2.4 foundation design

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exemplu calcul fundatie pe grinzi

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  • 39

    2.4 FOUNDATION DESIGN NP 112-2014

    2.4.1 Borehole RecordThe borehole record it has been done in order to establish the imposed conditions by the

    ground. This data are used for the design, execution and service of the studied constructionwhich is built on a site located in city Suceava county Suceava.2.4.1.1 The research targetThe borehole record determine the following information:

    - identification of the ground stratification and of the physical and geologicalcharacteristics of the layers which compose the foundation ground in the active zone ofthe foundation;- establishment of the bearing capacity of the foundation ground;- determination of the possible effects in time of the underground water which affectsthe foundation ground;- classification of the site according to P100-1-2013 design code;- determination of the frost depth of the site according to NP 112-2014;- establishment of the geotechnical category of the ground according to NP 074-2014;- classification of the natural grounds in the classes provided by the cost estimating worknorms for the excavations and embankments works.

  • 40

    2.4.1.2 The condition and amount of the research:The borehole record has been done according to NP 074/2014 and is composed with

    the lithological observations correlated with the previous researches and with the execution of2 boreholes in the active zone of the foundations which provided samples which have beenanalyzed in the laboratory.2.4.1.3 The building characteristics:

    - block of flats used for dwelling with height P+4E;- reinforced concrete frame structure;- continuous beam foundations under columns;- the building do not have special service conditions like thermic processes, chemicalprocesses or dynamical ones which can influence the foundation ground.

    2.4.1.4 The ground stratification:After the analyzing of the layering files and based on the observations from the site thefollowing specifications return :

    - the foundation will reach a layer of gray-brown silty clay with medium plasticity,plastic hard.

    2.4.1.5 Underground water, flooding:The underground water has not been reached when the boreholes were made, so that

    the water level is supposed to be around 10 meter depth.The site is not sensitive to flooding and does not allow the accumulation and

    stagnation of water after heavy rainfall periods having a slight slope.

  • 41

    2.4.1.6 Ground characteristics:The physical and mechanical characteristics for the direct foundation layer, are given

    in the following table, according to NP 124-2010:gray-brown silty clay with medium plasticity

    Nr.Crt. Name Symbol UM Design values1. Inferior plasticity limit PW % 30,00 35,72. Superior plasticity limit LW % 17,5 17,73. Plasticity index PI % 12,3 18,24. Moisture W % 23,0 18,2

    5. Consistency index CI - 0,75 0,896. Clay A % 21 337. Silt P % 54 568. Sand N % 13 239. Normal unit weight kN/m 3 18,55

    10. Dry state unit weight d kN/m 3 14,9711. Porosity n % 43,4412. Void ratio e - 0,7513. Moisture degree rS - 0,77

    14. Linear deformationmodulusnatural moisture E kPa 14262

    15. Linear deformationmodulus submerged E kPa 13335

    16. Internal friction angle Grade 14,76 13,5617. Cohesion c kPa 20,8 16,08

    Table 2.4.1 Physical and mechanical characteristics of the soil

  • 42

    2.4.2 Transversal Section Preliminary DesignFor this building it has been adopted the continuous beam footing system designed at servicelimit state according NP 112-2014 in drained condition.The verification condition is :

    pef,med ppl (2.4.1)where: pef,med is the medium effective pressure at the foundation base;

    ppl is the plastic pressure.The partial resistance coefficients of the soil materials are M = 1,00.

    The structure is loaded in the fundamental group:SLU1: 1.35DL + 1.5LL+ 1.05SL

    2.4.2.1 Preliminary design of the transversal cross-section:

    cmmhc 606,0

    plteref ppLBNp

    2,1)( 321 NcNqNBmp epl

    em - work conditions coefficient;- natural volumetric weight;B foundation width;q design overload at the level of the foundation footing, lateral to foundation (kPa)c- cohesion;

    321 ;; NNN - coefficients used to determine the plastic pressure plp

  • 43

    = 13,56 = interpolation : 27,01 N

    12,22 N63,43 N

    3/55,18 mkN ;q = 2/97,2010,155,18 mkNDf ;

    fD - the foundation depth will be establish taken in account the following criteria:- the foundation depth to be larger with minimum 20 cm than the frost depth:- the foundation footing to enter with at least 20 cm in the good foundation soil:Frost depth for Suceava = 100 cm;

    - the foundation depth will be ;10,1 mD f Longitudinal foundation beam B-B:B will be neglected in the design relation of the plastic pressure.

