Hedaya Hamaideh - Structural Design by hedaya abd

Course Name: Structures for Architect Course Tutor: Eng. Mariam Al-Azzeh

Project Name: Semi-Villa, Residential Project DWG. NO.: Site Plan

+6.80

Scale:

ROOF SLAB LEVEL

1- 100

North Direction:

+9.80

Date: 10/ 2 / 2021

+00.00 GROUND LEVEL

Student Name:

Arch. Hedaya Hamaideh 19340001

D Course Name:

Structures for Architect Course Tutor: Eng. Mariam Al-Azzeh Project Name:

Semi-Villa, Residential Project DWG. NO.: Ground Floor Level Scale:

1- 100 North Direction:

4 Date: 10/ 2 / 2021 Student Name:

Arch. Hedaya Hamaideh 19340001

Course Name: Structures for Architect Course Tutor: Eng. Mariam Al-Azzeh

Project Name: Semi-Villa, Residential Project DWG. NO.: First Floor Level Scale:

1- 100 North Direction:

4 Date: 10/ 2 / 2021

Student Name: Arch. Hedaya Hamaideh 19340001

Course Name: Structures for Architect

Course Tutor: Eng. Mariam Al-Azzeh Project Name: Semi-Villa, Residential Project DWG. NO.: North Elevation Scale: 1- 100 North Direction:

Date: 10/ 2 / 2021 Student Name: Arch. Hedaya Hamaideh 19340001

Course Name:

Structures for Architect Course Tutor: Eng. Mariam Al-Azzeh Project Name: Semi-Villa, Residential Project DWG. NO.: West Elevation Scale: 1- 100 North Direction:

Date: 10/ 2 / 2021 Student Name: Arch. Hedaya Hamaideh 19340001

Course Name: Structures for Architect

D Course Tutor: Eng. Mariam Al-Azzeh Project Name: Semi-Villa, Residential Project DWG. NO.: West Elevation Scale: 1- 100 North Direction:

Date: 10/ 2 / 2021 Student Name: Arch. Hedaya Hamaideh 19340001

Course Name: Structures for Architect

Eng. Mariam Al-Azzeh

Course Tutor:

Project Name: Semi-Villa, Residential Project DWG. NO.: West Elevation Scale: 1- 100 North Direction:

Date: 10/ 2 / 2021 Student Name: Arch. Hedaya Hamaideh 19340001

Course Name: Structures for Architect Course Tutor: Eng. Mariam Al-Azzeh Project Name: Semi-Villa, Residential Project DWG. NO.: Section A-A Scale: 1- 100 North Direction:

Date: 10/ 2 / 2021 Student Name: Arch. Hedaya Hamaideh 19340001

Section

Beam 1

Course Name:

Beam 3 - 1

Beam 2

Structures for Architect 2- way RS 31 cm

Course Tutor: Eng. Mariam Al-Azzeh

Cantilever

Beam 3 - 2

C.R

Beam 6 - 1

Beam 7 - 3

Beam 7 - 2

Beam 7 - 1

C.R

1- way RS 31 cm

Scale:

2Y 16 (TOP)

Beam 8 - 1

Beam 8 - 2

Semi-Villa, Residential Project DWG. NO.: Ground Slab

2Y 16 (TOP)

Project Name:

1- 200

Beam 8 - 3

2Y 18 (BOT)

Beam 4 - 2

Beam 6 - 3

Beam 4 - 1

North Direction:

Beam 3 - 3

C.R

Beam 6 - 2

2Y 18 (BOT)

Beam 4 - 3

Date:

2- way RS 31 cm

10/ 2 / 2021

Beam 5

Student Name: Arch. Hedaya Hamaideh 19340001

2- way RS 31 cm

Course Name: Structures for Architect

Beam 3 - 1

Beam 2

Beam 1

Course Tutor: Eng. Mariam Al-Azzeh

Beam 4 - 2

Beam 3 - 2

DWG. NO.: First Floor Slab

Beam 8 - 3

Project Name: Semi-Villa, Residential Project

Beam 3 - 3

Beam 8 - 2

Beam 4 - 1 Beam 6 - 3

1- way RS 31 cm

C.R

Beam 6 - 2

Beam 8 - 1

C.R

Beam 6 - 1

Beam 7 - 3

Beam 7 - 2

Beam 7 - 1

C.R

Scale: 1- 100 North Direction:

Beam 4 - 3

2- way RS 31 cm

Date: 10/ 2 / 2021

Beam 5

Student Name: Arch. Hedaya Hamaideh 19340001

Course Name: Structures for Architect Course Tutor: Eng. Mariam Al-Azzeh

Project Name: Semi-Villa, Residential Project DWG. NO.: Roof Slab

Scale: 1- 100 North Direction:

2Y 14 (TOP)

2Y 12 (BOT)

Date: 1- way RS 31cm

10/ 2 / 2021

Student Name: Arch. Hedaya Hamaideh 19340001

Ø12@20cm

Course Name: Structures for Architect Course Tutor: Eng. Mariam Al-Azzeh

2Ø12

Project Name: Semi-Villa, Residential Project

B8 30X31

DWG. NO.: Beams Details Scale: 1- 200 North Direction:

3Y 12

3Y 18

3Y 12

Date: 10/ 2 / 2021 Student Name: Arch. Hedaya Hamaideh 19340001

B6-1 30X31

1 C1

Course Name:

Structures for Architect Course Tutor: Eng. Mariam Al-Azzeh

2 C1

Project Name: Semi-Villa, Residential Project DWG. NO.: Columns Scale:

