Abstract:This project aims to evaluate the ophthalmic hospital in Nablus, which has a mixed structural system of braced

(stone) frames and shear walls .

The evaluation will include static, dynamic, and soil structure interaction analysis, in 3D view, in order to

approach the actual model as much as possible.

Two Methods will be used:

1 .Manual analysis: assuming 1D structural elements, using recommendations of ACI -05, as done in most

engineering offices, (trying to match the reality) .

2 .Program analysis: (assuming 3D views, and 1D, 2D structural elements, using structural analysis program sap

2000 v.12, more close to reality) .

After the checks on the model, by manual analysis and by program, the results will be compared with the as built design details, evaluate of the as built detail will be performed, first for the gravity loads and then for both,

gravity and seismic loads .

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What is evaluation?What is design?

1.1 What is design ?

1.1.1 Introduction:

Reinforced concrete, as a composite material, has occupied a special place in the modern construction of different types of structures due to its several advantages. Italian architect Ponti once remarked that concrete liberated us from the rectangle. Due to its flexibility in form and superiority in performance, it has replaced, to a large extent, the earlier materials like stone, timber and steel. Further, architect's scope and imaginations have widened to a great extent due to its mould ability and monolithicity. Thus, it has helped the architects and engineers to build several attractive shell forms and other curved structures. However, its role in several straight-line structural forms like multistoried frames, bridges,

foundations etc. is enormous.

The design of these modern reinforced concrete structures may appear to be highly complex. However, most of these structures are the assembly of several basic structural elements such as beams, columns, slabs, walls and foundations. Accordingly, the designer has to learn the design of these basic reinforced concrete elements. The

joints and connections are then carefully developed.

Design of reinforced concrete structures started in the beginning of last century following purely empirical approach. Thereafter came the so-called rigorous elastic

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theory where the levels of stresses in concrete and steel are limited so that stress-deformations are taken to be linear. However, the limit state method, though semi-empirical approach, has been found to be the best for the design of reinforced concrete structures. The constraints and applicabilities of the limit state method will be

considered in the design of this project.

1.1.2 Design objectives:

Every structure has got its form, function and aesthetics. Normally, we consider that the architects will take care of them and the structural engineers will be solely responsible for the strength and safety of the structure. However, the roles of architects and structural engineers are very much interactive and a unified approach of both will only result in an "Integrated" structure, where every material of the total structure takes part effectively for form, function, aesthetics, strength as well as safety and durability. This is possible when architects have some basic understanding of structural design and the structural engineers also have the basic knowledge of architectural requirements.

Both the engineer and the architect should realize that the skeletal structure without architecture is barren and mere architecture without the structural strength and safety is disastrous. Safety, here, includes consideration of reserve strength, limited deformation and durability. However, some basic knowledge of architectural and structural requirements would facilitate to appreciate the possibilities and limitations of exploiting the reinforced concrete material for the design of innovative structures.

Before proceeding to the design, one should know the objectives of the design of concrete structures. The objectives of the design are as follows:

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1. The structures so designed should have an acceptable probability of performing satisfactorily during their intended life.This objective does not include a guarantee that every structure must perform satisfactorily during its intended life. There are uncertainties in the design process both in the estimation of the loads likely to be applied on the structure and in the strength of the material. Moreover, full guarantee would only involve more cost. Thus, there is an acceptable probability of performance of structures as given in standard codes of practices of different countries.

2 .The designed structure should sustain all loads and deform within limits for construction and use. Adequate strengths and limited deformations are the two requirements of the designed structure. The structure should have sufficient strength and the deformations must be within prescribed limits due to all loads during

construction and use.

However, sometimes structures are heavily loaded beyond control. The structural engineer is not responsible to ensure the strength and deformation within limit under such situation. The staircases in residential buildings during festival like marriage etc., roof of the structures during flood in the adjoining area or for buildings near some stadium during cricket or football matches are some of the examples when structures get overloaded. Though, the structural designer is not responsible for the strength and deformations under these situations, he, however, has to ensure that the failure of the structures should give sufficient time for the occupants to vacate. The structures, thus, should give sufficient warning to the occupants and must not fail suddenly.

3. The designed structures should be durable. The materials of reinforced concrete structures get affected by the environmental conditions. Thus, structures having sufficient strength and permissible deformations

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may have lower strength and exhibit excessive deformations in the long run. The designed structures, therefore, must be checked for durability. Separate checks for durability are needed for the steel reinforcement and concrete. This will avoid problems of frequent repairing of the structure. 4. The designed structures should adequately resist to the effects of misuse and fire. Structures may be misused to prepare fire works, store fire works, gas and other highly inflammable and/or explosive chemicals. Fire may also take place as accidents or as secondary effects during earthquake by overturning kerosene stoves or lantern, electrical short circuiting etc. Properly designed structures should allow sufficient time and safe route for the persons inside to vacate the structures before they actually collapse.

1.1.3 How to fulfill the objectives? All the above objectives can be fulfilled by understanding the strength and deformation characteristics of the materials used in the design as also their deterioration under hostile exposure. Out of the two basic materials concrete and steel, the steel is produced in industries. Further, it is available in form of standard bars and rods of specific diameters. However, sample testing and checking are important to ensure the quality of these steel bars or rods. The concrete, on the other hand, is prepared from several materials (cement, sand, coarse aggregate, water and admixtures, if any). Therefore, it is important to know the characteristic properties of each of the materials used to prepare concrete. These materials and the concrete after its preparation are also to be tested and checked to ensure the quality. The necessary information regarding the properties and characteristic strength of these materials are available in the standard codes of practices of different countries. It is necessary to follow these clearly defined standards for materials, production,

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workmanship and maintenance, and the performance of structures in service .

1.2 What is evaluation?The notion of evaluation has been around a longtime-in fact, the Chinese had large functional evaluation systems in place for their civil servants as long ago as 2000 B.C. Not only does the idea of evaluation have a long history, but it also has varied definitions. Evaluation means different things to different people and takes place in

different contexts.

Thus, evaluation can be synonymous with tests, description, documentations, or management. Many definitions have been developed, but a comprehensive definition is presented by the joint committee on

standards for educational evaluation (1981) :

“Systematic investigation of the worth or merit of an object”...

This definition centers on the goals of using evaluation for a purpose. Evaluations should be conducted for action-related reasons, and the information provided should

facilitate deciding a course of action.

Over the years, evaluation has frequently been viewed as an adversarial process. Its main use has been to provide “thumbs-up” or “thumbs-down” about a program or project. In this role, it has all too often been considered by program or project Directors as an external imposition

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that is threatening, disruptive, and not helpful to project staff. Our contention is that while this may be true in some situations, this is not the case in all, or even in most, evaluation efforts. In addition, today in contrast to two decades or three ago, the view is gaining around that evaluation should be a tool that not only measures, but

also can contribute to success .

Evaluation can serve many different needs and provide critical data for decision-making at all steps of project development and implementation. Although some people feel that evaluation is an act that is done to a project, if

done well, an evaluation is really done for the project.

It is important to remember that evaluation is not a single thing, it is a process, when done well, evaluation can help inform the managers of the project as it progresses, can serve to clarify goals and objectives, and can provide important information on what is, or is not, working, and

why .

8Construction materialand Load combinationCodes andLoading

2.1 Introduction:Hospitals are considered as one of the most important buildings in the public sector, because of huge number of people, and expensive, important equipments that may exist in hospitals. So, many precautions should be taken before, during, and after design of hospitals in order to avoid the failure caused by additional loads and the natural disastrous circumstances, to achieve health and

economical safety .

In this project, we will evaluate (Sulama Bint Butti) charitable ophthalmic hospital in three stages:

First: static design, considering gravity loads, live and dead. Second: dynamic loads, considering horizontal loads, (such as seismic loads). Third: soil structure interaction, considering the relationship between reactions on columns, and

settlement in footings until convergence .

By these stages, we hope to achieve integrated design depending on scientific knowledge, engineering

judgment, and practical experience .

On the other hand, the engineering offices depend on static design only, and 1D analysis, which neglect the effect of lateral loads, and make inaccurate modeling that is far from the actual representation for the structures,

and can only be justified for conceptual understanding.

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2.2 Description:Evaluation of any project requires providing full information about it, to be able to evaluate in a good way, achieving justice as much as possible, and any lack of information will affect negatively on evaluation. The

following provisions describe the project.

2.2.1 Location description :

Geography: the ophthalmic hospital lies beside Aseera Street, in the northern mountain, in Nablus city. For more details about the location, see location map in appendix

(G1).

Topography: the site nature is mountainous, moderate steep and high variant in elevations. For more details

about hospital land topography, see appendix (G3).

Geology: the site investigation reports for the project are not found. So, the reports of the neighborhood location are considered (reports of the Orphan School). Depending on the mentioned reports, the project is expected to be constructed on hard to medium hard limestone soil with variable bearing capacity depending on the depth of the foundations. As a general recommendation depending on site investigation use bearing capacity of (3kg/cm²), as

shown in appendix (G5) .

2.2.2 Structure description:

Structural description:

The hospital has approximately a uniform grid with spans lengths between (6.5 – 7.5)m, which are constructed by two-way ribbed slab system, (to reduce the effect of large deflections, and economical saving). It has a structural system of braced (stone) frames and shear walls, with a structural separator in the middle of the

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hospital (to divide the building into two structures, increasing the seismic resistance for every one), besides, it is constructed on two levels (see appendix (S1), and

appendix (S2)).

