For those who took out time making fun of his accent, get a life guys! Thanks SL Khan for this wonderful video. Just what I needed. One small suggestion, please refer to original codes and clauses from codes if possible rather than the book. That would make our lives easier and your slides more interesting. Thanks once again.
hey. thank you dear for such a great appreciation............. happy to see your comment //// next time i will try to mention the codes .....................welcome to my channel
Hi Sir, after watched and learnt this video, can i check with you a question? Q1) with referenced to your final design (as shown at time: 17:36) , the Main Steel is #4@9"C/C. This means the whole RC Slab (10ft long) estimated having 15 units of #4 Main Steel Bar with 9" spacing between each bar? My Calculation: Span length : 10ft x 12 = 120 inch Main Steel C/C Spacing = 9 inch Number of Main Steel Bar in 10ft Span Length = (120 inch / 9 inch) = 13.333 + 2 (Most Left & Most Right steel bars) = 15 units Main Steel Bar Please correct me if my calculation is wrong. Thanks
Hon Tin Chin hi dear . thanks for asking. remember main bars are not going along 10 ft side. so your calculation is for distribution bars. which are #3@10" c/c. for total numbers. divide 10ft by 10" and add only one bar for one end not two bars for both ends. . #4@9"c/c mean that #4 bar will be placed after every 9". i mean distance b/w two bars will be 9". the total length is not a problem.
Hello sir, Thank you for the video. I am using Australian/New Zealand stabdard for the design of slab, and it uses the same method as you have explained. Could you please explain what im supposed to do when cover is larger than half depth of slab. Here is the problem: The slab is 100mm thick and located near the coastal area. According to AS/NZS, the cover should be 55 mm. After calculation, I found that I needed to use D20mm steel bars. My question is, how will this reinforcement be located in the slab? is it even possible to put reinforcement in slab with such deep cover? Thank you
Nice Video! Keep it up. Btw, Im just gonna ask if designing reinforced concrete slab is the "same" as the designing of reinforced concrete beams? What I mean of the same is, Beams are like trial and error process related to it reinforcement ratio right? Is that the same with slabs???
I m confuse (longer dimension / shorter dimension =2) then one way or two way ? What is it please clear Because mostly teacher considered two way slab when dimension ratio equal to 2 but u r considered one way slab
A reinforced concrete floor system consists of a 100 mm thick slab cast monolithically with beams having an effective span of 6 m and spaced 3 m centre to centre. The superimposed load on floor is 6 kN/m2. Design the beam using M 20 grade concrete and Fe 250 grade of stee (working stress method) ? Isko solution dikha do krke
Hi, Is there a maximum span for a one way slab, can I consider a 36f x12f a one way slab, and can I build a 9 feet high wall in top and along the span without beams under neath it? Thanks
Sir what should we do when we have to teach the deisgn of slab for international students. I am teacher for malaysian students but now i am confuse which textbook i should follow. I wanted to know alry from code book the tect books are same for every country or not sir please guide me
Hi. Sir it depends on your country’s institutions, like in my country we follow ACI codes. If your country follows ACI or whatever, then there are enough books available on google
hello misshollywood.... yes i have reference . i have taken this formula from my text book named as "Design of Concrete Structures,14th by Nilson" you can check this book.... thank you for joining us..
Very good lecture I have a request I am in my second year of civil engineering in London I have exam next week Can you tell me what are the basic things to learn for exam ? Thanks Deera manana wrora
No. There can be little different conditions. This method is for one way slab only. If you want to know whats oneway and two way? Watch uploaded video about oneway and twoway slab
salam aleykum sir. Awe thanks for your videos information and about books H.Nilson.They're so useful. Can u explain pls when u calculate D.L how did u solve 150 ?
