G+4 SolidSlabSystem

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ETHIOPIAN CIVIL SERVICE COLLEGE

INSTITUTE OF URBAN DEVELOPMENT STUDIES

DEPARTMENT OF RBAN ENGINEERING

ANALYSIS, DESIGN, AND CONSTRUCTION COST COMPARISON SOLID

VERSUS RIBED SLAB SYSTEM OF A G+4 COMMON RESIDENTIAL BUILDING

BY : WONDIMAGEGN SOLOMON UED_1064/98

MESFIN GIRMA UED_1041/98

TARIKU HAILE UED_1050/98

TEKLEGERIMA HAILU UED_1055/98

YINURSELAM SHIFAW UED_1067/98

TERFE SHIFERAW UED_1056/98

YIMER MESHESHA UED_1066/98

ADVISOR: MELESE YOHANES

A senior project submitted to:

Ethiopian Civil Service College Institute of Urban Development Studies Department

of Urban Engineering in partial fulfillment of the requirement for the degree of

Bachelor of Science in Urban Engineering.

JULY, 2010

ADDIS ABABA

ETHIOPIA

Table of Contents Page

PART -I: SOLID SLAB STRUCTURAL ANALYSIS AND DESIGN


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1. Introduction

1.1. General introduction

1.2. The Structural System

1.3. Building Layout

1.4. Quality of Construction and Method of Analysis

1.4. Construction Materials

1.5. Loading

1.6. Codes and References

1.7. Design Aids

2. Wind Load Analysis

2.1. Roof Wind Load Analysis

2.2. Truss Analysis and Design

2.3. Wind load analysis on external and internal surfaces of building

3. Solid slab analysis

4. Stair analysis and design

5. Earth quake analysis

5.1. Mass, mass-center and center of rigidity determination

5.2. D-value computation

5.3. Story shear distribution

5.4. Lateral load distribution among frame elements

6. 2D Frame Analysis

7. Beam Design

7.1. Flexural design

7.2. Design for shear

7.3. Development length and bar cutoff

8. Column Analysis and Design


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9. Footing design

PART -2: RIBBED SLAB STRUCTURAL ANALYSIS AND DESIGN

1. Ribbed Slab Analysis and Design

2. 3D frame analysis

2.1. Loading

2.2. Method of analysis

2.3. Analysis result (Output)

3. Girder design

4. Column design

5. Footing design

PART -3: SOLID SLAB VERSUS RIBBED SLAB

1. Comparison

2. Conclusion and recommendation


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Acknowledgement

Successful completion of this analysis and design project is last chapter in five

years study and hard work, which has been made possible today because of

assistance and support of our families, instructors, fellow students, and friends.

We would like to appreciate and thank our advisor Ato Melesse Yohannes(MSc)

who provide us considerate project type and above all for his guidance and supportthroughout the development of the project.

Our special thanks also go to our families and particularly to our parents who have

been all along with us assisting and encouraging.

Lastly, but not least we would like to express our heartfelt best gratitude to W/ro

Rahel Yohannes and Ato Yimer Mohamed for their effort for the modification

made and realization of our new curriculum as we are the first urban engineering

graduates in bachelor degree in the country.


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1. Introduction

1.1. General introduction

This booklet contains a static calcucalation of structural analysis and design ofa

common residential building located in Addis Ababa. The structure has 6stories one basement, one ground, and G + 4 stories with a solid/ribbed slab floo

system.

The solid slab system is modeled in 2D while the ribbed slab system is in 3D, just

for the academic interest there by to experience different scenarios.

Analysis and design comprises roofing (EGA sheet), purlin, truss members, beams,

columns, stairs, slabs and footings.

Loads for analysis include dead (self & transferred) loads, live loads, wind load, and

earth quack (lateral) loads.

Analysis of frames in 2D/3D involves:

Modeling based on architectural drawing.

Stiffness computation

Mass calculation

Center of mass & center of gravity determination.

Story shear distribution

Lateral load distribution according to stiffness of frame elements(columns)

Modeling on SAP 2000-V11/ ETABS -V9:

o Definition of material, frame section, load cases, analysis cases,

& combination.

o Drawing the model


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o Assigning joint constraints (joint restraints, diaphragm) lateral &

gravity loads.

o Set analysis options

Run analysis options & export relevant analysis report.

Using result of analysis of each of the cases, design of structural elements (slabs,

beams, columns, stair, &footing) for flexure, lateral & axial load and shear

followed. In addition design for development length is carried out.

In the mean time, comparison of the two systems with respect to cost of

construction material is made. However, only typical floor slab are considered for

cost analysis, tacking in to account its major proportion than other more or less

similar structural and none structural elements.

Finally, representative structural detail drawing is produced.

1.2. Structural System

In general the structural system consists of a solid/ribbed slab floor system with a

beam-column frame system. The slab system has a thickness of 130mm/200mm

according to the structural importance and deflection requirement. 130mm section

was used for the interior span two way slabs and 200mm for cantilever and one

way slabs. 100mm for the ground floor slab as this slab is resting on the 250mm

hardcore laid on the selected material fill on the ribbed mat. Square Reinforced

concrete columns with 300mm and 400mm sections were used to take all vertica

and lateral loads. Beams with 400mm x 400mm section were used to support al

the slab systems. Independent footings with 300/400/600mm slab section and2500mm tall foundation column below the natural ground level were used as a

foundation.

1.3. Building Layout


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a. Three Dimensional Layout of The Building

b. Ground Floor Plan


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c. Typical Floor Plan

1.4. Quality of Construction and Method of Analysis

o Class I work is assumed

o Ultimate limit state (ULS) method for loading and serviceability limit state

(SLS) for analysis

For concrete (gc), persistent and transit gc=1.5

Accidental gc=1.3

For steel (gs), persistent and transit gs=1.15

Accidental gs=1.00

1.5. Construction materials

Taking in to account availability of quality construction materials and skilledworkmanship concrete of quality C25 and steel of grade S300 are used.

Concrete C-25:

Fck=20Mpa,fcu =25Mpa

Fcd=0.85fck/1.5=11.33Mpa

Fck=0.7fctm,fctm=0.3*fck2/3

=0.21* fck2/3=1.5473/1.5=1.0315Mpa

Fctd=fctk/g=1.5473/1.5=1.0315Mpa

Ecm=9.5*(fck+8)1/3=9.5*(20+8)1/3=28.85Gpa29 Gpa

gc=24KN/m3

steel S300:

fyk=300Mpa

fyd=fyk/gs=300/1.15=260.87Mpa

gs=77KN/m3

Es=200Gpa

Timber

Esu=0.6Mpa

g=6KN/m3

n=

flooring

marble tile 27KN/m3

pvc covering 16KN m3

wall


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HCB gHCB=14KN/ m3

Glazing

Glass 25 KN/m3

Aluminum 27 KN/m3


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1.6. Loading

Since the site is located in Addis Ababa, which is in the area of seismic zone 2

according to EBCS-8/95, in addition to vertical loading Earthquake loading was

considered. The major loadings considered are:

Vertical Loading: Dead Load (DL) - (Self Weight, Wall Load and Finishing Load,

Roof loading)

Live load (LL)

Lateral Loading: Earthquake load in X and Y direction (EQX & EQy)

Wind Load

The above loadings make up a total of nine different combinations.

No. Combination Name Factored Loading Combination

1 Comb1 1.3DL+1.6LL

2 Comb2,3 0.75(1.3DL+1.6LL) EQx

3 Comb4,5 0.75(1.3DL+1.6LL) EQy

4 Comb6,7 0.75(1.3DL+1.6LL) EQx 5%Eccentricity in y dir

5 Comb8,9 0.75(1.3DL+1.6LL) EQy 5%Eccentricity in x dir

Out of the five combinations the critical case was taken for the analysis and design

of beams, slabs and columns. The footing was analyzed and designed for Combo1

(Vertical Loading).

1.7. Codes and References

EBCS 1-1995 EBCS 2-1995 and associated tables & charts

EBCS 5-1995 EBCS 6-1995

EBCS 7-1995 EBCS 8-1995 and

Euro Code 2-1992 (as used by the software), almost similar to EBCS-2/95

1.8. Design aids


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Sap 2000 -V11for modeling and analysis of solid slab G+4 building as 2D

frame

Etabs-V9 for modeling and analysis of ribbed slab same building as 3D

frame

Ms-Word, Ms-Excel are also used to facilitate computation and edition of this

booklet

Arch-CAD12 and AutoCAD 2007 for 3D modeling, working dawning, and bar

schedule

2. Wind Load Analysis

2.1. Roof wind load analysis

ROOF LOADING

LIVE LOAD

ROOF CATEGORY (EBCS-1, 1995 TABLE 2.13)

Roof not accessible except for normal maintenance, repair, painting &minor repairs

are under category H.

Imposed load on roof for slope category H

Roof is qk=0.25 KN/m

2

Qk= 1KN

WIND LOAD

DETERMINATION OF EXTERNAL WIND PRESSURE (We)

The wind pressure acting on external surface of a structure.

We, shall be obtained from

( ) peeeref CzCqWe = ---------------------EBCS-1, 1995 Eqn 3.1

Where qref = reference mean wind velocity pressure derived from reference wind

velocity.

Ce = exposure coefficient accounting for the terrain & height above the ground up

to a height z.


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(Ze) = reference height for the relevant pressure coefficient.

B. DETERMINATION OF REFERENCE WIND PRESSURE (qref)

refref vq2

2

= EBCS-1, 1995 sec. 3.7.1

Where:

r=air density (kg/m3) considering altitude of the place above mean sea level

Addis Ababa is found above sea level of 2000m

r = 0.94 kg/m3 EBCS-1,1995 table 3.1

refv = reference wind velocity

refv = CDIRCTEMCALT refv , 0 ..EBCS-1, 1995 sec. 3.7.2(2)

Mean return period 50 years..EBCS-1,1995 sec. 3.7.2(1)

Where

CDIR =is the direction factor taken as 1

CTEM = is the temporary (seasonal) factor to be taken as 1

CALT = is the altitude factor to be taken as 1

refv , 0 = is the basic value of reference wind velocity to be taken as 22m/s.

refv = CDIRCTEMCALT refv ,0 EBCS-1,1995- eqn. 3.7

=1*1*1*22m/se

=22m/se

Reference wind pressure ( refq )

refref

vq2

2

=

22*294.0=refq

2 =227.48

DETERMINATION OF EXPOSSURE COEFFICIENT, Ce (z).

The exposure coefficient taken in to account, (Ce, z). The effect of

Terrain roughness


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Topography and

Height above ground on the mean wind speed and turbulence

..EBCS-1, 1995 sec.3.8.5 (1)

For codification purpose it has been assumed that the quasi-static gust load is

determined from. EBCS-1, 1995 sec. 3.8.5(2)

( ) ( ) ( ) ( ) ( )

+=

zCzC

kzCzCzC

tr

Ttre

71

22

.EBCS-1, 1995 eqn.3.15

Where:

kT = is the terrain factor defined as (for terrain category IV) urban areas in which

at least 15% of the surface is covered with building & their average height exceeds

15m.

kT =0.24 ...EBCS-1, 1995 tab.3.2

Cr (z) = the roughness coefficient account for the variability of mean wind

velocity at the site of the structure due to

The height above the ground level (z)

The roughness of the terrain depending on the wind direction

...EBCS-1, 1995 sec.3.8.2

Cr (z), at height z is defined by logarithmic profile

One of the two cases must satisfy

Case-1 ( )

=

0

lnz

zKzC Tr for zmin


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And height of building z = 16.50m

zmin= 16m < z = 16.50m .satisfies case 1

Roughness coefficient

= 0.24 ln (16.50/1) = 0.673

Topography coefficient accounts for the increase of mean wind speed over isolated

hill & ridges.

