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GH. ASACHI TECHNICAL UNIVERSITY OF IASI CIVIL ENGINEERING AND BUILDING SERVICES FACULTY CIVIL ENGINEERING PROJECT II Civil constructions D E S I G N S U B J E C T : REINFORCED CONCRETE SHARE WALL MULTI-STOREY BUILDING, (Ug. + G + 4Fl . . . . 10 Fl.) AUTHOR, STUDENT: Blanariu DRAGOS Year: III – group: 3307 PROJECT ADVISOR: Sl. Dr. Ing. Radu PESCARU Sl. Dr. Ing. Irina BARAN Fetesti, pur.12 cm 1
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Page 1: Proiect constructii civile

GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

PROJECT IICivil constructions

D E S I G N S U B J E C T :

REINFORCED CONCRETE SHARE WALLMULTI-STOREY BUILDING,

(Ug. + G + 4Fl . . . . 10 Fl.)

AUTHOR, STUDENT: Blanariu DRAGOSYear: III – group: 3307

PROJECT ADVISOR: Sl. Dr. Ing. Radu PESCARU Sl. Dr. Ing. Irina BARAN

Fetesti, pur.12 cm

--- IAŞI 2012 – 2013 ---

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

PROJECT THEME

The project consists in some elements of design for a building that can have one of the following destinations: - block of flats; hotel; students accommodation building.

The building has UG+G+4Fl and it is placed in urban area on a land with the natural slope of 2% and connected to the city services facilities.

The structure will have:- pedestrian or non pedestrian reinforced concrete flat roof;- monolithic or precast reinforced concrete shear walls combined with

lamellar frames;- reinforced concrete floor slabs and stairs;- elastic cross beam or mat concrete foundations under the load bearing

walls and columns.

The envelope of the building consists of:- load-bearing walls of reinforced concrete or non load-bearing walls of

hollow brick or cellular concrete block masonry protected on the external face with al layer of efficient thermo insulating materials (fireproofed expanded polystyrene for face wall, mineral wool rigid plates, extruded polystyrene etc.);

- ventilated or un-ventilated could (insulated) terrace roof;- the ground floor plate over unheated basement will be insulated on the

ceiling of the basement;- the doors and windows can be of multi-layer wood or PVC profiles with

thermal insulating double or triple glassed window panels with Low CTE float glass. The windows will be provided with ventilating systems or with three opening positions in order to provide a adequate thermal protection and natural ventilation of the indoor space;

- the staircase is insulated and heated.

The project must contain information and details about:- the type and the composition of structural and for envelope elements

( walls, columns, floors, foundations, and roof);- the joints of the main structural and for envelope elements;- the heat conservation capacity of the building;- the structural performance under the load combination that contain

dead, snow, live loads and seismic action.The thermal and structural design and the constructive details of the elements

already mentioned.

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

The project consists of the following elements:

A. Written part :1. The project theme;2. Project content;3. Technical report;4. The design of the walls and roof for hygrothermal conditions

- the general factor of heat loos G,- estimation of water vapour condensation risk in the envelop structure;

5. The assessment of loads and design load combination;6. The computation of the cross shear walls sectional characteristics;7. Stress and strain analysis for the cross shear walls.

B. Drawings :

1. Current floor plan, scale 1/50,2. Ground floor plan (building access detail), scale 1/50,3. Cross section through the staircase, scale 1/50,4. Roof plan, scale 1/100,5. Constructive details, plan, scale 1/5, 1/10.

Technical Report

This project consists mainly of designing a block of flats with basement, groung floor and 4 floors. Now we shall determine the inner space area.

A dwelling could accomplish many functions, being determined by biological and psychological needs of people, which could be grouped in 4 categories:

- biological needs: food, hygiene assurance, rest, sleep;- social and professional activities: study, lecture;- entertainment activities;- auxiliary needs: traffic, cleaning.

To assure the optimal surfaces for rooms, we must take into account the purpose of every room. In the case of a villa, which is ment to assure a higher degree of confort, bigger surfaces must be used.

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

Building geometrical characteristics determination

1. Inferior slab area (A1), superior slab area (A2) are:

A1 = A2 = A3 = A4 = A5 = A6 = 328.06 m2

2. The perimeter:

P = 83.2 m

3. The free height of the ground floor (measured between the upper side of the inferior slab and the lower side of the superior slab) is: H = 2,80 m.

