chapter 1 rc ii columns design notes

8/10/2019 Chapter 1 RC II Columns design Notes

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School of Civil & Urban Eng., IOTec. Hawassa University

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RC II (CEng 3111) Chapter 1

1

Chapter 1 DESIGN OF COLUMNS

Columns are the members that take axial compressive load and bending moments. The bending

effect may be due to the lateral loads, end moments, and/or due to eccentricity of the axial loads.

Reinforced concrete columns are classified in EBCS 2, as un-braced (sway) or braced (non-sway), based on how the horizontal loads are transmitted by the super structure to the foundation.

Un-braced Columns (in sway frames): An un-braced structure is one in which frame action is

used to resist horizontal loads (lateral loads due to wind or earthquake). In such structure, beam

and column members may be designed to act together as a rigid frame in transmitting the lateral

forces down to the foundations through bending action in the beams and columns. In such an

instance the columns are said to be un-braced and must be designed to carry both the vertical

(compressive) and lateral (bending) loads. Moments in the columns can substantially reduce the

vertical load carrying capacity. The frame as a whole may exhibit significant lateral

displacement. The bending moment can increase due to second order effect.

Braced columns (in non-sway frames): If the lateral loads in a frame are transmitted to the

foundation through a system of bracing or shear walls, the column member in such a frame is

said to be braced column and consequently carry only vertical loads. In such a case, second

order effect will be negligible.

a). Braced columns in a non sway b). Unbraced columns in Sway type

Second order effect or P- effect: Consider a slender column subjected only to equal and

opposite end moment,M, as shown in figure. The column is bent into a single curvature with a

maximum deflection at the mid height as shown. If the axial compression is applied at the ends

of the column now, additional bending moment is caused due to the axial load acting on the

deformed shape. This additional bending moment causes additional deflection and so on, until

the final maximum deflection is reached at the stage of equilibrium under combined axial force

and bending moments. This is referred to asP- effects. It should be observed that due to P-

effects the maximum moment in the column is larger than the externally applied end moment M.


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2

If the column is short, P- effect is negligible. If the column is slender, P- effect is to be

considered.

(Section 4.4.4.4 of EBCS 2) Isolated columns: Columns may be considered as isolatedcolumns when they are isolated compression members (such as individual isolated columns and

columns with articulations in a non-sway structure), or compression members which are integral

parts of a structure but which are considered to be isolated for design purposes (such as slender

bracing elements considered as isolated columns, and columns with restrained ends in a non-

sway structure).

(Section 4.4.5 of EBCS 2) Slenderness ratio:

On the basis of the slenderness ratio columns may be classified as short or long (slender).

- For isolated columns, the slenderness ratio is defined by

WhereLeis the effective buckling length

i is the minimum radius of gyration of the concrete section only.

(Section 4.4.6 of EBCS 2) Limits of slenderness ratio:

Generally, the slenderness ratio of concrete columns shall not exceed 140.

Second-order effects in compressive members need not be taken into account in the following

cases:

(a).

For sway frames, the greater of the two

d

15

25

(b).For nonsway frames

2

12550M

M

WhereM1andM2are the first-order (calculated) moments at the ends, M2being always positive

and greater in magnitude than M1, and M1being positive if member is bent in single curvatureand negative if bent in double curvature.

i

Le

M

M

M

M

P

P

Max. Moment = M + PMax. Moment = M

A

Ii


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3

loadaxialdesignNAf

Nsd

ccd

sdd

Effective length of compression membersThe effective height (length) of a column is the distance between the two consecutive points of

contra flexure or zero bending moments. The figure shown below may serve this purpose.

In accordance with EBCS-2, 1995, the effective lengthLefor an RC Column is given as,

a. Non-sway mode 7.08.0

4.0

m

me

L

L

b. Sway mode

15.15.7

6.145.7

21

2121

L

Le

Or Conservatively .15.18.01 me

L

L

For the theoretical model shown below:

2

21

2221

22

1211

11

m

c

c

KK

KK

KK

KK

where K1 and K2 are column stiffness coefficients (EI/L) for the lower and the upper column

respectively.Kcis the stiffness coefficient (EI/L) of the column being designed.

