Proiect zid de sprijin:

Date:

qn=21.0 KN/m2 B=(12

÷23

¿◦H c=0.6◦a θ=¿10°

h=2.7m Df=1.40m

A F1:stratificatia S6:

±0.00………………………………….- 0.30m → Sol vegetal

γ=¿17.0 KN/m3

-0.30…………………………………..-3.30m → Argila prafoasa galbena plastic vartoasa

γ=20.0 KN/m3 , Ic=0.80, φ=22°,

c=27 KN/m3, E=12300 KN/m2

-3.30…………………………………….-6.50m→ Nisip mijlociu cu indesare mijlocie γ=¿18.5 KN/m3 , Id=0.65, φ=28°,

E=24800 KN/m3

-6.50……………………………………..-9.70→ Nisip mare cu pietris indesat

γ=¿19.2 KN/m3 , Id=0.90, φ=35°,

E=32200 KN/m3

Rezolvare:

① Predimensionare:

Df=1.40m Dfmin=0.5m θ=¿10° b=1m

a=0.9m H=Df+h=2.7m+1.4m=4.1m⟹ H=4.1m

h=2.7m c=0.6◦a⟹ c=0.6◦0.9 ⟹c=0.54m

B=(12

÷23

¿◦H ⟹ B=(2.05…………………..2.73) ⟹B=2.5m

② Calculul coeficientului Ka

∂i=(12…………………..

23¿◦φ

Kai=f(φi, θ,∂I,β

=0)

k a=cos2(ɸ+θ)

cos2θ×cos (δ−θ )× ¿¿

Strat1:

γ=20.0 KN/m3

φ=22°

θ=¿10°

∂1=(11……………..14.66)⟹∂1=12.5◦

k a1 =0.345

Strat2:

γ=¿18.5 KN/m3

φ=28°

θ=¿10°

∂2=(14……………..18.66)⟹∂2=17.5◦

k a2 =0.254

③ Determinarea greutatii zidului de sprijin

Determinarea pozitiei centrelor de greutate

G1=2.5cm=1.25m

G3=3.2cm=1.6m

G4⟹ b=1,4cm=0.7m

h=8.2cm=4.1m

G4=13 ∙8.2=2.73cm=1.36m

G2⟹ ⟹ b=3.3cm=1.65m

h=6.4 cm=3.2m

G2=23 ∙6.4=4.26cm=2.13 m

G1=(2.5m∙0.9m)∙24KN/m3=54 KN γ b=24 KN/m3

G2=(1.65∙3.22 )∙24KN/m3=63.36 KN

G3=(0.3∙3.2)∙24KN/m3=23.04 KN

G4=(0.7 ∙4.12 )∙24KN/m3=34.44 KN

④ Trasarea diagramelor de presiuni din impingerea pamantului

γ 2=¿18.5 KN/m3 k a1 =0.345 hstrat1=3.30m

γ 1=20.0 KN/m3 k a2 =0.254 hstrat2=0.8m

qn=21.0 KN/m2

Strat 1:

q=γ 1∙i⟹ i=qγ 1⟹ i=1.05m

PA=γ 1∙i∙ ka1 ⟹ PA=7.245 KN

P1=γ 1∙(i+ hstrat1)∙ ka1 ⟹P1=30.01 KN

Pa1=P A+P1

2∙ h strat 1⟹ Pa1=53.139 KN

Strat 2:

q+γ 1∙ hstrat1=γ 2∙hec1⟹ hec1=q+γ1 ∙h strat1

γ 2⟹ hec1=4.702m

P1,=γ 2∙ hec1∙ ka2⟹ P1

,=22.094 KN

P2=γ 2∙( hec1+ hstrat2)∙ ka2⟹ P2=25.853 KN

Pa2=P1,+P2

2∙ hstrat2⟹ Pa2=21.388 KN

Calculul componentelor

Pa1=53.139 KN ∂1=12.5◦ θ=¿10°

Pa2=21.388 KN ∂2=17.5◦

Strat1:

Pa1=53.139 KN → Pa1h= Pa1∙cos(∂1-θ) KN

→ Pa1V= Pa1∙sin(∂1-θ) KN

Pa1h=53.139∙cos2.5 KN

Pa1V=53.139∙sin2.5 KN

Strat2 :

