Estimation of Loads: Loads: Before we design any structure, it is very necessary to understand the complete structural system of the structure. How the load act on them, how are they transmitted from one element to the other, etc., must be visualized initially. If a loading diagram of the beam is given, it may be easy to design it, however, the first part of the problem, i.e. finding out the loads on the beam requires thorough knowledge of the structural system and the way in which the load transfer takes place. The buildings are designed to carry some live loads, functional loads and many other loads during its life span. To carry the loads, we provide the structural systems like slabs, beams, columns etc., which also have dead loads. The functional loads may be dead loads like tiles, walls, partitions etc., and other loads include wind, earthquake, impact etc. The loads are usually first carried
by the slabs although the beams may be
subjected to direct direct loads. The slabs will transfer the loads on the beams will be transferred to the columns by bending and shear. Columns will resist loading by axial compression with bending and will transfer them to the foundations. Foundations will also resist the loads by bending and shear and ultimately transfer them to the good soil. Before calculating loads on the structure, it is necessary to ascertain preliminary size of various structural components such as slab, beam, column and foundation type. PRELIMINARY SIZE OF RCC BUILDING COMPONENTS: COMPONENTS: 1.Slab:
Thumb rule: ½” per foot run Or 1 cm per 1 foot run (Convert 1 m multiply by 3 to get feet). Example for 5.0m span i.e 15’, the slab thickness required is 15x1=15cm.
Oneway slab: slab: As per Thumb rule: (i) Simply supported slab : 40 to 45mm/ m span (ii) Continuous slab
: 40mm/m span
Based on L/d ratio: (i) Simply supported slab = 25 (ii) Continuous slab
= 30
MAXIMUM SPAN FOR ONEWAY SLABS SLABS WITH RESPECT TO TOTAL DEPTH Total Depth of slab
Simply supported slab
Maximum Permissible Span for One edge Both ends Continuous Continuous
Cantilever slab
100 mm(4”)
2.03 m ( 6’ - 8”)
2.44 m ( 8’ - 0”)
2.84 m ( 9’ - 4”)
1.02 m ( 3’ - 4”)
115 mm(4½”)
2.29 m ( 7’ - 6”)
2.74 m ( 9’ - 0”)
3.20 m ( 10’ - 6”)
1.14 m ( 3- 9”)
125 mm(5”)
2.54 m ( 8’ - 4”)
3.05 m ( 10’ - 0”)
3.56 m ( 11’ - 8”)
1.27 m ( 4’ - 2”)
138 mm(5½”)
2.79 m ( 9’ - 2”)
3.35 m ( 11’ - 0”)
3.91 m ( 12’ - 10”)
1.40 m ( 4- 7”)
150 mm(6”)
3.05 m ( 10’ - 0”)
3.66 m ( 12’ - 0”)
4.27 m ( 14’ - 0”)
1.52 m ( 5’ - 0”)
162.5 mm(6½”)
3.30 m ( 10- 10”)
3.96 m ( 13- 0”)
4.62 m ( 15- 2”)
1.65 m ( 5- 5”)
175 mm(7”)
3.56 m ( 11’ - 8”)
4.27 m ( 14’ - 0”)
4.98 m ( 16’ - 4”)
1.78 m ( 5’ - 10”)
187.5 mm(7½”)
3.81 m ( 12’ - 6”) L/20
4.57 m ( 15’ - 0”) L/24
5.33 m ( 17’ - 6”) L/28
1.91 m ( 6’ - 3”) L/10
Reference
Where L = effective span of slab.
