Grade Slab CSB Building

PROJECT:DOCUMENT NO DATE

TITLE:DESIGNED CHECKED SHEET

DESIGN OF GRADE SLAB:

The Grade Slab is designed using Portland Cement Association (PCA ) method.

BASIC DATAGrade of Concrete = M 30Grade of Steel = 420Type of Sub-base below the Slab (SAND) = 200 mmLength of the slab between ends = 12000 mm = 39.4 ft

LOADING DATA

Type of Load = Uniformly DistributedLoad on Grade Slab,W = 4.2 kN/sq.mDESIGN OF GRADE SLABModulus of Subgrade Reaction on top of Subgrade, k = 2.40(For sand filling below grade slab refer page 2, Reference 1) = 87.0 pci

= 30= 4350 psi

Flexural Strength of Concrete (Modulus of Rupture), MR =(Ref page 2, Reference 1) = 593.6 psi at 28 days

As per reference 1, table 4 (page 13)

Assuming 175 thick grade slab,(7 inch) Allowable uniform distributed load on floor = 1045.0 psf

= 50.0Applied Floor load on grade slab = 4.2

Hence Provide 175 mm thick grade slab

REFERENCE

1 )Slab Thickness Design for Industrial Concrete Floors on Grade - Robert G. Packard- PCA(Enclosed in Annexure - 1)

mm2

The floor slab is subjected to both uniformly distributed load as well as concentrated Mechanical Equipement loads.

( DL = 0.2 kN/m2 and LL = 4.0kN/m2)

kg/cm3

Compressive Strength of Concrete,fc N/mm2

9 * √ fc

kN/m2

kN/m2

PROJECT:DOCUMENT NO DATE

TITLE:DESIGNED CHECKED SHEET

Uniform Load caseAllowable Load on Grade Slab, w =(From Ref (1) pg. 13) Where, = Allowable Working Stress, psi

h = Slab Thickness, in.k = Modulus of Subgrade Reaction,pci

= MR / FOSAssume FOS = 2

= 296.80 psiw = 4.20

= 86.1 psfThickness of Slab required, h =

=(86.1^2)/((0.123^2)×(296.80^2)×87) = 0.06 in= 1.62 mm

Provide 200 mm thick Grade slab

REINFORCEMENT CALCULATIONAmount of steel per linear foot of slab width =( As per Ref (2) pg. 23) F = Coeff. Of subgrade friction

= 1.5 for grade slabL = Slab length b/w free ends, ftw = Weight of Slab, psffs = Allowable Working Stress of Steel

= 0.67 times yield stress of steel in psi

=(1.5×12.5×(200/25.4)×(39.37)/(2×0.67×420×145)) = 0.07 = 150.76Spacing of 8 mm dia bar required = 330 mm C/C

Provide 8 mm dia bar at 200 mm C/C both ways .

REFERENCE

1 )Slab Thickness Design for Industrial Concrete Floors on Grade - Robert G. Packard- PCA2 )Concrete Floors on Ground - Portland Cement Association

0.123 f t√(hk)f t

f t

f t

kN/m2

w 2 / ( 0.1232 * f t2 * k )

A st F * L * w / 2 * f s in2/ft

A st in2/ft mm2/m

LARSEN & TOUBRO LIMITED GLOBAL ENGINEERING SERVICES

PROJECT: SIDRA MEDICAL RESEARCH CENTREDOCUMENT NO DATE

22/4/2023

TITLE:SAINT GOBAIN,EGYPT. DESIGNED CHECKED SHEET

DESIGN OF GRADE SLAB BAL CSR OF

Design of Grade slab-PCA method (PIN LOAD)

Basic dataGrade of Concrete = M 30Type of Soil below the Slab = Sandy

Loading Data

Type of Load = Point load(concentrated)Maximum Point Load = 200 kNLoading Area = 900

Pin Load

Concentrated Load = 200.0 kN = 45.0 kipsSlab Depth t = 175 mm = 6.9 in

= 30.0 = 4350 psi

= 593.6 psi

(I)Check as Post Load (Pg. 12 example problem, Ref. 1)

