there is no general formula for the calculation fo steel
(reinforcement) its depend upon the design of the structur
,for whih load and moment acting on the structure are
considered
for calculating the area of steel in rcc slabs,
ast = ( M ) / (.87*fy*(d-.42*xu))
where
M= moment corresponding to entire loads of span in one
meter length
fy= characteristic strength of steel
d= depth of slab
xu=lever arm depth depending upon the type of fy given
in Is:456-2000
Two Methods, First from Calculation, 2nd using SP16
1st Method Compute K= Mu/(fcu x b x d^2)
for Fy = 415 if K<= 0.138 Singly Reinforced
Calculate Z/d = 0.5 + sqrt { 0.25 - K/0.865}
Area of Steel = Mu / { 0.87 x fy x Z }
Z is Lever Arm Distance Between CG of Tension and Compression Force
2nd Method
Calculate Mu /{b x d^2}
Check Table 2 to 5 of SP16 based on Concrete Grade
It gives Percentage of Steel Pt for given Grade of Concrete,steel and Mu/{bxd^2}
using code book to find out is one method is456-2000
sp16
another method is design of rcc slab by limit state method
u refer any author
u check one way or two slab slab
what is ur liveload , dead load , seismic condition and
factor of safety
then u go to table or manual both will be same or it may be
slightly different
dont ask basic questions
It depends on type of loading on the slab
normally we go for ly/lx ratio after to find out one or two
way slab
depending on the loading condition u can find out IS
456-2000 Code book to
find the rft depends on moment acting on the slab
if it is two way slab provide shear reinforcement on all
four corners
better u can study design of concrete structures in slab
there is
calculation arrived by some model examples
but field condition is difference follow depends on
structural designer
u cannot ask direcly what is the formula for design the slab?
u have to read more book then u know
all...............................
regrds
Devendran
Saif Bin Darwish
Abu Dhabi
Design steps of one way slab
1. Depth of slab
d= lx/20 * M.F. (M.F. = Modification factor)
Assume the dia. of bar (8, 10, 12) and
Final overall depth
D= d+15+0/2
d= d-15-0/2
2. Effective Span
This is taken as the lesser of the followings
a) Center to center distance between followings
L= lx + t
L= lx + d
3. Load
Consider 1m width of the slab and find out the total udl
a) W = (Self weight + Live Load + Floor finish)
b) Find factor load
Wu = 1.5 * w
4. Factor bending moment
Mu = Wu * l²/8
5. Equating Mu limit to Mu find the depth req. if it is
less thand than O.K otherwise
Increase D & repeat 2, 3, and 4
Mulimit = Mu
6. Area of main steel
Pt = 50fck/fy (1-√1 4.6 Mu / fck bd²)
Ast = Pt/100 *100* d
7. Spacing of main steel
S = Area one bar * 1000/ Ast
Check: - a) 3d
b) 450 mm
8. Distribution Steel
Astd = 0.15 % of Ag -Fe 250 (mild)
= 0.12% of Ag Fe 415/5000 (torque)
Ag = b.D
9. Spacing of distribution steel
S= Area one bar * 1000/ Astd
Check: - a) 5d
b) 450 mm
10. Check for depth
a) v = V/bd
b) Pt Zc
c) Zv < Zc .. No shear reinforcement req.
1. why more stirrups are provided near the support?
2.where more stirrups is provided in a column and why?
3.why extra bars are provided at the top of beam?
4.what is the difference between overlap and lap lenght?
5.what is development lenght
Kindly tell me how much extra steel bars are provided in a
RCC roof and distance of bends in different sizes of roofs.
Please also tell me that bars provided in chajja are taken
seperate or main bars or short bars are extended in chajja.
please send me the answer@ suhail_akhter@hotmail.com
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