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    5/25/2010

    CIVILCONSTRUCTION PROJECT

    DESIGN CALCULATIONFOR 2-STOREY

    DWELLING MADE BY

    MASONRY CROSS WALLSACCORDINGTO EC6

    Student: MIRON Marcel-Cristian

    group 3211 ICE

    Coordinators: Prof Dr. Ing. Magda

    Broteanu

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    CONTENTS WRITTEN PARTS

    1. Tehnical Memoir

    1.1. General Data

    1.2. Functional Architectural Solution

    1.3. Construction Solution

    1.4. Urbanism Conditions

    2. Loads Computation

    2.1. General Notes About the Symbols/NotationsUsed

    2.2. Calculus of the Specific Thermal Resistance R

    2.3. Roof (chosen solution for roof covering &timber volume)

    2.4. Snow Load

    2.5. Characteristic Values of Dead Loads

    2.6. Design of a masonry structural memberaccording to EC6

    Interior wall

    Exterior wall

    Axial force diagram for exterior wall

    Eccentricity calculus

    2.7. Design of a foundation block

    GRAPHICAL PARTS

    Ground floor

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    First floor

    Section Plan

    Roof Plan

    Main Facade

    Details

    1.1.General Data

    CONSTRUCTION: 2-STORY DWELLING HOUSE

    Designer: MIRON Marcel-Cristian

    Height Regime: D+P+M

    Site characteristics:The site is situated in Craiova, in the western suburbs of

    the city. The site has an insignificant slope. The field has a localand general stability assured by respecting the instructionsfrom the geotechnic study. The destination of the building: asingle family dwelling, 2-storey height. The placement hasaccess to public utilities (water, sewerage, electricity).

    Maximum depth of frost is 0.90 m cf. STAS6054/77;

    Climatic zone III / summer temperatures 280C ; wintertemperatures -140C (STAS 6472/82)

    Climatic zone C / snow / (STAS 10101/21-90)Climatic zone C /wind / (STAS 10101/20-92)

    1.2. Functional Architectural Solution

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    Serving the clients needs, the architect and the designerelaborated a dwelling house consisted of 3 levels : partial

    basement (underground floor), ground floor, inner garret. Allthe rooms are properly oriented such that the mandatoryventilation and illumination conditions are fulfilled. Therefore,the living room and the dining room are oriented towards theSouth-East.

    The basement has a height of 2.30 m, the ground floor2.60 m, and the upper floor (inner garret) has a varying height

    between 1.80 m and 2.60 m.

    Concerning the interior staircase, the length of the stairs is1.96 m, having the counter-steps of 18 cm and the steps of 28cm.

    The house has the bearing walls made of a full brick of 25cm and a full white brick of 20 cm, thermo-isolated withexpanded polystyrene with the thickness of 4.8 cm. Above the

    partial basement, the ground floor and the first floor, reinforcedconcrete slabs are used.

    Sistematization of the Inner Space:

    PARTIAL BASEMENT (UNDERGROUND FLOOR):

    Staircase: 8.90 m2

    Depositing Space: 21.48 m2

    Total: 30.38 m2

    GROUND FLOOR:

    Living Room: 23.30 m2

    Dining Room: 9.05 m2

    Kitchen: 9.97 m2

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    Deposit: 1.07 m2

    Lobby: 5.68 m2

    Staircase: 10.85 m2

    Bathroom: 2.90 m2

    Total: 62.82 m2

    UPPER FLOOR:

    1st Bedroom: 15.34 m2

    SAS: 2.25 m2

    2nd Bedroom: 13.84 m2

    Master Bedroom: 17.27 m2

    1st Bathroom: 4.13 m2

    2nd Bathroom: 7.42 m2

    Lobby: 7.90 m2

    Total: 68.15 m2

    Sistematization of the Entire Surface:

    EFFECTIVE AREA:

    Underground Floor + Ground Floor + UpperFloor = 161.38 m2

    CONSTRUCTION AREA: 130.98 m2

    UNCOILED AREA: 270.26 m2

    P.O.T. = 12.2 %

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    C.U.T. = 0.28

    1.3.Construction Solution

    - Infrastructure:

    The infrastructure will be realized under the form of acontinuous foundation under the walls with a concrete shoemade up of concrete ( C 6/7.5) and concrete elevation ( C

    12/15), 90 cm thick under the first-floor walls, 100 cm under theunder-ground floor. Also, the infrastructure will be reinforcedwith girdles on the whole lenght. The reinforcement will bemade with PC52 and OB37 steel.

