Design of Column in Limit State Method

Design Of Column In Limit State Method The Design of R.C.C. Columns is very important, as in the Framed Structured Building that is in the modern day construction the Frames are formed by networs of Columns and Beams. ! Column is a Structural "ember which is vertical and a#ially loaded and sub$ected to Compressive forces, and having it%s &ffective 'ength ()eight* Three times greater than the least cross+sectional dimension of itself. To To Satisfactorily Complete Complet e the Design of Columns The Following Steps must followed.

STEP 1 :- Calculation of the Inuence Inue nce Aea of the Columns : The rst step is to nd out the -nuence !rea of the Column to be Designed. The -nuence !rea of a column is the area of which load is being transferred to the column to be designed for. For this purpose in a framed structure small and medium building the design of column is done for the column whose -nuence !rea is the largest hence the load coming on the column will be so the greater of the any other column in that building hence all the other column having lesser -nuence !rea hence lesser 'oads if provided with the same Designed parameters that re/uired for the column having largest -nuence !rea, then the whole Structure will automatically become safe against the 'oads.

DETERMINATION DETERMINATION OF INFLUENCE INFLUENCE AREAS FOR LOAD DISTRIBUTION ON COLUMNS

STEP !:- Calculation of the Loads Coming on Columns Colu mns fom the Inuence Aea : -n this step the 'oad Calculation is being done. This is done by calculating all the loads acting within the inuence area.

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The 'oads acting are broadly classied as Dead 'oad (D'* and 'ive 'oad (''*. Dead 'oads are the load of ob$ects which cannot be moved from on place to another lie the loads of Bric 0or, Beams, Slabs etc. and the 'ive 'oads are the loads coming from movable ob$ects such as )umans, Chair, Table etc. Thus 0e 1eed to Calculate the Dead 'oads as well as 'ive 'oads within the -nuence !rea, these are as follows in the general case of a Building 2+ A)Dead Loads : Due to weight of Slab 345666 17m3 8 Due to weight of Floor Finish 3566 17m28 Due to weight of Bric "asonry 39:466 17m38 Due to weight of Beam 345666 17m38 Due to weight of Self 0eight of Column 345666 17m38 B) Live Load : -t depends upon the 1ature of the Structure, and it values for different structural nature are given in the concerned Code of 5+
COLUMN FORM WORK OF BUILDING

1ow after correct calculation of above loads the Total 'oad is Calculated by, Total 'oad on each oor A Dead 'oad  'ive 'oad 1ow this the actual load which will be acting on column for each oor, now if the building say 5 storied, then $ust multiply the value with the nos. of oors, lie for ve storied building multiply the Total 'oad on each story with 5. 1ow thus the Total load acting on column at Column Base is btained and it is denoted with <%. )ence
STEP " :- #inding The $oss Coss-Sectional Aea %e&uied #o The Columns : This is the one of the most important and main step of the Design of Columns. First in the 'imit State "ethod of Design we must increase the load acting on the column with a 'oad Factor so that if there will be any accidental increase of loads the column will be still safe to resist the load without a failure. The Factor of Safety for Dead 'oad  'ive 'oad Combination is 9.5, hence we must multiply the load action on column (<* with the 9.5 to obtain the ltimate 'oad that is the Factored 'oad of the Column that is < u )ence Factored 'oad, < u A 9.5 E < For Design we will wor with this value of load. 1ow before going on -%m here to say that - will design according to the Code f .f#.!s! 3&/uation -8 0here, .f#.!s! 3&/uation --8 1ow !ssume the
Loading 'Pu( in ) Satisfacto* Design

Pecentage Of Steel  fo

Below 456666 ++++++++++++++++++++++++++++++++++++++++++++6.=K 456,666 to 566,666 ++++++++++++++++++++++++++++++++++++++9.6K 566,666 to >56,666 ++++++++++++++++++++++++++++++++++++++9.5K >56,666 to 9666,666 +++++++++++++++++++++++++++++++++++++4.6K 9666,666 to 9566,666 +++++++++++++++++++++++++++++++++++4.5K 9566,666 to 4666,666 +++++++++++++++++++++++++++++++++++L.6K !nd so on, with increase of each 456,666 1 increasing the .f#.6.69!$ Therefore, if we now the Jrade of Concrete and Jrade of Steel to be used and Factored 'oad coming on the Column and !ssuming the
!lso Rectangular Column Sections Can be made by using different proportion say b 2 D A 9 2 4 , )ence DA4b , Therefore, ! $ A b E D A b E 4b A 4b 2 or bA

)ence D can be also determined as DA4b after Calculating the b. "ost of the times after calculating the sides of a column it will give results such as 9:H.59mm or L4L.H4 etc. values, which practically cannot be provided at eld, hence we must increase those values to the nearest greater multiple of 45mm (i.e. 9 inch*. For e#amples a value of 9:H.59mm may be increased to 466mm or 445mm or 456 mm even, and a value of L4L.H4mm may be increased to L56mm. more it will be increased the more it will be safer, but it is uneconomical to increase by a very high amount, it should not be increased more than by >5mm to consider the economical factor.