1- 100 North Direction:

Date: 10/ 2 / 2021 C1

Student Name: Arch. Hedaya Hamaideh 19340001

Course Name: Structures for Architect Course Tutor: Eng. Mariam Al-Azzeh

C1 6Ø16

C2 Ø10 @20cm at mid Ø10 @10cm at end

6Ø16

C3 Ø10 @20cm at mid Ø10 @10cm at end

6Ø16

Project Name: Semi-Villa, Residential Project

Ø10 @20cm at mid Ø10 @10cm at end

DWG. NO.: Column Details Scale: 1- 100 North Direction:

Inner Columns

Edge Columns

Corner Columns

Date: 10/ 2 / 2021 Student Name: Arch. Hedaya Hamaideh 19340001

1 C1

Course Name:

Structures for Architect Course Tutor: Eng. Mariam Al-Azzeh

2 C1

Project Name: Semi-Villa, Residential Project DWG. NO.: Foundation Scale:

1- 200

North Direction:

Date: 10/ 2 / 2021 Student Name:

Arch. Hedaya Hamaideh 19340001

Hedaya Hamaideh 19340001 Architectural Engineering Eng. Mariam Al- Azzeh

STRUCTURE FOR ARCHITECT I Residential Project

Structure for Architect

Contents Introduction ..................................................................................................................................... 2 The Project ...................................................................................................................................... 2 Material ........................................................................................................................................... 2 Ductile Material........................................................................................................................... 2 1.

Rebar Steel ....................................................................................................................... 3

Aluminum Window Frames ............................................................................................. 3

Brittle Material ............................................................................................................................ 4 1.

Concrete ........................................................................................................................... 4

Glass ................................................................................................................................. 4

Concrete Bricks ................................................................................................................ 5

Determining Slab Type ................................................................................................................... 6 Loads ............................................................................................................................................. 12 Loads Acting on Slabs............................................................................................................... 12 Loads Acting on Beams ............................................................................................................ 15 Statically Determinate Beams ............................................................................................... 18 Statically Indeterminate Beams ............................................................................................. 21 Loads Acting on Columns ......................................................................................................... 22 1.

Inner Column.................................................................................................................. 22

Edge Column .................................................................................................................. 23

Corner Column ............................................................................................................... 24

Conclusion .................................................................................................................................... 25 References ..................................................................................................................................... 25

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Structure for Architect

Introduction This report represents the structural analysis and design of a residential project, it discusses the basic behavior of structural materials, especially the ductile materials, such as structural steel, and the brittle materials such as concrete. The material mechanical behavior is governed by a set of mechanical properties such as: yield strength, ultimate strength, failure point. Following the load and discussing the mechanical properties and their role in materials selection, a suitable structural system is suggested based on studying different possibilities to achieve the most efficient load path. Evaluating the forces acting on the structural elements and have the knowledge to structurally analyses these elements, choosing the suitable structural system to meet certain requirements and to estimate the reasonable dimensions for structural elements and producing structural technical drawing and extract necessary information.

The Project The project that was selected is a residential project, a semi-villa two floor for a small family including (Guest room, Living room, Kitchen, Maid room, 3 Master bed room and one balcony). The architectural drawings were prepared and ready to the next level which is structural drawing. Therefore, this report discusses the structural drawing process that were made during this project. As each civil engineer when start design they care for three main things, which are: safety, cost and functionality.

Material Construction material is an essential part during design as combining all material together is what makes the building stand. Therefore, the material within building need to be selected carefully. Each material has its own behavior and they vary in their capacity to sustain different kinds of loads, for examples some material can handle compression and tension, while other can handle only compression, in the latter case, a combination of two materials can be used to resist the expected loads like reinforced concrete. This demonstrates the importance of understanding and identifying the mechanical properties of materials in choosing the right material that would be used. In general, material could be classified into two main behaviors ductile and brittle material.

Ductile Material These materials are identified as material that can handle tensile stress, which mean their ability to be stretched and handle strains without breaking and ruptures. Ductile materials have high Young’s moduli; which mean their ability to resistance material elastic deformation, which mean their shape can change slightly under elastic loads. In addition, within the high Young’s moduli that mean that ductile material can handle high loads to permanently deform. Which can be the benefit of using it in construction site. In this project the material that were used as ductile material is:

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Structure for Architect

1. Rebar Steel One of the most important material that being used in construction and is used in this project is rebar steel. It has the ability to be blended and perform a variety of shapes within connecting structural elements with other structural elements, such as; beams, columns, shear wall and slabs. As for rebar steel it is generally produced to meet ASTM or AASHTO specifications as for mechanical properties, it has a lot of properties that need to be taken into consideration, such as; 1. Elastically, the modulus is 200'000 MPA and it’s a used value within design. also it has similar elastic properties under both tensile and compression loads. 2. Elongation, it can handle a significant elongation (%15) and to be stretched before breaking. 3. Yield: rebar steel it could be consider high; as its 420 MPa, which mean the point that steel can handle the elastic deformation with huge force before breaking. Rebar steel has different advantages such as; cost saving, no effect on the working speed, aesthetic reasons and the resistant to rough conditions during transport, storage, bundling and placing on construction site. Applying forces on rebar steel especially in beams causes internal stresses to develop where it can handle high tensile stress and make it strong and tight, which going to benefit us in our project; as it used in each structural element and sometime its combined with brittle material like reinforced concrete, to guarantee the resistance needed to support loads required and used as backup for brittle material to prevent cracks or deflections. 2. Aluminum Window Frames As aluminum considered ductile material, it behaves like steel, but its weaker than steel and stronger that brittle material. Ultimate Tensile Strength (310 MPa), tensile Yield strength (276 MPa) and elongation at break 12 %. Aluminum is used for window framing because of different reasons, such as; 1. Level up the security, insulation or aesthetics for the building 2. It weight is light, durable and easy to maintain. 3. Its strength allows to maximizes the glass area within the window 4. Its durable which meat it’s not effected by UV rays, it will not rot, rust or bend. 5. Aluminum can handle high levels of wind, rain and air-tightness, which effect in-house energy efficiency and having warmer house which effect on having less energy bills. 6. Aluminum can be recycled, which also have already low carbon footprint. Therefore, it's a huge benefit of using aluminum for window framing and combine this ductile material with brittle material; glass. Also, these reasons are the reasons for not using iron, as it rusts, get resonate and heavy. Using Ductile Material within the Project As mentioned above aluminum and steel are used in our project as a ductile material. They are chosen because of their mechanical properties. As one huge advantages of them is tensile P a g e 3 | 25

