4.3 Wind Load [WL] Manual Calculations By Force Coefficient Method as per IS 875 (Part-3): 2015
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- เผยแพร่เมื่อ 6 พ.ค. 2020
- Lecture: 4.3 Wind Load WL Calculations By Force Coefficient Method as per IS 875 Part 3 2015
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[Load Calculations and Considerations as per IS 875 and 1893 Part-1 2016] drive.google.com/uc?export=do...
[Excel File: Design of RCC Building as per Indian Standard by ETABS] drive.google.com/uc?export=do...
[Free] Course: Complete Design of G+3 Storey RCC Building as per IS Codes by ETABS
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Course Description: In this course, you will learn how to plan structure with respect to architectural planning and as per codal provisions of Indian Standard codes then you will learn how to use IS Codes to calculate the loads on the structure and modelling and analysis of building structure in computer software ETABS, and most importantly you will go through the design procedure of each type of structural element manually as well as with the help of MS Excel Sheets and at last how to detail your structural sections and preparation of drawings for Construction-Site Execution purpose.
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Akshay Thakur
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You really deserve my salute. People can learn from your video without even taking Etabs course from some institutions. You have explained it all from beginners to mid-level session. Hats off to you...
Thank You...!
Welcome... And thank you for providing design files too.
Happy Learning!
Great Work !!
Thanks!
Nice explanation ❤️ loved it
welcome dear!
what is this (for frames) 1 for cladding in excel file near Ka reading. it is not mentioned in lecture.
for frame structure ka = 1
and when there is cladding then ka = 0.8
Good explanation sir but how to calculate the tributary area. for considering value of Ka
length at x and y dirn is taken upto the cantilever top or upto the plinth level .thanx
It is max length in X and Y Direction at repective storey, for respective floor calculation ...
At top you need to only consider head room dimensions ... but for be on conservative side I took max at all floors respectively ...
sir my building height is 8 m then in that case, I have to consider wind force?
If structure is RCC with 8m height with regular shape, no need to define Wind Load, but you can define for being on conservative side...
😍😍... Hello sir u r really good sir.. And ur way of explaining is very excellent... Sir u r saying everything in details according to codal provisions.... Sir can u please tell about some MS excel.... Sir ur really deserves a salute❤️❤️❤️
welcome dear...
to learn excel for structural design you need to understand whole design procedure by manual calculation.....
sometime you need to take some values from tables and charts that forecasting values, use of various excel functions to view your design results, use of formulas that thing you need to know to learn more about excel....
although some other software also you can use to prepare design calculations such as MathCAD...
Hello sir
In excel sheet what is that " for udl calculations on columns"?
N at last " X C/C distance" what is that sir?
that udl value can be use with different methodology... ignore that for the force coefficient method...
Thank you sir
Sir can I get your mail id?
check the about section of channel....
why have you considered the dimension b as 10.965 even for head room roof ? is b base dimension in y dir or floor dimension in y dir?? Since wind strikes on each floor, so isn't the dimension of that floor be considered?
yes you are right, but at top Wind Load is at it extreme level and from perpendicular direction also wind in acting so to became on safer side I Consider whole length for Head Room Slab as respective bottom storey ....
in Load Calculations, always care about wind load at top storey because major chances of collapse/damage in windstroms is at top because of no respective support over it ... this type of safety may trouble you while analysis to satisfy codal requirements (that easy to resolve) but not affect much more in design.
as designer most important thing is safety ... so always give more importance to lateral loads ....
And please make video on how to transfer the design reaction to staad foundation. And which loads needs to create and define and apply to design footings in Staad foundation. If you can make this video please upload it. That would be appreciated.
Thanks for Suggestion...
