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ττττττττ
ZEM�NDE DO�AL DURUMZEM�NDE DO�AL DURUM
σσσσσσσσ
�������������� ������ ������������������������� ���� � ��� �� ��������������� ��� ������������������������������������ ����������� ���� � ����� �� �������������������������
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0 = x
z
Kσσ
0 1=
−K
νν
0( zeminde 3)<K
SÜKUNETTE TOPRAK BASINCI KATSAYISI SÜKUNETTE TOPRAK BASINCI KATSAYISI (K(K00))
Etki Eden Faktörler:Etki Eden Faktörler:Dane da�ılımı (Dmax, Cu, Cr)Kayma direnci açısı (φφφφ)Kumda çimentolanma A�ırı konsolidasyon (OCR)
Kilin aktivitesi
Jeolojik ko�ullar (birikme, olu�ma, tektonizma)Drenaj ko�ulları (k)
( )%= Pc
IA C
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Zemin CinsiZemin Cinsi KK00 ReferansReferans
Temiz Kum (NL) 1-Sinφ� (Jaky, 1945)
Kum (OC) KoNL × OCRSinφ� (Mayne ve Kulhawy, 1982)
NL Kil 0.19+0.23×log IP
OC Kil 0.7+0.1(OCR-1.2)
SÜKUNETTE TOPRAK BASINCI KATSAYISI SÜKUNETTE TOPRAK BASINCI KATSAYISI (K(K00))
Kum sıkı K0=0.37, gev�ek K0=0.46Kil, yu�rulmu� K0=0.69�2.00�eker K0=0.50
OC Kil 0.7+0.1(OCR-1.2)
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KK0 0 için Tipik De�erleriçin Tipik De�erler0 1
=−
Kν
νν �!� �����" ���
Zemin CinsiZemin Cinsi νννννννν için Tipik De�erleriçin Tipik De�erler KK00
KilKilDoygun 0.40-0.50 0.67-1.00
Doygun Olmayan 0.10-0.30 0.11-0.42Doygun Olmayan 0.10-0.30 0.11-0.42
Kumlu KilKumlu Kil 0.20-0.30 0.25-0.42
SiltSilt 0.30-0.35 0.42-0.54
KumKumSıkı 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
KayaKaya 0.10-0.40 0.11-0.67
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5 19 4 9.81 55.76′ = ⋅ − ⋅ =v kPaσ
0
0
1 1 28 0.531
55.76 0.531 29.61
= − = − =′ ′= ⋅ = ⋅ =
NC
h z
K Sin Sin
K kPa
φσ σ
280 0
0
0.531 5 1.13
55.76 1.13 63.01
= ⋅ = ⋅ =′ ′= ⋅ = ⋅ =
OC NC Sin Sin
h z
K K OCR
K kPa
φ
σ σ
������������ ����� ��������������������� ����� ��������� � ����������� ������ ���� ��� ����������� ������ ���� ��
ÖRNEK: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).
