Compression

Members
Version 2 CE IIT, Kharagpur

Lesson

ShortAxially
Loa
Compression

Members
Version 2 CE IIT, Kharagpur

instructional Objectives:
At the end of this lesson, the student should be able to:

o state additional assumptions regarding the strengths of concrete and steel
for the design of short axially loaded columns,
o specify the values of design strengths of concrete and steel,
o derive the governing equation for the design of short and axially loaded
tied columns,

o derive the governing equation for the design of short and axially loaded
spiral columns,
o derive the equation to determine the pitch of helix in spiral columns,
o apply the respective equations to design the two types of columns by
direct computation,
o use the charts of SP-16 to design these two types of columns subjected to
axial loads as per IS code.

10.22.1

Introduction

Tied and helically bound are the two types of columns mentioned in
sec.10.21.3 of Lesson 21. These two types of columns are taken up in this
lesson when they are short and subjected to axially loads. Out of several types of
plan forms, only rectangular and square crosssections are covered in this lesson
for the tied columns and circular cross-section for the helically bound columns.
Axially loaded columns also need to be designed keeping the provision of
resisting some moments which normally is the situation in most of the practical
columns. This is ensured by checking the minimum eccentricity of loads applied
on these columns as stipulated in IS 456. Moreover, the design strengths of
concrete and steel are further reduced in the design of such columns. The
governing equations of the two types of columns and the equation for
determining the pitch of the helix in continuously tied column are derived and
explained. The design can be done by employing the derived equation i.e., by
direct computation or by using the charts of SP-16. Several numerical examples
are solved to explain the design of the two types of columns by direct
computation and using the charts of SP-16.

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10.22.2

Further Assumptions Regarding the Strengths of

Concrete

and Steel

All the assumptions required for the derivation of the governing equations
are given in sec.10.21.11 of Lesson 21. The stress-strain diagrams of mild steel
(Fe 250) and cold worked deformed bars (Fe 415 and Fe 500) are given in
Figs.1.2.3 and 4, respectively of Lesson 2. The stress block of compressive part
of concrete is given in Fig.3.4.1.9 of Lesson 4, which is used in the design of
beam by limit state of collapse. The maximum design strength of concrete is
shown as constant at 0.446 fck when the strain ranges from 0.002 to 0.0035. The
maximum design stress of steel is 0.87 fy.
Sections 10.21.4 and 12 of Lesson 21 explain that all columns including
the short axially loaded columns shall be designed with a minimum eccentricity
(cls. 25.4 and 39.2 of IS 456). Moreover, the design strengths of concrete and
steel are further reduced to 0.4 fckand 0.67 fy, respectively, to take care of the
minimum eccentricity of 0.05 times the lateral dimension, as stipulated in c|.39.3
of IS 456. It is noticed that there is not attempt at strain compatibility. Also the
phenomenon of creep has not been directly considered.
exmin 2 greater of (I/500 + D/30) or 20 mm

(10.3)
eymin
2greater
of(I/500
+b/30)
or20mm
The maximum values of lax/D and lay/b should not exceed 12 in a short column
as per c|.25.1.2 of IS 456. For a short column, when the unsupported length l=
lax(for the purpose of illustration), we can assume l= 12 D (or 12b when b is
considered). Thus, we can write the minimum eccentricity = 12D/500 + D/30 =
0.057D, which has been taken as 0.05D or 0.05b as the maximum amount of

eccentricity of a short column.
It is, therefore, necessary to keep provision so that the short columns can
resist the accidental moments due to the allowable minimum eccentricity by
lowering the design strength of concrete by ten per cent from the value of
0.446fck, used for the design of flexural members. Thus, we have the design
strength of concrete in the design of short column as (0.9)(0.446fck)= 0.4014fck,
say 0.40 fck.The reduction of the design strength of steel is explained below.
For mild steel (Fe 250), the design strength at which the strain is 0.002 is
fy/1.15 = 0.87f,,. However, the design strengths of cold worked deformed bars (Fe
415 and Fe 500) are obtained from Fig.1.2.4 of Lesson 2 or Fig.23A of IS 456.
Table A of SP-16 presents the stresses and corresponding strains of Fe 415 and
Fe 500. Use of Table A of SP-16 is desirable as it avoids error while reading from
figures (Fig.1.2.4 or Fig.23A, as mentioned above). From Table A of SP-16, the

corresponding design strengths are obtained by making linear interpolation.
These values of design strengths for which the strain is 0.002 are as follows:
(i) Fe 415: {o.9r,., + o.o5r,.,(o.oo2
0.789f,,

0.00192)/(o.oo241

o.oo192)} = o.9o8r,., =

(ii) Fe 500: {o.85f,., + o.o5ry,,(o.oo2 0.00195)/(0.00226
0.746f,,

o.oo195)} = 0.859f,, =

A further

reduction

in each

of three

values

is made

to take

care

of the

minimum eccentricity as explained for the design strength of concrete. Thus, the
acceptable design strength of steel for the three grades after reducing 10 per
cent from the above mentioned values are 0-783fy, 0.710f,, and 0.671f,, for Fe
250, Fe 415 and Fe 500, respectively. Accordingly, cl. 39.3 of IS 456 stipulates
0.67fy as the design strength for all grades of steel while designing the short
columns. Therefore, the assumed design strengths of concrete and steel are
0.4fckand 0.67fy, respectively, for the design of short axially loaded columns.

