Soil Mechanics II Problem

EFFECTIVE STRESS

PROBLEM-1 Calculate total stress, porewater pressure and effective stress on a horizontal plane at levels A, B, C, D and draw the depth versus stress diagrams (z-s).

 n  17 kN/m 3  k  18 kN/m 3  d  20 kN/m 3

 d  18.5 kN/m3

SOLUTION-1 s  s  uw POINT-A

s A   n . z  17  3  51 kN/m3 uwA   w . z  0 s A  s A  uw  51 kN/m 3

 n  17 kN/m

3

 k  18 kN/m 3  d  20 kN/m 3

POINT-B

s B  17  3  18  2  87 kN/m 3 uwB   w . z  0 s B  s B  uw  87 kN/m3 POINT-C

s C  17  3  18  2  20  1  107 kN/m 3 uwC   w . z  9.81  1  9.81 kN/m3

 d  18.5 kN/m3

s C  s C  uw  107  9.81  97.19 kN/m3 POINT-D

s D  17  3  18  2  20  1  18.5  2.5  153.25 kN/m 3 uwD   w . z  9.81  3.5  34.34 kN/m 3

s D  s D  uw  153.25  34.34  118.91 kN/m3 or briefly, using buoyant unit weights  '

s D  17  3  18  2  (20  9.81)  1  (18.5  9.81)  2.5  118.91 kN/m3

VERTICAL STRESS DIAGRAM

SHEARING RESISTANCE of SOILS

PROBLEM-1 Diagnose thr soil type with the data from three shearbox tests.

s

100

200

300

max

58

115

175

 200

Kesme Kutusu Kırılma Zarfı

100

100

200

300

s

c=0 =30 Zemin Türü KUM

PROBLEM-2 Zemin örnekleri üzerinde yapılan kesme kutusu deneyinde aşağıdaki sonuçlar elde edildiğine göre; bu zeminlerin türünü tayin ediniz. s (kPa)

10

20

30

Simge

t1 (kPa)

6.25

12.49

18.76

•

t2 (kPa)

18.54

22.67

26.51

o

PROBLEM-3 Results of four triaxial shear tests on a c- are given. Calculate the shearing resistance if this soil is tested in the shear box at a normal stress of sn=200 kPa s3 (kPa)

50

100

200

300

s1 (kPa)

150

300

600

700

Graphical solution must be used for tests more than two,because circles are variable

c=0 =30?? or

c=20kPa =20

s=max=c+stan=20+200tan20=

ELASTIC - PLASTIC EQUILIBRIUM

PROBLEM-1 Calculate the shearing resistance of the sand at a normal stress of 100 kPa, given the result of triaxial tests. Test No

Cell pressure (kPa)

Deviator stress

Major principal stress, (kPa)

1

100

277.5

377.5

2

200

524.0

724.0

3

300

770.0

1070.0

Here, you have a choice: use graphical or analytical method.

c=0 and =34°: SAND

s  c  s  tan   0  100  tan 34  67.5 kPa

s1f=s3N+2cN0.5

377.5=100N+2cN0.5 1070 =300Nf+2cN0.5

PROBLEM-2 The following results have been obtained from a triaxial (UU) test on a sample of clay. Determine the type of soil and its parameters. Test No

Cell pressure s3 (kPa)

Deviator Stress sd (kPa)

Major principal stress s1 (kPa)

1

100

140

240

2

200

135

335

3

300

150

450

We draw a horizontal envelope regardless what the results show

Type: Normally loaded clay

c

70  67.5  75  71 kPa 3

PROBLEM-3 The uncunfined compressive stregth qu of a soil is given as 130 kPa olarak veriliyor. The same soil was tested in the triaxial cell with the cell pressure s3=qu . The deviator stress at failure was measured as 230 kPa. Determine the shearing resistance parameters and guess its type. s3=0 s3=130

s 1  qu  130 kPa s 1  s 3 s d  130  230  360 kPa

Graphic solution

Analytical solution

s1  s 3  N  2  c  N 130  0  N  2  c  N 360  130  N  2  c  N N 

230  1.769 130

  N   tan 2  45    1.769 2  =16 From the stress circle c=50 kPa =16 c- soil (overconsolidated)

130  2  c  N  c  49 kPa

PROBLEM-4 Serbest basma dayanımı qu=100 kPa olan yumuşak kil ve kayma direnci açısı =30 olan kum için s- diyagramı çizerek kırılma zarflarını işaretleyiniz. Bu zeminlerin d=20 kN/m3 ise yüzeyden 5 m derinlikte kayma direnci denklemlerini yazınız.

