okay so substitutive values this is the Q U for the W the L okay for the Q U for the fu3 one half the base is q u Max the height is W1 so l okay so we have a few three taking moment about the face of the column so for the f u two its distance is okay centroid of triangle panel um if Futo that's one third of w then for f u three that's two turns of w so that's geometry plus so fu2 again one third of w then fu3 two-thirds W okay okay so substitutive values so you have the MU critical sorry for the critical section two so 929 whereas this one 900 okay so much um okay so I hope this is clear to us guys one is okay anyway guys I I moment I critical moment moment about the critical section reinforce reinforcement and footing problem foreign situation five all right so in situation five uh due to the presence of ground water table and very weak soil strength the structural engineer will use footing on piles five files will be utilized one directly below the column analyze the foundation okay so columns actually load four thousand pile cups moment load Ayan uh 600 kilonewton meter for the column 500 by 500 for the pile cup guys for so for the pile cup 3.4 meter by 3.4 meter with thickness of 0.75 effective depth 600 mm spacing of piles 1.6 meter ion okay compute the one-way shear stress compute the two-way shear stress compute the critical moment okay let's talk about Pilots guys Pilots okay so so by the way so for the force being carried by the column negative compression for the moment compression is area of one pile only area of a pile so my times number of piles moment of inertia of group piles centroidal plus area D Squared we can consider the centroidal moment of inertia that is just Pi D to the fourth over 64. not over 32. over 32 Jae young polar moment of inertia over 64. we can neglect it so that the remaining will be summation of area D Squared the lamb okay so for this summation area foreign stress pile area pile so it will pile load so the file load the load being carried by a pile is equal to this expression okay I hope this is clear to us guys okay by the way area D Squared um distance times four okay see you uh 1.4 is a problem spacing 1.4 meters so 1.4 squared times four so we have 7.84 all right okay wide Beam share stress okay white beam sure is the distance from the face of the column right so from the face of the column as you can see distance this is the effective depth of the file cup provided regions a problem effective depth 600 mm or 0.6 meter okay I hope this is clear to us guys now buckets foreign Okay so actually nothing is CMU foreign 4 negative pu Over N negative Casey compression and mu compression so negative red m-u-s bucket s can see from the axis of the moment going to the centroid of piles one and two the distance is s so m u s over the I okay provided return problem then divided by five piles foreign so we have this all right now guys [Music] so negative meaning okay I hope this is clear to us okay so for the pile five R5 so zero over summation discrete okay now okay summation of forces okay I hope this is clear to us okay so okay so for the one-way share okay not punching sorry white beam or one-way Shear Force summation of forces vertical obvious that v u is equal to R1 and R2 R1 and R2 okay [Music] you know 3.4 Ang effective depth 600. so times so times one thousand para Newtons then times two reduction Factor okay and B three four six hundred for the effective depth okay so we have now the answer 1.186 okay okay so for the punching here is this Solution One summation of forces vertical equals zero so plus effective depth so that's 500 plus 600 so one one one one salvation of forces vertical so downward equals upward uh and downward equals R1 R2 R3 R4 plus High R5 okay so the one R1 or R1 times two iron four n times two n so we have three two more so summation of force says vertical equals zero and upward i v u at Chaka R5 foreign [Music] okay 500 600 times 4 iron 0.75 okay so Maritime answer okay I hope okay now critical moments okay so for the critical moment at the moment towards that uh young R1 R2 times lever arms forces then okay so Pandora okay okay so s n spacing 500 or 0.5 in meter so we have now the Z so for the moment that's R1 plus R2 times the Z um okay so this is the critical movement so I hope this is clear to us guys okay uh are you right so for the combined footing an exterior column with service okay board exterior column so we thought service that load 800 service live load 580 so extract 800 580 and an interior column with service deadload one one service live load 750 are to be supported on a combined rectangular footing as shown in the figure the center to Center spacing of the columns is six meters and has entered to Center 6 meters the base of the footing is two meters below the ground surface which carries a surcharge of 5 kPa in exception half the thickness of the footing is 1.3 the allowable bearing pressure is 300 unit weight of concrete unit rate of soil iron okay so number 20 calculate the effective bearing soil pressure calculate the distance of the resultant load from the outer face of the exterior column okay anyway effective soil bearing pressure so as you can see 800 580 ion 11750 ion okay sore charge 5 kPa thickness of footing bottom Nito I don't know two meters below the ground surface from 1.3 2.7 QA 2418 okay all right so effective pressure so QA is composed of one soil concrete sore charged and effective pressure ayos now for the QA it's 300 for the soil unit with soil depth of soil for the concrete unit weight of concrete depth of concrete for the sore charge five then effective so we have now the answer okay I hope this is clear to us 21. for 21 location of the resultant load guys variation of the resultant load from the outer face of the exterior column so anything exterior column Point distance between the column Center to Center guys six meters stated a while ago in the problem um for P1 it's on the lungs exterior so 800 580. foreign moment go from Statics the moment produced by the resultant is also the moment produced by its components so by the time will take anywhere moment produced by the resultant is equal to the moment produced by its components it's a point where P1 is acting no P1 so this is is acting going to this as you can see is x sub D time minus point two eating lever arm equally position [Music] and lever arm six meters okay substitute values so we have XT resultant load from the exterior face of the column so this is our answer I hope this is clear to us guys all right so number 22. okay so minimum dimension of the footing okay see that another one actually is [Music] foreign [Music] paper for the length is 2 times XD Cayenne Ayan okay now how about for the wind for the wind if you if we will use the QE QE is equal to the r the force being carried divided by the area of The 14. W Times l so we have the QE the effective soil pressure this is the resultant load the the the the L then the W so we have no W so dimensions these are the dimensions of our combined footing I hope this is clear to us foreign anyone problems ultimate impressions are you over area who in the first place ultimate [Music] okay so I hope this is clear to us [Music] all right okay continuous wall loading red okay file loading button okay so a continuous a continuous wall there is a load of 7 570 kilo Newton per meter which coincides with the piles situated in the middle row so as you can see um this wall is supported by three rows of files dimensions are in centimeter okay calculate the pi load in row a Kyle calculate the pi load in row C again calculate the pile load in row b okay locate the centroid of the group of iOS since continuity so when we say tributary and reference foreign foreign uh foreign guys foreign okay anyway okay so again baby area so 200 200 yeah yeah so 200. all right now uh if you will consider this Ayana Let's uh locate the centroid so using varignon's theorem total area times x c g uh CG stands for center of gravity so A1 y1 Plus A2 Y2 plus A3 Y3 continuous Okay so so five times area pile you know for the A1 distance is okay so for the area I am sorry 2A times 100 okay distance okay so components equals resultant then as you can see we can cancel out the A and the tall area of piles x sub G as 80 cm okay atcm all right now 80 cm in meter a non-equivalent known point eight meter so home um 1.2 neutral axis as you can see um in foreign [Music] [Music] area [Music] distance I point eight squared okay tapos is the I okay now for the stresses um so for the force a moment I 570 times 0.2 so results okay so as you can see 570 times 0.2 results okay I hope this is clear to us now for the stress a that is for the stress a this is the load divided by the area so anong total area limang piles okay then compression I positive tensions so the moment M then young distance [Music] okay okay all right now as you can see uh multiplied more stress times area is equal to force okay stress okay Lang I hope this is clear to us so actually foreign foreign so I am 114 divided by five then since then so moment then young distance 1.2 then young area my area can sell so this is the the pilots see then for the bee we are taking compression as positive distance 0.2 as you can see so times area cancel so my Force connected so again guys I am so I hope this is fair to us guys Foundation problems guys so all right thank you