55.76 29.6113.08
2−= =NC kPaτ 55.76 63.01
3.632−= = −OC kPaτ
1 3
2
′ ′−=
σ στ
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ττττττττ KUMKUM
KUMDA ELAST�K DENGEKUMDA ELAST�K DENGE
kum
Temiz Kum (φ�300)
K0 = ( 1 - Sin φ )
K0<1 � σz sa� tarafta
σσσσσσσσσσσσσσσσzzσσσσσσσσhh
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K�LDE ELAST�K DENGEK�LDE ELAST�K DENGE
ττττττττ K�LK�L
K�L
���� ����� � � �#$%&��������� #$ $�&'($�)*�����+,
� ��� �� ��� � � �#$-&� � #$ $�.($�& �" /01&�)�
σσσσσσσσσσσσσσσσhh σσσσσσσσzz σσσσσσσσhh
���� �����
� ��� �� ��� �#$-&
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2����� �� � �� ���� � �
��3��4�
ZEM�NDE EK GER�LME DURUMUZEM�NDE EK GER�LME DURUMU
2����� ��
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45+φ/2
ττττττττ
σσσσσσσσσh=K0σz σz=ρ.z
SükunetteSükunettec
φφφφ
ττττττττ
AktifAktif
φφφφ
45+φ/2
45-φ/2
σσσσσσσσσh=Kaσz σz=ρ.z
AktifAktif
c
σσσσσσσσσh=Kpσz
σz=ρ.z
PasifPasif
c φφφφ45-φ/2
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� ����� �! ���� " #� ! $� ����� �! ���� " #� ! $
PLAST�K DENGE PROBLEMLER�PLAST�K DENGE PROBLEMLER�
%��&�� ' ( ��) ��* ��) �����! ��# %��&�� ' ( ��) ��* ��) �����! ��#
σσσσσσσσσh=Kpσz
σz=ρ.z
PasifPasif
c φφφφ45-φ/2
σh=Kaσz
AktifAktif
SükunetteSükunette
σh=K0σz
AktifAktif
PasifPasif
SükunetteSükunette
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� � ( � + $! �* , �,
% ��( � &#� �%� ' $��$
C. C. PLAST�K PLAST�K DENGE DENGE
%� ' $��$
����� ! � -) . #� ��$�$/ $
DENGE DENGE PROBLEMLER�PROBLEMLER�
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PLAST�K DENGEPLAST�K DENGE
AKT�FAKT�F--PAS�F TOPRAK BASINÇLARIPAS�F TOPRAK BASINÇLARI
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PLAST�KPLAST�KDENGEDENGE
σσσσ3 σσσσ3
σσσσ1
σσσσ1
ττττ
��OBOB�=Yenilme Yok�=Yenilme Yok B=YenilmeB=Yenilme
cσ3 σ1
φττ
σ σ
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KIRILMA (KIRILMA (yenilme veya akma)’yenilme veya akma)’ NIN TEMS�L�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
max f( )σ = τ = σ τ
σ1σ3σ
φ
c
c tanτ = + σ ⋅ φ
1 3N 2c Nφ φσ = σ +
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AKT�F TOPRAK BASINCIAKT�F TOPRAK BASINCI
. .=a aP z Kρ
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AKT�F TOPRAK BASINCININ OLU�UMUAKT�F TOPRAK BASINCININ OLU�UMU
σ1
σ1
σ3σ3
τ
φh
0.001h gev�ek kum0.04 h yumu�ak kil
σσσσσσσσoo= �4�������� �������σσσσσσσσaa= ��� ������������= �4�������� ���,��������5���������������= ��� ����,��������5�����������
σσ1σ3=σa σ3=σo
σ3=Kaρ’z σ3=Koρ’z σ1=ρ’z
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PAS�F TOPRAK BASINCIPAS�F TOPRAK BASINCI
. .=p pP z Kρ
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PAS�F TOPRAK BASINCININ OLU�UMUPAS�F TOPRAK BASINCININ OLU�UMU
σσ
σ1
h
0.01 h gev�ek kum 0.05 h yumu�ak kil
σσσσσσσσoo= �4�������� �������σσσσσσσσPP= ,�� ��������KKoo= �4�������� ���,��������5�����������KKpp= ,�� ����,��������5�����������
τ
σ1
σ3σ3
σσ1 σ3=σpσ3=σo
φ
τ
σ3=Kpρ’zσ3=Koρ’z σ1=ρ’z
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TOPRAK BASINCININ OLU�UMU için TOPRAK BASINCININ OLU�UMU için GEREKL� HAREKET M�KTARIGEREKL� HAREKET M�KTARI
Kp ye gelebilmek için yakla�ık Ka nın 4 katı bir hareket gerekiyor
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45 / 2+φ45 /2−φ
tan=τ σ φ
τKUMDA PLAST�K DENGE KUMDA PLAST�K DENGE
1 3
2
1tan 5
41 2
+ � �= =
=
+ �− �
⋅
��
SinN
N
Sin
φ
φ
σφ
σφφ
AKT�FAKT�F PAS�FPAS�F
2
2
1tan 45
2
1
tan 452
= ⋅
� �= = −� �� �
= ⋅ ⋅
� �= ⋅ ⋅ −� �� �
x z a
a
x
x
K
KN
zN
z
φ
φ
σ σφ
σ ρ
φσ ρ
2
2
tan 452
tan 452
= ⋅
� �= = +� �� �
= ⋅ ⋅
� �= ⋅ ⋅ +� �� �
x z p
p
x
x
K
K N
z N
z
φ
φ
σ σφ
σ ρφσ ρ
σ
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KUMDA KUMDA AKT�FAKT�F DURUMDA BASINÇ DA�ILIMIDURUMDA BASINÇ DA�ILIMI
H H
452
+ φ
ρP
#+0+67 8#8 7 8 9 +
0=c
ρφ
. .aK Hρ
Pa
H/3
2. .1. . . .
