# Designing a Slender Concrete Column to Eurocode

A column is considered slender when second-order effects ( i.e. additional effects caused by structural deformations) cannot be safely ignored in the analysis and design of the column. Generally, a column is slender if it cross-sectional dimensions is small compared to its length.

Reinforced concrete columns have been the subject of a previous post. However, in that post, the design of slender columns was deliberately ignored in order not to excessively complicate the subject. This post presents a simplified approach to the design of slender concrete columns using the Eurocode.

According to the Eurocode, when the actual slenderness is less than the limiting slenderness, the column is slender and will fail by buckling rather than crushing which is the case of short columns. When a column is slender, it deflects with the application of axial loads, and as it does so, an additional bending moment known as the second-order moment is induced which causes the column to buckle.

To be clear, the procedure for designing a slender column is basically the same as that of a stocky column. The only concern here is taking into consideration the second-order moment due to the buckling effect when calculating design moments.

Kindly review the previous post on concrete columns, if you are new to designing concrete columns to Eurocode. Some of these principles have been laid down already and will not be repeated in this post. Designing a Concrete Column to Eurocode

#### Design Moments

As stated in the preceding section, the only difference and perhaps extra complexity in designing slender columns arise from the need to consider second-order effects. The methods of analysis available include a general method, for 2nd order effects based on non-linear second-order analysis and the following two simplified methods

• Nominal stiffness method
• Nominal curvature method

The nominal curvature method is considered primarily more suited for isolated members with defined effective length and a constant axial force. The method is based on applying a second-order moment due to buckling by calculating the deflection of the column which depends on the effective length and an estimated curvature. Refer to clause 5.8.8.3 of BS EN 1992-1-1 for more guidance.

Generally, the design value of bending moment in a slender column about any axis is defined as

{ M }_{ Ed }=max\left\{ { M }_{ 02 };{ M }_{ 0e }+{ M }_{ 2 };{ M }_{ 01 }+{ 0.5M }_{ 2 } \right\} \ge { M }_{ nom }

Where:

M02 and M01 are the first order moments defined as

{ M }_{ 02 }=max\left| { M }_{ bot };{ M }_{ top } \right| +{ e }_{ i }{ N }_{ Ed }\quad { e }_{ i }=\frac { { l }_{ o } }{ 400 }
{ M }_{ 01 }=min\left| { M }_{ bot };{ M }_{ top } \right| -{ e }_{ i }{ N }_{ Ed }\quad { e }_{ i }=\frac { { l }_{ o } }{ 400 }

M0e is the moment at mid height defined as

{ M }_{ 0e }=\left( 0.6{ M }_{ 02 }+0.4{ M }_{ 01 } \right) \ge 0.4{ M }_{ 02 }

The second order moment is defined as

{ M }_{ 2 }={ N }_{ Ed }{ e }_{ 2 }

Where: e2 is the second-order eccentricity, calculating this eccentricity using the method given in BS EN 1992-1-1 can be sometimes tedious. Therefore the institution of Structural Engineers manual for concrete design provided a simpler expression for calculating this eccentricity. It is defined as.

{ e }_{ 2 }={ f }_{ yk }\frac { { l }_{ o }^{ 2 } }{ d } \cdot { 10 }^{ -6 }

Where:

• fyk is the characteristic strength of the reinforcement in MPa
• lo is the effective length of the column in the direction considered
• d is the effective depth of section in the direction considered

The expression was derived using fck of 30mpa. BS EN 1992-1-1 requires an increase in the second order eccentricity for higher strength concrete to account for creep. As a result for an edge or corner column e2 may be increased by:

• 10% for fck = 35MP
• 20% for fck = 40MPa

The method provided here is a simplified method of calculating the second-order eccentricity, the detailed method in BS EN 1992-1-1 is different and slightly more accurate. For more details on this kindly refer to the texts in the further reading section of this post.

Once the design moments have been obtained by considering the second-order effect, the design of the column is then carried out the same way as for a stocky column.

### Worked Example

A 4.0m high 225×225 mm square column is required to support an axial load of 715kN and a bending moment of 32kN.m at the top and bottom in the y-y axis and 15kN.m at the top and bottom in the z-z axis. 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. Take the effective length factor = 0.85

Cover to concrete cnom =25mm

#### Slenderness Verification

{ l }_{ oy }={ l }_{ oz }=0.85\times 4000=3400mm
{ i }_{ yy }={ i }_{ zz }=\frac { h }{ 3.46 } =\frac { 225 }{ 3.46 } =65.03mm
{ \lambda _{ y } }={ \lambda }_{ z }=\frac { 3400 }{ 65.03 } =52.28
{ \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 } } =1.7-\frac { -32 }{ 32 } =2.7
n=\frac { { N }_{ Ed } }{ { A }_{ c }{ f }_{ cd } } =\frac { 715\times { 10 }^{ 3 } }{ { 225 }^{ 2 }\times \frac { 0.85\times 25 }{ 1.5 } } =0.997
{ \lambda }_{ lim }=\frac { 20\times 0.7\times 1.1\times 2.7 }{ \sqrt { 0.997 } } =41.64
\lambda >{ \lambda }_{ lim }\quad (52.28>41.64)

Since actual slenderness exceeds the limiting slenderness, the column is slender about the y-y and z-z axis.

