Thesis,Final Review

ADDIS ABABA UNIVERSITY ADDIS ABABA INSTITUTE OF TECHNOLOGY SCHOOL OF CIVIL AND ENVIRONMENTAL ENGINEERING

STRUCTURAL DESIGN OF A G+4 HOTEL AT ADDIS ABABA A Thesis in CIVIL ENGINEERING By:-

NAME

ID NUMBER

DANIEL ABERA

ENR/6603/04

GOYTOM KEBEDEW

ENR/3273/04

MEHARI TSEGAY

ENR/3484/04

YOHANNES AREFE

ENR/6397/04

A Thesis

Submitted in Partial Fulfillment of the Requirements for the Degree of Master of Science

STRUCTURAL DESIGN OF G+4 BUILDING 2016

The undersigned have examined the thesis entitled Structural Design of a G+4 hotel at Addis Ababa presented by

DANIEL ABERA, GOYTOM KEBEDEW, MEHARI TSEGAY, YOHANNES AREFE , a candidate for the degree of Bachelor of Science and hereby certify that it is worthy of acceptance.

Fresenay Zerabruk Advisor

Signature

Date

Internal Examiner

Signature

Date

External Examiner

Signature

Date

Chair person

Signature

Date

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STRUCTURAL DESIGN OF G+4 BUILDING 2016

UNDERTAKING We certify that research work titled Structural Design of a G+4 Hotel Building at Addis Ababa is my own work. The work has not been presented elsewhere for assessment. Where material has been used from other sources it has been properly acknowledged / referred.

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STRUCTURAL DESIGN OF G+4 BUILDING 2016

ABSTRACT This project is mainly concerned with the structural design and analysis of a G+4 building intended for the purpose of Hotel. The Analysis is computed using ETABs V9.6 software and the Design is done based on the Limitations and standards listed on EBCS 1995.

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STRUCTURAL DESIGN OF G+4 BUILDING 2016

ACKNOWLEDGMENTS Primarily, we want to specially thank the Almighty GOD for giving us the inspiration to start and patience to finalize this project work. Secondly, we want to extend our sincere appreciation to our advisor, Ato FIRESENAY for his valuable advice, constant support, dedication, encouragement and precious guidance, creative suggestions and critical comments, and for being everlasting enthusiastic from the beginning to the end of the project. Moreover, it will not be out place here to express special thanks to our dearest family, for their consistent and continuous advises, support, and encouragements valuable not only for the academic achievement but also for life lasting successfulness forwarded me during my stay in student life starting from high school to the university level. Thirdly, we are also grateful to all staffs‟ of Addis Ababa University, Addis Ababa Institute Of Technology (AAIT) for their heartily cooperation during our stay of five year in the University. Last, but not least, we would like to thanks for all our friends, and colleagues, for their love, encouragement, patience and support throughout the project study.

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STRUCTURAL DESIGN OF G+4 BUILDING 2016

Contents 1 INTRODUCTION ....................................................................................................................................................1 1.1 General................................................................................................................................................................................. 1 1.2 Limit States.......................................................................................................................................................................... 2 1.3 Overview of The Project ..................................................................................................................................................... 3 1.4 Design Consideration .......................................................................................................................................................... 4 1.5 Material Properties ............................................................................................................................................................. 5

2 SLAB ANALYSIS AND DESIGN ..........................................................................................................................7 2.1 General................................................................................................................................................................................. 7 2.2 Depth Determination .......................................................................................................................................................... 7 2.3 Load Calculations ............................................................................................................................................................... 9 2.4 Analysis .............................................................................................................................................................................. 11 2.5 Design ................................................................................................................................................................................. 14 2.6 Load Transfer to Beams ................................................................................................................................................... 17

3 DESIGN OF STAIR CASE ................................................................................................................................... 20 3.1 Depth Determination ........................................................................................................................................................ 20 3.2 Loading .............................................................................................................................................................................. 20 3.3 Design Moment Calculation ............................................................................................................................................. 21 3.4 Reinforcement Calculation ............................................................................................................................................... 22

4 LATERAL LOAD ANALYSIS ............................................................................................................................. 23 4.1 Earthquake Analysis ......................................................................................................................................................... 23 4.2 Base Shear Calculation ..................................................................................................................................................... 26 4.3 Story Shear Calculation.................................................................................................................................................... 27

5 FRAME ANALYSIS AND MODELING ............................................................................................................. 30 5.1 Modeling for 3D Frame Analysis Using ETABS 2013 ................................................................................................... 30 5.2 Load Combinations ........................................................................................................................................................... 33 5.3 Drift Analysis..................................................................................................................................................................... 34

6 BEAM ANALYSIS AND DESIGN ....................................................................................................................... 38 6.1 General............................................................................................................................................................................... 38

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6.2 Flexure Theory .................................................................................................................................................................. 38 6.2.1 Design of Beam for Flexure ............................................................................................................................................................ 39

6.3 Shear .................................................................................................................................................................................. 48 6.3.1 Design for Shear .............................................................................................................................................................................. 50 6.3.2 Minimum Shear Rebar ................................................................................................................................................................... 54

6.4 Bond and Development Length ........................................................................................................................................ 57 6.5 Serviceability ..................................................................................................................................................................... 59 6.5.1 Check for Crack .............................................................................................................................................................................. 61 6.5.2 Check for Deflection........................................................................................................................................................................ 68

7 COLUMN DESIGN AND ANALYSIS ................................................................................................................ 70 7.1 Introduction ....................................................................................................................................................................... 70 7.2 Design of columns ............................................................................................................................................................. 71 7.2.1 Design procedure............................................................................................................................................................................. 71 Slenderness.................................................................................................................................................................................................. 76

8 FOUNDATION ANALYSIS AND DESIGN ..................................................................................................... 112 8.1 Design philosophy ........................................................................................................................................................... 114 8.2 Design of Isolated Footing .............................................................................................................................................. 115

9 REINFORCEMENT DETAILS ......................................................................................................................... 125 9.1 Slab Detailing .................................................................................................................................................................. 125 9.2 Typical Beam Detailing .................................................................................................................................................. 129 9.3 Ground Beam .................................................................................................................................................................. 131 9.4 Roof Beam ....................................................................................................................................................................... 132 9.5 Foundation Detailing ...................................................................................................................................................... 133 9.6 Column Detailing ............................................................................................................................................................ 139

10 CONCLUSION ................................................................................................................................................... 141 11 RECOMMENDATION ..................................................................................................................................... 142 12 APPENDICES .................................................................................................................................................... 143 12.1 Determination of total building weight and Center mass ........................................................................................... 143

13 REFFERANCE ................................................................................................................................................... 156

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STRUCTURAL DESIGN OF G+4 BUILDING 2016

Tables Table 1 Depth determination of slab ............................................................................................... 8 Table 2 Slab self weight and finishing ............................................................................................ 9 Table 3 Support and field moments of slab .................................................................................. 12 Table 4 Design of slab support moments in the X direction......................................................... 16 Table 5 Design of slab support moments in the Y direction......................................................... 16 Table 6 Design of slab field moments in the X & Y direction ..................................................... 17 Table 7 Load transfer to beams ..................................................................................................... 18 Table 8 Stair design result............................................................................................................. 22 Table 9 Total building weight ....................................................................................................... 28 Table 10 Story shear distribution .................................................................................................. 28 Table 11 Center of mass of floors ................................................................................................. 29 Table 12 Drift analysis result ........................................................................................................ 35 Table 13 Typical floor beams design result .................................................................................. 44 Table 14 Typical floor beams shear design result ........................................................................ 55 Table 15Column 4C design result............................................................................................... 100 Table 16 Column 5D design result ............................................................................................. 103 Table 17 Column 5E design result .............................................................................................. 107 Table 18 Column design result ................................................................................................... 110 Table 19 Footing 4C,5D & 5E design result............................................................................... 123 Table 20 Isolated footing design result ....................................................................................... 124 Table 21 Typical Floor Weight and Center Mass ....................................................................... 143 Table 22 Ground Floor Weight and Center Mass ....................................................................... 147 Table 23 Roof Weight ................................................................................................................. 151

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Figures Figure 1 Typical floor slabs ............................................................................................................ 7 Figure 2 Stiffness modifier for beam  ....................................................................................... 31 Figure 3 Stiffness modifier for column ......................................................................................... 31 Figure 4 3-D Model ..................................................................................................................... 37 Figure 5 Beam on axis 4 ............................................................................................................... 39 Figure 6 Moment envelop of beam on axis 4 ............................................................................... 39 Figure 7 Shear force diagram of beam on axis 4 b/n A&B .......................................................... 50 Figure 8 Bending moment diagram of beam for serviceability .................................................... 62

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1 INTRODUCTION 1.1 General Structures shall be designed appropriately so that they will sustain all actions and influences likely to occur during their intended life. It is practical to choose types of structural members for different criteria especially with regards to economy after assuring safety. They have to remain fit for their intended purpose with adequate durability. It must also optimize the cost expended in building the structure and for maintenance. If damages occur, they shall be minimized or avoided by providing appropriate solutions such as: - avoiding, eliminating or reducing the hazards which the structure is to sustain - selecting a structural form which has low sensitivity to the hazards considered - selecting a structural form and design that can survive adequately the accidental removal of an individual element - tying the structure together Design of a certain structure involves determination of cross sectional dimensions, area of steel and their distribution and the area and spacing of transverse bars satisfying all strength and service equipment. In case the structure fails, it must be in such a way it will minimize risks. It must extend the time for evacuation of people inside a building. This requirement of structural design is accomplished by a principle called ductility. Ductility allows yielding of steel reinforcement prior to the collapse of the building. Yielding of steel bars warns the start of failure of a structure or its part. Therefore, structures are designed to be under reinforced by certain percent to assure ductility mode of failure if it happens. The design of any structure is categorized in to functional design and structural design. Functional design: the structure to be constructed should primarily serve the basic purpose for which it is to be used and must have a pleasing look.

