proiect constructii civile 2
TRANSCRIPT
“Gh. Asachi” Technical university of iasicivil engineering
Contents
Written part:
1. Project Theme;
2. Technical Report;
3. Hydrothermal design of the envelope elements :
3.1. Estimation of general factor of heat loss ”G”;
3.2. Checking for condensation in one of the envelope element;
4. Estimation of the loads and the loads combination
5. Diagrams for shear wall section.
Drawing part:
1. Floor plane current (1:50);
2. Building access detail (1:50);
3. Roof plane (1:100);
4. Transversal cross section through the stair case (1:50);
5. Construction details (1:5,1:10);
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“Gh. Asachi” Technical university of iasicivil engineering
1. Project Theme
The project consists in elements of design for a building that can have one of the
following destinations:
Collective dwelling
Hotel
Student accommodation
Hostel
The structure of the building is composed of monolithic reinforcement concrete,
shear walls, combined with frame structures, reinforced concrete floor slabs and
cross foundation under the walls or columns. The envelope of the building must be
covered with protective layer of efficient thermal resistance materials.
2. Technical report
This project combines all the technical documentation needed for the erection of a hotel with Ug+Gr+4Fl. , having the structure made of masonry walls, structural exterior walls, structural interior walls, reinforced concrete pillars and girdles.
Foundation placement The building is situated in Suceava, at the main street. The building terrain is
leveled and has the local and general stability assured.According to the geological study the terrain has the following characteristics:
At the surface vegetal soil and soil fill at about 80 cm depth; Underground water is present below 7.00 m; Seismic zone “C” conf P100/92: Tc=0.7s – corner period;
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Construction solutionInfrastructureThe infrastructure will be realized under the form of a continues foundation under
the walls with a concrete skid made up of concrete (C 6/7.5) and concrete elevation (C 16/20), 90 cm thick under the first-floor walls, 105 cm under the under-ground floor. Also, the infrastructure will be reinforced with girdles on the whole length. The reinforcement will be made with PC52 and OB37 steel.
SuperstructureThe strength structure is consisted of structural walls, masonry walls( mark 75 cal I
and masonry plaster MZ 50 ), and girdles. Exterior walls are 25 cm thick, interior walls are 20 cm thick and exterior and interior structural walls are 20 cm thick. The materials used in pillars, girdles and beams will be: concrete C16/20 (Bc20) and reinforcements PC52 and OB37. The masonry will be consisted of BCA.
The stairs are made up of monolith reinforced concrete.The slab over the ground floor will be made up of reinforced concrete (C16/20). The
slab will be 13 cm thick and it will be reinforced with steel STNB nets.The roof over the garret is consisted of infusible fir (resinous wood).
FinishingExterior paintings will be carefully chosen function of the volumetric of the
house:White and amber plaster, PVC windows, wood doors, wood handrail, PVC ditch
and down-comers.Interior paintings will be in accordance with the hygienic-sanitary requirements
that are imposed by the active normative and by the authorization given by D.S.P.So, interior paintings will be chosen such that the exterior-interior chromatic
sensation to be a fine one. For the living rooms we will use a white, creamy washable lime and for the bathrooms, kitchen and stores we’ll use sandstone assorted with the wall paintings.
Functional solutionInner space sistematization:
BASEMENT: 265.64 [m2]TERRACE AREA: 296.82[m2]GROUND FLOOR:
4 x Kitchen:45.88 [m2] 8 x Bathroom: 30.09[m2]
7 x Bedroom: 91.68 [m2]
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8 x Lobby: 48.68 [m2]
Staircase: 14.68 [m2] Entrance Lobby: 14.77[m2]
TOTAL: 245.78 [m2]UPPER FLOORS:
4 x Kitchen:45.88 [m2] 8 x Bathroom: 30.09[m2]
8 x Bedroom: 106.44 [m2]
8 x Lobby: 48.68 [m2]
Staircase: 14.68 [m 2 ] TOTAL: 245.77*4=1494.5 [m2]
Urbanism conditions Water supply will be assured by S.C. RAJAC S.R.L. Electricity supply will be assured by „E.ON Moldova S.A.” The sewerage will be made by joining the sewerage network from the
neighborhood. The heating will be made individually by central heating. The fuel (gas) supply will be made joining to the local network.