10.22.3 Governing Equation for Short Axially Loaded Tied
Columns
Factored concentric load applied on short tied columns is resisted by
concrete of area Ac and longitudinal steel of areas Asaeffectively held by lateral
ties at intervals (Fig.10.21.2a of Lesson 21). Assuming the design strengths of
concrete and steel are 0.4fck and 0.67f,,, respectively, as explained in sec.
10.22.2, we can write

Pu = 0.4fckAc+ 0.67fyAsc
(10.4)
where Pu = factored axial load on the member,

fck = characteristic compressive strength of the concrete,
Ac = area of concrete,

fy = characteristic strength of the compression reinforcement, and
A50= area of longitudinal reinforcement for columns.
The above equation, given in cl. 39.3 of IS 456, has two unknowns Ac and A50to
be determined from one equation. The equation is recast in terms of Ag, the
gross area of concrete and p, the percentage of compression reinforcement
employing

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A5,, = pAg/100
(10.5)

(10.6)
Accordingly, we can write
Pu/Ag = o.4r,,k+ (p/100) (o.67r,o.4r,,k)
(10.7)
Equation 10.7 can be used for direct computation of Ag when Pg, fckand fy are
known by assuming p ranging from 0.8 to 4 as the minimum and maximum
percentages of longitudinal reinforcement. Equation 10.4 also can be employed
to determine Ag and p in a similar manner by assuming p. This method has been
illustrated with numerical examples and is designated as Direct Computation
Method.

On the other hand, SP-16 presents design charts based on Eq.10.7. Each
chart of charts 24 to 26 of SP-16 has lower and upper sections. In the lower
section, P,/Ag is plotted against the reinforcement percentage p(= 100As/Ag)for
different grades of concrete and for a particular grade of steel. Thus, charts 24 to
26 cover the three grades of steel with a wide range of grades of concrete. When
the areas of crosssection of the columns are known from the computed value of
Pu/Ag,the percentage of reinforcement can be obtained directly from the lower
section of the chart. The upper section of the chart is a plot of Pu/Ag versus Pu
for different values of Ag. For a known value of Pu, a horizontal line can be drawn
in the upper section to have several possible Ag values and the corresponding
Pu/Agvalues. Proceeding vertically down for any of the selected Pu/Agvalue, the
corresponding percentage of reinforcement can be obtained. Thus, the combined
use of upper and lower sections of the chart would give several possible sizes of
the member and the corresponding Asc without performing any calculation. It is
worth mentioning that there may be some parallax error while using the charts.
However, use of chart is very helpful while deciding the sizes of columns at the
preliminary design stage with several possible alternatives.
Another advantage of the chart is that, the amount of compression
reinforcement obtained from the chart are always within the minimum and
maximum percentages i.e., from 0.8 to 4 per cent. Hence, it is not needed to
examine if the computed area of steel reinforcement is within the allowable range
as is needed while using Direct Computation Method. This method is termed as
SP-16 method while illustrating numerical examples.

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10.22.4

Governing Equation of Short Axially Loaded

Columns

with

Columns

Helical
with

helical

Ties
reinforcement

take

more

load

than

that

of

tied

columns due to additional strength of spirals in contributing to the strength of
columns. Accordingly, cl. 39.4 recommends a multiplying factor of 1.05 regarding
the strength of such columns. The code further recommends that the ratio of
volume

of helical

reinforcement

to the volume

of core

shall

not be less than

0.36

(Ag/Ac 1) (fck/fy),in order to apply the additional strength factor of 1.05 (cl.
39.4.1). Accordingly, the governing equation of the spiral columns may be written
as

P, = 1.05(0.4 r,,kAc+ 0.67 r,A,,,)
(10.8)
All the terms have been explained in sec.10.22.3.
Earlier observations of several investigators reveal that the effect of
containing holds good in the elastic stage only and it gets lost when spirals reach
the yield point. Again, spirals become fully effective after spalling off the concrete
cover over the spirals due to excessive deformation. Accordingly, the above two
points should be considered in the design of such columns. The first point is
regarding the enhanced load carrying capacity taken into account by the
multiplying factor of 1.05. The second point is maintaining specified ratio of
volume of helical reinforcement to the volume of core, as specified in c|.39.4.1
and mentioned

earlier.

The second point, in fact, determines the pitch p of the helical
reinforcement, as explained below with reference to Fig.10.21.2b of Lesson 21.