    B = 1,00 m = 2321

    /32,173)63,408,1612,297,2027,055,18(4,1)(

    mkNNcNqNmp epl

    Axial efforts for the pre-dimensioning will be the reactions for the fundamental group:kNNNN 1795max,23max

  • 44

    - Due to the symmetry of the structure the effects of the bending moments will notbe taken in consideration.

    plpLNB 2,1 mLLL yx 3,10)2

    0,625,3(55.5

    mLpNB

    pl83,03,1032,1732,1

    17952,1

    - It is recommended to rise the value of the foundation width up to 20% due to theinteraction between the statically undetermined beam and the foundation ground,the contact pressure diagram have a nonlinear distribution, with effortsconcentrations in the higher rigidity zones, commonly under columns.

    mB 00,12,183,0 The value of foundation width will be B = 1.30 mmHc )0,1.....0,2(0,6)6/1......3/1(

    The height of the cross-section of the foundation height: mHc 60,1mcmHt 50,030 ; mmmhb c 70.010.060.0)100.....50(

    mHH tC 10.150.060.1

  • 45

    2.4.2.2 Checking:Checking ppl without bending moments :

    plteref ppLBNp 2,1

    2/93,1423,103,11795 mkNpef

    2321

    /42,190)63,408,1612,297,2527,030,155,18(4,1)(

    mkNNcNqNBmp epl

    2/42.198 mkNp pl

    142,93 < 228.5 kN/m 2

  • 46

    Checking with the bending moments due to earthquaqe forces:

    plterxy

    ef ppLBMy

    LBMx

    LBNp 2,1

    6622

    222 /99.19955.53,1

    167675.43,1

    20063,103,1

    1795 mkNpef

    2/50.228 mkNppl 199.99 kPa < 228.5 kPa

  • 47

    Checking - pconv

    convxy

    ef pLBMy

    LBMx

    LBNp 22

    66

    pef = 199.99 kN/m2

    ][kPaCCpp DBconvconv

    ][256 kPap conv ; for e = 0.75 ; 75.0cI ; Ip = 12.3 %

    convp - conventional pressure characteristic valueBC - width correction kPa;

    DC - depth correction kPa;

    For B 5m width correction is determined with the following relation:])[1(1 kPaBKpC convB

    1K - coefficient for cohesive soils 05,01 K][56,2)12,1(05,0256 kPaC B

    Depth correction is determined with the following relation:

    For mD f 2 : ][4,384240.12564

    2 kPaDpC fconvD

    ][16.2204.3856.2256 kPaCCpp DBconvconv 199.99 kPa < 220.16 kPa

  • 48

    2.4.3 Reinforcement designNP 112-2014 and SR EN 1992-1-1-2004

    The Design Efforts will be computed in Axis VM.The ground which sustain the foundation is a cohesive soil with the following elastic

    ground coefficient ks :

    Ic consistency index ; Ic = 0.75 it results ks = 63000 (kN/m3)kz = k x B =63000 x 1.2 = 75600 kPa

    2.4.3.1 The static scheme overvie:

    The studied foundation beam :AXIS 1-1

  • 49

    2.4.3.2 Loads acting on foundation (Robot to AxisVM):

    Axes Nr Fz (kN) Mx (kNm) My (kNm)max (-) min (-) max (+) min (-) max (+)

    A-A

    1-A 695 174 151 155 1712-A 977 181 150 187 1803-A 978 119 89 112 1244-A 1292 187 147 174 1745-A 979 118 88 124 1126-A 976 181 149 181 1867-A 697 173 150 171 155

    B-B

    1-B 949 178 190 152 1732-B 1324 178 196 188 1793-B 1450 178 198 172 1904-B 1770 177 200 174 1745-B 1449 177 197 190 1726-B 1324 177 195 178 1877-8 951 177 189 173 151

    C-C

    1-C 737 191 191 152 1692-C 1014 194 194 185 1783-C 1111 195 195 175 1904-C 1317 195 195 171 1695-C 1085 193 193 185 1716-C 1013 193 192 177 1847-C 741 190 190 169 151

    D-D

    1-D 947 191 177 149 1712-D 1324 196 178 185 1763-D 1492 197 178 177 1994-D 1795 200 173 167 1605-D 1450 197 176 185 1686-D 1321 195 177 175 1847-D 951 190 176 170 148

    E-E

    1-E 693 151 174 146 1632-E 975 150 181 177 1713-E 978 89 119 106 1184-E 1292 147 187 165 1655-E 978 88 118 117 1056-E 973 149 180 170 1767-E 696 150 173 162 146

    For the efforts maximum design computation the min values where taken in account.