Pa2=21.388 KN → Pa2h=Pa2∙cos(∂2-θ) KN

→ Pa2V= Pa2∙sin(∂2-θ) KN

Pa2h=21.388∙cos7.5 KN

Pa2V=21.388∙sin7.5 KN

Calculul eforturilor N,T,M

G1=(2.5m∙0.9m)∙24KN/m3=54 KN γ b=24 KN/m3

G2=(1.65∙3.22 )∙24KN/m3=63.36 KN

G3=(0.3∙3.2)∙24KN/m3=23.04 KN

G4=(0.7 ∙4.12 )∙24KN/m3=34.44 KN

Pa1h=53.139∙cos2.5 KN Pa2h=21.388∙cos7.5 KN

Pa1V=53.139∙sin2.5 KN Pa2V=21.388∙sin7.5 KN

T=53.139∙cos2.5+21.388∙cos7.5=74.293 KN ⟹ T=74.293 KN

N=53.139∙sin2.5+21.388∙sin7.5+54+63.36+23.04+34.44=179.94 KN

⟹ N=179.94 KN

M-53.139∙sin2.5∙1.25+53.139∙cos2.5 ∙2.1-21.388∙sin7.5∙1.25+

+21.388∙cos7.5∙0.38-34.44∙1.5-23.04∙1.1-63.36∙0.5=0

⟹ M=4.47 KN∙m

Calculul pozitiei centrelor de greutate al diagramelor de eforturi

ZG=2∙ x+ yx+ y

∙z3

PA=7.245 KN=x

P1=30.01 KN=y ⟹ ZG1=1.31m

h=3.3m=z

P1,=22.094 KN=x

P2=25.853 KN=y ⟹ ZG2=0.38m

h=0.8m=z

⑤ Verificari:

a) Verificarea presiunii pe teren

N=179.94 KN M=4.47 KN∙m

P1,2=NA ± M

w →P1≥1.2∙Ppl γ 2=¿18.5 KN/m3

→P2>0 γ 1=20.0 KN/m3

A=B∙1 ⟹ A=2.5m2

W=B2 ∙16

⟹ W=1.04m2

Ppl=m∙(γ 2∙B∙N1+q∙N2+C∙N3)

m=1.5 N1=0.98

γ 2=¿18.5 KN/m3 N2=4.93 ⇒Din tabelul 11.5 N1,N2,N3

B=2.5m N3=7.40 Din tabelul 11.6 m

q=0.5∙ γ 1+(1.4-0.5)∙ γ 2 ⇒q=26.65

Ppl=1.5∙(18.5∙2.5∙0.98+26.56∙4.93)⟹Ppl=264.398

P1=179.942.5 +4.471.04≤1.2∙264.398 ⟹76.27≤317.277

P2=179.942.5 -4.471.04>0 ⟹67.67>0

b) Verificarea la rasturnare

Mr≤0.8 ∙Ms

G1=(2.5m∙0.9m)∙24KN/m3=54 KN γ b=24 KN/m3

G2=(1.65∙3.22 )∙24KN/m3=63.36 KN

G3=(0.3∙3.2)∙24KN/m3=23.04 KN

G4=(0.7 ∙4.12 )∙24KN/m3=34.44 KN

Pa1h=53.139∙cos2.5 Pa2h=21.388∙cos7.5

Pa1V=53.139∙sin2.5 Pa2V=21.388∙sin7.5

Mr=53.139∙cos2.5∙2.1+21.388∙cos7.5∙0.38=119.543 KN

Ms=53.139∙sin2.5∙2.5+21.388∙sin7.5∙2.5+34.44∙2.75+23.04∙2.35+

+63.36∙1.75+54∙1.25=340 KN ⟹ 119.543≤0.8∙340

c) Verificarea la lunecare

T≤0,8∙Ff T=74.293 KN N=179.94 KN

Ff=μ ∙N μ ⟶ din tabelul 8.1 pag 185

Ff=80,973 Nu verifica

a. Ff=N∙tg∂ ∂=17.5°

Ff=56.734 Nu verifica

b. Ff=N∙tg28 ° φ=28°

Ff=179.94∙tg28 ° ⟹ Ff=95.67 ⟹ 74.293 ≤ 0.8∙95.67

74.293 ≤ 76.536 Verifica

d) Verificarea rezistentei materialului in sectiunea 1-1

Ϭ1-1=N 1−1

A1−1±

M1−1

W 1−1 Ϭ1≤fcd Ϭ2>0

Sau: Ϭ2<0 atunci Ϭ2≤fcd

A1−1=B∙1=2.1m2 W 1−1=B2 ∙16

⟹ W 1−1=0.735m2

Calculul pozitilor centrelor de greutate

ZG=2∙ x+ yx+ y

∙z3

ZG1 ⟹ x=0.4m

Y=2.1m ⟹ZG1=1.23m

z=3.2m

ZG2=23

∙3.2 ⟹ ZG2=2.13m

G1=(0.4+2.1)∙3.22

∙24⟹G1=96 KN Pa1h=53.139∙cos2.5 KN

G2=0.6 ∙3.22

∙24⟹G2=23.04 KN Pa1V=53.139∙sin2.5 KN

Ϭ1-1=N 1−1

A1−1±

M1−1

W 1−1 fcd=rezistenta la compresiune a betonului

N1−1=G1+G2+53.139∙sin2.5 =121.35 KN

M1-1-53.139∙sin2.5∙1.05+53.139∙cos2.5∙1.31-G2∙1=0

M1-1=3.75KN

fcd=acc∙fckγc acc=1 , fck in fuctie de clasa betonului C16/20,γc=1.2

fcd= 161.2 ⟹ fcd=13.33 N/mm2

Ϭ1max=62.88 KN/m2≤13330 KN/m2

Ϭ1min=52.68 KN/m2>0