How to find the column size | How to calculate the load on column | Initial size of column | Column
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- เผยแพร่เมื่อ 24 ก.พ. 2022
- Hello friends,
This video explains how to find the column size, how to calculate the loads coming on the column, the safe load-carrying capacity of the column as per IS-456:2000.
Keep watching!!! Happy learning!!!
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It will be 0.008Ag instead of 0.08Ag @ 12.16.
Only to find out the gross area we use the formula 0.008Ag. So the gross area will become 23809.52mm^2 and d=23809.52/240 = 99.206 mm instead of 34.04 mm. This is to check the depth of the column. It was an error by mistake. Sorry for the inconvenience.
very nicely explained ma'am thankyou
This is a very beautiful video which clearly shows the load on the column and has a lot of information related to the building.
Thank you very much!
Very clear ❤ all concepts are define very well
This is an amazing ppreseentation
Love this
Thank You Very Much maam for so detailed explanation 🙏🙏🙏
Most welcome 😊
This video is too good
It will be better to consider 3 bays along X axis and 2 bays along Y axis with G plus 3 floors will be good to explain to arrive size of column.In addition to this thumb rule followed in practice also useful to assume size of column initially.
Thank you for explaining neatly with sketch with detailed calculation.
Thank you.
Please refer to the below link for load calculation of G+2 building part-1
th-cam.com/video/q8-puhJThI4/w-d-xo.html
Part-2
www.youtube.com/watch?v=XSptm...
Part-3
th-cam.com/video/7cpyZ8S666E/w-d-xo.html
Part-4
th-cam.com/video/0RXEcu5URYo/w-d-xo.html
Part-5
th-cam.com/video/vjSE1W3tgvE/w-d-xo.html
Very good
so much help full video thanks
Most welcome
Thank you so much madam 🙏🙏🙏
Most welcome 😊
Thank you.. Good explanation for new student as l am.
You are welcome
I will like you to make a video of design of a primary with a secondary beam attached to the primary beam making the primary beam to have an additional point load from the secondary beam to the UDL of the secondary beam
Superb 👌👌 And useful video for structural designing👍
Thank you! Cheers!
good
Short ,brief, very nicely explained❤
Glad you liked it.
@@CivilEngineeringMastery your lectures are extraordinary Mam
Thank you.
Thanks
Welcome.
Dear Ms. Architect, you explained it very well, thank you very much for your valuable lesson video.
Are there earthquake and wind loads in the calculations explained in this video?
Thank you very much.
Hakan
thank you so much mam
Most welcome 😊
Very well explained. What if mam it is a uniaxially or biaxially loaded ?
Similar to explain by madam you can calculate load on column for vertical load (DL+LL) for typical 2 or 3 columns by using tributary area method.
Multiplying factors
1. For beam spans up to 6.0m multiply by 1.2 and for beam spans greater than 6.0m multiply by 1.30 to arrive gravity loads.
2. Multiply 1.3, 1.4 and 1.5 for EQ zones 3, 4 and 5 respectively for including seismic loads.
By considering first floor steel as 2.50% determine size of column for assumed concrete grade and steel grade from formula Pu=0.4fck Ac+0.67 fy Asc. MInimum size of column as per IS13920-2016 will be 300mm or 20x beam dia bars assumed.This will be followed by Architect for arriving preliminary size of rcc column in building design.
This is suggestion. Thank you
Thank you.
How will I find the foundation depth? Please Make a video.
Why you take Asc as 0.08ag ?
In roof beam load calculation, self weight of beam is not added to self weight of parapet. Considered 6 kn /m only.
. Please correct me if iam wrong
Yes, you are right. self-weight of the beam needs to be added with the self-weight of the parapet wall which is 2.81 + 6.0 = 8.81 kN/m.
It was an error. Kindly ignore and thanks for the notification.
Slab load will act on beam like triangular motion only..but you consider like square motion, can clarify regarding this??
Clarify it, please
Discuss beams.
Madam vedio is good, please vedio few slowly is better
nice topic
Excellent
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Can explain about is this a one way or two way slab?
11/11=1
I think you have to explain how did tou arrive the load of floor finish to 1.5 kN/m2 , the live load, and the 25 that you multiply to the thickness of the slab. We have to know it by your explaination. Whare did you get them, and or how did you determine it? Its not just enough to introduce it, but how come?
