Structural Design of G+5 Building (Final Year Project for BSC in Civil Engineering)

MEKELLE UNIVERSITY

FACULTY OF SCIENCE AND TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING In Partial Fulfillment of B.Sc. Degree in Civil Engineering Structural Design of A G+4 Commercial Building with Solid & Pre-cast Slab and Cost Comparison Prepared by:Keralem Adane Osman Giragn Samuel Tesfaye Tewodros Kassa Tigistu Fisseha

Advisor:- Kibrealem Mebratu

June 2008

Senior project, structural design & cost comparison

June 2008

Table of content Acknowledgment…………………………………………………………..3 Introduction…………………………………………………………….…..4 Specification………………………………………………………………..5 1. Roof design...............................................................................................6 1.1. Wind load analysis………………………………………………...6 1.2. Analysis of lattice purlin………………………………………… 14 1.3. Design of truss……………………………………………………15 1.4. Design of lattice purloin………………………………………… 21 1.5. Slab roof design………………………………………………… 23 1.6. Weld design……………………………………………………… 25 2. Design of slab…………………………………………………………...26 2.1 Solid slab design…………………………………………………. 26 2.1.1 Depth determination……………………………………… 26 2.1.2 Loading…………………………………………………… 28 2.1.3 Analysis…………………………………………………… 29 2.1.4 Reinforcement design……………………………………… 38 2.2 Pre-cast slab design……………………………………………….39 2.2.1 Loading…………………………………………………… 40 2.2.2 Analysis and design……………………………………… 41 3. Design of stair………………………………………………………… 48 4. Frame analysis………………………………………………………… 55 4.1 Vertical load analysis…………………………………………… 55. 4.1.1 Solid slab………………………………………………… 55 4.1.2 Pre-cast slab……………………………………………… 59 4.2 Lateral load analysis…………………………………………… 62. 4.2.1 Earth quake analysis……………………………………… 62 4.2.2 Wind load analysis……………………………………… 64 4.3 Distribution of storey shear…………………………………… 67 4.4 Load combination……………………………………………… 83 5.Design of beam and column…………………………………………… 85 5.1 Beam design……………………………………………………85 5.1.1 Solid slab beams…………………………………………86 5.1.2 Pre-cast beams………………………………………… 93 5.1.3 Design of beams for shear and torsion………………… 99 5.2 Column design…………………………………………………104 5.2.1 Design procedure………………………………………. 104 5.2.2 Design of isolated columns…………………………… 111 5.2.3 Reinforcement design………………………………… 120 5.2.3.1 Solid slab column……………………………… 120 5.2.3.2 Pre-cast slab column…………………………… 121 6. Foundation design…………………………………………………… 124 Mekelle University, Department of Civil Engineering

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6.1 Footing design……………………………………………… 125 6.2 Mat foundation design…………………………………………128 7. Cost estimation…………………………………………………………135 Conclusion and recommendation…………………………………… 138 References…………………………………………………………… 139

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Acknowledgment We would like to express our sincere gratitude to our advisor Ato Kibrealem Mebratu for giving us critical advices throughout the project work. We would also like to thank our friend Fasil Meles for his material support. At last but not the least, we would like to thank to our parents for giving their countless material & moral support.

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Introduction Now a day’s it is being practical to choose types of structural members for different criteria’s especially economy after assuring safety. So in this particular project we have determined the cost variation for solid and pre-cast floor system. To do this we have passed many steps. This paper is prepared in partial fulfillment for the B.Sc. degree in civil engineering. The project is a structural design of a G+4 commercial building with solid and pre-cast slab systems with cost estimation & comparison for each staff. The building is located in Mekelle city. The structural design consists of the design of roof truss, Slabs, Staircase, Beams, Columns and foundation. The cost estimation comprises of cost for slabs, beams, columns and footings with their respective formwork. As can be seen from the architectural drawing, the floor arrangement is typical for all floors except some modification on the cantilever parts. For the solid type we first determine depth for deflection and made analysis for dead and live load as EBCS recommends. During the design of beams, columns and footings we have grouped them in reference to the stress they are bearing i.e. each group is taking relatively similar stress. Limit state design method has been adopted for the whole components. Ethiopian building code of standard EBCS 1, EBCS 2, EBCS 7 and EBCS 8, are referred for the design of the building. The roof truss, ribs and frame are analyzed using SAP 2000 V 9 for different combination of loads. And a combination with a critical effect is taken for sizing members and determination of rebar. Working drawings for beams, columns, footings stair case and floor slabs are prepared. And finally Bill of Quantity for Concrete work, Rebar and Footing are prepared. Generally in this project we have shown the basic steps for analysis and design of frame structures and we believe we have done a fabulous work which is almost accurate and we deserve a big hug.

