R.C.C design of slabs and beams manually
Design of slabs
1) Find the loadings.
a) Live load = LL
b) Dead load = DL
c) Floor finish = FF
2) Find the design load.
Design load = 1.5(LL + DL + FF)
3) Find the edge conditions.
Edge conditions = Ly/Lx
4) Do a deflection check
4) Do a the moment and steel calculation
Coefficients (x and y) are computed as per clause D-1.1 of IS 456:2000:
Mx = x x W x Lx x Lx:
My = y x W x Lx x Lx:
where
Mx,My = moments on strips of unit width spanning Lx and Ly respectively,
W = Total Design Load
Lx = Shorter span
Minimum steel check
As per clause 26.5.2.1 of IS 456:2000:
Minimum area of steel For Main Steel = 0.12 x C/S Area
Notes :
Extra steel at Top support is computed considering
the bent-ups,if any,coming from the adjoining span.
It is the maximum of the extra steel required
for each slab at that common support.
Design of beams
5) Find the loadings.
a) Self weight of beam
b) Slab reaction
Slab reaction = dead load slab reaction and live load slab reaction
c) Wall load
d) Staircase load (if present)
6) Find the total load
Total load = Self weight of beam + Slab reaction + Wall load + Staircase load (if present)
7) Find the area of steel required = Ast
8) Do a check for maximum and minimum steel required
9) Minimum and Maximum Steel Checks :
a)Minimum Steel Check as per IS 456:2000 Clause 26.5.1.1(a)
Ast = (0.85 x b x d) / fy
b) Maximum Steel Check as per IS 456:2000 Clause 26.5.1.1(b)
Ast = 0.04 x b x D
Where,
Fy = strength of steel
B = width of beam
D = depth of beam
10) Do a check for Shear Reinforcement(Stirrups)
Design Shear = (Equivalent S.F.)-(Vuc)
VusT = ( Tu/b1 + Vu/2.5)(d/d1)
Where,
Tu = Torsional moment,
Vu = Ultimate shear force
b1 = Center to center distance between corner bars in the direction of width
d1 = Center to center distance between corner bars in the direction of depth
(As per Clause 41.4.3 of IS 456:2000)
If Design Shear < VusT then Design Shear = VusT
VusT = Minimum shear for torsional transverse Reinforcement
Provide Shear Steel (Stirrups)
Provide minimum Shear Reinforcement of 0.75 d
(As per Clause 26.5.1.5 of IS 456:2000)
Wednesday, June 3, 2009
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