Unit-4
Design of Columns by Limit State Design Method
Column
Column is a compression member; the effective length is exceeding 3 times the least lateral dimension.
Left> 3 x least lateral dimension
Types of Columns (As per Clause 25.1.2 of IS456: 2000)
Pedestal
Short column
Long column
These above types are explaining follows,
1) Pedestal
If the length of column is less than 3 times its least lateral dimension lex/ D <3 or ley / b< 3, it is called a pedestal.
The other horizontal dimension D shall not exceed four times of b
2) Short Column
If both the slenderness ratios lex / D and ley / b ≤ 12, the column is a short column.
3) Long Column
If for a column lex/ D and ley/b > 12, it is a slender column in respect of major or minor axis, respectively
If both slenderness ratios are greater than 12, it is slender column in respect of both axes.
Here,
L ex= Effective length of column in respect of the major axis,
D = Depth in respect of the major axis,
L e y = Effective length of column in respect of minor axis, and
b = Width of the member.
Strain and stress variation diagrams
Effective Length (Clause 25.2 of IS456: 2000)
The effective length of a column (effective) is not equal to lo unsupported length (l) in all cases.
Effective length of column for different end conditions (clause 25.2 and table 28 of IS456 – 2000)
Effective length
- Effectively held in position and restrained against rotation in both ends – 0.65 l
- Effectively held in position at both ends, restrained against rotation in one ends – 0.80 l
- Effectively held in position at both ends, but not restrained against rotation – 1.00 l
- Effectively held in position and restrained against rotation in one end and at other restrained against rotation but not held in position – 1.20 l
- Effectively held in position and restrained against rotation in one end and at other partially restrained against rotation but not held in position – 1.5 l
- Effectively held in position at one end but not restrained against rotation and at other end restrained against rotation but not held in position – 2.00 l
- Effectively held in position and restrained against rotation in one ends but not held in position nor restrained against rotation at the other end – 2.00 l
Key takeaways:
Types of Columns (As per Clause 25.1.2 of IS456: 2000)
Pedestal
Short column
Long column
The assumptions for strength style of columns may be summarized as follows.
1. Strain of steel and concrete is proportional to distance from basic axis
2. Most usable compression strain of concrete is zero.
3. Stress, psi, in longitudinal reinforcing bars equals steel strain s times twenty nine, 000,000 for strains below yielding, and equals the steel yield strength ƒy, tension or compression, for larger strains
4. Durability of concrete is negligible.
5. Capability of the concrete in compression that is assumed at a most stress of, should be in line with check results. An oblong stress distribution zero.85ƒc could also be used. Depth of the parallelogram could also be taken as a 1c, wherever c is that the distance from the neutral axis to the acute compression surface and one zero.85 for ƒc 4000 psi and zero.05 less for every a thousand psi that ƒc exceeds4000 psi, however one mustn't be taken but zero.65.
Minimum Eccentricity (Clause 25.4 of IS456: 2000)
It is noted that in practical construction, columns are rarely truly concentric.
Even a theoretical column loaded axially which have the accidental eccentricity due to incorrect accuracy in construction or variation of materials etc.
Accordingly, all axially loaded columns which designed considering the minimum eccentricity which shows in Clause 25.4 of IS456 and given below.
E x min ≥ greater of (+) or 20 mm
E y min ≥ greater of (+) or 20 mm
Where l, D and b are the unsupported length, larger lateral dimension and least lateral dimension, respectively
Maximum Eccentricity (Clause 39.3 of IS456: 2000)
For the column to be short, as per Clause 39.3 the eccentricity shall not be more than 0.O5D or 0.05b.
Assumptions (Clause 39.1 of IS456: (000)
The maximum compressive strain in form of the concrete in axial compression is taken as 0.002.
Strain in concrete and steel are equal.
The maximum compressive strain at the highly compressed extreme fiber which is form in concrete subjected to axial compression and bending and If is no tension on the section which is 0.0035 minus 0.75 times the strain at the least compressed extreme fiber.
Stress in steel is governed by characteristic stress-strain curve of steel in compression. The stress-strain curve of steel in compression is the same as in tension.
The column is short.
Equation for Short axially Loaded Tied Columns (Clause39.3 of IS456: 2000)
According to Clause 39.3 of IS456: 2000, when the eccentricity does not exceed 0.05 times the lateral dimension, the member may be designed by the following equation
P u = 0.4 f c k Ac + 0.67 f y A s c
Where,
Pu = factored axial load on the member,
F c k= characteristic compressive strength where P. Of the concrete,
A c=area of concrete,
F y= characteristic strength of the compression reinforcement, and
A s c= area of longitudinal reinforcement for columns.
Equation of Short Axially Loaded Columns with Helical Ties (Clause39.4 of IS 456: 2000)
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) (f c k / fy), in order to apply the additional strength factor of 1.05 (Clause 39.4.1).
Accordingly, the governing equation of the spiral columns may be written as
P u = 1.05 (0.4 f c k Ac + 0.67 fy A s c)
= 0.4 f c k Ac + 0.67 fy A s c
The pitch of helical reinforcement is determined using the following relation as per CI 39.4.1 of IS456: 2000.
P <
Where,
Dc = Diameter of concrete core,
D sp = Diameter of spiral reinforcement,
A sp =Area of cross-section of spiral reinforcement,
p= Pitch of spiral reinforcement.
Note:
The pitch and diameter of the spiral reinforcement should also satisfy Clause 26.5.3.2 of 19456: 2000.
