Proračun vlačnog temelja
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Transcript of Proračun vlačnog temelja
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Proračun vlačnog temelja
-vježbe-
Geotehničko inženjerstvo
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Generalno, sustave prijenosa vlačnih sila u tlo možemo podijeliti u tri skupine:• Plitki vlačni temelji (D/B<4)• Duboki vlačni temelji (D/B>4)• Geotehnička sidra
Proračun plitkih vlačnih temelja:
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1. Metoda opterećenja i metoda otpora tla
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2. Metoda analize pasivnog loma u tlu (metoda Dž. Sarača)
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Ravninski model s aluminijskimvaljčićimaRavninski model s pijeskom(Sarač 1975.)
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osnovna podloga za proračun vlačnog temelja
D
B
b
S
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Riješeni primjer
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Metoda Dž. Sarača
B=1,0 m
b=0,4 m
h=0,3 m
α1=15°
γbet=25,0 kN/m3
Odrediti minimalnu dubinu temeljenja,ako je zadana vlačna sila: P=55 kN.
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tlatemcP2
m'
P3 WWKDcKDS ++⋅⋅+⋅⋅γ= ϕ
Karakteristike temelja:
B=1,0 m
b=0,4 m
h=0,3 m
α1=15°
γbet=25,0 kN/m3
( ) ( )
( ) ( )
D16,033,0V
08,03,0D4,04,04,011308,03,01V
vDbbbBB3vhBV
m 08,02
4,0115tg2
bBtgv
2222
2222
1
⋅+=
−−++⋅++⋅=
−++⋅++⋅=
=⎟⎠⎞
⎜⎝⎛ −°=⎟
⎠⎞
⎜⎝⎛ −⋅α= D (m)
Wtem (kN)
12,25
1 1,510
20
15
Težina temelja:
Wtem=V*γbet
Wtem=(0,33+0,16·D)·25,0=8,25+4·D
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Wtla=V⋅γtla
( ) ( )
( ) ( )
( )054,0D84,019W054,0D84,0V
08,0D4,04,04,011308,0D1V
vDbbbBB3vDBV
tla
2222
2222
−⋅⋅=−⋅=
−−+⋅+−⋅=
−−+⋅+−⋅=
Težina tla iznad temelja:γtla=19 kN/m3 za D<4,0 m
D (m)
Wtla (kN)
14,9
1 1,510
25
20Svojstva tla:
γtla=19 kN/m3
c=5 kPa; cm=c/2=2,5 kPa
ϕ=26°
λ=D/de
m 13,114F4d te =
π⋅
=π
=
Ft=L×B - za pravokutni temelj
- za kružni temelj: de = B = 2r, λ=D/B
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FS=1,5P=55 kN
Smax=FS⋅P=82,5 kNtlatemcP
2m
'P
3
tlatemcP2
m'
P3
tla
WWKDcKD19S
WWKDcKDS
++⋅⋅+⋅⋅=
++⋅⋅+⋅⋅γ=
ϕ
ϕ
157,731,4716,232,5120,5592,5261,772,0
143,729,7715,832,5780,5742,5261,681,9
130,428,1815,432,6760,5882,5261,591,8
119,326,5815,032,7690,6182,5261,51,7
108,324,9914,632,8670,6472,5261,421,6
99,823,3914,263,0000,7062,5261,331,5
7519,4013,263,3560,7942,5260,111,25
S (kN)Wtla(kN)Wtem(kN)KcPK’ϕPCm(kPa)ϕ°λD(m)
λ=D/de=D/1,13