Text Esanu

8/11/2019 Text Esanu

1/65

TECHNICAL UNIVERSITY OF CIVIL ENGINEERING BUCHAREST

FACULTY OF ENGINEERING IN FOREIGN LANGUAGES

Diploma Project

Reinforced concrete structure

B+GF+5S

SUPERVISOR

Asist.Dr.Ing.Andrei ZybaczynskiSTUDENT

Eanu Andrei Alin

BUCHAREST

2014


2/65

Esanu Andrei Alin Diploma project

Civil Engineering

- 2 -

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 -


3/65


Civil Engineering

- 3 -

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

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 -

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

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

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

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

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


4/65


Civil Engineering

- 4 -

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 faade ..................................................................................................... - 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 -


5/65


Civil Engineering

- 5 -

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 AM-N interaction diagram ............................................................. - 54 -

Figure 9 Wall 1 Zone BM-N interaction diagram .............................................................. - 56 -

Figure 10 Stair Etabs Diagram ............................................................................................... - 61 -
http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073424http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073424http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073425http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073425http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073426http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073426http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073427http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073427http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073433http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073433http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073433http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073427http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073426http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073425http://c/Users/MrPain/Dropbox/Licenta%20Ramona/Licenta/Text%20Esanu.docx%23_Toc399073424


6/65


Civil Engineering

- 6 -

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 m2

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 naturalstone 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 anextra 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 reinforcedmonolithic 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 /mm

2

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 aspects

Designing 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


7/65


Civil Engineering

- 7 -

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

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 propercomputational 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 site

The building will be located in Bucharest. As a result: The class of importance and exposure III with 1=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=300kPa

3 - 6.5m Plastic stiff clay e=0.8 pconv=300kPa

6.5 - 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.


8/65


Civil Engineering

- 8 -

2 LOAD EVALUATION

Table 2-1 Load on current floor

Load

Nominal

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 2.3625

Live load 2 0.4 0.8 1.5 3

Interior walls 1.5 1 1.5 1.35 2.025

qDV=13.125kN

Table 2-2 Load from terrace

LoadNominal

value

Coefficient

value

Long term

value

Coefficient

value

Design

value

Self-weight 4.25 1 4.25 1.35 5.7375

Terrace 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 2-3 Load from faade

ElementThickness

(m)

Height

(m)

(kN/m3)LTV

Coefficient

valueDV

Colored plastering 0.01 3 18 0.54 1.35 0.729

Thermo insulation 0.1 3 0.32 0.096 1.35 0.1296

Interior plastering 0.01 3 18 0.54 1.35 0.73

qDV= 1.5886 kN

Snow load

The 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= IsiCeCts0,k (2.3)

Where:

Is - the exposure-importance coefficient for the snow load

i- 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=1

Due to the small slope of the terrace roof 00300, i=0.8Due to its location, the structure has partial exposure, Ce=1

Due to good insulation of the roof, Ct=1


9/65


10/65


Civil Engineering

- 10 -

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 m

hpl= 5 m / 30= 0.166 m = 17 cm

3.2 Beams

3.2.1 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

length

Computed

height

Round up

height

Computed

width

Round

up width

(m) (m) (cm) (cm) (cm)

1 4.30 0.43 55 28 30

2 5.20 0.52 55 28 30

3 5.20 0.52 55 28 30

4 3.75 0.38 55 28 30

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

3.2.2

Tr ansversal beams

The 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

length

Computed

height

Round up

height

Computed

width

Round

up width

(m) (m) (cm) (cm) (cm)

1 4.20 0.42 55 28 30

2 3.90 0.39 55 28 30

3 5.00 0.50 55 29 30Same 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 faade, nonstructural

walls, and also self-weight of each column.

qfacade=1.59 kN/m2- Load from the weight of the facade

qter=10.16 kN/m2- Load from the terrace

qcl=13.13 kN/m2- Load from the current level


11/65


Civil Engineering

- 11 -

3.3.1 Column 1 - corner column

The tributary areas are A1=2.05 m2 and A2=3.02m

2, the dimensions of the beams are

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

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

Nter =83.13 kN-the axial load from the terrace Ncf =120.32 kN-the axial load from the terrace

N=5*Ncf+Nter N=684.73 KN 3.3.2 Column 2 - marginal column

The 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 =162.45 kN-the axial load from the terrace Ncf =227.16 kN-the axial load from the terrace

N=5*Ncf+Nter N=1365.76 KN 3.3.3

Column 3 - central column

The 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 =191.35 kN-the axial load from the terrace

Ncf =263.95 kN-the axial load from the terraceN=5*Ncf+Nter N=1511.08 KN


12/65


Civil Engineering

- 12 -

4 WALLS

The total area of the web on one direction will be at least: Where: is the sum of the horizontal sections of walls, laid parallel with the direction of theaction of the lateral loads is the importance factor of the structure according to 4.4.5 from P100-1:2013

ks=ag/g the ratio between the peak value of the acceleration and the gravitational

acceleration

n is the number of slabs of the buildingis the area of the slab in m2The 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.

