Reinforced Concrete Structure

Diploma Project

Reinforced concrete structure

B+GF+5S

BUCHAREST2014

Table of Contents

1 Technical report..........................................................................................................................- 6 -

1.1 Building description..........................................................................................................- 6 -

1.2 Architectural characteristics..............................................................................................- 6 -

1.3 Structural characteristics and materials used.....................................................................- 6 -

1.4 Structural computational aspects.......................................................................................- 6 -

1.5 Data regarding the building site........................................................................................- 7 -

2 Load evaluation...........................................................................................................................- 8 -

3 Predimensioning.......................................................................................................................- 10 -

3.1 Slab..................................................................................................................................- 10 -

3.2 Beams..............................................................................................................................- 10 -

3.2.1 Longitudinal beams................................................................................................- 10 -

3.2.2 Transversal beams..................................................................................................- 10 -

3.3 Columns...........................................................................................................................- 10 -

3.3.1 Column 1 - corner column.....................................................................................- 11 -

3.3.2 Column 2 - marginal column.................................................................................- 11 -

3.3.3 Column 3 - central column.....................................................................................- 11 -

4 Walls.........................................................................................................................................- 12 -

5 Evaluation of the seismic loads and load combination.............................................................- 14 -

5.1 Equivalent seismic force..................................................................................................- 14 -

5.2 Computational model for lateral and vertical load..........................................................- 15 -

5.3 Vibration modes..............................................................................................................- 16 -

6 Design of rigidity to lateral forces............................................................................................- 20 -

6.1 Serviceability Limit State Check (SLS)..........................................................................- 20 -

6.2 Ultimate Limit State Check (ULS)..................................................................................- 21 -

7 Dimensioning of the structural elements..................................................................................- 23 -

7.1 Slab..................................................................................................................................- 23 -

7.1.1 General computation scheme.................................................................................- 23 -

7.1.2 Static computation in the elastic-range slabs reinforced on one direction.............- 23 -

7.1.3 Static computation in the elastic-range slabs reinforced on two directions...........- 24 -

7.1.4 Dimensioning the reinforcement............................................................................- 29 -

7.2 Beams..............................................................................................................................- 35 -

7.2.1 Design of the longitudinal reinforcement..............................................................- 35 -

7.2.2 Design of the transversal reinforcement of the beams...........................................- 41 -

7.3 Columns...........................................................................................................................- 46 -

7.3.1 Geometrical characteristics....................................................................................- 46 -

7.3.2 Provisions regarding the materials used.................................................................- 46 -

7.3.3 Longitudinal reinforcement provisions..................................................................- 46 -

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7.3.4 Longitudinal reinforcement design........................................................................- 46 -

7.3.5 Transversal reinforcement provisions....................................................................- 47 -

7.3.6 Transversal reinforcement design..........................................................................- 47 -

8 Structural Walls........................................................................................................................- 51 -

8.1 General considerations for the computations of the structural walls..............................- 51 -

8.2 The values of the design sectional efforts in the walls....................................................- 51 -

8.3 The longitudinal and transversal reinforcement..............................................................- 52 -

8.4 Material provisions for structural walls...........................................................................- 59 -

8.5 Geometric Exigencies......................................................................................................- 59 -

9 Foundation................................................................................................................................- 60 -

9.1 General Information........................................................................................................- 60 -

9.2 Predimensioning..............................................................................................................- 60 -

10 Stairs.........................................................................................................................................- 61 -

10.1General issues..................................................................................................................- 61 -

10.2Loads...............................................................................................................................- 61 -

10.3Reinforcement.................................................................................................................- 62 -

11 Bibliography.............................................................................................................................- 65 -

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Table of Tables

Table 1-1 Geotechnical Conditions..........................................................................................- 7 -

Table 2-1 Load on current floor...............................................................................................- 8 -

Table 2-2 Load from terrace.....................................................................................................- 8 -

Table 2-3 Load from façade.....................................................................................................- 8 -

Table 5-1 Modal Participating Mass Ratios...........................................................................- 19 -

Table 6-1 SX drift check........................................................................................................- 21 -

Table 6-2 SY drift check........................................................................................................- 21 -

Table 6-3 SX drift check........................................................................................................- 21 -

Table 6-4 SY drift check........................................................................................................- 22 -

Table 7-1 Moment computation for the current floor slabs....................................................- 26 -

Table 7-2 Moment computation for the terrace slabs.............................................................- 28 -

Table 7-3 Current floor slab reinforcement............................................................................- 31 -

Table 7-4 Terrace slab reinforcement.....................................................................................- 33 -

Table 7-5 Field reinforcement for beams in Frame C-C........................................................- 37 -

Table 7-6 Support reinforcement for beams in Frame C-C....................................................- 38 -

Table 7-7 Field reinforcement for beams in Frame 6-6..........................................................- 39 -

Table 7-8 Support reinforcement for beams in Frame 6-6.....................................................- 40 -

Table 7-9 Transversal reinforcement for beams in frame C-C...............................................- 42 -

Table 7-10 Transversal reinforcement for beams in frame 6-6..............................................- 44 -

Table 7-11 Longitudinal reinforcement for Central Column.................................................- 49 -

Table 7-12 Transversal reinforcement for Central Column...................................................- 49 -

Table 8-1 Longitudinal reinforcement for the Structural Wall...............................................- 58 -

Table 8-2 Transversal reinforcement for the Structural Wall.................................................- 58 -

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Table of Figures

Figure 1 Ground level snow load - CR-1-1-3-2012.................................................................- 9 -

Figure 2 Peak ground acceleration IMR 225 years - P100-1-2013........................................- 14 -

Figure 3 Tc response spectrum zoning - P100-1-2013...........................................................- 14 -

Figure 4 Normalized elastic acceleration spectrum - P100-1-2013.......................................- 14 -

Figure 5 First mode of vibration.............................................................................................- 16 -

Figure 6 Second mode of vibration........................................................................................- 17 -

Figure 7 third mode of vibration.............................................................................................- 18 -

Figure 8 Wall 1 Zone A – M-N interaction diagram..............................................................- 54 -

Figure 9 Wall 1 Zone B – M-N interaction diagram..............................................................- 56 -

Figure 10 Stair Etabs Diagram...............................................................................................- 61 -

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1 TECHNICAL REPORT

1.1 Building descriptionThe purpose of the current project is to compute, design and detail the structural elements

for a B+GF+5S dual type concrete structure. The structure will be located on Grigore Manolescu street nr 10-14, District 2, Bucharest.

The building will have a residential function. Moreover the basement will be used for parking space.

The height regime of the structure is of 18 m. Each story has a height of 3 m, including the base one. The basement is also 3 m height. The surface of the ground floor is of 274.56 m 2

and of the current floors is of 307.17. As a result the construction has a developed area of 1810.41 m2.

In plane the building has a regular shape.

1.2 Architectural characteristics Partitioning made of vertical hollow brick walls finished with paint or natural stone,

according to the room. Ceilings are executed from gypsum boards on metallic structure, with a paint cover

over the support layer. Flooring made from stratified parquet in the living rooms and bedrooms, and natural

stone in kitchens, bathrooms and entrance halls. Outdoor swimming pool on the top terrace. Installations have separate meters for water, heating and electricity, better to value the

under floor heating system connected to the buildings common heating system. Preinstalled phone, internet and TV cable connections mean no cables hanging from the walls.

The parking access with remote control gate and the video surveillance system add an extra layer of security to the building.

Defrosting system on the access ramp for the parking lot and on the terraces from upper floors increase the comfort of the inhabitants during the cold season.

1.3 Structural characteristics and materials usedThe superstructure is designed based on a series of functional and architectural

requirements, but also by taking into consideration the seismic behavior. The design solution is based on the dual structural system: structural system in which the vertical loads are taken over mainly by spatial frames, while the lateral load is partially carried by the frames and partially by the structural walls. The floors are reinforced concrete slabs with hp=17cm. The structural walls, the columns, the beams and the slabs are made from reinforced

monolithic concrete Concrete used for superstructure: reinforced concrete C20/25 with fcd=13.3 N/mm2

For leveling and cross slope C8/10 Steel: PC52 with fyd=300 N/mm2, OB37 with fyd=210 /mm2

The foundation solution is mat foundation made from reinforced concrete C20/25. The infrastructure consists of concrete walls and frames. The basement is enclosed by a 30

cm thick reinforced concrete walls. The slab over the basement is 17 cm thick.

1.4 Structural computational aspectsDesigning structures with structural walls must satisfy all the exigencies (functional,

structural, esthetics, placement in the urban layout, execution, maintenance and repair /consolidation) based on the location conditions (geotechnical, climatic, seismic, influences on neighborhood buildings) and the importance of the building. In this way a favorable behavior can be assured during exploitation, at an acceptable safety level. The satisfaction of the structural

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exigencies imply taking over the different actions, especially the seismic ones. This is achieved by applying the structural mechanism of elastic-plastic deformation which incorporates also the mechanism of dissipating energy.

The adequate modeling with respect to the real behavior and using the proper computational models for determining the efforts and dimensioning the structural elements is also essential.

The computations are performed in accordance with the design codes EUROCODE, the corresponding National Annex, but also National Regulations. The structural modeling is achieved using software based on finite elements methods: ETABS 9. 7. The design is done in order to assure a proper seismic behavior. The actions have been considered from the fundamental combination, but also from the special combinations. The seismic force is established according to P100-1/2013.

1.5 Data regarding the building siteThe building will be located in Bucharest. As a result:

The class of importance and exposure III with γ1=1. The ground acceleration for design MRI 225 years is ag=0.3g. The control period of vibration is Tc=1.6s. Ductility class H. Snow characteristic value so,k=2kN/m2

Reference wind pressure, measured at a mean of 10 min, at 10m above ground level, for a mean recurrence interval of 50 years, is 0.5kPa; also the building is situated in a urban area with high density of buildings.

Table 1-1 Geotechnical ConditionsDepth Type of soil

0 - 0.7m Infill materials (pavements, 50 cm sand bed, broken and unbroken bricks)

0.7 - 3m Dark brown plastic silt sand e=0.5 pconv=300kPa3 - 6.5m Plastic stiff clay e=0.8 pconv=300kPa6.5 - 12m Sandy silt e=0.5 pconv=350kPa

Due to the soil stratification and the building layout the chosen foundation solution is a mat foundation.

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2 LOAD EVALUATION

Table 2-2 Load on current floor

LoadNominal value

Coefficient value

Long term value

Coefficient value

Design value

Self-weight 4.25 1 4.25 1.35 5.7375Flooring load 1.75 1 1,75 1.35 2.3625Live load 2 0.4 0.8 1.5 3Interior walls 1.5 1 1.5 1.35 2.025

qDV=13.125kN

Table 2-3 Load from terrace

LoadNominal value

Coefficient value

Long term value

Coefficient value

Design value

Self-weight 4.25 1 4.25 1.35 5.7375Terrace insulation 1.5 1 1.5 1.35 2.025Snow load 1.6 0.4 0.4 1.5 2.4

qDV=10.1625 kN

Table 2-4 Load from façade

ElementThicknes

s (m)Height

(m)γ

(kN/m3)LTV

Coefficient value

DV

Colored plastering 0.01 3 18 0.54 1.35 0.729

Thermo insulation 0.1 3 0.32 0.096 1.350.1296

Interior plastering 0.01 3 18 0.54 1.35 0.73qDV= 1.5886 kN

Snow loadThe snow load on the roof takes into account the quantity of snow depending on the roof shape, and the variation of snow height due to wind and snow melt.

sk = Isi Ce Ct s0,k (2.3)

Where: Is - the exposure-importance coefficient for the snow load μi - the shape coefficient s0,k - the charcteristic vale of snow load at the ground level [kN/m2], at the

location; Ce - the exposure coefficient of the locatiom Ct - the termic coefficient

Due to the importance of the structure, Is=1Due to the small slope of the terrace roof 00 300, μi=0.8Due to its location, the structure has partial exposure, Ce=1Due to good insulation of the roof, Ct=1

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The values of the characteristic value of snow at ground level in Romania are presented on the map:

The building is located in Bucharest so s0,k =2.0

sk=1*0.8*1*1*2=1.6kN/m2

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Figure 1 Ground level snow load - CR-1-1-3-2012

3 PREDIMENSIONING

3.1 SlabThe slab taken as basis of computation is the most unfavorable one, one way slab with

dimensions of L1 = 5 m and L2 = 14.7 mThe height of the slab: hpl = (/ 30)hpl = thickness of the slab Lmin = short length of the slab = 5 mhpl = 5 m / 30= 0.166 m = 17 cm

3.2 Beams3.2.1 Longitudinal beams

The dimensions of the longitudinal beams will be Hb.long and Wb.long, where:Hb.long = the height of the longitudinal beam Hb.long = t / (10..12)Wb.long = the width of the longitudinal beam Wb.long = Hb.long / (2..3)Where t represents the bays.

BeamBay

lengthComputed

heightRound up

heightComputed

widthRound

up width(m) (m) (cm) (cm) (cm)

1 4.30 0.43 55 28 302 5.20 0.52 55 28 303 5.20 0.52 55 28 304 3.75 0.38 55 28 30

Same size beams are chosen for ease of execution and symmetry.

