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Words 12700

Pages 51

2012

INTRODUCTION

The objective of this project is to examine the techniques used in the design of a building.

The structural design of this lecture room & office building involves design of floor slabs, stairs, beams, columns, shear walls, foundation and analysis of frames.

The building is composed of solid slab, inclined solid slabs & slabs with hole. Live load and dead load analysis is made according to EBCS-1, 1995.After the minimum depth of slab for serviceability limit state were determined, the slabs were designed for partition load, floor finish using self-weight load and live loads according to EBCS-1, 1995 using different combinations and analyzed for the worst load condition. Stairs and landings were designed as one-way slab.

For the analysis of frames, the restrained conditions at the foundation level are assumed fixed. Loads acting on beams from slab reactions and walls directly resting on beams were added to self-weight of beams to find total load acting on beams. These were inserted and analyzed using SAPv14 analysis for five load combinations.

The design of beams and columns is done for the critical moment’s shears and axial loads obtained from the dead and live load combinations mentioned above of the selected axis. Beams were designed according to EBCS-2, 1995 provisions.

The size of the footing was determined from assumed bearing capacity of the soil; the thickness of the footing is determined from punching and wide beam shear. Finally the footing was designed for flexure using design tables.

The limit state design method has been adopted for the design of the components. The Ethiopian Building code of standards, EBCS-1, EBCS-2, EBCS-2 part two, EBCS-7, and EBCS-8 1995 design charts are used.

CHAPTER ONE

GENERAL

Design data and material properties

Concrete

Class I workmanship and ordinary loading condition is used.

Sub structure

Concrete grade C-25 Class I

Partial safety factor,

= 1.5 (Ordinary loading) [EBCS – 2, 1995 table 3.1]

Characteristic strength [compression],

[Tension]

[EBCS -2, 1995 table 2.4 & 2.5]

Superstructure

Concrete grade C-25

(Ordinary loading) [EBCS – 2, 1995 table 3.1]

[EBCS -2, 1995 table 2.4 & 2.5]

Reinforcement Steel

Design load

Where: - Fk= Characteristic load = Partial safety factor for loads

1.6 For live load [EBCS – 2, 1995 table 3.3]

1.3 For dead load

Design method and reference

Limit state design method is adopted in this project. The reference (codes and aids) used are:-

* EBCS – 1, 1995 * EBCS – 2,1995 * EBCS – 2,1995, part 2 * EBCS – 7,1995 * EBCS – 8,1995

CHAPTER TWO

STRUCTURAL LAYOUT

1st floor Plan

2nd floor plan

3rd floor plan

Top level

Tracer Slab

CHAPTER THREE

Wind load analysis

According to EBCS1, Art 3.5.2, the wind pressure acting on the external surface of a structure shall be: We=qref*Ce(Zi)*Cpe

And the internal pressure acting on the internal surfaces becomes,

Wi=qref*Ce(Zi)*Cpi

Where:qref=reference mean wind velocity Ce(Ze)= exposure coefficients for external pressure Ce(Zi)=exposure coefficients for internal pressure Ze,Zi= reference height defined in Appendix A of EBCS1

Where: ρ= air density in Kg/m3 Vref=reference wind velocity

Vref=Cdir*Ctemp*Calt*Vref,0

For Ethiopia Vref,0= 22m/s Cdir=Ctemp=Calt=1

Vref=Cdir*Ctemp*Calt*Vref,0

=1*1*1*22m/s

=22m/s

For altitude of 2,840m ρ=0.94 Kg/m3

Roof slab Panel layout PANEL-1 Reference height, Ze= h =19.342m e= b 2h whichever is smaller e=min (17.501, 2*19.342) = min. (17.501, 38.684) =17.501m Debre Birhan is in terrain category is a sub-urban area, therefore we can read Kt ,Z0 and Zminfrom table 3.2 [EBCS 1] Kt= 0.22 Z0=0.3m and Zmin=8m For Z= 19.342m >Zmin = 8m, the roughness coefficient Cr will be Cr(34.5m)=KrlnZZ0

Cr(19.342m)=0.22*ln19.3420.3

Cr(34.5) =0.9166

Considering Ct=1

Ce(34.5)=CrZ2Ct(Z)21+7KTCr(Z)Ct(Z)

Ce(34.5)=0.91662*121+7*0.220.9166*1

Ce(34.5)=2.251 Calculation of External & Internal(Cpe &Cpi) pressure coefficient Cpe = Cpe,1....................................for A<1 Cpe=Cpe,1+Cpe,10-Cpe,1log10A…………for 1<A<10 Cpe = Cpe,10….………………………….for A>10 =100 Θ=00

For closed buildings with internal partitions and openings

Cpi = 0.8 or –0.5 Segment | Area | Cpe1 | Cpe 10 | Cpe | We | Cpi | Wi | Wnet | F1 | 7.6571 | -2.25 | -1.3 | -1.41 | -0.722 | 0.8 | 0.4096 | -1.13165 | F2 | 7.6571 | -1.15 | -0.75 | -0.796 | 0.407598 | -0.5 | -0.25603 | 0.663626 | G1 | 15.314 | -1.75 | -1 | -1 | -0.51206 | 0.8 | 0.409646 | -0.9217 | G2 | 15.314 | -0.9 | -0.5 | -0.5 | -0.25603 | 0.8 | 0.409646 | -0.66567 | H1 | 284.21 | -0.75 | -0.45 | -0.45 | -0.23043 | 0.8 | 0.409646 | -0.64007 | H2 | 284.21 | -0.5 | -0.2 | -0.2 | -0.10241 | 0.8 | 0.409646 | -0.51206 | I | 30.629 | -0.35 | -0.35 | -0.35 | -0.17922 | 0.8 | 0.409646 | -0.58887 | J | 284.21 | -0.9 | -0.65 | -0.65 | -0.33284 | 0.8 | 0.409646 | -0.74248 |

Wweig=-0.63564KN/m2

Segment 1 Vref=22m/s

Direction 2 qref=0.22748KN/m2

Ce(Z)=2.251

Θ=900 b=17.501 d=35.98 e=17.501 Segment | Area | Cpe 1 | Cpe 10 | Cpe | We | Cpi | Wi | Wnet | F1 | 7.6571 | -2.25 | -1.45 | -1.343 | -0.7901 | 0.8 | 0.40965 | -1.19975 | F2 | 7.6571 | -2.25 | -1.45 | -1.543 | -0.7901 | 0.8 | 0.40965 | -1.19975 | G1 | 15.314 | -2.1 | -1.3 | -1.3 | -0.66567 | 0.8 | 0.40965 | -1.07532 | G2 | 15.314 | -2.1 | -1.3 | -1.3 | -0.66567 | 0.8 | 0.40965 | -1.07532 | H1 | 284.21 | -1.2 | -0.65 | -0.65 | -0.33284 | 0.8 | 0.40965 | -0.74248 | H2 | 284.21 | -1.2 | -0.65 | -0.65 | -0.33284 | 0.8 | 0.40965 | -0.74248 | I | 30.629 | -0.85 | -0.55 | -0.65 | -0.28163 | 0.8 | 0.40965 | -0.69128 |

Wweig=-0.76671KN/m2

Segment 1 Vref=22m/s

Direction 2 qref=0.22748KN/m2

Ce(Z)=2.251

Θ=00 b=35.98 d=17.501 e=35.98 e10=3.598 e2=17.99 e4=8.995 Segment | Area | Cpe 1 | Cpe10 | Cpe | We | Cpi | Wi | Wnet | F1 | 32.36 | -2.25 | -1.3 | -1.3 | -0.66567 | 0.8 | 0.40965 | -1.07532 | F2 | 32.36 | -1.15 | -0.75 | -0.75 | -0.38404 | 0.8 | 0.40965 | -0.79369 | G1 | 64.73 | -1.75 | -1 | -1 | -0.51206 | 0.8 | 0.40965 | -0.9217 | G2 | 64.73 | -0.9 | -0.5 | -0.5 | -0.25603 | 0.8 | 0.40965 | -0.66567 | H1 | 185.39 | -0.75 | -0.45 | -0.45 | -0.23043 | 0.8 | 0.40965 | -0.64007 | H2 | 185.39 | -0.5 | -0.2 | -0.2 | -0.10241 | 0.8 | 0.40965 | -0.51206 | I | 129.46 | -0.35 | -0.35 | -0.35 | -0.17922 | 0.8 | 0.40965 | -0.58887 | J | 185.39 | -0.9 | -0.65 | -0.65 | -0.46085 | 0.8 | 0.40965 | -0.8705 |

Wweig=-2.44382KN/m2

Both +ve and –ve values are given for zones F,G and H, the most critical value should be selected considering combinations with internal pressure coefficient.

Θ=900

=300

Segment | Area | Cpe 1 | Cpe 10 | Cpe | We | Cpi | Wi | Wnet | F1 | 32.36 | -2.25 | -1.45 | -1.45 | -0.74248 | 0.8 | 0.409646 | -1.15213 | F2 | 32.36 | -2.1 | -1.45 | -1.45 | -0.74248 | 0.8 | 0.409646 | -1.15213 | G1 | 64.73 | -2 | -1.3 | -1.3 | -0.66567 | 0.8 | 0.409646 | -1.07532 | G2 | 64.73 | -2 | -1.3 | -1.3 | -0.66567 | 0.8 | 0.409646 | -1.07532 | H1 | 185.39 | -1.2 | -0.65 | -0.65 | -0.33284 | 0.8 | 0.409646 | -0.74248 | H2 | 185.39 | -1.2 | -0.65 | -0.65 | -0.33284 | 0.8 | 0.409646 | -0.74248 | I | 129.46 | -0.5 | -0.5 | -0.5 | -0.25603 | 0.8 | 0.409646 | -0.66567 |

Maximum suction pressure& positive pressure will be taken for design.

