application theme

8/3/2019 Application Theme

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"Gh. Asachi" Technical University of IasiFaculty of Civil Engineering

P R O J E C T

C iv i l const ruc t ions

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

IN DI VI DU AL DW EL LI NG : Ug + G + IG.

S t u d .

d i g i t a l

P o r t r a i t

Year: 2 Group: 3212

PROJECT ADVISORS:

I A I - 2 0 0 9

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APPLICATION THEME

The topic is to design an individual dwelling, which will be

used as a residence for one family, having three levels:

underground level, ground-floor and first floor, (Ug+G+IG). Thecalculations are made taking into account the settlement in the

Euro Codes and the Romanian Codes, and also they are made

using the ultimate limit state.

The placement of the building belongs to an urban area; the

connections with the electricity, natural gases, sewerage and

potable water systems are ensured.

The slope of the terrain will be less than 5% and the drains for

rain water are already traced. The level of the ground waters will

be provided from the geological survey office, and will be much

deeper than the maximum foundation depth. The allowable

pressure on the foundation ground will be less than 2.5 daN/ .

The reinforcement structure of the dwelling will be

composed by:

-main walls made by brickwork masonry with concrete portionsespecially above the windows and doors;

-slabs made by reinforced concrete;

-the roof will be designed in the hypothesis of pitched roof made

of timber and ceramic tiles;

-the foundations will be continuous under the main walls (rigid

foundations).

-the stairs are made by reinforced concrete, without

reinforcements inside steps.

- the secondary walls can be made of brickwork masonry, burnt

plaster plates or rigips

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The aspects which must be followed in the project:

functional distribution of the available space;

the composition of the reinforcement structure of different

elements;

the composition of the principal elements (like main exterior

walls, roof and slab over basement) and the design of these,

according to the thermal insulation condition. Also the

thickness of the insulation layer must be calculated, in order

to obtain the minimum loss of heat.

roof framework which must ensure enough space for

depositing;

the calculus of the main walls;

foundation calculus;

The written part of the project will consist of:

application theme;

technical report of the project

the design of the roof elements, the preliminary assumptions

from a static point of view, the design of the foundation

walls.

drawn parts: first floor (ground floor), scale 1:50

-first floor

-second floor(inner garret)

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-elevation

-the roof section and view from above

-foundation

-details, scale 1:10

Technical report

Project notes:

Destination :

- Individual dwelling for one family

- The building has 3 levels: Ug+G+IG;

Available space

-Projected area: 133 m2

- Used area: ground floor area: 76.85 m2

-First floor: 74.08m2

- Total area: 151 m2

Placement of the building

The climatic zone is the third one, having a ground snow

action that equals 2.5 , Iasi city.

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Guiding marks - the location is near the neighbors:

- N X street

- S Smith Family- E - Johnson Family

- W - Y street.

Design solutions

Design of the roof elements (made by wood)

load normative from EC-1 wood standing normative

There are proposed 2 hypothesis of loading:

a) the own weight + the snow weight action

b) the own weight of the roof + the concentrated

force P, which is a temporary load.

Calculi of the walls

Three steps have to be followed in the design of the

masonry walls:

a) Predesigned of the walls;

b) Checking the conditions of loading;

c) Redesign, if is necessary;

Floors

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Are predesigned, without making any checking for

them

Foundation design

Some details about the solution for foundation are

mentioned in the previously. We must remind here

that the method of calculations for the internal and

external foundation are done separately, considering

only a linear meter in the most loaded portion of the

wall.

Thermal insulation design

The thermal insulation layer is calculated using a

specialized soft ;

Computation of the roof elements

prop purlin2

purlin1

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=600kg/m3

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Element L[m]

B*h[cm]

Volume[m3]

Numberof pieces

Totalvolume

Pillars 3.25 15x15 0.073 10 0.73

Rafter 6.87 10x12 0.082 33 2.72

Rafter 5.87 10x12 0.071 6 0.423

Purlin 16.35 15x15 0.368 2 0.736

Purlin 15.35 15x15 0.345 2 0.691Slat net. 11.35 3.8x5.

80.025 42 1.051

=6.351m3


8/17

The roof is covered by tiles with 330x420x35mm.

Loads discharged by the own weight of the roof elements:

The minimum value for , = .

= => slope p=tan()=57.73%.

The roof framework load:

Gtimber=V* =600*6.351 = 3810.39 kg;

The weight of the tile layer:

Weight of a piece: 4.5 kg;

Necessary per m2: almost 10 buc;

The total weight of the tiles:

10*4.5*185.57=8350.763kg;

The total weight: (3810.39+8350.763)= 12161.153kg;

The total load : 12161.153*10/92.132 =121611.53/92.132=1319.97N/m2;

gk=1.319 kN/m2Snow loads:

Slope,


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0.8+0.8* 1.6 -

;

=1;

=1;

=1.6*2*1*1=3.2kN/m2;

=0.8;

=1.6;

Sd=1.5*Sk=4.8 kN/m2;

Sk-L=Sk*L0=Sk*cos =3.2*cos300=2.85kN/m2;

= => slope p= tan()=57.73%.

