Designing a Concrete Column to Eurocode

Introduction

The subject of this post is the design of reinforced concrete columns to BS EN 1992-1-1 Eurocode 2: Design of Concrete Structures-Part 1-1: General Rules for Buildings

In the last post, we analysed a concrete column in a frame structure for vertical actions: Axial loads and Bending Moments. However, the column was not designed because the design of columns has not been presented before coupled with the fact that some readers might not be familiar with the terms presented. In this post, we shall consider the aspects of designing concrete columns of all cross-section profiles, typically square, rectangular and circular columns to the Eurocode.

Principles

The general procedure to be adopted when designing any concrete column is as follows

  • Analyse the column for the required actions
  • Check that cover and concrete comply with requirements for durability
  • Check the slenderness of the column
  • Design section and reinforcement
  • Check biaxial bending

Slenderness

Columns are vertical elements designed to resist compression forces, and as with most vertical elements subjected compression forces, slenderness check is a compulsory check that must be carried out. Concrete columns are indifferent in this regard and are defined by the following expression.

\lambda =\frac { { l }_{ o } }{ i }
  • lo = the effective length of the column
  • i = radius of gyration of the column

The radius of gyration is defined as

i=\sqrt { \frac { I }{ A } } =\frac { h }{ 3.46 } for\quad recatngular\quad section\quad \& \frac { { d } }{ 4 } for\quad circular\quad columns

The value of the effective length may be determined by multiplying by the factors from table 1 which is based on the end restraints of the columns. However, BS-EN 1992-1-1 provides a more rigorous approach which is based on the relative stiffness’s of the beam and column. Please see equation 5.15 of BS EN1992-1-1.

Table 1: Effective length of column based on support conditions

Before we can design any concrete column, we must classify the column with respect to slenderness i.e. either short or slender.

Short Columns

A column is considered to be “short” if the second order moments are small enough to be neglected i.e. if they are less than 10% of the first order moments. This is likely to be the case in most braced structures with normal floor-floor distance except the column’s dimensions are particularly small in both directions.

In-order to determine whether a column is short, BS EN 1992-1-1 gives an expression that must be satisfied. This expression is defined as

{ \lambda }_{ lim }=\frac { 20\cdot A\cdot B\cdot C }{ n }

Where λlim is the limiting slenderness of the column ; λ is as defined in the preceding section.

Estimating Limiting Slenderness
A\quad is\quad defined\quad as\quad \frac { 1 }{ (1+0.2{ \varphi }_{ f }) } \quad if\quad not\quad known\quad A=0.7
B\quad is\quad defined\quad as\quad \sqrt { 1+2\left( \frac { { A }_{ s }{ f }_{ yd } }{ { A }_{ c }{ f }_{ cd } } \right) } if\quad not\quad known\quad B=1.1\\
C=1.7-\frac { { M }_{ min } }{ { M }_{ max } } \quad if\quad not\quad known\quad C=0.7\\

When designing columns the value of C has a very huge impact on the limiting slenderness and to take the default value might result in oversizing the columns. It is recommended that we always calculate this factor since it is unlikely for us not to have the bending moment values at the time of design.

n=\frac { { N }_{ Ed } }{ { A }_{ c }{ f }_{ cd } }

Where:

  • φef is the effective creep ration
  • Ac & As is the area of concrete and area of steel respectively
  • fcd & fyd is the design strength of concrete and steel respectively
  • M01 and M02 are smaller and larger end moments at the respectively
  • n is the relative normal force
  • Ned is the design axial loads on the column

Slender Columns

When the limiting slenderness is less than the actual slenderness, the column is slender and therefore subject to buckling failure as opposed to crushing which is the case of short columns. When a column is slender a bending moment is applied unto it as the axial load is applied. The column deflects due to the axial load and as it does so, this additional bending moment causes the column to buckle. The design of slender columns will not be covered in this post; however, it will be the subject of a later post.

Axial Actions & Bending Moment

Now, that the slenderness of the column has been established and the column classified. The next thing is determining the design values of the forces that will be used in designing the column. The axial force is the same as that obtained from the analysis however the bending moment must be modified to account for the effect of imperfections and nominal moments.

The design value of bending moment in a short column about any axis is defined as

{ M }_{ ed }=Max\left[ { M }_{ 02 };{ M }_{ nom } \right]

Where Mnom is the minimum moment that can be applied to a concrete column about an axis Irrespective of whether the column is axially, uniaxially or biaxially loaded. This moment is due to geometric imperfections and its defined as :

{ M }_{ nom }={ e }_{ o }{ N }_{ ed }\quad where\quad { e }_{ o }=\frac { { h }}{ 30 } \ge 20mm

M02 is defined as

{ M }_{ 02 }=max\left| { M }_{ bottom };{ M }_{ top } \right| +{ e }_{ i }{ N }_{ ed }\quad { e }_{ i }=\frac { { l }_{ o } }{ 400 }