Volumeof helical reinforcementin one loop = 7r(Dc- ¢sp)asp
(10.9)

Volumeof core = (7:/4)Df p
(10.10)
where Dc = diameter of the core (Fig.10.21.2b)

¢,p

diameterof the spiral reinforcement(Fig.10.21.2b)

asp = area of crosssection of spiral reinforcement
p

= pitch of spiral reinforcement (Fig.10.21.2b)

To satisfy the condition of c|.39.4.1 of IS 456, we have
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{7r(Dc
-¢xp)axp}/{(7r/4)
Df p} 2 0.36(Ag
/Ac-1)(fck/fy)
which finally gives

p s 111(1),-¢,p)a,p
fy/(D2-D3)f,,
(10.11)
Thus, Eqs.10.8 and 11 are the governing equations to determine the diameter of
column, pitch of spiral and area of longitudinal reinforcement. It is worth
mentioning that the pitch p of the spiral reinforcement, if determined from
Eq.10.11, automatically satisfies the stipulation of c|.39.4.1 of IS 456. However,
the pitch and diameter of the spiral reinforcement should also satisfy cl. 26.5.3.2
of IS 456:2000.

10.22.5 Illustrative Examples
Problem

1:

Design the reinforcement in a column of size 400 mm x 600 mm subjected
to an axial

load

of 2000

kN under

service

dead

load

and

live

load.

The

column

has an unsupported length of 4.0 m and effectively held in position and
restrained against rotation in both ends. Use M 25 concrete and Fe 415 steel.
Solution

1:

Step 1: To check if the column is short or slender
Given l= 4000 mm, b = 400 mm and D = 600 mm. Table 28 of IS 456 = lex= ley=
0.65(l) = 2600 mm. So, we have
[ex/D = 2600/600

= 4.33 < 12

lay/b = 2600/400 = 6.5 < 12
Hence, it is a short column.

Step 2: Minimum eccentricity
ex min

Greater of (lax/500+ D/30) and 20 mm = 25.2 mm

eymin= Greater of (lay/500+ b/30) and 20 mm = 20 mm
0.05 D = 0.05(600) = so mm > 25.2 mm (= exmin)

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0.05 b = 0.05(400) = 20 mm = 20 mm (= eym,-,,)
Hence, the equation given in cl.39.3 of IS 456 (Eq.10.4) is applicable for the
design here.
Step 3: Area of steel
Fro Eq.10.4, we have
Pu = 0.4 fckAc+ 0.67 f,,Asc

(10.4)

3000(103)= 0.4(25){(400)(600)
Asc} + 0.67(415)A5,,
which gives,

A5,,= 2238.39mm2

Provide6-20 mm diameterand 2-16 mm diameterrods giving 2287 mm2(>
2238.39mm2)and p = 0.953per cent,whichis morethan minimumpercentage
of 0.8 and less than maximum percentage of 4.0. Hence, o.k.
Step 4: Lateral ties
The diameter of transverse reinforcement (lateral ties) is determined from
cl.26.5.3.2 C-2 of IS 456 as not less than (i) ¢/4 and (ii) 6 mm. Here, ¢ = largest
bar diameter used as longitudinal reinforcement = 20 mm. So, the diameter of
bars

used

more

than

as lateral

ties = 6 mm.

The pitch of lateral ties, as per cl.26.5.3.2 C-1 of IS 456, should be not
the least

of

(i)

the least lateral dimension of the column

400 mm

(ii)

sixteen times the smallest diameter of longitudinal reinforcement
bar to be tied = 16(16) = 256 mm

(iii)

300 mm

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__f.:?&#39;3T

er

21:31.21

writ

g53r«..,.___,
7
%

g4«»2t}"?

ix.

is

Fig. "tt}.22.1:Prnsbflssm
1

Let us use p = pitch of lateral ties = 250 mm. The arrangement of
longitudinal and transverse reinforcement of the column is shown in Fig. 10.22.1.
Problem

2:

Design the column of Problem 1 employing the chart of SP-16.
Solution

2:

Steps 1 and 2 are the same as those of Problem 1.
Step 3: Area of steel

Pu/Ag 3000(103)/(600)(400)
= 12.5N/mm2
From the lower section of Chart 25 of SP-16, we get p = 0.95% when Pu/Ag=

12.5N/mm2andconcretegradeis M 25. This gives Asa= 0.95(400)(600)/100
=
2288 mm2.The resultsof boththe problemsare in goodagreement.Marginally
higher value of A50while using the chart is due to parallax error while reading the
value from the chart. Here also, 6-20 mm diameter bars + 2-16 mm diameter

bars(Aseprovided= 2287mm2)is o.k.,thoughit is 1 mm2less.
Step 4 is the same as that of Problem 1. Figure 10.22.1, thus, is also the
figure showing the reinforcing bars (longitudinal and transverse reinforcement) of
this problem (same column as that of Problem 1).
Problem

3:

Design a circular column of 400 mm diameter with helical reinforcement
subjected to an axial load of 1500 kN under service load and live load. The
column has an unsupported length of 3 m effectively held in position at both ends
but not restrained against rotation. Use M 25 concrete and Fe 415 steel.
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Solution

3:

Step 1: To check the slenderness ratio
Given data are: unsupported length l= 3000 mm, D = 400 mm. Table 28
of Annex E of IS 456 gives effective length /8= I= 3000 mm. Therefore, I9/D= 7.5
< 12 confirms

that it is a short

column.