Two way slab: slab: Based on Thumb rule: Support condition Simply supported slab Continuous slab Based on L/d ratio
Overall slab depth 40mm/m span 35mm/m span
Case 1: For span L x upto 3.5m and Live load up to 3.0KN/m 2 Support condition Simply supported slab Continuous slab Case 2: For span L x > 3.5m and Live load > 3.0KN/m 2
L/D ratio 28 32
Support condition Simply supported slab Continuous slab 2.Beams:
L/d ratio 25 30
Width: Width of beam= width of wall or 1/3 to 2/3 depth of beam for all beams 1/3 to ½ of overall depth for Tee beams b= 3.24(Lx)1/3 (As per per Swedish Regulations) Regulations) b, Lx are in cms. Depth: For simply supported and continuous beam=1/10 to 1/12 of clear span Tee beams=1/12 to 1/15 of clear span Cantilever beams=1/5 to 1/6 of clear span. Dp /Dt =1.67 Ly/Lx but not less than 2.5 where Dp=Depth of beam Dt=Depth of slab
Thumb rule: 1” per 1 foot run Or 7 to 8 cm per metre run Support condition Singly Reinforced beam
Doubly reinforced beam
Section Type Rectangular Flanged Cantilever
Depth 100mm/m span 80mm/m span 120 to 150mm/m span 2/3 D of Singly reinforced section
Simply supported Doubly reinforced beam=L/15 or 2/3 D of sSngly reinforced beam Continuous doubly reinforced beam= L/12 to L/15 Based on l/d ratio Span Loading Type 3 to 4m Light 5 to 10m Medium >10m Heavy Relation between Slab & Beam Beam Slab depth D in inches
Beam Depth ‘h’
4” 4½” 5” 5½” 6” 6½” 7”
L/d ratio 15 to 20 12 to 15 12
Maximum Span of beams Simply One End supported Continuous
Both End Continuous
12”
16’
18’ -6”
21’ -0”
14”
18’ -8”
21’ -7”
24’ -6”
16”
21’ -4”
24’ -9”
28’ -0”
18”
24’ -0”
27’ -7”
31’ -5”
18”
24’ -0”
27’ -9”
31’ -6”
20”
26’ -9”
30’ -10”
33’ -0”
22”
29’ -4”
34’ -0”
38’ -6”
h x 16 12
h x 18.5 12
h x 21 12
Reference is depth of beam in inches. Where ‘h’ is
Column Design: Design: Size of column shall be chosen on the higher side and richer mixes and age factor shall be used to the lower storeys. It shall be advantage to use minimum mix as M25 concrete and and Fe 500 TMT bars for Multistoreyed Multistoreyed building design. For achieving economy in shuttering, column column size can be kept the same throughout the height of building(or in steps of a few storeys at the least) for carrying the reinforcement. Moments in columns change sign in each storey, so that, we generally provide symmetrical bar arrangement in a column section and the steel area is kept constant throughout a given storey. In General steel reinforcement is equally on all faces, in case of columns subjected to only axial load where as in case of column subjected to bending also in addition to axial load the longitudinal bars are arranged at the faces faces in the plane of bending. Generally in top storey the axial load is less but the bending moment will be more. In construction of Multi storey building design if the building is designed for higher floors and construction is limit to less than designed floors, design should be checked for top most floor and accordingly the column reinforcement shall be adopted. The design of column necessitates determination of load transferred from beam at different floor levels. Loads are transferred transferred slabs to beams and then to columns. Hence, slabs and beams are normally designed prior to the design of columns. This method enables one to asses the loads on columns more accurately and design of column becomes realistic and economical. However, in practice, many times situation arise which require the design of columns and footings to be given to the client prior to the design of slabs and beams. In such situations, loads on columns and footings foo tings are required to be b e assessed using judgement based on past experience or using approximate methods. The loads on the columns can be determined approximately on the basic floor area shared by each column. These loads are normally calculated on the higher side so that they are not less than the actual loads transferred from slabs/ beams. This method of design of
column is likely to be uneconomical. However as per official procedure, one should should design slabs and beams first in order to know the actual design loads on columns. Only in case of emergency for need of giving the design of column and footings prior to the design of beams, the approximate method of assessing the load on column based on the floor area shared by the columns or approximate beam shears method can be used. 1.Approximate 1.Approximate Method for e for estimation stimation of Load on column by thumb rule: rule: Load on column = No. of floors x Tributary area of column x Load/m2 /Floor. Office/commercial building
