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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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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
2Ψ
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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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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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
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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
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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
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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 = 0.24 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
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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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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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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
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