Kijis the effective beam stiffness coefficient (EI/L)

= 1.0 (EI/L) for opposite end elastically or rigidly restrained.

= 0.5 (EI/L) for opposite end free to rotate.

= 0.0 (EI/L) for a cantilever beam.

For a nonsway frame 1.0cr

sd

N

N

Where: Nsdthe design value of the total vertical load.Ncrcritical vertical load for failure in a sway mode given as

Lc2

Lc

Lc1

Ib11Ib12

Ib22 Ib21

Ic

Ic1

Le 5.1

0.7L 1.2L

Le

L0.5L

Le 65.0 Le 8.0 Le 2.1

0.8L

Le 8.0

Ic2


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4

2

2

e

ecr

L

EIN

EIe= 0.2EcIc+ EsIs (or conservatively EIe= ccbal

bal IEr

M4.0

1 )

Ec 310511100 dr

Curvaturefbal

cd

I c= Moment of inertia of the concrete sections of the substitute column w.r.to centre

Is = Moment of inertia of reinforcement sections of the conc. section

A frame may be classified as braced if its sway resistance is supplied by a bracing system which

is sufficiently stiff to assume that all horizontal loads are resisted by the bracing system. (Not

more than 10% of the horizontal loads are attracted by the frame)

Reinforcement arrangement & Minimum Code Requirements

Functions of Lateral Rein forcement

Rules for the arrangement:

- Diameter of ties, t t 6mm or 4

- C/C spacing

mm

ensionlateralleastb

barsallongitudinofdiameterimumml

300

dim

)(min12

- Pitch of spiral 100mm

- Ties shall be arranged such that every bar or group of bars placed in a corner and

alternate longitudinal bar shall have lateral support provided by the corner of a tie

with an included angle of not more than 135oand no bar shall be farther than

150mm clear on each side along the tie from such a laterally supported bar.

Main or Longitudinal rein forcement

- Area of longitudinal reinforcement, As.

l

15t 300mm

At center

Equal or less

than 150mm

May be greater than 150mm

No intermediate tie is required

At center

135

15t

3

00mm

t

l = longitudinal bars

t = main ties


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5

0.008AcAs0.08Ac or 0.008 08.0c

s

A

A

- Min. # of bars

.6

tan4

tarrangemenCircularin

tarrangemengularrecin

- The diameter of longitudinal bars, .12mml

- The minimum lateral dimension of a column shall be at least 150mm- The Min. cover to reinforcement should never be less than

(a) ormmor n ),40(

(b) .3255 mmdifmmOrmm gn

n = ,55mmn n= no. of bars having the same diameter

dgthe largest nominal maximum aggregate size.

Minimum reinforcements are provided:

Example 4.1. (Classif ication of columns)

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.

Solution:

Moments of inertia

Girders: 483

105412

600300mmx

xIg

Columns: 483

1016

12

400300mmx

xIDE

.1071875.1012

350300 483

mmxx

IEF

F

E

I

H

G

E

D

C

B

A

M2= 45 KNm

M1= 30 KNm

7.5 m9 m

3.80 m

3.80 m

400

600

600

350

300 x 400

300 x 350

525 KNF

300300


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6

Stiffness Coefficients:

.102.77500

1054

.1069000

1054

:5

8

58

ExxE

KK

ExxE

KK

L

EIKGirders

FIEH

cFBE

g

g

g

Columns:

Exx

EK

Exx

xEK

L

EIK

EF

DE

c

cc

5

3

8

5

3

8

1082.2108.3

1071875.10

1021.4108.3

1016

The column being considered is column EF.