Structure for Architect

strength and their ability to stretched and handle huge strains without breaking and ruptures. Which give a major safety consideration and their durability for construction project without fracture. In addition, ductile material that are mentioned before have a variety in cost, also in our project using steel bar and aluminum for window framing, give the speed needed on the site, they are prefabricated and ready to be used immediately, which affect the speed of construction due to eliminating the need to place forms or to wait for curing. Also, a huge advantage of using ductile material is combining them with brittle material, such as; rebar with concrete and aluminum framing with glass, which their properties covers the weakness of brittle material and protect brittle materials form cracking.

Brittle Material These materials are identified as material that can handle high compression stress and low tensile test, which mean they exhibit good ability to be under high compressive stresses without breaking. Also, brittle materials have low Young’s moduli and ultimate stresses, which mean their ability to handle high loads before breaking. On the other hand, it can fail suddenly without warning and fracture at much lower strains. which means their shape can't change slightly under elastic loads as it has little elastic deformation and without worthy plastic deformation. In this project the material that were used as brittle material is: 1. Concrete One of the most important material were being used in construction and are going to be used in this project; is concrete, as it gives the construction the ability to handle high loads. as concrete is much stronger in compression 20 - 40 MPa than in tension: 2 - 5 MPa, giving it a higher strength to weight ratio in compression. The mechanical properties of concrete basically depend on the used amount of water, therefore, it's critical thing when adding water on concrete. In addition, mechanical properties for concrete is shrinkage, creep, flexural strength, and modulus of elasticity 14 - 41 GPa. As shrinkage relates to losing water within concrete, as "drying shrinkage and creep strains increase with the increasing age and volume of cement replacement. (2)". Also, "addition mineral admixtures slightly improves the tensile strength, flexural strength, and modulus of elasticity of concrete with the increase in replacement level. (2) Furthermore, the advantages of using concrete is it cost-effectively, lower carbon footprint and it doesn't burn, rust or rot. In this project, concrete in general is used in every structural element to be reinforced concrete. While plain concrete is used for pour hygiene under foundation as its resistant to salts and bacteria. 2. Glass As for glass, which going to be used in window within the project, it considers as brittle material. The compressive strength is very high (1000 MPa) which means that it requires a load of some 10 tons. Hence, the compressive strength is high therefore the tensile stress is very low. Also, glass consider as non-elastic material, as it’s not show their permanent deformation however, it breaks suddenly if loads were very high on it. However, in our project glass don't carry any load, as ductile material like the aluminum is there to carry the load and iron frame above the window is there to carry the load.

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Structure for Architect

3. Concrete Bricks Bricks is a main component at construction, as its used to support loads in non-bearing wall and partitions, therefore bricks are being used to transverse loads, weather in interior or exterior wall Also, it's mechanical properties is very important, brick have a high compressive strength and sufficient resistance for force, its tested by crushing until they fail or crumble to know the exact value. Also, ceramic tiles and stones are used in this project, which also consider as brittle material and have the same behaver as glass, however some numbers are different. Even if all these material consider as brittle one, each one has different behaver on specific value for compressive and tensile strength and different percentage for elastically. Using Brittle Material within the Project As mentioned above concrete, bricks, glass, tiles and stone consider as brittle material and used within the project like, bricks for wall, glass for window, tiles for flooring and so on. As they handle vertical load. The reason for using it is for their mechanical properties. As a huge advantage for using them is their ability to handle compressive load such as bricks and stone for wall, concrete for slab, tiles for flooring. All of these material carry all kinds of loads, stone, brick carry dead and wind loads. While tiles carry live load concrete carry dead and live loads. Which all of these loads caused an internal stress to develop. Also, as using specially concrete within residential and low-rise building support the same loads that ductile material such as steel can handle, therefore concrete is the economical choice. However, concrete take time for using and forming, other brittle material; such as glass, bricks, tiles and stone, came ready to use without losing time for forming, which can help in reducing time, reworking and increasing quality without having thermal damages, as they are fire resistance which increase the safety level. Reinforced Concrete As RC is the combination of concrete and steel reinforcing bars (rebar), and as its mentioned above, concrete consider as brittle material and steel consider as ductile material. Therefore, combining these elements together can enhance each other mechanical properties to give the material that needed to handle the building. As internal stresses come from the top of the building and transfer to the end of it " foundation", elements that have mechanical properties like reinforced is used to transfer the load, as loads come from slab, wall, beams need to have the ability to handle high loads. Acting force on the RC is much stronger than acting force on only concrete, as failure happens slower and crack forming fails slowly which help to get noticed before it collapses suddenly, which going to help in safety, the reasons concrete failure slowly and acts somehow as ductile material is because the use of steel rebar that prevent it to collapse suddenly. Therefore, RC is going to be used within this project in slabs, beams, columns and shear wall. Hence, concrete have the ability to handle compressive strength and steel have the ability to handle tensile forces, therefore, RC can handle both tensile and compressive loads. Which also, increase in ductility and elasticity of the material. In construction it’s difficult to use

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Structure for Architect

only brittle material such as, concrete which can suddenly collapse on overload. Therefore, it’s commonly to use RC.