On x direction: On y direction: The area of the flanges situated at the margins of the walls will respect the rule: Where, NEdis the axial compression stress.

qfacade=1.59 kN/m2- Load from the weight of the facade

qter=10.16 kN/m2- Load from the terraceqcl=13.13 kN/m

2- 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.

This check is performed to identify the cases in which the sections need strengthening by

means of flanges.

Wall 2:



The dimensions of the corresponding beams: L2=2.6m, L3=1.4m, T3=5m.


13/65


Civil Engineering

- 13 -

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. 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. Wall 5:



The dimensions of the corresponding beams: L17=2.6m, L18=1.4m, T14=4.05m,

T15=4.05m.


14/65


15/65


Civil Engineering

- 15 -

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 structuralelements, 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 seismicforces 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 verticalassociated 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 performedby 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 inthe international units: m, KN, KNm, seconds depending on each case.


16/65


17/65


Civil Engineering

- 17 -

Figure 6 Second mode of vibration


18/65


Civil Engineering

- 18 -

Figure 7 third mode of vibration


19/65


Civil Engineering

- 19 -

Table 5-1 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


20/65


Civil Engineering

- 20 -

6 DESIGN OF RIGIDITY TO LATERAL FORCES

Is 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 equipmentfor 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): d- inter-story drift associated to the conventional elastic analysis

- reduction factor which considers the shorter recurrence period of the earthquake

- the value ofis: 0.5 for the buildings in the importance classes III

q -behavior factor specific to the structures type- the relative displacement, determined by static computation under the seismic loads- allowable value of the relative story displacement 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 suchcircumstances 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:

Where:EcElasticity modulus of the concrete

Icinertia moment of the cross section

From 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.


21/65


Civil Engineering

- 21 -

Table 6-1 SX drift check

Level Direction Combination dre q d,r,sls dadm

Story 5 SX PSXPSY 0.000378 0,5 5,75 0.002174 0,015 OK


Story 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 OK


Base SX PSXPSY 0.000165 0,5 5,75 0.000949 0,015 OK

Table 6-2 SY drift check


Story 5 SY PSYNSX 0.001219 0,5 5,75 0.007009 0,015 OK





Base 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. - relative story displacement under the seismic action associated to ULSq -behavior factor specific to the structures type

dr -the relative displacement, determined by static computation under the seismic loads

c- the displacement amplification factor, which takes into account that for T


22/65


Civil Engineering

- 22 -

Table 6-4 SY drift check




Story 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 OK


Base SY PSYNSX 0.00049 0,5 5,75 0.01127 0,075 OK


23/65


Civil Engineering

- 23 -

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 therespective support and in that alternate openings.

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

The slabs are computed on one direction if where = 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. one way slab (which is reinforced on the shorter direction)Positive moment in field:

Negative moment in support: Slab 2 q(kN/m2) l(m) M field(KN*m) M support (KN*m)

Current floor 13.13 5 13.67708 27.35417

Terrace 10.16 5 10.58333 21.16667


24/65


Civil Engineering

- 24 -

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

The slabs are computed on two directions if , where .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: , where:g is the permanent load

p is the variable load

In 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:

Slab 1

Scheme 1 Scheme 2a Scheme 2b Scheme 2c


25/65


Civil Engineering

- 25 -

Example:

Scheme 1 Scheme 2a

Scheme 2b

Scheme 2c Total moment


26/65


Civil Engineering

- 26 -

Table 7-1 Moment computation for the current floor slabs

q+p/2= 11.63 L1= 4.05

p/2= 1.50 L2= 5.00

Slab1

1 2 1 2 M1(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

Slab5

1 2 1 2 M1(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

Slab6

1 2 1 2M1

(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

1 2 1 2M1

(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


27/65


Civil Engineering

- 27 -

q+p/2= 11.63 L1= 4.20

p/2= 1.50 L2= 4.60

Slab10

1 2 1 2M1

(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

Slab11

1 2 1 2M1

(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

Slab12

1 2 1 2M1

(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

Slab13

1 2 1 2

M1

(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


28/65


Civil Engineering

- 28 -

Table 7-2 Moment computation for the terrace slabs

q+p/2= 7.76 L1= 3.90

p/2= 1.20 L2= 4.60

Slab5

1 2 1 2 M1(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

Slab6

1 2 1 2M1

(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

Slab7

1 2 1 2M1

(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

Slab10

1 2 1 2M1

(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


29/65


Civil Engineering

- 29 -

q+p/2= 7.76 L1= 3.00

p/2= 1.20 L2= 4.20

Slab11

1 2 1 2M1

(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

Slab12

1 2 1 2M1

(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

Slab13

1 2 1 2M1

(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 rein forcement

Based 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: Where: b=1000 mm - the unitary strip

x - the depth of the neutral axis

Where: d - the width without the concrete coveringfcd - the strength of concrete in compression

fyd - the yielding strength of steel


30/65


Civil Engineering

- 30 -

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.

For hp0.20%.

Perpendicular on the direction of the strength reinforcement, found bycomputations, 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 thepouring and the vibrating of concrete.