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Specification Purpose – commercial building Approach- Limit state design method Material – Concrete – 25, class – I works Steel S – 300 deformed bars RHS for roof truss and purlin EGA- 300 for roof cover is used. Partial safety factors – concrete γc=1.5 Steel γs=1.15 Unit weight of concrete γc=24KN/m3 Supporting ground condition = sandy gravel with bearing capacity of 560KPa Design Data and Materials Concrete fck = 0.8*25MPa =20MPa fctk= 0.21*fck 2/3 =1.547MPa fcd= 0.85*fck =11.33MPa γc fctd= fctk=1.032MPa γc Steel fyk= 300MPa fcd= 260.87MPa Design loads Fd= γf*Fk Where Fk = characteristics loads γf = partial safety factor for loads = 1.3 for dead loads = 1.6 for live loads Seismic condition Mekelle – Zone 4

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1. Roof Design Roof layout

1.1 Wind load Analysis The roof is categorized according to EBCS-1,1995 table 2.1.3 under category-H :- roof not accessible except normal maintenance, repair, painting and minor repairs. From table 2.1.4 the roof will be sloping roof with category-H qk=0.25KN/m2 Qk=1.0KN

Characteristics wind load Wind pressure (EBCS-1, 1995 Art. 3.5)

I. External wind pressure Wind pressure acting on external surface of the structure will be obtained from We =qref.Ce(ze) (Cpe) Where qref =reference wind pressure 2

= ρ/2 Vref …………… (Art. 3.7.1) ρ= air density (from site with altitude above sea level) > 2000m (Mekelle) ρ=0.94 Kg/m3 Vref = reference wind velocity Cdir .Ctem .Calt .Vref o = 1*1*1*22m/s 2 qref = ρ/2 Vref =0.5* 0.94Kg/m3*222 =227.48N/m2 Ce=pressure coefficient that accounts the effect of terrain roughness, topography and height above the ground on the mean wind speed is defined as

Vref

=

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  7 * KT 2 Ce(z) = C r ( Z ) * C t ( Z ) * 1 +   C r ( Z ) * Ct ( Z )  Where KT = terrain factor - For urban area in which at least 15% of the surface is covered with buildings by their average height is 15m. KT = 0.24 Cr(ze)= roughness coefficient = KT.ln.(z/z0) for Z min≤ Z Cr(ze)=Cr(Z min) for Z22.78K* OK ! Therefore use ST-30 with thickness 2mm for lower and upper chords

Top members Nsd=8.14KN use Fe430

fy=275 Determination of buckling length L=0.41m Selection of trial section by assuming trial value of reduction factor x=0.45 8.14=0.45*1*A*275/1.1 A=72.36mm2 Nbrd = χβAfy γM1 Trial section use ST-20 with thickness t=1mm A=0.73cm2 I=0.73 cm4 20mm r=0.77cm t=1.2mm 20mm Class of x-section d/t ≤ 90 E2 17/1.2 = 14.17 ≤ 90*0.922= 76.17 Therefore the class is at least class-3 and has no problem of local buckling thus βA=1 Determine λλ1= 86.4 λx= λz= l/r =0.41*102/0.77 =53.25 λ-x= λ-z= (53.25/86.4) * 11/2 =0.616 for cold formed RHS we use curve C and the value of the reduction factor x=0.7757 calculate the design buckling resistance Nbrd = χβAfy = 0.7757*1*73*275/1.1 = 14.16K* >8.14K* OK! γM1

Bottom members Nsd=8.25KN(tension) take asection of ST-20 with thickness t=1mm A=0.73 I=0.73 Mekelle University, Department of Civil Engineering

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r=0.77 Nu,Rd=0.9Aefffy/ γM2 =0.9*73*275/1.25 =14.45KN>8.25KN

OK!

DESIG* ACTIO*S FOR THE DIAGO*AL MEMBERS Nsd=1.99KN Ncd=1.9KN fy=260.8 Material Ø8 ,S-300 Check is done for both compression and tension actions .