Design Procedure for Axially Loaded Short Columns
Step 1: Find effective length.
Step 2: Find the minimum eccentricity and check the type of column.
Step 3: Find the area of steel
Use, P u = 0.4 f c k Ac + 0.67 fy A s c
Step 4: Design of lateral ties
The diameter of transverse reinforcement (lateral ties) determined from Clause 26.5.3.2 C-2 of IS456 as not less a (1) d/2 (2) 6 mm.
Where, d = diameter of bar
The pitch of lateral ties, as per Clause26.5.3.2 C-1 of IS45 should be not more than the least of
The least lateral dimension of the column
Sixteen times the smallest diameter of longitudinal reinforcement bar which can tied.
300 mm
Step 5: Draw the cross section
Problems
1) Design an axially loaded column 400 mm x 450 mm pinned at both ends with an unsupported length of 3 m for carrying axial service load of 2000 kN. Use M20 Fe415.
Answer
Given:
b = 400 mm
D = 450 mm
L = 3m
P = 2000 KN
F c k = 20 N/mm2
F y = 415 N/mm2
To find: Reinforcements
Solution:
Step 1: Check for minimum eccentricity
e min = ( + )
= ( + )
= 21 mm or 20 mm ………….whichever is greater
e min = 21 mm
e max = 0.05 D = 0.05 x 450 = 22.5 mm
e min < e max , the formula given for axial load I applicable
Step 2: Factored load P u = d x p = 1.5 x 2000 = 3000 KN
Step 3: Longitudinal reinforcement:
Pu = 0.4 fck Ac + 0.67 fyAsc
Ag = 450 x 400 = 180000 mm2
Ac = Ag – Asc
= 180000 –Asc
3000 x 103 = 0.4 x 20 x (180000 – As c) + 0.67 x 415 x Asc
= 1.44 x 106 – 8 Asc + 278.05 Asc
270 Asc = 1.56 x 196
Asc = 5777.8 mm2
Using 32 mm diameter of bar
Area of one bar Ad = • d2 = • 322 = 804.2 mm2
Number of bars n = 7.2 = 8 bar
Provide 8 – 32 mm bar as longitudinal steel
A s c provided = 8 x 322 = 6434 mm2
Pt = x 100
= x 100
= 3.6 % < 6 % ……………………OK
Use cover 40 mm
Steel spacing = 144 mm
Clear spacing between bars = 144 – 32 = 112 mm < 300 mm
Step 4: Transverse steel :
Diameter of link or lateral ties should not less not x diameter of largest longitudinal bar
= x 32 = 8 mm
Diameter should not less not 6 mm
Dt = 8 mm ……………………..whichever is greater
Spacing for lateral ties
The least lateral dimension b = 400 mm
16 x diameter of longitudinal reinforcement bar = 16 x 32 = 512 mm
300 mm
Spacing = 300 mm ……….....whichever is least
Provided lateral ties of 8 mm diameter of bar at 300 mm c/c
2) Design a square column with the following data Factored load = 3000 k N concrete grade = M20 Steel = Fe415 Unsupported length of column = 3 m Check for minimum eccentricity
Answer.
Given:
Pu = 3000 x 103 N
F ck = 20 N/mm2
F y = 415 N/mm
L = 3000 mm
To find: Size of column and steel requirement
Solution:
Step 1: Size of column
Assume area of steel is 1% of grass area
Asc = 0.01 Ag
Area of concrete = Ag – Asc
= Ag (1 – 0.008)
= 0.992 Ag
Factored load, Pu = 0.4 fck Ac + 0.67 fyAsc
3000 x 103 = 0.4 x 20 x 0.992 Ag + 0.67 x 415 x 0.008 Ag
A g = 295264 mm2
For square column side,
= 544 mm = say 560 mm
Dimension of column = 560 m > 400 mm
The formula given for axial load I applicable
Step 2: Check for minimum eccentricity
e min = ( + )
= ( + )
= 24.67 mm or 20 mm ………….whichever is greater
e min= 24.67 mm
e max = 0.05 D = 0.05 x 560 = 28 mm
e min< e max , ……………………………….OK
Step 3: Slenderness ratio
= 5.36 < 12 ……………………..Short column
Step 4: Longitudinal steel
Asc = 0.008 x Ag = 0.008 x 5602
= 2508.8 mm2
Using 20 mm diameter of bar
Area of one bar Ad = • d2 = • 202 = 314.15 mm2
Number of bars n = 7.99 = 8 bar
Provide 8 – 20 mm bar as longitudinal steel
As c provided = 8 x 202 = 2513.3 mm2
Pt = x 100
= x 100
= 0.802 % > 0.8 % ……………………OK
Step 5: Transverse steel:
Diameter of link or lateral ties should not less not x diameter of largest longitudinal bar
= x 20 = 5 mm
Diameter should not less not 6 mm
Dt = 6 mm ……………………..whichever is greater
Spacing for lateral ties
The least lateral dimension b = 560 mm
16 x diameter of longitudinal reinforcement bar = 16 x 20 = 320 mm
300 mm
Spacing = 300 mm ……….....whichever is least
Provided lateral ties of 6 mm diameter of bar at 300 mm c/c
Step 6: Reinforcement details:
3) Design a circular column to carry an axial load 1600 kN using concrete M 20 and Fe415 steel. The unsupported length of column is 3.75 m.