Column Position
Residential building
Interior column
12KN/m2
14KN/m2
Side or end column
17KN/m2
19KN/m2
Corner column
22KN/m2
24KN/m2
Note: Add extra 2KN/m2 in toilet and staircase areas. For example for residential flats (GF+3 Floors) for 4m x 4m panel Load on interior column = 4 x 16 m2 x 12 KN/m 2 =768 KN say 780KN or 78 tonnes. 2. Speedy calculation calculation for Load on column column (Alternate Method) Following approximate loads may be considered for various types of buildings. Type of Building
Load /m2 per floor in KN/m2
Residential Flats
16
Education/Medical hospitals building
18
Library/Godowns/printing press building
22
Industrial Building
20
Steel Roof ACC sheet building
7
Staircase/ Escalators
18
Car parking (Silt Floor)
8
Roof terrace floor
10
Load on the column= Intensity of loading x Area to be covered under the influence of the column. This area area can be arrived by bisecting the distance between the columns on all four sides (2 sides on corner columns). Note: 1. Extra load for sitout & Balcony: This can be separately computed and apportioned to the column concerned. 2. For Lift & Machine room: extra loads can be arrived from the lift & Machine room data. 3. Overhead Tank: extra load can be easily calculated Dead wt. of OHT+weight of water to be stored and apportioned to the column concerned. For example 10,000 litres capacity of OHT the dead weight of tank may be taken as 100% weight of water i.e 10m 3 or 10tonnes. Total weight is 10t+10t =20t. Load on each column assuming supporting water tank column column as 4, =20t/4=5t. Moment factors to be considered for calculating loads on columns as per Reynolds hand book: Position of column
Moment factor
Interior Middle column
1.10
Exterior intermediate column
1.30
Corner column
1.80
For example interior intermediate column with 5m x 4m grid with (silt +4Floors) Load on column= Equivalent load x Reynolds moment factor. Equivalent load = Tributary area x Total load intensity. Tributary area = (5m x 4m) =20 m2 Total Load intensity=load intensity for silt Floor+ load intensity for residential flat for 3 Floors +Load intensity for Roof area = (8 +3x16+10)=66KN/m 2 Equivalent load on interior column =20 x 66=1320KN.
Actual axial Load on interior column= Equivalent load x Moment Factor =1320 x 1.10=1452KN. Arriving size of Column: Column: Based on load on column, size of the Column can be calculated by any one of the following method. Method 1 : Based on ultimate load/m2 (Approximate) (Approximate) Ag (Gross area) required = 80 to 100 mm2 per every 1KN ultimate load carried by the column depending on the grade grade of concrete used. Ac= cross sectional area of column D= large dimension of column =Ac/b b=width of column. For 1.5 x 1452= 2178 KN ultimate load, area required= 80x 2178 =174240 mm 2 If b=230mm, D=174240/230=758mm say 750mm. Provide Size of column as 230 x 750mm. Assume 1 to 2% of steel and M25 concrete. If we provide 2% of steel, Ast required= 2/100 x(230 x750) =3450mm2 . Provide 8 Nos. 25 dia steel. Safe Load carrying capacity of column P=(2.6805p+10)bD/1500=(2.7005x2.275+10)230x750/1500 =1851KN. >1452KN . Factored Load =1.5x1851= 2776.5KN> 2178KN. Method 2 : Based on Axial load on Column
Equivalent axial column Load with moment factor (KN)
Column size (mm)
Up to 500
230 x 230
Above 500 to 800
230 x 300
Above 800 to 1200
230 x 450
Above 1200 to 1500
230 x 600
Above 1500 to 1950
300 x 750
Above 1950
300 x 830
As per table for axial load of 1452 KN , approximate size of column =
230
x600mm. The reinforcement shall be taken as 1 to 2% of cross sectional area provided. Method 3: 3: (Based on tributary area) Area required is mm2 /m2 of area covered by the column Grade of concrete
External column
Internal column
M20
2000
1500
M25
1800
1200
M30
1600
900
M35
1400
600
M40
1200
300
For example for (Silt +4Floors) residential residential building interior column with M20 concrete with 5m by 4m grid Area required = (5 storey) x tributary area (5x4)x 1500 mm 2 =150000mm2 Assuming width of column b=230 mm Depth required
D = 150000/230 =652mm say 600mm
The size of column to be adopted is 230 x 600 mm.
