VI Proiectarea fundatiilor pe retele de grinzi

Se cere pentru proiecterea fundatiilor unui loc de locuinte avind regimul de inaltime p+10 etaje cu o structura de rezistenta tip cadre din beton armat.Ca sistem de fundare se adopta solutia de fundatii pe retele de grinzi.

Dupa repatizarea sarcinilor axiale din noduri pe directia longitudinala si transversala se vor calcula fundatiile pentru o grinda longitudinala sau transversala in urmatoarele ipoteze:· Metoda aproximativa - se vacalcula grinda longitudinala cu forte si momente· Metoda Winckler (in ipoteza deformatiilor elastice) - pentru o grinda transversala

incarcata cu forte si momente· Metoda Jemoskin (ipoteza deformatiilor elastice generala) - pentru o grinda

longitudinala cu forte dar fara momente

Sectiunea stilpilor vor avea sectiunea in forma de T

Date numerice ale grinzilor de fundare:· grinda longitudinala marginala A-A, D-D are latimea Bx=1,4m· grinda teansversala interioara 2-2, 3-3, 4-4 are latimea By=1,2m· grinda longitudinala interioara B-B, C-C are latimea Bx=1,4m· grinda transversala laterala 1-1, 5-5 are latimea By=1,1m· dimensiunile stilpilor 40x50cm