Step 2: Minimum eccentricity
em

= Greater of (I/500 + D/30) or 20 mm = 20 mm

0.05 D= 0.05(400) = 20 mm
As per cI.39.3 of IS 456, eminshould not exceed 0.05D to employ the equation
given in that clause for the design. Here, both the eccentricities are the same.
So, we can use the equation given in that clause of IS 456 i.e., Eq.10.8 for the

design.
Step
3:Area
ofsteel
From Eq.10.8, we have
Pu = 1.05(0.4 fckAc+ 0.67 f,,Asc)
Ac =

Asc

=

(10.8)

&#39;
Asc

Substituting the values of Pu, fck,Ag and fy in Eq.10.8,

1.5(1500)(103)
= 1.05{0.4(25)(125714.29
Asa)+ 0.67(415)Asa}
we get the valueof A5,;= 3304.29mm2.Provide11 nos.of 20 mmdiameterbars
(= 3455mm&#39;)
as longitudinalreinforcement
giving p = 2.75%.Thisp is between
0.8 (minimum) and 4 (maximum) per cents. Hence o.k.
Step 4: Lateral ties
It has been mentioned in sec.10.22.4 that the pitch p of the helix
determined from Eq.10.11 automatically takes care of the c|.39.4.1 of IS 456.
Therefore, the pitch is calculated from Eq.10.11 selecting the diameter of helical
reinforcement from c|.26.5.3.2 d-2 of IS 456. However, automatic satisfaction of
c|.39.4.1

of IS 456 is also checked

here for confirmation.

Diameter of helical reinforcement (c|.26.5.3.2 d-2) shall be not less than
greater of (i) onefourth of the diameter of largest longitudinal bar, and (ii) 6 mm.

Therefore, with 20 mm diameter bars as longitudinal reinforcement, the diameter
of helical

reinforcement

= 6 mm.

From Eq.10.11, we have

Pitchof helixp s 11.1(D,,¢,p)as,,
fy/(D2&#39;
03);,

(10.11)

whereDc = 400 4o 40 = 320mm,¢,p= 6 mm,asp= 28 mm2,fy= 415
N/mm2,D = 400 mm and rck= 25 N/mm2.

So,

p s 11.1(3206) (28)(415)/(4002-3202)
(25) 3 28.125mm

As per c|.26.5.3.2 d1, the maximum pitch is the lesser of 75 mm and 320/6 =
53.34 mm and the minimum pitch is lesser of 25 mm and 3(6) = 18 mm. We
adopt pitch = 25 mm which is within the range of 18 mm and 53.34 mm. So,
provide 6 mm bars @ 25 mm pitch forming the helix.

Checking of cl. 39.4.1 of IS 456
The values of helical reinforcement and core in one loop are obtained from
Eqs.10.8 and 9, respectively. Substituting the values of Dc, ¢ , aspand pitch p in

theabove
twoequations,
wehave

Sp

Volumeof helicalreinforcement
in oneloop = 27632mm3and
Volumeof corein one loop = 2011428.571
mm3.
Their

ratio

= 27632/2011428.571

= 0.0137375

0.36(Ag/Ac 1) (rck/ry) = 0.012198795
It is, thus, seen that the above ratio (0.0137375) is not less than 0.36(Ag/Ac 1)
(fck/fy).

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t}? rm:

16.22.22

Pretailern 3

Hence, the circular column of diameter 400 mm has eleven longitudinal
bars of 20 mm diameter and 6 mm diameter helix with pitch p = 25 mm. The
reinforcing bars are shown in Fig.10.22.2.

10.22.6

Practice

Questions

and Problems

with Answers

Q.1: State and explain the values of design strengthsof concrete and steel to be
considered in the design of axially loaded short columns.
A.1:

Q.2:

Seesec.10.22.2.

Derive the governing equation for determining the dimensions of the
column and areas of longitudinal bars of an axially loaded short tied
column.

A.2:

See sec.

10.22.2.

Q.3: Derive the governing equation for determining the diameter and areas of
longitudinal bars of an axially loaded circular spiral short column.
A.3:

First and second paragraph of sec. 10.22.4.

Q.4: Derive the expression of determining the pitch of helix in a short axially
loaded spiral column which satisfies the requirement of IS 456.
A.4:

See third paragraph onwards up to the end of sec. 10.22.4.

Q.5: Design a short rectangular tied column of b = 300 mm having the maximum
amount of longitudinal reinforcement employing the equation given in
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cl.39.3 of IS 456, to carry an axial load of 1200 kN under service dead
load and live load using M 25 and Fe 415. The column is effectively held
in position at both ends and restrained against rotation at one end.
Determine the unsupported length of the column.

