How to find a number of bars in column | column'ல் எத்தனை கம்பி போடுவது என்று கண்டுபிடிப்பது எப்படி?
ฝัง
- เผยแพร่เมื่อ 22 ก.ค. 2024
- As per Indian standard (ISCODE 456:2000)
There are limitations and regulations for column design.
Limitations
1. A column must have 4 numbers of bars.
2. Minimum 12mm bars should be used in column
3. stirrups should follow the rules below
a. Least lateral length of the column
b. 16 times dia of smallest longitudinal bar
c. 300mm
4. stirrups dia should be 1/4 of the longitudinal bar.
5. Minimum 5mm bar should be used as a stirrup.
In this video we deal with column reinforcements
To find number of bars in column, we need to know how much load acting on the column and dimension of the column.
Source:
• How to decide the numb...
www.iitk.ac.in/ce/test/IS-cod...
civilsite.in/reinforcement-de...
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chapters
00:00 - starting
00:35 - purpose of the column
00:48 - Load and its types
02:29- Area of steel
03:49 - Example
04:29- Axial load
04:59- Gross area
06:09- Main reinforcement equations
08:33- Limitations
09:11- How to find area of one bar
10:32- Number of bars
12:36 - Distribution bars
13:06 - Spacing of stirrups
14:26 - Contact details
Vera level la teach panringa sir.ivlo naal unga videos lam TH-cam la recommend eh pannala
Thanks for the feedback my dear friend 🥳
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நன்றி தம்பி
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Kindly do videos related to rebar drawings, like how to draw rebar in pile cap, pier, etc
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Footing design videos podunga
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Storage of building roll in column pl explain.
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Oru slab oda dead load yeppudy calculate pandrathu athuku column yeppudy Design pandrathu
How to calculate p value
Structural drawing epdi podanum nu A to Z fulla solli kuduga anna
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Bro 12"×9" column la 15 feet distance vidalama..?
15" x9" safe ha irrukum bro!
Spacing for beam also same method bro????
column ly spacing kandupudika innum silla methods irruku heavy structures vera mathiri dhan spacing provide pannanum.
Bro main rod ahh column la entha direction la kudukanumnu sollunga bro..
Example ahh ippo oru 1x.75 post kku 6number of 12 mm kudukuromna atha 0.75 pakkam kudukanuma illa.... 1 pakkam 3 rod kudukanuma sollunga bro... Plz
Perpendicular to Longer span (unsupported length). 4 side la entha side la support oda distance adhigamo athuku perpendicular ah Rebar vara mathiri pathukalam.
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Steel reinforcement for column

Calculate the diameter and number of steel reinforcement required for a column size 230 × 300mm carrying an ultimate axial load 800 kN and the great of the concrete is m20 and Fe550d
To calculate the diameter and number of steel reinforcement required for the given column size and load, we need to perform the following steps:
Step 1: Calculate the cross-sectional area of the column.
The cross-sectional area of the column can be calculated as follows:
Area = Width x Depth Area = 230 mm x 300 mm Area = 69,000 mm²
Step 2: Determine the compressive strength of the concrete.
The grade of concrete given is M20. According to the IS 456:2000, the characteristic compressive strength of M20 grade concrete is 20 N/mm².
Step 3: Determine the yield strength of the steel reinforcement.
The grade of steel reinforcement given is Fe550d. According to the IS 1786:2008, the characteristic yield strength of Fe550d grade steel reinforcement is 550 N/mm².
Step 4: Calculate the area of steel required to resist the given axial load.
The area of steel required to resist the given axial load can be calculated using the following formula:
Ast = (Pu x 1000)/(0.87 x fy)
Where,
Ast = Area of steel required (mm²) Pu = Ultimate axial load (kN) fy = Characteristic yield strength of steel reinforcement (N/mm²)
Plugging in the values, we get:
Ast = (800 x 1000)/(0.87 x 550) Ast = 1687.29 mm²
Step 5: Calculate the diameter and number of steel bars required.
Assuming that we will use only one steel bar, we can calculate the diameter of the steel bar required as follows:
Area of one steel bar = (π/4) x d²
where d is the diameter of the steel bar.
Plugging in the values, we get:
Area of one steel bar = (π/4) x d² 1687.29 = (π/4) x d² d² = (4 x 1687.29)/π d² = 2151.05 d = 46.37 mm (approx.)
Therefore, we can use a steel bar with a diameter of 46 mm.
Now, we can calculate the number of such steel bars required as follows:
Number of steel bars = Ast/Area of one steel bar
Plugging in the values, we get:
Number of steel bars = 1687.29/((π/4) x 46²) Number of steel bars = 3.52
Since we cannot use fractional steel bars, we need to use 4 numbers of steel bars.
Therefore, the diameter of steel reinforcement required for the given column size and load is 46 mm and the number of steel bars required is 4.