Kayma direnci denklemleri NL Kil Kum

sc

qu 100   50 kPa 2 2

s  s  tan     z  tan   5  20  tan 30  57.7 kPa

PROBLEM-5 Aşağıda kesiti verilen zeminden alınan numuneler üzerinde üç eksenli kesme deneyi yapılmıştır. Buna göre kesitte a) efektif gerilmenin, b) kayma direncinin derinlikle değişimini gösteren s-z doğrularını çiziniz. YASS

5

SP

n =18.5

2

d =22.1

3

a) EFEKTİF GERİLME

KUMDA KİLDE

8

s 5m  37  3   22.1  9.81  73.9 kPa

s13m  73.9  8  18.7  9.81  145.1 kPa

b) KAYMA DİRENCİ

d =18.7

CH

s 2 m  2  18.5  37

KUMDA

s 1  s 3  s deviatör max s1  s 3  N  2  c  N

UU Deney Sonuçları ( sdeviatörmax ) ZEMİN

s3 = 100

s3 = 200

SP

70

140

CH

59

55

170  100  N 

N  1.70 

  tan 2  45    1.70 2 

s  c veya

cu

s0

 0.11  0.037  I P

s  s  tan15

0 YASS

37 SP

74

5

59  30 kPa 2

   15 

0

0 2

KİLDE

c

340  200  N 

9.9 19.8

30

s

30

CH

s  c  30 kPa 13

s

145

PROBLEM-6

PROBLEM-7 Draw the total stress diagrams for the soil profile shown

GWL SP

=32 OCR=3 d= 21 kN/m3

CH OCR= 1 IP= 40 d= 19 kN/m3

EARTH PRESSURES

PROBLEM 1:

Determine the active lateral earth pressure on the frictionless surface shown in figure. Calculate the resultant force and its location from the base of the wall. STEP-1: Calculate Ka H=5 m

d  20 kN / m3    30o c0

1  Sin 1  Sin(30o ) 1 Ka    veya o 1  Sin 1  Sin(30 ) 3  30 1 2 2 K a  tan (45  )  t an (45  )  2 2 3

STEP-2: Calculate the vertical effective stress (s’z) Yüzeyde: s z  0

uw  0

Tabanda: s z   .H o  (20  9.81)  5  51 kPa uw   w .H o  9.81 5  49 kPa

STEP-3: Calculate the lateral effective stress (s’x)a

1 (s x ) a  K a .s z   51  17 kPa 3

PROBLEM 1: (continued) STEP-4: Sketch the lateral earth pressure distributions Lateral Earth Yanal Toprak Pressure Basıncı

Su BasıncıPressure Hydrostatic

H=5 m

17 kPa

49 kPa

STEP-5: Calculate the active thrust

Pa  Psu basıncı  Ptoprak basıncı where (Pwater pressure) is the due to the pore water and (Pearth pressure) is the lateral force due to the soil

Pa 

1 1 1 1  (s x ) a  H o   u  H o  (  17  5)  (  49  5)  165 kN/m 2 2 2 2

STEP-6: Determine the location of the resultant Since both lateral pressure and the pore water (hydrostatic)pressure distributions are triangular over the whole depth, the resultant is at the centroid of the triangle, that is;

z  H o / 3  5/ 3  1.67 m

from the base of the wall

PROBLEM 2:

Determine the active earth pressure on the frictionless surface shown 15 kPa

GWT

STEP-1: Calculate Ka and Kp

H=4 m

 d  19 kN/m

 1  sin  1  sin 26 K a  tan 2 (45  )    0.39 2 1  sin  1  sin 26  1  sin  1  sin 26 K p  tan 2 (45  )    2.56 2 1  sin  1  sin 26

3

   26o c  8 kPa

STEP-2: Calculate the vertical effective stress (s’z) Yüzeyde: s z  0 uw  0 Tabanda: s z   .H o  (19  9.81)  4  36.76 kPa uw   w .H o  9.81 4  39.24 kPa

STEP-3: Calculate the lateral effective stress at the base (s’x)a (s x )a  Ka .s z  0.39  36.76  14.34 kPa

STEP-4: Calculate the lateral effective stress due to external load at the base(s’q)a (s q )a  K a .q  0.39 15  5.85 kPa

STEP-5: Calculate the lateral tension effect due to cohesion (s’c)a (s c ) a  

2c N



28 26 tan 2 (45  ) 2

 10 kPa

PROBLEM 2: (continued) STEP-6: Sketch the distributions of lateral earth pressure 15 kPa