2 2= = a
a a
K HP K H H
ρρ
#�
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KUMDA KUMDA PAS�FPAS�F DURUMDA BASINÇ DA�ILIMIDURUMDA BASINÇ DA�ILIMI
HH
452
− φ
ρ
#����#����
H
0=c
φ
. .pK Hρ
Pp
H/3
2. .1. . . .
2 2= = p
p p
K HP K H H
ρρ
#�
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Herhangi bir derinlikte zeminde efektif yatay gerilme � ′= ⋅ ⋅p z Kρ
KUMDA AKT�F ve PAS�F DURUMDA BASINÇ DA�ILIMLARIKUMDA AKT�F ve PAS�F DURUMDA BASINÇ DA�ILIMLARI
Herhangi bir derinlikte zeminde efektif yatay gerilme � ′= ⋅ ⋅hp z Kρ
2
0
1 12
′= ⋅ = ⋅ ⋅ ⋅�H
a hP p dz HNφ
ρ
2
0
12
′= ⋅ = ⋅ ⋅ ⋅�H
p hP p dz H Nφρ
AKT�FAKT�F
PAS�FPAS�F
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Batık durumda toplam yatay kuvvetBatık durumda toplam yatay kuvvet
,= +xx a p wP P P
Aktif durumAktif durum
( )2 2 21 1 12 2 2
′ ′= ⋅ ⋅ ⋅ + ⋅ ⋅ = ⋅ ⋅ ⋅ +a a w a wP H K H H Kρ ρ ρ ρ( )2 2 2
= ⋅ ⋅ ⋅ + ⋅ ⋅ = ⋅ ⋅ ⋅ +a a w a wP H K H H Kρ ρ ρ ρ
Pasif durumPasif durum
( )2 2 21 1 12 2 2
′ ′= ⋅ ⋅ ⋅ + ⋅ ⋅ = ⋅ ⋅ ⋅ +p P w P wP H K H H Kρ ρ ρ ρ
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ρ
YAYILI YÜKÜN YANAL BASINCA ETK�S�YAYILI YÜKÜN YANAL BASINCA ETK�S�(q)(q)
( ). ' .+H h Kaρ( ). ' .+H h Kpρ( ). ' .+H h Koρ
.q K . .H Kρ
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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
K�LDE AKT�F BASINÇK�LDE AKT�F BASINÇ
kadar çekme gerilmesi alıyor dolayısı ile 2×z0 derinli�e kadar teorik olarak kazılabilir.
