Design of Slabs Part - II

[@sahilofficial.7193]

No need of coaching and wasting money if u get these types👌👌👌👌👌 of good lectures free.

[@ahmedabdi56]

please which method the profferse is using is it limit state

[@vignesharavinthan7696]

very thank you to iit for their kind online teaching

[@denning2781]

sir u just saved my career

[@rajeshpritijain]

EXCELLENT

[@premchand-ww4ie]

good explanation of slab design

[@edsnarsi]

Very Helpful, thank you Professor

[@constructionsknowledgefrom361]

please explain step 5 calculation ofarea of steel by another method

[@gilitato697]

Pls, see the balanced singly reinforcement moment formula (Mu) of the beam.

[@LeXpedit]

M.R = 0.87 fy Ast d[1-(fyAst)/(fckbd)] Use this formula to calculate Ast . Or u can generally solve this formula to get the equation in your question. M.R=Moment of Resistance Other notations are as usual

[@thesarcasticengineer7]

@18-037 Sabaz Mazumder they have assumed spacing is *150mm* instead of 165.05 When u will 150 in spacing formula *u will get area 523.33mm2* instead of 475.6 mm2 Which they used as *provided area*

[@WahranRai]

Hi, How you will modelize slab (4.30 x 5.00) based on 9 steel beams (space between beams 0.60m) on 2 walls. Shall we assume that we have continuous beam supported simply by 9 supports (span 0.60m).

[@joebosah2727]

Kudos to Prof

[@syedalifahadzaidi]

Why are bars cranked when there is no negative moment?

[@chinnareddy2151]

Why we are taking beam length 1000mm always ? Can u explain sir

[@mirzasalim5902]

Because we are providing steel per meter

[@kgpathy9534]

This is design of slab and not beam. In slab design always consider 1000mm which is perpendicular to span. if you could design for 1000mm, then the same is extended to 10m, 100m and so on...

[@yashsahu1]

in step 4;- check for shear, how you can find out Tc without Ast

[@Inquisite1031]

u can assume x/d=0.48 or u can calculate x/d then calculate z, then calculate Ast from Mu formula

[@chipuul]

am interested in RCC Structures

[@swatideo9323]

sir in dead load slab 0.16(depth) × 25 ( what is 25 here ? ) = 4.0 kn/m 2

[@41titheekulkarni90]

It's the density of concrete

[@kgpathy9534]

unit weight of concrete is 2500kg/m^3 or 25KN/m^3

[@AbdulQadirqadirjames]

Why we are taking beam length as 1000 mm always?? Please explain that.. thanku..

[@akshaysingh733]

by taking 1 m width design of slab problem converges to design f beam abdul bhai..

[@lgledesma211]

On sample problem about one way slab where did Mr. Dang got the value 523 mm for Ast(provided)? just for clarification.ty

[@bekiroztrk]

Luisito Ledesma Area provided by bars which is taken 523 by prof.Dhang is calculated by 1000mm/150mm(spacing between bars)=6.666 and Ast will become 6.666*78.5=523 which is wrong. Because we can't put 0.6 bar. If we put 7 bars then Ast will become 549.5mm2

[@engahmedjigjigaawi2426]

as the slab is same in the design for the beam it is the same process that is why is taken 1m length and the slab will be calculated per meter of the result

[@praveenprakashprakash3895]

Bekir öztürk hello sir , dhang sir check for shear by tou c by (Ast*100)/bd from table no. 19 .but actual value comes 0.39 if i take Ast=523 and b=1000and d=140 and sir take the value for

[@abithar9682]

super

[@shreydon35]

can somebody explain to me how he got ast provided?

[@animemaths3901]

AST provided is decided according to the dia of steel choosen

[@kumaranselvaraju6345]

=Area of single bar x (number of bars) =(pi x sq(d)/4) x (B/Spacing) =523 sqmm

[@chinnareddy2151]

Shreyas S same problem bro ,u clear that one So explain me bro

[@gilitato697]

In problem, we have width 1000 mm, area of one steel bar 78.50 mm^2, spacing 150 mm. Ast provided= width x area of single bar/ spacing = 523 mm^2

[@ishwarchandra84]

@@gilitato697 how do we got 150mm spacing

[@yohannesshewarega5639]

Thanks 😘

[@ananyanair2640]

can some one explain why it is said in the concentrated wheel load part, moment decreases with increased dispersion? isn't moment directly proportional to length?

[@Inquisite1031]

u are distributing the same force over a larger area/length, so the resultant force will be less

[@aniketkesarkar9635]

It is limit state method ??

[@thesarcasticengineer7]

Can someone explain why he is taking span by depth ratio is 25 initially why not 20 or 26?

[@kgpathy9534]

The IS:456 says for simply supported case the Basic Value is 20. So you should take a value more than 20. Sir is taking 25. If you want you take 22,23 or 24. nothing wrong in it.

[@civilstudent5846]

Can some one explain how the relation for spacing is obtained please?

[@thesarcasticengineer7]

It's too easy first find no. Of Bar in founded area of steel with respect to Dia which is want to use Then divide 1(1000mm)m by no. Of Baar u will get spacing be th baar

[@myganitbydinker9977]

Where i can get the formula used in step 3,4,5

[@shikhapandey7428]

sir please tell me,how we get 523 as Ast provided?

[@samael5095]

(breadth x area of the bars)/spacing= ast provided

[@gilitato697]

Breadth 1000 mm x area of Single bar 78.50 / provided spacing in problem ( 150 mm) = 523

[@Bangon2024]

How you taken WL/8 ?

[@902Jr]

He used the total load (kN) and not uniformly distributed load (kN/m)

[@902Jr]

Med = 13.5*3.64^2/8= 22.36kNm

[@aamirabdulsalam]

@@902Jr which formula is this

[@902Jr]

​@@aamirabdulsalam​ Moment using total load WL/8 or m0ment using UDL wl^2/8.

[@RaghavSharma]

how did u calculate ast????

[@parag27865]

thank sir me an student svits indore,civil 4 year

[@91sk64]

8:56 What's the reference

[@thesarcasticengineer7]

Sir what is x¹ @51:40

[@kishanpatel3546]

for bending moment=w*L(square)/12

[@mahaveerprasad6431]

wL when w is udl here W is total load that's why it's W/2

[@mahaveerprasad6431]

WL is takes when w is udl but here w total load so

[@somnathbera2681]

🕉️👑🔱💎🪔💝💐🙏🙏🙏

[@jagadeeshamjuri7907]

sorry sir for say to this, your teach good but not soo good, we want how they toe V and Mu formulas are discoverd in table book.