Method 4: Based on thumb rule rule : (i) Based on height or span of the beam Column depth is 3 to 5% of total height of building For example (Silt +4 Floors) 5 storeyed building with 3m height The depth of column is (5x3=15m+0.6m(Basement height)) 15.6x3/100= 0.47m say 480mm. If the beam beam span is 4.0m, along along transverse direction, width (b) = 1/12 of span of beam b= 1/12x4.00 =0.333m say 300 mm. Column size is 300 x 480 (ii) Based on storey/ span of beam If building height is 3 storeys or less: If beam span is < 6m, D=300mm; If beam span is between 6.0 to 9m, D=350mm If the beam span is more than 12.0m, D=400mm. If the building height is 4 to 9 storeys: If beam span is < 6m,
D=400mm;
If beam span is between 6.0 to 9m,
D=500mm
If the beam span is more than 12.0m,D=600mm In this example, the span of beam is 5.0m and No. of storeys =5 Size of column is 300 x 400mm. (iii) Based on Seismic and Non seismic areas: For Seismic areas : Assume Pu/fck bD=0.35 for side column and 0.30 for corner column. In case of non seismic areas, areas, the ratio will be 0.40 for side side column and 0.35 for corner column.
For example if Pu= 2178KN, fck=20, column area for non seismic zone for side column=2178x1000/0.40x20=272250mm2 . The size of column is 300 x 900mm. Assume 0.8% of C.S area=2160mm2. Provide 8 Nos. 20mm dia bars. (2512mm2.2160mm2) Method 5: 5: Using Column load & Moment Based on Column load & Moment; Assume 2% of C.S area for fy=250N/mm2 (i) If the line of action of the eccentric eccentric Load is outside C.S.area = Pu /0.4 fck (ii) If the line of action of the eccentric Load is inside (within the section) C.S.area=Pu /0.45 fck Example : Pu=2460KN
Mu=91 KNm fck=20N/mm2 fy=415 N/mm2
Calculate eccentricity of Load =Mu/Pu = 91/2460=0.037m Assume that line of action of axial load is inside the section and check this later. C.S.area required =2460 x10 3 /0.45 x20x10 6 =0.273m2 If one dimension is 460mm, the other needs to be= 0.273/0.46=0.59m say 0.60m Therefore section of column=460mm x 600mm Area of steel reinforcement -=0.20 x 0.273 x(250/415) x 10 6 =3289.16mm2 100
Method 6: 6: Safe load carrying carrying capacity of column based on % of steel and known known column section for various grade of concrete and Steel.
Steel Grade Fe 415
Fe 500
Concrete Grade M20
P=(2.7005 p+ 8) bD/1500
P=(3.27p + 8) bD/1500
M25
P=(2.6805 p+ 10) bD/1500
P=(3.25p +10) bD/1500
M30.
P=(2.6605 p+ 12) bD/1500
P=(3.23p + 12) bD/1500
M35
P=(2.6405 p+ 14) bD/1500
P=(3.21p + 14) bD/1500
M40
P=(2.6205 p+ 16) bD/1500
P=(3.19p + 16) bD/1500
Where P is Axial Load carrying capacity of column in KN. p = % of steel reinforcement reinforcement (say 2% is 2) b = Breadth of Column in mm D = Depth of Column in mm. Method 7: Based on formula formula given in IS 456-2000 STEP 1 :- Calculation of the Influence Area of the Column : The first step is to find out the Influence Area of the Column to be Designed. The Influence Area of a column is the area of which load is being transferred to the column to be designed for. For this purpose in a framed structure small and medium building the design of column is done for the column whose Influence Area is the largest hence the load coming on the column will be so the greater of the any other column in that building hence all the other column having lesser Influence Area hence lesser Loads if provided with the same Designed parameters that required for the column having largest Influence Area, then the whole Structure will automatically become safe against the Loads.