GWT GWT

Lateral Earth Yanal Toprak Pressure Basıncı

Yayılı Load Yük

Hydrostatic Su Basıncı

Cohesion Kohezyon

H=4 m

(+) 14.34 kPa

(+) 5.85 kPa

(+) 39.24 kPa

STEP-7: Calculate the total lateral force (thrust)

Pa  Pearth  Pload  Pwater  Pcohesion 1 1  (s x ) a  H o  (s q ) a  H o   u  H o  (s 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

PROBLEM 3: Determine the active earth pressure on the frictionless surface shown. 10 kPa SW

 n  17 kN/m3

H=3 GWT

SP H=3

  32o SW

 n  19.31 kN/m 3   30o SP

STEP-1: Calculate Ka values

 1  sin  1  sin 32 ( K a ) SW  tan 2 (45  )    0.307 2 1  sin  1  sin 32  1  sin  1  sin 30 ( K a ) SP  tan 2 (45  )    0.333 2 1  sin  1  sin 30

PROBLEM 3: (continued) 10 kPa

H=3

SW  n  17 kN/m3 GWT

  32 SW SP  d  19.31 kN/m 3

Yayılı Load Yük

I

o

H=3

  30 SP

Yanal Toprak Lateral Earth Basıncı Pressure SW SW

III

1

II

Yanal Toprak Lateral Earth Basıncı Pressure SP SP

3

IV 4

5

p1  q  K aSW  10  0.307  3.07 kPa p2  q  K aSP  10  0.333  3.33 kPa p3  K aSW   nSW  H SW  0.307  17  3  15.66 kPa p4   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   SP

6

FORCES

PRESSURES

p6  W  HW  9.81  3  29.43 kPa

VI

V

o

2

Hydrostatic Su Basıncı Pressure

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

PROBLEM 4 Bir SW kumda yapılan kesme kutusu deneyinde aşağıdaki sonuçlar alınmış. Kum 5 m derinlikte büyük bir kutuya ( B=10 m, L=30 m ) dolduruluyor. Kapağın a) kuru durumda sükunette, b) kuru durumda aktif, c) kutu su ile dolu iken pasif durumda alacağı toplam itkiyi bulunuz. 3 3 H=5m

X

s

(  k  19 kN / m ,  d  22.5 kN / m ) Results from Shearbox tests

max

m

L=30m

10 = B

1 ) Calculate shearing resistance of sand:

a)

113 229 tan       37 150 300

Test No

1

2

s

150

300



113

229

s   max  s  tan   s  tan37

K0  1  Sin  1  sin 37  0.398 p0  K 0  z  k  0.398  5  19  37.81 kPa

 tabanda 

1 1  P0    k  H 2  K 0   B   19  52  0.398  10  945.3 kN 2 2 

sw

pa  K a  z  k  0.249  5  19  23.66 kPa

Aktif Durum (Kuru Kum)

5m

  1  Sin 1  b ) K a  tan 2  45      tan 2 26.5  0.249 2  1  Sin N 

 tabanda 

1 1  Pa     k  H 2  K a   B   19  52  0.249  10  591.4 kN 2 2 

63.5 23.66 kPa

  1  Sin  c) K p  tan 2  45     N  tan 2 63.5  4.023 2  1  Sin  p p  K p  z     z   w  4.023  5  12.69  5  9.81  255.26  49.05  304.31 kPa  taban 

sw

Pasif Durum (Batık Kum)

5 m

1  1  Pp       H 2  K p   B     w  H 2   B 2  2  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

255.26 kPa

49.05 kPa

PROBLEM 5 Calculate the force F required to keep the door in active equilibrium. 30  1  K a SP  tan 2  45    2  3  35   K a SW  tan 2  45    0.271 2  

FORCES (THRUST)

PRESSURES p1  18  2 

1  12 kPa 3

1 p2  18  2   20  9.81  1   15.4 kPa 3

p3  p4  18  2  10.19  1  0.271  12.5 kPa p5  0 p6   22  9.81  3  0.271  9.91 kPa p7  0 p8  9.81  4  39.24 kPa

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 

MOMENTS

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 

 M0  0

2 3 1  6  F  12   3  1    13.7  3.4  37.5  1.5  14.9   78.4  4   F  46 kN / m 3 3 3 

PROBLEM 6 Construct the active pressure diagram on vertical plane XX and calculate the thrust Pa. Dry sand Submerged CH