1. Calculul mamentelor de inertie ale grinzilor

pe directia X h

H

B

b

Ib H

cmx

x

x zx

xx

0 35

1 200 29

1 4

0 43 5

1 69712

1 69740 120

129774720

3 34

,

,,

,

,,

, ,

pe directia Y h

H

B

b

Ib H

cmy

y

y zy

y

y

0 35

1 200 29

1 2

0 52 4

1 47312

1 47350 120

1210605600

3 34

,

,,

,

,,

, ,

2. Calculul lungimilor elastice ale grinzilor

LE I

K Bcmex

zx

s x

4 4 270000 9774720

1 6 140511 684 4

,,

LE I

K Bcmez

zz

s z

4 4 270000 10605600

1 6 120494 214 4

,,

B L

B Lx ex

y ey

140 511 68

120 494 211 208 1 2

,

,, ,

3. Calculul incarcarilor in nodurile centrale (7,8,9,12,13,14)

p P tf

p P tf

x

y

1

1 208

1 1 208278 832 152 55

1

1

1

1 1 208278 832 126 28

7

7

,

,, ,

,, ,

4. Calculul incarcarilor in nodurile curente ale grinzii transversale (6,10,11,15)

Lungimile consolelor pe cele doua directii adoptate sint: l1=1,8m, l2=0,9m

l

L

l

L

ex

ey

1

2

180

511 6800 352 2 05

90

494 1820 182 2 87

,, ,

,, ,

p P tf

p P tf

x

y

16

1

16

1 208

1 208 2 05157 994 58 336

2 05

1 208 2 05157 994 99 658

,

, ,, ,

,

, ,, ,

4. Calculul incarcarilor in nodurile curente ale grinzii longitudinale(2,3,4,17,18,19)

p P tf

p P tf

x

y

2

22

22

1

1 208 2 87

1 1 208 2 87179 984 139 484

1

1

1

1 1 208 2 87179 984 40 5

, ,

, ,, ,

, ,, ,

4. Calculul incarcarilor in nodurile din colturile retelei(1,5,16,20)

l

L

l

L

excolt

eycolt

1

2

180

511 6800 352 2 8

90

494 1820 182 3 36

,, ,

,, ,

p P tf

p P tf

x

y

2

1 21

1

1 22

1 208 3 36

2 8 1 208 3 3693 90 55 416

2 8

2 8 1 208 3 3693 90 38 484

, ,

, , ,, ,

,

, , ,, ,

1. Metoda aproximativa

1. Calculul rezultantei incarcarilor

R Pi P P P P P KNi

n

16 17 18 19 20 529 284,

2. Verificarea presiunii terenului de sub fundatie

2.1 Calculul greutatii grinziiG L B H KNf T x med 27 6 1 4 1 2 20 927 36, , , ,

2.2 Claculul incarcarilor totaleR R G KNt f 529 284 927 36 1456 644, , ,

2.3 Pozitia rezultantei incarcarilor

eM

Rmt

t

3 13 5 2 6 7505

1456 6440 037

, ,

,,

2.4 Presiunea efectiva sub talpa fundatiei

pR

L B

e

L

p KN m

p KN mt

t x t1 2

12

22

16 1456 644

27 6 1 41

6 0 037

27 6

37 698

37 395,

,

, ,

,

,

, /

, /

p m B N q N c N

P KN m

pl l x

pl

1 2 3

1 5 19 48 1 4 0 325 21 758 2 3 40 4 848 379 240, , , , , , , , /

p pG

L BKN m

q p B KN m

ef

f

t x

x

' ,,

, ,, /

' , , , /

37 698927 36

27 6 13 69813 698

13 698 1 4 19 177

2

3. Calculul fortei taietoareQ tf

Q tf

st

dr

16

16

18 19 77 34 519

34 519 55 416 20 897

, , ,

, , ,Q tf

Q tf

st

dr

17

17

7 8 19 77 55 416 94 165

94 165 319 484 45 319

, , , ,

, , ,Q tf

Q tf

st

dr

18

18

6 19 77 45 319 69 743

69 743 139 484 69 741

, , ,

, , ,Q tf

Q tf

st

dr

19

19

6 19 77 69 741 45 319

45 819 139 484 94 165

, , ,

, , ,Q tf

Q tf

st

dr

20

20

6 19 77 94 615 20 897

20 897 55 416 34 519

, , ,

, , ,

4. Calculul momentelor in reazeme

M tf m

M tf m

st

dr

16

2

16

19 77 18

231 06

31 06 6 7505 24 316

, ,,

, , ,

M tf m

M tf m

st

dr

17

2

17

19 77 7 8

26 7505 55 416 6 244 117

244 117 13 5 230 618

, ,, , ,

, , ,

M tf m

M tf m

st

dr

18

2

18

19 77 13 8

26 7505 55 416 12 13 5 139 484 6 303 887

303 887 13 5 290 387

, ,, , , , ,

, , ,

M

tf m

M tf m

st

dr

19

2

19

19 77 19 8

26 7505 55 416 18 13 5 139 484 12

13 5 139 484 6 217 125

