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TECHNOLOGICAL INSTITUTE OF THE PHILIPPINES 938 Aurora Boulevard Cubao, Quezon City

A Project in Partial Fulfilment for the Requirements in

CE473 (REINFORCED CONCRETE DESIGN)

Entitled as

A DESIGN OF A FIVE STOREY REINFORCED CONCRETE SEMINARY MAIN BUILDING

Submitted by LAZO, EMMANUEL M.

Submitted to Engr. Rhonnie C. Estores

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October, 2015

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APPROVAL SHEET

The design project entitled “A Design of a Reinforced Concrete Seminary Main Building” prepared by Emmanuel M. Lazo of the Civil Engineering Department was examined and evaluated by designer himself, and is hereby recommended for approval.

Engr. Rhonnie Estores Adviser

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ABSTRACT This project is entitled as “A Design of a Five Storey Reinforced Concrete Seminary Main Building” is presented by Emmanuel M. Lazo, as partial fulfillment for the requirements for CE 473 (Reoinforced Concrete Design. The project was about structural analysis and design of identified parts of a five storey reinforced concrete seminary main building utilizing special moment resisting space frames. Design specifications from NBCP and NSCP were utilized in the design process. The parts analysed and designed included: beams, columns, and slabs. The parts of the building chosen were considered to be the most critical due to the highest result computed throught STAAD pro considering all load combinations. Design schedule and member details of the structure were then created for the design proper.

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TABLE OF CONTENTS TITLE PAGE....................................................................................................................................................1 APPROVAL SHEET..........................................................................................................................................2 ABSTRACT.......................................................................................................................................................3 LIST OF TABLES..............................................................................................................................................6 LIST OF FIGURES............................................................................................................................................7 CHAPTER I - PROJECT BACKGROUND........................................................................................................8 1.1 Introduction.............................................................................................................................................8 1.2 The Project..............................................................................................................................................9 1.3 Project Location....................................................................................................................................10 1.4 Project Objectives.................................................................................................................................12 1.5 The Client..............................................................................................................................................12 1.6 Project Scope and Limitation................................................................................................................12 1.7 Project Development.............................................................................................................................13 CHAPTER 2: DESIGN INPUTS......................................................................................................................15 2.1 Description of the Structure..................................................................................................................15 2.2 Classification of the Structure...............................................................................................................18 2.3 Architectural Plans................................................................................................................................18 CHAPTER 3: DESIGN CONSTRAINTS, TRADE-OFFS, AND STANDARDS...............................................21 3.1 Design Constraints................................................................................................................................21 3.2 Tradeoffs...............................................................................................................................................22 3.2.1 One Way Slab................................................................................................................................22 3.2.2 Two Way Slab................................................................................................................................23 3.3 Significance of Chosen Tradeoffs to the Quantitative Design Constraints...........................................23 3.4 Method of Measurements for Quantitative Constraints........................................................................24 3.5 Ranking Scale.......................................................................................................................................24 3.6 Initial Estimate and Ranking Computation............................................................................................25 3.7 Raw Designer’s Ranking and Assessment...........................................................................................27 3.8 Design Standards.................................................................................................................................29 CHAPTER IV: DESIGN OF STRUCTURE.....................................................................................................30 4.1 Design Methodology.............................................................................................................................30 4.1.1 Structural Plans..............................................................................................................................32 5

4.1.2 Design Specifications....................................................................................................................33 4.1.3 Material Properties.........................................................................................................................33 4.1.4 Structural Models...........................................................................................................................33 4.1.5 Load Models..................................................................................................................................34 4.1.6 Structural Analysis.........................................................................................................................39 4.1.7 Structural Design...........................................................................................................................41 4.2 Raw Ranking Validation, Comparison of Results, and Final Ranking Assessments...........................49 4.2.1 Final Estimates of Tradeoffs..........................................................................................................49 4.2.2 Validation of Raw Designer’s Ranking...........................................................................................49 4.2.3 Final Designer’s Ranking...............................................................................................................50 4.2.4 Designer’s Final Ranking Assessment..........................................................................................51 CHAPTER V: FINAL DESIGN.........................................................................................................................52 5.1 Design Schedules.................................................................................................................................52 5.1.1 Design Schedule of Slabs..............................................................................................................52 5.1.2 Design Schedule of Beams...........................................................................................................53 5.1.3 Design Schedule of Columns........................................................................................................55 5.1.4 Beam Details..................................................................................................................................56 5.1.5 Column Details...............................................................................................................................57 APPENDICES.................................................................................................................................................59 APPENDIX A: CODES AND STANDARDS................................................................................................59 APPENDIX B: RESULTS OF STRUCTURAL ANALYSIS..........................................................................69 APPENDIX C: DESIGN OF BEAMS...........................................................................................................73 APPENDIX D: DESIGN OF ONE-WAY SLAB............................................................................................85 APPENDIX E: DESIGN OF TWO-WAY SLAB............................................................................................91 APPENDIX F: SAMPLE DESIGN OF COLUMNS....................................................................................104 APPENDIX G: COST ESTIMATE.............................................................................................................108 APPENDIX H: ESTIMATE OF MAN HOURS...........................................................................................110 APPENDIX I: PERCENTAGE DEFLECTION FROM ALLOWABLE.........................................................110 APPENDIX H: REFERENCES..................................................................................................................110

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LIST OF TABLES

Table 1. Total Floor Areas and Functions per Floor........................................................................................15 Table 2. Summary of Initial Estimate of Values...............................................................................................26 Table 3. Wind Intensity (One-Way).................................................................................................................36 Table 4. Wind Intensity (Two-Way).................................................................................................................38 Table 5. Final Estimate of Tradeofs................................................................................................................49 Table 6. Comparison of Initial and Final Estimate of Tradeoffs......................................................................49 Table 7. Final Designer’s Ranking..................................................................................................................51 Table 8. Slab Schedule...................................................................................................................................52 Table 9. Beam Schedule.................................................................................................................................53 Table 10. Column Schedule............................................................................................................................55

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LIST OF FIGURES Figure 1. Perspective of the Proposed Seminary...........................................................................................10 Figure 2. Distance of the Nearest Fault Line to the Structure........................................................................11 Figure 3. Vicinity Map of the Seminary...........................................................................................................11 Figure 4. Project Development Process.........................................................................................................13 Figure 5. Structural Model of the Structure.....................................................................................................15 Figure 6. Floor Plans of the Building...............................................................................................................20 Figure 7. One Way Slab System.....................................................................................................................22 Figure 8. Two Way Slab System.....................................................................................................................23 Figure 9. Ranking Scale for Importance Factor..............................................................................................25 Figure 10. Ranking Scale for Satisfactory Factor...........................................................................................25 Figure 11. Design Methodology......................................................................................................................30 Figure 12. One Way Slab Framing Plan.........................................................................................................32 Figure 13. Two Way Slab Framing Plan.........................................................................................................32 Figure 14. Geometric Modelling of One Way Slab.........................................................................................33 Figure 15. Geometric Modelling of Two-Way Slab.........................................................................................34 Figure 16 Load Diagrams for One-Way Slab.................................................................................................36 Figure 17. Load Diagrams for Two-Way Slab.................................................................................................38 Figure 18. Result of Structural Analysis for One-Way Slab............................................................................39 Figure 19. Result of Structural Analysis for Two-Way Slab............................................................................40 Figure 20. Stress-Strain Diagram for Singly Reinforced Beam......................................................................41 Figure 21. Design of Singly Reinforced Beam................................................................................................42 Figure 22. Design of Doubly Reinforced Beam..............................................................................................43 Figure 23. Design for Spacing of Stirrups for Beams.....................................................................................44 Figure 24.Beindg of Slab................................................................................................................................45 Figure 25. Two Way Slab Strips Considered in EFM......................................................................................45 Figure 26. Two Way Slab Design (EFM).........................................................................................................46 Figure 27. Determining the Steel Area of a Column.......................................................................................47 Figure 28. Column Check for Compression or Tension Controls...................................................................48 Figure 29. Beam Details..................................................................................................................................56 8

Figure 30. Column Details..............................................................................................................................57 Figure 31. Beam-Column Interaction Detail....................................................................................................58

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CHAPTER I - PROJECT BACKGROUND 1.1 Introduction Structural analysis and design is a very old art and is known to human beings since early civilizations. The Pyramids constructed by Egyptians around 2000 B.C. stands today as the testimony to the skills of master builders of that civilization. Many early civilizations produced great builders, skilled craftsmen who constructed magnificent buildings such as the Parthenon at Athens (2500 years old), the great Stupa at Sanchi (2000 years old), Taj Mahal (350 years old), Eiffel Tower (120 years old) and many more buildings around the world. These monuments tell us about the great feats accomplished by these craftsmen in analysis, design and construction of large structures. Today we see around us countless houses, bridges, fly-overs, high-rise buildings and spacious shopping malls. Planning, analysis and construction of these buildings is a science by itself. In the early periods houses were constructed along the riverbanks using the locally available material. They were designed to withstand rain and moderate wind. Today structures are designed to withstand earthquakes, tsunamis, cyclones and blast loadings. These have been made possible with the advances in structural engineering and a revolution in electronic computation in the past 50 years. The construction material industry has also undergone a revolution in the last four decades resulting in new materials having more strength and stiffness than the traditional construction material. Combinations of some materials were also utilized so as to have a better performance in the maintenance of the structure. One good example are the reinforced concrete structures which are one of the most popular structural systems today. The combination of concrete and a steel reinforcement gives advantages that make a structure maintain its form for a long time. Concrete is one of the most popular materials for buildings because it has high compressive strength and flexibility in its form and it is widely available. The history of concrete usage dates back for over a thousand years. Contemporary cement concrete has been used since the early nineteenth century with the development of Portland cement. Despite the high compressive strength, concrete has limited tensile strength, only about ten percent of its compressive strength and zero strength after cracks develop. In the late nineteenth century, reinforcing materials, such as iron or steel rods, began to be used to increase the tensile strength of concrete. Today steel bars are used as common reinforcing material. Usually steel bars have over 100 times the tensile strength of concrete; but the cost is higher than concrete. Therefore, it is most economical that concrete resists compression and steel provides tensile strength. Also it is essential that concrete and steel 10

deform together and deformed reinforcing bars are being used to increase the capacity to resist bond stresses. Advantages of reinforced concrete can be summarized as follows (Hassoun, 1998). 1. It has a relatively high compressive strength. 2. It has better resistance to fire than steel or wood 3. It has a long service life with low maintenance cost. 4. In some types of structures, such as dams, piers, and footing, it is the most economical structural material. 5. It can be cast to take any shape required, making it widely used in precast structural components. Disadvantages of reinforced concrete can be summarized as follows: 1. It has a low tensile strength (zero strength after cracks develop). 2. It needs mixing, casting, and curing, all of which affect the final strength of concrete. 3. The cost of the forms used to cast concrete is relatively high. The cost of form material and artisanry may equal the cost of concrete placed in the forms. 4. It has a lower compressive strength than steel (about 1/10, depending on material), which requires large sections in columns of multi-storey buildings. 5. Cracks develop in concrete due to shrinkage and the application of live loads.

1.2 The Project The project is a seminary constituted of five-storeys containing all the necessary rooms for the residents of the building. It is intended to be built in Antipolo, Rizal. As a city with many public and private schools, constructing a seminary is appropriate. This will be very important for the Antipoleneos since the city contains the National Shrine of the Philippines and thus needs training areas for students who want to become priests someday.

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Figure 1. Perspective of the Proposed Seminary

The building is rectangular shaped and has a total area of 700 m 2 with dimensions of 50 m x 14 m. The structure to be constructed will be the main building of a seminary. The first floor contains the refectory (dining), chapel, lobby, infirmary (clinic), recreation area, kitchen and staff room. The second and third floors contain class rooms, laboratories, library, and offices. The fourth and fifth floor contain the study area

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and dormitories. It has a main stair, fire exit, ramps, and an elevator. The height of each floor is 3 m having a total of 15 m. The covering of the building will be a roof deck.

1.3 Project Location The project area is located in Antipolo, Rizal, which is included in the areas under seismic zone 4. The figure below shows the distance of the planned structure from the nearest fault line which is the Makati Valley Fault System.

Figure 2. Distance of the Nearest Fault Line to the Structure

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Figure 3. Vicinity Map of the Seminary Address: Lot 6 Blk.1, Sampaguita St. Bermuda Hts. Subd., Brgy. San Luis, Antipolo City Nearest Fault Line Distance: 16.3 km

1.4 Project Objectives The main objective of this project is to analyse and design a reinforced concrete structure in accordance with the principles written in NSCP 2010. Other objectives of the project are as follows: a. To design a seminary that will have an acceptable probability of performing satisfactorily during its intended life time. b. To provide all the necessary architectural plans, structural plans, and the estimate of the building cost. c. To plan the structure considering balanced constraints, trade-offs and standards on the design.

1.5 The Client

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The client of this structure is a set of religious people lead by Mrs. Sharon Umayam. She is a businesswoman and at the same time the president of the lectors in Our Lady of Peace and Good Voyage Church (National Shrine of the Philippines).

1.6 Project Scope and Limitation The following were the scope covered by the design project: 1.) The project was designed in accordance to the National Building Code of The Philippines and the National Structural Code of the Philippines. 2.) Analysis of the loads and moments was done using STAAD Pro. 3.) All architectural plans such as floor plan and perspective of the apartment were provided. The following were the limitations of the design project: 1.) 2.) 3.) 4.)

Only beams, slabs, and columns were considered in the design. The cost estimates for the mechanical, plumbing and architectural plan were not included. The plumbing and electrical plans are not included in this design. The interior design of the structure was not considered.

1.7 Project Development PLANNING/CONCEPTUALIZATION

IDENTIFICATION OF DESIGN STANDARDS AND PARAMETERS

PRESENTATION OF ARCHITECTURAL AND STRUCTURAL PLANS WITH INITIAL ESTIMATE

IDENTIFICATION OF DESIGN CONSTRAINTS, TRADE-OFF 15

LOAD IDENTIFICATION, STRUCTURAL ANALYSIS, AND FINAL DESIGN

Figure 4. Project Development Process The project development process started with the planning/conceptualization. In this stage, the identification of client was the most important so as to know the structure to be build. In this case, the structure requested by the client was a seminary. It also included the identification of the location where the structure was intended to be built. The next stage was the identification of design standards. Knowing the structure to be constructed, the next part was to know the specific design standards that are required before coming up to the design (i.e., minimum dimension of a classroom, minimum size of an elevator shaft, etc.). These will set the parameters in the creation of the architectural and floor plans which is the next stage in the process. In the third stage, the plans will be presented to the client so that alterations could be made. After all has been settled, constraints can now be identified, which is the next stage. In this, the constraints that were projected will then be classified as either qualitative or quantitative. Knowing the quantitative tradeoffs will pave the way to the determination of the trade-offs for the structure. In the last stage, geometric design, computation, and final estimation for each trade-offs will be made. Then, all of these will be presented to the client. The client will then rate each trade-off. The one which has the most favorable rating among all will then be chosen for the design of the structure.

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CHAPTER 2: DESIGN INPUTS

2.1 Description of the Structure As what was said, the structure will be a seminary. The structure contains five floors with each floor having different function from the other. The structure has two access stairs, set of ramps, and an elevator. The structure contains five floors with each floor having different functions. The structure has special moment reinforced concrete frames both in longitudinal and transverse axes. The figure below shows the model of the structure.

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This model shows

the

structural

members of

the

structure.

The blue

members

beams are

the

and columns while the violet members

the

total floor

are

slabs that form the building. The table below shows the area and the different areas of the rooms contained in each floor. Figure 5. Structural Model of the Structure

Table 1.

Total Floor Areas and Functions per Floor AREA (m2)

FUNCTION 1ST Floor Ramps and Elevator

49

Stairs

25

C.R.

22.5

Chapel

168

Refectory

168

Staff Room

63

Clinic

49

Lobby

70

Kitchen

63

Hallway

22.5

TOTAL

700 2nd Floor

Ramps and Elevator

49

Stairs

25 18

C.R.

22.5

Offices

3(45)

Class Rooms & Laboratories

4(63)

Other Rooms

32.5

Lounge

35

Hallway

79

TOTAL

700 3rd Floor

Ramps and Elevator

49

Stairs

25

C.R.

22.5

Offices

45

Class Room

2(63)

Other Rooms

133

Faculty Room

65

Library

94.5

Hallway

73.5

Sisters’ Room

66.5

TOTAL

700 4th Floor

Ramps and Elevator

49

Stairs

12.5

C.R.

22.5

Study Area

178.5

Dormitory

255.5

Vice Rector’s and Prefect’s Room

66.5

Toilet & Bath

59.5

Laundry

28

Hallway

28 19

TOTAL

700 5th Floor

Ramps and Elevator

49

Stairs

25

Hallway

28

Dormitory (1)

201

Dormitory (2)

196

Toilet & Bath

2(59.5)

Laundry

28

Rector’s Room

66.5

TOTAL

700

TOTAL FLOOR AREA

3500

2.2 Classification of the Structure

Using the National Structural Code of the Philippines (NSCP) 2010, the designer was able to classify the structure. With respect to the occupancy category, the building is classified as an Essential Facility. With respect to the structural members, the building will have special moment resisting frames. These data will help in designing the structure especially in the determination of the seismic forces acting on the structure.

2.3 Architectural Plans Height is 3 m per floor, for a total of 15 m.

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21

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Figure 6. Floor Plans of the Building

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CHAPTER 3: DESIGN CONSTRAINTS, TRADE-OFFS, AND STANDARDS 3.1 Design Constraints Constraint based design takes the parameters associated with a design problem and links them to the attributes of the formal components and relationships of a solution. The forms that compose a building are defined by a set of attributes. Constraints have to be managed effectively throughout the decision making process, and also could be reduced or eliminated. In this project, the design constraints were divided into two types, namely, quantitative and qualitative. Quantitative constraints are those constraints that can be measured using engineering methods (estimation). The qualitative constraints are those which cannot be measured but are ranked through the designer’s perception and experience. The following are the constraints that are considered in the design of the structure. 3.1.1 Quantitative Constraints 1. Economic. Cost is always an integral part of marketing and livelihood even in ancient times. Thus, this constraint is considered in this design. The cost of the structure is highly significant both to the designer and the client. 2. Constructability. This constraint refers to the ratio of the number of workers that will be hired for the construction, to period of time for the structure to be built. One of these two will be considered as a constant so as to measure an accurate difference between them. 3. Safety/Serviceability. Structures always meet some limitations but sometimes some part of it could be accidentally damaged. The safety of a structure is the one which must have the most outstanding consideration among all. Safety makes a structure function effectively overtime. 3.1.2 Qualitative Constraints 1. Aesthetics. The beauty of the structure lies upon its final output. This constraint depends on the taste of a person therefore it is considered as a qualitative constraint. It depends on a person’s perception which design is more presentable. 24

2. Sustainability. In civil engineering, sustainability refers to the conditions under which a building is still considered useful. Should these limit states be exceeded, a structure that may still be structurally sound would nevertheless be considered unfit. In the process, only the quantitative constraints was focused by the designer. Tradeoffs were enumerated next in this section which will then be ranked and assessed. 3.2 Tradeoffs Design trade-off strategies are always present in the design process. Considering design constraints, trade-offs that have a significant effect on the structural design of the structure were provided by the designer. As a trade-off, the designer will have to evaluate which of the two is more effective considering each constraint. The following are the tradeoffs that were chosen by the designer because they are the most fitted to the said constraints.

3.2.1 One Way Slab

Figure 7. One Way Slab System One-way slabs are those slabs with an aspect ratio in plan of 2:1 or greater, in which bending is primarily about the long axis. In heavily loaded slabs, the thickness is often governed by shear or flexure, while in lightly-loaded slabs, the thickness is generally chosen based on deflection limitations. Both lightly and heavily loaded slabs are typically dimensioned so that no shear reinforcement is required, as placing 25

stirrups in slabs is perceived to be difficult and costly. One-way slabs are designed for flexure and shear on a per meter width basis, assuming that they act as a series of independent strips. Thus one-way shear in slabs is often referred to as beam shear, and design for flexure and shear is carried out using a beam analogy

3.2.2 Two Way Slab

Figure 8. Two Way Slab System When a rectangular slab is supported on all the sides and the length-to-breadth ratio is less than two, it is considered to be a two-way slab. The slab spans in both the orthogonal directions. In general, a slab which is not falling in the category of one-way slab is considered to be a two-way slab.

3.3 Significance of Chosen Tradeoffs to the Quantitative Design Constraints In this section, the constraints enlisted in the beginning of the chapter will be related to the tradeoffs chosen by the designer. The final decision of choosing the tradeoff that will be used for the structure lies on the client. Thus, the significance of the tradeoffs to the constraints is needed. Economic. For the cost effectiveness of the structure, the tradeoffs chosen will be designed to be compared whether of the two will be more economical. Clients do not have the same state of living and 26

thus might give priority to this constraint. Some might choose the tradeoff that have lower price but might not give way to the positivity of other tradeoffs. Constructability. Time measures is significant in the construction of the structure. Knowing which of the difference in the period of construction two tradeoffs might be significant for a client. Some clients need shorter period of time and thus give priority to this constraint. Safety/Serviceability. The magnitudes of deflection for concrete members are also important. Any structure used by the people should be quite rigid and relatively-vibration free so as to provide security. Designing these two tradeoffs will give different results. Thus, one tradeoff might be safer than the other. A safer structure known to a client might be given priority. Through the consideration of multiple constraints, the designer will have to choose what particular design among the tradeoffs will be used. The tradeoff is very significant in the design for it will solve the problem regarding the concern of client considering the constraints.

3.4 Method of Measurements for Quantitative Constraints The main method of measurement that will be used in this design is estimation. For the economic constraint, the cost of the whole building. This includes the materials that will be used for the construction of the beams, slabs, and columns. It also includes the cost of the reinforcements that will be used for the structure. For the constructability of the structure, the period of time that will be utilized to construct the building will be estimated, together with the number of workers that will work on that period of time. The number of workers will be constant for both tradeoffs. The difference between the days will give the result for each tradeoff. For the last constraint, the deflection of the most critical beam will be computed for each tradeoff and will then be compared.

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3.5 Ranking Scale The ranking scale that will be used in this design is based on the model on tradeoff strategies formulated by Otto and Antonsson (1991). The importance factors in each constraint is scaled from 0 to 5, while the ability to satisfy the constraint is scaled from -5 to 5, 5 being the highest for both. After obtaining the results, the product of the importance and ability to satisfy the criteria will be summed of from each constraint. The result will then be the overall ranking of the tradeoff.

Figure 9. Ranking Scale for Importance Factor

Figure 10. Ranking Scale for Satisfactory Factor

Computation of ranking for ability to satisfy criterion of materials:

Difference( )=

Higher value−Lower value ×100( ) Lower value

Subordinate rank =Governing rank −(

difference ) 10

Equation 1 Equation 2

The above equations will be used for the manipulation of the rankings of each constraint given to the tradeoffs. The governing rank is the highest possible value set by the designer. The subordinate rank in second equation is a variable that corresponds to its percentage difference from the governing rank along the ranking scale.

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3.6 Initial Estimate and Ranking Computation To determine the difference between the two tradeoffs, certain methods were used by the designer. For the economic constraint, a cost estimate was provided. For the constructability constraint, an estimate of the number of working days was provided, given that there will be 50 workers. For the safety/serviceability constraint, the deflection of the most critical beam was considered. In this part, a rough computation of the estimates was utilized. The values written in the table below were just assumed by the designer whose basis came from experience. Table 2. Summary of Initial Estimate of Values Estimated Value One-Way Slab Two-Way Slab Php 9,000,000 Php 8,000,000 500 days 450 days 4 % of allowable 5 % of allowable

Constraint Economic Constructability Safety/Serviceability

Computation of ranking for Economic Constraint % difference=

higher value−lower value ×100 higher value

% difference=

9000000−8000000 ×100 9000000

%difference=11.11

Subordinate rank =Governing rank −(

difference ) 10

Subordinate rank =5−1.11

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Subordinate rank =3.89

Computation of ranking for Constructability Constraint

% difference=

higher value−lower value ×10 higher value

% difference=

500−400 ×100 500

%difference=20

Subordinate rank =Governing rank −

difference 10

Subordinate rank =5−2 Subordinate rank =3

Computation of ranking for Safety/Serviceability Constraint

% difference=

higher value−lower value ×10 higher value

% difference=

30

5−4 × 10 5

%difference=20

Subordinate rank =Governing rank −

difference 10

Subordinate rank =5−2

Subordinate rank =3

3.7 Raw Designer’s Ranking and Assessment After making an initial estimate of the structure considering the constraints, the design came up with the raw rankings on the one-way slab and two-way slab. The values computed in the latter section is tabulated. Table 1. Raw Designer’s Ranking CONSTRAINT (CRITERIA)

ABILITY TO SATISFY THE CRITERION IMPORTANCE

One-Way Slab

Two-Way Slab

1. Economic

5

4

5

2. Constructability

4

3

5

3. Maintenance

2

5

3

42

51

Over All Ranking

These tabulated values are just subjective, especially the importance factors. This values will still go on with the validation after making a final estimate and final ranking. Knowing the significance of the constraints to the tradeoffs, the ranks in its importance are given as 5, for economic, 4, for constructability, and 2, for maintenance.