    - Suprastructure:

    The strength structure is consisted of mansory walls,reinforced concrete pillars, and girdles. Exterior walls are 45 cm

    thick and the interior bearing walls are 25 cm thick. Thematerials used in pillars (25x25 cm), girdles and beams will be:concrete C12/15 and reinforcements PC52 and OB37. Themasonry will be consisted of 25 cm full bricks and 20 cm whitebricks.

    The stairs are mare up of monolith reinforced concrete. Theslab over the ground floor will be made up of reinforcedconcrete (C12/15). The slab will be 13 cm thick and it will be

    reinforced with steel STNB nets.

    The roof over the garret is consisted of fir wood.

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    - Trims:

    The builing will have modern, high quality trims asfollows:

    Exterior paintings will be carefully chosen function of thevolumetry of the house:White and beige plaster, PVC windows,wood doors, wood handrails, pvc ditches and down-comers(water pipes), TONDACH tiles, brick revetment, pergola and

    ornamental flower pots.

    Interior paintings will be in accordance with the hygienic-sanitary requirements that are imposed by the activenormative. Interior paintings will be chosen such that theexterior-interior chromatic sensation to be a fine one. For theliving rooms we will use a white, creamy washable lime and forthe bathrooms, kitchen and stores well use sandstone assorted

    with the wall paintings and faience.

    1.4.Urbanism Conditions

    Water supply will be made by joining the water networkhaving the authorization obtained from S.C. RAJAC S.R.L.

    Electricity supply will be made having the autorizationobtained from E.ON, Moldova S.A.

    The sewerage will be made by joining the seweragenetwork from the neighbourhood .

    The heating will be made individually by central heating.

    2. Loads Computation

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    2.1. Design according to the Ultimate Load State

    method:

    Symbols:

    Fk characteristic value of a load (force);

    Fd= (f)*(Fk) design value of a load;

    f =1.35(1.40 in USA, Canada, Japan, Australia, etc)for Gk and 1.50 (1.60 in USA, Canada, Japan, Australiaetc) for Qk partial safety coefficients for loads;

    Dead (Permanent) Loads (EC1Part 2.1):

    Gk or Wk or SWk concentrated permanent load, or

    weight, or self-weight load, in N or daN or kN;

    gk or wk weight per unit area in N*m -2, orweight per unit length in N *m -1;

    = ()(g) unit weight in N.m-3 or kN.m-3 ;

    = m*V-1 unit mass or density of material inkg.m-3 ;

    g =9.81(~10) m.s-2 gravitational acceleration;

    )(40.1)(35.1 USAEU ff == (overloading) partial safetyfactor for dead load;

    Loading Effects Grouping:

    ikikjk QQG ,,01,, 5.15.135.1 ++ , where:

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    1.35 Gk, j Design Dead Load;

    f=1.35 Partial Safety Coefficient for DeadLoad for unfavorable exception;

    1.5 Qk,1 Design Dominant Variable Load;

    f=1.5 Partial Safety Coefficient for Variable Loadfor unfavorable exception;

    1.5 0,i Qk, i Design Variable Load;

    0,i=0,7 Concomitance (Reduction) Factor forcombination value of a variable action;

    2.2.Calculus of the Specific ThermalResistance R

    For the purpose of computing these values, the followingsymbols are used:

    R thermal resistance of the element (m2K/W);

    Rsi inner surface thermal resistance (m2K/W);

    Rse outer surface thermal resistance (m2K/W);

    Rmin the minimum overall thermal resistance (m2K/W);

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    a) Outer wall

    No.crt.

    Material layerThickness

    d (m)

    Thermalconductivity

    coefficient [W/(mK)]

    1. Inner plaster 0.02 0.872. 1 full brick 0.25 0.83. 1 white brick (BCA) 0.20 0.283. Expanded polystyrene X 0.0444. Plaster cement mortar 0.03 0.03

    C107/05 Design Code states that in the case of an

    exterior wall, , and

    .

    Therefore, the thickness of the expand polystyrene is

    . Because the material is found only with thicknessesmultiple of 2.4, a 4.8 cm thick polystyrene is chosen.

    b) Slab above the basement

    No.crt.