For the regular slabs the repartition reinforcement should be at least 15%of the

load strengthening reinforcement, but minimum 46.

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.


31/65


Civil Engineering

- 31 -

Table 7-3 Current floor slab reinforcement

SLAB TYPE

M

[kN

m]

hs

[mm]

fcd

[N/mm2]

d

[mm]

x

[mm]

fyd

[N/mm]

Anec[mm]

ReinforcementAeff

[mm]

Steel

percentage

[%)

1

Mid-spanx direction 8.62 170 13.33 165 3.97 300 176 38+212/m 377.00 0.23

y direction 5.79 170 13.33 155 2.83 300 126 38+212/m 377.00 0.24

Supportx direction 19.17 170 13.33 165 8.96 300 398 48+212/m 427.26 0.26

y direction 13.36 170 13.33 155 6.61 300 294 38+212/m 377.00 0.24

2Mid-span y direction 13.68 170 13.33 155 6.77 300 301 38+212/m 377.00 0.24

Supporty direction up 27.35 170 13.33 155 13.86 300 616 612/m 678.58 0.44

y direction down 27.35 170 13.33 155 13.86 300 616 612/m 678.58 0.44

5


y direction 6.45 170 13.33 155 3.15 300 140 38+212/m 377.00 0.24

Support

x direction left 7.70 170 13.33 165 3.54 300 157 38+212/m 377.00 0.23

x direction right 7.70 170 13.33 165 3.54 300 157 38+212/m 377.00 0.23

y direction up 20.21 170 13.33 155 10.11 300 449 512/m 565.49 0.36


6


y direction 1.91 170 13.33 155 0.93 300 41 38+212/m 377.00 0.24

Support



y direction up 6.71 170 13.33 155 3.28 300 146 38+212/m 377.00 0.24

y direction down 6.71 170 13.33 155 3.28 300 146 38+212/m 377.00 0.24

7


y direction 6.68 170 13.33 155 3.27 300 145 38+212/m 377.00 0.24

Support



y direction up 19.38 170 13.33 155 9.68 300 430 512/m 565.49 0.36



32/65


Civil Engineering

- 32 -

10


y direction 6.03 170 13.33 155 2.95 300 131 38+212/m 377.00 0.24

Support



y direction up 19.46 170 13.33 155 9.72 300 432 512/m 565.49 0.36


11


y direction 2.22 170 13.33 155 1.08 300 48 38+212/m 377.00 0.24

Support



y direction up 7.78 170 13.33 155 3.81 300 169 38+212/m 377.00 0.24


12


y direction 1.57 170 13.33 155 0.76 300 34 38+212/m 377.00 0.24

Support



y direction up 5.24 170 13.33 155 2.56 300 114 38+212/m 377.00 0.24


13


y direction 7.74 170 13.33 155 3.79 300 169 38+212/m 377.00 0.24

Support


x direction right 13.35 170 13.33 165 6.19 300 275 38+212/m 377.00 0.23y direction up 22.47 170 13.33 155 11.29 300 501 512/m 565.49 0.36



33/65


Civil Engineering

- 33 -

Table 7-4 Terrace slab reinforcement

SLAB TYPEM

[kNm]

hs

[mm]

fcd

[N/mm2]

d

[mm]

x

[mm]

fyd

[N/mm]

Anec[mm]

ReinforcementAeff

[mm]

Steel

percentage

[%)

2

Mid-span y direction 1.58 170 13.33 155 0.77 300 34 38+212/m 377.00 0.24

Supporty direction up 21.17 170 13.33 155 10.61 300 471 512/m 565.49 0.36


5


y direction 4.45 170 13.33 155 2.17 300 96 38+212/m 377.00 0.24

Support



y direction up 13.80 170 13.33 155 6.83 300 303 38+212/m 377.00 0.24


6


y direction 1.29 170 13.33 155 0.63 300 28 38+212/m 377.00 0.24

Support



y direction up 4.70 170 13.33 155 2.29 300 102 38+212/m 377.00 0.24


7


y direction 4.64 170 13.33 155 2.26 300 101 38+212/m 377.00 0.24

Support


x direction right 7.72 170 13.33 165 3.55 300 158 38+212/m 377.00 0.23y direction up 13.26 170 13.33 155 6.56 300 291 38+212/m 377.00 0.24


10


y direction 4.22 170 13.33 155 2.06 300 91 38+212/m 377.00 0.24

Support



y direction up 13.30 170 13.33 155 6.58 300 292 38+212/m 377.00 0.24


34/65


35/65


Civil Engineering

- 35 -

7.2 Beams

The 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 r ein forcement

MEd = Mdesignthe 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 . As a result beff isinvolved in the computations for the beams which are done on a rectangular section.

The necessary area of reinforcement is: Where: is the maximum value of the effective bending moment from ETABS.c is the concrete cover

Based on the chosen diameters of reinforcement the area given by the actual reinforce

layout is computed: As1

The reinforcement percentage is:

The capable moment of the section is: For the top reinforcement we assume that x


36/65


Civil Engineering

- 36 -

The minimum diameter is 12.

