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DESIGN OF AXIALLY LOADED COLUMN

Wherever columns and beams are rigidly connected, the columns will be subjected tobending moments. Even when the beams are simply supported over the columns, attimes, bending moments are developed in columns due to partial fixity. When the effectof all the beams connected to the column results in zero net bending moment, thecolumn will be acted upon by only axial load with no moment. Such columns are calledas “ Axially loaded columns”.In any structure the columns under the following situations are treated as Axially loadedcolumns.Case (i):

In the fig above, the column C1 is connected to beams B1,B2, B3 and B4 as shown. If the bending moments on the column due to beams B1 and B3 are equal then there willbe no net moment on the column along Y – Y axis. Similarly, if the bending moments onthe column due to beam B2 and B4 are equal, then the net moment along X – X axis willalso be zero. Therefore the column will be acted upon by only axial load with nomoment, such columns are designed as axially loaded columns.In general internal columns in a structure come under this category.Case (ii):

In the fig the column C2 is connected to beams B1 and B2 as shown. If the bendingmoments on the column due to beams B1 and B2 are equal, then there will be no netmoment. Such columns are designed as axially loaded columns.

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The above eqn. accounts for minimum eccentricity of (i) 20 mm (ii) ( l/500 + D/30)whichever is larger but not to exceeding 0.05 times lateral dimension of the column.In case of short columns with helical reinforcements are used instead of laterals, thestrength value given by eqn. may be multiplied by 1.05, provided in such case the ratioof the volume of helical reinforcement to the volume of the core shall not be less than0.36 ( Ag - 1) f ck

Ac f y Where Ag = gross area of the section of column

Ac= area of the core of helically reinforced column measured to the outsidediameter of the helix.

Type I Problem :

Data known:Unsupported length of column =4.0 mEdge condition : Both ends fixed (lef = 0.65 l)Concrete Grade : M25

Steel Grade : Fe415Column Section : 400 x 600 mmSteel Reinforcement : 8 Nos. of 20 Φ

Design (What to be obtained ) : To find Ultimate Load carrying Capacity of Column

Step 1: Check whether the column is short or slender l = 4000 mmlef = 0.65 l (as per Table 28 of IS 456 – 2000)

= 0.65 x 4000 = 2600 mmFor column to be short lex = 2600 = 4.33 < 12

D 600

ley = 2600 = 6.50 < 12b 400

Hence the given section is a short column.

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Step 2 : Check for minimum eccentricityexmini. = Greater of ( lex + D ) and 20 mm = 2600 + 600

500 30 500 30= 5.2 + 20 = 25.2 mm and 20 mm

exmini = 25.2 mm.

eymini. = Greater of ( lex + b ) and 20 mm = 2600 + 400 500 30 500 30

= 5.2 + 13.33 = 18.53 mm and 20 mmeymini = 20 mm.

0.05 D = 0.05 x 600 = 30 mm > 25.2 mm ( = exmin)0.05 b = 0.05 x 400 = 20 mm = 20 mm (= eymin)Hence, the equation given in Clause 39.3 of IS 456 – 2000 is applicable for the design.Step 3 : Ultimate Strength of Column

METHOD 1 : USING EQUATION GIVEN IN IS 456 - 2000

Formula used : Pu= ( 0.4 f ck Ac+ 0.67 f y Asc ) - ---- (1)

Pu = { 0.4 x 25 x ( 600 x 400 – 2512) + ( 0.67 x 415 x 2512) } x 10-3

= ( 2374880 + 698461.6 ) x 10-3

= 3073.34 KN

METHOD 2 : USING CHART 25 OF SP – 16 or EQUIVALENT TABLE C5Pt = 2512 x 100 = 1.047

600 x 400From Table C5 for M25 Pu = 0.5122

Ag f ckPu = 0.5122 x 25 x 600 x 400 x 10-3 = 3073.2 KN

METHOD 3 : DIRECT TABLE (READY RECONER )

From Table C10 for Column 400 x 600 mm with 8 Nos. 20 ΦUltimate Load carrying capacity of Column Pu = 3074 KN

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Type II Problem : To design the column section for the given Ultimate Load

Data given :Ultimate Load Pu = 3000 KNUnsupported length of Column = l = 4000 mmEffective length of Column lef = 0.65 l =2600 mmConcrete Grade = M25

Steel Grade = Fe415

Design (what to be obtained ) :

Alternate 1 : Assuming Column section and to find steel reinforcement

Step 1 & 2 same as above

Step 3 : Area of Steel

METHOD I : USING EQUATION GIVEN IN IS 456 – 2000

Formula used : Pu= ( 0.4 f ck Ac+ 0.67 f y Asc ) - ---- (1)Pu= 0.4 f ck {Ag - Asc}+ 0.67 f y Asc )

