0 4 Reinforced Concrete Slab Design Example 1 Unfilled

RC Slab Design Example 1

Reinforced Concrete Slab Design Example - 1 Design a simply supported, one way internal RC slab in a residential accommodation block. Refs

EC2

Output

DATA Effective Span of Slab, L = 4.8m; Concrete Compressive Strength Class = C25/30; fck = N/mm2; fyk = N/mm2 ; b = 1000mm; h = 175mm Characteristic Dead Load on slab = 0.78kN/m2 (finishes) Characteristic Imposed Load on slab = 1.50kN/m2 (resi) Concrete Density = 25 kN/m3 Main Bar φ = 12mm,

DURABILITY & FIRE RESISTANCE T 4.1

Exposure Class = Required Fire Resistance = 60 mins T NA2 Nominal Cover for Durability = mm T2 Axis Dist, a = mm, Minimum slab depth = mm CEC2 Nominal Cover+φ /2 = mm, a < Nominal Cover +φ /2, h > Minimum slab depth, > mm Effective Depth of Slab, d = h – cover – ½ main barφ =

<

mm

a & h OK Cover = mm d=

LOADING @ ULS - for a 1m width of slab Characteristic Dead Load, kN

Characteristic Imposed Load, kN

Slab Loads Self Weight

EC 0

Total Characteristic Dead Load = kN Gk Total Characteristic Imposed Load = kN Qk Design Ultimate Load, F = 1.35DL + 1.5 IL F=

(= 1.35×Gk + 1.5×Qk)

F=

kN

MOMENT @ ULS Design Ultimate Moment, MEd = K = MEd/bd2fck =

MEd = K ≤ K’ < 0.167

z/d = 0.5 + √( 0.25 – 0.882K) =

kNm

∴ No compression steel required z/d = 0.82 ≤ z/d ≤ 0.95 0.82 ≤ 0.95

∴ z/d OK

Lever Arm, z = z/d × d =

mm

z=

mm

MAIN REINFORCEMENT Required Area of steel, As = MEd / 0.87fykz = Use H12 bars ( mm2)

mm2 H12

As = NA9.2.

mm2

0.0013bd = 0.00016 fck2/3bd = 0.04bh = As

1

∴As OK

RC Slab Design Example 1

SECONDARY REINFORCEMENT Required Area of Secondary Reinforcement = 0.2As = 0.2 Use @ bars ( mm2/m)

mm2/ m @ Area =

bars mm2/m Area OK

CRACKING Main bar Clear Spacing = Main Bar spacing =

∴ Cracking

Secondary Bar spacing = OK

SHEAR @ ULS Design value of the applied Shear Force, VEd = F/2=

(support reactions)

N/mm2

Design Shear Stress, vEd = VEd/bd = Provided tension reinforcement ratio,

ρ l = (As/bd) ×100 =

d<

ρl

vRd,c

0.25

0.54

0.50

0.59

N/mm2

T 7.1 Concrete Shear Stress Capacity, vRd,c = CEC2

vEd ≤ vRd,c

1.0

∴No Shear steel required

DEFLECTION @ SLS Actual span-to-effective-depth ratio, L/d = Required tension reinforcement ratio, ρ = (As,req /bd)×100 =( fck==

T.15.10 Basic span-to-effective-depth ratio, N = CEC2

ρ

T.15.11 Factor K = CEC2 Factor F1 = Factor F2 = Service stress, σ s = 435×(Gk + 0.3Qk /1.35×Gk + 1.5×Qk)×(As req /As prov) = = N/mm2 Factor F3 = 310/σ s =

N/mm2 N

0.30

32.2

0.40

22.4 1.0

Allowable L/d= N x K x F1 x F2 x F3 = Actual L/d ≤ Allowable L/d Deflection OK

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