Awais Karni this design is for self weight as dead load and live load. if your slab has all these types of dead loads you can add them. thanks for comments
Thanks for the reply. Another thing, for suspended slab that you illustrated, the main re-bar is located 3/4" from the bottom of the slab. What about if the slab is on grade, where do we locate the re-bar?
hello.#3 bar means 3/8 inches dia bar #4 means 4/8 inches dia bar #6 means 6/8 inches dia bar #7 means 7/8 inches dia bar #8 means 8/8 inches dia bar or 1 inche #9 means 9/8 inches dia bar etc
If you are designing a slab and a beam would you design the slab first and then the beam, and for dead load and live would be same from beam, let’s say you have slab of 16” and b 48 the building is a theater how would you find the dead load and live load
First you must know , For design purpose you have to follow some steps given in your country code ,, this code is giving you live load and the dead load you can know it from the dimenstion of your slab plan. for each type of building there is a different live load value. Building design must start from top to down you need to design the slab first then take loads from slab to design your beam , from beam you again design the column
Hi sir, if you are free, could you post some lectures on the following design? Q1)design a 20ft x 10ft RC Ground Slab without supporting beam. (This Ground Slab shall withstand the weight of 20tons lorry driving over it.) Let me know what are the Reinforcement Bar or BRC required for such design Q2) design a 50' run ft long concrete drainage (Width ? x Depth ? x Thickness of concrete ? x Reinforcement Bar require? ) which shall be able to withstand the soil force & side force without crack.
Q2) I mean the 50' Run Feet drainage is rectangular "U -shape"instead of "round-shape". I would like your advice on how to design it so that it can withstand the soil forces (up force & side force) without crack & slanted. All unknownsto be assumed by sir. Thanks sir and I will wait for your upcoming lecture videos
Hon Tin Chin for size design check a lecture name as design of a rectangular open channel. how ever for forces capacity i will try to bring new lecture. thanks
hi dear. you right about that student problem. i checked my mistake. thanks for catching it. check that video tomorrow i am going to edit that. how ever please do not delete your comments. that can help other people to know about the mistake. thanks dear
Can you solve this problem please sir? Size :8×3m l.l=2 kN and a floor finish of 25 mm thick terrazzo flooring width of the wall is 300 mm grade material is fe415 and M20
I can design it but only if u let me know how to calculate the floor finish load because it is not given ,it is given that floor thickness is25 mm terrazzo flooring
DESIGN OF TWO-WAY SLAB BY ACI-COEFFICIENTS FF SLAB DESIGN PANEL No : Input Data: Slab Dimensions Short Span = 10 ft Long Span = 20 ft Span Ratio = 10 / 20 = 0.5 Slab Thickness (reqd) = 2 x (10+20) x 12 / 180 = 4. in Slab Thickness (prvd) = 6. in in Edge Conditions Short Span Edge Condition = One Edge Continuous Long Span Edge Condition = One Edge Continuous Applied Loads Slab self wt = 75.00 psf Service Dead Load = 0.075 psf Service Live Load = 220 psf Material Strengths Concrete Strength = 4 ksi Steel Yield Strength = 60 ksi Design Moments Loads Total Ultimate DL =1.4(75 + 0.075) = 105.105 psf Total Ultimate LL =1.7(220) = 374 psf Total Ultimate Load =105.105 + 374 = 479.105 psf Short Span Moments Postive Short span Moments = 1/11 x 479.105 x 10^2 = 4.356 k-ft Negative Moment on Discontinuous Edge = 1/24 x 479.105 x 10^2 = 1.996 k-ft Negative Moment on Continuous Edge = 1/9 x 479.105 x 10^2 = 5.323 k-ft
Long Span Moments Postive Long Span Moments = N.A. Negative Long Span Moments = N.A.