At least 2 bars 14 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 (for C20/25 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-C

The chosen diameter is 112 and 216 so the The reinforcement percentage is The chosen diameter is 412 + 216 so the =25.45


37/65


Civil Engineering

- 37 -

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

Story FieldMEd+[kNm]

fyd[N/mm2]

as[mm]

bb[mm]

hb[mm]

hs=hb-2as

[mm]

As2nec

[mm2]

Diameter[mm]

Aeff =n2/4[mm2]

MRd+[kNm]

GF

1 4.75 300 35 300 550 480 33 112+216 515.22 0.0033 74.192 63.76 300 35 300 550 480 443 112+216 515.22 0.0033 74.19

3 12.01 300 35 300 550 480 83 112+216 515.22 0.0033 74.19

4 58.81 300 35 300 550 480 408 112+216 515.22 0.0033 74.19

5 4.68 300 35 300 550 480 33 112+216 515.22 0.0033 74.19

1

1 7.88 300 35 300 550 480 55 112+216 515.22 0.0033 74.19

2 63.67 300 35 300 550 480 442 112+216 515.22 0.0033 74.19

3 14.30 300 35 300 550 480 99 112+216 515.22 0.0033 74.19

4 58.76 300 35 300 550 480 408 112+216 515.22 0.0033 74.19

5 17.68 300 35 300 550 480 123 112+216 515.22 0.0033 74.19

2

1 10.89 300 35 300 550 480 76 112+216 515.22 0.0033 74.19

2 64.06 300 35 300 550 480 445 112+216 515.22 0.0033 74.19

3 17.02 300 35 300 550 480 118 112+216 515.22 0.0033 74.19

4 58.98 300 35 300 550 480 410 112+216 515.22 0.0033 74.19

5 17.21 300 35 300 550 480 120 112+216 515.22 0.0033 74.19

3

1 12.70 300 35 300 550 480 88 112+216 515.22 0.0033 74.19

2 64.42 300 35 300 550 480 447 112+216 515.22 0.0033 74.19

3 18.49 300 35 300 550 480 128 112+216 515.22 0.0033 74.19

4 59.17 300 35 300 550 480 411 112+216 515.22 0.0033 74.19

5 20.20 300 35 300 550 480 140 112+216 515.22 0.0033 74.19

4

1 12.54 300 35 300 550 480 87 112+216 515.22 0.0033 74.19

2 64.03 300 35 300 550 480 445 112+216 515.22 0.0033 74.19

3 19.88 300 35 300 550 480 138 112+216 515.22 0.0033 74.19

4 59.09 300 35 300 550 480 410 112+216 515.22 0.0033 74.19

5 20.56 300 35 300 550 480 143 112+216 515.22 0.0033 74.19

5

2 53.36 300 35 300 550 480 371 112+216 515.22 0.0033 74.19

3 14.71 300 35 300 550 480 102 112+216 515.22 0.0033 74.19

4 48.60 300 35 300 550 480 338 112+216 515.22 0.0033 74.19


38/65


Civil Engineering

- 38 -

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

Story SupMEd

-

[kNm]fyd

[N/mm2]as

[mm]bb

[mm]hb

[mm]

hs=hb-2as

[mm]

As2nec

[mm2]Diameter

[mm]

Aeff=n

2/4

[mm2]

xn[mm]

MRd

-

[kNm]

GF

112.67 300 35

300550 480 88

112+216515.22 0 0.0033 74.19

2 60.31 300 35 300 550 480 419 112+216 515.22 0 0.0033 74.19

3 32.94 300 35 300 550 480 229 112+216 515.22 0 0.0033 74.19

4 110.47 300 35 300 550 480 767 12+216 854.51 25 0.0055 123.05

5 11.03 300 35 300 550 480 77 112+216 515.22 0 0.0033 74.19

1

1 14.88 300 35 300 550 480 103 112+216 515.22 0 0.0033 74.19

2 60.31 300 35 300 550 480 419 112+216 515.22 0 0.0033 74.19

3 44.52 300 35 300 550 480 309 112+216 515.22 0 0.0033 74.19

4 114.15 300 35 300 550 480 793 12+216 854.51 25 0.0055 123.05

5 17.68 300 35 300 550 480 123 112+216 515.22 0 0.0033 74.19

2

1 17.60 300 35 300 550 480 122 112+216 515.22 0 0.0033 74.192 61.14 300 35 300 550 480 425 112+216 515.22 0 0.0033 74.19