Floor finish thickness 75mm, unit weight of floor finish 20 kN/m3.
0.075 x 20 = 1.5 kN/m2.
Unit weight of RCC is 25 kN/m3. This has to be multiplied with the thickness of the slab to get the self weight of the slab.
Live load need to be considered as per IS-875-Part-2.
@@CivilEngineeringMastery Thanks a lot my dear Sir. Now i know it.
In load for second floor why self weight of beam and parapet wall weight is not add.
How you taken area of steel as 804 mm^2 when finding load bearing capacity of column.
The required steel area was 720 mm^2, whereas the provided steel is 4T16, so the provided area is 804 mm^2. The area of one 16mm bar is 201mm^2.
Mam I have column to column distance 12.10 feet span for g+1 building i have use 4 T12 for column .Can it be safe
Please check what is the load coming on that column and adequate size is provided or not and then check this reinforcement is suffice.
Just increase the no. Of stirrups rings along to suffice column strengthen
Load on column
How did you get the 804mm² area of steel?
804mm^2 is the area of steel provided. 4 nos of 16mm (4T16) dia bar provided, if you calculate the area that will come to around 804mm^2.
Madam you forgot to say about how to calculate the lateral ties dia
Niceeeeeee explain mam
Thank you.
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Kindly use the below link to avail the full course.
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Madam please contact number
I am a diploma in civil engineering. Me any helpful this course
Sis 0.8% in calcuion is 0.08which is 8 %insted of .8 pls explain
At last moments while calculating the area of column
I.e, Pu...
Why you added 10^3 in the equation?
300×10^3.....!!!???
I didnt understand that
What is ( 4 T 12 ) or ( 4 T 16 )
How you taken area of steel as 452mm^2 or 804mm^2
Please explain.
4 T 12 - 4 nos of 12mm dia bar. 4 T 16 - 4 nos of 16mm dia bar.
The area of one 12mm bar is 113 mm^2, Area of 4 T 12 is 452 mm^2.
The area of one 16mm bar is 201 mm^2, Area of 4 T 16 is 804 mm^2.
How did u derived Ac =0.992
Ag = Ac+0.008Ag
Ag-0.008Ag = Ac
Ac = 0.992Ag.
IS THE LOAD CALCULATION OF SLAB IS CORRECT?
Only the calculation of slab is wrong and you identified that
I feel like subscribing thrice after watching this. Can someone help me out?
Thank you. Kindly share with your friends.
.008g ki jagah .08g likha hai formule mein , isliye 34.04mm column ki one arm ki length aayi hai.weight mein kn kya hai?
Thank you for the notification. It was an error. Sorry for the inconvenience. It will be 0.008Ag instead of 0.08Ag @ 12.16.
Only to find out the gross area we use the formula 0.008Ag. So the gross area will become 23809.52mm^2 and d=23809.52/240 = 99.206 mm instead of 34.04 mm. Other than this everything will be ok. This is to check the depth of the column. Hope this helps.
How we can contact u...
civilengineeringmastery@gmail.com
Area of Steel kase aaya mam
Please reply
The area of steel has to be calculated as per the % of steel which is given in IS-456:2000.
For columns % steel is 0.8 to 6.0%.
While calculating Pu you take equation 300X10^3.
Why you take like that pl kindly tell us.
The total unfactored load calculated was 200 kN and factored load Pu is 300 kN.
Please check the load calculation.
300 kN is 300 x 10^3 N. Just a unit conversion.
@@CivilEngineeringMastery Thank you☺️
YOU FORGOT TO ADD THE SELF WEIGHT OF 2ND FLOOR BEAM AND SELF WEIGHT OF PARAPET WALL COMING ON COLUMN C1. AFTER ADDING THOSE TWO THE TOTAL LOAD WILL BE {(2.81+6)X3.35}/2 = 14.75KN
I have a doubt maam that 0.8% should been written at 0.008 na
0.8 / 100 = 0.008
So you need to write 0.008Ag insteed of 0.08Ag
Then Ag = 23809.52 mm2
Ast = 423.20 mm2
Yes mam he is right it should be 0.008 in place of 0.08
Madar there are 2 mistakes
Explain