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CONCRETE STRUCTURES 2011 17

DESIGN OF REINFORCED CONCRETE COLUMNS

UNDER CENTRIC LOAD ACCORDING TO

EUROCODE 2

Bernt Csuka - Lszl P. Kollr

The paper presents a very simple method for the design and analysis of centric loaded, symmetrically rein-

forced concrete columns with rectangular or circular cross-sections. The concept of the capacity reduction

factor (or instability factor, buckling coefcient) is introduced, which was applied for steel, timber

and masonry columns in Eurocode 3, 5 and 6, respectively. The capacity reduction factor is determined

on the basis of Eurocode 2. It is shown numerically that the method is always conservative and reasonably

accurate. The usage of the method is demonstrated through numerical examples.Keywords:Reinforced concrete column, concentric compression, capacity reduction factor, simplified design, parametric calculation,Eurocode 2.

1. INTRODUCTIONConcentricly loaded RC columns are common structural

elements of braced building structures. In Eurocode 2 (2004)

unlike the previous Hungarian code MSZ these columns

must be designed the same way as eccentricly loaded columns,

with the only difference that the eccentricity of the load is set

equal to zero. In EC 2 there are two methods of calculation:

(1) Nominal Stiffness, (2) Nominal Curvature. The National

Annex (NA) has to decide which method must be used in a

country. In Hungary both methods are accepted.

There are several articles in the literature which deal with

the design procedures of columns according to Eurocode 2

(Bonet et al. (2007), Bonet et al. (2004), Mirza and Lacroix

(2002), Aschheim et al. (2007)), however none of these

treats the centric loaded columns separately. Other parts of

Eurocode contain simple methods, which can be used for

the calculation of centric loaded columns. For example, the

buckling coefcient, , the instability factor, kc,y

or kc,z

or

the capacity reduction factor, m,s

are introduced for steel-

(EC 3, 2004), timber- (EC 5, 2004) and masonry structures(EC 6, 2006), respectively.

Our aim in this paper is to derive a similarly simple design

method for centric loaded RC columns. We wish to use the

following expression:

NRd

= Nu (1)

whereNRd

is the ultimate load of a column, is the capacityreduction factor, and N

u is the plastic ultimate load of the

cross-section:

u cd c s yd

,N f A A f= + (2)

where Ac is the cross-sectional area of concrete, A

s is the

total cross-sectional area of the reinforcement, fcd

is the

design compressive strength of concrete, andfyd

is the design

yield strength of steel. The concrete cross-section and the

arrangement of the reinforcing bars is doubly symmetric

and hence the center of gravity of the reinforcement and the

concrete are at the same point.

In a previous paper Kollr (2004) presented a solution for

this problem applying an approximate interaction diagram ,

however the presented method has the following shortcomings:

it is valid only for rectangular columns and according to the

applied approximations the accuracy is not satisfactory.

2. THE PROCEDURE OFEUROCODE 2

Here we present briefly the method of the Eurocode for

calculating the design value of eccentricities for concentric

loading applying the method of Nominal Curvature.

These eccentricities are used in the cross-sectional design of

columns.

In the analysis the original eccentricity of the normal force

on the undeformed column (ee= 0), the eccentricities due to

the imperfections (ei) and the (second order) eccentricities

due to the deformations of the column (e2) must be taken into

account (Fig. 1).

When the rst order bending moment along the column is

uniform, the cross-section of the column must be designed for

the eccentricity etot

:

e i 2

tot

0

sum of eccentricitiesmax

minimal value of eccentricities

e e ee

e

+ +=

(3)

where ee=M

0e/N

Edis the rst order eccentricity (for uniform

bending moment), ei is due to the imperfections and e2 isthe second order imperfection. The expression for e

i is as

follows:

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18 2011 CONCRETE STRUCTURES

0

0i

0

, if 4 m400

2, if 4 m 9 m

400

2, if 9 m

3 400

ll

le l

l

ll

=