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Structural design: it is an art and science of understanding the behavior of structural members subjected to loads and designing them with economy and elegance to give a safe, serviceable and durable structure. Design situations The severe conditions which can be foreseen to occur in the life time of the building include: 1. Persistent and transient situations 2. Seismic situations 3. Accidental situation This project is executed based on the Ethiopian Building Code Standard (EBCS) prepared in 1995 E.C, which follows the Limit State design approach.

1.2 Limit States When a structure or structural element becomes unfit for its intended use, it is said to have reached a limit state. Limit states can be divided into three basic groups. a) Ultimate limit states: involve a structural collapse of part or all of the structure. Its main concern is the safety of structure and people. Such a limit state should have a very low probability of occurrence, because it may lead to loss of life and major financial losses. The major ultimate limit states are:  loss of equilibrium  rupture  progressive collapse  formation of a plastic mechanism  instability  fatigue b) Serviceability limit states: involve disruption of the functional use of the structure, but not collapse per se. Serviceability Limit states are those associated to conditions beyond for which a structure does not accomplish specified service requirements. It is mainly concerned about the function of construction works, comfort of people, and appearance of the building. Because there is less danger of loss of life, a higher probability of

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occurrence can generally be tolerated than in the case of an ultimate limit state. The major serviceability limit states include:  excessive deflections  excessive crack widths  undesirable vibrations c) Special limit states: involves damage or failure due to abnormal conditions or abnormal loadings and includes:  damage or collapse in extreme earthquakes,  structural effects of fire, explosions, or vehicular collisions,  structural effects of corrosion or deterioration, and  long-term physical or chemical instability

1.3 Overview of The Project This project deals about the structural design and analysis of a G+4 building located at Addis Ababa with soil class B, without a basement floor, considering live load and dead load analysis and all the external effects according to EBCS, 1995. The building has typical floor of 252 squared meter area (18m*14m). All slabs including the ground floor have a typical Hotel function and identical partitions. As a result of the natural stability of the ground below the foundation it will be designed with simple isolated footings. The structural design of this typical building involves design of solid slab for the floors, stairs, frames analysis and lateral load analysis beams, columns and foundation. In the design process, first the minimum depth of slab for serviceability limit state was determined. The slabs were designed for partition load, floor finish load along with its selfweight and live loads. The ground and typical floor slabs were designed using coefficient method. Stairs and landings were designed as one-way slab. 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 of the selected axis. Beams and columns were designed according to EBCS-2, 1995 provisions.

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To simplify the design procedure, calculations were done using designed MS-Excel spreadsheets. The size of the footing was determined from the bearing capacity of the soil; the thickness of the footing is determined from punching and wide beam shear. For the analysis of frames, the restrained conditions at the foundation level are assumed fixed. Loads acting on beams from slab reactions and partition walls directly resting on beams and lateral load acting on the frame were added to self-weight of beams to find total load acting on beams. All the significant loads are inserted and analyzed for nine load combinations using ETABS Nonlinear V9.6 Design criteria To analyze or design a structure, it is necessary to establish criteria for determining whether a given structure is acceptable for use in a specified circumstance or for use directly as a design objective that must be met.

1.4

Design Consideration

Safety:

Safety implies the likelihood of partial or total collapse of the structure is acceptably

low not only under normal expected loads (service loads), but also under abnormal but probable overloads (such as due to earthquake or extreme wind). Collapse may occur due to various possibilities such as exceeding the load bearing capacity, overturning, sliding, buckling, fatigue etc. Serviceability: Serviceability implies satisfactory performance of the structure under service loads, without discomfort to the user due to excessive deflection, cracking, vibration etc. Other considerations that come under the preview of serviceability are durability, acoustics and thermal insulation. The structure must be able to carry the design load safely without excessive material distress and with deformations with in an acceptable range. The ability of a structure to carry loads safely and without material distress is achieved by using safety factors in the design of the

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element. By altering the size, shape, and choice of material, stresses in a structure can be maintained at safe levels and such that material distress.

1.5 Material Properties Concrete The main measure of structural quality of concrete is its compressive strength. Our code EBCS2 -1995 recommends concrete grade based on a test of 150mm cube at the age of 28 days in terms of its characteristic compressive strength (f cu ). Class I workmanship and ordinary loading condition is used. In which values are measured by weight and using mixer. It requires less safety factor. Concrete grade C-25 C denotes the characteristic compressive strength in MPa. Partial safety factor

= 1.5

Compressive strength: fck = 0.8*25 = 20MPa where fck is the characteristic compressive strength of cylinder tests.

=

= 11.33MPa where the design characteristic compressive strength of cylinder tests.

Characteristic tensile strength As it is difficult to obtain accurate data because of hardening problems empirical relations are used to obtain tensile strength. = 0.21(fck) 2/3 =1.5 MPa = 0.21

= 0.21

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= 1.03MPa where

= tensile strength of concrete fck = characteristic cylindrical compressive strength in MPa

The ultimate stress in concrete for design purpose is: = 0.85fck The ultimate strain in concrete for design purpose is taken as 0.0035. The unit weight of concrete is 24 KN/m 3. Reinforcement steel Characteristic properties of reinforcement bar are expressed using its yielding strength

and is

given as: Steel grade S-300

,where S - characteristic strength of steel in MPa.

Partial safety factor:

= 1.15

= 300Mpa where

=

fyk 300 =  s 1.15

- characteristic tensile strength of steel

= 260.87MPa.

Where fyd is the design tensile strength of steel Es = 200GPa

where Es = Modulus of elasticity of steel.

Partial safety factor for actions in building structure for persistent and transient design situation is taken for unfavorable condition. Factor of safety for permanent and variable loading condition are 1.3 and 1.6 respectively (EBCS 2 table 3.3). Generally, a Hotel ground plus four (G+4) building will be designed in the pages that follow with a solid slab and frame combining beams and columns, and foundation with isolated footing. AAiT:BSc thesis

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2 SLAB ANALYSIS AND DESIGN 2.1

General

Slabs are horizontal structural elements which transfer service loads to the frame elements. There are two types of slabs based on the load transferring mechanisms. These are one way and two way slabs. One-way slabs transmit their load in one direction while two way slabs resist applied two directions. These types of slabs are composed of rectangular panels supported at all four edges by walls or beams stiff enough to be treated as unyielding. And their design should follow procedures. 

Figure 1 Typical floor slabs

2.2 Depth Determination The minimum depth required for the slab can be calculated from the minimum depth required for deflection. The effective depth requirement for deflection can be calculated using the following formula (EBCS – 2 – 1995 Article 5.2.3) d ≥ (0.4 + 0.6*

AAiT:BSc thesis

fyk Le ) 400  a

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STRUCTURAL DESIGN OF G+4 BUILDING 2016

Where: fyk – is the characteristic strength of the reinforcing bars. Le – is the effective span. For two-way solid slabs it is the shorter span βa - is the appropriate constant which depends on the support condition of the slab Note: For the purpose of construction simplicity and monolithic construction the governing overall depth has been taken. Table 1 Depth determination of slab

Support Panel

Ly(mm) Le(mm) Ly/Lx

condition

Provided "d" Βa

d(mm)

(mm)

C1

4500

1500

3 Cantilever

12

106.25

107

C2

4500

1500

3 Cantilever

12

106.25

107

C3

4500

1500

3 Cantilever

12

106.25

107

C4

4500

1500

3 Cantilever

12

106.25

107

C5

4500

2000

2.74 Cantilever

12

141.67

142

C6

4500

2000

2.74 Cantilever

12

141.67

142

C7

4500

1500

3 Cantilever

12

106.25

107

Support Panel

Ly(mm) Le(mm) Ly/Lx

condition

Provided "d" Βa

d(mm)

(mm)

P1

6500

450

1.44 Edge span

35.6

107.44

108

P2

6500

450

1.44 Edge span

35.6

107.44

108

P3

6500

450

1.44 Edge span

35.6

107.44

108

P4

4500

450

1 Edge span

40

95.6

96

P5

4500

450

1 Interior span

40

95.6

96

P6

4500

450

1 Interior span

45

85

85

P7

4500

450

1 Edge span

40

95.6

96

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STRUCTURAL DESIGN OF G+4 BUILDING 2016 Using ∅12 rebar and concrete cover of 15mm ,Total depth (D) is calculated as: D=Cover + = 15+

 +d 2

12 + 142 2

= 163mm

Use D=165mm for C4,C5,C6 & C7

For C1,C2,C3 and P1-P7 D= 15+

12 + 108 = 129mm 2

Use D=150mm for ease of electrical installation

2.3 Load Calculations a. Dead load (DL) The dead load is composed of the self-weight of the slab itself, weights of the partition walls, weight of the finishing and other considerable permanent loads. Self-weight of the slab is equal to the overall depth times unit weight of concrete. Partition loads are distributed over the slab if they are not large enough to cause localized effects. However, for line load effect Renold method should be checked. For C1, C2, C3 and P1-P7

For C4, C5, C6 & C7 Table 2 Slab self weight and finishing

Depth (m) Own weight Cement screed Plastering Ceramic tile

0.15

Unit weight (KN/m3) 25

load Depth

(KN/m2)

