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Sample Slab Computation

This document summarizes the design of a two-way concrete slab (S1) using the coefficient method. Key details include a 4.38m long by 3.98m short slab with 12mm reinforcing bars spaced 200mm. Loads include 1.9kPa live and 2.45kPa dead. Moments are calculated at short and long spans and edges. Reinforcement ratios are checked to satisfy design requirements with a 125mm slab thickness.

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Allan Bautista
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525 views7 pages

Sample Slab Computation

This document summarizes the design of a two-way concrete slab (S1) using the coefficient method. Key details include a 4.38m long by 3.98m short slab with 12mm reinforcing bars spaced 200mm. Loads include 1.9kPa live and 2.45kPa dead. Moments are calculated at short and long spans and edges. Reinforcement ratios are checked to satisfy design requirements with a 125mm slab thickness.

Uploaded by

Allan Bautista
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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DESIGN OF TWO-WAY SLAB (BY COEFFICIENT METHOD)

Slab Designation : S1
Design Specifications
Compressive strength of concrete, f'c = 20.7 MPa
Yield strength of reinforcement, fy = 276 MPa

Loadings
Unfactored Live Load = 1.9 kPa
Unfactored Dead Load (excluding self wt) = 2.45 kPa

Slab Detail Reinforcement: Phils. Standards


Long Span, L = 4.383 m Bar Diameter = 12 mm
Short Span, S = 3.975 m
Slab Height = 125 mm ≥ 121mm
Concrete Cover = 20 mm
Slab Support Condition = Case #
2 === "all four edges continues"

Computation

Slab Self Weight = unit wt. of concrete x thickness = 2.943 kPa


Total Dead Load = DL + self weight = 5.393 kPa
Factored Dead Load = 6.471 kPa
Total Live Load = 1.900 kPa
Factored Live Load = 3.040 kPa
Total Factored Load = 1.2DL + 1.6LL = 9.511 kPa

m = S/L = 0.91
0.95 0.90
0.04 0.01
therefore use, m = 0.90
Consider 1 m strip : ω = 9.511 kN/m

At Short Direction "S"


Negative Moment at Continues Edges
Negative Cs = 0.055
MS = -CsωS2 = 8.265 kN-m

Positive Moment at Midspan


Positive Dead Cs = 0.022
MS = +CsωDLS2 = 2.249 kN-m
Positive Live Cs = 0.034
MS = +CsωLLS2 = 1.633 kN-m

Total Postive Moment at Midspan = 3.882 kN-m

Negative Moment at Discontinues Edge = 1/3(Total Positive Moment)


MS = -1/3( +Msmidspan ) = 1.294 kN-m

At Long Direction "L"


Negative Moment at Continues Edges
Negative Cl = 0.037
MS = -ClωL2 = 6.76 kN-m

Positive Moment at Midspan


Positive Dead Cl = 0.014
MS = +CsωDLL2 = 1.74 kN-m

Positive Live Cs = 0.022


MS = +CsωLLL2 = 1.285 kN-m

Total Postive Moment at Midspan = 3.025 kN-m

Negative Moment at Discontinues Edge = 1/3(Total Positive Moment)


MS = -1/3( +Msmidspan ) = 1.008 kN-m

Value of β1 ;
β1 = 0.85 (max)
β1 = 0.85 - 0.008(f'c - 30) = 0.9244
β1 = 0.65 (min)
Final Value of β1 = 0.85

0.85f'cβ1600
ρb = = 0.03711
fy(600 + fy)
ρmax = 0.75ρb = 0.02784

1.4
ρmin = = 0.00507
fy
Short Span Long Span
Top Bar Bot. Bar Bot. Bar Top Bar Bot. Bar Bot. Bar
Cont. Discont. Cont. Discont.
Midspan Midspan
Edge Edge Edge Edge
Mu 8.265 3.882 1.294 6.76 3.025 1.008
ω 0.06079 0.02799 0.00923 0.04938 0.02173 0.00718
ρactual 0.00456 0.0021 0.00146 0.00146 0.00163 0.00054
ρsupplied 0.00507 0.00507 0.00507 0.00507 0.00507 0.00507
As =ρbd 441.304 441.304 441.304 441.304 441.304 441.304
RSB Spacing 250 250 250 250 250 250
min. S = 3t 375 375 375 375 375 375
min. S = 450 mm 450 450 450 450 450 450
S, mm c to c 12 mm 200 200 200 200 200 200

Check for Adequacy of Slab Thickness Mu = Φbd2f'cω(1-0.59ω)


Max Mu = 8.265 kN-m trequired = 121 mm
ω= 0.06763 tsupplied = 125 mm
dmin = 82.6568 mm tmin = P/180 = 92.87 mm
dactual = 87 mm

trequired < tsupplied ok

t = 125 mm

Use 12 mm RSB spaced at 200 mm O.C.