Answer:
Given:
Pu = 1600 x 103 N
F ck = 20 N/mm2
F y = 415 N/mm
L = 3750 mm
To find: Diameter of column and steel requirement
Solution:
Step 1: Factored load Pu = 1.5 P = 1.5 x 1600 = 2400 KN
Step 2: Diameter of column
Assume area of steel is 1% of grass area
Asc = 0.01 Ag
Area of concrete = Ag – Asc
= Ag (1 – 0.008 )
= 0.992 Ag
Using equation,
Pu = 0.4 fck Ac + 0.67 fyAsc
2400 x 103 = 0.4 x 20 x 0.992 Ag + 0.67 x 415 x 0.008 Ag
Ag = 236211 mm2
D = 548 = 550 mm > 400 mm
Or 0.12 x L = 0.12 x 3750 = 450 mm
Step 3: Slenderness ratio
= 6.82< 12 ……………………..Short column
Step 4: Longitudinal steel
Asc = 0.008 x Ag = 0.008 x 5502
= 1900 mm2
Using 18 mm diameter of bar
Area of one bar Ad = • d2 = • 182 = 254.5 mm2
Number of bars n = 7.5 = 8 bar
Provide 8 – 18 mm bar as longitudinal steel
Step 5: Transverse steel:
Diameter of link or lateral ties should not less not x diameter of largest longitudinal bar
= x 18 = 4.5 mm
Diameter should not less not 6 mm
Dt = 6 mm ……………………..whichever is greater
Spacing for lateral ties
The least lateral dimension D = 550 mm
16 x diameter of longitudinal reinforcement bar = 16 x 18 = 288 mm
300 mm
Spacing = 288 mm = 280mm ….....whichever is least
Provided lateral ties of 6 mm diameter of bar at 280 mm c/c
Step 6: Reinforcement details:
Short Compression Members under Axial Load with Biaxial Bending
Beams and girders transfer their end moments near the corner columns of a building frame in two perpendicular planes. Interior columns have biaxial moments if the layout of the columns is irregular. Accordingly, such columns are design considering axial load with biaxial bending.
Key takeaways:
Design steps
Step 1: Find effective length.
Step 2: Find the minimum eccentricity and check the type of column.
Step 3: Find the area of steel
Use, P u = 0.4 f c k Ac + 0.67 f y A s c
Step 4: Design of lateral ties
The diameter of transverse reinforcement (lateral ties) determined from Clause 26.5.3.2 C-2 of IS456 as not less a (1) d/2 (2) 6 mm.
Where, d = diameter of bar
The pitch of lateral ties, as per Clause26.5.3.2 C-1 of IS45 should be not more than the least of
The least lateral dimension of the column
Sixteen times the smallest diameter of longitudinal reinforcement bar which can tied.
300 mm
Step 5: Draw the cross section
Types of Column Reinforcement
A column has two types of reinforcements as mention as below,
- Longitudinal or main reinforcement
- Transverse or lateral reinforcement
- Helical reinforcement
Longitudinal or Main Reinforcement (Clause 26.5.3.1 of IS456: 2000)
The longitudinal reinforcement resists axial load and bending moment.
The transverse reinforcement resists shear force, shares a small fraction of axial load if provided in the form of helical reinforcement, keeps main reinforcement in position and prevents longitudinal bars from buckling.
The longitudinal reinforcing bars in the form of vertical steel are used to carry the compressive loads along with the concrete. The following are the IS requirement for design of column:
IS requirement for design of column
- Areas of longitudinal steel
- Minimum number of longitudinal steel
- Diameter of longitudinal steel
- Spacing of longitudinal bars
- Nominal cover to longitudinal bars
(A) Areas of longitudinal steel
Minimum amount of steel should not be less than 0.80 % of gross cross-sectional area
Amount of area should not be more than 4 % of the gross cross-sectional area of the column
In no case, it does not exceed 6 % when bars from column below have to be lapped with those in the column under consideration.
(B) Minimum Number of longitudinal steel
For rectangular column shall be 4
For circular columns shall be 6
(C) Diameter of longitudinal steel
The diameter of the longitudinal bars shall not be less than 12 mm
(D) Spacing of longitudinal bars
The minimum horizontal spacing between two parallel main bars shall not be less than diameter of larger bar or maximum size of coarse aggregate plus 5 mm.
The maximum spacing of longitudinal bars measured along the periphery of the column shall not be more than 300 mm.
(E) Nominal cover to longitudinal bars
Nominal cover in case of a column equals to (i) 40 mm (ii) the diameter of longitudinal bar whichever is greater.
In case of columns of width 200 mm or less whose longitudinal bar not exceed 12 mm diameter, the nominal cover of 25mm may be used.
Transverse or Lateral Reinforcement (Clause 26.5.3.2 of IS456: 2000)
Clause 26.5.3.2(b) of IS 456: 2000 stipulates the guidelines of the arrangement of transverse reinforcement as under.
As per Clause 26.5.3.2 (b)(1), If the longitudinal bars are not spaced more than 75 mm on either side, transverse reinforcement shall only go round corner and alternate bars for the purpose of providing effective lateral supports.
As per Clause 26.5.3.2 (b)(2),If longitudinal bars spaced at a distance not exceeding 48 times the diameter of the tie are effectively tied in two directions, additional open ties are provided in between longitudinal bars
As per Clause 26.5.3.2(b)(3),.For longitudinal bars in compression members are placed in more than one row
As per Clause 26.5.3.2(b) (3(i), Transverse reinforcement is provided for the outer-most row in accordance with (a) above, and
As per Clause 26.5.3.2 (b)(3(ii),no bar of the inner row is closer to the nearest compression face than three times the diameter of the largest bar in the inner row.
