PPT DESIGN AND COMPARISON OF FLAT SLAB USING IS 456 – 2000

PRESENTATION ON

DESIGN AND COMPARISON OF FLAT SLAB USING

IS 456 – 2000 AND ACI 318 – 08

PRESENTED BY

TUSHAR MANOJ MADAMWAR

(11001063)

GUIDED BY

DR. K. N. KADAM

CONTENTS

1. INTRODUCTION

2. ADVANTAGES OF FLAT SLAB

3. LITERATURE REVIEW

4. DESIGNOF FLAT SLAB USING IS 456-2000

5. DESIGN OF FLAT SLAB USING ACI 318-08

6. RESULT AND COMPARISION

INTRODUCTION



_{DEFINATION-Slab and column without beams}



_COMPONENTS

- Drops-increase shear strength against punching.

Advantages of Flat Slab

1.

Flexibility in room layout

Saving in building height

Shorter construction time

Ease of installation

5.

Pre-fabricated welded mesh

Buildable score

PROBLEM STATEMENT

Design a flat slab of size 6.6m×5.6m using direct design method by IS

456-2000 and ACI 318-08 .It is subjected to live load of 7.75 kN/m

.The

grade of steel used is Fe 415. The exposure condition is mild. Height of

floor is 4m.size of column is 400mm × 400mm.

DESIGN OF FLAT SLAB USING IS 456-2000

 Column Strip

A design strip having a width of 0.25l₂ not greater than 0.25l₁ on each side of column center-line.

For longer span=1.4m (not greater than 1.65) For shorter span=1.65m(not greater than 1.4)

 Middle strip

A design strip bounded on each of its opposite sides by the column strip. For longer span=2.8m

For shorter span=3.8m

 Panel

The part of a slab bounded on-each of its four sides by the line of a Column or center-lines of adjacent-spans.

 Drop

The drops when provided shall be rectangular in plan and have a length in each direction not less than one- third of the panel length in that direction

For longer span =2.2m For shorter span=1.866m Provide drop of 2.2m×2.2m

 Column Head

For longer span=1.65m For shorter span=1.4 Adopting 1.3m

 Depth of Flat Slab

The thickness of the flat slab up to spans of 10 m shall be generally controlled by considerations of span ( L ) to effective depth ( d ) ratios given as below:

Cantilever 7; simply supported 20; Continuous 26 For longer span=260mm

For shorter span= 220mm

Loads acting on slab

Dead load(𝑊_𝑑1)=0.285× 25=6.25kN/m2

Floor finish(𝑊_𝑑2)=1.45kN/m2

Live load(𝑊_𝑙)=7.75kN/m2

Total design moment on slab

M_o=𝑾𝒍𝒏

𝟖

M_o=total moment

W = design load on an area l₁ l₂

l_n = clear span extending from face to face of columns, capitals, brackets or walls, but not less than 0.65 l₁

l₁ = length of span in the direction of M_o.

l₂ = length of span transverse to l₁.

A=𝝅

𝟒 𝒅𝟐= 𝜋

4 1.32 = 1.152𝑚

Clear span along long span =l_n =6.6-1

2 (1.152)-1

2(1.152)=5.448 >4.29

(Should not be less than 0.65 l₁ )

Clear span along long span =l_{n =}5.6-1

2 (1.152)-1

2(1.152)=4.44 >3.64

Total Design Load

For longer span(l _n=5.448 m , l₂ =5.6 m)= W = w l ₂ l _n =471.36kN For shorter span(l _n=4.44 m , l₂ =6.6 m)=W = w l₂ l_n= 452.74k N

Absolute Sum of Negative and Positive Moments

For longer span=M_o=𝑤𝑙𝑛

8 =

471.36×5.448

8 = 320.99 𝑘𝑁

For shorter span=M_o=𝑊𝑙𝑛

8 = 452.74×4.44 8 = 251.2 𝑘𝑁

Stiffness Calculations

Distribution of Bending Moment Across

The Panel Width



_{Longer Span}

1.Column strip

Negative B.M at exterior support=−121.34𝑘𝑁𝑚 positive span B.M =90𝑘𝑁𝑚

Negative span BM at interior support =-166.5 kNm

2.Middle strip

Negative BM at exterior support =0𝑘𝑁𝑚 Positive span BM =59.96𝑘𝑁𝑚

Effective Depth of The Slab

Maximum positive BM occurs in the column strip (long span) = 90.91 kNm factored moment = 1.50 x 90.91 = 136.36 kNm