3 51.02 300 35 300 550 480 354 112+216 515.22 0 0.0033 74.19

4 117.35 300 35 300 550 480 815 12+216 854.51 25 0.0055 123.05

5 21.60 300 35 300 550 480 150 112+216 515.22 0 0.0033 74.19

3

1 20.23 300 35 300 550 480 140 112+216 515.22 0 0.0033 74.19

2 61.73 300 35 300 550 480 429 112+216 515.22 0 0.0033 74.19

3 53.64 300 35 300 550 480 373 112+216 515.22 0 0.0033 74.19

4 119.68 300 35 300 550 480 831 12+216 854.51 25 0.0055 123.05

5 25.51 300 35 300 550 480 177 112+216 515.22 0 0.0033 74.19

4

1 14.91 300 35 300 550 480 104 112+216 515.22 0 0.0033 74.19

2 62.35 300 35 300 550 480 433 112+216 515.22 0 0.0033 74.19

3 54.70 300 35 300 550 480 380 112+216 515.22 0 0.0033 74.19

4 120.92 300 35 300 550 480 840 12+216 854.51 25 0.0055 123.05

5 19.70 300 35 300 550 480 137 112+216 515.22 0 0.0033 74.19

5

2 49.06 300 35 300 550 480 341 112+216 515.22 0 0.0033 74.19

3 46.89 300 35 300 550 480 326 112+216 515.22 0 0.0033 74.19

4 101.90 300 35 300 550 480 708 12+216 854.51 25 0.0055 123.05


39/65


Civil Engineering

- 39 -

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

Story FieldMEd+[kNm]

fyd[N/mm2]

as[mm]

bb[mm]

hb[mm]

hs=hb-2as

[mm]

As2nec

[mm2]

Diameter[mm]

Aeff =n2/4[mm2]

MRd+[kNm]

GF1 55.65 300 35 300 550 480 386 112+216 515.22 0.0033 74.192 69.06 300 35 300 550 480 480 112+216 515.22 0.0033 74.19

3 34.59 300 35 300 550 480 240 112+216 515.22 0.0033 74.19

1

1 54.59 300 35 300 550 480 379 112+216 515.22 0.0033 74.19

2 69.22 300 35 300 550 480 481 112+216 515.22 0.0033 74.19

3 34.8 300 35 300 550 480 242 112+216 515.22 0.0033 74.19

2

1 54.67 300 35 300 550 480 380 112+216 515.22 0.0033 74.19

2 69.4 300 35 300 550 480 482 112+216 515.22 0.0033 74.19

3 35.1 300 35 300 550 480 244 112+216 515.22 0.0033 74.19

3

1 54.46 300 35 300 550 480 378 112+216 515.22 0.0033 74.19

2 69.68 300 35 300 550 480 484 112+216 515.22 0.0033 74.19

3 35.19 300 35 300 550 480 244 112+216 515.22 0.0033 74.19

4

1 54.64 300 35 300 550 480 379 112+216 515.22 0.0033 74.19

2 69.3 300 35 300 550 480 481 112+216 515.22 0.0033 74.19

3 35.64 300 35 300 550 480 248 112+216 515.22 0.0033 74.19

5

1 32.88 300 35 300 550 480 228 112+216 515.22 0.0033 74.19

2 51.11 300 35 300 550 480 355 112+216 515.22 0.0033 74.19

3 20.08 300 35 300 550 480 139 112+216 515.22 0.0033 74.19


40/65


Civil Engineering

- 40 -

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

Story SupMEd

-

[kNm]fyd

[N/mm2]as

[mm]bb

[mm]hb

[mm]

hs=hb-2as

[mm]

As2nec

[mm2]Diameter

[mm]

Aeff=n

2/4

[mm2]

xn[mm]

MRd

-

[kNm]

GF 185.03 300 35

300550 480 590

112+316716.28 15 0.0046 81.52

2 69.23 300 35 300 550 480 481 112+316 716.28 15 0.0046103.14

3 62.02 300 35 300 550 480 431 112+316 716.28 15 0.0046103.14

1

1 91.84 300 35 300 550 480 638 112+316 716.28 15 0.0046103.14

2 70.14 300 35 300 550 480 487 112+316 716.28 15 0.0046103.14

3 75.32 300 35 300 550 480 523 112+316 716.28 15 0.0046103.14

2

1 94.44 300 35 300 550 480 656 112+316 716.28 15 0.0046103.14

2 73.25 300 35 300 550 480 509 112+216 515.22 0 0.0033 74.19

3 82.39 300 35 300 550 480 572 112+316 716.28 15 0.0046103.14

3

1 96.87 300 35 300 550 480 673 112+316 716.28 15 0.0046103.14

2 73.16 300 35 300 550 480 508 112+216 515.22 0 0.0033 74.193 84.96 300 35 300 550 480 590 112+316 716.28 15 0.0046103.14

4

1 98.26 300 35 300 550 480 682 112+316 716.28 15 0.0046103.14

2 72.66 300 35 300 550 480 505 112+216 515.22 0 0.0033 74.19

3 85.11 300 35 300 550 480 591 112+316 716.28 15 0.0046103.14

5

2 57.33 300 35 300 550 480 398 112+216 515.22 0 0.0033 74.19

3 46.81 300 35 300 550 480 325 112+216 515.22 0 0.0033 74.19

4 60.49 300 35 300 550 480 420 112+216 515.22 0 0.0033 74.19


41/65


Civil Engineering

- 41 -

7.2.2 Design of the transversal r einforcement of the beams

The 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 ofthe 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

coefficient which takes into account a possible over resistance due to increased

rigidity of the steel to deformation = 1. 2Provisions for computing the transversal reinforcement:

wmin= 6 mm for h800mm 0 1 ,-

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

modulus

For = or || because =45and cot=1

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

and || are needed inclinedreinforcement 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 =