DETERMINATION OF INFLUENCE AREAS FOR LOAD DISTRIBUTION ON COLUMNS STEP 2:- Calculation of the Loads Coming on Column from the Influence Area : In this step the Load Calculation is being done. This is done by calculating all the loads acting within the influence area. The Loads acting are broadly classified as Dead Load (DL) and Live Load (LL). Dead Loads are the load of objects which cannot be moved from on place to another like the loads of Brick Work, Beams, Slabs etc. and the Live Loads are the loads coming from movable objects such as Humans, Chair, Table etc. Thus We Need to Calculate the Dead Loads as well as Live Loads within the Influence Area, these are as follows in the general case of a Building : A)Dead Loads : I. II.
Due to weight of Slab
[25000 N/m3 ]
Due to weight of Floor Finish
[500 N/m2]
III.
Due to weight of Brick Masonry
IV.
Due to weight of Beam
V.
Due to weight of Self Weight of Column
[19200 N/m3] [25000 N/m3] [25000 N/m3]
B) Live Load : It depends upon the Nature of the Structure, and it values for different structural nature are given in the concerned Code of Practice, like in India these are given in I.S.: 875-Part II. For Residential Buildings it is generally considered @ 2KN/m 2 = 2000 N/m2 Now after correct calculation of above loads the Total Load is Calculated by, Total Load on each floor = Dead Load + Live Load Now this the actual load which will be acting on column for each floor, now if the building say 5 storied, then just multiply the value with the nos. of floors, like for five storied building multiply the Total Load on each story with 5. Now thus the Total load acting on column at Column Base is Obtained and it is
denoted with ‘P’. Hence P= Total Load on each Floor X Nos. of Stories = (Dead Load + Live Load) X Nos. of Stories. Now we shall move to the actual Designing to determine suitable Column sections and its Reinforcements so that the above load is safely resisted by the column Designed. It can be done by Three main Methods of Design : a) Working Stress Method b) Ultimate Load Method and c) Limit State Method. The Modern Practice is to use Limit State Method for all types of Designing, Hence we discuss here the Limit State Method Of Design Of Column. STEP 3 :- Finding The Gross Cross-Sectional Area Required For The Column This is the one of the most important and main step of the Design of Column. First in the Limit State Method of Design we must increase the load acting on the column with a Load Factor so that if there will be any accidental increase of loads the column will be still safe to resist the load without a failure. The Factor of Safety for Dead Load + Live Load Combination is 1.5, hence we must multiply the load action on column (P) with the 1.5 to obtain the Ultimate Load that is the Factored Load of the Column that is Pu. Hence Factored Load, Pu = 1.5 X P For Design we will work with this value of load.
Now before going on we here to say that we will design according to the Code Of Practice of I.S.: 456-2000 The Ultimate Load of a Column is given by, Pu = 0.4.f ck .A c + 0.67.f y.A sc ck .A sc [Equation I] Where, Pu = Ultimate Load of the Column in N/mm2 2 f ck ck = Yield Strength of Concrete in N/mm
A c = Area of Concrete (Cross-Sectional Area) of Column in mm 2 f y = Yield Strength Of Steel in N/mm2 2 A sc sc = Area of Steel (Cross-Sectional Area) in Column in mm
Now the column consists of Concrete and as well as Steel in the form of Reinforcements hence the Total Cross-Sectional Area of Column is made of Area of Concrete and Area of Steel. The Total Cross-Sectional area of Column can be also termed as Gross CrossSectional Area of Column and it’s denoted by A g. Hence, Gross Cross-Sectional Cross-Sectional Area of Column Column = C/S Area of Concrete + C/S Area of Steel Therefore, A g = A c + A sc sc And hence, A c = A g - A sc sc Now putting the above obtained value in the original equation (Equation I) we get, Pu = 0.4.f ck .(A g-A sc ck .(A sc) + 0.67.f y.A sc sc [Equation II] Now Assume the Percentage of Steel you want to use ranging anywhere from 0.8% to 6% with Respect to Gross Cross-Sectional Area of the Column (A g). Say Assuming Steel as 1% of A g it means Area of Steel A sc sc = 1% of A g = 0.01A g The higher will be the percentage of steel used the lower will be A g and thus lesser will be the cross-sectional dimension of the column. But the as the Price of Steel is very high as compared to the Concrete hence it is desirable to use as less as steel possible to make the structure economical, again if the percentage of steel is lowered then the A g will increase at higher rate, about 30% with decrease of just 1% of steel and so each lateral dimension of the column will increase and will cause a gigantic section to be provided to resist the load. Therefore both the factors are to be considered depending upon the amount of loadings.
