ZEM NDE DO AL DURUM τ
σ
SÜKUNETTE TOPRAK BASINCI KATSAYISI (K0) σx K0 = ( zeminde K 0 < 3) σz K0 =
ν 1 −ν
Etki Eden Faktörler: Dane da ılımı (Dmax, Cu, Cr) Kayma direnci açısı (φ φ) Kumda çimentolanma A ırı konsolidasyon (OCR) Kilin aktivitesi
( A = I %C ) c
P
Jeolojik ko ullar (birikme, olu ma, tektonizma) Drenaj ko ulları (k)
SÜKUNETTE TOPRAK BASINCI KATSAYISI (K0) Zemin Cinsi
K0
Temiz Kum (NL)
1-Sinφ
Kum (OC)
KoNL × OCRSinφ
NL Kil
0.19+0.23×log IP
OC Kil
0.7+0.1(OCR-1.2)
Kum Kil, yu rulmu eker
Referans (Jaky, 1945) (Mayne ve Kulhawy, 1982)
sıkı K0=0.37, gev ek K0=0.46 K0=0.69 2.00 K0=0.50
K0 için Tipik De erler K0 =
ν 1 −ν
ν !
"
ν için Tipik De erler
K0
Doygun
0.40-0.50
0.67-1.00
Doygun Olmayan
0.10-0.30
0.11-0.42
Kumlu Kil
0.20-0.30
0.25-0.42
Silt
0.30-0.35
0.42-0.54
0.20-0.40
0.25-0.67
ri (e=0.4-0.7)
0.15
0.18
nce (e=0.4-0.7)
0.25
0.33
0.10-0.40
0.11-0.67
Zemin Cinsi Kil
Sıkı Kum Kaya
ÖRNEK: ρd = 19 kN/m3 ve φ’ = 280 olan bir zeminin 5 m derinli inde OCR=1 ve OCR=5 durumları için dü ey ve yatay efektif gerilmeler ile maksimum kayma gerilmelerini bulunuz (Su seviyesi yüzeyden 1 m a a ıda ve zemin tüm kesitte doygundur).
σ v′ = 5 ⋅ 19 − 4 ⋅ 9.81 = 55.76 kPa K 0NC = 1 − Sinφ = 1 − Sin28 = 0.531
K 0OC = K 0NC ⋅ OCR Sinφ = 0.531 ⋅ 5Sin 28 = 1.13
σ h′ = σ z′ ⋅ K 0 = 55.76 ⋅ 0.531 = 29.61 kPa
σ h′ = σ z′ ⋅ K 0 = 55.76 ⋅ 1.13 = 63.01 kPa
τ= τ NC
55.76 − 29.61 = = 13.08 kPa 2
σ 1′ − σ 3′ 2
τ OC
55.76 − 63.01 = = −3.63 kPa 2
KUMDA ELAST K DENGE τ
kum
KUM
σh
σz
Temiz Kum (φ 300) K0 = ( 1 - Sin φ ) K0<1
σz sa tarafta
σ
K LDE ELAST K DENGE τ
KL
KL
σh
σz
σh #$-&
#$%&
#$ $ &' ($ ) *
+,
#$-&
#$ $ . ($ & " / 0 1& )
σ
ZEM NDE EK GER LME DURUMU 3 4
2
τ
φ Sükunette c σh=K0σz
σ
σz=ρ.z
τ
φ Aktif
45+φ/2
c σh=Kaσz
σ
σz=ρ.z
Pasif 45-φ/2
c
φ
45-φ/2
σz=ρ.z
σh=Kpσz
σ
PLAST K DENGE PROBLEMLER !
%
&
'(
"# ! $
)
*
)
!