The critical suction wind pressure is found to bewhen the wind is at Ө=00 Wnet (-ve) = -1.19975KN/m2

And critical positive wind pressure is found to bewhen the wind is at Ө=00 Wnet (+ve) = 0.663626KN/m2

Wind load for walls

b=35.48m, d=17m, h=17.30m

For buildings, whose height h is less than b, shall be considered to be one parts. (EBCS 1 1995 Appendix A 2.2) Therefore the building shall be considered to be one part

Reference height, Ze Ze= h =17.3m e= b 2h whichever is smaller e=min (35.48, 2*17.3m) = min. (35.48m, 34.6m) =34.6m for, d>e

Debrebrihan is in terrain category –iiisuburban areas, therefore we can read Kt ,Z0 and Zminfrom table 3.2 [EBCS 1] Kt= 0.22Z0=0.3m and Zmin=8m For Z= 17.3m>Zmin = 8m, the roughness coefficient Cr will be Cr(34.6)=KrlnZZ0

Cr(34.6)=0.22*ln34.60.3

Cr(34.6) =1.045 Cr(17.3)=KrlnZZ0

Cr(17.3m)=0.22*ln17.30.3

Cr(17.3) =0.52

Considering Ct=1

Ce(34.6)=CrZ2Ct(Z)21+7KTCr(Z)Ct(Z)

Ce(34.6)=1.0452*121+7*0.221.045*1

Ce(34.6)=2.7

Ce(17.3)=CrZ2Ct(Z)21+7KTCr(Z)Ct(Z)

Ce(17.3)=0.522*121+7*0.220.52*1

Ce(17.3)=1.0712 External pressure

We=Ce*Cpe*qref

Area of all zone are greater than 10m2so, we use Cpe10 d÷h=17÷17.3=0.98 qref=0.2275

Zone | | D | A | B | E | Cpe | | 0.8 | -1.0 | -0.8 | -0.3 | Ce | Cpe,h | 1.0712 | 1.0712 | 1.0712 | 1.0712 | | Cpe,b | 0.52 | 0.52 | 0.52 | 0.52 | We | We,h | 0.195 | -0.2437 | -0.195 | -0.07311 | | We,b | 0.09464 | -0.1183 | -0.09464 | -0.3549 | Area | A,b | 1,227.61 | 119.716 | 468.484 | 1,227.61 | We,force | We,b | 116.18KN | -14.16KN | -44.337KN | -435.68KN |

INTERNAL PRESSURE

Zi=17.30 for the bottom zone

Zi=34.60 for the top zone

Roughness coefficient

Cr(17.30)=KrlnZZ0 Cr(17.30)=0.22*ln17.300.3

Cr(17.30) =0.892 Cr(34.60)=KrlnZZ0

Cr(34.60)=0.22*ln34.600.3

Cr(34.60) =1.0445

Considering Ct=1 Ce(17.30)=CrZ2Ct(Z)21+7KTCr(Z)Ct(Z)

Ce(17.30)=0.8922*121+7*0.220.892*1

Ce(17.30)=2.169

Ce(34.60)=CrZ2Ct(Z)21+7KTCr(Z)Ct(Z)

Ce(34.60)=1.04452*121+7*0.221.0445*1

Ce(34.60)=2.6995

To get the maximum effect we use Cpi=+0.8 for –ve external pressure and -0.5 for +ve external pressure.

Zone | | D | A | B | E | Cpi | | -0.5 | +0.8 | +0.8 | +0.8 | Ce | Ce,h | 2.6995 | 2.6995 | 2.6995 | 2.6995 | | Ce,b | 2.169 | 2.169 | 2.169 | 2.169 | Wi | Wi,h | -0.30707 | 0.4913 | 0.4913 | 0.4913 | | Wi,b | -0.24672 | 0.39476 | 0.39476 | 0.39476 | Area | A,b | 1,227.608 | 119.716 | 468.484 | 1,227.61 | Wi,force | Wi,b | -302.875KN | 47.259KN | 184.939KN | 484.611KN |

CHAPTER FOUR

Roof truss design

Dupich roof

Live load [L.L]

The roof is categorized as H roof not accessible except normal maintenance, repair and painting.[EBCS-1,1995 section 2.6.3.4.5]

There for the characteristic dead load is

Qk=1KN,qk = 0.25KN/m2 -sloping roof

Wind load [W.L]

From the calculation done in the above section, we get the suction is found to be when the wind is found to be when the wind is at 900

Wnet (-ve) = -1.19KN/m2

And critical positive wind pressure is found to be when the wind is at Ө=00 Wnet (+ve) =0.6636 KN/m2

Design load

Dead load transferred from the roof cover (iron sheet in aluminum framing) will be approximated as:

Ribbed iron sheet= 0.008*77 =0.616KN/m2 Rig cap =0.008*77KN/m2 =0.616KN/m2

Total= 1.232KN/m2

Load combinations

After calculating loads on the structure it has to be decided that which combination of loading may be made to get the maximum stress in any member .However these combination of loads should involve their safety factors for the ultimate limit state for persistent and transient design situation.

Safety factor for favorable and unfavorable load condition indicated below are adopted from [EBCS-1, 1995 section 1.9.4.3 table 1.2]

I. Unfavorable Distributed live load + dead load 1.3DL+1.6LL II. Unfavorable Concentrated live load +dead load 1.3DL+1.6LL III. Favorable load combination Wind load + dead load 0.9DL +1.6 WL IV. Wind load +concentrated live load +dead load 1.3DL +1.35(LL+WL) V. Wind load +distributed live load +dead load 1.3DL +1.35(LL+WL) In the above combinations wind load will be checked for both max suction & positive pressure

Live load for sloping roof will be Distributed live load =0.25KN/m2 Concentrated live load =1KN (table 2.14of EBCS-1)

And Separated analysis should be carried for both cases.

Design of purlins Purlin c/c spacing = 1.49m Truss c/c spacing = 1.8m

Loading

Load transfer to purlin

Select EGA 500, 0.4

Dead load

Due to EGA sheet =1.232KN/m2*1.49m=1.835KN/m

Due to self-weight of purlin (Assume 60 x 40x3 mm purlin) =77KN/m3 x 0.06 x 0.04=0.1848KN/m Total DL=2.0198KN/m

Load perpendicular to the rafter =2.0198KN/m cos100 = 1.99KN/m

Load parallel to the rafter

=2.0198KN/m sin100 =0.35KN/m

Uniformly distributed live load per meter length of purlin.

=0.25*1.49=0.3725KN/m

Ө=100

WL

DL &LL

Load perpendicular to the rafter

=0.3725 cos100 =0.367KN/m

Load parallel to the rafter

=0.3725 sin100 =0.065KN/m

Concentrated live load Load perpendicular to the rafter

=1KN cos100 = 0.985 KN Load parallel to the rafter

=1KN sin100 =0.174 KN

Wind load Suction pressure

-1.19 KN/m2 *1.49 m = -1.77 KN/m Positive pressure

0.6636 KN/m2*1.49m= 0.99 KN/m

Design load for suction pressure

Pd=1.3DL +1.35(LL+WL) =(1.3*2.0198)+1.35(0.895-1.77) =1.17KN/m

Design load for positive pressure

Pd=1.3DL +1.35(LL+WL) =(1.3*1.807)+1.35(0.895+0.6636) =4.453KN/m

We take the larger value of design loads i.e Pd=4.453KN/m * The max moment is on combination –IV(Wind load +concentrated live load +dead load )

= =1.301KNm

The max shear is on combination –IV (Wind load +concentrated live load +dead load Vmax= =3.447KN

For Fe360 steel grade fy=235KN/mm2. Thus e= =

For RT106 (100x60x3)

Sx=24.11cm3 Sy= 18.22cm3

Wply= 30714mm3 Wplx= 21474mm3

Tf=3mm= tw d=h-3t =100-3*3=91mm b=60mm

Internal element of compression flange: b/tf=60/3=20≤42e (42*1) class 1

Web subjected to bending only: d/tw=91/3=30.33≤79e(79*1) class 1

The section is class one (plastic) section.

Check the section for pure bending

Mc,Rd= Wply*fy /gM0 =30714mm3*235/1.1

=6.56KNm> 2.6324 KNm………safe!

Check the section for shear [EBCS-3,1995 Section 4.6.1.2,4.6.4]

d/tw =91/3=30.33 69e=69√(235/235)=69>>30.33

Hence, shear buckling resistance need to be verified

Vpl,Rd= (fy/√3)*Av,eff/gMo =(fy/√3)*(Ah/(b+h))/gMo =(235/√3)*((1039*100)/(60+100))/1.1 =80.0KN>>5.06 KN………………………………safe!

The effect of shear force on the resistance moment Vsd/Vpl,rd*100=5.508/80*100=6.885%<50%

Reduction of design resistance moment not required * Check for deflection

For purlin the recommended limiting value for vertical deflection is

LWL4

384 EI

Since the roof is supporting non-flexible or brittle finish [table 5.1 EBCS3] =1840/350=5.257 mm

From SAP the vertical deflection for this purlin becomes 3.536mm which is safe! * Interaction equation

For biaxial bending the interaction equation must satisfy

NsdNplrd+MxsdMpxrd+MysdMpyrd<1

Mpyrd=Wply*fyϒmo=21474*235/1.1=4.587 KNm>0.0415KNm…………safe!!

5.0680+2.63246.656+0.0454.587=0.468<1

………hence safe!!!!

Truss analysis and design

Since the type of truss is specified on the architectural drawing which fulfills structural strength.

Loading on the truss The maximum reactions determined from the purlin can be used for loading at each joint of the member-purlin location and self-weight of the truss can be conceded during SAP analysis.

* The max reaction is on combination –IV(Wind load +concentrated live load +dead load 0

For purlin the recommended limiting value for vertical deflection is

LWL4

384 EI since the roof is supporting non-flexible or brittle finish [table 5.1 EBCS3]

* 3.64 KN per purlin and 2 purlins rest on single joint of a truss so,2*3.64=7.28KN 2*0.174=0.348KN

After analyzing using SAP for

(Self-weight of the truss *1.3) + Maximum reaction coming from the purlin

The values of tensile & compressive force will become FOR TRUSS -1 Reaction at the support * R1=3.03 KN * R2=19.50 KN * R3=2.5 KN The maximum tensile force =16.32 KN The maximum compressive force =29.76 KN The maximum shear force 5.72KN The maximum moment 2.923 KNm FOR TRUSS -2 Reaction at the support * R1=0.72 KN * R2=3.67 KN * R3=0.72 KN The maximum tensile force =6.553 KN The maximum compressive force =22.698 KN

Section provided for truss members is the same and is RT64 (60*40*2)

h=60mm, b=40mmm,tw=tf=2mm The design tensile force =16.32 KN The design compressive force =29.76 KN The design shear force 5.72 KN The design moment, 5.35 KNm * Classification Flange C=b-3*tw =40-3*2=36mm e=√ (235/fy)= √(235/235)=1 c/tf=36/2=18<26e=26*1=26……so, the flange is class 1 web,with neutral axis at mid depth d/tw=h-3*tf/tw=(60-3*2)/2=27<79e=79…….so, the web is class 1

The section is class one (plastic) section.

* Check for compression member Nt,sd= 39.42 KN(max.)