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As for economic constraint, it turned out that the initial cost for the two-way slab is cheaper than the one-way slab, considering only the volume of concrete that will be used. As for the constructability constraint, it turned out that the labor constituting of 50 workers will have to work for longer time for the construction of the one way slab. As for the safety/serviceability constraint, the deflection of the critical member in the two-way slab is quite greater than that of the one-way slab. Overall, it turned out that the two-way slab tradeoff outranked the one-way slab tradeoff for the raw designer’s ranking.

3.8 Design Standards To come up with the final design of the structure, the designer utilized the codes and standards written in the following: 1 2

National Building Code of the Philippines National Structural Code of the Philippines (NSCP) 2010 The National Building Code of the Philippines (PD 1096). The National Building Code of the

Philippines, also known as Presidential Decree No. 1096 was formulated and adopted as a uniform building code to embody up-to-date and modern technical knowledge on building design, construction, use, occupancy and maintenance. The Code provides for all buildings and structures, a framework of minimum standards and requirements to regulate and control

location, site, design, and quality of materials,

construction, use, occupancy, and maintenance. The National Structural Code of the Philippines 2010. This code provides minimum standards to safeguard life or limb, property and public welfare by regulating and controlling the design, construction, quality of materials pertaining to the structural aspects of all buildings and structures within its jurisdiction. The provision of this code shall apply to the construction, alteration, moving, demolition, repair, maintenance and use of any building or structure within its jurisdiction, except work located primarily in a

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public way, public utility towers and poles, hydraulic flood control structures, and indigenous family dwellings.

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CHAPTER IV: STRUCTURAL ANALYSIS AND DESIGN

4.1 Design Methodology The design was done in accordance with the codes and standards appropriate for a reinforced concrete structure. The figure below shows the step by step process of the design of the building. FRAMING PLANS

STRUCTURAL PLANS

DESIGN SPECIFICATIONS

NSCP NBCP

MATERIAL PROPERTIES

COMPRESSIVE STRENGTH MODULUS OF ELASTICITY STRUTURAL MEMBER DIMENSIONS

GEOMETRIC MODELLING

STRUCTURAL MODEL

LOAD MODELS

DEAD AND LIVE LOAD SEISMIC AND WIND LOAD LOAD COMBINATIONS

STRUCTURAL ANALYSIS

SHEAR DIAGRAMS MOMENT DIAGRAMS REACTIONS AND DEFLECTIONS

STRUCTURAL DESIGN

DESIGN SCHEDULES DETAILING

Figure 11. Design Methodology 34

The first process in design methodology was the creation of structural plans. The structural plans included the framing plans of the two trade-offs. The next step was to know the design specifications. These specifications are the codes and standards needed for the structure’s classification and description. The National Building Code and National Structural Code of the Philippines are the main books used for design specifications. The third step in the process was the identification of the material properties. The compressive stresses and modulus of elasticity of the concrete and steel to be used were determined. Also, the structural member dimensions (b, d, etc.) were assumed. The fourth step was the creation of the structural model. These models included geometric modelling, which showed the positioning of the structural members (beams, columns, slabs) in 3D form. The fifth step was the presentation of load models. In this part, the loads acting on the structure were computed. These loads were the dead load, live load, wind load, and seismic (earthquake) load, applying also the load combinations. After computing for these loads, load models was presented also in 3D form. The sixth step was the structural analysis. In structural analysis, member (beams and columns) forces and reactions were determined. The member forces included were the axial force, shear force, and moment acting on the member. The last part was the structural design. The structural design did not include the design of footings. The values from the structural analysis was utilized to design the structural members of the structures, mainly the beams and columns. The maximum moment acting on a beam was used to design the beam, and the maximum value of the axial force acting on a column was used to design the column. To design the slab, the total load on the floors was utilized.

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4.1.1 Structural Plans

Figure 12. One Way Slab Framing Plan

Figure 13. Two Way Slab Framing Plan

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4.1.2 Design Specifications The all the design specifications coming from NBCP and NSCP for the structure is stated Appendix A of the project.

4.1.3 Material Properties The material properties that were utilized are 20.7 MPa for the compressive strength of the concrete (f’c) and 415 MPa for the compressive strength of steel (fy). The modulus of elasticity of steel (Es) is 200000 MPa, while the modulus of elasticity of concrete (Ec) was solved through the following formula, Ec =4700 √ f ' c

4.1.4 Structural Models

Figure 14. Geometric Modelling of One Way Slab

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Figure 15. Geometric Modelling of Two-Way Slab

4.1.5 Load Models

The loads considered in this project are the dead load, live load, wind load and seismic loads. Load combinations were also applied to these loads. The load combinations that were utilized were those that are written in Section 203 of NSCP 2010.

38

The figure below show the preliminary loads acting on the structure.

Dead Load Self Weight = 2.8272 kPa

Floor Loads = 3.49 kPa

39

Figure 16a. Dead Load

Live Load Floor Loads = 1.9 kPa

Figure 16b. Live Load Wind Load

Table 3. Wind Intensity (One-Way) Height (m)

Intensity (kPa)

3

1.34496

6

1.40150

9

1.4884

12

1.55674

15

1.5999

Figure 16c. Wind Load (+X)

40

Figure 16

Figure 16d. Earthquake Load (-Z) Figure 16. Load Diagrams for One-Way Slab

Dead Load Self Weight = 2.8272 kPa

Floor Loads = 3.49 kPa

41

Figure 17a. Dead Load

Live Load Floor Loads = 1.9 kPa

Figure 17b. Live Load Wind Load

Table 4. Wind Intensity (Two-Way)

Figure 17c. Wind Load (+Z) 42

Height (m)

Intensity (kN)

3

1.34496

6

1.40150

9

1.4884

12

1.55674

15

1.5999

Figure 17d. Earthquake Load (+X) Figure 17. Load Diagrams for Two-Way Slab

4.1.6 Structural Analysis For the structural analysis of the members, the results considered are those that came from the load combination which gave the maximum values of member forces and reactions. A summary of values of the member forces is presented in the appendices. The following figures show the results of the structural analysis done through the software STAAD.

Maximum Value of Axial Forces in Columns F = 1337.143 kN

43

Figure 18a. Axial Forces and Shear Forces

Maximum Value of Bending Moments (Z) in Beams +M = 198.802 kN-m -M = 103.972 kN-m

Figure 18b. Moment along Z and Y axes. Figure 18. Result of Structural Analysis for One-Way Slab

44

Maximum Value of Axial Force in Columns F = 2424.714 kN

Figure 19a. Axial and Shear Forces

Maximum Value of Bending Moments (Z) in Beams +M = 310.203 kN-m -M = 164.303 kN-m

Figure 19b. Moments along Z and Y axes

Figure 19. Result of Structural Analysis for Two-Way Slab 4.1.7 Structural Design In this section, the beams, columns, and slabs were designed. The main goal of the structural design of the members is to know the number of bars and their spacing, and check if the assumed dimensions are adequate for the structure 45

. For beams and columns, only the most critical parts were designed. For one-way slab, only one slab was considered both in longitudinal and transverse directions was designed. For two-way slab, only one strip was designed also considering both longitudinal and transverse directions. For convenience, a sample procedure of computation for a structural member will be shown. The manual computations of the members is shown in the appendices.

4.1.7.1 Design of Beams Due to forces acting on the beam, the whole structure experiences flexure, and thus the whole length of the beam have moments within them. Also due to these forces, the beam experiences a shearing stress, which makes a part of the beam to be compressed (top), and another part to be tensed (bottom) .

To design the beams

of the entire structure, the beam which had the highest moment was picked and the resulting design for that beam will be applied to all other beams in the structure. The dimensions of the beam (b,t) and the stresses (f’c,fy) were provided by the designer.

Figure 20. Stress-Strain Diagram for Singly Reinforced Beam

The parts of the beam to be designer are the supports, which

experience negative moment, and the midspan, which experience positive moment. Moreover, the stressstrain diagram of the cross-sectional of the beam was used for the design. The following flow charts present the step by step process of designing a beam.

46

Given b, d, f’c, fy, and Mu

ρmin =

Ru =

ρ=

ρb =

ρmax = 0.75ρb

ρ< ρmax

NO

YES

As = ρminbd N=

YES

ρ> ρmin

NO

As = ρbd N=

eams Figure 21. Design of Singly Reinforced Beam 47

DOUBLY REINFORCED

=

=

=

=

As = As1 + As2 =

Y =

N N =

48

Figure 22. Design of Doubly Reinforced Beam

49

Given Vu and other properties

Vc =

4.1.7.2 Design of Slabs To design a slab, we always consider the longer and shorter span of the slab

since

bending is experience by the whole. For One-way slabs, the process is quite N same in designing a singly reinforced concrete beam. The only

the the

N Vu > ΦVc

Vu < 0.5ΦVc

NO NEED STIRRUPS difference is FOR that we assume that we get a strip from the whole

length of the slab. The width of that strip is 1 meter with thickness provided by the designer.

Y

Following the procedure of solving for the reinforcement of singly reinforced beams, the desired number of bars for one-way slab was computed. For the Vs = Vn - Vc

Y

Vn = Vu/Φ

spacing of bars, the width, b (1 m) was divided by the diameter of the bar times the quantity of bars. Since the two-way slab transmits the load to the supports in trapezoidal form, the method used for one-way slab is not

Figure 24.Beindg of Slab applicable. For the two-way slab, the equivalent frame method was used. The two-way slab was designed N

Y

S =The flow considering REDESIGN the positive and negative moments passedVs through < the column strip and middle strip.

chart below shows the procedure of equivalent frame method.

N Smax = d/4 or 300 (get smaller)

Vs <

Y

Use the smaller value between S and Smax

Smax = d/2 or 600 (get smaller)

Figure for Spacing of Stirrups in forEFM Beams Figure 25.23. TwoDesign Way Slab Strips Considered 50

Given E and L1, L2, t as dimensions of slab

Given E and a, b, lcolumn as dimensions of column

PROCEED WITH THE MDM TO OBTAIN THE MOMENTS

Figure 26. Two Way Slab Design (EFM)

SLAB AS SINGLY REINFORCED BEAM USING THE MOMENTS OBTAINED 51

4.1.7.3 Design of Columns From the structural analysis, the column that experienced the greatest axial forces was designed. The designer started the design of the column in determining the number of bars and its positioning within the gross area of the column. Knowing the position of bars in the column, the designer then computed for the axial force capacity column due to the eccentric load. The flow chart below shows the step by step process done by the designer. The second flow chart is applicable only in this design (eccentricity on one side only).

Given P, My, f’c, fy, b, and t

Assume a value of ρg between 0.02 – 0.04

As = ρgAg

N = As/Abar Redesign Determine actual As and ρg

P < Pcap Pcap = Φ(0.8)Ag(0.85f’c(1-ρg)+ρgfy)

Check If Compression or Tension Controls

Figure 27. Determining the Steel Area of a Column

52

Given e, As and other properties

fy = 600(d-c)/c , solve for c

Pb = 0.85f’cab

Pb(eb+x) = C1(d-.5a) + C2(d-d’), solve for eb

e > eb

Compression Controls

fs = 600(d-c)/c, in terms of c

Tension Controls

f’s = 600(c-d’)/c, in terms of c

Pcap + T = C1 + C2 Pcap(e+x) = C1(d-.5βc) + C2(d-d’) Solve for c and Pcap, then check.

Figure 28. Column Check for Compression or Tension Controls

53

4.2 Raw Ranking Validation, Comparison of Results, and Final Ranking Assessments In this section, the raw designer’s ranking was validated through the gathered results of the design. The initial and final estimated values was then be compared. With the help of the final designer’s ranking, the final ranking assessments was concluded.

4.2.1 Final Estimates of Tradeoffs The table below shows the result of the estimation of construction cost, man days, and cost of maintenance for each tradeoff. Table 5. Final Estimate of Tradeoffs TRADEOFFS

CONSTRAINT

One-Way Slab

Two-Way Slab

Economic (Construction Cost)

Php 10,778,163.00

Php 8,735,033.00

Constructability

435 days

375 days

Safety/Serviceability

1.6 %

5.7 %

4.2.2 Validation of Raw Designer’s Ranking

Table 6. Comparison of Initial and Final Estimate of Tradeoffs CONSTRAINT Economic Constructability Safety/Serviceabilit y

Initial Estimate One Way Slab Two Way Slab Php 9,000,000 Php 8,000,000 500 days 450 days 4 % of allowable

5 % of allowable

Final Estimate One Way Slab Two Way Slab Php 10,778,163 Php 8,735,033 435 days 375 days 1.6 % of allowable

5.7 % of allowable

Looking at the table, there are large discrepancies between the assumed values and the computed values. However, the results of the final estimate of values has almost the same outcome with the initial estimate. It turned out that the two way slab is better than the one way slab in terms of both economic and constructability constraint, while one way slab is better than two way slab in terms of safety/serviceability

54

constraint. These results are the same as what was said in the raw ranking, which makes raw design to be quite certain in this project.

4.2.3 Final Designer’s Ranking Computation of ranking for Economic Constraint % difference=

10778163−8735033 × 100 10778163

%difference=18.9562

Subordinate rank =5−1.8956

Subordinate rank =3.10438

Computation of ranking for Constructability Constraint

% difference=

435−375 ×100 435

%difference=13.79

Subordinate rank =5−1.379 Subordinate rank =3.621

Computation of ranking for Safety/Serviceability Constraint

55

% difference=

5.7−1.6 ×10 5.7

%difference=71.93

Subordinate rank =5−7.193

Subordinate rank =−2.193

Table 7. Final Designer’s Ranking CONSTRAINT (Criteria)

Ability to Satisfy the Criterion

Importance

One-Way Slab

Two-Way Slab

Economic

5

3.10438

5

Constructability

4

3.621

5

Safety/Serviceability

2

5

-2.193

40.0059

40.614

Overall Rank

4.2.4 Designer’s Final Ranking Assessment In terms of economic constraints, the two-way slab got the rank of 5 considering both the concrete works and rebar works. As for the constructability constraints, the number of man hours needed to construct the structure in one-way slab is larger rather than the two-way slab, thus making the two-way slab gets the rank of 5. For safety/serviceability constraint, the percentage of deflection from allowable in the one way slab is smaller than the two way slab making this trade get a rank of 5 in this constraint. After gathering all data and making the designer’s overall final ranking assessment. Overall, the two tradeoffs almost had the same rank, but then the two-way slab outranked the one way slab tradeoff with a nail biting difference. With these information, the designer concluded that the governing tradeoff is still two-way slab in contrast with the raw designer’s ranking.

56

57

CHAPTER V: FINAL DESIGN As what was proven from the previous chapters, the governing tradeoff was the two-way slab. After going through all the design processes, the designer can now conclude the final design of the structure which includes the design schedule of the structural members.

5.1 Design Schedules The design schedule of the structural members included the investigated dimensions and designed number of bars with spacing. The following tables below show the design schedule of the project.

5.1.1 Design Schedule of Slabs Table 8. Slab Schedule SLAB (2F-Roof)

t (mm)

S-1 S-2

150 150

S-1 S-2

150 150

Spacing (mm)

Φ bar

Column Strip Middle Strip LONGITUDINAL DIRECTION 12 300 300 12 300 300 TRANSVERSE DIRECTION 12 300 300 12 300 300

58

Φ tie

10 10 10 10

5.1.2 Design Schedule of Beams Table 9. Beam Schedule 2F-Roof Beam B1 B2 B3 B4 B5 B6 B7 B8 B9 B10 B11 B12 B13 B14 B15 B16 B17 B18 B19 B20 B21 B22 B23 B24 B25 B26 B27 B28 B29 B30 B31 B32 B33 B34 B35 B36

Dimensions b (mm) t (mm) 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500

Top (Left) 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 59

Number of Bars Bottom (Mid) Top (Right) 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ

B37 B38 B39 B40 B41 B42 B43 B44 B45 B46 B47 B48 B49 B50 B51 B52 B53 B54 B55 B56

350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350

500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500

5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ

60

4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ

5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ

5.1.3 Design Schedule of Columns Table 10. Column Schedule 2F-Roof Column C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 C17 C18 C19 C20 C21 C22 C23 C24 C25 C26 C27 C28 C29 C30 C31 C32 C33

Dimensions b (mm) t (mm) 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550

# of Bars 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ

61

Tie Wires Φtie Spacing 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm

5.1.4 Beam Details

Figure 29. Beam Details

5.1.5

Column

Details

62

Figure 30. Column Details

Figure 31. Beam-Column Interaction Detail

63

APPENDICES APPENDIX A: CODES AND STANDARDS

National Building Code of the Philippines (NBC) The following are the sections and codes that are followed in conceptualizing and designing the structural plan of the apartment building:

Section 401. Types of Construction Type I. The structural elements may be any of the materials permitted by this Code.

Section 701. Occupancy Classified. Group B. Residentials, Hotels and Apartments

Section 805. Ceiling Heights. Habitable rooms provided with artificial ventilation have\ ceiling heights not less than 2.40 meters measured from the floor to the ceiling; Provided that for buildings of more than one-storey, the minimum ceiling height of the first storey shall be 2.70 meters and that for the second storey 2.40 meters and succeeding storeys shall have an unobstructed typical head-room clearance of not less than 2.10 meters above the finished floor. Above stated rooms with a natural ventilation shall have ceiling height not less than 2.70 meters.

Section 806. Size and Dimensions of Rooms. Minimum sizes of rooms and their least horizontal dimensions shall be as follows: 1. Rooms for Human Habitations. 6.00 square meters with at least dimensions of 2.00 2. Kitchens. 3.00 square meters with at least dimension of 1.50 meters; 3. Bath and toilet. 1.20 square meters with at least dimension of 0.90 meters.

Section 808. Window Openings. Every room intended for any use, not provided with artificial ventilation system as herein specified in this Code, shall be provided with a window or windows with a total free area of openings equal to at least ten percent of the floor area of room, and such window shall open directly to a court, yard, public street or alley, or open water courses.

Section 1207. Stairs, Exits and Occupant Loads. General. The construction of stairs and exits shall conform to the occupant load requirements of buildings, reviewing stands, bleachers and grandstands:

a. Determinations of Occupant Loads. The Occupant load permitted in any building or portion thereof shall be determined by dividing the floor area assigned to that use by the unit area allowed per occupant as determined by the Secretary. 64

b. Exit Requirements. Exit requirements of a building or portion thereof used for different purposes shall be determined by the occupant load which gives the largest number of persons. No obstruction shall be placed in the required width of an exit except projections permitted by this Code.

National Structural Code of the Philippines (NSCP) 2010 Notation Ag

= gross area of section, mm2.

As

= area of nonprestressed tension reinforcement, mm 2.

A s , min = minimum amount of flexural reinforcement, mm 2. A st

= total area of nonprestressed longitudinal reinforcement (bars and steel shapes), mm 2.

Av

= area of shear reinforcement within a distance s, mm2.

A vf

= area of shear-friction reinforcement, mm 2.

A ' s = area of compression reinforcement, mm 2. b

= width of compression face of member, mm.

bw

= web width, mm.

c

= distance from extreme compression fiber to neutral axis, mm.

cc

= clear cover from the nearest surface in tension to the surface of the flexural tension reinforcement, mm.

Cm = a factor relating actual moment diagram to an equivalent uniform moment diagram. D

= dead loads, or related internal moments and forces.

d

= distance from extreme compression fiber to centroid of tension reinforcement, mm.

d'

= distance from extreme compression fiber to centroid of compression reinforcement, mm. 65

db

= nominal diameter of bar, wire, or prestressing strand, mm.

dc

= thickness of concrete cover measure from extreme tension fiber to center of bar or wire located closest thereto, mm.

ds

= distance from extreme tension fiber to centroid of tension reinforcement, mm.

dt

= distance from extreme compression fiber to extreme tension steel, mm.

E

= load effects of earthquake, or related internal moments and forces.

Ec

= modulus of elasticity of concrete, MPa.

Es

= modulus of elasticity of reinforcement, MPa.

EI

= flexural stiffness of compression member, N-mm 2.

F

= loads due to weight and pressures of fluids with well defined densities and controllable maximum heights, or related internal moments and forces.

f 'c

= specified compressive strength of concrete, MPa.

fy

= specified yield strength of nonprestressed reinforcement, MPa.

f yt = specified yield strength fy H

= loads due to weight and pressure of soil, water in soil, or other materials, or related internal moments and forces.

h

= overall thickness of member, mm.

I

= moment of inertia of section beam about the centroidal axis, mm 4.

I cr

= moment of inertia of cracked section transformed to concrete, mm 4.

Ie

= effective moment of inertia for computation of deflection, mm 4.

Ig

= moment of inertia of gross concrete section about centroidal axis, neglecting reinforcement, mm 4.

L

= live loads, or related internal moments and forces. 66

Ld

= development length, mm.

ln

= length of clear span measured face-to-face of supports, mm.

M a = maximum moment in member at stage deflection is computed. M cr = cracking moment. Pb

= nominal axial load strength at balanced strain conditions

Pn

= nominal axial load strength at given eccentricity.

Vc

= nominal shear strength provided by concrete

W

= wind load, or related integral moments and forces.

wc

= unit weight of concrete, kN/m 3.

wu

= factored load per unit length of beam or per unit area of slab.

αf

= ratio of flexural stiffness of beam section to flexural stiffness of a width of slab bounded laterally by center line of adjacent panle, if any on each side of beam.

α fm = average value of α f

for all beams on edges of a panel.

β1

= factor

εt

= net tensile strain in extreme tension steel at nominal strength.

λ

= modification factor reflection the reduced mechanical properties of lightweight concrete.

λΔ

= multiplier for additional long-time deflection

reinforcement =

ρ

= ration of nonprestressed tension

A s /bd

ρ'

= ratio of nonprestressed compression reinforcement =

ρb

= reinforcement ratio producing balanced strain conditions

Φ

= strength-reduction factor. 67

A ' s /bd

The following are the sections and codes that are followed in conceptualizing and designing the structural plan of the apartment building:

Section 203 - Combination of Load a.Minimum densities for design loads from materials b.Minimum design loads c. Minimum uniform and concentrated live loads Section 206 - Other Minimum Loads a.206.3 Impact loads b.206.3.1 Elevators c. 206.3.2 Machinery

Section 207 - Wind Load a. 207.5.10 Velocity Pressure b. 207.5.6.6 Velocity Pressure Exposure Coefficient c. 207.5.7.2 Topographic Factor d. 207.5.4.4 Wind Directionality Factor e. 207.5.6 Exposure

Section 208 - Earthquake Loads a. 208.5.1.1 Earthquake Loads b. 208.5.2.1 Design Base Shear c. 208.5.2.2 Structure Period

Wind Load Section 207.5.4 Wind Directionality Factor The wind directionality factor, Kd, shall be determined form Table 207-2. This factor Shall only be applied when used in conjunction with load combinations specified in Section 203.3 and 203.4.

Section 207.5.5 Importance factor An importance factor Iw, for the building or other structure shall be determined from Table 207-3 based on building and structure categories listed in Table 103-1.

Section 207.5.6 Exposure For each wind direction considered, the upwind exposure category shall be based on ground surface roughness that is determined from natural topography, vegetation, and constructed facilities.

Section 207.5.7 Topographic factor The wind speed up effect shall be included in the calculation of design wind loads by using the factor kzt. If site conditions and locations of structures do not meet all the conditions specified in Section 207.5.7.1 the kzt= 1.0

68

Section 207.5.8 Gust Effect factor The gust effect factor shall be calculated as permitted in Sections 207.5.8.1 to 207.5.8.5, using appropriate values for natural frequency and damping ratio as permitted in Section 207.5.8.6.

Section 207.5.9 Enclosure Classifications For the purpose of determining internal pressure coefficients, all buildings shall be classified as enclosed, partially enclosed, or open as defined in Section 207.2. Section 207.5.10 Velocity Pressure Velocity pressure, qz, evaluated at height z shall be calculated by the following equation qz= 47.3x10-6 kz kzt kd V2 Iw.

Section 207.5.11 Pressure and Force Coefficients Internal Pressure Coefficients, GCpi, shall be determined from fig. 207-5 based on building enclosure classifications determined from Section 207.5.9

Section 207.5.12 Rigid Building for all heights Design wind pressures for the MWFRS of a buildings of all heights shall be determined by the following equation; P= qGCP – qi(GCPi)

Section 207.5.13 Design Wind Loads on Open Buildings with Monoslope, Pitched, or Troughed Roofs Plus and minus signs signify pressure acting toward and away from the top surface of the roof, respectively.