#
Aktif Sükunette
Pasif
Aktif c
Sükunette
Pasif
φ
45-φ/2
σh=Kaσz σh=K0σz
σz=ρ.z
σh=Kpσz
σ
( + $! *, ,
C. PLAST K DENGE PROBLEMLER
% ( &# % '$
$
! ).# $ $/ $
PLAST K DENGE AKT F F--PAS F TOPRAK BASINÇLARI
PLAST K DENGE
σ1 σ3
τ
σ3
σ1
τ
OB =Yenilme Yok
B=Yenilme
τ
σ
c
σ3
σ1
φ
σ
KIRILMA ((yenilme yenilme veya akma)’ NIN TEMS L • Malzemenin direnci uygulanan gerilmeden dü ükse yenilme meydana gelir ve PLAST K DENGE olarak isimlendirilir. • Bu durum gerilme dairelerinin kırılma zarfına te et olmaları ile temsil edilir. • Kırılma, gerilme dairelerinin kırılma zarfına te et olması ile gerçekle ti ine göre bu durumda gerilmeler arasında bir ba ıntı olmalı. τ
σ = τmax = f( σ )
τ σ
τ = c + σ ⋅ tan φ
φ
c
σ3
σ1
σ1 = σ3Nφ + 2c Nφ
σ
AKT F TOPRAK BASINCI
Pa = ρ .z.K a
AKT F TOPRAK BASINCININ OLU UMU 0.001h gev ek kum 0.04 h yumu ak kil
σ1
τ σ3
σ3
φ h
σ1
σ3=σa σ3=Kaρ’z
σ3=σo σ3=Koρ’z
σ
σ1 σ1=ρ’z
σ o= 4 σ a= = 4 =
, ,
5 5
PAS F TOPRAK BASINCI
Pp = ρ .z.K p
PAS F TOPRAK BASINCININ OLU UMU 0.01 h gev ek kum 0.05 h yumu ak kil
σ o= 4 σP= , K o= 4 K p= ,
h
, ,
5 5
σ1
τ
σ3
σ3
φ
σ1
σ3=σo
σ1
σ3=σp
σ3=Koρ’z
σ1=ρ’z
σ3=Kpρ’z
σ
TOPRAK BASINCININ OLU UMU için GEREKL HAREKET M KTARI
Kp ye gelebilmek için yakla ık Ka nın 4 katı bir hareket gerekiyor
τ
KUMDA PLAST K DENGE τ = σ tan φ
σ 1 = σ 3 ⋅ Nφ Nφ = 45 + φ / 2
45−φ /2
σ
AKT F
PAS F
σ x = σ z ⋅ Ka
σx =σz ⋅ Kp
1 = tan 2 Ka = Nφ
σx = ρ ⋅ z⋅
45 −
φ 2
1 Nφ
σ x = ρ ⋅ z ⋅ tan
1 + Sinφ φ = tan 2 45 + 1 − Sinφ 2
K p = Nφ = tan
2
45 +
φ 2
σ x = ρ ⋅ z ⋅ Nφ 2
45 −
φ 2
σ x = ρ ⋅ z ⋅ tan
2
45 +
φ 2
KUMDA AKT F DURUMDA BASINÇ DA ILIMI 45 +
φ 2
#+0 +67 8 #8 7 8 9 +
H #
ρ φ
H
Pa
c=0
H/3
K a .ρ .H
1 K a . ρ .H 2 Pa = K a .ρ .H .H . = 2 2
KUMDA PAS F DURUMDA BASINÇ DA ILIMI 45 −
φ 2
# # H #
ρ φ
H
P
c=0
p
H/3
K p .ρ .H 2 1 K p .ρ .H Pp = K p .ρ .H .H . = 2 2
KUMDA AKT F ve PAS F DURUMDA BASINÇ DA ILIMLARI
Herhangi bir derinlikte zeminde efektif yatay gerilme
AKT F
Pa =
H
0
PAS F
Pp =
H
0
ph = ρ ′ ⋅ z ⋅ K
1 1 2 ph ⋅ dz = ⋅ ρ ′ ⋅ H ⋅ 2 Nφ
1 2 ′ ph ⋅ dz = ⋅ ρ ⋅ H ⋅ Nφ 2
Batık durumda toplam yatay kuvvet
Pxx = Pa , p + Pw Aktif durum
1 1 1 2 2 2 ′ Pa = ⋅ H ⋅ ρ ⋅ K a + ⋅ H ⋅ ρ w = ⋅ H ⋅ ( ρ ′ ⋅ K a + ρ w ) 2 2 2 Pasif durum
1 1 1 2 2 Pp = ⋅ H ⋅ ρ ′ ⋅ K P + ⋅ H ⋅ ρ w = ⋅ H 2 ⋅ ( ρ ′ ⋅ K P + ρ w ) 2 2 2
YAYILI YÜKÜN YANAL BASINCA ETK S (q)
ρ
ρ . ( H + h ') .Ka ρ . ( H + h ') .Kp ρ . ( H + h ') .Ko
q.K
ρ .H .K
K LDE AKT F BASINÇ Zeminde c (kohezyon) varsa bundan dolayı teorik olarak bir çekme gerilmesi alabilir denilebilir. Yani belli bir yere kadar bu zemin kendini desteksiz tutabilir. Kil z0 derinli e kadar çekme gerilmesi alıyor dolayısı ile 2×z0 derinli e kadar teorik olarak kazılabilir.