Architecture description : The hospital consists of five stories: a basement (area = 900m²), three floors (total area = 3545m²), and a roof (area = 830m²), with 4.16 m height per storey, as shown in the appendices: for elevations (A1) to (A4) and for plans

(S5) to (S9).

2.3 Construction material :The properties of materials used in this project are not available on a hard copy, so, they will be assumed depending on practice in Palestine and the approximate

known used ones, See appendix (S21).

2.3.1 Physical properties :

These properties are related to the material its self in any situation the material is, such as material densities.

1 .Reinforced and plain concrete: Reinforced concrete is mainly used in slabs, shear walls, columns, stair, and footings. Density of R.C= 2.5 ton/m³. Plain concrete is mainly used in masonry walls between stone and blocks. Density of plain concrete= 2.3

ton/m³ .

2. Stone: Used in external walls for buildings to give beautiful

aspect. Density of stone = 2.6 ton/m³ .

3. Blocks: Used for internal dividing in buildings, external walls, and

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in ribbed slabs. Density of blocks used in ribbed slab = 1.2 ton/m³. Density of blocks

used in internal dividing and external walls = 1.4 ton/m³ .

4 .Steel: Used in concrete reinforcing, especially in tension stresses (where concrete is very weak), and as ties for confinement of concrete and bonding steel. Density of

steel = 7.86 ton/m³ .

2.3.2 Mechanical properties:

These properties are related to behavior of material under forces, (such as, modulus of elasticity, compressive

strength (concrete), and yielding stress (steel)).

1 .Reinforced concrete: Compressive strength of concrete used in slabs, shear walls, columns, and beams is B300 kg/cm². Compressive strength of concrete used in masonry walls B150 kg/cm². Compressive strength of concrete used in cleanness concrete B200 kg/cm².

2 .Steel: Yielding stress (fy) = 4200 kg /cm². Modulus of elasticity (Es) = 200Gpa = 2*106 kg/cm². (ACI 318-05 8.5.2).

Note: concrete investigation has not been found, so no checks of the actual strength of structural members have been done.

2.4 Codes and analysis methods:Engineering is a renewable science depends on physical laws and experimental observations. So, during the last decades, many methods were used to explain the

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behavior of structures, and these methods are still used until a failure in structures occurs, then a new method

appears and so on .

Now, after many developments on engineering methods during the last decades, more accurate methods are

appeared, which have less error than provisions methods.

In the last decades, the experimental knowledge and experience, have been documented and arranged in a uniform way, called codes, so, when these codes depend on the local practice and experience, every country or nation has its private codes, despite, these codes could be used in other nations with carefulness with difference between local practices in different countries. In this project, we will use the following methods and codes for

evaluating :

1. American Concrete institute (ACI-2005).

2. Ultimate design method

2.5 Loadings:Humans were, and are still trying to overcome the natural phenomena in order to have a comfortable life, and one of these phenomena is loading (due to gravity, lateral, etc). Loading controls and affects the affecting of humanitarian activities, due to resulting stresses, which may be larger than the stress capacity of the structure elements and materials, so, loading affect the structure should be considered. Types of loadings considered in design in this

project are as follows :

2.5.1 Vertical loads :

Dead loads:

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Dead loads are constant loads in place, including the weight of the structural elements, (such as slabs, beams, columns, and foundations), and super imposed loads,

(such as filling, tiles, and walls).

Live loads:

Live loads are changeable loads in place like weight of people, furniture, and the house content. Live loads are changeable with the variation of type of structures, for example, in residential buildings, density of people and thus live load is less than in public buildings, (such as

schools, universities, and hospitals) .

2.5.2 Lateral loads:

Earthquake loads: this load affects structure vertically and horizontally. The horizontal component is more dangerous than vertical component because it makes shear forces on the structures. So, it should be considered in analysis to

achieve safety.

Earthquake load effect varies with the magnitude of the earthquake, structural system, soil type, site seismicity,

and structure period .

2.6 Load combinations:The method used in this project is the ultimate design method. This method uses factors of safety for different loads, and these factors are changeable depending on the

type of load.

The load factors are: (ACI 318-05 9.2.1)Wu= 1.4D.L

Wu= 1.2D.L+ 1.6L.L

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Wu= 1.2D.L +1.0L.L ±1.0EWu= 0.9D.L ±1.0E

Where: D.L: Dead load. L.L: live load. E: Earthquake load.

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3.1 Introduction:

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IntroductionDesign Inputs.

Modeling is the language between the engineer and the computer, which helps the engineer to represent the actual structure as he understands, and making the calculations and analysis on the model depending on the

engineering laws and theories.

Since, the model the engineer builds should be as it is in the mind of the engineer, achieving the laws and theories the engineer follows in his work. So, it is necessary to know if the model on the computer achieves the required limitations and considerations, or not, by considering the

following conditions :

1. Compatibility. 2. Equilibrium. 3. Stress-Strain relationships .

3.2 Design Inputs:3.2.1 Introduction:

The first step to build a good model for a structure is to identify the following:

1. Properties of the structure: Such as, grid (columns centers and shear walls), and

geometry .

3. Properties of materials :

A. Physical properties: such as materials densities (plain concrete, reinforced concrete, stone, sand, and tiles). See

2.3.1

B. Mechanical properties: such as compressive strength of concrete, modulus of elasticity, Poisson’s ratio, yields

stress of steel). See 2.3.2

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2. Properties of structural elements (slabs, columns, walls, and beams): Such as, cross-sections and

dimensions .

3.2.2 Grid of center columns:

The first step must be done to represent the structure on sap program is to define grid of columns centers and shear walls in three dimensions (X.Y.Z) as shown in the

table (3-1) below and appendix (S0):

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Table 3-1: (Grid of columns centers and shear walls) Z- grid dataY- grid dataX- grid dataGrid No. Ordina

teGrid IDOrdina

teGrid IDOrdina

teGrid ID

00.00Z10.00170.00A104.16Z20.65160.70B208.32Z31.60151.10C312.48Z42.35146.00D416.64Z53.45138.40E520.80Z64.15129.90F624.13Z76.051111.05G7

--7.351011.40H8--8.05913.50I9--14.85821.00J10--22.35721.60K11--29.10623.70L12--30.00525.375M13--31.30427.00N14--31.85329.10O15--32.25236.60P16--34.05138.75Q17----39.80R18----40.90S19----41.55T20----41.90U21----42.75V22----43.05W23----44.60X24

3.2.3 Structural members:

3.2.3.1 Slabs:

Given the difficulty of representation of ribbed slab using the sap program, as it is constructed, it is better to use an (equivalent solid slab) in order to simplify representation

in away close to reality, see appendix (S11).

► Slab properties:

Concrete strength (fc) = B300 kg/cm² = 300*0.8 = 240 kg/cm².

Yield strength of steel (fy) = 4200 kg/cm².

Elasticity for concrete = 15100* √ fc = 15100 * √240 = 2.34*105 kg/cm².

Density of reinforced concrete (ρ) = 2500 kg/m3.

See figure (3-1) that represents cross section for ribbed slab.

For strip of ribbed slab, figure (3-2):

Y = ∑ A*X / ΣA .

Y = {((0.15)*(0.24)*(0.24/2)) + ((0.06)*(0.55)*(0.06/2+0.24))} = 0.1917m.

{0.15*0.24+ 0.06*0.55}

I rib = ∑ I + ∑ Ad².

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Figure 3-2: (Strip of ribbed slab) Figure 3-1: (Cross section for ribbed slab)

{ = 0.15(*0.24)³( / 12{ + })0.55(*0.06)³( / 12{ + })0.15*0.24((*0.24/2 – )0.1917)²{ + }0.06*0.55((*0.06/2( + )0.24 – 0.1917 ))² .}

I rib = 5.7 *10 -4m4.

I rectangular = bh³/12.

I rib = I rect. 5.7*10 -4 = 0.55*h³ (equivalent) /12 h = 0.2317m .

See figure (3-3) that represents cross section for equivalent solid

slab .

o.w for rib = o.w for rectangular.

o.w for rib = (0.06* 0.55* 2.5) + (0.15 *0.24 *2.5) + (0.24 *0.4 *1.2)

= 0.2877 ton/m.

o.w for slab (ton/m²) = 0.2877/0.55

= 0.523 ton/m².

h (equivalent) * γ (modified) = 0.523 γ (modified) = 2.257 ton/m³.

Thickness and density of slab that had been used in the sap program as result of the use of

the equivalent solid slab is 0.2317m and 2.257 ton/m³, respectively. See figure (3-4) .

► Loads on the slab:

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Figure 3-3: (Cross section for equivalent solid slab)

Figure 3-4: (Definition of area section for slab on sap)

Live loads = 400 kg/m². From table (4-1) in ASCE STANDARD, (Table 4-1 in appendix) .

See figure (3-5) that represents cross section for superimposed load on slab.

Superimposed loads /m² = t * γ tiles + t* γ concrete + t* γ sand. = 0.02*2.6 +

0.05* 2.3 + 0.1* 2 = 0.367 ton/m² .

∴ Superimposed loads = 0.367 ton/m².

Total dead loads on slab = o.w + S.I.D = 0.523 +0.367 = 0.89 ton/m².

Where; t: thickness. , γ: density of material.

3.2.3.2 Masonry walls:

As we know in the process of construction, the masonry walls are made usually of the stone, concrete and block. It is usual practice in our country not to link the stone or the block with concrete (which is located between the two) in

carrying out the masonry wall.

Therefore, the representation of the masonry walls is as follows :

1. The thickness of the masonry walls is the same of the thickness of the concrete layer (which is located between the stone and blocks), and therefore, the

concrete layer in the masonry walls is a carrier layer.

