Design of Long Coumn | RCC Structures | Mumbai University | IOE , TU , PU

hello guys and welcome to another video in the video series of RCC and in this video we'll be discussing about a numerical related to long column so we have a question here design a column with the following data that is B into D data dimension 500 into 500 that is a square column clear respond L is 8 meter le X and Lu a is given as 6 7 meter and 6 meters respectively that is effective length along x axis and y axis similarly moment along both axis is given as 120 and 170 respectively X 0 tu is given as 2500 and f-ck and FY are given as 25 and 500 Newton per mm square respectively so solution step 1 calculation of slenderness ratio obviously for a long column the column must be slender l affected by least lateral dimension and we have to check it for both effective length along major axis and minor axis that is x and y respectively so for lambda x le x divided us draw a figure here at drawing the axis x and y so along x axis will be keeping D and along y axis will be keeping B so both are 500 so that doesn't make any difference here and le X is given as 7 divided by 0.5 so this comes as 14 and this is greater than 2 well and lambda y is equal to le y by B that is dimension of that axis and we have 6 divided by 0.5 this comes as 12 and which is less or equal to 2 well hence what we find is that it is long column or it is slender along x axis but not along Y axis hence it is designed as a long axis a long column along major axis and short column for the minor axis that is y axis now for the eccentricity okay we have to check for eccentricity and that is calculated as e^x and ey both so we have here e^x as L by 500 plus d by 30 so which comes as thirty two point six seven and both who are same because the value of D is same for both X's and the value of 0.05 D and 0.05 B is calculated which is 25mm and what we find here is that e^x and ey both are greater than 0.05 B and 0.053 respectively hence a centrist should be considered since e ex and ey a both are greater than 0.05 G and 0.05 B respectively hence it is a biaxial loaded column now moment due to eccentricity any X that is calculated as PU into ey and m ey is calculated as PU into ax and both of them comes as a 2-1 point 675 kilo Newton meter therefore M X and M ey is equal to 81 point 675 kilonewton meter also mu x is equal to 120 kilonewton meter which is greater than M E X and mu y is equal to 170 kilo Newton meter which is greater than M e Y hence will be taking greater moment for designing purpose hence design moment due to axial loading is mu X is equal to 120 kilonewton meter and mu Y is equal to 170 kilo Newton meter also for long column additional moment due to buckling is considered okay but in this case we have long column along x axis only so additional moment along x axis about Y axis is considered so we have to calculate ma X that is equal to PU D divided by 2,000 le X by D whole square substitute the value and this value is calculated as one twenty two point five kilo Newton meter then for total moment as we have already discussed in long Coulomb introduction M DX is equal to design moment along X and design moment along y mu X plus K into ma Y and mu y plus K into ma x 4k K is equal to PU at minus PU by PU at minus PB and for PB which is a new term here in long column so from SP 16 page 171 you can find the value of PB by using that equation so we get P by f-ck BD is equal to k1 plus k2 P by f-ck then for P is equal to percentage of still taking 2.5 percent as assumed value D dash by D where D dash is the effective cover and D is the effective depth so 50 by 500 is equal to 0.1 meter providing D dash as 50 mm and we get k1 is equal to for a rectangular section having the ratio of D dash by D as 0.1 the value of k1 is taken as zero point two zero seven so P 1 is equal to zero point two zero seven and providing a for fist Coulomb k2 will be calculated as four D dash by D the value of 0.1 and a four phase rectangular column F I as 500 a 2 will be taken as zero point four two five now PV is equal to from the above equation just arranging the variables f-ck B into D into K 1 plus K 2 into P by f-ck substitute the value 25 500 square because B and a are same K 1 is zero point two zero seven k2 is zero point 4 2 5 and P is taken as two point five percent f-ck is 25 so this value comes as one five five nine point three seven kilonewton from is code four five six puz as we have already done P 0.45 f-ck AC plus 0.75 FY is 0.45 into 25 into 100 minus 2.5 by 100 into 500 square okay plus 0.75 500 into 500 square because area so into 2.5 percentage so this comes as five thousand eighty five point three nine seven kilonewton also K is equal to P you Z minus PU by PU Z minus PV and this is five thousand eighty five point nine three seven minus twenty