The diameter of such transverse reinforcement need not, however, exceed 20 mm
Pitch and Diameter of Lateral Ties (Clause 26.5.3.2(c) of IS456 : 2000)
1) Pitch
The pitch of transverse reinforcement shall be the least of the following,
The least lateral dimension of the compression 9.5 members
Sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied; and
300 mm.
2) Diameter
The diameter of the polygonal links or lateral ties is not less than one-fourth of the diameter of the largest longitudinal bar, and in have no case less than 6 mm.
Key takeaways:
A column has two types of reinforcements as mention as below,
- Longitudinal or main reinforcement
- Transverse or lateral reinforcement
- Helical reinforcement
IS requirement for design of column
- Areas of longitudinal steel
- Minimum number of longitudinal steel
- Diameter of longitudinal steel
- Spacing of longitudinal bars
- Nominal cover to longitudinal bars
Helical Reinforcement (Clause 26.5.3.2 (d) of IS456: 2000)
1) Pitch
Helical reinforcement which the regular formation with the turns of the helix spaced having evenly and its ends anchored properly by providing one and a half extra turns of the spiral bar.
The pitch of the helical turns shall not be the more than 75 mm, nor more than one sixth or the core diameter of the column, not less than 25 mm or less than three times the diameter of the steel bar forming the helix
2) Diameter
The diameter of the helical reinforcement shall be as mentioned
Interaction Diagrams
The interaction diagram is the curve drawn with )as an abscissa and ) ordinate for the various values of p ranging from 0 to 0.26 for distribution of reinforcement such as equal reinforcement on two parallel faces or on all faces of rectangular sections.
This diagram is also called as Pu – Mu interaction curve.
This diagram is drawn for square, rectangular and circular column for different grades of concrete and steel.
This diagram is advantageously used for the column subjected to axial load and bending moment. This diagram provides for the different modes of failures namely compression failure, balanced failure and tension failure.
The interaction curve is drawn for various load-moment (Pu and Mu) combinations for all possible eccentricities of loading. For design purposes, Pu and Mu are calculated on the basis of design stress-strain curves (including partial safety factors).
The design interaction curve represents the failure envelope and the point given by the co-ordinates (Pu and Mu) falling within the interaction curve indicates the safe values of the combination of load and moments.
The salient points on the interaction curve are as under.
Point-1 corresponds to the axial loading with zero moment (P u) and e = 0. This indicates, the column is subjected to axial load only.
Point-1' corresponds to the axial load with the minimum eccentricity prescribed in Clause 254 of 5456 the corresponding ultimate load is represented as P u
As the eccentricity increases, the moment increases with the neutral axis xumoving from outside towards the extreme fiber.
Point-3 on the interaction diagram indicates the balanced failure state. The design strength values for the balanced failure condition are denoted as Pub and Mub
Point-4 on the interaction curve refers to the pure flexural state (Pu= 0) with the ultimate moment of resistance Muo
Interaction diagrams or Design Charts of SP-16
SP-16 has three sets of design charts prepared by IS456 published in SP-16 for rectangular and circular types of cross- sections of columns. These three sets are as follows:
The charts 27 to 38 are the first set of twelve charts for rectangular columns having symmetrical longitudinal steel bars in two rows for three grades of steel (Fe250, Fe415 and Fe500) and each of them has four values of d'/D ratios (0.05, 0.10, 0.15 and 0.20).
Charts 39 to 50 are the 2nd set of 12 charts for rectangular columns which have the symmetrical longitudinal steel bars (twenty numbers) distributed equally on four sides (in six rows, for three grades of steel (Fe250, Fe415 and Fe500) and each of them having four values of d/D ratios (0.05, 0.10, 0.15 and 0.20).
The 3rd set of 12 charts have the numbering from 51 to 62, are for circular columns having eight longitudinal steel bar equal diameter and uniformly spaced circumferentially for three grades of steel (Fe250, Fe415 a Fe500) and each of them haves four values of d'/D ratios (00s 0.10,0.15 and 0.20).
All the 36 charts which can prepared for M 20 grade of concrete only.
Chart 1: Interaction diagram for combined bending and compression rectangular section-equal reinforcement on opposite sides
Chart 2: Interaction diagram for combined bending and compression rectangular section- equal reinforcement on opposite sides
Chart 3: Interaction diagram for combined bending and compression rectangular section- equal reinforcement on opposite sides
Chart 4: Interaction diagram for combined bending and compression rectangular section-equal reinforcement on opposite sides
Chart 5: Interaction diagram for combined bending and compression rectangular section-equal reinforcement on all sides
Chart 6: Interaction diagram for combined bending and compression rectangular section-equal reinforcement on all sides
Chart 7: Interaction diagram for combined bending and compression rectangular section-equal reinforcement on all sides
Chart 8: Interaction diagram for combined bending and compression rectangular section-equal reinforcement on all sides
Approximations and Limitations of Design Charts of SP-16 9.14.2.1
Approximations
The following are the approximations of the design charts of SP-16:
(a) Grade of concrete
(b) The d'/D ratio
(c) Equal distribution of twenty longitudinal steel bars on four sides of rectangular columns
(d) Longitudinal bars in circular columns
This are approximations are explain s follow
Grade of concrete
As mentioned in earlier section, all the design chars of SP-16 assume the constant grade M 20 of concrete.