DESIGN OF TWO-WAY SLAB BY ACI-COEFFICIENTS FF SLAB DESIGN PANEL No : Reinforcement Calculations Along Short Span Positive Steel Mu=4.36 k-ft ; b=12 in ; d=5.313 in ; fc'=4ksi ; fy= 60 ksi As+ = 0.85 x 4 / 60 x [1 - [1-2 x 12 x4.356/(0.9x0.85x4x12x5.313^2) ]^0.5 ] x 12 x 5.313 = 0.187 sq.in. Asmin = 0.0018 x 12 x 6 = 0.130 sq in use As+ = 0.187 sq.in. (#3 @ 7 in ) Negative Steel Mu=2 k-ft ; b=12 in ; d=5.313 in ; fc'=4ksi ; fy= 60 ksi As- = 0.85 x 4 / 60 x [1 - [1-2 x 12 x1.996/(0.9x0.85x4x12x5.313^2) ]^0.5 ] x 12 x 5.313 = 0.084 sq.in. Asmin = 0.0018 x 12 x 6 = 0.130 sq in use As- = 0.130 sq.in. (#3 @ 10 in ) Negative Steel Mu=5.32 k-ft ; b=12 in ; d=5.313 in ; fc'=4ksi ; fy= 60 ksi As- = 0.85 x 4 / 60 x [1 - [1-2 x 12 x5.323/(0.9x0.85x4x12x5.313^2) ]^0.5 ] x 12 x 5.313 = 0.230 sq.in. Asmin = 0.0018 x 12 x 6 = 0.130 sq in use As- = 0.230 sq.in. (#3 @ 5.5 in ) Along Long Span Positive Steel Mu=0 k-ft ; b=12 in ; d=4.938 in ; fc'=4ksi ; fy= 60 ksi As+ = 0.85 x 4 / 60 x [1 - [1-2 x 12 x0/(0.9x0.85x4x12x4.938^2) ]^0.5 ] x 12 x 4.938 = 0.000 sq.in Asmin = 0.0018 x 12 x 6 = 0.130 sq in use As+ = 0.130 sq.in. (#3 @ 10 in ) Negative Steel Mu=0 k-ft ; b=12 in ; d=5.313 in ; fc'=4ksi ; fy= 60 ksi As- = 0.85 x 4 / 60 x [1 - [1-2 x 12 x0/(0.9x0.85x4x12x5.313^2) ]^0.5 ] x 12 x 5.313 = 0.000 sq.in Asmin = 0.0018 x 12 x 6 = 0.130 sq in use As- = 0.130 sq.in. (#3 @ 10 in ) Negative Steel Mu=0 k-ft ; b=12 in ; d=5.313 in ; fc'=4ksi ; fy= 60 ksi As- = 0.85 x 4 / 60 x [1 - [1-2 x 12 x0/(0.9x0.85x4x12x5.313^2) ]^0.5 ] x 12 x 5.313 = 0.000 sq.in Asmin = 0.0018 x 12 x 6 = 0.130 sq in use As- = 0.130 sq.in. (#3 @ 10 in )
Shear Check Total load on the panel = 10 x 20 x 479.105 = 95.8 kips Shear Capacity of the section fVc = 0.85 x 2 x ( 4000 )^½ x 12 x 5.25 / 1000 = 6.77 k/ft Shear along short side Vua = ( 0.06 x 95.8 ) / (2 x 10 ) = 0.28 k/ft
Expertise civil engineering here
thank you rora.... abad osa ... zra de rala khushala kro
welcome rora, daira manana
For those who took out time making fun of his accent, get a life guys!
Thanks SL Khan for this wonderful video. Just what I needed. One small suggestion, please refer to original codes and clauses from codes if possible rather than the book. That would make our lives easier and your slides more interesting.
Thanks once again.
hey. thank you dear for such a great appreciation............. happy to see your comment //// next time i will try to mention the codes .....................welcome to my channel
Right dear
Dear Sir, Very good lecture thank lot for
welcome dear..............
Very good explanation! Thank you for posting it!
Glad it was helpful!
Thank you genius for viral your video on design of beam on youtube and solve my problem in some steps
Allah give u to long life tnx
Sir Can You Please explain that why you did not consider 10' when you was calculating volume for Dead Load. Please tell me Sir..
Because we design unit width e.g 12 inches..
Mashallah sir g der zbr10. ao der simple teaching
welcome dear
GREAT SIR PROUD OF U.....
thank you dear
Hi Sir, after watched and learnt this video, can i check with you a question?
Q1) with referenced to your final design (as shown at time: 17:36) , the Main Steel is #4@9"C/C. This means the whole RC Slab (10ft long) estimated having 15 units of #4 Main Steel Bar with 9" spacing between each bar?
My Calculation:
Span length : 10ft x 12 = 120 inch
Main Steel C/C Spacing = 9 inch
Number of Main Steel Bar in 10ft Span Length = (120 inch / 9 inch) = 13.333 + 2 (Most Left & Most Right steel bars) = 15 units Main Steel Bar
Please correct me if my calculation is wrong. Thanks
Hon Tin Chin
hi dear . thanks for asking.
remember main bars are not going along 10 ft side. so your calculation is for distribution bars. which are #3@10" c/c.
for total numbers. divide 10ft by 10" and add only one bar for one end not two bars for both ends.
.