3.2.2 Transversal beamsThe dimensions of the longitudinal beams will be Hb.trans and Wb.trans, where:Hb.trans = the height of the longitudinal beam Hb.trans = t / (10..12)Wb.trans = the width of the longitudinal beam Wb.trans = Hb.long / (2..3)Where t represents the bays.

BeamBay

lengthComputed

heightRound up

heightComputed

widthRound

up width(m) (m) (cm) (cm) (cm)

1 4.20 0.42 55 28 302 3.90 0.39 55 28 303 5.00 0.50 55 29 30

Same size beams are chosen for ease of execution and symmetry.

3.3 ColumnsEach column will have its axial load computed. For this task we will consider the loads

supported by the slabs, which are carried to the beams, and the beams then transfer the loads to the column. We will also take into consideration the loads coming from façade, nonstructural walls, and also self-weight of each column.

qfacade=1.59 kN/m2 - Load from the weight of the facadeqter=10.16 kN/m2 - Load from the terraceqcl=13.13 kN/m2 - Load from the current level

3.3.1 Column 1 - corner columnThe tributary areas are A1=2.05 m2 and A2=3.02m2, the dimensions of the beams are

Hb=0.55m and Wb=0.3 and dimensions of the slab are L1=4.05m, T1=5m.

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The chosen cross section will be a rectangle of 30 cm x 60 cm with Area = 0.18 m2.

Nter=Hb∗Wb∗γ∗L 1+T 12

+0.9m∗qfac∗L1+T 2

2+q ter∗( A 1+A 2 )

Nter =83.13 kN-the axial load from the terrace

Ncf=Hb∗Wb∗γ∗L1+T 12

+Hs∗q fac¿L1+T 2

2+q

CL

∗( A 1+A 2 )+0.3∗0.6∗Hs∗γ

Ncf =120.32 kN-the axial load from the terraceN=5*Ncf+Nter N=684.73 KN

b 1∗b 2≤N

fc∗0.35→b 1∗b 2=0.18 m2 ≥

Nfc∗0.35

=0.147 m2

3.3.2 Column 2 - marginal columnThe tributary areas are A1=1.90 m2, A2=3.46m2, A3=3.75 m2, A4=3.13 m2, the

dimensions of the beams are Hb=0.55m and Wb=0.3 and dimensions of the slab are L1=5.2m, T1=5m, T2=3.9.

The chosen cross section will be a square of 60 cm x 60 cm with Area = 0.36 m2.

Nter=Hb∗Wb∗γ∗L 1+T 1+T 22

+0.9m∗q fac∗T 1+T 2

2+q ter∗( A 1+A 2+A 3+A 4 )


Ncf=Hb∗Wb∗γ∗L1+T 1+T 22

+Hs∗q fac ¿T 1+T 2

2+q

CL

∗( A 1+A 2+A 3+A 4 )+0.6∗0.6∗Hs∗γ

Ncf =227.16 kN-the axial load from the terraceN=5*Ncf+Nter N=1365.76 KN

b 1∗b 2≤N

fc∗0.35→b 1∗b 2=0.36 m2 ≥

Nfc∗0.35

=0.293 m2

3.3.3 Column 3 - central columnThe tributary areas are A1=2.62 m2, A2=2.59m2, A3=1.9 m2, A4=1.8 m2, A5=1.12 m2,

A6=1.12m2, A7=2.04 m2, A8=2.2 m2, the dimensions of the beams are Hb=0.55m and Wb=0.3 and dimensions of the slab are L1=4.3m, T1=4.05m, L2=3, T2=4.2.

The chosen cross section will be a square of 60 cm x 60 cm with Area = 0.36 m2.

Nter=Hb∗Wb∗γ∗L 1+T 1+L2+T 22

+q ter∗( A 1+A 2+A 3+A 4+A 5+A 6+A 7+A 8 )


Ncf= Hb∗Wb∗γ∗L1+T 1+L 2+T 22

+qCL∗( A 1+A 2+A 3+A 4+A 5+A 6+A 7+A 8 )+0.6∗0.6∗Hs∗γ

Ncf =263.95 kN-the axial load from the terrace

N=5*Ncf+Nter N=1511.08 KN

b 1∗b 2≤N

fc∗0.35→b 1∗b 2=0.36 m2 ≥

Nfc∗0.35

=0.324 m2

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4 WALLS

The total area of the web on one direction will be at least:

∑ Abi≥α∗k s∗n∗A pl

120

Where:bi is the sum of the horizontal sections of walls, laid parallel with the direction of the

action of the lateral loadsα is the importance factor of the structure according to 4.4.5 from P100-1:2013ks=ag/g the ratio between the peak value of the acceleration and the gravitational

accelerationn is the number of slabs of the buildingApl is the area of the slab in m2

The thickness of the walls will be at least 15 cm. For structures that have less than 12 stories it is recommended to keep constant dimensions of the wall sections on the whole height.

α∗k s∗n∗Apl

120=1∗0.3∗6∗287

120=4.30 m2

On x direction:

∑ Abi=2∗6.5∗0.3+2∗6.1∗0.3+2∗2.6∗0.3+10∗0.6∗0.3=10.92 m2 On y direction: ∑ Abi=0.3∗2.6+0.3∗4.5+0.3∗3.1+5∗0.6∗0.6=4.86 m2

The area of the flanges situated at the margins of the walls will respect the rule:

no≤ 1.5A f

Ac

+0.35=1.5∗2∗0.3∗0.35.9∗0.3

+0.35=0.50 no=

N Ed

A c∗f cd

Where, “NEd” is the axial compression stress.qfacade=1.59 kN/m2 - Load from the weight of the facadeqter=10.16 kN/m2 - Load from the terraceqcl=13.13 kN/m2 - Load from the current level

Wall 1:Dimensions of the slab: L=9.675m, T=2.5m.The tributary area is A=24.56m2.The dimensions of the corresponding beams: L1=3.75m, L2=2.6m, T2=5m.

Nter=Hb∗Wb∗γ∗L 1+ L2+T 22

+L∗0.9 m∗q facade+qter∗A=287.09 KN

Ncf= Hb∗Wb∗γ∗L1+L 2+T 22

+ L∗Hs∗q facade+qCL∗A+0.3∗Hs∗6.5∗γ=583.59 KN

N=5∗Ncf +Nter=3205.04 KN

no=N Ed

A c∗f cd

= 3205.040.3∗5.9∗13.33∗1000

=0.136<0.5 →OK

This check is performed to identify the cases in which the sections need strengthening by means of flanges.

Wall 2:Dimensions of the slab: L=8.25m, T=2.5m.The tributary area is A=20.625m2.The dimensions of the corresponding beams: L2=2.6m, L3=1.4m, T3=5m.

12

Nter=Hb∗Wb∗γ∗L 2+ L3+T 32

+L∗0.9 m∗q facade+q ter∗A=239.87 KN

Ncf= Hb∗Wb∗γ∗L2+L 3+T 32

+L∗Hs∗q facade+qCL∗A+0.3∗Hs∗6.1∗γ=465.90 KN


no=N Ed

A c∗f cd

= 2569.390.3∗5.5∗13.33∗1000

=0.117<0.5 → OK

Wall 3:Dimensions of the slab: L=4.175m, T=5.825m.The tributary area is A=24.32m2.The dimensions of the corresponding beams: L6=2.6m, L7=3m, T7=1.45m.

Nter=Hb∗Wb∗γ∗L 6+L 7+T 72

+q ter∗A=261.63 KN

Ncf= Hb∗Wb∗γ∗L6+ L7+T 72

+qCL∗A+0.3∗Hs∗2.6∗γ=392.36 KN


no=N Ed

A c∗f cd

= 2223.440 .3∗2∗13.33∗1000

=0.278<0.5 → OK

Wall 4:Dimensions of the slab: L=8.575m, T=2.1m.The tributary area is A=18.01m2.The dimensions of the corresponding beams: L16=1.55m, L17=2.6m, T13=4.05m,

T14=4.05m.

Nter=Hb∗Wb∗γ∗L 16+L17+T 13+T 142

+ L∗0.9 m∗qfacade+q ter∗A=220.52 KN

Ncf= Hb∗Wb∗γ∗L16+L 17+T 13+T 142



no=N Ed

A c∗f cd

= 2464.970 .3∗5.9∗13.33∗1000

=0.104<0.5 →OK

Wall 5:Dimensions of the slab: L=8.1m, T=2.1m.The tributary area is A=17.01m2.The dimensions of the corresponding beams: L17=2.6m, L18=1.4m, T14=4.05m,

T15=4.05m.

Nter=Hb∗Wb∗γ∗L 17+L18+T 14+T 152

+L∗0.9 m∗q facade+q ter∗A=209.37 KN

Ncf= Hb∗Wb∗γ∗L17+L 18+T 14+T 152



no=N Ed

A c∗f cd

= 2330.290 .3∗5.5∗13.33∗1000

=0.106<0.5→ OK

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5 EVALUATION OF THE SEISMIC LOADS AND LOAD COMBINATION

5.1 Equivalent seismic forceThe seismic loads are determined with the method of seismic equivalent force according

to P100-1/2013Fb=γ I ,e∗Sd (T1 )∗λ∗m Where:

γ I=1 is the importance factor of the building

Sd (T 1 ) is the design acceleration spectrum which takes into account the characteristics of ground motion and the dynamic properties of the structural system.

Sd (T 1 )=Se (T1 )

q

T 1 is the fundamental vibration period of the structural system

q is the behavior factor Se (T1 ) represents the elastic acceleration

spectrum Se (T1 )=ag∗β (T1 )=0.3∗2.5 ag is the design acceleration of the peak

ground acceleration β (T 1) is the normalized elastic

acceleration spectrum

q=5∗αu

α 1

=5∗1.15=5.75

αu

α1

takes into account the over strength and redundancy of the structural system

λ is a correction factor that takes into account the contribution of the fundamental vibration mode through its modal mass making the equivalence between real multi degree of freedom(MDOF) and a single degree of freedom system.

λ =0.85 if T 1≤ T c and the building has two levels or λ =1 otherwise

Fb=γ I∗ag∗β (T 1)

qλ∗m

Fb=c∗m

c=

γ I∗ag

g∗β (T1)

qλ

c=1∗0.3∗2.55.75

∗0.85=0.111

Fb=c∗G G=16628 KN−¿ETABS Fb=0.111∗16628 Fb=1845.7 kN T=0.3331TB< T <TC

14

Figure 2 Peak ground acceleration IMR 225 years - P100-1-2013

Figure 3 Tc response spectrum zoning - P100-1-2013

Figure 4 Normalized elastic acceleration spectrum - P100-1-2013

5.2 Computational model for lateral and vertical load

The computation of the efforts is done with ETABS. The superstructure is considered fixed at the level of the slab above the basement.

In order to define the model the following steps are followed: Choose the units of measurement KN and m and define the geometry of the structure

(axes, spans, bays, story height); Define the materials and the linear cross sections (beam, columns), the plane elements

(walls, slabs); define the geometrical characteristics, the materials and the rigidity of the structural elements;

Define the load cases due to gravitational loads: from the self-weight of the structural elements, terrace layers, floorings, partitioning elements, attic, facade, but also from variable load: live load and snow;

Define the horizontal load from the design seismic situation. The equivalent seismic forces are defined as a fraction of the weight of the superstructure. The response spectrum is defined. The additional eccentricities are considered 5% from the length of the building on each direction, on one side and the other of the center of the masses.

Define the load combinations which contain the action of the earthquake and the vertical associated loading.

Define the mass source for computing the basic seismic force. Position the elements in the structure, define the piers, define the slab diaphragm Define the location and value of the loads associated to different loading hypothesis Define the supports Choose the type of analysis-elastic analysis After having the spatial model for computation, the structural computation is performed

by determining the first 3 vibration modes (translation on X, translation on Y and rotation on Rz) and the values of displacements and efforts. The displayed results have the units in the international units: m, KN, KNm, seconds depending on each case.

15

5.3 Vibration modes

Figure 5 First mode of vibration

16

Figure 6 Second mode of vibration

17

Figure 7 third mode of vibration

18

Table 5-5 Modal Participating Mass Ratios

Mode Period UX UY UZ SumUX SumUY SumUZ RX RY RZ SumRX SumRY SumRZ

1 0.333133 0.0003 71.1639 0 0.0003 71.1639 0 97.4479 0.0001 1.0634 97.4479 0.0001 1.0634

2 0.200045 0.2923 1.0544 0 0.2926 72.2183 0 1.5182 0.3966 71.5823 98.9661 0.3967 72.6456

3 0.18462 71.5095 0.0006 0 71.8021 72.219 0 0.0057 98.4545 0.3292 98.9719 98.8512 72.9749

4 0.08479 0.0001 18.2615 0 71.8022 90.4804 0 0.7802 0.0094 0.1851 99.752 98.8606 73.16

5 0.052368 0.0086 0.2771 0 71.8107 90.7576 0 0.009 0.0003 18.2446 99.761 98.8609 91.4046

6 0.046056 19.2178 0.0161 0 91.0286 90.7736 0 0.0022 0.9207 0.0337 99.7632 99.7816 91.4383

7 0.040308 0.0057 5.5695 0 91.0343 96.3431 0 0.1973 0.0007 0.0949 99.9604 99.7823 91.5332

8 0.026245 0.0001 2.4338 0 91.0344 98.7769 0 0.0266 0 0.0026 99.987 99.7824 91.5357

9 0.025652 0.0119 0.0038 0 91.0463 98.7807 0 0.0002 0.0008 5.6091 99.9871 99.7832 97.1448

10 0.02198 5.8886 0.0018 0 96.9349 98.7825 0 0 0.1903 0.012 99.9872 99.9734 97.1568

11 0.020003 0.0015 0.9669 0 96.9364 99.7494 0 0.0112 0.0001 0.0074 99.9984 99.9735 97.1643

12 0.01731 0.0035 0.0195 0 96.94 99.7689 0 0 0 1.9627 99.9984 99.9736 99.127

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6 DESIGN OF RIGIDITY TO LATERAL FORCESIs considered the check at two limit states, to the Serviceability Limit State (SLS) and the

Ultimate Limit State (ULS) (Appendix E, P100/2013)

6.1 Serviceability Limit State Check (SLS)The aim of checking is to limit the degradation of nonstructural elements and equipment

for frequent seismic action (recurrence period of 30 years). This check also aims to reduce the cost of the post-earthquake repair works.