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3. Hydrothermal design of the envelope elements
3.1. Estimation of general factor of heat loss ”G”;
Calculus of the distinct areas for the thermal design of the building
exterior structural wall
- total area: 403.61m2
- gap area: 32.4m2
- solid area: 371.21m2
exterior wall
- total area: 565.896m2
- gap area: 148.8m2
- solid area: 417.10m2
Determination of the specific thermal resistance(R)
a. For structural external wall:
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NO. Layer d[m]
λ[w/mk]
R[m2k/w]
Interior surface - - Ri=1/81. Lime cement mortar 0.01 0.87 R1=0.0112. Reinforced concrete 0.20 1.74 R2=0.1143. Polystyrene 0.10 0.04 R3=2.5004. External plaster 0.005 0.52 R4=0.009
External surface - - Re=1/24TOTAL R=2.825
R=R i+∑d i
λi
+Re=1α i
+∑ d i
λ i
+ 1λe
R=18+ 0.001
0.87+ 0.2
1.74+ 0.1
0.04+ 0.005
0.52+ 1
24=2.825[m2 K /W ]
b. For B.C.A. masonry external wall:
NO. Layer d[m]
λ[w/mk]
R[m2k/w]
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Interior surface - - Ri=1/81. Lime cement mortar 0.01 0.87 R1=0.0112. B.C.A. 0.25 0.27 R2=0.9263. Polystyrene 0.10 0.04 R3=2.5004. External plaster 0.005 0.52 R4=0.009
External surface - - Re=1/24TOTAL R=3.636
R=18+ 0.03
0.87+ 0.25
0.27+ 0.10
0.04+ 0.005
0.52+ 1
24=3.636 [m2 K /W ]
c. For floor over basement:
NO. Layer d[m]
λ[w/mk]
R[m2k/w]
Interior surface - - Ri=1/61. Cold flooring 0.03 1.16 R1=0.025
2. Equalization screed (M100) 0.03 0.91 R2=0.0323. Reinforced concrete slab 0.13 1.74 R3=0.0744. Polystyrene 0.10 0.04 R4=2.5005. Lime cement mortar 0.012 0.87 R5=0.013
External surface - - Re=1/12TOTAL R=2.894
R=16+ 0.03
1.16+0.03
0.91+ 0.13
1.74+ 0.10
0.04+ 0.012
0.87+ 1
12=2.894 [m2 K /W ]
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d. For terrace:
NO. Layer d[m]
λ[w/mk]
R[m2k/w]
Interior surface - - Ri=1/81. Lime cement mortar 0.01 0.87 R1=0.0112. Reinforced concrete slab 0.13 1.74 R2=0.0053. Equalization screed 0.03 0.91 R3=0.0324. Vapor barrier - - -5. Protection layer - - -6. B.C.A. 0.08 0.27 R6=0.2967. Equalization screed 0.03 0.91 R7=0.0328. Diffusion layer - - -9. Polystyrene 0.25 0.04 R9=6.25010. Protection screed 0.03 0.91 R10=0.03211. Waterproof membrane - - -12. Gravel - - -
External surface - - Re=1/24TOTAL R=6.898
R=18+ 0.01
0.87+ 0.13
1.74+ 0.03
0.91+ 0.08
0.27+ 0.03
0.91+ 0.25
0.04+ 0.03
0.91+ 1
24=6.898
Determination of specific thermal adjusted resistance (R’)
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Element Thermal bridge Ψ[w/m]
l[m]
Ψ*l[w]
A[m2]
R[m2k/w]
U’ R’[m2k/w]
Structural external wall
V. j. corner out 0.1 55.48 5.548
371.217 2.825 0.520 1.923V. j. current ‘T’ 2*0.14 83.22 23.31V. j. at cornice 0.15 29.1 4.36H. j. at basement 0.18 29.1 5.23H. j. current 0.11 116.4 12.8H. j. current 0.13 116.4 15.13
Masonry external wall
V. j. corner out 0.06 55.48 3.33
417.10 3.636 0.453 2.207
V. j. current ‘T’ 2*0.06 55.48 6.657V. j. current ‘T’ 2*0.04 83.22 6.657H. j. at cornice 0.14 40.8 5.72H. j. at basement 0.15 40.8 6.12H. j. current 0.09 110.4 9.93H. j. current 0.11 110.4 12.14H. balcony 0.15 66 9.9
0.21 66 13.86Floor over
BasementH. j. at basement 0.27 29.1 7.85
296.82 2.894 0.498 2.01H. j. at basement 0.25 40.8 10.2H. j. int. wall 2*0.11 123.5 27.17
Terrace H. j. at cornice 0.19 29.1 5.53 296.82 6.898 0.188 5.31H. j. at cornice 0.18 40.8 7.34
Window Woodwork j. 0.27 471.6 127.33 181.2 - - 0.55
U '= 1R
+Σ (Ψ∗l )
A
R – specific resistance
Ψ – specific linear coefficient of the thermal transfer process
l – the length of the thermal bridge
A – area of the element
Determination of global coefficient of thermal insulation (G)
9
R '= 1U '
G=1V
⋅∑( A⋅τ
R ' )+0. 34⋅n[ W /(m3 K ) ]
“Gh. Asachi” Technical university of iasicivil engineering
Vheated – interior heated volume of the building
A- area of the outer cover element
n - the natural ventilation speed of the building, number of air changes per hour
n =0.6
τ -correction factor of the exterior temperature
τ=T i−T u
T i−T e
τ basement=20−520+21
=0.6097
Vheated=Abasement¿ l
l=5¿hfloor - hslab
l=5¿2.8 - 0.13=13.87 [m]
Vheated=296.82¿13.87 [m3]
∑ ( A⋅τ
R' )= Aexterior wallI
⋅τ
Rexterior wallI
'+