#4@9"c/c mean that #4 bar will be placed after every 9". i mean distance b/w two bars will be 9". the total length is not a problem.
SL Khan : noted with thanks sir. I found that your videos are very comprehensive and easy for understanding, and very useful to me.
Please make a video on design of two way slabs.
v good sir.bohat acha hain.
wow great explanation
Glad you liked it
Sir kindly upload a lecture on design of two way slab by aci codes
Will try
this video made me learn somrthing with lots of cheers....simply awsome
Thank you for watching dear ❤️❤️
Take my salam sir,
Can you make a video of design one way slab with beams (50*80 hall)
Thank you sir for this videos you know it's a very helpful especially when you have exam in the next day and being so lost.
Nice work
the same was happened to me when i was student of 5th semester.... that why i have posted this video... enjoy dear
Good
Thank alot❤️❤️❤️
Hello sir,
Thank you for the video. I am using Australian/New Zealand stabdard for the design of slab, and it uses the same method as you have explained. Could you please explain what im supposed to do when cover is larger than half depth of slab.
Here is the problem:
The slab is 100mm thick and located near the coastal area. According to AS/NZS, the cover should be 55 mm. After calculation, I found that I needed to use D20mm steel bars.
My question is, how will this reinforcement be located in the slab? is it even possible to put reinforcement in slab with such deep cover? Thank you
Thanks for sharing such good information regarding reinforcement plan.
welcome dear....
this video really helpful, but don't we need to check the required depth??
Very well explained
thank you dear, enjoy
@@Sulimankhanacademy have you made any video regarding shear reinforcement or design of doubly reinforced beam ...
yes, check the same playlist
Thank you Sir.
Right
Simple & perfect presentation
can you do two way slab?
thank you so much . i will try it soon
Very nice. Great job.
thank you for appreciation
thank u sir its was very intresting lectre for me
welcome dear.. enjoy
By your video, i got 93 number on RCC ,without using any code.
My teacher also shocked , because they was gave me 21/50 number.
❤
enjoy dear
enjoy dear
Thank You Sir, that is really useful.
most welcome dear
Execellent explanation and video. how do you make your images, or what program you use?
thank you. Its PPT
See here more
How can she slab?
Sir
Thank you very much for making videos.
you are strongly welcome
Pls share the design of cantilever slab by ACI code. Your video is so helpful.
Will upload soon
Well done. Thank you.
Thank you too!
Nice Video! Keep it up. Btw, Im just gonna ask if designing reinforced concrete slab is the "same" as the designing of reinforced concrete beams? What I mean of the same is, Beams are like trial and error process related to it reinforcement ratio right? Is that the same with slabs???
very nice video
Enjoy dear
I m confuse (longer dimension / shorter dimension =2) then one way or two way ?
What is it please clear
Because mostly teacher considered two way slab when dimension ratio equal to 2 but u r considered one way slab
y=longer dimension=20ft
x=shorter dimension=10ft
y
__=2
x
x 1
x
thanks for these nice lesson.
welcome dear. i hope you will invite your friends here too.
A reinforced concrete floor system consists of
a 100 mm thick slab cast monolithically with
beams having an effective span of 6 m and
spaced 3 m centre to centre. The superimposed
load on floor is 6 kN/m2. Design the beam
using M 20 grade concrete and Fe 250 grade
of stee (working stress method)
? Isko solution dikha do krke
JAZAKALLAH
very nice sir
abdul malik
thanks
"Sleb" :D
what?
IN UR LANGUAGE HOW WE CALL IT?
Thank you sir
Most welcome dear ❤️
I think i would have chosen the #3 bar for both main and temp for ease of manufacture and less chance of mistake.
thats good, i appreciated,,,
Hi, Is there a maximum span for a one way slab, can I consider a 36f x12f a one way slab, and can I build a 9 feet high wall in top and along the span without beams under neath it? Thanks
Sir what should we do when we have to teach the deisgn of slab for international students. I am teacher for malaysian students but now i am confuse which textbook i should follow. I wanted to know alry from code book the tect books are same for every country or not sir please guide me
Hi. Sir it depends on your country’s institutions, like in my country we follow ACI codes. If your country follows ACI or whatever, then there are enough books available on google
@@Sulimankhanacademy thanqqq i hope you cluld upload the rest of the.process and also upload two way slab
8:44 you said H = 6 but in equation wrote down 5?