Considering the criteria of interaction between the structural and nonstructural elements, the present structure is in the situation where the nonstructural walls interact with the structure, so the admissible inter story drift is 0.005.

The displacement check is done based on the expression (relation E.1-appendix E, P100-1-2013):

drSLS=ν∗q∗dℜ≤ dr , a

SLS

ddrSLS- inter-story drift associated to the conventional elastic analysis

ν - reduction factor which considers the shorter recurrence period of the earthquake - the value of ν is: 0.5 for the buildings in the importance classes IIIq - behavior factor specific to the structure’s typedℜ- the relative displacement, determined by static computation under the seismic loadsdr ,a

SLS - allowable value of the relative story displacement

dr ,aSLS=0.005 h−Table E .2P 100−1−2013

Displacement values dr are calculated using calculation assumptions of structural elements rigidity in accordance with the actual cracking condition, depending on the degree of interaction between structural elements and the nonstructural elements (partitioning and closures). At the action of a moderate intensity earthquake is assumed that connections between closures and partitioning elements and columns and beams will not be compromised and damage of nonstructural elements to be insignificant as a result of limiting lateral movement. In such circumstances it is justified the contribution of nonstructural elements in the global stiffness of the structure. Because it cannot build rigorous models , but sufficiently simple structure - elements of partitioning for design practice, it is permitted in a simplified way, the global evaluation of structural rigidity by considering the deformation properties of not cracked sections (stage I behavior) of the structural elements and neglect the contribution of non-structural elements. In the case where the nonstructural elements are not deforming in the same way with the structure, the rigidity of the structure is evaluated considering the deformation properties of structural elements in the cracked stage.

So, in my case the values of dr are estimated in the hypothesis of the sectional stiffness of the structural elements in the not cracked stage:

(EI )conv=Ec∗Ic Where:Ec– Elasticity modulus of the concreteIc– inertia moment of the cross sectionFrom the table below, the structure with the dimensions of the elements, obtained from

preliminary design, obey the lateral displacement check corresponding to SLS.The admissible displacement is 0.5% of Hs.

20

Table 6-6 SX drift checkLevel Direction Combination dre ν q d,r,sls dadm

Story 5 SX PSXPSY 0.000378 0,5 5,75 0.002174 0,015 OKStory 4 SX PSXPSY 0.000407 0,5 5,75 0.00234 0,015 OKStory 3 SX PSXPSY 0.000422 0,5 5,75 0.002427 0,015 OKStory 2 SX PSXPSY 0.000399 0,5 5,75 0.002294 0,015 OKStory 1 SX PSXPSY 0.000323 0,5 5,75 0.001857 0,015 OKBase SX PSXPSY 0.000165 0,5 5,75 0.000949 0,015 OK

Table 6-7 SY drift checkLevel Direction Combination dre ν q d,r,sls dadm

Story 5 SY PSYNSX 0.001219 0,5 5,75 0.007009 0,015 OKStory 4 SY PSYNSX 0.001219 0,5 5,75 0.007009 0,015 OKStory 3 SY PSYNSX 0.001268 0,5 5,75 0.007291 0,015 OKStory 2 SY PSYNSX 0.00121 0,5 5,75 0.006958 0,015 OKStory 1 SY PSYNSX 0.000986 0,5 5,75 0.00567 0,015 OKBase SY PSYNSX 0.00049 0,5 5,75 0.002818 0,015 OK

6.2 Ultimate Limit State Check (ULS)The aim is to avoid the loss of human life during a major earthquake (with a long

recurrence period) by preventing the collapse of the closures and partitioning elements, limiting the structural degradation and the second order effects. The earthquake associated to this limit state is considered when designing the resistance to lateral forces of the building. All the horizontal seismic action will be resisted by the reinforced concrete walls and frames, which will be heavily cracked so their stiffness is equal to 0.5EcIc.

The check is made based on the expression E.2., Appendix E, P100/2013. dr

ULS=c∗q∗dr<draULS

drULS - relative story displacement under the seismic action associated to ULS

q - behavior factor specific to the structure’s typedr -the relative displacement, determined by static computation under the seismic loadsc- the displacement amplification factor, which takes into account that for T<Tc the

seismic displacements computed in the inelastic domain are larger than those corresponding to elastic seismic response.

1 ≤ c=3−2.5∗T1

T c

≤ 2 ≤ dr ,admisULS =2.5 %

c=3−2.5∗T 1

T c

=3−2.5∗0.3331.6

=2.48=2

dr ,admisULS - allowable value of the relative story displacement, equal to 2.5% of Hs

Table 6-8 SX drift checkLevel Direction Combination dre ν q d,r,sls dadm

Story 5 SX PSXPSY 0.000378 0,5 5,75 0.008694 0,075 OKStory 4 SX PSXPSY 0.000407 0,5 5,75 0.009361 0,075 OKStory 3 SX PSXPSY 0.000422 0,5 5,75 0.009706 0,075 OKStory 2 SX PSXPSY 0.000399 0,5 5,75 0.009177 0,075 OKStory 1 SX PSXPSY 0.000323 0,5 5,75 0.007429 0,075 OKBase SX PSXPSY 0.000165 0,5 5,75 0.003795 0,075 OK

21

Table 6-9 SY drift checkLevel Direction Combination dre ν q d,r,sls dadm

Story 5 SY PSYNSX 0.001219 0,5 5,75 0.028037 0,075 OKStory 4 SY PSYNSX 0.001219 0,5 5,75 0.028037 0,075 OKStory 3 SY PSYNSX 0.001268 0,5 5,75 0.029164 0,075 OKStory 2 SY PSYNSX 0.00121 0,5 5,75 0.02783 0,075 OKStory 1 SY PSYNSX 0.000986 0,5 5,75 0.022678 0,075 OKBase SY PSYNSX 0.00049 0,5 5,75 0.01127 0,075 OK

22

7 DIMENSIONING OF THE STRUCTURAL ELEMENTS

7.1 Slab

7.1.1 General computation schemeIn the case of continuous beams and slabs the permanent load together with the

temporary load can determine several loading schemes. The efforts are determined for the most unfavorable loading scheme corresponding to each section.

To determine the maximum positive moment in a field, the permanent load is considered to be applied in all the spans and the temporary load in the respective opening, and in alternative spans.

To determine the maximum negative moment in a field, the permanent load is considered to be applied in all openings and the temporary load in the adjacent openings, but also in alternate openings.

For determining the maximum negative moment in the support, the permanent load is considered to be applied in all openings and temporary load in the adjacent openings of the respective support and in that alternate openings.

7.1.2 Static computation in the elastic-range slabs reinforced on one direction

The slabs are computed on one direction if λ¿0. 5∨ λ>2 where λ=Ll

The computation of the efforts in the section of the slabs reinforced on one direction due to permanent or temporary loads, applied uniform, is done like in the case of isolated or continuous beams, computed in the elastic range for a strip of slab with the width equal to the unit.

The slab is loaded with the sum of the temporary and permanent loads.

λ=Ll=14.6

5=2.92 →one way slab (which is reinforced on the shorter direction)

Positive moment in field: M= 124∗q∗l2

Negative moment in support: M= 112∗q∗l2

Slab 2 q(kN/m2) l(m) M field(KN*m) M support (KN*m)Current floor 13.13 5 13.67708 27.35417Terrace 10.16 5 10.58333 21.16667

23

7.1.3 Static computation in the elastic-range slabs reinforced on two directions

The slabs are computed on two directions if 0.5 ≤ λ ≤ 2, where λ=Ll

.

The maximum and minimum moments are computed based on two loading schemes, based on the support type of the slab.

In the first load scheme the panels are considered fixed supported or simply supported based on their contour. The load applied is:

q=g+ p2

, where:

g is the permanent loadp is the variable loadIn the second conventional scheme the panels are considered simply supported on all the

sides. On all the panels is applied a conventional load directed up-down and respectively down-

up in all the possible ways. This load is: q'= p

2

Slab 1

q=qselfweight+q flooring+qpartitionwalls=5.74+2.36+2.03=10.13kN

m2−permanent load

P=qlive=3kN

m2−variable load

Scheme 1 Scheme 2a Scheme 2b Scheme 2c

q=q+P2

λ=Ll

M 1=α 41∗q∗l12

M 2=α 42∗q∗l22

q1=β41∗q

M 1' =

−−q1∗l12

8

q= P2

λ=Ll

M 1=α 11∗q∗l12

M 2=α 12∗q∗l22

q= P2

λ=Ll

q1=β21∗q

M 1' =

−−q1∗l12

8

q= P2

λ '= lL

q2=β22∗q

M 2' =

−−q2∗l22

8

24

q2=β42∗q

M 2' =

−−q2∗l22

8

25

Example:

Scheme 1

q=q+P2=10. 13+1. 5=11.63 kN /m2

λ=Ll= 5

4.05=1.23

M 1=0.03832∗11.63∗4.052=7.31 kN∗m

M 2=0.01688∗11.63∗52=4.91 kN∗m

q1=0.6945∗11.63=8.08 kN /m2

M 1' =−−8.08∗4.052

8=16.57 kN∗m

q2=0.3055∗11.63=3.55 kN /m2

M 2' =−−3.55∗52

8=11.09kN∗m

Scheme 2a

q= P2=1.5 kN /m2

λ=Ll= 5

4.05=1.23

M 1=0.05362∗1.5∗4.052=1.32 kN∗m M 2=0.02354∗1.5∗52=0.88 kN∗m Scheme 2b

q= P2=1.5 kN /m2

λ=Ll= 5

4.05=1.23

q1=0.84997∗1.5=1.275 kN /m2 M 1

' =−−1.275∗4.052

8=2.61 kN∗m

Scheme 2c

q= P2=1.5 kN /m2

λ '= lL=0.81

q2=0.48206∗1.5=0.723 kN /m2 M 2

' =−−0.723∗52

8=2.26 kN∗m

Total momentM midspan shortside=7.31+1.32=8.63 kN∗m M support shortside=16.57+2.61=19.18 kN∗m M midspanlongside=4.91+0.88=5.79 kN∗m M support longside=7.28+2.26=9.54 kN∗m

26

Table 7-10 Moment computation for the current floor slabs

q+p/2= 11.63 L1= 4.05

p/2= 1.50 L2= 5.00

Slab 1

λ α1 α2 β1 β2M1

(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.23 0.038 0.017 0.695 0.306 7.31 16.56 4.91 11.10

a 1.23 0.054 0.024 - - 1.32 - 0.88 -

b 1.23 - - 0.850 - - 2.61 - -

c 0.81 - - - 0.482 - - - 2.26

8.62 19.17 5.79 13.36

q+p/2= 11.63 L1= 3.90

p/2= 1.50 L2= 4.60

Slab 5


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.18 0.03 0.012 0.806 0.194 5.31 17.82 2.93 5.98

a 1.18 0.05 0.025 - - 1.14 - 0.79 -

b 1.18 0.04 - 0.838 - - 2.39 - -

c 0.85 - 0.04 - 0.434 - - - 1.72

6.45 20.21 3.72 7.70

q+p/2= 11.63 L1= 3.00

p/2= 1.50 L2= 3.90

Slab 6


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.30 0.03 0.008 0.761 0.232 3.14 9.96 1.49 5.12

a 1.30 0.06 0.019 - - 0.81 - 0.42 -

b 1.30 0.05 - 0.892 - - 1.51 - -

c 0.77 - 0.052 - 0.558 - - - 1.59

3.95 11.47 1.91 6.71

q+p/2= 11.63 L1= 3.90

p/2= 1.50 L2= 5.50

Slab 7


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.28 0.03 0.008 0.761 0.232 5.31 16.83 2.96 10.18

a 1.28 0.06 0.019 - - 1.37 - 0.84 -

b 1.28 0.05 - 0.892 - - 2.54 - -

c 0.78 - 0.052 - 0.558 - - - 3.17

6.68 19.38 3.80 13.35

q+p/2= 11.63 L1= 4.20

p/2= 1.50 L2= 4.60

27

Slab 10


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.10 0.02 0.02 0.667 0.333 4.10 17.10 4.87 10.25

a 1.10 0.04 0.037 - - 1.06 - 1.16 -

b 1.10 0.03 - 0.714 - - 2.36 - -

c 0.91 - 0.027 - 0.286 - - - 1.13

5.16 19.46 6.03 11.39

q+p/2= 11.63 L1= 3.00

p/2= 1.50 L2= 4.20

Slab 11


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.40 0.03 0.008 0.761 0.232 3.14 9.96 1.72 5.94

