Proiectarea unui zid de sprijin din pământ armat cu geogrile

1 = Verifica

Calcul rambleu pamant armat R h γ b := = 0.359 Nγ H:=:=1.58 Greutate 3 +pamant 0.1 ⋅ 18 =armat 4.8 Rv

he := 1.3 Kn γ := 18 Evaluarea greutatii si suprasarcinii Stabilitatea a := 1.5 la alunecarea armat mc la contactul armata/teren NqH:=:=5.9 i := 0.9 − 2.05 ⋅ γ b = 0.165 H + aN gama = 6.3 Nq Ncγ Kn t ⋅ 1 phi G1fal := :=H ⋅ x ⋅ γ = 41.472 pentru alunecare pe talpa 1.3 15 armat 0.70 - factorul 3.9 partial 11 2 20 := 6 1.80 ml Larmare 6.4 Nc := 14.04 iq14.8 := 0.95 − 1.7 ⋅ γ b = 0.34 Evaluarea presiunii pamantului 19 armaturilor 1.58 5.9 14.04 Verificarea la smulgere Kn de frecare dintre armatura si teren - unghiul δa := 0.95 0.443 0.8⋅ ( L⋅ tan G2tan :=D γf armat − x( )φ⋅ 1H) == 435.456 q1 := γ 1 ⋅ Df = 15.04 Ti ic := iml q = 0.34 H L := Larmare Pi ≥x f:= ⋅ =δ⋅ tan 0.96 ⋅+γ cKn sm tan + q) + ca ⋅ Lpi vγa ⋅⋅L(δpi a −( 21⋅⋅e⋅Lh) i⋅ N G3fPal :=⋅ R:= Gh1 ≤ =5R41.472 0.9 ⋅ L  cr ml γ ⋅ iγ + q1 ⋅ Nq ⋅ iq + c2 ⋅ Nc ⋅ ic = 143.356 Verifica !  2

Kn Lpi⋅ R1 lungimea KN = 347.067 armaturilor G4fal := h⋅ 2.25 ⋅ 1.5 pasiva ⋅ γ 1 = a31.725 2 ml

Calculul rezistentei laφ intindere a armaturilor π 1   ⋅ H Kn KN := L − tan − G5L 4.25 ⋅ 1.5 ⋅ γ = 119.85 Verifica ! R:= ⋅ tan δ + c ⋅ L = 347.665 pv a 114 2  KN ml Se alege o geogrila cu rezistenta : T := 33

m Kn perimetrul Pi q0 ⋅ 3.0361 Q := = armaturii 74.202 pe un metrul liniar de zid KNml Tcalcul := 18.75 Verificarea stabilitatii la rasturnare Kn Pi := 2 + 2 ⋅ L = 14 m G := G1 + G2 + G3 + G4 + G5 = 669.975 Momentul tuturor fortelor fata de centrulml bazei, Mo : tanδ a = 0.443

Dispunerea armaturilor

2 nG := 1.2 L KNm M fsm := P1.3 0 := aq ⋅ 2.75 + PaH ⋅ 1.83 + G4 ⋅  2 − x − 2.25 ⋅ 3  = 578.079   Tcalcul nG.1 sv :=:= 0.8 = 0.331 m H ⋅ γ armat + q0 ⋅ ka ⋅ 1.2 cRa := 0 KN Kn v = 744.177 G := = 803.97 q0 :=defhe ⋅ γ 1G=⋅ nG 24.44 ml  πM φ 1  - se intre armaturi egala cu 0.3 m  0−o distanta  = 0.637 tanalege 4 G⋅ n=2 0.777 :=⋅:=  = 535.98m Kn β := 34 deg Gefav Rv G.1 ml β 1 := 9 ⋅ deg = 0.157 ⋅ rad Calculul fortei de intindere in armaturi L 1 = Verifica e< =1 2 2 6 cos β − cos β ( ( ( ) 1) 1) ) − ( cos ( φpe 1) talpa Verificarea stabilitatii la alunecarea Forta datorata greutatii proprii a umpluturii ka := cos de = 0.426armate si suprasarcinii ce ( β 1intindere )⋅ 2 2 actioneaza la suprafata terenului. Ti1 cosPai( β 1) + Paqi ( cos ( βMoi Cota Gi Rvi ViTi1 1) ) − (eicos ( φLi-2*eiSigma 1) )

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Stabilitatea la alunecarea la contactul umplutura armata/teren Caracteristici teren : Verificarea presiunii pe teren

0.3 32.4 191.98 14.66 20.00 25.20 0.13 5.74 33.46 4.21 0.6 64.8 224.38 Rvi19.96 23.33 37.29 0.17 5.67 39.59 4.98 0.9 97.2 256.78 26.07 26.67 52.61 T0.20 :=5.59 σ vi := ka ⋅ σ 45.93 i1 vi ⋅ svi 5.77 L − 2.ei R := P + P = 266.975 1.2 129.6 289.18 32.99 30.00 71.54 0.25 5.51 52.53 i aq actiunii pentru impingerea pamantului 6.60 coeficientul n := 1.2 h aH 1.5 162 321.58 40.73 27.78 88.87 0.28 5.45 59.03 7.42 Presiunea de baza 49.28 1.8 194.4 353.98 36.67 121.65 0.34 5.31 66.63 8.38 σv 2.1 226.8 386.38 58.65 40.00 153.57 0.40 5.21 74.23 9.33 Kn Rv := Q2=4.8m 744.177 1 G h+ := γφ 2 := 19 + 0.1 ⋅ 18 2.4 259.2 418.78 68.83 43.34 190.55 0.46 5.09 82.28 10.34 0.349 1 2 =11.42 := ⋅ γ451.18 ⋅ Ht ⋅ ka79.82 ⋅ nR = 190.644 2.7Pa291.6 46.67 232.97 ml 0.52 4.97 90.83 1 v 50.00 281.18 Kn 2 483.58 12.57 3 324 91.64 0.58 4.84 99.97 := γ ⋅σ := 17 + 0.1 ⋅ 18= 167.366 = 18.8 φ 2 := 20 ⋅ deg v R ≤ R tan φ = 3.3 356.4 515.98 104.26 0.65 4.70 109.80 13.80 Kn mc h v1 2L − 21⋅ e 53.34 335.56 := q0548.38 ⋅ Ht ⋅ ka ⋅ n117.70 = 78.678 3.6Paq 388.8 56.67 396.48 0.72 4.55 120.42 15.14 P 421.2 Rv:= ⋅ tan Pa580.78 φc⋅ cos :==β325 ⋅ deg = 188.296 tan 131.96 φ c22 := =16.59 0.364 22 ml 0.80 4.40 121 := 1270.858 3.9 aH 131.95 60.00 464.29

Stratul 1

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Stratul de fundatie

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= 20.8