The design plastic resistance of the gross section Npl,Rd=A*fy/ gM1

=374*235/1.1

=79.9KN >>39.42KN……………….safe! * Check for moment resistance in member M,plsd= 5.35 KNm(max.)

The design plastic resistance of the gross section Mpl,Rd=wply*fy/ gM0

=7776*235/1.1

=30.081KN >>5.35 KN……………….safe! * Check for shear resistance in member V,plsd= 7.67 KNm(max.)

Vpl,Rd=(Av*fy /√3)/ gM0 , Av=AH/(B+H) =384*60/(60+40)=230.4mm2

Vpl,Rd=(230.4*235/√3)/ 1.1 =28.42 KN>7.67 KN………………….safe!!

* Buckling resistance of axially loaded compression member

Nb,Rd=X* βA* A* fy gM1 For class one section βA =1

Fy =235MPa, gM1=1.1

For all cold formed RHS buckling curve b (table 4.11of EBCS-3) is used

Fyb =fy

Le (for the longest compressive member) =1.15 m

Radius of gyration (i) =ry=22.4 mm

l = Le /i = 1150/22.4 =51.34 For Fe360, l1=93.3e=93.3

ly=(ly/l1)√(bA) =(51.34/93.3) √(1) =0.55

From table 4.9 of EBCS-3 the reduction factor X for curve b and ly=0.55 X=0.8842 for l=0.5and

X=? for l=0.55

X=0.8371 for l=0.6

X=0.86065(after interpolation)

Nb,Rd=X* βA* A* fy gM1 =0.86065*1*384*235/1.1 =70.604KN > 39.42 KN ………………………safe!

* Check for tension member

This section is class 1so no buckling takes place Nt,sd= 20.065 KN(max.)

The design plastic resistance of the gross section

Npl,Rd=A*fy/ gM1 =384*235/1.1 =82.04KN >>20.065KN………………………..safe!

First let’s try a single section with dimensions 60x40x3mm for the purlin

tf= tw=3mm,h=60mm, b=40mm

This section is class 1so no buckling takes place Nt,sd= 20.065 KN(max.)

The design plastic resistance of the gross section

Npl,Rd=A*fy/ gM1 =384*235/1.1 =82.04KN >>20.065KN………………………..safe!

CHAPTER FIVE

DESIGN OF SOLID SLAB

In the building there are: * Slabs with four sides support * Inclined slab

According to EBCS-1, section 5-2-3 the minimum effective depth of slab from deflection point of view. d=0.4+0.6*fyk400*leβa Where, d is the effective depth of the slab fyk is the characteristic strength of reinforcement in MPA, 300 MPA le is the effective span, in two way slab the shorter of the two sides βais constant taken from table 2-5 of EBCS-1

Sample calculation for depth determination, for s-1, which is an end span two way slab

For 2:1 span ratio end span,βa=30

1:1 span ratio end span,βa=40

For Ly/Lx=1.21,by interpolation we get βa=37.4 ,Le=Lx=7 d= (0.4+0.6*300/400)*7000/37.4 =158.96mm

We have calculated the depth accordingly as tabulated below

Depth Determination slab | Lx[cm] | Ly[cm] | Ly/Lx | support condition | βa | d[cm] | | Second floor | | | S1 | 700 | 880 | 1.257143 | End span | 37.428571 | 15.89695 | S2 | 600 | 880 | 1.466667 | >> | 35.333333 | 14.43396 | S3 | 300 | 880 | 2.933333 | >> | 30 | 8.5 | S4 | 700 | 427.5 | 0.610714 | >> | 40 | 14.875 | S5 | 300 | 682.5 | 2.275 | >> | 30 | 8.5 | S6 | 525.5 | 880 | 1.674596 | >> | 33.254044 | 13.4322 | S7 | 577.5 | 880 | 1.52381 | >> | 34.761905 | 14.12106 | S8 | 690 | 880 | 1.275362 | >> | 37.246377 | 15.7465 | S9 | 577.5 | 880 | 1.52381 | >> | 34.761905 | 14.12106 | S10 | 577.5 | 880 | 1.52381 | >> | 34.761905 | 14.12106 | S11 | 700 | 820 | 1.171429 | >> | 38.285714 | 15.54104 | S12 | 600 | 820 | 1.366667 | Intermediate | 41.333333 | 12.33871 | S13 | 1000 | 410 | 0.41 | End span | 40 | 8.7125 | S14 | 300 | 682.5 | 2.275 | Intermediate | 35 | 7.285714 | S15 | 525.5 | 820 | 1.560419 | End span | 34.395814 | 12.98632 | S16 | 577.5 | 820 | 1.419913 | >> | 35.800866 | 13.71126 | S17 | 690 | 820 | 1.188406 | >> | 38.115942 | 15.38726 | S18 | 690 | 820 | 1.188406 | >> | 38.115942 | 15.38726 | S19 | 577.5 | 820 | 1.419913 | >> | 35.800866 | 13.71126 | S20 | 577.5 | 820 | 1.419913 | >> | 35.800866 | 13.71126 | S21 | 727.5 | 880 | 1.209622 | >> | 37.90378 | 16.31434 | S22 | 727.5 | 880 | 1.209622 | >> | 37.90378 | 16.31434 | S23 | 690 | 880 | 1.275362 | >> | 37.246377 | 15.7465 | S24 | 727.5 | 880 | 1.209622 | >> | 37.90378 | 16.31434 | S25 | 675.6 | 880 | 1.302546 | >> | 36.974541 | 15.53123 | S26 | 325 | 880 | 2.707692 | interior | 35 | 7.892857 | S27 | 1000 | 397.5 | 0.3975 | End span | 40 | 8.446875 | S28 | 600 | 880 | 1.466667 | inter | 40.333331 | 12.64463 | S29 | 700 | 880 | 1.257143 | End span | 37.428571 | 15.89695 | S30 | 727.5 | 820 | 1.127148 | End span | 38.728522 | 15.96691 | S31 | 727.5 | 820 | 1.127148 | End span | 38.728522 | 15.96691 | S32 | 690 | 820 | 1.188406 | >> | 38.115942 | 15.38726 | S33 | 727.5 | 820 | 1.127148 | >> | 38.728522 | 15.96691 | S34 | 675.6 | 820 | 1.213736 | >> | 37.862641 | 15.16693 | S35 | 300 | 550 | 1.833333 | >> | 31.666667 | 8.052632 | S36 | 800 | 550 | 0.6875 | inter | 45 | 10.38889 | S37 | 200 | 820 | 4.1 | End span | 30 | 5.666667 | S38 | 600 | 820 | 1.366667 | End span | 36.333333 | 14.0367 | S39 | 700 | 820 | 1.171429 | End span | 38.285714 | 15.54104 | | | | | | | |

d = 163.1434mmgoverns using Ø10mm & cover 15 mm, the span.

Clear cover is 15mm EBCS 2 ART. Table 7.2 for Dry environment D = 163.1434+15+5 = 183.5524mm USE D=190mm dx=190-15-5=170mm dy=190-15-10-5=160mm

Load on slabs

Marble floor

* Floor finish -0.02*27=0.54 KN/M2 * Cement screed-0.02*23=0.46 KN/M2 * 190mm RC slab-0.19*25=4.75 KN/M2 * Ceiling plaster-0.02*23=0.46 KN/M2 ∑6.21 KN/M2 PVC floor

* Floor finish -0.002*16=0.032 KN/M2 * Cement screed-0.02*23=0.46 KN/M2 * 190mm RC slab-0.19*25=4.75 KN/M2 * Ceiling plaster-0.02*23=0.46 KN/M2 ∑ 5.702 KN/M2

Terrazzo floor

* Floor finish -0.02*23=0.46 KN/M2 * Cement screed-0.02*23=0.46 KN/M2 * 190mm RC slab-0.19*25=4.75 KN/M2 * Ceiling plaster-0.02*23=0.46 KN/M2 ∑6.13 KN/M2

Ceramic floor

* Floor finish -0.006*16=0.096 KN/M2 * Cement screed-0.02*23=0.46 KN/M2 * 190mm RC slab-0.19*25=4.75 KN/M2 * Ceiling plaster-0.02*23=0.46 KN/M2 ∑5.766 KN/M2

* Live Load Live load on slab is considered to be 4KN/m2 on the lecture hall & using 3KN/m2 on the other offices and corridors(EBCS 2.6.3 category C3)

Two way slab αxs = 0.063 αys = 0.045 αxf = 0.047 αyf = 0.034

Mxs = 0.063*14*7.2752 =46.67 KNm/m Mys = 0.045*14*7.2752 = 33.34 KNm/m Mxf = 0.047*14*7.2752 = 34.82 KNm/m Myf = 0.034*14*7.2752 = 25.19 KNm/m Panel-2

= 1.21 Two way slab αxs = 0.055 αys = 0 αxf = 0.041 αyf = 0.034

Mxs = 0.055*13.96*7.2752 =40.62 KNm/m Mys = 0*13.96*7.2752 = 0 KNm/m Mxf = 0.041*13.96*7.2752 = 30.28 KNm/m Myf = 0.034*13.96*7.2752 = 25.11 KNm/m

Ly = 8.80= 4 One way slab Lx 2.2

M=Pd*lx2

12

=13.81*2.22 12 =5.57KNm

Panel-4

Ly = 8.80= 1.87 Two way slab Lx 4.7

αxs = 0.065 αys = 0.037 αxf = 0.049 αyf = 0.028

Mxs = 0.065*13.81*4.72 =19.83 KNm/m Mys = 0.037*13.81*4.72 = 11.29 KNm/m Mxf = 0.049*13.81*4.72 = 14.95 KNm/m Myf = 0.028*13.81*4.72 = 8.54 KNm/m

Design load

From EBCS 1,1995 Table 2.1 ,we have the design load calculation as follows:-

qd= 1.6LL+1.3DL Design load calculation for slab 1 Dead Load Own weight = 0.19*25= 4.75KN/m2 Cement screed = 23*0.02= 0.46 KN/m2 Plastering and painting = 23*0.02= 0.46 KN/m2 PVC tile = 16*0.002= 0.032 KN/m2 Total dead load = 5.702 KN/m2 Live Load=4 KN/m2 EBCS-1, 1995 Table 2.1 Pd= 1.6LL+1.3DL = 13.81 KN/m2