4. The area of the windows and doors are:

Ag = 196.8 m2

The block of flats has a of 4x80x90 cm bathroom windows, 6x120x120 cm windows, 10x150x120 cm windows, one door, 6 doors H = 205 cm, L = 80 cm.

5. The area of exterior walls (A = P * 5H – Ag):

A = 83.2 * 2,80 * 5 – 196.8 = 968 m2

6. The envelope of the building (A);

Ae = 328.06 + 328.06 + 968 = 1624.12 m2

7. The heated volume of the building (V); V = A1 * 5H

V = 328.06 x 2.80 x 5 = 4592.84 m3

8. Heated height of the building;

Hw = 2.81 * 5 – 0.14 = 13.86 m

Unidirectional specific thermal resistance determination (“R”)

R - the unidirectional specific thermic resistance, in the area without thermal bridges,where the element is composed of one or more layers of material considered homogenious(without thermal bridges),including layers of unventilated air, all disposed perpendicular to the direction of the air flow.

R=Ri+∑d/ +Rλ e [m2 *K/W]

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

Ri = 0.123 (outer walls) , 0.167(inferior slab) , 0.125(superior slab)

Re=0.06(outer walls,inferior slab) , 0.04(superior slab);

R'= 11R

+∑(Ψ ×l)

A ;

R=unidirection specific thermal resistance(in the area without thermic bridges;

Ψ= linear specific coefficient of thermal transport;

l = the length of the thermal bridge;

A = the area;

Unidirectional Specific Thermal Resistance R

Nr. Layer type d λ R=d/λCrt. [m] [W/m*K] [mp*k/W]

Rei

nfor

ced

conc

rete

ex

teri

or w

all

Interior surface - - 0.1251 Interior plaster M5 0.01 0.9 0.0112 Reinforced concrete 0.2 1.74 0.1143 Thermal isolation

EPS200.1 0.044 2.272

4 Exterior plaster M10 0.005 0.93 0.005Exterior surface - - 0.042

TOTAL 2.571

Mas

onry

ext

erio

r w

all Interior surface - - 0.125

1 Interior plaster M5 0.01 0.9 0.0112 Masonry GVP 0.25 0.7 0.3573 Thermal isolation

EPS200.1 0.044 2.272

4 Exterior plaster M10 0.005 0.93 0.005Exterior surface - - 0.042

TOTAL 2.813

Supe

rior

sla

b

Interior surface - - 0.1251 Ceiling plaster M5 0.01 0.9 0.0112 Reinforced concrete 0.14 1.74 0.0803 Equalizing slab M10 0.03 0.93 0.0324 Vapour barier - - -5 Slope slab 0.08 0.34 0.2356 Equalizing slab M10 0.03 0.93 0.0327 Difusion layer - - -

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

8 Thermal isolation 0.25 0.04 6.2509 Equalizing slab M10 0.03 0.93 0.032

10 Waterproof isolation - - -11 Gravel infilling - - -

Exterior surface - - 0.042TOTAL 6.840

Infe

rior

sla

b

Interior surface - - 0.1671 Mosaic 0.03 1.74 0.0172 Equalizing slab M10 0.03 0.93 0.0323 Reinforced concrete 0.14 1.74 0.0804 Mineral wool 0.1 0.04 2.5005 Cardboard plaster 0.01 0.4 0.025

Exterior surface - - 0.042TOTAL 2.863

Preliminary Calculus for the determination of the Rectified Thermic Resistance R’

Thermal Bridge

n L

[m]

Ψ

[W/m]

n* *lΨ R U’ R’