217 125 13 5 230

, ,, , , ,

, , ,

, ,

M

tf m

M tf m

st

dr

20

2

20

19 77 25 8

26 7505 55 416 24 13 5 139 484 18

13 5 139 484 12 13 5 139 484 6 10 831

10 831 13 5 24 33

, ,, , , ,

, , , , ,

, , ,

5. Calculul momentelor injcovoietoare din cimp

M tf ma

19 77 4 8

26 7505 55 416 3 47 9

2, ,, , ,

M tf mb

19 77 10 8

26 7505 55 416 9 13 5 139 484 3 180 956

2, ,, , , , ,

M tf mc

19 77 16 8

26 7505 55 416 15 13 5 139 484 9 13 5 139 484 3 167 460

2, ,, , , , , , ,

M

tf m

d

19 77 22 8

26 7505 55 416 21 13 5 139 484 15 13 5 139 484 9

13 5 139 484 3 34 43

2, ,, , , , , ,

, , ,

6.Corectarea momentelor incovoietoare

reazem 1

x m

yx

LM m

M tf m

M tf m

t

st

dr

1

1

1

1

18

18

27 613 496 0 88

31 067 0 88 31 947

24 316 0 88 25196

,

,

,, ,

, , ,

, , ,

reazem 2

x m

yx

LM m

M tf m

M tf m

t

st

dr

2

2

2

2

7 8

7 8

27 613 496 3 814

244 117 3 814 247 931

230 618 3 814 234 432

,

,

,, ,

, , ,

, , ,

reazem 3

x m

yx

LM m

M tf m

M tf m

t

st

dr

3

3

1

1

13 8

13 8

27 613 496 6 748

303 887 6 748 310 635

290 387 6 748 297 138

,

,

,, ,

, , ,

, , ,

reazem 4

x m

yx

LM m

M t fm

M tf m

t

st

dr

4

4

4

4

18

19 8

27 613 496 9 682

217 125 9 682 226 807

230 620 9 682 240 302

,

,

,, ,

, , ,

, , ,

reazem 5

x m

yx

LM m

M t fm

M tf m

t

st

dr

5

5

5

5

18

25 8

27 613 496 12 616

10 831 12 616 23 447

24 33 12 616 25196

,

,

,, ,

, , ,

, , ,

2.Metoda Winkler

L cm

L cm

L

L

L

ex

ey

ex

ey

511 68

494 21

4 8 3 2 15

511683 4

1 1

4 94210 202

,

,

, ,

,,

,,

x x

x x

x x m

x x m

1 4 1 4

2 3 1 3

2 2

1

2 0 2027 763

4 4

1

4 0 2023 8815

,,

,,

Grinda incarcata cu forteDeterminarea V1 si V2

forta x f3(x) M=Pf3/4 f4(x) Q=-Pf4/2V1 /2 -0.2079 -V10.2573 0 0V2 /4 0 0 0.3224 -V20.1612P3=40.5 0.303 0.4888 24.5 0.7077 -14.3309P8=126.128 1.273 -0.1807 -28.2071 0.061 -3.8469P13=126.128 2.242 -0.1482 -23.1339 -0.067 4.2252P18=40.5 3.212 -0.0383 -1.9197 -0.0407 0.8242

M V V V tf

M V V V tf

B

A

0 0 2573 28 7607 111779

0 01612 131284 814417

1 1 4

2 2 3

. . .

. . .

Sectiunea A=Bforta x f1 p=Pf1/2b f3 M f4 Q=-pf4/2V1=-111.779 /2 0.2079 -1.9559 -0.2079 28.7610V2=-81.442 /4 0.6448 -4.4199 0.3224 13.1285P3=40.5 0.303 0.9267 3.1589 0.4888 24.5005 0.7077 -14.3309P8=126.128 1.273 0.3627 3.8503 -0.18065 -28.1992 0.061 -3.8469P11=126.128 2.242 0.0162 0.1720 -0.1482 -23.1339 -0.066 4.1622P18=40.5 3.212 -0.0431 -0.1466 -0.0383 -1.9197 -0.0407 0.8242total 0.6588 0.0087 -0.062

Sectiunea 3=18forta x f1 p=Pf1/2b f3 M f4 Q=-pf4/2V1=-111.779 1.8738 0.0932 -0.8768 -0.1899 26.2708 -0.0484 -2.7051V2=-81.442 1.0883 0.4476 -3.0682 -0.1457 14.6858 0.1510 6.1489P3=40.5 0 1 3.4088 1 50.1238 0.1 +-20.25P8=126.128 0.9696 0.5398 5.7304 -0.1283 -20.0275 0.1988 -12.5371P13=126,12 1.9392 0.0932 0.9894 -0.1899 -29.6463 -0.0484 3.0523P18=40.5 2.9088 -0.0403 -0.1374 -0.0666 -3.3382 -0.0534 1.0814V3=-81.442 3.9972 -0.0.258 0.1769 0.0139 -1.4010 -0.0120 -0.4887V4=-111.779 4.7826 -0.0085 0.08 0.0089 -1.2312 0.0007 0.0391

6.3031 -25.659214.8408

Sectiunea a=fforta x f1 p=Pf1/2b f3 M f4 Q=-pf4/2V1=-111.779 2.197 0.0244 -0.2296 -0.1548 21.4151 -0.0652 -3.644V2=-81.442 1.4116 0.2849 -1.9529 -0.2011 20.2698 0.0419 1.7062P3=40.5 0.3232 0.9267 3.1589 0.4888 24.5005 0.7077 -14.3309P8=126.128 0.6464 0.7628 8.0977 0.4431 22.3378 0.4530 -28.568