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K�LDE AKT�F BASINÇK�LDE AKT�F BASINÇX
X’
1 0=σ
3σ
H
nρK�LK�L
( + )( + ) ( ( -- ))
( ( -- ))
( + )( + )n zKρ
2− cNφ
2cNφ
z0
X’n zKρ
1 3 2= ⋅ + ⋅ ⋅N c Nφ φσ σ
1
3
3
. 0
0 2
2
� = =
= ⋅ + ⋅ ⋅
⋅= −
nyüzeyde z
N c N
c
N
φ φ
φ
σ ρ
σ
σ
3
0
0
lg 0
0 2
2
� =
⋅ = + ⋅ ⋅
⋅= ⋅n
çekme bö esi
z c N
cz N
φ
φ
σ
ρ
ρ
0
4Kritik kazı derinli�i 2
⋅� = ⋅ =c
n
cH z Nφρ
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3
2
o
(kN/m )
(kPa = kN/m )
( )
c
ρ
φ
28 2+6+�2: # #" ; <�2" = > ��8 ? +06+#
K�L�N PAS�F BASINCIK�L�N PAS�F BASINCI
( )2 = ⋅ + ⋅ ⋅ + ⋅ ⋅ =h pp q N c N H N N Kφ φ φ φρ21
22
= ⋅ ⋅ + ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅pP q N H c N H H Nφ φ φρ
452
− φ
.q Nφ 2 .c Nφ . .H Nφρ
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> ���@A �� ���� ��4�4��� ����B�������3������ ����,���������������B����,����������4�4�����,����,���������4�4����� ��������������� ��� � ��������B ��� ��4�������
320 /
300
=′ ==
d
o
kN m
c
ρφ
H=5 m
1) Ka
2 2
1 1 (30 ) 1 veya
1 1 (30 ) 330 1
tan (45 ) an (45 )
− −= = =+ +
= − = − =
o
a o
a
Sin SinK
Sin Sin
K t
φφ
φtan (45 ) an (45 )
2 2 3= − = − =aK t
2) Dü�ey efektif gerilmeler (σ’z) ve su basınçları (Uw)
0 0
. (20 9.81) 5 51
.
Tabanda:
Yüzeyd
9.
8
e
1 5 49
: ′ = =′ ′= = − × =
= = × =
z w
z o
w w o
u
H kPa
u H kPa
σσ ρ
ρ
3) Tabanda yanal efektif gerilme (σ’x)a�1
( ) . 51 173
′ ′= = × =x a a zK kPaσ σ
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0� 2�������,��������5������������
H=5 m
17 kPa 49 kPa
2�����C �,��D����5�
9 ��D����5�
1� 2�������BB���� ����,���1� 2�������BB���� ����,���
su basıncı toprak basıncı= +aP P P
1 1 1 1( ) ( 17 5) ( 49 5) 165 kN
2 2 2 2′= ⋅ ⋅ + ⋅ ⋅ = × × + × × =a x a o oP H u Hσ
2� 2�������BB�� ����� ����
; ����������,��������5����������������5��� ��������4�������������E���� � � � ����,�����������BB�� ����� ���� �4���� ������������� ��������5�����
/ 3 5/ 3 1.67 m= = =oz H �C ��������&�F.����������
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3 ���43 ���49 4�4��� ����B�������3������ ����,��������5��������,��������
1) Ka, Kp
2
2
1 sin 1 sin 26tan (45 ) 0.39
2 1 sin 1 sin 261 sin 1 sin 26
tan (45 ) 2.562 1 sin 1 sin 26
− −= − = = =+ ++ += + = = =− −
a
p
K
K
φ φφ
φ φφ
2) �43�������� ���� ���(σ’z)
319 kN/m
268 kPa
=′ ==
d
o
c
ρφ
H=4 m
GWT15 kPa
0 0
. (19 9.81) 4 36.76 kPa
Yüzeyde:
Taba
nd
. 9.81 4 39.