=45o=

Are chosen 210 at s=100mm =1.57 so it verifies


42/65

- 42 -

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

Story Field(|MRd

+|+|MRd

-|)

*1.2/ L [KN]

qltv*L/

2 [KN]

VEdmax

[kN]

VEdmin

[kN]

=VEdmin

/VEdma

x

Incline

d Bars

d=hb-as

[mm]

bb

[mm]

=0.6(1-

fck/250)

fcd

[N/mm2]

|VEdmax

| [kN]

0.5bbd

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 < 566.36 464

2-3 41.41 91.60 133.01 50.19 0.38 >-0.5 No 515 300 0.55 13.33 133.01 < 566.36 464

3-4 62.48 39.59 102.07 -22.89 -0.22 >-0.5 No 515 300 0.55 13.33 102.07 < 566.36 464

5-6 44.24 94.98 139.22 50.74 0.36 >-0.5 No 515 300 0.55 13.33 139.22 < 566.36 464

6-7 142.45 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 < 566.36 464

2-3 41.41 91.10 132.51 49.69 0.37 >-0.5 No 515 300 0.55 13.33 132.51 < 566.36 464

3-4 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 < 566.36 464

6-7 142.45 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 < 566.36 464

2-3 41.41 90.63 132.04 49.22 0.37 >-0.5 No 515 300 0.55 13.33 132.04 < 566.36 464

3-4 62.48 45.76 108.24 -16.72 -0.15 >-0.5 No 515 300 0.55 13.33 108.24 < 566.36 464

5-6 44.24 97.33 141.57 53.09 0.37 >-0.5 No 515 300 0.55 13.33 141.57 < 566.36 464

6-7 142.45 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 < 566.36 464

2-3 41.41 90.26 131.67 48.85 0.37 >-0.5 No 515 300 0.55 13.33 131.67 < 566.36 464

3-4 62.48 47.22 109.70 -15.26 -0.14 >-0.5 No 515 300 0.55 13.33 109.70 < 566.36 464

5-6 44.24 98.16 142.40 53.92 0.38 >-0.5 No 515 300 0.55 13.33 142.40 < 566.36 464

6-7 142.45 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 < 566.36 464

2-3 41.41 90.17 131.58 48.76 0.37 >-0.5 No 515 300 0.55 13.33 131.58 < 566.36 464

3-4 62.48 48.05 110.53 -14.43 -0.13 >-0.5 No 515 300 0.55 13.33 110.53 < 566.36 464

5-6 44.24 98.55 142.79 54.31 0.38 >-0.5 No 515 300 0.55 13.33 142.79 < 566.36 464

6-7 142.45 43.80 186.25 -98.65 -0.53 >-0.5 No 515 300 0.55 13.33 186.25 < 566.36 464

5

2-3 41.41 71.64 113.05 30.23 0.27 >-0.5 No 515 300 0.55 13.33 113.05 < 566.36 464

3-4 62.48 39.31 101.79 -23.17 -0.23 >-0.5 No 515 300 0.55 13.33 101.79 < 566.36 464

5-6 44.24 81.90 126.14 37.66 0.30 >-0.5 No 515 300 0.55 13.33 126.14 < 566.36 464


43/65

- 43 -

fywd

[N/mm2]

smax[mm]

smin[mm]

cot

Choosen

stirrups

S

[mm]

Choosen

S [mm]pw [%]

210 137.5 100 45 1 2 10 114 100 0.5 >0.2%

210 137.5 100 45 1 2 10 115 100 0.5 >0.2%

210 137.5 100 45 1 2 10 150 100 0.5 >0.2%

210 137.5 100 45 1 2 10 110 100 0.5 >0.2%

210 137.5 100 45 1 2 12 134 100 0.8 >0.2%

210 137.5 100 45 1 2 10 111 100 0.5 >0.2%

210 137.5 100 45 1 2 10 115 100 0.5 >0.2%

210 137.5 100 45 1 2 10 145 100 0.5 >0.2%

210 137.5 100 45 1 2 10 109 100 0.5 >0.2%

210 137.5 100 45 1 2 12 126 100 0.8 >0.2%

210 137.5 100 45 1 2 10 108 100 0.5 >0.2%

210 137.5 100 45 1 2 10 116 100 0.5 >0.2%

210 137.5 100 45 1 2 10 141 100 0.5 >0.2%

210 137.5 100 45 1 2 10 108 100 0.5 >0.2%

210 137.5 100 45 1 2 12 122 100 0.8 >0.2%

210 137.5 100 45 1 2 10 107 100 0.5 >0.2%

210 137.5 100 45 1 2 10 116 100 0.5 >0.2%

210 137.5 100 45 1 2 10 139 100 0.5 >0.2%

210 137.5 100 45 1 2 10 107 100 0.5 >0.2%

210 137.5 100 45 1 2 12 117 100 0.8 >0.2%

210 137.5 100 45 1 2

10 108 100 0.5>0.2%210 137.5 100 45 1 2 10 116 100 0.5 >0.2%

210 137.5 100 45 1 2 10 138 100 0.5 >0.2%

210 137.5 100 45 1 2 10 107 100 0.5 >0.2%

210 137.5 100 45 1 2 12 118 100 0.8 >0.2%

210 137.5 100 45 1 2 10 135 100 0.5 >0.2%

210 137.5 100 45 1 2 10 150 100 0.5 >0.2%

210 137.5 100 45 1 2 10 121 100 0.5 >0.2%


44/65

- 44 -

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

Story Field(|MRd

+|+|MRd

-|)