2. The block and stone are considered an additional load (superimposed load) located on the masonry walls.

3. The thickness of the isolation is neglected.

► Masonry wall properties :

Concrete strength (fc) = B150 kg/cm² = 0.8* 150 = 120 kg/cm².

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Figure 3-5: (Cross section for superimposed load on slab)

Elasticity for concrete (Ec) =15100 *√ fc = 15100 * √120 = 1.65*105 kg/cm² .

Density of concrete (ρ) =2300 kg/m³.

Thickness of concrete layer = 12cm.

See figure (3-6) that represents cross section for masonry walls. (Return to appendix (S20)).

► Loads on masonry walls :

γ block= 1.2 ton/m³ , γ stone = 2.6 ton/m³.

Superimposed load (resulting from the stone and block)

=0.05(t )*2.6( γ for stone) + 0.1(t) * 1.2( γ for Block) = 0.25ton/m² .

∴ Superimposed load on masonry walls = 0.25ton/m² .

3.2.3.3 Shear walls and stair walls :

The rest of the walls in this building are shear walls (including stair walls), which is different in thickness and

can be categorized into the following :

1 .Shear walls #1 (s.w.1), thickness = 20cm.

2. Shear walls #2 (s.w.2), thickness = 25cm.

3 .Shear walls #3 (s.w.3), thickness = 30cm.

Note: most of the shear walls in this building of (s.w.1).

See figure (3-7) that represents cross section for shear walls .

► Shear walls properties:

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Figure 3-6: (Cross section for masonry walls)

Concrete strength fc = B300 = 300*0.8= 240kg/cm².

Yield strength of steel (fy) = 4200kg/cm².

Elasticity for concrete (Ec) = 15100 * √ fc = 15100 * √240 = 2.34*105 kg/cm² .

Note: the thickness of plaster is neglected.

3.2.3.4 Columns:

This building contains (33) columns, divided into three forms, as it is distributed in regular grid as shown in

appendices (S0) and (S4).

Note: since it is difficult to represent the column form (C2) as frame section, it has been representing as an area

section .

See figure 3-8, that represents cross section for columns.

3.2.3.5 Beams:

The construction planes (which were obtained by an engineering office who has designed this building) contain

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Figure 3-7: (Cross section for shear walls)

Figure 3-8: (Cross section for columns)

C1 C2 C3

several forms of beams as shown in appendices: (S13), (S14), (S15), (S16), and (S18).

In sap, beams are represented as shown in the table (3-2) and as in figure (3-9) below:

Density (ton/m³)

Width(m)

Depth( m)

Name

2.50.50.3B/30/502.50.60.3B/30/602.50.70.3B/30/702.50.400.45B/45/402.50.600.45B/45/602.50.200.50B/50/202.50.300.60B/60/302.50.400.60B/60/402.50.600.60B/60/602.50.301B/100/30

3.2.4 Load cases:

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Table 3-2: (Properties of beams sections that represented on sap)

Figure 3-9: (Beams and columns properties that represented on sap)

The sap program is able to calculate the weight of construction elements that are represented, and it’s based on the properties of structural elements that are introduced to the program, and we can do this through define loads list and put number 1 for dead case in

self weight multiplier as shown in Figure (3-10):

3.2.5 Load combinations:

Buildings and other structures, and parts thereof, shall be designed and constructed to support safely the factored loads in load combinations defined

in ACI-05code. The load combination that has been used in manual and program solutions is:

Wu= 1.2D.L+ 1.6L.L

Moreover, this combination can be defined on sap program as shown in Figure

(3-11):

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Figure 3-11: (Load combination)

Figure 3-10: (Load cases)

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CompatibilityStress-Strain RelationshipsEquilibrium

4.1 Introduction :

To make sure that the model is close to the actual structure and satisfies the physical and engineering laws and theories, we have to check on the conditions of

modeling mentioned in 2.1, which are as follow:

4.2 Compatibility :

This requires that the structure behave as one block, since, when the structure in reality is supposed to loads, its structural elements move with each others, and do not move each one in different direction, so the computerized model should achieve compatibility, to be more approach

to reality .

For checking compatibility, run the computerized model, press start animation button on the down right side of the

window as Figure (4-1) below :

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4.3 Equilibrium:

4.3.1 Introduction:

As known in the physical, engineering, and creation laws, the bodies tend to be in the lowest energy position, which leads to the equilibrium equations, such as for the static bodies, the summation of forces in any direction must be zero. So, for the model, the summation of the loads applied on the structure must equal the reaction from the columns to achieve the equilibrium, but because of the errors occur from human and the computer, accuracy

greater than or equal 95% is allowed .

4.3.2 Manual solution:

A. Dead loads calculations : Area of structure that is entered to sap = 1042.23 m²

1. Weight of one-meter strip of the rib:

0. 06 (t for slab) *0.55( h for slab ) *2.5( γ for slab ) + 0.15( t for rib )*0.24( h for

rib )*2.5( γ for rib ) + 0.24( t for block )*0.4( h for block )*1.2( γ for block ) = 0.2877 ton / m. 0.2877 / 0.55 = 0.523 ton / m².

Where; t: thickness, / h: height of section, / γ: density of material .

∴ Total own weight for the slab = 0.523*1042.23 = 545 ton.

2. Super imposed weight:

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Figure 4-1: (Compatibility)

Own weight of 1 m² of slab = 0.02( t ) * 2.6( γ for tiles ) + 0.05( t ) * 2.3( γ for concrete ) + 0.1( t ) * 2( γ for

sand ) = 0.367 ton / m².

Where; t: thickness. / γ: density of material .

∴ Total super imposed weight = 0.367*1042.23 = 382.5 ton.

3. The own weight of beams:

Weight (ton)

Density(ton/m³)

length(m)

Width(m)

Depth(m)

BeamID

7.2672.519.380.50.3B/30*50

2.7672.506.150.60.3B/30*60

15.7502.530.000.70.3B/30*70

8.7752.519.500.400.45B/45*40

10.1252.515.000.600.45B/45*60

1.8752.507.500.200.50B/50*20

6.7502.515.000.300.60B/60*30

181.6802.5302.800.400.60B/60*40

4.8602.505.400.600.60B/60*60

4.8002.506.400.301B/100*30

∑= 244.65

∴ Total own weight for the beams = 244.65 ton .

WeightDensitNumbDimensiShapeSectionColu

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Table 4-1: (Properties and own weight of beams sections represented on sap)

(ton)y(

ton/m³)

erons(m)

mn ID

108.57602.5290.60*0.60SquareC1

15.91202.531*1*0.3L-sectionC2

3.122.511*0.30Rectangular

C3

=∑127.608

4. The own weight of columns:

∴ Total own weight for the columns = 127.608ton.

5. The own weight of shear walls:

Weight(ton)

Density(ton / m³)

Net area(m²)

Thickness(m)

Wall ID

144.9762.5289.9520.2S.W.147.36162.563.14880.3S.W.3

=∑192.3286

∴ Total own weight for the shear walls = 192.3286 ton.

6 .The own weight of masonry walls:

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Table 4-2: (Properties and own weight of columns sections represented on sap)

Table 4-3: (Properties and own weight of shear walls sections represented on sap)

Own weight for one-squared meter including stone, block, and concrete

=0.05 (t for stone) *2.6( γ for stone ) + 0.12( t for plan concrete )*2.3( γ plan concrete )

+ 0.1( t for block )*1.2( γ for block ) = 0.526 ton / m².

Where; t: thickness. / γ: density of material .

Net area = 394.85 (without the area of windows included in stone walls).

∴ Total own weight for the masonry walls = 394.85 * 0.526 = 207.69 ton.

Summation of dead loads:

∑ D.L = weight of slab + weight of super imposed on slab + weight of beams + weight of columns + weight of shear walls + weight of masonry walls. = 545 + 382.5 + 244.65 + 127.61 + 192.33 + 207.69 = 1699.78 ton.

B. Live loads calculations : Live loads on slab = 0.4ton/m².

∴ Total live loads weight = 0.4*1042.23 = 416.89ton .

4.3.3 Sap program solution:

The values obtained from sap program for dead and live loads as figure (4-2):

31

Figure 4-2: (Results of sap program)

4.3.4 Percentage of errors:

D.L (manual) = 1699.78 ton , D.L (sap) = 1705.59 ton

%of error = (1705.59 - 1699.78)/ 1705.59 = 0.34%

L.L (manual) = 416.89 ton , L.L (sap) = 416.23 ton

%of error = (416.89 - 416.23)/416.23 = 0.16%

4.4 Stress-Strain relationships:

4.4.1 Introduction:

According to the physical meaning, for the integration of applied load (as a function of distance) on a structural element, gives the function of shear, and the integration of the shear (as a function of distance) gives the function of moment. In addition, the stress strain diagram for a reinforced concrete section, a formula for moments could be formed, which is considered as the nearest formula to

32

the actual status. Therefore, the moment values given from the computerized model should be closely equal the moment given from the formula, with allowed error not

greater than 5% .

4.4.2 Frame along y-direction as Figure (4-5) :

I slab (in figure 2-14) = (b) * (h) ³/ (12) = (7.5) * (0 .2317) ³/ 12 = 0.007774 m4 .

Y = ∑AX / ∑A.

={ 0.54 * 0.4 (* 0.27 + ) 1.48 * 0.06 (* 0.54 .+ 03 }) { 0.54*0.4 + 1.48*

0.06 }

Y = 0.357 m.

I beam = ∑ I + Ad ².