Rotational stiffnesses at joints E and F.

effgf

col

effgf

col

LI

LI

LEI

LEI

/

/

/

/

Joint E: 53.0102.7106

1082.21021.455

55

xx

xx

KK

KK

EHBE

DEEFE

Joint F: 21.0102.7106

1082.255

5

xx

x

KK

K

FICF

EFF

37.02

21.053.0

2 FEm

For a braced column (Non sway structure ) for design

7.066.08.037.0

4.037.0

8.0

4.0

m

me

L

L

Le= (0.7) (3.8) = 2.66m = 2660mm

The slenderness ratio: 3503001071875.10

2660

8 xxAI

L

I

L ee

.

!66.6645

302550

.327.26

shortiscolumThe

ok


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7

Design of short Columns subjected to Axial compression

The ultimate capacity of an axially loaded short column can be determined by

Pdu = fcd(Ag - As) + fydAst, = fcdAg (1) + Agfyd

Where, =g

st

A

Aor Ast = Ag

Pdu = Ag [fcd(1) +fyd]

Ag= Gross concrete area; Ast = Area of main reinforcement

A column may be classified as long when the slenderness factor is defined as

;12b

Le

For long (Slender) columns a load reduction factor Cr is introduced in such a way that theconcrete concentric design axial load capacity can be given by

Pduc= CrPduWhere Cr= 1.25Le/48b

Le= Effective height; b= least lateral dimension

Example: 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 (b) 0.02. 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) +fyd]

(a)For = 0.01 and Pd = 1000 kN,

])1([ ydcd

dg

ff

PA

S2=)87.260(01.0)01.01(6.13

10*1000 3

S = 249 mm

Use 250 mm 250 mm cross section

Ast =Ag = 0.01 (250)2= 625 mm2

Use 4 numbers of 16 mm dia rods; Astprovided = 804 mm2

Ties: d 6 mm (or) S 12* dia of main bar = 192 mm

Dia of main bar/4 = 16/4 = 4 mm Least lateral dimension = 250 mm

300 mm

Therefore, use 6 mm dia rods at 190 mm center to center

(b)For = 0.02 and Pd = 1000 kN,


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])1([ ydcd

dg

ff

PA

S2=)87.260(02.0)02.01(6.13

10*1000 3

S = 232 mm

Use 240 mm 240 mm cross section

Ast =Ag = 0.02 (240)2= 1125 mm2

Use 4 numbers of 20 mm dia rods; Astprovided = 1256 mm2

Ties: d 6 mm (or) S 12* dia of main bar = 240 mm

Dia of main bar/4 = 20/4 = 5 mm Least lateral dimension = 240 mm

300 mm

Therefore, use 6 mm dia rods at 240 mm center to center

Design of Columns with moments (eccentrically Loaded Columns)When a member is subjected to combined axial load and bending moment it is more convenient

to replace the axial load and moment with an equivalent load applied at an eccentricity e.

I nteraction diagram:It is a plot of axial load capacity of a column against the bending moment

it sustains. To illustrate conceptually the interaction between axial load and moment in a

column, an idealized, homogeneous and linearly elastic column with compressive strength fcu

equal to its tensile strengthftuis considered. This type of column fails in compression when

max=fcu.

;cufI

My

A

P or 1

If

My

Af

P

cucu

-------------------------(1)

ButPmax= fcuA & Mmax=y

Ifcu

Equation (1) becomes, 1ma xma x

MM

PP ; This equation is called interaction equation.

Pd

Md

e

Pd


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It shows the interaction of (or) relationship between P and M at failure.

Reinforced concrete is not elastic, and it hasft


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10

Where ee=equivalent constant first-order eccentricity of the design axial load.

ea = additional eccentricity in account of geometric imperfections.

ea= mmLe 20300

ee = eo,for eoequal at both ends of a column

For first order moments varying linearly along the length, the equivalent eccentricity is the higherof the following two values. eo1and eo2are first order eccentricities at the ends with eo2being positive and greater in

magnitude than eo1and eo1 is the smaller and positive for single curvature and negative for

double curvature.

e2 = Second-order eccentricity and is ignored if column is short.