a:
24 kPa
′ = =′ ′= = − × =
= = × =
z w
z o
w w o
u
H
u H
σσ ρ
ρ3) C ������������������ ���� �� (σ’x)a
( ) . 0.39 36.76 14.34 kPa′ ′= = × =x a a zKσ σ
4) Yüzeyde ve t���������������4�4����� � �������3�������������� ���� �� (σ’q)a
( ) . 0.39 15 5.85 kPa′ = = × =q a aK qσ
5) # � ������������������3��������������� � (σ’c)a
2
2 2 8( ) 10 kPa
26tan (45 )
2
×′ = − = − =+
c a
c
Nφ
σ
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2�2� 2�������,��������5����������� �������
H=4 m
14.34 kPa 39.24 kPa
2�����C �,���D����5� 9 ��D����5�
GWT15 kPa
5.85 kPa
2�����2 4� #�������
-10 kPa
(+)(+)(+)(+) (+)(+) ((--))
5�5� C �,�����������BB��
zemin su= + + −a kohezyonyayılı yükP P P P P
1 1( ) ( ) ( )
2 21 1
( 14.34 4) 5.85 4 ( 39.24 4) 10 42 2
28.68 23.4 78.48 40 90.56 kN
′ ′ ′= ⋅ ⋅ + ⋅ + ⋅ ⋅ − ⋅
= × × + × + × × − ×
= + + − =
a x a o q a o o c a o
a
a
P H H u H H
P
P
σ σ σ
39.24 kPa5.85 kPa -10 kPa
![Page 33: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/33.jpg)
3 ���43 ���4A �� ���� ��4�4��� ����B������� ����,�������������������,��������
317 kN/m
32
=′ =n
oSW
ρφ
H=3GWT
10 kPa
SW
SP 319.31 kN/m=nρH=3
1) Kesitteki tüm zeminlerin Ka’ları
2
2
1 sin 1 sin 32( ) tan (45 ) 0.307
2 1 sin 1 sin 321 sin 1 sin 30
( ) tan (45 ) 0.3332 1 sin 1 sin 30
− −= − = = =+ +− −= − = = =+ +
a SW
a SP
K
K
φ φφ
φ φφ
30′ =n
oSPφ
H=3
![Page 34: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/34.jpg)
317 kN/m
32
=′ =n
oSW
ρφ
H=3GWT
10 kPa
SW
SP319.31 kN/m
30
=′ =n
oSP
ρφ
H=3
2�����C �,��D����5�9 G
9 ��D����5�2�����2 4�
II
IIII
1
2
IIIIII
IVIV
3
4VIVIVV
65
2�����C �,��D����5�
SP
%� ' $�-�� #%� ' $�-�� # �. " " �( ��#�. " " �( ��#
1
2
3
4
5
6
10 0.307 3.07
10 0.333 3.33
0.307 17 3 15.66
17 3 0.333 16.98
(19.31 9.81) 3 0.333 9.50
9.81 3 29.43
= × = × =
= × = × =
= × × = × × =
= × = × × =′= × × = − × × =
= × = × =
aSW
aSP
aSW nSW SW
nSW SW aSP
SP SP aSP
W W
p q K kPa
p q K kPa
p K H kPa
p H K kPa
p H K kPa
p H kPa
ρρρρ
%� ' $�-�� #%� ' $�-�� # �. " " �( ��#�. " " �( ��#3.07 3 9.21 /
3.33 3 9.99 /
115.66 3 23.49 /
216.98 3 50.94 /
19.50 3 14.25 /
21
29.43 3 44.15 /2
= × =
= × =
= × × =
= × =
= × × =
= × × =
aI
aII
aIII
aIV
aV
aVI
P kN m
P kN m
P kN m
P kN m
P kN m
P kN m
152.05 / =P kN m
![Page 35: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/35.jpg)
> ���@�D �9 G ��������,���������������������� �����3������ ��������������3��#��H���� �� �����4�4��� ����������D &$���6 *$����������������#�,�������������������4������������������������� ���5���������� �������� ����,�� �����������5������,��� �� � ��������� 3 3( 19 / , 22.5 / )= =k dkN m kN mρ ρ
L=30m
H=5
m
B=10m
X
τmax
σKesme Kutusu Deney Sonuçları
Test No 1 2
σ 150 300
τ 113 229
L=30mB=10m