K LDE AKT F BASINÇ
σ1 = 0
−
X
σ3
KL
2c Nφ
2c Nφ
(-)
ρn
z0
H (-)
(+) X’
(+)
ρ n zK
σ 1 = σ 3 ⋅ N φ + 2 ⋅ c ⋅ Nφ yüzeyde
σ 1 = ρ n .z = 0
0 = σ 3 ⋅ Nφ + 2 ⋅ c ⋅ Nφ
σ3 = −
2⋅c Nφ
Kritik kazı derinli i
çekme bö lg esi
σ3 = 0
ρ ⋅ z0 = 0 + 2 ⋅ c ⋅ Nφ z0 =
H c = 2 ⋅ z0 =
2⋅c
ρn
4⋅c
ρn
⋅ Nφ
Nφ
K L N PAS F BASINCI 2 8 2 +6+ 2 : #
#" ; < 2 " =
>
8 ? +0 6+#
ρ (kN/m3 ) c
(kPa = kN/m 2 )
φ (o ) 45 −
φ 2
q.Nφ
ph = q⋅ Nφ + 2⋅ c ⋅ Nφ + ρ ⋅ H ⋅ Nφ
2c. Nφ
ρ .H .Nφ
( Nφ = Kp )
1 2 Pp = q ⋅ Nφ ⋅ H + 2 ⋅ c ⋅ Nφ ⋅ H + ⋅ ρ ⋅ H ⋅ Nφ 2
>
@
A
4 4 4 4
B ,
,
3
,
B
4 4
, B
4 1) Ka H=5 m
ρ d = 20 kN / m3 φ ′ = 30o c=0
1 − Sinφ 1 − Sin(30o ) 1 veya Ka = = = o 1 + Sinφ 1 + Sin(30 ) 3 30 1 φ K a = tan 2 (45 − ) = t an 2 (45 − ) = 2 2 3
2) Dü ey efektif gerilmeler (σ’z) ve su basınçları (Uw) Yüzeyde: σ z′ = 0
uw = 0
Tabanda: σ z′ = ρ ′.H o = (20 − 9.81) × 5 = 51 kPa uw = ρ w .H o = 9.81 × 5 = 49 kPa
3) Tabanda yanal efektif gerilme (σ’x)a
1 (σ x′ ) a = K a .σ z′ = × 51 = 17 kPa 3
0 2
,
5 2
C , D 5
9
D
5
H=5 m
17 kPa
1 2
BB
49 kPa
,
Pa = Psu basıncı + Ptoprak basıncı Pa =
1 1 1 1 ⋅ (σ x′ ) a ⋅ H o + ⋅ u ⋅ H o = ( × 17 × 5) + ( × 49 × 5) = 165 kN 2 2 2 2
2 2
BB
;
, BB
5
5 4
z = H o / 3 = 5 / 3 = 1.67 m
4 5
E
C
& F.