(m) 3.75

Own weight

0.03

23

0.69

Cement

0.02

23

0.46

screed

0.008

23

0.184

Ceramic

5.084

tile

Plastering

Total

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Unit weight (KN/m3)

load (KN/m2)

0.165

25

4.125

0.03

23

0.69

0.02

23

0.46

0.008

23

0.184

Total

5.459

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Self-Weight and finishing Partition Load From HCB wall = height * thickness * unit weight = 3.05m * 0.2m * 10 KN/m3= 6.1 KN/m For P1 and P3, by taking partition length and panel area from the architectural drawing we get 6.1 KN/m *partition length/Panel area= 6.1 KN/m*

8.64m =1.8 KN/m2 29.28m2

Similarly; For P2 …………………1.47 KN/m2

,

P4, P5, P6 & P7 ………………….0.82 KN/m2

C1-C7……………….0.99 KN/m2 For P1 & P3 since 1.8KN/m2 > 1.5 KN/m2

use 1.8 KN/m2

For the rest use 1.5 KN/m2 which is minimum because they are less b. Live load (LL) Since the building is multifunctional the live loads are different depending on the function of the building. According to EBCS 1(1995) we have live loads as following: P1-P7 ……………………. 2 KN/m2 C1-C7 …………………….. 4 KN/m2 c. Design load (Pd) The design load is factored according to the following formula Pd  1.3Gk  1.6Qk

Where,

Pd = design load Gk = total dead load on slab Qk = total live load on slab

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For P1 & P3

Pd  1.3 5.084  1.8  1.6*2   12.15 KN / m2 For P2,P4,P5,P6 & P7

Pd  1.3 5.084  1.5  1.6*2   11.76 KN / m2 For C1-C3

Pd  1.3 5.084  1.5  1.6*4   14.96 KN / m2 For C4-C7

Pd  1.3 5.459  1.5  1.6*4   15.45 KN / m2

2.4 Analysis Analysis of the design moment will be done as per the EBCS-2-1995 Art A.3.2 for two-way solid slabs and for one way solid slabs the calculation will be performed as 1m wide beam. The analysis of slab moments of two way slabs is accomplished by coefficient method using the formula: Mi  α i Pd Lx 2

Where, Mi = the design moment per unit width at the point of reference α i = the coefficient given in Table A-1 in EBCS2-1995. Pd = the design load Lx = the shorter span of the panel Ly = is the longer span of the panel s = support f = span x = direction of shorter span y = direction of longer span

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STRUCTURAL DESIGN OF G+4 BUILDING 2016 

Reading αi for each panel from EBCS 2, support and span moments of P1-P7 are shown below: Table 3 Support and field moments of slab

For the cantilevers the moments are calculated using equilibrium equation. For C1-C3 6.1KN 14.96 KN/m Mxs

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1.5m

14.96 1.5  Mxs   6.1*1.5  2

2

Mxs = 25.98 KNm

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For C4 & C7 6.1KN

15.45 1.5  Mxs   6.1*1.5  + 2

15.45 KN/m Mxs

2

Mxs  26.53 KNm

1.5m

For C5 & C6 6.1KN 15.45 KN/m Mxs

2m

15.45  2  Mxs   6.1*2  + 2

2

Mxs  43.1 KNm

Moment Adjustment 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. According to EBCS 2, 1995 A.3.3.2, there are two cases A. If ΔM < 20% of the larger moment, the design moment is the average of the two B. If ΔM ≥ 20% then the unbalanced moment is distributed based on their stiffness, use Moment Distribution Method. When using this method: 

Unbalanced moment is distributed using the moment distribution method



Relative stiffness of each panel shall be taken proportional to its gross moment of inertia divided by the smaller span

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2.5 Design For design, use C-25 concrete and S-300 steel which means fcu= 25MPa & fyk= 300 MPa fcd=

0.68* f cu 0.68*25 MPa =  11.33 MPa c 1.5

d=D-cover-

fyd=

f yk s

=

300 = 260.87 MPa 1.15

 12 =150-15- = 129mm 2 2

Sample is shown For Panel 1 and then summarized in a tabular form for the rest. For P1

Mxs1 = Mxs2 = Mys2 = 0 Mys1 = 12.732 KNm

Step 1 Evaluate μsds

μsds 

=

M sd fcd  b  d 2

12.732 KNm 11.33MPa *1000 mm 129 mm  *1000 2

= 0.067508669  sds*  0.295 Step 2 Read Kz from the design chart of EBCS 1995, using sds  0.0675 Kz  0.955

Step 3 determine Z from the following equation Z  Kz * d  0.955*0.129m

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STRUCTURAL DESIGN OF G+4 BUILDING 2016 Z  0.123m

Step 4 Calculate area of steel (As)

M sd 12.732*1000 Nm = Z * f yd 0.123m * 260.87 MPa

As =

As  396.169 mm2

Step 5 Calculate spacing of rebars

b * as S= As S=

as =

 D2 4

=

 10  4

2

= 78.54 mm2

1000mm *78.54 mm2 396.169mm2

S  198.15 mm

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Design for support moments In the X direction Table 4 Design of slab support moments in the X direction

As slab

d

Mxs1

p1

0.129

0

p2

0.129

p3

Μsds

Kz

Z

(mm2)

S in mm

0

1

0.129

0

18.399

0.097557

0.94

0.12126

581.6386

134.963526

0.129

18.399

0.097557

0.94

0.12126

581.6386

134.963526

p4

0.129

12.732

0.067509

0.952

0.122808

397.4171

197.525468

p5

0.129

9.287

0.049242

0.96

0.12384

287.469

273.072879

p6

0.129

10.24

0.054295

0.96

0.12384

316.9681

247.658967

p7

0.129

11.379

0.060335

0.96

0.12384

352.2246

222.86913

In the Y direction Table 5 Design of slab support moments in the Y direction

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Design for field moments Both in the X and Y direction Table 6 Design of slab field moments in the X & Y direction

2.6 Load Transfer to Beams To transfer load from slab to the beams un factored slab DL and LL are computed separately and transferred to the beam as shear using the following formula

Vx  vx  gd  qd  Lx

Vy  vy  gd  qd  Lx Where, gd and qd are un factored dead and live loads respectively βvx and βvy are read from table based on their support condition and span ratio

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Dead load transfer to beam, from P1-P7 Table 7 Load transfer to beams

Live load transfer to beam , from P1-P7

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Live and dead load transfer to beam From cantilevers

Live and dead load transfer to beam From Stair

Slab

Lx

gd

qd

stair 1

7.074

3

B1=33.332 LB=54.509

stair 2

8.074

3

B1=31.668 LB=64.623 B1=16.384 LB=26.224

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Load transferred B1=20.24

LB=22.364

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STRUCTURAL DESIGN OF G+4 BUILDING 2016

3 DESIGN OF STAIR CASE

Stair 1 is between ground floor and REFPL1(first flight) Stair 2 is between REFPL1 and fourth floor

3.1 Depth Determination For Stair 1 Le=6.5m ßa=30 d=0.85Le/ßa =>d=184.1mm D=184.1+21 =206.1mm use D=210mm ForStair 2 Le=6.5m ßa=25 d=0.85Le/ßa =>d=221mm D=221+21 =242mm use D=250mm 3.2

Loading

Live load take LL=3KN/m2 according to EBCS2 1995

,

Dead load on the stair (taking 1m strip) Own weight=25*0.21=5.25 KN/m Cement Screed=23*0.03=0.69 KN/m Plastering=23*0.02=0.46 KN/m Marble=27*0.025=0.674 KN/m AAiT:BSc thesis

Page 20

STRUCTURAL DESIGN OF G+4 BUILDING 2016

Total Dead load on the stair 1 =7.074KN/m Total Dead load on the stair 2 =8.074KN/m Partition Load 1.5 KN/m2 for shaded area B HCB=0.2m*3.05m*10KN/m3=6.1KN/m Factoed HCB load=6.1KN/m*1.3=7.93 Design Load On stair 1=1.3*7.074+1.6*3 = 14KN/M On stair 2=1.3*8.074+1.6*3= 15.3KN/m

3.3 Design Moment Calculation 7.93KN 14KN/m

6.5m

16.77KN/m

1.5m

14KN/m

6.5m

M at Stair 1=78.45 16.77KN/m

7.93KN

1.5m M at landing for stair 1 and 2=16.77*(1.52)/2+7.93*1.5=30.76 16.77KN/m

7.93KN

1.8m M at landing for stair at section B=16.77*(1.82)/2+7.93*1.8=41.44 AAiT:BSc thesis

Page 21

STRUCTURAL DESIGN OF G+4 BUILDING 2016

15.33KN/m

16.77KN/m

6.5m

7.93KN

1.5m

M at stair 2

15.3KN/m

6.5m M at Stair 2=(15.3*6.52)/8=80.8KNm

3.4 Reinforcement Calculation Table 8 Stair design result

Notation

d

D

mxs1

Μsds

Kz

Z

As in mm^2

S in mm

Stair 1

0.189

210

78.45

0.19378122

0.886

0.167454

1795.866327

85.67452802

Stair 2

0.229

250

80.8

0.1359511

0.922

0.211138

1466.971049

104.8827788

AAiT:BSc thesis

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STRUCTURAL DESIGN OF G+4 BUILDING 2016

4 LATERAL LOAD ANALYSIS In designing a building, various forces which act on the structures in different modes should be considered. Lateral loads are one of the modes of forces and they include: 1. Wind load 2. Earthquake 3. Earth pressure Wind load Analysis Wind actions are fluctuating with time. They act directly on the external surfaces of enclosed structures and, through porosity of the external surface, also act indirectly on the internal surfaces. They may also directly affect the internal surface of open structures. Pressures act on areas of the surface providing forces normal to the surface for the structure or for individual cladding components. Additionally, when large areas of structures are swept by the wind, frictional forces acting tangentially to the surface, may be significant. For this project, only Earthquake load is considered since it is clearly the dominant one at our site. Plus Earth quake is not likely to occur simultaneously with wind or maximum flood or maximum sea waves, so we design for earthquake alone.

