Slab Examples

Design constant

Table of Contents

1. introduction

2. Basic Assumptions

3. DESIGN OF SOLID TOW WAY SLAB3.1. SLAB3.2. BEAM3.3. FORM WORK

3.4. BAR SCHEDULE

4. DESIGN OF FLAT SSLAB4.1. Slab

4.2. FORM WORK4.3. BAR SCHEDULE5. DESIGN OF RIBBED SLAB5.1. DESIGN OF RIB5.2. DESIGN OF TOPPING SLAB5.3. Design of GIRDER5.4. FORM WORK5.5. BAR SCHEDULE6. CONCLUSION and SUMMERY1. INTRODUCTION

This structural design document presents all the assumptions analysis and design Calculations done in the design Material In the design of the building material properties of concrete C25 and steel S300 are used for structural elements.

Slab and beam design and method of analysisThe slab and beam is designed according to EBCS-1 and EBCS-2 1995 using direct design method. The partition wall load on each slab panel is also considered properly for each panel area. After calculation of the design actions, shear fore and bending moments, the proper reinforcement is provided for moment and the slab panel against shear.

Objective:- comparision between solid tow way slab, flat slab and ribbed slab To help and identify least cost and most economical To know which is ease to construct and needed minimum construction material Help the designer to identify and understand the baste preferable in terms of design, economy and way of construction To provide the first estimation for the designer2. BASIC ASSUMPTIONDead Load:

- Reinforced concrete =25KN/m3- Cement Screed =23KN/m3 - HCB- 200mm thick =14KN/m3- Terrazzo tile =23N/m3Partial Safety Factor for Dead Load =1.30Live Load:

LL= 4KN/m2

Partial Safety Factor for Live Load =1.50

Material Properties

Concrete:

Grade C-25 Fck = 20 Mpa

Fctk =1.5 Mpa

Partial Safety Factor = 1.5 Fcd = 0.85[20/1.5] = 11.33 Mpa

Fctd = 1.5/1.5 = 1 Mpa

Ecm = 29Gpa Reinforcing Steel:

Fyk = 300 Mpa

Partial Safety Factor =1.15 Fyd = 300/1.15 = 260.87 Mpa

Es = 200Gpa

Spacing of column: Lx= 5.0m Ly= 6.0m

Size of Column: 0.5m X 0.5m

3. DESIGN OF SOLID TOW WAY SLAB

3.1 SLAB

Design constant

For S-300, C-25, = 5m = 6m , Clear cover = 20mm

Live Load LL = 4 KN/m2

method Depth determination by Serviceability Limit State

,

The value of can be determined: using the ratio of length is end span. And from EBCS-2 , ratio 2 = 30, and ratio 1 = 40For ratio 1.2 by interpolation

, Use d = 115mm

Total depth D = 115 + 20 + 5 = 140 mmDesign Load:

Dead Loads

Selfe weight = 0.14 * 25 = 3.5 KN/m2 Ceiling plaster(15mm) =23 *0.015= 0.345KN/m2

Cement scread(30mm) = 23*0.03 = 0.69KN/m2

Terrazzo tile(20mm) = 23*0.02 = 0.46 KN/m2

Partition wall = 2 KN/m2

Total Gk = 7 KN/m2

Moment alculate

=0.063*15.49 * 52=24.40

=0.047*5.49 * 52=18.20

=0.047*5.49* 52=18.20

=0.036*5.49* 52=13.94

=0.056*15.49* 52=21.69

=0.042*5.49 * 52=16.27

=0.039*5.49* 52=15.10

=0.030*5.49* 52=11.62

=0.048*15.49* 52=18.59

=0.036*5.49* 52=13.94

=0.039*5.49* 52=15.10

=0.029*5.49* 52=11.23

=0.042*15.49* 52=16.27

=0.032*5.49* 52=12.39

=0.032*5.49* 52=12.39

=0.024*5.49* 52=9.30

Depth check

Mmax.= 24.40KNm

Use = 115mmCalculate Moment Redistribution for longer span1. Section 1-1

18.20, 13.94 .. for S1

15.04, 11.62 .. for S1

We take average moment method to adjusts support moment.

Adjustment of moment for span (field) For Panel S1: Moment adjustment coefficients from the table by using .

Cx = 0.338, cy = 0.172

Adjusted span moments

For Panel S2:The support moment is increases, according to EBCS-2 no adjustment requires for span moments.Therefore

2. Section 2-2

15.10, 11.23 .. for S3

12.39, 9.29 .. for S4

We take average moment method to adjust support moment.

