two way slab design

PetrofacPetrofac International Limited

PROJECT: Kashagan Development Experimental ProgramDOCUMENT No. DATE

28-05-2006

TITLE : Design Calculation for Two-Way SlabDESIGNED CHECKED SHEET

JJX MNR

Grade of concrete = 25

Grade of reinforcement = 390

Density of concrete = 25

Clear cover to reinforcement = 25 mm

= 3500 mm

= 5500 mm

1.571 < 2. Hence, the slab is two-way.

Thickness of slab = 150 mm

Type of panel (Edge Condition) =

3.1 Dead Load:

Self-weight of slab = 3.75

Load due to floor finish = 1.25

Load due to insulation, if any = 0.5

Other loads, if any = 0

= 5.5

3.2 Live Load:

Live load on slab = 7.5

3.3 Snow Load:

Snow load on slab, if any = 1.5

3.4 Design Load:

Load factor for dead load = 1.4

Load factor for live load = 1.6

Load factor for snow load = 1.6

= 1.4 x 5.5 + 1.6 x 7.5 + 1.6 x 1.5

= 22.1

DESIGN OF TWO-WAY SLAB: (as per BS 8110:Part 1 - 1997)

1.0 GENERAL DATA:

N/mm2

N/mm2

kN/m3

2.0 SIZE OF SLAB:

Length of slab in shorter direction, lx

Length of slab in longer direction, ly

\ ly/lx =

One short edge discontinuous

3.0 LOADING:

kN/m2

kN/m2

kN/m2

kN/m2

\ Total dead load kN/m2

kN/m2

kN/m2

\ Total design load, n

kN/m2



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Refer Clause 3.5.3.4 of BS 8110-1:1997.

Maximum design ultimate moments (for unit width),

Refer Table 3.14 of BS 8110-1:1997 for bending moment coefficients.

Bending Moment (kN-m/metre width)

Direction Location Coefficient Moment, kN-m

X directionSupport 0.059 15.973

Mid span 0.044 11.912

Y directionSupport 0.037 10.017

Mid span 0.028 7.58

Refer Clause 3.5.3.7 of BS 8110-1:1997.

Maximum design ultimate shear force (for unit width),

Refer Table 3.15 of BS 8110-1:1997 for bending moment coefficients.

Shear Force (kN/metre width)

Direction Edge Coefficient Shear Force, kN

X directionContinuous 0.479 37.051

Discontinuous 0.000 0

Y directionContinuous 0.360 27.846

Discontinuous 0.240 18.564

4.0 CALCULATION OF BENDING MOMENTS:

for span lx, msx = bsx n lx2

for span ly, msy = bsy n lx2

msxS =

msxM =

msyS =

msyM =

5.0 CALCULATION OF SHEAR FORCE AT EDGES:

for span lx, vsx = bvx n lx

for span ly, vsy = bvy n lx

vsxC =

vsxD =

vsyC =

vsyD =



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6.1 Design for Bending Moment:

Refer 3.4.4.4 of BS 8110:Part 1 - 1997 K' = 0.156 (no redistribution of moments)

Moment redistribution ratio, 1

Considering unit width of slab, b = 1000 mm

Minimum reinforcement required, 0.0013 x 1000 x 150 (0.13%, Refer Table 3.25

= 195 of BS 8110:Part 1 - 1997)

a) Calculation of Bottom Reinforcement in X direction:

Design Bending Moment = 11.912 kN-m

d = 120 mm

K =

= 0.033 (Compression reinf. is not required)

z = d {0.5 + sqrt(0.25 - K/0.9)}

= 114 mm

= 283

283

Diameter of reinforcement = 10 mm

Required spacing of reinforcement = 277 mm

Maximum spacing of reinforcement allowed = Refer 3.12.11.2.3 of

= 190 BS 8110:Part 1 - 1997

Provide 10 T @ 175 mm c/c (at the bottom of slab along X direction).