Section 207.5.14 Design Wind Loads on Solid Freestanding Walls and Solid Signs The design wind force for solid freestanding walls and solid signs shall be determined by the following formula: F= qhGCfAs Section 207.5.15 Design Wind Loads on other Structures The design wind force for other structures shall be determined by the following equation: F=qzGfCfAf Earthquake Load

Section 208.5.1 Earthquake Loads and Modeling Requirements Structures shall be designed for ground motion producing structural response and seismic forces in any horizontal direction. The following earthquake loads shall be used in the load combinations set forth in Section 203: E= ρEh + Eb Section 208.5.2 Static Force Procedure Section 208.5.2.1 The total design base shear in a given direction shall be determined form the following equation: V= CvI (W) 69

RT The total design base shear need not exceed the following: V= 2.5CaI (W) R The Base Shear shall not be less than the following: V= .11CaIW

Section 208.5.2.2 The value of T shall be determined using the following method: Determine the structure period T using Method A T = Ct (hn)3/4

The following are the tables used in each design computations:

Stone Concrete Fill 1.53 Kpa Gypsum Board 0.2 Kpa Suspended Steel Channel 0.1 Kpa Mechanical Duct Allowance 0.2 Kpa Terrazo 1.53 Kpa Grout 0.11 Kpa CHB 1.65 Kpa Clay Dry 0.6435 Kpa Water Proofing 0.05 Kpa Cement Finish 1.53 pa Table 204-1 Minimum Densities for Design Loads from Materials

Material Density (KN/m3) Masonry, Concrete 16.5 Table 204-2 Minimum Design Dead Loads

Basic Floor Area Roof Live Load

1.9 Kpa 1.9 Kpa 70

Table 205-1 Minimum Uniform Concentrated Live Loads

Occupancy Category Seismic Importance Factor I Seismic Importance Factor Ip I. Essential facilities 1.5 1.5 II. Hazardous facilities 1.25 1.5 III. Special Occupancy Structures 1 1 IV. Standard Occupancy Strutures 1 1 V. Miscellaneous Structures 1 1 Table 208-1 Seismic Importance Factors

Soil Profile

Soil Profile Name

Ave. Properties for Top 30 m Soil Profile Shear Wave Velocity SPT Undrained Shear Strenght

SA SB Sc SD SE SF

Hard Rock >1500 Rock 760 to 1500 Very Dense Soil 360 to 760 >50 >100 Stiff Soil Profile 180 to 360 15 to 50 50 to 100 Soft Soil Profile .5*ΦVc Stirrups needed * If Vu < .5*Φ*Vc, stirrups are not needed

Vc ΦVc .5ΦVc

Step 2. Calculate the Shear Strength by Stirrup (Vs) Vn = Vu/Φ

Vn

189.3352

kN

Vs parameter

73.22246 466.7734

kN

Si = Av*fy*d/Vs, Av = pi*(Φtie^2)/4 220 Av For Smax, 110 Si * If Vs < 0.33*sqrt(f'c)*b*d, Smax = d/2 or 600mm (get smaller) parameter * If Vs > 0.33*sqrt(f'c)*b*d, Smax = d/4 or 400mm (get smaller) Smax1 Smax2 Epoxy Light Sf Zinc Normal

78.53982 200 229.9033 220 600 200

mm2 mm mm mm mm mm

Vs = Vn - Vc * If Vs < 0.67*sqrt(f'c)*b*d, go to Step 3. * If Vs > 0.67*sqrt(f'c)*b*d, redesign.

116.1128 104.5015 52.25075 Stirrups Needed

kN kN kN

Step 3. Spacing of Stirrups

Uncoated Part 3. Development Length The following are the supplementary data. Cc 40 Mm Bar Coat

Epoxy

Step 1. Determine the Value of the Coefficients (ψt,ψe,ψs,λ) ψt = 1.0 for all other situations ψe = 1.5 for epoxy-coated bars with cover less than 3d or 6d = 1.2 for all other epoxy-coated bars = 1 for uncoated or zinc coated bars ψt = .8 for 20 mm bars and smaller = 1 for 25 mm bars and larger λ = 1 for normal weight concrete 79

RESULTS ψt ψe

1 1.2

ψs λ

1 1

Step 2. Compute for the development length ld = (fy*ψt*ψe*ψs*d)/(1.1*λ*sqrt(f'c)*((cc+Ktr)/d)) Ktr = 40*Atr/(S*N), Atr = 2*pi*(Φtie^2)/4

Atr

157.079 6

Ktr

7.853982

ld

80

1299.611

mm2 mm

For Midspan The following are the given data Mu 103.972 Vu f'c fy b t d' d Φbar Φtie

89.11886 20.7 415 350 500 62.5 437.5 25 10

kN-m

Es

kN MPa MPa mm mm mm mm mm mm

Ec n L

Part 1. Computation of Steel Area and Number of Bars Step 1. Solve for ρmax and Mu(max) ρb = (0.75*0.85*f'c*β1*600)/(fy*(600+fy) β = 0.85, for f'c < 28 MPa ρmax = ρ = 0.75ρb ω = ρ*fy/f'c Mu(max) = Φ*f'c*ω*b*(d^2)*(1-.59ω) Φ = 0.9 * If Mu < Mu(max), design is Singly Reinforced * If Mu > Mu(max), design is Doubly Reinforced

200000 21383.7 1 10 7

MPa MPa m

RESULTS β

0.85 0.01597 ρb 7 0.01198 ρmax 3 ω 0.24024 Φ 0.9 Mu(max) 257.336 SINGLY

kN-m

Step 2. Using Singly Reinforcement. Solving As and N bars As1 = ρmax*b*d N = As/Abar, Abar = pi*(Φbar^2)/4

As N'

1834.90 6 4

mm2 pcs

Part 2. Designing the Vertical Stirrup Step 1. Calculate the Shear Strength by Concrete (Vc) Vc = sqrt(f'c)*b*d/6

RESULTS

* If Vu > ΦVc, stirrups needed, go to Step II * If Vu < ΦVc, but Vu > .5*ΦVc Stirrups needed

Vc ΦVc 81

116.112 8 104.501

kN kN

5 52.2507 .5ΦVc 5 kN Stirrups Needed

* If Vu < .5*Φ*Vc, stirrups are not needed Step 2. Calculate the Shear Strength by Stirrup (Vs) Vn = Vu/Φ Vs = Vn - Vc * If Vs < 0.67*sqrt(f'c)*b*d, go to Step 3. * If Vs > 0.67*sqrt(f'c)*b*d, redesign.

Vn

99.0209 5

Vs paramete r

17.0918 466.773 4

Av Si paramete r

78.5398 2 230 229.903 3

Smax1 Smax2 Sf

220 600 220

kN kN

Step 3. Spacing of Stirrups Si = Av*fy*d/Vs, Av = pi*(Φtie^2)/4 For Smax, If Vs < 0.33*sqrt(f'c)*b*d, Smax = d/2 or 600mm (get * smaller) If Vs > 0.33*sqrt(f'c)*b*d, Smax = d/4 or 400mm (get * smaller) Epoxy Zinc Uncoated

220 110

Light Normal

Part 3. Development Length The following are the supplementary data. cc 40 mm Bar Coat

Epoxy

Step 1. Determine the Value of the Coefficients (ψt,ψe,ψs,λ) ψt = 1.0 for all other situations ψe = 1.5 for epoxy-coated bars with cover less than 3d or 6d = 1.2 for all other epoxy-coated bars = 1 for uncoated or zinc coated bars ψt = .8 for 20 mm bars and smaller = 1 for 25 mm bars and larger λ = 1 for normal weight concrete 82

RESULTS

ψt ψe

1 1.2

ψs λ

1 1

mm2 mm mm mm mm mm

Atr Step 2. Compute for the development length

Ktr

ld = (fy*ψt*ψe*ψs*d)/(1.1*λ*sqrt(f'c)*((cc+Ktr)/d)) Ktr = 40*Atr/(S*N), Atr = 2*pi*(Φtie^2)/4

ld

157.079 6 7.13998 3 1319.29 5

mm2

mm

Part 4. Checking the Beam in Deflection Step 1. Calculate the Gross Moment of Inertia and the Cracking Moment of the Beam RESULTS Ig = b(t^3)/12 Mcr = Ig*fr/ϒt, fr = 0.62*λ*sqrt(f'c), yt = t/2

Ig

mm4

Mcr

3645833333 2.82082966 5 250 41.1370992 8

kN-m

Icr

91438412.2 9

mm4

Ie

311587799

mm4

fr yt

MPa mm

Step 2. Calcualte the Moment of Inertia of the Cracked Section Icr = b*(c^3)/12 + nAs(d-c)+nAs'(c-d') Step 3. Calculate the Effective Moment of Inertia Ie = ((Mcr/Mu)^3)*Ig + ((1-(Mcr/Mu)^3)*Icr) Step 4. Determine and Check the Deflection Mu = W(L^2)/8, W=____ δ = 5*W*(L^4)/(384*Ec*Ie)

W δ

δmax = L/360

δmax

83

16.9750204 1 0.30585036 19.4444444 4 OK

kN/m mm mm

Beam with Maximum Moment was Designed (Two-Way Tradeoff) For Support

The following are the given data Mu 310.203

kN-m

Es

Vu

265.89

kN

Ec

200000 21383.7 1

f'c fy b t d' d

20.7 415 350 500 62.5 437.5

MPa MPa mm mm mm mm

N L

10 7

Φbar Φtie

25 10

mm mm

Part 1. Computation of Steel Area and Number of Bars Step 1. Solve for ρmax and Mu(max) ρb = (0.75*0.85*f'c*β1*600)/(fy*(600+fy) β = 0.85, for f'c < 28 MPa

MPa MPa M

RESULTS β

0.85 0.01597 ρb 7 0.01198 ρmax 3 ω 0.24024 Φ 0.9 Mu(max) 257.336 DOUBLY

ρmax = ρ = 0.75ρb ω = ρ*fy/f'c Mu(max) = Φ*f'c*ω*b*(d^2)*(1-.59ω) Φ = 0.9 * If Mu < Mu(max), design is Singly Reinforced * If Mu > Mu(max), design is Doubly Reinforced

kN-m

Step 2. Using Doubly Reinforcement. Solving As1, Mu1, Mu2, and As2 As1 = ρmax*b*d Mu1 = Mu(max)

As1 Mu1

Mu2 = Mu - Mu1

Mu2

Mu2 = Φ*As2*fy*(d-d'), Solve for As2 Step 3. Solve for the Stress of the Compression Steel 84

As2

1834.90 6 257.336 52.8669 6 377.452 6

mm2 kN-m kN-m mm2

C=T 0.85*f'c*a*b = As1*fy, Solve for a

a

a = βc, Solve for c

c 377.452 6 457.722 6

f's = 600*(c-d')/c * If f's > fy, A's = As2 * If f's < fy, A's = As2*fy/f's

A's

123.653 145.474 1 342.222 2 457.722 6

N N'

5 1

f's

mm mm MPa mm2

Step 4. Determine the Number of Bars As = As1 + As2 For Tension N = As/Abar, Abar = pi*(Φbar^2)/4 For Compression N' = A's/Abar, Abar = pi*(Φbar^2)/4

pcs pcs

Part 2. Designing the Vertical Stirrup Step 1. Calculate the Shear Strength by Concrete (Vc) Vc = sqrt(f'c)*b*d/6

RESULTS 116.112 Vc 8 kN 104.501 ΦVc 5 kN 52.2507 .5ΦVc 5 kN Stirrups Needed

* If Vu > ΦVc, stirrups needed, go to Step II * If Vu < ΦVc, but Vu > .5*ΦVc Stirrups needed * If Vu < .5*Φ*Vc, stirrups are not needed Step 2. Calculate the Shear Strength by Stirrup (Vs) Vn = Vu/Φ

Vs paramete r

295.433 3 179.320 6 466.773 4

Av

78.5398 2

Vn

Vs = Vn - Vc * If Vs < 0.67*sqrt(f'c)*b*d, go to Step 3. * If Vs > 0.67*sqrt(f'c)*b*d, redesign.

kN kN

Step 3. Spacing of Stirrups Si = Av*fy*d/Vs, Av = pi*(Φtie^2)/4

220 85

mm2

For Smax, If Vs < 0.33*sqrt(f'c)*b*d, Smax = d/2 or 600mm (get * smaller) If Vs > 0.33*sqrt(f'c)*b*d, Smax = d/4 or 400mm (get * smaller) Epoxy Zinc Uncoated

110

Light Normal

Si paramete r

80 229.903 3

mm

Smax1 Smax2 Sf

220 600 80

mm mm mm

mm

Part 3. Development Length The following are the supplementary data. Cc 40 mm Bar Coat

Epoxy

Step 1. Determine the Value of the Coefficients (ψt,ψe,ψs,λ) ψt = 1.0 for all other situations ψe = 1.5 for epoxy-coated bars with cover less than 3d or 6d = 1.2 for all other epoxy-coated bars = 1 for uncoated or zinc coated bars ψt = .8 for 20 mm bars and smaller = 1 for 25 mm bars and larger λ = 1 for normal weight concrete

RESULTS

ψt ψe

1 1.2

ψs λ

1 1

Atr Step 2. Compute for the development length

Ktr

ld = (fy*ψt*ψe*ψs*d)/(1.1*λ*sqrt(f'c)*((cc+Ktr)/d)) Ktr = 40*Atr/(S*N), Atr = 2*pi*(Φtie^2)/4

ld

86

157.079 6 78.5398 2 524.647 1

mm2

mm

For Midspan

The following are the given data Mu 164.303 Vu f'c Fy B T d' D Φbar Φtie

kN-m

Es

kN MPa MPa mm mm mm mm mm mm

Ec n L

143.83 20.7 415 350 500 62.5 437.5 25 10

Part 1. Computation of Steel Area and Number of Bars Step 1. Solve for ρmax and Mu(max) ρb = (0.75*0.85*f'c*β1*600)/(fy*(600+fy) β = 0.85, for f'c < 28 MPa ρmax = ρ = 0.75ρb ω = ρ*fy/f'c Mu(max) = Φ*f'c*ω*b*(d^2)*(1-.59ω) Φ = 0.9 * If Mu < Mu(max), design is Singly Reinforced * If Mu > Mu(max), design is Doubly Reinforced

200000 21383.7 1 10 7

MPa MPa M

RESULTS β ρb ρmax ω Φ Mu(max)

0.85 0.01597 7 0.011983 0.24024 0.9 257.336 SINGLY

kN-m

Step 2. Using Singly Reinforcement. Solving As and N bars As1 = ρmax*b*d N = As/Abar, Abar = pi*(Φbar^2)/4

As N

1834.90 6 4

mm2 pcs

Part 2. Designing the Vertical Stirrup Step 1. Calculate the Shear Strength by Concrete (Vc) Vc = sqrt(f'c)*b*d/6

RESULTS

* If Vu > ΦVc, stirrups needed, go to Step II * If Vu < ΦVc, but Vu > .5*ΦVc Stirrups needed

Vc ΦVc 87

116.1128 104.5015

kN kN

.5ΦVc 52.25075 Stirrups Needed

* If Vu < .5*Φ*Vc, stirrups are not needed Step 2. Calculate the Shear Strength by Stirrup (Vs) Vn = Vu/Φ Vs = Vn - Vc * If Vs < 0.67*sqrt(f'c)*b*d, go to Step 3. * If Vs > 0.67*sqrt(f'c)*b*d, redesign.

kN

Vn

159.8111

kN

Vs parameter

43.69833 466.7734

kN

Av Si parameter Smax1 Smax2 Sf

78.53982 330 229.9033 220 600 220

Step 3. Spacing of Stirrups Si = Av*fy*d/Vs, Av = pi*(Φtie^2)/4 220 For Smax, 110 * If Vs < 0.33*sqrt(f'c)*b*d, Smax = d/2 or 600mm (get smaller) * If Vs > 0.33*sqrt(f'c)*b*d, Smax = d/4 or 400mm (get smaller) Epoxy Zinc

Light Normal

mm 2

mm mm mm mm mm

Part 3. Development Length The following are the supplementary data. Cc 40 mm Bar Coat

Epoxy

Step 1. Determine the Value of the Coefficients (ψt,ψe,ψs,λ) ψt = 1.0 for all other situations ψe = 1.5 for epoxy-coated bars with cover less than 3d or 6d = 1.2 for all other epoxy-coated bars = 1 for uncoated or zinc coated bars ψt = .8 for 20 mm bars and smaller = 1 for 25 mm bars and larger λ = 1 for normal weight concrete Step 2. Compute for the development length ld = (fy*ψt*ψe*ψs*d)/(1.1*λ*sqrt(f'c)*((cc+Ktr)/d)) Ktr = 40*Atr/(S*N), Atr = 2*pi*(Φtie^2)/4

88

RESULTS Ψt Ψe

1 1.2

Ψs Λ

1 1

Atr

157.079 6

mm2

Ktr Ld

7.13998 1319.3

mm

Part 4. Checking the Beam in Deflection Step 1. Calculate the Gross Moment of Inertia and the Cracking Moment of the Beam RESULTS Ig = b(t^3)/12 Mcr = Ig*fr/ϒt, fr = 0.62*λ*sqrt(f'c), yt = t/2

Ig

mm4

Mcr

3645833333 2.82082966 5 250 41.1370992 8

kN-m

Icr

93279189.2 4

mm4

Ie

149036936. 7

mm4

fr yt

MPa mm

Step 2. Calcualte the Moment of Inertia of the Cracked Section Icr = b*(c^3)/12 + nAs(d-c)+nAs'(c-d') Step 3. Calculate the Effective Moment of Inertia Ie = ((Mcr/Mu)^3)*Ig + ((1-(Mcr/Mu)^3)*Icr) Step 4. Determine and Check the Deflection Mu = W(L^2)/8, W=____ δ = 5*W*(L^4)/(384*Ec*Ie)

W δ

δmax = L/360

δmax

89

26.8249795 9 1.01047278 19.4444444 4 OK

kN/m mm mm

APPENDIX D: DESIGN OF ONE-WAY SLAB Considering Longer Side Design of S-1 The following are the given data. Dead Loads (kPa) Weight of Slab 3.537 Stone Concrete 1.53 Fill Gypsum Board 0.2 Total 5.267 Live Load (kPa) Basic Floor 1.9 Area

f'c fy

20.7 415

MPa MPa

L

7

m

t b Φbar

150 1000 12

mm mm mm

Φtie

10

mm

d

134

mm

Step 1. Calculate the Factored Loads and the Moment in the Slab W = 1.2DL + 1.6LL For Midspan, M = W*(L^2)/14 For Continuous Edge, M = W*(L^2)/10

RESULTS W

9.36040

kN/m

Mmid

32.76140

kN/m

Mc.e.

45.86596

kN/m

R

1.82454

ρi

0.00465

ρmax

0.01598

ρmin

0.00337

ρf

0.00465

Step 2. Calculate the ρ and check for the Midspan R = Mu/(b*(d^2)) ρ = (0.85*f'c/fy)*(1-(sqrt(1-((2*R)/ (0.85*f'c)))) ρmax = 0.75*0.85*f'c*β*600/ (fy*(600+fy)) ρmin = 1.4/fy If ρ > ρmax, * redesign * If ρmin < ρ < ρmax, ok If ρmin > ρ, use * ρmin

90

Step 3. Calculate the Steel Area and Spacing of Bars As

697.74823

mm2

S = b*Abar/As, Abar = pi*(Φbar^2)/4

Abar S

113.09734 162.08903

mm2 mm

Step 4. Calculate the ρ and check for the Continuous Edge

R Ρi ρmax ρmin Ρf

2.55435 0.00668 0.01598 0.00337 0.00668

As

1002.23229

mm2

Abar S

113.09734 112.84543

mm2 mm

As = ρ*b*d

Step 5. Calculate the Steel Area and Spacing of Bars

Considering Shorter Side Dead Loads (kPa) Weight of Slab Stone Concrete Fill Gypsum Board Total Live Load (kPa) Basic Floor Area

3.537 1.53 0.2 5.267 1.9

f'c fy L t b Φbar Φtie d

20.7 415 3.5 150 1000 12 10 134

Step 1. Calculate the Factored Loads and the Moment in the Slab W = 1.2DL + 1.6LL For Midspan, M = W*(L^2)/14 For Continuous Edge, M = W*(L^2)/10 91

MPa MPa m mm mm mm mm mm RESULTS

W

9.36040

kN/m

Mmid

8.19035

kN/m

Mc.e.

11.46649

kN/m

Step 2. Calculate the ρ and check for the Midspan R = Mu/(b*(d^2)) ρ = (0.85*f'c/fy)*(1-(sqrt(1-((2*R)/ (0.85*f'c)))) ρmax = 0.75*0.85*f'c*β*600/(fy*(600+fy)) ρmin = 1.4/fy * If ρ > ρmax, redesign * If ρmin < ρ < ρmax, ok * If ρmin > ρ, use ρmin Step 3. Calculate the Steel Area and Spacing of Bars As = ρ*b*d S = b*Abar/As, Abar = pi*(Φbar^2)/4

Step 4. Calculate the ρ and check for the Continuous Edge

Step 5. Calculate the Steel Area and Spacing of Bars

92

R

0.45613

ρi

0.00111

ρmax ρmin ρf

0.01598 0.00337 0.00337

As

506.02410

mm2

Abar

113.09734

mm2

S

223.50188

mm

R ρi ρmax ρmin ρf

0.63859 0.00157 0.01598 0.00337 0.00337

As

506.02410

mm2

Abar

113.09734

mm2

S

223.50188

mm

Design of S-2 The following are the given data. Dead Loads (kPa) Weight of Slab 3.537 Stone Concrete Fill 1.53 Gypsum Board 0.2 Total 5.267 Live Load (kPa) Basic Floor Area 1.9

f'c Fy

20.7 415

MPa MPa

L T B Φbar

7 150 1000 12

M mm mm mm

Φtie D

10 134

mm mm

Step 1. Calculate the Factored Loads and the Moment in the Slab W = 1.2DL + 1.6LL For Midspan, M = W*(L^2)/14 For Continuous Edge, M = W*(L^2)/10 Step 2. Calculate the ρ and check for the Midspan R = Mu/ (b*(d^2)) ρ = (0.85*f'c/fy)*(1-(sqrt(1-((2*R)/(0.85*f'c)))) ρmax = 0.75*0.85*f'c*β*600/ (fy*(600+fy)) ρmin = 1.4/fy * If ρ > ρmax, redesign * If ρmin < ρ < ρmax, ok * If ρmin > ρ, use ρmin

RESULTS

W

9.36040

kN/m

Mmid Mc.e.

32.76140 45.86596

kN/m kN/m

R

1.82454

ρi ρmax ρmin ρf

0.00465 0.01598 0.00337 0.00465

As

697.74823

mm2

Abar S

113.09734 162.08903

mm2 mm

R

2.55435

Step 3. Calculate the Steel Area and Spacing of Bars As = ρ*b*d S = b*Abar/As, Abar = pi*(Φbar^2)/4

Step 4. Calculate the ρ and check for the Continuous Edge 93

Step 5. Calculate the Steel Area and Spacing of Bars

ρi ρmax ρmin ρf

0.00668 0.01598 0.00337 0.00668

As

1002.23229

mm2

Abar S

113.09734 112.84543

mm2 mm

Considering Shorter Side Dead Loads (kPa) Weight of Slab Stone Concrete Fill Gypsum Board Total Live Load (kPa) Basic Floor Area

3.537 1.53 0.2 5.267 1.9

f'c fy L t b Φbar Φtie d

20.7 415 3 150 1000 12 10 134

Step 1. Calculate the Factored Loads and the Moment in the Slab W = 1.2DL + 1.6LL For Midspan, M = W*(L^2)/14 For Continuous Edge, M = W*(L^2)/10 Step 2. Calculate the ρ and check for the Midspan R = Mu/(b*(d^2)) ρ = (0.85*f'c/fy)*(1-(sqrt(1-((2*R)/ (0.85*f'c)))) ρmax = 0.75*0.85*f'c*β*600/ (fy*(600+fy)) ρmin = 1.4/fy * If ρ > ρmax, redesign * If ρmin < ρ < ρmax, ok * If ρmin > ρ, use ρmin 94

MPa MPa m mm mm mm mm mm

RESULTS

W

9.36040

kN/m

Mmid Mc.e.

