Answer / md gulam moinuddin

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

Answer / mahendra gaikwad

For Tender purpose, just take valume of RCC conc.X 110 =
steel in rcc

Answer / shashwati nag

Ast=0.5fck/fy(1-(1-4.6Mu/fck.b.d.d))bd

Answer / k.srinivasa rao

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

Answer / aanand babasaheb chougule

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 than‘d’ 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) Žv = V/bd
b) Pt —› Zc
c) Zv < Zc …….. No shear reinforcement req.

Answer / vinesh

Ast=(0.5Fck/Fy) into(1-(in sqrt(1-((4.6Mu)/Fck.bd^2))into bd
Fck=grade of cement i.e M25, M20
Fy=grade of steel i.e.Fe415,Fe250
Mu=moment
b=width
d=depth

Answer / hemant gor

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}

Answer / shafiullah

ast = ( M ) / (.87*fy*(d-.42*xu))

Answer / dinesh chahal

it will be depend on the area. just calculate the area then
spacing.,.,after than total length/ spacing = no of bar. use
the formula d*d/162 = total steel waight per meter.,.,.
multiply total to total length= total steel in the rcc slab

Answer / engr shabbir

Short formulas.
for coloums=1% t0 5% of concrte volume
for beams=.7% t0 1%
for beams=.5% .8%

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