However, each chart has fourteen plots having different values of the parameter p/f ranging from zero to 0.26 at an interval of 0.02. The designer which can formed to be use of the actual grade of concrete by multiplying the p/fck which can obtained from the plot with the actual fck for the particular grade of concrete to partially compensate the approximation.
The d'/D ratio
The three sets of charts have four fixed values of d'/D ratios (0.05, 0.10, 0.15 and 0.20). However, in the practical design, the d'/D ratio may be different from those values. In such situations intermediate values are determined by making linear interpolations.
Equal distribution of 20 longitudinal steel bars on four sides of rectangular columns
In above consideration, the design charts can used without significant error for any number of bars greater than eight provided the bars which can be distributed equally on four sides.
Longitudinal bars in circular columns
Though the design charts are prepared considering ight bars uniformly placed circumferentially, they may generally be used for any number of bars greater than six, uniformly placed circumferentially.
Limitations
Limitations of the design chars of SP-16 are as follow
Longitudinal bars equally distributed on four sides of rectangular columns
Unsymmetrical arrangement of longitudinal bars in rectangular cross-sections Non-uniform placing of longitudinal bars in circular cross-sections
Cross-sections other than rectangular or circular like, I, T, H, X etc
It is known that the design of columns by direct computations involves several trials for exact determination of steel percentage in which the percentage is assumed by comparing the strength calculated from the assumed steel with the given load on the column and hence, time taking. Thus, design charts prepared by IS456 published in SP-16 are used in getting several alternative solutions quickly.
Further, these design charts are also used for the analysis of columns for safety etc.
However, there are limitations of using the design charts, which are mentioned in the succeeding sections.
The Analysis of Short Columns under Axial Load with Uniaxial Bending
In many situations, it becomes necessary to assess the safety of a column with known cross-section dimensions, and longitudinal and transverse steel reinforcing bars. The objective is examining if the column can resist some critical values of Pu or Mu or pairs of Pu and Mu as may be expected to be applied on the column.
This is done by comparing if the given values of pair of Pu and Mu Pu and Mu are less than the respective strength capacities pair of Pu and Mu.
The word "given" shall be used in the suffix of pairs of Pu and Muto indicate that they are the given values for which the column has to be examined. The strength capacity of the column, either Pu or Mu alone or pair of Pu and Mu will not have any suffix. Thus, the designer shall assess
(Pair of Pu and Mu) given< Pair of P u and Mu, as strength capacities
This type of problem is known as analysis types of problem the three steps are given below are used to design the column using design charts of SP-16.
Step 1: Select the respective design chart,
Specified by the chart number, from the known value of d'/D and the grade of steel for circular columns; and considered the distribution of longitudinal steel bars equally on two or four sides for the rectangular columns.
Step 2: Select the particular curve
Select the particular curve out of the family of 14 curves in the chart selected in Step 1. The selection of the curve shall be made from the value of p/fck parameter which is known.
Step 3: Assessment of the column
This can be done in any of the three methods selecting two of the three parameters as known and comparing the third parameter to satisfy. The parameters are Pu /fckbD. Mu/fckbD2and p/fck for rectangular columns. For circular columns the breadth b shall be replaced by D (the diameter of the column).
Design of Short Columns under Axial Load with Uniaxial Bending
The design of columns is involving with the determination of percentage of longitudinal steel p, assuming or knowing the dimensions b and D, grades of concrete and steel, distribution of longitudinal bars in two or multiple rows and d'/D from the analysis or elsewhere. This can be minimized by using the design charts prepared by IS456 in SP-16.
The procedure to design the column using design charts of SP-16 is explained below:
Step 1: Assume the size of the column
Assume the size of the column and calculate d'/D.
Step 2: Calculate P u / f c k b D and M u / f c k b D2
Step 3: Selection of the design chart
Select suitable the design chart from SP-16 for the corresponding to d'/D, grade of concrete and distribution of steel.
Step 4: Determination of the percentage of longitudinal steel
Locate the point of intersection of P u /f c k b D and M u / f c k b D2 and find the suitable value of p/fck
Thus, amount of longitudinal steel is obtained by using the following relation.
Asc = p b D /100 fck
Step 5: Design of transverse reinforcement
Design the transverse ties as per the section 9.3.2.1,
IS Code Method for Design or Columns under Axial Load and Biaxial Bending (Clause 39.6 of IS456: 2000)
A Clause 39.6 of IS456 recommends the following Simplified formula, based on "Breslers load Contour Method", for the of biaxial loaded columns. The relationship between Muxx and Muyz for a particular value of Pu = Puz expressed dimensional form is
(M u x / Mu x 1)+ (M u y / M u y 1) ≤ 1
Where,
Mux and Muy= moment about x and y axes due to design load
Mux1and Mu y 1= maximum uniaxial moments the column can take under the actual load P by bending about x and y axes.
n is related to Pu,/Puz where
Puz= design load on the column
Puz= 0.45 fck Ac + 0.75 fyAsc (I e value Pu of when M = 0)
Puz= 0.45 fc k Ag (0.75 fy – 0.45 fck )Asc
n = Exponent whose value is to be taken as follow:
Where,
Ag= gross area of the section, and
Asc = total area of steel in the section
It is worth mentioning that the quantities Mux ,Muyand Pu are due to external loadings applied on the structure and are available from the analysis
Whereas Mu x ,Muyand Pu are the capacities of the column section to be considered for the design.