I beam = {0.4*(0.54) ³/12 + 1.48*(0.06) ³/12 + (0.54*0.4)*(0.357-0.27) ² + 1.48*0.06*(0.57-0.357) ²} =

0.0109m4 .

∝ = I beam / I slab = 0.0109 / 0.007774 = 1.4.

Dead load on slab = o.w + S.I.D = 0.523 + 0.367 = 0.89ton/m².

Live load on slab = 0.4ton/m².

Wu = 1.2 D.L + 1.6 L.L = 1.2*(0.89) + 1.6*(0.4) = 1.71ton/m².

Mo = Wu * l2 * (ln) ²/ 8 = 1.71 * 7.5 * (6.9) ²/ 8 = 76.325ton .m.

33

Figure 4-5: (Frame along y-direction)

Figure 4-3: (Cross section for slab) Figure 4-4: (Cross section for beam)

∎ For M (negative) :

M beam = 0.85*0.75*0.65*Mo = 0.85*0.75*0.65*76.325 = 31.63 ton .m.

M col.strip = 0.15*0.75*0.65*Mo = 0.15*0.75*0.65*76.325 = 5.58 ton .m.

∴ M col.strip = 5.58 / (3.75-1.48) = 2.46 ton.m /m.

M mid.strip = 0.25*0.65*Mo = 0.25*0.65*76.325 = 12.4 ton.m.

∴ M mid.strip = 12.41 / 3.75 = 3.31 ton.m / m.

∎ For M (positive) :

M beam = 0.85*0.75*0.35*Mo = 0.85*0.75*0.35*76.325 = 17 ton.m.

M col.strip = 0.15*0.75*0.35*Mo = 0.15*0.75*0.35*76.325 = 3 ton.m.

∴ M col.strip = 3 / (3.75-1.48) = 1.324 ton.m / m.

M mid.strip = 0.25*0.35*Mo = 0.25*0.35*76.325 = 6.68 ton.m.

∴ M mid.strip = 6.68 / 3.75= 1.8 ton.m / m.

Results from sap program:

∎ For M (negative) :

M beam = (30.64+33.7) /2 = 32.17 ton.m.

M col.strip = 5.1 ton.m.

M mid.strip = 4.1 ton.m.

See Figure 4-6.

∎ For M (positive) :

34

30.6433.7

15.2

M beam = 15.2 ton.m.

M col.strip = 2 ton.m.

M mid.strip = 2.3

I slab (in figure 2-18) = (b) * (h) ³/ (12)

( =7.5( * )0. 2317) ³ /12

=0.007774 m4.

I beam (in figure 2-19) = ∑ I + Ad ²

Y = ∑AX / ∑A.

={ 0.54 * 0.4 (* 0.27 + ) 1.48 * 0.06 (* 0.54 .+ 03 }) { 0.54*0.4+

1.48*0.06}

Y = 0.357 m.

I beam = ∑ I + Ad ².

35

Figure 4-9: (Frame along x-direction)

Figure 4-7: (Cross section for slab) Figure 4-8: (Cross section for beam)

Figure 4-6: (Moment values for beam from sap)

Figure 2-16: (frame along y-direction)

I beam = {0.4*(0.54) ³/12 + 1.48*(0.06) ³/12 + (0.54*0.4)*(0.357-0.27) ² + 1.48*0.06*(0.57-0.357) ²}

= 0.0109 m4 .

∝ = I beam / I slab = 0.0109 / 0.007774 = 1.4.

Dead load on slab = o.w + S.I.D = 0.523 + 0.367 = 0.89 ton / m ².

Live load on slab = 0.4 ton / m ².

Wu = 1.2 D.L + 1.6 L.L = 1.2*(0.89) + 1.6*(0.4) = 1.71 ton /m ².

Mo = Wu * l2 * (ln) ²/ 8 = 1.71 * 7.5 * (6.9) ²/ 8 = 76.325 ton .m.

∎ For M (negative) :

M beam = 0.85*0.75 *0.65*Mo = 0.85*0.75*0.65*76.325 = 31.63 ton .m.

M col.strip = 0.15*0.75*0.65*Mo = 0.15*0.75*0.65*76.325 = 5.58 ton .m.

∴ M col.strip = 5.58 / (3.75 -1.48) = 2.46 ton .m/m.

M mid.strip = 0.25*0.65*Mo = 0.25*0.65*76.325 = 12.4 ton .m.

∴ M mid.strip = 12.4 / 3.75 = 3.3 ton .m / m.

∎ For M (positive) :

M beam = 0.85*0.75*0.35*Mo = 0.85*0.75*0.35*76.325 = 17 ton .m.

M col.strip = 0.15*0.75 *0.35*Mo = 0.15*0.75*0.35*76.325= 3 ton .m.

∴ M col.strip = 3 / (3.75-1.48) = 1.32 ton .m / m.

M mid.strip = 0.25*0.35*Mo = 0.25*0.35*76.325 = 6.68 ton .m.

36

∴ M mid.strip = 6.68 / 3.75 = 1.8 ton.m/m.

Results from sap program:

∎ For M (negative) :

M beam = (34.24+29) /2 = 31.6 ton.m.

M col.strip = 4.1 ton.m.

M mid.strip = 3.27 ton.m.

See Figure 4-10.

∎ For M (positive) :

M beam = 15.75 ton.m.

M col.strip = 2 ton.m.

M mid.strip = 2.26 ton.m.

4.5 Discussion:

Checks on the model have been done, to ensure that the model is close to reality as much as possible, but some relatively high percents of errors appeared, and some

37

Figure 4-10: (Moment values for beam from sap)

34.2429

15.75

decisions must be made, so, these things will be discussed in this section .

1 .Compatibility:

The model indicates a good compatibility, and its members move with each other, so, percent of error is so

little .

2 .Equilibrium:

Equilibrium check of the model indicates an error of 0.34% and 0.16% in dead and live loads respectively, which are considered as small and excepted values, which means

that the model is close to reality.

3 .Stress-Strain relationships:

Stress –Strain relationship is more difficult check compared with others, because of the large difference between values of 1D and 3D model, which usually

appears during check.

Engineering judgment is an effective factor in this check, as an engineer, you have to decide whether trust the results or not, according to the engineering sense, so, decisions depend on the sense, which means somehow , a

difficult decision for most of engineers.

Making decisions should be carefully, trusting a model or not, may mean trusting of life of citizen or not, so, modeling trust is a life trust, so, engineer should take all

the precautions and care during modeling check.

38

5.1 Introduction

:Many years ago, the manual static analysis and design was the basic method that followed, until the new technology arose and changed many methodologies, so, in this chapter classical manual static analysis and design

39

Slabs And FootingsColumns And BeamsStairs

will be covered, not as a basic, but as manual checks on sap modeling and planning designs .

5.2 Compatibility:

5.3 Equilibrium:Since there are two separate models, checks on both are required separately; in this section the equilibrium including dead and live loads will be covered, in order to check sap modeling, and shows the results acceptance due to actual status and

modeling.

Weight of live

Weight of

masonr

Weight of

shear

Weight of colu

Weight of

beam

Weight of

super

Weight of slab

AreaStorey

40

Figure 5-2: (Compatibility for block B {left building})Figure 5-3: (Compatibility for block A {right building})

Figure 5-1: (Compatibility for building {block A + B})

loadsy wallswallsmnssimposed

(ton)(ton)(ton)(ton)(ton)(ton)(ton)(m2)

305.80109.25325.2791.75180.10280.50399.80764.50B416.89207.69192.33127.61244.65382.50545.001042.2

3G

416.89207.69192.33127.61244.65382.50545.001042.23

F1

416.89207.69192.33127.61244.65382.50545.001042.23

F2

310.10167.53224.80101.40170.220.00405.45775.25Roof1866.5

7899.851127.06575.98

1084.27

1428.00

2440.254666.4

4Sum

5.3.1 Manual solution:

A. Dead loads calculations: ∑ D.L = weight of slab + weight of super imposed on slab + weight of beams + weight of columns + weight of shear walls + weight of masonry walls. = 2440.25 + 1428.00 + 1084.27+ 575.98 + 1127.06 + 899.85 = 7555.40 ton.B. Live loads calculations : Live loads on slab = 0.4ton/m².

∴ Total live loads weight = 0.4 * 4666.44 = 1866.57 ton .

5.3.2 Sap program solution:

41

Table 5-1: (Dead and live loads for building storeys)

The values obtained from sap program for dead and live loads as shown in figures (5-4) and (5-5):

5.3.3

Percentage of errors:

D.L (manual) = 7555.4 ton. , D.L (sap) = (3683.79 + 3845.25) = 7529 ton.

% of error = (7555.4 - 7529) / 7529 = 0.35%.L.L (manual) = 1866.57 ton. , L.L (sap) = (888.281 +

960.078) = 1848.36 ton.% of error = (1866.57 – 1848.36) /1848.36 = 0.98%.

5.4 Design and evaluation of slabs:5.4.1 Introduction:

42

Figure 5-5: (Results of dead and live loads from sap program for block B {left building})

Figure 5-4: (Results of dead and live loads from sap program for block A {right building})

Floor slabs are the main horizontal elements that transmit the moving live loads as well as the stationary dead loads to the vertical framing supports of a structure.They can be slabs on beams, waffle slabs, slabs without beams (flat plates), resting directly on columns, or composite slabs on joists.They can be proportioned such that they act in one direction (one-way slab), or proportioned so that they act in two perpendicular direction (two-way slabs and flat plates).