6.01740 8.42436

kN/m kN/m

R

0.33512

ρi ρmax ρmin ρf

0.00082 0.01598 0.00337 0.00337

Step 3. Calculate the Steel Area and Spacing of Bars As = ρ*b*d S = b*Abar/As, Abar = pi*(Φbar^2)/4

Step 2. Calculate the ρ and check for the Continuous Edge

Step 3. Calculate the Steel Area and Spacing of Bars

95

As

506.02410

mm2

Abar

113.09734

mm2

S

223.50188

mm

R ρi ρmax ρmin ρf

0.46917 0.00115 0.01598 0.00337 0.00337

As

506.02410

mm2

Abar

113.09734

mm2

S N

223.50188 5

mm pcs

APPENDIX E: DESIGN OF TWO-WAY SLAB EQUIVALENT FRAME METHOD

Figure. Design

Strips for Equivalent Frame Method Considering Longitudinal Frame

The following are the given data.

ts B l1 l2 DL LL W f'c fy E

Slab 150 1000 4.5 7 General 5.267 1.9 9.3604 20.7 415 21383.7

mm mm m m

c1 c2 lc

Column 550 550 3

mm mm m

kPa kPa MPa MPa MPa

96

Summary of Values for Slabs SAMPLE RESULTS (AB) Slab1. Determine l1 l2the MOIc1/l1 c2/l2 FEM COF K Step and Stiffness of Columnk 1 7 a*(b^3)/12 7 0.078571 0.078571 4.11 269.6918 0.5086 Ic 7.6E+09 2.47E+10 mm4 Ic = 2 7 7 0.078571 0.078571 4.11 269.6918 0.5086 2.47E+10 Determine lu/lc lc 3 mm 3 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 Determine a/b lu 2.85 mm 4 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 Find kab in the tables. lu/lc 0.95 5 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 a/b 1 K4.5 c = kab*E*Ic/h 6 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 C = (1-0.63(t/a))*((t^3)(a)/3) kab 4.55 7 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 K t = 9*E*C/(l2*(1-b/l2)^2) K c 2.5E+11 8 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 K ec = 4Kt*Kc/(2*Kt+2*Kc) C 5.1E+08 9 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 Kt 1.7E+10 10 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 Step 2. Determine the MOI and Stiffness of Slab Is = b*(t^3)/12 Compute a/l1 & b/l1 Find ks in the tables. Find FEM and COF in the tables. Ks = ks*E*Is/l1

Kec

3.1E+10

Is a/l1 b/l2 ks FEM COF Ks

2E+09 0.12222 0.12222 4.4 111.454 0.526 4.1E+10

Figure. Design Strip for Longitudinal Frame

Table. Summary of Coefficients and Stiffness for Slabs in Longitudinal Frame

97

mm4

kN-m

Step 3. Determine the Distribution Factors DFxy = Kxy/ΣK Table. Summary of Distribution Factors and FEM

Member AB BA BC CB CD DC DE ED EF FE FG

K 2.5E+10 2.5E+10 2.5E+10 2.5E+10 4.12E+1 0 4.12E+1 0 4.12E+1 0 4.12E+1 0 4.12E+1 0 4.12E+1 0 4.12E+1 0

SUMMARY OF VALUES FOR DF AND FEM DF FEM Member K 0.28438 -269.69 GF 4.1E+10 0.22141 269.6918 GH 4.1E+10 0.22141 -269.69 HG 4.1E+10 0.19299 269.6918 HI 4.1E+10

DF 0.28481 0.28481 0.28481 0.28481

FEM 111.454 -111.45 111.454 -111.45

0.32138

-111.454

IH

4.1E+10

0.28481

111.454

0.28481

111.4543

IJ

4.1E+10

0.28481

-111.45

0.28481

-111.454

JI

4.1E+10

0.28481

111.454

0.28481

111.4543

JK

4.1E+10

0.28481

-111.45

0.28481

-111.454

KJ

4.1E+10

0.39824

111.454

0.28481

111.4543

0.28481

-111.454

98

Step 4. Continue with the Moment Distribution Method (MDM) Table. MDM for Longitudinal Frame

Member DF FEM COF Balance CO1 Balance CO2 Balance CO3 Balance CO4 Balance TOTAL

A B AB BA BC 0.284 0.22 0.22 -270 270 270 0.509 0.51 0.51 -76.7 0 0 0 -39 15.5 0 -5.2 6.88 -2.64 0 0 -0.75 0 0 0 0.38 0.74 0 0.08 0.08 0.04 0 0 0.011 0 0 -350 225 246

C CB CD 0.19 0.32 270 0.51 30.5 0 0 3.5 1.45

-111 0.53 50.9 0 0 4.01 2.41

MDM FOR LONGITUDINAL FRAME D E F G H DC DE ED EF FE FG GF GH HG HI IH 0.285 0.285 0.28 0.285 0.285 0.285 0.28 0.285 0.285 0.285 0.285 111.5 -111 111 -111 111.5 -111 111 111.5 111.5 -111 111.5 0.526 0.526 0.53 0.526 0.526 0.526 0.53 0.526 0.526 0.526 0.526 0 0 0 0 0 0 0 0 0 0 0 26.75 0 0 0 0 0 0 0 0 0 0 7.619 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 0 0 0 0 0.996

0 0 1.269 0 0 0 0 0 0 0.361 0.361 0 0 0 0.04 0.19 0 0 0.19 0 0 0.04 0.07 0 0 0.05 0.054 0 305 53.9 147.5 -111 112 -111 111.5

0 0 0 0 -111

J IJ JI JK 0.285 0.28 0.285

KJ KJ 0.4

-111 111 -111 111 0.526 0.53 0.526 0.53 0 0 0 44.4 0 0 23.35 0 0 6.65 13.3 0 3.498 0 0 7 0.996 0 0 2.79

0 0 0 0.524 0 0 0.52 1.465 0 0 0 0.149 0.149 0 0 0.57 0.567 0 0 0.078 0 0 0.078 0.298 0 0 0.3 0.02 0.022 0 0 0.107 0.107 0 0 0.12 111 111.4 111.6 -111 112.6 -107 119 -72.8 166

Figure. Result of Moment Distribution Method (Moment Diagram)

99

I

Table. Summary of Moments for Longitudinal Frame Sla b

W

1

401.3272

2

401.3272

3

165.8546

4

165.8546

5

165.8546

+ + + + + -

Summary of Moments Sla M b W 349.73093 113.87016 6 165.8546 225.18306 246.46759 125.4633 7 165.8546 305.26012 53.916191 65.170221 8 165.8546 147.45254 111.09292 54.458879 9 165.8546 111.69849 111.40015 54.427373 10 165.8546 111.45428

Step 5. Get the Design Moments For Exterior Slabs S-1 M(+) 113.8702 M(-) 349.7309 Moments (%) 0.85M(+) 96.78963 0.70M(-) 244.8117 Moment (%) Passed Column Strip

(+) (-)

(+) (-)

Beam (85%) 82.27119 208.0899

Middle Strip

Slab(15%) 14.5184 36.7217

17.0805 104.919

Design Moments CS MS 7.259222 8.54026 18.36087 52.4596 100

+ + + + + -

M 111.45428 54.389126 111.47664 111.35343 54.376115 111.60352 110.78107 54.145967 112.63617 106.5553 52.979831 119.19421 72.777157 46.447589 166.03684

Slab-10 M(+) 46.44759 M(-) 166.0368 Moment (%) 0.85M(+) 39.48045 0.70M(-) 116.2258 Moment (%) Passed Column Strip Beam (85%) Slab(15%) (+) 33.55838 5.92207 (-) 98.79192 17.4339

(+) (-)

Design Moments CS MS 2.961034 3.48357 8.716934 24.9055

For Interior S-2 M(+) 125.4633 M(-) 305.2601 Moments (%) 0.85M(+) 106.6438 0.70M(-) 213.6821 Moment (%) Passed Column Strip Beam Slab(15%) (85%) (+) 90.64723 15.9966 (-) 181.6298 32.0523

(+) (-)

Middle Strip 6.96714 49.8111

Middle Strip 18.8195 91.578

Design Moments CS MS 7.998285 9.40975 16.02616 45.789

101

S-3 to S-9 54.3891 3 111.454 M(-) 3 Moments (%) 0.85M( 46.2307 +) 6 0.70M( 78.018 -) M(+)

Moment (%) Passed

(+) (-)

Column Strip Beam Slab(15 (85%) %) 39.2961 6.9346 4 1 11.702 66.3153 7

Middle Strip 8.15837 33.4363

Design Moments CS MS 3.46730 4.0791 (+) 7 8 16.718 (-) 5.85135 1

102

Step 5. Determine the Steel Areas and Number of Bars per meter of Width

SAMPLE DESIGN f'c fy B T D Φbar Φtie cc

GIVEN DATA MPa 20.7 MPa 415 mm 1000 mm 150 mm 114 mm 12 mm 10 mm 20

Design for Exterior (S-1) Column Strip (+) R = Mu/(Φb*(d^2)) ρ = (0.85*f'c/fy)*(1-(sqrt(1-((2*R)/(0.85*f'c)))) ρmax = 0.75*0.85*f'c*β*600/(fy*(600+fy)) ρmin = 1.4/fy * If ρ > ρmax, redesign * If ρmin < ρ < ρmax, ok * If ρmin > ρ, use ρmin

As = ρ*b*d S = b*Abar/As, Abar = pi*(Φbar^2)/4

103

M R ρi ρmax ρmin ρf

RESULTS 7.25922243 0.62064 0.00152 0.01598 0.00337 0.00337

As Abar S

384.57831 113.09734 294.08142

kN-m

mm2 mm2 mm

Table. Summary of Steel Areas and Number of Bars for Two-Way Slabs

104

+ CS Ext

S-1 M S

+ +

CS S-2 M S

+ -

Int + CS -

S-(39) M S

+ +

CS Ext

S-10 M S

+ -

Summary of Design for Two-Way Slabs (Longitudinal ρ(initi ρ(fina M R al) ρ(max) ρ(min) l 7.259 0.620 0.001 0.0160 0.0034 0.0034 2 6 5 18.36 1.569 0.004 0.0160 0.0034 0.0040 09 8 0 8.540 0.730 0.001 0.0160 0.0034 0.0034 3 2 8 52.45 4.485 0.012 0.0160 0.0034 0.0127 96 1 7 7.998 0.683 0.001 0.0160 0.0034 0.0034 3 8 7 16.02 1.370 0.000 0.0160 0.0034 0.0034 62 2 1 9.409 0.804 0.002 0.0160 0.0034 0.0034 7 5 0 45.78 3.914 0.010 0.0160 0.0034 0.0108 90 8 8 3.467 0.296 0.000 0.0160 0.0034 0.0034 3 4 7 5.851 0.500 0.001 0.0160 0.0034 0.0034 3 3 2 4.079 0.348 0.000 0.0160 0.0034 0.0034 2 8 0 16.71 1.429 0.003 0.0160 0.0034 0.0036 81 3 6 2.961 0.253 0.000 0.0160 0.0034 0.0034 0 2 6 8.716 0.745 0.001 0.0160 0.0034 0.0034 9 3 8 3.483 0.297 0.000 0.0160 0.0034 0.0034 6 8 7 24.90 2.129 0.005 0.0160 0.0034 0.0055 55 3 5 105

Frame) As 384.578 3 452.390 5 384.578 3 1449.36 4 384.578 3 384.578 3 384.578 3 1232.54 5 384.578 3 384.578 3 384.578 3 410.030 5 384.578 3 384.578 3 384.578 3 625.384 4

Abar 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7

S 294.081 4 249.999 4 294.081 4 78.0324 294.081 4 294.081 4 294.081 4 91.7591 8 294.081 4 294.081 4 294.081 4 275.826 7 294.081 4 294.081 4 294.081 4 180.844 5

Considering Transverse Frame The following data are given. (Slab AB as an Example) Slab Column ts 150 mm c1 550 b 1000 mm c2 550 l1 7 m lc 3 l2 7 m General DL 5.267 kPa LL 1.9 kPa W 9.3604 f'c 20.7 MPa fy 415 MPa E 21383.7 MPa

mm mm m

SAMPLE RESULTS (AB) Step 1. Determine the MOI and Stiffness of Column Ic = a*(b^3)/12 Determine lu/lc Determine a/b Find kab in the tables. Kc = kab*E*Ic/h C = (1-0.63(t/a))*((t^3)(a)/3) Kt = 9*E*C/(l2*(1-b/l2)^2) Kec = 4Kt*Kc/(2*Kt+2*Kc)

Step 2. Determine the MOI and Stiffness of Slab Is = b*(t^3)/12 Compute a/l1 & b/l1 Find ks in the tables. Find FEM and COF in the tables. Ks = ks*E*Is/l1

106

Ic lc lu lu/lc a/b kab Kc C Kt Kec

7.6E+09 3 2.85 0.95 1 4.55 2.5E+11 5.1E+08 1.7E+10 3.1E+10

mm4 mm mm

Is a/l1 b/l2 ks FEM COF Ks

2E+09 0.07857 0.07857 4.11 269.692 0.5086 2.5E+10

mm4

kN-m

Figure. Design Strip for Transverse Frame Step 3. Determine the Distribution Factors DFxy = Kxy/ΣK SUMMARY OF VALUES FOR DF Membe r K DF FEM AB 2.5E+10 0.28438 -269.69 269.691 BA 2.5E+10 0.22141 8 BC 2.5E+10 0.22141 -269.69 269.691 CB 2.5E+10 0.28438 8

Step 4. Continue the MDM To obtain the moments in the midspan of the slabs, M+ = FEM - (-Mave)

MDM FOR TRANSVERSE FRAME Membe A B C r AB BA BC CB 0.22141 0.22141 DF 0.28438 4 4 0.28438 269.691 269.691 FEM 269.692 8 -269.692 8 107

COF

0.5086 Balance 76.6949

0.5086

0.5086

0

CO1 Balance

-39.007 0 230.684 8

0 39.0070 4 0

TOTAL

Slab 1

2

0 0 346.387

-230.685

For Interior 112.7914 346.3868

Moments (%) 0.85M(+) 95.87266 0.70M(-) 242.4707 Moment (%) Passed Column Strip Beam Slab(15%) (85%) (+) 81.49176 14.3809 (-) 206.1001 36.37061

(+) (-)

0 0 346.386 8

Summary of Moments W M 346.3868 401.3272 + 112.7914 230.6848 230.6848 401.3272 + 112.7914 346.3868

For Exterior M(+) M(-)

0.5086 76.6949 2

M(+) M(-)

Middle Strip 16.9 104

112.7914 230.6848

Moments (%) 0.85M(+) 95.87266 0.70M(-) 161.4794 Moment (%) Passed Column Strip Beam Slab(15%) (85%) (+) 81.49176 14.38 (-) 137.2575 24.22

Design Moments CS MS 7.190449 8.459352 18.18531 51.95801

(+) (-)

108

Design Moments CS MS 7.190449 8.459 12.11095 34.6

Middle Strip 16.918705 69.205443

109

Table. Summary of Steel Areas and Number of Bars for Two-Way Slabs

CS Exterior

MS CS

Interior

MS

+ + + + -

Summary of Design of Two-Way Slab (Transverse Frame) M R ρ(initial) ρ(max) ρ(min) ρ(final 7.19 0.6147575 0.00151 0.015977 0.003 0.00337 18.2 1.554778 0.00393 0.015977 0.003 0.00393 8.46 0.7232441 0.00178 0.015977 0.003 0.00337 52 4.4422228 0.01257 0.015977 0.003 0.01257 7.19 0.6147575 0.00151 0.015977 0.003 0.00337 12.1 1.0354427 0.00257 0.015977 0.003 0.00337 8.46 0.7232441 0.00178 0.015977 0.003 0.00337 34.6 2.9584078 0.00786 0.015977 0.003 0.00786

110

As 385 448 385 1433 385 385 385 896

Abar 113.097 113.097 113.097 113.097 113.097 113.097 113.097 113.097

S 294.0814 252.5375 294.0814 78.94669 294.0814 294.0814 294.0814 126.2729

APPENDIX F: SAMPLE DESIGN OF COLUMNS One-Way Slab Tradeoff The column with the maximum axial force was designed. The following are the given data. P My b t cc d f'c fy Φbar Φtie

1337.143 277.02 550 550 40 484 20.7 415 32 10

kN kN-m mm mm

MPa MPa mm mm

Part 1. Determine the Steel Area and Positioning of Bars Step 1. Determine the Steel Area and N bars ρg = ____, assumed value from 0.02 0.04

Results

ρg

0.02

As = ρgAg

Ag

302500

mm2

N = As/Abar then determine actual As

As

mm2

Get actual ρg Pcap = Φ*0.8*Ag(0.85*f'c*(1-ρg)+fy*ρg) If Pcap > P, the dimensions are * adequate * If Pcap < P, redesign

Abar N

6050 804.247 7 8 0.02126 9 0.65 4097.28 OK

111

actual ρg Φ Pcap

mm2 pcs

kN

Step 2. Determine the position of the bars.

Part 2. Checking of Capacity due to Eccentric Load Step 1. Determine if Tension or Compression Controls fy = 600*(d-c)/c, Solve for c a = βc Pb = 0.85*f'c*a*b Pb*(eb+x) = A's*(d-d')+0.85*f'c*a*b*(d-a/2) Solve eb in this equation ex = My/P Solve ex in this equation * If eb > e, Compression Controls, solve for fs * If eb < e, Tension Controls, solve for f's Step 2. Solve and check for Pcap Since tension conrtols, f's = 600(c-d')/c Solve for c and Pcap in the following equations Pcap(ex+(d-c)) = A'sf's(d-d')+0.85f'c(βc)b(d-βc/2) Pcap + Asfy = A'sf's + 0.85f'c(βc)b * If Pcap > P, The dimensions are adequate * If Pcap < P, Redesign For Two-Way Slab Tradeoff The column with the maximun axial forces was designed. The following are the given data.

112

Results c β a Pb eb ex

c Pcap

286.1084 mm 0.85 243.1921 mm 2353431 kN 33.039 mm 207.173 mm Tension Controls

272.163 2247.189

mm kN

P My b t Cc D f'c Fy Φbar Φtie

2424.17 363.257 550 550 40 484 20.7 415 32 10

kN kN-m mm mm

MPa MPa mm mm

Part 1. Determine the Steel Area and Positioning of Bars Step 1. Determine the Steel Area and N bars ρg = ____, assumed value from 0.02 - 0.04

Results ρg

0.02

As = ρgAg

Ag

302500

mm2

N = As/Abar then determine actual As

As

6050

mm2

Get actual ρg Pcap = Φ*0.8*Ag(0.85*f'c*(1-ρg)+fy*ρg) * If Pcap > P, the dimensions are adequate * If Pcap < P, redesign

Abar N actual ρg Φ Pcap

804.2477 8 0.021269 0.65 4097.28 OK

mm2 pcs

Step 2. Determine the position of the bars.

113

kN

Part 2. Checking of Capacity due to Eccentric Load Step 1. Determine if Tension or Compression Controls fy = 600*(d-c)/c, Solve for c a = βc Pb = 0.85*f'c*a*b Pb*(eb+x) = A's*(d-d')+0.85*f'c*a*b*(d-a/2) Solve eb in this equation ex = My/P Solve ex in this equation * If eb > e, Compression Controls, solve for fs * If eb < e, Tension Controls, solve for f's Step 2. Solve and check for Pcap Since tension conrtols, f's = 600(c-d')/c Solve for c and Pcap in the following equations Pcap(ex+(d-c)) = A'sf's(d-d')+0.85f'c(βc)b(d-βc/2) Pcap + Asfy = A'sf's + 0.85f'c(βc)b * If Pcap > P, The dimensions are adequate * If Pcap < P, Redesign

114

Results c β a Pb eb ex

286.1084 0.85 243.1921 2353431 33.039 149.848 Tension Controls

c Pcap

268.6011 2873.398

mm mm kN mm mm

mm kN

APPENDIX G: COST ESTIMATE COST ESTIMATE - ONE WAY TRADEOFF

Member B-1 B-2 B-3 C-1 Slab S-1 S-2 S-3 S-4

L (m) 7 3.5 3 3

b (m) 0.35 0.35 0.35 0.55

L (m) 7 7 3.5 3.3

b (m) 3.5 3 3.3 3

ITEM

TOTAL

CEMENT

9662.96

SAND

536.831

GRAVEL

1073.66

B-1 B-2 B-2 C-1 Slonger Sshorter

BAR Ø (mm) 25 26 25 32 12 12

ITEM

TOTAL

Steel

98654.5

Member

CONCRETE WORKS t V pcs (m) (m3) 0.5 170 208.25 0.5 50 30.625 0.5 180 94.5 0.55 260 235.95 TOTAL t V pcs (m) (m3) 0.15 30 110.25 0.15 120 378 0.15 5 8.6625 0.15 5 7.425

CEMENT (bags) 1874.25 275.625 850.5 2123.55 5123.93 CEMENT (bags) 992.25 3402 77.9625 66.825

TOTAL

4539.038

PRICES ITEM LABOR bags 2415741 966296.3 26841.5 m3 50 10736.63 6 m3 800 858930 343572 TOTAL PRICE 3301512 1320605 REBAR WORKS As L NN bars 2 members (mm ) (m) 490.873 7 6 170 530.929 3.5 6 50 490.873 3 6 180 804.247 3 8 260 113.097 50 224 113.097 14 500 PRICE per kg ITEM LABOR TOTAL 513003 52 1026008 6156046 8 per pc 250

TOTAL COST 115

10778163

TOTAL 4622117 Total W (kg) 27162.5 4320.436 12325.84 38893.42 9816.849 6135.53

SAND (m) 104.125 15.3125 47.25 117.975 284.66 SAND (m) 55.125 189 4.33125 3.7125 252.168 8

GRAVEL (m) 208.25 30.625 94.5 235.95 569.325 GRAVEL (m) 110.25 378 8.6625 7.425 504.3375

COST ESTIMATE - TWO WAY TRADEOFF

Member B-1 B-2 B-3 C-1 Slab S-1 S-2 S-3 S-4 S-5

L (m) 7 3.5 4.5 3

b (m) 0.35 0.35 0.35 0.5

L1 (m) 7 7 3.7 4.5 3.7

L2 (m) 7 4.5 3.5 3.3 1

ITEM

TOTAL

CEMEN T

7894.12 5 438.562 5 877.125

SAND GRAVEL

Member

BAR Ø (mm)

B-1

25

B-2

26

B-2

25

C-1 Slonger Sshorter

32 12 12

ITEM

TOTAL

CONCRETE WORKS t V pcs (m) (m3) 0.5 140 171.5 0.5 20 12.25 0.5 120 94.5 0.5 165 123.75 TOTAL t V pcs (m) (m3) 0.15 10 73.5 0.15 80 378 0.15 5 9.7125 0.15 5 11.1375 0.15 5 2.775

per pc bags

TOTAL PRICES ITEM LABOR

1973531 789412.5 21928.1 m3 50 3 8771.25 3 m 800 701700 280680 TOTAL PRICE 2697159 1078864 REBAR WORKS As L NN bars 2 members (mm ) (m) 490.873 9 7 7 140 530.929 2 3.5 6 20 490.873 9 4.5 6 120 804.247 8 7 3 165 215 113.0973 50 113.0973 14 425 PRICE per kg ITEM LABOR TOTAL

CEMENT (bags) 1543.5 110.25 850.5 1113.75 3618 CEMENT (bags) 661.5 3402 87.4125 100.2375 24.975 4276.125

250

116

TOTAL 3776023 Total W (kg) 26097.31 1728.174 12325.84 24682.36 9422.422 5215.201

SAND (m) 85.75 6.125 47.25 61.875 201 SAND (m) 36.75 189 4.85625 5.56875 1.3875 237.562 5

GRAVEL (m) 171.5 12.25 94.5 123.75 402 GRAVEL (m) 73.5 378 9.7125 11.1375 2.775 475.125

Steel

79471.3 1

52

4132508

826501. 6

TOTAL COST

117

4959010 8735033

APPENDIX H: ESTIMATE OF MAN HOURS

For Tradeoff 1 (One Way Slab)

B-1 B-2 B-3 C-1 S-1 S-2 S-3 S-4

b 350 350 350 550 t 150 150 150 150

T L 500 7 500 3 500 3.5 550 3 S l 3.5 7 3 7 3.3 3.5 3 3.3 TOTAL VOLUME

Quantity 170 180 50 250

Volume 208.25 94.5 30.625 226.875

25 115 5 5

91.875 362.25 8.6625 7.425 1030.463

Assuming that 500% of Total Volume of Concrete Works is equal to Total Man Days, Adding 200% For Rebar Works and 350% For Finishing TOTAL MAN DAYS = 5(1030.463) + 2(1030.463) + 3.5(1030.463) TOTAL MAN DAYS = 10820 days Given that there will be 25 workers TOTAL MAN DAYS = 435 Days

118

For Tradeoff 1 (Two Way Slab)

B-1 B-2 B-3 C-1 S-1 S-2 S-3 S-4

b 350 350 350 550 t 150 150 150 150

t 500 500 500 550 s 7 4.5 3.3 3 TOTAL VOLUME

L 7 4.5 3.5 3 l 7 7 3.5 3.3

Quantity 135 120 20 165

Volume 165.375 94.5 12.25 149.7375

10 80 5 5

73.5 378 8.6625 7.425 889.45

Assuming that 500% of Total Volume of Concrete Works is equal to Total Man Days, Adding 200% For Rebar Works and 350% For Finishing TOTAL MAN DAYS = 5(889.45) + 2(889.45) + 3.5(889.45) TOTAL MAN DAYS = 9940 days Given that there will be 25 workers TOTAL MAN DAYS = 375 Days

119

APPENDIX I: PERCENTAGE DEFLECTION FROM ALLOWABLE

Tradeoff 1 (One Way Slab) (Beam with maximum moment was used) Beam Deflection at Midspan 0.30585 mm Allowable Deflection 19.4444 4 mm Percentage of Computed Deflection from Allowable % = (LV/HV)*100% % = 1.5729 %

Tradeoff 2 (Two Way Slab) (Beam with maximum moment was used) Beam Deflection at Midspan 1.10473 mm Allowable Deflection 19.4444 4 mm Percentage of Computed Deflection from Allowable % = (LV/HV)*100% % = 5.68148 %

120

APPENDIX H: REFERENCES Manuals

Choi K. K. (2002). Reinforced Concrete Structure Design Assistant Tool. California, USA Dahlgren A., & Svensson L. (2013). Guidelines and Rules for Detailong of Reinforcement in Concrete Structures. Goteborg, Sweden. Al-Shamma A. K. (2013). Novel Flowchart for Design of Concrete Rectangular Beams. International Journal of Scientific & Engineering Research. Manual for Design and Detailing of Reinforced Concrete to the Code of Practice for Structural Use of Concrete, 2013

Books

Everrad & Tanner (1996). Theory and Problems of Reinforced Concrete Design. New York: Schaum Publishing Company. McCormac, J.C., & Brown, R. H. (2014). Design of Reinforced Concrete 9 th Edition. United States: John Wiley & Sons, Inc. Association of Structural Engineers of the Philippines. National Structural Code of the Philippines 2010. Quezon City, Philippines: Association of Structural Engineers of the Philippines, Inc. National Building Code of the Philippines (1977). Philippines.

Websites

www.google.com www.wikipedia.com http://www.bca.gov.sg/publications/BuildabilitySeries/others/prh_s2.pdf http://elearning.vtu.ac.in/P6/enotes/CV61/Beams-GS.pdf

121

View more...
A Project in Partial Fulfilment for the Requirements in

CE473 (REINFORCED CONCRETE DESIGN)

Entitled as

A DESIGN OF A FIVE STOREY REINFORCED CONCRETE SEMINARY MAIN BUILDING

Submitted by LAZO, EMMANUEL M.