My suggestion is to use the following Percentage of steel for the Design, Which I’ve found to be effective and to produce economical and safe section of Column. Loading (Pu) in N
Percentage Of Steel for Satisfactory Design
Below 250000 --------------------------------------------0.8% 250,000 to 500,000 --------------------------------------1.0% 500,000 to 750,000 --------------------------------------1.5% 750,000 to 1000,000 -------------------------------------2.0% 1000,000 to 1500,000 -----------------------------------2.5% 1500,000 to 2000,000 -----------------------------------3.0% And so on, with increase of each 250,000 N increasing the Percentage of Steel as 0.5%. Now input the value of the A sc sc in the form of A g in the Equation I. For example suppose 1% Steel is used then the equation will be like the one below :Pu = 0.4.f ck .(A g-0.01A g) + 0.67.f y.0.01A g ck .(A Therefore, if we know the Grade of Concrete and Grade of Steel to be used and Factored Load coming on the Column and Assuming the Percentage of steel required appropriately then we can Very Easily Calculate the Gross-Sectional Area (A g) of the Column required from the above form of the equation. Now as the Ag is obtained thus the Lateral Dimensions of the Column that are the sides of the column can be easily determined. The A g or Gross-Sectional Area of the Column means that it is the product of the two lateral sides of a column [i.e. Breadth (b) X Depth (D)], hence reversely knowing the A g we can determine the Lateral Dimensions. For making a Square Section just Determine the Root Value of the A g. Like if the Value of A g is 62500 mm 2 Then considering square section of a column colum n we can get each side
Also Rectangular Column Sections Can be made by using different proportion say b : D = 1 : 2 , Hence D=2b , Therefore, A g = b X D = b X 2b = 2b2 or b=
Hence D can be also determined as D=2b after Calculating the b. Most of the times after calculating the sides of a column it will give results such as 196.51mm or 323.62 etc. values, which practically cannot be provided at field, hence we must increase those values to the nearest greater multiple of 25mm (i.e. 1 inch). For examples a value of 196.51mm may be increased to 200mm or 225mm or 250 mm even, and a value of 323.62mm may be increased to 350mm. more it will be increased the more it will be safer, but it is uneconomical to increase by a very high amount, it should not be increased more than by 75mm to consider the economical factor. STEP 4 : Check For Long/Short Column Depending upon the ratio of Effective Length to the Least Lateral Dimension of a column, a column may be classified as Long Column and Short Column. If the value
of this ratio is less than 12 then it’s called as a short column and if the value is more than 12 then it’s called as a Long Column. A short column mainly fails by direct compression and has a lesser chance of failure by buckling. And in the case of a long column the failure mainly occurs due to the buckling alone. Long column being slender, that is being thin like stick as compared with its length it grows a tendency to get bended by deviating from its verticality under the action of loads. Due to this tendency of long column to get buckled (bended) a long column of all same properties and dimensions that of a short column will be able to carry much lesser load safely than that of the short column. Suppose a 400mmx400mm short column can take a load of 1000KN , then a long column of 400mmx400mm having same grade of concrete, same amount of reinforcement and same workmanship will be able to carry a lesser load like say about 800KN only, hence we get a loss of 200KN which is 20% loss of load carrying capacity. So the above formula used in Step 3 holds good only for the Short Column. For using it in long column a little modification is needed, which I will update it later when I will get hands on this article again. For now let us concentrate on Short Column. First of all we need to find out the effective length of a column, which can be obtained by multiplying a factor with the actual unsupported length of the column. The factor depends upon the end condition of the column. In most general cases we use a Both End Fixed Column for which The Factor is 0.65. Therefore, Effective Length = Effective Length Factor (0.65) x Unsupported Length (l). suppose a column has a unsupported length of 2.7m = 2700mm, hence the effective length will be lef = 0.65x2700 = 1755mm. Least lateral dimension means the shorter of the two dimensions of column that is length and breadth. But in case of a circular column as there is only diameter, hence we will use the diameter.