A.5:
Step
1:Dimension
Dand
area
ofsteel
A5,,
Substituting the values of Pu = 1.5(1200) = 1800 kN and A50= 0.04(300)D
in Eq.10.4, we have

1800(103)= 0.4(25)(300D)(10.04)+ 0.67(415)(0.04)(300D)
we getD = 496.60 mm. Use 300 mm x 500 mm column.

Asa= 0.04(300)(500)= 6000mm2,provide4-36mmdiameter+ 4-25mm
diameterbarsto give4071+ 1963= 6034mm"&#39;
> 6000mm2.
Step 2: Lateral ties

t{.:Ti 31:10

.359 356*.

tawrilfr

Fig. E{f,.22.v3i::
Diameter of lateral ties shall not be less than the larger of (i) 36/4 = 9 mm
and (ii) 6 mm. Use 10 mm diameter bars as lateral ties.
Pitch of the lateral ties p shall not be more than the least of (i) 300 mm, (ii)
16(25) = 400 mm and (iii) 300 mm.
So, provide 10 mm diameter bars @ 300 mm c/c. The reinforcement bars
are shown in Fig.10.22.3.

The centre to centre distance between two corner longitudinal bas along
500 mm direction is 500 2(4) + 10 + 18) = 364 mm which is less than 48
(diameter of lateral tie). Hence, the arrangement is satisfying Fig.9 of cl. 26.5.3.2
b2 of IS 456.

Step 3: Unsupported length
As per the stipulation in cl. 25.1.2 of IS 456, the column shall be
considered as short if lax= 12(D) = 6000 mm and I8, = 12(300) = 3600 mm. For
the type of support conditions given in the problem, Table 28 of IS 456 gives
unsupported length is the least of (i) l= lax/0.8 = 6000/0.8 = 7500 mm and (ii)
ley/0.8= 3600/0.8 = 4500 mm. Hence, the unsupported length of the column is
4.5 m if the minimum eccentricity clause (cl. 39.3) is satisfied, which is checked
in the next step.
Step 4: Check for minimum eccentricity
According to cl. 39.4 of IS 456, the minimum eccentricity of 0.05b or 0.05D
shall not exceed as given in cl. 25.4 of IS 456. Thus, we have
(i)

0.05(500) = I/500 + 500/30 giving I: 4165 mm

(ii)

0.05(300) = I/500 + 300/10 giving I: 2500 mm

Therefore, the column shall have the unsupported length of 2.5 m.
Q.6: (a) Suggest five alternative dimensions of square short column with the
minimum longitudinal reinforcement to carry a total factored axial load of
3000 kN using concrete of grades 20, 25, 30, 35 and 40 and Fe 415.
Determine the respective maximum unsupported length of the column if it
is effectively held in position at both ends but not restrained against
rotation. Compare the given factored load of the column with that obtained
by direct computation for all five alternative columns.
(b) For each of the five alternative sets of dimensions obtained in (a),
determine the maximum factored axial load if the column is having
maximum longitudinal reinforcement (i) employing SP-16 and (ii) by direct
computation.

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A.6:

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Solution of Part (a):

Step 1: Determination of Ag and column dimensions b (= D)
Chart 25 of SP-16 gives all the dimensionsof five cases. The two input
data are Pu= 3000 kN and 100 145/149
= 0.8. In the lower sectionof Chart 25, one
horizontalline AB is drawn startingfrom A where p = 0.8 (Fig.10.22.4) to meet
the linesfor M 20, 25, 30, 35 and 40 respectively.In Fig.10.22.4, B is the meeting
point for M 20 concrete. Separate vertical lines are drawn from these points of
intersectionto meet another horizontalline CD from the point C where Pu= 3000
kN in the upper section of the figure. The point D is the intersectingpoint. D

happens
tobeonlinewhenAg= 3000cm2.Otherwise,
itmaybeinbetween
two
lienswithdifferent
values
ofAg.ForM 20,Ag= 3000cm2.However,
incasethe
point is in between two lines with differentvalues of Ag, the particularAg has to
be computedby linear interpolation.Thus, all five values of Agare obtained.
The dimension b = D = 4300000 = 550 mm. Other four values are
obtained similarly.Table 10.1 presentsthe values of Ag and D along with other
resultsas explainedbelow.

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Step 2: Unsupported length of each column
The unsupportedlengthlis determinedfrom two considerations:
(i) Clause 25.1.2 of IS 456 mentionsthat the maximumeffectivelength lgx
is 12 times b or D (as b = D here for a square column).The unsupportedlengthis
related to the effective length dependingon the type of support. In this problem
Table 28 of IS 456 stipulates l= lgx.Therefore, maximumvalue of l= 12 D.
(ii) The minimumeccentricityof cl. 39.3 shouldbe more than the same as
given in cl. 25.4. Assumingthem to be equal, we get I/500 + D/30 = D/20, which
gives l= 8.33D. For the columnusing M 20 and Fe 415, the unsupportedlength
= 8.33(550) = 4581 mm. All unsupportedlengths are presented in Table 10.1
usingthe equation
I = 8.33