    Material layerThickness

    d (m)

    Thermalconductivity

    coefficient [W/(mK)]

    1. Sandstone 0.007 2.032. Concrete layer 0.01 1.62

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    3. Equalizing layer 0.03 0.934. Reinforced concrete slab 0.13 1.62

    5. Expanded polystyrene X 0.044

    6.Plaster cement and lime

    mortar0.02 0.87

    C107/05 Design Code states that in the case of a slab

    above the basement, , and

    .

    Therefore, the thickness of the expanded polystyrene

    is . Because the material is found only withthicknesses multiple of 2.4, a 7.2cm thick polystyrene is chosen.

    2.3.Roof

    Chosen solution for roof

    covering

    TONDACH Castel tiles

    Material: sand, cement, water,inorganic pigments.

    Surface: even, with a tintlesscoat.

    Dimensions: 43 x 27 cm

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    Active length: 43 cm

    Active width: 21.5 cm

    Profile height: 2.2 cm

    Weight: 3.0 kg / piece

    Necessary/ m2 13.1 pieces

    Positioning: network mesh

    Timber volume used for the roof

    ElementNo.

    Dimensions (cm) Slope(degre

    es)

    Totalvolume(cm3)

    b h L/B/H

    Purlins111

    121212

    121212

    136592

    127526

    19656013248

    183600

    Rafters172

    7.57.5

    1010

    557393

    2671017558950

    Boarding 1area

    2.4 area 26 1914389

    Props56

    14

    14

    10140

    --

    267773936945

    Longitudinalbattens

    172

    55

    33

    557393

    2614203511790

    Transversebattens 11 5 5 1365 26 375375Ridge 1 12 12 1302 - 187488

    Tongs2x3

    4 15 460 - 165600

    Purlins11

    1212

    1212

    5001254

    2272000

    180576Rafters 17 7.5 10 462 22 589050

    Boarding 1area

    2.4 area 22 639418

    Props 1 1 194 - 22 59728

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    4Longitudinal

    battens 17 5 3 462 22 117810Transverse

    battens15 5 5 1365 22 511875

    Total volume of wood: 9244351 cm3= 9.24 m3

    2.4.Snow Load

    The computation was done according to CR 1-1-3-2005, Design Code. The following notations are used:

    snow density of 235350400 kgm-3 depending onthe breaking state and unfavorable snow falling

    Sk snow characteristic value on roof horizontal projectionin kNm-2

    g gravitational constant of 10 ms-2

    shape coefficient

    Ce exposure coefficient of placement

    Ct thermal coefficient

    S0,k characteristic value of snow load on the earth level

    Sk,L snow characteristic value on pitched roof

    Sk0,L - snow characteristic value on pitched roof per unitlength

    Sk0,Lx - snow characteristic value on pitched roof per unit

    length perpendicular on the roof

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    Sk0,Ly - snow characteristic value on pitched roof per unitlength tangent to the roof

    Fk characteristic value of snow pushing force in kNm-1

    cfr friction coefficient between snow and tiles (cfr = 0.05)

    Se characteristic value of snow load hanged down at roofeaves and distributed on roof length

    k coefficient depending on unsteady snow falling

    The building being situated in Iasi, will have s0,k=2.5kNm-2. The presence of other buildings around it doesnt notallow an important blowing of the snow by the wind, therefore,a partial exposure will be considered (Ce=1). The thermalcoefficient Ct is considered to be 1.0.

    The snow weight:

    For the uncrowded snow loading, the distribution is:

    14

    1 2

    1(1)2(2)

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    In this case, 1=26 and 2=22. Because 1

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    where: - L01 is the horizontal projection of the width of the roofwith slope 1=26

    - L1 is the width of the roof with the slope 1=26

    Geometrical features of the roof

    Fig.4.3. Snow characteristic value on pitched roof per unitlength

    16

    Sk0,L

    Sk0,Lx

    Sk0,Ly

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    The characteristic value of snow pushing force will be:

    For the part of the part of the roof with 2=22

    The characteristic value of snow pushing force will be:

    2.5. Characteristic Values of Dead Loads

    Roof with Thermal Insulation

    No

    .

