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İki farklı yapı sisteminin malzeme miktarları bakımından kıyaslanması

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  1. Tez No: 66783
  2. Yazar: S. SİNAN CERRAHOĞLU
  3. Danışmanlar: PROF. DR. METİN AYDOĞAN
  4. Tez Türü: Yüksek Lisans
  5. Konular: İnşaat Mühendisliği, Civil Engineering
  6. Anahtar Kelimeler: Belirtilmemiş.
  7. Yıl: 1997
  8. Dil: Türkçe
  9. Üniversite: İstanbul Teknik Üniversitesi
  10. Enstitü: Fen Bilimleri Enstitüsü
  11. Ana Bilim Dalı: İnşaat Mühendisliği Ana Bilim Dalı
  12. Bilim Dalı: Belirtilmemiş.
  13. Sayfa Sayısı: 138

Özet

ÖZET Bu çalışmada on katlı ve beş katlı kiriş plak bir yapı ile on katlı ve beş katlı tünel kalıp bir yapı olmak üzere iki farklı sistemdeki yapının yeni deprem yönetmeliğine, yatay ve düşey yüklere göre statik ve betonarme hesaplan yapılıp, malzeme miktarları bakımından karşılaştırılmıştır. Yatay yüklere göre yapılan statik hesaplarda Prof.Dr. Adnan Çakıroğlu ve Prof. Dr. Günay Özmen tarafından geliştirilen bir yaklaşık yöntem kullanılmıştır. Düşey yüklere göre statik hesaplarda ise SAP 90 programı kullanılmıştır. Kiriş plak yapılarda kat çerçeveleri her bir aksta çıkarılarak tek tek çözülmüştür. Tünel kalıp yapıda ise kirişsiz plak sonlu elemanlarla çözülmüştür. Düşey yüklerde TS 500' de belirtilen yük kombinezonları kullanılmış ve en elverişsiz durum dikkate alınmıştır. Temelde kirişli radye kullanılmıştır. Malzeme miktarları karşılaştırılmalarında ise beton, beton çeliği, tuğla ve kalıp miktarları açısından hesap yapılmıştır. xıı

Özet (Çeviri)

SUMMARY One of the leading aims of civil engineering is to use the knowledge gained from theoretical and practical studies in the field of strengthening structures by changing codes according to financial condition of countries. As an example four different structure compare each other according to financial condition. First structure is ten stories and built by tunnel form. Second structure is five stories and built by tunnel form. Third structure is ten stories and built standard system. Fourth structure is five stories and built standard system. According to vertical loads static analysis of the structure is performed by computer program SAP 90 which is a module specially improve for building type structure. According to horizontal loads static analysis of the structure is performed by approximately method developed by Prof. Dr. Adnan Çakıroğlu and Prof. Dr. Günay Özmen and new Turkish earthquake standard and Turkish code for reinforced concrete structure (TS 500). TS 500 five different load cases are concerned: Load Case 1 : Static vertical loads ; 1.4G+1.6Q Load Case 2: Static lateral loads applied along x axis in positive directions;. 1.0G+1.0Q+1.0E Load Case 3:Static lateral loads applied along x axis in negative directions; 1.0G+1.0Q-1.0E Load Case 4: Static lateral loads applied along y axis in positive directions; xiiil.OG+l.OQ+l.OE Load Case 5 : Static lateral loads applied along y axis in negative directions; 1.0G+1.0Q-L0E As told before the static analysis of the structure is done by SAP 90. The input, output and numerical solution techniques are specifically designed, taking into consideration specific characteristics that are unique to building type structures, there by giving the engineer an analytical tool that offers significant savings in the time associated with data preparation, output interpretation and execution throughput over general purpose computer programs. The standard system building is idealized as an assemblage of vertical frame and shear wall systems interconnected by horizontal floor diaphragm slabs which are rigid in their own plane. The basic frame geometry is defined with reference to a three dimensional grid system and, with special modeling techniques, very complex framing situations may be considered. The tunnel form system building is idealized as an assemblage of shear wall system interconnected by horizontal floor diaphragm slabs which are rigid in their own plane. By importing SAP 90 output files to MS Excel that is an electronic table pro- gram, reinforcements are calculated. As told before the horizontal loads analysis of the structure is done by approximately method. This method is; 5(i,i-l)X(i-l) + 5(i,i)Xi + 5(i,i+l)X(i+l) + 8(i,o)=0 in there 5(i,i-l) = f(i-l)-F(i-l) 5(i,i) = 2fi+2f(i-l)+Fi+F(i-l) XIV5(i,i+l) = fi-Fi 8(i,o) = fıM(i+l,o) + 2[fı+f(i-l)]Mi,o + f(i-l)M(i-l),o fi = hi/6(SIf)i Fi = l/(ZD)i i : Number of floor X : Unknown 5 : Displacements of floor E : Elasticity module If : Inertia moment of shear walls D : rigidly of columns According to the new seismic code, total equivalent seismic load can be calculated by the belove formula, F=m*A/R A is defined as coefficient of spectral acceleration and can be calculated by the belove formula, A=Ao*I*S Ao is defined as coefficient of effective ground acceleration. In this example, seismic area is 1 so Ao is taken 0.40.. I is importantly coefficient of structure, in this example structure is house so I is taken 1. S is coefficient of spectrum and can be calculated by the belove formulas, xvIf 0TB S=2.5(TB/T)A0.8 TA and TB are spectrum characteristic spectrum period. T is natural vibration period of structure, T=27t[Z(midiA2)/I(Fidi)]Al/2 mi=wi/g.Compare financial condition of four structure; For the all structure the concrete is chosen as C25. Floor of standard system structure steel is chosen as St I.Floor of tunnel form system structure the steel is chosen StIV. The other part of all structures steel is chosen St III. At the ten stories standard system structure, total volume of concrete needed is 1 169.5 mA3 and the total mass of steel needed is 164.05 ton. At the ten stories tunnel form system structure, total volume of concrete needed is 2081.39 mA3 and the total mass of steel needed is 242.623 ton. At the five stories standard system structure, total volume of concrete needed is 59 1. 1 73 mA3 and the total mass of steel needed is 77.675 ton. At the five stories tunnel form system structure, total volume of concrete needed is 1084.026 mA3 and the total mass of steel needed is 1 16.687 ton In order to examine the changes in the cost of the structural system when different structures are used, unit prices of the year 1997 are used. To take lost of material, occupation of worker, profit of cocnrator and factors like those into account, the analyses of prices are examined and the most realistic ones are chosen. But since the analyses are not so much parallel to Turkish Code for Reinforced Concrete Buildings (TS 500) approximately price of an upper class of material is accepted. xviCivil engineering aims to use the knowledge gained from theoretical and practical studies in the field of strengthening structures by changing codes according to those and construct earthquake resistant structures. Although economical design is desired it is obvious that saving human life is the leading aim. According to total amount of material of the structural system; Sum of them according to ten stories; tunnel form system is cheaper than standard form. According to five stories ; tunnel form system is more expensive than standard form. xvu

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