Upload
davidescu-daniel
View
133
Download
16
Embed Size (px)
Citation preview
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