P13=126,128 1.616 0.1959 2.0796 -0.2077 -32.4218 -0.0059 0.3721P18=40.5 2.5856 -0.0254 -0.0866 -0.1019 -5.1076 -0.0636 1.2879V3=-81.442 3.674 -0.0341 0.2337 -0.0079 0.7963 -0.0210 -0.8551V4=-111.779 4.4594 -0.0132 0.1242 0.0085 -1.1759 0.0023 -0.1285

11.425 49.8975 -43.3907

Sectiunea b=Bforta x f1 p=Pf1/2b f3 M f4 Q=-pf4/2V1=-111.779 2.5202 -0.0166 0.1562 -0.1149 15.8953 -0.0658 -3.6775V2=-81.442 1.7348 0.1576 -1.0803 -0.2047 20.6326 -0.0235 -0.9569P3=40.5 0.6464 0.7628 2.6002 0.1431 7.1727 0.4530 -9.1733P8=126.128 0.323 0.9267 9.8376 0.4888 76.3012 0.7077 -44.6304P13=126,12 1.2928 0.3355 3.5616 -0.1897 -29.6120 0.0129 -0.8135P18=40.5 2.2624 0.0080 0.0273 -0.1416 -7.0975 -0.0668 1.3527V3=-81.442 3.3508 -0.0408 0.2797 -0.0237 2.3888 -0.0323 -1.3153V4=-111.779 4.1362 -0.0231 0.2173 0.0040 -0.5531 -0.0095 -0.5309

15.5996 85.1277 -59.7351

Sectiunea 8=13forta x f1 p=Pf1/2b f3 M f4 Q=-pf4/2V1=-111.779 2.8434 -0.369 3.4716 -0.0771 10.666 -0.0523 -3.2025V2=-81.442 2.058 0.0439 -0.301 -0.1675 16.8831 -0.0618 -2.5166P3=40.5 0.9696 0.5083 1.7327 -0.1108 -5.5537 0.1988 -4.0257P8=126.128 0 1 10.6158 1 156.099 1 -+63.064P13=126,12 0.9696 0.5063 5.396 -0.1108 -17.2958 0.1988 -12.537P18=40.5 1.9392 0.0932 0.3177 -0.1899 -9.5185 -0.0484 0.9801V3=-81.442 3.0276 -0.0423 0.29 -0.0563 5.6747 -0.0493 -2.0075V4=-111.779 3.813 -0.0314 0.2954 -0.0040 0.5534 -0.0177 -0.9892

21.8182 157.508 -87.362538.7655

Sectiunea c=dforta x f1 p=Pf1/2b f3 M f4 Q=-pf4/2V1=-111.779 3.1666 -0.0431 0.4055 -0.0383 5.2984 -0.0407 -2.2747V2=-81.442 2.3812 -0.0056 0.0384 -0.1282 12.9219 -0.0669 -2.7242P3=40.5 1.2929 0.3355 1.1436 -0.1897 -9.5058 0.0129 -0.2612P8=126.128 0.3232 0.9267 9.8376 0.4888 76.3012 0.7077 -44.6304P13=126,12 0.6464 0.7628 8.0977 0.1431 22.3378 0.4530 -28.568P18=40.5 1.616 0.1959 0.6678 -0.2077 -10.4107 -0.0059 0.1195V3=-81.442 2.7044 -0.0320 0.2194 -0.0895 98.0211 -0.0608 -2.4758V4=-111.779 3.4898 -0.0389 0.366 -0.0177 2.4486 -0.0283 -1.5817

20.776 104.1426 -82.3965

Grinda incarcata cu momente

x x

x xx x

x x

1 4

2 3

1 4

2 3

3

4

20 202

11 6643

7 7723,

,

,

Determinarea momentelor m1, m2, m3, m4

Sect AMomentul µx f1(µx) T=Mµf1/2 f4(µx) M0f4/2m1 3/4 0 - -0.0670 m10.0335m2 /2 0.2079 m2*0.021 - -M3=4.7885 0.303 0.9267 0.4482 0.7077 -1.6944M8=9.577 1.2726 0.3355 0.3245 0.0129 -0.0618M13=9.577 2.2422 0.0244 0.0236 -0.0652 -0.3122M18=4.7885 3.2118 -0.0431 0.2084 -0.0407 -0.0974Total 0