X

3

k=19 kN/m3 SP =300 YASS

d=18 kN/m3 3 =100 CH c=10 kPa 2

d=22 =450

kN/m3

cohesion

I

1

VI 4

2

IV

III X

3

pressures 1 p1  19  3   19 kPa 3 p2  19  3  0.704  40.13 kPa

p3  19  3  0.171  9.75 kPa p4  18  9.81  3  0.704  17.30 kPa p5  18  9.81  3  0.171  4.20 kPa 2c 2 10 p6     16.78 kPa 1.192 N

p7   22  9.81  2  0.171  4.17 kPa

p8  9.81 5  49.05 kPa

(- )

10   K aCH  tan 2  45    0.704 2 

6

V 5

30  1  K a SP  tan 2  45    2  3 

45   K aGW  tan 2  45    0.171 2  

CH

II

GW

submerged GW water

VIII

7 VII 8

FORCES

10   N CH  tan  45    1.192 2 

1 PaI   3 19  28.47 kN / m 2

PaII  3  40.13  120.39 kN / m Pa III  2  9.75  19.5 kN / m 1 PaIV   3 17.30  25.95 kN / m 2

PaV  2  4.20  8.40 kN / m PaVI  3  16.78  50.34 kN / m 1 PaVII   2  4.17  4.17 kN / m 2 1 PaVIII   5  49.05  122.63 kN / m 2

 P  279.17 kN / m

Örnek:

XX düzleminde aktif toprak basıncı diyagramlarını çizip toplam aktif kuvveti Batık (Pa) hesaplayınız. Batık 3

CH

Kuru kum

X

k=19 kN/m3 SP =300 YASS

d=18 3 =100 c=10 kPa

CH

d=22 =450

GW

kohezyon CH

I

1

Su

GW

kN/m3

2

VI

II 4

2

kN/m3

IV

III X

3

Basınçlar

(-)

6

V

VIII

7 VII

5

8

Kuvvetler

p1  19  3 

1 PaI   3  18.98  28.47 kN / m 2 PaII  3  40.13  120.39 kN / m

p3  19  3  0.171  9.75 kPa

Pa III  2  9.75  19.5 kN / m

p4  18  9.81  3  0.704  17.30 kPa

PaIV 

1  18.98 kPa 3 p2  19  3  0.704  40.13 kPa

p5  18  9.81  3  0.171  4.20 kPa p6  

2c N



2  10  16.78 kPa 1.192

p7   22  9.81  2  0.171  4.17 kPa p8  9.81  5  49.05 kPa

30  1  K a SP  tan 2  45    2  3  45   K aGW  tan 2  45    0.171 2   10   K aCH  tan 2  45    0.704 2  10   N CH  tan  45    1.192 2 

1  3  17.30  25.95 kN / m 2 PaV  2  4.20  8.40 kN / m 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

KISA SINAV-1 • Aşağıdaki kesitte sürtünmesiz duvara etkiyen toplam aktif kuvveti hesaplayıp, bu kuvvetin etkime noktasını bulunuz. 33 kPa

1  Sin N  1  Sin 1 Ka  N Kp  N

SP n=17 kN/m3 = 330

2.5 m YASS

CI d=19.81 kN/m3 =110 c=22 kPa

3.5 m

Sonuçları Ka virgülden sonra 3 hane, yükler ve basınçlar için ise 2 hane olacak şekilde yuvarlayınız.

KISA SINAV-1 (Çözüm) 33 kPa SP YASS

CI

n=17 kN/m3 = 330

2.5 m

d=19.81 3.5 m kN/m3 0 =11 c=22 kPa

YÜK

SP

CI

CIKOHEZYON

SU

I III

1

3

2

II (+)

4

IV (+)

V (+)

VI 5

6

(-)

1  Sin N  1  Sin

KaSP

1 1  Sin    0.295 N 1  Sin

1 1  Sin KaCI    0.680 N 1  Sin

VII (+)

7

KISA SINAV-1 (Çözüm) 33 kPa SP YASS

P

CI

x=?

n=17 kN/m3 = 330

2.5m

d=19.81 3.5 m kN/m3 0 =11 c=22 kPa

YÜK

SP

CI

SU

I III

1

3

2

II (+)

4

IV (+)

p1  q  KaSP  33  0.295  9.74kPa p2  q  KaCI  33  0.680  22.44kPa p3   SP H SP KaSP  17  2.5  0.295  12.54kPa p4   SP H SP KaCI  17  2.5  0.680  28.90kPa  H CI KaCI  (19.81  9.81)  3.5  0.680  23.80kPa p5  CI p6  