,
3
4
94 4
B
3
15 kPa
,
5
,
GWT
1) Ka, Kp
H=4 m
ρ d = 19 kN/m
φ 1 − sin φ 1 − sin 26 K a = tan 2 (45 − ) = = = 0.39 2 1 + sin φ 1 + sin 26 φ 1 + sin φ 1 + sin 26 = = 2.56 K p = tan 2 (45 + ) = 2 1 − sin φ 1 − sin 26
3
φ ′ = 26o c = 8 kPa
2) 43 (σ’z) Yüzeyde: σ z′ = 0 uw = 0 Tabanda: σ z′ = ρ ′.H o = (19 − 9.81) × 4 = 36.76 kPa uw = ρ w .H o = 9.81 × 4 = 39.24 kPa
(σ’x)a
3) C
(σ x′ ) a = K a .σ z′ = 0.39 × 36.76 = 14.34 kPa
4) Yüzeyde ve t
4 4
(σ’q)a
3
(σ q′ ) a = K a .q = 0.39 × 15 = 5.85 kPa
5) #
(σ’c)a
3 (σ c′ ) a = −
2c Nφ
=−
2×8 26 tan 2 (45 + ) 2
= 10 kPa
2 2
,
5
15 kPa
GWT
2 D
C , 5
2 24
(+)
(+)
9
D
5
#
H=4 m
14.34 kPa
5 C ,
5.85 kPa
(+) 39.24 kPa
BB
Pa = Pzemin + Pyayılı yük + Psu − Pkohezyon 1 1 ⋅ (σ x′ ) a ⋅ H o + (σ q′ ) a ⋅ H o + ⋅ u ⋅ H o − (σ c′ ) a ⋅ H o 2 2 1 1 Pa = ( × 14.34 × 4) + 5.85 × 4 + ( × 39.24 × 4) − 10 × 4 2 2 Pa = 28.68 + 23.4 + 78.48 − 40 = 90.56 kN Pa =
(-) -10 kPa
3 A
4 4 4
B
,
,
10 kPa SW H=3
ρ n = 17 kN/m3 GWT
SP H=3
′ = 32o φSW
ρ n = 19.31 kN/m3 ′ = 30o φSP
1) Kesitteki tüm zeminlerin Ka’ları
φ 1 − sin φ 1 − sin 32 ( K a ) SW = tan 2 (45 − ) = = = 0.307 2 1 + sin φ 1 + sin 32 φ 1 − sin φ 1 − sin 30 ( K a ) SP = tan 2 (45 − ) = = = 0.333 2 1 + sin φ 1 + sin 30
10 kPa
H=3
2 24
SW ρ n = 17 kN/m3 GWT
′ = 32 φSW SP ρ n = 19.31 kN/m3 o
H=3
′ = 30 φSP
% '$ -
2
o
2
C , D 5 9G
I
III
1
2
#
p1 = q × K aSW = 10 × 0.307 = 3.07 kPa p2 = q × K aSP = 10 × 0.333 = 3.33 kPa p3 = K aSW × ρ nSW × H SW = 0.307 × 17 × 3 = 15.66 kPa p4 = ρ nSW H SW × K aSP = 17 × 3 × 0.333 = 16.98 kPa ′ × H SP × K aSP = (19.31 − 9.81) × 3 × 0.333 = 9.50 kPa p5 = ρ SP p6 = ρW × HW = 9.81 × 3 = 29.43 kPa
9
D
5
3
IV
II
C , D 5 SP
4
V
VI 5
. "" (
6