a 1.40 0.06 0.019 - - 0.81 - 0.49 -

b 1.40 0.05 - 0.892 - - 1.51 - -

c 0.71 - 0.052 - 0.558 - - - 1.85

3.95 11.47 2.22 7.78

q+p/2= 11.63 L1= 2.60

p/2= 1.50 L2= 4.20

Slab 12


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.62 0.05 0.006 0.893 0.107 3.93 8.78 1.31 2.74

a 1.62 0.08 0.01 - - 0.81 - 0.26 -

b 1.62 0.06 - 0.954 - - 1.21 - -

c 0.62 - 0.073 - 0.755 - - - 2.50

4.74 9.99 1.57 5.24

q+p/2= 11.63 L1= 4.20

p/2= 1.50 L2= 5.50

Slab 13


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.31 0.03 0.008 0.761 0.232 6.15 19.52 2.96 10.18

a 1.31 0.06 0.019 - - 1.59 - 0.84 -

b 1.31 0.05 - 0.892 - - 2.95 - -

c 0.76 - 0.052 - 0.558 - - - 3.17

7.74 22.47 3.80 13.35

28

Table 7-11 Moment computation for the terrace slabs

q+p/2= 7.76 L1= 3.90

p/2= 1.20 L2= 4.60

Slab 5


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.18 0.03 0.012 0.806 0.194 3.54 11.89 1.97 3.98

a 1.18 0.05 0.025 - - 0.91 - 0.63 -

b 1.18 0.04 - 0.838 - - 1.91 - -

c 0.85 - 0.04 - 0.434 - - - 1.38

4.45 13.80 2.61 5.36

q+p/2= 7.76 L1= 3.00

p/2= 1.20 L2= 3.90

Slab 6


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.30 0.03 0.008 0.761 0.232 2.10 6.64 0.94 3.42

a 1.30 0.06 0.019 - - 0.65 - 0.35 -

b 1.30 0.05 - 0.892 - - 1.20 - -

c 0.77 - 0.052 - 0.558 - - - 1.27

2.74 7.85 1.29 4.70

q+p/2= 7.76 L1= 3.90

p/2= 1.20 L2= 5.00

Slab 7


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.28 0.03 0.008 0.761 0.232 3.54 11.23 1.55 5.63

a 1.28 0.06 0.019 - - 1.10 - 0.57 -

b 1.28 0.05 - 0.892 - - 2.04 - -

c 0.78 - 0.052 - 0.558 - - - 2.09

4.64 13.26 2.12 7.72

q+p/2= 7.76 L1= 4.20

p/2= 1.20 L2= 4.60

Slab 10


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.10 0.02 0.02 0.667 0.333 2.74 11.41 3.28 6.83

a 1.10 0.04 0.037 - - 0.85 - 0.94 -

b 1.10 0.03 - 0.714 - - 1.89 - -

c 0.91 - 0.027 - 0.286 - - - 0.91

3.58 13.30 4.22 7.74

q+p/2= 7.76 L1= 3.00

p/2= 1.20 L2= 4.20

29

Slab 11


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.40 0.03 0.008 0.761 0.232 2.10 6.64 1.10 3.97

a 1.40 0.06 0.019 - - 0.65 - 0.40 -

b 1.40 0.05 - 0.892 - - 1.20 - -

c 0.71 - 0.052 - 0.558 - - - 1.48

2.74 7.85 1.50 5.45

q+p/2= 7.76 L1= 2.60

p/2= 1.20 L2= 4.20

Slab 12


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.62 0.05 0.006 0.893 0.107 2.62 5.86 0.82 1.83

a 1.62 0.08 0.01 - - 0.65 - 0.21 -

b 1.62 0.06 - 0.954 - - 0.97 - -

c 0.62 - 0.073 - 0.755 - - - 2.00

3.27 6.82 1.03 3.83

q+p/2= 7.76 L1= 4.20

p/2= 1.20 L2= 5.50

Slab 13


(kN*m)M1'

(kN*m)M2

(kN*m)M2'

(kN*m)

1.31 0.05 0.006 0.893 0.107 6.84 15.28 1.41 3.14

a 1.31 0.08 0.01 - - 1.69 - 0.36 -

b 1.31 0.06 - 0.954 - - 2.52 - -

c 0.76 - 0.073 - 0.755 - - - 3.43

8.54 17.80 1.77 6.57

7.1.4 Dimensioning the reinforcementBased on the mean of the maximum values of the bending moment in each support the

reinforcement can be determined for each slab. The computation is performed by considering a rectangular cross section with the length

equal to unit. If the length of the strip is considered 1 m, the resulting reinforcement will be laid on 1 m of slab. On the drawing the representation of the reinforcement is done by the number of bars on meter.

The necessary reinforcement is:Anecc=b∗x∗f cd / f yd Where: b=1000 mm - the unitary stripx - the depth of the neutral axis

x=d (1−√1−2∗M max

b∗d2∗fcd)

Where: d - the width without the concrete coveringfcd - the strength of concrete in compressionfyd - the yielding strength of steel

The constructive rules which have to be respected are: The minimum diameter of the bars is 6 mm for the bottom part and 8 mm(OB37)

or 6 mm(PC52) for the upper part and for the inclined bars.

30

For hp<30, the minimum number of bars per meter is 5 and the maximum number is 12.

The minimum reinforcement area (SR-EN-1992-1-1-2004).

A s ,min=0.26∗f ctm

f yk

∗b∗d, is equal to 443.3 mm2 for the computed slab.

From the Romanian codes, the minimum reinforcement percentage should be ρ>0.20%.

Perpendicular on the direction of the strength reinforcement, found by computations, repartition reinforcement has to be placed. This reinforcement has the following purposes: constructive because it takes over bending moments from the lower part, from the vicinity of the support and serves to ensure a good behavior in case of concentrated loads; for mounting the reinforcement nets because it holds the reinforcement bars from migrating before and during the pouring and the vibrating of concrete.

For the regular slabs the repartition reinforcement should be at least 15%of the load strengthening reinforcement, but minimum 4Ø6.

The number of bars should be multiple of 2.5 and 3. The length of the bars on one support is the minimum between l1/4, l2/4.

31

Table 7-12 Current floor slab reinforcement

SLAB TYPEM

[kN·m]

hs

[mm]fcd

[N/mm2]d

[mm]x

[mm]

fyd

[N/mm²]

Anec

[mm²]Reinforcement

Aeff

[mm²]

Steel percentage

[%)

1Mid-span

x direction 8.62 170 13.33 165 3.97 300 176 3φ8+2φ12/m 377.00 0.23y direction 5.79 170 13.33 155 2.83 300 126 3φ8+2φ12/m 377.00 0.24

Supportx direction 19.17 170 13.33 165 8.96 300 398 4φ8+2φ12/m 427.26 0.26y direction 13.36 170 13.33 155 6.61 300 294 3φ8+2φ12/m 377.00 0.24

2Mid-span y direction 13.68 170 13.33 155 6.77 300 301 3φ8+2φ12/m 377.00 0.24

Supporty direction up 27.35 170 13.33 155 13.86 300 616 6φ12/m 678.58 0.44y direction down 27.35 170 13.33 155 13.86 300 616 6φ12/m 678.58 0.44

5

Mid-spanx direction 3.72 170 13.33 165 1.70 300 76 3φ8+2φ12/m 377.00 0.23y direction 6.45 170 13.33 155 3.15 300 140 3φ8+2φ12/m 377.00 0.24

Support

x direction left 7.70 170 13.33 165 3.54 300 157 3φ8+2φ12/m 377.00 0.23x direction right 7.70 170 13.33 165 3.54 300 157 3φ8+2φ12/m 377.00 0.23y direction up 20.21 170 13.33 155 10.11 300 449 5φ12/m 565.49 0.36y direction down 20.21 170 13.33 155 10.11 300 449 5φ12/m 565.49 0.36

6


Support

x direction left 11.47 170 13.33 165 5.30 300 235 3φ8+2φ12/m 377.00 0.23x direction right 11.47 170 13.33 165 5.30 300 235 3φ8+2φ12/m 377.00 0.23y direction up 6.71 170 13.33 155 3.28 300 146 3φ8+2φ12/m 377.00 0.24y direction down 6.71 170 13.33 155 3.28 300 146 3φ8+2φ12/m 377.00 0.24

7


Support


10 Mid-span x direction 5.16 170 13.33 165 2.36 300 105 3φ8+2φ12/m 377.00 0.23

32

y direction 6.03 170 13.33 155 2.95 300 131 3φ8+2φ12/m 377.00 0.24

Support


11


Support


12


Support


13


Support


33

Table 7-13 Terrace slab reinforcement

SLAB TYPEM

[kN·m]hs

[mm]fcd

[N/mm2]d

[mm]x

[mm]fyd

[N/mm²]Anec

[mm²]Reinforcement

Aeff

[mm²]

Steel percentage

[%)

2Mid-span y direction 1.58 170 13.33 155 0.77 300 34 3φ8+2φ12/m 377.00 0.24

Supporty direction up 21.17 170 13.33 155 10.61 300 471 5φ12/m 565.49 0.36y direction down 21.17 170 13.33 155 10.61 300 471 5φ12/m 565.49 0.36

5


Support


6


Support


7


Support


10


Supportx direction left 7.74 170 13.33 165 3.56 300 158 3φ8+2φ12/m 377.00 0.23x direction right 7.74 170 13.33 165 3.56 300 158 3φ8+2φ12/m 377.00 0.23y direction up 13.30 170 13.33 155 6.58 300 292 3φ8+2φ12/m 377.00 0.24y direction down 13.30 170 13.33 155 6.58 300 292 3φ8+2φ12/m 377.00 0.24

34

11Mid-span


Support

x direction left 7.85 170 13.33 165 3.61 300 160 3φ8+2φ12/m 377.00 0.23

x direction right 7.85 170 13.33 165 3.61 300 160 3φ8+2φ12/m 377.00 0.23y direction up 5.45 170 13.33 155 2.66 300 118 3φ8+2φ12/m 377.00 0.24y direction down 5.45 170 13.33 155 2.66 300 118 3φ8+2φ12/m 377.00 0.24

12


Support


13Mid-span


Supportx direction left 6.57 170 13.33 165 3.01 300 134 3φ8+2φ12/m 377.00 0.23x direction right 6.57 170 13.33 165 3.01 300 134 3φ8+2φ12/m 377.00 0.23y direction up 17.80 170 13.33 155 8.87 300 394 5φ12/m 565.49 0.36y direction down 17.80 170 13.33 155 8.87 300 394 5φ12/m 565.49 0.36

35

7.2 BeamsThe bending moments were obtained through automatic computation in ETABS. To

obtain the needed moments, the building was loaded with the envelope of the load combinations. The results displayed the maximum value of the bending moment as shown in the figure below:

7.2.1 Design of the longitudinal reinforcement MEd = Mdesign the effective bending moment from the envelope.In the present project the beams were declared as T-shape but in computation the cross

section is considered rectangular because ¿>¿ so A s−¿>A s

+¿ ¿¿. As a result beff is involved in the computations for the beams which are done on a rectangular section.

The necessary area of reinforcement is:

A s ,necc

+¿=M Ed

+¿

(d−c )∗f yd

¿¿

Where:M Ed

+¿¿ is the maximum value of the effective bending moment from ETABS.

c is the concrete coverBased on the chosen diameters of reinforcement the area given by the actual reinforce

layout is computed: As1The reinforcement percentage is:

ρ=A s 1

bw∗d

The capable moment of the section is:M Rb

+¿=A s 1∗f yd∗d ¿ For the top reinforcement we assume that x<2*cAs a result:

A s ,necc

−¿=M Ed

−¿

(d−c )∗f yd

¿¿

Then x is computed for the effective reinforcement:

x=( A s 2

eff−A s 1eff )∗f yd

b∗f cd

If x<2a the area of reinforcement is computed correctly, otherwise the section is computed as a double reinforced section.

36

M Rd=bw∗f cd∗(d− x2 )+A s 1

eff∗f yd∗(d−c)

Provisions for the longitudinal reinforcement: The minimum diameter is Ø12. At least 2 bars Ø14 should be placed in the bottom and top parts of the beams. The compressed reinforcement at the ends of the beams should be at least 50% of

the tensioned reinforcement. At the upper part of the beam at least one quarter of the top reinforcement should

be continuous.

The reinforcement ratio ρ=A s

eff

bw∗d≤ ρmin=0.5

f ctm

f yk

(for C20/25 ρmin=¿ 0. 32%)

Only 2 or maximum 3 nonconsecutive diameters are used. The minimum clear distance between rebars is 30 mm, and at least one space of

50 mm for the top rebars. For the bottom rebars the distance should be 25mm. The maximum clear distance is 200 mm.

