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6.0 Column

6.1 Introduction

Is usually refers to the structural member that placed

vertically.

Is normally subject to compression and bending forces.

A small compression member, such as that in a framed

structure, is known as a strut.

A larger member, such as the main support for a beam in a

building, is known as a column, or more traditionally a

stanchion.

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Axially loaded compression members can fail in two

principal ways:

• Short fat members fail by crushing or splitting of the

material. This is a strength criterion.

Load subjected to a column and the possible failure mode

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Axially loaded compression members can fail in two

principal ways:

• Long thin members fail by sideways buckling. This is

stiffness criterion.

Figure 6-1: Load subjected to a column and the possible failure mode

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6.2 Basis of Design

i) Short axially loaded columns

When its height to minimum width ratio is less than 15 if the

top restrained against lateral movement and less than 10 if

unrestrained.

In practice it is usually only reinforced concrete or brickwork

columns fall into this category.

The ultimate compressive load capacity, N, to be the sum of

the strength of both the concrete and steel components.

N= 0.4 fcu Ac + 0.75 fy Asc

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Where

fcu = characteristic concrete cube crushing strength

Ac = area of concrete

fy = characteristic yield stress of steel

Asc = area of steel

The cross-section

of a reinforced

concrete column

Ac

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Example 1

A short reinforced concrete column is to support the

following axial loads:

Characteristic dead load = 758 kN

Characteristic imposed load = 630 kN

If the column is to measure 325 mm x 325 mm and the

concrete characteristic strength is 30 N/mm2, determine

the required size of high yield reinforcing bar, and specify

suitable links.

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Example 1

Solution:

Design load = 1.4Gk + 1.6 Qk

= 1.4 x758 + 1.6 x 630

= 2069 kN

Using N= 0.4 fcu Ac + 0.75 fy Asc

2069 x 103 = 0.4 x 30 x (325 x 325) + 0.75 x 460 x Asc

Asc = 2323 mm2

Using 4 bars areas/bar = 581 mm2

From Table 6.1

Area of 32 mm dia. bar = 804 mm2

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Example 1

Bar dia. (mm) 6 8 10 12 16 20 25 32 40

C/s area(mm2)

28 50 79 113 201 314 491 804 1256

Table 6.1

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Example 1

Solution:

Steel percentage = 4 x 804 x 100%

(325 x 325)

= 3.04 %

This is between 0.4 % and 6% and therefore satisfactory

Minimum diameter of links = 32/4 = 8 mm

Maximum spacing of links = 32 x 12 = 384 mm

Answer : Use four 32 mm diameter bars with 8 mm

diameter links at 350 mm spacing

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6.2 Basis of Design

ii) Slender Columns

A slender column fails by side-ways buckling and the

length of column makes a significant difference.

The load at which a slender column buckles is known as

its critical buckling load, Pcrit .

The value of the critical buckling load for a slender

column is given by the Euler buckling formula.

Pcrit = 2EI

L2

Where E = modulus of elasticity

I = second moment of area

L = length between pins

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Real strut tend to be design using design chart which take

into account imperfections in the member. The strength pc is

based on the slenderness ratio of the member.

- λ, LE/r, the slenderness

where LE is the effective length of the strut

r is the radius .of gyration of the cross-section about

the axis of buckling (r = √(I/A))

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The pc is determined based on Perry strut formula.

This is based on elastic analysis of pin ended centrally

loaded strut which has an initial lateral deflection.

Perry assumed that failure would occur when the

maximum stress was equal to the yield (py in limit state

design).

To simplify the calculation of pc including consideration

on the member's imperfection (e.g. out-of-straightness,

particularly for residual stresses arising from rolling or

welding) could be summarized in the strut curve (see

Figure 6-2).

In BS, the pc is presented in tabulated form in Table 23

and Table 24.