#### Design Moments

##### Moment in the y-y direction
{ M }_{ Ed,y }=max\left\{ { M }_{ 02,y };{ M }_{ 0e,y }+{ M }_{ 2,y };{ M }_{ 01,y }+0.5{ M }_{ 2,y } \right\} \ge { M }_{ nom }
{ M }_{ 02,y }=max\left| { M }_{ top };{ M }_{ bot } \right| +{ e }_{ i }{ N }_{ Ed }
max\left( 32+\frac { 3400 }{ 400 } \times 715 \right) =38.08kN.m
{ M }_{ 01,y }=min\left| { M }_{ top };{ M }_{ bot } \right| -{ e }_{ i }{ N }_{ Ed }
min\left( 32-\frac { 3400 }{ 400 } \times 715 \right) =23.92kN.m
{ M }_{ 0e,y }=(0.6{ M }_{ 02,y }+{ 0.4M }_{ 01,y })\ge 0.4{ M }_{ 02,y }
0.6(38.08)+0.4(-23.92)\ge 0.4(38.08)
{ M }_{ 0e,y }=15.23kN.m
{ M }_{ 2,y }={ N }_{ Ed }{ e }_{ 2 }
{ e }_{ 2 }={ f }_{ yk }\frac { { l }_{ o }^{ 2 } }{ d } \cdot { 10 }^{ -6 }=410\times \frac { { 3400 }^{ 2 } }{ 179.5} \cdot { 10 }^{ -6 }=26.4mm
{ M }_{ 2,y }=715\times 26.4\times { 10 }^{ -3 }=18.88kN.m
{ M }_{ nom }={ N }_{ Ed }{ e }_{ o }
{ e }_{ o }=\frac { h }{ 30 } =\frac { 225 }{ 30 } =7.5\ge 20mm
{ M }_{ nom }=715\times 20\times { 10 }^{ -3 }=14.3kN.m
{ M }_{ Ed,y }=max\left\{ 38.08;15.23+18.88;23.92+0.5(18.60) \right\} \ge 14.3
{ M }_{ Ed,y }=38.08kN.m
##### Moment in the z-z axis

The effect of imperfections have been considered in the y-y axis, therefore the effect will be ignored in the z-z axis.

{ M }_{ Ed,z}=max\left\{ { M }_{ 02,z };{ M }_{ 0e,z }+{ M }_{ 2,z };{ M }_{ 01,z }+0.5{ M }_{ 2,z } \right\} \ge { M }_{ nom }
{ M }_{ 02,z }=max\left| { M }_{ top };{ M }_{ bot } \right| =15kN.m
{ M }_{ 01,z }=min\left| { M }_{ top };{ M }_{ bot } \right| =15kN.m
{ M }_{ 0e,z }=(0.6{ M }_{ 02,y }+{ 0.4M }_{ 01,y })\ge 0.4{ M }_{ 02,y }
0.6(15)+0.4(-15)\ge 0.4(15)
{ M }_{ 0e,z }=6kN.m
{ M }_{ 2,z }={ N }_{ Ed }{ e }_{ 2 }
{ e }_{ 2 }={ f }_{ yk }\frac { { l }_{ o }^{ 2 } }{ d } \cdot { 10 }^{ -6 }=410\times \frac { { 3400 }^{ 2 } }{ 179.5 } \cdot { 10 }^{ -6 }=26.4mm
{ M }_{ 2,z }=715\times 26.4\times { 10 }^{ -3 }=18.88kN.m
{ M }_{ nom }={ N }_{ Ed }{ e }_{ o }
{ e }_{ o }=\frac { h }{ 30 } =\frac { 225 }{ 30 } =7.5\ge 20mm
{ M }_{ nom }=715\times 20\times { 10 }^{ -3 }=14.3kN.m
{ M }_{ Ed,z }=max\left\{ 15;6.0+18.88;15+0.5(18.88) \right\} \ge 14.3
{ M }_{ Ed,z }=24.44kN.m

Haven obtained the design moment in both directions, we will design the column in the most critical direction y-y and then conclude by checking the detailing requirements.

#### Design using Charts

{ d }_{ 2 }={ c }_{ nom }+\frac { \phi }{ 2 } +links\quad =25+\frac { 25 }{ 2 } +8=45.5mm
\frac { { d }_{ 2 } }{ h } =\frac { 45.5 }{ 225 } =0.20
\frac { { N }_{ Ed } }{ bh{ f }_{ ck } } =\frac { 715\times { 10 }^{ 3 } }{ { 225 }^{ 2 }\times 25 } =0.56
\frac { { M }_{ Ed } }{ { bh }^{ 2 }{ f }_{ ck } } =\frac { 38.08\times { 10 }^{ 6 } }{ 225^{ 3 }\times 25 } =0.13
\frac { { A }_{ s }{ f }_{ yk } }{ bh{ f }_{ ck } } =0.52
{ A }_{ s }=\frac { 0.52bh{ f }_{ ck } }{ { f }_{ yk } } =\frac { 0.52\times { 225 }^{ 2 }\times 25 }{ 410 } =1605.18{ mm }^{ 2 }

#### Detailing Requirements & Containment Links

##### Minimum Area of Steel
{ A }_{ s,min }=\frac { 0.1{ N }_{ Ed } }{ 0.87{ f }_{ yk } } \ge 0.002{ A }_{ c }
=\frac { 0.1\times 715\times { 10 }^{ 3 } }{ 0.87\times 410 } \ge 0.002\times { 225 }^{ 2 }
{ A }_{ s,min }=200.45{ mm }^{ 2 }
##### Maximum Area of Steel
{ A }_{ s,max }=0.04{ A }_{ c }
0.04\times 225\times 225={ 2025mm }^{ 2 }
{ \phi }_{ req }=0.25{ \phi }_{ bar }=0.25\times 25=6.25mm\quad use\quad 8mm