*1.2/ L [KN]

qltv*L/2

[KN]

VEdmax

[kN]

VEdmin

[kN]=VEd

min/VEd

max

Inclined

Bars

d=hb-as

[mm]

bb

[mm]

=0.6(1-

fck/250)

fcd

[N/mm2]

|VEdmax

|

[kN]

0.5bbd

fcd[kN]

z

[mm]

GF

1-2 46.14 87.65 133.79 41.51 0.31 >-0.5 No 515 300 0.55 13.33 133.79 < 566.36 464

2-3 59.11 135.67 194.78 76.56 0.39 >-0.5 No 515 300 0.55 13.33 194.78 < 566.36 464

3-4 43.88 68.62 112.50 24.74 0.22 >-0.5 No 515 300 0.55 13.33 112.50 < 566.36 464

1

1-2 52.54 89.92 142.46 37.38 0.26 >-0.5 No 515 300 0.55 13.33 142.46 < 566.36 464

2-3 59.11 135.76 194.87 76.65 0.39 >-0.5 No 515 300 0.55 13.33 194.87 < 566.36 464

3-4 43.88 71.52 115.40 27.64 0.24 >-0.5 No 515 300 0.55 13.33 115.40 < 566.36 464

2

1-2 52.54 90.99 143.53 38.45 0.27 >-0.5 No 515 300 0.55 13.33 143.53 < 566.36 464

2-3 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

3

1-2 52.54 91.89 144.43 39.35 0.27 >-0.5 No 515 300 0.55 13.33 144.43 < 566.36 464

2-3 49.46 135.41 184.87 85.95 0.46 >-0.5 No 515 300 0.55 13.33 184.87 < 566.36 464

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

4

1-2 52.54 92.51 145.05 39.97 0.28 >-0.5 No 515 300 0.55 13.33 145.05 < 566.36 464

2-3 49.46 135.40 184.86 85.94 0.46 >-0.5 No 515 300 0.55 13.33 184.86 < 566.36 464

3-4 43.88 75.57 119.45 31.69 0.27 >-0.5 No 515 300 0.55 13.33 119.45 < 566.36 464

5

1-2 43.97 52.90 96.87 8.93 0.09 >-0.5 No 515 300 0.55 13.33 96.87 < 566.36 464

2-3 49.46 89.53 138.99 40.07 0.29 >-0.5 No 515 300 0.55 13.33 138.99 < 566.36 464

3-4 36.71 44.03 80.74 7.32 0.09 >-0.5 No 515 300 0.55 13.33 80.74 < 566.36 464


45/65

- 45 -

fywd

[N/mm2]

smax[mm]

smin[mm]

cot

Choosen

stirrups

S

[mm]

Choosen

S [mm]pw [%]

210 137.5 100 45 1 2 10 114 100 0.5 >0.2%

210 137.5 100 45 1 2 12 113 100 0.8 >0.2%

210 137.5 100 45 1 2 10 136 100 0.5 >0.2%

210 137.5 100 45 1 2 10 107 100 0.5 >0.2%

210 137.5 100 45 1 2 12 113 100 0.8 >0.2%

210 137.5 100 45 1 2 10 132 100 0.5 >0.2%

210 137.5 100 45 1 2 10 107 100 0.5 >0.2%

210 137.5 100 45 1 2 12 119 100 0.8 >0.2%

210 137.5 100 45 1 2 10 131 100 0.5 >0.2%

210 137.5 100 45 1 2 10 106 100 0.5 >0.2%

210 137.5 100 45 1 2 12 119 100 0.8 >0.2%

210 137.5 100 45 1 2 10 129 100 0.5 >0.2%

210 137.5 100 45 1 2 10 105 100 0.5 >0.2%

210 137.5 100 45 1 2 12 119 100 0.8 >0.2%

210 137.5 100 45 1 2 10 128 100 0.5 >0.2%

210 137.5 100 45 1 2 10 158 100 0.5 >0.2%

210 137.5 100 45 1 2 10 110 100 0.5 >0.2%

210 137.5 100 45 1 2 10 189 100 0.5 >0.2%


46/65


Civil Engineering

- 46 -

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 Geometr ical characteristics

The 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 mater ials used

The 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 STNBcan 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 provisions

The total reinforcement coefficient 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: * +Where, hc is the largest dimension of the column and lcl is the story clear height.