Suppose a column is of 400mmx200mm section and has an unsupported length of 2700mm, then the Ration of Effective length t the Least Lateral Dimension will be as follows :(Effective Length/Least Lateral Dimension) = (lef/b) = (1755/200) = 8.775 8. 775 which is less than 12 and hence is a Short Column. Co lumn. STEP 5 :Check For Eccentricity : Eccentricity means deviating from the true axis. Thus an Eccentric Load refers to a load which is not acting through the line of the axis of the column in case of column design. The eccentric load cause the column to bend towards the eccentricity of the loading and hence generates a bending moment in the column. In case of eccentric loading we have to design the column for both the Direct Compression and also for the bending moment also. Practically all columns are eccentric to some extent which may vary from few millimetres to few centimetres. In practical field it is almost impossible to make a perfectly axially loaded column, as a reason we have to consider a certain value of eccentricity for safety even though if we are designing for a axially loaded column. The conditions of considering eccentricity and its value may differ from code to code according to the country. Here I will tell you what I.S. : 456-2000 says. According to it the eccentricity which we have to consider for design must be taken as the greater of the followings :i) 20mm. ii) (lef/500) + (b/30) Where, lef = Effective Length of the Column b = Lateral Dimension of the Column (We have to calculate two separate values for two sides in case of rectangular column) Permissible Eccentricity :- 0.05b where b is the dimension of a side of a column, we have to check for two sides separately in case of rectangular column. The Permissible eccentricity must be greater than t han or equal to the actual eccentricity of the column. Or else we have to design it for bending also. STEP 6 : Calculating The Area Of Steel Required Now the Area of Steel Required Asc is to be calculated from the Ag as the predetermined percentage of Ag. For example if the Gross-Sectional Area of the Column is 78600 mm2 and at the starting of calculation of Ag it was assumed that 1% Steel is used then we get,
Asc = 1% of Ag = 0.01Ag = 0.01 X 78600 = 786 mm2 Now we shall provide such amount of Reinforcements that the Cross-Sectional Area of the Reinforcement provided is Equal to or Greater than the Cross-Sectional Area of Steel required above. Hence in the above case we shall Provide 4 Nos. of 16mm Diameter Bars Hence, The Actual Area of Steel Provided, Hence the Area of Steel Provided is Greater than Area Of Steel Required, Hence the Structure will be Safe. NOTE : The minimum of 4 Nos. of Bars to be provided at the four corners of a rectangular or Square Column and minimum diameter of o f Bars that to be used is 12mm Diameter. Hence 4 Nos. of 12mm Diameter Bars are must in any Column irrespective of their necessities. STEP 7 : Determining The Diameter and Spacing Of The Lateral Ties In this step we will Determine the Diameter and the Spacing of the Lateral Ties or Transverse Links or Binders. The Diameter of the Ties shall not be lesser than the Greatest of the following two values 1.
6mm
2.
1/4th of the Diameter of the Largest Diameter Bar
For an example if a Column has 16mm and 20mm both types of bar as Longitudinal Bars or main Reinforcement then 1/4th of 20mm = 5mm Hence we shall provide 6mm diameter Ties. But in practice we use 8 dia RTS only. The Spacing of Ties shall not exceed the least of the followings three values 1.