4.1

Earthquake Analysis

Plate tectonics theory visualizes the earth as consisting of a viscous, molten magma core with a number of lower-density rock plates floating on it. Earthquakes result from the sudden movement of these tectonic plates in the earth‟s crust. Once this movement has started, energy is released rapidly, causing intense vibrations to propagate out from the fault. As a result, the effects of these seismic waves and local soil conditions will lead to different ground motions at various sites. Earthquakes may involve regions of slip and/or offsets along surface faults. Earthquake ground motions impart vertical and horizontal accelerations, a, to the base of a structure. If the structure was completely rigid, forces of magnitude F=ma would be generated in it, where m is the mass of the structure. Because real structures are not rigid, the actual forces generated will differ from this value depending on the period of the building and the dominant periods of the earthquake ground motions. The determination of the seismic force, E, is made more complicated because recorded earthquake ground motions contain a wide range of frequencies and maximum values of base acceleration. Lateral seismic forces are closely related AAiT:BSc thesis

Page 23

STRUCTURAL DESIGN OF G+4 BUILDING 2016

to the fundamental period of vibration of the building. As the ground moves, inertia tends to keep structures in place, resulting in the imposition of displacements and forces that can have catastrophic results. -the closer the frequency of the ground motion is to one of the natural frequencies of a structure, the greater the likelihood of the structure experiencing resonance, resulting in an increase in both displacement and damage. -the taller a structure, the more susceptible it is to the effects of higher modes of vibration. The purpose of seismic design is to proportion structures so that they can withstand the displacements and the forces induced by the ground motion. Design for earthquakes differs from design for gravity and wind loads in the relatively greater sensitivity of earthquake-induced forces to the geometry of the structure. Without careful design, forces and displacements can be concentrated in portions of a structure that are not capable of providing adequate strength or ductility. Steps to strengthen a member for one type of loading may actually increase the forces in the member and change the mode of failure from ductile to brittle. Earthquake force determination To calculate Base shear force (Fb) we take height of the building:i) From the bottom of the basement if the building have a basement. ii) From the bottom of the ground floor if the building does not have a basement. iii) From the bottom of the foundation if the building foundation have different elevation difference that are cover and exposed to the surface. Ductility Class (KD) have three different values. i) KD= 1.0 is high ductility which implies that (1) The structure is in plastic range. (2) It require low reinforcement while it is constructed with small cross sectional area of frame structure. (3) It is most economical design but it also requires one of the most

complicated

Reinforcement detailing. ii) KD= 1.5 is medium ductility iii) KD= 2.0 is low ductility which implies that (1) The structure is in elastic range. AAiT:BSc thesis

Page 24

STRUCTURAL DESIGN OF G+4 BUILDING 2016

(2) It require high reinforcement while it is constructed with large cross sectional area of frame structure. (3) It is not economical design and have normal reinforcement detailing. Importance Factor (I) i) The magnitude of importance factor that we will use in earth quake design depends on building purpose or extent of building failure after anticipated earth quake hits. ii) We use high value of importance factor (I) for hospitals and nuclear plant because we do not allow any structural failure to these types of structures. After the earthquake hits, these buildings should remain fully functional for their intended purposes. iii) Using High value of importance factor (I) put the structure into elastic range with larger columns. Moment resisting factor (C1) i) Have two values, for Reinforced Concrete C1=0.075 and for Steel C1=0.085. Higher value of moment resisting factor gives higher Fundamental Period of Building (T1). ii) We use lower value for Reinforced concrete because the connection of beams to columns is homogeneously casted. It has to be higher value for steel structure because the connection is relatively weaker as it is bolt or welding connection. Soil Type (S) i) It is found between the foundation of the building and bed rock. ii) If the soil acceleration right below the foundation is too close or equal to bed rock acceleration the magnitude of Earth Quake to the structure will be magnified. To avoid such type of problem for example in Dubai where an exposed to earth quake foundation engineers set soil acceleration to design standard. First by digging out and changing soil below the foundation with different type of soil and second by compacting the soil to the required density. Wiping Effect (Ft) i) Such force come from second mode (Dynamic Analysis) which descends from top of the building to the foundation. ii) It has to be added at the top because it is the only larger value to be considered. Other second mode forces to different building floors are insignificant to be taken to calculation. iii) The whole structure of the building is resisted from Earth quake by its frame system. The buildings lateral load distribution is computed as follows. AAiT:BSc thesis

Page 25

STRUCTURAL DESIGN OF G+4 BUILDING 2016

4.2 Base Shear Calculation According to EBCS-8, 1995 static method of analysis, the seismic base shear force Fb for each main direction is determined by the equation: Fb=Sd*(T1)*W Where:Sd*(T1) = Ordinate of the design spectrum T1= the fundamental period of vibration of the structure (in seconds) for translational motion in the direction of the motion. W= seismic dead load computed. The influence of local ground conditions, soil type, Seismic zonal subdivision of the building in accordance to EBCS-8 1995 Article 1.4 shall be considered for the design. For linear analysis, the design spectrum Sd*(T1) normalized by the acceleration of gravity g is defined by the following expression:Sd*(T) = αβγ

(EBCS-8, 1995 Art 1.4.2.2 (4))

Where:α= the ratio of the design bed rock acceleration to the acceleration of gravity g given by α =αo*I αo = bed rock acceleration ratio for the site and depend on the Seismic zone and I = the importance factor (EBCS-8, 1995 Table 1.1 and Table 2.4 respectively) Since our building is located at Addis Ababa which is zone 2, we use the following values αo = 0.05 and use I = 1.0 α= 0.05*1.0 =0.05 β= the design response factor for the site and is given by:β=1.2*S/ (T2/3) ≤2.5 S= site coefficient for the soil characteristics. In our case our soil was Soil class B S=1.2 For building up to 80m height the value of T1 may be approximated by the formula:T1 = C1 *H3/4 (sec) Where:H = Height of the building (m) = 16m C1= 0.075 AAiT:BSc thesis

Page 26

STRUCTURAL DESIGN OF G+4 BUILDING 2016 T1 = 0.075*(163/4) = 0.6 β= 1.2*1.2/ (0.62/3)=2.0242 As ,+ve = 351.69 mm2 100

Therefore take As = 628.3185 mm2 to calculate No of rebar for the positive reinforcement i.e. No of bars =

As As 628.3185 = 2= = 4.082 ……….. Use 5 ∅14 rebars 49 as  r

Similarly for the negative moment on the left (129.75 KNm) , we get 3 ∅20 rebars  design for flexure of the rest moments is summarized in the table below, which is done on excel using template.