Reinforced concrete exterior walls

1.R.V. Outer corner

20 13.86 0.05 13.86

2.R.O. Footing

2 10.6 0.225 4.77

3.R.O. Curent

10 27.2 0.08 21.76

4. R.V. Curent

30 13.86 0.02 8.31

5.R.O. Eave 2 10.6 0.145 6.14A = 316.12 m2 54.85 2.57 0.55 1.79

Masonry exterior

walls

1.R.V. Curent

60 13.86 0.02 16.632

2.R.O. Footing

2 31 0.225 13.95

3.R.O. Balcony

8 3.2 0.15 3.84

4.R.O. Balcony

16 3.4 0.15 8.16

5.R.O. Eave

2 31 0.145 17.98

6.Carpent 1 653 0.25 163.25

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

ry contour

A = 687.72 m2 223.812 2.81 0.92 1.07Superior

slabR.V. Eave 2 31 0.225 13.95

2 10.6 0.225 4.77A = 328.06 m2 18.72 6.84 0.24 4.02

Inferior slab

1.R.O. Footing

2 10.6 0.225 4.772 31 0.225 13.95

2.R.O.Interior walls

2 262 0.2 104.8

A = 328.06 m2 123.52 2.86 1.02 0.97Carpentry A = 196.8 mp - - 0.55

Outer Wall no. 1

Outer Wall no. 2

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

Roof Floor

1 – waterproof membrane protection (mortar 3 cm + gravel or concrete boards);2 – waterproof membrane;3 – diffusive layer;4 – waterproof membrane support (mortar 3 cm);5 – polystyrene insulation 15 cm ;6 – vapor barrier;7 – diffusion layer;

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

8 – sloping layer (b.c.a.) 8 cm;9 – reinforced concrete flor slab – 12-15 cm;10 – ceiling plaster – 1,5 cm.

Slab Over Basement

1 – ceiling plaster – 1.5 cm;2 – thermal insulation of polystyrene – 10 cm;3 – reinforced concrete floor slab – 12-15 cm;4 – mortar bead – 3 cm;5 – ceramic floor tile – 1.2 cm.

Thermal bridges

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Page 10: Proiect constructii civile

GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

A thermal bridge represents a area of the buildings envelope where the thermic flow is in one direction and it is influenced by:

- the partial or total penetration of the building elements with different thermic conductivity;

- thickness alternation of the construction element;- alternation between the interior and exterior areas of the surfaces.

Types of thermal bridges:a) Vertical thermal bridges:

- bridge of outer corner - R.V. outer corner (symmetric bridge)

interiorR.V. - Outer corner

Bridge length= Hheated

Intinrasds 1 = 2 = 0,05 W/mΨ Ψ

Exterior

- brigde of inner corner – R.V. inner corner (symmetric bridge)

exterior

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

interior

R.V. – Inner corner

Bridge length= Hheated

1 = 2 = 0,01 W/mΨ Ψ

b) Horizontal thermal bridges- current bridge R.V. current (T)

Exterior1Ψ 2Ψ

interior interior

R.V. current (T)

Bridge length= Hheated

1 = 2 - 0,02 W/mΨ Ψ

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

- R.O. nonsymmetric eave

exterior interior

R.O. eave

Bridge length=Pslab

Pslab= the perimeter of the superior slab at the interior face of the walls

1= 0,14 W/mΨ

2=0,24 W/mΨ

R.O. footing Exterior Ti=+20o C

interior

Ts=+5o C

Hthermal isolation = min 70 cm

R.O. footing

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

Bridge length=Pslab

Pslab= the perimeter of the inferior slab at the interior face of the walls

1= 0,225 W/mΨ

2=0,30W/mΨ

- R.O. interior walls – inferior slab Ti=+20o C Ti=+20o C

Ts=+5o C Ts=+5o C

RO footing

Bridge length= Lwall

Lwall – interior wall length

1 = 2 = 0,2 W/mΨ Ψ

- R.O. interior walls – superior slab Nonheated tehnic bridge

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

1Ψ 2Ψ

interior interior

R.O. slab

Bridge length = L

L - interior wall length

1 = 2 = 0,24 W/mΨ Ψ

- thermal bridge on the carpentry contourexterior 1Ψ

interior

R.V. current (T)

Bridge length = Pcarpentry

Pcarpentry = 2*(L+H)

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

1=0.25 W/mΨ

The general factor of thermal insulation G

This factor of thermal insulation reflects the sum of all thermal loss trough all the building envelope elements for a thermal difference of 1 Kelvin ( ) plus the losses Δθbecause of ventilation and air infiltration.