The right side of the equation is the additional moment sue to effects of global imperfections. Please see clause C6.1(4) of BS EN 1992-1-1. If the column is biaxially loaded, the effect of global imperfections is only evaluated in the critical direction only

Detailing Requirement

There are very strict guidelines with regards to reinforcement detailing of columns. A minimum of 4 bars are required in a square column and 6 bars for a circular column unless it is quite small. The minimum size of bars is 12mm

The minimum area of a column is defined in BS EN 1992-1 as

{ A }_{ smin }=\frac { 0.1{ N }_{ Ed } }{ 0.87{ f }_{ y } } \ge 0.002{ A }_{ c }

The maximum area of steel is defined as

{ A }_{ smax }\le 0.04{ A }_{ c }

Links in columns are there to primarily hold the bars together as they work to resist the axial load. They should be no less than 0.25 times the diameter of the compression reinforcement. Spacing is limited to 20 times that of the diameter of compression reinforcement or the smaller column dimension b or 400mm, whichever is smallest. However, for a distance h above and below the junction with a beam or slab, the spacing should be closed up to a factor of 0.6 of the spacing required for the rest of the column. All compression reinforcement must be restrained by transverse bars and cannot be more than 150mm from a restrained bar

Designing columns for axial forces and bending moment

Now that the forces in the column have been established and the detailing requirements are understood, the required reinforcement in the column can be calculated using design charts.

The two expressions that need to be used in order to read from the charts are as follows

\frac { { N }_{ Ed } }{ bh{ f }_{ ck } } \quad \& \quad \frac { { M }_{ Ed } }{ b{ h }^{ 2 }{ f }_{ ck } }

These values are read against the design chart. Each chart is based on the ratio of d2/h and is therefore selected using this value. These charts can be downloaded from the Concrete Centre’s website at: www.eurocode2.info

Biaxial Bending

Finally, having done the analysis and design one way, you have to do it in the other direction and check biaxial bending. Often it will be non-critical by inspection but one should check. The expression given below must be satisfied.

\left( \frac { { M }_{ Edz } }{ { M }_{ Rdz } } \right) ^{ a }+\left( \frac { { M }_{ Edy } }{ { M }_{ Rdy } } \right) ^{ a }\le 1
NEd/NRd 0.10.71.0
a1.01.52.0

The table given above is used for rectangular cross-sections, for circular sections a=2

where\quad { N }_{ Rd }={ A }_{ c }{ f }_{ cd }+{ A }_{ s }{ f }_{ yd }

Worked Example

A 3.5m high 350x350mm square column is required to support an axial load of 1687.06kN and a bending moment of -46.86kN.m at the top and 46.86 at the bottom about the y-y axis only. The column is to have a fire rating of 1 hour and to be constructed from C25/30 concrete with 410 mpa bars. It is unlikely to be exposed to water. Determine the compression reinforcement and containment links required in this column.

This example is taken from the post: Derivation of Actions on Concrete Columns

Cover to reinforcement Cnom =25mm

Slenderness verification

{ \lambda \le \lambda }_{ lim }
from\quad table\quad 1\quad { l }_{ e }=0.75\times 3500=2625mm
since\quad column\quad is\quad square\quad i=\frac { h }{ 3.46 } =\frac { 350 }{ 3.46 } =101.15
{ \lambda }_{ y }={ \lambda }_{ z }=\frac { { l }_{ e } }{ i } =\frac { 2625 }{ 101.15 } =25.96
{ \lambda }_{ lim }=\frac { 20\cdot A\cdot B\cdot C }{ \sqrt { n } } \quad A=0.7,\quad B=1.1
C=1.7-\frac { { M }_{ min } }{ { M }_{ max } } =1-\frac { -46.86 }{ 46.86 } =2.7
n=\frac { { N }_{ ed } }{ { A }_{ c }{ f }_{ cd } } =\frac { 1687.06\times { 10 }^{ 3 } }{ 350\times 350\times \frac { 0.85\times 25 }{ 1.5 } } =0.97
{ \lambda }_{ lim }=\frac { 20\times 0.7\times 1.1\times 2.7 }{ \sqrt { 0.97 } } =42.2
\lambda \le { \lambda }_{ lim };\quad (23.96\le 42.2)

Therefore the column can be classified as a short column

Design Moments

Moment in the y-y direction
{ M }_{ ed,y }=max({ M }_{ 02 }\quad ;{ \quad M }_{ nom })
{ M }_{ 02 }=max\left| { M }_{ bot }\quad ;{ M }_{ top } \right| +{ e }_{ i }{ N }_{ Ed }
{ e }_{ i }=\frac { { l }_{ 0 } }{ 400 } =\frac { 2625 }{ 400 } =6.56mm
{ M }_{ 02 }=46.86+(6.56\times 1687.06)\times { 10 }^{ -3 }=58kN.m
{ M }_{ nom }={ e }_{ o }{ N }_{ ed }
{ e }_{ o }=\frac { h }{ 30 } =\frac { 350 }{ 30 } =11.67\ge 20=20mm
{ M }_{ nom }=20\times 1687.06\times { 10 }^{ -3 }=33.74kN.m
{ M }_{ Ed,y }=max\left( 54\quad ,\quad 33.74 \right) =54kN.m
Moment in the z-z direction
{ M }_{ Ed,z }=max({ M }_{ 02 }\quad ;{ \quad M }_{ nom })\quad ;{ M }_{ 02 }=0
{ M }_{ Ed,z }={ e }_{ o }{ N }_{ ed }
{ M }_{ Ed,z }=33.74kN.m

Now that the design moments has been obtained, we will design the column in the most critical direction which is the y-y direction and then conclude by checking biaxial bending.