Submitted to Engr. Rhonnie C. Estores

1

October, 2015

2

APPROVAL SHEET

The design project entitled “A Design of a Reinforced Concrete Seminary Main Building” prepared by Emmanuel M. Lazo of the Civil Engineering Department was examined and evaluated by designer himself, and is hereby recommended for approval.

Engr. Rhonnie Estores Adviser

3

ABSTRACT This project is entitled as “A Design of a Five Storey Reinforced Concrete Seminary Main Building” is presented by Emmanuel M. Lazo, as partial fulfillment for the requirements for CE 473 (Reoinforced Concrete Design. The project was about structural analysis and design of identified parts of a five storey reinforced concrete seminary main building utilizing special moment resisting space frames. Design specifications from NBCP and NSCP were utilized in the design process. The parts analysed and designed included: beams, columns, and slabs. The parts of the building chosen were considered to be the most critical due to the highest result computed throught STAAD pro considering all load combinations. Design schedule and member details of the structure were then created for the design proper.

4

TABLE OF CONTENTS TITLE PAGE....................................................................................................................................................1 APPROVAL SHEET..........................................................................................................................................2 ABSTRACT.......................................................................................................................................................3 LIST OF TABLES..............................................................................................................................................6 LIST OF FIGURES............................................................................................................................................7 CHAPTER I - PROJECT BACKGROUND........................................................................................................8 1.1 Introduction.............................................................................................................................................8 1.2 The Project..............................................................................................................................................9 1.3 Project Location....................................................................................................................................10 1.4 Project Objectives.................................................................................................................................12 1.5 The Client..............................................................................................................................................12 1.6 Project Scope and Limitation................................................................................................................12 1.7 Project Development.............................................................................................................................13 CHAPTER 2: DESIGN INPUTS......................................................................................................................15 2.1 Description of the Structure..................................................................................................................15 2.2 Classification of the Structure...............................................................................................................18 2.3 Architectural Plans................................................................................................................................18 CHAPTER 3: DESIGN CONSTRAINTS, TRADE-OFFS, AND STANDARDS...............................................21 3.1 Design Constraints................................................................................................................................21 3.2 Tradeoffs...............................................................................................................................................22 3.2.1 One Way Slab................................................................................................................................22 3.2.2 Two Way Slab................................................................................................................................23 3.3 Significance of Chosen Tradeoffs to the Quantitative Design Constraints...........................................23 3.4 Method of Measurements for Quantitative Constraints........................................................................24 3.5 Ranking Scale.......................................................................................................................................24 3.6 Initial Estimate and Ranking Computation............................................................................................25 3.7 Raw Designer’s Ranking and Assessment...........................................................................................27 3.8 Design Standards.................................................................................................................................29 CHAPTER IV: DESIGN OF STRUCTURE.....................................................................................................30 4.1 Design Methodology.............................................................................................................................30 4.1.1 Structural Plans..............................................................................................................................32 5

4.1.2 Design Specifications....................................................................................................................33 4.1.3 Material Properties.........................................................................................................................33 4.1.4 Structural Models...........................................................................................................................33 4.1.5 Load Models..................................................................................................................................34 4.1.6 Structural Analysis.........................................................................................................................39 4.1.7 Structural Design...........................................................................................................................41 4.2 Raw Ranking Validation, Comparison of Results, and Final Ranking Assessments...........................49 4.2.1 Final Estimates of Tradeoffs..........................................................................................................49 4.2.2 Validation of Raw Designer’s Ranking...........................................................................................49 4.2.3 Final Designer’s Ranking...............................................................................................................50 4.2.4 Designer’s Final Ranking Assessment..........................................................................................51 CHAPTER V: FINAL DESIGN.........................................................................................................................52 5.1 Design Schedules.................................................................................................................................52 5.1.1 Design Schedule of Slabs..............................................................................................................52 5.1.2 Design Schedule of Beams...........................................................................................................53 5.1.3 Design Schedule of Columns........................................................................................................55 5.1.4 Beam Details..................................................................................................................................56 5.1.5 Column Details...............................................................................................................................57 APPENDICES.................................................................................................................................................59 APPENDIX A: CODES AND STANDARDS................................................................................................59 APPENDIX B: RESULTS OF STRUCTURAL ANALYSIS..........................................................................69 APPENDIX C: DESIGN OF BEAMS...........................................................................................................73 APPENDIX D: DESIGN OF ONE-WAY SLAB............................................................................................85 APPENDIX E: DESIGN OF TWO-WAY SLAB............................................................................................91 APPENDIX F: SAMPLE DESIGN OF COLUMNS....................................................................................104 APPENDIX G: COST ESTIMATE.............................................................................................................108 APPENDIX H: ESTIMATE OF MAN HOURS...........................................................................................110 APPENDIX I: PERCENTAGE DEFLECTION FROM ALLOWABLE.........................................................110 APPENDIX H: REFERENCES..................................................................................................................110

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LIST OF TABLES

Table 1. Total Floor Areas and Functions per Floor........................................................................................15 Table 2. Summary of Initial Estimate of Values...............................................................................................26 Table 3. Wind Intensity (One-Way).................................................................................................................36 Table 4. Wind Intensity (Two-Way).................................................................................................................38 Table 5. Final Estimate of Tradeofs................................................................................................................49 Table 6. Comparison of Initial and Final Estimate of Tradeoffs......................................................................49 Table 7. Final Designer’s Ranking..................................................................................................................51 Table 8. Slab Schedule...................................................................................................................................52 Table 9. Beam Schedule.................................................................................................................................53 Table 10. Column Schedule............................................................................................................................55

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LIST OF FIGURES Figure 1. Perspective of the Proposed Seminary...........................................................................................10 Figure 2. Distance of the Nearest Fault Line to the Structure........................................................................11 Figure 3. Vicinity Map of the Seminary...........................................................................................................11 Figure 4. Project Development Process.........................................................................................................13 Figure 5. Structural Model of the Structure.....................................................................................................15 Figure 6. Floor Plans of the Building...............................................................................................................20 Figure 7. One Way Slab System.....................................................................................................................22 Figure 8. Two Way Slab System.....................................................................................................................23 Figure 9. Ranking Scale for Importance Factor..............................................................................................25 Figure 10. Ranking Scale for Satisfactory Factor...........................................................................................25 Figure 11. Design Methodology......................................................................................................................30 Figure 12. One Way Slab Framing Plan.........................................................................................................32 Figure 13. Two Way Slab Framing Plan.........................................................................................................32 Figure 14. Geometric Modelling of One Way Slab.........................................................................................33 Figure 15. Geometric Modelling of Two-Way Slab.........................................................................................34 Figure 16 Load Diagrams for One-Way Slab.................................................................................................36 Figure 17. Load Diagrams for Two-Way Slab.................................................................................................38 Figure 18. Result of Structural Analysis for One-Way Slab............................................................................39 Figure 19. Result of Structural Analysis for Two-Way Slab............................................................................40 Figure 20. Stress-Strain Diagram for Singly Reinforced Beam......................................................................41 Figure 21. Design of Singly Reinforced Beam................................................................................................42 Figure 22. Design of Doubly Reinforced Beam..............................................................................................43 Figure 23. Design for Spacing of Stirrups for Beams.....................................................................................44 Figure 24.Beindg of Slab................................................................................................................................45 Figure 25. Two Way Slab Strips Considered in EFM......................................................................................45 Figure 26. Two Way Slab Design (EFM).........................................................................................................46 Figure 27. Determining the Steel Area of a Column.......................................................................................47 Figure 28. Column Check for Compression or Tension Controls...................................................................48 Figure 29. Beam Details..................................................................................................................................56 8

Figure 30. Column Details..............................................................................................................................57 Figure 31. Beam-Column Interaction Detail....................................................................................................58

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CHAPTER I - PROJECT BACKGROUND 1.1 Introduction Structural analysis and design is a very old art and is known to human beings since early civilizations. The Pyramids constructed by Egyptians around 2000 B.C. stands today as the testimony to the skills of master builders of that civilization. Many early civilizations produced great builders, skilled craftsmen who constructed magnificent buildings such as the Parthenon at Athens (2500 years old), the great Stupa at Sanchi (2000 years old), Taj Mahal (350 years old), Eiffel Tower (120 years old) and many more buildings around the world. These monuments tell us about the great feats accomplished by these craftsmen in analysis, design and construction of large structures. Today we see around us countless houses, bridges, fly-overs, high-rise buildings and spacious shopping malls. Planning, analysis and construction of these buildings is a science by itself. In the early periods houses were constructed along the riverbanks using the locally available material. They were designed to withstand rain and moderate wind. Today structures are designed to withstand earthquakes, tsunamis, cyclones and blast loadings. These have been made possible with the advances in structural engineering and a revolution in electronic computation in the past 50 years. The construction material industry has also undergone a revolution in the last four decades resulting in new materials having more strength and stiffness than the traditional construction material. Combinations of some materials were also utilized so as to have a better performance in the maintenance of the structure. One good example are the reinforced concrete structures which are one of the most popular structural systems today. The combination of concrete and a steel reinforcement gives advantages that make a structure maintain its form for a long time. Concrete is one of the most popular materials for buildings because it has high compressive strength and flexibility in its form and it is widely available. The history of concrete usage dates back for over a thousand years. Contemporary cement concrete has been used since the early nineteenth century with the development of Portland cement. Despite the high compressive strength, concrete has limited tensile strength, only about ten percent of its compressive strength and zero strength after cracks develop. In the late nineteenth century, reinforcing materials, such as iron or steel rods, began to be used to increase the tensile strength of concrete. Today steel bars are used as common reinforcing material. Usually steel bars have over 100 times the tensile strength of concrete; but the cost is higher than concrete. Therefore, it is most economical that concrete resists compression and steel provides tensile strength. Also it is essential that concrete and steel 10

deform together and deformed reinforcing bars are being used to increase the capacity to resist bond stresses. Advantages of reinforced concrete can be summarized as follows (Hassoun, 1998). 1. It has a relatively high compressive strength. 2. It has better resistance to fire than steel or wood 3. It has a long service life with low maintenance cost. 4. In some types of structures, such as dams, piers, and footing, it is the most economical structural material. 5. It can be cast to take any shape required, making it widely used in precast structural components. Disadvantages of reinforced concrete can be summarized as follows: 1. It has a low tensile strength (zero strength after cracks develop). 2. It needs mixing, casting, and curing, all of which affect the final strength of concrete. 3. The cost of the forms used to cast concrete is relatively high. The cost of form material and artisanry may equal the cost of concrete placed in the forms. 4. It has a lower compressive strength than steel (about 1/10, depending on material), which requires large sections in columns of multi-storey buildings. 5. Cracks develop in concrete due to shrinkage and the application of live loads.

1.2 The Project The project is a seminary constituted of five-storeys containing all the necessary rooms for the residents of the building. It is intended to be built in Antipolo, Rizal. As a city with many public and private schools, constructing a seminary is appropriate. This will be very important for the Antipoleneos since the city contains the National Shrine of the Philippines and thus needs training areas for students who want to become priests someday.

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Figure 1. Perspective of the Proposed Seminary

The building is rectangular shaped and has a total area of 700 m 2 with dimensions of 50 m x 14 m. The structure to be constructed will be the main building of a seminary. The first floor contains the refectory (dining), chapel, lobby, infirmary (clinic), recreation area, kitchen and staff room. The second and third floors contain class rooms, laboratories, library, and offices. The fourth and fifth floor contain the study area

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and dormitories. It has a main stair, fire exit, ramps, and an elevator. The height of each floor is 3 m having a total of 15 m. The covering of the building will be a roof deck.

1.3 Project Location The project area is located in Antipolo, Rizal, which is included in the areas under seismic zone 4. The figure below shows the distance of the planned structure from the nearest fault line which is the Makati Valley Fault System.

Figure 2. Distance of the Nearest Fault Line to the Structure

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Figure 3. Vicinity Map of the Seminary Address: Lot 6 Blk.1, Sampaguita St. Bermuda Hts. Subd., Brgy. San Luis, Antipolo City Nearest Fault Line Distance: 16.3 km

1.4 Project Objectives The main objective of this project is to analyse and design a reinforced concrete structure in accordance with the principles written in NSCP 2010. Other objectives of the project are as follows: a. To design a seminary that will have an acceptable probability of performing satisfactorily during its intended life time. b. To provide all the necessary architectural plans, structural plans, and the estimate of the building cost. c. To plan the structure considering balanced constraints, trade-offs and standards on the design.

1.5 The Client

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The client of this structure is a set of religious people lead by Mrs. Sharon Umayam. She is a businesswoman and at the same time the president of the lectors in Our Lady of Peace and Good Voyage Church (National Shrine of the Philippines).

1.6 Project Scope and Limitation The following were the scope covered by the design project: 1.) The project was designed in accordance to the National Building Code of The Philippines and the National Structural Code of the Philippines. 2.) Analysis of the loads and moments was done using STAAD Pro. 3.) All architectural plans such as floor plan and perspective of the apartment were provided. The following were the limitations of the design project: 1.) 2.) 3.) 4.)

Only beams, slabs, and columns were considered in the design. The cost estimates for the mechanical, plumbing and architectural plan were not included. The plumbing and electrical plans are not included in this design. The interior design of the structure was not considered.

1.7 Project Development PLANNING/CONCEPTUALIZATION

IDENTIFICATION OF DESIGN STANDARDS AND PARAMETERS

PRESENTATION OF ARCHITECTURAL AND STRUCTURAL PLANS WITH INITIAL ESTIMATE

IDENTIFICATION OF DESIGN CONSTRAINTS, TRADE-OFF 15

LOAD IDENTIFICATION, STRUCTURAL ANALYSIS, AND FINAL DESIGN

Figure 4. Project Development Process The project development process started with the planning/conceptualization. In this stage, the identification of client was the most important so as to know the structure to be build. In this case, the structure requested by the client was a seminary. It also included the identification of the location where the structure was intended to be built. The next stage was the identification of design standards. Knowing the structure to be constructed, the next part was to know the specific design standards that are required before coming up to the design (i.e., minimum dimension of a classroom, minimum size of an elevator shaft, etc.). These will set the parameters in the creation of the architectural and floor plans which is the next stage in the process. In the third stage, the plans will be presented to the client so that alterations could be made. After all has been settled, constraints can now be identified, which is the next stage. In this, the constraints that were projected will then be classified as either qualitative or quantitative. Knowing the quantitative tradeoffs will pave the way to the determination of the trade-offs for the structure. In the last stage, geometric design, computation, and final estimation for each trade-offs will be made. Then, all of these will be presented to the client. The client will then rate each trade-off. The one which has the most favorable rating among all will then be chosen for the design of the structure.

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CHAPTER 2: DESIGN INPUTS

2.1 Description of the Structure As what was said, the structure will be a seminary. The structure contains five floors with each floor having different function from the other. The structure has two access stairs, set of ramps, and an elevator. The structure contains five floors with each floor having different functions. The structure has special moment reinforced concrete frames both in longitudinal and transverse axes. The figure below shows the model of the structure.

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This model shows

the

structural

members of

the

structure.

The blue

members

beams are

the

and columns while the violet members

the

total floor

are

slabs that form the building. The table below shows the area and the different areas of the rooms contained in each floor. Figure 5. Structural Model of the Structure

Table 1.

Total Floor Areas and Functions per Floor AREA (m2)

FUNCTION 1ST Floor Ramps and Elevator

49

Stairs

25

C.R.

22.5

Chapel

168

Refectory

168

Staff Room

63

Clinic

49

Lobby

70

Kitchen

63

Hallway

22.5

TOTAL

700 2nd Floor

Ramps and Elevator

49

Stairs

25 18

C.R.

22.5

Offices

3(45)

Class Rooms & Laboratories

4(63)

Other Rooms

32.5

Lounge

35

Hallway

79

TOTAL

700 3rd Floor

Ramps and Elevator

49

Stairs

25

C.R.

22.5

Offices

45

Class Room

2(63)

Other Rooms

133

Faculty Room

65

Library

94.5

Hallway

73.5

Sisters’ Room

66.5

TOTAL

700 4th Floor

Ramps and Elevator

49

Stairs

12.5

C.R.

22.5

Study Area

178.5

Dormitory

255.5

Vice Rector’s and Prefect’s Room

66.5

Toilet & Bath

59.5

Laundry

28

Hallway

28 19

TOTAL

700 5th Floor

Ramps and Elevator

49

Stairs

25

Hallway

28

Dormitory (1)

201

Dormitory (2)

196

Toilet & Bath

2(59.5)

Laundry

28

Rector’s Room

66.5

TOTAL

700

TOTAL FLOOR AREA

3500

2.2 Classification of the Structure

Using the National Structural Code of the Philippines (NSCP) 2010, the designer was able to classify the structure. With respect to the occupancy category, the building is classified as an Essential Facility. With respect to the structural members, the building will have special moment resisting frames. These data will help in designing the structure especially in the determination of the seismic forces acting on the structure.

2.3 Architectural Plans Height is 3 m per floor, for a total of 15 m.

20

21

22

Figure 6. Floor Plans of the Building

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CHAPTER 3: DESIGN CONSTRAINTS, TRADE-OFFS, AND STANDARDS 3.1 Design Constraints Constraint based design takes the parameters associated with a design problem and links them to the attributes of the formal components and relationships of a solution. The forms that compose a building are defined by a set of attributes. Constraints have to be managed effectively throughout the decision making process, and also could be reduced or eliminated. In this project, the design constraints were divided into two types, namely, quantitative and qualitative. Quantitative constraints are those constraints that can be measured using engineering methods (estimation). The qualitative constraints are those which cannot be measured but are ranked through the designer’s perception and experience. The following are the constraints that are considered in the design of the structure. 3.1.1 Quantitative Constraints 1. Economic. Cost is always an integral part of marketing and livelihood even in ancient times. Thus, this constraint is considered in this design. The cost of the structure is highly significant both to the designer and the client. 2. Constructability. This constraint refers to the ratio of the number of workers that will be hired for the construction, to period of time for the structure to be built. One of these two will be considered as a constant so as to measure an accurate difference between them. 3. Safety/Serviceability. Structures always meet some limitations but sometimes some part of it could be accidentally damaged. The safety of a structure is the one which must have the most outstanding consideration among all. Safety makes a structure function effectively overtime. 3.1.2 Qualitative Constraints 1. Aesthetics. The beauty of the structure lies upon its final output. This constraint depends on the taste of a person therefore it is considered as a qualitative constraint. It depends on a person’s perception which design is more presentable. 24

2. Sustainability. In civil engineering, sustainability refers to the conditions under which a building is still considered useful. Should these limit states be exceeded, a structure that may still be structurally sound would nevertheless be considered unfit. In the process, only the quantitative constraints was focused by the designer. Tradeoffs were enumerated next in this section which will then be ranked and assessed. 3.2 Tradeoffs Design trade-off strategies are always present in the design process. Considering design constraints, trade-offs that have a significant effect on the structural design of the structure were provided by the designer. As a trade-off, the designer will have to evaluate which of the two is more effective considering each constraint. The following are the tradeoffs that were chosen by the designer because they are the most fitted to the said constraints.

3.2.1 One Way Slab

Figure 7. One Way Slab System One-way slabs are those slabs with an aspect ratio in plan of 2:1 or greater, in which bending is primarily about the long axis. In heavily loaded slabs, the thickness is often governed by shear or flexure, while in lightly-loaded slabs, the thickness is generally chosen based on deflection limitations. Both lightly and heavily loaded slabs are typically dimensioned so that no shear reinforcement is required, as placing 25

stirrups in slabs is perceived to be difficult and costly. One-way slabs are designed for flexure and shear on a per meter width basis, assuming that they act as a series of independent strips. Thus one-way shear in slabs is often referred to as beam shear, and design for flexure and shear is carried out using a beam analogy

3.2.2 Two Way Slab

Figure 8. Two Way Slab System When a rectangular slab is supported on all the sides and the length-to-breadth ratio is less than two, it is considered to be a two-way slab. The slab spans in both the orthogonal directions. In general, a slab which is not falling in the category of one-way slab is considered to be a two-way slab.

3.3 Significance of Chosen Tradeoffs to the Quantitative Design Constraints In this section, the constraints enlisted in the beginning of the chapter will be related to the tradeoffs chosen by the designer. The final decision of choosing the tradeoff that will be used for the structure lies on the client. Thus, the significance of the tradeoffs to the constraints is needed. Economic. For the cost effectiveness of the structure, the tradeoffs chosen will be designed to be compared whether of the two will be more economical. Clients do not have the same state of living and 26

thus might give priority to this constraint. Some might choose the tradeoff that have lower price but might not give way to the positivity of other tradeoffs. Constructability. Time measures is significant in the construction of the structure. Knowing which of the difference in the period of construction two tradeoffs might be significant for a client. Some clients need shorter period of time and thus give priority to this constraint. Safety/Serviceability. The magnitudes of deflection for concrete members are also important. Any structure used by the people should be quite rigid and relatively-vibration free so as to provide security. Designing these two tradeoffs will give different results. Thus, one tradeoff might be safer than the other. A safer structure known to a client might be given priority. Through the consideration of multiple constraints, the designer will have to choose what particular design among the tradeoffs will be used. The tradeoff is very significant in the design for it will solve the problem regarding the concern of client considering the constraints.

3.4 Method of Measurements for Quantitative Constraints The main method of measurement that will be used in this design is estimation. For the economic constraint, the cost of the whole building. This includes the materials that will be used for the construction of the beams, slabs, and columns. It also includes the cost of the reinforcements that will be used for the structure. For the constructability of the structure, the period of time that will be utilized to construct the building will be estimated, together with the number of workers that will work on that period of time. The number of workers will be constant for both tradeoffs. The difference between the days will give the result for each tradeoff. For the last constraint, the deflection of the most critical beam will be computed for each tradeoff and will then be compared.

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3.5 Ranking Scale The ranking scale that will be used in this design is based on the model on tradeoff strategies formulated by Otto and Antonsson (1991). The importance factors in each constraint is scaled from 0 to 5, while the ability to satisfy the constraint is scaled from -5 to 5, 5 being the highest for both. After obtaining the results, the product of the importance and ability to satisfy the criteria will be summed of from each constraint. The result will then be the overall ranking of the tradeoff.

Figure 9. Ranking Scale for Importance Factor

Figure 10. Ranking Scale for Satisfactory Factor

Computation of ranking for ability to satisfy criterion of materials:

Difference( )=

Higher value−Lower value ×100( ) Lower value

Subordinate rank =Governing rank −(

difference ) 10

Equation 1 Equation 2

The above equations will be used for the manipulation of the rankings of each constraint given to the tradeoffs. The governing rank is the highest possible value set by the designer. The subordinate rank in second equation is a variable that corresponds to its percentage difference from the governing rank along the ranking scale.

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3.6 Initial Estimate and Ranking Computation To determine the difference between the two tradeoffs, certain methods were used by the designer. For the economic constraint, a cost estimate was provided. For the constructability constraint, an estimate of the number of working days was provided, given that there will be 50 workers. For the safety/serviceability constraint, the deflection of the most critical beam was considered. In this part, a rough computation of the estimates was utilized. The values written in the table below were just assumed by the designer whose basis came from experience. Table 2. Summary of Initial Estimate of Values Estimated Value One-Way Slab Two-Way Slab Php 9,000,000 Php 8,000,000 500 days 450 days 4 % of allowable 5 % of allowable

Constraint Economic Constructability Safety/Serviceability

Computation of ranking for Economic Constraint % difference=

higher value−lower value ×100 higher value

% difference=

9000000−8000000 ×100 9000000

%difference=11.11

Subordinate rank =Governing rank −(

difference ) 10

Subordinate rank =5−1.11

29

Subordinate rank =3.89

Computation of ranking for Constructability Constraint

% difference=

higher value−lower value ×10 higher value

% difference=

500−400 ×100 500

%difference=20

Subordinate rank =Governing rank −

difference 10

Subordinate rank =5−2 Subordinate rank =3

Computation of ranking for Safety/Serviceability Constraint

% difference=

higher value−lower value ×10 higher value

% difference=

30

5−4 × 10 5

%difference=20

Subordinate rank =Governing rank −

difference 10

Subordinate rank =5−2

Subordinate rank =3

3.7 Raw Designer’s Ranking and Assessment After making an initial estimate of the structure considering the constraints, the design came up with the raw rankings on the one-way slab and two-way slab. The values computed in the latter section is tabulated. Table 1. Raw Designer’s Ranking CONSTRAINT (CRITERIA)

ABILITY TO SATISFY THE CRITERION IMPORTANCE

One-Way Slab

Two-Way Slab

1. Economic

5

4

5

2. Constructability

4

3

5

3. Maintenance

2

5

3

42

51

Over All Ranking

These tabulated values are just subjective, especially the importance factors. This values will still go on with the validation after making a final estimate and final ranking. Knowing the significance of the constraints to the tradeoffs, the ranks in its importance are given as 5, for economic, 4, for constructability, and 2, for maintenance.

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As for economic constraint, it turned out that the initial cost for the two-way slab is cheaper than the one-way slab, considering only the volume of concrete that will be used. As for the constructability constraint, it turned out that the labor constituting of 50 workers will have to work for longer time for the construction of the one way slab. As for the safety/serviceability constraint, the deflection of the critical member in the two-way slab is quite greater than that of the one-way slab. Overall, it turned out that the two-way slab tradeoff outranked the one-way slab tradeoff for the raw designer’s ranking.