G≤GN

G= 1V

∗∑ A∗τR 'm

+0.34∗n [W/m3K];

τ – The temperature correction factor

The area chosen is Fetesti city situated on thermal map in zone II with Te = -150 C.

τ=(T i−Tu )(T i−Te )

= 20−520+15

=0.42

A – The area of the element with the specific main thermal resistance R’m;A[m2];

V – represents the inside volume of the building V[m3] ;R’m - represents the average corrected thermal resistance of the building element,[m2K/W];n – represents the ventilation rate: n = 0.5

AV

=1856.764592.84

=0.40 => GN = 0.59

G= 14592.84

∗(316.12∗0.421.79+ 687.72∗1

1.07+328.06∗1

4.02+328.06∗1

0.97 )+0.34∗0.5=0 .41 <

GN

Calculul higrotermic

Checking the risk of condensation ( C107/6; C107/3 )

1. On the face of the element;

T si=T i−(T i−T e)

R∗Ri=20−

20+152.57

∗0.125=18.29≫τn = 9.3 => no risk of

condensation.

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

2. In the structure of the element.

T 1=T i−(T i−T e)

R∗( 1α i

+d1λ1 )=20−20+152.57

∗(0.125+ 0.010.9 )=18 .14

T 2=T i−(T i−T e)

R∗( 1α i

+d1λ1

+d2λ2 )=20−20+152.57

∗(0.125+ 0.010.9 + 0.21.74 )=16 .58

T 3=T i−(T i−T e)

R∗( 1α i

+d1λ1

+d2λ2

+d3λ3 )=20−20+152.57

∗(0.125+ 0.010.9 + 0.21.74

+ 0.10.044 )=−14 .37

T 4=T i−(T i−T e)

R∗( 1αi

+d1λ1

+d2λ2

+d3λ3

+d4λ4 )=20−20+152.57

∗(0.125+ 0.010.9 + 0.21.74

+ 0.10.044

+ 0.0050.93 )=−14 .44

T se=T i−(T i−Te )

R∗(R1+R2+R3+R4+R5+R6 )=20−20+152.57

∗(0.125+0.011+0.114+2.272+0.005+0.042 )=−14 .96

Tsi = 18.29 => Ps,si = 2105 pa T1 = 18.14 => Ps1 = 2009 pa T2 = 16.58 => Ps2 = 1937 pa T3 = -14.37 => Ps3 = 181 pa T4 = -14.44 => Ps4 = 170 pa Ts,se = -14.96 => Ps,se = 167 pa

Pvi=Φ i∗P si

100=55∗2105 pa

100=1157 .75 pa

Pve=Φe∗Pse

100=85∗167 pa

100=141.95 pa

R v1=d1∗μ1∗M 1=0.01∗7.1∗54∗108=3.83∗108

R v2=d2∗μ2∗M 2=0.2∗21.3∗54∗108=230∗108

R v3=d3∗μ3∗M3=0.1∗30∗54∗108=162∗108

R v4=d4∗μ4∗M 4=0.005∗7.1∗54∗108=1.91∗108

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

Determining the moments of inertia for the reinforced concrete elements

Element No. 1

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

A1 = 2.22 m2

A2 = A3 = 0.05 m2

A4 = 0.28 m2

xG=x1 A1+x2 A2+x3 A3+x4 A4

A1+A2+A3+A4=−0.20∗0.05−0.20∗0.05−0.8∗0.28

2.6=−0 .09m

Ix = Ix1 + Ix2 + Ix3 + Ix4 = 22.79 + 2.22 * ( -0.09 )2 + 0.00026 + 0.05 * ( -0.19 )2 + 0.00026 + 0.05 * ( -0.19 )2 + 0.00093 + 0.28 * ( -0.79 )2 = 22.979 m4

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

Iy = Iy1 + Iy2 + Iy3 + Iy4 = 0.0074 + 0.00016 + 0.05 * ( -5.4 )2 + 0.00016 + 0.05 * ( -5.4)2 +0.045 = 2.968 m4

Active sections to shear force

Am, t=b∗hk

=0.20∗11.11.1 = 2.01 m2 k = 1.1 for “T” and “L” sections;