Sir In sidl load 50 mm 0.21*5 ( what convert met/inch/ what5)how in wall load internality mortar
for 10 mm it is 0.21 kN/m2 right
so for 50 mm
it is 10×5=50 :: 0.21×5=1.05
hence for 50 mm it is 1.05 kN/m2
Thanks you very much sir
Sir 6 mm Ceiling plaster is 456:2000 page no.? Nominal size and mass/weight?
for dead load refer IS 875 Part-1 1987
in which you get mortar weight in kN/m2 i.e., 0.21 kN/m2 for 10 mm
so for 6 mm it is 0.6 × 0.21 = 0.126 kN/m2
SIR A DOUBT, AFTER WE FOUND OUT THAT THE WIND LOAD IN Y DIR IS GREATER, WHERE AND HOW DO WE ENTER THOSE VALUES FOR DIFFERENT STORIES IN ETABS?
All of this are discussed in Section 6 ... video 6.11 ...
Great...👍🏼
thanks...
Can you share the excel sheet also and also why we considered ht of 1st storey as h/2?
PFA Excel in Decscription Box...
Effective Area for Exposed Wind is Half of Storey Above + Half of Storey Below!!!
how you calculate area avg factor Ka.In code tributary area is given.
basically tributory area is the areq that causes loading and so here in this case area of 1 panel of floor is 3.25×17.105=55.59 m2
for exact actual "ka" value you have to interpolate this area value to value shown in table 4 of IS 875-Part 3
sorry but while preparing the course i won't consider too minute things to include so i don't mention that...
I hope you get the point, that how to consider ka value as per tributory area!
Sir kindly can you give of this lacture in pdf copy.....in giving file doesn't show...
check description box for download link of pdf
Dear sir,
I downloaded the excel sheet, but they are emply..
Plz guide, how to get that excel sheet.
which one you find empty can you please specify?? so that i can fix it!
tell me your email id i'll mail you
Wonderfully done....i can the efforts put into this....please also incude design usin IS 16700 at some point if possible
IS 16700 is only for Structures more than height 50m and upto 250m ... for small building project IS 16700 is not used ....
your suggestion, I will consider for future video creation ...
@@AkshayThakur thanks... this code needs detailed explanation and very little material is available to understand this code....if you have any material source on this code please suggest
Book: Brief Review of Structural Aspects of IS 16700 www.springerprofessional.de/en/a-brief-review-of-structural-aspects-of-is-16700-2017/16895598
Article: SEFINDIA
www.sefindia.org/forum/viewtopic.php?t=19035
also go through some research papers on sciencedirect and books on springer over this for crystal clear understanding on structural safety of tall concrete building structures
sir how effective height of 1st floor is calculated?
effective height is the half from below storey and half from above storey and in the way we need to consider ....
by mistake, I have not mentioned ground floor so the height from below the storey to 1st floor not consider ... it is ok for small rc frame building but for high-rise building structure you need to consider that as well ...
you can consider that for more effective results ... 👍🏻
Edited: download the updated excel sheet through the same link for all things set with formulas and correct effective height....
How to find value for wind pressure ????? Couldn't understand.
@Bhushan Patil
pz = Wind Pressure = 0.6 Vz^2
whereas Vz depends on vb, k1, k2, k3, k4 you can also make whole calculations manually...
for only multiplication function [mathematics], I have used excel sheet ....
take a copy of IS 875-3:2015 and go step by step likewise in video .....
I hope your doubt will now resolve ...
sir how to get your excel sheel thanku sir ...
check the description box for download links...
Can we get these excel sheets plz
check the description box for download link
The excel that you have provided. In that excel fine the effective height of floor 1 is 3.25m and in the utube video tutorial it's 1.625. Which one is correct
And what is that table For UDL calculation on column for?
sorry for my mistake, in the excel sheet what i provide is correct...
1.625m is the effective height for head room slab
3.25m is the effective height for all the intermediate floors
that udl is the same for the given input... btw it is another approach to apply wind loads which i won't mention or teach in the course...
@@AkshayThakur can I remove that udl part. I could be asked what it is for. I have given a task to design and analyze a modal in 3 days I have to complete that
yes you can 👍🏻
Why is there no slab at plinth level sir??
here building not has basement floor and here in the example taken in video considered earth filling upto plinth level so that....