Solution of Problems using IS Code Method
18456: 2000, suggest the following steps to design the columns subjected to load and biaxial moment.
1) Selection of trial section for the design type of problems
The preliminary dimensions are assumed during the analysis of structure Assume the percentage of longitudinal steel from the given Mux , Muy, Pu, fy and fck
Note:
Pillai and Menon suggested a simple way of considering a moment of approximately 15 per cent in excess (lower percentage up to 5 per cent if Pu/Puzis relatively high) of the resultant moment
Mu = 1.15 x (M u x 2 + M u y 2) .
As the uniaxial moment for the trial section with respect to the major principal axis xx, if Mu x≥ Muy otherwise, it should be with respect to the minor principal axis.
The reinforcement should be assumed to be distributed equally on four sides of the section.
2) Checking the eccentricities ex and ey for the minimum eccentricities
Clause 25.4 of IS256 stipulates the amounts of the minimum eccentricities and are given in Equation s follows,
(M u x /M u x 1)+ (M u y /M u y 1) ≤ 1
Puz= 0.45 fck Ag +( 0.75 fy – 0.45 fck )Asc
Where 1, b and D are the unsupported length, least lateral dimension and larger lateral dimension, respectively. The clause further stipulates that for the biaxial bending, it is sufficient to ensure that the eccentricity exceeding the minimum value about one axis at a time.
3) Verification of eccentricities
It is to be done determining ex = Mux/Pu andey = Muy/Pu from the given data of Mu x,Muy and Pu and Equation above from the assumed b and D and given leff.
4) Assuming a trial section including longitudinal reinforcement
This step is needed only for the design type of problem, which is to be done as explained in (a) above.
5) Determination of Mux1 and Muy1
Use of design charts should be made for this. Mux1 and Muy1 corresponding to the given Pu should be significantly greater than Mux and Muy respectively.
6) Determination of Puz and n
The values of Puz and n can be determined from Equations as
Puz= 0.45 fck Ag + (0.75 fy – 0.45 fck )
. Alternatively, Puz can be obtained from Chart 63 of SP-16.
7) Checking the adequacy of the section
This is done either using Equation (9.16.1) or using Chart 64 of SP-16.
Design of isolated column footing for axial load and uniaxial bending.
Key takeaways:
Design steps
1) Selection of trial section for the design type of problems
2) Checking the eccentricities ex and ey for the minimum eccentricities
3) Verification of eccentricities
4) Assuming a trial section including longitudinal reinforcement
5) Determination of Mux1 and Muy1
6) Determination of P u z and n
7) Checking the adequacy of the section
Problems:
1) Design an uniaxial square, short column by limit state method with material M 25 and Fe 415 to carry ultimate load of 800 kN and working moment of 80 k Nm about major axis bisecting the depth of column. The unsupported length of column is 3.6m. The column is fixed at one end and hinged at the other. Also design the footing for this column only for flexure and punching shear. Take SBC = 250 kN/m2. Show detailed design calculations and reinforcement details in plan and sectional elevation
Answer:
Given: Uniaxial square column
M 25,fck = 25 N/mm2
Fe 415 fy= 415 N/mm2
Ultimate load P = 800 kN
Working Moment M = 80 k Nm
Unsupported length L = 3.6 m
End condition: One end fixed and other hinged
Solution
A) Design of uniaxial square column
Factored Load Pu = 1.5 x P = 1.5 x 800 = 1200 kN
Mu = 1.5 x M = 1.5 x 80 = 120 k N m
Effective length for column Le = 0.8 L = 0.8 x 3600 = 2880 mm
Min. Size of short square column
L e / L L D1 ≤ 12
2880 / b ≤ 12
b > 240 mm
D = 400 mm
Eccentricity
Mu = P u x e
12 x 106 = 1200 x 103 e
e = 100m
e min = unsupported length/500 + lateral dimension /30
= 360/500 + 400/30
= 20.53 mm or 20 mm
e min = 20.5 mm
e > e min, the short column is designed for uniaxial bending.
Determine d/D
Assume cover 60 mm, d/D =50 /400 = 0.15
Percentage of steel from design chart (refer sp – 16, chart 32)
Refer design chart for fy= 415 N/mm2 and d/D = 0.15
P u /f c k b D = 120 * 103 /25 * 400 *400 = 0.3
M u /f c k b D2= 120 * 106 /25 * 400 * 4002 = 0.075
From chart, (refer chart 2 )
P u/ f c k = 0.03
P = 0.03 fck = 0.03 x 25 = 0.75 %
Required percentage of steel is less than minimum percentage of steel as per IS 456 i.e. 0.8 %
Interaction Diagram for Combined Bending and Compression Rectangular Section-Equal Reinforcement on All sides .