    Layer Thickn

    ess[m]

    Unit

    Weight

    Characteris

    tic Valuegk=d*

    Ultimate

    Valuegd=F*gk=1.3

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    =p*g[N/m3]

    [N/m2] 5*gk [N/m2]

    1 roof tile - - 393 530.55

    2longitudinal

    strip0.03 6000 180 243

    3transverse

    strip0.05 6000 300 405

    4water

    proofing0.005 - 60 81

    5 roof baten 0.024 6000 144 194.4

    6mineral wool

    1 0.14 3500 490 661.5

    7mineral wool

    20.05 3500 175 236.25

    8vapourbarrier

    0.01 - 60 81

    9insulating

    board0.01 11000 104.5 141.075

    Total Load 1906.5 2115.45

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    Outer Wall

    No.

    Layer Thickness[m]

    UnitWeig

    ht=p*g

    [N/m3]

    Characteris

    tic Valuegk=d*[N/m2]

    Ultimate

    Valuegd=F*gk=1.35*gk [N/m2]

    1 inner plaster 0.021900

    0380 513

    2 1 full brick 0.251850

    04625 6243.75

    3 1 white brick(BCA) 0.20 4500 900 12154 vapour barrier 0.005 - 60 81

    5expanded

    polystyrene(EPS)

    0.05 200 10 13.5

    6 outer plaster 0.032200

    0660 891

    Total Load 6635 8957.25

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    Inner Wall

    No.

    LayerThickn

    ess[m]

    UnitWeig

    ht=p*

    g[N/m

    3]

    Characteristic Valuegk=d*

    [N/m2]

    UltimateValue

    gd=F*gk=1.3

    5*gk [N/m2]

    1 Inner plaster 0.021900

    0380 513

    2 1 full brick 0.251850

    04625 6243.75

    3 Inner plaster 0.021900

    0380 513

    Total Load 5385 7449.3

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    Basement Wall

    No.

    LayerThickn

    ess[m]

    Unit

    Weight

    =p*g

    [N/m3]

    Characteristic Valuegk=d*[N/m2]

    UltimateValue

    gd=F*gk=1.35*gk [N/m2]

    1 inner plaster 0.021900

    0380 513

    2 1 full brick 0.251850

    04625 6243.75

    3 concrete 0.20 2400 4800 6480

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    0

    4

    cement and

    lime mortar 0.03

    1700

    0 510 688.55 waterproofing 0.005 - 175 236.25

    6Protectingbrick wall

    0.151850

    02775 3746.25

    Total Load 13265 17907.75

    Floor above Basement

    No.

    LayerThickn

    ess[m]

    UnitWeig

    ht=p*

    g[N/m

    3

    ]

    Characteristic Valuegk=d*[N/m2]

    UltimateValue

    gd=F*gk=1.35*gk [N/m2]

    1 Parquet 0.02 8000 176 237.6

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    2 False boarding 0.01 8000 80 108

    3

    Acousting

    insulation 0.03 3500 105 141.75

    4 Concrete slab 0.132500

    03250 4387.5

    5 Vapour barrier 0.005 - 60 81

    6Expanded

    polystyrene0.075 200 14.4 19.44

    7 Plaster 0.021700

    0340 459

    Total Load 4025.4 5434.29

    Floor above First Floor (including Staircase)

    No.

    Layer Thickness[m]

    UnitWeig

    ht=p*

    g[N/m

    Characteristic Valuegk=d*[N/m2]

    UltimateValue

    gd=F*gk=1.35*gk [N/m2]

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    Technical University Gh. Asachi, IaiFaculty of Civil Engineering and Building Services

    3]

    1 Sandstone 0.007 24000

    168 226.8

    2 Concrete layer 0.022500

    0500 675

    3 Vapour barrier 0.005 - 60 81

    4 Equalizer layer 0.031900

    0570 769.5

    5 Concrete slab 0.132500

    03250 4387.5

    6 Plaster 0.02 17000

    340 459

    Total Load 4888 6598.8

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    2.6.Design of a masonry structural member

    according to EC6

    The computation will be made on a section of 1 meterwidth from an exterior wall and from an interior wall.

    We use M50 binder and 100daN/cm2 full bricks.