Ct (z) = 1 .................EBCS-1, 1995 sec. 3.8.4(2)

Exposure coefficient

( ) ( ) ( )( ) ( )

+=

zCzC

kzCzCzC

tr

Ttre

71

22

= 0.673

2

*1

2

(1+(7*0.24)/(0.673*1)) = 1.5836

D. DETERMINATION OF EXTERNAL PRESSURE COEFFICIENT (Cpe

(z))

External pressure coefficient for hipped roofs.

For wind direction = 0

0


0

( )

=

0

lnz

zKzC Tr

20.70 23.90

Wind

=00

Wind

=900

a=00

a=900


0

For wind direction

= 90

0


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o= tan-1(1.5*2/20.7)=8.250 90= tan-

1(1.5*2/10.45)=8.170

Reference height: ze=h=16.50m Reference

height: ze=h=16.50m

e=b or 2h whichever is smaller

b=cross wind dimension

FOR WIND DIRECTION =900

Imposed areas

F = 2.07*3.175=6.75=2*(6.75+2.12)=17.74m2

o.5*2.07*2.05=2.12

G = 2.07*10.25=21.22 m2

H = 0.5*16.60*8.38=69.55 m2

I = 0.5*16.56*8.33=68.97 m2

J = (0.5*20.70*10.45)-68.97=39.19 m2

L = 2.07*10.35=21.42 m2

M=0.5*8.28*8.28=34.28 m2*2=68.56m2

N=0.5*(21.83+3.10)*10.35-

34.28=94.73 m2*2

=189.466 m2

FOR WIND DIRECTION =00

F=3.562*2.39=8.51

=0.5*2.413*2.39=2.88

=2(8.51+2.88)=22.78 m2

G=11.95*2.39=28.56 m2

H=0.5(18.38+3)*7.96=85.01 m2

I=0.5(0.74+19.12)*7.96=79.04 m2

J=2.39*7.96=19.02*2=38.05 m2

K=0.5(3+5.52)*2.39=10.18 m2

L=0.5(10.45*20.70)-

84.49=23.67*2=47.34 m2

M=0.5(9.24*18.31)=84.59*2=169.18

m2


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Excel values are required from

wondmagegn

1.3 DESIGN OF ROOF COVER (DESIGN OF EGA)

Maximum wind load (critical windload) is

WL=-0.521 KN/m2from

computation

Live load from .EBCS-1,

1995

Distributed live load qk=0.25 KN/m2

Concentrated live load Qk=0.25 1KNPurlin spacing=1.5m

Take EGA-300 from KMPF manual

Thickness t=0.5mm, section modulus

Sx=1970

Load W=3.92kg/m

Uniform load carrying capacity

=1.36KPa=ULC

Load computation for roof cover

EGA width is 0.90m

EGA load in KN/m2

DLEGA=(3.92Kn/m)/(0.90m)

2/044.01000

10*

92.0

/92.3mKN

m

mKNDLEGA ==

Case -1Pd=LLq+DL

=1.6 qk+1.3 Gk

=1.6*0.25+1.3(0.044)

=1.6*0.25*cos(8.25)+1.3(0.044)

=0.453KN/m2


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mKNmWlPL

M /609.08

5.1*0572.0

4

5.1*58.1

84

22

=+=+=

Section modulus, SX

Allowable carrying capacity of roof is 250N/mm2

32

6

4.2434/250

/10*609.0 mmmmN

mNmmMSall

x ===

From table of KMPF manual selec EGA sheet having section of modulus Sx

>2434

Take section of modulus Sx =2749mm

3

Section property

Thikness t=0.7mm

Weight w=5.49kg\m

Capacity=1.90kpa

Dead Load of EGA=

2/061.00.9*1000

10*5.49mkn=

Perpendicular dead load of EGA Ro0f is

DL EGA =O.O61 cos8.25

=0.0604

Load computation at 0=8.25

Case 1

Pd=LLq+DL

=1.6qk+1.3Gk

=1.6*0.25COS8.25+1.3

(0.0604)

=0.474KN/m2

Case 2

Pd=WL+DL

=1.6WL+0.9GK

=1.6*O.521+0.9(0.060

4)

=0.834KN/m2

Case 3

Pd=LLQK+DL

=1.6QK+1.3GK

=1.6*1*cos8.25+1.3(0.0604)

=1.583KN+0.0785KN/

m2

Moment Computation


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Case1

M=Pd

L2/8=0.474*1.52/8=0.1

33KNm/m

Case 2

M=PdL2/8=0.834*1.52/8

=0.235kNm/m

Case 3

M=Pd*L/4+PdL2/8

=1.583*1.5/4+0.0785*

1.52/8

=0.616kNm/m

Section modulus computation

SX=M/rall=0.616*106Nmm/m/250N/mm2

2464.31mm3


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Reaction forceKN

WLR 344.0

2

5.1*459.0

2===

Exterior reaction force Re=0.344KN at 0.9m spacing,

Interior reaction force Ri=0.344KN at 0.9m spacing on rafter of the truss.

LOAD FROM TRUSS

Material property:-property of equliptus.

TRUSS TOTAL LENGTH COMPUTITION

Truss 1

Major horizontal =10.45m

>> Vertical =1.50m

>> Diagonal (rafter) =10.56m

Total length =22.51m

Vertical & diagonal members

Spacing of vertical i=1.20m

Slope of roof in (%)=14.36%

No of verticals =9

No of diagonals =8

No of joints =10

Total length of vertical and diagonal

members =17.35

TOTAL LOAD COMPUTATION

Factor loads:

wt(10)=16.6m*0.087 KN/m=1.44 KN

wt(12)=22.62m*0.125

KN/m=2.83 KN

total load of truss

=4.27 KN

No of joints =10

Unit wt.

(KN/m3)

diameter(

m)

area(m2

)

wt.

(KN/m)

remark

10 8.5 0.10 0.0079 0.067 V&D members

12 8.5 0.12 0.0113 0.096 major

V,H&rafter


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0.614

0.6141.22

1.22 1.221.22

1.221.22

1.221.22

0.103

R1 R2

12 11

R3

0.10

3

0.20

60.20

6

0.20

6

0.20

63

0.20

6

0.20

6

0.20

6

0.20

6

0.10

3

A B C D E F G H I

I

I

I

I

I

J

12 3

45

67

89

Reactions on truss

Truss Stretching Axes A-B-C

Analysis Result

Truss Stretching Axes 1-2-3

Analysis Result

2.3. Truss Analysis and Design

Internal joints =8

External joints =2

Loads on truss joints

KNsofvertical

totalloadsernaljo 474.0

9

27.4

#intint ===

KNloadsernaljo

sexternaljo 237.02

474.0

2

intintint ===

DEAD LOAD OF CHIP WOOD CEILING (CHIP BOARD)

Properties of chip board

Unit wt., =8KN/m3

Thickness, t=8mm=0.008m

Truss spacing =1.5m

Length =10.45m

Wt=*t*l*c/c spacing

=8KN/m3*0.008m*10.45m*1.5m

=1.0032KN

Factored load=1.3*1.0032KN

=1.30416 KN

DEAD LOAD OF CEILING BATTENS (30mmx40mm)

Spacing c/c=60mm

In the truss direction # of ceiling

battens=10.45/0.6=17.4

Use 18 battens

Other direction # of ceilingbattens=1.5/0.6=2.5,use 3 battens

Length of ceiling battens is

=18*1.5+3*10.45=58.35m

Unit wt, = 6kN/m3

Wt of

battens=6*0.03*0.04*38.35=0.42012

KNFactored battens load =0.546156kN

Total load at the bottom of the truss

Joints

Load

=1.30416+0.546156=1.850kN

# Of Joints=10

Exterior =2

Interior =2

Joint load Int = kN206.09

850.1=

Joint load ext = kN103.02

206.0=

Total load at the top of the truss joints

Interior joint load =0.688+0.51=1.188

KN

Exterior joint load

=0.344+0.26=0.604 KN


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I

2.3. Wall Wind Load Analysis on External and Internal Surfaces of

Building

I. Wind Load External Surfaces

We =qrefce(ze)cpe


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qref=5/2v2req ,

=

=

3/94.0

sec/22

mkg

mrefv

since

Addis Ababa >2500 amsl

= 0. 94/2kg/m3 *22m/sec =2/48.227 mN

Ce(z)=Cr2(z) Ct2(z)

+

)()(

71

zCzzC

kg

r

- Terrain category IV for urban

areas

- Kt=0.24 zo(m)=1 zmin(m)=16

- Crz=crz(min)=ktln(zmin/zo

since z=15


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C)

h=15m and b=19.9

he=19.9m

We,A=227.48*1.575*-1=-0.358kN/m2

We,D=227.48*1.575*-

0.764=0.274kN/m2

We,B=227.48*1.575*-0.8=-

0.287kN/m2

We,E=22.48*1.575*-0.3=-0.108kN/m2

WeC=227.48*0.5=-0.5=-0.175kN/m2

II) Wind load on internal surfaces

These pressures are neglectedbecause the openings on both the wind ward and

leeward directions are the same, and the net effect on the entire structure is zero.

3. Solid slab analysis

Coefficient and strip methods of solid slab analysis are used here. Since the

building is symmetric in two directions we considered quarter of the building for

design of slabs in each typical panel.

zone A B C D Ed/n Cpe,10 Cpe,10 Cpe,10 Cpe,10 Cpe,101 -1 -08 -0.5 +0.8 -0.34 -1 -08 -0.5 +0.6 -0.3234=1.5

4

15

-1 -08 -0.5 +0-

764

-0.3

B=19.

9

A

B

C

E

D

H=2

3.1


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Quarter of typical floor layout

3.1. Water Tank Seat Slab

Design considerations

5 people per household

100 liter reserve water requirement per head (individual) for

residential building

4 households in each floor except ground floor

Two water tanks

Total volume of water; VT=4floors*4households*5individuals per

household*100l=800l

Each of tanks expected to hold V=800/2=400l=4m3

Assume: A 2.5m diameter cylindrical Roto reservoir.

V=D2h/4h=4V/D2=(4*4m3)/(2.5)2=0.815m

Take h=1m (standard size)


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V=(D2h)/4=(*2.52*1)/4=4.91m3

Wwater=w*Vw=4.91m3*10KN/m3=49.1KN.

W=49.1KN/(D2/4)=10KN/m2

Depth determination ;Ly/Lx=1.815 Ly/Lx a

d(0.4 +0.6*300/400)2700/a 2 25

0.85*2700/26.85=85.475mm 1.815 ? a=25+(35-25)

((2-1.815)/(2-1))

1 35

=26.85

Take d=86mm

D=86+12/2+15=107mm

Take D=110mmd=89mm

Design Load :

w=1.3*[0.11*25+10]+1.6*0.5; Assuming no occupancy except for casual

maintenance.