Aexterior wallII
⋅τ
Rexteriorwall II
'+
Abasement⋅τ
Rbasement'
+Aterrace⋅τ
Rterrace'
+Awoodwork⋅τ
Rwoodwork'
[w/m3k]
G= 14116.89
+0.34*0.6=0.412 [w/m3k]
Ah eated
V h eated
=1563.1674116.85
=¿0.37=¿GN=0.574 [w /m3 k ]
G=0.412<GN
Checking up the risk of condensation for a structural external wall according
to C107/3
10
∑ ( A⋅τ
R' )=371 .217⋅11 . 923
+417 .1⋅12 . 207
+296.82⋅0.6097
2. 01+
296 .82⋅15 .31
+181.2⋅1
0. 55
∑ ( A⋅τ
R' )=857 . 412
“Gh. Asachi” Technical university of iasicivil engineering
T Si=T i−T i−T e
R∗Ri=20−
20+212.825
∗1
8=18.1℃→ PS , Si=2079[ Pa]
T 1=T i−T i−T e
R∗( 1
α i
+d1
λ1)=20−20+21
2.825∗( 1
8+ 0.03
0.87)=17.6℃→
PS 1=2014 [Pa]
T 2=T i−T i−T e
R∗( 1
α i
+d1
λ1
+d2
λ2)=20−20+21
2.825∗( 1
8+ 0.03
0.87+ 0.2
1.74)=16.0℃→
PS 2=1818[ Pa]
T 3=T i−T i−T e
R∗( 1
α i
+d1
λ1
+d2
λ2
+d3
λ3)=20−20+21
2.825∗( 1
8+ 0.03
0.87+ 0.2
1.74+ 0.1
0.04)
T 3=−20.2℃→ PS 3=101[ Pa]
T Se=T i−T i−T e
R∗( 1
αi
+d1
λ1
+d2
λ2
+d3
λ3
+d4
λ4)=¿
¿20−20+212.825
∗( 18+ 0.03
0.87+ 0.2
1.74+ 0.1
0.04+ 0.005
0.52 )=−20.4℃→ PS , Se=99[Pa ]
Dew temperature
φi=55%
Ti= +20 OC => τr=10.7 OC
TSi=18.3 OC > τr
RVi=di*μi*M [m/s]
RV1=0.03*8.5*54*108=13.77*108 [m/s]
RV2=0.2*21.3*54*108=230.04*108 [m/s]
RV3=0.1*30*54*108=162*108 [m/s]
RV4=0.005*4.7*54*108=1.269*108 [m/s]
PVi=φ i∗PSi
100=55∗2340
100=1287
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PVe=φe∗PSe
100=85∗94
100=79.9
Checking for the water accumulation risk as a consequence of vapors condensation in the structure elements.
4. Estimation of the loads and the loads combination
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4.1The geometrical characteristics of calculus of structural walls
External Structural Wall
Data for the vertical beam
Vertical beam 1.
Active section at eccentric compression
Am ,1=4.4∗0.2+0.2∗0.2=0.92[m2]
Active section at shear force
Am1 , t=4.4∗0.2=0.88 [m2 ]The section centroid position
Y G, 1=Y 1 A1∗Y 2 A2
A1+ A2
=0∗0.88+2.1∗0.040.92
=0.0913 [m ]
Moment of inertia
I m1=[ 4.43∗0.212
+0.09132∗4.4∗0.2]+[ 0.23∗0.212
+2.00872∗0.22]=1.588 [m4 ]
Vertical beam 2.
Active section at eccentric compression
Am ,2=0.85∗0.2+0.65∗0.2=0.30[m2]
Active section at shear force
Am ,t=b∗h
k
k =1 for rectangular sections;
k=1.1 for sections in T and L ;
k=1.0 for sections in I and C;
Am2 , t=0.85∗0.2
1.1=0.154 [ m2 ]
The section centroid position
Y G, 2=Y 1 A1∗Y 2 A2
A1+ A2
=0∗0.17+0.025∗0.130.30
=0.0108 [ m ]
Moment of inertia
I m2=[ 0.853∗0.212
+0.01082∗0.17 ]+[ 0.23∗0.06512
+0.1358∗0.13]=0.0148 [m4 ]
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Vertical beam 3.
Active section at eccentric compression
Am ,3=0.8∗0.2+0.2∗0.2=0.20 [m2]
Active section at shear force
Am3 , t=0.8∗0.2
1.1=0.145 [m2 ]
The section centroid position
Y G, 3=Y 1 A1∗Y 2 A2
A1+ A2
=0 [m ]
Moment of inertia
I m3=[ 0.83∗0.212 ]+[ 0.23∗0.2
12 ]=0.00866 [m4 ]
Vertical beam 4.
Active section at eccentric compression
Am ,4=0.85∗0.2+0.65∗0.2=0.30 [m2]
Active section at shear force
Am 4 ,t=0.85∗0.2
1.1=0.154 [m2 ]
The section centroid position
Y G, 4=Y 1 A1∗Y 2 A2
A1+ A2
=0∗0.17+0.025∗0.130.30
=0.0108 [ m ]
Moment of inertia
I m4=[ 0.853∗0.212
+0.01082∗0.17]+[ 0.23∗0.06512
+0.1358∗0.13]=0.0148 [m4 ]
Vertical beam 5.
Active section at eccentric compression
Am ,5=4.4∗0.2+0.2∗0.2=0.92[m2]
Active section at shear force
Am1 , t=4.4∗0.2=0.88 [m2 ]The section centroid position
Y G, 1=Y 1 A1∗Y 2 A2
A1+ A2
=0∗0.88+2.1∗0.040.92
=0.0913 [m ]
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Moment of inertia
I m1=[ 4.43∗0.212
+0.09132∗4.4∗0.2]+[ 0.23∗0.212
+2.00872∗0.22]=1.588 [m4 ]
Data for the cross-beam
Distance between centroids
L1−2=3.605 [ m ] L2−3=1.635 [ m ] L3−4=1.635 [ m ] L4−5=3.605 [ m ] L1−5=10.68 [ m ]
Section depth of the cross-beam constraining
a=0.35∗hr ≤ 0.40 [ m ]
a=0.35∗1.6=0.56 [ m ]>0.40 → a=0.40 [ m ]
Length of the gap
l0=0.90 [m ]
Calculus length of the cross-beam
l=l0+2a=0.90+2∗0.40=1.70 [m ]
The floor plate does not pull together with the cross-beam
I r=0.2∗1.63
12=0.0682 [ m ]
Active section at shear force
Ar ,t=b∗hr
k=0.2∗1.6
1.2=0.267[m ]
The rigidities calculus considering the structural wall a cantilever equal with the building height.