ST = 0 m20.021-0.9575 = 0SM = 0 m10.035=2.1658 = 0

Sectiunea Ax f1 Q=Mf1/2 f2 p=M2f2/2b f4 M=-mf4/2

m1=61.88 3/4 0 0 0.6070 0.0705 -0.0670 2.073m2=-45.5952 /2 1.2079 -0.9574 0.2709 -0.1612 0M3=4.7885 0.303 0.9267 0.4482 0.2189 0.0178 0.7077 -1.6944M8=9.577 1.2726 0.3355 0.3245 0.2626 0.0428 0.0129 -0.0618M13=-9.577 2.2422 0.0080 -0.0077 0.0784 -0.01277 -0.0668 -0.3199M18=-4.7885 3.2118 -0.0431 0.0208 -0.0024 0.0002 -0.0407 0.0974total -0.1709 -0.0427 -0.1005

Sectiunea 3-18x f1 Q=Mf1/2 f2 p=M2f2/2b f4 M=-mf4/2

m1=61.88 2.6592 -0.0320 -0.2 0.0287 0.6302 -0.0608 1.8812m2=-45.5952 1.8738 0.0932 -0.4292 0.1415 -0.1097 -0.0484 -1.1034M3=4.7885 0 1 0.4836 0 1 -+.3943M8=9.577 0.9696 0.5083 0.4917 0.3096 0.0504 0.1988 -0.952M13=-9.577 1.9392 0.0932 -0.0902 0.1415 -0.023 -0.0484 -0.2318M18=-4.7885 2.9088 -0.0403 0.0195 0.0132 -0.0011 -0.0534 -0.1279m3=45.5952 4.7826 -0.0085 -0.0391 -0.0082 -0.0064 0.0007 -0.016m4=-61.88 5.568 0.0005 0.0031 0.0005 -0.0005 0.0029 0.0897total 0.2332 -0.059 -2.8545

1.9341

Sectiunea a=fx f1 Q=Mf1/2 f2 p=M2f2/2b f4 M=-mf4/2

m1=61.88 2.9824 -0.0423 -0.2644 0.0070 0.0074 -0.0493 -1.5253m2=-45.5952 2.1966 0.0244 -0.1124 0.0895 -0.0694 -0.0652 -1.4864M3=4.7885 0.3232 0.9267 0.4482 0.2189 0.0178 0.7077 -1.6944M8=9.577 0.6464 0.7628 0.7378 0.3099 0.0505 0.4530 -2.1692M13=-9.577 1.616 0.1959 -0.1895 0.2018 -0.0329 -0.0059 -0.0283M18=-4.7885 2.5856 -0.0254 0.0123 0.0383 -0.0031 -0.0636 -0.1523m3=45.5952 4.458 -0.0132 -0.0608 -0.0108 -0.0084 -0.0023 0.0524m4=-61.88 5.2438 -0.0023 0.0144 0.0049 -0.0052 0.0026 0.0080total 0.5854 -0.0433 -3.8729

Sectiunea b=ex f1 Q=Mf1/2 f2 p=M2f2/2b f4 M=-mf4/2

m1=61.88 3.3056 -0.0422 -0.2637 -0.0058 -0.0061 -0.0364 1.1262m2=-45.5952 2.5198 -0.0166 0.0764 0.0492 -0.0381 -0.0658 -1.5001M3=4.7885 0.6464 0.7628 0.3689 0.3099 0.0252 0.4530 -1.0846M8=9.577 0.3232 0.9267 0.8964 1.2189 0.0356 0.7077 -3.3888M13=-9.577 1.2928 0.3355 -0.3245 0.2626 -0.0428 0.0129 0.0618M18=-4.7885 2.2624 0.0080 -.0039 0.0748 -0.0061 -0.0688 -0.1599m3=45.5952 4.1348 -0.0231 -0.1064 -0.0136 -0.0105 -0.0095 0.2166m4=-61.88 4.9206 -0.0059 0.0369 -0.0073 0.0077 0.0014 0.0433total 0.6801 -0.0351 -4.6855