After convert ''H'' from inches to feet, we divide per 12
6/12=0.5 in
0.5 feet not inches .المهدي حسن سالم الزوي
Why we convert into inches first??
sir if slab is 8feet*17feet where as supports are at 8feet span so can it be designed as one way with main rod at longer span
love u .lala
manana ror
Hello, can you reference where you got the long formula to find the "required reinforced concert ratio" please.
hello misshollywood.... yes i have reference . i have taken this formula from my text book named as "Design of Concrete Structures,14th by Nilson" you can check this book.... thank you for joining us..
Question: Consider a slab with L/B=2. And you are to support it on 2 edges only. Where do you put those supports or beams? on L or B?
Provide beam, supports along L
well done
nice and good lecturers
Fakhar Zaman
thank you dear. and welcome back
design table available in is 456:2000
Very good lecture
I have a request
I am in my second year of civil engineering in London
I have exam next week
Can you tell me what are the basic things to learn for exam ?
Thanks
Deera manana wrora
Fazel Khan
thank you rore for joining SLK's tuts.
rore exam da sa da ? i mean da kam subject?
Is it the same approach for calculating composite slab?
No. There can be little different conditions. This method is for one way slab only. If you want to know whats oneway and two way? Watch uploaded video about oneway and twoway slab
thanx sulaiman khan
Are you using eurocode 2 ??
achref Messaoud
i follow ACI codes
salam aleykum sir. Awe thanks for your videos information and about books H.Nilson.They're so useful. Can u explain pls when u calculate D.L how did u solve 150 ?
w.salam. 150 lbs/cft is the unit weight of RCC.
lbs= pound
Manam de Engr Sahb (Y) Thumbs up for you.
taso dira dira manana g.... welcome to SLK's tuts...
hy man what about roof treatment in dead load like brick tiles , mud , plaster.? .u did'nt consider these things in dead load :)
Awais Karni
this design is for self weight as dead load and live load.
if your slab has all these types of dead loads you can add them. thanks for comments
well its ok btw i meant there should be a glimpse of these things so other can get it anyways nice video
Thank so much sir!!
Enjoy dear
I like your straggle
thank you dear for joining us...
Don't you need any rod in slab top? You are placing only in bottom layer.
thnqq sir. this is which country code book. ...
Awesome information
Please brother tell me also the name of slab?? if slab supported on 4 sides and the length of slab greater thenL/2...
hello dear. i will bring a video soon. thank you for joining us again
sir kindly use SI units like m, N. these will make ur video more student friendly
dear method is the same.... you can convert my values into SI............ i will try to make videos but that will take much time
Thank you sir! ❤️
What would be the benefit, if I choose to provide some re-bars 20 mm below the top of the slab?
depends upon load............... but in my conditions there is no need to provide them,,,
Thanks for the reply.
Another thing, for suspended slab that you illustrated, the main re-bar is located 3/4" from the bottom of the slab. What about if the slab is on grade, where do we locate the re-bar?
welcome dear.... in suspended slab rebars can be located in both zones (top and bottom) and the concrete cover may be assume as 3/4" etc
Thanks but I am asking about the re-bars location for the slab on grade/ground.
thanks, how can i calculate the size and number of reinforcement, require and provide reinf?
welcome. check the channel in detail you will find video lectures on your demand
Kindly upload two way slab in Feet inches System...There is no auch video on Whole TH-cam
not uploaded yet. will upload soon
idont understand what means ksi and psf
and how to conversation inches to meters
snow Heart
ksi is strength unit which means kips per square inch.
psf is load which is pound per square foot
thanks very much sir
snow Heart
welcome
KSI means Kilopound per square inch
sir kindly if u can explain it in urdu it will be good for it
What about the design of continuous one way slab ?
we will try bring this approach ....
sir what is the #4 and #3 can u pls tell me dia of bars
hello.#3 bar means 3/8 inches dia bar
#4 means 4/8 inches dia bar
#6 means 6/8 inches dia bar
#7 means 7/8 inches dia bar
#8 means 8/8 inches dia bar or 1 inche
#9 means 9/8 inches dia bar
etc
can u pls tell me book name
If you are designing a slab and a beam would you design the slab first and then the beam, and for dead load and live would be same from beam, let’s say you have slab of 16” and b 48 the building is a theater how would you find the dead load and live load
Thats a good question but due to long answer i can not type in comment
First you must know , For design purpose you have to follow some steps given in your country code ,, this code is giving you live load and the dead load you can know it from the dimenstion of your slab plan. for each type of building there is a different live load value. Building design must start from top to down you need to design the slab first then take loads from slab to design your beam , from beam you again design the column
Slab 12in.. What do you mean
We design slab for 12 inches then apply final design to overall slab
What´s the software you use for make this video,please?