AAiT:BSc thesis

Page 43

STRUCTURAL DESIGN OF G+4 BUILDING 2016

For the typical beam Table 13 Typical floor beams design result Beam on Axis

As b/n

μsd,s

Moment(KN) +ve

b/n 2&3

0

b/n 3&4

91.4 6

-ve 136.09

+ve 0

136.09 0.031 152.85

A b/n 4&5

66.2 7

152.85 0.017 152.85

b/n 5&6

0

b/n 2&3

0

b/n 3&4

79.0 3

0

65.8 4

C

0

b/n

0

AAiT:BSc thesis

+ve 0

-ve 0

+ve

-ve 0

positive 0

0

944.974

0

1256.637

0

0

0

0

0

0.164

0

944.974

0

1256.637

0

0

0

0

0

588.697

0

615.7522

0

628.3185

5

769.6902

0.185

0

1075.597

0

1256.637

0

0

0.185

0

1075.597

0

1256.637

0

0

423.072

0

461.8141

0

628.3185

0.185

0

1075.597

0

1256.637

0

1075.597

0

-ve 0.164

0.185

No. bar 0

Ф 0

1256.637

4

20

0

1256.637

4

20

5

14

0

0

0

0

0

0

1256.637

4

20

0

0

0

0

1256.637

4

20

5

769.6902

5

14

0

0

0

0

0

0

0

0

1256.637

4

20

1256.637

0

0

0

0

0

1256.637

4

20

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0.19

0

1113.11

0

1256.637

0

0

0

0

0

1256.637

4

20

0.19

0

1113.11

0

1256.637

0

0

0

0

0

1256.637

4

20

0

507.123

0

615.7522

0

628.3185

5

769.6902

5

14

0

0

0

157.65

0.19

0

1113.11

0

1256.637

0

0

0

0

0

1256.637

4

20

157.65

0.19

0

1113.11

0

1256.637

0

0

0

0

0

1256.637

4

20

0

420.327

0

461.8141

0

628.3185

5

769.6902

5

14

0

0

0

0.188

0

1102.377

0

1256.637

0

0

0

0

0

1256.637

4

20

0.188

0

1102.377

0

1256.637

0

0

0

0

0

1256.637

4

20

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

00

0

0 157.65 0.026

0.017 156.13

b/n 5&6

As provided for reversal effect As No. positive bar Ф Negative 0 0 0 0

No. bar 0

0

157.65

B b/n 4&5

152.85

As provided

156.13

0 0

Page 44

STRUCTURAL DESIGN OF G+4 BUILDING 2016

2&3

b/n 3&4

74.5 9

153.16

0.185

0

1077.778

0

1256.637

0

0

0

0

0

1256.637

4

20

153.16

0.185

0

1077.778

0

1256.637

0

0

0

0

0

1256.637

4

20

0

478.632

0

615.7522

0

942.4778

7

1077.566

7

14

0

0

0

0.263

0

1632.668

0

1884.956

0

0

0

0

0

1884.956

6

20

0.263

0

1632.668

0

1884.956

0

0

0

0

0

1884.956

6

20

0.025 217.74

b/n 4&5

122. 08

217.74 0.031 217.74

b/n 5&6

0

b/n 2&3

0

b/n 3&4

79.8 7

217.74

0

b/n 4&5

116. 33

E

0

b/n 2&3

0

b/n 3&4

90.9 4

b/n 4&5

63.3 3

923.6282

0

942.4778

7

1077.566

7

14

0

0

0

1632.668

0

1884.956

0

0

0

0

0

1884.956

6

20

0.263

0

1632.668

0

1884.956

0

0

0

0

0

1884.956

6

20

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

1144.371

0

1256.637

0

0

0

0

0

1256.637

4

20

161.35

0.195

0

1144.371

0

1256.637

0

0

0

0

0

1256.637

4

20

0

513.04

0

615.7522

0

942.4778

7

1077.566

7

14

0

0

0

0.26

0

1608.139

0

1884.956

0

0

0

0

0

1884.956

6

20

0.26

0

1608.139

0

1884.956

0

0

0

0

0

1884.956

6

20

0

748.007

0

769.6902

0

942.4778

7

1077.566

7

14

0

0

0

0.26

0

1608.139

0

1884.956

0

0

0

0

0

1884.956

6

20

0.26

0

1608.139

0

1884.956

0

0

0

0

0

1884.956

6

20

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0.163

0

937.134

0

942.4778

0

0

0

0

0

942.4778

3

20

0.163

0

937.134

0

942.4778

0

0

0

0

0

942.4778

3

20

0

584.748

0

615.7522

0

628.3185

5

769.6902

5

14

0

0

0

145.85

0.176

0

1020.63

0

1256.637

0

0

0

0

0

1256.637

4

20

145.85

0.176

0

1020.63

0

1256.637

0

0

0

0

0

1256.637

4

20

0

403.89

0

461.8141

0

628.3185

5

769.6902

5

14

0

0

0

0.027 214.98 0.029 214.98

0 0

135.11

AAiT:BSc thesis

0

0

0.195

0

214.98 b/n 5&6

785.788

161.35

214.98

D

0 0.263

135.11 0.03

0.016

Page 45

STRUCTURAL DESIGN OF G+4 BUILDING 2016

145.85 b/n 5&6

0

b/n A&B

60.3 2

145.85

0

130.88 0.02 133.67

b/n B&C

55.4 1

133.67 0.018 133.97

3 b/n C&D

55.2 5

133.97

60.3 9

b/n A&B

55.0 9

b/n B&C

88.4 8

b/n C&D

53.5

b/n D&E

57.8 4

4

133.97 0.02 130.77 129.75 0.018 167.91 167.91 0.03 167.91 167.91 0.018 124.68 124.68 0.019 124.68

5

b/n A&B

71.7 1

AAiT:BSc thesis

0

1020.63

0

1256.637

0

0

0

0

0

1256.637

4

20

0.176

0

1020.63

0

1256.637

0

0

0

0

0

1256.637

4

20

0

0

0

0

0

0

0

0

0

0

0

0

0

0.158

0

905.797

0

942.4778

0

0

0

0

0

942.4778

3

20

0

385.875

0

461.8141

0

471.2389

4

615.7522

4

14

0

0

0

0.161

0

926.125

0

942.4778

0

0

0

0

0

942.4778

3

20

0.161

0

926.125

0

942.4778

0

0

0

0

0

942.4778

3

20

0

353.741

0

461.8141

0

471.2389

4

615.7522

4

14

0

0

0

0.162

0

929.227

0

942.4778

0

0

0

0

0

942.4778

3

20

0.162

0

929.227

0

942.4778

0

0

0

0

0

942.4778

3

20

0.018 133.97

b/n D&E

0.176

141.26

0.024

352.72

0

461.8141

0

471.2389

4

615.7522

4

14

0

0

0

0.162

0

929.227

0

942.4778

0

0

0

0

0

942.4778

3

20

0.162 0 0.158 0.157 0 0.203

0 386.323 0 0 351.698 0

929.227 0 905.036 896.99 0 1197.649

0 461.8141 0 0 461.8141 0

942.4778 0 942.4778 942.4778 0 1256.637

0 471.2389 0 0 628.3185 0

0 4 0 0 5 0

0 615.7522 0 0 769.6902 0

0 4 0 0 5 0

0 14 0 0 14 0

942.4778 0 942.4778 942.4778 0 1256.637

3 0 3 3 0 4

20 0 20 20 0 20

0.203 0 0.203 0.203 0 0.151

0 568.93 0 0 341.547 0

1197.649 0 1197.649 1197.649 0 859.108

0 615.7522 0 0 461.8141 0

1256.637 0 1256.637 1256.637 0 942.4778

0 628.3185 0 0 628.3185 0

0 5 0 0 5 0

0 769.6902 0 0 769.6902 0

0 5 0 0 5 0

0 14 0 0 14 0

1256.637 0 1256.637 1256.637 0 942.4778

4 0 4 4 0 3

20 0 20 20 0 20

0.151

0

859.108

0

942.4778

0

0

0

0

0

942.4778

3

20

0 0.151

369.632 0

0 859.108

461.8141 0

0 942.4778

471.238 0

4 0

615.7522 0

4 0

14 0

0 942.477

0 3

0 20

0.171

0

891.345

0

942.4778

0

0

942.477

3

20

0

459.679

4

14

0

0

461.8141

471.238

4

Page 46

615.7522

STRUCTURAL DESIGN OF G+4 BUILDING 2016

141.26 274.42 b/n B&C

133. 95

b/n C&D

69.6 3

b/n D&E

54.3 3

0 0 780.073

891.345 1875.501

0 0 445.889 0

1875.501 871.91

0.167

0

871.91

0

346.846

0.153

0

0.036 274.42 138.18 0.023 138.18 138.18 0.018 127.16

AAiT:BSc thesis

0.171 0.271

0.271 0.167 0 0.167

871.91

0 0 804.2477

942.4778 2261.947

0 0 461.8141 0

2261.947 942.4778

0

942.4778

1130.973

942.4778

461.8141 802.374

0

6

1206.372

0 471.2389 0

4

471.2389

4

942.4778

Page 47

615.7522

615.7522

0 0 6

0 0 16

942.477 2261.94

3 5 0

20 24 0

0 0 4 0

0 0 14 0

2261.947 942.477 942.477

5 3 0 3

24 20 0 20

0

0

942.477

3

20

4

14

0

0

0

0

3

20

942.4778

STRUCTURAL DESIGN OF G+4 BUILDING 2016

6.3 Shear Failure due to shear is sudden and brittle compared to flexure failure, it should be provided with enough stirrups. The following five modes of failure due to shear are identified. Diagonal tension failure: an inclined crack propagates rapidly due to inadequate shear reinforcement.

Figure: diagonal tension failure Shear Compression Failure: There is crushing of the concrete near the compression flange above the tip of the inclined crack.

Figure: shear compression failure Shear Tension Failure: Due to inadequate anchorage of the longitudinal bars, the diagonal

cracks propagate horizontally along the bars.

Figure: shear tension failure

AAiT:BSc thesis

Page 48

STRUCTURAL DESIGN OF G+4 BUILDING 2016

Web Crushing Failure: The concrete in the web crushes due to inadequate web thickness.

Figure: web crushing failure Arch Rib Failure: For deep beams, the web may buckle and subsequently crush. There can be anchorage failure or failure of the bearing.

Figure: arch rib failure The occurrence of a mode of failure depends on the span-to-depth ratio, loading, crosssection of the beam, amount and anchorage of reinforcement. Shear failure starts at the neutral axis and extends in both directions. It depends on the a/𝑑 ratio. Shear failure is very explosive and brittle. That is why flexural failure is preferred over shear failure. In order to guarantee flexural failure over shear • Make the beam slender • Over design the beam for shear by 20-25%

AAiT:BSc thesis

Page 49

STRUCTURAL DESIGN OF G+4 BUILDING 2016

6.3.1 Design for Shear Below is shown sample calculation for shear design of beam on axis 4 b/n A&B Step 1 Material property C-25 concrete ………………….. fcd= 11.33 MPa S-300 steel …………..............fyd= 260.87 MPa Step 2 Determine design shear force (Vsd*) For beam A

Vsd=121.69 KN

From comb 5

Figure 7 Shear force diagram of beam on axis 4 b/n A&B

44.44 121.69 = 4.5  x x

44.44x = 121.69 (4.5-x) 44.44x = 547-121.69x X = 3.296 m For d = 460mm

For d = 457mm

AAiT:BSc thesis

,

Vsd*=

300 = 104.70 KN 1.15

Vsd*=

121.69  3.296  0.457  =104.81 KN 3.296

Page 50

STRUCTURAL DESIGN OF G+4 BUILDING 2016

From comb 4

40.36 125.78 = 4.5  x x 40.36x = 125.78(4.5-x) X = 3.41 m Vsd * 125.78 = x 3.41  d