#���������� ��5 ������@max
113 229tan 37 tan tan37
150 300= ≈ � = ° = = ⋅ = ⋅sφ φ τ σ φ σ
( )0
0 0
2 20 0
1 1 sin 37 0.398
0.398 5 19 37.81 tabanda
1 119 5 0.398 10 945.3
2 2
= − = − =
= ⋅ ⋅ = ⋅ ⋅ =
� �= ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ =� �� �
k
k
K Sin
p K z kPa
P H K B kN
φρ
ρ
���
![Page 36: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/36.jpg)
( )
2 2
2 2
1 1tan 45 tan 26.5 0.249
2 1
0.249 5 19 23.66 tabanda
1 119 5 0.249 10 591.4
2 2
b ) −� �= − = = = =� � +� �
= ⋅ ⋅ = ⋅ ⋅ =
� �= ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ =� �� �
a
a a k
a k a
SinK
Sin N
p K z kPa
P H K B kN
φ
φ φφ
ρ
ρ
( )
2 21tan 45 tan 63.5 4.023
2 1)
+� �= + = = = =� � −� �p
SinK N
Sinc φ
φ φφ
sw
5 m
63.523.66 kPa
� 6 7) 8�8����8�8��8��
( )2 2
2 2
4.023 5 12.69 5 9.81 255.26 49.05 304.31 taban
1 12 2
1 112.69 5 4.023 10 9.81 5
2 2
′= ⋅ ⋅ + ⋅ = ⋅ ⋅ + ⋅ = + =
� � � �′= ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅� � � �� � � �
� �= ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅� �� � �
p p w
p p w
p
p K z z kPa
P H K B B
P
H
ρ ρ
ρ ρ
10
6381.5 1226.3 7607.8
⋅ �= + =pP kN
sw
5 m
26.5255.26 kPa49.05 kPa
&�� 7) 8�8���%�6���8��
![Page 37: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/37.jpg)
3 ���43 ���4A �� ���� ���B������ ���������������������� � ������� �I ���BB�� � �����,��������
2 30 1tan 45
2 3� �= − =� �� �
a SPK
2 35tan 45 0.271
2� �= − =� �� �
a SWK
%����9���%����9��� �8::�6�����8::�6����%����9���%����9���
( )
( )
( )
1
2
3 4
5
6
7
8
118 2 12
31
18 2 20 9.81 1 15.43
18 2 10.19 1 0.271 12.5
0
22 9.81 3 0.271 9.91
0
9.81 4 39.24
= ⋅ ⋅ =
= ⋅ + − ⋅ ⋅ = �� �
= = ⋅ + ⋅ ⋅ ==
= − ⋅ ⋅ ==
= ⋅ =
p kPa
p kPa
p p kPa
p
p kPa
p
p kPa
( )
1
2
3
4
5
12 12 12 /
21
12 15.4 1 13.7 /212.5 3 37.5 /
13 9.91 14.9 /
21
39.2 4 78.4 /2156.5 /
= ⋅ ⋅ =
= ⋅ + ⋅ =
= ⋅ =
= ⋅ ⋅ =
= ⋅ ⋅ =
=
a
a
a
a
a
P kN m
P kN m
P kN m
P kN m
P kN m
P kN m
! ����6����:��* ���� ��8::�6 ��) �;�� ! ����6����:��* ���� ��8::�6 ��) �;�� 0 0 =M
2 3 16 12 3 1 13.7 3.4 37.5 1.5 14.9 78.4 4
3 3 3� �⋅ = × + + + × + × + × + × ×� �� �
F
� 46 /� =F kN m
![Page 38: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/38.jpg)
3 ���43 ���4J J ��4��� ������� ����,��������5��� ����������� � ,���,������ ����BB�� ��!�� ����,��������
2
2
2
30 1tan 45
2 3
45tan 45 0.171
2
10tan 45 0.704
2
10tan 45 1.192
2
� �= − =� �� �
� �= − =� �� �
� �= − =� �� �
� �= + =� �� �
a SP
aGW
aCH
CH
K
K
K
Nφ
BasınçlarBasınçlar KuvvetlerKuvvetler
ρk=19 kN/m3
φ=300
ρd=22 kN/m3
φ=450
ρd=18 kN/m3
φ=100
c=10 kPa
SP
GW
YASS
3
3
2
CH
X
X
2
1
3
4
5
6
78
I
II
III
IV
V
VI
VIIVIII
Kuru kum
BatıkCH
kohezyonCH
BatıkGW Su
( - )
BasınçlarBasınçlar
( )( )
( )
1
2