5.4.2 Design for flexure:In this project the two way ribbed slabs is used, see appendices S5 to S9.Mo = 76.325ton .m. (See page 29)

∎

For M (negative) :

M col.strip = 2.46 ton .m/m. (See page 29)∴ M col.strip / rib = 0.55*2.46 = 1.353 ton.m / rib.

ρ = 0.85 fcfy [1−√1−2.61∗10 ⁵∗Mu

0.9∗b∗d ²∗fc ]ρ = 0.85∗240

4200 [1−√1− 2.61∗10 ⁵∗1.3530.9∗15∗26 ²∗240 ] = 0.0041 < 0.002(ρ min) →

Use ρ

43

Figure 5-6: (Section for two way ribbed slab)

As (-ve) = ρ * b * d = 0.0041 * 15 * 26 = 1.6 cm2. → (Use 2Φ14mm/rib)

M mid.strip = 3.31 ton.m/m. (See page 29)∴ M mid.strip / rib = 0.55 * 3.31 = 1.82 ton.m / rib.

ρ = 0.85∗2404200 [1−√1− 2.61∗10 ⁵∗1.82

0.9∗15∗26²∗240 ] = 0.0056 > 0.002 → Use ρ

As (-ve) = ρ * b * d = 0.0056 * 15 * 26 = 2.184 cm2. (Use 2Φ14mm/rib)

∎ For M (positive) : M col.strip = 1.324 ton.m. (See page 29)

∴ M col.strip / rib = 0.55 * 1.324 = 0.73 ton.m/rib.

ρ = 0.85∗2404200 [1−√1− 2.61∗10 ⁵∗0.73

0.9∗55∗26²∗240 ] = 0.00058 < ρ min → Use ρ min

As (+ve) = ρ * b * d = 0.002 * 55 * 26 = 2.86 cm2. → Use 2Φ14mm/rib

M mid.strip = 1.8 ton.m. (See page 29)

∴ M mid.strip / rib = 0.55 * 1.8 = 0.99 ton.m/rib.

ρ = 0.85∗2404200 [1−√1− 2.61∗10 ⁵∗0.99

0.9∗55∗26²∗240 ] = 0.00079 < ρ min → Use ρ minAs (+ve) = ρ * b * d = 0.002 * 55 * 26 = 2.86 cm2. → Use 2Φ14mm/rib

The area of steel needed for moment on slab from planning, (See appendix S11)

44

∎ For col. strip: ∎ For mid. strip :

As (-ve) = 1Φ16mm/rib As (-ve) = not needed .

As (+ve) = 1Φ16mm/rib. As (+ve) = 1Φ16mm/rib .

........................................................................

................................5.4.3 Check for shear:Wu = 1.2 * {(0.367 * 0.55) + (0.24 * 0.4 * 1.2) + 2.5 * (0.15 * 0.24 + 0.06 * 0.55)} + 1.6 * (0.4 * 0.55) = 0.94 ton/m. Max. shear (Vu at the center of support) = 4.23 ton. Vu at x from the center of support = Vu – x *Wu = 4.23-(0.26+0.3)*0.94= 3.7ton.Vn = Vu / Φ = 3.7 / 0.75 = 4.93 ton. Vc = {1.1*0.53*√ fc*bw * d}/ 1000 = {1.1*0.53*√240* 15*26}/1000=3.52 ton. Vc < Vn → shear reinforcement is required.

5.5 Design and evaluation of beams:5.5.1 Introduction:Beams are the structural elements that transmit the tributary area loads from floor slabs to vertical supporting columns. They are normally cast monolithically with the slab and are structurally reinforced on one face, the lower

45

Figure 5-7: (Shear force diagram for rib)

tension side, or both the top and bottom faces. As they are cast monolithically with the slab, they form a T-beam section for interior beams or an L-beam at the building exterior. The plan dimensions of a slab panel determine whether the floor slab behaves essentially as a one-way or two-way slab.

5.5.2 Design for flexure:We use direct design method for manual design of beam:Dimensions for this beam = (60*40 * 750) cm.

∎

For M (negative):

(See page 29)

M beam = 0.85*0.75*0.65*Mo = 0.85*0.75*0.65*76.325 = 31.63 ton .m.

∎ For M (positive): (See page 29)M beam = 0.85*0.75*0.35*Mo = 0.85*0.75*0.35*76.325 = 17.03 ton.m.

See figure 5-9 that display the bending moment and shear force diagrams from sap program.

46

Figure 5-8: (Location of selected beam in ground floor)

∎ Percentage of errors:Negative moment on beam (manual) = 31.63 ton.m. Negative moment on beam (sap) = (32.43+29)

2 = 30.72 ton.m.

% of error = (31.63−30.72)30.72 = 2.96 %.

Positive moment on beam (manual) = 17.03 ton.m. Positive moment on beam (sap) = 16 ton.m.

% of error = (17.03 –16.00)16.00 = 6.43 %.

ρ = 0.85 fcfy [1−√1− 2.61∗10 ⁵∗Mu

0.9∗b∗d ²∗fc ]

ρ (+ve) = 0.85∗2404200 [1−√1− 2.61∗10 ⁵∗17.03

0.9∗40∗52 ²∗240 ]=0.00484 .

ρ (min) = max. of { 14fy

= 144200

=0.003

0.8√ fcfy

=0.8 √2404200

=0.0029

As (-ve) = ρ*b*d = 0.00951 * 40 * 52 = 19.8 cm2. → Use 10Φ16mm.As (+ve) = ρ*b*d = 0.00484 * 40 * 52 = 10 cm2. → Use 5Φ16mm.......................................................................................................................................Area of steel for this beam from planning:

47

Figure 5-9: (Bending moment and Shear force diagrams from sap program)

As (-ve) → Use 4Φ16mm and 7Φ12mm. okAs (+ve) → Use 9Φ16mm. okSee Figure 5-10 that display section in beam.Area of steel for this beam from sap program:As (-ve) → Use 12Φ14mm. As (+ve) → Use 6Φ16mm.

Positive

moment

Negative moment

Beams in

horizontal

direction

Positive

moment

Negative moment

Beams in

vertical directio

n

RightLeftRightLeft

1.10000.0000.00AH 11.4800.1400.200AV 1 1.9002.44000.00B1.7900.3601.150B11.3010.5611.75C1.6600.3300.950C0.4601.29000.00D0.5600.1504.410AV 26.41011.7614.00AH 215.6019.8027.00B16.3833.9728.80B14.9029.0030.00C14.8530.1629.72C14.9225.0027.67D

6.40011.6314.34AH 3 W17.3531.0027.71AV 3

16.4032.2729.41B13.5222.0026.14B14.9530.9428.83C16.3026.1433.00C

1.1000.19000.00AH 4 W00.0012.8000.00D

2.7003.3701.000B10.8317.2017.30AV 40.81000.000.410C6.51010.3210.62B

0.5200.36000.00AH 4 E9.70014.6017.37C

0.5400.3500.310B00.009.30000.00D0.8301.35000.00C10.4417.3816.88AV 5

18.2033.5529.50AH 3 E8.91615.56B

15.9529.4631.96B9.8515.115.9C10.9517.2519.60C17.9530.6433.71AV 6

18.0533.9027.50AH 2 E16.0029.0032.43B

13.5527.3028.90B15.9529.9025.62C

48

Figure 5-10: (Section in beam)

11.6515.7322.25C1.3309.3101.860D

3.3107.3002.550AH 1 E16.3027.9930.40AV 7

13.7125.9622.87B14.6826.3228.19B10.429.47021.16C9.35018.9518.26C

3.4109.2504.720D0.7400.6400.260AV 81.2700.7300.220B0.3100.2102.440C

0.37000.000.150B 33.1403.7404.360B1-A00.002.00000.00B2-E2.7004.0004.000B1-B00.001.0001.060B2-F0.7901.3701.600B2-A0.21000.000.960B2-G0.16000.0000.00B2-B2.2001.6803.200B1-C0.26000.001.160B2-C3.0302.2003.700B1-D0.2100.48000.00B2-D

See table 5-3 and appendices from S13 to S19 for more details about area of steel and distribution of bars from sap and plans.

#of bars (plannin

g)

#of bars (sap)

As min

Beams in

horizontal

direction

#of bars (plannin

g)

#of bars (sap)

As min

Beams in

vertical

direction +(

ve)-(ve)

+(ve)

-(ve)

+(ve)

-(ve)

+(ve)

-(ve)

9ф16

12ф12

4ф16

7ф12

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9ф16

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5ф14

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5ф14

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9ф16

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9ф16

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9ф16

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5ф16

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5ф14

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9ф16

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49

Table 5-2: (Beams analysis results from sap program for ground floor)

Table 5-3: (Area of steel for beams from sap and planning for ground floor)

14ф16

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50

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4.29

B2-C12ф14

7ф12

4ф14

6ф12

6.006

B1-D7ф12

5ф124ф1

2

4ф12

4.29

B2-D

5.5.3 Check for shear:Dimensions for this beam = (60*40 * 750) cm.See Figure 5-11 that display the tributary area for this beam.D.l on beam = o.w for beam + D.l for slab.D.l on slab = (super imposed + o.w for 1m strip of slab)*tributary distance for beam = (0.367 + 0.523) * 7.5/2 = 3.3375 ton/m.

o.w for beam = 0.6 * 0.4 * 2.5 = 0.6 ton/m.Total D.l = 0.6 + 3.3375 = 3.94 ton/m.L.L on beam = 0.4 * 3.75 = 1.5 ton/m.Wu = 1.2D.l + 1.6L.L = 1.2 * 3.94 + 1.6 * 1.5 = 7.128 ton/m.