Furthermore, from the reasons of using RC is combining ductile and brittle material have a huge advantage, as ductile material like rebar give the brittle material like concrete the protection to control it from cracking caused by loads, while concrete protect reinforcing bar from fire and corrosion. Also, rebar has the ability to expand and contract within temperatures changing, therefore, concrete won't be effected from the structure area because of having a degree of cohesiveness with each other and the ability for these two material for expand and contract in the same rate. Therefore, it’s hard to construct huge building concrete beams, slabs, columns and share wall without rebar, because of their mechanical properties alone and the huge safety effect that they cause of suddenly collapse on overload. In addition, sometime within the building material with low Young’s moduli require a backup with ductile material as each structural element in the building need to be light and capable of bending and flexing of sudden loads and weather changes, Therefore, columns, share wall and beams need to contain rebar as having only concrete will make it heavy and collapse easily. Also, for concrete slabs using rebar can necessitate thinner concrete slabs, which result for cost saving and time of mixing huge amount of concrete.

Determining Slab Type An important step to design a structural system is determining slab type and how it’s going to be supported, options were made to know the suitable slab to be chosen, as determining slab type help to know its thickness and loads acting on it. When designing a building its important to know that loads are transferred from top to bottom and designing start from slab then beams then columns then foundation, based on loads path design are made, therefore, it’s important to know how the loads path will act and how it will be supported to handle it. As it’s important for the structural support elements to provide the stiffness and strength to resist the load coming from above and safely transferred it to the ground. Based on the slab and loads acting on it structural supporting elements are designed based on it, which influences the behavior of the whole system within the building. Different structural systems used to support slab, such as; P a g e 6 | 25

Structure for Architect

1. Beam supported slab; This slab type is supported by beam and beam is supported by column, while column is supported by foundation and foundation is supported by the soil. This type is one of the best in determining slab type based on its supporting, as it carries heavy load, larger span and give a greater fire resistance, also it could be attractive exposed ceiling as beams could be hidden in the wall, but sometime it could be drop and shown which could be used in artistic way. Therefore, its more popular and used. However, using a lot of supporting structural system take time to build it and cost increase because of using different and a lot of material. 2. Flat plate; In this type the flat slab is supported directly by the column without using beams, which has a low cost for formwork and don’t need a lot of time to constructed it. However, it has a structural issue, as having low stiffness slab lead of noting deflection, low shear capacity and column could simply punch through the slab under high loads 3. Flat slab; In this type column punching through the slab is solved by adding panels and columns capital and similar with the advantages of flat plate like low cost for formwork and don’t need a lot of time to constructed it. However, it need more time in form working for capital and panels. On the other hand, different structural systems used to transfer mechanism such as; 1. Solid Slab; Is a full of pre-stressed concrete that have cross section, its used in a variety of application; such as slab, bridge, flooring, and so on. While its manufactured by using high tensile strength. Solid slab is used where its requires a high level of loading, thermal insulation, fire resistance. Its framework is simpler and faster but, its self-weight is heavy, could be costly for material and using a lot of steel, and bear somehow short span. 2. Ribbed Slab: Is slab where concrete, block and steel rib is used, where load could be acted on one or two way. Its suitable for heavy loads and bear a reasonable distance for spans. Ribbed slab provides stiff slab and sometime a have similar weight to solid slab, as block have light weight. 3. Waffle Slab: Its name come from the grid pattern it made by the rib and model of concrete within two-way joist slab. while it can handle longer span, heavy load. However, its costly for using perfect shape model. P a g e 7 | 25

Structure for Architect

Determining slab type based on slab thickness. Options were made and based on the result the suitable slab were chosen.

Option 1:

Solid Slab One way – two way

Strip 1: 3M

3.5 M

Strip 2: 4.5 M

Strip 3: 5M Strip 4: 5M Strip 5: 4M

Strip Num.

Distance

Type

Calculation

Height

2 way - One end counties

= 3*0.87 / 45

0.05 cm

Strip 1

3.5 3.5 1.2

1 way - Both end counties 1 way - One end counties Cantilever

= 3.5 / 28 = 3.5 / 28 = 1.2 / 10

0.12 cm 0.12 cm 0.12 cm

Strip 2

4.5

2 way - Simply Supported

= 1*4.5 / 35

0.12 cm

Strip 3

1 way - Simply Supported

= 5 / 20

0.25 cm

Strip 4 Strip 5

5 4

2 way - Simply Supported 2 way - Simply Supported

= 1*5 / 35 =1 *4 / 35

0.14 cm 0.11 cm

As shown in the table each part has different slab thickness, the biggest slab thickness was 25 cm in strip 3 and the smallest thickness is 5 cm in strip 1. Also, in strip 1 beside strip 3 had 12cm slab thickness which is huge different and could cause difficulties within construction site and would costs a lot if we treated strip 1 like strip 3 and increase concrete.