3
4
5
6
7
8
119 3 18.98
319 3 0.704 40.13
19 3 0.171 9.75
18 9.81 3 0.704 17.30
18 9.81 3 0.171 4.20
2 2 1016.78
1.192
22 9.81 2 0.171 4.17
9.81 5 49.05
= ⋅ ⋅ =
= ⋅ ⋅ =
= ⋅ ⋅ =
= − ⋅ ⋅ =
= − ⋅ ⋅ =⋅ ⋅= − = − = −
= − ⋅ ⋅ =
= ⋅ =
p kPa
p kPa
p kPa
p kPa
p kPa
cp kPa
N
p kPa
p kPa
φ
KuvvetlerKuvvetler1
3 18.98 28.47 /23 40.13 120.39 /
2 9.75 19.5 /
13 17.30 25.95 /
22 4.20 8.40 /
3 16.78 50.34 /
12 4.17 4.17 /
21
5 49.05 122.63 /2
= ⋅ ⋅ =
= ⋅ =
= ⋅ =
= ⋅ ⋅ =
= ⋅ =
= ⋅ − = −
= ⋅ ⋅ =
= ⋅ ⋅ =
aI
aII
a III
aIV
aV
aVI
aVII
aVIII
P kN m
P kN m
P kN m
P kN m
P kN m
P kN m
P kN m
P kN m
279.17 / =P kN m
![Page 39: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/39.jpg)
( )2cos 1.1= ⋅ ⋅b hσ σ χ
TEK�L YÜK ETK�S�N�N UZUNLAMASINA DE����M�TEK�L YÜK ETK�S�N�N UZUNLAMASINA DE����M�
) . " � # �� #�� ' $�) � ��, �( ���) . " � # �� #�� ' $�) � ��, �( ���* �# �! ��) � / $�$! $* �# �! ��) � / $�$! $
![Page 40: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/40.jpg)
q:q: çizgi yükQ:Q: nokta yük
= ⋅= ⋅
x m H
z n H
TEK�L ve Ç�ZG� YÜKLERDEN YANAL BASINÇTEK�L ve Ç�ZG� YÜKLERDEN YANAL BASINÇ
q:q: Çizgi YükÇizgi Yük
( )2
22 21.27 0.4
0.203 0.4
⋅= = ⋅ ⋅ >+
= = ⋅ ⋅ ≤
h h
q m np m
H m n
q np m
σ
σ
z
xa
( )
( )
2 2
2 32 2
2
2 32
1.77 0.4
0.28 0.40.16
⋅= = ⋅ ⋅ >+
= = ⋅ ⋅ ≤+
h h
h h
Q m np m
H m n
Q np m
H n
σ
σ
Q:Q: Nokta YükNokta Yük
( )220.203 0.4
0.16= = ⋅ ⋅ ≤
+h h
q np m
H nσ
hσ
H
![Page 41: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/41.jpg)
PROBLEM: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.
= ⋅ � =
= ⋅ � =
xx m H m
Hz
z n H nH
( )
( )
2 2
2 32 2
2
2 32
1.77 0.4
0.28 0.40.16
⋅= = ⋅ ⋅ >+
= = ⋅ ⋅ ≤+
h h
h h
Q m np m
H m n
Q np m
H n
σ
σ
a)a) AA ekseni boyunca gerilme da�ılı�ı2
0.286 0.47
= = = ≤xm
H( ikinci formül kullanılacak )
zemin özellikleri ve yer altı su seviyesi hesaplarda etkili de�il!!!
z (m) n n2 0.16+n2 (0.16+n2)3 n2/(0.16+n2)3 0.28Q/H2 σh (kPa)
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
0
1
2
3
4
5
6
7
0 10 20 30 40
σσσσh (kPa)
Der
inlik
(m)
AA
![Page 42: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/42.jpg)
b)b) y=2 m (BB), y=5 m (CC) yatay uzaklıkta yatay gerilme artı�ları σK
( )2 1.1= ⋅ ⋅Q h Cosσ σ ψ
y=2 m y=2 m ( )21 1
2tan 1 45 1.1 45 0.422
2� = = � = ° � ⋅ =Cosψ ψ
y=5 m y=5 m ( )22 2
5tan 2.5 68.199 1.1 68.199 0.067
2� = = � = ° � ⋅ =Cosψ ψ
00 5 10 15
sh (kPa)
2
Bhσ
Chσ
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
1
2
3
4
5
6
7D
erin
lik (
m)
BB
CC
![Page 43: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/43.jpg)
KISA SINAVKISA SINAV--11• A�a�ıdaki kesitte sürtünmesiz duvara etkiyen
toplam aktif kuvveti hesaplayıp, bu kuvvetin etkime noktasını bulunuz.