ν0=βs
θ s 4 αs
; I 0 S=ηs∗I 0
I eS=ηS∗I 0
1+ν0∗ηS∗I 0
H2∗∑ Am,t
ν0=the number of levels coeffcient
I 0=the global moment of inertia of the whole sectionηs=a cofficient depending on the number of levels∧λ , γ .
λ=the ratiobetwe nthe distorsion rigidities of the vertical∧cross−beams
γ=the coeficient wich introduces the axial deformability15
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Kme=∑
1
n
Kmi
2; K re
=∑1
n−1
K r ; λ=2∑
1
n−1
K r
∑1
n
K m
λ=K r
K m
;γ=1+∑
1
n
I m
L1 ,n2 ( 1
Am1
+ 1Am,n
)
Rigidity of the cross-beam
K r=6 E r∗I r
L ( Ll )
3
∗μ
μ= 1
1+30 I r
A r . t∗l2
= 1
1+30∗0.0682
0.2667∗1.702
=0.273
K r ,1=6∗6.75∗0.0682∗109
3.605∗( 3.605
1.70 )3
∗0.273=1.994∗109 [ N∗m ]
K r ,2=6∗6.75∗0.0682∗109
1.635∗( 1.635
1.70 )3
∗0.273=0.41∗109 [ N∗m ]
K r ,3=6∗6.75∗0.0682∗109
1.635∗( 1.635
1.70 )3
∗0.273=0.41∗109 [ N∗m ]
K r ,4=6∗6.75∗0.0682∗109
3.605∗( 3.605
1.70 )3
∗0.273=1.994∗109 [ N∗m ]
∑1
n−1
K r=4.808∗109[N∗m ]
Rigidity at distortion
Km=Em∗I m
he
Km, 1=27∗109∗1.588
2.8=15.312∗109[ N∗m ]
Km, 2=27∗109∗0.0148
2.8=0.142∗109 [N∗m ]
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Km, 3=27∗109∗0.00866
2.8=0.0835∗109[N∗m ]
Km, 4=27∗109∗0.0148
2.8=0.142∗109[ N∗m]
Km, 5=27∗109∗1.588
2.8=15.312∗109[ N∗m]
∑1
n
K mi=30.991∗109[N∗m ]
Calculus of the coefficients λ and γ of the structural wall
λ=2∑
1
n−1
K r
∑1
n
Km
=2∗4.808∗109
30.991∗109 =0.310 .
γ=1+∑
1
n
I m
L1 ,n2 ( 1
Am1
+ 1Am,n )=1+ 3.214
10.682 ( 10.92
+ 10.92 )=1.061 .
{λ=0.310γ=1.061
→ηs=0.185
Global Moment of Inertia
Is computed for each case.
I 0=I m,1+ Am ,1∗c12+ I m,2+ Am, 2∗c2
2+ I m,3+ Am ,3∗c32+ I m, 4+Am, 4∗c4
2+ I m,5+ Am ,5∗c52 .
I 0=2∗[1.588+(5.3412∗0.92 )+0.0148+(1.712∗0.3 ) ]+0.00866=57.542 m2
Equilibrium Moment of Inertia
I 0 S=ηs∗I 0=0.185∗57.542=10.645 [m4 ]Equivalent Moment of Inertia
∑ Am,t=0.88+0.154+0.145+0.154+0.88=2.054 [m¿¿2]¿
I eS=ηS∗I 0
1+ν0∗ηS∗I 0
H 2∗∑ Am,t
= 0.185∗57.542
1+ 10.87∗0.185∗57.542142∗2.054
=8.269 [ m4 ]
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Internal Structural Wall
Data for the vertical beam
Vertical beam 1.
Active section at eccentric compression
Am ,1=4.375∗0.2+0.6∗0.2=0.995[m2]
Active section at shear force
Am1 , t=4.575∗0.2=0.915 [m2 ]The section centroid position
Y G, 1=Y 1 A1∗Y 2 A2
A1+ A2
=0∗0.0875+2.2875∗0.0120.995
=0.275 [ m ]
Moment of inertia
I m1=[ 4.3753∗0.212
+0.2752∗0.875 ]+[ 0.23∗0.612
+2.01252∗0.12]=1.948 [m4 ]
Vertical beam 2.
Active section at eccentric compression
Am ,2=0.8∗0.2+0.3∗0.2=0.28[m2]
Active section at shear force
Am ,t=b∗h
k
k =1 for rectangular sections;
k=1.1 for sections in T and L ;
k=1.0 for sections in I and C;
Am2 , t=0.8∗0.2
1.1=0.145 [m2 ]
The section centroid position
Y G, 2=Y 1 A1∗Y 2 A2
A1+ A2
=0∗0.16+0.1∗0.06+0.1∗0.060.28
=0.0428 [ m ]
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Moment of inertia
I m2=[ 0.83∗0.212
+0.04282∗0.16 ]+[ 0.23∗0.312
+0.05722∗0.06]∗2=0.0398 [m4 ]
Vertical beam 3.