STEP + Chec, #o LongShot Columns:

Depending upon the ratio of &ffective 'ength to the 'east 'ateral Dimension of a column, a column may be classied as 'ong Column and Short Column. -f the value of this ratio is less than 94 then it%s called as a short column and if the value is more than 94 then it%s called as a 'ong Column. ! short column mainly fails by direct compression and has a lesser chance of failure by bucling. !nd in the case of a long column the failure mainly occurs due to the bucling alone. 'ong column being slender, that is being thin lie stic as compared with its length it grows a tendency to get bended by deviating from its verticality under the action of loads. Due to this tendency of long column to get bucled (bended* a long column of all same properties and dimensions that of a short column will be able to carry much lesser load safely than that of the short column. Suppose a G66mm#G66mm short column can tae a load of 9666@1 , then a long column of G66mm#G66mm having same grade of concrete, same amount of reinforcement and same wormanship will be able to carry a lesser load lie say about =66@1 only, hence we get a loss of 466@1 which is 46K loss of load carrying capacity. So the above formula used in Step L holds good only for the Short Column. For using it in long column a little modication is needed, which - will update it later when - will get hands on this article again. For now let us concentrate on Short Column. First of all we need to nd out the effective length of a column, which can be obtained by multiplying a factor with the actual unsupported length of the column. The factor depends upon the end condition of the column. -n most general cases we use a Both &nd Fi#ed Column for which The Factor is 6.H5. Therefore, &ffective 'ength A &ffective 'ength Factor (6.H5* # nsupported 'ength (l*. suppose a column has a unsupported length of 4.>m A 4>66mm, hence the effective length will be lef A 6.H5#4>66 A 9>55mm. 'east lateral dimension means the shorter of the two dimensions of column that is length and breadth. But in case of a circular column as there is only diameter, hence we will use the diameter. Suppose a column is of G66mm#466mm section and has an unsupported length of 4>66mm, then the Ration of &ffective length t the 'east 'ateral Dimension will be as follows 2+ (&ffective 'ength7'east 'ateral Dimension* A (lef7b* A (9>557466* A =.>>5 which is less than 94 and hence is a Short Column.

STEP .Chec, #o Eccenticit* : &ccentricity means deviating from the true a#is. Thus an &ccentric 'oad refers to a load which is not acting through the line of the a#is of the column in case of column design. The eccentric load cause the column to bend towards the eccentricity of the loading and hence generates a bending moment in the column. -n case of eccentric loading we have to design the column for both the Direct Compression and also for the bending moment also.

 T%e &e'(issi*e e!!e+,'i!i,# (us, e $'ea,e' ,%a+ o' e-ua* ,o ,%e a!,ua* e!!e+,'i!i,# o. ,%e !o*u(+ O' e*se /e %ave ,o desi$+ i, .o' e+di+$ a*so

STEP / :- Calculating The Aea Of Steel %e&uied : 1ow the !rea of Steel Re/uired !s! is to be calculated from the ! $ as the predetermined percentage of !$. For e#ample if the Jross+Sectional !rea of the Column is >=H66 mm2 and at the starting of calculation of !g it was assumed that 9K Steel is used then we get, !s! A 9K of !$ A 6.69!$ A 6.69 E >=H66 A >=H mm2 1ow we shall provide such amount of Reinforcements that the Cross+Sectional !rea of the Reinforcement provided is &/ual to or Jreater than the Cross+Sectional !rea of Steel re/uired above. )ence in the above case we shall
)ence the !rea of Steel
)OTE : The minimum of + )os0 of as to 2e 3o4ided at the fou cones of a ectangula o S&uae Columns and minimum diamete of as that to 2e used is 1!mm Diamete0 5ence + )os0 of 1!mm Diamete as ae must in an* Columns ies3ecti4e of thei necessities0

STEP 6 :- Detemining The Diamete and S3acing Of The Lateal Ties: -n this step we will Determine the Diameter and the Spacing of the 'ateral Ties or Transverse 'ins or Binders. The Diameter of the Ties shall not be lesser than the Jreatest of the following two values 0 1(( 2 0,% o. ,%e Dia(e,e' o. ,%e La'$es, Dia(e,e' Ba' For an e#ample if a Column has 9Hmm and 46mm both types of bar as 'ongitudinal Bars or main Reinforcement then 97G,% of 46mm A 5mm )ence we shall provide 5mm diameter Ties. The Spacing of Ties shall not e#ceed the least of the followings three values 0 Leas, La,e'a* Di(e+sio+ 2 04 Ti(es o. ,%e Dia(e,e' o. ,%e S(a**es, Dia(e,e' Lo+$i,udi+a* Ba'

3 5 Ti(es o. ,%e Dia(e,e' o. Ties For an e#ample ! Column of 456mm E L56mm Dimension having 46mm and 9Hmm Diameter 'ongitudinal Bars and 5mm Diameter Ties we get, •

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Leas, La,e'a* Di(e+sio+ 6 217(( 04 Ti(es o. ,%e Dia(e,e' o. ,%e S(a**es, Dia(e,e' Lo+$i,udi+a* Ba' 6 04 8 04 6 214(( 5 Ti(es o. ,%e Dia(e,e' o. Ties 6 5 8 1 6 227((

)ence