For nonsway frames, e2 = 10

121 rLK e

.350.1

351575.020

1

1

forK

forK

3

2 105

sec1

dK

tioncriticaltheatCurvaturer

Where d = the Column dimension in the buckling plane less the cover to the center of the

longitudinal reinforcement.

K2=bal

sd

bal

sd

MM

Msd = design moment at the critical section including second-order effects.

Mbal = balanced moment capacity of the column.

Design of columns for uni-axial bending

A column is said to be bending uniaxially if it is loaded with a bending moment only in one

direction, in addition to axial force. For the design of such a column interaction charts are

prepared using non-dimensional parameters, and , in which,bhf

N

cd

sd and2bhf

M

cd

sd

A chart showing the interaction diagrams are prepared (and compiled in EBCS 2 - part 2) for

different values of d' and h as given in the sample chart. Ascan be calculated by the formula,

yd

cdcs

f

fAA

In using these charts for design, the following procedure may be adopted.Given: axial load and bending moment, (BM = axial load total eccentricity)

* Assume the cross section dimensions band h.


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* Assume d' and evaluate d'/hto choose appropriate chart number

* Calculate and ,

* The coordinate (,) gives the value of from the appropriate chart chosen

* Determineyd

cdcs

f

fAA

Sample design chart


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Example 1

Design a slender braced (non-sway) column subjected to uniaxail bending.

Given:- factored load=1650KN

-factored 1storder 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; long = 20mm and lat. = 10mm

400

40'

h

d= 0.1 and d = 400-40 = 360mm

ea>=300

eL=

3007000*7.0

= 16.33 or 20mm

Therefore; ea=20mm

Check for second order effect

- =

A

I

Le =

12

400

4900

2= 42.4

- max = 50-25( 21

M

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

Therefore; max= 50-25=25

As > max, 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

For C-30 concrete; fck= 24; fcd=s

ckf

85.0

=5.1

24*85.0

= 13.6MPa

fyd =s

ykf

=15.1

460

= 400MPa

sd =cdc

sd

fA

N

=6.13*400

10*16502

3

= 0.76

hfA

M

cdc

sdsd =

400*6.13*400

10*1632

6

=0.187

Using chart no- 2; forsd = 0.76 and sd = 0.187; = 0.32; bal = 0.25


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K2 =bal

sd

=25.0

187.0= 0.75,

r

1

= K2(

d

5) 10-3 = 0.75( 310*

360

5

= 10.42*10-6

e2 = (10

2

1LeK

r

1) here K1= 1 for > 35

= )10*42.10(10

)4900(1 62

= 25mm

e tot = ee+ ea+e2 = mm8.12325201650

10*130 3

Msd = 1650*1000

8.123= 204.3kNm ,

6.13*400

10*3.2043

3

= 0.236 implies =0.45

Recalculating k2, bal=0.3

k2=

3.0

235.0= 0.78 ,

r

1

= 0.78( 310*)

360

5 = 10.8*10-6

e2= 26mm

etot = 124.8

Msd = 1650*1000

8.124

= 205.09 kNm ,

hfA

M

cdc

sdsd =

400*6.13*400

10*1.2052

6

= 0.236

= 0.45

Interaction can be stopped.

400

6.13*400*45.0 2stA= 2448mm2

Use 8 number of 20mm diameter rods.