Since the slab is inclined, the design load is divided by the inclined angle i.e 100. Therefore Pd=13.81KN/m2/cos100 =14KN/m2 Panel | Floor finish used | Floor finish | Screed | Self weight | plaster | Partition load | Total dead load | Live load | Design load(Pd) | P1 | PVC tile | 0.032 | 0.46 | 4.75 | 0.46 | | 5.702 | 4 | 14.00 | P2 | Cement screed | | 0.46 | 4.75 | 0.46 | | 5.67 | 4 | 13.96 | P3 | PVC tile | 0.032 | 0.46 | 4.75 | 0.46 | | 5.702 | 4 | 13.81 | P4 | >> | 0.032 | 0.46 | 4.75 | 0.46 | | 5.702 | 4 | 13.81 | P5 | Cement screed | | 0.46 | 4.75 | 0.46 | | 5.67 | 4 | 13.96 | P6 | >> | | 0.46 | 4.75 | 0.46 | | 5.67 | 4 | 13.96 | P7 | Ceramic tile | 0.096 | 0.46 | 4.75 | 0.46 | 2.62 | 8.39 | 3 | 15.71 | P8 | Marble | 0.54 | 0.46 | 4.75 | 0.46 | 1.14 | 7.35 | 3 | 14.36 | P9 | >> | 0.54 | 0.46 | 4.75 | 0.46 | 1.07 | 7.28 | 3 | 14.27 | P10 | >> | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P11 | >> | 0.54 | 0.46 | 4.75 | 0.46 | 2.49 | 8.70 | 3 | 16.11 | P12 | >> | 0.54 | 0.46 | 4.75 | 0.46 | 1.25 | 7.46 | 3 | 14.50 | P13 | >> | 0.54 | 0.46 | 4.75 | 0.46 | 1.02 | 7.23 | 3 | 14.19 | P14 | marble | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P15 | >> | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P16 | >> | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P17 | >> | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P18 | >> | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P19 | Terrazzo tile | 0.46 | 0.46 | 4.75 | 0.46 | | 6.13 | 3 | 12.77 | P20 | >> | 0.46 | 0.46 | 4.75 | 0.46 | | 6.13 | 3 | 12.77 | P21 | marble | 0.54 | 0.46 | 4.75 | 0.46 | 2.62 | 8.83 | 3 | 16.28 | P22 | PVC tile | 0.032 | 0.46 | 4.75 | 0.46 | | 5.70 | 3 | 12.21 | P45 | marble | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P23 | PVC tile | 0.032 | 0.46 | 4.75 | 0.46 | 1.54 | 7.24 | 3 | 14.21 | P24 | marble | 0.54 | 0.46 | 4.75 | 0.46 | 2.99 | 9.20 | 3 | 16.76 | P25 | >> | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P26 | >> | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P27 | >> | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P28 | Terrazo tile | 0.46 | 0.46 | 4.75 | 0.46 | | 6.13 | 3 | 12.77 | P29 | >> | 0.46 | 0.46 | 4.75 | 0.46 | | 6.13 | 3 | 12.77 | P30 | marble | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P31 | >> | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P32 | >> | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P33 | PVC tile | 0.032 | 0.46 | 4.75 | 0.46 | | 5.70 | 4 | 14.02 | P34 | >> | 0.032 | 0.46 | 4.75 | 0.46 | | 5.70 | 4 | 14.02 | P35 | >> | 0.032 | 0.46 | 4.75 | 0.46 | | 5.70 | 4 | 14.02 | P36 | >> | 0.032 | 0.46 | 4.75 | 0.46 | | 5.70 | 4 | 13.81 | P37 | >> | 0.032 | 0.46 | 4.75 | 0.46 | | 5.70 | 4 | 14.02 | P38 | >> | 0.032 | 0.46 | 4.75 | 0.46 | | 5.70 | 4 | 14.02 | P39 | marble | 0.54 | 0.46 | 4.75 | 0.46 | 1.06 | 7.27 | 3 | 14.25 | P40 | >> | 0.54 | 0.46 | 4.75 | 0.46 | 1.12 | 7.33 | 3 | 14.33 | P41 | >> | 0.54 | 0.46 | 4.75 | 0.46 | | 6.21 | 3 | 12.87 | P42 | >> | 0.54 | 0.46 | 4.75 | 0.46 | 0.81 | 7.02 | 3 | 13.92 | P43 | >> | 0.54 | 0.46 | 4.75 | 0.46 | 1.12 | 7.33 | 3 | 14.33 | P44 | >> | 0.54 | 0.46 | 4.75 | 0.46 | 1.08 | 7.29 | 3 | 14.28 |

Sample calculations

Design load

Live load-5 KN/m2

Dead load –ceramic floor finish -5.838KN/m2

Wall load =(1.8+0.82)*2.19=5.7378 KN

When distributed over the area =0.3243KN/m2

Design load(w)=1.3*Gk+1.6*Qk =1.3*(5.838+0.3243)+1.6*5=16.011KN/m2

Balance of Support Moments

For each support over which the slab is continuous there will thus generally be two different support moments. The difference may be distributed between the panels on either side of the support to equalize their moments, as in the moment distribution method for frames.

Two methods of differing accuracy are given here for treating the effects of this redistribution on moments away from the support.

EBCS 2, 1995 article A.3.3.1 No.3 and 4

When differences between initial support moments are less than 20 percent of the larger moment only

EBCS 2, 1995 article A.3.3.2 No.1

Section 1-1

For panel S19& S20

Calculated support moments

For S19-M= 40.67KNM , For S20-M= 40.62KNM = (40.67 -40.62)/40.67 = 0.0013<0.2 25.28% > 20% we use larger moment for design

Msd=40.67KNM

For panel 20 &21

Calculated support moment

For S20-M=40.62KNM

For S21-M=5.57KNM

ΔM=(40.62-5.57)/40.62=0.863

0.863>0.2

Moment distribution →For the support

Moment Adjustment → For the span

ΔM=35.05KNM

DF for panel S20=EI/Lx= 1/7.275 =0.232 (1/7.275 +1/2.2)

DF for panel S21=EI/Lx= 1/2.2 =0.768 (1/7.275 +1/2.2)

S21

S20

DF

0.77

0.2328

-5.57

40.62

0.77*(40.62-5.57) =26.99

0.23*(40.62-5.57)=8.06 ==4.875

+32.56

-32.56

Balanced moment

Span moment Adjustments

Coefficients

Ly =8.8 = 1.21

Lx 7.275

Cx=0.344 , Cy=0.364

Mxf=30.28 + ΔM =30.28 + 35.05x0.344 =42.34KNM

Myf=25.11 + 35.05x0.364 =37.87KNM

According to the above EBCS recommendations we balanced the support moments respectively and tabulated as follows.

Balancing of support moment | | | | | | | | | Panel No | Length | K | DF | Unbalanced Moment | | Balanced Moment | S-19&S-20 | S19 | 7.275 | 0.1375 | 0.5 | 46.67 | Take Average | 43.670 | | S20 | 7.275 | 0.1375 | 0.5 | 40.67 | | 43.670 | S20&S21 | S20 | 7.275 | 0.1375 | 0.76781 | 40.67 | Take Larger | 40.670 | | S21 | 2.2 | 0.4545 | 0.23219 | 5.57 | | 40.670 | S21&S22 | S21 | 2.2 | 0.4545 | 0.318841 | 5.57 | Take Larger | 8.540 | | S22 | 4.7 | 0.2128 | 0.681159 | 8.54 | | 8.540 | S22&S23 | S22 | 4.7 | 0.2128 | 0.392484 | 8.54 | Take larger | 25.110 | | S23 | 7.275 | 0.1375 | 0.607516 | 25.11 | | 25.110 | S23&S24 | S23 | 7.275 | 0.1375 | 0.665173 | 25.11 | Take larger | 25.110 | | S24 | 3.662 | 0.2731 | 0.334827 | 15.60 | | 25.110 | S24&S25 | S24 | 3.662 | 0.2731 | 0.542117 | 15.60 | Take larger | 15.600 | | S25 | 3.093 | 0.3233 | 0.457883 | 12.53 | | 15.600 | S29&S30 | S29 | 7.275 | 0.1375 | 0.5 | 43.32 | Take Average | 40.160 | | S30 | 7.275 | 0.1375 | 0.5 | 37.00 | | 40.160 | S30&S31 | S30 | 7.275 | 0.1375 | 0.513228 | 37.00 | Take Average | 26.105 | | S31 | 6.9 | 0.1449 | 0.486772 | 15.21 | | 26.105 | S31&S32 | S31 | 6.9 | 0.1449 | 0.486772 | 15.21 | Use Distribution | 26.854 | | S32 | 7.275 | 0.1375 | 0.513228 | 39.13 | | 26.854 | S32&S33 | S32 | 7.275 | 0.1375 | 0.489339 | 39.13 | Take Average | 26.738 | | S33 | 7.592 | 0.1317 | 0.510661 | 14.35 | | 26.738 |

Adjustments of Field Moments

According to EBCS 2, 1995 article A.3.3.3 Method II 1. In this method consideration of the effects of changes of support moments is limited to the adjacent spans. Since no effects on neighboring support sections need to be considered; only a simple balancing operation is required at each edge and no iterative process is involved. 2. The procedure for applying method II, is as follows a. Support and span moments are first calculated for individual panels by assuming each panel to be fully loaded. This is done by using the coefficients given in Table A-1 as described in Section A.3.2 b. The unbalanced moment is distributed using the moment distribution method. The relative stiffness of each panel shall be taken proportional to its gross moment of inertia divided by the smaller span. c. If the support moment is decreased, the span moments mxf and myfare thenincreased to allow for the changes of support moments. This increase is calculated as being equal to the change of the support moment multiplied by the factors given in Table A-2. If a support moment is increased, no adjustments shall be made to the span moments.

Un Balanced supportmoment

Balanced fieldmoment

Balanced supportmoment

Unbalanced Field

Moment

Balanced supportmoment

Un Balanced supportmoment

Un Balanced supportmoment

Adjusted Span Moments

The change in the respective moments will be described as follows:

Mxf = CxM Myf = CyM Adjusted field moment: Mxf = Mxf+ (Cx*M) Myf = Myf+ (Cy*M)

Reinforcement calculations

fyk = 300Mpa, γc = 1.15 fyd= fyk = 300 = 260.87 Mpa γc 1.15

As min = ρmin bd

Smax = 2D which ever is smaller. 350

Slab Reinforcement /for the 2nd floor/

Checking adequacy of depth

Since D = 190mm from serviceability requirements.