,

Vsd*=

125.78 3.41  d  3.41

For d = 460mm Vsd*=

125.78 3.41  0.46  = 108.81 KN 3.41

For d = 457mm Vsd*=

125.78 3.41  0.457  =108.92 KN 3.41

Step 3 Determine diagonal compression failure For d = 460mm VRD = 0.25* fcd*bw*d VRD = 0.25* 11.33

N *350mm * 460mm mm 2

VRD = 456.03 KN

, for ∅14

For d = 457mm VRD = 0.25* fcd*bw*d VRD = 0.25* 11.33 AAiT:BSc thesis

N *350mm * 457mm mm 2 Page 51

STRUCTURAL DESIGN OF G+4 BUILDING 2016 VRD = 453.06 KN , for ∅20 Step 4 Calculate shear capacity of concrete (Vc) Vc = 0.25*fctd*k1*k2*bw*d 2/3 f ctk  0.21 f ck  fctd = = = 1.5 c

 0.21 *20  = 1.0315 MPa 2/3

1.5

k1= 1+50ρ ≤ 2 ρ = As/bd For 5∅14 As = 5*49π = 769.69 mm2 ρ=

769.69 = 0.0057 300* 460

k1= 1+50*0.0057= 1.28 For 3∅20 As = 3*100π = 942.478 mm2 ρ=

942.478 = 0.00687 300* 457

k1= 1+50*0.00687=1.34 k2 = 1.6 – d k2 = 1.6 – 0.46 = 1.14

, for +ve moment (bottom)

k2 = 1.6 – 0.457 = 1.143

, for –ve moment (top)

taking minimum k1 & k2 Vc = 0.25*1.0315 Vc < Vsd AAiT:BSc thesis

N *1.28*1.14*300*460 = 51.93 KN mm 2

, provide shear reinforcement Page 52

STRUCTURAL DESIGN OF G+4 BUILDING 2016

Step 5 Determine spacing Vsd = Vc + Vs Vs = Vsd – Vc using Vsd = 121.69 KN Vs =Vsd – Vc =121.69KN – 58.642KN Vs = 63.05 KN S=

S=

Asv * f yd * d

, Asv=2*16π = 100.531 (two legs of ∅8 stirrups)

Vs

100.531*260.87*460 = 156.98mm 76.849*1000

,use S=150mm

using Vsd*= 106.55 Vs* = Vsd* – Vc = 106.55KN – 44.841KN Vs* = 61.709 KN S=

Asv * f yd * d Vs *

=

100.531*260.87*460 = 195.49mm 61.709*1000

,use S=190mm

2 2 (VRD) = (345.848) = 230.565 KN 3 3

Vsd <

AAiT:BSc thesis

2 (VRD) 3

Page 53

STRUCTURAL DESIGN OF G+4 BUILDING 2016

6.3.2 Minimum Shear Rebar For beam on axis 4 b/n A&B Comb 5

Comb 4

Vc= 51.93 KN

121.69 51.93  3.296 y 51.93*3.296 y  1.41mm 121.69 S min 

Vsd



Asv bw *  min



125.78 51.93  3.41 y 51.93*3.41 y  1.41mm 125.78

100.531 100.531   251.33mm 0.4 0.4 300* 300* fyk 300

2 (VRD ) 3

AAiT:BSc thesis

Page 54

STRUCTURAL DESIGN OF G+4 BUILDING 2016

Beam Shear design result from excel template; For the typical floor Table 14 Typical floor beams shear design result Beam Axis on

Spacing for sttirup b/n

Vsd

b/n 2&3

17.36

b/n 3&4

96.09

A b/n 4&5

17.36

b/n 2&3

17.36

b/n 3&4

134.12

B

17.36

b/n 2&3

24.17

b/n 3&4

129.07

C

D

Asv

VRD

Vc

769.69

14

8

100 .53

391

51.884

769.69

14

8

100 .53

391

769.69

14

8

100 .53

769.69

14

8

769.69

14

769.69

S max

S used

251.3

230

230

51.884

125.5

125.55

120

391

51.884

114.2

114.17

110

100 .53

391

51.884

251.3

230

230

8

100 .53

391

51.884

251.3

230

230

14

10

157 .08

389.3

51.798

139.9

139.93

130

769.69

14

8

100 .53

391

51.884

113.1

113.05

110

769.69

14

8

100 .53

391

51.884

251.3

230

230

942.48

20

8

100 .53

388.5

54.302

251.3

228.5

220

942.48

20

10

157 .08

386.8

54.22

144.5

144.45

140

1077.6

14

10

157 .08

389.3

56.332

92.34

92.342

90

1077.6

14

8

100 .53

391

56.409

251.3

230

230

1077.6

14

8

100 .53

391

56.409

251.3

230

230

1077.6

14

10

157 .08

389.3

56.332

138.5

138.54

130

1077.6

14

10

157 .08

389.3

56.332

96.97

96.97

90

203.24

b/n 5&6

17.36

b/n 2&3

17.36

b/n 3&4

135.47

b/n 4&5

Ф(sttirup)

106.71

b/n 5&6

b/n 4&5

Ф(longt)

105.66

b/n 5&6

b/n 4&5

As

193.54

AAiT:BSc thesis

S,cal

Page 55

STRUCTURAL DESIGN OF G+4 BUILDING 2016

b/n 5&6

17.36

b/n 2&3

17.36

b/n 3&4

95.72

E b/n 4&5

17.36

b/n A&B

127.88

b/n D&E

b/n A&B

b/n B&C

b/n D&E

b/n A&B 5

b/n B&C/ LANDI NG BEAM

100 .53

391

56.409

251.3

230

230

769.69

14

8

100 .53

391

51.884

251.3

230

230

769.69

14

8

100 .53

391

51.884

126

126.03

120

769.69

14

8

100 .53

391

51.884

122.6

122.62

120

769.69

14

8

100 .53

391

51.884

251.3

230

230

615.75

14

10

157 .08

389.3

49.531

146.8

146.76

140

615.75

14

10

157 .08

389.3

49.531

139.1

139.13

130

615.75

14

10

157 .08

389.3

49.531

139.1

139.05

130

615.75

14

10

157 .08

389.3

49.531

146.7

146.73

140

769.69

14

10

157 .08

389.3

51.798

149.2

149.22

140

769.69

14

10

157 .08

389.3

51.798

89.57

89.566

80

769.69

14

10

157 .08

389.3

51.798

161

161.04

160

615.75

14

10

157 .08

389.3

49.531

155.2

155.19

150

615.75

14

10

157 .08

389.3

49.531

136.3

136.28

130

1206.4

14

10

157 .08

389.3

58.228

70.06

70.057

70

134.97

127.91

125.77

209.54

4 b/n C&D

8

134.89

3 b/n C&D

14

98.38

b/n 5&6

b/n B&C

1077.6

116.54

120.93

137.71

267.89

AAiT:BSc thesis

Page 56

STRUCTURAL DESIGN OF G+4 BUILDING 2016

b/n C&D b/n D&E

133.29

615.75

14

10

157 .08

389.3

49.531

140.8

140.8

140

615.75

14

10

157 .08

389.3

49.531

150.8

150.83

150

124.43

6.4 Bond and Development Length Bond: is adhesion between reinforcing steel and surrounding concrete. It is responsible for the transfer of axial force from a reinforcing bar to the surrounding concrete providing strain compatibility, composite action. Here it is important to note that the fundamental theory of flexure, i.e. plane section remain plane after bending becomes valid only if there is perfect bending. Mechanism of bond resistance The mechanism of bond resistance arises from the following factors.  Chemical adhesion  Frictional resistance between concrete and steel  Mechanical interlocking Bond stress Is achieved by the development of tangential (shear) stress components along the interface (contact surface) between the reinforcing bar and surrounding concrete. Bond failure mechanism 

Break up of adhesion between bar and concrete



Longitudinal splitting of concrete around the bar



Crushing of concrete in front of the bar ribs



Shearing of the concrete found between the ribs along cylindrical surface surrounding the ribs.

Bond strength can be enhanced by: 

Using deformed bars (ribbed) bars instead of plain bars.



Using smaller diameter bars



Using higher grade of concrete ( improved tensile strength)



Using Increased cover is provided around each bar



Increased length of embedment



Stirrups with increased areas, reduced spacing and higher grade

AAiT:BSc thesis

Page 57

STRUCTURAL DESIGN OF G+4 BUILDING 2016

Development length 

Basic development length (lb)

lb 

  fyd    4  fbd  fbd  fctd ……………. Plain bars

For good bond condition,

fbd  2 fctd …………….. Deformed bars Other bond conditions , 

fbd  0.7 (good bond condition)

Required anchorage length ( lbnet )

lbnet 

a * lb * As , cal As , provided

a =1 ………….straight bar a = 0.7 ………..hook anchorage

Sample calculation  Basic anchorage length

lb 

fyd 20*347.826 6956.52    844.24MPa 4 fbd 4*2* fctd 4*2*1.03  Required anchorage length

lbnet  lbnet 

a * lb * As , cal As , provided

 =1 for deformed bar

1*844.25*896.99  803.5mm 942.478

lb, min 

0.3(lb)  0.3*844.24  253.27mm 10*∅ = 10*20 = 200mm 200mm

Take

lb, min  253.27 mm

AAiT:BSc thesis

Page 58

STRUCTURAL DESIGN OF G+4 BUILDING 2016 Vc  0.25 * fctd * k 1 * k 2 * bwd  50*942.478  Vc  0.25*1.03* 1  *(1.6  0.457)*(300* 457) 300* 457   Vc  54.22 KN Vsd  121.69KN  2(Vc)  108.44

al  0.5d  0.5*457  228.5mm Total length  1540 mm  803.5mm  228.5mm Total length  2572mm