Active section at eccentric compression
Am ,3=0.6∗0.2+0.2∗0.2=0.16[m2]
Active section at shear force
Am3 , t=0.6∗0.2
1.1=0.109 [m2 ]
The section centroid position
Y G, 3=Y 1 A1∗Y 2 A2
A1+ A2
=0 [m ]
Moment of inertia
I m3=[ 0.63∗0.212 ]+[ 0.23∗0.2
12 ]=0.00373 [m4 ]
Vertical beam 4.
Active section at eccentric compression
Am ,4=0.8∗0.2+0.3∗0.2=0.28 [m2]
Active section at shear force
Am 4 ,t=0.8∗0.2
1.1=0.145 [m2 ]
The section centroid position
Y G, 4=Y 1 A1∗Y 2 A2
A1+ A2
=0∗0.16+0.1∗0.06+0.1∗0.060.28
=0.0428 [ m ]
Moment of inertia
I m4=[ 0.83∗0.212
+0.04282∗0.16]+[ 0.23∗0.312
+0.05722∗0.06 ]∗2=0.0398 [m4 ]
Vertical beam 5.
Active section at eccentric compression
Am ,5=4.375∗0.2+0.6∗0.2=0.995 [m2]
Active section at shear force
Am5 , t=4.575∗0.2=0.915 [m2 ]19
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The section centroid position
Y G, 5=Y 1 A1∗Y 2 A2
A1+ A2
=0∗0.0875+2.2875∗0.0120.995
=0.275 [ m ]
Moment of inertia
I m5=[ 4.3753∗0.212
+0.2752∗0.875]+[ 0.23∗0.612
+2.01252∗0.12]=1.948 [m4 ]
Data for the cross-beam
Distance between centroids
L1−2=3.72 [ m ] L2−3=1.642 [m ] L3−4=1.642 [ m ] L4−5=3.72 [ m ] L1−5=10.725 [ m ]
Section depth of the cross-beam constraining
a=0.35∗hr ≤ 0.40 [ m ]
a=0.35∗0.7=0.21 [m ]
Length of the gap
l0=0.90 [m ]
Calculus length of the cross-beam
l=l0+2a=0.90+2∗0.21=1.32 [ m ]
The floor plate does not pull together with the cross-beam
I r=0.2∗0.73
12=5.716∗10−3 [ m ]
Active section at shear force
Ar ,t=b∗hr
k=0.2∗0.7
1.2=0.116 [m]
The rigidities calculus considering the structural wall a cantilever equal with the building height.
ν0=βs
θ s 4 αs
; I 0 S=ηs∗I 0
I eS=ηS∗I 0
1+ν0∗ηS∗I 0
H2∗∑ Am,t
ν0=the number of levels coeffcient
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I 0=the global moment of inertia of the whole sectionηs=a cofficient depend ingon the number of levels∧λ , γ .
λ=the ratiobetwenthe distorsion rigidities of the vertical∧cross−beams
γ=the coeficient wich introduces the axial deformability
Kme=∑
1
n
Kmi
2; K re
=∑1
n−1
K r ; λ=2∑
1
n−1
K r
∑1
n
K m
λ=K r
K m
;γ=1+∑
1
n
I m
L1 ,n2 ( 1
Am1
+ 1Am,n
)
Rigidity of the cross-beam
K r=6 E r∗I r
L ( Ll )
3
∗μ
μ= 1
1+30 I r
A r . t∗l2
= 1
1+ 30∗5.716∗10−3
0.116∗1.322
=0.541
K r ,1=6∗6.75∗109∗5.716∗10−3
3.72∗( 3.72
1.32 )3
∗0.541=0.753∗109 [ N∗m ]
K r ,2=6∗6.75∗109∗5.716∗10−3
1.643∗( 1.643
1.32 )3
∗0.541=0.147∗109 [ N∗m ]
K r ,3=6∗6.75∗109∗5.716∗10−3
1.643∗( 1.643
1.32 )3
∗0.541=0.147∗109 [ N∗m ]
K r ,4=6∗6.75∗109∗5.716∗10−3
3.72∗( 3.72
1.32 )3
∗0.541=0.753∗109 [ N∗m ]
∑1
n−1
K r=2.735∗109[ N∗m ]
Rigidity at distortion
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“Gh. Asachi” Technical university of iasicivil engineering
Km=Em∗I m
he
Km, 1=27∗109∗1.948
2.8=18.784∗109[N∗m ]
Km, 2=27∗109∗0.0398
2.8=0.383∗109[N∗m ]
Km, 3=27∗109∗0.00373
2.8=0.03645∗109[N∗m ]
Km, 4=27∗109∗0.0398
2.8=0.383∗109[ N∗m]
Km, 5=27∗109∗1.948
2.8=18.784∗109[N∗m ]
∑1
n
K mi=38.698∗109[N∗m ]
Calculus of the coefficients λ and γ of the structural wall
λ=2∑
1
n−1
K r
∑1
n
Km
=2∗2.735∗109
38.698∗109 =0.141 .
γ=1+∑
1
n
I m
L1 ,n2 ( 1
Am1
+ 1Am,n )=1+ 4.013
10.7252 ( 10.995
+ 10.995 )=1.070 .