Effective depth X - direction : dx = 190 - 10 - 15 = 170 2 y - direction : dy = 190 - 10- 10-15 = 160 2 ρmin = 0.5= 0.5 = 0.00167 fyx300 Asmin= ρmin b*d = 0.00167*1000*170 = 283.9mm2 (for X-direction)

= 0.00167*1000*160 = 267.2mm2(for y- direction)

Panel S-30 X -direction : Mxs = 40.16KNm/m

ρ= [ 1 – √ ( 1- 2Mmax ) ] * fcd bd2*fcdfyd

ρ = [ 1 - √ ( 1- 2*40.16*106 ) ] * 11.33 1000*1702*11.33 260.87

ρ = 0.0057

As = ρbd

As = 0.0057* 1000*170 = 969.176mm2

Using ø10mm bar

Spacing

S = as* b As

S = 78.54*1000 = 81.04mm 969.176

Spacing provided = 80mm

Use ø10mm bars c/c 80mm

As provided = 969.176mm2

y - direction : Mys = 25.67 KNm/M

ρ = [ 1 – √ ( 1- 2Mmax ) ] * fcd bd2*fcd fyd

ρ = [ 1 - √ ( 1- 2*25.67*106 ) ] * 11.33 1000*1602*11.33 260.87 ρ = 0.004

As = ρbd As = 0.004* 1000*160 = 640mm2

Using ø10mm bar

Spacing S = as* b As

S = 78.54*1000 = 122.72mm 640

Spacing provided= 120mm Use ø10mm bars c/c 120mm

As provided = 640mm2

Reinforcement Calculations for the rest of the Slab is done in the spread sheet. See the spread sheets of sample slab design for second floor.

Panel | Design moment | Fcd | Fyd | d | m | C1 | C2 | b | ρ | ρmin | Area | Space | Space provided | As provided | Panel 19 | Mxs | 46.67 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.0067 | 0.00167 | 1,139.00 | 68.95 | 60 | 1308.96 | | Mys | 34.2 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00546 | 0.00167 | 873.6 | 89.9 | 80 | 981.72 | | Mxf | 34.82 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00489 | 0.00167 | 831.3 | 94.48 | 90 | 872.64 | | Myf | 25.19 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00395 | 0.00167 | 632 | 124.27 | 120 | 654.48 | Panel 20 | Mxs | 40.62 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.0067 | 0.00167 | 1139 | 68.95 | 60 | 1308.96 | | Mxs | 32.56 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00455 | 0.00167 | 773.5 | 101.54 | 100 | 785.38 | | Mys | - | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0 | 0.00167 | - | | | | | Mxf | 42.09 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00599 | 0.00167 | 1018.3 | 77.13 | 70 | 1121.96 | | Myf | 30.99 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00491 | 0.00167 | 785.6 | 99.97 | 90 | 872.64 | Panel 21 | Msd | 32.56 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00455 | 0.00167 | 773.5 | 101.54 | 100 | 785.38 | | Msd | 15.26 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00234 | 0.00167 | 374.4 | 209.77 | 200 | 392.69 | Panel 22 | Mxs | 32.51 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00454 | 0.00167 | 771.8 | 101.76 | 100 | 785.38 | | Mxs | 15.26 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00207 | 0.00167 | 351.9 | 223.18 | 100 | 785.38 | | Mys | 12.46 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.0019 | 0.00167 | 304 | 258.35 | 250 | 314.15 | | Mxf | 18.7 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00255 | 0.00167 | 433.5 | 181.17 | 180 | 436.32 | | Myf | 9.42 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00143 | 0.00167 | 228.8 | 343.26 | 340 | 230.99 | Panel 23 | Mxs | 32.51 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00454 | 0.00167 | 771.8 | 101.76 | 100 | 785.38 | | Mxs | 32.33 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00452 | 0.00167 | 768.4 | 102.21 | 100 | 785.38 | | Mys | 12.46 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.0019 | 0.00167 | 304 | 258.35 | 250 | 314.15 | | Mxf | 18.7 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00255 | 0.00167 | 433.5 | 181.17 | 180 | 436.32 | | Myf | 9.42 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00143 | 0.00167 | 228.8 | 343.26 | 340 | 230.99 | Panel 24 | M | 32.33 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00452 | 0.00167 | 768.4 | 102.21 | 100 | 785.38 | | M | 15.6 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.0024 | 0.00167 | 384 | 204.52 | 200 | 392.69 | Panel 25 | M | 15.6 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.0024 | 0.00167 | 384 | 204.52 | 200 | 392.69 | Panel 29 | Mxs | 43.32 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00618 | 0.00167 | 1050.6 | 74.75 | 70 | 1121.96 | | Mys | 34.2 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00546 | 0.00167 | 873.6 | 89.9 | 80 | 981.72 | | Mxf | 32.68 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00457 | 0.00167 | 776.9 | 101.09 | 100 | 785.38 | | Myf | 25.84 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00405 | 0.00167 | 648 | 121.2 | 120 | 654.48 | Panel 30 | Mxs | 22.46 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00308 | 0.00167 | 523.6 | 150 | 150 | 523.58 | | Mys | - | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0 | 0.00167 | - | | | | | Mxf | 28.7 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00398 | 0.00167 | 676.6 | 116.08 | 110 | 713.98 | | Myf | 25.67 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00402 | 0.00167 | 643.2 | 122.1 | 120 | 654.48 | Panel 31 | Mxs | 12.55 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.0017 | 0.00167 | 289 | 271.76 | 270 | 290.88 | | Mys | 23.24 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00363 | 0.00167 | 580.8 | 135.22 | 130 | 604.13 | | Mxf | 14.48 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00196 | 0.00167 | 333.2 | 235.71 | 230 | 341.47 | | Myf | 6.72 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00102 | 0.00167 | 163.2 | 481.23 | 350 | 224.39 | Panel 32 | Mxs | 23.24 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.0032 | 0.00167 | 544 | 144.37 | 140 | 560.98 | | Mys | - | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0 | 0.00167 | - | | | | | Mxf | 29.15 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00405 | 0.00167 | 688.5 | 114.07 | 110 | 713.98 | | Myf | 26.09 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00409 | 0.00167 | 654.4 | 120.01 | 120 | 654.48 | Panel 33 | Mxs | 41.46 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00589 | 0.00167 | 1001.3 | 78.44 | 70 | 1121.96 | | Mys | 29.15 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.0046 | 0.00167 | 736 | 106.71 | 100 | 785.38 | | Mxf | 31.1 | 11.33 | 260.87 | 170 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00433 | 0.00167 | 736.1 | 106.69 | 100 | 785.38 | | Myf | 22.03 | 11.33 | 260.87 | 160 | 28.78 | 0.087 | 3,003.04 | 1000 | 0.00343 | 0.00167 | 548.8 | 143.11 | 140 | 560.98 |

Design of slabs resting on the ground

Since the ground floor slab rests on the ground directly, no bending moment is created and therefore, no structural design is required.

But minimum reinforcement in both directions is provided to prevent shrinkage of the slab, stress due to temperature variation and crushing of slab due to concentrated loads.

[EBCS-2, 1995] Art 7.2.2.2]

Use D = 100mm

F8 diameter reinforcement 25 mm clear cover d = 100-25-4 = 71 mm

Spacing (s)

[EBCS-2, 1995 Art 7.2.2.2]

Use F8 c/c 200mm in both directions.

CHAPTER SIX

DESIGN OF STAIR CASE 1. Stair case between axis 15 and 16 from ground to the roof

Lay out

Sectional view

2.2 Material properties Steel grade fyk=300Mpa Concrete class fck = 25Mpa

2.3 Depth determination

Span ratio

[EBCS-2, 1995 Art5.2.3

[EBCS-2,1995 Table 5.1]

= 153mm

Using F 12 reinforcement bar and concrete cover of 20 mm D = 153 + 20 + 6= 179mm Use D = 180mm d = 180 – 20 - 6 = 154mm

2.4 Stair dimension Thread = 340mm Riser = 150mm

Flight length =

= 371.62mm

2.5 Load on stair

i) Dead load on flight

a. Own weight of concrete

b. Steps

c. Base mortar plaster = d. Mortar for thread = 0.02*23 = 0.46 KN/m2

e. mortar for riser = f. Marble thread = g. Marble riser = TOTAL = 8.624KN/m2

ii) Dead load on landing

a. Own weight of concrete =0.19 * 25 = 4.75 KN/m2 b. Base mortar plaster = 0.02 * 23 = 0.46KN/m2 d. Mortar for floor finish = 0.02*23 = 0.46 KN/m2 e. Weight of floor finish= 0.03 * 27 =0.81KN/m2 TOTAL = 6.48KN/m2

iii) Live load According to EBCS-2, 1995 stairs are classified as category A Live load taken as 3.0KN/m2

iv) Design load on flight

Live load on flight = 1.6 * 3.0/cos 27 = 5.387KN/m2 Dead load on flight = 1.3 *8.624 =11.2112KN/m2 16.60KN/m2 iv) Design load on landing

Live load on flight = 1.6 * 3.0/cos 27 = 5.387KN/m2 Dead load on flight = 1.3 *6.48 = 8.424KN/m2 13.811KN/m2

2.6 Moment determination and steel reinforcement i) Maximum bending moment

ii) Check the adequacy of the section for flexure d =Mmax0.2952*b*fcd =122.5367*1060.2952*2350*11.33 =124.86 mm < 154 mm……………………..OK! iii) Reinforcement a. Longitudinal reinforcement

Span

[EBCS-2, 1995 Art 7.2.2.2] = 0.00167 ρ=1-1-2Mmaxbd2*fcd*fcdfyd ρ=1-1-2*66.7158*1062350*1542*11.33*11.33260.87 ρ =0.00486 As= ρ*b*d = 0.00486*2350*154 As= 1759.11 mm2 Using F 12 bar

Spacing provided 150mm

Smax = 2D=2*180=360mm >150mm…….. OK! 350mm

Use F 12 bars c/c 150mm As provided = 1771.9 mm2

Support

[EBCS-2, 1995 Art 7.2.2.2] = 0.00167 ρ=1-1-2Mmaxbd2*fcd*fcdfyd ρ=1-1-2*122.5367*1062350*1542*11.33*11.33260.87 ρ =0.0095 As= ρ*b*d = 0.0095*2350*154 As= 3422.84 mm2

Using F 14 bar

Spacing provided 100mm

Use F 14 bars c/c 100mm As provided = 3617.59 mm2

Smax = 2D=2*180=360mm >100mm…….. OK!