Sectiunea 8=13x f1 Q=Mf1/2 f2 p=M2f2/2b f4 M=-mf4/2

m1=61.88 3,6288 -0,0366 -2,2648 -0,0121 -0,0127 -0,0245 0,7580m2=-45.5952 2,842 -0,369 1,6993 0,0204 -0,0158 -0,0573 -1,3063M3=4.7885 0,9696 0,5083 0,2458 0,30936 0,0852 0,1988 -0,476M8=9.577 0 1 0,9673 0 0 1 -*4,7885M13=-9.577 0,9696 0,5083 -0,4917 0,3096 -0,0504 0,1988 0,0994M18=-4.7885 1,9392 0,0932 -0,0451 0,1415 -0,0115 -0,0484 -0,1159m3=45.5952 3,8116 -0,0314 -0,1446 -0,0137 -0,0106 -0,0177 0,4035m4=-61.88 4,5974 -0,0111 0,0694 0,0100 0,0105 -0,0011 -0,340total 0,0356 -0,0653 -5,4598

4,1172

Sectiunea c=dx f1 Q=Mf1/2 f2 p=M2f2/2b f4 M=-mf4/2

m1=61.88 3,952 -0,0258 -0,1612 -0,0139 -0,0146 -0,0120 0,3713m2=-45.5952 3,1662 -0,0431 0,1985 -0,0024 0,0019 -0,0407 -0,92979M3=4.7885 1,2928 0,3355 0,1623 0,2626 0,0214 0,0129 -0,0309M8=9.577 0,3232 0,9267 0,8964 0,2189 0,0356 0,7077 -3,3888M13=-9.577 0,6464 0,7628 -0,7378 0,3099 -0,0505 0,4530 2,1692M18=-4.7885 1,616 0,1959 -0,0947 0,2018 -0,0164 -0,0059 -0,0141m3=45.5952 3,4884 -0,0389 -0,1791 -0,0106 -0,0082 -0,0283 0,6452m4=-61.88 4,2742 -0,0179 0,1119 -0,0125 0,0132 0,0054 -0,1671total 0,1963 -0,0176 -1,3431

sectiunea A=Bp M(tfm) Q y(m)

F 0,6588 0 0 0,02928M 90,0427 0 0 -0,0019total 0,6161 0 0 0

Sectiunea 3=18p Mstg Mdr Qstg Qdr y

F 6,3031 35,4393 14,8408 -25,6598 0,2801M -0,059 1,9338 -2,8548 0,2332 0,0026total 6,2441 37,3731 32,5845 15,074 25,4263 0,2827

sectiunea a=fp M(tfm) Q y(m)

F 11,425 49,8975 -43,3907 0,5078M 90,0433 -3,8729 0,5854 -0,0019total 11,3817 46,0246 -42,8053 0,5085

sectiunea b=ep M(tfm) Q y(m)

F 15,5996 85,1277 -59,7351 0,6933M -0,0351 -4,6885 0,6801 -0,0016total 15,5645 80,4392 -59,055 0,6918

Sectiunea 8=13p Mstg Mdr Qstg Qdr y

F 21,8182 157,5082 38,7655 -87,3625 0,9697M -0,0653 4,1172 -5,4598 0,0356 -0,0028total 21,2529 161,625 152,0484 38,8011 -87,3269 0,9668

sectiunea c=dp M(tfm) Q y(m)

F 20,776 104,1426 -82,3965 0,9234]M -0,0176 -1,3431 0,1963 0,0008total 20,7584 102,7995 -82,2002 0,9226

Armare agrinzii longitudinale:

Se cunosc dimensiunile: H=1,20m si Bx=1,40m

h h a mbeton0 1 20 0 03 117 , , ,

sectiunea 3=18 Mmax=373731daNm

BM

b h Rp

A pb h

cm

c

ao

max . ,

, ,

02 2

2

373731

140 117 27000 00007 0 05

1000 05

140 117

1008 19

sectiunea a=f Mmax=498975daNm

BM

b h Rp

A pb h

cm

c

ao

max . ,

, ,

02 2

2

498975

140 117 27000 00009 0 05

1000 05

140 117

1008 19

sectiunea b=e Mmax=851277daNm

BM

b h Rp

A pb h

cm

c

ao

max . ,

, ,

02 2

2

851277

140 117 27000 00016 0 05

1000 05

140 117

1008 19

sectiunea 8=13 Mmax=1616250daNm

BM

b h Rp

A pb h

cm

c

ao

max . ,

, ,

02 2

2

1616250

140 117 27000 0003 0 05

1000 05

140 117

1008 19

sectiunea c=d Mmax=1041426daNm

BM

b h Rp

A pb h

cm

c

ao

max . ,

, ,

02 2

2

104126

140 117 27000 0002 0 05

1000 05

140 117

1008 19

3. Metoda Jemoskin

1.Caracteristicile grinzii si terenului de fundare

h

H

B

b

Ib H

m

E c

E I

y

zy

b zy

0 35

1 20 2917

1 4

0 43 5

1 680

121 680

40 120

120 1

6 1

314 2300 2 8

6 2700000 0 1 1 0 30 3

3 34

4

2

4

2

,

,,

,

,,

,

, ,

, ,

, ,,

2. Calculul coeficientilor si termenilor liberi

wki

k\i 1 2 3 4 5 6 7 8 9 101 0.25 1 1.75 2.50 3.25 4. 4.75 5.50 6.25 72 3.75 13.5 20.25 27 33.75 40.50 47.25 54 60.75.3 31.25 50 68.75 87.5 106.25 125 143.75 162.54 85.75 122.5 159.25 196 232.75 269.5 306.255 182.25 245 303.75 364.5 425.25 4866 332.75 423.5 514.25 605 695.757 549.25 676 802 929.58 843.75 1012.5 1181.259 1228.25 144510 1714.75

Fki

1 2 3 4 5 6 7 8 9 101 2.406 0.929 0.490 0.330 0.249 0.199 0.166 0.143 0.125 0.1112 2.406 0.929 0.490 0.33 0.249 0.199 0.166 0.143 0.1253 2.406 0.929 0.490 0.330 0.249 0.199 0.166 0.1434 2.406 0.929 0.490 0.330 0.249 0.199 0.1665 2.406 0.929 0.490 0.330 0.249 0.1996 2.406 0.929 0.490 0.330 0.2497 2.406 0.929 0.490 0.3308 2.406 0.929 0.4909 2.406 0.92910 2.406

dki

1 2 3 4 5 6 7 8 9 101 2.481 1.229 1.015 1.018 1.224 1.399 1.591 1.793 2 2.2112 3.531 4.979 6.565 8.43 10.374 12.349 14.341 16.343 18.353 11.781 15.929 21.115 26.58 32.124 37.699 43.29 48.8994 25.965 37.679 48.265 59.13 70.074 81.049 92.0415 54.915 74.429 91.615 109.68 127.824 145.9996 100.065 127.98 154.765 181.83 208.9747 165.015 203.729 241.09 279.188 255.53 304.679 354.8659 340.881 434.42910 516.831

3. Calculul termenilor liberi

11

P ij ji

n

P d

d

ij ij

j

ja

c

a

c

a

c

1

1 13

Pentru j=1...5 avem:a1=2m a2=8m a3=14m a4=20m a5=26m

wij

1 2 3 4 5 1P

1 0.411 2.018 3.625 5.232 6.839 616.7762 1.446 15.911 30.375 44.839 59.304 6631.6403 -2.232 37.946 78.125 118.304 158.482 17056.6884 -16.625 62.125 140.875 224 298.375 309925 -47.732 82.446 212.625 342.804 472.982 46421.4816 -101.554 92.911 287.375 481.839 676.304 627417 -184.089 87.518 359.125 630.732 902.339 78406.1818 -301.339 60.268 421.875 783.482 1145.089 92106.11259 -459.304 5.161 469.625 934.089 1398.554 102531.16010 -663.982 -83.804 496.375 1076.554 1656.732 106913.499