=17.375KN/m2

Design Moment:

Ly/Lx xf/xs

1.75 0.103

1.815 ? x=(0.111-0.103)[(1.82-1.75)/(2.1)]

+0.103=0.105

2 0.111

ys=yf=0.056

Mxs=0.105*17.375*2.72=13.3 KNm =Mxf

Mys=0.056*17.375*2.72=7.09KNm=Myf

Design for Flexure


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A) For Span Moment

Check for ductility

Depth check:

!.......8908.63

33.11*1000*295.0

/10*3.13

**

26

OKmmmmmmKN

fb

Md

cd

f min

As, cal= 0.007*1000*89mm= 623.06mm2

i) Middle Strip

Reinforcement

51.54/12

06.623

4/12#

2

2

2

, ===

mm

D

Abars

cals

Provide 6 12 bars per meter width

Spacing:

mm

mm

mmh

mmAA

S

rebarcals

52.181

350

220110*22

52.181

4/12*06.6231000

/1000

2,

max =

==

==

Apply 12 C/C 180mm


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ii) Edge Strip

Reinforcement

=4.28[minimum reinforcement]


Spacing

Apply 8 C/C 230mm

B) For Support Moment

Depth check

!.......77128906.4633.11*1000*295.0

/10*09.7

**

26

OKmmmmmmKN

fb

Md

cd

f =min

As, cal= 0.00486*1000*77mm= 373.86mm2

31.34/12

86.3734/

12#2

2

2, ===

mmDAbars cals

Apply 4 12 bars per meter width


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mm

mm

mmh

mmAA

S

rebarcals

220

350

220110*22

51.302

4/12*86.373

1000

/

1000

2,

max =

==

==

Provide 12

C/C 220mm

3.2 Ground floor slab

The design of ground floor slab is similar to that of the other floor slab. In theground floor slab design we have two alternatives. The first alternative is that two

lay the slab on the ground when the stability & strength of the ground soil is good.

Theether alternative is to suspend the ground floor slab on the tie-beames like the

other floor.

In our case we madethe slab lay on the soil for our academic interest

Slab depth determination

The minimum slab depth determined from minimum depth require for deflection

according to EBCS 2 this depth has to be checked for flexural requirement after

the computation of bending moment

)4006.04.0(

ykfd +=

ba

le

Where fyk = the characteristics strength of the rebar

Le = effective span &for two way slab the

Be = constant from table

Depth for cantilever slab

One way slab

Le = 2.15m

Ba = 10


29/182

d =10

21500.85= 182.75mm

Clear louver for internal structure

D = 182.75mm +15+7=204.75mm

Slab loading

The loads involved on designing the slab are

1. Dead weight of slab & partition wall carried by the slab

2. Live load to each floor is obtained according to the function of the room as

specified EBCS 1-1995 table.

Slab design moment (for slab)

The bending moment that are required for the determination of reinforcement

can be computed for the design loads computed for each panel in previoussection according to EBCS 2 1995

Check for slab depth

For flexure

The slab total depths have provided was from the requirement of deflection.

This depth has to be checked for flexure requirement has to be checked for

flexure requirement

mfcdbmd m= max moment in the slab

M

fcd= concrete strength

b= liner width taken

for shear

The shale is checked for the highly loaded panel w/c is --------------------

Slab reinforcement

the area of steal required for a section with given the acting moment ,the depth

&width can be calculated according limit design aids for reinforced concrete

member issued by BDE km= 2bdm ,k3 is value from table ab=

d

ksmd


30/182

3.3. Typical Floor Slab

a) Pannel-1

One way end span ( which can be treated as beam ) a=24

from serviceability requirement

mmlef

da

yk170

24

4800*)

400

300*6.04.0()

4006.04.0( =+=+>=

Use d=170mm

b) Pannel-2

One way interior span a=28

mmlef

da

yk

57.17828

5000*)400

300*6.04.0()4006.04.0( =+=+>=

Use d=179mm

c) Pannel-3

Two way end span

Lx=2150-400/2=1950mm=Le

(Assuming a 400mm width strong band along the unsupported side)

Ly=2700mm

mmL

L

x

y39.1

1950

2700==

1.36)12

39.12(*)3040(3039.1

21

239.1

3040

30=

+===>=

=

xmmx

mmlef

da

yk914.45

1.36

1950*)

400

300*6.04.0()

4006.04.0( =+=+>=

Use d=46mm

Ly/Lx a2 30

1.02 ?1 40

Ly/Lx a

2 30

1.39 ?

1 40


31/182

d) Pannel-4

Two way end span

Lx=4800mm=Le

Ly=4900mm

mmL

L

x

y021.1

4800

4900==

79.39)12

021.12(*)3040(3039.1

21

2021.1

3040

30=

+===>=

=

xmmx

mmlef

da

yk54..102

79.39

4800*)

400

300*6.04.0()

4006.04.0( =+=+>=

Use d=46mm

e) Pannel-5

Two way end span

Lx=4900mm=Le

Ly=5000mm

mmLL

x

y02.1

49005000 ==

8.39)12

02.12(*)3040(3039.1

21

202.1

3040

30=

+===>=

=

xmmx

mmlef

da

yk65.104

8.39

4800*)

400

300*6.04.0()

4006.04.0( =+=+>=

Use d=105mm

f) Pannel-6

Two way end span

Lx=2300-400/2=1900mm=Le

Ly/Lx a2 30

1.02 ?1 40

Ly/Lx a2 30

1.42 ?1 40


32/182

(Assuming a 400mm width strong band

along one side)

Ly=2700mm

mm

L

L

x

y421.1

1900

2700==

75.35)12

421.12(*)3040(3039.1

21

2421.1

3040

30=

+===>=

=

xmmx

mmlef

da

yk18.45

75.35

1900*)

400

300*6.04.0()

4006.04.0( =+=+>=

Use d=46mm

g) Pannel-7

Two way end span

Lx=2900mm=Le

Ly=4800mm

mmL

L

x

y66.1

2900

4800 ==

4.33)12

66.12(*)3040(3039.1

21

266.1

3040

30=

+===>=

=

xmmx

mmlef

da

yk802.73

4.33

42900*)

400

300*6.04.0()

4006.04.0( =+=+>=

Use d=74

h) Pannel-8

Two way end span

Lx=2900mm=Le

Ly/Lx a2 30

1.72 ?1 40

Ly/Lx a2 30

1.66 ?1 40


33/182

Ly=5000mm

mmL

L

x

y724.1

2900

5000==

76.32)12

724.12(*)3040(3039.1

21

2724.1

3040

30=

+===>=

=

xmmx

mmlef

da

yk24.75

76.32

2900*)

400

300*6.04.0()

4006.04.0( =+=+>=

Use d=76mm

For one way slabs

The maximum effective depth so far was 170mm assuming 12 main

reinforcement bars and 15mm clear cover

D=170+12/2+15=191mm

Take D=200mm==> d=200-12/2-15=179mm

For one way slabs

The maximum effective depth so far was d=105mm assuming 12 main

reinforcement bars and 15mm clear cover

D=105+12/2+15=126mm

Take D=130mm==> d=130-12/2-15=109mm

3.2 LOADING

i. DEAD LOAD

SELF WEIGHT

130mmRCslab=0.13*25=3.25KN/m2

200mmRCslab=0.20*25=5KN/m2

30mm cement screed=0.03*23=0.69KN/m

20mm floor finish(terrazzo til,ceramic, or pvc)

20mm selling=0.02*23=0.46KN/m2

Partition wall with finishing distributed over the respective panels:

Pannel-1

((4.80+2.15)*0.2*14+(4.80+2.15)*0.04*23)*3=77.562KN


34/182

==> 2/52.715.2*80.4

562.77mKN

mm

KN=

Pannel-2

((5.00*0.2+2.15*0.1)*14+(5.00+2.15)*0.04*23)*3=70.764KN

==> 2/58.615.2*00.5

764.70mKN

mm

KN=

Panel-4

((4.80+1.45)*0.1*14+(4.80+1.45)*0.04*23)*3=43.5KN

==> 2/85.7190.4*80.4

5.43mKN

mm

KN=

Pannel-5

((4.90*0.1*14+(4.90*0.04*23)*3=34.104 KN

==> 2/392.190.4*00.5

104.34 mKNmm

KN =

Pannel-7

(0.2*14+*0.04*23)*2.90*3=32.364 KN

==> 2/325.280.4*90.2

364.32mKN

mm

KN=

Pannel-8

(0.1*14+0.04*23)*(2.90+5)*3=54.984 KN

==> 2/792.300.5*90.2

984.54mKN

mm

KN=

Note: wall along beams (on top of beam) and reasonable near beams are

intentionally left to be loaded later on the respective beams.

ii. LIVE LOAD

Residential building==>category A

I. Panel 3&6

Longer side unsupported two way slabs

(Strip method of analysis)2.70

2.15(1-)b

b

K1

(1-K1)

-K2 (1+K 2)


35/182

P=1.3(5.09)+1.6(2)=9.82 KN/m2

D=130mm, ==> d=130-12/2-15=109 mm

fcu=25Mpa ==>fcd=11.33 Mpa

fyk=300Mpa ==> fyk =260.87 Mpa

Let b=0.4m ==>=0.4/2.15=0.186

To apply at least minimum reinforcement bar in the longer direction

i.e. rmin=o.5/fyk=0.5/300=1.67*10-3

==>Amin= rmin*b*d=1.67*10-3*103*109=181.67mm2/m width

Using 8 bars:

Spacing S= ( )m

mm

mmh

A

a

s

s

69.276

250

300150*2267.181

4

8**1000

*1000

2

=

==

=

take S=250mm

Therefore 23

2

min 08.201250

10*4

8*

mms

ba

As

s ===

3

310*85.1

109*10

08.201

===

bd

As


36/182

We know that

=

yd

cd

cd

sd

f

f

fdb

M*

**

211

2

Rearranging terms

= *5.01***2

cd

yd

cd

yd

sd f

f

fdbf

f

M cd

mkNMsd

=

= 612.510*85.1*

33.11

87.2605.0110*85.1*33.11*109*1000*

33.11

87.260 332

==>select k1=0.5 since the panel is newly squared.

==>k1=0.5*9.82=4.91 kN/m2

mkNwl

kMys === 67.42

95.1*91.4*5.0

22

1 22

1

( )

( )

( ) ( )( )

mkNwb

M

kk

ys

=

+

=

+= 372.0

186.02186.0

15.2*82.9

67.4*2186.01

*5.02

21 2

2

2

2

12

==>k2=0.372*9.82=3.654 KN/m2

Check!

Mys=4.91*1.752/2-3.654*0.4*1.95=4.47 kN-mOK!