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1.2M

Structure for Architect

Option 2:

Solid Slab One way – two way Strip 1: 3M

3.5 M

Strip 2:

1.2M

4.5 M

Strip 3+6: 5M

Strip 4:

Strip 5:

Strip Num.

Distance

Type

Calculation

3.5 M

Height

2 way -One end counties

= 3*0.87 / 45

0.05 cm

Strip 1

3.5 3.5 1.2

1 way - Both end counties 1 way - One end counties Cantilever

= 3.5 / 28 = 3.5 / 28 = 1.2 / 10

0.12 cm 0.12 cm 0.12 cm

Strip 2

4.5

2 way - Simply Supported

= 1*4.5 / 35

0.12 cm

Strip 3 + 6

2 way - Simply Supported

= 5*1 / 35

0.14 cm

Strip 4

5 4 3.5 3.5

2 way - Simply Supported 2 way -One end counties 2 way – Both end counties 2 way -One end counties

= 1*5 / 35 =0.87 *4 / 45 = 0.76 * 3.5 / 45 =0.87 *3.5 / 45

0.14 cm 0.07 cm 0.06 cm 0.06 cm

Strip 5

The differences in this option is made to reduce the previous thickness (20 cm). Therefore, as shown in the table each part has different slab thickness, the biggest slab thickness is 14 cm in strip 3+6+4, and the smallest thickness is 12 cm which is a logical difference. However, here we need to take into consideration that beam length has increased, yes the thickness is more logical than the previous option but the cost is higher.

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3.5 M

Structure for Architect

Option 3:

Solid Slab One way – two way

Strip 1: 3M

3.5 M

Strip 2:

1.2M

3.5 M

3.5M

Strip 3: 4M

3.5 M

Strip 4: 5M Strip 6+5: 4.5 M

4.5 M

Strip 7: 4.5 M

Strip Num. Strip 1

Strip 2 Strip 3 Strip 4 Strip 5+6 Strip 7

Distance 3 3.5 3.5 1.2 3.5 3.5 4 3.5 3.5 5 4.5 4.5 5 4.5

Type 2 way -One end counties 2 way -Both end counties 2 way - Both end counties Cantilever 2 way -One end counties 2 way -One end counties 2 way -One end counties 2 way -Both end counties 2 way -One end counties 2 way - Simply Supported 2 way -One end counties 2 way – Both end counties 2 way -One end counties 2 way - Simply Supported

Calculation = 3*0.87 / 45 = 3.5*0.76 / 45 = 3.5*0.76 / 45 = 1.2 / 10 = 3.5*0.87 / 45 = 3.5*0.87 / 45 = 0.87*4 / 45 = 0.76*3.5 / 45 = 0.87*3.5 / 45 = 1*5 / 35 = 0.87*4.5 / 45 = 0.76*4.5 / 45 = 0.87 *5 / 45 = 0.87*4.5 / 45

Height 0.05 cm 0.06 cm 0.06 cm 0.12 cm 0.06 cm 0.06 cm 0.07 cm 0.06 cm 0.07 cm 0.14 cm 0.08 cm 0.07 cm 0.09 cm 0.08 cm

The differences in this option is made to reduce the differences between previous thickness. Therefore, as shown in the table each part has different slab thickness, the biggest slab thickness is 14 cm in strip 4, and the smallest thickness is 7 cm which is not really logical difference comparing it with the previous choose. All strips thickness between 7-9 cm expect strip 4 which can be dealt with differently, yes is the best options with the thinner thickness. However, it’s the worse option with the high cost of the length of beams. Therefore, one last option will be made with different slab type. P a g e 10 | 25

Structure for Architect

Option 4:

Ribbed Slab One way – two way

= 3.5 / 21

Strip 1: 3M

3.5 M

Strip 2: 4.5 M

Strip 3: 5M

Strip 4:

Strip Num.

Distance

Type

Calculation

Height

2 way - One end counties

= 3/ 18.5

0.16 cm

Strip 1

3.5 3.5 1.2

1 way - Both end counties 1 way - One end counties Cantilever

= 3.5 / 21 = 3.5 / 21 = 1.2 / 8

0.16 cm 0.16 cm 0.15 cm

Strip 2

4.5

2 way - Simply Supported

= 4.5 / 16

0.28 cm

Strip 3

2 way - Simply Supported

= 5 / 16

0.31 cm

Strip 4

2 way - Simply Supported

= 4 / 16

0.25 cm

The differences in this option is made to be more logical and traditional, as shown in the table each part has different slab thickness, the biggest slab thickness is 31 cm in strip 3, and the smallest thickness is 16 cm, it’s the first option that ribbed slab is used and less length of beams. Also, one column is added but it keeps the building handle more, this is one of the best options as less beams are used and thickness can be dealt with to be the same. Therefore, this is the chosen one.