Example for beam 4 frame C-CM Ed

+¿=58.81 kN∗m¿

A s ,necc

+¿= 58.81∗1000000(550−2∗35)∗300

=408.40 mm2¿

The chosen diameter is 1Ø12 and 2Ø16 so the A s 1eff=515.22 mm2

M Rd

+¿=515.22∗300∗4801000000

=74.19 kN∗m¿

The reinforcement percentage is ρ=74.19

300∗(550−35)=0.33%

M Ed−¿=110.47kN∗m¿

A s ,necc

−¿= 110.47∗1000000(550−2∗35)∗300

=767.15 mm2¿

The chosen diameter is 4Ø12 + 2Ø16 so the A s 2eff=854.51 mm2

x=(854.51−515.22)∗300

300∗13.13=25.45 <2*c

M Rd=300∗13.33∗(550−25.452 )+515.22∗300∗(550−35 )=81.75 kN∗m

ρ= 854.51300∗515

=0.55 %

37

Table 7-14 Field reinforcement for beams in Frame C-C

Story FieldMEd+ [kN·m]

fyd [N/mm2]

as [mm]

bb [mm]

hb [mm]

hs=hb

-2as

[mm]

As2nec

[mm2]Diameter

[mm]

Aeff = n·π·φ2/4 [mm2]

ρMRd+ [kN·m]

GF

1 4.75 300 35 300 550 480 33 1φ12+2φ16 515.22 0.0033 74.192 63.76 300 35 300 550 480 443 1φ12+2φ16 515.22 0.0033 74.193 12.01 300 35 300 550 480 83 1φ12+2φ16 515.22 0.0033 74.194 58.81 300 35 300 550 480 408 1φ12+2φ16 515.22 0.0033 74.195 4.68 300 35 300 550 480 33 1φ12+2φ16 515.22 0.0033 74.19

1

1 7.88 300 35 300 550 480 55 1φ12+2φ16 515.22 0.0033 74.192 63.67 300 35 300 550 480 442 1φ12+2φ16 515.22 0.0033 74.193 14.30 300 35 300 550 480 99 1φ12+2φ16 515.22 0.0033 74.194 58.76 300 35 300 550 480 408 1φ12+2φ16 515.22 0.0033 74.195 17.68 300 35 300 550 480 123 1φ12+2φ16 515.22 0.0033 74.19

2

1 10.89 300 35 300 550 480 76 1φ12+2φ16 515.22 0.0033 74.192 64.06 300 35 300 550 480 445 1φ12+2φ16 515.22 0.0033 74.193 17.02 300 35 300 550 480 118 1φ12+2φ16 515.22 0.0033 74.194 58.98 300 35 300 550 480 410 1φ12+2φ16 515.22 0.0033 74.195 17.21 300 35 300 550 480 120 1φ12+2φ16 515.22 0.0033 74.19

3

1 12.70 300 35 300 550 480 88 1φ12+2φ16 515.22 0.0033 74.192 64.42 300 35 300 550 480 447 1φ12+2φ16 515.22 0.0033 74.193 18.49 300 35 300 550 480 128 1φ12+2φ16 515.22 0.0033 74.194 59.17 300 35 300 550 480 411 1φ12+2φ16 515.22 0.0033 74.195 20.20 300 35 300 550 480 140 1φ12+2φ16 515.22 0.0033 74.19

4

1 12.54 300 35 300 550 480 87 1φ12+2φ16 515.22 0.0033 74.192 64.03 300 35 300 550 480 445 1φ12+2φ16 515.22 0.0033 74.193 19.88 300 35 300 550 480 138 1φ12+2φ16 515.22 0.0033 74.194 59.09 300 35 300 550 480 410 1φ12+2φ16 515.22 0.0033 74.195 20.56 300 35 300 550 480 143 1φ12+2φ16 515.22 0.0033 74.19

52 53.36 300 35 300 550 480 371 1φ12+2φ16 515.22 0.0033 74.193 14.71 300 35 300 550 480 102 1φ12+2φ16 515.22 0.0033 74.194 48.60 300 35 300 550 480 338 1φ12+2φ16 515.22 0.0033 74.19

38

Table 7-15 Support reinforcement for beams in Frame C-C

Story SupMEd

-

[kN·m]fyd

[N/mm2]as

[mm]bb

[mm]hb

[mm]

hs=hb -2as

[mm]

As2nec

[mm2]Diameter

[mm]

Aeff=n·π ·φ2/4

[mm2]

xn

[mm]ρ

MRd-

[kN·m]

GF

1 12.67 300 35 300 550 480 88 1φ12+2φ16 515.22 0 0.0033 74.19

2 60.31 300 35 300 550 480 419 1φ12+2φ16 515.22 0 0.0033 74.193 32.94 300 35 300 550 480 229 1φ12+2φ16 515.22 0 0.0033 74.194 110.47 300 35 300 550 480 767 4φ12+2φ16 854.51 25 0.0055 123.055 11.03 300 35 300 550 480 77 1φ12+2φ16 515.22 0 0.0033 74.19

1

1 14.88 300 35 300 550 480 103 1φ12+2φ16 515.22 0 0.0033 74.192 60.31 300 35 300 550 480 419 1φ12+2φ16 515.22 0 0.0033 74.193 44.52 300 35 300 550 480 309 1φ12+2φ16 515.22 0 0.0033 74.194 114.15 300 35 300 550 480 793 4φ12+2φ16 854.51 25 0.0055 123.055 17.68 300 35 300 550 480 123 1φ12+2φ16 515.22 0 0.0033 74.19

2

1 17.60 300 35 300 550 480 122 1φ12+2φ16 515.22 0 0.0033 74.192 61.14 300 35 300 550 480 425 1φ12+2φ16 515.22 0 0.0033 74.193 51.02 300 35 300 550 480 354 1φ12+2φ16 515.22 0 0.0033 74.194 117.35 300 35 300 550 480 815 4φ12+2φ16 854.51 25 0.0055 123.055 21.60 300 35 300 550 480 150 1φ12+2φ16 515.22 0 0.0033 74.19

3

1 20.23 300 35 300 550 480 140 1φ12+2φ16 515.22 0 0.0033 74.192 61.73 300 35 300 550 480 429 1φ12+2φ16 515.22 0 0.0033 74.193 53.64 300 35 300 550 480 373 1φ12+2φ16 515.22 0 0.0033 74.194 119.68 300 35 300 550 480 831 4φ12+2φ16 854.51 25 0.0055 123.055 25.51 300 35 300 550 480 177 1φ12+2φ16 515.22 0 0.0033 74.19

4

1 14.91 300 35 300 550 480 104 1φ12+2φ16 515.22 0 0.0033 74.192 62.35 300 35 300 550 480 433 1φ12+2φ16 515.22 0 0.0033 74.193 54.70 300 35 300 550 480 380 1φ12+2φ16 515.22 0 0.0033 74.194 120.92 300 35 300 550 480 840 4φ12+2φ16 854.51 25 0.0055 123.055 19.70 300 35 300 550 480 137 1φ12+2φ16 515.22 0 0.0033 74.19

52 49.06 300 35 300 550 480 341 1φ12+2φ16 515.22 0 0.0033 74.193 46.89 300 35 300 550 480 326 1φ12+2φ16 515.22 0 0.0033 74.194 101.90 300 35 300 550 480 708 4φ12+2φ16 854.51 25 0.0055 123.05

39

Table 7-16 Field reinforcement for beams in Frame 6-6

Story FieldMEd+ [kN·m]

fyd [N/mm2]

as [mm]

bb [mm]

hb [mm]

hs=hb

-2as

[mm]

As2nec

[mm2]Diameter

[mm]

Aeff = n·π·φ2/4 [mm2]

ρMRd+ [kN·m]

GF1 55.65 300 35 300 550 480 386 1φ12+2φ16 515.22 0.0033 74.192 69.06 300 35 300 550 480 480 1φ12+2φ16 515.22 0.0033 74.193 34.59 300 35 300 550 480 240 1φ12+2φ16 515.22 0.0033 74.19

11 54.59 300 35 300 550 480 379 1φ12+2φ16 515.22 0.0033 74.192 69.22 300 35 300 550 480 481 1φ12+2φ16 515.22 0.0033 74.193 34.8 300 35 300 550 480 242 1φ12+2φ16 515.22 0.0033 74.19

21 54.67 300 35 300 550 480 380 1φ12+2φ16 515.22 0.0033 74.192 69.4 300 35 300 550 480 482 1φ12+2φ16 515.22 0.0033 74.193 35.1 300 35 300 550 480 244 1φ12+2φ16 515.22 0.0033 74.19

31 54.46 300 35 300 550 480 378 1φ12+2φ16 515.22 0.0033 74.192 69.68 300 35 300 550 480 484 1φ12+2φ16 515.22 0.0033 74.193 35.19 300 35 300 550 480 244 1φ12+2φ16 515.22 0.0033 74.19

41 54.64 300 35 300 550 480 379 1φ12+2φ16 515.22 0.0033 74.192 69.3 300 35 300 550 480 481 1φ12+2φ16 515.22 0.0033 74.193 35.64 300 35 300 550 480 248 1φ12+2φ16 515.22 0.0033 74.19

51 32.88 300 35 300 550 480 228 1φ12+2φ16 515.22 0.0033 74.192 51.11 300 35 300 550 480 355 1φ12+2φ16 515.22 0.0033 74.193 20.08 300 35 300 550 480 139 1φ12+2φ16 515.22 0.0033 74.19

40

Table 7-17 Support reinforcement for beams in Frame 6-6

Story SupMEd

-

[kN·m]fyd

[N/mm2]as

[mm]bb

[mm]hb

[mm]

hs=hb -2as

[mm]

As2nec

[mm2]Diameter

[mm]

Aeff=n·π ·φ2/4

[mm2]

xn

[mm]ρ

MRd-

[kN·m]

GF1 85.03 300 35 300 550 480 590 1φ12+3φ16 716.28 15 0.0046 81.52

2 69.23 300 35 300 550 480 481 1φ12+3φ16 716.28 15 0.0046 103.143 62.02 300 35 300 550 480 431 1φ12+3φ16 716.28 15 0.0046 103.14

11 91.84 300 35 300 550 480 638 1φ12+3φ16 716.28 15 0.0046 103.142 70.14 300 35 300 550 480 487 1φ12+3φ16 716.28 15 0.0046 103.143 75.32 300 35 300 550 480 523 1φ12+3φ16 716.28 15 0.0046 103.14

21 94.44 300 35 300 550 480 656 1φ12+3φ16 716.28 15 0.0046 103.142 73.25 300 35 300 550 480 509 1φ12+2φ16 515.22 0 0.0033 74.193 82.39 300 35 300 550 480 572 1φ12+3φ16 716.28 15 0.0046 103.14

31 96.87 300 35 300 550 480 673 1φ12+3φ16 716.28 15 0.0046 103.142 73.16 300 35 300 550 480 508 1φ12+2φ16 515.22 0 0.0033 74.193 84.96 300 35 300 550 480 590 1φ12+3φ16 716.28 15 0.0046 103.14

41 98.26 300 35 300 550 480 682 1φ12+3φ16 716.28 15 0.0046 103.142 72.66 300 35 300 550 480 505 1φ12+2φ16 515.22 0 0.0033 74.193 85.11 300 35 300 550 480 591 1φ12+3φ16 716.28 15 0.0046 103.14

52 57.33 300 35 300 550 480 398 1φ12+2φ16 515.22 0 0.0033 74.193 46.81 300 35 300 550 480 325 1φ12+2φ16 515.22 0 0.0033 74.194 60.49 300 35 300 550 480 420 1φ12+2φ16 515.22 0 0.0033 74.19

41

7.2.2 Design of the transversal reinforcement of the beamsThe shearing forces for computations were found using equilibrium of the beam under the

action of the transversal loading due to earthquake and the moments acting on the ends of the beam. The design is performed in order to place the plastic hinges at the ends of the beam.

The principle of computation is based on computing two values of the shear force corresponding to the maximum and minimum positive moments developing in the two ends of the beams. Since the shear failure is brittle we have to amplify the shear force produced by the earthquake with a safety coefficient and finally the design shear forces will be computed with the following equation

V Ed−¿=−γ Rb∗¿¿ ¿

V Ed+¿=γ Rb∗¿ ¿¿

qbeam¿ =

qslab¿ ∗A trib

L+bw∗(hw−hs )∗γ Rd

γ Rb coefficient which takes into account a possible over resistance due to increased rigidity of the steel to deformation = 1. 2

Provisions for computing the transversal reinforcement:Øw

min= 6 mm for h≤ 800mm

s≤[ hb

4;15∗d long ;200 mm]= [137.5 ;180 ;200 ]=137.5

Choose Øw and the spacing s which can have values between 75. . . 200mm with a 25 mm modulus

For ς=V Ed , min

❑

Vedmax>0.5−¿ or ς←0. 5∧|V Edmax|< (2+ς )∗bw∗d∗f cdt

A sw

s≥

V Ed , maxel

0. 9∗d∗f ywd

because θ=45b and cotθ=1

A sw

s≥ 0.

2∗bw100

The diameter and the bars are chosen in order to comply with the conditions 1 and 2.

For ς=V Ed ,max❑

0. 9∗d∗f yd

←0. 5 and |V Edmax|<(2+ς )∗bw∗d∗f cdt are needed inclined

reinforcement in two orthogonal directions and half of the design shear force should be resisted by stirrups and half by the inclined reinforcement.