The maximum positive moment in the y-direction strips will be located atpoint of zero shear with y1 as the distance of that point from the free edge to

the zero shear location:

o.4*3.654-4.91*(y-0.4)=0

Y1=0.7m

The maximum positive moment, found at that location is:

Myf= k2 *0.4*( Y1-0.4/2)- k1 *((Y1-0.4)/2)2

Myf=3.654*0.4*(0.7-0.4/2)-4.91*((0.7-0.4)/2)2=0.51KNm

For latter reference in cutting of bars, the point of inflection is located a

distance y2 from the free edge:

4.91

Myf

Y

1

3.6

54

0.4


37/182

3.654*0.4*(y2-0.4/2)-(4.91/2)*(y2-0.4)2 =0

For the x-direction slab strips, the cantilever moment is:

(1-k1)*=(1-0.5)*9.82=4.91 KN/m2

The cantilever Mx=4.91*2.72/2=17.9 kNm

A ratio of negative to positive moments of 2.0 will be chosen here:Negative Mxs=17.9*2/3=11.93 KNm

Positive Mxf=17.9 *1/3=5.97 KN/m

The unit load on the strong band in the x-direction is:

(1+k2)*=(1+0.372)*9.82=13.473 KN/m2

So for the 0.4m wide band the load per meter is

0.4*13.473=5.39 KN/m

(Reaction from the stair is included here)

The cantilever negative and positive strong band moments are respectively:

Cantilever Mx=5.39*2.72/2=19.65 KN-m

Negative Mxs=19.65*2/3=13.10 KN-m

Positive Mxf=19.65 *1/3=6.55 KN-m

With negative moment of 14.32 kN and support reaction of 5.39*2.7/2=7.28

KN The point of inflection in the strong band is found as follows:

13.473+7.28x-5.39x2/2=0

II. Panel -6

Longer side unsupported two way slabs

(Solid slab analysis)

2.70

2.0(1-)b

b

K1

(1-K1)

-K2 (1+K

2)


38/182

P=1.3(5.09)+1.6(2)=9.82 KN/m2

D=130mm, ==> d=130-12/2-15=109 mm

fcu=25Mpa ==>fcd=11.33 Mpa

fyk=300Mpa ==> fyk =260.87 Mpa

Let b=0.4m ==>=0.4/2.15=0.186

To apply at least minimum reinforcement bar in the longer direction

i.e. rmin=o.5/fyk=0.5/300=1.67*10-3

==>Amin= rmin*b*d=1.67*10-3*103*109=181.67mm2/m width

Using 8 bars:

Spacing S=

==

==

mm

mmh

mmA

a

s

s

250

300150*22

69.27667.181

4

8**1000

*1000

2

Take S=250mm

Therefore 23

2

min 08.201250

10*4

8*

mms

baA ss ===

3

310*85.1

109*10

08.201

===

bd

As

We know that

=

yd

cd

cd

sd

f

f

fdb

M*

**

211

2

Rearranging terms


39/182

= *5.01*** 2

cd

yd

cd

yd

sdf

ffdb

f

fM cd

mkNMsd

=

= 612.510*85.1*

33.11

87.2605.0110*85.1*33.11*109*1000*

33.11

87.260 332

==>select k1=0.5 since the panel is newly squared.

==>k1=0.5*9.82=4.91 kN/m2

mkNwl

kMys === 98.32

8.1*91.4*5.0

22

1 22

1

( )

( )

( ) ( )( )

mkNwb

M

kk

ys

=

+

=

+= 326.0

21.022.0

2*82.9

981.3*22.01

*5.02

21 2

2

2

2

12

==>k2=0.326*9.82=3.2 KN/m2

Check!

Mys=4.91*1.62/2-3.2*0.4*1.8=3.981 kN-mOK!

The maximum positive moment in the y-direction strips will be located at

point of zero shear with y1 as the distance of that point from the free edge to

the zero shear location:

o.4*3.2-4.91*(Y1-0.4)=0

Y1=0.661m

The maximum positive moment, found at that location is:

Myf= k2 *0.4*( Y1-0.4/2)- k1 *((Y1-0.4)/2)2

Myf=3.2*0.4*(0.661-0.4/2)-4.91*((0.661-0.4)/2)2

=0.423 KN-m

For latter reference in cutting of bars, the point of inflection is located a

distance y2 from the free edge:

3.2*0.4*(y2-0.4/2)-(4.91/2)*(y2-0.4)2 =0

For the x-direction slab strips, the cantilever moment is:

(1-k1)*=(1-0.5)*9.82=4.91 KN/m2

4.91

Y1

3.

2

0.

4

Myf


40/182

The cantilever Mx=4.91*2.72/2=17.9 kNm

A ratio of negative to positive moments of 2.0 will be chosen here:

Negative Mxs=17.9*2/3=11.93 KNm

Positive Mxf=17.9 *1/3=5.97 KN/m

The unit load on the strong band in the x-direction is:(1+k2)*=(1+0.326)*9.82=13.02 KN/m2

So for the 0.4m wide band the load per meter is

0.4*13.02=5.21 KN/m

(Reaction from the stair is included here is 5.13KN-m)

The total load perimeter width on strong band is

5.13+5.21 =10.34 KN/m

The cantilever negative and positive strong band moments are respectively:

Cantilever Mx=10.34*2.72/2=37.69 KN-m

Negative Mxs=37.69*2/3=25.13 KN-m

Positive Mxf=37.69 *1/3=12.56 KN-m

With negative moment of 25.13 KN and support reaction of

10.34*2.7/2=13.96 KN

The point of inflection in the strong band is found as follows:25.13+13.96x-10.34x2/2=0

III. Panel -4

Type1 two way slab

Lx=4.8 and Ly=4.9

02.18.4

9.4==

x

y

L

L

1.0 1.1 1.02

1

xs 0.03

2

0.037 ?

xf 0.02

4

0.028 ?


41/182

ys=0.032 yf=0.024

x=0.032+(0.037-0.032)*((1.021-1)/(1.1-1))=0.0331

x=0.024+(0.028-0.024)*((1.021-1)/(1.1-1))=0.025

P=1.3DL+1.6LL

=1.3*(3.25+0.69+2*0.46+1.89)+1.6*2=11.923 KN/m2

Mxs=0.0331*11.923*4.82=9.093 KN-m

Mxf=0.025*11.923*4.82=6.868 KN-m

Mys=0.032*11.923*4.82=8.79 KN-m

Myf=0.024*11.923*4.82=6.593 KN-m

IV. Panel -5

Type 2 two way slab

Lx=4.9 and Ly=5.0

021.19.4

0.5==

x

y

L

L

ys=0.039 yf=0.029

x=0.039+(0.044-0.039)*((1.021-1)/(1.1-1))=0.04

x=0.029+(0.033-0.029)*((1.021-1)/(1.1-1))=0.03

P=1.3DL+1.6LL

1.0 1.1 1.02

1

xs 0.03

9

0.044 ?

xf 0.02

9

0.033 ?

1.5 1.75 1.655

xs 0.05

8

0.063 ?

xf 0.04

3

0.047 ?


42/182

=1.3*(3.25+0.69+2*0.46+1.392)+1.6*2=11.3276 KN/m2

Mxs=0.04*11.3276*4.92=10.879 KN-m ?

Mxf=0.03*11.3276*4.92=8.159 KN-m?

Mys=0.039*11.3276*4.92=10.61 KN-m

Myf=0.029*11.3276*4.92=7.89 KN-m

V. Panel -7

Type 2 two way slab

Lx=2.9 and Ly=4.8

655.19.2

8.4==

x

y

L

L

ys=0.039 yf=0.029

x=0.058+(0.063-0.058)*((1.655-1.5)/

(1.75-1.5))

=0.0611

x=0.043+(0.047-0.043)*((1.655-1.5)/

(1.75-1.5))

=0.046

P=1.3DL+1.6LL

=1.3*(3.25+0.69+2*0.46+2.325)+1.6

*2

=12.54 KN/m2

Mxs=0.0611*12.54*2.92=6.44 KN-m

Mxf=0.046*12.54*2.92=4.85 KN-m

Mys=0.039*12.54*2.92=4.11 KN-m

Myf=0.029*12.54*2.92=3.06 KN-m

VI. Panel -8Type 2 two way slab

Lx=2.9 and Ly=5.0

724.19.2

0.5==

x

y

L

L

ys=0.039 yf=0.029

1.5 1.75 1.72

xs 0.05 0.06 ?

xf 0.04 0.04 ?

x=0.058+(0.063-0.058)*((1.724-1.5)/

(1.75-1.5))

=0.0625

x=0.043+(0.047-0.043)*((1.724-1.5)/

(1.75-1.5))

=0.0466

P=1.3DL+1.6LL

=1.3*(3.25+0.69+2*0.46+3.792)+1.6*2

=14.45 KN/m2

Mxs=0.0625*14.45*2.92=7.6 KN-m

Mxf=0.0466*14.45*2.92=5.66 KN-m


43/182

Mys=0.039*14.45*2.92=4.74 KN-m Myf=0.029*14.45*2.92=3.52 KN-m


44/182

VII. Panels- 1,2&3


45/182

(One way cantilever slab)

W1=1.3*(5+0.69+2*0.46+7.52)+1.6*4KN/m=24.769KN/m

W2=1.3*(5+0.69+2*0.46+6.58.)+1.6*4KN/m=23.547KN/m

Assuming 20% moment redistribution:

2.045.66

45.66=

x==>x2=53.16 2.0

98.1

98.1=

x==>x2=1.584

Let us find maximum span moment, where shear force is zero.

Consider the following section:

BM

D

1.584

? ?

53.16KN-m

0.40mm

24.769

KN/m

4.8m

53.16 KN-m

RB1

RA

Mx

24.769

KN/m

20.347

KN/m

BM

D

1.98 KN-

m

40.71 KN-

m

34.46 KN-

m

66.45

KNm

5

m0.40

mm

4.8

m


46/182

Taking moment about point B

53.16+4.8RA=24.769*5.22/2

RA=58.531KN

RB1=70.268KN

Now, let x be the distance of zero shear form left edge in a span AB.

24.769*x=58.531

==>x=2.37m

Mx=(2.37-0.4)*58.531-

24.769*2.372/2

=45.744KN

Consider again the following section:

Taking moment about point B.

53.16+5RC=20.347*5 2/2

RC=40.236 KN

==>RB2=5*20347-40.236=61.5KN

RB=RB1+RB2=131.77KN

Let y be the distance of form right edge zero shear in a span BC.

5.0

m RC

RB1

Mx

20.347

24.769

KN/m

x

53.16 KN-m

xR

A

53.16 KN-m


47/182

20.347*y=40.236

==>y=1.98m

My=RC*y-20.347*y2/2

=40.236*1.98-20.347*1.982/2

=39.783 KN

After redistribution:

i. SECTION A-A

Cx=0.338+(0.325-0.338)*(1.256-1.2)/(1.3-

1.2)=0.331

Cy=0.172+(0.135-0.172)*(1.256-1.2)/(1.3-1.2)=0.1513

0.4

k=0.37

11.93 BMD1.584

45.744

39.783

53.16 KN-

m

5.97

k=0.2

33

2

1

4.8 5.0 2.7

Lx/Ly Cx Cy1.2 0.33 0.171.25 ? ?1.3 0.32 0.13

BMD1.584

46.059 KN-m 39.72

53.16 KN-

m

My

20.347

KN/m

y

53.16 KN-m

RC

4.187

-7.743-4.187

-4.187

0.628

4

0.649


48/182

Mxf=0+0.33*11.93=3.95

Myf=5.97+0.1513*11.93=7.775

ii. SECTION B-B

Cx=0.28+(0.314-0.28)*(1.02-1)/(1.1-1)=0.287

Cy=0.38+(0.374-0.38)*(1.02-1)/(1.1-1)=0.379

Mxf= Mxf+ CxM=6.868+0.287*(9.093-4.0314)=8.320

11.93

8.71

4.187

-7.7430.9232

-4.187

-4.0374 0.649

8.71

11.2711.27

10.61 11.9

39.852 9.852

9.093 10.61

k=0.208

0.4

k=0.3711.934.0374

6.868 7.89 KN-m

9.093 KN-m

5.97

k=0.2

3321

4.8 5.0 2.7

10.61 KN-m

Lx/Ly Cx Cy1 0.28 0.38

1.02 ? ?1.1 0.31 0.37

BMD


49/182

Myf= Myf+ CxM=6.593+0.379*(9.093-4.0314)=8.510

iii. SECTION C-C

iv. SECTION D-D

BMDk=0.208

3.06 3.52 KN-

m

4.11KN

mk=0.2

31

4.8 5.0

4.74 KN-

m

4.425KN

m

4.425KN

m

2


50/182

a) left side

Cx=0.38+(0.356-0.38)*(1.02-1)/(1.1-1)=0.375

Cy=0.28+(0.22-0.28)*(1.02-1)/(1.1-1)=0.268

Mxf= Mxf+ CxM=8.320+0.375*(8.79-6.44)=9.201

Myf= Myf+ CxM=8.510+0.268*(8.79-6.44)=9.140

b) Right side

Mxf= Mxf+ CxM=9.201+0.375*(8.79-0)=12.497

Myf= Myf+ CxM=9.14+0.268*(8.79-0)=11.496

v.