Asprovided = 2512, compare the result with minimum and maximum code requirements

>0.008*4002=1280


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14

Soln: Assume d= 40mm;h

d' =

400

40= 0.1 use uniaxail chart no-2

ee 0.6eo2+0.4eo1 or 0.4eo2

eo2= 1000*1280

155=121.1mm

eo1= 1000*1280

82=-64.1mm

ea

300

le=

300

7500*75.0=18.75mm or 20mm; use ea =20mm

Check for e2; =

12

400

7500*75.02

=48.7 ; max=50-25(155

82)=63.2

< max; therefore; neglect second order eccentricity

etot=eo2+ea =121.1+20=141.1mm

Msd= Nsd*etot=1280*1000

1.141=180.6kNm; fcd= 13.6 ; fyd=347.8

=400*300*6.13

10*1280 3

bhf

N

cd

sd =0.78 and =2

6

2 400*300*6.13

10*61.180

bhf

M

cd

sd =0.28

implis =0.6

As=

yd

cdc

f

fA **=

8.347

6.13*300*400*6.0= 2815.4mm; use 822mm bar

Aspov= 8*4

*22 2 =3041mm2

< Asmax= 0.08*Ag=9600mm2

>Asmin=0.008Ag=960mm2

Lateral reinforcement

6 or 22/4 S12* 22 =264 or 300

Use 6mm ties at 260mm spacing.


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15

Design of columns for biaxial bending

A column may receive moments from beams and grade framing to it, in addition to the axial

loads. This situation of a biaxial loaded rectangular section is shown below with the

corresponding interaction curves.

Uniaxial loading about y-y.

Uniaxil loading about X-X

Biaxial loading about a diagonal axis x-x where: r=arc tandy

dx

e

e=arc tan

dxM

Mdy

Three-dimensional interaction diagramInteraction surface

Any combination of Pdn, Mdx,and Mdylying inside the surface can be applied safety.

Any point lying outside the surface would represent failure.

Due to the mathematical complication arising from the construction of interaction surface,

in practice a simpler approximation methods are used of which the one developed by

Bresler is satisfactory.


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16

It is given as reciprocal load equation.

dodnydnxdn PPPP

1111 , which simplifies to

dnydnxdnydnxdo

dnydnxdo

dn

PPPPP

PPPP

)(

Where: Pdnis the desing axial load capacity when applid at edxand edy

simultaneously.(biaxial bending ).

Pdnxand Pdny- design axial load capacity when edx and edy are only acting ( case of

uniaxal loading).

PdoDesign axial force capacity for concentric load case.

However, in EBCS 1995, interaction charts are prepared for this purpose and they can be easly

used for actual design where the following procedures need be followed. Given : Pd; Mband Mh

Assume a cross section and evaluateh

h

b

b ',

Calculate b , h, and

ccd

d

Af

N ,

bAf

M

ccd

bb and

bAf

M

ccd

hh

Select suitable chart, which nearly satisfy the calculatedh

h

b

b ', and

Enter the chart for suitable value of ( 0.0,0.2,0.41.4)

Note: - > 1.4 shows very small concrete cross section

For intermediate value of ,use interpolation

Select corresponding to,b , h

Compute As, totyd

cd

f

bhf

Check minimum and maximum requirements.

Example 1

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.

Soln: fck= 24MPa ;fcd=13.6MPa; fyd=260.87MPa

Assume b*h = 400*600mm andb

b' =

h

h'=0.1, Nsd= 900kN

Mh=Mdx=270kNm

Mb=Mdy=180kNm

= 600*400*6.13

10*900 3

=0.28(between0.2 and0.4)


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17

bAf

M

ccd

bb = 2

6

400*600*6.13

10*180=0.14 and

hAf

M

ccd

hh = 2

6

600*400*6.13

10*270=0.14

Using biaxial chart no- 9 thus:

for =0.2; h =0.14 , b =0.14; =0.4

for =0.4; h =0.14, b =0.14; =0.4

By interpolation for =0.28; =0.4

As=yd

cdc

f

fA **=

87.260

6.13*600*400*4.0=5005mm2

< Asmax= 0.08*Ag=19200mm2

>Asmin=0.008Ag=1920mm2

use 830mm bar

Lateral reinforcement

6 or 30/4

{

Use 8mm ties at 300mm spacing.

chapter 1 rc ii columns design notes

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