Example beam 1 frame C-C

V Ed

−¿=−1. 20∗74.19+74.191 .60

± 28.8∗1.62

=134.33kN ¿

V Ed

−¿=−1. 20∗74.19+74.191 .60

± 28.8∗1.62

=−88.25 kN ¿

qbeam¿ =13.13∗3.16

1.6+0. 3∗(0.55−0.17 )∗25=28.8 kN /m2

ς=V Ed , min

❑

Vedmax=−88.25

134.33=−0.66>−0.5 θ=45o

ς=V Ed ,max❑

0.9∗d∗f yd

= 134.330.9∗0.515∗210∗1000

=1.38mm

Are chosen 2Ø10 at s=100mmAss

=1.57 so it verifies

42

Table 7-18 Transversal reinforcement for beams in frame C-C

Story Field(|MRd

+|+|MRd-|)

*1.2/ L [KN]qltv*L/2 [KN]

VEdmax

[kN]VEd

min

[kN]ξ=VEd

min/VEdmax Incline

d Barsd=hb-as

[mm]bb

[mm]ν=0.6(1-fck/250)

fcd

[N/mm2]|VEd

max| [kN]

0.5·bb·d·ν·fcd[

kN]

z [mm]

GF

1-2 111.29 23.04 134.33 -88.25 -0.66 >-0.5 No 515 300 0.55 13.33 134.33 < 566.36 4642-3 41.41 91.60 133.01 50.19 0.38 >-0.5 No 515 300 0.55 13.33 133.01 < 566.36 4643-4 62.48 39.59 102.07 -22.89 -0.22 >-0.5 No 515 300 0.55 13.33 102.07 < 566.36 4645-6 44.24 94.98 139.22 50.74 0.36 >-0.5 No 515 300 0.55 13.33 139.22 < 566.36 4646-7 142.45 21.29 163.74 -121.16 -0.74 >-0.5 No 515 300 0.55 13.33 163.74 < 566.36 464

1

1-2 111.29 26.39 137.68 -84.90 -0.62 >-0.5 No 515 300 0.55 13.33 137.68 < 566.36 4642-3 41.41 91.10 132.51 49.69 0.37 >-0.5 No 515 300 0.55 13.33 132.51 < 566.36 4643-4 62.48 42.80 105.28 -19.68 -0.19 >-0.5 No 515 300 0.55 13.33 105.28 < 566.36 4645-6 44.24 96.19 140.43 51.95 0.37 >-0.5 No 515 300 0.55 13.33 140.43 < 566.36 4646-7 142.45 31.86 174.31 -110.59 -0.63 >-0.5 No 515 300 0.55 13.33 174.31 < 566.36 464

2

1-2 111.29 29.66 140.95 -81.63 -0.58 >-0.5 No 515 300 0.55 13.33 140.95 < 566.36 4642-3 41.41 90.63 132.04 49.22 0.37 >-0.5 No 515 300 0.55 13.33 132.04 < 566.36 4643-4 62.48 45.76 108.24 -16.72 -0.15 >-0.5 No 515 300 0.55 13.33 108.24 < 566.36 4645-6 44.24 97.33 141.57 53.09 0.37 >-0.5 No 515 300 0.55 13.33 141.57 < 566.36 4646-7 142.45 37.79 180.24 -104.66 -0.58 >-0.5 No 515 300 0.55 13.33 180.24 < 566.36 464

3

1-2 111.29 32.21 143.50 -79.08 -0.55 >-0.5 No 515 300 0.55 13.33 143.50 < 566.36 4642-3 41.41 90.26 131.67 48.85 0.37 >-0.5 No 515 300 0.55 13.33 131.67 < 566.36 4643-4 62.48 47.22 109.70 -15.26 -0.14 >-0.5 No 515 300 0.55 13.33 109.70 < 566.36 4645-6 44.24 98.16 142.40 53.92 0.38 >-0.5 No 515 300 0.55 13.33 142.40 < 566.36 4646-7 142.45 45.65 188.10 -96.80 -0.51 >-0.5 No 515 300 0.55 13.33 188.10 < 566.36 464

4

1-2 111.29 30.62 141.91 -80.67 -0.57 >-0.5 No 515 300 0.55 13.33 141.91 < 566.36 4642-3 41.41 90.17 131.58 48.76 0.37 >-0.5 No 515 300 0.55 13.33 131.58 < 566.36 4643-4 62.48 48.05 110.53 -14.43 -0.13 >-0.5 No 515 300 0.55 13.33 110.53 < 566.36 4645-6 44.24 98.55 142.79 54.31 0.38 >-0.5 No 515 300 0.55 13.33 142.79 < 566.36 4646-7 142.45 43.80 186.25 -98.65 -0.53 >-0.5 No 515 300 0.55 13.33 186.25 < 566.36 464

52-3 41.41 71.64 113.05 30.23 0.27 >-0.5 No 515 300 0.55 13.33 113.05 < 566.36 4643-4 62.48 39.31 101.79 -23.17 -0.23 >-0.5 No 515 300 0.55 13.33 101.79 < 566.36 4645-6 44.24 81.90 126.14 37.66 0.30 >-0.5 No 515 300 0.55 13.33 126.14 < 566.36 464

43

fywd

[N/mm2]smax

[mm]smin

[mm]θ

cot θ

Choosen stirrups

S [mm

]

Choosen S [mm]

pw [%]

210137.

5100 45° 1 2

φ10 114 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 115 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 150 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 110 100 0.5

>0.2%

210137.

5100 45° 1 2

φ12 134 100 0.8

>0.2%

210137.

5100 45° 1 2

φ10 111 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 115 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 145 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 109 100 0.5

>0.2%

210137.

5100 45° 1 2

φ12 126 100 0.8

>0.2%

210137.

5100 45° 1 2

φ10 108 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 116 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 141 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 108 100 0.5

>0.2%

210137.

5100 45° 1 2

φ12 122 100 0.8

>0.2%

210137.

5100 45° 1 2

φ10 107 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 116 100 0.5

>0.2%210 137. 100 45° 1 2 φ 10 139 100 0.5 >0.2%

44

5

210137.

5100 45° 1 2

φ10 107 100 0.5

>0.2%

210137.

5100 45° 1 2

φ12 117 100 0.8

>0.2%

210137.

5100 45° 1 2

φ10 108 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 116 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 138 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 107 100 0.5

>0.2%

210137.

5100 45° 1 2

φ12 118 100 0.8

>0.2%

210137.

5100 45° 1 2

φ10 135 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 150 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 121 100 0.5

>0.2%

Table 7-19 Transversal reinforcement for beams in frame 6-6

StoryFiel

d(|MRd

+|+|MRd-|)

*1.2/ L [KN]qltv*L/2 [KN]

VEdmax

[kN]VEd

min

[kN]ξ=VEd

min/VEdmax Inclined

Barsd=hb-as

[mm]bb

[mm]ν=0.6(1-fck/250)

fcd

[N/mm2]|VEd

max| [kN]

0.5·bb·d·ν·fcd

[kN]

z [mm]

GF1-2 46.14 87.65 133.79 41.51 0.31 >-0.5 No 515 300 0.55 13.33 133.79 < 566.36 4642-3 59.11 135.67 194.78 76.56 0.39 >-0.5 No 515 300 0.55 13.33 194.78 < 566.36 4643-4 43.88 68.62 112.50 24.74 0.22 >-0.5 No 515 300 0.55 13.33 112.50 < 566.36 464

11-2 52.54 89.92 142.46 37.38 0.26 >-0.5 No 515 300 0.55 13.33 142.46 < 566.36 4642-3 59.11 135.76 194.87 76.65 0.39 >-0.5 No 515 300 0.55 13.33 194.87 < 566.36 4643-4 43.88 71.52 115.40 27.64 0.24 >-0.5 No 515 300 0.55 13.33 115.40 < 566.36 464

21-2 52.54 90.99 143.53 38.45 0.27 >-0.5 No 515 300 0.55 13.33 143.53 < 566.36 4642-3 49.46 135.54 185.00 86.08 0.47 >-0.5 No 515 300 0.55 13.33 185.00 < 566.36 4643-4 43.88 73.22 117.10 29.34 0.25 >-0.5 No 515 300 0.55 13.33 117.10 < 566.36 464

31-2 52.54 91.89 144.43 39.35 0.27 >-0.5 No 515 300 0.55 13.33 144.43 < 566.36 4642-3 49.46 135.41 184.87 85.95 0.46 >-0.5 No 515 300 0.55 13.33 184.87 < 566.36 464

45

3-4 43.88 74.55 118.43 30.67 0.26 >-0.5 No 515 300 0.55 13.33 118.43 < 566.36 464

41-2 52.54 92.51 145.05 39.97 0.28 >-0.5 No 515 300 0.55 13.33 145.05 < 566.36 4642-3 49.46 135.40 184.86 85.94 0.46 >-0.5 No 515 300 0.55 13.33 184.86 < 566.36 4643-4 43.88 75.57 119.45 31.69 0.27 >-0.5 No 515 300 0.55 13.33 119.45 < 566.36 464

51-2 43.97 52.90 96.87 8.93 0.09 >-0.5 No 515 300 0.55 13.33 96.87 < 566.36 4642-3 49.46 89.53 138.99 40.07 0.29 >-0.5 No 515 300 0.55 13.33 138.99 < 566.36 4643-4 36.71 44.03 80.74 7.32 0.09 >-0.5 No 515 300 0.55 13.33 80.74 < 566.36 464

46

fywd

[N/mm2]smax

[mm]smin

[mm]θ

cot θ

Choosen stirrups

S [mm

]

Choosen S [mm]

pw [%]

210137.

5100 45° 1 2

φ10 114 100 0.5

>0.2%

210137.

5100 45° 1 2

φ12 113 100 0.8

>0.2%

210137.

5100 45° 1 2

φ10 136 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 107 100 0.5

>0.2%

210137.

5100 45° 1 2

φ12 113 100 0.8

>0.2%

210137.

5100 45° 1 2

φ10 132 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 107 100 0.5

>0.2%

210137.

5100 45° 1 2

φ12 119 100 0.8

>0.2%

210137.

5100 45° 1 2

φ10 131 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 106 100 0.5

>0.2%

210137.

5100 45° 1 2

φ12 119 100 0.8

>0.2%

210137.

5100 45° 1 2

φ10 129 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 105 100 0.5

>0.2%

210137.

5100 45° 1 2

φ12 119 100 0.8

>0.2%

210137.

5100 45° 1 2

φ10 128 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 158 100 0.5

>0.2%

210137.

5100 45° 1 2

φ10 110 100 0.5

>0.2%

47

210137.

5100 45° 1 2

φ10 189 100 0.5

>0.2%

48

7.3 Columns

The columns were dimensioned taking into account that the beams were of medium ductility M, so the behavior of the columns will be in the class of ductility M. The rules and principles for design are found in P 100-2013.

7.3.1 Geometrical characteristicsThe minimum dimension of the section is 350 mm. The dimensions of the cross-section must be a multiple of 50 mm. The ratio between hc and bc ≤ 2. 5, where hc is the height of the column and bc is the

width of the column.

7.3.2 Provisions regarding the materials usedThe minimum concrete class used for the main structural elements is C16/20 for medium

ductility buildings. The structural elements will be reinforced only with profiled steel type PC52, PC60. The

closed stirrups and the hooks are be made of plain steel type OB37. Non ductile steel, like STNB can be used, only if, by design, an elastic behavior can be assured.

In the critical zones of the main structural elements, steel types with a minimum strain of 7. 5% corresponding to the maximum stress will be used.

7.3.3 Longitudinal reinforcement provisionsThe total reinforcement coefficient ρ will be at least 0.01 and maximum 0.04.The minimum reinforcement percentage on each side is pmin,side ≥0 . 20 %. Between the corner reinforcement bars, at least 1 intermediate bar will be placed. For circular columns, the minimum number of reinforcement bars is 6. The maximum distance between the axes of the reinforcement bars is 250 mm. The length of the critical zones at the ends of the column lcr is:

lcr ≥ max {1,5 hc ;lcl6

;600 mm }

Where, hc is the largest dimension of the column and lcl is the story clear height.

If lclh<3, the entire column section is considered critical and will be reinforced

accordingly.The minimum diameter of the reinforcement bars is 12 mm. The maximum diameter of the reinforcement bars is 28mm. For good adherence under normal conditions of solicitation the anchorage length will be

40 diameters. Splicing will usually be placed in zones of minimum effort in reinforcement bars; for the

vertical bearing elements, the splicing is admitted above the level of each floor plate. For the reinforcement bars spliced by overlapping, overlapping length will be 30d for elements with concrete class < C20/25 and 20d in case of elements with concrete class > C20/25.

7.3.4 Longitudinal reinforcement designThe equation below will be computed on the two main directions of the seismic action.

Always consider both directions of the actions of the moments in the beams in the joints. The values of the design shear force are obtained from the column equilibrium at each level.

M Ed=γ Rd

M Ed, ∗Σ M Rb

Σ M Ed ,b

Where:M Ed is the design bending moment in the considered section of the column.

49

M Ed, is the bending moment taken from ETABS by applying the envelope combination.

Σ M Rb is the sum of the capable bending moments in the beams from all the spans of the considered level associated to the direction of the seismic action.

ΣM Ed ,b is the effective bending moment in the beam resulted after the computation of the reinforcement

γ Rd is a factor taking into consideration the effect of steel and concrete consolidation in the compressed areas and is considered 1.3 at the base of the building and 1.2 for the rest of the levels.