Check for compression resistance

Φ8 deformed bar buckling length for pin –pin support l= 289.91mm class determination :-the class of the section is taken as class -2⇒βa=1 determination of slenderness ratio (λ)






λ1=93.9є=93.9*0.92=86.4

√

λx= λy=l/r,r= (I/A), A=πD2/4=π*0.82/4=0.503cm2 I=πr4/4=0.02cm4 r=

√(0.02/0.503)=0.199≅0.2cm

λx= λy=28.91cm/0.2cm=144.55 λ-=[144.55/86.4]* βa1/2=1.673 using curve C , χ =0.2667
determination of buckling resistance Nbrd = χβAfy γM1

=0.2677*0.503*10-4m2*260.87*106N/m2 1.1 Nbrd=3.193KN > Ncd=1.9KN




OK

!

Check for tensile capcity ,Nsd=1.99KN Aeff=Agross Npl,rd=Afy =0.503*10-4m2*260.87*106N/m2 γM1 1.1

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=11.93KN >1.99KN OK! Nu,rd= 0.9Aeff*fy = 0.9*0.503*10-4m2*260.87*106N/m2 γM2 1.25 =9.45 KN > 1.99 KN OK! So the section can carry both tensile and compressive action

1.5 Slab roof Design For the design purpose we take the max positive pressure. The suction pressure can be easily counter balanced by the weight of the slab. Net positive wind pressure = 0.269 KN

pc

ps

Changing into equivalent rectangular section using Bare’s equation Equivalent rectangular section c/a= 3.9/5= 0.78 > 0.25 ar =2/3[(2c +a)*a/(a +c)] = 4.79

{

4.79

}

br = b- a*(a – c) 6*(a + c) br= 2.599≅2.6

2.6  Depth determination [From EBCS 2-1995 Art. 5.2.3] for ps d=

[0.4 + 0.6*f ]le yk

=0.85*2600/24= 108.32 mm

400 βa For pc d= 0.85*1650 =116.87 mm 12 Use for both slabs overall depth, D= d +15 +5= 116.87 + 15 +5=136.87 Use D= 150 mm  Loading :From water tank having a capacity of 5000 lit = 5m3 Mekelle University, Department of Civil Engineering

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unit wt of water= 9.81KN/ m3 P= 5*9.81= 49.05 KN Distributed load =49.05 KN =4.08 KN/ m2 12.015 m2 Self weight wt of slab = 0.15* 24KN/ m3=3.6 KN/m2 Live load =0.5KN/ m2 Wind load =0.269 KN/ m2 Pd=1.25DL+ LL + 0.8WL = 1.25*7.68 +0.5 + 0.8*.269 Pd =10.31 KN/ m2  Design of slab For ps ly/lx= 1.84 s.c =8 αxf =0.0922 αys= 0.0584 αyf = 0.045

Mi = αi*Pd*lx2 Mys =0.0584* 10.31*2.72= 4.36 KN-m Myf =0.044*10.31*2.72 =3.31 KN-m Mxf = 0.0922*10.31*2.72 = 6.85 KN-m Moment for cantilever slab ps ,pd=5.21KN/m M= 0.5*wl2= 5.21*1.65^2*0.5 =7.099 KN-m Moment adjustment

Reinforcement calculation Moment 7.091

km 20.35

ks 3.96

As 216

Asmin =ρmin*b*d= 0.002*1000*130=260mm2 Asminthe needed Ls=11.84mm So use fillet weld with length Ls=30mm and also use connected plate with Fe-430 and thickness 2mm. and also use this conection for all other joints since their design action is less than this.

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2. DESIG* OF SLAB 2.1-SOLID SLAB DESIG*

Sample floor layout 2.1.1 Depth determination The effective depth requirement for deflection can be calculated by using the following formula (EBCS- 2; 1995 Art 5.2.3) D ≥ (0.4+0.6 fyk) Le 400 βa Where fyk = is the characteristics strength of the reinforcement Le = is the effective span and for two span the shorter span Βa = is the appropriate constant from table 5.1

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Fourth floor slab Panel 1

Ly/Lx = 1 Lx = 5000mm Ly = 5000 mm βa= 45 d ≥ (0.4+0.6*300) 5000 400 45

= 94.44mm

Panel 4

Panel 8

Lx = 5000mm Ly = 5000 mm Ly/Lx = 1 βa= 40 d ≥ (0.4+0.6*300) 5000 400 40

= 106.25mm

Ly/Lx = 5/1.65 = 3.03 Mekelle University, Department of Civil Engineering

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βa= 12 d ≥ (0.4+0.6*300) 1650 400 12