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1.2M

Structure for Architect

Among the four options which all of them appropriate and can be applied. However, I tried to find the lowest cost and the suitable one for this project, by considering different aspect: 1. Slab type according to supporting system: As were discussed before there are several type of slab according to its supporting system. However, the best choice is slab supported by beam, yes it may be the one that cost more, however it’s the one that much safer and can handle the required loads, without supporting it with panel and column capital. Therefore, in this project the slab is supported by beam. 2. Slab type according to the load transfer mechanism: The three first option, were I used solid slab has required a lot of beam length, sometime huge slab thickness and a lot of differences between different area on the same slab, none of them were appropriate to be used in this project. On the other hand, in Option 4 were I used ribbed slab it uses less beam length, not huge slab thickness taking into consideration that ribbed slab thickness means block and concrete topping, also, all of these factors means less costs on structural elements. Therefore, ribbed slab in option 4 was the appropriate one. In addition, the reasons for choosing less thickness, is for economic reasons which were mentioned before beside of having less concrete won’t cost a lot and architectural reasons, as using thinner thickness slab will give the opportunity to have suitable space down and put and hide all mechanical and structural elements. Therefore, to have the same level on, ribbed slab with 31cm thickness is used among all area in the slab and it also won’t cost a lot because of not having different situation to deal with. Ribbed slab based on option 4 will be chosen in this project its advantages is much more than solid slab in our case, also taking into consideration the long span we have, ribbed slab can deal with it much better that solid slab. In addition, as its residential building and live load is generally less than 3 KN/m2 therefore ribbed slab with thickness 31cm is used.

Loads Structural system is very important to serve some main purpose which is to make the building stand and handle loads acting on it which called the structural loads as it can be defined "forces, deformations, or accelerations applied to a structure or its components."(3) Structural loads are divided into four categories; dead loads, live loads, impact loads, and environmental loads. These loads need to be calculated and get structural analysis to know how to handle them within the building. The following is showing the acting loads on the project.

Loads Acting on Slabs 1. Dead Loads Dead loads are identified as constant load over time on specific area of the structural elements or the building, as they cause stresses, deformations and displacements in the structure. Such as the own weight of slabs, columns, beams, walls, roof and building material.

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Structure for Architect

As the slab was chosen is ribbed slab and it thickness is 31cm concrete. Therefore, the dead load acting on the one way- ribbed slab is: P.S.1: unit weight based on Jordanian building codes P.S.2: Block and rib is taking for width of 55 cm and depth of 1 m.

Material Tiles Mortar Aggregate Concrete 2% Block Rib Plaster

Unit Weight KN/m3 25 22 14.5 25 0.18

Thickness (m)

Unit weight * Thickness ( KN/m2) 0.03 = 25 * 0.03 0.02 = 22 * 0.02 0.05 = 14.5 * 0.05 0.07 = 25 * 0.07 0.24 = 5* 0.18 * 0.24 /0.55 25 0.24 = 1* 25 * 0.24 *0.15 /0.55 22 0.02 = 22 * 0.02 Total Dead load for one way ribbed slab is

Dead Load ( KN/m2) 0.75 0.44 0.72 1.75 0.39 1.6 0.44 = 5.64 KN/m2

Also, partitions are defined therefore their load will be added to the dead load which is ~2 RS dead load = 5.64+ 2 = 7.64 KN/m2. “this number is for all one way slabs within the project ground slab, first slab and roof slab as all of them is the same”

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As for dead load acting on the two way- ribbed slab is: P.S.1: unit weight based on Jordanian building codes P.S.2: Block and rib is also taking for width of 55 cm and depth of 1 m Material Tiles Mortar Aggregate Concrete 2% Block Rib Plaster

Unit Weight KN/m3 25 22 14.5 25 0.18

Thickness (m)

Unit weight * Thickness 0.03 = 25 * 0.03 0.02 = 22 * 0.02 0.05 = 14.5 * 0.05 0.07 = 25 * 0.07 0.24 = 3* 0.18 * 0.24 /0.55 25 0.24 = 3 * 0.24 *0.15 * 25 /0.55 22 0.02 = 22 * 0.02 Total Dead load for one way ribbed slab is

Dead Load ( KN/m2) 0.75 0.44 0.72 1.75 0.23 4.9 0.44 = 9.32 KN/m2

Also, partitions are defined therefore their load will be added to the dead load which is ~2 RS dead load = 9.32+ 2 = 11.32 KN/m 2. “this number is for all two way slabs within the project ground slab, first slab and roof slab as all of them is the same”

2. Live Loads Live loads are identified as load can change over time on specific area in the building, as they depend on usage and capacity, such as; people and furniture. Building codes has determined equivalent loads based on the building types, therefore, for this project the live load is: Live load for building usage: 2 KN/m 2. Live load for roof slab = 1.5 KN/m2

3. Environmental Loads Environmental loads are identified as load caused by natural forces and can change over time on specific area or direction in the building, such as; wind, rain, snow and earthquake. Building codes has determined equivalent loads based on the building location, which civil engineer refer to when calculate environmental loads.

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After knowing the main loads acting on the project, Ultimate load = 1.2 (D.L) + 1.6 (L.L) For one way ribbed slab = 1.2 (7.64) + 1.6 (2) = 12.3 KN/m2 Roof load for one way ribbed slab = 1.2 (7.64) + 1.6 (1.5) = 11.5 KN/m2

For two way ribbed slab = 1.2 (11.32) + 1.6 (2) = 16.78 KN/m2 Roof load for two way ribbed slab = 1.2 (11.32) + 1.6 (1.5) = 15.98 KN/m2

Loads Acting on Beams As for the loads acting on each beam:

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Structure for Architect

10 cm block = 1.94 KN/m2 **Sketch wall section** 15 cm block = 1.5 KN/m2 5 cm stone = 2.7 KN/m2 Stone wall = (wall height * block weights) + (stone height * stone width * stone weight) Stone wall = 3 * (1.94 + 1.5) + (3 * 0.05 * 27) = 18.87 KN/m