Part of the Table 23 is shown in Table 5-1 below.Jan-May 2016

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Figure 6-2: Strut Curve

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Type of sectionMaximum

thickness

Axis of buckling

x-x y-y

Rolled I-section<40mm

≥40mm

a)

b)

b)

c)

Rolled H-section<40mm

≥40mm

b)

c)

c)

d)

Welded I- or H-section (see not 2 and

4.7.5)

<40mm

≥40mm

b)

b)

c)

d)

NOTE 2 For welded I- or H-section with their flanges thermally cut by machine without

subsequent edge grinding or machining, for buckling about the y-y axis, strut curve b)

may be used for flange up to 40mm thick and strut curve c) for flange over 40mm thick

Allocation of strut curve

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Effective Length, LE

It is representing the length of the part of column

which is susceptible to flexural deformation ( see

Figure 6-3). The effective length of column without

pin ends must be adjusted before the slenderness ratio

is evaluated.

Figure 6-3: Effective length of various end connection

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a) non-sway mode

Restraint ( in the plane under consideration) by other pats of

the structure

LE

Effective held in

position at both

ends

Effectively restrained in direction at both

ends

0.7L

Partially restrained in direction at both ends 0.85L

Restrained in direction at one end 0.85L

Not restrained in direction at either ends 1.0L

b) Sway mode

One end Other end LE

Effectively held

in position and

restrained in

direction

Not held in

position

Effectively restrained in

direction

1.2L

Partially restrained in direction 1.5L

Not restrained in direction 2.0L

Excluding angle, channel or T-section struts designed in accordance with 4.7.10.

Nominal effective length LE for a compression member.

Table 22 (pg 79) – BS5950

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6.3 Design Procedure for Columns in Simple Constructions

6.3.1General

The simple construction is refers to the structural form in which

the joints assumed not to develop moments (pin jointed).

For a column in simple construction design, it is not necessary to

consider the effects on column of pattern loading.

All beams supported by a column at any level of the structure

should be assumed to be fully loaded.

The considerations taken in a column design for simple construction include:

- the axial compressive resistance of the column

- moment due to eccentricity

- moment distribution in multi-storey continuous columns

- the effects due to interaction of axial load and nominal moments

- buckling resistance of the columnJan-May 2016

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Example 2

Determine the critical Euler buckling load of a solid

aluminium rod of diameter l2 mm if it has a length of l.0

m between pin- ended supports.

Solution :

Pcrit = 2EI where E = 70 000 N/mm2

L2 I = r4/4 = x 64/4 =1018 mm2

= 2 x 70 000 x 1018 = 703 N

10002

Answer : Critical Euler buckling load = 703 N

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Example 3

A pin-ended column in a building is 4 m long and is subjected

to the following axial loads :

Characteristic dead load = 350 kN

Characteristic imposed load = 300 kN

Determine the dimension of a suitable standard universal

column section.

Solution :

Step 1

Design load = 1.4Gk + 1.6 Qk

= 1.4 x350 + 1.6 x 300

= 970 kNJan-May 2016

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Solution :

Step 2

Approximate area = design load

estimated stress

= 970 x 103 = 9 700 mm2 = 97 cm2

100

Step 3 (Trial 1)

Try 254 x 254 x 89 kg/m universal column

(A = 114 cm2 , r min = 6.52 cm)

Step 4

Slenderness ratio = LE = 4000 = 61.3

r min 65.2Jan-May 2016

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Solution :

From figure given

Compressive strength, pc = 189 N/mm2

Actual stress = Design load = 970 x 103

Area 114 x 102

= 85 N/mm2

Step 5

The above column would be safe but not very economic

as the loads produce a stress which is less than half the

compressive strength of the member. It is worth trying a

smaller section in this case.

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Solution :

Step 6 (Trial 2)

Try 203 x 203 x 52 kg/m universal column

(A = 66.4 cm2 , r min = 5.16 cm)

Step 7

Slenderness ratio = LE = 4000 = 77.5

r min 51.6

From figure 8.6

Compressive strength, pc = 165 N/mm2

Actual stress = Design load = 970 x 103

Area 66.4 x 102

= 146 N/mm2Jan-May 2016

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Solution :

As 146 N/mm2 is less than 165 N/mm2 this section is

satisfactory.

Answer : Use 203 x 203 x 52 kg/m universal column.

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