350mm

b. Transverse reinforcement

Transverse reinforcement must be provided to account for shrinkage. The ratio of secondary reinforcement to the main reinforcement shall be at least equal to 0.2

At the support [EBCS-2, 1995 Art 7.2.2.2] = 0.2 *3617.59= 723.518 mm2 Using F 8 bar

= 163.28mm [EBCS-2, 1995 Art 7.2.2.2]

Use F 8 c/c 160 mm

At the span [EBCS-2, 1995 Art 7.2.2.2] = 0.2 *1771.9=354.38 mm2 Using F 8 bar

= 333.36mm [EBCS-2, 1995 Art 7.2.2.2]

Use F 8 c/c 330 mm

3.6Shear reinforcement

Checking flexural shear at a distance d from the face of the support(at critical section)

Vsd =78.23KN

The shear force at the critical section (Vsd)= 78.23KN

The shear force Vc carries the concrete in members without significant axial forces shall be taken as [EBCS-2, 1995 Art4.5.3.1]

=1.245< 2

=1.245

[EBCS-2,1995 Art3.5.3.1] [EBCS-2, 1995 Art2.4]

= 1.032 Mpa

Vc = 0.25 * 1.032 * 1.466 * 2350 * 154 = 136.88KN

Because Vc> Vu, no shear reinforcement is required. But it is recommended to use F 8 c/c 300 mm to hold the main reinforcement.

CHAPTER SEVEN FRAME ANALYSIS

EARTHQUAKE ANALYSIS The building is to be constructed in Debre Birhan town, which is located in latitude 07-04’ N and 38-31’ E .It, is categorized seismic hazard zone four

The seismic base shear force Fb for each main direction can be computed from

[EBCS-8, 1995 Art 2.3.3.2.2.(1)]

Where

, Ordinate of the design spectrum at period T1. T1, fundamental period of vibration of the structure for Translational motion in the direction considered. W, seismic dead load computed in accordance with clause 1.4.3(3)

For linear analysis the design spectrum normalized by the acceleration of gravity g is by the following expression,

[EBCS-2, 1995 Art.1.4.2.2.(4)]

The parameter in the above expression is the ratio of the design bed rock acceleration to the acceleration of gravity g and is given by

Where

, the bed rock acceleration ratio for Debre Birhan Zone -4 = 0.01 [EBCS-8, 1995 Table 1.1] I, the importance factor given table 2.4 for ordinary building, Importance category III, the importance factor I = 1.0

= 0.01*1 = 0.01

The parameter b is design response factor for the site and given by

Where b is the coefficient for soil characteristics given in table 1.2, S = 1. For building with height up to 80 m to a value of T, may be Approximated by

Where T1 = Fundamental period of building, in second H = height of building above the basin in meters C1 = 0.075 (for reinforced concrete moment – resisting frames and eccentrically braced steel frame)

H = 33.5 m

Then T1 = 0.075 * 33.5 = 1.044 sec

The behavior factor γ to account for energy dissipation: [EBCS-8, 1995 Art 3.3.2.1(1)]

Where = Basic value of the behavior factor dependent on the structural type given in table 3.2 EBCS-8 page 37 = 0.2 for RC structure frame. KD = factor reflecting the ductility class = 1.5 for RC structure medium ductile. KR = structure regularity factor. = 1.25 for RC irregular structure. KW = factor reflecting the prevailing failure made in structural system with walls = 1.0 frame structure

U = 0.2 * 1.5 * 1.25 * 1.0 =0.375 <0.7 = =0.00437

The effect of the seismic action shall be evaluated using seismic dead load.

CHAPTER EIGHT

FRAME ELEMENT DESIGN

BEAMS DESIGN

DESIGN OF AXIS 1 BEAM

(2ndFLOOR )

D= 700mm b= 300mm fcd = 11.33MPa fyd= 260.87 MPa

Concrete cover -25mm (EBCS2-art7.1.3.6)

Reinforcement provided φ 20mm rods Stirrup diameter 6

From serviceability (EBCS2-1995, art 5.2.3): d =0.4+0.6fyk400Leβa = 0.85* 727528 =220.85 dprov =700-25 - 202 – 6 =659mm

DESIGN FOR FLEXURAL REINFORCEMENT

Fig: Bending moment diagram for combination 3 @ axis B (2th floor )

Design of beams is done using design table

Assume 20% moment distribution

Support 1

Md =15.112 KNm

Km=Mdbd = 15.110.30.659 =10.77

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced Ks=4.36

As= KsMdd = 4.36*10.770.659 = 71.26mm2

Use 2 φ20mm rods at the top

Span1

Md =39.10 KNm

Km=Mdbd = 39.100.30.659 =17.32

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced Ks=4.36

As= KsMdd = 4.36*39.100.659 =258.696mm2

Use 2 φ20mm rods at the bottom

Support 2

Md =87.62KNm

Km=Mdbd = 87.620.30.659 =25.933

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced Ks=4.36

As= KsMdd = 4.36*87.620.659 = 579.70mm2

Use 2 φ20mm rods at the top

Span 2

Md =17.2KNm

Km=Mdbd = 17.200.30.659 =11.49

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced Ks=4.36

As= KsMdd = 4.36*17.200.659 = 113.79mm2

Use 2 φ20mm rods at the bottom

Support 3

Md =46.10KNm

Km=Mdbd = 46.100.30.659 =18.81

K*m=48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced Ks=4.36

As= KsMdd = 4.36*46.100.659 = 305.00mm2

Use 2 φ20mm rods at the top

Span 3

Md =16.15KNm

Km=Mdbd = 16.150.30.659 =11.13

K*m=48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced Ks=4.36

As= KsMdd = 4.36*16.150.659 = 106.85mm2…………………………Use 2 φ20mm rods at the bottom

Support 4

Md =50KNm

Km=Mdbd = 500.30.659 =19.60

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced

Ks=4.36

As= KsMdd = 4.36*500.659 = 330.80mm2

Use 2φ20mm rods at the bottom

Span4

Md =17.05KNm

Km=Mdbd = 17.050.30.659 =11.43

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced Ks=4.36

As= KsMdd = 4.36*17.050.659 = 112.80mm2

Use 2 φ20mm rods at the bottom

Support 5

Md =62.50KNm

Km=Mdbd = 62.500.30.659 =21.90

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced

Ks=4.36

As= KsMdd = 4.36*62.500.659 = 413.51mm2

Use 2φ20mm rods at the bottom

Span 5

Md =17.20KNm

Km=Mdbd = 17.200.30.659 =11.49

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced

Ks=4.36

As= KsMdd = 4.36*17.200.659 = 113.80mm2

Use 2φ20mm rods at the bottom

Support 6

Md =54.95KNm

Km=Mdbd = 54.950.30.659 =20.54

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced

Ks=4.36

As= KsMdd = 4.36*54.950.659 = 363.55mm2

Use 2φ20mm rods at the bottom

Span 6

Md =23.57KNm

Km=Mdbd = 23.570.30.659 =13.45

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced

Ks=4.36

As= KsMdd = 4.36*23.570.659 = 155.94mm2

Use 2φ20mm rods at the bottom

Support 7

Md =49.98KNm

Km=Mdbd = 49.980.30.659 =19.59

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced

Ks=4.36

As= KsMdd = 4.36*49.980.659 = 330.67mm2

Use 2φ20mm rods at the bottom

Span 7

Md =23.60KNm

Km=Mdbd = 23.600.30.659 =13.45

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced

Ks=4.36

As= KsMdd = 4.36*23.600.659 = 155.94mm2

Use 2φ20mm rods at the bottom

Support 8

Md =7.07KNm

Km=Mdbd = 7.070.30.659 =7.36

K*m= 48.23 for 20% moment redistribution

Km< K*m,Section is singly reinforced

Ks=4.36

As= KsMdd = 4.36*7.360.659 = 48.69 mm2

Use 2φ20mm rods at the bottom

Shear reinforcement design

Fig: Shear force diagram

Vsd=68.02KN

Critical section for shear is considered at‘h’ distance from face of support, for interior span of a continuous beam.

Shear force at critical section (Vsd) =48KN

Vrd =0.25*fcdbwd =0.25*11.33*300*659*10-3 =559.985KN > 48KN OK

At the support there are 2φ20mm rods

Fctd= 0.21*fck2/31.5 =0.21*202/31.5 = 1.03MPa

Ρ=Astbw*d=0.000429

k1=1+50 =1+50*0.000429= 1.02145≤ 2.0, k1=1.02145 k2=1.6-d = 1.6 – 0.659 = 0.941< 1.0

Vc= 0.25fctdk1k2bwd = 0.25*1.03*1.02145*1*300*659*10-3 = 52 KN

Vsd<Vc

Vs =Vc –Vsd = 52 – 48= 4KN

Consider φ8mm stirrups with 2 legs

Av = 2 * 82 * π/4 = 100.48mm2

Spacing S= Avfydd /Vs =100.48*260.87*3594*103 = 4318.46 mm

Smax= 0.5d= 0.5* 659= 329.5 mm

Use φ8 stirrups c/c 320mm

Development length The reinforcement shall be properly anchored at each end to prevent bond failure.

Basic anchorage length

The basic anchorage length lb diameter d of the bar is given as lb =φ4fydfbd [EBCS-2, 1995 Art 7.1.6.1] , The design bond strength for deformed compression bar of good bond condition.

lb = 204260.872.06= 633.179mm

Required anchorage length [EBCS-2, 1995 Art7.1.6.2]

Where: - Ascal = theoretical area of reinforcement required by the design Aseff = Area of reinforcement actually provided

For span1

Ascal = 258.696mm2 Aseff = 628.32mm2 lb.net=1.0*633.179*258.696628.32=261.70mm 0.3lb = 0.3* 633.179 = 189.54mm

Lbmin = 10* 20 =200mm

Use lnet = 270mm

Curtailment of longitudinal flexure reinforcement

Reinforcement shall extend beyond the point of zero moments for length lbnet calculated above at point of zero moment. [EBCS-2, 1995 Art 7.1.7] Rest of reinforcement of positive moment shall extend through span min of two bars also extend through span.

Result is shown in structural drawings. Points of zero moments are directly read from the ETABS v 9.2.

COLUMN DESIGN The lateral loads in a frame are transmitted to the foundation through a system of shear walls ,the column in such a frame is said to be braced(or non-sway frame ) and second order effect can be neglected.

Effective length of compression members

In accordance with EBCS-1995, the effective length lefor on RC column is given as,

For non-sway mode,

LeL=am+0.4am+0.8

Where, αm= ( αc1 + αc2)/2 and, In this project column at the intersection of Axis-B’ and Axis-3 is selected for design. On the choosen frame, column cross section through out the floors are 500 x 500mm is selected.