= 116.88mm

First and second floor

A = ½ ЛD2 = 0.5*Л*3.82 = 5.67m 2 4 4 βa= 12 Equivalent rectangle

d ≥ (0.4+0.6*300) 1530 = 108.37mm 400 12 Comparing all the above critical panels d ≥ 116.88mm Over all depth D = 116.88 + 10 + 15 = 141.88mm Use D = 150 mm Actual depth d =150 – 10 – 15 = 125mm

For first and second floor slabs 2.1.2 Loading Unit weight -pvc – 16 KN/m3 -Cement – 23 KN/m3 -Terrazzo - 23 KN/m3 Mekelle University, Department of Civil Engineering

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- Concrete - 24 KN/m3 Dead load -pvc tile = 16 *0.008 = 0.128 KN/m2 -Cement screed = 23 *0.03 = 0.69 KN/m2 -terrazzo tile = 23 *0.02 = 0.46 KN/m2 -slab concrete = 24*0.15 = 3.6 KN/m2 -plaster = 23*0.03 = 0.69 KN/m2 -partition load = 1,2 KN/m2 DL1 = (PVC floor finish) = (0.128 + 0.69 + 3.6 + 0.69 + 1.2) KN/m2 = 6.31 KN/m2 DL2 = (Terrazzo tile floor finish = (0.46 + 0.69 + 3.6 + 0.69 + 1.2) KN/m2 = 6.64 KN/m2 Live load (from EBCS-1-1995 Table 2.9 and 2.10) Category D1 (Areas in general retail shops) = 5 KN/m2 For stairs = 3 KN/m2 Balconies = 4 KN/m2

2.1.3 Analysis Design moment calculation The support and span moment of simply supported (external edges) or fully fixed (contentious edges) are calculated as Mi = αi Pd LX2 Where; Mi = the design moment per unit width at point of reference Pd = the uniformly distributed load αi = the coefficient given in table A-1 as function of aspect ratio (Ly/Lx) and Support condition Lx = shorter span of the panel Ly = longer span of the panel Fig. Where; s = support f = field (span) x = direction of shorter span y = direction of longer span

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Panel 1

Ly/Lx =1 αxs = 0.039 αys = 0.039 αxs = 0.029 αyf = 0.029 2 DL1 = 6.31 KN/m LL = 5 KN/m2 Pd = ((1.3* DL1) + (1.6*LL)) ((1.3* 6.31) + (1.6*5)) KN/m2 = 16.203 KN/m2 Mi = αi Pd LX2 Mys = Mxs = 0.039*16.203 KN/m2 *(5m)2 = 15.8 KN.m Myf = Mxf = 0.029*16.203 KN/m2 *(5m)2 = 11.75 KN.m Panel 2,3 and 4

Ly/Lx =1 αxs = 0.032 αys = 0.032 αxs = 0.024 αyf = 0.024 2 DL1 = 6.31 KN/m LL = 5 KN/m2 Pd =16.203KN/m2 Mys = Mxs = 0.032*16.203 KN/m2 *(5m)2 = 12.96 KN.m Myf = Mxf = 0.024*16.203 KN/m2 *(5m)2 = 9.72 KN.m Mekelle University, Department of Civil Engineering

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Panel 5

Ly/Lx =1 αxs = αys = 0.039 αxs = αyf = 0.03 Pd =16.203KN/m2 Mys = Mxs = 0.039*16.203 KN/m2 *(5m)2 = 15.8 KN.m Myf = Mxf = 0.03*16.203 KN/m2 *(5m)2 = 12.15 KN.m Panel 6

Equivalent rectangular section

Let’s take equivalent rectangular area Fig Ly/Lx = 5/0.905 = 5.52m one way slab Pd =16.203KN/m2 * 1m = 16.203KN/m Fig Mxs = Pd Lx2 2 Mekelle University, Department of Civil Engineering

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= 16.203KN/m * (0,905m)2 = 6.64KN.m 2 Panel 7

Equivalent rectangular section 2

2

2

A = ½ ЛD = 0.5*Л*3.8 = 5.67m 4 4 Equivalent rectangular area Ly*Lx = 5.67m 2 Lx = 5.67m 2 = 1.49m 3.8m Ly/Lx = 3.8/1.49 = 2.55  one way slab Pd = 16 203 KN/m Fig Mxs = Pd Lx2 = 16.203KN/m* (1.49m)2 = 17.986KN.m 2 2 Panel 8