3.8 Design Standards To come up with the final design of the structure, the designer utilized the codes and standards written in the following: 1 2

National Building Code of the Philippines National Structural Code of the Philippines (NSCP) 2010 The National Building Code of the Philippines (PD 1096). The National Building Code of the

Philippines, also known as Presidential Decree No. 1096 was formulated and adopted as a uniform building code to embody up-to-date and modern technical knowledge on building design, construction, use, occupancy and maintenance. The Code provides for all buildings and structures, a framework of minimum standards and requirements to regulate and control

location, site, design, and quality of materials,

construction, use, occupancy, and maintenance. The National Structural Code of the Philippines 2010. This code provides minimum standards to safeguard life or limb, property and public welfare by regulating and controlling the design, construction, quality of materials pertaining to the structural aspects of all buildings and structures within its jurisdiction. The provision of this code shall apply to the construction, alteration, moving, demolition, repair, maintenance and use of any building or structure within its jurisdiction, except work located primarily in a

32

public way, public utility towers and poles, hydraulic flood control structures, and indigenous family dwellings.

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CHAPTER IV: STRUCTURAL ANALYSIS AND DESIGN

4.1 Design Methodology The design was done in accordance with the codes and standards appropriate for a reinforced concrete structure. The figure below shows the step by step process of the design of the building. FRAMING PLANS

STRUCTURAL PLANS

DESIGN SPECIFICATIONS

NSCP NBCP

MATERIAL PROPERTIES

COMPRESSIVE STRENGTH MODULUS OF ELASTICITY STRUTURAL MEMBER DIMENSIONS

GEOMETRIC MODELLING

STRUCTURAL MODEL

LOAD MODELS

DEAD AND LIVE LOAD SEISMIC AND WIND LOAD LOAD COMBINATIONS

STRUCTURAL ANALYSIS

SHEAR DIAGRAMS MOMENT DIAGRAMS REACTIONS AND DEFLECTIONS

STRUCTURAL DESIGN

DESIGN SCHEDULES DETAILING

Figure 11. Design Methodology 34

The first process in design methodology was the creation of structural plans. The structural plans included the framing plans of the two trade-offs. The next step was to know the design specifications. These specifications are the codes and standards needed for the structure’s classification and description. The National Building Code and National Structural Code of the Philippines are the main books used for design specifications. The third step in the process was the identification of the material properties. The compressive stresses and modulus of elasticity of the concrete and steel to be used were determined. Also, the structural member dimensions (b, d, etc.) were assumed. The fourth step was the creation of the structural model. These models included geometric modelling, which showed the positioning of the structural members (beams, columns, slabs) in 3D form. The fifth step was the presentation of load models. In this part, the loads acting on the structure were computed. These loads were the dead load, live load, wind load, and seismic (earthquake) load, applying also the load combinations. After computing for these loads, load models was presented also in 3D form. The sixth step was the structural analysis. In structural analysis, member (beams and columns) forces and reactions were determined. The member forces included were the axial force, shear force, and moment acting on the member. The last part was the structural design. The structural design did not include the design of footings. The values from the structural analysis was utilized to design the structural members of the structures, mainly the beams and columns. The maximum moment acting on a beam was used to design the beam, and the maximum value of the axial force acting on a column was used to design the column. To design the slab, the total load on the floors was utilized.

35

4.1.1 Structural Plans

Figure 12. One Way Slab Framing Plan

Figure 13. Two Way Slab Framing Plan

36

4.1.2 Design Specifications The all the design specifications coming from NBCP and NSCP for the structure is stated Appendix A of the project.

4.1.3 Material Properties The material properties that were utilized are 20.7 MPa for the compressive strength of the concrete (f’c) and 415 MPa for the compressive strength of steel (fy). The modulus of elasticity of steel (Es) is 200000 MPa, while the modulus of elasticity of concrete (Ec) was solved through the following formula, Ec =4700 √ f ' c

4.1.4 Structural Models

Figure 14. Geometric Modelling of One Way Slab

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Figure 15. Geometric Modelling of Two-Way Slab

4.1.5 Load Models

The loads considered in this project are the dead load, live load, wind load and seismic loads. Load combinations were also applied to these loads. The load combinations that were utilized were those that are written in Section 203 of NSCP 2010.

38

The figure below show the preliminary loads acting on the structure.

Dead Load Self Weight = 2.8272 kPa

Floor Loads = 3.49 kPa

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Figure 16a. Dead Load

Live Load Floor Loads = 1.9 kPa

Figure 16b. Live Load Wind Load

Table 3. Wind Intensity (One-Way) Height (m)

Intensity (kPa)

3

1.34496

6

1.40150

9

1.4884

12

1.55674

15

1.5999

Figure 16c. Wind Load (+X)

40

Figure 16

Figure 16d. Earthquake Load (-Z) Figure 16. Load Diagrams for One-Way Slab

Dead Load Self Weight = 2.8272 kPa

Floor Loads = 3.49 kPa

41

Figure 17a. Dead Load

Live Load Floor Loads = 1.9 kPa

Figure 17b. Live Load Wind Load

Table 4. Wind Intensity (Two-Way)

Figure 17c. Wind Load (+Z) 42

Height (m)

Intensity (kN)

3

1.34496

6

1.40150

9

1.4884

12

1.55674

15

1.5999

Figure 17d. Earthquake Load (+X) Figure 17. Load Diagrams for Two-Way Slab

4.1.6 Structural Analysis For the structural analysis of the members, the results considered are those that came from the load combination which gave the maximum values of member forces and reactions. A summary of values of the member forces is presented in the appendices. The following figures show the results of the structural analysis done through the software STAAD.

Maximum Value of Axial Forces in Columns F = 1337.143 kN

43

Figure 18a. Axial Forces and Shear Forces

Maximum Value of Bending Moments (Z) in Beams +M = 198.802 kN-m -M = 103.972 kN-m

Figure 18b. Moment along Z and Y axes. Figure 18. Result of Structural Analysis for One-Way Slab

44

Maximum Value of Axial Force in Columns F = 2424.714 kN

Figure 19a. Axial and Shear Forces

Maximum Value of Bending Moments (Z) in Beams +M = 310.203 kN-m -M = 164.303 kN-m

Figure 19b. Moments along Z and Y axes

Figure 19. Result of Structural Analysis for Two-Way Slab 4.1.7 Structural Design In this section, the beams, columns, and slabs were designed. The main goal of the structural design of the members is to know the number of bars and their spacing, and check if the assumed dimensions are adequate for the structure 45

. For beams and columns, only the most critical parts were designed. For one-way slab, only one slab was considered both in longitudinal and transverse directions was designed. For two-way slab, only one strip was designed also considering both longitudinal and transverse directions. For convenience, a sample procedure of computation for a structural member will be shown. The manual computations of the members is shown in the appendices.

4.1.7.1 Design of Beams Due to forces acting on the beam, the whole structure experiences flexure, and thus the whole length of the beam have moments within them. Also due to these forces, the beam experiences a shearing stress, which makes a part of the beam to be compressed (top), and another part to be tensed (bottom) .

To design the beams

of the entire structure, the beam which had the highest moment was picked and the resulting design for that beam will be applied to all other beams in the structure. The dimensions of the beam (b,t) and the stresses (f’c,fy) were provided by the designer.

Figure 20. Stress-Strain Diagram for Singly Reinforced Beam

The parts of the beam to be designer are the supports, which

experience negative moment, and the midspan, which experience positive moment. Moreover, the stressstrain diagram of the cross-sectional of the beam was used for the design. The following flow charts present the step by step process of designing a beam.

46

Given b, d, f’c, fy, and Mu

ρmin =

Ru =

ρ=

ρb =

ρmax = 0.75ρb

ρ< ρmax

NO

YES

As = ρminbd N=

YES

ρ> ρmin

NO

As = ρbd N=

eams Figure 21. Design of Singly Reinforced Beam 47

DOUBLY REINFORCED

=

=

=

=

As = As1 + As2 =

Y =

N N =

48

Figure 22. Design of Doubly Reinforced Beam

49

Given Vu and other properties

Vc =

4.1.7.2 Design of Slabs To design a slab, we always consider the longer and shorter span of the slab

since

bending is experience by the whole. For One-way slabs, the process is quite N same in designing a singly reinforced concrete beam. The only

the the

N Vu > ΦVc

Vu < 0.5ΦVc

NO NEED STIRRUPS difference is FOR that we assume that we get a strip from the whole

length of the slab. The width of that strip is 1 meter with thickness provided by the designer.

Y

Following the procedure of solving for the reinforcement of singly reinforced beams, the desired number of bars for one-way slab was computed. For the Vs = Vn - Vc

Y

Vn = Vu/Φ

spacing of bars, the width, b (1 m) was divided by the diameter of the bar times the quantity of bars. Since the two-way slab transmits the load to the supports in trapezoidal form, the method used for one-way slab is not

Figure 24.Beindg of Slab applicable. For the two-way slab, the equivalent frame method was used. The two-way slab was designed N

Y

S =The flow considering REDESIGN the positive and negative moments passedVs through < the column strip and middle strip.

chart below shows the procedure of equivalent frame method.

N Smax = d/4 or 300 (get smaller)

Vs <

Y

Use the smaller value between S and Smax

Smax = d/2 or 600 (get smaller)

Figure for Spacing of Stirrups in forEFM Beams Figure 25.23. TwoDesign Way Slab Strips Considered 50

Given E and L1, L2, t as dimensions of slab

Given E and a, b, lcolumn as dimensions of column

PROCEED WITH THE MDM TO OBTAIN THE MOMENTS

Figure 26. Two Way Slab Design (EFM)

SLAB AS SINGLY REINFORCED BEAM USING THE MOMENTS OBTAINED 51

4.1.7.3 Design of Columns From the structural analysis, the column that experienced the greatest axial forces was designed. The designer started the design of the column in determining the number of bars and its positioning within the gross area of the column. Knowing the position of bars in the column, the designer then computed for the axial force capacity column due to the eccentric load. The flow chart below shows the step by step process done by the designer. The second flow chart is applicable only in this design (eccentricity on one side only).

Given P, My, f’c, fy, b, and t

Assume a value of ρg between 0.02 – 0.04

As = ρgAg

N = As/Abar Redesign Determine actual As and ρg

P < Pcap Pcap = Φ(0.8)Ag(0.85f’c(1-ρg)+ρgfy)

Check If Compression or Tension Controls

Figure 27. Determining the Steel Area of a Column

52

Given e, As and other properties

fy = 600(d-c)/c , solve for c

Pb = 0.85f’cab

Pb(eb+x) = C1(d-.5a) + C2(d-d’), solve for eb

e > eb

Compression Controls

fs = 600(d-c)/c, in terms of c

Tension Controls

f’s = 600(c-d’)/c, in terms of c

Pcap + T = C1 + C2 Pcap(e+x) = C1(d-.5βc) + C2(d-d’) Solve for c and Pcap, then check.

Figure 28. Column Check for Compression or Tension Controls

53

4.2 Raw Ranking Validation, Comparison of Results, and Final Ranking Assessments In this section, the raw designer’s ranking was validated through the gathered results of the design. The initial and final estimated values was then be compared. With the help of the final designer’s ranking, the final ranking assessments was concluded.

4.2.1 Final Estimates of Tradeoffs The table below shows the result of the estimation of construction cost, man days, and cost of maintenance for each tradeoff. Table 5. Final Estimate of Tradeoffs TRADEOFFS

CONSTRAINT

One-Way Slab

Two-Way Slab

Economic (Construction Cost)

Php 10,778,163.00

Php 8,735,033.00

Constructability

435 days

375 days

Safety/Serviceability

1.6 %

5.7 %

4.2.2 Validation of Raw Designer’s Ranking

Table 6. Comparison of Initial and Final Estimate of Tradeoffs CONSTRAINT Economic Constructability Safety/Serviceabilit y

Initial Estimate One Way Slab Two Way Slab Php 9,000,000 Php 8,000,000 500 days 450 days 4 % of allowable

5 % of allowable

Final Estimate One Way Slab Two Way Slab Php 10,778,163 Php 8,735,033 435 days 375 days 1.6 % of allowable

5.7 % of allowable

Looking at the table, there are large discrepancies between the assumed values and the computed values. However, the results of the final estimate of values has almost the same outcome with the initial estimate. It turned out that the two way slab is better than the one way slab in terms of both economic and constructability constraint, while one way slab is better than two way slab in terms of safety/serviceability

54

constraint. These results are the same as what was said in the raw ranking, which makes raw design to be quite certain in this project.

4.2.3 Final Designer’s Ranking Computation of ranking for Economic Constraint % difference=

10778163−8735033 × 100 10778163

%difference=18.9562

Subordinate rank =5−1.8956

Subordinate rank =3.10438

Computation of ranking for Constructability Constraint

% difference=

435−375 ×100 435

%difference=13.79

Subordinate rank =5−1.379 Subordinate rank =3.621

Computation of ranking for Safety/Serviceability Constraint

55

% difference=

5.7−1.6 ×10 5.7

%difference=71.93

Subordinate rank =5−7.193

Subordinate rank =−2.193

Table 7. Final Designer’s Ranking CONSTRAINT (Criteria)

Ability to Satisfy the Criterion

Importance

One-Way Slab

Two-Way Slab

Economic

5

3.10438

5

Constructability

4

3.621

5

Safety/Serviceability

2

5

-2.193

40.0059

40.614

Overall Rank

4.2.4 Designer’s Final Ranking Assessment In terms of economic constraints, the two-way slab got the rank of 5 considering both the concrete works and rebar works. As for the constructability constraints, the number of man hours needed to construct the structure in one-way slab is larger rather than the two-way slab, thus making the two-way slab gets the rank of 5. For safety/serviceability constraint, the percentage of deflection from allowable in the one way slab is smaller than the two way slab making this trade get a rank of 5 in this constraint. After gathering all data and making the designer’s overall final ranking assessment. Overall, the two tradeoffs almost had the same rank, but then the two-way slab outranked the one way slab tradeoff with a nail biting difference. With these information, the designer concluded that the governing tradeoff is still two-way slab in contrast with the raw designer’s ranking.

56

57

CHAPTER V: FINAL DESIGN As what was proven from the previous chapters, the governing tradeoff was the two-way slab. After going through all the design processes, the designer can now conclude the final design of the structure which includes the design schedule of the structural members.

5.1 Design Schedules The design schedule of the structural members included the investigated dimensions and designed number of bars with spacing. The following tables below show the design schedule of the project.

5.1.1 Design Schedule of Slabs Table 8. Slab Schedule SLAB (2F-Roof)

t (mm)

S-1 S-2

150 150

S-1 S-2

150 150

Spacing (mm)

Φ bar

Column Strip Middle Strip LONGITUDINAL DIRECTION 12 300 300 12 300 300 TRANSVERSE DIRECTION 12 300 300 12 300 300

58

Φ tie

10 10 10 10

5.1.2 Design Schedule of Beams Table 9. Beam Schedule 2F-Roof Beam B1 B2 B3 B4 B5 B6 B7 B8 B9 B10 B11 B12 B13 B14 B15 B16 B17 B18 B19 B20 B21 B22 B23 B24 B25 B26 B27 B28 B29 B30 B31 B32 B33 B34 B35 B36

Dimensions b (mm) t (mm) 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500 350 500

Top (Left) 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 59

Number of Bars Bottom (Mid) Top (Right) 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ 4 - 25Φ 5 - 25Φ

B37 B38 B39 B40 B41 B42 B43 B44 B45 B46 B47 B48 B49 B50 B51 B52 B53 B54 B55 B56

350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350 350

500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500 500

5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ

60

4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ 4 - 25Φ

5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ 5 - 25Φ

5.1.3 Design Schedule of Columns Table 10. Column Schedule 2F-Roof Column C1 C2 C3 C4 C5 C6 C7 C8 C9 C10 C11 C12 C13 C14 C15 C16 C17 C18 C19 C20 C21 C22 C23 C24 C25 C26 C27 C28 C29 C30 C31 C32 C33

Dimensions b (mm) t (mm) 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550 550

# of Bars 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ 8-32mmΦ

61

Tie Wires Φtie Spacing 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm 10 mm 480 mm

5.1.4 Beam Details

Figure 29. Beam Details

5.1.5

Column

Details

62

Figure 30. Column Details

Figure 31. Beam-Column Interaction Detail

63

APPENDICES APPENDIX A: CODES AND STANDARDS

National Building Code of the Philippines (NBC) The following are the sections and codes that are followed in conceptualizing and designing the structural plan of the apartment building:

Section 401. Types of Construction Type I. The structural elements may be any of the materials permitted by this Code.

Section 701. Occupancy Classified. Group B. Residentials, Hotels and Apartments

Section 805. Ceiling Heights. Habitable rooms provided with artificial ventilation have\ ceiling heights not less than 2.40 meters measured from the floor to the ceiling; Provided that for buildings of more than one-storey, the minimum ceiling height of the first storey shall be 2.70 meters and that for the second storey 2.40 meters and succeeding storeys shall have an unobstructed typical head-room clearance of not less than 2.10 meters above the finished floor. Above stated rooms with a natural ventilation shall have ceiling height not less than 2.70 meters.

Section 806. Size and Dimensions of Rooms. Minimum sizes of rooms and their least horizontal dimensions shall be as follows: 1. Rooms for Human Habitations. 6.00 square meters with at least dimensions of 2.00 2. Kitchens. 3.00 square meters with at least dimension of 1.50 meters; 3. Bath and toilet. 1.20 square meters with at least dimension of 0.90 meters.

Section 808. Window Openings. Every room intended for any use, not provided with artificial ventilation system as herein specified in this Code, shall be provided with a window or windows with a total free area of openings equal to at least ten percent of the floor area of room, and such window shall open directly to a court, yard, public street or alley, or open water courses.

Section 1207. Stairs, Exits and Occupant Loads. General. The construction of stairs and exits shall conform to the occupant load requirements of buildings, reviewing stands, bleachers and grandstands:

a. Determinations of Occupant Loads. The Occupant load permitted in any building or portion thereof shall be determined by dividing the floor area assigned to that use by the unit area allowed per occupant as determined by the Secretary. 64

b. Exit Requirements. Exit requirements of a building or portion thereof used for different purposes shall be determined by the occupant load which gives the largest number of persons. No obstruction shall be placed in the required width of an exit except projections permitted by this Code.

National Structural Code of the Philippines (NSCP) 2010 Notation Ag

= gross area of section, mm2.

As

= area of nonprestressed tension reinforcement, mm 2.

A s , min = minimum amount of flexural reinforcement, mm 2. A st

= total area of nonprestressed longitudinal reinforcement (bars and steel shapes), mm 2.

Av

= area of shear reinforcement within a distance s, mm2.

A vf

= area of shear-friction reinforcement, mm 2.

A ' s = area of compression reinforcement, mm 2. b

= width of compression face of member, mm.

bw

= web width, mm.

c

= distance from extreme compression fiber to neutral axis, mm.

cc

= clear cover from the nearest surface in tension to the surface of the flexural tension reinforcement, mm.

Cm = a factor relating actual moment diagram to an equivalent uniform moment diagram. D

= dead loads, or related internal moments and forces.

d

= distance from extreme compression fiber to centroid of tension reinforcement, mm.

d'

= distance from extreme compression fiber to centroid of compression reinforcement, mm. 65

db

= nominal diameter of bar, wire, or prestressing strand, mm.

dc

= thickness of concrete cover measure from extreme tension fiber to center of bar or wire located closest thereto, mm.

ds

= distance from extreme tension fiber to centroid of tension reinforcement, mm.

dt

= distance from extreme compression fiber to extreme tension steel, mm.

E

= load effects of earthquake, or related internal moments and forces.

Ec

= modulus of elasticity of concrete, MPa.

Es

= modulus of elasticity of reinforcement, MPa.

EI

= flexural stiffness of compression member, N-mm 2.

F

= loads due to weight and pressures of fluids with well defined densities and controllable maximum heights, or related internal moments and forces.

f 'c

= specified compressive strength of concrete, MPa.

fy

= specified yield strength of nonprestressed reinforcement, MPa.

f yt = specified yield strength fy H

= loads due to weight and pressure of soil, water in soil, or other materials, or related internal moments and forces.

h

= overall thickness of member, mm.

I

= moment of inertia of section beam about the centroidal axis, mm 4.

I cr

= moment of inertia of cracked section transformed to concrete, mm 4.

Ie

= effective moment of inertia for computation of deflection, mm 4.

Ig

= moment of inertia of gross concrete section about centroidal axis, neglecting reinforcement, mm 4.

L

= live loads, or related internal moments and forces. 66

Ld

= development length, mm.

ln

= length of clear span measured face-to-face of supports, mm.

M a = maximum moment in member at stage deflection is computed. M cr = cracking moment. Pb

= nominal axial load strength at balanced strain conditions

Pn

= nominal axial load strength at given eccentricity.

Vc

= nominal shear strength provided by concrete

W

= wind load, or related integral moments and forces.

wc

= unit weight of concrete, kN/m 3.

wu

= factored load per unit length of beam or per unit area of slab.

αf

= ratio of flexural stiffness of beam section to flexural stiffness of a width of slab bounded laterally by center line of adjacent panle, if any on each side of beam.

α fm = average value of α f

for all beams on edges of a panel.

β1

= factor

εt

= net tensile strain in extreme tension steel at nominal strength.

λ

= modification factor reflection the reduced mechanical properties of lightweight concrete.

λΔ

= multiplier for additional long-time deflection

reinforcement =

ρ

= ration of nonprestressed tension

A s /bd

ρ'

= ratio of nonprestressed compression reinforcement =

ρb

= reinforcement ratio producing balanced strain conditions

Φ

= strength-reduction factor. 67

A ' s /bd

The following are the sections and codes that are followed in conceptualizing and designing the structural plan of the apartment building:

Section 203 - Combination of Load a.Minimum densities for design loads from materials b.Minimum design loads c. Minimum uniform and concentrated live loads Section 206 - Other Minimum Loads a.206.3 Impact loads b.206.3.1 Elevators c. 206.3.2 Machinery

Section 207 - Wind Load a. 207.5.10 Velocity Pressure b. 207.5.6.6 Velocity Pressure Exposure Coefficient c. 207.5.7.2 Topographic Factor d. 207.5.4.4 Wind Directionality Factor e. 207.5.6 Exposure

Section 208 - Earthquake Loads a. 208.5.1.1 Earthquake Loads b. 208.5.2.1 Design Base Shear c. 208.5.2.2 Structure Period

Wind Load Section 207.5.4 Wind Directionality Factor The wind directionality factor, Kd, shall be determined form Table 207-2. This factor Shall only be applied when used in conjunction with load combinations specified in Section 203.3 and 203.4.

Section 207.5.5 Importance factor An importance factor Iw, for the building or other structure shall be determined from Table 207-3 based on building and structure categories listed in Table 103-1.

Section 207.5.6 Exposure For each wind direction considered, the upwind exposure category shall be based on ground surface roughness that is determined from natural topography, vegetation, and constructed facilities.

Section 207.5.7 Topographic factor The wind speed up effect shall be included in the calculation of design wind loads by using the factor kzt. If site conditions and locations of structures do not meet all the conditions specified in Section 207.5.7.1 the kzt= 1.0

68

Section 207.5.8 Gust Effect factor The gust effect factor shall be calculated as permitted in Sections 207.5.8.1 to 207.5.8.5, using appropriate values for natural frequency and damping ratio as permitted in Section 207.5.8.6.

Section 207.5.9 Enclosure Classifications For the purpose of determining internal pressure coefficients, all buildings shall be classified as enclosed, partially enclosed, or open as defined in Section 207.2. Section 207.5.10 Velocity Pressure Velocity pressure, qz, evaluated at height z shall be calculated by the following equation qz= 47.3x10-6 kz kzt kd V2 Iw.

Section 207.5.11 Pressure and Force Coefficients Internal Pressure Coefficients, GCpi, shall be determined from fig. 207-5 based on building enclosure classifications determined from Section 207.5.9

Section 207.5.12 Rigid Building for all heights Design wind pressures for the MWFRS of a buildings of all heights shall be determined by the following equation; P= qGCP – qi(GCPi)

Section 207.5.13 Design Wind Loads on Open Buildings with Monoslope, Pitched, or Troughed Roofs Plus and minus signs signify pressure acting toward and away from the top surface of the roof, respectively.