SECTION E-E

3.52 KN-

m4.8 5.0

0.628

4

0.3716

BMDk=0.208

3.06KN

m

4.11KN

mk=0

.2

321

4.74 KN-m

Lx/Ly Cx Cy1 0.38 0.28

1.02 ? ?1.1 0.35 0.22

7.9167

0.8733-1.4767

7.916

7


51/182

c) left side

Cx=0.28+(0.314-0.28)*(1.02-1)/(1.1-

1)=0.287

Cy=0.38+(0.374-0.38)*(1.02-1)/(1.1-

1)=0.379

Mxf= Mxf+

CxM=8.189+0.287*(10.879-7.6)=9.1

Myf= Myf+ CxM=7.89+0.379*(10.879-

7.6)=9.133

d) Right side

Mxf= Mxf+ CxM=9. 1+0.287*(10.879-0)=12.22

Myf= Myf+ CxM=9.133+0.379*(10.879-0)=13.256

Panel-5

One way cantilever slab P=1.3Gk+1.3gk+1.6qk , Gk=wall load

qk=2 ==>P2=1.6*2=3.2 KN/m

gk=(13.25+0.69+2*0.46)

==>P3=1.3*4.86=6.38 KN

P1=8.4*1.3=10.92 KN

M=0.3*10.92+0.42*9.518/2=4.0374

KN

R=0.4*9.518+10.92=14.73 KN

10.879 KN-

m

7.6 KN-m

7.6 KN-m

9.676

-1.2-2.076

9.676

BMDk=0.345

5.66 KN-m8.159 KN-m

k=0.2

AB

C

2.9 2.1

5

10.879 KN-m

0.6334

0.3716

0 KN-m

k=0.465

4.9

7.615

7.615

-7.615

0.3 0.7-

3.264

M

0.10

mm

24.769

KN

0.3

m

9.518 KN


52/182

4. Stair analysis and design

3.3 STAIR ANALYSIS

I. Lay out and effective depth

From serviceability requirement:

mmlef

da

yk170

24

4800*85.0)

4006.04.0( ==+>=

D=170+12/2+15=191mm

Putting on strong band to make the depth the same as the rest of slabs and avoid

tipping effect.

Load Area qk

(KN/m2)

general 2

stairs 3

balconies 4

1.80

m3.00

m

1.15

m

1.50m


53/182

Serviceability Check:

mmlef

da

yk25.106

24

3000*)

400

300*6.04.0()

4006.04.0( =+=+>=

D=106.25+12/2+15=127.25 mm

Take D=130mm ==> d=109 mm

01 565.26)3

5.1(tan

3

5.1tan ====>=

II. LOADING

A. Dead Load

1. Flight

Slab self weight + plastering mKNCos

/405.4565.26

23*03.025*13.0=

+=

steps mKN/625.1565.26cos

23*03.025*13.0=

+=

2cm marble thread + 2cm cement mortar =0.02*(23+27)=1KN/m

13cm marble riser + 2cm cement mortar =

mKN/5.03

)2327(*15.0*02.0*10 =

+=

Total load = 4.405+1.625+1+0.5= 7.53 KN/m

2. Landing

13cm RC slab = 0.13*25 = 3.25 KN/m

2cm marble = 0.02*27 = 0.54 KN/m

3cm cement screed = 0.03*23 = 0.69KN/m

3cm plastering = 0.03*23 = 0.69KN/m


54/182

Total = 5.17 KN/m

B. Live Load

=3KN/m2*1m(width) = 3KN/m

C. Design load

1. Flight=1.3*7.53+1.6*3=14.589KN/m

2. Landing=1.3*5.17+1.6*3=11.521KN/m

III. Analysis

1.14 KN/m

BMD

7.84 KN/m

7.62 KN/m9.43 KN/m

11.521

KN/m

14.589

KN/m11.521

KN/m

1.80m 3.00m 1.15m

21.28 KN

-22.49 KN5.13 KN 13.25 KN

15.61 KN

SFD

Reactio

n

A B C

5.13 KN 38.09

KN

34.53 KN

KN


55/182

Fig. SAP Output of Moment, Shear force, and Support Reactions diagrams

i. Design Loadsdesign moment

span BC=7.84 KN-m

support B=9.43 KN-m

design shear

span AB=15.61 KN

span BC=22.49 KN

ii. Check for Deflection

Where

W=Flight unfactored design load =7.53+3=10.53 KN/m

L=length of stair =3m

E=29GPa

0003.0*10*29*384

3*1000*53.10*59

4

max = =1.28mm

According to EBCS-2, 1995 sec. 5.2.2.

The final deflection shall not be exceed the value

where Le effective length

Le=3000 mm


56/182

mm15200

3000==

Here max=1.28mmVsd,max=22.49KN . . . .OK!

iv. Reinforcement Calculation

For Flight, span BC

Check for ductility

Depth check:

D = it is ok!

As,min= smin*b*d=0.5/300*1000*109=181.17mm2

b= 1000mm (unit width)

As,cal=bd


57/182

= [1- ]

[ 1-

As, cal= 0.00261*1000*109mm= 284.25mm2

Reinforcement

=2.51


Spacing:

Apply 12 C/C 260mm

v. Secondary Reinforcement

According to EBCS-2, 1995 sec. 7.2.2.2.

The ratio of the secondary reinforcement to the main reinforcement shall be at

least equal to 0.2.

Thus, the transverse reinforcement:

As,t=0.2*As=0.2*341.388=68.28mm2/m< As,mi=181.17mm2

use 10mm as=78.54mm2

spacing,s=as*b/As=78.54*1200/181.17=433.52mm

Use 10mm c/c 430mm


58/182

Fig Reinforcement Distribution design

vi. Load transform from stair to beam

Reaction force at support A,RA=5.13KN

Reaction force at support B, RB=38.09KN

Reaction force at support C, RC=34.53KN

5. Earth Quake Analysis

5.1. Mass, Mass-center and Center of Rigidity Determination

I. Mass

Weight of a typical floor

i) weight of walls = 2,082.02KN

ii)weight of beams =

5*0.4*0.27*23.1*25-

2.5*0.4*027*25

= 305.1

2*6*[1.95+4.5+2.5]*0.4*0.27*25

= 289.98

iii) slab and finishing

4*7.8*9.8*0.13*25 = 993.72

4*2.15*9.8*0.2*25 = 421.40

2*2.7*4.15*0.13*25 = 72.83

= 1487.95

Finishing = [19.9*23.1-

12*2.7]*0.07*23

= 687.94

Total = 2,175.89KN

iV. Columns:

30*13-0.13*2) *0.4*0.3*25 = 246.6KN


59/182

v. Stair = 2*2.4*(3*7.53+2.95*5.17) =

181.64KN

Total = 5,281.23KN Take

5,300KN

Weight of 5th floor [water tank

seat]

Slab

2*0.11*0.3*5.3*25 = 87.45KN

Columns

2*4*0.3*0.4*(1/2) *1.5*25 =

18KN

beams

4*(0.4-0.11)*0.4*[2.7+ (4.9-

0.8)]*25

= 78.88KN

Total =184.33KN

Take 185KN

Weight of ground floor

Wall

2*3*[4.5*0.2**3+4.7*0.2*3]*14

+2*4*0.10*14*[4.5+2.5]

+2*2*0.2*14*[4.5+2.5]=

620.48KN

column

30*(2.5/2+3/2)*0.3*0.4*25 =

247.5KN

Stair half stair

[181.64] = 90.82KN

Total = 958.8KN take

960KN

Weight of foot story

Columns

[8(2/2) +(3/2)30]*0.4*0.3*25 =

195KN

Beams (top tie beams)

[5*23.1+12*(1.95+4.5+2.5]*0.4

*0.4*25 = 891.6KN

Roofing= (GIS + Zigba purlin

and truss]

purlin +EGA = 0.-76KN/m

truss 3.08 per 1.5m

= 3.08*2*[17+3+8*71]*0.076

= 240.056KN

Total = 1290.656 take

1291KN


60/182

II. Mass Center

Only typical floor mass center determination is presented here, others are done in

similar fashion and results are used.


61/182


62/182


63/182


64/182


65/182

5.2. D-value computation

Assuming 30X40 columns and 40X40 beams with similar E value

Typical column typical beam

Icy= (0.4*0.33)/12=0.9*10-3===kcy= (0.9*10-3)/Lc and Icx=

(0.3*0.43)/12=1.6*10-3Kcx= (1.6*10-3)/Lc

Ib=(0.4*0.43)/12=2.133*10-3kb=(2.133*10-3)/Lb

k

ka

kc

kbk

+=

=

2

2

D=13.8

16

FRAME ON AXIS A-A

D=4.74

5

D=10.1

24

Kb=7.90

1

40

3

0

=

4.

4

4

4

x

y

40

4

0

y

k

ka

kc

kbk

2

5.0

+

+=

=

D=10.1

24

D=10.1

24

D=10.1

24

Kb=4.44

4

Kb=4.26

7

Kc=4.

5 a=0.527D=2.372Kb=7.90

1

=2.23

D=1.93

a=0.536

=1.234

a=0.661D=2.378

=2.42

D=2.6a=0.721

=3.38

a=0.67D=2.009

=4.056

D=1.776Kb=4.26

7

=2.904

a=0.426

D=1.27

7

=1.481

Kc=3.