Sir can you tell how to find cf for cylinder or pole
To calculate wind moment you can use following formulae as per AASHTO
(½ x circumference x wind pressure) x (pole height^2)/2
this formula might gave you some higher moment than in general if you check with NESC or with other codes....
Indian Code IS 875-3, i am not sure where actually code mentioned this or not....
in such structures like light pole Force-Coefficient method is not used instead of that Pressure Coefficient Method is used.
@@AkshayThakur thank you sir
Also sir can you make a video how to design a flag pole manually of a height of 60 ft and in staad
@@divyanshsharma5538 I am not working in steel structures in major, anything related to high-rise rcc I am open to discussion....
Sir, How did you get 1.63 as effective height @1st floor? while you have taken 3.25 as effective height for the rest of the floors in excel sheet!
sorry... for mistake... there should be 3.25 meter because half of below and above height.....
thing is that, for first floor half of ground floor height and half of 1st floor need to be taken, & in sheet I don't mention Ground Floor, so sheet did't get thing in formula for calculation.... that's the problem... but now i have updated the sheet in which you need to input efffective height for 1st floor and for other floors sheet will calculate by itself from input data.... download the excel file again... for updated things....
@@AkshayThakur thanks a million sir, Appreciated 👏👏👏
@@AkshayThakur sir can you make the similar video of shear wall?
@@jamescarter3090 welcome dear & thanks for suggestion...
how k2 value will be same? please check
k2 is different as per table that i mention.... it is found by interpolation from table... please check the full video...
While calculating Pd= Kd.Ka.Kc.Pz
Can you explain little how to read value of Ka & Kc from IS CODE. You skipped it in the video.
Thanks
ka is area averaging factor that depends on tributary areas of columns
Ka = 1.0 for area less than or equal to 10 sq.m.
Ka = 0.9 for area equal to 25 sq.m.
Ka = 0.8 for area greater than or equal to 100 sq.m.
kc is combination factor that is depends on structures's framing action against wind
combined effect of wind loads on the frame, a reduction factor of Kc = 0.90 may be used over the building envelope when roof is subjected to pressure and internal pressure is suction, or vice-versa.
Value of Ka depends upon tributary area. What is tributary area?
tributary area is a loaded area that contributes to the load on the
member supporting that area
Clause 7.2.2 of IS 875 Part-3 (2015)
@@AkshayThakur Yes..I referred to IITk GSDMA explanatory examples, there I could see lot of examples for framed and unframed buildings. Now it's clear. Thanks.
Can I get excel sheet??
Check the Description Box for Download Link.
why force is not changing wrt height of the building?
it has changed at first storey only. why not for remaining?
wind force is depends upon varoius factors.... most importantly k2 is increase as height increase so....
effective height also matters....
@@AkshayThakur in your example.. Why it is not changing as you said in beginning of the video.. Force will increase with the height.
WL increases if your structure is tall structure... in this case buiilding is of height 15 m only & you can see in IS 875.3-2015 for particular considered case in lecture is terrain category 4 in which k2 value till 20m is 0.8 hence here no sudden change in increment of wind is not possible...
but if you are calculating wind load of 10 storey or above there you will ge to see this....
No sudden wind load increment for low rise building structure....
@@AkshayThakur ok.. I agree that there will not be any change in the wind force up to 20m height.
But in your example you are showing 11 & 21in x direction and 19 & 38 in y direction....how?
Why did you perform the wind analysis ..
as per NBC we only perform wind analysis when the structure height is above 15m
this is not any guideline that should we not apply for lower than 15m structures by NBC or IS, this is general conecption that wind load is not domniating for low-rise RC structures and it somewhat true also ... but if designer wants to apply the loads governing for better stability and safety it is a better thing ....
& to give the overall procedure, I made this ...
@@AkshayThakur I want ask. One thing. I have read in online articles and answers from Structural Engineers. That if earthquake load exceeds Windload Windload need not to be considered ... But I couldn't find the reference in is-875-p3 .. do u know about this.. will be very grateful for your reply...
@@omarbhati5038 to do not consider is not mentioned at all in code...
Oooo
!