The final load, transmitted to the exterior walls by the roof:

Fk=4.8+1.319=6.119 kN/m2

Loadings on rafters

There are 2 hypothesis of loading for the design of the roof

elements:

a) G+Z, the own weight together with the snow load;

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b) G+P, the own weight together with the concentrated force;

Load combination a):

Z=snow load=2.85kN/m2;

=>2.85*1.575/2.25=1.995kN/m;

P=the weight of the tiles= 1.575*45*10=0.70875kN;

=>0.315kN/

m;

The vertical load,V=1.995+0.315=2.31kN/m;

=>N=2.31*2.25*sin=2.31/2=1.155kN/m;

Q=2.31*cos=2.05kN/m;

=2.05*2.252/8=0.577075kN*m;

Loadin combination b):

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The only forces that produce moments are the

forces projected perpendicular to the longitudinal

axis of the beam, so they will be considered.

P=800*cos=0.712kN;

q=0.315cos*2.25=0.6315kN;

=0.712*1.125+0.6315*1.125=1.5115kN*m;

=max(1.5115;0.57707)= 1.5115kN*m;

=1.5115kN*m;

= = =151.15daN*m/0.00024m3=629791.666667daN/

m2=

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62.979daN/cm2,< ,taking into account that

=100daN/cm2;

W=b*h2/6=0.1*0.12*0.12/6=0.00024m3;

Dead loads acting on building

Warm floorNo Layer Thickness

D [m]Unitweight

Characteristicvalue

Ultimate value

1 Parquet 0.015 5886 88.29 119.1935

2 Counter floor 0.025 5886 147.35 198.6525

3 Vapour barrier foil 0.001 0 0 0

4 Glass-wool 0.074 142.245 10.52613 14.2202735 Glass-wool

insulation0.02 1128.15 22.563 30.46005

6 Reinforcedconcrete

0.1 23544 2354.4 31.7844

3565

Outer wall

No Layer ThicknessD [m] Unitweight Characteristicvalue Ultimate value

1 Floor tiles fixed toplaster

0.02 17658 353.16 476.766

2 Plaster 0.015 16677 250.155 337.70925

3 Brick 0.25 12556.8 3139.2 432.92

4 Polystyreneinsulation

0.1 194.3 29.43 39.7305

5 Faade plaster 0.005 16677 83.385 112.56975

3855 5205

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Inner wallNo Layer Thickness

D [m]Unitweight

Characteristicvalue

Ultimate value


0.02 17658 353.16 476.766

2 Plaster 0.015 16677 250.155 337.70925

3 Brick 0.25 12556.8 3139.2 4237.92

4 Plaster 0.015 16677 250.155 337.70925


0.02 17658 353.16 476.764

4345 5867

Floor above basementNo Layer Thickness

D [m]Unitweight

Characteristicvalue

Ultimate value

1 Panel parquet 0.015 5886 88.29 119.1915

2 Counter floor 0.025 5886 147.15 198.6525

3 Vapour barrier foil 0.001 0 0 0

4 Glass-wool 0.074 142.245 10.52613 14.2102735

5 Glass woolinsulation

0.02 1128.15 22.563 30.46005

6 Reinforcedconcrete slab

0.1 23544 2345.4 3178.44

7 Glass wool 0.1 142.245

14.2245 19.203

8 Plaster board 0.015 12262.

5

183.9375 248.315

3808

Basement walls

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No Layer ThicknessD [m]

Unitweight

Characteristicvalue

Ultimate value

1 Plaster 0.015 16647 250.155 337.70

2 Brick 0.25 12556.8 3139.2 4237.92

3 Polystyreneinsulation

0.1 249.3 29.43 39.7305

4 Watter resistantfoil

0.002 0 0 0

3619 4611

Loads on exterior walls

For checking the walls it is considered a linear meter of the

biggest wall, at the midspan;

=3.296+8.76=12.056kN;

Roof self-weight: 1.35* *A=1.35* 1.338*1.825= 3.296kN

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gk= =1.1922kN/m2;

= gk/cos 30=1.338kN/m2;

Snow load: 1.5* *A=1.5*3.2*1.825=8.76kN;

Last floor inner garret:

Floor sw: 3.908*1.825=7.1321kN;

Live load: 1.5*1.500*1.825=4.106kN;

Area, 1*3.65/2=1.825m2;

Design of an interior wall

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Foundation design

The necessary width is obtained as follows:

;

;

;

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;

, we will adopt 1.05m;

, we will adopt 1.35 m;

tan = 1.1 .... 1.3 condition ;

;

a = 0.4 m;

tan = 1.25 for outer wall, =57.04;

a = 0.55 cm;

tan = 0.909 for inner wall, =46.96;

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