Vera level la teach panringa sir.ivlo naal unga videos lam KZfaq la recommend eh pannala

Supera solli kuduthinga gurunatha nee valga

Great brother

Super explain bro thank you so much

Thanks bro hru 🎉❤ kovilpatti garpender❤

💆‍♀️finally i got your videos

நன்றி தம்பி

Tq you bro.

Bro....Elevation low budget la super ah irukanum athuku ethuna tips video podunga

Good explanation bro..oru building manual ah structural design panni podunga bro

Bro super bro

Super teach

spr bro very useful keep posting these kind of quality contents

Andha load calculation pathiyum podunga bro .........detail ah explain panni

Super bro Load calculation podunka

Kindly do videos related to rebar drawings, like how to draw rebar in pile cap, pier, etc

Centre line method estimate video podunga bro

Bro same steel design beam ku oru video podunga bro

Storage of building roll in column pl explain.

Oru slab oda dead load yeppudy calculate pandrathu athuku column yeppudy Design pandrathu

Clg la kuda ipdi explain pannala. 👏👏👏

Bro 800kn load eppudi one column ku varuthunu oru video podunga bro

How to calculate p value

Very very useful vedio sir thanks how we can find the loud in the colum

Bro complete structural design for residential building notes...pdf..book .... Ethachu iruka bro..?

Super bro Footing design videos podunga

Structural drawing epdi podanum nu A to Z fulla solli kuduga anna

Bro job vacancy eruntha solunga

Do more videos

bro estimation series video podunga bro neraiya bbs laam pathi solli kudunga bro

Spacing for beam also same method bro????

Bro 12"×9" column la 15 feet distance vidalama..?

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

Sir teach column position

Roof concrete podrathu...Ready mix concrete(RMC) best'ah illa Hand mix concrete best'ah. Which is economical??? Video podunga bro...its usefull to everyone

One KN என்பது எத்தனை கிலோ

Bro 6* 9* inches column podalamnu sollirukanga appadi podalama illa 6* 12 podalama 6 rod sollirukanga (4x12mm 2x10mm ) Hollow bricks la katturom

Ninga use pannirukka formula axial loaded column kku mattum than use akuma

Bro chat gpt la intha answer varuthu bro kongam clarification Pannu ga bro 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.

ப்ரோ , காலத்தில் stirrups கொடுக்கும்போது நடுப்பகுதியில் அதிகமான spacing,கீழ் மற்றும் மேல் பகுதியில் குறைவான spacing ஏன் கொடுக்கிறார்கள்

Boss i will pay ur fees can u teach me design