Simplified engineering for architecture students

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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.
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Figure 6-2: Strut Curve
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Type of section
Maximum
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 kN
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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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