6

Kc=3

Kb=4.4

44

a=0.592

Kb=7.90

1

1.277

1.277

1.277

1.277

1.776

1.776

1.776

1.776

2.009

2.009

2.009

2.009

a=0.426

D=1.27

7

=1.481

Kb=4.4

44

D=1.776Kb=4.26

7

=2.904

a=0.592

D=1.93

a=0.536

=1.234

a=0.661D=2.378

=2.42

1.2771.776

1.277

1.277

1.277

1.776

1.776

1.776


66/182

D=10.1

24

11.57

FRAME ON AXIS B-B

11.57

11.57

D=11.5

7

D=13.8

16

kb=7.9

01

D=4.745

D=1.27

7

D=1.77

6 D=2.00

9

D=2.378

=2.42

a=0.661D=1.93

Kc=3.6k=1.234

a=0.536

D=1.277

Kb=4.44

4

kc=3a=0.426

D=1.776Kb=4.26

7

=2.904a=0.592 D=2.019

Kb=4.26

7

kc=6a=1.014

D=2.6

=3.38

a=0.721

kc=4.5

kb=7.9

01kc=3

=2.23

a=0.527

D=2.37

2

=1.481 a=0.455kc=6

Kb=7.90

1

=4.056

kb=4.2

67

5 6

Kb=4.44

4

kc=3a=0.426

=1.481

D=1.277D=1.776

Kb=4.26

7

=2.904a=0.592

D=2.378

=2.42

a=0.661 D=1.93

Kc=3.6k=1.234

a=0.536

D=1.776

D=1.776

D=1.776

D=1.77

6

D=1.776

D=1.776

D=1.776

D=1.27

7

D=1.27

7

D=1.27

7

D=1.27

7

D=1.27

7

D=1.27

7

D=1.27

7

D=2.019

D=2.019

D=2.019

4


67/182

D=9.574

D=8.6

D=8.6

D=8.6

D=9.57

D=13.8

FRAME ON AXIS C-C

5 64

D=0.00

D=1.27

7

D=1.77

6 D=1.73

4

D=2.378

=2.42

a=0.661D=1.93

Kc=3.6k=1.234

a=0.536

D=1.277

Kb=4.44

4

kc=3a=0.426

D=1.776

Kb=4.26

7

=2.420

a=0.592

D=1.734

Kb=7.90

1

=2.739

a=0.578

D=2.596

=3.38

a=0.721

kb=7.901

a=0.57

8

=1.481

=2.739kb=4.2

67

D=1.27

7

D=1.27

7

D=1.27

7

D=1.77

6

D=1.77

6

D=1.77

6

1.247

1.247

=1.422

a=0.416

D=1.247

D=1.77

6

D=1.77

6

D=1.77

6

D=1.77

6

D=1.27

7

D=1.27

7

D=1.27

7

D=1.27

7

D=2.378

=2.42

a=0.661

D=1.776

Kb=4.26

7

=2.420

a=0.592

D=1.93

Kc=3.6k=1.234

a=0.536

D=1.277

Kb=4.44

4

kc=3

=1.481

a=0.426


68/182

D=17.5

89

FRAME ON AXIS 1-1 and 2-2

3.092

D=0

11.766

D=11.7

66

D=11.7

66

2.7911.546

Kc=5.333

=0.816

a=0.29D=1.546kb=4.354

Kc=6.4=0.68

a=0.44

=2.196a=0.523

D=2.791Kb=7.356

=1.83a=0.608D=3.893

=2.759a=0.580

D=3.092Kb=7.356

=2.99a=0.651D=4.167

1.546 2.791 3.092

1.546

1.546

1.546

1.546

1.546

1.546

3.0922.791

3.0922.791

Kc=5.333

=0.816

a=0.29

kb=4.354

Kc=6.4=0.68

a=0.44

D=11.7

66

D=11.7

66

D=2.818 D=2.818


69/182

FRAME ON AXIS 3

kc=8

D=17.5

89

D=11.7

66

D=8.44

D=11.7

66

D=11.7

66

D=11.7

66

D=11.7

66

=1.004

a=0.33

4D=2.67

4

D=1.54

6

2.00

2.90 4.90

kc=5.3

33=0.816

a=0.29D=1.54

6kb=4.354kc=6.4

03a=0.44

D=2.81

8

=0.68

=2.196

a=0.52

3D=2.79

1kb=7.356=1.82

a=0.60

8D=3.89

3

=2.759

a=0.58D=3.09

2

=2.299

a=0.65

1D=4.16

7

kc=6.4

03a=0.44

D=2.81

8

=0.68

kc=5.3

33=0.816

a=0.29D=1.54

6kb=4.354

D=1.54

6

D=1.546

D=1.54

6

D=1.54

6

D=1.54

6

D=1.54

6

D=1.54

6

D=1.54

6

D=2.79

1

D=3.09

2

D=2.79

1

D=3.092

D=2.79

1

D=3.09

2

D=2.79

1

D=3.09

2


70/182


71/182


72/182


73/182

(10.124)

(10.124)

(10.124)

(9.574)

1 2 3 3 2 1

A

(11

.7

6

6)

(11

.7

66

) (11

.7

66

)(11

.7

669

(11

.7

6

6

1.

54

6 1.

54

6

2.

79

1 2.

79

1

1.

54

6

2.

79

1

3.

09

2 3.

09

2 3.

09

2

2.7

9

1 2.7

9

12

.7

9

1

1.

54

6 1.

54

61.

54

6

(11

.7

66)9

1.77

62.00

9

2.00

91.27

622.

3

17.

5

15.

2

9.84.8

1.27

7

1.27

7

1.9

3

1.77

6

1.776

2.00

9

2.00

91.77

6

1.77

6

1.73

4

1.73

4

1.27

7

1.277

1.277

1.277

D 4.9

C 7.8

7.8

B 10.7

A 15.66

(10.124)

4. 4th floor story


74/182


75/182

Shear center

1.

a. Basement floor

=

x

iy

sD

xDX

6.11589.17*6

3.22*589.175.17*589.172.15*589.178.9*589.178.4*589.170*589.17=

+++++=

=

x

ix

sD

yDY

8.7816.13*5

6.15*816.1370.10*816.138.7*816.139.4*816.130*816.13=

++++=

b. Ground floor

Xs

766.11*6

3.22*766.115.17*766.112.15*766.118.9*766.118.4*766.110*766.11 +++++=

Ys 8.7)574.957.11*2124.10*2(

6.15*124.107.10*57.118.7*574.99.4*57.110*124.10=

++++++

=

c. Typical 1st 2nd and 3rd floor

Xs

766.11*6

3.23*766.115.17*766.112.15*766.118.9*766.118.4*766.110*766.11 +++++=

Ys 8.760.82*57.11124.10*2

6.15*124.107.10*57.118.7*60.89.4*57.110*124.10=

++++++

=

d. 4th floor

Xs

6.11766.11*6

3.22*766.115.17*766.112.15*766.118.9*766.118.4*766.110*766.11=+++++=

Ys 8.7)574.957.11*2124.10*2(

6.15*124.107.10*57.118.7*574.99.4*57.110*124.10=

++++++

=

e. 5th floor


76/182

Xs 5.1244.8*2

3.22*05.17*02.15*44.88.944.88.4*00*0=

++++++=

Ys m8.7745.4*4

6.15*745.47.10*745.48.7*09.4*745.4000*745.4=

++++=

5.3. Story shear distribution

Method response spectrum

since the building is regular both in plan and in elevation static method

of frame analysis will be adopted here

We know that seismic base shear in the main direction is given by:

Fb = Sd(Ti)w

Ti = CiH3/4, Ci = 0.075 (RC moment resisting frame)

= 0.075(195)3/4 H = 19.5m above the base

= 0.696

Sd(To) = , =oI, o=0.05,for zone 2

=0.05*1 I=1, for ordinal building

=0.05

= 1.2S/T2/32.5 ,S=1.5 for sub soil class-c

= !5.2292.23/2)696.0(

5.1*2.1ok=

g= g okDkekN0.70

go=0.2 for frame system

Kd = 1.5 for DC"M"

Ke=1 regular in elevation

KN= 1 for frame

= g =0.2*1.5*1*1=0.30.7 0k!


77/182

Sd (Ti)=0.05*2.292*0.3=0.03440****************

W= Gk(no need for live load allowance)

Fb = sd (T1)*w= 0.0344*23,636= 813.08 kN

Ft = 0.07 T1Fd= 0.07*0.696*813.08 = 39.61 kN

Fb-Ft = 775,47 kN

Story Wi(KN

)

hi(m

)

wihi Fi(KN)

5th 185 19.5 3607.5 11.60+39.6

1

Roof 1,291 17.5 22,592.5

72.63

4th 5,300 14.5 .76,850 247.0543rd 5,300 11.5 60,950 195.942nd 5,300 8.5 45,050 144.831st 5,300 5.5 29,150 93.71Ground 960 2.5 2,400 7.72

23,63

6

240.600 773.484

5.4. Lateral load distribution among frame elements

Story shear in each of the floors is distributed to each of 2D-frames

according to their stiffness in the following tables. In addition to actual

eccentricity accidental torsion effect is also considered

+

+=22

1

1

)()(

1

xDyyDx

yeD yxx

++=22

1

2

)()(1 xDyyDx

yeDyx

x

+

+=22

1

1

)()(

1

xDyyDx

yeD xyy

+

+=22

2

2

)()(

1

xDyyDx

xeD xyy

Qy=),max(*

VD

21

yy

yyyD

Qx=

),max(*VD

21xx

xx

xD

Eyi=ey0.05Lx


78/182

Exi=ex0.05Ly

Where:

Dx, Dy: d- value of frames in the X-

Y directions

X, Y: co-ordinate distance from

origin of the axis

YX, : Distance of frame from

shear center

ey y-direction eccentricity

ex x-direction eccentricity


79/182

1) 5th

Floor

X-direction

Qx6= 51.21KN

Axis Dx y Dx2

ax1 ax2 Q6x

A 1.380 0.000 7.800 83.959 1.205 0.795 15.317

B 1.400 4.900 2.900 11.774 1.076 0.924 13.878

C 0.000 7.800 0.000 0.000 1.000 1.000 0.000

B 1.400 10.700 -2.900 11.774 0.924 1.076 13.878

A 1.380 15.600 -7.800 83.959 0.795 1.205 15.317

5.560 191.466 58.390

eui1 eui2 eli1 eli2

x 11.15 12.5 -1.35 23.1 1.155 -1.155 -0.195 -2.505

y 7.8 7.8 0 19.1 0.955 -0.955 0.955 -0.955

Y-direction

Qy6= 51.21KN

Axis Dy x xb Dyxb2

ay1 ay2 Q6y

1 0 0 12.5 0 0.983 0.777 0.000

2 0 4.8 7.7 0 0.989 0.863 0.000

3 0.718 9.8 2.7 5.23422 0.996 0.952 25.509

3 0.718 15.2 -2.7 5.23422 1.004 1.048 26.837

2 0 17.5 -5 0 1.007 1.089 0.000

1 0 22.3 -9.8 0 1.014 1.175 0.000

1.436 10.468 52.346

Total Eccentricity

Action axis

mass

center

shear

centerActual

eccentricity Li

Accidental Eccentricity


80/182

2) 4thFloor

X-direction

Qx6= 72.63 KN

Axis Dx y Dx2

ax1

ax2

Q6x

A 13.616 0.000 7.800 828.397 1.228 0.772 18.119

B 13.614 4.900 2.900 114.494 1.085 0.915 16.002

C 12.574 7.800 0.000 0.000 1.000 1.000 13.624

B 13.614 10.700 -2.900 114.494 0.915 1.085 16.002

A 13.616 15.600 -7.800 828.397 0.772 1.228 18.119

67.034 1885.782 81.865

eui1 eui2 eli1 eli2

x 11.15 12.5 -1.35 23.1 1.155 -1.155 -0.195 -2.505

y 7.8 7.8 0 19.1 0.955 -0.955 0.955 -0.955

Y-direction

Qy6= 72.63 KN

Axis Dy x xb Dyxb2

ay1

ay2

Q6y

1 0.8616 0 12.5 134.625 0.994 0.926 12.035

2 0.8616 4.8 7.7 51.084264 0.996 0.954 12.062

3 0.8616 9.8 2.7 6.281064 0.999 0.984 12.090

3 0.8616 15.2 -2.7 6.281064 1.001 1.016 12.298

2 0.8616 17.5 -5 21.54 1.002 1.030 12.463

1 0.8616 22.3 -9.8 82.748064 1.005 1.058 12.807

Action axis

mass

center

shear

centerActual

eccentricity Li

Accidental Eccentricity Total Eccentricity


81/182

3) 3rdFloor

X-direction

Qy5= 247.054 KN

Axis Dx y Dx2

ax1

ax2

Q6x

A 13.616 0.000 7.800 828.397 1.218 0.782 64.491

B 12.668 4.900 2.900 106.538 1.081 0.919 53.247

C 10.980 7.800 0.000 0.000 1.000 1.000 42.687

B 12.668 10.700 -2.900 106.538 0.919 1.081 53.247

A 13.616 15.600 -7.800 828.397 0.782 1.218 64.491

63.548 1869.871 278.162

eui1 eui2 eli1 eli2

x 11.15 11.6 -0.45 23.1 1.155 -1.155 0.705 -1.605

y 7.8 7.8 0 19.1 0.955 -0.955 0.955 -0.955

Y-direction

Qx5= 247.054 KN

Axis Dy x xb Dyxb2

ay1

ay2

Q6y

1 0.8616 0 11.6 115.936896 1.019 0.956 41.979

2 0.8616 4.8 6.8 39.840384 1.011 0.974 41.646

3 0.8616 9.8 1.8 2.791584 1.003 0.993 41.300

3 0.8616 15.2 -3.6 11.166336 0.994 1.014 41.743

2 0.8616 17.5 -5.9 29.992296 0.990 1.023 42.105

1 0.8616 22.3 -10.7 98.644584 0.982 1.041 42.862


Action axis

mass

center

shear

centerActual

eccentricity Li


82/182

4) 2ndFloor

X-direction

Qy5= 195.94 KN

Axis Dx y Dx2

ax1

ax2

Q6x

A 13.616 0.000 7.800 828.397 1.218 0.782 51.148

B 12.668 4.900 2.900 106.538 1.081 0.919 42.230

C 10.980 7.800 0.000 0.000 1.000 1.000 33.855

B 12.668 10.700 -2.900 106.538 0.919 1.081 42.230

A 13.616 15.600 -7.800 828.397 0.782 1.218 51.148

63.548 1869.871 220.612

eui1 eui2 eli1 eli2

x 11.15 11.6 -0.45 23.1 1.155 -1.155 0.705 -1.605

y 7.8 7.8 0 19.1 0.955 -0.955 0.955 -0.955

Y-direction

Qx5= 195.94 KN

Axis Dy x xb Dyxb2

ay1 ay2 Q6y

1 0.8616 0 11.6 115.936896 1.019 0.956 33.293

2 0.8616 4.8 6.8 39.840384 1.011 0.974 33.030

3 0.8616 9.8 1.8 2.791584 1.003 0.993 32.755

3 0.8616 15.2 -3.6 11.166336 0.994 1.014 33.107

2 0.8616 17.5 -5.9 29.992296 0.990 1.023 33.394

1 0.8616 22.3 -10.7 98.644584 0.982 1.041 33.994

Total Eccentricity

Action axis mass center

shear

centerActual

eccentricity Li



83/182

5) 1stFloor

X-direction

Qy5= 144.83 KN

Axis Dx y Dx2

ax1

ax2

Q6x

A 13.616 0.000 7.800 828.397 1.218 0.782 37.807

B 12.668 4.900 2.900 106.538 1.081 0.919 31.215

C 10.980 7.800 0.000 0.000 1.000 1.000 25.024

B 12.668 10.700 -2.900 106.538 0.919 1.081 31.215

A 13.616 15.600 -7.800 828.397 0.782 1.218 37.807

63.548 1869.871 163.067

eui1 eui2 eli1 eli2

x 11.15 11.6 -0.45 23.1 1.155 -1.155 0.705 -1.605

y 7.8 7.8 0 19.1 0.955 -0.955 0.955 -0.955

Y-direction

Qx5= 144.83 KN

Axis Dy x xb Dyxb2

ay1 ay2 Q6y

1 0.8616 0 11.6 115.936896 1.019 0.956 24.609

2 0.8616 4.8 6.8 39.840384 1.011 0.974 24.414

3 0.8616 9.8 1.8 2.791584 1.003 0.993 24.211

3 0.8616 15.2 -3.6 11.166336 0.994 1.014 24.471

2 0.8616 17.5 -5.9 29.992296 0.990 1.023 24.683

1 0.8616 22.3 -10.7 98.644584 0.982 1.041 25.127

Action axis

Total Eccentricity

mass center

shear

centerActual

eccentricity Li



84/182

6) GroundFloor

X-direction

Qy5= 93.71 KN

Axis Dx y Dx2

ax1

ax2

Q6x

A 13.616 0.000 7.800 828.397 1.224 0.776 23.971

B 12.668 4.900 2.900 106.538 1.083 0.917 19.740

C 12.574 7.800 0.000 0.000 1.000 1.000 18.088

B 12.668 10.700 -2.900 106.538 0.917 1.083 19.740

A 13.616 15.600 -7.800 828.397 0.776 1.224 23.971

65.142 1869.871 105.510

eui1 eui2 eli1 eli2

x 11.15 11.6 -0.45 23.1 1.155 -1.155 0.705 -1.605

y 7.8 7.8 0 19.1 0.955 -0.955 0.955 -0.955

Y-direction

Qx5= 93.71 KN

Axis Dy x xb Dyxb2

ay1 ay2 Q6y

1 0.8616 0 11.6 115.936896 1.019 0.956 15.923

2 0.8616 4.8 6.8 39.840384 1.011 0.974 15.797

3 0.8616 9.8 1.8 2.791584 1.003 0.993 15.666

3 0.8616 15.2 -3.6 11.166336 0.994 1.014 15.833

2 0.8616 17.5 -5.9 29.992296 0.990 1.023 15.971

1 0.8616 22.3 -10.7 98.644584 0.982 1.041 16.258



shear

centerActual

eccentricity Li


85/182

7) Basment Floor

X-direction

Qy5= 7.72 KN

Axis Dx y Dx2

ax1

ax2

Q6x

A 20.660 0.000 7.800 1256.954 1.236 0.764 1.908

B 20.660 4.900 2.900 173.751 1.088 0.912 1.679

C 20.660 7.800 0.000 0.000 1.000 1.000 1.544

B 20.660 10.700 -2.900 173.751 0.912 1.088 1.679

A 20.660 15.600 -7.800 1256.954 0.764 1.236 1.908

103.300 2861.410 8.718

eui1 eui2 eli1 eli2

x 11.15 11.6 -0.45 23.1 1.155 -1.155 0.705 -1.605

y 7.8 7.8 0 19.1 0.955 -0.955 0.955 -0.955

Y-direction

Qx5= 7.72 KN

Axis Dy x xb Dyxb2

ay1 ay2 Q6y

1 1.17 0 11.6 157.4352 1.018 0.960 1.309

2 1.17 4.8 6.8 54.1008 1.010 0.977 1.300

3 1.17 9.8 1.8 3.7908 1.003 0.994 1.290

3 1.17 15.2 -3.6 15.1632 0.995 1.012 1.303

2 1.17 17.5 -5.9 40.7277 0.991 1.020 1.313

1 1.17 22.3 -10.7 133.9533 0.984 1.037 1.334

Total Eccentricity


shear

centerActual

eccentricity Li



86/182

1. 2D Frame Analysis

Gravity loads on the frame are transferred from slabs (own weight of slabs

plus additional loads from partition walls and finishing), walls reasonably

close to support beams, and own weight of the frame elements. The lateral

loading is a result of analysis of earth quake loading which is distributedaccording to capacity of each frame. The loading of each frame is shown in

the following pages each of the frames are analyzed using SAP 2000

For sap analysis three combinations were applied.

Comb 1 : 1.3DL+1.6LL

Comb 2 : 0.75(1.3DL+1.6LL)+EQleft

Comb 2 : 0.75(1.3DL+1.6LL)+EQright

The output is shown only for severe maximum and sever minimum, which

were used as basis for design of the frame elements. Hence output shows

the envelope for desired action.

For sake of clarity and readability the loading and analysis results are

presented graphically.

7. Beam analysis and Design

7.1 Analysis

I. Load Transfer to Beams

Coefficient and other conventional load transfer method is used for two way

slabs.

a. Panel-1,2,& 3

Reaction R=58.531KN on cantilever beam on axis 1-1



Reaction from two way one long edge unsupported on cantilever

beams.


87/182

i. Non-strong band portion

W1= (1-k1)w

R=

ii. Strong band portion

W2=(1+k2)*wR=

Reaction on beam 3-4 on axis A-A

R=k1w*(1-)*b-k2*w b =0.5*9.82*1.75-

0.372*9.82*0.4=7.13KN/m

b. Panel -6 type-1

Bvx=0.336+ (0.39-0.36)*(1.02-1)/(1.1-1)=0.336

Bvy=0.33

Vx=0.336*11.923*4.8=19.23KN/m

Vy=0.336*11.923*4.8=18.23KN/m

c. Panel -7 type-2

Vx,c=0.336*(0.39-0.36)*(1.02-1)/(1.1-1)=0.336

Ly/Lx Bvx1.0 1.001.021 ?1.1 0.36


88/182

Vx,d=____

Vy,c=0.36

Vy,d=0.24

Vy,c=0.336*11.3276*4.9=18.65KN/m

Vx,d=____

Vy,c=0.36*11.3276*4.9=20KN/m

Bvy,d=0.24*11.3276*4.9=20KN/m

d. Panel-8: Longer side unsupported two way slabs

-Reaction on beam AB on axis 3-3

i. Non-strong band portion

R=

ii. Strong band portion and reaction from stair

R=(10.34*2.7)/2=13.96

-Reaction on beam 3-3 on axis A-A

.

e. Panel-9 type -2 two way slab

Ly/Lx Bvx1.0 0.361.02 ?1.1 0.39


89/182

Vx,c=0.47+(0.5-0.47)*(1.655-1.5)/

(1.75-1.5)=0.49

Vx,d=____

Vy,c=0.36

Vy,d=0.24

Vx,c=0.49*12.54*2.9=17.77KN/m

Vx,d=____

Vy,c=0.36*12.54*2.9=13.092KN/m

Bvy,d=0.24*12.54*2.9=13.092KN/

m

f. Panel -10 type 2 two way slab

Vx,c=0.47+(0.5-0.47)*(1.724-

1.5)/(1.75-1.5)

=0.497

Vx,d=____

Vy,c=0.36

Vy,d=0.24

Vx,c=0.49*14.45*2.9=20.83KN/m

Vx,d=____

Vy,c=0.36*

14.45*2.9=15.09KN/m

vy,d=0.24*14.45*2.9=10.06K

N/m

g. Panel-5

R=0.4*9.518+10.92=14.727KN/m on beam A-B on axis 1-1

Load Transfer to Grade Beams

Slab on grade type of construction of ground floor slab is assumed and hence

loads on grade beam are from self and wall load.