CIKOHEZYON

2c 2  22   36.29kPa N 1.47

p7   w H w  9.81 3.5  34.34kPa

VI

V

5

(+)

6

(-)

VII

7

(+)

PI  9.74  2.5  24.35 kN / m PII  22.44  3.5  78.54 kN / m 12.54  2.5  15.68 kN / m 2 PIV  28.90  3.5  101.15 kN / m PIII 

23.80  3.5  41.65 kN / m 2 PVI  36.29  3.5  127.02 kN / m PV 

34.34  3.5  60.10 kN / m 2 P  194.45 kN / m PVII 

194.45 kN / m  X  24.35  4.75  78.54 1.75  15.68  4.33  101.15 1.75  41.65 1.17  127.02 1.75  60.10 1.17  X  2.03 m

RETAINING WALLS

PROBLEM-2: Check the equilibrium of the wall. b20

DOLGU

 duvar= 23 kN/m3

H=3.5 m

  20

Df=1m O

 = 21 kN/m3  = 300

a88 B=2.5m

DOĞAL ZEMİN

n = 18 kN/m3  = 150 c =12 kN/m2

Problem 2 (Solution) - Aktif ve Pasif Kuvvetlerin Hesabı b20

Pv

Pp

O

Ph

DOĞAL ZEMİN

a  88, b  20,   20,   30

Pa 

n = 18 kN/m3  = 150 c =12 kN/m2

a88 B=2.5m

 K a  0.436

 = 21 kN/m3  = 300

Pa

H=3.5 m Df=1m

DOLGU

 Coulomb 

1  21  3.52  0.436  56.1 kN / m 2

    15, b  0  K p  tan 2  45    1.7 2 

Pp 



1  18  12  1.7  15.3 kN / m 2

Aktif Kuvvetin Yatay Bileşeni  Ph  56.1  cos  90  88  20   52 kN / m Aktif Kuvvetin Düşey Bileşeni  Pv  56.1  sin  90  88  20   21 kN / m

Problem 2 (Solution) - Duvarın Ağırlığı B-2x x

x

H=3.5 m

Duvar üst genişliği tan a 

3.5 3.5 x x tan a

 3.5  2.50  2     2.50  2  0.122  2.25 m  tan 88 

 duvar= 23 kN/m3 Df=1m

a88

O B=2.5m

x

M b, taş  Alantaş  taş 

1   2.5  2.25  3.5  23  191.19 kN / m 2

Mtaş=191.2

H=3.5 m

Problem 2 (Solution) - Sliding check

Pv=21

DOLGU  = 21 kN/m3  = 300

Pa=56.1 Ph=52

a88

Pp=15.3 B=2.5m

n = 18 kN/m3,  = 150, c =12 kN/m2

Düşey   Fz  Mb, taş +Pv  191.2  21  212.2 kN / m Resisting Forces   FR   Fz  f  B  c  Pp

2  2  F  212.2  tan  15  2.5   12  R      15.3  37.42  20  15.3 3  3   FR  72.2 kN/m

GS KAYMA

F  

R

Ph

72.2   1.39  1.50 KAYMAZ, GÜVENSİZ 52

Df=1m

Mtaş=191.2

H=3.5 m

Problem 2 (solution) - Sliding check DOLGU  = 21 kN/m3  = 300

Pv=21

Pa=56.1 Ph=52

a88

Pp=31.5

n = 18 kN/m3,  = 150, c =12 kN/m2

B=2.5m

    30, b  0  K p  tan 2  45    3.0 2 

Pp 



1  21  12  3  31.5 kN / m 2

KONTROLDE G.S.=1.00 ARANIR

2  2  F  212.2  tan  15  2.5   12  R      31.5  37.42  20  31.5  88.92 kN / m 3  3 

FSSLIDING

F  

R

Ph

88.92   1.71  1.50 52

Does not slide, safe

Problem 2 (solution) - Overturning control :