* For the 2nd floor column

Effective length calculation axis –B or x-axis

1

2

Ib1

Ib2

Ib3

Ib4

centerline of nearby column

Lb1

Lb2 column elevation | | | | Lc2 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | h | | | | | | | | | Lc1 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Lc3 | | | | | | | | | | | | | | | | | | | | | | | | |

K11=(bh3/12)/L

K11=((0.3*0.73)/12)/7.275 , K12=((0.3*0.73)/12)/6.9 =12.4x10-4

K21=((0.3*0.73)/12)/7.275 , K22=((0.3*0.73)/12)/6.9

Kc=((0.5*0.53)/12)/3.6

K2 αc1 = (KC +K1)/(K11+K12) =413.82 αc2= (KC+K2)/(K21+K22) =70.85 αm= ( αc1 + αc2)/2

=(413.82+70.85)/2

=242.33

For isolated columns, the slenderness ratio is defined by

V=NuAcfcd

=606.48 x 103 500x500x13.6 =0.18 µsd=Msd=60.90 x 106 =0.036

Asfcd 5002x13.6x500 λ=lei le –effective buckling length i -min radius of gyration of the concrete section only

i = √I/A=h/√12 = 0.144 λ=lei=7.275 *0.7 =35.36 > 35 K1=1 0.144

K2=µsd=0.036=0.17

µsdbal 0.21 ω=0.05 µbal=0.2

LeL=242.33+0.4242.33+0.8=0.9998≥0.7

So Lex=0.9998*3.6=3.59m

Effective length calculation axis–2 or Y-axis

1

2

Ib1

Ib2

Ib3

Ib4

centerline of nearby column

Lb1

Lb2 column elevation | | | | Lc2 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | h | | | | | | | | | Lc1 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Lc3 | | | | | | | | | | | | | | | | | | | | | | | | |

K11=(bh3/12)/L

K11=((0.3*0.73)/12)/8.2 , K12=((0.3*0.73)/12)/8.8

K21=((0.3*0.73)/12)/8.2 , K22=((0.3*0.73)/12)/8.8

Kc=((0.5*0.53)/12)/3.6

K2=0.17

K3

αc1 = (KC +K1)/(K11+K12) =497.78 αc2= (KC+K2)/(K21+K22)

=84.71

αm= ( αc1 + αc2)/2

=(497.78+84.71)/2

=289.75

For isolated columns, the slenderness ratio is defined by

V=NuAcfcd

=606.48 x 103 500x500x13.6 =0.18 µsd=Msd=60.90 x 106 =0.036

Asfcd 5002x13.6x500 λ=lei le –effective buckling length i -min radius of gyration of the concrete section only

i = √I/A=h/√12 = 0.144 λ=lei=4.73 =39.86 > 35 K1=1 0.144

K2=µsd=0.036=0.17

µsdbal 0.21 ω=0.05 µbal=0.2

LeL=289.75+0.4289.75+0.8=0.9986≥0.7

So Lex=0.9986*3.6=3.59m

* Slenderness ratio [EBCS-2,1995 section 4.4.5]

For a non-sway frame λ=50-25M1M2 Where, M2-is first order calculated moment at the end and is always positive and greater in magnitude than M1

M1--is first order calculated moment at the end and is positive if it is in the opposite direction than that of M2

* In the x-axis

M1=-60.89 KNm

M2=69.08 KNm

λ=50-2560.8969.08 , λx=lexix=3.640.144=25

λ≤28.20 * In the y-axis

M2=69.08 KNm

M1=60.89KNm

λ=50-2560.8969.08 ,λy=leyiy=4.40.173=25.43

λ≤27.96

Hence, the column is not slender in both directions and second order effect can be ignored.

* Design action [EBCS-2,1995,section 4.4.10] etot=ee+ea+e2 ee=0.6eo2+0.4eo1 whichever is greater =0.4eo2 ea-additional eccentricity due to geometric imperfection eax=lex/300=7275*0.7/300=16.975mm<20mm eay=ley/300=8800*0.7/300=20.53 mm>20mm

So take eax=20mm=0.02m eay=20.53mm=0.02053 m e2=0, because secondary effect is ignored

In the x-axis etotx=eex+eax eo2x=M2x/Psd=69.08/769.31=0.0898m eo1x=M1x/Psd=60.89/769.13=0.079m eex=0.0844m

In the y-axis etoty=eey+eay eo2y=M2y/Psd=69.08/7369.51=0.0898m eo1y=M1y/Psd=60.89/7369.51=0.079m eey=0.0844m

so etotx=0.0844+0.02=0.01044m etoty=0.0844+0.02053=0.105m The design loads are:

Msdx-x=Psd*etotx

=769.301*0.01044 =11.08KNm

Msdy-y=Psd*etoty

=769.301*0.0844 =11.08KNm

* Reinforcement calculations

Vsd=Nsdfcd*Ac=769.30111.33*1000*0.5*0.5=0.27

μsdx=Msdxfcd*Ac*h=11.0811.33*1000*(0.5*0.5)*0.5=0.0078

μsdy=Msdyfcd*Ac*d=11.0811.33*1000*(0.5*0.5)*0.5=0.0078

From EBCS-2, 1995 part-2 biaxial chart number-18

Astot=w*Ac*fcdfyd w=0.25 so

Astot=0.25*0.5*0.5*11.33*10^6260.87*10^6=0.00271m2

Using 16 number of bars

Astot=16*ᴨ*d24 wegetd=10mm So we use 16# of 10 mm bars

Asprovided=16*ᴨ*4024=20,106.2mm2 * Lateral ties

For reinforcing concrete section to resist shear force ties are used as lateral reinforcement. The diameter of ties shall not be less than 6mm or one quarter of the diameter of the longitudinal bars.

The center to center spacing of lateral reinforcement shall not exceed: * 12 times the minimum diameter of longitudinal bars * Least dimension of column * 300 mm

Ø ≥ 6mm 10/4=3mm c/c spacing ≤ 12*10 =120mm b(least lateral dimension)= 500mm 300mm

Use Ø10 c/c spacing of 300mm

Lap Length

[EBCS-1, 1995, section 7.1.5.1&4]

The length of lap, lo, shall be at least equal to α1 is a function of the percentage of reinforcement lapped

Required anchorage length lb, net depends on the type of anchorage and on the stress in the reinforcement and calculated as

For bar in tension For bar in compression

Where As, cal = the theoretical area of reinforcement required by the design As, eff = the area of reinforcement actually provided a = 1.0 for straight bar anchorage in tension or compression = 0.7 for anchorage in tension with standard hook lb, min = minimum anchorage length

The basic anchorage length is the embedment length required to develop the full design strength of a straight reinforcing bars

Basic anchorage length, lb, for a bar diameter Ø is

For good bond condition the design bond strength of deformed bars may be taken as twice the value for plain bars: fbd= 2fctd = 2*1.0 = 2.0MPa the percentage of lapped bars in compression in any one section may be 100% of the total steel cross section

Therefore, α1 = 1.4

α=1.0 for straight bar anchorage

lb=(260.87*10)/4*2*1=360.09mm

For bar in compression

lb,min=0.6*360.09=216.05 ≥10*10=100 mm

α1 = 1.4 α= 1.0 lo,min =0.3*1*1.4*360.09=151.24<15*10=150mm

lo,min=150mm

Lb ,net=485.34mm>300mm

The length of lap, lo lo≥ 1.4*485.34=679.48mm

Provide lap length lo=690mm The rest of columns are designed like above.

CHAPTER NINE

FOUNDATION DESIGN

Axially loaded footing at intersection of axis B&2 on structure which has a compression force of 2512.45KN, is selected as a representative sample for isolated footing design

From ETABS analysis output results the following loading data are obtained.

P =2512.45KN Mx =140.67KN-m My = 140.67KN-m

a)Proportioning of footings ex=My/p=140.67/2512.45=0.56 ey=Mx/p=140.67/2512.45=0.56

Assume the size of footing B=2m ,L=2m

Fs =2

Effective dimensions L’=L-2el =2-2(0.56)=0.88 B’=L-2eB=2-2(0.56)=0.88

Foundation depth assumed to be 3m below natural ground surface.

Bearing capacity The soil must be capable of carrying the loads from an engineered structure placed up on it without a shear failure and with the resulting settlements being tolerable for the structure. The design bearing resistance is calculated from drained condition. Using Meyerhof bearing capacity equation.

Drained conditions [EBCS-7, 1995 Art 6.5.2.2]

With the design value of dimensionless factors for the bearing resistance

The value Nq ,Nc ,NU are given in tabulated as a function of F

Also C, ϕ andγ values are 0, 260 and 18.5 respectively (assumed). Nq = 12.24 Nγ = 8.58 B=1.2m, q’=γD=18.8*3=56.40KN/m2

For vertical loads ic ,iq,iU = 1.0

Bearing capacity of the soil is given as

Qu=1.3CNc+q’Nq+0.4γBNγ

= (56.40*18.5*12.24) + (0.4*18.50*1.5*8.58) = 12,866.45 KN/m2 A recommendation for the allowable bearing capacity qa to be used for design is based on the ultimate bearing capacity divided by a suitable safety factor (SF)

=12,866.45/2=6,433.225KN/m2 The safety factor is based on the type of soil (cohesive or cohesion less) reliability of the soil parameter, & the structure information, possible safety factor 2-3 is recommended.

The factored design load is 2,512.45KN……………→from SAP output.

Mx = 140.69 KN-m

My = 140.69 KN-m σ σmax=2738.57

Check ;-σmax < qall =2,738.57 < 6,433.225---------------------------OK

L’ = L-2el =0.88

B’ = B-2eB =0.88

Contact formula using flexure σ ave=746.76Kpa

Structural Design fctm=0.3fck(2/3)=0.320(2/3)=2.21Mpa fctk=0.7fctm=1.55Mpa fctd=0.35√25/1.5=1.167Mpa fcd=0.85fck/yc=11.33Mpa fyd=fyk/ys=260.87Mpa a)Depth determination

Punching shear (two way shear)

Vacting=Pcolomn-(a’+d)(b’+d) σ ave =2074.33-(0.25+d+d2)σave

Vresisting=3((a’+d)+(b’+d)Vupd

=(2+6d)Vupd Where, Vupd=0.5fctd(1+50ρ) ρmin=0.5/300=0.00167

Vup=632.22Kpa

Vresisting=(2+4d)632.22d

Vacting =Vresisting d=0.42 d from punching=0.50m wide beam shear (one way shear)

Critical at ’d’

a)along section (1)-(1) Vacting=3*x*σaverage =(b/2 –(b’/2 +d))*3*σav = (1.25-d)3σave Vresisting=3*d*Vvw ,vvw=0.3*fctd(1+50ρ)=379.33kpa

Vacting =Vresisting d=0.83m b)along section (2)-(2) Vacting=y*3*σave =(0.5*3-0.5*0.5-d)*3*σave

=(1.25-d)*3*σave

Vresisting=3*d*379.33

Vacting =Vresisting d=0.83 Take d=0.90 from wide beam shear

Compare trhe above three and take the larger d=0.90m

Assuming we will use Ø16 reinforcement and a clear cover of 75mm, overall thickness required will be:- =0.9+0.028/2+0.075=0.989mm, provide D=1.00m, d= 0.911mm

Determination of design moments

The critical section for bending moment is as shown in the figure. Section 1-1 & 2-2. Since the proposed footing is square & the effective area is the same as provided the moment acting in both section is identical.