Area of steel in compression
Asc = p b D /100 = 1 * 400 * 400 = 1600 mm2
Provided 4 bars of 20 mm and 4 bars of 16 mm
As c provided = 4 * π/2 * 202 + 4 * π/2 * 162 = 2061 mm2> 1600 mm2
Design of lateral ties:
Diameter of link or lateral ties should not less not ¼ x diameter of largest longitudinal bar
= ¼ x 20 = 5 mm
Diameter should not less not 6 mm
Dt = 6 mm ……………………..whichever is greater
Spacing for lateral ties
The least lateral dimension b = 400 mm
16 x diameter of longitudinal reinforcement bar = 16 x 20 = 320 mm
300 mm
Spacing = 300 mm ……….....whichever is least
Provided lateral ties of 6 mm diameter of bar at 300 mm c/c
Design of footing
Assume self wt. Of footing = 10% of load on column
Total load on footing = 1.1 p = 1.1 x 800 = 880kN
Area of footing = total weight on footing /safe bearing capacity of soil
= P u /200 = 880/ 200
= 3.52 m2
Size of footing
Assume square footing L = B = 
= 1.879 = 2 m
Provided size of footing = 2 m x 2 m
Since the column is subjected to eccentricity of 100mm, the C.G. Of footing is taken at 100 mm away from the axis column as shown in fig as below,
Upward soil pressure = load on column /size of footing provided
= 800 /2 *2
= 200 KN/m2< SBC
Depth of footing based on BM:
Critical section for maximum bending moment occur at the face of column Projection beyond face of column
a = (B-d /2) + e = (2000 – 400/2 ) + 100
= 900 mm
Bending at face of column
BM = po B a2 /2 = 200 * 2 * 0.92/2 = 162 K Nm
Factored BM,
Mu = 1.5 x M = 1.5 x 162 = 243 K Nm
For Fe415
Mu.lim = 0.138fck bd2
243 x 106= 0.138x 25 x 1000 x d2
d = 265.4 mm
Depth of footing based on punching shear:
Critical section for two way shear or punching shear is considered at a distance d/2 from the column face periphery.
Share width at critical Section b0 = b+ d = (0.4 + d)
Factored shear force
Vu= 1.5 p o (B2 - b02)
= 1.5 x 200 (22 - (0.4 + d2)
= 300 (4 – (0.16 + 0.8 d + d2)
= 300 (3.84 + 0.8d + d2) ………(1)
Shear capacity of concrete,
Vu = v (4 bo do)
But, 
v= Ks
c = 1 x 0.25
= 1.25 N/mm2
= 1250 KN/m2
= 500 (0.4 + d) d
= 500 (0.4d + d2) …………..(2)
Equating equation (1) and (2), we get,
300 (3.84 +0.8d + d2) = 500 (0.4d + d2)
3.84 +0.8 d + d2 = 16.67 (0.4d + d2)
3.84 +0.8d + d2 = 6.67d+ 16.67 d2
15.67d2 + 5.87 d - 3.84 = 0
Solving quadratic equation by using calculator,
d = 0.342 mm = 0.35 mm.
Assume cover to footing 0.5 m
Overall depth D = d +c = 0.35 + 0.5 = 0.4 m = 400 mm
Provide Depth for footing is 400 mm, which is greater value based on BM and punching shear.
Steel reinforcement
A s t = 0.5 f c k /f y (1 -
1 – ((4.6 M u)/ (f c k b d2) b d
As t = 2020.8 mm2
Using 12 mm diameter of bar
Area of one bar Ad =
• d2 = 
• 122 = 113.09 mm2
Number of bars n = A s t/ A d = 2020.8/ 113.09 = 17.9 = 18 bar
Provide 18 - 12 mm dia. Of bars at an equal distance
Check for development length:
Ld = 47 x d = 47 x12 = 564 mm
2) Design a bi-axial rectangular short column by limit state method with material M25 and Fe 415 to carry a working load of 900 kN. Working moment of 90 kN-m about major axis bisecting the depth of column and 35 kN-m about minor axis bisecting the width of column. The unsupported length of column about major and minor axis is 3.6 m and 3.2 m. The column is fixed at one end and hinged at the other. Show detailed design calculations and reinforcement details.
Answer:
Given:
Working load = 900 kN
Lex= 3.6 m, fck = 25 N/mm2
Ley = 3.2 m, fy = 415 N/mm2
Solution
Minimum eccentricities
Assume b = 300 mm, D = 500
Ex min = greater of ( 3600/500+ 300/30 ) and 20 mm
= 17.2 mm and 20 mm
E y min = greater of (3200/500 +500/30) and 20 mm
= 23.06 mm and 20 mm
P u = ultimate load= 900 x 1.5= 1350 kN
Mux = ultimate moment @ major axes
2x90 = 180 kN-m
Muy= ultimate moment @major axes
2x35 = 70KN-m
E x = M u x/ P u = 133.33
E y = M u y/ P u = 25.93
Ex and ey are greater than ex min and eymin
2 )Assuming trial section,
b = 300 mm, d = 500 mm
Mu = 1.15 (mu x2 + muy2) ½
= 1.15 (1802 + 702) ½
=222.10 K Nm
P u /f c k b D=1350 * 103 /25 *300 *500 = 0.36
M u/ f c k b D2= 222.10 * 106 /25 * 300 * 5002 = 0.12
d /D =50/ 500 = 0.1
= 0.08
P = 0.08fck = 0.08 x 25 = 2 %
Area of steel in compression
Asc =p b D/ 100 = 2 * 300 *500 /100 = 3000 mm2
Use 20 mm diameter of bar
Number of bar = 3000/ π/2 202 = 7.55 = 8
Provided 8 bar of 20 mm diameter of bar
Asc provided = = 8 x π/2 x 202 = 2513.3 mm2< 3000 mm2
Pt % (provided) =100 * 2513.3 /300* 500 = 1.68 %
3) Determine of Mux 1 and Muy 1
P u / f c k b D =1350 * 103 /25 * 300 * 500 = 0.36
p/ f c k = 1.68/25 = 0.0672
Determine,
M u /f c k b D2= 222.10 * 10 6/ 25 * 300 * 500 2= 0.12
Mux 1 = 0.11 x 25 x 300 x 5002
= 206.25 x 106 N mm > 180 N mm
P u /f c k b D = 0.36 and = 0.0672
d/b = 50 / 300 = 0.166
Determine value Muy1/ f c k b D2
d/b = 0.15 and d/b =0.2
For d/b = 0.15 = 0.11
For d/b = 0.2 = 0.09
For d/b = 0.166
Mu y 1/f c k b D2= 0.09 +((0.11 – 0.09)/(0.2 – 0.15)) x ( 0.2 – 0.166 ) = 0.104
Muy1 = 0.104 x 25 x 3002 x 500
= 117 x 106KN m =117 KN m
As Mux1 and Muy1 are significantly greater than Mux and Muy respective. Redesign of section is not needed.