Sistemul va fi de forma:X y a

X P

X a P a

i ij P o i oi

ii

ii

i ii

i ii

d 11

10

1

10

1

10

1

10

1

10

0

X1=28,79 X2=33,441 X3=71,265 X4=153.690 X5=103.680 X6=94.238 X7=121.052 X8=61.092 X9=33.900 X10=27.900

In acest caz presiunile sub talpa fundatiei vor fi:

P1=7.344 P2=8.531 P3=18.180 P4=39.207 P5=26.449 P6=24.040 P7=30.881 P8=15.585 P9=8.648 P10=7.117

Daca yo=150.702 si o=0,45211 atunci:

y

y

E cm

E cm

o

oo

0

2

2

10 0062

10 0000187

' ,

' ,

Determinarea fortei taietoare:Q0=0 Q1

st=X10=27,9 KNQ1

dr=-X1-P1=27,9-93,90=-66Q2

st=-66+X9+X8=28.992Q2

dr=28.992-179.984=-150.992Q3

st=-150.992+X7+X6-64.298

Q3dr=64.298-179.984=-115.686

Q4st=-115.686+X5+X4=141.684

Q4dr=141.684-179.984=-38.3

Q5st=-38.3+X3+X2=66.406

Q5dr=66.406-93.9=28.79

Q6=28.79-X1=0

Determinarea momentelor:Mx1=0Mp1=11,516KNmMx2=Mx3=Mp2=Mx4=Mx5=Mp3=Mx6=Mx7=Mp4=Mx8=Mx9=Mp5=Mx10=

VII Calculul fundatiilor

Se adopta sistemil de fundatii sub forma de talpi continue, rigide, din beton simplu sub peretii portanti.

Calculul se va face separat pentru fundatiile peretilor exteriori si interiori, pe o portiune de 1m lungime de talpa situata sub fisia cea mai incarcata.

Pentru simplificare dimensiunile rezultate se adopta pentru toate talpile similare (interioare si exterioare). Calculul se efectueaza cu valori normate ale incarcarilor, in gruparea fundamentala.

Datorita modului de rezemare a peretilor pe fundatie(articulatie), momentul incovoietor se atenueaza pe inaltime, astfel ca pe fundatie actioneaza numai forte verticale provenite din infrastructura.

Stratul de protectie termica se poate considera atasat la peretii exteriori si se descarca impreuna la peretele de la subsol si in continuare la fundatie.

In aceste conditii incarcarile se transmit axial la fundatie, iar presiunea pe teren se distribuie uniform.

Calculul fundatie se realizeaza in doua etape:· dimensionarea latimii talpii si verificarea presiunii sub talpa· calculul de rigiditate

1. Calculul de rezistenta

Calculul de rezistenta are la baza conditia ca presiunea efectiva pe teren sa nu dapaseasca valoarea admisibila:

pef<pn

Se impune o inaltime a fundatiei de hf=40cm

G b h

N N G

A b

pN b h

bp b

N

p hcm

f f f b

totn

f

f f

ef

nf f b

f

nfnec o

n

nf b

1

1

140

0

0

Valoarea obtinuta pentru bfnec se rotunjeste cu 5cm in plus abtinind bf

ef.Se urmareste ca fundatia sa permita si plasarea peretelui de protectie a

hidroizolatiei, deci:

bd

cm cmfef s

2

210 2 5,

Daca nu este satisfacuta conditia se va majora valoarea tot la 5cm si se obtine valoarea finala bf

ef.

2. Calculul de rigiditate

Acest calcul urmareste respectarea urmatoarei conditii:

tg tg

a b d

tgh

a

f s

f

min

1

22

BC50 BC70

tg 1,3 1,1tg 1,6 1,3

Daca nu este indeplinita conditia se impune majorarea inaltimii fundatiei hf.