#
PaI = 3.07 × 3 = 9.21 kN / m PaII = 3.33 × 3 = 9.99 kN / m 1 = 23.49 kN / m 2 = 16.98 × 3 = 50.94 kN / m
PaIII = 15.66 × 3 × PaIV
1 = 14.25 kN / m 2 1 = 29.43 × 3 × = 44.15 kN / m 2
PaV = 9.50 × 3 × PaVI
P = 152.05 kN / m
@ 9G
,
3
4 4
D &$
4 ,
5
#
6 *$
H
# , ,
( ρk = 19 kN / m3 , ρd = 22.5 kN / m3 ) X
L=30m
3 #
5
H=5m
> D
10 = B
σ
Kesme Kutusu Deney Sonuçları
τmax
m
5
@
113 229 tanφ = ≈ 150 300
Test No
1
2
σ
150
300
τ
113
229
φ = 37° s =τmax =σ ⋅ tanφ =σ ⋅ tan37
K 0 = 1 − Sinφ = 1 − sin 37 = 0.398 p0 = K 0 ⋅ z ⋅ ρ k = 0.398 ⋅ 5 ⋅ 19 = 37.81 kPa
( tabanda )
1 1 2 ⋅ ρ k ⋅ H ⋅ K 0 ⋅ B = ⋅ 19 ⋅ 52 ⋅ 0.398 ⋅ 10 = 945.3 kN P0 = 2 2
1 − Sinφ 1 45 − = = = tan 2 26.5 = 0.249 2 1 + Sinφ Nφ
pa = K a ⋅ z ⋅ ρ k = 0.249 ⋅ 5 ⋅ 19 = 23.66 kPa Pa =
( tabanda )
1 1 ⋅ ρ k ⋅ H 2 ⋅ K a ⋅ B = ⋅ 19 ⋅ 52 ⋅ 0.249 ⋅ 10 = 591.4 kN 2 2
c) K p = tan 2 45 +
φ 2
=
sw
67 )8 8 8 8 8
5m
b ) K a = tan
2
φ
63.5
23.66 kPa
1 + Sinφ = Nφ = tan 2 63.5 = 4.023 1 − Sinφ
p p = K p ⋅ z ⋅ ρ ′ + z ⋅ ρ w = 4.023 ⋅ 5 ⋅ 12.69 + 5 ⋅ 9.81 = 255.26 + 49.05 = 304.31 kPa ( taban ) Pp =
1 1 ⋅ ρ′⋅ H 2 ⋅ Kp ⋅ B + ⋅ ρw ⋅ H 2 ⋅ B 2 2
sw 5m
1 1 Pp = ⋅ 12.69 ⋅ 52 ⋅ 4.023 ⋅ 10 + ⋅ 9.81 ⋅ 52 ⋅ 10 2 2 Pp = 6381.5 + 1226.3 = 7607.8 kN
26.5
& 7 )8 8 % 6 8
255.26 kPa 49.05 kPa
3
4
A
B
I
BB
,
K a SP = tan 2 45 −
K a SW = tan 2 45 −
%
30 1 = 2 3
35 = 0.271 2
8:: 6
9
1 ⋅ 2 ⋅ 12 = 12 kN / m 2 1 Pa 2 = ⋅ (12 + 15.4 ) ⋅ 1 = 13.7 kN / m 2 Pa 3 = 12.5 ⋅ 3 = 37.5 kN / m Pa1 =
1 p1 = 18 ⋅ 2 ⋅ = 12 kPa 3 1 = 15.4 kPa 3 p3 = p4 = (18 ⋅ 2 + 10.19 ⋅ 1) ⋅ 0.271 = 12.5 kPa p2 = 18 ⋅ 2 + ( 20 − 9.81) ⋅ 1 ⋅
1 ⋅ 3 ⋅ 9.91 = 14.9 kN / m 2 1 Pa 5 = ⋅ 39.2 ⋅ 4 = 78.4 kN / m 2 P = 156.5 kN / m
Pa 4 =
p5 = 0
p6 = ( 22 − 9.81) ⋅ 3 ⋅ 0.271 = 9.91 kPa p7 = 0 p8 = 9.81 ⋅ 4 = 39.24 kPa
!
