Column Design Examples EBCS

School of Civil & Urban Eng., IOTec.

Hawassa University

Example 4.1. (Classification of columns as short or long) The frame shown in figure below is composed of members with rectangular cross sections. All members are constructed of the same strength concrete (E is the same for both beams and columns). Considering bending in the plane of the frame only, classify column EF as long or short if the frame is braced. All girders are 300 x 600 mm. 600 C

525 KN F

I F

M1 = 30 KNm

E

M2 = 45 KNm

300

350 300

3.80 m

300 x 350

600 B

H

E 400

3.80 m 300 x 400 A

D 9m

G 7.5 m

Solution: Moments of inertia 3

300 x600  54 x10 8 mm 4 12 300 x 400 3  16 x10 8 mm 4 Columns: I DE  12 3 300 x350 I EF   10 .71875 x10 8 mm 4 . 12

Girders: I g 

Stiffness Coefficients:









 E  54 x108  6 x105 E. K  K  cF  BE EI g 9000 Girders : K g    Lg   E  54 x108  7.2 x105 E.  K EH  K FI  7500 





 E  16 x108  4.21 x105 E K   DE EI 3.8 x103 Columns: K c  c    Lc E 10.71875 108  2.82 x105 E  K EF  3.8 x103 

 

The column being considered is column EF. RC II (CEng 3111)

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column design examples Page 1


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Rotational stiffnesses at joints E and F.  EIcol / L I col / L     f EIg / Leff   f I g / Leff  Joint E:

E 

K EF  K DE 4.21x105  2.82x105   0.53 K BE  K EH 6 x105  7.2 x105

Joint F:

F 

K EF 2.82x105   0.21 KCF  K FI 6 x105  7.2 x105

E  F

0.53  0.21  0.37 2 2 For a braced column (Non sway structure ) for design Le  m  0.4 0.37  0.4    0.66  0.7 L  m  0.8 0.37  0.8 Le= (0.7) (3.8) = 2.66m = 2660mm L Le 2660 The slenderness ratio:   e   i I A 10.71875 x108 300 x 350    26.327.

m 







  30    66.66  ok!  45   The colum is short .

  50  25

Example 2: A column resting on an independent footing supports a flat slab. The super imposed factored load transferred from the slab is 1000 kN. Design the column assuming a gross steel ratio of (a) 0.01. Use concrete C30, steel S300 and class I works. Assume column height h = 4 m. Solution:

fcd = 13.6 MPa;

fyd = 260.87 MPa

Pdu = Ag [ fcd (1 – ρ) + ρ fyd] (a) For ρ = 0.01 and Pd = 1000 kN, Ag 

Pd [ f cd (1   )  f yd ]

S2 =

1000 * 10 3 13 .6(1  0.01)  0.01(260 .87 )

S = 249 mm

Use 250 mm × 250 mm cross section Ast

=

ρ Ag = 0.01 (250)2 = 625 mm2

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Ast provided = 804 mm2

Use 4 numbers of 16 mm dia rods; Ties: d ≥ 6 mm (or) ≥ Dia of main bar/4 = 16/4 = 4 mm

S ≤ 12* dia of main bar = 192 mm ≤ Least lateral dimension = 250 mm ≤ 300 mm Therefore, use 6 mm dia rods at 190 mm center to center Example 3 Design a slender braced (non-sway) column subjected to uniaxail bending. Given: - factored load=1650KN -factored 1st order equivalent constant Moment=130KNm -Geometric length: L=7m and Le=0.7L -Material data; C-30, S-460 class I work -Assume Column size b = 400mm; h = 400mm; Required: - quantity of reinforcement. Solution Assume cover = 20mm; ølong = 20mm and ølat. = 10mm d ' 40 = 0.1 and d = 400-40 = 360mm  h 400

ea >=

Le 0.7 * 7000 = = 16.33 300 300

or

20mm

Therefore; ea=20mm Check for second order effect

- λ=

Le I A

=

- λmax = 50-25( MM

1

4900 4002 12

= 42.4

) ; here first order moment is constant throughout the column.