{λ=0.141γ=1.070
→ ηs=0.197
Global Moment of Inertia
Is computed for each case.
I 0=I m,1+ Am ,1∗c12+ I m,2+ Am, 2∗c2
2+ I m,3+ Am ,3∗c32+ I m, 4+Am, 4∗c4
2+ I m,5+ Am ,5∗c52 .
I 0=2∗[1.948+(5.36252∗0.995 )+0.0398+ (1.64882∗0.28 ) ]+0.00373=63.715 m2
Equilibrium Moment of Inertia
I 0 S=ηs∗I 0=0.197∗63.715=12.55 [m4 ]
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“Gh. Asachi” Technical university of iasicivil engineering
Equivalent Moment of Inertia
∑ Am,t=0.915+0.145+0.109+0.145+0.915=2.229 [m¿¿2]¿
I eS=ηS∗I 0
1+ν0∗ηS∗I 0
H 2∗∑ Am,t
= 0.197∗63.715
1+ 10.87∗0.197∗63.715142∗2.229
=9.645 [m4 ]
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“Gh. Asachi” Technical university of iasicivil engineering
4.2 Gravitational load evaluation
External structural wall
25
“Gh. Asachi” Technical university of iasicivil engineering
No. Layer typeThickness
[m]
[daN/m3]gn
[daN/m2]1 Lime cement mortar 0.03 1800 542 Reinforced concrete 0.20 2500 5003 Polystyrene 0.10 20 24 Exterior plaster 0.005 2100 10.5
Total 566.5
External masonry wall
No. Layer typeThickness
[m]
[daN/m3]gn
[daN/m2]1 Lime cement mortar 0.03 1800 542 BCA masonry 0.25 700 1753 Polystyrene 0.10 20 24 Exterior plaster 0.005 2100 10.5
Total 241.5
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Internal structural wall
No. Layer typeThickness
[m]
[daN/m3]gn
[daN/m2]1 Lime cement mortar 0.01 1800 542 Reinforced concrete 0.20 2500 5003 Lime cement mortar 0.01 1800 54
Total 608
Internal masonry wall
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“Gh. Asachi” Technical university of iasicivil engineering
No. Layer typeThickness
[m]
[daN/m3]gn
[daN/m2]1 Lime cement mortar 0.01 1800 542 BCA masonry 0.20 700 1403 Lime cement mortar 0.01 1800 54
Total 248
Partition wall
No. Layer typeThickness
[m]
[daN/m3]gn
[daN/m2]1 Lime cement mortar 0.01 1800 542 BCA masonry 0.10 700 703 Lime cement mortar 0.01 1800 54
Total 178
Attic
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“Gh. Asachi” Technical university of iasicivil engineering
No. Layer typeThickness
[m]
[daN/m3]gn
[daN/m2]1 Lime cement mortar 0.03 1800 542 Reinforced concrete 0.10 2500 2503 Polystyrene 0.10 20 24 Exterior plaster 0.005 2100 10.5
Total 316.50
Plate over basement – cold flooring
No. Layer typeThickness
[m]
[daN/m3]gn
[daN/m2]1 Ceramic plates 0.01 2400 242 Screed 0.03 1800 543 Reinforced concrete slab 0.13 2500 3254 Polystyrene 0.10 20 25 Lime cement mortar 0.012 1800 21.6
Total 426.6
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Plate over basement – warm flooring
No. Layer typeThickness
[m]
[daN/m3]gn
[daN/m2]1 Parquet 0.02 600 122 Screed 0.03 1800 543 Reinforced concrete slab 0.13 2500 3254 Polystyrene 0.10 20 25 Lime cement mortar 0.012 1800 21.6
Total 414.60
Plate over floor – cold flooring
No. Layer typeThickness
[m]
[daN/m3]gn
[daN/m2]1 Ceramic plates 0.01 2400 242 Screed 0.03 1800 543 Sound insulation 0.03 1600 48
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4 Reinforced concrete slab 0.13 2500 3255 Lime cement mortar 0.01 1800 18
Total 469
Plate over floor – warm flooring
No. Layer typeThickness
[m]
[daN/m3]gn
[daN/m2]1 Parquet 0.02 600 122 Screed 0.03 1800 543 Sound insulation 0.03 1600 484 Reinforced concrete slab 0.13 2500 3255 Lime cement mortar 0.01 1800 18
Total 457
Stairs
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“Gh. Asachi” Technical university of iasicivil engineering
No. Layer typeThickness
[m]
[daN/m3]gn
[daN/m2]1 Lime cement mortar 0.01 1800 542 Reinforced concrete 0.13 2500 3253 Concrete 0.08 2400 1924 Mosaic 0.02 2400 48
Total 619
Terrace
No. Layer typeThickness
[m]
[daN/m3]gn
[daN/m2]1 Lime cement mortar 0.01 1800 182 Reinforced concrete 0.13 2500 325
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3 Screed 0.03 1800 544 Vapors barrier - - -5 Definition layer - - -6 BCA masonry 0.08 700 567 Equalization screed 0.03 1800 548 Polystyrene 0.25 20 59 Screed 0.03 1800 5410 Water proof membrane - - -11 Gravel - - -
Total 566
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Internal, external walls and plates distribution
34
“Gh. Asachi” Technical university of iasicivil engineering
4.3 Walls load evaluation
Elem.type
gsolid
[daN/m2]gwoodworkl
[daN/m2]
Surfaces Weight No. ofsimilar elem.