    The loading surface for an exterior wall is Se = 2.235 m2

    The loading surface for an interior wall is Si = 1.85 m2

    Calculus for the interior wall

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    1. Load from the roof being an inner garret house, the axial

    force at the upper side is considered 0 for the interior wall

    2. Self weight of the wall - Nsw :

    Nd, wall = 7449.3 x 2.60 = 19368.18 N

    3. Load from plate Nd, plate :

    Live load -> 1.5 x 1500 x 1.85 = 4162.5 N

    Self weight -> 6598.8 x 1.85 = 12207.78 NNd, plate = 16370.28 N

    4. Self weight of the wall - Nsw :

    Nd, wall = 7449.3 x 2.60 = 19368.18 N

    5. Load from plate - Nd, plate :Live load -> 1.5 x 1500 x 1.85 = 4162.5 N

    Self weight -> 5434.29 x 1.85 = 10053.44 NNd, plate = 14215.94 N

    6. Self weight of the basement wall Nd,basement wall :

    Nd, basement wall= 17907.75 x 2.33 = 41725.06 N

    Nd,0 = 111047.64 N

    Calculus for the exterior wall

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    1. Load from the roof Nd, roof:

    Dead load -> 2115.45 x 1 = 2115.45 NSnow load -> 9260 x 1.5 x 1 = 13890 NNd, roof = 3913.35 x 1.9 = 16005.45 N

    2. Self weight of the wall - Nsw :

    Nd, wall = 8957 x 1.58 x 1 = 14152.06 N

    3. Load from the plate Nd, plate :

    Self weight -> 6598.8 x 2.235 = 14748.318 NLive load -> 1.5 x 1500 x 2.235 = 5028.75 NNd, plate = 19777.068 N

    4. Self weight of the wall - Nsw :

    Nd, wall = 8957 x 2.60 = 23288.20 N

    5. Load from plate Nd, plate :

    Live load -> 1.5 x 1500 x 2.235 = 5028.75 NSelf weight -> 5434.29 x 2.235 = 12145.64 NNd, plate = 17174.39 N

    6. Self weight of the basement wall Nd,basement wall :

    Nd, basement wall= 17907.75 x 2.33 = 41725.06 N

    Nd,0 = 132122.16 N

    Axial force diagram for external wall

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    Nd,0=132122.16 N => Md,f=Nd,fx d= Nd,fx (t/2-t/6)= Nd,fx 0.075

    Md,f1=19777.068 x 0.075 = 1483.28 NmMd,f2=17174.39 x 0.075 = 1288.08 Nm

    Eccentricity calculus

    We take the x-x section at a height of 2 meters.

    Mx-x / 1288.08 = 2 / 2.60 => Mx-x=990.83 Nm

    Nx-x / 73222.778 = 2 / 2.60 => Nx-x = 56325.21 N

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    => e = 990.83 / 56325.21 +ead=0.0176+0.02=0.0376cm

    0.0376 (e) < 0.075 (t/6) => all the section works incompression

    => Ac = t x b = 0.45 m2

    2.7.Design of foundation block

    We impose hf = 50 cm and we verify that pef pn

    External block

    N0,d=131023 N

    pef = Ntotal / Af pn we impose this equation at limit and we have

    Ntotal = Af x pn N0,d + Vg = pn x 1 x bf N0,d + hfbf = pnx 1 x bf => N0,d / bf +hf = pn =>

    bfnec = N0,d / (pn - hf) = 132.122 / (159-20 x 0.5 x 1)= 0.887 m

    => we will adopt 90 cm

    tg C 3.5 / 5 C 5 / 7.5

    Pn 2 daN / cm2 1.3 1.1

    Pn > 2 daN / cm2 1.6 1.3

    Stiffnes calculus:

    tg tgmin

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    tg = hf/ a , a= x (bfc - d) => tg = 2 x 50 / (90-45)=2.2222> tgmin

    Internal block

    N0,d=111047 N

    pef = Ntotal / Af pn we impose this equation at limit and we have

    Ntotal = Af x pn N0,d + Vg = pn x 1 x bf N0,d + hfbf = pn

    x 1 x bf => N0,d / bf +hf = pn =>

    bfnec = N0,d / (pn - hf) = 111047/ (159-20 x 0.5 x 1)= 0.7453 m

    => we will adopt 80cm

    Stiffnes calculus:

    tg tgmin

    tg = hf/ a , a= x (bfc

    - d) => tg = 2 x 50 / (80-45)=2.86 >tgmin