Design of Column to BS 8110 using Design Chart – Worked Example

Column design charts are created to simplify the design of reinforced concrete columns. These charts are published by British Standard Institution (BSI) and are available in part 3 of BS 8110. They are applicable to rectangular columns with symmetrical arrangement of reinforcement. Each chart is suitable to certain columns with peculiar parameters which are:

 

• Characteristic strength of concrete (fcu)
• Characteristics strength of reinforcement (fy)
• Ratio between the effective depth and column height. (d/h)

The charts are available for columns of concrete grade 25, 30, 35, 40, 45, and 50. They are available in d/h ratio between 0.75 and 0.95 at 0.05 increments. The reinforcement grade for which they are applicable is 460N/mm2.

 

Worked Example

This article presents a worked example on how to design a column using these charts. The column that shall be designed is column EF which has been previously analysed as part of a frame using moment distribution method. The moment in the column has been obtained already and it has been evaluated to be 11.2KNm. We are left with evaluating the axial load on the column.

The axial force of a column in a structural frame may be calculated on the assumption that the beams and slabs transmitting force into the particular column are simply supported. By this, the reaction from the column can be calculated which in turn serves as the load on the column. The reaction on the column shall be calculated as follows:

 

We shall first provide the subframe below for easy reference.

 

From basic load take down technique, column EF supports half of the load acting on both Beams BE and EH. Since the load acting on Beams BE and EH are 207.5 and 117.6 respectively, then the column reaction is:

207.5/2 + 117.6/2 = 162.55KN

Having calculated the axial load on the column, we can now proceed to write down the loads and other design parameters necessary for the column design.

Moment on the column (M2) = 11.2KNm

Axial load on the column (N) = 162.55 KN

Length = 4m

b = 300mm

h = 350mm

fcu = 25N/mm2

cover = 25mm

Steps in designing the column

• Check the Slenderness of the column.

The slenderness ratio of the column shall be compared against the limiting slenderness, if the slenderness ratio is greater than the limiting slenderness then the column is declared “slender”. The column is declared “short” if otherwise. This is shown in the below steps

1. Calculate the effective height of the column

Since the column is braced, the Effective height (lo) is calculated thus:

le = β lo

a)     Calculate the clear height (lo) of the column

lo (clear height of the column) = 4000 – 600 x 0.5 – 600 x 0.5 = 3400mm

(The clear height of the column is the obvious height of the column between end restraints. In this case, the clear height of the column is taken to midway of the depth of both top and bottom beams framing into the column)

b)     Evaluate the relative stiffness at the column ends

β is a coefficient dependent on the end condition of the column. Its value can be obtained from table 3.19 and 3.20 of BS 8110-1:1997, however a rigorous approach will be adopted here according to clause 2.5 of BS 8110-2: 1985.

According to BS 8110 part 2, for braced column B is adopted as the lesser of:

β = 0.7 + 0.05(αc1 + αc2)

β = 0.85 + 0.05αcmin

αc2 and αc1 are the relative flexibilities of the top and bottom ends respectively

αc2 =    $
\frac{\frac{\varSigma I_c}{l}}{\frac{\varSigma EI_b}{l}}
$

$
\frac{\frac{\frac{bh^3}{12}}{l}}{\frac{\frac{bh^3}{12}}{l}\,\,+\,\,\frac{\frac{bh^3}{12}}{l}}
$

$
\frac{\frac{\frac{300 X 350^3}{12}}{4000}}{\frac{\frac{300 X 600^3}{12}}{6000}\,\,+\,\,\frac{\frac{300 X 600^3}{12}}{6000}}
$

= 31525.35/(900000 + 900000)

= 0.175

We do the same thing all over again to get αc1

αc1 =    $
\frac{\frac{\varSigma I_c}{l}}{\frac{\varSigma EI_b}{l}}
$

$
\frac{\frac{\frac{bh^3}{12}}{l}}{\frac{\frac{bh^3}{12}}{l}\,\,+\,\,\frac{\frac{bh^3}{12}}{l}}
$

$
\frac{\frac{\frac{300 X 350^3}{12}}{4000}}{\frac{\frac{300 X 600^3}{12}}{6000}\,\,+\,\,\frac{\frac{300 X 600^3}{12}}{6000}}
$

= 31525.35/(900000 + 900000)

= 0.175

β = 0.7 + 0.05(0.175 + 0.175) =  0.7175

β = 0.85 + 0.05 x 0.175 = 0.8588

Adopt the lesser as

β = 0.7175

 

c.   Determine the effective height

              l = β lo

 l = 0.7175 x 2439.5 =   2439.5

2.     Compare the slenderness ratio against the limiting slenderness.

lex/h and ley/b < 15 – braced column

lex/h = 2439.5/350 = 6.97

ley/b = 2439.5/300 = 8.13

Since lex/h and ley/b are less than 15 then the column is short and additional moment due to slenderness is not required.

 

3.    Compute the design moment (M)

 

The design moment is the greater between the end moment (M2) and the moment due to nominal eccentricity (Nemin)

emin = 0.05 X h (≤ 20mm)

emin = 0.05 X 350 (≤ 20mm)

emin = 17.5mm (≤ 20mm)

Since 17.5mm is less than 20mm, the minimum eccentricity is taken as 17.5mm

Nominal moment= 0.00175 X 162.55 = 0.28KNm

Since the minimum moment (0.28KNm) is less than M2(11.2KNm), hence the critical moment to be adopted for the design is 11.2KNm

4.   Use the design Chart to determine the area of longitudinal reinforcement required

 

In order to use the design chart, few parameters have to be calculated;

 

a.    Determine N/bh

N/bhf = 162.55 X 10^3 /300 X 350   = 1.55

 

b.     Determine M/bh²

M/bh² =    14.451 X 10^6/300 X 350²   = 0.4

 

c.    Calculate d/h

d/h determines the exact graph that shall be used to determine the reinforcement area.

d = h- c + ϕ/2 + ϕ

= 350 – 25 – 10 – 8

d/h   = 307/350 = 0.88

This is approximated to 0.9 and the design chart with parameters which conforms with d/h = 0.9, fcu = 25N/mm2, and fy = 460N/mm2 is used. Chart 25 in BS 8110 Part 3 matches this. The chart is reproduced below.

From the design chart, the intersection of M/bh² = 0.4 and N/bh = 1.55 falls below the 0.4 mark. This implies there is no need to provide reinforcement for the column. However, to keep to BS 8110 directives minimum area of reinforcement shall be provided.

100Asc/Acol = 0.4

As = 0.004 X 300 X 350

As = 420mm2

Hence provide 4Y12 (449.856mm2) as the longitudinal reinforcement of the column

 

5.   Determine the size of links required and spacing

 

a.  Link diameter

The link diameter is the greater of;

•       ¼ x maximum diameter of longitudinal bar
•       6mm

¼ x maximum diameter of longitudinal bar = ¼ x 12 = 3mm

Hence use 6mm as the link diameter.

b.  Link Spacing

The link spacing should not be greater than 12 x minimum diameter of longitudinal bar

 

• 12 x minimum diameter of longitudinal bar = 12 x 12 = 144mm

 Provide 6mm links at 125mm spacing.