M_o= 0.138f_{ck 𝑏𝑑}2 _{= (𝑏 = 2800𝑚𝑚}

d= 136.36×106

0.138×20×2800 𝑀20 𝑔𝑟𝑎𝑑𝑒 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒

d=132.83 mm ≅140 mm

Using 12 mm ∅ (diameter) main bars. Overall thickness of slab =140+15+ 12

2 =161 mm ≅170 mm

Depth (along longitudinal direction) =170-15-12

2 = 150𝑚𝑚

Shear In Flat Slab

The critical section for shear shall be at a distance d/2 from the periphery of the column/capital/ drop panel, perpendicular to the plane of the slab

Check for shear stress developed in slab

The critical section for shear for the slab will be at a distance d/2 from the face of drop. Perimeter of critical section =4× 2340=9340mm

V_o=1.5× 15.45 × 𝐿₁ × 𝐿₂ − 2.34 (2.34) =729.78kN Nominal shear stress =T_v=𝑉𝑢

𝑏𝑑 =

729.78×103

9340×140 = 0.55𝑁/𝑚𝑚2

shear strength of concrete =T_c=0.25 𝑓_𝑐𝑘=0.25 20 =1.1N/𝑚𝑚2 Permissible shear stress =𝑇_𝑣 < 𝑘_𝑠𝑇_𝑐

K_s=(0.5+𝛽_𝑐) ,β_c=0.848 K_s=1.348 > 1

Reinforcement

Longer span

Negative exterior reinforcement: 𝑀_𝑢=0.87 𝑓_𝑦A_st 𝑑 − 0.42𝑥_𝑢

1.5× 121.34 × 106 = 0.87 × 415 × 𝐴_𝑠𝑡 𝑑 − 0.42 × 0.48 × 150

DESIGN OF FLAT SLAB USING ACI 318-08

 Drop of flat slabs

Drop panel shall extend in each direction from centerline of support a distance not less than one-sixth the span length measured from center -to center of supports in that direction.

Span of panel in longer direction = 16.76 ft length of drop panel=1

6 × 16.76 × 2=5.58ft

With half width on either side of the center line of support = 0.85 m Thickness of drop =1

4 × 6 = 1.5 𝑖𝑛 = 38.1𝑚𝑚

 Thickness of the slab

the minimum thickness shall be in accordance with (a) Slabs without drop panels as ... 5 in. (b) Slabs with drop panels as defined... 4 in.

Depth of the slab from deflection criteria =𝑙𝑛

36 (the yield stress 𝑓𝑦𝑖=60,000psi,≅ 415𝑁/𝑚𝑚2

Minimum depth of slab=max of (16.76×12

36 ,

14.22×12

36 )=max of (5.58in,4.74in)=5.58in

Providing a slab of thickness 6 in or 152.4 mm.

 Design strips

Column strip is a design strip with a width on each side of a column centerline equal to 0.25 l₂ or 0.25 l_1,

whichever is less.

Middle strip is a design strip bounded by two column strips.

Designed Load on Slab

Dead load on the slab = 6

12 × 150 = 75𝑝𝑠𝑓 = 3.6𝑘𝑁/𝑚3

Live load on the slab = 161.80 psf = 7.75 KN /m2

Design load on the slab = (1.2 x 7.5 + 1.6 x 161.80) = 348.88 » 350 psf= 16. 765 KN /m2

Total factored static moment for a span

M_o=𝑾𝒖𝒍𝟐𝒍𝒏

𝟐

𝟖

W_u=load per unit area acting on the slab panel

l_n=Clear span l_n shall extend from face to face of columns, capitals ,brackets, or walls. Value of l_n shall not be

less than 0.65 l₁ .

l₂=When the span adjacent and parallel to an edge is being considered, the distance from edge to panel centerline shall be substituted for l₂ .