Mxx = Myy= L * X2/2 * s

X1 = 1.25m

L = 3.0m s = 746.76KN/m2

Mxx = Myy= 3.0*1.25*1.25*0.5*746.76 =1750.22KNm

Reinforcement

Mxx= Myy= 1750.22KNm

Km=Mb/d=537.673/0.911=14.03

Ks=3.95

As=M*Ksd=537.67*3.950.911=2,331.28mm2

Asmin=ρmin*sectional area =0.00167*1000*911=1521.37mm2 Ascal > Ascal

Provide As-cal

Use ø20 Spacing=s=b*as/As=1000*314.16/2331.18=134.69 Provide Ø20 c/c 130mm Distribute uniformly

Check the bond length (development length)

required=111.18cm

required<available

No bending Detail

CHAPTER TEN

RETAING WALLS

These structures are used to retain a mass of earth or any other material where prevailing conditions do not allow the masses to assume its natural slope. They commonly support vertical slopes of soil located at the north direction of the building.

It assumed that the cantilever types of retaining wall are used for the design.

Step 1

Select the tentative dimension

Step 2

Compute all the vertical & horizontal loads acting on.

Earth pressure is computed by using columb’s theory Use:-wall inclination (α)=90˚ =1.57Rad Angle of friction (ø) =26˚ =0.56 rad Wall frication (δ) =21.30˚ =0.37 rad Surface inclination (β) =0˚ =0.00rad Then Ka=sin2(∝+∅)sin2∝sin∝-δ1+sin∅+δsin(∅-β)sinα-δsin(α+β)2 =0.21 Kp=sin2(∝-∅)sin2∝sin∝-δ1+sin∅+δsin(∅-β)sinα+δsin(α+β)2=4.71 a) | Vertical forces | | | | | Item | Forces | Moment arm from the toe | Moment about the toe | | WI | (KNM) | (M) | (KNM/M) | | ws | 103.60 | 3.00 | 310.80 | | w1 | 35.00 | 1.75 | 61.25 | | w2 | 17.50 | 1.33 | 23.28 | | w3 | 70.00 | 2.00 | 140.00 | | Total | 226.10 | | 535.33 | b) | lateral force | | | | | Pa | 23.80 | 1.17 | 27.76 | | Pp | 98.03 | 0.50 | 49.01 | | surcharge effect of live load | | | | | Po2 | 5.00 | 1.75 | 8.75 | | Total | 28.80 | | 36.51 | Step 3 Stability Checks a)Stability Against Overturning Stabilizing Moment =535.33KNM/M Overturning Moment =36.81KNM/M Factor of safety against Overturning =Stabilizing moment Overturning moment =535.33 36.81 ⇒F.s=14.54 > F.s allowable=1.5 ---------------------OK! b)Stability against sliding Coefficient of friction (μ)=tan=tan21.3˚=0.39 Base adhesion Ca=0.6C= 0 ΣF resisting=Σvtanø+bca=88.179 ΣF sliding =28.80 (By ignoring passive resistance) Factor of safty against sliding=ΣF resisting =3.06 > 1.5 -------------------OK! ΣF acting Note:-if FS<FS allowable consider the effect of passive resistance. c)Stability against bearing

Σ vertical forces=226.10

Σ clockwise moment =516.31

Bearing capacity existing under the wall qall=6,433.225KPa x=MomentVertical force=516.10226.10=2.283 (the distance from the toe to the point of application of R)

The eccentricity=0.5B-x=-0.283⇒ The resultant is with the middle of 1/3

⇒max=vB1+6ebB=226.141+6*0.2834=80.52< qall=6,433.225KPa ------------OK

⇒ min=vB1-6ebB=226.141-6*0.2834=32.53 >0--------------------OK

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...of air conditioning namely central chilled water air conditioning system. An all-air system provides complete sensible and latent cooling capacity in the cold air supplied by the system. Heating can be accomplished by the same air stream, either in the central system or at a particular zone. All-air systems can be classified into 2 categories namely single duct systems and dual duct systems. The single duct system as shown in figure 1 is used for a large room such as an atrium, a banking hall, a swimming pool, or a lecture, entertainment or operating theatre. It can be applied to groups of rooms with a similar demand for air conditioning, such as offices facing the same side of the building. A terminal heater coil under the control of temperature sensor within the room can be employed to provide individual room conditions. A variable value (AV) system has either an air volume control damper or a centrifugal fan in the terminal unit to control the quantity of following into the room in response to signals from a room air temperature...

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...phase of a final year BSc (Hons) computing science project aiming to investigate biometric technologies and develop a fingerprint recognition application to allow logging of student attendance at lectures. The introduction offers some background to the project and establishes the aims and objectives of the project overall. Following on from the introduction, the literature review presents a critique of research material that provides the basis for the project. This material includes a number of texts, journals and research papers as well as additional information sourced from the Web. As drawn from the literature, the subject areas covered include; history and background to modern biometrics; technological, social, organisational and environmental influences; key technologies in the market today; design issues, including security, performance and testing. In chapter 3, attention turns to requirements analysis for the development of a fingerprint recognition system. The process follows a requirements engineering approach to development by formally establishing user requirements and allowing continuous requirements assessment throughout the project life-cycle. The design approach and methodology used to model the problem are also addressed here. Chapter 4 deals with high level design issues such as requirements engineering in the solution domain; assessment and selection of technology options; project management and implementation strategy and evaluation of user......

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...The work of this year’s laureate, Shigeru Ban, has also been displayed at Vitra. Huddled on a lawn, his structures, three fifty-dollar tents sheathed in standard-issue plastic tarps from the U.N., intended for the refugees of the Rwandan civil war, looked as if any minute they might be loaded on a pallet and removed. Ban’s work lay underneath the plastic: a simple skeleton of recycled-paper tubes, fitted together with plastic joints and braced with ropes describing the pattern of an unfinished star. Ban, who has built museums, mansions, corporate headquarters, and a golf-course clubhouse in South Korea, takes pleasure in distinguishing himself from his peers, and in pointing up their excesses: not much of their work could fit into a kit that comprises eleven elements (Paper Tube A, Paper Tube B, plastic peg), including the bag. “This company has the most expensive collection of architecture,” he says. “My tents became their cheapest collection.” In a profession often associated with showmanship and ego, Ban’s work appears humble, and appropriate to a historical moment that celebrates altruism, or its posture. The Supreme Court Justice Stephen G. Breyer, a member of the Pritzker jury, told me that he was moved by Ban’s commitment to the dispossessed. “The world is filled with billions of people, and most of them live in conditions where they will never see an architect or an architect-designed space,” he said. “To have a first-rate architect pay attention to those in need......

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...PARSONS 2012–2013 CATALOG ACADEMIC CALENDAR 2012-2013 ABOUT PARSONS PROGRAMS OF STUDY PROGRAM REQUIREMENTS FACULTY ADVISING ACADEMIC POLICIES AND PROCEDURES REGISTRATION FINANCIAL INFORMATION STUDENT LIFE OTHER UNIVERSITY POLICIES ABOUT THE NEW SCHOOL 1 ACADEMIC CALENDAR 2012-2013 FALL 2012 Registration April 2–27 (Registration (Most new for continuing students) students register over the Aug. 20–24 (Registration summer for the fall term) for transfer students and late registration for continuing students) . Classes Begin Mon., Aug. 27 Last Day to Add a Class Mon., Sept. 10 Last Day to Drop a Class Tues., Sept. 18 Last Day to Withdraw From a Class With a Grade of W Undergraduate Fri., Oct. 12 Students Parsons Graduate Fri., Oct. 12 Students All Other Graduate Tues., Dec. 18 Students Holidays Labor Day Sat., Sept. 1–Mon., Sept. 3 weekend: Rosh Hashanah: Sun., Sept. 16 eve*–Mon., Sept. 17 Yom Kippur: Tues., Sept. 25 eve*–Wed., Sept 26 *Sunday and Tuesday classes scheduled for 3:50 p.m. or later do not meet. No classes meet on Monday and Wednesday. See rescheduled days below. Thanksgiving: Wed., Nov. 21–Sun., Nov. 25 Winter Break: Wed., Dec. 19–Fri., Jan. 25 Makeups and On Tuesday, Nov. 20, Rescheduled Days classes will follow the Wednesday schedule. On Tuesday, Dec. 18, daytime classes will not meet. Spring 2013 Registration Nov. 5–30 Juries Arranged by program Classes and Exams End Tues., Dec. 18 Online Session A Aug. 27–Dec. 18 Online Session B Aug. 27–Oct. 26......

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...Rachel Dillie SOS-360 Final Project March 25, 2014 Game Theory “Game theory is the study of strategic interaction among rational players in an interactive setting” (Stevens, Lecture 1). Elements of a game consist of common knowledge, the players, strategies and payoffs. Every other Friday night my wife and I play cards at our friend’s house and it generally gets more intense as the game and the drinks progress. We mainly play a card game called spades and we play as partners. Common knowledge in this game is every player knows that in order to win you have to bid the correct number of books, meaning you have to guess how many rounds you and your partner can win based off what cards you are holding and what you think the other players are holding in their hands. Each player also is assumed to be an experienced player unless otherwise stated and even if they are it’s easier to guess how the hand will play out. In this card game there is not strictly a pure strategy at play, meaning it doesn’t not involve an act of randomness but a game of mixed strategy. Mixed strategy means using a pure strategy with acts of randomness. I have an idea what the other players will do when it is their turn in the card game but sometimes to force their hand, it is necessary to randomly sacrifice one of your cards to make them display their higher ranked cards. The payoff to a player reflects what that player cares about, not what another player thinks they should care about. Being......

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