Equivalent moment of inertia

I0 = Im = Ix = 22.979 m4

I e=η s∗I 0

1+ν s∗ηs∗I 0Am, t∗H2

= 1∗22.979

1+10.87 1∗22.9792.01∗142

=14 .063m4

Element No. 2

A1 = 0.15 m2

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

A2 = 0.68 m2

A3 = 0.04 m2

xG=x1 A1+x2 A2+x3 A3

A1+A2+A3=0.20∗0.04

0.87=0 .009m

yG=y1 A1+ y2 A2+ y3 A3

A1+A2+A3=0.125∗0.15+2.30∗0.68+3.75∗0.04

0.87=1 .99m

Ix = Ix1 + Ix2 + Ix3 = 0.00078 + 0.15 * ( 0.009 )2 + 1.14 + 0.68 * ( 0.009 )2 + 0.00013 + 0.04 * ( 0.291 )2 = 1.1443 m4

Iy = Iy1 + Iy2 + Iy3 = 0.0045 + 0.15 * ( -1.865 )2 + 0.0027 + 0.68 * ( 0.31 )2 + 0.00013 + 0.04 * ( 1.76 )2 = 0.7182 m4

Active sections to shear force

Am, t=b∗hk

=0.20∗4.351.1

=0 .79m2 k = 1.1 for “T” and “L” sections;

Equivalent moment of inertia

I0 = Im = Ix = 1.443 m4

I e=η s∗I 0

1+ν s∗ηs∗I 0Am, t∗H2

= 1∗1.443

1+10.87 1∗1.4430.79∗142

=1 .3102m4

Element No.3

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

A1 = 0.04 m2

A2 = 0.12 m2

yG=y1 A1+ y2 A2

A1+A2= 0.20∗0.12

0.16=0 .15m

Ix = Ix1 + Ix2 = 0.00013 + 0.0004 = 0.00053 m4

Iy = Iy1 + Iy2 = 0.00013 + 0.04 * ( -0.15 )2 + 0.0036 + 0.12 *( 0.05 )2 = 0.0049 m4

Active sections to shear force

Am, t=b∗hk

=0.20∗0.401.1

=0 .07m2 k = 1.1 for “T” and “L” sections;

Equivalent moment of inertia

I0 = Im = Iy = 0.0049 m4

I e=η s∗I 0

1+ν s∗ηs∗I 0Am, t∗H2

= 1∗0.0049

1+10.87 1∗0.00490.07∗142

=0 .0048m4

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

Element No.4

A1 = 0.60 m2

A2 = 0.34 m2

yG=y1 A1+ y2 A2

A1+A2=0.95∗0.34

0.94=0 .34m

Ix = Ix1 + Ix2 = 0.002 + 0.0818 = 0.0838 m4

Iy = Iy1 + Iy2 = 0.45 +0.60 * ( 0.34 )2 + 0.0011 + 0.34 * ( 0.61 )2 = 0.6446 m4

Active sections to shear force

Am, t=b∗hk

=0.20∗1.901.1

=0 .34m2 k = 1.1 for “T” and “L” sections;

Equivalent moment of inertia

I0 = Im = Iy = 0.6446 m4

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

I e=ηs∗I 0

1+ν s∗ηs∗I 0Am, t∗H2

= 1∗0.6446

1+10.87 1∗0.64460.34∗142

=0 .5832m4

Element No.5

A1 = 0.08 m2

A2 = 0.48 m2

A3 = 0.15 m2

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

xG=x1 A1+x2 A2+x3 A3

A1+A2+A3=−0.10∗0.08

0.71=−0 .011m

yG=y1 A1+ y2 A2+ y3 A3

A1+A2+A3= 0.10∗0.08+1.40∗0.48+2.725∗0.15

0.71=1.53m

Ix = Ix1 + Ix2 + Ix3 = 0.00026 + 0.08 * ( -0.089 )2 + 0.2304 + 0.48 * ( 0.011 )2 + 0.00078 + 0.15 * ( 0.011 )2 = 0.2320 m4

Iy = Iy1 + Iy2 + Iy3 = 0.0010 + 0.08 * ( -1.43 )2 + 0.0016 + 0.48 * ( -0.13 )2 + 0.0036 + 0.15 * ( 1.195 )2 = 0.3920 m4