449

Hence, percentage of reinforcement provided 0.299 %

b) Calculation of Bottom Reinforcement in Y direction:


d = 111 mm

K =


6.0 DESIGN OF SLAB:

bb =

Astmin =

mm2

Mu/bd2 fcu

(zmax = 0.95d)

Ast = Mu/(0.95 fy z)

mm2

Comparing with Astmin,

\ Area of reinforcement required Astreq = mm2

70000 bb/fy + dia. of bar

\ Area of reinforcement provided, Astprovided = mm2

Mu/bd2 fcu



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z = d {0.5 + sqrt(0.25 - K/0.9)}

= 105.45 mm

= 195

195




= 188 BS 8110:Part 1 - 1997

Provide 8 T @ 175 mm c/c (at the bottom of slab along Y direction).

287


c) Calculation of Top Reinforcement in X direction:


d = 120 mm

K =


z = d {0.5 + sqrt(0.25 - K/0.9)}

= 113.81 mm

= 379

379




= 190 BS 8110:Part 1 - 1997

Provide 10 T @ 175 mm c/c (at the top of slab along X direction).

449


(zmax = 0.95d)


mm2





Mu/bd2 fcu

(zmax = 0.95d)


mm2







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d) Calculation of Top Reinforcement in Y direction:


d = 111 mm

K =


z = d {0.5 + sqrt(0.25 - K/0.9)}

= 105.45 mm

= 257

257




= 188 BS 8110:Part 1 - 1997

Provide 8 T @ 175 mm c/c (at the top of slab along Y direction).

287


6.2 Check for Shear Stress:

Refer 3.5.5 of BS 8110:Part 1 - 1997.

Maximum shear force, V = 37.051 kN

Average effective depth, d = 115.5 mm

Design shear stress, v = V/bd

= 0.321

Refer 3.5.5.2 of BS 8110:Part 1 - 1997.

Maximum allowable shear stress =

= 4 Hence Ok.

Refer Table 3.8 of BS 8110: Part 1 - 1997.

Mu/bd2 fcu

(zmax = 0.95d)


mm2





N/mm2

0.8 (fcu)0.5 or 5 N/mm2, whichever is less

N/mm2



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=

Factor for characteristic strength of concrete = where,

= 1

0.191

1.364 (without shear reinforcement)

Partial safety factor for material 1.25 (Refer 2.4.4.1 of BS 8110:Part 1 - 1997)

0.496

Shear stress is less than the design concrete shear stress, hence Ok.

6.3 Check for Deflection:

Refer 3.5.7 of BS 8110:Part 1 - 1997.

Span/effective depth, 3500/120

= 29.167

Refer Table 3.9 & Cl. 3.4.6.4 of BS 8110: Part 1 - 1997.

Basic span/effective depth ratio = 26 (Continuous Support)

Refer equations 7 & 8 of BS 8110: Part 1 - 1997.

Modification factor for tension reinforcement =

Design service stress, 1

Refer 3.4.6.2 of BS 8110: Part 1 - 1997.

1

163.88

0.827

2

Refer equation 9 of BS 8110: Part 1 - 1997.

Modification factor for compression reinf. =

= 1.091

Hence, modified span/effective depth ratio = 2 x 1.091 x 26

= 56.732 > 29.167, hence Ok.

Design concrete shear stress, nc 0.79 {100 As/(bv d)}1/3 {400/d}1/4 /gm

(fcu/25)1/3 fcumax = 40 N/mm2

100 As/bv d =

{400/d}1/4 =

gm =

\ nc = N/mm2

lx/dx =

0.55 + (477 - fs) < 2

120 [0.9 + M/bd2]

fs = 2 fy As req

3 As prov bb

As no redistribution of moment is done, bb =

\ fs = N/mm2

Mu/bd2 =

\ Modification factor for tension reinforcement =

1 + 100A's prov/bd < 1.5

3 + 100A's prov/bd



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CONTENTS

S. No. Description Page No.

1.0 General Data 1

2.0 Size of Slab 1

3.0 Loading 1

3.1 Dead Load 1

3.2 Live Load 1

3.3 Snow Load 1

3.4 Design Load 1

4.0 Calculation of Bending Moments 2

5.0 Calculation of Shear Force at Edges 2

6.0 Design of Slab 3

6.1 Design for Bending Moment 3

6.2 Check for Shear Stress 5

6.3 Check for Deflection 6



28-05-2006


JJX MNR

two way slab design

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