ω=Σ M Rb

Σ M Ed ,b

The depth of the active zone is computed using the value of the axial loading given by the LTV loading,N Ed .

x=N Ed

bc ¿ f cd

If x≤ 2c then A

s ,necc=¿M Ed−

N Ed∗(b−c)2

f yd∗(b−c)¿

If x>2 c then A

s ,necc=¿M Ed+

NEd∗(b−c )

2−b∗x∗f cd∗(b−c−0.5∗x )

f yd∗b¿

Based on the necessary area of reinforcement the reinforcement layout is chosen and the capable bending moment can be computed:

If x>2 c M Rd=b∗x∗f cd∗(b−c−0.5∗x )+A s∗f yd∗(b−c )−N Ed∗0.5(b−c)If x≤ 2c M Rd=A s∗(b−c )∗f yd+N Ed∗0.5∗(b−c )

ρ=A steel

b∗(b−c )∗100

7.3.5 Transversal reinforcement provisionsThe minimum transversal reinforcement coefficient must be at least: - 0.005 in the critical zone situated at the bottom part of the ground story. - 0.0035 for the other critical zones. - 0.0015 for the rest of the column. The minimum diameter of stirrups is: - d/4, where d is the maximum diameter of the longitudinal reinforcement; - 6 mm; - 8 mm, for perimetral stirrups. The distance in section between consecutive bars situated at a stirrup corner or tied with

hooks will not be greater than 200 mm. For the first 2 stories of buildings with more than 5 stories and for the first storey of

buildings with fewer stories, the stirrups will be placed beyond the critical zone on a length equal to 0. 5lcr.

The minimum distance between stirrups is:For the plastic zones: ae ≤ { b0/3; 7 dmin (6 dmin for the bottom part of the ground

column); 125 mm}, where b0 is the minimum dimension of the stirrup and dmin is the minimum diameter of the longitudinal reinforcement.

Outside the plastic zones: ae ≤ { 3hc/4; 15 dmin; 200 mm}.

50

7.3.6 Transversal reinforcement designThe design values of the shear forces are obtained from the equilibrium of the column at

each story, considering the action of the bending moments at each end, corresponding, for each sense of the seismic action, to the formation of plastic hinges which can occur either in the beams or in the columns connected to the nodes.

M i ,d=γ Rd∗M Rc ,i∗(1 ,Σ M Rb/Σ M Rc) Where:M Rc ,i -value of the capable bending moment at the end iΣ M Rb-sum of the capable moments of the beams entering the jointΣ M Rc-sum of the capable moments of the columns entering the jointThe design value of the shear force is:

V Ed=M 1 d+M 2d

ld

Q=V Ed , max

bc∗d∗f ctd

p=A s

bc∗d∗100

pe=Q2∗f ctd∗100

3. 2∗√ p∗f yd

sd=√ 100∗√ p∗f ctd

pe∗0.8∗f yd

The reinforcement is chosen and the area of reinforcement is computed taking into account ne which refers to the number of stirrups in one section, but also their position.

A sw

s≥

V ed

z∗cotθ∗f yd

A sw

s≥

pwmin∗b

100

In order to assure a ductile behavior, plastic and elastic zones have to be decided. The plastic zone is found at the extremities of the column, and the elastic zone in the middle part.

Example for column 4 - intersection of axes C and 6 - the first level

M Ed=1.2∗50.27∗411.917

352. 9=562. 348kN*m

x= 1852.88600∗13.13

=227. 94 mm

x>2 c so the Asteel=-1713.84 mm2

The chosen reinforcement is 10Ø16 so the effective area of steel is 2009.6. The computations are performed also on the perpendicular direction and the results show

that the same amount of reinforcement is needed.

ρ= 2∗2009. 6600∗(600−35 )

∗100=1.1 %

For the transversal reinforcement

V Ed=2∗1.3∗498.99

3=432.46 kN

A sw

s≥

432.46509∗1∗210

A sw

s≥ 5.54

51

A sw

s≥ 0.5

Chosen bars of Ø10 OB 37 spaced at 100mm: A sw

s=6.11

Both conditions are fulfilled.

52

Table 7-20 Longitudinal reinforcement for Central ColumnFrame 6-6

Story Supportfyd

[N/mm2]fcd

[N/mm2]as

[mm]b

[mm]h

[mm]h0=h-as

[mm]hs=h-2as

[mm]xb

[mm]pTOTAL%

|ME'| [kN·m]

|NE'| [kN]

ω γRd|Med|

[kN·m]|Ned|

[kN]

GFBottom 300 13.33 35 600 600 565 530 311

1.1950.26 1852.88 1.023 1.3 66.84 1852.88

Top 300 13.33 35 600 600 565 530 311 8.13 1823.11 1.023 1.3 10.81 1823.11

1Bottom 300 13.33 35 600 600 565 530 311

1.1955.63 1504.75 1.023 1.2 68.29 1504.75

Top 300 13.33 35 600 600 565 530 311 23.13 1474.98 1.007 1.2 27.96 1474.98

2Bottom 300 13.33 35 600 600 565 530 311

1.1957.94 1172.51 1.007 1.2 70.03 1172.51

Top 300 13.33 35 600 600 565 530 311 31.53 1142.74 1.052 1.2 39.81 1142.74

3Bottom 300 13.33 35 600 600 565 530 311

1.1958.03 851.14 1.052 1.2 73.28 851.14

Top 300 13.33 35 600 600 565 530 311 36.89 821.37 1.030 1.2 45.58 821.37

4Bottom 300 13.33 35 600 600 565 530 311

1.1953.76 540.00 1.030 1.2 66.43 540.00

Top 300 13.33 35 600 600 565 530 311 33.95 510.23 1.057 1.2 43.05 510.23

5Bottom 300 13.33 35 600 600 565 530 311

1.1959.56 228.44 1.057 1.2 75.53 228.44

Top 300 13.33 35 600 600 565 530 311 55.24 198.67 1.046 1.2 69.32 198.67

Table 7-21 Transversal reinforcement for Central Column

StoryMcap

[kN·m]HS

[m]γ VEd [kN]

0.5·b·d·ν·fcd [kN]

h [mm] b [mm]d=h-as

[mm]ν=0.6(1-fck/250)

fcd

[N/mm2]z [mm]

fywd

[N/mm2]θ cot θ

smin

[mm]

GF 498.99 3 1.3 432.46 < 1242.69 600 600 565 0.55 13.33 509 210 45° 1 1001 466.33 3 1.2 373.06 < 1242.69 600 600 565 0.55 13.33 509 210 45° 1 1002 421.03 3 1.2 336.82 < 1242.69 600 600 565 0.55 13.33 509 210 45° 1 1003 364.08 3 1.2 291.26 < 1242.69 600 600 565 0.55 13.33 509 210 45° 1 1004 295.06 3 1.2 236.04 < 1242.69 600 600 565 0.55 13.33 509 210 45° 1 1005 212.49 3 1.2 169.99 < 1242.69 600 600 565 0.55 13.33 509 210 45° 1 100

53

Frame 6-6 Frame C-C

x [mm]

Asnec

[mm2]Nr of bars

φ [mm]Aeff=n·π·φ2/4

[mm2]p%

MRd

[kN·m]|ME'|

[kN·m]|NE'| [kN]

ω k|Med|

[kN·m]|Ned| [kN]

x [mm]

Asnec

[mm2]Nr of bars

φ [mm]

Aeff=n·π·φ2/4 [mm2]

p%MRd

[kN·m]

228 -1713 5 16 1005.31 0.30 498.997.38 1852.88 1.072 1 10.29 1852.88

228 -2068 5 16 1005.31 0.30 498.9917.93 1823.11 1.072 1 25.00 1823.11

184 -1498 5 16 1005.31 0.30 466.3332.15 1504.75 1.072 1 44.82 1504.75

184 -1646 5 16 1005.31 0.30 466.3320.87 1474.98 1.052 1 28.55 1474.98

143 -1202 5 16 1005.31 0.30 421.0330.74 1172.51 1.052 1 42.05 1172.51

143 -1378 5 16 1005.31 0.30 421.0321.43 1142.74 1.107 1 30.73 1142.74

103 -824 5 16 1005.31 0.30 364.0833.78 851.14 1.107 1 48.63 851.14

103 -979 5 16 1005.31 0.30 364.0822.71 821.37 1.103 1 34.29 821.37

64 -580 5 16 1005.31 0.30 295.0634.82 540.00 1.103 1 49.93 540.00

64 -536 5 16 1005.31 0.30 295.0621.34 510.23 1.162 1 32.22 510.23

25 105 5 16 1005.31 0.30 212.4937.77 228.44 1.162 1 57.03 228.44

25 28 5 16 1005.31 0.30 212.4928.28 198.67 1.219 1 44.82 198.67

smax

[mm] Chosen stirrups

s [mm]S

[mm]lppz

pw

[%]nsl φ125 5.4 φ 10 105 100 900 0.71 >0.5%125 5.4 φ 10 121 100 600 0.71 >0.35%125 5.4 φ 10 134 100 600 0.71 >0.35%125 5.4 φ 10 155 150 600 0.47 >0.35%125 5.4 φ 10 192 150 600 0.47 >0.35%125 5.4 φ 10 266 200 600 0.35 >0.35%

54

8 STRUCTURAL WALLS

8.1 General considerations for the computations of the structural wallsThe main idea is to design the structural walls in order to favor the development of a

structural mechanism for dissipating energy in a favorable manner for the structure as a whole and to assure enough ductility to the structural elements.

The dimensioning is done according to P-100-1-2013 for the class of high ductility H, and according to the Romanian Code for designing buildings with structural reinforced concrete walls CR 2 – 1 – 1. 1 : 2011 .

The critical zone is at the base of the wall having the length: hcr=max (lw , Hw /6)≤ 2hs because the building has 6 levels.

Where:Hw-the height of the wallhs-the clear height of the storylw-the length of the wall

In the case of multistory buildings this height is rounded up to a full number of stories if the limit of the computed plastic zone is bigger or smaller with 0.2hs. In this way the base becomes Area A with specific design performances, and the rest of the wall becomes Area B, with smaller efforts.

Wall 1hcr=max(3.3¿, 18/6)≤2∗2.45¿ hcr=max(3.3¿, 35)≤ 4. 9¿ hcr=3.3 m - the height of Area A is 3m

8.2 The values of the design sectional efforts in the wallsThe values of the design bending moments MEd in the horizontal sections of the wall for the

structures which are in the classes of ductility DCH:MEd=MEd,o for the zone AMEd=km*ω*MEd'< ω*MEd,o' for the zone BWhere:MEd' is the bending moment due to seismic loadingMEd,o' is the value of MEd' at the base

55

km is the correction coefficient of the bending moments in the walls which is km=1 in zone A and km=1.30 in zone B for the ductility class DCH

ω is the ratio between the capable moment and the design moment in each section

ω=Σ M Rd ,o

Σ M Ed ,o; ≤q

Where:M Rd ,o is the capable moment at the base of the wallq is the behavior factor considered for designing the structure

8.3 The longitudinal and transversal reinforcementIn the case of longitudinal reinforcement the overlapping is in zone A:45d in the case of horizontal bars PC5245d in the case of vertical bars PC52 with d<20mmBars with d>20mm need weldingIn the case of longitudinal reinforcement the overlapping in zone B is: the minimum

overlapping lengths are with 10d less than those from the table. Also, in the zone B it is not necessary the overlapping by welding of the reinforcement with d 16(20) mm.

For the flexural reinforcement for the web the minimum web reinforcement are given in the table below. For zone B the minimum web percentages are the ones for as<0. 12g.

Seismic zoneMinimum reinforcement percentageHorizontal bars Vertical barsOB 37 PC 52, PC 60 OB 37 PC 52, PC 60

ag≥0.12g 0,30% 0,25% 0,35% 0,30%ag<0.12g 0,25% 0,20% 0,25% 0,20%The minimum bar diameter is 8 mm for horizontal bars and 10 mm for vertical bars. The

maximum distance between bars is 350 mm for horizontal bars and 250 mm for vertical bars.

For the local reinforcement at the ends on each wall one has to take into account the following:

56

The reinforcement is realized as for the columnsIf the wall has no flanges the portions with the length 0. 1*lw at the ends of the beams are

considered flangesThe minimum vertical percentage is:

Seismic zoneMinimum reinforcement percentageOB 37 PC 52, PC 60zone A zone B zone A zone B

ag≥0. 12g 0,7% 0. 5% 0,6% 0. 5%ag<0. 12g 0. 4% 0,4%

The minimum diameter is 12 mmThe layout rules are the ones in the pictures below:

The anchorage of the vertical reinforcement of the flanges can be done by overlapping and the length is 50dBL for the DHL class in zone A, and in zone B the lengths are smaller with 10 dbl. In the area A should be avoided if possible the anchorage of the bars situated at the extremities. The horizontal reinforcement should be overlapped for at least 50dBT for the class DCH.

The maximum distances between stirrups for class DCH with ag>0. 12g are:Zone A 8dbl<125mmZone B 10dbl<200mmFor computing the longitudinal reinforcement the section was modeled in Response 2000.

The value of the effective bending moment was taken from the envelope of load combinations from ETABS.

Based on the results ω is computed for the wall.

57

Wall 1 - zone A

Figure 8 Wall 1 Zone A – M-N interaction diagram

58

From response were taken the values of the bending moments corresponding the axial range and interpolation is performed in MathCAD.