Active sections to shear force

Am, t=b∗hk

=0.20∗2.851.1

=0 .51m2 k = 1.1 for “T” and “L” sections;

Equivalent moment of inertia

I0 = Im = Iy = 0.3920 m4

I e=η s∗I 0

1+ν s∗ηs∗I 0Am, t∗H2

= 1∗0.3920

1+10.87 1∗0.39200.51∗142

=0 .3959m4

Element No.6

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GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

A1 = 0.15 m2

A2 = 0.76 m2

A3 = 0.12 m2

xG=x1 A1+x2 A2+x3 A3

A1+A2+A3= 0.30∗0.12

1.03=0.034m

yG=y1 A1+ y2 A2+ y3 A3

A1+A2+A3=0.125∗0.15+2.15∗0.76+4.15∗0.12

1.03=2 .08m

Ix = Ix1 + Ix2 + Ix3 = 0.00078 + 0.15 * ( -0.034 )2 + 0.9145 + 0.76 * ( -0.034 )2 + 0.0004 + 0.12 * ( 0.226 )2 = 0.9250 m4

Iy = Iy1 + Iy2 + Iy3 = 0.0045 + 0.15 * ( -1.955 )2 + 0.0025 + 0.76 * ( 0.07 )2 + 0.0036 + 0.12 * ( 2.17 )2 = 1.5019 m4

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Page 26: Proiect constructii civile

GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

Active sections to shear force

Am, t=b∗hk

=0.20∗4.251.1

=0 .77m2 k = 1.1 for “T” and “L”

sections;

Equivalent moment of inertia

I0 = Im = Iy = 1.5019 m4

I e=η s∗I 0

1+ν s∗ηs∗I 0Am, t∗H 2

= 1∗1.5019

1+10.87 1∗1.50190.77∗142

=1 .3553m4

Element No.7

A1 = 0.04 m2

A2 = 0.12 m2

A3 = 0.04 m2

Ix = Ix1 + Ix2 + Ix3 = 0.00013 + 0.0004 + 0.00013 = 0.00066 m4

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Page 27: Proiect constructii civile

GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

Iy = Iy1 + Iy2 + Iy3 = 0.00013 + 0.04 * ( -0.20 )2 + 0.0036 + 0.00013 + 0.04 * ( 0.20 )2 = 0.00706 m4

Active sections to shear force

Am, t=b∗hk

=0.20∗0.601.1

=0 .10m2 k = 1.1 for “T” and “L” sections;

Equivalent moment of inertia

I0 = Im = Iy = 0.00706 m4

I e=η s∗I 0

1+ν s∗ηs∗I 0Am, t∗H 2

= 1∗0.00706

1+10.87 1∗0.007060.10∗142

=0 .00703m4

Element No.8

27

Page 28: Proiect constructii civile

GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

A1 = 0.08 m2

A2 = 0.48 m2

A3 = 0.15 m2

xG=x1 A1+x2 A2+x3 A3

A1+A2+A3=−0.10∗0.08

0.71=−0 .011m

yG=y1 A1+ y2 A2+ y3 A3

A1+A2+A3= 0.10∗0.08+1.40∗0.48+2.725∗0.15

0.71=1.53m

Ix = Ix1 + Ix2 + Ix3 = 0.00026 + 0.08 * ( -0.089 )2 + 0.2304 + 0.48 * ( 0.011 )2 + 0.00078 + 0.15 * ( 0.011 )2 = 0.2320 m4

Iy = Iy1 + Iy2 + Iy3 = 0.0010 + 0.08 * ( -1.43 )2 + 0.0016 + 0.48 * ( -0.13 )2 + 0.0036 + 0.15 * ( 1.195 )2 = 0.3920 m4

Active sections to shear force

28

Page 29: Proiect constructii civile

GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

Am, t=b∗hk

=0.20∗2.851.1

=0 .51m2 k = 1.1 for “T” and “L” sections;