Adjustment of moment for span (field)

For Panel S3:

Moment adjustment coefficients from the table by using

Cx = 0.338, cy = 0.172


For Panel S2:

The support moment is increases, according to EBCS-2 no adjustment requires for span moments.

Therefore

Calculate Moment Redistribution for shorter span

1. Section A-A

24.40, 18.20 .. for S1

18.59, 13.94 .. for S3

Therefore we use Moment Distribution Method Relative stiffness:

Distribution factors:

Adjusted support Moment:

DF 0.5

0.5 0.50.5

FEM -24.4018.59 -18.5924.40

Adj. 2.9052.905 -2.905-2.905

Madj -21.49521.495 -21.49521.495

Adjusted support Moment 21.495


For Panel S1:


Cx = 0.344, cy = 0.364


For Panel S3:


Therefore

1. Section B-B

21.69, 16.27 .. for S2

16.27, 12.39 .. for S4

Therefore we use Moment Distribution Method

Relative stiffness:

Distribution factors:

Adjusted support Moment:

DF 0.5

0.5 0.50.5

FEM -21.6916.27 -16.2724.40

Adj. 2.7122.712 -2.712-2.712

Madj -18.97818.978 -18.97818.978

Adjusted support Moment 18.978


For Panel S2:


Cx = 0.344, cy = 0.364


For Panel S4:


Therefore

Maximum Moments on the panelPanel -S1

21.495KNm 19.20KNm

16.652KNM 14.997KNm

Panel- S2

18.978KNm 17.203KNm

16.652KNM 12.607KNm

Panel- S3

21.495KNm 14.409KNm

13.75KNM 11.472KNm

Panel- S4

18.978KNm 13.749KNm

13.75KNM 9.295KNmReinforcementb= 1000mm, d = 115mm

Minimum Reinforcement:

Calculated reinforcement:

PANELMOMENTAscal (mm2)SPACING (mm)SPACING PROVIDED (mm)

TYPEVALUE

PANEL1Mxs21.500.0068776.92145.63145

Mxf19.390.0060694.64162.88160

Mys16.650.0051589.83191.82190

Myf15.150.0046533.49212.08210

PANEL2Mxs18.980.0059678.79166.68165

Mxf17.360.0054616.74183.45180

Mys16.650.0051589.83191.82190

Myf12.730.0039444.07254.78250

PANEL3Mxs21.500.0068776.92145.63145

Mxf14.540.0044510.78221.51220

Mys13.750.0042481.55234.96230

Myf11.580.0035402.19281.32280

PANEL4Mxs18.980.0059678.79166.68165

Mxf12.510.0038436.03259.48255

Mys13.750.0042481.55234.96230

Myf9.380.0028323.12350.16350

Load transfer to beam:

Panel S-1:Shear force coefficients for uniformly loaded rectangular panel is: Lx = 5m Pd=, =15.491KN

EMBED Equation.3

Panel S-2:

Shear force coefficients for uniformly loaded rectangular panel is:

Lx = 5m , Pd = 15.491KN

EMBED Equation.3

Panel S-3:


Lx = 5m , Pd = 15.491KN

EMBED Equation.3

Panel S-2:


Lx = 5m , Pd = 15.491KN

EMBED Equation.3

PanelLy- longer directionLx- Shorter direction

axisVcxVdxVcyVdy

S - 11-discontineous-24.011-discontineous-20.14

2-contineous36.4-2-contineous30.98-

S - 21-discontineous-22.462-contineous27.88-


S - 32-contineous32.53-1-discontineous-18.59


S - 42-contineous30.21-2-contineous25.56-


3.2 Beam design

design constant

Longer Direction

b = 250mm, D = 400mm, cc=25mm, main=20mm, stirrup =8mm ,d= 357mm

check depth:

Use d = 357mm

Beam span AB & CD , Msd= 79.79 KNm

From GDC for sd = 0.22,

Reinforcment

Provide 4 20 bar

Beam span BC, Msd= 20.98 KNm


Reinforcment

Provide 2 20 bar

For support B & C, Msd = 95.84


Reinforcment

Provide 4 20 barShear design

By similarity of triangle: design shear

EMBED Equation.3

EMBED Equation.3 Resistance shear

Shear Capacity

At support

For beam section 2,3,4,5

Design shear for span AB & CD Vsd,max = 88.07 Vc,max = 48.42

S max =

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