In an interior span, total static moment M_o shall be distributed as follows: Negative factored moment ...0.65

Positive factored moment ...0.35 M_o= 350

8×1000 × 170762 × 14.22 = 174.75𝑓𝑡 − 𝑘𝑖𝑝𝑠 = 237𝑘𝑁𝑚

Negative design moment = 237 x 0. 65 = 154 ft-kips = 208.89 kNm Positive design moment = 237 x 0.35 = 83.00 ft -kips = 113.22 kNm

 Bending moment for column strip

Negative moment for column strip = 75 % of total negative moment in the pannel = 0.75 x 154.00 = 115.50 ft -kips = 157.66 KNm

Positive moment for column strip = 60 % of total positive moment in the panel. = 0.60 x 83.00 = 49.8 ft -kips = 67.977 KNm

Max moment (+ve or –ve ) along shorter span = 72.18 ft -kips Max moment (+ve or –ve) along longer span = 115.50 ft –kips

Shear Provision

 For flat slabs V_c =Nominal shear strength of concrete, V_c Shall be smallest of the

following: [Where β_c is the ratio of long side to short side of the column,

concentrated load or reaction area and where a_s is 40 for interior columns, 30 for

edge columns,20 for corner columns] V_c== 2 + 4 𝛽_𝑐 𝑓𝑐 ′_𝑏 𝑜𝑑 V_c== 2 + 𝑎_𝑏𝑠𝑑 0 𝑓𝑐 ′_𝑏 𝑜𝑑 V_c=4 𝑓_𝑐′𝑏_𝑜𝑑

V_u= factored shear, acting at distance d/2 from face of the support. (assuming column of size 400 mm by 400 mm)

V_u=350 16.76 × 14.22 − 1.31 + 0.5 1.31 + 0.5 =350 238.32 − 1. . 812 =82265.365 lb=365.91kN

𝑓_𝑐′𝑏_𝑜𝑑= 4000 × 4 × 21.72 × 6 = 32968.64𝑙𝑏 (𝛽_𝑐 = 1.17) The nominal stress of concrete will be smallest of the following

V_c== 2 + 4

1.17 × 32968.64 = 178650.57 𝑙𝑏

V_c== 2 + 40×6

4×21.72 × 32968.64 = 157010.87 𝑙𝑏

V_c= 𝑓_𝑐′𝑏_𝑜𝑑=4× 32968.64 = 131874.56 𝑙𝑏 V_c > V_u section safe in punching shear \safe.

Reinforcement

For negative moment in column strip

R= 𝑀𝑢 𝑏(324) = 115.5×103 16.22×32.4=219.77 Reinforcement ratio = 0.00375 Area of reinforcement = 0.00375 x 16.22 x 6 x 12 = 4.38 in2_/ft

Results and Comparisons

CODE IS 456 -2000 ACI 318-08

Shape of test specimen for concrete strength (mm) Cube 150x150x150 Cylinder 152.4x304.8 Grade of concrete(N/mm²) 20 20 Grade of steel (N/mm²) 415 413.7 Negative moment (KN-m) 188.5 208.89 Positive moments (KN-m) 90 113.22 Area of reinforcement(mm²) 4290 2829

Thickness of slab for Serviceability criteria(mm)

170 150

Conclusions

 By comparing with different codes we concluded that ACI 318 code is more effective in designing of flat slabs.

 As per Indian code we are using cube strength but in international standards cylinder strength is used which gives higher strength than cube.

 Drops are important in increasing the shear strength of the slab.

 Flat slabs enhance resistance to punching failure at the junction of concrete slab & column.

 By incorporating heads in slab, we are increasing rigidity of slab.

Conclusions

 The negative moment’s section shall be designed to resist the larger of the two interior negative design moments for the span framing into common supports.

 According to Indian standard (IS 456) for RCC code has recommended characteristic strength of concrete as 20, 25, and 30 and above 30 for high strength concrete. For design purpose strength of concrete is taken as 2/3 of actual strength this is to compensate the difference between cube strength and actual strength of concrete in structure. After that we apply factor of safety of 1.5. So in practice Indian standard actually uses 46% of total concrete characteristic strength. While in International practice is to take 85% of total strength achieved by test and then apply factor of safety which is same as Indian standard so in actual they use 57% of total strength.

References

 Bureau of Indian Standards, New Delhi, “IS 456:2000, Plain and Reinforced

Concrete - Code of Practice”, Fourth Revision, July (2000).

 American Concrete Institute, “ACI 318-08, Building Code Requirements for

Structural Concrete and Commentary”, January (2008).

 Dr. V. L. Shaha & Dr. S. R. Karve “Limit State Theory and Design of Reinforced Concrete” Sixth edition

 Amit A. Sathwane , R. S. Deoalate (IJERA)“Analysis and Design of Flat Slab and Grid Slab and Their cost omparision”