Equivalent moment of inertia

I0 = Im = Iy = 0.3920 m4

I e=η s∗I 0

1+ν s∗ηs∗I 0Am, t∗H2

= 1∗0.3920

1+10.87 1∗0.39200.51∗142

=0 .3959m4

Element No.9

29

Page 30: Proiect constructii civile

GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

A1 = 0.15 m2

A2 = 0.76 m2

A3 = 0.24 m2

xG=x1 A1+x2 A2+x3 A3

A1+A2+A3=−0.50∗0.24

1.15=−0 .10m

yG=y1 A1+ y2 A2+ y3 A3

A1+A2+A3= 0.125∗0.15+2.15∗0.76+4.15∗0.24

1.15=2.30m

Ix = Ix1 + Ix2 + Ix3 = 0.00078 + 0.15 * ( -0.10 )2 + 0.9145 + 0.76 * ( 0.10 )2 + 0.0008 + 0.24 * ( -0.40 )2 = 0.9635 m4

Iy = Iy1 + Iy2 + Iy3 = 0.0045 + 0.15 * ( -2.175 )2 + 0.0025 + 0.76 * ( -0.15 )2 + 0.028 + 0.24 * ( 1.825 )2 = 1.7373 m4

Active sections to shear force

30

Page 31: Proiect constructii civile

GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

Am, t=b∗hk

=0.20∗4.251.1

=0 .77m2 k = 1.1 for “T” and “L” sections;

Equivalent moment of inertia

I0 = Im = Iy = 1.7373 m4

I e=η s∗I 0

1+ν s∗ηs∗I 0Am, t∗H 2

= 1∗1.7373

1+10.87 1∗1.73730.77∗142

=1 .5441m4

Element No.10

A1 = 0.44 m2

A2 = 1.04 m2

A3 = 0.15 m2

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Page 32: Proiect constructii civile

GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

yG=y1 A1+ y2 A2+ y3 A3

A1+A2+A3=2.7∗1.04+5.425∗0.15

1.63=2 .22m

Ix = Ix1 + Ix2 + Ix3 = 0.0014 + 2.3434 + 0.00078 = 2.3455 m4

Iy = Iy1 + Iy2 + Iy3 = 0.1774 + 0.44 * ( -2.22 )2 + 0.0034 + 1.04 * ( 0.48 )2 + 0.0045 + 0.15 * ( 3.205 )2 = 4.1341 m4

Active sections to shear force

Am, t=b∗hk

=0.20∗5.651.1

=1 .02m2 k = 1.1 for “T” and “L” sections;

Equivalent moment of inertia

I0 = Im = Iy = 4.1341 m4

I e=ηs∗I 0

1+ν s∗ηs∗I 0Am, t∗H2

= 1∗4.1341

1+10.87 1∗4.13411.02∗142

=3 .3756m4

Weight per floor. Total weight of the structure

Snow Load

qs=μi∗C e∗Ct∗qs0

C e=1

C t=1

α∈ (0 ,30° )→μ i=0.8

qs0=2.5 [ KNm2 ]=250[ daNm2 ]qs=0.8∗1∗1∗250=200[ daNm2 ]

Ce -> coefficient = 1

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Page 33: Proiect constructii civile

GH. ASACHI TECHNICAL UNIVERSITY OF IASICIVIL ENGINEERING AND BUILDING SERVICES FACULTYCIVIL ENGINEERING

Ct -> thermal coefficient = 1 = μ μi = 0.8

Sd=γ A∗Sk

Sd=1 .5∗200=300daN {m¿2

Glast floor=Gterrace+12Gwalls+max (q terrace , qsnow )∗φ=58968+1

2∗564649.65+1.6∗0.4=341293 .46daN

Gfloor=Gslab+Gwalls+φ∗(Q slab+Qinner walls)=156778.2+564649.65+0.4∗115385.18=767581 .922daN

Gstructure=4∗Gfloor+Glast floor=341293.46+4∗767581.922=3411621 .148daN

Seismic load

Tc = 1.6 (for Buzău) T1 = 0.36 T1 < Tc => = 0.85λ

β (T )=β0∗T c

T1=2.75∗1.6

0.36=12 .22

Gt=G4+4∗G0=3411621 .148daN

m=Gt

g=3411621.148

10=341162 .11kg

Se=ag∗β (T )=2.74∗12.22=33 .48

Sd=Se∗[ T 12π ]2

=33.48∗[0.362π ]2

=0 .109

Fb=φ∗Sd∗m∗λ=1∗0.109∗341162.11∗0.85=31608 .66 .39daN

33


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