Section 207.5.14 Design Wind Loads on Solid Freestanding Walls and Solid Signs The design wind force for solid freestanding walls and solid signs shall be determined by the following formula: F= qhGCfAs Section 207.5.15 Design Wind Loads on other Structures The design wind force for other structures shall be determined by the following equation: F=qzGfCfAf Earthquake Load

Section 208.5.1 Earthquake Loads and Modeling Requirements Structures shall be designed for ground motion producing structural response and seismic forces in any horizontal direction. The following earthquake loads shall be used in the load combinations set forth in Section 203: E= ρEh + Eb Section 208.5.2 Static Force Procedure Section 208.5.2.1 The total design base shear in a given direction shall be determined form the following equation: V= CvI (W) 69

RT The total design base shear need not exceed the following: V= 2.5CaI (W) R The Base Shear shall not be less than the following: V= .11CaIW

Section 208.5.2.2 The value of T shall be determined using the following method: Determine the structure period T using Method A T = Ct (hn)3/4

The following are the tables used in each design computations:

Stone Concrete Fill 1.53 Kpa Gypsum Board 0.2 Kpa Suspended Steel Channel 0.1 Kpa Mechanical Duct Allowance 0.2 Kpa Terrazo 1.53 Kpa Grout 0.11 Kpa CHB 1.65 Kpa Clay Dry 0.6435 Kpa Water Proofing 0.05 Kpa Cement Finish 1.53 pa Table 204-1 Minimum Densities for Design Loads from Materials

Material Density (KN/m3) Masonry, Concrete 16.5 Table 204-2 Minimum Design Dead Loads

Basic Floor Area Roof Live Load

1.9 Kpa 1.9 Kpa 70

Table 205-1 Minimum Uniform Concentrated Live Loads

Occupancy Category Seismic Importance Factor I Seismic Importance Factor Ip I. Essential facilities 1.5 1.5 II. Hazardous facilities 1.25 1.5 III. Special Occupancy Structures 1 1 IV. Standard Occupancy Strutures 1 1 V. Miscellaneous Structures 1 1 Table 208-1 Seismic Importance Factors

Soil Profile

Soil Profile Name

Ave. Properties for Top 30 m Soil Profile Shear Wave Velocity SPT Undrained Shear Strenght

SA SB Sc SD SE SF

Hard Rock >1500 Rock 760 to 1500 Very Dense Soil 360 to 760 >50 >100 Stiff Soil Profile 180 to 360 15 to 50 50 to 100 Soft Soil Profile .5*ΦVc Stirrups needed * If Vu < .5*Φ*Vc, stirrups are not needed

Vc ΦVc .5ΦVc

Step 2. Calculate the Shear Strength by Stirrup (Vs) Vn = Vu/Φ

Vn

189.3352

kN

Vs parameter

73.22246 466.7734

kN

Si = Av*fy*d/Vs, Av = pi*(Φtie^2)/4 220 Av For Smax, 110 Si * If Vs < 0.33*sqrt(f'c)*b*d, Smax = d/2 or 600mm (get smaller) parameter * If Vs > 0.33*sqrt(f'c)*b*d, Smax = d/4 or 400mm (get smaller) Smax1 Smax2 Epoxy Light Sf Zinc Normal

78.53982 200 229.9033 220 600 200

mm2 mm mm mm mm mm

Vs = Vn - Vc * If Vs < 0.67*sqrt(f'c)*b*d, go to Step 3. * If Vs > 0.67*sqrt(f'c)*b*d, redesign.

116.1128 104.5015 52.25075 Stirrups Needed

kN kN kN

Step 3. Spacing of Stirrups

Uncoated Part 3. Development Length The following are the supplementary data. Cc 40 Mm Bar Coat

Epoxy

Step 1. Determine the Value of the Coefficients (ψt,ψe,ψs,λ) ψt = 1.0 for all other situations ψe = 1.5 for epoxy-coated bars with cover less than 3d or 6d = 1.2 for all other epoxy-coated bars = 1 for uncoated or zinc coated bars ψt = .8 for 20 mm bars and smaller = 1 for 25 mm bars and larger λ = 1 for normal weight concrete 79

RESULTS ψt ψe

1 1.2

ψs λ

1 1

Step 2. Compute for the development length ld = (fy*ψt*ψe*ψs*d)/(1.1*λ*sqrt(f'c)*((cc+Ktr)/d)) Ktr = 40*Atr/(S*N), Atr = 2*pi*(Φtie^2)/4

Atr

157.079 6

Ktr

7.853982

ld

80

1299.611

mm2 mm

For Midspan The following are the given data Mu 103.972 Vu f'c fy b t d' d Φbar Φtie

89.11886 20.7 415 350 500 62.5 437.5 25 10

kN-m

Es

kN MPa MPa mm mm mm mm mm mm

Ec n L

Part 1. Computation of Steel Area and Number of Bars Step 1. Solve for ρmax and Mu(max) ρb = (0.75*0.85*f'c*β1*600)/(fy*(600+fy) β = 0.85, for f'c < 28 MPa ρmax = ρ = 0.75ρb ω = ρ*fy/f'c Mu(max) = Φ*f'c*ω*b*(d^2)*(1-.59ω) Φ = 0.9 * If Mu < Mu(max), design is Singly Reinforced * If Mu > Mu(max), design is Doubly Reinforced

200000 21383.7 1 10 7

MPa MPa m

RESULTS β

0.85 0.01597 ρb 7 0.01198 ρmax 3 ω 0.24024 Φ 0.9 Mu(max) 257.336 SINGLY

kN-m

Step 2. Using Singly Reinforcement. Solving As and N bars As1 = ρmax*b*d N = As/Abar, Abar = pi*(Φbar^2)/4

As N'

1834.90 6 4

mm2 pcs

Part 2. Designing the Vertical Stirrup Step 1. Calculate the Shear Strength by Concrete (Vc) Vc = sqrt(f'c)*b*d/6

RESULTS

* If Vu > ΦVc, stirrups needed, go to Step II * If Vu < ΦVc, but Vu > .5*ΦVc Stirrups needed

Vc ΦVc 81

116.112 8 104.501

kN kN

5 52.2507 .5ΦVc 5 kN Stirrups Needed

* If Vu < .5*Φ*Vc, stirrups are not needed Step 2. Calculate the Shear Strength by Stirrup (Vs) Vn = Vu/Φ Vs = Vn - Vc * If Vs < 0.67*sqrt(f'c)*b*d, go to Step 3. * If Vs > 0.67*sqrt(f'c)*b*d, redesign.

Vn

99.0209 5

Vs paramete r

17.0918 466.773 4

Av Si paramete r

78.5398 2 230 229.903 3

Smax1 Smax2 Sf

220 600 220

kN kN

Step 3. Spacing of Stirrups Si = Av*fy*d/Vs, Av = pi*(Φtie^2)/4 For Smax, If Vs < 0.33*sqrt(f'c)*b*d, Smax = d/2 or 600mm (get * smaller) If Vs > 0.33*sqrt(f'c)*b*d, Smax = d/4 or 400mm (get * smaller) Epoxy Zinc Uncoated

220 110

Light Normal

Part 3. Development Length The following are the supplementary data. cc 40 mm Bar Coat

Epoxy

Step 1. Determine the Value of the Coefficients (ψt,ψe,ψs,λ) ψt = 1.0 for all other situations ψe = 1.5 for epoxy-coated bars with cover less than 3d or 6d = 1.2 for all other epoxy-coated bars = 1 for uncoated or zinc coated bars ψt = .8 for 20 mm bars and smaller = 1 for 25 mm bars and larger λ = 1 for normal weight concrete 82

RESULTS

ψt ψe

1 1.2

ψs λ

1 1

mm2 mm mm mm mm mm

Atr Step 2. Compute for the development length

Ktr

ld = (fy*ψt*ψe*ψs*d)/(1.1*λ*sqrt(f'c)*((cc+Ktr)/d)) Ktr = 40*Atr/(S*N), Atr = 2*pi*(Φtie^2)/4

ld

157.079 6 7.13998 3 1319.29 5

mm2

mm

Part 4. Checking the Beam in Deflection Step 1. Calculate the Gross Moment of Inertia and the Cracking Moment of the Beam RESULTS Ig = b(t^3)/12 Mcr = Ig*fr/ϒt, fr = 0.62*λ*sqrt(f'c), yt = t/2

Ig

mm4

Mcr

3645833333 2.82082966 5 250 41.1370992 8

kN-m

Icr

91438412.2 9

mm4

Ie

311587799

mm4

fr yt

MPa mm

Step 2. Calcualte the Moment of Inertia of the Cracked Section Icr = b*(c^3)/12 + nAs(d-c)+nAs'(c-d') Step 3. Calculate the Effective Moment of Inertia Ie = ((Mcr/Mu)^3)*Ig + ((1-(Mcr/Mu)^3)*Icr) Step 4. Determine and Check the Deflection Mu = W(L^2)/8, W=____ δ = 5*W*(L^4)/(384*Ec*Ie)

W δ

δmax = L/360

δmax

83

16.9750204 1 0.30585036 19.4444444 4 OK

kN/m mm mm

Beam with Maximum Moment was Designed (Two-Way Tradeoff) For Support

The following are the given data Mu 310.203

kN-m

Es

Vu

265.89

kN

Ec

200000 21383.7 1

f'c fy b t d' d

20.7 415 350 500 62.5 437.5

MPa MPa mm mm mm mm

N L

10 7

Φbar Φtie

25 10

mm mm

Part 1. Computation of Steel Area and Number of Bars Step 1. Solve for ρmax and Mu(max) ρb = (0.75*0.85*f'c*β1*600)/(fy*(600+fy) β = 0.85, for f'c < 28 MPa

MPa MPa M

RESULTS β

0.85 0.01597 ρb 7 0.01198 ρmax 3 ω 0.24024 Φ 0.9 Mu(max) 257.336 DOUBLY

ρmax = ρ = 0.75ρb ω = ρ*fy/f'c Mu(max) = Φ*f'c*ω*b*(d^2)*(1-.59ω) Φ = 0.9 * If Mu < Mu(max), design is Singly Reinforced * If Mu > Mu(max), design is Doubly Reinforced

kN-m

Step 2. Using Doubly Reinforcement. Solving As1, Mu1, Mu2, and As2 As1 = ρmax*b*d Mu1 = Mu(max)

As1 Mu1

Mu2 = Mu - Mu1

Mu2

Mu2 = Φ*As2*fy*(d-d'), Solve for As2 Step 3. Solve for the Stress of the Compression Steel 84

As2

1834.90 6 257.336 52.8669 6 377.452 6

mm2 kN-m kN-m mm2

C=T 0.85*f'c*a*b = As1*fy, Solve for a

a

a = βc, Solve for c

c 377.452 6 457.722 6

f's = 600*(c-d')/c * If f's > fy, A's = As2 * If f's < fy, A's = As2*fy/f's

A's

123.653 145.474 1 342.222 2 457.722 6

N N'

5 1

f's

mm mm MPa mm2

Step 4. Determine the Number of Bars As = As1 + As2 For Tension N = As/Abar, Abar = pi*(Φbar^2)/4 For Compression N' = A's/Abar, Abar = pi*(Φbar^2)/4

pcs pcs

Part 2. Designing the Vertical Stirrup Step 1. Calculate the Shear Strength by Concrete (Vc) Vc = sqrt(f'c)*b*d/6

RESULTS 116.112 Vc 8 kN 104.501 ΦVc 5 kN 52.2507 .5ΦVc 5 kN Stirrups Needed

* If Vu > ΦVc, stirrups needed, go to Step II * If Vu < ΦVc, but Vu > .5*ΦVc Stirrups needed * If Vu < .5*Φ*Vc, stirrups are not needed Step 2. Calculate the Shear Strength by Stirrup (Vs) Vn = Vu/Φ

Vs paramete r

295.433 3 179.320 6 466.773 4

Av

78.5398 2

Vn

Vs = Vn - Vc * If Vs < 0.67*sqrt(f'c)*b*d, go to Step 3. * If Vs > 0.67*sqrt(f'c)*b*d, redesign.

kN kN

Step 3. Spacing of Stirrups Si = Av*fy*d/Vs, Av = pi*(Φtie^2)/4

220 85

mm2

For Smax, If Vs < 0.33*sqrt(f'c)*b*d, Smax = d/2 or 600mm (get * smaller) If Vs > 0.33*sqrt(f'c)*b*d, Smax = d/4 or 400mm (get * smaller) Epoxy Zinc Uncoated

110

Light Normal

Si paramete r

80 229.903 3

mm

Smax1 Smax2 Sf

220 600 80

mm mm mm

mm

Part 3. Development Length The following are the supplementary data. Cc 40 mm Bar Coat

Epoxy

Step 1. Determine the Value of the Coefficients (ψt,ψe,ψs,λ) ψt = 1.0 for all other situations ψe = 1.5 for epoxy-coated bars with cover less than 3d or 6d = 1.2 for all other epoxy-coated bars = 1 for uncoated or zinc coated bars ψt = .8 for 20 mm bars and smaller = 1 for 25 mm bars and larger λ = 1 for normal weight concrete

RESULTS

ψt ψe

1 1.2

ψs λ

1 1

Atr Step 2. Compute for the development length

Ktr

ld = (fy*ψt*ψe*ψs*d)/(1.1*λ*sqrt(f'c)*((cc+Ktr)/d)) Ktr = 40*Atr/(S*N), Atr = 2*pi*(Φtie^2)/4

ld

86

157.079 6 78.5398 2 524.647 1

mm2

mm

For Midspan

The following are the given data Mu 164.303 Vu f'c Fy B T d' D Φbar Φtie

kN-m

Es

kN MPa MPa mm mm mm mm mm mm

Ec n L

143.83 20.7 415 350 500 62.5 437.5 25 10

Part 1. Computation of Steel Area and Number of Bars Step 1. Solve for ρmax and Mu(max) ρb = (0.75*0.85*f'c*β1*600)/(fy*(600+fy) β = 0.85, for f'c < 28 MPa ρmax = ρ = 0.75ρb ω = ρ*fy/f'c Mu(max) = Φ*f'c*ω*b*(d^2)*(1-.59ω) Φ = 0.9 * If Mu < Mu(max), design is Singly Reinforced * If Mu > Mu(max), design is Doubly Reinforced

200000 21383.7 1 10 7

MPa MPa M

RESULTS β ρb ρmax ω Φ Mu(max)

0.85 0.01597 7 0.011983 0.24024 0.9 257.336 SINGLY

kN-m

Step 2. Using Singly Reinforcement. Solving As and N bars As1 = ρmax*b*d N = As/Abar, Abar = pi*(Φbar^2)/4

As N

1834.90 6 4

mm2 pcs

Part 2. Designing the Vertical Stirrup Step 1. Calculate the Shear Strength by Concrete (Vc) Vc = sqrt(f'c)*b*d/6

RESULTS

* If Vu > ΦVc, stirrups needed, go to Step II * If Vu < ΦVc, but Vu > .5*ΦVc Stirrups needed

Vc ΦVc 87

116.1128 104.5015

kN kN

.5ΦVc 52.25075 Stirrups Needed

* If Vu < .5*Φ*Vc, stirrups are not needed Step 2. Calculate the Shear Strength by Stirrup (Vs) Vn = Vu/Φ Vs = Vn - Vc * If Vs < 0.67*sqrt(f'c)*b*d, go to Step 3. * If Vs > 0.67*sqrt(f'c)*b*d, redesign.

kN

Vn

159.8111

kN

Vs parameter

43.69833 466.7734

kN

Av Si parameter Smax1 Smax2 Sf

78.53982 330 229.9033 220 600 220

Step 3. Spacing of Stirrups Si = Av*fy*d/Vs, Av = pi*(Φtie^2)/4 220 For Smax, 110 * If Vs < 0.33*sqrt(f'c)*b*d, Smax = d/2 or 600mm (get smaller) * If Vs > 0.33*sqrt(f'c)*b*d, Smax = d/4 or 400mm (get smaller) Epoxy Zinc

Light Normal

mm 2

mm mm mm mm mm

Part 3. Development Length The following are the supplementary data. Cc 40 mm Bar Coat

Epoxy

Step 1. Determine the Value of the Coefficients (ψt,ψe,ψs,λ) ψt = 1.0 for all other situations ψe = 1.5 for epoxy-coated bars with cover less than 3d or 6d = 1.2 for all other epoxy-coated bars = 1 for uncoated or zinc coated bars ψt = .8 for 20 mm bars and smaller = 1 for 25 mm bars and larger λ = 1 for normal weight concrete Step 2. Compute for the development length ld = (fy*ψt*ψe*ψs*d)/(1.1*λ*sqrt(f'c)*((cc+Ktr)/d)) Ktr = 40*Atr/(S*N), Atr = 2*pi*(Φtie^2)/4

88

RESULTS Ψt Ψe

1 1.2

Ψs Λ

1 1

Atr

157.079 6

mm2

Ktr Ld

7.13998 1319.3

mm

Part 4. Checking the Beam in Deflection Step 1. Calculate the Gross Moment of Inertia and the Cracking Moment of the Beam RESULTS Ig = b(t^3)/12 Mcr = Ig*fr/ϒt, fr = 0.62*λ*sqrt(f'c), yt = t/2

Ig

mm4

Mcr

3645833333 2.82082966 5 250 41.1370992 8

kN-m

Icr

93279189.2 4

mm4

Ie

149036936. 7

mm4

fr yt

MPa mm

Step 2. Calcualte the Moment of Inertia of the Cracked Section Icr = b*(c^3)/12 + nAs(d-c)+nAs'(c-d') Step 3. Calculate the Effective Moment of Inertia Ie = ((Mcr/Mu)^3)*Ig + ((1-(Mcr/Mu)^3)*Icr) Step 4. Determine and Check the Deflection Mu = W(L^2)/8, W=____ δ = 5*W*(L^4)/(384*Ec*Ie)

W δ

δmax = L/360

δmax

89

26.8249795 9 1.01047278 19.4444444 4 OK

kN/m mm mm

APPENDIX D: DESIGN OF ONE-WAY SLAB Considering Longer Side Design of S-1 The following are the given data. Dead Loads (kPa) Weight of Slab 3.537 Stone Concrete 1.53 Fill Gypsum Board 0.2 Total 5.267 Live Load (kPa) Basic Floor 1.9 Area

f'c fy

20.7 415

MPa MPa

L

7

m

t b Φbar

150 1000 12

mm mm mm

Φtie

10

mm

d

134

mm

Step 1. Calculate the Factored Loads and the Moment in the Slab W = 1.2DL + 1.6LL For Midspan, M = W*(L^2)/14 For Continuous Edge, M = W*(L^2)/10

RESULTS W

9.36040

kN/m

Mmid

32.76140

kN/m

Mc.e.

45.86596

kN/m

R

1.82454

ρi

0.00465

ρmax

0.01598

ρmin

0.00337

ρf

0.00465

Step 2. Calculate the ρ and check for the Midspan R = Mu/(b*(d^2)) ρ = (0.85*f'c/fy)*(1-(sqrt(1-((2*R)/ (0.85*f'c)))) ρmax = 0.75*0.85*f'c*β*600/ (fy*(600+fy)) ρmin = 1.4/fy If ρ > ρmax, * redesign * If ρmin < ρ < ρmax, ok If ρmin > ρ, use * ρmin

90

Step 3. Calculate the Steel Area and Spacing of Bars As

697.74823

mm2

S = b*Abar/As, Abar = pi*(Φbar^2)/4

Abar S

113.09734 162.08903

mm2 mm

Step 4. Calculate the ρ and check for the Continuous Edge

R Ρi ρmax ρmin Ρf

2.55435 0.00668 0.01598 0.00337 0.00668

As

1002.23229

mm2

Abar S

113.09734 112.84543

mm2 mm

As = ρ*b*d

Step 5. Calculate the Steel Area and Spacing of Bars

Considering Shorter Side Dead Loads (kPa) Weight of Slab Stone Concrete Fill Gypsum Board Total Live Load (kPa) Basic Floor Area

3.537 1.53 0.2 5.267 1.9

f'c fy L t b Φbar Φtie d

20.7 415 3.5 150 1000 12 10 134

Step 1. Calculate the Factored Loads and the Moment in the Slab W = 1.2DL + 1.6LL For Midspan, M = W*(L^2)/14 For Continuous Edge, M = W*(L^2)/10 91

MPa MPa m mm mm mm mm mm RESULTS

W

9.36040

kN/m

Mmid

8.19035

kN/m

Mc.e.

11.46649

kN/m

Step 2. Calculate the ρ and check for the Midspan R = Mu/(b*(d^2)) ρ = (0.85*f'c/fy)*(1-(sqrt(1-((2*R)/ (0.85*f'c)))) ρmax = 0.75*0.85*f'c*β*600/(fy*(600+fy)) ρmin = 1.4/fy * If ρ > ρmax, redesign * If ρmin < ρ < ρmax, ok * If ρmin > ρ, use ρmin Step 3. Calculate the Steel Area and Spacing of Bars As = ρ*b*d S = b*Abar/As, Abar = pi*(Φbar^2)/4

Step 4. Calculate the ρ and check for the Continuous Edge

Step 5. Calculate the Steel Area and Spacing of Bars

92

R

0.45613

ρi

0.00111

ρmax ρmin ρf

0.01598 0.00337 0.00337

As

506.02410

mm2

Abar

113.09734

mm2

S

223.50188

mm

R ρi ρmax ρmin ρf

0.63859 0.00157 0.01598 0.00337 0.00337

As

506.02410

mm2

Abar

113.09734

mm2

S

223.50188

mm

Design of S-2 The following are the given data. Dead Loads (kPa) Weight of Slab 3.537 Stone Concrete Fill 1.53 Gypsum Board 0.2 Total 5.267 Live Load (kPa) Basic Floor Area 1.9

f'c Fy

20.7 415

MPa MPa

L T B Φbar

7 150 1000 12

M mm mm mm

Φtie D

10 134

mm mm

Step 1. Calculate the Factored Loads and the Moment in the Slab W = 1.2DL + 1.6LL For Midspan, M = W*(L^2)/14 For Continuous Edge, M = W*(L^2)/10 Step 2. Calculate the ρ and check for the Midspan R = Mu/ (b*(d^2)) ρ = (0.85*f'c/fy)*(1-(sqrt(1-((2*R)/(0.85*f'c)))) ρmax = 0.75*0.85*f'c*β*600/ (fy*(600+fy)) ρmin = 1.4/fy * If ρ > ρmax, redesign * If ρmin < ρ < ρmax, ok * If ρmin > ρ, use ρmin

RESULTS

W

9.36040

kN/m

Mmid Mc.e.

32.76140 45.86596

kN/m kN/m

R

1.82454

ρi ρmax ρmin ρf

0.00465 0.01598 0.00337 0.00465

As

697.74823

mm2

Abar S

113.09734 162.08903

mm2 mm

R

2.55435

Step 3. Calculate the Steel Area and Spacing of Bars As = ρ*b*d S = b*Abar/As, Abar = pi*(Φbar^2)/4

Step 4. Calculate the ρ and check for the Continuous Edge 93

Step 5. Calculate the Steel Area and Spacing of Bars

ρi ρmax ρmin ρf

0.00668 0.01598 0.00337 0.00668

As

1002.23229

mm2

Abar S

113.09734 112.84543

mm2 mm

Considering Shorter Side Dead Loads (kPa) Weight of Slab Stone Concrete Fill Gypsum Board Total Live Load (kPa) Basic Floor Area

3.537 1.53 0.2 5.267 1.9

f'c fy L t b Φbar Φtie d

20.7 415 3 150 1000 12 10 134

Step 1. Calculate the Factored Loads and the Moment in the Slab W = 1.2DL + 1.6LL For Midspan, M = W*(L^2)/14 For Continuous Edge, M = W*(L^2)/10 Step 2. Calculate the ρ and check for the Midspan R = Mu/(b*(d^2)) ρ = (0.85*f'c/fy)*(1-(sqrt(1-((2*R)/ (0.85*f'c)))) ρmax = 0.75*0.85*f'c*β*600/ (fy*(600+fy)) ρmin = 1.4/fy * If ρ > ρmax, redesign * If ρmin < ρ < ρmax, ok * If ρmin > ρ, use ρmin 94

MPa MPa m mm mm mm mm mm

RESULTS

W

9.36040

kN/m

Mmid Mc.e.

