This lecture was indeed the best among all the other lectures i came across on youtube....!! Hats off to u sir and huge respect to your amazing explanations skills!!
Good job .. but u didnt explained the requied depth of beam. Which most important part in single reinforced beam. And please tell me that when the requied depth is greater than the assumed depth , what to do then next pls..
If the question is specificly asking for singley reinforced bram design, than go with greater 'd' and continue steps mentioned in video. If there is no limitation ( or limitation of depth) , then design doubly reinforced beam
Design a beam using limit state of 300 × 600 mm (overall) to support ultimate bending moment 50 kNm, shear force 50 kN and torsion 50 kNm with M 25 grade of concrete and Fe 415 grade of steel
This lecture was indeed the best among all the other lectures i came across on youtube....!! Hats off to u sir and huge respect to your amazing explanations skills!!
Thankss you explain with english so its very clear for us
Thanks and welcome
Where did the support diagram and width come from, it was never mentioned. If it eas assumed do we really need to find the effective length?
sir, how did we get the ratio of length by depth as 12?
Assumed
Though the speed was a bit fast, but was totally useful! Thank you so much sir!
Glad it helped you.
Do like and subscribe
AMAZING SIR.. new subscriber
Cheers
Really nice Explanation. Thanks a lot...
thank you
How is l/d = 12. Explain. In IS code l/d = 20 for simply supported beam
20 is max
Do a video on anchorage lenght numeriacal too..
Will give a try. Cheers
Good job .. but u didnt explained the requied depth of beam. Which most important part in single reinforced beam. And please tell me that when the requied depth is greater than the assumed depth , what to do then next pls..
Adopt the greater depth and compute rebar with respect to it.
If the question is specificly asking for singley reinforced bram design, than go with greater 'd' and continue steps mentioned in video.
If there is no limitation ( or limitation of depth) , then design doubly reinforced beam
Sir near 7.07 min how did you use x= and find the value
Thank u sir very helpful...New subscriber
Cheers
Unit weight nikalne ke 24kN/m hi aayega kya rcc ke liye to 25 lete hai🙄
Very good explanation ...but sir can u please explain all the rest examples in hindi....
I wish I could, but i can’t speak hindi
@@civilconst 😅no problem sir.....
How bearing width 230 came?
Support width
you should use "Leff" to find d, not 4m. After finding 4.2, you should have repeated your calc
L/d how.....?
Nice but had to watch in 0.75 speed 😀
Sorry for the inconvenience.
Ye kesa pta chla k under reinforced section h?
MOR and moment load comparison.
Design a beam using limit state of 300 × 600
mm (overall) to support ultimate bending
moment 50 kNm, shear force 50 kN and torsion
50 kNm with M 25 grade of concrete and
Fe 415 grade of steel
Thankyou 👍🏿👍
Welcome 👍
How to calculate ast
Sir how l/d become 12???!!!
Thumb rule
Total width of supporting is 230....mm assumption or code...pls somebody tell...😭
General assumption.
@@civilconst thank you
The design of beam are not properly do
udl load is not considered while calculating MOR
All load shall be included.
Please make diagram also
Bro Ast find karne ke liya tumne (0.5fck)/fy* [1-......]bd ye formulae Q ni use kiya
👍tq
Cheers
C/c distance explain kr do
How। 4+(.23/2)+(.23/2)
230mm generally wall width hoti hai
it was good bt you were too fast 🤐
Sorry for inconvenience.
Clear nahi hai aur bahoot fast video 😐😐
You can rewind and watch it again.
Sir clear nazar ñii aa rha h
Bro tospeed bro i do not understand I am a poor
Sorry.
Pdane bhi ata hi kya smaj aa raha isme
Sorry for inconvenience.
I took l/d ratio as 20,but MOR>MAX is not satisfy. why.....?
Go for greater depth or rebar area.
Afaile ghokya hoki padhako ho.. 🙄
ho ho aafai le ghokyaa ho.. aru lai ni help huncha ki ali ali bhanera upload gardyaa ho
How to take support width 230mm
Considering brick wall of width 230 mm