DESIGN OF TWO-WAY SLAB (BY COEFFICIENT METHOD)
Slab Designation : S1
Design Specifications
Compressive strength of concrete, f'c = 20.7 MPa
Yield strength of reinforcement, fy = 276 MPa

Loadings
Unfactored Live Load = 1.9 kPa
Unfactored Dead Load (excluding self wt) = 2.45 kPa

Slab Detail Reinforcement: Phils. Standards


Long Span, L = 4.383 m Bar Diameter = 12 mm
Short Span, S = 3.975 m
Slab Height = 125 mm ≥ 125mm
Concrete Cover = 20 mm
Slab Support Condition = Case #
9 === "one short edge discontinues"

Computation

Slab Self Weight = unit wt. of concrete x thickness = 2.943 kPa


Total Dead Load = DL + self weight = 5.393 kPa
Factored Dead Load = 6.471 kPa
Total Live Load = 1.900 kPa
Factored Live Load = 3.040 kPa
Total Factored Load = 1.2DL + 1.6LL = 9.511 kPa

m = S/L = 0.91
0.95 0.90
0.04 0.01
therefore use, m = 0.90
Consider 1 m strip : ω = 9.511 kN/m

At Short Direction "S"


Negative Moment at Continues Edges
Negative Cs = 0.068
MS = -CsωS2 = 10.219 kN-m

Positive Moment at Midspan


Positive Dead Cs = 0.026
MS = +CsωDLS2 = 2.658 kN-m
Positive Live Cs = 0.036
MS = +CsωLLS2 = 1.729 kN-m

Total Postive Moment at Midspan = 4.387 kN-m

Negative Moment at Discontinues Edge = 1/3(Total Positive Moment)


MS = -1/3( +Msmidspan ) = 1.462 kN-m

At Long Direction "L"


Negative Moment at Continues Edges
Negative Cl = 0.025
MS = -ClωL2 = 4.568 kN-m

Positive Moment at Midspan


Positive Dead Cl = 0.015
MS = +CsωDLL2 = 1.865 kN-m

Positive Live Cs = 0.022


MS = +CsωLLL2 = 1.285 kN-m

Total Postive Moment at Midspan = 3.15 kN-m

Negative Moment at Discontinues Edge = 1/3(Total Positive Moment)


MS = -1/3( +Msmidspan ) = 1.05 kN-m

Value of β1 ;
β1 = 0.85 (max)
β1 = 0.85 - 0.008(f'c - 30) = 0.9244
β1 = 0.65 (min)
Final Value of β1 = 0.85

0.85f'cβ1600
ρb = = 0.03711
fy(600 + fy)
ρmax = 0.75ρb = 0.02784

1.4
ρmin = = 0.00507
fy
Short Span Long Span
Top Bar Bot. Bar Bot. Bar Top Bar Bot. Bar Bot. Bar
Cont. Discont. Cont. Discont.
Midspan Midspan
Edge Edge Edge Edge
Mu 10.219 4.387 1.462 4.568 3.15 1.05
ω 0.07587 0.0317 0.01043 0.03304 0.02264 0.00748
ρactual 0.00569 0.00238 0.00146 0.00146 0.0017 0.00056
ρsupplied 0.00569 0.00507 0.00507 0.00507 0.00507 0.00507
As =ρbd 495.023 441.304 441.304 441.304 441.304 441.304
RSB Spacing 225 250 250 250 250 250
min. S = 3t 375 375 375 375 375 375
min. S = 450 mm 450 450 450 450 450 450
S, mm c to c 12 mm 200 200 200 200 200 200

Check for Adequacy of Slab Thickness Mu = Φbd2f'cω(1-0.59ω)


Max Mu = 10.219 kN-m trequired = 125 mm
ω= 0.07587 tsupplied = 125 mm
dmin = 87 mm tmin = P/180 = 92.87 mm
dactual = 87 mm

trequired < tsupplied ok

t = 125 mm

Use 12 mm RSB spaced at 200 mm O.C.


Slan Designation: S2

Unfactored Loads Factored Loads


Self Weight = 2.94 kPa U = 1.2DL + 1.6LL
Superimposed DL= 2.45 kPa U = 1.2(5.39) + 1.7(1.9)
Total Dead Load = 5.39 kPa U= 9.70 kPa
Superimposed LL = 1.90 kPa wu = 9.70 kN/m (Consider 1-m Strip)

2
Mu =wuS^2/10 (9.7)(3.8) Vu(max) = 1.15(wuS/2)
Mu =
10 Vu(max) = 9.7(3.783)/2
Mu = 13.883 kN - m Vu(max) = 21.1 kN
Mu = ϕbd2f'cω(1-0.59ω)
13883000 = 0.90(1000)(99)^2(20.7ω(1-0.59ω)
ω = 0.079776
ρ = ωf'c/fy
ρ = 0.005983
Check the Value of ρ(actual)
Value of b1 = 0.85
0.85β1f'c(600) 0.85(0.85)(20.7)(600)
ρ(bal) = =
fy (600 + fy) 276(600 +276)
ρ (bal) = 0.0371147 ρ max =0.75 x ρ(bal) = 0.027836
ρ min = 1.4/fy = 0.005072
ρ max > ρ > ρ min OK
Therefore, ρ = 0.005983
As = ρbd = 0.005983(1000)(99)
2
As = 592.32 mm
Shrinkage and Temperature Reinforcements
2
As = 0.002bt = 0.002(1000)(125) = 250 mm

Temperature
Spacing of Rebar Requirements Main Bars Bars
Actual = Area of 1 Bar (b/As) , mm 190.94 314.16
Min S ≤ 3t , mm 375 -
Min S ≤ 5t , mm - 625
Min S ≤ 450 , mm 450 450
Provided Spacing mm 175 200

Check for Shear: Vu ≤ ØVc/2


Vu = Vu(max) - wud = 21.1 - 9.7(0.099)
Vu = 20.14 kN
Vc =1/6√f'c bd = 1/6 √(20.7) (1000)(99)
Vc = 75.07 kN
ØVc/2 = 28.15 kN > Vu = 20.14 kN OK

l/3 l/3
1.26 m 1.26 m

Temperature & Shrinkager bar Main Bars


10 mm @ 200 mm O.C. 12 mm @ 175 mm O.C.

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