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Structure for Architect

Calculation beam loads Beams

Q ultimate One way / two way

(KN/m2) 1 2 3-1 3-2 3-3

4-1 4 4-2 4-3 5 6

Two way = 16.78 One way = 12.3 One way = 12.3 One way = 12.3 Two way = 16.78

6-1 6-2 6-3 7-1 7-2 7-3

Two way = 16.78 Two way = 16.78 Two way = 16.78 0 0

8-1 8-2 8-3

0 0 Two way = 16.78 One way = 12.3 One way = 12.3 One way = 12.3 Two way = 16.78 One way = 12.3 One way = 12.3 One way = 12.3

Qult * half distance Some are trapezoid + triangle ½ * (3+5) * 16.78 ( ½ * 4 * 16.78) (½ * 4 * 16.78) 0 0 ( ½ * (2.5+4) * 16.78) (12.3 * 1.75) (12.3 * 1.75) (12.3 * 1.75) ( ½ * (2.5+4) * 16.78) 0 0 ( ½ * 4.5 * 16.78) (12.3 * 1.75) (12.3 * 1.75) (12.3 * 1.75) ( ½ * (3+5) * 16.78) (12.3 * (1.75+1.75)) (12.3 * (1.75+1.75)) (12.3 * (1.75+1.75))

Stone wall

Loads

(18.87 KN/m) 18.87

(KN/m) 67.12 41.96 33.568 0 0

Loads acting on Beams

(KN/m) 85.99 60.83 52.438

18.87

18.87 18.87 18.87

18.87 18.87

54.535

21.525 21.525 21.525

18.87 18.87 18.87

40.39 40.39

54.535

18.87

73.405

0 0 37.76 21.525 21.525 21.525 67.12 43.05 43.05 43.05

18.87 18.87 18.87 18.87 18.87

18.87 18.87

88.645

0 0 0

43.05 43.05 43.05

94.93

56.6 40.39 40.39

To calculate the H optimum to know the height of the beam, Beam 7 + 8 + 4 are the same length= 4.5 M

One end cont. 4.5/ 18.5 24 cm

4.5 M

Both end cont. --

4.5 / 21

21 cm

---

One end cont. 5/18.5 27 cm

The highest high is 27 cm which also the minimum height for beams, however as our slab thickness is 31cm therefore, 7, 8, 4 beams will be hidden beam with thickness of 31cm.

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Structure for Architect

Statically Determinate Beams A statically determinate beams its where the reaction can be calculated using equilibrium equations alone, for beams that is statically determinate in this project: Beam 1 is simply supported. Distributed loads according to previous calculation = 85.99 KN/m Point load = ½ * (85.99) * (3+5) Point load (R)= 343.96 𝜀fx = 0

Bx = 0

𝜀fy = 0

Ay+ By = 343.96

𝜀M= 0

-(343.96)(2.5) + Ay (5) = 0

By = 171.98KN Ay = 171.98 KN

Sare force diagrm (A1+) = ½ *2.5* 171.98 = + 214.975 KN Sare force diagrm (A2-) = ½ * 2.5* 171.98 = -214.975 KN

And based on share force and bending moment diagrams, Civil eng. Will know where is the weakest point to increase Steel bar and kanat. Based on the figure the disturbution of Kanat will be organized, also the moment is positive.

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Beam 6-1 is continued with cantilever in first floor. Distributed loads according to previous calculation = 18.87 KN/m Point load ( R) = 18.87* 4.7 = 88.689 1. Support Reaction 𝜀fx = 0

Ax =0

𝜀fy = 0

Ay + By = 88.689

𝜀M= 0

Ay = 29.141 KN

+(88.689)(2.35) - By (3.5) = 0

By = 59.548 KN 2. Share Force Diagram For A1

X/3.5 = 29.141/ 66.04

X = 1.54

A1 = ½ * 1.54 * 29.141

A1 = 22.43 KN/m

A2 = ½ * 1.96 * - 43.61

A2 = - 42.73 KN/m

A3 = ½ * 1.2 * 13.59

A3 = 8.154 KN/m

And based on share force and bending moment diagrams, Civil eng. Will know where is the weakest point to increase Steel bar and kanat, also, where is the water will be collected.

As based on the figure the disturbution of Kanat won’t be organized and will be increased in the beginning of the cantiliver and reflectiong point will be in the middle. Also, to make sure of my data I used Prokon software, to see eleastic deflection and try put steel bar with appropriate width.

Figure 1: Elastic Deflection

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Structure for Architect

Figure 2: Long-term deflection

Figure 3: Reinforcement Bar

Figure 4: Main Reinforcement

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Structure for Architect

Statically Indeterminate Beams Statically indeterminate beams, is beams where it has a lot of reaction to calculate, therefore, in some cases code simplification are used to calculate the reaction, share and moment diagram, based on ACI 2014 codes. Beam 8, is simply supported, in one way ribbed slab, according to the slab based on deflection the moment diagram is showed, where when its positive or negative its calculated by W ult 2 / (Certain num based on where the span is).

