nm405_12b
DESCRIPTION
L H d d φ φ ( ) tan 45 tan 45 2 2 ′ ′ = + ⋅ − + ⋅ + 0 ANKRAJ LEM D free earth < D fixed earth Serbest Toprak Destei Yönteminde minimum gömme derinlii Ankrajlı Palplan Perdede ekil Deitirme ve Moment Diyagramı b) Optimum position of plate or block ΣM=0 at C PALPLAN PERDEDE GÖÇME (AKSARAY) KUMDA SERBEST TOPRAK DESTE LE ÇÖZÜMTRANSCRIPT
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0( ) tan 45 tan 452 2a zL H d dφ φ′ ′� � � �= + ⋅ − + ⋅ +� � � �
� � � �
ANKRAJ ��LEM�
Ankrajlı Palplan� Perdede �ekil De�i�tirme ve Moment Diyagramı
a) Serbest Toprak Deste�i Yöntemi b) Ankastre Durum
Dfree earth < Dfixed earth
Serbest Toprak Deste�i Yönteminde minimum gömme derinli�i
Appropriate anchor zone: Place block/root beyond intersection ofactive and passive areas .
b) Optimum position of plate or block
ΣM=0 at C
PALPLAN� PERDEDE GÖÇME (AKSARAY)
KUMDA SERBEST TOPRAK DESTE�� �LE ÇÖZÜM
KUMDA SERBEST TOPRAK DESTE�� �LE ÇÖZÜM
= → = ⋅ ⋅� � ��������� � � � �ρ
′= + → = ⋅ + ⋅ ⋅� � � � ��������� � � � � � � � �ρ ρ
( )ρ=
′ ⋅ −�
�
�
� �
= + +� � �� � � �Kazı tabanı altında
= + + +� � � �� � � � � derinli�inde net gerilme ρ ′= ⋅ − ⋅� �� � � � �
DEF çizgisinin e�imi 1 dü�eye, yatayρ ′ ⋅ −� � � �
derinli�inde net gerilme = 0
Denge için Fx=0ΣMO’=0
ΣMO’=0
ρ′� �− ⋅ + + − + + ⋅ ⋅ − ⋅ ⋅ + + + ⋅ =� � �
� � � � � � � � �
� �� � � � � � � � �
� � � � � � � � � � � � � � �
ρ⋅ ⋅ + + − +� � � + ⋅ ⋅ + + − =
′ ⋅ −� � � � � �
� � � � �
� � � � ���� � � �
� �
� � � � � �� � � � �
� �
E�itlik deneme-yanılma yöntemiyle çözülerek L bulunur
ΣFx=0 � AACDE-AEBF-T=0
− ⋅ ⋅ − =� �
��
�� � �
ρ ′� �= − ⋅ ⋅ − ⋅ �
�
�� �
� � � � � �
E�itlik deneme-yanılma yöntemiyle çözülerek L4 bulunur
Dteorik=L3+L4 Dgerçek=1.3 to 1.4 Dteorik
E�er Kp* kullanılırsa, teorik derinli�i 30%-40% arttırmaya gerek yok
Max. teorik moment z=L1 and z=L1+L2 derinlikleri arasında olu�acak.
Kesme kuvvetinin 0, momentin maksimum oldu�u derinlik,
ρ ′⋅ ⋅ − + ⋅ − + ⋅ ⋅ ⋅ − =�
� � � � �
� �� � � � �
� � � � � � � � �
z-L1=x olsun ρ ′⋅ ⋅ − + ⋅ + ⋅ ⋅ ⋅ =�
� � �
� ��
� � � � � � �
Maksimum moment, 0 kesme kuvvetinin Maksimum moment, 0 kesme kuvvetinin oldu�u yerde moment alınarak bulunabilir.
ANKRAJLI ÇEL�K PERDE DUVARDA MOMENT AZALTMA YÖNTEM� (ROWE)
PALPLAN� R�J�TL���
′ = + +� � ���� � � �
1. Uygun kesitte palplan� seç
ρ − � �′= ⋅ ⋅ � ⋅
������� ��
�
� �
MN/m2 m4/m
1. Uygun kesitte palplan� seç2. Bu kesitin kesit modülünü (S) bul ve atalet momentini (I) hesapla.3. ρ yu hesapla4. Kesitin direnç momentini hesapla5. Kesit için e�risini noktala.
6. 1−5 i�lemlerini birkaç kesit için tekrarla. E�rinin altında kalan kesitler güvensizdir. E�rinin üstündekilerden seçim yap. (Md<Mmax)
σ= ⋅ ��! "
ρ −#�
�$% !
!
Problem: L1=3.05 m, L2=6.1 m, l1=1.53 m, l2=1.52 m, c=0, φ=30�, ρ=16 kN/m3, ve ρsat=19.5 kN/m3.
a)Teorik ve gerçek gömme derinliklerini bulunuz.b) Ankraj kuvvetini hesaplayınız..c) Maksimum momenti hesaplayınız, Mmax.d) Rowe Moment Azaltma Yöntemiyle uygun palplan� kesitini bulunuz.