five hundred divided by thousand eighty five point nine three seven minus fifteen fifty nine point three seven this comes at zero point seven thirty three less or equal to one okay now design moment M DX is equal to MU X plus K into ma Y which is equal to zero so we'll be taking the value of MU X only which is 120 kilonewton and MD Y is equal to MU y plus K into ma X and we have calculate the value of M ax 1 1 is 170 plus zero point seven 33 into two one twenty two point five this comes as two fifty nine point seven nine kilonewton meter now from interaction diagram taking D dash by D as 0.1 pu by fckbd 2500 by twenty five into 500 square we get this value as 0 point 4 P by f-ck two point five by 25 we have zero point one so from the interaction diagram for FY as five hundred and D dash by D is at zero point one we can use chart 48 and we have the value for the ordinate that is pu by fckbd as zero point four and for P dash by P by f-ck is zero point one we can find the value for the abscissa that is a value of moment ratio so this value coincide at this point and from the graph we get the value of mu X comma one f-ck B Square D mu Y comma one also same so BD square that will the difference 0.135 so mu X comma 1 and mu y comma 1 this comes as 4 106.3 5 kilonewton meter for by Xillia loaded Coulomb the condition has to satisfied that is mu x comma 1 by mu X to the power alpha N plus mu y by mu a comma 1 alpha n should be less or equal to 1 and for alpha we have to find the ratio of P you by puz and this comes at 0.49 1 and from the code the value of alpha ranges from 1 2 to 4 bu by P use it as zero point two to zero point eight so we get the value of alpha as one point four eight five from the interpolation then mu x by mu X comma 1 to the power alpha n plus mu Y by L naming Y comma 1 alpha less or equal to 1 so substrate the value that is four hundred twenty one point six seven please correct here for twenty one point six seven this comes at zero point six four one which is less or equal to one okay now for reinforcement P percentage is as two point five percent ast percentage two point five percent of BD this is calculator six two five zero mm square area of Steel is much more greater so we have to arrange a different type of bar that is various sizes of bar so providing four numbers of 32 mm bar four numbers of okay eight number of 28 mm bar so let us take 25 mm okay as there is a greater value so area of steel is 6 2 5 0 that has to be provided and the area of steel that will be provided with the arrangement of 4 and 8 number of bars 4 into PI into 32 is square by 4 plus 8 into PI into 20 a v square by 4 so this comes as 7 1 4 3 point 9 8 mm square which is greater than 6 2 5 0 so it is okay the longitudinal bar has been done now for lateral Tai's diameter of lateral type should be greater or equal to 1 by 4 times of Phi L max so taking the greater dimension of bar the diameter of bar so that is 32 this comes as 8mm should be greater or equal to 6 mm so according to the code the greater value is taken so providing PI T as 8 mm for peach should be less or equal to 16 times of Phi L taking the minimum size of bar that has been provided that is 25 so this comes as 400 mm so do less or equal to least lateral dimension that is 500 mm should be less or equal to 300 mm so taking the list value taking peach as 300 mm now let us draw the reinforcement designing detailing so for numbers of bar taking the for number of 32 mm bar at the corner and the other bar at middle 8 number of 25 mm bar providing a cover and for exercise Twp check along x axis spacing between bars that is 500 along x axis and effective cover of 50 so we have to find the spacing between bar so 500 minus 2 into 50 okay divided by there are 3 number of spacing so this comes at 1:33 point 3 3 which is greater or equal to 75 mm then extra stirrup is required now for the type of steel I will directly provide a closed type of steed up okay you can make a check whether to provide a closed or open I'm directly providing a closed I was stated I have done in the previous videos regarding the type of Street up to be provided in spacing between bars in y axis same which is greater than 75 mm hence a closed stearic will be provide along y axis also thank you

Channel: Parash Joshi - Civil Construction and Tutor

Views: 10,455

Keywords: IOE, Tu, Pulchowk, Nepal, Engineering, Design, TOS, Structural, Analysis, Autocad, Etabs, Course, Survey, Camp, Numerical, Methods, Irrigation, Civil, nepal, TCES, civil, er, tutor, design, of, rcc, Singlyreinforcedbeam, doubly, reinforced, beam, over, reinofrced, column, biaxially, loaded, bracedcolumn, thecivilguy, machenlink, mumbaiuniversity, long column numerical, design of long column, rcc long column, slender column

Id: twjxLqeK4fA

Length: 12min 28sec (748 seconds)

Published: Tue Apr 23 2019