MS powerpoint
so temp bars are distribution bars ?
yes
multiple span example required pl...one way slab
Yes dear
Thanks sir
Welcome
Sr where do you get 12 inches in slab?
we design slab only for 12 inches and then apply the 12 inches design into total slab
Can't you make a video for designing two way slab
yes dear. but this will take enough time . bec next video is about double RC beam
SL Khan thank you 🙏
How does it become 8 in the denominator of wL²/8?
this is max moment at mid span of beam having UDL. to derive it , just take sum of all moment at mid equal zero then solve it
U SHOULD EXPLAIN IN INTERNATIONAL SYSTEM OF UNIT
why unit is important ... just learn the method . you can do it with any system
Hi sir, if you are free, could you post some lectures on the following design?
Q1)design a 20ft x 10ft RC Ground Slab without supporting beam. (This Ground Slab shall withstand the weight of 20tons lorry driving over it.) Let me know what are the Reinforcement Bar or BRC required for such design
Q2) design a 50' run ft long concrete drainage (Width ? x Depth ? x Thickness of concrete ? x Reinforcement Bar require? ) which shall be able to withstand the soil force & side force without crack.
Q2) I mean the 50' Run Feet drainage is rectangular "U -shape"instead of "round-shape". I would like your advice on how to design it so that it can withstand the soil forces (up force & side force) without crack & slanted. All unknownsto be assumed by sir.
Thanks sir and I will wait for your upcoming lecture videos
Hon Tin Chin
for size design check a lecture name as design of a rectangular open channel. how ever for forces capacity i will try to bring new lecture. thanks
hi dear.
you right about that student problem.
i checked my mistake. thanks for catching it. check that video tomorrow i am going to edit that.
how ever please do not delete your comments. that can help other people to know about the mistake.
thanks dear
Could you explain the one end continuous and both end continuous?
Yes i will to bring a separate video on this
Sir waiting for Ur video 🤲😭
Also , why we design the slab for 1 feet breadth.
we design for 1feet because to make the calculate little easy ... the one foot design is then applied to whole slab..
V nice
turn you notification on for more...
please upload the video to calculate the unit weigth of equal angle iron.
haris iqbal
good and easy question . i will try to make a video on it. thanks
Thank you
Welcome dear
Thank you brother
welcome dear..........
can u pls tell me book name sir
design of concrete structures by nillson 14th edition
nice
but how to design dia of a bar.
already explained dear, watch again
Can you solve this problem please sir? Size :8×3m l.l=2 kN and a floor finish of 25 mm thick terrazzo flooring width of the wall is 300 mm grade material is fe415 and M20
but what is the problem? you have only mentioned the data
SL Khan design it
Design it as a simply supported slab
I can design it but only if u let me know how to calculate the floor finish load because it is not given ,it is given that floor thickness is25 mm terrazzo flooring
excillent
DESIGN OF TWO-WAY SLAB BY ACI-COEFFICIENTS
FF SLAB DESIGN
PANEL No :
Input Data:
Slab Dimensions
Short Span = 10 ft
Long Span = 20 ft
Span Ratio = 10 / 20 = 0.5
Slab Thickness (reqd) = 2 x (10+20) x 12 / 180 = 4. in
Slab Thickness (prvd) = 6. in in
Edge Conditions
Short Span Edge Condition = One Edge Continuous
Long Span Edge Condition = One Edge Continuous
Applied Loads
Slab self wt = 75.00 psf
Service Dead Load = 0.075 psf
Service Live Load = 220 psf
Material Strengths
Concrete Strength = 4 ksi
Steel Yield Strength = 60 ksi
Design Moments
Loads
Total Ultimate DL =1.4(75 + 0.075) = 105.105 psf
Total Ultimate LL =1.7(220) = 374 psf
Total Ultimate Load =105.105 + 374 = 479.105 psf
Short Span Moments
Postive Short span Moments = 1/11 x 479.105 x 10^2 = 4.356 k-ft
Negative Moment on Discontinuous Edge = 1/24 x 479.105 x 10^2 = 1.996 k-ft
Negative Moment on Continuous Edge = 1/9 x 479.105 x 10^2 = 5.323 k-ft
Long Span Moments
Postive Long Span Moments = N.A.