D

(1)
Step 3: Area of longitudinal steel
Step 1 showsthat the area providedfor the firstcase is 550 mm x 550 mm

= 302500mm2,slightlyhigherthanthe requiredarea of 300000mm2for the
practical aspects of construction.However, the minimum percentage of the
longitudinalsteel is to the calculated as 0.8 per cent of area required and not
area provided (vide cl. 26.5.3.1 b of IS 456). Hence, for this case Asc =

0.8(300000)/100
= 2400 mm2.Provide4-25 mm diameter+ 4-12 mm diameter
bars(area= 1963+ 452 = 2415mm2).Table10.1presents
thisandotherareas
of longitudinalsteel obtainedin a similarmanner.
Step 4: Factored load by direct computation
Equation10.4 is employedto calculatethe factoredload by determiningAc
from Ag and A5,. With a view to comparingthe factored loads, we will use the
values of Ag as obtained from the chart and not the Ag actually provided. From
Eq.10.4, we have
Pufrom directcomputation= 0.4-(fck)(0.992
Ag)+ 0.67(fy)(0.008)Ag
or
(2)

Pu = Ag(0.3968 fck+0.00536 fy)

ForthefirstcasewhenAg= 300000mm2,fck= 20 N/mm2,
and fy= 415N/mm2,
Eq.(2) gives Pu = 3048.12 kN. This value and other values of factored loads
obtainedfrom the directcomputationare presentedin Table 10.1.

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Table 10.1 Results of Q.6a (Minimum Longitudinal Steel), given factored P =
3000

kN

Gross area of

Area of steel (Asg)
Require

Provide

Bars

d cmz

d cmz

M 20

3000

3025

55

24

24.15

4-25

M 25

2500

2500

50

20

20.60

4-20

M 30

2200

2209

47

17.60

17.85

2.25

M 35

1800

1806

42.5

14.40

14.57

2-28

M 40

1600

1600

40

12.80

13.06

2-20

Puby direct
computation

I
(m)

+

3048.12

4.58

+

3036.10

4.16

+

3108.25

3.91

+

2900.23

3.54

+

2895.42

3.33

4-12

1

4-16

5

4-16

5

2-12

0

6-12

Solution

2

of Part (b):

Step 1: Determination of Pu
Due to the known dimensions of the column section, the Ag is now known.
With known Ag and reinforcement percentage 100/45/Agas 4 per cent, the
factored Pu shall be determined. For the first case, when b = D = 550 mm, Ag =

302500mm2.InChart25,wedrawa horizontal
lineEFfromE,where100/45/Ag
=
4 in the lower section of the chart (see Fig.10.22.4) to meet the M 20 line at F.
Proceeding vertically upward, the line FG intersects the line Ag = 302500 at G. A
horizontal line towards left from G, say GH, meets the load axis at H where Pu =
5600 kN. Similarly, Pu for other sets are determined and these are presented in
Table 10.2, except for the last case when M 40 is used, as this chart has ended
at p = 3.8 per cent.

Step 2: Area of longitudinal steel
The maximum area of steel, 4 per cent of gross area of column =

0.04(550)(550)= 12100 mm2.Provide12-36 mm diameterbars to have the
actualareaof steel= 12214mm2> 12100 mmz,as presentedin Table10.2.
Step 3: Factored Pu from direct computation
From Eq,10.4, as in Step 4 of the solution of Part (a) of this question, we
have

Pu = 0.4+

A50

(3)

Version 2 CE IIT, Kharagpur

Substitutingthe values of Ag and A50actually provided,we get the maximum P
of the same column when the longitudinalsteel is the maximum. For the first

casewhenAg= 302500mm2,Asa= 12214mm2,1:,= 20 N/mm2and fy= 415
N/mm2,we get Pu= 5718.4kN. Thisvaluealong withotherfourvaluesare
presentedin Table 10.2.
Remarks:
Tables

10.1

and

10.2

reveal

that two sets of results

obtained

from charts

of SP-16 and by direct computationmethods are in good agreement. However,
values obtained from the chart are marginallydifferent from those obtained by
direct computationboth on the higher and lower sides. These differences are
mainly due to personal error (parallax error) while reading the values with eye
estimation

from the chart.

