Effective length nahi diya hoga to clear length diya hoga and support ka co dition bataega, ki support dono side se fixed hai ya rotational hai, uske hisab se IS 456 2000 me 1 table hai shayad page 90/91 pe effective length of columns k naam se, uske hisab se 0.65 x L ya 0.8 x L ya 1.2 x L karna parega
Sir, AAP concept wale video me, e min. Ke formula me, unsupported length/500 + D/30 liye hn. Pr hr numerical me aap l.effec/500 + D/30 liye hn. Kyu?? Pls clear the confusion sir,
Unsupported length hi hota hai, agar question me unsupported length nahi given ho to effective length se check kar skte hain, jaise Q.1 me kiye hain, but Q.2 me mistake ho gya hai usme unsupported length given tha, galti se effective length le liye hain. I hope confusion clear ho gya hoga.😊
For 1st question Sir agar column ki dimension pehle se diye hue hai to hamara Ag= 400x550= 220000 mm² aayega Agar humne 1% steel of Ag assume kiya To 0.01x220000= 2200 mm² Ac= Ag-Asc = 220000-2200=217800 Ye bhi to sahi hain naa Aise bhi solve kr skte hai kya?
Practical problems me kar skte hain but theoretically Karna nahi chahiye, kyuki aapko kaise pata ki 1% steel hi aaega, apne to assume kiya na, to load kam hai ya jyada iska to koi role aaya hi nahi apke formula me, kya pata requirememt 0.8% ka hi hota, isliye is tarah se nahi solve karte hain. Load ko conaider karne hue load k hisab se Asc reqd nikalna is the correct process. Agar isi numerical me load increase kar denge 1.5 times to aap imagine kar skte hain ki Asc reqd bhi jyada aaega.
Nice explained sir
Good morning Sir ☀️
52:04 pe Ac aapne 500*900 kyu liya?... Wo to Ag hota hein na?
Very very good explanation
Sir yadi effective length question mai nhi diya ho to kaise nikale ge , please reply
Effective length nahi diya hoga to clear length diya hoga and support ka co dition bataega, ki support dono side se fixed hai ya rotational hai, uske hisab se IS 456 2000 me 1 table hai shayad page 90/91 pe effective length of columns k naam se, uske hisab se 0.65 x L ya 0.8 x L ya 1.2 x L karna parega
Good morning sir 🙏🏽🙏🏽
Sir isme 1% of gross area se steel nhi provide kr skte h kya ? Jo ki assume kiye the pehle
thnkuh sir
if column length is not provided and one side of the column given then what
Sir check for minimum % of steel is necessary for exam
Yes
Sir, AAP concept wale video me, e min. Ke formula me, unsupported length/500 + D/30 liye hn. Pr hr numerical me aap l.effec/500 + D/30 liye hn. Kyu??
Pls clear the confusion sir,
Unsupported length hi hota hai, agar question me unsupported length nahi given ho to effective length se check kar skte hain, jaise Q.1 me kiye hain, but Q.2 me mistake ho gya hai usme unsupported length given tha, galti se effective length le liye hain. I hope confusion clear ho gya hoga.😊
Gd moning sr
For 1st question
Sir agar column ki dimension pehle se diye hue hai to hamara
Ag= 400x550= 220000 mm² aayega
Agar humne 1% steel of Ag assume kiya
To
0.01x220000= 2200 mm²
Ac= Ag-Asc
= 220000-2200=217800
Ye bhi to sahi hain naa
Aise bhi solve kr skte hai kya?
Practical problems me kar skte hain but theoretically Karna nahi chahiye, kyuki aapko kaise pata ki 1% steel hi aaega, apne to assume kiya na, to load kam hai ya jyada iska to koi role aaya hi nahi apke formula me, kya pata requirememt 0.8% ka hi hota, isliye is tarah se nahi solve karte hain. Load ko conaider karne hue load k hisab se Asc reqd nikalna is the correct process. Agar isi numerical me load increase kar denge 1.5 times to aap imagine kar skte hain ki Asc reqd bhi jyada aaega.
2%assume kar sakte h area of steel
Karne me to kuch b kar skte hain, but exam k numericals me load agar light to medium ho to 1% assume kar k start karna chahiye numerical.
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