Pv=21

H=3.5 m

1.25m

Mb, taş=191.2 B/3=1.17m

Df=1m

O 2.46 m

B=2.5m

GSOVERTURN

Ph=52

M   M

RES

OVER



880 X

tan a 

1.17 1.17 x  0.04 m x tan 88

191.2  1.25 239   26.03 52  1.17  21  2.46 60.84  51.66

Does not overturn, safe

Problem 2 (solution) - eccentricity

Mb, taş=191.2

Pv=21

H=3.5 m

1.25m

F

z

x

Df=1m

 212.2

Ph=52

e

O

880 2.46 m

F

z

 x   M0

B=2.5m

212.2 x  191.2  1.25  21  2.46  52  1.17 M  x F

o

z

229.82   1.08 m 212.2

B 2.5 e x   1.08  0.17 m 2 2

Problem 2 (solution) - eccentricity

Mtaş=191.2

Pv=21

F

z

x  1.08

Df=1m

O

 212.2 e=0.17

H=3.5 m

1.25m

Ph=52

880

2.46 m

B=2.5m

smin=50.3

smax=119.5

s

min

 0, s max  qemn 

s

max min

max s min

F  

 6e  1   BL  B 212.2  6  0.17    1   2.5  1  2.5   119.5 kPa z

s max s min  50.3 kPa

PROBLEM-3: Determine the dimensions of the wall against sliding, overturning and eccentricity.

Dimensions selected b20

 duvar= 24 kN/m3

DOLGU   21 c0   30

8.5 m

Df=1.3 m

DOĞAL ZEMĠN    16 c  12   18

PROBLEM-3 (solution) - Ön Boyutlandırma b=20-30cm

100

2

Seçilen Boyutlar

Konsol

H

0.3 m

b20

0.08H-0.1H 0.33B

B=0.4H-0.7H

7.7 m 8.5 m

2.2 m

0.8 m

Df=1.3 m B=6.5 m

(Gerekirse Diş)

0.8 m

PROBLEM-3 (solution) – Yanal Toprak Basınçları Dolguda

  0, b  20,   30

Try B=6.5m (0.4-0.7H!) b20

(Rankine)

H=8.5 m

K a  cos b  Pa 

cos b  cos b  cos  2

2

 0.397

1  21  8.52  0.397  301.2 kN / m 2

Pa Pa

Pv

Doğal Zeminde    18

b=200

Ph Df=1.3

cos b  cos 2 b  cos 2 

Pp b20

18   K p  tan 2  45    1.894 2  1 Pp   16  1.32  1.894  25.6 kN / m 2

Ph  Pa  cos 20  283.04 kN / m Pv  Pa  sin 20  103.02 kN / m

PROBLEM-3 (solution) – Ağırlıklar 0.3 m

b20

3.7 m

 dolgu= 21 kN/m3  duvar= 24 kN/m3

Ms Gövde

7.7 m 8.5 m

Mb

2.2 m

0.8 m

3.5 m

Df=1.3 m B=6.5 m

Temel

0.8 m

0.8  0.3  7.70  24  124.8  101.6  226.4 kN / m 2 3.5  3.7 Ms   7.70  21  582.1 kN / m 2

M b  6.5  0.8  24 

PROBLEM-3 (solution) Sliding Check

DOLGU   21 c0   30

Ms=582.1

H=8.5 m

Mb=226.4

b20

Pv  103.02 b=200

Ph  283.04 Pp=25.6

Df=1.3

H/3=m B=6.5 m

Doğal Zemin    16 c  12   18

F

R

F F F

z

 M b  Pv  M s

z

 226.4  103.02  582.1

z

 911.52 kN / m

2 2  18 2  12 2    Fz  tan     B   c  Pp  911.52  tan  6.5   25.6  271.4 kN / m 3 3 3 3 

FSkayma

F  

R

Ph



271.4  0.96  1.25 283.03

Wall slides! Assuming GS=1.25 is enough for sliding, make a key to obtain the required additional passive resistance.

Mb=226.4

Wanted total passive resistance :

  21 c0

  30

Pv  103.02

Ms=582.1

Ph  283.04 H/3=m

Df=1.3

D=?

Pp=??

PROBLEM-3 (solution) Kayma Kontrolu Diş Uygulaması

B=6.5 m

Doğal Zemin    16 c  12   18

Ddiş=?

Re quired additional passive res. : FSsliding 

F

R

Ph

271.4  PpEK



283.03

Key depth :1.4 m Then ;

 1.25  Pp EK  283.03  1.25  271.4  82.39 kN / m

Wanted total passive resistance : PpToplam  25.6  82.39  107.99 107.99 

FSoverturn

1  16  D2  1.89  D  2.67 m  Ddiş  D  D f  Ddiş  2.67  1.3  1.4 m 2

F 

R

Ph

Key depth : 1.4m, Then : D f  2.7 m 1 Pp   16  2.72  1.894  110.46 kN / m 2 193.75  52  110.46   1.259  1.25 283.03

√

(Dişten doğan beton ağırlığı lehimize hesaba katılmadan).