M1 13510.69M2 11439.61N1 7025.288N2 4711.667N 5018

x M2 N N2( )M1 M2( )

N1 N2( )

x 1.171 104

MRd xMedb 11379

MRd

Medb

1.029

59

Wall 1 - zone B

Figure 9 Wall 1 Zone B – M-N interaction diagram

60

Dimensioning of the Transversal Reinforcement for Structural Walls

Three checks are needed:Check the web of the concrete section to resist to compression Check the horizontal reinforcement from the webCheck the joints For the horizontal reinforcement:1.5V Ed

' ≤V Ed=kq∗ω∗V Ed, ≤ 5 V Ed

' Where:V Ed is the design shear forceV Ed

, is the shear force from ETABSk qis the correction factor V Rd=V Rd ,c+V Rd , s>V Ed

Where:V Rd , s is the shear due to the participation of steelV Rd , s=0. 8∗A s∗f yd V Rd ,c is the shear due to the participation of concrete V Rd ,c=0.3∗bw∗hw∗σ o<0.6∗bw∗hw∗f cdt for zone AV Rd ,c=0.2∗bw∗hw∗σo+0.7∗bw∗hw∗f cdt>0 for zone BWhere:bw is the width of the sectionhwis the length of the section

σ o=NLTV

bw∗hw

If Hw

lw

≤ 1 then V Ed≤ V Rd ,c+Σ A sh∗f yd ,h+lw−Hw

lw

∗Σ A sv∗f yd , v

Where:Σ A sv is the sum of the vertival reinforcement from the webΣ A sh is the sum of the horizontal reinforcement from the webf yd ,v is the design value of the yielding limit of the vertical reinforcement

61

Table 8-22 Longitudinal reinforcement for the Structural WallMs ω kM M N b h Flange Aefffl ρflange Field Aefffield ρfield Mcap

zone A 11379,01 1,03 1 11379,01 5018,57 400 3300 3ϕ25+2ϕ25+3ϕ25 3924,96 2,4531 24ϕ16 4823,04 0,4823 14180

zone B

7579,5 1,03 1,3 10139,1 4311,34 400 3300 3ϕ25+2ϕ25+3ϕ25 3924,96 2,23009 12ϕ12 2712,96 0,25691 103205057,7 1,03 1,3 5209,431 3594,13 400 3300 3ϕ12+3ϕ12+3ϕ12 904,32 0,68509 12ϕ12 2712,96 0,25275 58973406,01 1,03 1,3 3508,19 2873,2 400 3300 3ϕ12+3ϕ12+3ϕ12 904,32 0,68509 12ϕ12 2712,96 0,25275 58972253,78 1,03 1,3 2321,393 2149,41 400 3300 3ϕ12+3ϕ12+3ϕ12 904,32 0,68509 12ϕ12 2712,96 0,25275 58971533,52 1,03 1,3 1579,526 1424,32 400 3300 3ϕ12+3ϕ12+3ϕ12 904,32 0,68509 12ϕ12 2712,96 0,25275 58971206,44 1,03 1,3 1242,633 398,52 400 3300 3ϕ12+3ϕ12+3ϕ12 904,32 0,68509 12ϕ12 2712,96 0,25275 5897

Table 8-23 Transversal reinforcement for the Structural Wall

Ved' kq Ved Diameter Asteel ρ Vrd,s bw hw Nltv σo Vrd,c0,6*bw*hw*fcdt

VRd

1569,66 1,2 2354,49 1,5*Ved'<Ved<5*Ved' ϕ18/10 7630,2 1,92682 1831,248 0,4 3,3 2476,8 1876,348 743,034 1425,6 2574,282

1345,98 1,2 2018,97 1,5*Ved'<Ved<5*Ved' ϕ10/20 1334,5 0,33699 320,28 0,4 3,3 2476,8 1876,348 1881,356 0 2201,6361160,61 1,2 1740,915 1,5*Ved'<Ved<5*Ved' ϕ10/20 1334,5 0,33699 320,28 0,4 3,3 1771,8 1342,273 1740,36 0 2060,641052,66 1,2 1578,99 1,5*Ved'<Ved<5*Ved' ϕ10/20 1334,5 0,33699 320,28 0,4 3,3 1417,3 1073,742 1669,468 0 1989,748949,93 1,2 1424,895 1,5*Ved'<Ved<5*Ved' ϕ10/20 1334,5 0,33699 320,28 0,4 3,3 1061,8 804,4242 1598,368 0 1918,648812,56 1,2 1218,84 1,5*Ved'<Ved<5*Ved' ϕ10/20 1334,5 0,33699 320,28 0,4 3,3 705,72 534,6364 1527,144 0 1847,424638,23 1,2 957,345 1,5*Ved'<Ved<5*Ved' ϕ10/20 1334,5 0,33699 320,28 0,4 3,3 348,5 264,0152 1455,7 0 1775,98

62

8.4 Material provisions for structural wallsThe minimum concrete class used for the main structural elements is C20/25 for high

ductility class buildings and C16/20 for medium ductility buildings. The structural elements will be reinforced only with profiled steel type PC52, PC60. The

closed stirrups and the hooks can be made of plain steel type OB37. Non ductile steel, like STNB can only be used as constructive reinforcement.

In the critical zones of the main structural elements, steel types with a minimum strain of 7. 5% corresponding to the maximum stress will be used.

The ratio between the strength of the steel and its yielding limit must not be excessively large (<1. 4).

8.5 Geometric ExigenciesThe minimum thickness of the wall bw0 ≥ max {150 mm, hs/20}, where hs is the story

clear height. For buildings up to 10-12 stories it is recommended to keep the same wall section on the entire height.

The height of the compressed wall section will not be greater than:

x u≤0 . 10 (Ω+2 ) lw Ω – the ratio between capable bending moment at the base, associated to the

plastification of the walls and the bending moment obtained from structural analysis in the seismic combination; the capable bending moments must be computed for γRd=1.1.

lw – is the length of the wall cross-section. In the potentially plastic zone, when the height of the compressed zone xu exceeds the

smallest of the values 5bw0 and 0.4lw, it is necessary to make the check for loss of stability. The same check must be made at the edges of the flanges, in the parts which exceed 4hp

on each part of the wall web. If the check can’t be fulfilled, bulbs will be added. It is recommended that for seismic zones I-IV, or for buildings with more than 6 stories,

the walls should have at their edges either bulbs or flanges. The minimum bulb thickness is 250mm and the minimum bulb width is bp ≥ 2 bw0. The coupling beams for the walls with doors will have the same thickness as the rest of

the wall. If by computation this thickness is not enough, the beams will be thickened on a length sufficient to allow the reinforcement anchorage.

63

9 FOUNDATION

9.1 General InformationThe purpose of the foundation is to transmit the loads from the superstructure to the

supporting ground. From the seismic point of view the foundation should have an elastic behavior.

The chosen foundation solution is mat foundation. This is a type of direct foundation which ensures a large support surface. The use of mat foundation is done in the following situations:

the foundation terrain has a small bearing capacity and the raft foundation can spread the loads on a larger area

very large loading which implies great surfaces for foundation so the loads have to be spread on a large area, usually the whole area of the structure

non-homogenous terrain where the risk of differential settlement appears because concrete slabs resist differential movement between loading positions

the presence of ground water above the level of the foundation closely spaced vertical elements for which the individual pad foundation would

interact in the case of high rise buildings

The chosen solution is raft foundation with constant width. For design of the raft foundation should be taken into consideration the compatibility

between the deformability of the terrain with the deformability of the structural elements. The sectional efforts are obtained in ETABS by modeling the interaction between the foundation and the terrain. In order to perform this task the whole structure has to be modeled including the basements. The maximum efforts obtained are used for designing the raft foundation. The ground is modeled as an elastic environment, a Winkler model and the efforts in each spring in computed. Based on these results the reinforcement can be computed.

The reinforcement is made with horizontal nets of horizontal reinforcement, laid at the upper and lower part of the element to resist the positive and negative moments. In addition to this it is necessary an additional reinforcement on the central area of the raft for preventing contraction phenomena. The minimum reinforcement percentage is 0.15% on each face. 9.2 Predimensioning

The height of the raft foundation: H r=( 18

;1

12 )∗Lmax

The length of the cantilever lc=(0.2 ;0.3)Lmax The dimensions in plan of the raft foundation: Br=2∗L1+L2+2∗lc

Lr=5∗T +2∗l c

H r=10. 07

12=0.83m so is chosen H r=0. 9 m

lc=0.2∗10. 07=¿0.2mBr=21.55+2∗0.2=21.95 m Lr=14.30+2∗0.2=14.60 m

64

10 STAIRS

10.1 General issuesStairs should be designed in order to not influence the rigidity of the structure or to

interfere with the elements which assure the rigidity. The design of the stairs must assure the following performance conditions:

Stability-avoid buckling or local deflection Resistance: limit the damage in case of severe actions and avoid collapse Ductility: no interference with the ductility of the adjacent vertical structural

elements which can be assured by avoiding uncontrolled links with the horizontal sub-ensembles

Rigidity: limit the displacements and deflection of the stairs and of the neighboring sub-ensembles; limit the cracking

Durability: assure a normal behavior during the lifetime of the structure under normal exploitation conditions and limit the premature deterioration of materials and elements

10.2 LoadsLandingplastering qplast=γplast*hplast=19*0. 015=0. 285kN/m2flooring qflooring=γflooring*hflooring=22*0. 05=1. 1kN/m2live load qlanding=3kN/m2FlightFor the structural elements a projection of loads has to be performed. For the steps Atriangle=Arectangleb∗h

2=hc∗√b2+h2

The height of a stair is computed considering the needed number of stairs, which is 18 in this case and the height of the level which is 3m.

htr=H level

n=3. 00

18=16. 67 cm

hc=0.26∗0. 1676

2∗√0.282+0.16762=0.0665 m

qsteps=hc∗γ PC=0.0665∗24=1.596 kN /m2 The finishingA1=b*0. 05=0. 26*0. 05=0. 013m2A2=0. 05*0. 05=0. 0025m2A3=h*0. 05=0. 1676*0. 05=0. 00838m2A1+A2+A3=0. 02388m2A1+A2+A3=hech

, ∗√b∗h so

hech. =

A1+A2+A3

√b2+h2

hech. = 0. 02388

√0.26∗0. 26+0.176∗0. 176

hech. =0.077 m

qfinishing=hp*γp=0. 015*19=. 0285kN/m2The variable load :qu=qu*cosα=0. 43kN/m2where α is the angle of the slope of the ramp.

65

Figure 10 Stair Etabs Diagram

tgα=1.50

2.065=0.726

10.3 Reinforcement Landing

L 2.5ml 1mL

l2.5

Due to the L/l ratio the landing will be reinforced on one directionFor computation we consider one strip with the length of 1 m. b 1000mm

fcd 13.33N

mm2

For concrete C20/25

fyd 300N

mm2

For steel PC52 which will be used for reinforcementMfield 0.9kN mMsupp 2.63kN md1 150mm

x d1 1 1 2Mfield

b d12 fcd

x 4.508 104 m

x

d1

3.005 103

bb 0.55

As1 b xfcd

fyd

As1 2.003 105 m

2

min0.2

100

Asmin minb d1

Asmin 3 104 m

2The chosen reinforcement is 6ϕ 8 with

Asteel 6 3.148mm( )

2

4

Asteel 3.014 104 m

2

Asteel 100

b d1

0.201OK

d1 150mm

x d1 1 1 2Msupp

b d12 fcd

66

x 1.321 103 m

x

d1

8.808 103

bb 0.55

As1 b xfcd

fyd

As1 5.87 105 m

2

min0.2

100

Asmin minb d1

Asmin 3 104 m

2

Asteel 6 3.148mm( )

2

4

Asteel 3.014 104 m


Asteel 100

b d1

0.201OK

On the long direction will be used repartition reinforcement computed from the minimum reinforcement percentage: 6ϕ 8

Flight

L 2.55ml 1mL

l2.55

Due to the L/l ratio the landing will be reinforced on one directionFor computation we consider one strip with the length of 1 m. b 1000mm

fcd 13.33N

mm2

For concrete C20/25

fyd 300N

mm2

For steel PC52 which will be used for reinforcementMfield 2.11kN mMsupp 2.90kN m

d1 130mm

x d1 1 1 2Mfield

b d12 fcd

x 1.223 103 m

x

d1

9.411 103

b

67

b 0.55

As1 b xfcd

fyd

As1 5.436 105 m

2

min0.2

100

Asmin minb d1

Asmin 2.6 104 m


Asteel 6 3.148mm( )

2

4

Asteel 3.014 104 m

2

Asteel 100

b d1

0.232OK

d1 150mm

x d1 1 1 2Msupp

b d12 fcd

x 1.457 103 m

x

d1

9.716 103

bb 0.55

As1 b xfcd

fyd

As1 6.476 105 m

2

min0.2

100

Asmin 2.6 104 m

2Asmin minb d1

Asteel 6 3.148mm( )

2

4

Asteel 3.014 104 m


Asteel 100

b d1

0.201OK

On the long direction will be used repartition reinforcement computed from the minimum reinforcement percentage: 6ϕ 8

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11 BIBLIOGRAPHY

P100-1-2013 SREN1992-1-1-2004 CR-1-1-3-2012-ZAPADA CR2-2006-Cod-de-Proiectare-Pereti-Structurali-Beton-Armat

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Reinforced Concrete Structure

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