3000 x 103 = 0.4 x 25 x {(400 x 600) – Asc} + 0.67 x 415 x Asc

= 2400000 – 10 Asc + 278.05 Asc Asc = (3000 x 103 – 2400 x 103) = 2238.39 mm2

268.05

Main Reinforcement : Provide 8 Nos. 20 Φ RTS all round the column

METHOD 2 : USING CHART 25 OF SP – 16

Assume column section as 600 x 400 mm. Ag = 2400 cm2

For Pu = 3000 KN & Ag = 2400 cm2 &

Pu = 3000 x 103 = 12.50 N/ mm2 Ag 600 x 400

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Pt from Chart 25 of SP – 16 = 0.95 %

Ast = 0.95 x 600 x 400 = 2288 mm2

100

Provide 8 Nos. 20 Φ RTS all round the column (Ast = 2512 mm2 )

METHOD 3 : TABLE C5 EQUIVALENT TO CHART 25 OF SP – 16

Pu = 3000 x 103 = 0.5Ag f ck 600 x 400 x 25

From Table C5 for M25 Pt = 0.90 + (1.0 – 0.90) x ( 0.50- 0.496)(0.507 – 0.496)

= 0.90 + 0.036 = 0.936 %

Ast = 0.936 x 600 x 400 = 2246.40 mm2

100

Provide 8 Nos. 20 Φ RTS all round the column (Ast = 2512 mm2 )

METHOD 4 : DIRECT TABLE (READY RECKNER )From Table C10 for Ultimate Load carrying capacity of Column Pu = 3000 KNColumn section : 400 x 600 mmReinforcement Steel : 8 Nos. 20 Φ

Step 4 : Lateral Ties

Dia of lateral ties is greater of the following: (Clause 26.5.3.2 C2 of IS 456 – 2000)(i) ¼ of largest main dia = ¼ x 20 = 5 mm(ii) Readily available in the market = 8 mmProvide 8 mm Φ RTS

Pitch : Pitch is least of the following (Clause 26.5.3.2 C1 of IS 456 – 2000)

(i) least lateral dimension of column = 400 mm(ii) 16 x least main dia = 16 x 20 =320 mm(iii) 300 mmProvide 8 Φ RTS @ 300 mm C/ c

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Sketch:

Alternative 2: Assuming % of reinforcement & find column cross sectionStep 1: METHOD 1 USING EQUATION GIVEN IN IS 456 – 2000

Assuming short columnPu= ( 0.4 f ck Ac+ 0.67 f y Asc ) - ---- (1) Assume % of steel as 1%, Ac = 0.99 % Ag ; Asc = 1% of Ag3000 x 103 = 0.4 x 25 x 0.99 Ag + 0.67 x 415 x 1 x Ag

1003000 x 103 = 9.90 Ag + 2.7805 Ag Ag = 3000 x 103 = 236583.73 mm2

12.6805

Assuming b= 400 mm D= 236583.73 = 591.46 mm say 600 mm400

Ast = 1 x 400 x 600 = 2400 mm2

100Provide 8 Nos. 20 Φ RTS al round the column ( Ast= 2512 mm2 > 2400 mm2)

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METHOD 2 USING TABLE C5 EQUIVALENT TO CHART 25 OF SP – 16

For Pt =1% for M25

Pu = 0.507Ag f ck

3000 x 103 = 0.507 Ag x 25

Ag = 3000 x 103 = 236686.39 mm2

25 x 0.507

Assume b=400 mm D= 236686.39 = 591.72 mm say 600 mm 400

Provide 8 Nos. 20 Φ RTS al round the column ( Ast= 2512 mm2 > 2400 mm2)

PROBLEM 2Data given: f ck= 20 N/mm2

f y = 415 N/mm2

Pu= 2100 KNlef = 2.75 m

Design (what to be obtained ) :Step 1: Check whether the column is less than or greater than 400 mm

To begin with, it will be decided whether the lateral dimensions of an axially loadedcolumn shall be less than 400 mm or greater than 400 mm. This is because, if the size of the column is less than 400 mm, the column is required to be designed for axial load andminimum moment. Mu.mini.= (= Pu x emini)Load carrying capacity of the column section (400 x 400 ) mm with 0.8% steel for M20 is1625 KN from Table 3 < Pu i.e. 2100 KNSince the size of column > 400 mmlef = 2.75 x 1000 = 6.9 < 12 Column is shortb 400 Assume 1.5 % steel p = 0.015

Ag = 2100 x 103 = 174263 mm2

[ 0.4 x 20+ ( 0.67 x 415 – 0.4 x 20) x 0.015]

Size of square column = √ 174263 = 417.4 mm

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Provide column size of 420 mm x 420 mmAg = 176400 mm2 > 174263 mm2

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