6.5 Serviceability Is the fitness of the structure to serve the desired function satisfactorily under service loads. It disrupts the use of the structure but do not cause total collapse. These are the major serviceability limits;  Excessive crack width  Excessive deflection  Undesirable vibrations etc. Limit state of cracking: Crack widths are concern for aesthetical appearance, leakage, corrosion, reduction in the stiffness of members. There are various types of cracks. Load induced cracks (i.e. axial, shear, flexural, torsion) imposed cracks and other various types of cracks are the common ones. Load induced cracks: Tensile stresses induced by loads cause distinctive crack patterns. These are also further divided into direct tension induced cracks which extends through the entire the cross section. They are characterized by their vertical nature. The other common type of load induced cracks are the flexural cracks that are when the section is subjected to moment. These are also vertical in nature like the direct tension cracks. They can extend up to the neutral axis. The shear cracks are the other types of load induced cracks caused when the section is subjected to shear. These are different from the previous ones with their orientation having an inclined orientation. They can extend high up to the neutral axis and sometimes into the compression zone. Torsion cracks are also in this group caused when the section is subjected to torsion. These types of cracks are characterized by their spiral orientation around the beam. Bond cracks are the last types of cracks in this category, which are characterized by splitting around the reinforcement. AAiT:BSc thesis

Page 59

STRUCTURAL DESIGN OF G+4 BUILDING 2016

Development of cracks due to loads Crack due to loads doesn‟t reach the final stage spontaneously rather it develops from one stage to the other gradually. Ones the final stage is reached, the crack pattern has stabilized and the further loading of the section merely widens the existing cracks. The distance between the stabilized cracks is the function of the following factors;  The overall member thickness  Concrete cover  Efficiency of the bond Limits on crack width Although cracking of concrete is inevitable, it is desirable to aim for a large number of well distributed fine hairy cracks than a few but wide cracks. The acceptable limits of cracks vary with the following factors;  The type of the structure  The environment There are two types of limit states;  Limit state of crack formation  Limit state of crack width limit state of crack formation The calculated stress shall not exceed δ =1.7*fctk, for flexure δ=fctk ,for direct tension limit state of crack width 0.4mm ---------------- mild 0.2mm----------------- moderate 0.1mm------------------ severe Factors influencing crack width Load induced cracks are influenced by the following factors;  Tensile stress in the steel bars  The thickness of concrete cover  Diameter and spacing of bars  Depth of members and location of neutral axis  Bond strength and tensile strength of concrete AAiT:BSc thesis

Page 60

STRUCTURAL DESIGN OF G+4 BUILDING 2016

6.5.1 Check for Crack Step 1 Material properties fck  20MPa

fctk  0.21 fck 2/3  1.547MPa

Cracking stress

 ct  1.7 fctk  1.7*1.5473  2.6303MPa Step 2 Moment of inertia for the uncracked section Section with 2 ∅20 and 5 ∅14 rebars

Ec  9.5( fck  8)1/3  28.85GPa  29GPa Es  200GPa

n

Es 200   6.89  7 Ec 29

Y

(b * D)( D / 2)  (n  1) As1 * Y 1  (n  1) As 2 * Y 2 b * D  (n  1) As1  (n  1) As 2

Y

(350*500)250  6*769.3*40  6*628*457 350*500  6*769.3  6*628

Y  248.97m

AAiT:BSc thesis

Page 61

STRUCTURAL DESIGN OF G+4 BUILDING 2016

n

n

i 1

i 1

Iuncr   Igi   Ai di2 Iuncr 

bD3  bD( D / 2  Y ) 2  (n  1) As1(Y 1  Y ) 2  (n  1) As 2(Y 2  Y ) 2 12

350(500)3 Iuncr   350*500(250  248.97)2  6*769.3(40  248.97)2  6*628(457  248.97)2 12 Iuncr  4.01*109 mm4

Step 3 Calculate cracking moment Mcr 

 cr * Iuncr



Y Mcr  42.36 KNm

2.6303MPa * 4.01*109 mm4 248.97mm

Step 4 Bending moment diagram for service load

Figure 8 Bending moment diagram of beam for serviceability For sample below are calculation of one positive and one negative moment For positive moment Mk  65.35  Mcr  42.36KNm , section cracked

Step 5 Calculate moment of inertia of the cracked section

AAiT:BSc thesis

Page 62

STRUCTURAL DESIGN OF G+4 BUILDING 2016 n

 AY

Y 

i

i 1 n

i

A

i

i 1

bY (Y / 2)  ( n) As1 * Y 1  ( n  1) As 2 * Y 2 bY  (n  1) As1  (n  1) As 2

Y 

350Y (Y / 2)  7 * 769.3* 460  6 * 628* 43 350Y  6 * 769.3  6 * 628 Y  99.41mm Y 

n

n

Icr   I i   Ai di2 i 1

i 1

3

bY  (n  1) As1(Y 1  Y ) 2  (n  1) As 2(Y 2  Y ) 2 3 350(99.41)3 Icr   6*769.3(460  99.41) 2  6*628(43  99.41) 2 Step 6 3 Calculate Icr  7.27*108 mm4

Icr 

mean strain of reinforcement ( sm )

 sm 

s 

 sr 2  s 1   1 2( )   0.4  Es  s  Es

 sr (due to cracked section) Triangular stress block

Mcr  T * Z

AAiT:BSc thesis

,

T  fs * As

Page 63

STRUCTURAL DESIGN OF G+4 BUILDING 2016

T  fs As 42.36*106 Nmm Mcr  sr   As * Z 769.3mm 2 *(460  99.41)mm 3  sr  128.99MPa

 sr 

 s (stress due to serviceability load moment) 65.35*106 Nmm Mk  As * Z 769.3mm2 *(460  99.41)mm 3  s  199MPa

s 

1  1

, high bond stress

 2  0.5

, service load

 sm 

199 200*103

128.99 2  199  1  1*0.5( 199 )   0.4 200*103

 sm  0.000786  0.000398 Step 7 Average distance between cracks ( Sm)

Sm  50  0.25* K1* k 2 *

r 

As

Aeff



As

2.5d 1 * b





r

769.3 2.5(500  460)*350

 r  0.02198 k1  0.8 , for deformed bar k 2  0.5 , for bending

Sm  50  0.25*0.8*0.5*

14 0.02198

Sm  113.69mm

AAiT:BSc thesis

Page 64

STRUCTURAL DESIGN OF G+4 BUILDING 2016

mean crack width ( wm )



wm  Sm *  sm  113.69mm *0.0006725  0.076mm crack width ( wk )



wk  1.7wm  1.7*0.076mm  0.129mm 0.7

= 0.81*L = 0.81 *3200 = 2592mm

• Radius of gyration ( in x- direction). =

=

=129.96mm

• Slenderness ratio in the x- direction.

=

=

= 19.94

• Critical 𝛌 in x- direction.

AAiT:BSc thesis

Page 79

STRUCTURAL DESIGN OF G+4 BUILDING 2016

= 50 - 25( •

<

) = 50- 25(

) = 69.57.

, The column is short in x- direction. Therefore

= 0.

calculate eccentricity of the load 

e1=

=

=2.17mm



e2=

=

=2.79mm

 emax =max emax=e2=2.79mm • Additional eccentricity in y- direction,

.

≥ 20mm

=

=8.64 ≤20 mm

=

= 20 mm. Therefore

=

+

= 20+ 2.79= 22.79mm.

• Design moment in y- direction, = P*(

) = 2102.11KN *22.79mm

= 47.91KNm •

=

=

= 0.046

• Stiffness coefficient at ground floor joint ( in y- direction 3-3) =

bh3 =

=

bh3 =

=

bh3 =

=

bh3 =

AAiT:BSc thesis

450*4503 = 3.42 * 109 mm4 300*5503 = 4.16 * 109 mm4 250*4503 = 1.90 * 109 mm4 350*5003 = 3.65 *109 mm4

Page 80

STRUCTURAL DESIGN OF G+4 BUILDING 2016

=

,

= 1, &

=0

=

= = 3.45 • Stiffness coefficient at 1st floor joint ( in y- direction) =

= 3.81

=

=

=3.63

• The effective length of buckling (

•

=

≥ 0.7

=

= 0.91 > 0.7

) of a column given by:

= 0.91*L = 0.91 *3200 = 2912mm

• Radius of gyration ( in y- direction). =

=

=129.96mm

• Slenderness ratio in (y- direction).

=

=

= 22.41

• Critical slenderness ratio in y- direction.

= 50 - 25( •

<

) = 50- 25(

) = .64.50

, The column is short in y- direction. therefore

= 0.

• Equivalent 1st order eccentricity in y- direction,

AAiT:BSc thesis

Page 81

STRUCTURAL DESIGN OF G+4 BUILDING 2016

emax

= max

= max

emax = max

emax = max

= 27.94mm. • Additional eccentricity in xi direction,

.

≥ 20mm

= =

=9.71 ≤20 mm

= 20 mm. Therefore

=

+

+

= 20+ 27.94= 47.94mm.

• Design moment in x- direction,

• •

= P*(

) = 2102.11KN *47.94mm = 100.775KNm

=

=

= 0.0976

=

AAiT:BSc thesis

Page 82

STRUCTURAL DESIGN OF G+4 BUILDING 2016

=

= 0.90

• Determine • •

.

= = 25+8+20/2 = 43mm = .= 43/450 = 0.096

• Using

0.1

= .=0.1 and

= 0.90 , choose biaxial chart No. 35.