51

Max. shear (manual) = 22 ton.Vu at center of support = 22 ton.Vu at distance d from the face of support = Vu - {d + (c/2)} * Wu = 22 – {0.52 + 0.3} * 7.128 = 16.15 ton.Vn = Vu/Φ = 16.15/0.75 = 21.54 ton.

Vc = 0.53∗√ fc∗bw∗d1000 = 0.53∗√240∗40∗52

1000 = 17 ton

Vn > Vc then we use shear reinforcement.Vs = Vn – Vc = 21.54 – 17 = 4.54 tonAvS

= Vsfy∗d

=4.54∗10004200∗52

=0.0208 .

{AvS }min = max. of { 3.5∗bwfy

=3.5∗404200

=0.0333

0.2√ fc∗bwfy

=0.2√240∗404200

=0.0295

{AvS }min = 0.0333 > 0.0208 → Use {AvS }min = 0.0333.

Use stirrups Φ8mm → Av = 2 * 0.5 = 1.00 cm2

{AvS }= 1S=0.0333 → S = 30 cm

S max. = min. of {d2=522

=26cm

60cm

S = 30 cm > S max. → use stirrups 1Φ8mm/25cm.∴ Spacing of stirrups for this beam from manual solution:

52

Figure 5-11: (Tributary area for beam)

Use stirrups 1Φ8mm/25cm.∴ Spacing of stirrups for this beam from planning:Use stirrups 1Φ8mm/17cm.

S used

(cm

)

S calc-

ulate

(cm)

S max

[ d2 ](

cm)

AvS(

min)

AvS

Beams in

horizontal

direction

S used(

cm)

S calc-

ulate (cm)

S max

[ d2 ](

cm)

AvS(

min)

AvS

Beams in

vertical

direction

172527

0.033

0.005AH 1172527

0.033

00.00AV 1

172527

0.033

00.00B

172527

0.033

00.00B

172527

0.033

0.004C

172527

0.033

00.00C

171213

0.050

0.014D

171213

0.050

00.00AV 2

1720200.00.03AH 21725270.030.0B

53

33333317

25270.033

0.049B

172327

0.033

0.043C

171220

0.050

0.078C

172527

0.033

0.033D

172020

0.033

0.033A

H 3 W

172746

0.025

0.036AV 3

172527

0.033

0.040B

172527

0.033

0.031B

171220

0.050

0.082C

172527

0.025

0.030C

172527

0.033

00.00A

H 4 W

172527

0.033

00.00D

172527

0.033

00.00B

173046

0.025

00.00AV 4

172527

0.033

00.00C

172527

0.033

00.00B

172527

0.033

00.00A

H 4 E

172527

0.025

00.00C

172527

0.033

00.00B

172527

0.033

00.00D

172527

0.033

00.00C

172527

0.033

00.00AV 5

172027

0.033

0.046A

H 3 E

172527

0.033

00.00B

172027

0.033

0.044B

172527

0.033

00.00C

172527

0.033

0.033C

172027

0.033

0.052AV 6

172027

0.033

0.046A

H 2 E

172027

0.033

0.049B

172527

0.033

0.033B

172527

0.033

0.035C

172527

0.033

0.026C

172527

0.033

00.00D

172027

0.050

00.00A

H 1 E

172527

0.033

0.037AV 7

172527

0.033

0.033B

172527

0.033

0.027B

172527

0.033

0.015C

171520

0.033

0.033C

172527

0.033

0.033D

1725270.033

00.00AV 8

1725270.033

00.00B

54

1725270.033

00.00C

1725270.016

00.00B 3

1712130.058

00.00B1-A

1712130.042

00.00B2-E

1712130.058

0.037B1-B

1712130.042

00.00B2-F

1712130.042

0.014B2-A

1712130.042

0.022B2-G

1712130.042

00.00B2-B

171213

0.058

00.00B1-C

1712130.042

00.00B2-C

171213

0.058

00.00

B1-D1712130.042

00.00B2-D

5.6 Design and evaluation of columns:5.6.1 Introduction:They are vertical compression members supporting the structural floor system and subjected in most cases to both bending and axial load and are of major importance in the safety considerations of any structure. They transmit loads from the upper floor to the lower levels and then to the soil throw the foundation. Since columns are compression elements failure of one column in a critical location can cause the progressive collapse of the adjoining floor and ultimate total collapse of the entire structure. If a structural system is also composed of horizontal compression members, such members would be considered as beam-columns.

5.6.2 Design critical column for axial load (Col. I-7):

55

Table 5-4: (Spacing of stirrups for beams from sap and planning for ground floor)

∎ Manual solution: (See appendix S0 that display the critical col. I-7)o.w for slab (ton/m²) = 0.523 ton/m². (See page 17)Superimposed loads = 0.367 ton/m². (See page 17)Tributary area for column without openings= 7.5 * 7.5 * 5(# of storey) = 281.25 m2.Area of openings = 21.98 m2.Net area = 281.25 – 21.98 = 259.27 m2.Tributary area for column from roof = 50.755 m2.Dead load due to tributary area for column from roof = 50.755 * 0.523 = 26.55 ton.Dead load due to tributary area for column from other floors = 208.515 * (0.523+0.367) = 185.58 ton.Dead load on column from own weight of slab = 26.55 + 185.58 = 212.13 ton.Dead load on column from beams = 51.715 ton.Dead load of column (own weight) = 0.6 * 0.6 * 2.5 * 4.16 * 5 = 18.72 ton.Total dead load on column = 212.13 + 51.715 + 18.72 = 282.57 ton.Total live load on column = net area * 0.4 = 259.27 * 0.4 = 103.7 ton.Service load = D.L + L.L = 282.57 + 103.7 = 386.3 ton.∴ Reaction Fz (Wu) = 1.2*D.L + 1.6*L.L = 1.2 * 282.57 + 1.6 * 103.7 = 505 ton.

56

∴ Reaction Fz from sap = 526 ton.∎Percentage of errors:Reaction Fz (manual) = 505ton. , Reaction Fz from sap = 526 ton. % of error = (526 - 505) / 526 = 4%.……………………………………………………………………………Since we have a continuity of column in vertical and horizontal direction, the fixed model shown is near to the actual one, see figure 5-12 that display different cases of effective length factor.But it is unable to have very stiff (fixed) restraints in practice, then take K= 0.7

Dimensions of column from plans and design drawings = (0.6 * 0.6 * 4.16) m.r =√( I /A ) = √{(0.6)∗(0.6) ³ /12¿/(0.6) ² }¿ = 0.173 m.

M1b/M2b ≈ 0 → the columns tributary area is approximately equal in the four directions.KL/r = (0.7 * 4.16) / (0.173) = 16.8 < 34.∴ Non-slender column (short column).By conceptual: 416/60 = 6.93 < 15 → short.

57

Figure 5-12: (Cases of effective length factor)

Pd = λ Φ {0.85*fc (Ag – As) + (As fy)} ACI code 05 10.3.6.2505*10³ = 0.8 * 0.65 * {0.85 * 240 * {(60 * 60) - As} + As (4200)}As from manual solution = 59.248 cm². As from plans = 24.120 cm². (See appendix S4)

Comment

Distribution

of steel(

planning)

Distribution

of steel(sap)

As from sap

program

(cm2)

Reaction or

Pu design (ton)

Columns(vertical-

horiz)

NOT OK10ф1615ф1630.0057.750D - 5NOT OK12ф1618ф1636.0071.000I - 5NOT OK12ф1618ф1636.0091.500J - 5NOT OK12ф1618ф1636.00106.60K- 5NOT OK12ф1618ф1636.00110.20O - 5NOT OK12ф1618ф1636.0082.600P - 5NOT OK12ф1618ф1636.00102.00A - 7NOT OK12ф1618ф1636.00369.65D - 7NOT OK12ф1618ф1636.00405.36I - 7NOT OK12ф1618ф1636.00201.68J - 7NOT OK12ф1618ф1636.00222.27K - 7NOT OK12ф1624ф1647.86441.35O - 7NOT OK12ф1618ф1636.00382.80P - 7NOT OK12ф1618ф1636.0089.410V - 7NOT OK12ф1618ф1636.00106.00A - 8 NOT OK12ф1618ф1636.00351.65D - 8NOT OK12ф1618ф1636.00359.70I - 8NOT OK12ф1618ф1636.00215.60J - 8NOT OK12ф1618ф1636.00221.85K - 8NOT OK12ф1618ф1636.00382.47O - 8NOT OK12ф1618ф1636.00317.43P - 8NOT OK12ф1618ф1636.00105.10V - 8NOT OK12ф1618ф1636.0070.320D - 10

58

NOT OK12ф1618ф1636.00105.02I - 10NOT OK12ф1618ф1636.00107.60J - 10NOT OK12ф1618ф1636.0037.850K - 10NOT OK12ф1618ф1636.00182.50O - 10NOT OK12ф1618ф1636.00215.10P - 10NOT OK12ф1618ф1636.0023.070O - 14NOT OK12ф1618ф1636.0024.240P - 14

As from sap program = 84.4 cm².

5.7 Design and evaluation of footing:5.7.1 Introduction:Foundations are the structural concrete elements that transmit the weight of the super-structure to the supporting soil. They could be in many forms, the simplest being the isolated footing. It can be viewed as an inverted slab transmitting a distributed load from to the column. Other forms of foundation are piles driven to rock, combined footing supporting more than one column, mat foundation, and rafts, which are basically inverted slab and beam construction.