2

Therefore; λmax= 50-25=25 As λ > λmax, second order effect has to be considered Msd = etot*Nsd=(ee+ea) Nsd =ee* Nsd+ ea* Nsd =first order moment + moment due to ea = 130+ (1650*0.02) =163kNm

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For C-30 concrete; fck= 24;

f yk

fyd = νsd =

s

=

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fcd=

0.85 f ck

s

=

0.85 * 24 = 13.6MPa 1.5

460 = 400MPa 1.15

N sd 1650 * 10 3 = = 0.76 Ac f cd 400 2 * 13 .6

 sd 

M sd 163 *10 6 = =0.187 Ac f cd h 400 2 *13 .6 * 400

Using chart no- 2; for νsd = 0.76 and  sd = 0.187; ω = 0.32;  bal = 0.25 K2 =

 sd 0.187 = = 0.75,  bal 0.25 e2 =

1 5 5 = K2 ( ) 10-3 = 0.75( * 10 3 = 10.42*10-6 r 360 d

K1 Le 2 1 ( ) here K1 = 1 for λ > 35 10 r

=

1(4900 ) 2 (10 .42 * 10 6 ) = 25mm 10

e tot = ee + ea+e2 = Msd =Nsd* etot = 1650*

130 *10 3  20  25  123 .8mm 1650

123.8 = 204.3kNm 1000

,



204 .3 *10 3 400 3 *13 .6

= 0.236 implies

ω=0.45 Recalculating k2, μbal=0.3 k2=

0.235 = 0.78 , 0.3

1 5 = 0.78( ) *10 3 = 10.8*10-6 r 360

e2= 26mm etot = 124.8 Msd = 1650*

124.8 = 205.09 kNm , 1000

 sd 

M sd 205 .1 *10 6 = = 0.236 Ac f cd h 400 2 *13 .6 * 400

ω = 0.45 Interaction can be stopped.

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Ast 

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0.45 * 400 2 * 13 .6 = 2448mm2 400

Use 8 number of 20mm diameter rods. As provided = 2512, compare the result with minimum and maximum code requirements >0.008*4002=1280 <0.08*4002=12800 Lateral ties: ø  6 or 20/4=5

Hence ok Hence use ø 10 bar

 {

Use 10mm diameter bar @240mm c/c. Example 4 A uni-axial column is to be constructed from a materials C-30, S-400 class I works. If the diagram for 1st order end moment and axial force are as shown, determine the area of reinforcement assuming non-sway frame system.(use b/h =300/400 and Le=0.75L, with L=7.5m)

d' 40 = = 0.1 h 400 ee  0.6eo2+0.4eo1 or 0.4eo2 155 eo2= *1000 =121.1mm 1280  82 eo1= *1000 =-64.1mm 1280

Soln: Assume d’= 40mm;

use uniaxail chart no-2

{ ea 

le 0.75 * 7500 = =18.75mm or 20mm; use ea =20mm 300 300

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Check for e2; λ=

0.75 * 7500

=48.7

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; λmax=50-25(

 82 )=63.2 155

4002 12 λ < λmax; therefore; neglect second order eccentricity etot=eo2 +ea =121.1+20=141.1mm 141.1 Msd= Nsd*etot=1280* =180.6kNm; fcd= 13.6 ; fyd=347.8 1000 N M sd 1280 *103 180.61*106 ν = sd  =0.78 and μ= =0.28  f cdbh 13.6 * 300 * 400 f cd bh2 13.6 * 300 * 4002

implis ώ=0.6 As=

 * Ac * f cd f yd

=

0.6 * 400 * 300 *13.6 = 2815.4mm; use 8ø22mm bar 347.8

22 2 *  =3041mm2 4 < Asmax= 0.08*Ag=9600mm2 >Asmin=0.008Ag=960mm2 Lateral reinforcement Ø  6 or 22/4 S  12* 22 =264 or 300

Aspov= 8*

Use 6mm Ø ties at 260mm spacing. Example 5 Design a column to sustain a factored design axial load of 900KN and biaxial moments of Mdx=270KNM and Mdy=180KNm including all other effects. Use C-30, S-300 class I works. Solution:

fck= 24MPa ;fcd=13.6MPa; fyd=260.87MPa b' h' Assume b*h = 400*600mm and  = =0.1, Nsd= 900kN b h Mh=Mdx=270kNm Mb=Mdy=180kNm 900 *10 3 ν= =0.28(between0.2 and0.4) 13 .6 * 400 * 600 Mb Mh 180 *10 6 270 *10 6 b  = =0.14 and  h  = =0.14 f cd Ac h 13 .6 * 400 * 600 2 f cd Ac b 13 .6 * 600 * 400 2 Using biaxial chart no- 9 thus: for  =0.2;  h =0.14 ,  b =0.14; ώ=0.4 for  =0.4;  h =0.14,  b =0.14; ώ=0.4 By interpolation for  =0.28; ώ=0.4

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 * Ac * f cd

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0.4 * 400 * 600 *13.6 =5005mm2 260.87 f yd < Asmax= 0.08*Ag=19200mm2 >Asmin=0.008Ag=1920mm2 use 8ø30mm bar Lateral reinforcement

As=

Ø  6 or 30/4

=

{

Use 8mm Ø ties at 300mm spacing.

RC II (CEng 3111)

Chapter 1


Column Design Examples EBCS

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