(σall=172500 kN/m2, E=207000 MN/m2)
φ� � � �= − = − =� � � �� � � �
� � �� �&� �� &� ��
� � ��
φ� � � �= + = + =� � � �� � ��
&� �� &� �� �� �
� = + = + =� � � �� � � �
&� �� &� �� �� �
�
− = − =� ����� ��''� � �
ρ ρ ρ′ = − = − = ����� ���� ��'� ()& * �+ #
ρ= ⋅ ⋅ = ⋅ ⋅ = �
� �
��'���� �'��� (
� � � �+ #
ρ ρ ′= ⋅ + ⋅ ⋅ = ⋅ + ⋅ ⋅ = �
� � �
�� � ��'���� ��'�'��� ����� (
� � � � �+ #
ρ= = =
′ ⋅ − ⋅�
�
���������
� � ��'���''�
� #
� �
= ⋅ ⋅ + ⋅ + ⋅ − ⋅ + ⋅ ⋅� � � � � � � � �
� � �� �
� � �� � � � �
= ⋅ ⋅ + ⋅ + ⋅ − ⋅ + ⋅ ⋅� � ��'��� ���� �'��� '��� ������ �'����'��� ����� ����
� � ��
= + + + =����� ����� '���� ���� ������ (� �+ #
� �⋅� � � � � � � �= = ⋅ + + + ⋅ + + ⋅ + + ⋅ ⋅ =� � � � � � � � � � � � � � � � �
� ���� '��� '��� � ���� ������ ���� '��� ����� ���� '���� ���� ���� ����
� � � � ������
�!� #
�
ΣMO’=0ρ
⋅ ⋅ + + − +� � � + ⋅ ⋅ + + − =′ ⋅ −
� � � �� �� � � � �
� � � � ���� � � �
� �
� � � � � �� � � � �
� �ΣMO’=0
ρ� + ⋅ ⋅ + + − =
′ ⋅ −� � � � ���� � � �� �
� � � � �� �
⋅ ⋅ + + − +� � � + ⋅ ⋅ + + − =⋅
� �
� �
� ������ ����� '��� ����� ����� �������� ����� '��� ����� �
��'����''��� �
� =� ���� # � = + = + = ≈� � ���� ��� ���� ����&,�$�-� � � # #
� = ⋅ = ⋅ =��� ��� ��� ����.&/� &,�$�-� � #
( )ρ ′= − ⋅ ⋅ − ⋅ �
�
�
� � � � � �
= − ⋅ ⋅ ⋅ = ≈�������� ��'���''� ��� ������ ( ��� (
�� �+ # �+ #
ρ ′⋅ ⋅ − + ⋅ − + ⋅ ⋅ ⋅ − =�
� � � � �
� �� � � � �
� � � � � � � � �
z-L1=x olsun ρ ′⋅ ⋅ − + ⋅ + ⋅ ⋅ ⋅ =�� � �
� ��
� � � � � � �
⋅ ⋅ − + ⋅ + ⋅ ⋅ ⋅ =�� � ��'��� ���� ��� �'��� ��'� �
� � �� �
� + ⋅ − = � =� ����� ����� � �� � � #
Max. teorik moment z=L1 and z=L1+L2 derinlikleri arasında olu�acak.
Kesme kuvvetinin 0, momentin maksimum oldu�u derinlik (z),
� + ⋅ − = � =� ����� ����� � �� � � #
� = + = + =� � ���� ����� � � #
ρ� � ′= − ⋅ ⋅ ⋅ + + ⋅ + − ⋅ − ⋅ ⋅ ⋅ ⋅� �� �
��
#� � � �
� ���� �� �����
� � � � �
� �! � � � � � �
� �= − ⋅ ⋅ ⋅ + + ⋅ + − ⋅ − ⋅ ⋅ ⋅ ⋅� �� �
��
#�
� ���� � � � ��'��� ���� � ����� ����� �'��� ��'��
� � � � � �!
= ����� (�+# #
Maksimum moment, 0 kesme kuvvetinin oldu�u yerde moment alınarak
′ = + + = + + =� � ���� '�� ���� �����.&/�� � � � #
σ= ⋅ ��! "
Section I (m4/m) H’ (m) ρ log ρ S (m3/m) Md (kNm/m)
Md/Mma
x
PZ-22 115.2*10-6 14.48 20.11*10-4 -2.7 97*10-5 167.33 0.485
σall=172500 kN/m2, E=207000 MN/m2ρ − � �′= ⋅ ⋅ � ⋅
������� ��
�
� �
MN/m2 m4/m
PZ-22 115.2*10-6 14.48 20.11*10-4 -2.7 97*10-5 167.33 0.485
PZ-27 251.5*10-
614.48 9.21*10-4 -3.04 162.3*10-
5284.84 0.826
PZ-27 yeterli
FREE EARTH SUPPORT METHOD FOR PENETRATION OF CLAY (having a granular soil backfill)
ρ ρ ′= ⋅ − ⋅ + ⋅' � �� � � . � �
= � − ⋅ =� � '��0 � � �
P1= area of the pressure diagram ACDT= anchor force per unit length of the sheet pile
′ =� �1!
� �⋅ + − − − ⋅ ⋅ + + =� �� �
� � � � � ' � �� � ��
�� � � � � � � �
( )� ⋅ + ⋅ ⋅ ⋅ + − − ⋅ ⋅ + − − =�
' ' � � � � � � � �� � � � � � � � � � � � � � �
This equation gives theoretical depth of penetration, D
Maximum moment occurs at depth L1<z<L1+L2. The depth of zero shear (and thus the maximum moment) may be determined from:
ρ ′⋅ ⋅ − + ⋅ − + ⋅ ⋅ ⋅ − =�� � � � �
� �� � � � �
� � � � � � � � �
Various types of anchoring for sheet pile walls: a) anchor plate or beam b) tie back c) vertical anchor pile d) anchor beam with batter piles
ULTIMATE RESISTANCE OF TIE BACKS
π σ φ′= ⋅ ⋅ ⋅ ⋅ ⋅&�/ 2� � �
σ ρ′ = ⋅2 �
�� � 3)4�5�4��6��)�-$�
��#�&� �& φ≈ −� �� "��
Kumda Ankraj Kapasitesi
��#�&� �& φ≈ −� �� "��
Kilde Ankraj Kapasitesi
π α= ⋅ ⋅ ⋅ ⋅/ /� � . α = >���� ��/. ��