5.7.2 Design for single footing:Dimensions of footing (for critical col. (I-7)) = (300 * 300 * 70) cm (See appendix S3)Service load on the footing F1 (Ps) = 386.2 ton.Allowable bearing capacity (Q all) = 30 ton/m2.Ultimate load (Pu) = 505 ton (See page 45)

59

Table 5-5: (Columns design results from sap program and planning for ground floor)

Thickness of footing = 70 cm → d = 63 cmRequired area of footing = Ps / Q all = 386.2 / 30 = 12.873 m2.Area of footing used in plans = 3 * 3 = 9 m → not ok.

5.7.2.1 Check for punching shear:bo = 4 * (c + d) → {for square column}.bo = 4 * (0.6+ 0.63) = 4.92 m = 492 cm. qu = Pu/Area = 505 / 9 = 56.11 ton/m2. Vu = qu [ (B∗H ) – (b+d ) (h+d ) ] = 56.11* [ (3∗3 )– (0.6+0.63 ) (0.6+0.63 ) ]= 420 ton.

βc = long side of columns hort side of column

=0.60.6

=1 ≤ 2

bod

=49263

=7.8≤20 → Ok.

Vc = 0.53 * (1+ 2βc ) √ fc * bo * d = 0.53∗(1+1)√240∗492∗63

1000 = 509

ton.Vc > Vu → ok (no need for shear reinforcement).

5.7.2.2 Check for wide beam shear:Vu = qu * b ( L2 −C

2−d )

= 56.11 * 3 * ( 32– 0.6

2–0.63)= 95.95 ton.

60

Figure 5-13: (Dimension of footing)

Figure 5-14: (Punching shear)

Vc = 0.53∗√ fc∗bw∗d1000

= 0.53∗√240∗300∗631000 = 155.18 ton.

Vc > Vu → no need shear reinforcement.

5.7.2.3 Design for flexure:Mu = qu *b*(0.5L-0.5C) ² / 2 = 56.11*3*1.2²/2 = 121.2 ton.m.Mn = Mu/Φ = 121.2/0.9 = 134.67 ton.m.

ρ = 0.85 fcfy [1−√1− 2.61∗10 ⁵∗Mu

0.9∗b∗d ²∗fc ]ρ = 0.85∗240

4200 [1−√1− 2.61∗10 ⁵∗121.20.9∗300∗63 ²∗240 ]

= 0.00308 > 0.002(As min) → Ok.As = ρ*b*d=0.00308*300*63= 58.212 cm² (Use 23Φ18/13cm).Since we have a square footing then the longitudinal reinforcement and the transverse reinforcement is the same.

5.7.3 Design for combined footing:See figure 5-17 that display the dimensions of footing. Ultimate load (PuA) form sap = 382.47 ton. Ultimate load (PuB) form sap = 317.43 ton.

61

Figure 5-15: (Wide beam shear)

Figure 5-16: (Flexure of footing)

Thickness of footing = 70 cm → d = 63 cm.

5.7.3.1 Top longitudinal steel:

ρ = 0.85 fcfy [1−√1−2.61∗10 ⁵∗Mu

0.9∗b∗d ²∗fc ]ρ = 0.85∗240

4200 [1−√1− 2.61∗10 ⁵∗3200.9∗200∗63 ²∗240 ]

ρ = 0.0138 > 0.002(As min)As = ρ*b*d = 0.0138 * 200 * 63 = 173.8 cm² (Use 55Φ18)

5.7.3.2 Bottom longitudinal steel:

ρ = 0.85∗2404200 [1−√1− 2.61∗10 ⁵∗27

0.9∗200∗63 ²∗240 ]ρ = 0.0018 < 0.002(As min)As = ρ*b*d = 0.002 * 200 * 63 = 25.2 cm² (Use 10Φ18)

5.7.3.3 Transverse steel:● Transverse steel for column A:Ultimate load (PuA) form sap = 382.47 ton/m²

62

Figure 5-17: (Dimension of footing)

27 26

320

Figure 5-18: (M11 for combined footing)

Soil pressure= 382.47[ (2 )∗(0.45+0.6+0.63) ]= 113.8ton.

Mu = [ (113.8)∗(1.68 )∗(0.7) ² ]2

= 46.84 ton.m

Mu (from sap) = 54 ton.m

ρ = 0.85∗2404200 [1−√1− 2.61∗10 ⁵∗54

0.9∗168∗63 ²∗240 ] = 0.00244 > 0.002(As min)As = ρ*b*d = 0.00244 * 168 * 63 = 25.8 cm² (Use 10Φ18)● Transverse steel for column B:Ultimate load (PuB) form sap = 317.43 ton.

Soil pressure= 317.43[ (2 )∗(0.45+0.6+0.63) ]= 94.47 ton/m².

Mu = [ (94.47 )∗(1.68 )∗(0.7) ² ]2

= 38.9 ton.m , Mu (from sap) = 49.5 ton.m.

As = ρ*b*d = 0.00222 * 168 * 63 = 23.6 cm² (Use 10Φ18)

Type length

width

depth

Rein. (manual)

Rein. (planning)

comment

m m mlong Dir.

short Dir.

long Dir.

short Dir.

F1 3.0 3.0 0.7 23Φ18

23Φ18

22Φ16

22Φ16

Not ok

F2 2.6 2.6 0.7 21Φ16

23Φ16

20Φ16

20Φ16

Not ok

F3 3.4 2.6 0.7 23Φ16

25Φ16

22Φ16

24Φ16

Not ok

63

Figure 5-19: (Dimensions of footing for transverse steel calculation)

Table 5-6: (Footing design results from manual solution and planning)

F4 3.4 2.4 0.7 21Φ16

22Φ16

22Φ16

24Φ16

ok

5.8 Design and evaluation of stairs:5.8.1 Introduction:Stairs is a continuous substructure expand along the vertex used for moving from a level to another vertically, so it requires a different types of supporting such as shear walls or beams which makes the modal of the stairs in some cases (such as shear walls) is complicated. Beside, the type of loading is different since it is not vertical on the member but inclined, so these things require more

care in modeling the stair .

In this project, we have four shear walls surrounding the stair, which means fixation in three sides for stair.

(See Figure 5-20)

There are many several types of models for stairs, according to the conditions of constructing and type of structural system surrounding stairs, but we will discuss only the model of this projects stairs.Required data for analysis and design of stair:

1) Live load on A, B and C slabs = 0.4 ton/m2.2) Super imposed on A and B = 0.367 ton/m2.

(See page 17)3) Superimposed loads /m² (see Figure 5-21) =

= t * γ tiles + t* γ concrete + t* γ sand + weight of one stair = 0.03*2.6 + 0.02* 2.3 + 0.05* 2 + ( 0.5∗0.3∗0.16∗2.5

0.34 ) = 0.4 ton /m2.

64

4) Thickness of A, B and C equal 0.2317, 0.2 and 0.15 m respectively.

5) Dimensions of beam = (60*30) cm, dimensions of column = (100*30*416) cm.

6) Thickness and height of shear walls equal 0.2m and 4.16m respectively.

5.8.2 Check for compatibility and equilibrium:∎ Check for compatibility:

65

Figure 5-21: (Cross section in stair)

∎ Check for equilibrium:o.w for slabs (A+B+C) = 4.4 + 2.85 + 5.613 = 12.86 ton.Total super imposed dead load for (A+B+C) = 2.79 + 1.425 + 6 =10.2 ton.o.w for shear walls = net area*thickness*unit weight = 82.1 * 0.2 * 2.5 = 41.1 ton.o.w for column = 0.3 * 1 *2.5 * 4.16 = 3.12 ton.o.w for beam = 0.6 * 0.3 * 2.5 * 3.8 = 1.71 ton.Total dead load = 12.86 + 10.2 + 41.1 + 3.12 + 1.71 = 69 ton.Live load on (A+B+C) = 2.28 + 3.04 + 6 = 11.32 ton.

66

Figure 5-22: (Compatibility for stair)

∎ Sap program results:The values obtained from sap program for dead and live loads as shown in Figure (5-23):

∎ Percentage of errors:

D.L (manual) = 69 ton. , D.L (sap) = 69.2 ton.

%of error = (69.2 - 69) / 69.2= 0.29%.

L.L (manual) = 11.32 ton. , L.L (sap) = 11.307 ton.

%of error = (11.32 – 11.307) / 11.307= 0.12%.

67

Figure 5-23: (Results of dead and live loads from sap program for stair)

5.8.3 Design for flexure:

We note that the moment transverse to shear walls because its stiffness is very large and so the moment of slab is very small.

Area of negative steel by conceptual =Mu∗30d = 0.28∗30

18 = 0.5cm2.

Area of positive steel by conceptual =Mu∗30d = 0.28∗30

18 = 0.5cm2.

68

Figure 5-24: (Bending moment values for stair in x-direction {M11})

We

observ

e that there is considered difference between (1-D) and (3-D), results from the shear walls, which carries the most of

the loads because of its high relative stiffness.

69

Response spectrum method Time history methodEquivalent method

6.1 Introduction:Structural dynamics is a subset of structural analysis that covers the behavior of structures subjected to dynamic loading. Dynamic loads include people, wind, waves, traffic, earthquakes, and blasts. Any structure can be subject to dynamic loading. Dynamic analysis can be used to find dynamic displacements, time history, and modal analysis.A static load is one, which does not vary. A dynamic load is one, which changes with time. If it changes slowly, the structure's response may be determined with static analysis, but if it varies quickly (relative to the structure's ability to respond), the response must be determined with a dynamic analysis.Dynamic analysis for simple structures can be carried out manually, but for complex structures, finite element analysis can be used to calculate the mode shapes and frequencies. An open-source, lightweight, free software can be used to solve basic structural dynamics problems.