a. Axis 1-1

1. Beam AB

Wall load

=(0.15*14+0.04*23)*3=9.06KN/m

Ly/Lx Bvx1.0 0.361.02 ?1.1 0.39


90/182

2. Beam BC

Wall load =9.06KN/m

b. Axis 2-2

1. Beam AB

Wall load

=(0.1*14+0.04*23)*3=6.96KN/m

2. Beam BC

Wall load

=(0.1*14+0.04*23)*3=9.96KN/m

c. Axis 3-3

Wall load

=(0.2*14+0.04*23)*3=11.19KN/m

d. Axis A-A

1. Beam 1-2

Wall load

=(0.15*14+0.04*23)*3=9.06KN/m

2. Beam2-3

Wall load=(0.15*14+0.04*23)*3=9.06KN/m

3. Beam3-3

Wall load

=(0.15*14+0.04*23)*3=9.06KN/m

e. Axis B-B

1. Beam 1-2, 2-3, and 3-3

No load

f. Axis c-c

1. Beam 1-2 and 2-3

Wall load

=(0.15*14+0.04*23)*3=9.06KN/m

Load transfer to beams of a typical floor

a. Axis 1-1

1. Cantilever beam

Reaction from panel -1 and panel

-2

= 58.531KN/m

2. beam A-B

Reaction from panel -5=

14.727KN/m

Shear transfer from panel -6 =

19.23KN/m

3. Beam B-C


8.73KN/m

b. Axis 2-2

1. Cantilever beam

Reaction from panel -2and panel -3

= 131.767KN/m

2. beam A-B


19.23KN/m


91/182


20KN/m

Total shear transfer

=39.23KN/m

Wall load =

(0.1*14+0.04*23)*3=6.96KN/m

3. Beam B-C


19.23KN/m


20KN/m

Total shear transfer

=28.182KN/m

c. Axis 3-3

1. Cantilever beam

Non strong band portion

Reaction from panel -3 =

40.236KN/m

Reaction from panel -4 = 6.63KN/m

Wall load =

(0.2*14+0.04*23)*3=11.16KN/m

Total =58.026KN/m

strong band portion


40.236KN/m

Reaction from panel -4 = 7.28KN/m

Wall load

=

(0.2*14+0.04*23)*3=11.16KN/

m

Total =58.676KN/m

2. Beam A-B

-Non strong band portion

Shear from panel -7 =

13.32KN/m

Wall load =11.16KN/m


6.63KN/m

Total =17.79KN/m

-strong band portion

Wall load = 11.16KN/m

Reaction from panel -8=

13.96KN/m

Total =25.12KN/m

Shear from panel -7 =

13.32KN/m

-Stair hole portion


Shear transfer from panel

-7=13.32KN/m

3. Beam B-C

Wall load 11.16KN/m

Shear transfer from panel -10=

10.06KN/m

d. Axis A-A


92/182

1. Cantilever beam

Wall load =

(10.2*14+0.04*23)*3=11.16KN/m

2. Beam 1-2


18.89KN/m

3. Beam 2-3


18.65KN/m

4. Beam 3-3


7.13KN/m


3.29KN/m

Total =10.42KN/m

e. Axis B-B

1. Cantilever beam


2. Beam 1-2


18.89KN/m


17.77KN/m

Total =36.66KN/m

Wall

load=(0.1*14+0.04*23)*3=6.96KN/

m

3. Beam 2-3


18.65KN/m


20.83KN/m

Total =39.48KN/m

f. Axis C-C

1. Cantilever beam


2. Beam 1-2

Shear transfer from two panel -9

slabs = 2*20.83=41.66KN/m


3. Beam 2-3

Shear transfer from two panel -10

slabs = 2*17.77=35.54KN/m


c. Load Transfer to Top-tier Beams

To be constructive and minimize confusion we analyzed the maximum case

of the wind loading and taken the reaction for analysis

A uniformly distributed load on top-tie beam on axis 1-1, 2-2, and 3-3 are

1.4KN, 11.18KN, and 0.02KN per 1.5m (truss spacing) respectively.


93/182

A uniformly distributed load on top-tie beams on A-A, B-B, and C-C are

11.31,1.72, and 2.39KN per 1.5m (truss spacing) respectively.

=+==>=+==>=+==>

=+==>=+==>=+==>

992.04.1*)1

9.23

5.1

9.23(....992.04.1*)1

9.23

5.1

9.23(....086.831.11*)1

7.20

5.1

7.20(

992.04.1*)1

9.23

5.19.23(.....992.04.1*)1

9.23

5.19.23(....992.04.1*)1

9.23

5.19.23(

Load transfer on 5th floor Beams

Type two way slabs Ly/Lx=4.9/2.7=1.815

Vx,d=0.48+(0.5-0.48)*(1.815-1.75)/(2-

1.75)=0.4852

Vy,d=33

i. Shear transfer to beam AB on axis 3-3

= vx=0.4852*17.375=8.43KN/m

ii. Shear transfer to beam 3-3 on axis A

= vy=0.33*17.375=5.734KN/m

Ly/Lx Bvx,d1.75 0.481.815 ?

2 0.5


94/182

Axis A-A Loading

AxisC-C Loading


95/182

Axis2-2 Loading

Axis3-3 Loading


96/182

AxisA-A Bending Moment


97/182

AxisC-C Shear Force and Bending Moment

A) Design for flexure

On Axis 2-2 (for grade beam)

b = 400mm, D = 400mm, cc=25mm, main=20mm, stirrup

=8mm ,d= 357mm

Check depth:

mmd

bf

Md

wcdsd

u

66.164

400*33.11*295.0

10*25.36

**

6

= Use d = 357mm

Beam span AB Msd= 21.86KNm

!.....295.0038.033.11*357*400

10*86.212

6

2ok

fbd

M

cd

sd

sd


98/182

Reinforcement2

6

5.187.260*357*1

10*14.0mm

dfk

MA

ydz

sd

s ===

2

minmin 6.285 mmuseAAAs =

#42.116 =

; provide 2 16 bars

For support AMsd =0 KN-m

No Negative reinforcement is required simply provideminimumreinforcement i.e. As=285.6mm2

# 42.116 = ; Provide 2 16 bars

For support B Msd=36.25

!.....295.0063.033.11*357*400

10*25.362

6

2 okfbd

M

cd

sd

sd


99/182

Shear force Max is at a distance of effective depth d from face of

each support

By similarity of triangles: design shear

( )KNVsd 64.17

82.1

357.082.194.211 =

=

( )KNVsd 05.32

08.3

357.008.325.362 =

=

( )KNVsd 10.16

68.1

357.068.145.203 =

=

Resistance shear

!05.32VsdKN48.404

357*400*11.33*0.2525.0

max OkKNVrd

bdfVrdctd

=>>=

==

Shear Capacity

bdkkfv ctdc 2125.0=

!.....1243.16.12 okdk ==

For beam section(1)

!.....21.1357*400

6.2855011 okk =+=

28.50357*400*243.1*1.1*03.1*25.0 ==cv

Vsd=17.64Vc>VsdThen provide minimum Design shear


100/182

==

==

mm

mmd

mmb

avfyk

S

300

5.178357*5.05.0

5.188400*4.0

300*4

*8*2

4.0

2

max

Provide 8 c/c 175mm

For beam section(2)

50.74357*400*243.1*63.1*03.1*25.0 ==cv

!.....215.1

357*400

08.4235011 okk =+=

56.52357*400*243.1*15.1*03.1*25.0 ==cv

Vsd=32.05Vc>Vsd;Then provide minimum Design shear

==

==

mm

mmd

mmb

avfyk

S

300

5.178357*5.05.0

5.188400*4.0

300*4

*8*2

4.0

2

max

Provide 8 c/c 175mm

For beam sections (3)

!.....215.1357*400

08.4235011 okk =+=

56.52357*400*243.1*15.1*03.1*25.0 ==cv



101/182

==

==

mm

mmd

mmb

avfyk

S

300

5.178357*5.05.0

5.188400*4.0

300*4

*8*2

4.0

2

max

Provide 8 c/c 175mm

For beam sections (4,5)

!.....21.1357*400

6.2855011 okk =+=

28.50357*400*243.1*1.1*03.1*25.0 ==cv


==

==

mm

mmd

mmb

avfyk

S

300

5.178357*5.05.0

5.188400*4.0

300*4

*8*2

4.0

2

max

Provide 8 c/c 175mm

On Axis 2-2 (for 1st -4th beam)

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

357mm

Check depth:

!61.486

400*33.11*295.0

10*57.316

**

6

NotOKmmd

bf

Md

wcdsd

u

>=

==

Shear Capacity

bdkkfv ctdc 2125.0= ; !.....109.16.12 okdk ==

For beam sections(1)

!.....273.1507*400

64.29555011 okk =+=

47.98507*400*09.1*73.1*03.1*25.0 ==cv


==

==

mm

mmd

mmb

avfyk

S

300

5.178357*5.05.0

5.188400*4.0

300*4

*8*2

4.0

2

max

Provide 8 c/c 175mm


!.....215.1507*400

22.6095011 okk =+=

46.65507*400*09.1*15.1*03.1*25.0 ==cv

Vsd=66.38Vs=Vsd-Vc=66.38-65.46=0.92KN


105/182

==

==

mm

mmd

mmb

avfyk

S

300

5.178357*5.05.0

5.188400*4.0

300*4

*8*2

4.0

2

max

mmv

fyddavS

s

21.1437692.0

87.260*507*100**===

S>Smax--------------------------Not OK!

Provide 8 c/c 175mm


!.....215.1507*400

22.6095011 okk =+=

46.65507*400*09.1*15.1*03.1*25.0 ==cv


==

==

mm

mmd

mmb

avfyk

S

300

5.178357*5.05.0

5.188400*4.0

300*4

*8*2

4.0

2

max

Provide 8 c/c 175mm


!.....207.1507*400

6.2855011 okk =+=

93.60507*400*09.1*07.1*03.1*25.0 ==cv



106/182

==

==

mm

mmd

mmb

avfyk

S

300

5.178357*5.05.0

5.188400*4.0

300*4

*8*2

4.0

2

max

Provide 8 c/c 175mm


!.....273.1507*400

67.29555011 okk =+=

4.98507*400*09.1*73.1*03.1*25.0

==cv

Vsd=225.04; Vs=Vsd-Vc=225.04-98.4=126.64KN

==

==

mm

mmd

mmb

avfyk

S

300

5.178357*5.05.0

5.188400*4.0

300*4

*8*2

4.0

2

max

mmv

fyddavSs

44.10464.126

87.260*507*100** ===

S

=

Use d = 357mm


107/182

Beam span AB Msd= 22.86KNm

!.....295.004.0

33.11*357*400

10*86.222

6

2ok

fbd

M

cd

sd

sd


108/182

# 7.120 = ; Provide 2 20bars

For support C,Msd=18.94

!.....295.003.033.11*357*400

10*94.182

6

2ok

fbd

M

cd

sd

sd


109/182

( )KNVsd 64.13

36.1

357.036.157.184 =

=

( )KNVsd 31.24

15.2

357.015.215.295 =

=

Resistance shear

!04.225VsdN48.404

357*400*11.33*0.2525.0

max OkKNKVrd

bdfVrdcd

=>>=

==

Shear Capacity

bdkkfv ctdc 2125.0=

!.....1243.16.12 okdk ==

For beam sections(1)

!.....222.1357*400

54.3505011 okk =+=

76.55357*400*243.1*22.1*03.1*25.0 ==cv


==

==

mm

mmd

mm

b

avfyk

S

300

5.178357*5.05.0

5.188

400*4.0

300*4

*8*2

4.0

2

max

Provide 8 c/c 175mm


!.....218.1357*400

73.5345011 okk =+=

26.54357*400*243.1*18.1*03.1*25.0 ==cv



110/182

==

==

mm

mmd

mmb

avfyk

S

300

5.178357*5.05.0

5.188400*4.0

300*4

*8*2

4.0

2

max

Provide 8 c/c 175mm


!.....218.1357*400

73.5345011 okk =+=

26.54357*400*243.1*18.1*03.1*25.0

==cv

Vsd=16Vc>Vsd;Then provide minimum Design shear

==

==

mm

mmd

mmb

avfyk

S

300

5.178357*5.05.0

G+4 SolidSlabSystem

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