Design using Charts

{ d }_{ 2 }={ c }_{ nom }+\frac { \phi }{ 2 } +links\quad =25+\frac { 16}{ 2 } +8=41mm
\frac { { d }_{ 2 } }{ h } =\frac { 41 }{ 350 } =0.12
\frac { { N }_{ Ed } }{ bh{ f }_{ ck } } =\frac { 1687.06\times { 10 }^{ 3 } }{ 350\times { 350\times }{ 25 } } =0.55
\frac { { M }_{ Ed } }{ b{ d }^{ 2 }{ f }_{ ck } } =\frac { 58\times { 10 }^{ 6 } }{ 350\times { 350 }^{ 2 }\times 25 } =0.05
Interpolating\quad between\quad 0.10\quad \& \quad 0.15\quad of\quad \frac { { d }_{ 2 } }{ h }
\frac { { A }_{ s }{ f }_{ yk } }{ bh{ f }_{ ck } } =0.125
{ A }_{ s }=\frac { 0.125\times { 350 }^{ 2 }\times 25 }{ 410 } =933.68{ mm }^{ 2 }
Try\quad 4Y16mm+4Y12mm\quad bars\quad ({ A }_{ s,prov }=1256mm^{ 2 })

Verify minimum area of steel & containment links

Minimum Area
{ A }_{ s,min }=\frac { 0.10{ N }_{ Ed } }{ 0.87{ f }_{ yk } } \ge 0.002{ A }_{ c }
{ A }_{ s,min }=\frac { 0.1\times 1687.06\times { 10 }^{ 3 } }{ 0.87\times 410 } \ge 0.002\times { 350 }^{ 2 }
{ A }_{ s,min }=473mm^{ 2 }\le 1256{ mm }^{ 2 }
Containment links
{ \phi }_{ required }=0.25{ \phi }_{ bar }=0.25\times 16=4mm\quad Use\quad Y8
spacing\quad =\quad lesser\quad of\quad 20d\quad 350mm\quad or\quad 400
Place\quad links\quad at\quad 200mm\quad c/c

Biaxial Bending Verification

\left( \frac { { M }_{ Edz } }{ { M }_{ Rdz } } \right) ^{ a }+\left( \frac { { M }_{ Edy } }{ { M }_{ Rdy } } \right) ^{ a }\le 1
{ M }_{ Rd,z }={ M }_{ Rd,y }={ M }_{ Rd }
Assuming\quad 4Y16mm + 4Y12mm\quad bars
\frac { { A }_{ s }{ f }_{ yk } }{ bh{ f }_{ ck } } =\frac { 1256\times 410 }{ 350\times 350\times 25 } =0.168
\frac { { N }_{ ed } }{ bh{ f }_{ ck } } as\quad before\quad =0.55
interpolating\quad from\quad charts\quad for\quad { M }_{ Rd }
\frac { { M }_{ Rd } }{ { bh }^{ 2 }{ f }_{ ck } } =0.069
{ M }_{ Rd }=0.069\times 350\times { 350 }^{ 2 }\times 25=74kN.m
{ N }_{ Rd }={ A }_{ c }{ f }_{ cd }+{ A }_{ s }{ f }_{ yd }
=\frac { 350\times 350\times \times 0.85\times 25 }{ 1.5 } +1256\times \frac { 410 }{ 1.15 } =2530.3kN
\frac { { N }_{ Ed } }{ { N }_{ Rd } } =\frac { 1687.06 }{ 2530.3 } =0.67
interpolating\quad from\quad table\quad a=1.46
\left( \frac { 54 }{ 74 } \right) ^{ 1.46 }+\left( \frac { 33.74 }{ 74 } \right) ^{ 1.46 }=0.64+0.31=0.94<1\quad O.k

Therefore Provide 4Y16mm + 4Y12mm bars and Y8 containment links at 200mm c/c.

Further Reading & References

  • The Institution of Structural Engineers (2006) Manual for the design of concrete building structures to Eurocode 2 London: The Institution of Structural Engineers
  • The Concrete Centre (2009) Worked Examples to Eurocode 2: Volume 1 [Online] Available at: www.concretecentre.com/ pdf/Worked_Example_Extract_Slabs.pdf (Accessed: February 2013)

THANKYOU!!!

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