proiect constructii civile 2

“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);

1


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;

2


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]

3


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.

4


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:

5


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]

6


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


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


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 ]

7


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 - - -


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’)

8


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 ) ]


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


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

11


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

12


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.


Am ,2=0.85∗0.2+0.65∗0.2=0.30[m2]


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 ]

13


Vertical beam 3.


Am ,3=0.8∗0.2+0.2∗0.2=0.20 [m2]


Am3 , t=0.8∗0.2

1.1=0.145 [m2 ]


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.


Am ,4=0.85∗0.2+0.65∗0.2=0.30 [m2]


Am 4 ,t=0.85∗0.2

1.1=0.154 [m2 ]


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.


Am ,5=4.4∗0.2+0.2∗0.2=0.92[m2]



Y G, 1=Y 1 A1∗Y 2 A2

A1+ A2

=0∗0.88+2.1∗0.040.92

=0.0913 [m ]

14


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 ]


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


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 ]

16


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 ]

17


Internal Structural Wall

Data for the vertical beam

Vertical beam 1.


Am ,1=4.375∗0.2+0.6∗0.2=0.995[m2]



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.


Am ,2=0.8∗0.2+0.3∗0.2=0.28[m2]


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 ]


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 ]

18


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.


Am ,3=0.6∗0.2+0.2∗0.2=0.16[m2]


Am3 , t=0.6∗0.2

1.1=0.109 [m2 ]


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.


Am ,4=0.8∗0.2+0.3∗0.2=0.28 [m2]


Am 4 ,t=0.8∗0.2

1.1=0.145 [m2 ]


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.


Am ,5=4.375∗0.2+0.6∗0.2=0.995 [m2]


Am5 , t=4.575∗0.2=0.915 [m2 ]19



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 ]


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

20


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

21


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 ]

22


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 ]

23


24


4.2 Gravitational load evaluation

External structural wall

25


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


[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

26


Internal structural wall


[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

27



[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


[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

28



[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


[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

29


Plate over basement – warm flooring


[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


[m]

[daN/m3]gn

[daN/m2]1 Ceramic plates 0.01 2400 242 Screed 0.03 1800 543 Sound insulation 0.03 1600 48

30


4 Reinforced concrete slab 0.13 2500 3255 Lime cement mortar 0.01 1800 18

Total 469

Plate over floor – warm flooring


[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

31



[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


[m]

[daN/m3]gn

[daN/m2]1 Lime cement mortar 0.01 1800 182 Reinforced concrete 0.13 2500 325

32


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

33


Internal, external walls and plates distribution

34


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

35


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

36


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 ]

37


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 ]

38


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


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


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


Shear force diagram

42


Axial force diagram

43


Bending force diagram

44

proiect constructii civile 2

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