Equivalent rectangular area using Bales theorem

so,

ar =2*((2c + a) * a ) 3 (a + c) br = b – a(a – c) b(a + c) ar =2*((2* 1.2m + 1.9m) * 1.9 ) = 1.756 m 3 (1.9m + 1.2m) br = 5 m – 1.9m(1.9m – 1.2m) = 4.57 m 5(1.9m + 1.2 m)

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Ly/Lx = 4.57/ 1.756 = 2.6 > 2  one way slab Pd = 16.203KN/m Mys = Pd Lx2 = 16.203KN/m *(1.756m)2 = 24.81KN.m 2 2 Panel 9

Equivalent rectangle ar =2*((2c + a) * a ) 3 (a + c) br = b – a(a – c) b(a + c) ) = 1.043 m so, ar =2*((2* 0.525m + 1.2m) * 1.2 3 (1.2m + 0.525m) br = 3.1 m – 1.2m(1.2m – 0.525m) = 2.95 m 3.1(1.2m + 0.525 m) Ly/Lx = 2.95m/ 1.043m = 2.8 > 2  one way slab Pd = 16.203KN/m Mys = Pd Ly2 = 16.203KN/m *(1.043m)2 = 8.8132KN.m 2 Panel 10

Lx/Ly = 3.4m/ 1.35m = 2.52 > 2  one way slab Pd = 16.203KN/m Mys = Pd Ly2 = 16.203KN/m *(1.35m)2 = 14.76KN.m 2 2 Panel 11

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Equvalent rectangular area Ly/Lx = 3.25m/ 1.4m = 2.62 > 2  one way slab Pd = 16.203KN/m Mys = Pd Ly2 = 16.203KN/m *(1.4m)2 = 15.88 KN.m 2 2 Panel 12’ 13,14 and 15

Ly/Lx = 5m/ 1.55m = 3.23 > 2  one way slab Pd2 = 16.632 KN/m2 Mys = Pd2 Ly2 = 16.632KN/m *(1.55m)2 = 19.98 KN.m 2 2 Panel 17

Pd2 = 16.632 KN/m2 Mys = Pd2 Ly2 2 Panel 18

= 16.632KN/m *(1.55m)2 = 19.98 KN.m 2

Ly/Lx = 4.65m/ 1.1m = 4.23 > 2  one way slab DL2 = 6.31 KN/m2 LL = 5 KN/m2 Mekelle University, Department of Civil Engineering

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Pd = ((1.3* DL1) + (1.6*LL)) ((1.3* 6.31) + (1.6*5)) KN/m2 = 16.203 KN/m2 For one meter strip Pd =16.203KN/m Mys = Pd Lx2 = 16.203KN/m *(1.1m)2 = 2.45 KN.m 8 8 Panel 1 Is cantilever slab with length of 1.03m Pd2 = 16.632 KN/m2 Mys = Pd Ly2 = 16.632KN/m *(1.03m)2 = 8.82KN.m 2 2

Adjustment of support and span moment Support adjustment Between 1&4 and 4&5 MR = 15.8 KN.m M = 12.96 KN.m ΛM = 15.8 – 12.96 = 2.84 ΛM = 2.84 *100% = 17.97% < 20% ; Then Md =15.8 +12.86 = 14.38KN/m 2 Between 5&6 MR = 6.64 KN.m ML = 15.8 KN.m Md = 15.8 KN. m ; Since it is cantilever Span moment Panel 1&5 Mxfd = Mxf + Cx ΛM = 11.75 + 0.38*(2.84) =12.83 KN.m Panel 5 Mxfd = Mxf + Cx ΛM = 12.15 + 0.38*(2.84) = 13.58 KN.m

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After moment adjustment check depth for flexure, taking the maximum moment Mmax =24.81 KN-m

From material used the design constants c1=0.0858 fcd=11.33Mpa c2=3074 fyd= 260.87Mpa m= 28.78 ρb= 0.8Ecu * fcd ( Ecu + Eyd ) fyd b= 1000mm, Ecu= 0.0035, Eyd= (260.87/(200*103) = 0.0013 ρb=0.8*0.0035*11.33*0.75 = ( 0.0035 + 0.0013) *260.87 Check for effective depth

d=79.89< 125 mm

OK!!