If , 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 thevertical 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 design

The 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. Where:is the design bending moment in the considered section of the column. is the bending moment taken from ETABS by applying the envelope combination.


47/65


Civil Engineering

- 47 -

is the sum of the capable bending moments in the beams from all the spans of theconsidered level associated to the direction of the seismic action.is the effective bending moment in the beam resulted after the computation of thereinforcement

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 thelevels.

The depth of the active zone is computed using the value of the axial loading given by the

LTV loading, If then

If then Based on the necessary area of reinforcement the reinforcement layout is chosen and the

capable bending moment can be computed:

If If

7.3.5 Transversal reinforcement provisions

The 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}.

7.3.6 Transversal reinforcement design

The 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 eachsense 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.


48/65


Civil Engineering

- 48 -

Where:-value of the capable bending moment at the end i-sum of the capable moments of the beams entering the joint

-sum of the capable moments of the columns entering the joint

The design value of the shear force is: 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. 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 =562. 348kN*m

so the Asteel=-1713.84 mm2The chosen reinforcement is 1016 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. For the transversal reinforcement Chosen bars of 10 OB 37 spaced at 100mm: Both conditions are fulfilled.


49/65


Civil Engineering

- 49 -

Table 7-11 Longitudinal reinforcement for Central Column

Frame 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'|

[kNm]

|NE'|

[kN] Rd

|Med|

[kNm]

|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

2 Bottom 300 13.33 35 600 600 565 530 311 1.19 57.94 1172.51 1.007 1.2 70.03 1172.51Top 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 7-12 Transversal reinforcement for Central Column

StoryMcap

[kNm]HS[m]

VEd [kN]0.5bd

fcd [kN]h [mm] b [mm]

d=h-as[mm]

=0.6(1-fck/250)

fcd[N/mm

2]

z [mm]fywd

[N/mm2]

cot smin

[mm]

GF 498.99 3 1.3 432.46 < 1242.69 600 600 565 0.55 13.33 509 210 45 1 100

1 466.33 3 1.2 373.06 < 1242.69 600 600 565 0.55 13.33 509 210 45 1 100

2 421.03 3 1.2 336.82 < 1242.69 600 600 565 0.55 13.33 509 210 45 1 100

3 364.08 3 1.2 291.26 < 1242.69 600 600 565 0.55 13.33 509 210 45 1 100

4 295.06 3 1.2 236.04 < 1242.69 600 600 565 0.55 13.33 509 210 45 1 100

5 212.49 3 1.2 169.99 < 1242.69 600 600 565 0.55 13.33 509 210 45 1 100


50/65


Civil Engineering

- 50 -

Frame 6-6 Frame C-C

x

[mm]

Asnec

[mm2]

Nr of

bars[mm]

Aeff=n2/4

[mm2]

p%MRd

[kNm]

|ME'|

[kNm]

|NE'|

[kN] k

|Med|

[kNm]

|Ned|

[kN]

x

[mm]

Asnec

[mm2]

Nr of

bars

[mm]

Aeff=n2/4

[mm2]

p%MRd

[kNm]

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 0.30 421.03 30.74 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 stirrupss [mm]

S[mm]

lppzpw[%]nsl

125 5.4 10 105 100 900 0.71 >0.5%

125 5.4 10 121 100 600 0.71>0.35%125 5.4 10 134 100 600 0.71 >0.35%

125 5.4 10 155 150 600 0.47 >0.35%

125 5.4 10 192 150 600 0.47 >0.35%

125 5.4 10 266 200 600 0.35 >0.35%


51/65


Civil Engineering

- 51 -

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 andto 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 211. 1 : 2011 .

The critical zone is at the base of the wall having the length: because the building has 6 levels.

Where:-the height of the wall-the clear height of the story

-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 0.2. In this way the base becomesArea A with specific design performances, and the rest of the wall becomes Area B, with smaller

efforts.

Wall 1

- the height of Area A is 3m8.2 The values of the design sectional efforts in the walls

The 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 A

MEd=km**MEd'< *MEd,o' for the zone B

Where:

MEd' is the bending moment due to seismic loading

MEd,o' is the value of MEd' at the base


52/65


Civil Engineering

- 52 -

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

is the ratio between the capable moment and the design moment in each section Where: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 reinforcement

In the case of longitudinal reinforcement the overlapping is in zone A:

45d in the case of horizontal bars PC52

45d in the case of vertical bars PC52 with d20mm need welding

In 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


53/65


Civil Engineering

- 53 -

If the wall has no flanges the portions with the length 0. 1*lw at the ends of the beams are

considered flanges

The minimum vertical percentage is:

Seismic zone

Minimum reinforcement percentage

OB 37 PC 52, PC 60

zone A zone B zone A zone B

ag0. 12g 0,7% 0. 5% 0,6% 0. 5%

ag0. 12g are:

Zone A 8dbl


54/65


Civil Engineering

- 54 -

Wall 1 - zone A

Figure 8 Wall 1 Zone AM-N in

Text Esanu

Documents

Transcript of Text Esanu