6.2 Design input:IE: seismic factor (importance factor) = 1.5 (See table 1 in appendix)R: response modification factor = 3 (See table 4 in appendix)

70

PGA: peak ground acceleration = 0.25 g according to seismic map for Palestine (Nablus). (See PGA diagram for Palestine in appendix)Soil type: SB (Rock)Ss: spectral curve at short period = 0.5 , S1: spectral curve at 1second period = 0.2Fa (site coefficient) =1 (See appendix) , Fv (site coefficient) =1 (See appendix)

SDS = 23Fa Ss = 2

3*1* 0.5= 0.3333 , SD1 = 23Fv S1 = 2

3*1* 0.2= 0.1333Area mass (super imposed load) = 0.367/9.81= 0.03741 ton/m2.

Scale factor = I∗gR = 1.5∗9.81

3 = 4.905

In this project a structural separator exist in the middle of the building, so the structure has been divided into two sections .

6.3 Response spectrum method:6.3.1 Introduction:Response spectra are very useful tools of earthquake engineering for analyzing the performance of structures and equipment in earthquakes, since many behave principally as simple oscillators (also known as single degree of freedom systems). Thus, if you can find out the natural frequency of the structure, then the peak response of the building can be estimated by reading the value from the ground response spectrum for the appropriate frequency. In most building codes in seismic regions, this value forms the basis for calculating the forces that a structure must be designed to resist (seismic analysis).

71

As mentioned earlier, the ground response spectrum is the response plot done at the free surface of the earth. Significant seismic damage may occur if the building response is 'in tune' with components of the ground motion (resonance), which may be identified from the response spectrum. This was observed in the 1985 Mexico City Earthquake where the oscillation of the deep-soil lake bed was similar to the natural frequency of mid-rise concrete buildings, causing significant damage. Shorter (stiffer) and taller (more flexible) buildings suffered less damage.

6.3.2 Representation of response spectrum

on sap:

72

Figure 6-1: (Determine of response spectrum function) Figure 6-2: (Define area mass for building)

73

Figure 6-3: (Definition of response spectrum function)

Figure 6-4: (Adding of load cases)

Figure 6-5: (Define of earthquake load case in x-direction)

6.4 Equivalent static method:● Equivalent static method for left building:

●Manual solution for equivalent static method (left building):

V = Cs*W

Cs = S D1∗IR∗T ≤ SDS∗I

R

Cs = 0.1333∗1.53∗0.344 ≤ 0.3333∗1.5

3 → Cs = 0.1937 is not less than or equal 0.166So, Cs = 0.166

74

Figure 6-7: (Definition of equivalent static method on sap for left building)

W = 3551.6 ton (from sap). → V = 0.166 *3551.6= 589.6 ton.

These values applied in x and y direction whether in sap or in manual solution, it is the same .

● Equivalent static method for right building:

●Manual solution for equivalent static method (right building):

V = Cs*W

Cs = S D1∗IR∗T ≤ SDS∗I

R

75

Figure 6-8: (Definition of equivalent static method on sap for right building)

Cs = 0.1333∗1.53∗0.244 ≤ 0.3333∗1.5

3 → Cs = 0.273 is not less than or equal 0.166So, Cs = 0.166 ton.

W = 3377.1 ton (from sap). → V = 0.166 *3377.1= 560.6 ton.

These values applied in x and y direction whether in sap or in manual solution, it is the same .

6.5 Time history method:6.5.1 Introduction:The basic mode superposition method, which is restricted to linearly elastic analysis, produces the complete time history response of joint displacements and member forces. In the past, there have been two major disadvantages in the use of this approach. First, the method produces a large amount of output information that can require a significant amount of computational effort to conduct all possible design checks as a function of time. Second, the analysis must be repeated for several different earthquake motions in order to assure that all frequencies are excited, since a response spectrum for one earthquake in a specified direction is not a smooth function. The recent increase in the speed of computers has made it practical to run many time history analyses in a short period. In addition, it is now possible to run design checks as a function of time, which produces superior results, since each member is not designed for maximum peak values as required by the response spectrum method.

6.5.2 Representation of time history functions on sap:

76

77

These function applied in x and y direction whether on left or right building.

78

Figure 6-10: (Data entry for time history function on sap)

6.6 Results from sap:Throw the dynamic analysis for the building we obtained on these results:

Base ShearPeriod

(T) Time HistoryEquivalent Static

Response Spectrum

Y-directio

n

X-directio

n

Y-directio

n

X-directio

n

Y-directio

n

X-directio

n(ton)(ton)(ton)(ton)(ton)(ton)(sec)

382.83301.34591.93591.93376.82402.20.344Left

Building

231.62246.25562.85562.85308.86357.880.244Right Buildi

ng

79

Figure 6-11: (Block B {left building})Figure 6-12: (Block A {right building})

Table 6-1: (Base shear and period values for buildings from sap)

● For left building:

80

81

Figure 6-13: (Base reaction for defined functions)

Figure 6-14: (Base reaction for equivalent static function)

Figure 6-15: (Values of modal mass participation ratio for Block B {left building})

● For right building:

82

Figure 6-16: (Base reaction for defined functions)

Figure 6-17: (Base reaction for equivalent static function)

83

Figure 6-18: (Values of modal mass participation ratio for Block A {right building})

84

Stiff Soil Actual SoilWeak Soil

7.1 Introduction:Most of the civil engineering structures involve some type of structural element with direct contact with ground. When the external forces, such as earthquakes, act on these systems, neither the structural displacements nor the ground displacements, are independent of each other. The process in which the response of the soil influences the motion of the structure and the motion of the structure influences the response of the soil is termed as soil-structure interaction (SSI).Conventional structural design methods neglect the SSI effects. Neglecting SSI is reasonable for light structures in relatively stiff soil such as low-rise buildings and simple rigid retaining walls. The effect of SSI, however, becomes prominent for heavy structures resting on relatively soft soils for example nuclear power plants, high-rise buildings and elevated-highways on soft soil.If a lightweight flexible structure is built on a very stiff rock foundation, a valid assumption is that the input motion at the base of the structure is the same as the free-field earthquake motion. This assumption is valid for a large number of building systems since most building type structures are approximately 90 percent voids, and, it is not unusual that the weight of the structure is excavated before the structure is built. However, if the structure is very massive and stiff, such as a concrete gravity dam, and the foundation is relatively soft, the motion at the base of the structure may be significantly different than the free-field surface motion. Even for this extreme case, however, it is apparent that the most significant interaction effects will be near the structure and, at some finite distance from the base of the structure, the

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displacements will converge back to the free-field earthquake motion.Because the difficulty of the representation of the soil under the footing on the sap program, we used a set of springs to operate the work of the soil, and we define the properties of these springs to be similar to the characteristics of the soil. Will study the impact of representation the footing on the mathematical model in this project and in accordance with the three steps as follows: 1. Representation of the footing as carried out in reality with the approximate value of bearing capacity for soil as it is in location = 3 kg/cm².2. Representation of the footing as carried out in reality with high bearing capacity for soil (soil is very strong) = 9 kg/cm².3. Representation of the footing as carried out in reality with low bearing capacity for soil (soil is poor) = 1 kg/cm².

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Figure 7-1: (Foundations-level one)

7.2 Stiff soil (bearing capacity = 9kg/cm²):

●For left building:

Period (T) = 0.497 sec

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Figure 7-2: (Foundations-level two)

Figure 7-3: (Definition of springs on sap for stiff soil case)

● For right building:Period (T) = 0.319 sec.

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Figure 7-4: (Bending moment diagram and area of steel in stiff soil case for left building)

7.3 Actual soil (bearing capacity = 3kg/cm²):

● For left building: Period (T) = 0.582 sec.

● For right building:

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Figure 7-6: (Definition of springs on sap for actual soil case)

Period (T) = 0.373 sec.

7.4 Weak

soil (bearing capacity = 1 kg/cm²):

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Figure 7-8: (Bending moment diagram and area of steel in actual soil case for right building)

● For left building: Period (T) = 0.717 sec.

● For right building:

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Figure 7-10: (Bending moment diagram and area of steel in weak soil case for left building)

Period (T) = 0.451 sec.

7.5 Fixation case:

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Figure 7-11: (Bending moment diagram and area of steel in weak soil case for right building)

● For left building: Period (T) = 0.315 sec.

●For right building:

Period (T) = 0.222 sec.

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Figure 7-12: (Bending moment diagram and area of steel in fixation case for left building)

Discussion:Through the results of the previous analysis (when the soil is stiff), we note that the values are very close with fixation case, and this demonstrates that the representation of the building on the program was accurate, and therefore we can trust with the results of

the program.If we compare the results of reinforced in the previous cases, we note that the results of the reinforced for columns may vary significantly, contrary to what was expected, it was expected that the difference of the values is small, and the differences in the values of the reinforced in beams is high, and the reason for this, to a more From the level of the footing, and this may lead to differential settlement.

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Figure 7-13: (Bending moment diagram and area of steel in fixation case for right building)

..……………………………………………………

.

References: Nawy, Edward G. (2005). Reinforced concrete: A

fundamental approach, 5th edition. NJ, Upper Saddle River: Pearson / Prentice Hall.

Wang, Chu-Kia (2007). Reinforced concrete design, 7th edition. NJ, Hoboken, John Wiley & Sons .

American concrete institute (2005). ACI Code 318-05. USA, MI: Farmington Hills.

International code council (2006). IBC Code 06. USA.

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APPENDIX

ReportsDrawings