MbTEMEL=124.8

MbBOYUN=101.6

PROBLEM-3 (solution) Check for overturning

  21 c0

  30

Pv  103.02

Ms=582.1

Df1=1.3

Ph  283.04 8.5/3

Df2=2.7

B=6.5 m

Pp=110.46 Ddiş=1.4 1

Doğal Zemin

  16 c  12   18

FSoverturn

M

2.6 3.25 4.7 Duvar temeli MbTEMEL

Boyun MbBOYUN

diş

Duvar üzerindeki molgu Ms

124.8  3.25  101.6  2.6  1.4  1  24  3.25  582.1  4.7   8.5 Ph  z  Pv  b 283.04   103.02  6.5 3 o

3514.8   26.6 132.3

MbTEMEL=124.8

MbBOYUN=101.6

PROBLEM-3 (solution) Eccentricity

  21

Pv  103.02

c0

  30

Ms=582.1

Ph  283.04

Df1=1.3

8.5/3

Df2=2.7

B=6.5 m

Pp=110.46 Ddiş=1.4

Doğal Zemin

1

  16 c  12   18

2.6

F

z

3.25

x   Mo

4.7

M  x F

boyun

temel

o

z



dolgu

Pv

Ph

405.6  264.16  109.2  2735.87  669.63  801.95 3382.51   3.58 m 911.52  33.6 945.12 M b  Pv  M s

M diş  1.4  1  24

B 6.5 x  3.58  3.25  3.58  0.33 m 2 2 Fz  6e  945.12  6  (0.33)  189.7   1    1   kPa s min  0, s max  qemn      101.1 BL  B  6.5  1  6.5  e

max s min

diş

PROBLEM-3 (solution) Check for bearing capacity (strip footing)

1 qd  c( N c )   D f ( N q )   B( N  ) 2  = 18  Nc= 15.5

Nq=6

N=3.9

qd  12  15.5  18  1.3  6  0.5  18  6.5  3.9  186  140  228  554kPa sem= qd/FS  185 kPa

smax=189kPa OK!

PROBLEM-5: Calculate the active pressure acting on the wall a)Horizontal fill state of Rankine b)Sloping (b) fill state by Rankine c)Sloped (b) fill state by Coulomb approaches.

  25

b20 a ) Rankine, horizontal



sw



H=6m

 = 350 k = 22kN/m3

K a  tan2  45  Pa 

35   0.271 2 

1  22  62  0.271  107.32 kN / m 2

b ) Rankine, b

a8

K a  cos b 

Pa 

cos b  cos 2 b  cos 2  cos b  cos b  cos  2

2

 0.303

1  22  62  0.303  119.99 kN / m 2

c ) Coulomb, b

Ka 

Pa 

sin 2 a     sin      sin   b   sin 2 a  sin a     1   sin a     sin a  b    

1  22  62  0.378  149.69 kN / m 2

2

 0.378

BEARING CAPACITY

PROBLEM 1: Calculate the highest axial column load (Qmax) for square and circular footings. Qmax

SC Df=3m

c=10 kPa  =19 kN/m3 = 20

For =20 Nc=17.69 Nq=7.44 N= 4.97

B=Df=1

Footing base at -3m But by definition Df  B →

For square footing

⁂

Df= 1m

qd  1.3  c  N c  1  D f  N q  0.4   2  B  N

=(1.3x10x17.69)+(19x1x7.44)+(0.4x19x1x4.97) =230+141+28= 410 kPa Ultimate load Pmax=qdxA=410x12 = 410 kN Circular footing

qd  1.3  c  N c  1  D f  N q  0.3   2  B  N 

= (1.3x10x17.69)+(19x1x7.44)+(0.3x19x1x4.97)=230+141+28=399 kPa Ultimate load Pmax=qdxA=399p0.52=313 kN This is 76% of the square footing !!

P

PROBLEM 2: Şekilde verilen daire temelde 2.5 güvenlik sayısı ile taşınabilecek yükü (Qem=P) hesaplayınız.