Also, based on codes simplification, in the figure it shows how in my case Beam 8 will be solved, to know how shear and moment diagram will be. As shown the moment is positive

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Loads Acting on Columns Loads that act on the columns come from, columns own weight, loads from slab and weight of non-structural vertical elements. To know the area that required to carry the calculated loads and to make sure of it. 1. Inner Column Q ultimate (one way) = 12.3 KN/m2 (two floor) H= 3m Pin – Pin connected Fc = 25 MPa for concrete / Fy = 420 MPa for steel bar A = (2.5 + 2.25) * (1.75 + 1.75) A= 16.875 m2 Pu = Q Altamont * Area * Num. floor * 1.1 Pu = 12.3 * 16.875 * 2 * 1.1 Pu = 456.63 KN (Amount of load acting on the column) Pn = 702 * 103

Pn = Pu / 0.65

702 * 10 3 = 0.8 ( 0.85 * 25 * ( Ag – 0.02 Ag) + 0.02 Ag * 420) 702* 10 3 = 23.38 Ag Ag = 30025 mm2 Therefore, according to the square root the columns dimension is 17 *17 cm, but I’m going the least dimension for the columns, which is 25 * 25 cm. To check the stability for the columns 𝛾 = KL/r 𝛾 = 1 * 3 /0.3 * 0.25 𝛾 = 40 (which is not stable and slender, therefore I’m going to check the P critical and buckling factor) P cr = π2 * EI / (kL)2 P cr = π2 * 4700 * √25 * 0.25 * bh 3 / 12 (KL)2 P cr = 41944 KN As P cr > Pn, therefore the design of the column safe and the dimension for all inner column within the project will be (25*25 cm) C3

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2. Edge Column Q ultimate (one way) = 12.3 KN/m2 (two floor) H= 3m Pin – Pin connected Fc = 25 MPa / Fy = 420 MPa A = (2.25 + 2.25) * (1.75) A= 7.875 m2 Pu = Q Altamont * Area * Num. floor * 1.1 Pu = 12.3* 7.875 * 2 * 1.1 Pu = 213.0975 KN (Amount of load acting on the column) Pn = 327 * 103

Pn = Pu / 0.65

327 * 10 3 = 0.8 ( 0.85 * 25 * ( Ag – 0.02 Ag) + 0.02 Ag * 420) 347* 10 3 = 23.38 Ag Ag = 10098 mm2 Therefore, I will use the square root, the columns dimension will be 31 *31 cm. To check the stability for the columns 𝛾 = KL/r 𝛾 = 1 * 3 /0.3 * 0.31 𝛾 = 32 (which is not stable and slender, therefore I’m going to check the P critical and buckling factor) P cr= π2 * EI / (kL)2 P cr= π2 * 4700 * √25 * 0.25 * bh 3 / 12 (KL)2 P cr= 99165 KN As P𝑐𝑟 > Pn , therefore the design of the column safe and the dimension for all edge columns within the project will be ( 31*31 cm ) C2

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3. Corner Column Q ultimate (two way) = 16.78 KN/m2 (two floor) H= 3m Pin – Pin connected Fc = 25 MPa / Fy = 420 MPa A = (2) * (2.5) A= 5 m2 Pu = Q Altamont * Area * Num. floor * 1.1 Pu = 16.78 * 5 * 2 * 1.1 Pu = 184.58 KN (Amount of load acting on the column) Pn = 283 * 103

Pn = Pu / 0.65

283 * 10 3 = 0.8 (0.85 * 25 * (Ag – 0.02 Ag) + 0.02 Ag * 420) 283* 10 3 = 23.38 Ag Ag = 12104 mm2 Therefore, I will use the square root, the columns dimension will be 35*35 cm. To check the stability for the columns 𝛾 = KL/r 𝛾 = 1 * 3 /0.3 * 0.35 𝛾 = 28.5 (which is not stable and slender, therefore I’m going to check the P critical and buckling factor) P cr= π2 * EI / (kL)2 P cr= π2 * 4700 * √25 * 0.25 * bh 3 / 12 (KL)2 P 𝑐𝑟 = 16113 KN As P cr > Pn , therefore the design of the column safe and the dimension for all corner columns within the building will be ( 35*35 cm) C1

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Conclusion As it is showed in the report and attached in the drawing, structural design need to go through different phases to be design and to know each structural element thickness, height and width. All of this depends on calculation and how materials act with each other. However, it’s important for architectural design to help and improve structural designs. One of the factors that affect structural design negatively is sometime architectural design require a lot of open spaces according to certain concept, however this has a huge effect on structural design, as thickness could be bigger, more material use and all effect on the final cost within the project, also it could cause a lot of money and working. On the other hand, architectural design could help structural design to be safe by using pure building shape, or use less of cantilever. When architect and structural engineer understand each other, the design would be in high quality. Low communication between them cause a lot problem, therefore, its important from early stage within the project architect and structural engineer collaborate and understand each other. Also, architect need to not only think of designing a great concept, but also need to think how structural engineer will design. All of that help in the project to become successfully structured and with beautiful concept, as also it will help in the high quality within the project. As it’s important for architect when design to classify their certain detail for structural engineer to understand. The high level of details in architectural drawing help to increase the quality in structural design, which all lead to perfectly structured building within the required quality, and time and cost isn’t wasted because of good details and communications.

References 1. Edmonton Structural Steel Fabrication. 2021. Structural Steel Fabricators & Manufacturing Companies in Edmonton. [online] Available at: <http://northernweldarc.com/> [Accessed 10 February 2021]. Edmonton Structural Steel Fabrication. 2021. Structural Steel Fabricators & Manufacturing Companies in Edmonton. [online] Available at: <http://northern-weldarc.com/> 2. : Downloads.hindawi.com. 2021. [online] Available at: <https://downloads.hindawi.com/journals/tswj/2014/875082.pdf. 3. Cement.org. 2021. Structural Loads. [online] Available at: https://www.cement.org/buildingcodes/structural-design/enhanced-resiliency/structuralloads.

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