GT
[daN]Ssolid
[m2]Swood
[m2]Stotal
[m2]Gsolid
[daN]Gwood
[daN]Gtotal
[daN]
E1 241.5 60 7.92 2.1610.0
81908.72 129.6 2038.32 4 8153.28
E2 241.5 60 6.27 2.97 9.24 1514.2 178.2 1692.41 4 6769.64
E3 241.5 60 8.28 2.5210.0
81999.62 151.2 2150.82 4 8603.28
E4 566.5 60 15.85 0.8116.6
68979.03 48.6 9027.63 4 36110.52
E5 566.5 60 3.95 0.81 4.76 2237.67 48.6 2286.27 4 9145.08
I1 248 60 13.44 1.8915.3
33333.12 113.4 3446.52 4 13786.08
I2 608 60 13.44 1.8915.3
38171.52 113.4 8284.92 4 33139.68
I3 248 60 15.33 015.3
33801.84 0 3801.84 2 7603.68
I4 248 60 4.76 0 4.76 1180.48 0 1180.48 4 4721.92I5 608 60 2.87 1.89 4.76 1744.96 113.4 1858.36 4 7433.44I6 608 60 7.77 1.47 9.24 4724.16 88.2 4812.36 4 19249.44
I7 608 60 20.16 020.1
612257.3 0 12257.2 2 24514.56
I8 248 60 7.14 0 7.14 1770.72 0 1770.72 2 3541.44I9 248 60 6.93 0 6.93 1718.64 0 1718.64 2 3437.28I10 608 60 5.04 1.89 6.93 3064.32 113.4 3177.72 4 12710.88
Total 198920.2
4.4 Floor load evaluation
Elem.type
Weight [daN/m2]
Surface [m2]
Weight [daN]
No. ofsimilar elem.
GTColdfloor
Warmfloor
Scold Swarm Stotal Gcold Gwarm Gtotal
P1 469 457 2.72 14.24 16.96 1275.68 6507.68 7783.36 4 31133.44P2 - 457 - 16.59 16.59 0 7581.63 7581.63 4 30326.52P3 - 457 - 14.77 14.77 0 6749.89 6749.89 4 26999.2P4 469 - 4.8 - 4.8 2251.2 0 2251.2 4 9004.8P5 - 457 - 4.65 4.65 0 2125.05 2125.05 1 8500.2
P6(stairs) 619 - 24.48 - 24.48 15153.1 0 15153.1 4 15153.12
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“Gh. Asachi” Technical university of iasicivil engineering
P7 469 - 6.435 - 6.435 3018.1 0 3018.1 4 12072.06Total 133189.34Pterrace 566 - - - 305.32 0 0 172811.12
4.5 Partition walls load
Elementtype
Weight of the wall Equivalent
weightge
[daN/m2]
Area of the plate
[m2]
Equivalent weight of the walls[daN/m2]
No. of similar
elements
GT
[daN]g
[daN/m2]Gh
[daN/m]
P1 178 475.26 150 16.96 2544 4 10176P2 178 475.26 150 16.59 2488.5 4 9954
Total 20130
Gh<150daN/m => ge=50 daN/m2
150<Gh<300daN/m => ge=100 daN/m2
300<Gh<500 daN/m => ge=150 daN/m2
4.4 Variable load
Type of plate
Area[m2]
qeffective
[daN/m2]Qeffective
[daN/m2]
No. of similar elemnts
Total
P1 16.96 200 3392 4 13568P2 16.59 200 3318 4 13272P3 14.77 200 2954 4 11816P4 4.8 200 960 4 3840P5 4.65 200 930 4 3720P6 24.48 300 7344 1 7344P7 6.435 300 1930.5 4 7722
Total 61282Pterrace 305.32 75 22899 1 22899
4.5 Snow load
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“Gh. Asachi” Technical university of iasicivil engineering
qz=μi ∙ C e ∙C t ∙ qz0=0.8∙1 ∙1 ∙250=200[ daN
m2 ]C e=1
C t=1
α∈ (0 ;30 ° )→ μi 0.8
qz0=250[ daN
m2 ](Suceava)
Qsnow=qz ∙ S terrace=200 ∙296.82=59364 [daN ]
4.6 Seismic load
Go=G1=G2=G3=G plate+Gwalls+0.4 (qeffective+q partition)=133189.34+198920.20+0.4 (61282+20130 )=364674.34[daN ]
G4=Gterrace+12
Gwalls+max ( qeff ,terrace , qeff ,snow ) ∙ 0.4+gattic=172811.12+12
198920.2+59364 ∙0.4+633 ∙ 73.75=342700.57 [daN ]
Suceava { land acceleration ag=0.16 gcontrol period of the responsespectrum T c=0.7[ s]
Fb=γ L ∙ Sd ( T1 ) m∙ λ;
γ L=1 λ={0.85 ( T1 ≤T c )1
;
T 1=CT ∙ H34 =0.361 CT=0.05 H=5 ∙ he=14 [m ]
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“Gh. Asachi” Technical university of iasicivil engineering
Sd=Se ( T1
2 π )=8.365 ∙( 0.3612π )
