NM405_03

45
τ τ τ τ τ ZEMNDE DOAL DURUM ZEMNDE DOAL DURUM σ σ σ σ σ

description

ZEMNDE DOAL DURUMZEMNDEDOALDURUM σσσσσσσσ ττττττττ 0 = x ν ν ( zeminde 3) < K K % = P Dane daılımı (D max , C u , C r ) Kayma direnci açısı (φ φφφ ) Kumda çimentolanma Aırı konsolidasyon (OCR) σ σ Etki Eden Faktörler:EtkiEdenFaktörler: 0 0 z c Zemin CinsiZeminCinsi KK 00 ReferansReferans Temiz Kum (NL) Kum sıkı K 0 =0.37, gevek K 0 =0.46 Kil, yurulmu K 0 =0.692.00 eker K 0 =0.50 NL Kil 0.19+0.23×log IP OC Kil 0.7+0.1(OCR-1.2) OC Kil 0.7+0.1(OCR-1.2) 1-Sin φ

Transcript of NM405_03

Page 1: NM405_03

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

=−

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

Page 20: NM405_03

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

Page 21: NM405_03

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

#�

Page 22: NM405_03

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

#�

Page 23: NM405_03

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

Page 24: NM405_03

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

Page 26: NM405_03

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.

Page 27: NM405_03

K�LDE AKT�F BASINÇK�LDE AKT�F BASINÇX

X’

1 0=σ

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

Page 28: NM405_03

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

Page 29: NM405_03

> ���@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

Page 30: NM405_03

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.����������

Page 31: NM405_03

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

σ

Page 32: NM405_03

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

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

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

> ���@�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

( )

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

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

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

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

( )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

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

H

Page 41: NM405_03

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

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

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

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

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