  • 50

    2.4.3.3 Efforts diagrams and design values:The bending moment diagram :

    The shear force diagram :

    The Design Bending Moment and Shear Force values:

    VEd,max= 395 kN MEd,max,span = 537 kNm in span, superior partMEd,max,sup = 336 kNm in support, inferior part

  • 51

    2.4.3.4 Foundation beam reinforcement:Materials:Concrete strength class C16/20

    2/16 mmNfck ;2/66.105,1/16/ mmNff bcdck

    Reinforcement strength class:

    PC 52; ;/30015,1/345/ 2mmNff sykyd

    OB 37; ;/2042558,08,0 2mmNff ywkywd i) Longitudinal reinforcement

    NP 112-2014 General rules for foundation beam transversal andlongitudinal reinforcement

    - the minimum reinforcement coefficient for all sections (superior/inferior) is= 0.50 for foundation beams designed for seismic action, but not less than 0.002.- minimum diameter of the longitudinal reinforcement is 14 mm;- on the lateral faces of the beam are mounted reinforcement with 10/300 mm OB37;- the stirrups are designed at shear force and twisting moment;

    - minimum transversal reinforcement percent is 0,1 % , but not less than , = 0.08 ;- minimum stirrups diameter is 8 mm. If the width of the beam (b) is greater than 400 mm,there are mounted double stirrups (4 legged).- the longitudinal reinforcement from the inferior part can be mounted on all the width of thebeams footing

  • 52

    Longitudinal reinforcement computation in support (inferior part) As1 :

    = = 0.035= 1 (1 2 ) = 40.26= = 1001= 0.50 = 0.27%= = 2929As1 = 2929 mm2

    dl = = 21.59 mm2dl = 22 mm

    As1r = 8dl2 /4 = 3041 mm2The longitudinal reinforcement in the support in the superior part is composed of :

    8 bars of 22

    hf = hc = 500 mmbw = b = 700 mmhw = 1600 mm

    beff = B = 1300 mma = 50 mm

    d = hw a = 1550 mm

    fcd = 16.66 N/mm2fyk = 345 N/mm2fyd = fyk/s = 300 N/mm2s = 1.15fctm = 1.9 N/mm2 = 0.27 %Asmin = 2135 mm2

    MEd = 336 kNm

    bw

    da

    beff

    hf

    x

    hw

    As1 fyd

    beff x fcd

    As1

    MEd d-0.5x

  • 53

    As2bw

    ad

    beff

    hf x

    hw

    As2 fyd

    beff x fcdMEd d-0

    .5x

    Longitudinal reinforcement computation in span (superior part) As2 :

    = = 0.056= 1 (1 2 ) = 66.31= = 1649= 0.50 = 0.27%= = 2929As2 = 1649 mm2

    dl = = 21.59 mm2dl = 22 mm

    As2r = 8dl2 /4 = 3041 mm2The longitudinal reinforcement in the support in the superior part is composed of :

    8 bars of 22

    hf = ht = 500 mmbw = b = 700 mmhw = 1600 mm

    beff = B = 1300 mma = 50 mm

    d = hw a = 1550 mm

    fcd = 10.66 N/mm2fyk = 345 N/mm2fyd = fyk/s = 300 N/mm2s = 1.15fctm = 1.9 N/mm2 = 0.27 %Asmin = 2156 mm2

    MEd = 537 kNm

  • 54

    ii) Transversal reinforcement dimensioning :

    maxcw w 1 cdRd,max Ed b zv fV = V (kN)(ctg+tg)acw= 1 coefficient considering the effort in the compressed area

    z = 0.9d = 990 mmv1 = 0,6 cracked concrete strength reduction coefficient at shear force

    max(ctg1; ctg2) 1 if the condition is not fulfilled the cross-section dimensions mustbe modified

    ctg 2.5 if it results a higher value in the computation will be consideredctg = 2.5

    s = 100 mm distance between stirrupsfywk = 255 N/mm2 the yielding strength of OB37

    fywd = 0.8 fywk = 204 N/mm2, = 0.0011 700 990 0.6 16.66+ 1 345 10+1/ 49.80= .

    maxSWRd,s ywd EdAV = zf ctg V (kN)sAsw = = 4 = 201.06 mm2

    VRd,s = 201.06/100x990x204x2.5 = 1015 kN > VEd,max= 345 kN

    hf = ht = 500 mmbw = 700 mmhw = 1600 mm

    beff = b = 1300 mma = 40 mm

    d = hw a = 1560 mm

    fcd = 10.66 N/mm2fyk = 255 N/mm2fyd = 204 N/mm2fctm = 1.9 N/mm2

    VEd = 395 kN

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    = 100 = 0.28 %The stirrups will have a diameter of 8 and will be mounted at 100 mm distance

    stirrups OB37 8/10cm/20cmThe stirrups will be mounted on first quarter of the span of the beam at 100 mm distance and

    on the rest of the span at 200 mm.There will be mounted 4 legged stirrups (double stirrups b > 400 mm)

    iii) Anchorage lengths :Support :

    lbd = 1 2 3 4 5 lbdr= 825 mm1 = 2 = 3 = 4 = 5 = 1= 4 = 825= = 300 /= 22= 2.25 = 2.00 /

    h1 = 1 , h2 = 1= , . = 0.9 /, . = 1.3 /lbd = 850 mm

    iv) Overlapping lengths :lbd = 850

    l0 = 6 lbd = 1.275 m6 = 1.5

    l0 = 1.30 m