6.01740 8.42436

kN/m kN/m

R

0.33512

ρi ρmax ρmin ρf

0.00082 0.01598 0.00337 0.00337

Step 3. Calculate the Steel Area and Spacing of Bars As = ρ*b*d S = b*Abar/As, Abar = pi*(Φbar^2)/4

Step 2. Calculate the ρ and check for the Continuous Edge

Step 3. Calculate the Steel Area and Spacing of Bars

95

As

506.02410

mm2

Abar

113.09734

mm2

S

223.50188

mm

R ρi ρmax ρmin ρf

0.46917 0.00115 0.01598 0.00337 0.00337

As

506.02410

mm2

Abar

113.09734

mm2

S N

223.50188 5

mm pcs

APPENDIX E: DESIGN OF TWO-WAY SLAB EQUIVALENT FRAME METHOD

Figure. Design

Strips for Equivalent Frame Method Considering Longitudinal Frame

The following are the given data.

ts B l1 l2 DL LL W f'c fy E

Slab 150 1000 4.5 7 General 5.267 1.9 9.3604 20.7 415 21383.7

mm mm m m

c1 c2 lc

Column 550 550 3

mm mm m

kPa kPa MPa MPa MPa

96

Summary of Values for Slabs SAMPLE RESULTS (AB) Slab1. Determine l1 l2the MOIc1/l1 c2/l2 FEM COF K Step and Stiffness of Columnk 1 7 a*(b^3)/12 7 0.078571 0.078571 4.11 269.6918 0.5086 Ic 7.6E+09 2.47E+10 mm4 Ic = 2 7 7 0.078571 0.078571 4.11 269.6918 0.5086 2.47E+10 Determine lu/lc lc 3 mm 3 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 Determine a/b lu 2.85 mm 4 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 Find kab in the tables. lu/lc 0.95 5 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 a/b 1 K4.5 c = kab*E*Ic/h 6 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 C = (1-0.63(t/a))*((t^3)(a)/3) kab 4.55 7 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 K t = 9*E*C/(l2*(1-b/l2)^2) K c 2.5E+11 8 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 K ec = 4Kt*Kc/(2*Kt+2*Kc) C 5.1E+08 9 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 Kt 1.7E+10 10 4.5 7 0.122222 0.122222 4.4 111.4543 0.526 4.12E+10 Step 2. Determine the MOI and Stiffness of Slab Is = b*(t^3)/12 Compute a/l1 & b/l1 Find ks in the tables. Find FEM and COF in the tables. Ks = ks*E*Is/l1

Kec

3.1E+10

Is a/l1 b/l2 ks FEM COF Ks

2E+09 0.12222 0.12222 4.4 111.454 0.526 4.1E+10

Figure. Design Strip for Longitudinal Frame

Table. Summary of Coefficients and Stiffness for Slabs in Longitudinal Frame

97

mm4

kN-m

Step 3. Determine the Distribution Factors DFxy = Kxy/ΣK Table. Summary of Distribution Factors and FEM

Member AB BA BC CB CD DC DE ED EF FE FG

K 2.5E+10 2.5E+10 2.5E+10 2.5E+10 4.12E+1 0 4.12E+1 0 4.12E+1 0 4.12E+1 0 4.12E+1 0 4.12E+1 0 4.12E+1 0

SUMMARY OF VALUES FOR DF AND FEM DF FEM Member K 0.28438 -269.69 GF 4.1E+10 0.22141 269.6918 GH 4.1E+10 0.22141 -269.69 HG 4.1E+10 0.19299 269.6918 HI 4.1E+10

DF 0.28481 0.28481 0.28481 0.28481

FEM 111.454 -111.45 111.454 -111.45

0.32138

-111.454

IH

4.1E+10

0.28481

111.454

0.28481

111.4543

IJ

4.1E+10

0.28481

-111.45

0.28481

-111.454

JI

4.1E+10

0.28481

111.454

0.28481

111.4543

JK

4.1E+10

0.28481

-111.45

0.28481

-111.454

KJ

4.1E+10

0.39824

111.454

0.28481

111.4543

0.28481

-111.454

98

Step 4. Continue with the Moment Distribution Method (MDM) Table. MDM for Longitudinal Frame

Member DF FEM COF Balance CO1 Balance CO2 Balance CO3 Balance CO4 Balance TOTAL

A B AB BA BC 0.284 0.22 0.22 -270 270 270 0.509 0.51 0.51 -76.7 0 0 0 -39 15.5 0 -5.2 6.88 -2.64 0 0 -0.75 0 0 0 0.38 0.74 0 0.08 0.08 0.04 0 0 0.011 0 0 -350 225 246

C CB CD 0.19 0.32 270 0.51 30.5 0 0 3.5 1.45

-111 0.53 50.9 0 0 4.01 2.41

MDM FOR LONGITUDINAL FRAME D E F G H DC DE ED EF FE FG GF GH HG HI IH 0.285 0.285 0.28 0.285 0.285 0.285 0.28 0.285 0.285 0.285 0.285 111.5 -111 111 -111 111.5 -111 111 111.5 111.5 -111 111.5 0.526 0.526 0.53 0.526 0.526 0.526 0.53 0.526 0.526 0.526 0.526 0 0 0 0 0 0 0 0 0 0 0 26.75 0 0 0 0 0 0 0 0 0 0 7.619 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 0 0 0 0 0.996

0 0 1.269 0 0 0 0 0 0 0.361 0.361 0 0 0 0.04 0.19 0 0 0.19 0 0 0.04 0.07 0 0 0.05 0.054 0 305 53.9 147.5 -111 112 -111 111.5

0 0 0 0 -111

J IJ JI JK 0.285 0.28 0.285

KJ KJ 0.4

-111 111 -111 111 0.526 0.53 0.526 0.53 0 0 0 44.4 0 0 23.35 0 0 6.65 13.3 0 3.498 0 0 7 0.996 0 0 2.79

0 0 0 0.524 0 0 0.52 1.465 0 0 0 0.149 0.149 0 0 0.57 0.567 0 0 0.078 0 0 0.078 0.298 0 0 0.3 0.02 0.022 0 0 0.107 0.107 0 0 0.12 111 111.4 111.6 -111 112.6 -107 119 -72.8 166

Figure. Result of Moment Distribution Method (Moment Diagram)

99

I

Table. Summary of Moments for Longitudinal Frame Sla b

W

1

401.3272

2

401.3272

3

165.8546

4

165.8546

5

165.8546

+ + + + + -

Summary of Moments Sla M b W 349.73093 113.87016 6 165.8546 225.18306 246.46759 125.4633 7 165.8546 305.26012 53.916191 65.170221 8 165.8546 147.45254 111.09292 54.458879 9 165.8546 111.69849 111.40015 54.427373 10 165.8546 111.45428

Step 5. Get the Design Moments For Exterior Slabs S-1 M(+) 113.8702 M(-) 349.7309 Moments (%) 0.85M(+) 96.78963 0.70M(-) 244.8117 Moment (%) Passed Column Strip

(+) (-)

(+) (-)

Beam (85%) 82.27119 208.0899

Middle Strip

Slab(15%) 14.5184 36.7217

17.0805 104.919

Design Moments CS MS 7.259222 8.54026 18.36087 52.4596 100

+ + + + + -

M 111.45428 54.389126 111.47664 111.35343 54.376115 111.60352 110.78107 54.145967 112.63617 106.5553 52.979831 119.19421 72.777157 46.447589 166.03684

Slab-10 M(+) 46.44759 M(-) 166.0368 Moment (%) 0.85M(+) 39.48045 0.70M(-) 116.2258 Moment (%) Passed Column Strip Beam (85%) Slab(15%) (+) 33.55838 5.92207 (-) 98.79192 17.4339

(+) (-)

Design Moments CS MS 2.961034 3.48357 8.716934 24.9055

For Interior S-2 M(+) 125.4633 M(-) 305.2601 Moments (%) 0.85M(+) 106.6438 0.70M(-) 213.6821 Moment (%) Passed Column Strip Beam Slab(15%) (85%) (+) 90.64723 15.9966 (-) 181.6298 32.0523

(+) (-)

Middle Strip 6.96714 49.8111

Middle Strip 18.8195 91.578

Design Moments CS MS 7.998285 9.40975 16.02616 45.789

101

S-3 to S-9 54.3891 3 111.454 M(-) 3 Moments (%) 0.85M( 46.2307 +) 6 0.70M( 78.018 -) M(+)

Moment (%) Passed

(+) (-)

Column Strip Beam Slab(15 (85%) %) 39.2961 6.9346 4 1 11.702 66.3153 7

Middle Strip 8.15837 33.4363

Design Moments CS MS 3.46730 4.0791 (+) 7 8 16.718 (-) 5.85135 1

102

Step 5. Determine the Steel Areas and Number of Bars per meter of Width

SAMPLE DESIGN f'c fy B T D Φbar Φtie cc

GIVEN DATA MPa 20.7 MPa 415 mm 1000 mm 150 mm 114 mm 12 mm 10 mm 20

Design for Exterior (S-1) Column Strip (+) R = Mu/(Φb*(d^2)) ρ = (0.85*f'c/fy)*(1-(sqrt(1-((2*R)/(0.85*f'c)))) ρmax = 0.75*0.85*f'c*β*600/(fy*(600+fy)) ρmin = 1.4/fy * If ρ > ρmax, redesign * If ρmin < ρ < ρmax, ok * If ρmin > ρ, use ρmin

As = ρ*b*d S = b*Abar/As, Abar = pi*(Φbar^2)/4

103

M R ρi ρmax ρmin ρf

RESULTS 7.25922243 0.62064 0.00152 0.01598 0.00337 0.00337

As Abar S

384.57831 113.09734 294.08142

kN-m

mm2 mm2 mm

Table. Summary of Steel Areas and Number of Bars for Two-Way Slabs

104

+ CS Ext

S-1 M S

+ +

CS S-2 M S

+ -

Int + CS -

S-(39) M S

+ +

CS Ext

S-10 M S

+ -

Summary of Design for Two-Way Slabs (Longitudinal ρ(initi ρ(fina M R al) ρ(max) ρ(min) l 7.259 0.620 0.001 0.0160 0.0034 0.0034 2 6 5 18.36 1.569 0.004 0.0160 0.0034 0.0040 09 8 0 8.540 0.730 0.001 0.0160 0.0034 0.0034 3 2 8 52.45 4.485 0.012 0.0160 0.0034 0.0127 96 1 7 7.998 0.683 0.001 0.0160 0.0034 0.0034 3 8 7 16.02 1.370 0.000 0.0160 0.0034 0.0034 62 2 1 9.409 0.804 0.002 0.0160 0.0034 0.0034 7 5 0 45.78 3.914 0.010 0.0160 0.0034 0.0108 90 8 8 3.467 0.296 0.000 0.0160 0.0034 0.0034 3 4 7 5.851 0.500 0.001 0.0160 0.0034 0.0034 3 3 2 4.079 0.348 0.000 0.0160 0.0034 0.0034 2 8 0 16.71 1.429 0.003 0.0160 0.0034 0.0036 81 3 6 2.961 0.253 0.000 0.0160 0.0034 0.0034 0 2 6 8.716 0.745 0.001 0.0160 0.0034 0.0034 9 3 8 3.483 0.297 0.000 0.0160 0.0034 0.0034 6 8 7 24.90 2.129 0.005 0.0160 0.0034 0.0055 55 3 5 105

Frame) As 384.578 3 452.390 5 384.578 3 1449.36 4 384.578 3 384.578 3 384.578 3 1232.54 5 384.578 3 384.578 3 384.578 3 410.030 5 384.578 3 384.578 3 384.578 3 625.384 4

Abar 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7 113.09 7

S 294.081 4 249.999 4 294.081 4 78.0324 294.081 4 294.081 4 294.081 4 91.7591 8 294.081 4 294.081 4 294.081 4 275.826 7 294.081 4 294.081 4 294.081 4 180.844 5

Considering Transverse Frame The following data are given. (Slab AB as an Example) Slab Column ts 150 mm c1 550 b 1000 mm c2 550 l1 7 m lc 3 l2 7 m General DL 5.267 kPa LL 1.9 kPa W 9.3604 f'c 20.7 MPa fy 415 MPa E 21383.7 MPa

mm mm m

SAMPLE RESULTS (AB) Step 1. Determine the MOI and Stiffness of Column Ic = a*(b^3)/12 Determine lu/lc Determine a/b Find kab in the tables. Kc = kab*E*Ic/h C = (1-0.63(t/a))*((t^3)(a)/3) Kt = 9*E*C/(l2*(1-b/l2)^2) Kec = 4Kt*Kc/(2*Kt+2*Kc)

Step 2. Determine the MOI and Stiffness of Slab Is = b*(t^3)/12 Compute a/l1 & b/l1 Find ks in the tables. Find FEM and COF in the tables. Ks = ks*E*Is/l1

106

Ic lc lu lu/lc a/b kab Kc C Kt Kec

7.6E+09 3 2.85 0.95 1 4.55 2.5E+11 5.1E+08 1.7E+10 3.1E+10

mm4 mm mm

Is a/l1 b/l2 ks FEM COF Ks

2E+09 0.07857 0.07857 4.11 269.692 0.5086 2.5E+10

mm4

kN-m

Figure. Design Strip for Transverse Frame Step 3. Determine the Distribution Factors DFxy = Kxy/ΣK SUMMARY OF VALUES FOR DF Membe r K DF FEM AB 2.5E+10 0.28438 -269.69 269.691 BA 2.5E+10 0.22141 8 BC 2.5E+10 0.22141 -269.69 269.691 CB 2.5E+10 0.28438 8

Step 4. Continue the MDM To obtain the moments in the midspan of the slabs, M+ = FEM - (-Mave)

MDM FOR TRANSVERSE FRAME Membe A B C r AB BA BC CB 0.22141 0.22141 DF 0.28438 4 4 0.28438 269.691 269.691 FEM 269.692 8 -269.692 8 107

COF

0.5086 Balance 76.6949

0.5086

0.5086

0

CO1 Balance

-39.007 0 230.684 8

0 39.0070 4 0

TOTAL

Slab 1

2

0 0 346.387

-230.685

For Interior 112.7914 346.3868

Moments (%) 0.85M(+) 95.87266 0.70M(-) 242.4707 Moment (%) Passed Column Strip Beam Slab(15%) (85%) (+) 81.49176 14.3809 (-) 206.1001 36.37061

(+) (-)

0 0 346.386 8

Summary of Moments W M 346.3868 401.3272 + 112.7914 230.6848 230.6848 401.3272 + 112.7914 346.3868

For Exterior M(+) M(-)

0.5086 76.6949 2

M(+) M(-)

Middle Strip 16.9 104

112.7914 230.6848

Moments (%) 0.85M(+) 95.87266 0.70M(-) 161.4794 Moment (%) Passed Column Strip Beam Slab(15%) (85%) (+) 81.49176 14.38 (-) 137.2575 24.22

Design Moments CS MS 7.190449 8.459352 18.18531 51.95801

(+) (-)

108

Design Moments CS MS 7.190449 8.459 12.11095 34.6

Middle Strip 16.918705 69.205443

109

Table. Summary of Steel Areas and Number of Bars for Two-Way Slabs

CS Exterior

MS CS

Interior

MS

+ + + + -

Summary of Design of Two-Way Slab (Transverse Frame) M R ρ(initial) ρ(max) ρ(min) ρ(final 7.19 0.6147575 0.00151 0.015977 0.003 0.00337 18.2 1.554778 0.00393 0.015977 0.003 0.00393 8.46 0.7232441 0.00178 0.015977 0.003 0.00337 52 4.4422228 0.01257 0.015977 0.003 0.01257 7.19 0.6147575 0.00151 0.015977 0.003 0.00337 12.1 1.0354427 0.00257 0.015977 0.003 0.00337 8.46 0.7232441 0.00178 0.015977 0.003 0.00337 34.6 2.9584078 0.00786 0.015977 0.003 0.00786

110

As 385 448 385 1433 385 385 385 896

Abar 113.097 113.097 113.097 113.097 113.097 113.097 113.097 113.097

S 294.0814 252.5375 294.0814 78.94669 294.0814 294.0814 294.0814 126.2729

APPENDIX F: SAMPLE DESIGN OF COLUMNS One-Way Slab Tradeoff The column with the maximum axial force was designed. The following are the given data. P My b t cc d f'c fy Φbar Φtie

1337.143 277.02 550 550 40 484 20.7 415 32 10

kN kN-m mm mm

MPa MPa mm mm

Part 1. Determine the Steel Area and Positioning of Bars Step 1. Determine the Steel Area and N bars ρg = ____, assumed value from 0.02 0.04

Results

ρg

0.02

As = ρgAg

Ag

302500

mm2

N = As/Abar then determine actual As

As

mm2

Get actual ρg Pcap = Φ*0.8*Ag(0.85*f'c*(1-ρg)+fy*ρg) If Pcap > P, the dimensions are * adequate * If Pcap < P, redesign

Abar N

6050 804.247 7 8 0.02126 9 0.65 4097.28 OK

111

actual ρg Φ Pcap

mm2 pcs

kN

Step 2. Determine the position of the bars.

Part 2. Checking of Capacity due to Eccentric Load Step 1. Determine if Tension or Compression Controls fy = 600*(d-c)/c, Solve for c a = βc Pb = 0.85*f'c*a*b Pb*(eb+x) = A's*(d-d')+0.85*f'c*a*b*(d-a/2) Solve eb in this equation ex = My/P Solve ex in this equation * If eb > e, Compression Controls, solve for fs * If eb < e, Tension Controls, solve for f's Step 2. Solve and check for Pcap Since tension conrtols, f's = 600(c-d')/c Solve for c and Pcap in the following equations Pcap(ex+(d-c)) = A'sf's(d-d')+0.85f'c(βc)b(d-βc/2) Pcap + Asfy = A'sf's + 0.85f'c(βc)b * If Pcap > P, The dimensions are adequate * If Pcap < P, Redesign For Two-Way Slab Tradeoff The column with the maximun axial forces was designed. The following are the given data.

112

Results c β a Pb eb ex

c Pcap

286.1084 mm 0.85 243.1921 mm 2353431 kN 33.039 mm 207.173 mm Tension Controls

272.163 2247.189

mm kN

P My b t Cc D f'c Fy Φbar Φtie

2424.17 363.257 550 550 40 484 20.7 415 32 10

kN kN-m mm mm

MPa MPa mm mm

Part 1. Determine the Steel Area and Positioning of Bars Step 1. Determine the Steel Area and N bars ρg = ____, assumed value from 0.02 - 0.04

Results ρg

0.02

As = ρgAg

Ag

302500

mm2

N = As/Abar then determine actual As

As

6050

mm2

Get actual ρg Pcap = Φ*0.8*Ag(0.85*f'c*(1-ρg)+fy*ρg) * If Pcap > P, the dimensions are adequate * If Pcap < P, redesign

Abar N actual ρg Φ Pcap

804.2477 8 0.021269 0.65 4097.28 OK

mm2 pcs

Step 2. Determine the position of the bars.

113

kN

Part 2. Checking of Capacity due to Eccentric Load Step 1. Determine if Tension or Compression Controls fy = 600*(d-c)/c, Solve for c a = βc Pb = 0.85*f'c*a*b Pb*(eb+x) = A's*(d-d')+0.85*f'c*a*b*(d-a/2) Solve eb in this equation ex = My/P Solve ex in this equation * If eb > e, Compression Controls, solve for fs * If eb < e, Tension Controls, solve for f's Step 2. Solve and check for Pcap Since tension conrtols, f's = 600(c-d')/c Solve for c and Pcap in the following equations Pcap(ex+(d-c)) = A'sf's(d-d')+0.85f'c(βc)b(d-βc/2) Pcap + Asfy = A'sf's + 0.85f'c(βc)b * If Pcap > P, The dimensions are adequate * If Pcap < P, Redesign

114

Results c β a Pb eb ex

286.1084 0.85 243.1921 2353431 33.039 149.848 Tension Controls

c Pcap

268.6011 2873.398

mm mm kN mm mm

mm kN

APPENDIX G: COST ESTIMATE COST ESTIMATE - ONE WAY TRADEOFF

Member B-1 B-2 B-3 C-1 Slab S-1 S-2 S-3 S-4

L (m) 7 3.5 3 3

b (m) 0.35 0.35 0.35 0.55

L (m) 7 7 3.5 3.3

b (m) 3.5 3 3.3 3

ITEM

TOTAL

CEMENT

9662.96

SAND

536.831

GRAVEL

1073.66

B-1 B-2 B-2 C-1 Slonger Sshorter

BAR Ø (mm) 25 26 25 32 12 12

ITEM

TOTAL

Steel

98654.5

Member

CONCRETE WORKS t V pcs (m) (m3) 0.5 170 208.25 0.5 50 30.625 0.5 180 94.5 0.55 260 235.95 TOTAL t V pcs (m) (m3) 0.15 30 110.25 0.15 120 378 0.15 5 8.6625 0.15 5 7.425

CEMENT (bags) 1874.25 275.625 850.5 2123.55 5123.93 CEMENT (bags) 992.25 3402 77.9625 66.825

TOTAL

4539.038

PRICES ITEM LABOR bags 2415741 966296.3 26841.5 m3 50 10736.63 6 m3 800 858930 343572 TOTAL PRICE 3301512 1320605 REBAR WORKS As L NN bars 2 members (mm ) (m) 490.873 7 6 170 530.929 3.5 6 50 490.873 3 6 180 804.247 3 8 260 113.097 50 224 113.097 14 500 PRICE per kg ITEM LABOR TOTAL 513003 52 1026008 6156046 8 per pc 250

TOTAL COST 115

10778163

TOTAL 4622117 Total W (kg) 27162.5 4320.436 12325.84 38893.42 9816.849 6135.53

SAND (m) 104.125 15.3125 47.25 117.975 284.66 SAND (m) 55.125 189 4.33125 3.7125 252.168 8

GRAVEL (m) 208.25 30.625 94.5 235.95 569.325 GRAVEL (m) 110.25 378 8.6625 7.425 504.3375

COST ESTIMATE - TWO WAY TRADEOFF

Member B-1 B-2 B-3 C-1 Slab S-1 S-2 S-3 S-4 S-5

L (m) 7 3.5 4.5 3

b (m) 0.35 0.35 0.35 0.5

L1 (m) 7 7 3.7 4.5 3.7

L2 (m) 7 4.5 3.5 3.3 1

ITEM

TOTAL

CEMEN T

7894.12 5 438.562 5 877.125

SAND GRAVEL

Member

BAR Ø (mm)

B-1

25

B-2

26

B-2

25

C-1 Slonger Sshorter

32 12 12

ITEM

TOTAL

CONCRETE WORKS t V pcs (m) (m3) 0.5 140 171.5 0.5 20 12.25 0.5 120 94.5 0.5 165 123.75 TOTAL t V pcs (m) (m3) 0.15 10 73.5 0.15 80 378 0.15 5 9.7125 0.15 5 11.1375 0.15 5 2.775

per pc bags

TOTAL PRICES ITEM LABOR

1973531 789412.5 21928.1 m3 50 3 8771.25 3 m 800 701700 280680 TOTAL PRICE 2697159 1078864 REBAR WORKS As L NN bars 2 members (mm ) (m) 490.873 9 7 7 140 530.929 2 3.5 6 20 490.873 9 4.5 6 120 804.247 8 7 3 165 215 113.0973 50 113.0973 14 425 PRICE per kg ITEM LABOR TOTAL

CEMENT (bags) 1543.5 110.25 850.5 1113.75 3618 CEMENT (bags) 661.5 3402 87.4125 100.2375 24.975 4276.125

250

116

TOTAL 3776023 Total W (kg) 26097.31 1728.174 12325.84 24682.36 9422.422 5215.201

SAND (m) 85.75 6.125 47.25 61.875 201 SAND (m) 36.75 189 4.85625 5.56875 1.3875 237.562 5

GRAVEL (m) 171.5 12.25 94.5 123.75 402 GRAVEL (m) 73.5 378 9.7125 11.1375 2.775 475.125

Steel

79471.3 1

52

4132508

826501. 6

TOTAL COST

117

4959010 8735033

APPENDIX H: ESTIMATE OF MAN HOURS

For Tradeoff 1 (One Way Slab)

B-1 B-2 B-3 C-1 S-1 S-2 S-3 S-4

b 350 350 350 550 t 150 150 150 150

T L 500 7 500 3 500 3.5 550 3 S l 3.5 7 3 7 3.3 3.5 3 3.3 TOTAL VOLUME

Quantity 170 180 50 250

Volume 208.25 94.5 30.625 226.875

25 115 5 5

91.875 362.25 8.6625 7.425 1030.463

Assuming that 500% of Total Volume of Concrete Works is equal to Total Man Days, Adding 200% For Rebar Works and 350% For Finishing TOTAL MAN DAYS = 5(1030.463) + 2(1030.463) + 3.5(1030.463) TOTAL MAN DAYS = 10820 days Given that there will be 25 workers TOTAL MAN DAYS = 435 Days

118

For Tradeoff 1 (Two Way Slab)

B-1 B-2 B-3 C-1 S-1 S-2 S-3 S-4

b 350 350 350 550 t 150 150 150 150

t 500 500 500 550 s 7 4.5 3.3 3 TOTAL VOLUME

L 7 4.5 3.5 3 l 7 7 3.5 3.3

Quantity 135 120 20 165

Volume 165.375 94.5 12.25 149.7375

10 80 5 5

73.5 378 8.6625 7.425 889.45

Assuming that 500% of Total Volume of Concrete Works is equal to Total Man Days, Adding 200% For Rebar Works and 350% For Finishing TOTAL MAN DAYS = 5(889.45) + 2(889.45) + 3.5(889.45) TOTAL MAN DAYS = 9940 days Given that there will be 25 workers TOTAL MAN DAYS = 375 Days

119

APPENDIX I: PERCENTAGE DEFLECTION FROM ALLOWABLE

Tradeoff 1 (One Way Slab) (Beam with maximum moment was used) Beam Deflection at Midspan 0.30585 mm Allowable Deflection 19.4444 4 mm Percentage of Computed Deflection from Allowable % = (LV/HV)*100% % = 1.5729 %

Tradeoff 2 (Two Way Slab) (Beam with maximum moment was used) Beam Deflection at Midspan 1.10473 mm Allowable Deflection 19.4444 4 mm Percentage of Computed Deflection from Allowable % = (LV/HV)*100% % = 5.68148 %

120

APPENDIX H: REFERENCES Manuals

Choi K. K. (2002). Reinforced Concrete Structure Design Assistant Tool. California, USA Dahlgren A., & Svensson L. (2013). Guidelines and Rules for Detailong of Reinforcement in Concrete Structures. Goteborg, Sweden. Al-Shamma A. K. (2013). Novel Flowchart for Design of Concrete Rectangular Beams. International Journal of Scientific & Engineering Research. Manual for Design and Detailing of Reinforced Concrete to the Code of Practice for Structural Use of Concrete, 2013

Books

Everrad & Tanner (1996). Theory and Problems of Reinforced Concrete Design. New York: Schaum Publishing Company. McCormac, J.C., & Brown, R. H. (2014). Design of Reinforced Concrete 9 th Edition. United States: John Wiley & Sons, Inc. Association of Structural Engineers of the Philippines. National Structural Code of the Philippines 2010. Quezon City, Philippines: Association of Structural Engineers of the Philippines, Inc. National Building Code of the Philippines (1977). Philippines.

Websites

www.google.com www.wikipedia.com http://www.bca.gov.sg/publications/BuildabilitySeries/others/prh_s2.pdf http://elearning.vtu.ac.in/P6/enotes/CV61/Beams-GS.pdf

121

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