21 SinN tan 45
1 Sin 21
K
φ+ φ φ� �= = +� �− φ � �
= ρρρρn=16 kN/m3 2 m
20 kPa
SPSP
a
p
1K
N
K Nφ
φ
=
=
ρρρρρρρρdd=17.81 kN/m=17.81 kN/m33
φφφφφφφφ=10=1000
c=12 kPac=12 kPa
ρρρρn=16 kN/m3
φφφφ= 300
YASS
2 m
4 m
CICI
Sonuçları Ka virgülden sonra 3 hane, yükler ve basınçlar için ise2 hane olacak �ekilde yuvarlayınız.
![Page 44: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/44.jpg)
KISA SINAVKISA SINAV--1 (Çözüm)1 (Çözüm)
1+ Sinφ
ρd=17.81 kN/m3
φ=100
c=12 kPa
ρn=16 kN/m3
φ= 300YASS2 m
4 m
20 kPa
SPSP
CICI
I
II IV VI
III
V VII
12
34
5 6 7
YÜK SP CI CIKOHEZYON SU
( + ) ( + ) ( + ) ( - ) ( + )
11
+=−
SinN
Sinφφφ
1 1 11 3
1 10.704
1
−= = =+
−= = =+
SP
CI
SinKa
N Sin
SinKa
N Sin
φ
φ
φφφφ
![Page 45: NM405_03](https://reader030.fdocuments.net/reader030/viewer/2022013123/568c4a731a28ab49169835a4/html5/thumbnails/45.jpg)
KISA SINAVKISA SINAV--1 (Çözüm)1 (Çözüm)
120 6.67= ⋅ = =p q Ka kPa
ρd=17.81 kN/m3
φ=100
c=12 kPa
ρn=16 kN/m3
φ= 300YASS2 m
4 m
20 kPa
SPSP
CICI
I
II IV VI
III
V VII
12
34
5 6 7
YÜK SP CI CIKOHEZYON SU
( + ) ( + ) ( + ) ( - ) ( + )
6.67 2 13.34 /= ⋅ =IP kN m
x=?
ΣP
1
2
3
4
5
6
7
120 6.67
320 0.704 14.08
116 2 10.67
316 2 0.704 22.53
(17.81 9.81) 4 0.704 22.53
2 2 1220.17
1.42
9.81 4 3
= ⋅ = =
= ⋅ = ⋅ =
= = ⋅ ⋅ =
= = ⋅ ⋅ =′= = − ⋅ ⋅ =
×= − = − = −
= = ⋅ =
SP
CI
SP SP SP
CI SP SP
CI CI CI
w w
p q Ka kPa
p q Ka kPa
p H Ka kPa
p H Ka kPa
p H Ka kPa
cp kPa
N
p Hφ
ρ
ρρ
ρ 9.24kPa
6.67 2 13.34 /
14.08 4 56.32 /
10.67 210.67 /
222.53 4 90.12 /
22.53 445.06 /
220.17 4 80.68 /
39.24 478.48 /
2213.43 /
= ⋅ == ⋅ =
⋅= =
= ⋅ =⋅= =
= − ⋅ = −⋅= =
Σ =
I
II
III
IV
V
VI
VII
P kN m
P kN m
P kN m
P kN m
P kN m
P kN m
P kN m
P kN m
203.87 / 5 13.34 2 56.32 4.67 10.67 2 90.12 1.33 45.06 2 80.68 1.33 78.481.93
⋅ = ⋅ + ⋅ + ⋅ + ⋅ + ⋅ − ⋅ + ⋅� =
kN m X
X m