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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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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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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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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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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:
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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
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Sk0,L
Sk0,Lx
Sk0,Ly
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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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=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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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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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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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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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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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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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