4) Determine P u z
Puz= 0.45 fck Ag +( 0.75 fy – 0.45 fck ) Asc
= 0.45 x 25 x 300 x 500 + (0.75 x 415 -0.45 x 25 ) x 2513.3
= 2441.49 x 103 N
Puz = 2441.49 KN
5 ) Check for adequacy of section
(M u x/ Mu x 1)+ (M u y/ Muy1) ≤ 1
0.88 ≤ 1
Design is safe
Spacing for lateral ties
The least lateral dimension b = 300 mm
16 x diameter of longitudinal reinforcement bar = 16 x 20 = 320 mm
300 mm
Spacing = 300 mm ……….....whichever is least
Provided lateral ties of 8 mm diameter of bar at 300 mm c/c
Design of footing
Load of column = 900 KN
Assume self wt. Of footing = 10% of load on column = 90 KN
Total load on footing = 900 + 90 = 990kN
Factored load = 1.5 x 990 = 1485 KN
Area of footing = factored load on footing/ ultimate bearing capacity of soil
= P u/200 = 1485 / 200 *2
= 3.71m2
Size of footing
Assume square footing L = B =
= 1.93 = 2 m
Provided size of footing = 2 m x 2 m
Upward soil pressure = load on column /size of footing provided
= 1.5 * 900/ 2* 2
= 337.5 KN/m2
Depth for BM
Determine eccentricity = e =M u/P u = 180/ (1.5 * 900) = 0.023 m = 23 mm
a1 = (L-b)/2 + e
= (2- 0.3)/2 + 0.023
= 0.873 m
a1 = (L-b)/2 - e
= (2- 0.3)/2 - 0.023
= 0.827 m
b1 = (B- a)/2
= (2- 0.5)/2
= 0.75 m
BM calculation
For BM calculation takes greater of a1 and a2
Mxd = Pu calx a12/2 = upward soil pressure a12/2
= 337.5 x 0.8732= 128.60 KN .m
Myd = Pu calx b12/2 = upward soil pressure b12/2
= 337.5 x 0.752/2 = 95 KN .m
Md= maximum of Mxd and My d= 128.60 KN m
For Fe415
Mu.lim = 0.138fck bd2
128.60 x 106= 0.138x 25 x 1000 x d2
d = 193.06 mm
v = should be less than ks 
c
K s = 0.5 + 
c = 0.5 +b/ D = 0.5 +300/ 500 = 1.1
c = 0.25 
= 0.25
= 1.25 N/mm2
v = 1.1x 1.25 = 1.375 N/mm2
Nominal shear stress 
v = 1375KN/m2
Vu = 2 [((2- 0.3)/ 2) – d ] x 337.5 ………1
v < ks 
c
Shear resisted by concrete
Vu = 
c B d
= 1375 x 2 x d
= 2750 d ……2
From equation 1 and equation 2
2 [ ((2- 0.3)/2 ) – d ] x 337.5 = 2750 d
573.75 – 675d = 2750d
d = 0.168m = 168 mm
Depth is safe against one way shear
Check for two way shear
Shear force on shaded area
Vu = [22 – (0.3 + d ) ( 0.5 +d) ]x 337.5…………Equation 1
Shear force resisted by concrete
Vu = 
c b0 d
b0 = perimeter of critical section
b0 = 2 [(D + d) + (b + d)]
= 2 [(0.3 + d) + (0.5 + d)]
= 2[0.8 + 2d]
= 1.6 + 4d
Vu = b0 d
= 1375 (1.6 + 4d) d …………………Equation 2
Equating Equation (1) and (2)
[22 – (0.3 + d) (0.5 +d)]x 337.5= 1375 (1.6 + 4d) d
3.85 – 0.8 d – d2= 4.07 (1.6d + 4d2)
15.3 d2 + 7.31 d – 3.85 = 0
d = 0.316 = 0.316 m
= 320 mm
Overall depth = D = d + c + d/2
= 320 + 50 + 16/2
= 378 mm = 390 mm
Main reinforcement
A s t = 0.5 f c k /f y (1 -
1 – ((4.6 M u)/ (f c k b d2) b d
As t = 1357.38 mm2
As t for 2 m width = 2 x 1357.38 = 2714.77 mm2
Using 16 mm diameter of bar
Area of one bar Ad = π/2• d2 = π/2• 162 = 201.01 mm2
Number of bars n =A s t/ A d = 92714.77/ 201.01 = 13.5 = 14 bar
Provided 14 bars of 16 mm diameter of bar
References
- Reinforced concrete design by S.N.Sinha
- SP- 16 Design aid to Is - 456