Contact Area = 900 = 140

Load Periphery = Perimeter of contact area = 120 cm

= 47 inCheck for Bearing stress

Allowable Bearing Stress: LOAD DISTRIBUTIONFor interior load = 4.2 MR = 2493 psiFor edge or corner load = 2.1 MR = 1247 psi

Actual bearing stress = Load / Area = 322.3 psiActual bearing stress = < Allowable Bearing Stress Hence Safe

The floor slab is subjected to both uniformly distributed load as well as concentrated wheel loads from trucks.

cm2

Compressive Strength of Concrete fc N/mm2

Modulus of Rupture MR = 9 * √ fc

cm2 in2

(For 900 cm2 loading area minimum perimeter is 120 cm)



22/4/2023



Check for Shear StressAllowable Shear Stress = 0.27 MR = 160.3 psi

Actual Shear Stress for Interior Load

=Post Load = 87.245 psi < 160.3

Slab Depth [load periphery + 4(slab depth)] SAFE

Actual Shear Stress for Edge Load

=Post Load = 132.61 psi < 160.3

Slab Depth [0.75(load periphery) + 2(slab depth)] SAFE

Actual Shear Stress for Corner Load

=Post Load = 213.89 psi < 160.3

Slab Depth [0.5(load periphery) + (slab depth)] UNSAFE

200 mm thick Grade Slab shall be provided.

Minimum reinforcement for shrinkage shall be provided.

Provide T- 10 at 150 c/c in both directions

1 )Slab Thickness Design for Industrial Concrete Floors on Grade - Robert G. Packard- PCA(Enclosed in Annexure -1)

'(0.262%)



22/4/2023



(II)Check as Column Load

= X A X (Sec 5.4, Ref. 2)Where,t = Slab thickness beneath column, in

= Factored Column Load, lbA = = 0.0005B = = 6E+07C = k * b^4 = 1E+05

= Compressive strength of Concrete, psi = 4350 psik = Modulus of Subgrade of Soil, pci = 200 pcib = Base Plate dimension, in = 5.12 in

Slab thickness 150 mmt = 150 mm = 5.91 in

= X A X

5.91 2 = X 0.0005 X 6.0.E+07 X 5.91 3 / 137236 )= 14119.9 lb = 64.1 kN < 70.0 kN Hence increase the thickness

Slab thickness 175 mmt = 175 mm = 6.89 in

= X A X

6.89 2 = X 0.0005 X 6.0.E+07 X 6.89 3 / 137236 )= 18470.4 lb = 83.86 kN > 70.0 kN Hence safe

Punching Shear CheckPunching shear check shall be done for slab thickness = 175 mm

b+d

Base plated/2 Grade slab100 263

263.0

= 4 X X X d (Sec. 5.4, Ref 2)

t2 Pu log 10 (B * t3 / C )

Pu

Constant - 0.03 / √ fc

Constant - 915000 * √ fc

fc

Ultimate load that can be carried by the slab is calculated substituting all the values in the above eq. except Pu

t2 Pu log 10 (B * t3 / C )

Pu log 10(Pu

t2 Pu log 10 (B * t3 / C )

Pu log 10(Pu

Perimeter at critical section

Punching Shear Capacity Vc √ fc bo



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d = Effective depth of the Slab = 175 - 8 - 8/2 = 163 mm = 6.42 inb = Base plate dimension = 100 mm = 3.94 in

= Perimeter of the critical section at distance d/2 from the base plate= 4 ( b + d ) = 1052 mm = 41 in

Vc = 4 X √ 3625 X 41 X 6.42 = 70120 lb = 318.3 kN > 70.0 kN Safe

175 mm thick Grade Slab shall be provided.REFERENCE1 Slab Thickness Design for Industrial Concrete Floors on Grade - Robert G. Packard.2 Designing Floor Slabs on Grade, Ringo and Anderson.

200 kN

Slope 1:188 mm

Slab Thick= 175 mm

Tributary area 30.5 cm( 250.0 x 13.0 ) cm

267.5 cmContact area after dispersion = 8159 = 1264.6

bo

cm2 in2

Documents

Grade Slab CSB Building