6
: *
6 ⋅ F = 12 × 3 + 1 +
8:: 6
) ;
M0 = 0
2 3 1 + 13.7 × 3.4 + 37.5 × 1.5 + 14.9 × + 78.4 × 4 × 3 3 3 F = 46 kN / m
3 JJ
4 4
, ,
! 3
ρ =22 2 d 0 φ=45
Batık GW
YASS
1
VI 4
2
IV
III
GW
3
Basınçlar
(-)
V
VIII
7 VII
5
Pa III = 2 ⋅ 9.75 = 19.5 kN / m
p5 = (18 − 9.81) ⋅ 3 ⋅ 0.171 = 4.20 kPa Nφ
=−
2 ⋅ 10 = −16.78 kPa 1.192
p7 = ( 22 − 9.81) ⋅ 2 ⋅ 0.171 = 4.17 kPa p8 = 9.81 ⋅ 5 = 49.05 kPa
K aCH = tan 2 45 −
10 = 0.704 2
Nφ CH = tan
45 +
10 = 1.192 2
Kuvvetler
p3 = 19 ⋅ 3 ⋅ 0.171 = 9.75 kPa
p4 = (18 − 9.81) ⋅ 3 ⋅ 0.704 = 17.30 kPa
45 = 0.171 2
8
p1 = 19 ⋅ 3 ⋅
1 = 18.98 kPa 3 p2 = 19 ⋅ 3 ⋅ 0.704 = 40.13 kPa
30 1 = 2 3
K aGW = tan 2 45 −
6
1 PaI = ⋅ 3 ⋅ 18.98 = 28.47 kN / m 2 PaII = 3 ⋅ 40.13 = 120.39 kN / m
2⋅c
BB
Su
kohezyon CH
I II
X
p6 = −
, K a SP = tan 2 45 −
ρd=18 kN/m3 3 φ=100 CH c=10 kPa kN/m3
,
Batık CH
Kuru kum
X
ρk=19 kN/m3 SP φ=300
5
1 ⋅ 3 ⋅ 17.30 = 25.95 kN / m 2 PaV = 2 ⋅ 4.20 = 8.40 kN / m PaIV =
PaVI = 3 ⋅ −16.78 = −50.34 kN / m 1 ⋅ 2 ⋅ 4.17 = 4.17 kN / m 2 1 = ⋅ 5 ⋅ 49.05 = 122.63 kN / m 2
PaVII = PaVIII
P = 279.17 kN / m
TEK L YÜK ETK S N N UZUNLAMASINA DE
M
σ b = σ h ⋅ cos 2 (1.1 ⋅ χ )
)." # * # !
# '$ ) ) / $ $! $
, (
TEK L ve Ç ZG YÜKLERDEN YANAL BASINÇ q: Çizgi Yük
x = m⋅H
q: çizgi yük Q: nokta yük
x a
z
z = n⋅H q m2 ⋅ n σ h = ph = 1.27 ⋅ ⋅ H ( m 2 + n 2 )2
σ h = ph = 0.203 ⋅
H
σh
q n ⋅ H ( 0.16 + n 2 )2
m > 0.4
m ≤ 0.4
Q: Nokta Yük
Q m2 ⋅ n2 σ h = ph = 1.77 ⋅ 2 ⋅ H ( m 2 + n 2 )3 Q n2 σ h = ph = 0.28 ⋅ 2 ⋅ H ( 0.16 + n 2 )3
m > 0.4 m ≤ 0.4
PROBLEM:
Yüzeydeki 1000 kN’luk Q (tekil) yükünden dolayı AA, BB (y=2m) ve CC (y=5m) dü ey eksenlerinde olu an yatay gerilme artı ını derinli e ba lı olarak hesaplayınız.
x H z n= H
x = m⋅H
m=
z = n⋅H
σ h = ph = 1.77 ⋅
Q m2 ⋅ n 2 ⋅ H 2 ( m 2 + n 2 )3
σ h = ph = 0.28 ⋅
Q n2 ⋅ H 2 ( 0.16 + n 2 )3
m ≤ 0.4
σh (kPa)
a) AA ekseni boyunca gerilme da ılı ı x 2 m= = = 0.286 ≤ 0.4 H 7
m > 0.4
0
( ikinci formül kullanılacak ) zemin özellikleri ve yer altı su seviyesi hesaplarda etkili de il!!! n2/(0.16+n2)3
0.28Q/H2
σh (kPa)
z (m)
n
0
0
0
0.160
0.0041
0
5.714
0
1
0.143
0.020
0.180
0.0058
3.448
5.714
19.7
2
0.286
0.082
0.242
0.0142
5.775
5.714
33.0
3
0.429
0.184
0.344
0.0407