Least Lateral Dimension
2.
16 Times of the Diameter of the Smallest Diameter Longitudinal Bar
3.
300 mm
[In this case our objective is to minimize the value to reduce the spacing and to make the structure more stable, hence we shall take least value and suitably in a multiple of 25mm]
4. Foundation: (i) Depth of foundation: Minimum depth of foundation: foundation: 500 mm from G.L. ( As per I.S. 1080-1962)
Minimum depth of foundation as per Rankine’s theory d = p / γ {1- sin Ø / 1+ sin Ø } 2 Where p = gross bearing capacity (SBC )
γ = density of soil. Ø= angle of repose of soil. But in practice, the foundation depth is kept at 0.90m or even more. The depth of foundation depends not only on the nature of soil strata but also on the height of building. It is customary customary practice to place the foundat foundation of a “simple footing “ at a minimum depth of 1.50 m from ground level or at least 1.50 times the width of footing. In cold climates the depth is kept at a minimum of 1.50 m below surface because of possible frost action. For low rise building (less than four storeys) a depth of foundation of 1.2m to 1.50m may be adequate. For taller buildings (6 to 12 storeys) 2.0m to 3.0m foundation depth will be adequate. For still taller buildings shallow foundation may not be suitable. In these case higher capacity of pile foundation shall be be adopted to suit the super structure of load and nature of soil condition. For tall isolated Structures, like water towers, the foundation depth may be at 3.0m also. A few practical requirements also may have to be considered while deciding the depth of foundation such as the existing foundation of nearby building, the possible influence of future expansion etc. If the height of building is more, the horizontal forces acting on the building such as wind force are large. As a thumb rule,
minimum depth of foundation may be selected as 5% to 10% of the height of building. Where the moisture content may vary and cause shrinkage, the depth must be considered with the minimum moisture moisture content content variation(1.50m variation(1.50m to 2.0m).In case of black cotton soils soils of of expansive nature, the zone of movement may be as deep as 3.0m to 3.50m. This is why ,the under reamed piles in expansive clays are taken to minimum depth of 3.50m. (ii) Size of Footing: Footing: For working out size of footing working load on foundation to be considered. The load to be increased for 10% on account for self weight of footing. Area of Footing required= Axial Load on column+10% for self weight SBC of Soil
For square square footing footing side of footing = √ Area = X in m PRACTICAL DIMENSIONS DIMENSIONS:: (i) Size of footing : 1000 mm to 3000 mm in multiples of 250 mm. (ii) Maximum depth near column face: 500 to 1000 mm in multiples of 50 mm. (iii) Depth of Footing: Depth of footing can be calculated based on bending moment, one way shear and two way shear shear consideration. For preliminary, depth of footing can can be calculated using any one of the following approximate methods. (a) Thumb rule based on projection of footing Depth of footing D =650 x a where a = projection of footing from face of column in metres and D iis
in
(b) Thumb rule based based on side of the footing for Fe 415 / Fe 500 steel.
mm.
Net upward soil pressure in t / m2
D / A value
5
1/7
10
1/5.5
15
1/5.0
20
1 /4.5
25
1 /4.0
30
1 /3.50
Note : 1)
Increase 20% for sloped and stepped footing.
2)
p = Net upward pressure in t / m2 D = Overall depth of footing in cm. A = Average side of footing in cm.
3)
Minimum depth at the edges =200 mm (150 mm according to IS 456-2000)
(iv) Reinforcement: Reinforcement: Minimum Dia. of bar : 8 Φ RTS
Preferable
: 10 Φ RTS
Thumb rule: Up to 2.0 m width Above 2.0m Up to 3.0 m width
: use 10 Φ RTS : use 12 Φ RTS
Above 3.0 m width
: use 16 Φ RTS
Minimum reinforcement : Not less than 0.15% of c.s area area for mild steel & 0.12% when HYSD bars are used. Maximum spacing spacing of bar : 200 mm (180 mm for Tor 40 as per IS 456-2000)