Calculate the 12mm diameter and number of steel reinforcement required for a column size 230 × 300mm carrying an ultimate axial load 800 kN and the great of the concrete is m20 and Fe550Sd
2 / 3
To calculate the number of 12mm diameter steel reinforcement required for the given column size and load, we need to perform the following steps:
Step 1: Calculate the cross-sectional area of the column.
The cross-sectional area of the column can be calculated as follows:
Area = Width x Depth Area = 230 mm x 300 mm Area = 69,000 mm²
Step 2: Determine the compressive strength of the concrete.
The grade of concrete given is M20. According to the IS 456:2000, the characteristic compressive strength of M20 grade concrete is 20 N/mm².
Step 3: Determine the yield strength of the steel reinforcement.
The grade of steel reinforcement given is Fe550Sd. According to the IS 1786:2008, the characteristic yield strength of Fe550Sd grade steel reinforcement is 550 N/mm².
Step 4: Calculate the area of steel required to resist the given axial load.
The area of steel required to resist the given axial load can be calculated using the following formula:
Ast = (Pu x 1000)/(0.87 x fy)
Where,
Ast = Area of steel required (mm²) Pu = Ultimate axial load (kN) fy = Characteristic yield strength of steel reinforcement (N/mm²)
Plugging in the values, we get:
Ast = (800 x 1000)/(0.87 x 550) Ast = 1687.29 mm²
Step 5: Calculate the number of 12mm diameter steel bars required.
We can calculate the area of one 12mm diameter steel bar as follows:
Area of one 12mm diameter steel bar = (π/4) x d² Area of one 12mm diameter steel bar = (π/4) x 12² Area of one 12mm diameter steel bar = 113.1 mm²
Now, we can calculate the number of 12mm diameter steel bars required as follows:
Number of 12mm diameter steel bars = Ast/Area of one 12mm diameter steel bar
Plugging in the values, we get:
Number of 12mm diameter steel bars = 1687.29/113.1 Number of 12mm diameter steel bars = 14.92
Since we cannot use fractional steel bars, we need to use 15 numbers of 12mm diameter steel bars.
Therefore, the number of 12mm diameter steel reinforcement required for the given column size and load is 15.
in video what did i explain sample the AI Repeats.
1. The load acting on the column,
2.Area of column
you will get ast.
with the help of ast
3. area of bar using.
proved the bar as per ast requirement
Bro 6* 9* inches column podalamnu sollirukanga appadi podalama illa 6* 12 podalama 6 rod sollirukanga (4x12mm 2x10mm )
Hollow bricks la katturom
Illa, puriyala..
1/2 * 3/4 size column podalamnu mester solluranga appadi podala illa 3/4 * 1 feet podalama column la 4*12mm rod 2*10mm rod podalamnu sollirukanga
@@Manidhan-om2ih minimum column size 9"x 9" ithuku keela poga kudathu.
@@Manidhan-om2ih athey maari column ku minimum vertical bar 12mm provide pananum.
@@Manidhan-om2ih Mullion ku matum 10mm use panalam athu wall length 3m ku mela pogum pothu Mulion provide panuvanga. Athu column illa but appdi act aagum for additional support.
Ninga use pannirukka formula axial loaded column kku mattum than use akuma
Apdi illa, ella column ku ore formula dhan, but corner illa side la vara column ku unga load la irrunthu some % of the amount load increase pannaum.
4 corner column mella vara load kooda, athanodala irrunthu 15 to 20 % load oda add panikanum , side la vara column ku mella vara load kooda athanoda load la irrunthu 11 to 15% ungaloda load kooda add panni kanum. Then atha P (load) ah unga formula la use pannanum.
@@civilonlycivil bro center la vara columnku thana load athikam varum corner la irukara columnku kammiyathana varum?
@@vigneshbalaji591 most of the case la apdi la dhan, but water tank la vara apo corner column hum athigamana load ah handle panna vendi irrukum. Corner column ku moment aagra thamai athigama irrukum adhanala antha load oda antha percentage of loads ah addition pannanrom.
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One KN என்பது எத்தனை கிலோ
ப்ரோ , காலத்தில் stirrups கொடுக்கும்போது நடுப்பகுதியில் அதிகமான spacing,கீழ் மற்றும் மேல் பகுதியில் குறைவான spacing ஏன் கொடுக்கிறார்கள்
Shear force supports la adhingavum, mid space la kammiyavum irrukum so apdi tharanga..
Ok bro, அந்த stirrups இடைவெளியை எவ்வாறு கணக்கிடுவது?
13:06
@@manikandanmani-zb1tl roof height la irrunthu l/4 split pannikonga antha length ku mattum stirrups spacing 100mm maintain panunga both floor and roof L/4 la balance area la stirrups spacing maximum aa irukalam but not exceeding 200mm.
Example floor to roof height 3m
3/4 is 0.75metre
So floor bottom to 0.75metre ku close spacing of stirrups provide pannanum.
Same roof side also.
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Actually I was making a course, soon I will announce. 💓