4.521
5.714
25.8
4
0.571
0.327
0.487
0.1155
2.831
5.714
16.2
5
0.714
0.510
0.670
0.3008
1.696
5.714
9.7
6
0.857
0.735
0.895
0.7169
1.025
5.714
5.9
7
1
1
1.160
1.5609
0.641
5.714
3.7
2
Derinlik (m)
(0.16+n2)3
20
30
0
1
0.16+n2
n2
10
3
4
5
6
7
A
40
b) y=2 m (BB), y=5 m (CC) yatay uzaklıkta yatay gerilme artı ları σK
σ Q = σ h ⋅ Cos 2 (1.1 ⋅ψ )
y=5 m y=2 m
5 = 2.5 ψ 2 = 68.199° Cos 2 (1.1 ⋅ 68.199 ) = 0.067 2 0 2 2 0 tanψ 1 = = 1 ψ 1 = 45° Cos (1.1 ⋅ 45 ) = 0.422 2
tanψ 2 =
sh (kPa) 5
B
1
σh
B
σh
C
z (m)
0
1
2
3
4
5
6
7
(y=2 m) (kPa)
0
8.3
13.9
10.9
6.8
4.1
2.5
1.6
(y=5 m) (kPa)
0
1.3
2.2
1.7
1.1
0.7
0.4
0.3
Derinlik (m)
2
3
4
5
6
7
10
C
15
KISA SINAV SINAV--1
• A a ıdaki kesitte sürtünmesiz duvara etkiyen toplam aktif kuvveti hesaplayıp, bu kuvvetin etkime noktasını bulunuz. Nφ =
1 + Sinφ φ = tan2 45 + 1 − Sinφ 2
Ka =
1 Nφ
K p = Nφ
20 kPa SP ρn=16 kN/m3 φ= 300
2m YASS
CI ρd=17.81 kN/m3 φ=100 c=12 kPa
4m
Sonuçları Ka virgülden sonra 3 hane, yükler ve basınçlar için ise 2 hane olacak ekilde yuvarlayınız.
KISA SINAV SINAV--1 (Çözüm) 20 kPa SP YASS
CI
ρn=16 kN/m3 φ= 300
2m
ρd=17.81 4m kN/m3 0 φ=10 c=12 kPa
YÜK
SP
I 1
II (+)
2
III
CI
3
IV (+)
CIKOHEZYON
SU
4
V
VI 5
(+)
6
(-)
1 + Sinφ 1+ Nφ = 1 − Sinφ
1 1 − Sinφ 1 KaSP = = = Nφ 1 + Sinφ 3 1 1 − Sinφ KaCI = = = 0.704 Nφ 1 + Sinφ
VII (+)
7
KISA SINAV SINAV--1 (Çözüm) 20 kPa SP YASS
ΣP x=?
CI
ρn=16 kN/m3 φ= 300
YÜK
2m
I 1
ρd=17.81 4m kN/m3 0 φ=10 c=12 kPa
SP
2
II (+)
III
CI
3
p3 = ρ SP H SP KaSP
IV (+)
p4 = ρCI H SP KaSP p5 = ρCI′ H CI KaCI = (17.81 − 9.81) ⋅ 4 ⋅ 0.704 = 22.53kPa p6 = −
2c Nφ
=−
2 × 12 1.42
= −20.17 kPa
p7 = ρ w H w = 9.81 ⋅ 4 = 39.24kPa
SU
4
1 p1 = q ⋅ KaSP = 20 = 6.67 kPa 3 p2 = q ⋅ KaCI = 20 ⋅ 0.704 = 14.08kPa 1 = 16 ⋅ 2 ⋅ = 10.67kPa 3 = 16 ⋅ 2 ⋅ 0.704 = 22.53kPa
CIKOHEZYON
V (+)
VI 5
6
(-)
VII
7
(+)
PI = 6.67 ⋅ 2 = 13.34 kN / m PII = 14.08 ⋅ 4 = 56.32 kN / m 10.67 ⋅ 2 = 10.67 kN / m 2 PIV = 22.53 ⋅ 4 = 90.12 kN / m PIII =
22.53 ⋅ 4 = 45.06 kN / m 2 PVI = −20.17 ⋅ 4 = −80.68 kN / m PV =
39.24 ⋅ 4 = 78.48 kN / m 2 ΣP = 213.43 kN / m PVII =
203.87 kN / m ⋅ X = 5 ⋅13.34 + 2 ⋅ 56.32 + 4.67 ⋅10.67 + 2 ⋅ 90.12 + 1.33 ⋅ 45.06 − 2 ⋅ 80.68 + 1.33 ⋅ 78.48 X = 1.93 m