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Reinforcement design Moment Km Ks As Asmin S calculated 19.98 35.76 4.108 656.62 212.5 119.55 12.83 28.66 4.03 413.64 212.5 189.78 17.986 33.92 4.09 588.50 212.5 133.39 24.81 39.85 4.16 825.68 212.5 95.07 14.38 30.34 4.05 465.91 212.5 168.49 9.72 24.94 4 311.04 212.5 252.38 12.96 28.8 4.04 418.87 212.5 187.41 14.76 30.73 4.06 479.40 212.5 163.74 15.88 31.88 4.07 517.05 212.5 151.82 13.58 29.48 4.05 439.99 212.5 178.41 2.45 12.52 3.87 75.85 212.5 236.71 8.813 23.75 3.982 280.75 212.5 179.16 6.64 20.65 3.961 210.41 212.5 236.71

S provided Φ10c/c110 Φ10c/c180 Φ10c/c130 Φ10c/c90 Φ10c/c160 Φ10c/c250 Φ10c/c180 Φ10c/c160 Φ10c/c150 Φ10c/c170 Φ8c/c220 Φ8c/c170 Φ8c/c220

Reinforcement detail

NB: The full reinforcement details are attached with AutoCAD files.

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2.2Pre-cast slab design General Precast beam slab system is a system of slab construction in which reinforced concrete precast beam elements, with their latticed reinforcement bars projected out, are used. During construction, these beam elements will be placed at certain intervals, to accommodate hollow concrete block. These blocks of specified dimensions are placed along these prefabricated beams and across the span of these elements in a similar fashion as in the case of ribbed slab construction. Concrete will be casted over the blocks and the beam elements. The projected reinforcement bars from the beam elements are used as an anchorage for the concrete, in addition to their main purpose, i.e. shear resistance. The beam elements, together with the blocks, act as formwork for the concrete casted. In addition the beam elements will acts as flexural members to carry the loads until the cast in-situ concrete attains its full strength The pre-cast beam span is 5m and it is only one type therefore from the GTZ technical manualII the section is recommended as follows

And also the cross-sectional dimensions of the HCB is given below

220mm

550mm

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Design constants and assumptions Material properties C-25 fcd = 0.85 fck in compression fctd=fck/gc in tension Therefore fcd=11.33 Mpa fctd=1.0 Mpa steel S-300 fyd=fyk/gs Hollow concrete block (HCB) g=14KN/m2

Design of pre-cast beam element For starting use Ø10 for top members and Ø8 for diagonal members

2.2.1Loading A) Initial condition

625mm Dead load – pre-cast beam and Hollow Block Pre-cast beam = 1.3*0.12*0.08*24=0.299KN/m Hollow block = 1.3*0.069*14=1.256 KN/m Live load = 1.6*2*0.625= 2KN/m qd= 0.299+1.256+2= 3.6K*/m

B)Final condition

hf

60 280 220mm

625mm Dead load – Mekelle University, Department of Civil Engineering

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Pre-cast beam 1.3*.12*.08*24 = .299 KN/m Hollow Block 1.3*.069*14 = 1.256 KN/m Cast In situ Concrete 1.3(.625*.06+2(.14*0.06))*24= 1.69KN/m Concrete Floor finishes and partition wall Partition load=0.625*1.2*1.3=0.975 KN/m Plastering = 1.3*23*0.03*0.625=0.561KN/m PVC floor finish = 1.3*16*0.002*0.625=0.104KN/m Terrazzo floor finish=1.3*23*0.02*0.625=0.374KN/m Live load =5*0.625*1.6=5KN/m qd1= 0.299+1.256+1.69+0.975+0.561+0.104+5=9.884K*/m (for PVC floor finish) qd2= 0.299+1.256+1.69+0.975+0.561+0.374+5=10.154K*/m (for terrazzo floor finish)

2.2.2Analysis and design Mmax=wl2/8=10.154*52/8=31.73KNm

Final condition 10.154K*/m

The minimum depth required for deflection d≥ (.4+.6*300/400)5000/20=212.5 Actual d=D-Ø/2 –cover = 280-16/2-15=257>212.5

OK!!!

be ≤ bw + le/5=120+5000/5=1120 actual=625 be = 625mm

Determination of neutral axis ρ= ½ c1-√(c12-4M/(c2bd2)) ρ= 0.00305 m=fyd/fcd*0.8=260.87/0.8*11.33=28.78 Mekelle University, Department of Civil Engineering

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x= ρmd=0.00305*28.78*257=22.56 y=0.8X=0.8*22.56=18.05