2
=0.0276 ;
Se=ag ∙ β (t );
β (t )=β0
T c
T 1
=2.75 ∙0.7
0.361=5.332 ;
β0=2.75 ;T c=0.7 [ s ] ; ag=0.16 g=1.569 [ ms];
Se=ag ∙ β (t )=1.569 ∙5.332=8.365 ;
m=GT
g;
GT=G0+G1+G2+G3+G 4=1801397.93 [ daN ]=18013979.30 [N ]
m=GT
g=1836287.4 [ Kg ] ;
Fb=1 ∙0.0276 ∙183542.44 ∙0.85=43079.31[daN ]
5. Diagrams for shear wall section.5.1. Calculus of sectional efforts
FbDi=Fb
I echDi
∑1
n
I echDn
;
FbD1=43079.31
8.2692(8.269+9.645)
=10043.49 [ daN ] ;
FbD2=43079.31
9.6452 (8.269+9.645)
=11368.61 [ daN ] ;
qreq ,i=2∗n∗Fb
D i
(n+1 ) H
H=5∗he=14.00 [m ]
n=5 floors
qreq1=2n ∙ Fb
D1
(n+1 ) H=2∙5 ∙10043.49
6 ∙ 14=1195.65 [d aN ]
qreq2=2n ∙Fb
D 1
(n+1)H=2 ∙5 ∙11368.61
6 ∙14=1353.40[daN ]
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“Gh. Asachi” Technical university of iasicivil engineering
External structural wall (D1)
M d=m s∗qreq∗H 2
100[ daN∗m ]
M m,i=M d∗I m,i
I 0
[ daN∗m ]
Section ms
qreq1
[daN/m]H2/100 MD
[daNm]Im1/I0 Mm1
[daNm] Im2/I0
Mm2
[daNm] Im3/I0
Mm3
[daNm]
0 1 2 3 4(1∙ 2∙ 3) 5 6 (4∙ 5) 7 8 (4∙7) 9 10(4∙9)
5-4 2.061
1195.65 1.96
4829.89
0.192
917.67
0.0018
8.69
0.00104
5.02
4-5 0.162 379.64 72.89 0.68 0.39
4-3 2.485 5823.53 1118.11 10.48 6.056
3-4 1.113 2608.28 500.78 4.69 2.71
3-2 1.624 3805.80 730.71 6.85 3.95
2-3 3.174 7438.18 1428.13 13.38 7.735
2-1 -0.703 -1647.46 -316.31 -2.96 -1.71
1-2 6.108 14313.93 2748.27 25.76 14.88
1-0 -4.079 -9559.03 -1835.3 -17.20 -9.95
0-1 10.163 23816.72 4572.81 42.87 24.76
Vertical beam 1 (Vertical beam 5)
39
λ 0.3331γ 1,061
SE
CT
ION
5-4 2.0614-5 0.1624-3 2.4853-4 1.1133-2 1.6242-3 3.1742-1 -0.7031-2 6.1081-0 -4.0790-1 10.163
“Gh. Asachi” Technical university of iasicivil engineering
V 5−41 =
M 5−4−M 4−5
he
=301.70¿
V 4−31 =
M 4−3−M 3−4
he
=220.47 [daN ]
V 3−21 =
M 3−2−M 2−3
he
=−249.07[daN ]
V 2−11 =
M 2−1−M1−2
he
=−868.55 [daN ]
V 1−01 =
M 1−0−M 0−1
he
=−2288.63[daN ]
Vertical beam 2 (Vertical beam 4)
V 5−42 =
M 5−4−M 4−5
he
=2 .96[daN ]
V 4−32 =
M 4−3−M 3−4
he
=1 .92[daN ]
V 3−22 =
M 3−2−M 2−3
he
=−2. 16[daN ]
V 2−12 =
M 2−1−M1−2
he
=−9 . 52[daN ]
V 1−02 =
M 1−0−M 0−1
he
=−1 9. 91 [daN ]
Vertical beam 3
V 5−43 =
M 5−4−M 4−5
he
=1 .71 [daN ]
V 4−33 =
M 4−3−M 3−4
he
=1 .11 [daN ]
V 3−23 =
M 3−2−M 2−3
he
=−1 . 25[daN ]
V 2−13 =
M 2−1−M1−2
he
=−5 . 56[daN ]
V 1−03 =
M 1−0−M 0−1
he
=−11 . 50[daN ]
40
“Gh. Asachi” Technical university of iasicivil engineering
Axial Force
A=(14.55+11.35) ∙1.6
2+15.25 ∙ 0.2+[(0.2∙ 0.2+ 0.2∙ 0.2
2) ∙ 2]=23.83 [m2]
I. TerraceDead Load from attic: 633∙ 15.2 = 9621.6 [daN]Snow Load: 200 *23.83 = 4766 [daN]
Total Weight: 15275.03[daN]Load from plate:
Live Load: 75*23.83=1787.25[daN]Self Weight: 566*23.83=13487.7[daN]
Total Weight: 15275.03[daN]
II. Floor 4Self Weight of the wall: 566.5*39.46=22354.09[daN]Load from plate:
Live Load: 200*23.83= 4766[daN]Partitions: 150*23.83= 3574.5[daN]Self Weight of the plate: 469*23.83=11181.75[daN]
Total Weight: 19522.25[daN]III. Floor 3 = Floor 2 = Ground Floor = Floor 4 = 41876.34[daN]
41
“Gh. Asachi” Technical university of iasicivil engineering
Shear force diagram
42
“Gh. Asachi” Technical university of iasicivil engineering
Axial force diagram
43
“Gh. Asachi” Technical university of iasicivil engineering
Bending force diagram
44