• Read from chart No. 35 with,

= 0.90 ,

=0.0976 and

=0.046 and read ω.

• ω = 0.17 5.The amount of reinforcement required by the substitute column is computed and the moment of inertia of the reinforcement with respect to the centroid of the concrete section is determined. Determine the amount of reinforcement, =

. =1126.465 mm2.

=

• Let's check minimum and maximum reinforcement requirement. = 0.008 *450mm*450mm = 1620mm2.≥

= 0.8% = 8% Since

= 0.08*450mm*450mm = 16200mm2.

<

ᶲ20, n =

=1126.465 mm2.

, we provide

.

= 5.16. n = 6,

For symetry the Numbers have to 8,

==8*

= 2513.3 mm2.

= 2513.3 mm2.

Use 8ᶲ20,

• let's check shear capacity of the column.

•

+ 0.1

= 0.25

=

=

k1 = 1+50ρ ≤ 2 ,ρ

=

*p

=

= 1.0315

=

= 0.0137

k1 = 1+50*0.0137= 1.685 ≤ 2 ok AAiT:BSc thesis

Page 83

STRUCTURAL DESIGN OF G+4 BUILDING 2016

k2 = 1.6-d ≥ 1 = 1.6-0.407 = 1.193 ≥1 0k = 0.25* 1.0315N/mm2 *1.685*1.193*450*407+0.1* = 132.579KN ≥ Vsd = therefore let's provided =

=

*2102.11*103

.

= 0.00133

• Lateral Reinforcement

=

> 6mm

=

= 5mm< 6mm.

• So let's take •

= 8mm >6mm.

spacing of the lateral reinforcement ,c/c given by:

C/C =

=

= 167.6mm, use C/C = 160mm

= min

160mm ≤ 240mm , therefore use ᶲ8C/C160mm.

• Use ᶲ8

240mm

8ᶲ20

• Detailing  Ties shall be arranged such that every bar or group of bars placed in a corner and alternate longitudinal bar shall have lateral support provided by the corner of a tie with an included angle of not more than 1350 and no bar shall be further than 150 mm clear on each side along the tie.Up to five longitudinal bars in each corner may be secured against lateral buckling by means of the main ties. The center-to--center distance between the outer most of these bars and the corner bar shall not exceed 15 times the diameter of the tie. Smax = 350 mm AAiT:BSc thesis

Page 84

STRUCTURAL DESIGN OF G+4 BUILDING 2016

The spacing between the longitudinal re-bar is; S = =

= 162mm>150mm

Therefore the number of longitudinal reinforcement bar has to increase until the spacing between two coscative re-bar alonge the tie is less than 150mm. = 3769.911 mm2.

Use 12 ϕ 20,

=

Checke spacing, S = S=101.33mm 0.7

= 0.91*L = 0.91 *3200 = 2912mm AAiT:BSc thesis

Page 88

STRUCTURAL DESIGN OF G+4 BUILDING 2016

• Radius of gyration ( in y- direction). =

=

=129.96mm

• Slenderness ratio in (y- direction).

=

=

= 22.41

• Critical slenderness in y- direction.

= 50 - 25( •

<

) = 50- 25(

) = 69.44

, The column is short in y- direction. Therefore

= 0.

• Equivalent 1st order eccentricity in y- direction,

= max 

=

=

= max

= max

= max

AAiT:BSc thesis

Page 89

STRUCTURAL DESIGN OF G+4 BUILDING 2016 e max, y

= 26.095mm.

• Additional eccentricity in x- direction,

=

.

≥ 20mm

=

=9.71 ≤20 mm

= 20 mm. =

Therefore

+

+

= 20 + 26.095=46.095mm.

• Design moment in x- direction, = P(

) = 1154.61KN *46.095mm

= 53.222KNm

=

•

•

=

= 0.052

= = = 0.50

.

• Determine •

= = 25+8+20/2 = 43mm

•

=

• Using

= 43/450 = 0.096

=

=0.1 and

• Enter to the chart with

0.1

= 0.50 , choose biaxial chart No. 34. = 0.50,

=0..052 and

=.052 and read ω.

•ω=0 • Provided the amount of reinforcement, AAiT:BSc thesis

. Page 90

STRUCTURAL DESIGN OF G+4 BUILDING 2016

•

= 0.008 *450mm*450mm = 1620mm2.

=

ᶲ20, n =

= 5.16. n = 8,

=8*

= 2513.3 mm2.

= 2513.3 mm2

Use 8ᶲ20,

• Let's check minimum and maximum reinforcement requirement. = 0.08* 450mm*450mm = 16200mm2.

= 8%

= 1620mm2) < (

(

= 2513.3 mm2.) < (

(16,200mm2)/2)

OK ᵎ

• let's check shear capacity of the column. •

+ 0.1

= 0.25

=

=

*p

=

k1 = 1+50ρ ≤ 2 ,ρ

=

= 1.0315 =

= 0.0137

k1 = 1+50*0.0137= 1.685 ≤ 2 ok k2 = 1.6-d ≥ 1 = 1.6-0.407 = 1.193 ≥1 0k = 0.25* 1.0315N/mm2 *1.685*1.193*450*407+0.1* = 115.614KN ≥ Vsd = therefore lets provide =

=

*1154.61*103

.

= 0.00133

• Lateral Reinforcement =

> 6mm

=

= 5mm< 6mm.

• So let's take

= 8mm >6mm.

• spacing of the lateral reinforcement ,c/c given by: C/C =

=

AAiT:BSc thesis

= 167.6mm, use C/C = 160mm

Page 91

STRUCTURAL DESIGN OF G+4 BUILDING 2016

= min 160mm ≤ 240mm , therefore use ᶲ8C/C160mm. 8ᶲ20

The spacing between the longitudinal re-bar is; S = =

= 162mm>150mm

Therefore the number of longitudinal reinforcement bar has to increase until the spacing between two coscative re-bar alonge the tie is less than 150mm. = 3769.911 mm2.

Use 12 ϕ 20,

=

Checke spacing, S = S=101.33mm 0.7

= 0.93*L = 0.93 *3200 = 2976mm

• Radius of gyration ( in y- direction). AAiT:BSc thesis

Page 95

STRUCTURAL DESIGN OF G+4 BUILDING 2016

=

=

= 173.21mm

• Slenderness ratio in (y- direction). =

=

= 17.18

• Critical in y- direction. = 50 - 25( •

<

) = 50- 25(

) = 65.55

, The column is short in y- direction. Therefore

= 0.

• Equivalent 1st order eccentricity in y- direction, = max 

=

=

= max

= max

= max

= 15.995mm.

• Additional eccentricity in xi direction, =

.

≥ 20mm

AAiT:BSc thesis

Page 96

STRUCTURAL DESIGN OF G+4 BUILDING 2016

= 9.92 ≤20 mm

=

= 20 mm. Therefore

=

+

+

= 20+15.995 =35.995mm.

• Design moment in x- direction, = P(

) = 3010.34KN *35.995mm

= 108.357KNm • •

=

=

= 0.0443

= = = 0.74

• Determine •

.

= = 25+8+20/2 = 43mm

• = .= 43/600 = 0.072

0.1

• Using

= 0.70 and

= .=0.1 and

= 0.80 , choose biaxial chart

No. 34.and 35. • Enter to the chart with biaxial chart No. 34 and

= 0.70 ,

=0.0353 and

= 0.80 ,

=0.0353 and

• ω1 = 0 , ω2 = .045 by interpolation for

= 0.74 between

=0.0443 and read ω1 from =0.0443 & read ω2 . &

,

ω = 0.0 • Determine the amount of reinforcement,

.

• Let's check minimum and maximum reinforcement requirement. = 0.008 *600mm*600mm = 2880mm2.

= 0.8%

= 0.08*450mm*450mm = 16200mm2.

= 8% provide ᶲ20 , n =

. = 9.17. n = 12,

AAiT:BSc thesis

= 12*

= 3,769.91 mm2. Page 97

STRUCTURAL DESIGN OF G+4 BUILDING 2016

Use 12ᶲ20. . • Check the 600mm dimension is enough for 12ᶲ20. 600-2*(25)-2*(8) = 534mm 534 = n(20) + (n-1)25 559 = 45n n = 12.42 > 12

OK.

• let's check shear capacity of the column. •

= 0.25 =

+ 0.1 =

=

*p = 1.0315

k1 = 1+50ρ ≤ 2 ,ρ =

=

= 0.0113

k1 = 1+50*0.0113= 1.564 ≤ 2 ok k2 = 1.6-d ≥ 1 = 1.6-0.557 = 1.043 ≥1 0k = 0.25* 1.0315N/mm2 *1.564*1.043*600*557+0.1* = 188.495KN ≥ Vsd = therefore let's provided =

=

*3010.34*103

.

= 0.00133

• Lateral Reinforcement =

> 6mm

=

= 5mm< 6mm.

• So let's take

= 8mm >6mm.

• spacing of the lateral reinforcement ,c/c given by: C/C =

=

= 125.98mm, use C/C = 120mm

= min 120mm ≤ 240mm , therefore use ᶲ8C/C120mm.

AAiT:BSc thesis

Page 98

STRUCTURAL DESIGN OF G+4 BUILDING 2016

• Detailing

12ᶲ20mm

The spacing between the longitudinal re-bar is; S=

=

=151.33mm>150mm

Therefore the number of longitudinal reinforcement bar has to increase until the spacing between two consecutive re-bar along the tie is less than 150mm. Use 16 ϕ 20,

= 5026.54mm2.

check the spacing; S=

=

=108.5mm