Table 10.2 Fesults of Q.6(b) (Maximum LongitudinalSteel) given the respective

Area of steel Asa
Provide

P = Factored load

Bars

SP-chart

(No.
)

(kN)

Direct

Computatio

(
36
M 25

50

2500

8-36
+ 4-

5200

5208.9

5000

5017.8

4500

4474.5

25
M 30

47

2209

88.36

88.97

M35

42.5

1806.25

72.25

73.69

M 40

40

64

64.46

8-32
+ 428
23
8-28

Not

+ 4-

availabl

32

4249.2

e

Q.7: Designa short, helicallyreinforcedcircularcolumnwith minimumamountof
longitudinalsteel to carry a total factored axial load of 3000 kN with the
same supportconditionas that of Q.6, using M 25 and Fe 415. Determine
its unsupportedlength. Compare the resultsof the dimensionand area of
longitudinalsteel with those of Q.6(a) when M 25 and Fe 415 are used.
A.7:

Step 1: Diameter of helically reinforced circular column
Version 2 CE IIT, Kharagpur

As per cl. 39.4 of IS 456, applicable for short spiral column, we get from
Eq.10.8
P,, = 1.05(0.4fc,,Ac+0.67 ryA,,,)

(10.8)

Givendataare: P, = 3000kN,Ac= 71/4(D2)(0.992),
A5,,= 0.008(7r/4)D2,rck= 25

N/mm2
andfy= 415N/mm2.
So,wehave
3000(103)= 1.05(12.1444)(7r/4)D2

givingD = 547.2mm and Ag= 235264.2252
mm2.Provide
diameter
of 550mm.
Step 2: Area of longitudinal steel
Providing 550 mm diameter, the required Ag has been exceeded. Clause
26.5.3.1b stipulates that the minimum amount of longitudinal bar shall be
determined on the basis of area required and not area provided for any column.
Accordingly, the area of longitudinal steel = 0.008 Ag = 0.008 235264.2252) =

1882.12mm2.Provide6-20mmdiameterbars(area= 1885mm ) as longitudinal
steel, satisfying the minimum number of six bar (cl. 26.5.3.1c of IS 456).
Step 3: Helical reinforcement

Fig. 1032.0: f?
Minimum diameter of helical reinforcement is greater of (i) 20/4 or (ii) 6
mm. So, provide 6 mm diameter bars for the helical reinforcement (cl. 26.5.3.2d
2 of IS 456). The pitch of the helix p is determined from Eq.10.11 as follows:

p s 11.1(D,,¢,,,)a,,,ry/(D2D3)r,,k

(10.11)
Version 2 CE IIT, Kharagpur

UsingDc= 550 4o 40= 470mm, ¢Sp
= 6 mm,asp= 28mm2,
D= 550mm,
rck= 25 N/mm2
andfy= 415N/mm2,
weget
p s 11.1(4706)(28)(415)/(55024702)(25)
3 29.34mm
Provide 6 mm diameter bar @ 25 mm c/c as helix. The reinforcement bars are

shown in Fig. 10.22.5. Though use of Eq.10.11 automatically checks the
stipulation of cl. 39.4.1 of IS 456, the same is checked as a ready reference in
Step 4 below.
Step 4: Checking of cl. 39.4.1 of IS 456

Volumeof helix in one loop = 7r(Dc- ¢Sp)asp

(10.9)

Volumeof core = (Ir/4) Df (P)

(10.10)

Theratioof Eq.10.9andEq.10.10= 4(Dc ¢Sp)
asp/Dfp
= 4(4706)(28)/(470)(470)(25)

= 0.009410230874

This ratio should not be less than 0.36(Ag/Ac 1)(fck/fy)

= 0.36{(D2/Df)1)}(fck/fy)
= 0.008011039177
Hence, the stipulation of cl. 39.4.1 is satisfied.
Step 5: Unsupported length
The unsupported length shall be the minimum of the two obtained from (i)
short column requirement given in cl. 25.1.2 of IS 456 and (ii) minimum
eccentricity requirement given in cls. 25.4 and 39.3 of IS 456. The two values are
calculated separately:
(i) I:

la = 12D = 12(550) = 6600 mm

(ii) I/500 + D/30 = 0.05 D gives I= 4583.3 mm
So, the unsupported length of this column = 4.58 m.
Step 6: Comparison of results
Table 10.3 presents the results of required and actual gross areas of
concrete and area of steel bars, dimensions of column and number and diameter

of longitudinal reinforcement of the helically reinforced circular and the square
Version 2 CE IIT, Kharagpur

columns of Q.6(a) when M 20 and Fe 415 are used for the purpose of
comparison.
Table 10.3 Comparison of results of circular and square columns with minimum
longitudinal steel (Pu = 3000 kN, M 25, Fe 415)
Gross

concrete

area

Area

Required

Provided

Dimension

(cm2)

(cm2)

D (cm)

of steel

Required

Provided

Bar dia.

(cm2)

(cm2)

and No.

18.82

18.85

mm,No.
2352.64

2376.78

2500

2500

o.7

Square

50

20

Q.6a

20.6
4-16

10.22.7

References

. ReinforcedConcreteLimit State Design,6" Edition,by Ashok K. Jain,
Nem Chand & Bros, Roorkee, 2002.

LimitStateDesignof ReinforcedConcrete,2 Edition,by P.C.Varghese,
PrenticeHa|I of India Pvt. Ltd., New Delhi, 2002.

Advanced Reinforced Concrete Design, by P.C.Varghese, Prentice-Hall of
India Pvt. Ltd., New Delhi, 2001.

ReinforcedConcreteDesign,2 Edition,by S.UnnikrishnaPillai and
Devdas Menon, Tata McGrawHiII Publishing Company Limited, New
Delhi, 2003.