Negative Long Span Moments = N.A.
DESIGN OF TWO-WAY SLAB BY ACI-COEFFICIENTS
FF SLAB DESIGN
PANEL No :
Reinforcement Calculations
Along Short Span
Positive Steel
Mu=4.36 k-ft ; b=12 in ; d=5.313 in ; fc'=4ksi ; fy= 60 ksi
As+ = 0.85 x 4 / 60 x [1 - [1-2 x 12 x4.356/(0.9x0.85x4x12x5.313^2) ]^0.5 ] x 12 x 5.313
= 0.187 sq.in.
Asmin = 0.0018 x 12 x 6 = 0.130 sq in
use As+ = 0.187 sq.in. (#3 @ 7 in )
Negative Steel
Mu=2 k-ft ; b=12 in ; d=5.313 in ; fc'=4ksi ; fy= 60 ksi
As- = 0.85 x 4 / 60 x [1 - [1-2 x 12 x1.996/(0.9x0.85x4x12x5.313^2) ]^0.5 ] x 12 x 5.313
= 0.084 sq.in.
Asmin = 0.0018 x 12 x 6 = 0.130 sq in
use As- = 0.130 sq.in. (#3 @ 10 in )
Negative Steel
Mu=5.32 k-ft ; b=12 in ; d=5.313 in ; fc'=4ksi ; fy= 60 ksi
As- = 0.85 x 4 / 60 x [1 - [1-2 x 12 x5.323/(0.9x0.85x4x12x5.313^2) ]^0.5 ] x 12 x 5.313
= 0.230 sq.in.
Asmin = 0.0018 x 12 x 6 = 0.130 sq in
use As- = 0.230 sq.in. (#3 @ 5.5 in )
Along Long Span
Positive Steel
Mu=0 k-ft ; b=12 in ; d=4.938 in ; fc'=4ksi ; fy= 60 ksi
As+ = 0.85 x 4 / 60 x [1 - [1-2 x 12 x0/(0.9x0.85x4x12x4.938^2) ]^0.5 ] x 12 x 4.938
= 0.000 sq.in
Asmin = 0.0018 x 12 x 6 = 0.130 sq in
use As+ = 0.130 sq.in. (#3 @ 10 in )
Negative Steel
Mu=0 k-ft ; b=12 in ; d=5.313 in ; fc'=4ksi ; fy= 60 ksi
As- = 0.85 x 4 / 60 x [1 - [1-2 x 12 x0/(0.9x0.85x4x12x5.313^2) ]^0.5 ] x 12 x 5.313
= 0.000 sq.in
Asmin = 0.0018 x 12 x 6 = 0.130 sq in
use As- = 0.130 sq.in. (#3 @ 10 in )
Negative Steel
Mu=0 k-ft ; b=12 in ; d=5.313 in ; fc'=4ksi ; fy= 60 ksi
As- = 0.85 x 4 / 60 x [1 - [1-2 x 12 x0/(0.9x0.85x4x12x5.313^2) ]^0.5 ] x 12 x 5.313
= 0.000 sq.in
Asmin = 0.0018 x 12 x 6 = 0.130 sq in
use As- = 0.130 sq.in. (#3 @ 10 in )
Shear Check
Total load on the panel = 10 x 20 x 479.105 = 95.8 kips
Shear Capacity of the section fVc = 0.85 x 2 x ( 4000 )^½ x 12 x 5.25 / 1000 = 6.77 k/ft
Shear along short side Vua = ( 0.06 x 95.8 ) / (2 x 10 ) = 0.28 k/ft
L/B=20/10=2
That means one way slab
thanks
Hey i need a help to you
سلام انجنير صاحب محترم ديره مننه ښايسته ويدوګاني مو کړى دى ،،مننه
Daira manana ror. Enjoy