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Shows a simple computation involving Design of a one way slab., Exercises of Reinforced Concrete Design

This is from an excel wherein we just input values and it will automatically compute the design for one way slab

Typology: Exercises

2023/2024

Uploaded on 03/28/2024

rudel-adrian-torno
rudel-adrian-torno 🇵🇭

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Slab I.D. 2S-1
LENGTH: 3500 mm Main barØ: 12 mm
Fc': 21 Mpa temp. barØ: 10 mm
Fy: 275 Mpa
min. "t" req'd = 138.75 mm
use slab thickness = 140.00 mm
INPUT DATA:
Super Imposed Dead Load (SDL) = 1.8 kPa
Live Load (LL) = 2.4 kPa
OUTPUT DATA:
Slab Selfweight = 3.36 kN/m
therefore, factored load on 1 meter strip,
W = 1.2*Self weight + 1.2*SDL + 1.6*LL 10.03 kN/m
Mmax. = 15.36 kN-m ρ min. =0.00509
Vmax. =17.56 kN ρ max =0.02837
ρ bal. =0.04 ρ reqd. =0.00497 use ρ = 0.00509
DESIGN:
ρ reqd. < ρ max therefore,
Check for Shear, Vu is critical at "d" distance from the support
Vu = 16.41 kN
actual V = 0.17 Mpa
allow. V = 0.78 Mpa
actual V < allow. V therefore,
DESIGN OF REINFORCEMENT:
As min. = 0.0020bt 280
mm2
As req'd. = ρbd 580.36
mm2NOTE:
For Main reinforcement, "s" should not be
greater than 3*t
use 12 mm dia. use 10 mm dia. ForTemp. reinforcement, "s" should not be
@190 @280 greater than 5*t nor
greater than 450 mm
78.53981634
3.565070725 280.499344
DESIGN OF ONE-WAY SLAB
"t" is safe for shear!!!
SPACING OF REINFORCEMENT
MAIN
TEMPERATURE
mm.o.c
mm.o.c
Simply Supported
L/20 (0.4 + Fy/700)
"t" is safe for flexure!!!
pf3

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Slab I.D. 2S- LENGTH: 3500 mm Main barØ: 12 mm Fc': 21 Mpa temp. barØ:^ 10 mm Fy: 275 Mpa min. "t" req'd = 138.75 mm use slab thickness = 140.00 mm INPUT DATA: Super Imposed Dead Load (SDL) = 1.8 kPa Live Load (LL) = 2.4 kPa OUTPUT DATA: Slab Selfweight = 3.36 kN/m therefore, factored load on 1 meter strip, W = 1.2Self weight + 1.2SDL + 1.6*LL (^) 10.03 kN/m Mmax. = 15.36 kN-m ρ (^) min. = 0. Vmax. = (^) 17.56 kN ρ (^) max = (^) 0. ρ (^) bal. = (^) 0.04 ρ (^) reqd. = (^) 0.00497 use ρ = 0. DESIGN:

ρ reqd. < ρ max therefore,

Check for Shear, Vu is critical at "d" distance from the support Vu = 16.41 kN actual V = 0.17 Mpa allow. V = 0.78 Mpa

actual V < allow. V therefore,

DESIGN OF REINFORCEMENT:

As min. = 0.0020bt 280 mm 2 As req'd. = ρbd 580.36 mm 2 NOTE: For Main reinforcement, "s" should not be greater than 3*t use 12 mm dia. use 10 mm dia. ForTemp. reinforcement, "s" should not be @ 190 @ 280 greater than 5*t nor greater than 450 mm

3.565070725 280. DESIGN OF ONE-WAY SLAB "t" is safe for shear!!! SPACING OF REINFORCEMENT MAIN TEMPERATURE mm.o.c mm.o.c Simply Supported L/20 (0.4 + Fy/700) "t" is safe for flexure!!!

Simply Supported Both Ends Cont. One End Cont Cantilever L/20 (0.4 + Fy/700) 20 L/24 (0.4 + Fy/700) 24 20 L/28 (0.4 + Fy/700) 28 L/10 (0.4 + Fy/700) 10 depth d = 114.00 10 ;β = 0.85 12 16 20 formulas: ρ (^) bal. = (^) 0.85fc'β600 / fy (600+fy) ρ (^) min. = (^) 1.4/fy ρ (^) max = 0.75*ρ bal. ρ (^) reqd. = (^) wfc'/fy β = 0.85 if fc' ≤ 30 β = 0.85 - 0.008 (fc'-30) if fc' > 30 but β must not be less than 0.

Mu = Øfc' bd

w (1-0.59w)

w = 0.

ρ reqd. <

ρ reqd. >

"t" is safe for flexure!!! increase slab thickness!!! actual V = Vu/Øbd ;Ø=0. allow. V = 0.17 (Fc')^0. actual V < actual V > "t" is safe for shear!!! increase slab thickness!!! 214 210.