Limit State Design of Reinforced Concrete Structures, by P.Dayaratnam,
Oxford & I.B.H. Publishing Company Pvt. Ltd., New Delhi, 2004.

ReinforcedConcreteDesign, 13 RevisedEdition,by S.N.Sinha,Tata
McGrawHiII Publishing Company. New Delhi, 1990.

ReinforcedConcrete,6" Edition,by S.K.Ma|Iickand A.P.Gupta,Oxford &
8.
9.

IBH Publishing Co. Pvt. Ltd. New Delhi, 1996.
Behaviour, Analysis & Design of Reinforced Concrete Structural Elements,
by |.C.Sya| and R.K.Ummat, A.H.Whee|er & Co. Ltd., Allahabad, 1989.

ReinforcedConcreteStructures,3 Edition,by |.C.Sya|and A.K.Goe|,
A.H.Whee|er & Co. Ltd., Allahabad, 1992.

10.Textbook of R.C.C, by G.S.Birdie and J.S.Birdie, Wiley Eastern Limited,
New Delhi, 1993.

11.Designof ConcreteStructures,13" Edition,by Arthur H. Nilson,David
Darwin and Charles W. Dolan, Tata McGraw-Hill Publishing Company
Limited, New Delhi, 2004.

12.Concrete Techno|oQY. by A.M.NeviIIe and J.J.Brooks, ELBS with
Longman, 1994.

13.Propertiesof Concrete,4 Edition, 13 Indian reprint, by A.M.NeviIIe,
Longman, 2000.
Version 2 CE IIT, Kharagpur

14.Reinforced
ConcreteDesignersHandbook,10"Edition,by C.E.Reyno|ds
and J.C.Steedman,

E & FN SPON, London, 1997.

15.lndianStandardPlain and ReinforcedConcrete Code of Practice(4"
Revision), IS 456: 2000, BIS, New Delhi.
16. Design Aids for Reinforced Concrete to IS: 456

10.22.8

Test

22 with

Maximum Marks = 50,

1978, BIS, New Delhi.

Solutions
Maximum Time = 30 minutes

Answer all questions carrying equal marks.
TQ.1: Derive the expression of determining the pitch of helix in a short axially
loaded spiral column which satisfies the requirement of IS 456.
(20 marks)
A.TQ.1: See third paragraph onwards up to the end of sec. 10.22.4.
TQ.2: Design a square, short tied column of b = D = 500 mm to carry a total
factored load of 4000 kN using M 20 and Fe 415. Draw the reinforcement
diagram.
(30 marks)
A.TQ.2:

Step 1: Area of longitudinal steel
Assuming p as the percentage of longitudinal steel, we have Ac =

(500)(500)(10.01p),Asc= (500)(500)(0.01p),
fck= 20 N/mm2and fy= 415
N/mm2.Usingthesevaluesin Eq.10.4
Pu = 0.4 fckAc+ 0.67 f,,As,
or

4000000 = 0.4(20)(250000)(1

(10.4)
0.01p) + 0.67(415)(250000)(0.01p)

we getp = 2.9624,whichgives A5,,= (2.9624)(500)(500)/100
= 7406mm&#39;.
Use
4-36+ 4-25+ 4-22mmdiameterbars(4071+ 1963+ 1520)= 7554mm2> 7406
mm2as longitudinalreinforcement.

Version 2 CE IIT, Kharagpur

75?

_yv.*.~wM»-rs:
f QT
,4

ESQ
5:5:

?..EE.5:§£7.f§.E

Step 2: Lateral ties
Diameter of tie is the greater of (i) 36/4 and (ii) 6 mm. Provide 10 mm
diameter

lateral

ties.

The pitch of the lateral ties is the least of (i) 500 mm, (ii) 16(22) = 352 mm,
and (iii) 300 mm. Provide 10 mm diameter @ 300 mm c/c. The reinforcement
bars are shown in Fig.10.22.6. It is evident that the centre to centre distance
between

two

corner

bars

= 364

mm

is less than

48 times

the diameter

of lateral

ties = 480 mm (Fig.9 of cl. 26.5.3.2b2 of IS 456).

1022.9 Summary of this Lesson
This lesson illustrates the additional assumptions made regarding the
strengths of concrete and steel for the design of short tied and helically
reinforced columns subjected to axial loads as per IS 456. The governing
equations for determining the areas of cross sections of concrete and longitudinal
steel are derived and explained. The equation for determining the pitch of the
helix for circular columns is derived. Several numerical examples are solved to
illustrate the applications of the derived equations and use of charts of SP-16 for
the design of both tied and helically reinforced columns. The results of the same
problem by direct computation are compared with those obtained by employing
the charts of SP-16 are compared. The differences of results, if any, are
discussed. Understanding the illustrative examples and solving the practice and
test problems will explain the applications of the equation and use of charts of
SP-16 for designing these two types of columns.

Version 2 CE IIT, Kharagpur