2m

  300

SP

n  19.5 kN / m3 YASS

3m

  200 SC

c  15 kPa

 d  20 kN / m3

qd  K1  c  Nc  Cw  1  Df  Nq  C'w  K2  2  B  N Daire Temel İçin Şekil Katsayıları  K1=1.3, K2=0.3 (D=B)

=30o  Nq  22.46 ve   20o  Nc  17.69, N  4.97 (Terzaghi)

YASS olmasa idi Cw=C’w=1

qd  1.3 15 17.69  119.5  2  22.46  1 0.3  20  3  4.97

P

qd  344.96  875.94  89.46  1309.36 kPa Son Taşıma Gücü 

SP

Qd 

p D2

p x32

 qd   1309.36  9255 kN 4 4 Güvenli Taşıma Gücü  Q 9255 Qemn  d   3702 kN GS 2.5

SC

YASS temel tabanında ise Cw=1, C’w=0.5

qd  1.3 15 17.69  1 19.5  2  22.46  0.5  0.3  20  3  4.97

P

qd  344.96  875.94  44.73  1265.63 kPa Son Taşıma Gücü 

SP YASS

SC

Qd 

p D2

 qd 

p x32

 1265.63  8946 kN

4 4 Güvenli Taşıma Gücü  Q 8946 Qemn  d   3579 kN GS 2.5

YASS yüzeyde ise Cw=C’w=0.5

P YASS

SP

SC

qd  1.3 15 17.69  0.5 19.5  2  22.46  0.5  0.3  20  3  4.97 qd  344.96  437.97  44.73  827.66 kPa Ultimate load  Qd 

p D2

Allowable load  Qemn

 qd 

p x32

 827.66  5850.4 kN

4 4 Q 5850.4  d   2340 kN GS 2.5

M=600 kNm P=1200 kN

1.75 m

  300

SP

k  20 kN / m3

footing to carry 1200 kN axial load and 600 kNm moment .

B

  200 2m

c  20 kPa

 d  19 kN / m3

PROBLEM 3: Design a square

SC YASS

B’

M 600   0.5 m B  B  2e P 1200 B Re quired e   try min imum size B  3 m 6 B  3  2  0.5  2 m

e

B

B

Şekil katsayıları yeni boyutlara göre bulunacak

B 2 B 2 K1  1  0.2  1  0.2   1.133 K 2  0.5  0.1  0.5  0.1  0.433 L 3 L 3 db 2 Su Seviyesi Düzeltme Katsayıları   0.67  Cw  0.82 (temelin gerçek boyutu kullanılıyor) B 3 q  K cN  C  D N  C  K  BN B  B' alınacak d

1

c

w

1

f

q

w

2

2



  30 için N q  22.46,   20 için N c  17.69, N  4.97 qd  1.133  20 17.69  1 20 1.75 17.69  0.82  0.433 19  2  4.97 qd  400.86  619.15  67.06  1087.07 kPa Qd  q d B  L  1087.07  2  3  6522.42 kN GS 

max s min 

Qd 6522.42   5.44 P 1200

P  6e  1200  6  0.5  267  1  1      0 kPa B L  B  3 3  3  (temelin gerçek boyutu kullanılıyor)

PROBLEM 4: A footing will be designed to carry an axial load of 20000 kN with

one basement, using properties measured in laboratory. The owner has accepted to use select material (c=0, f=40, =22kN/m3) up to the garden level. P=20000 kN

Select fill

GP  = 40  = 22 kN/m3 (backfill)

3.5m

B

CH (natural soil) c = 60 kPa  = 16  = 18.5 kN/m3

SOLUTION We shall use basement depth as -3.5m and use it as the footing dimension (B=Df). We shall then demonstrate the profound effect of Df on the bearing capacity. Bearing capacity factors from Table IV-1, page 97 CH   =16 : Nc =13.68 N=2.94 ; GP    40: Nq81.27 kitaptan

The ultimate bearing capacity of square footing by Terzaghi

qd  1.3cN c  1 D f N q  0.4  2 BN  =(1.3x60x13.68)+(22x3.0x81.27)+(0.4x18.5x3x2.94) =(1067+5364+65)= 6496 kPa The benefit of coarse material is observed in the high second term. Let us use a FOS like 3 to be on the safe side sall = qd/FOS = 6496/3 = 2165 kPa

The load to be carried by a footing 3.5 by 3.5: Qem = sem x A = 2165x32 = 19485 KN

which is sufficiently above the required load Q=20000kN !

PROBLEM 5: YASS I düzeyinde iken şekilde gösterilen yükleri (Mx= 280 kNm, My= 120 kNm, P= 1700 kN) 2.5 güvenlikle taşıyacak bir temeli Terzaghi yöntemi ile boyutlandırınız. Bulunan boyutlara göre su seviyesinin YASS II düzeyine düşmesi ile güvenlik sayısında oluşacak değişimi hesaplayınız.

Taşıma gücü katsayıları;  '  36 için Nq  47.16,  '  28 için Nc  31.61, N  15.15

Terzaghi

Çift yönde eksantriklik var; Mx =280 kNm, My =120 kNm  MB =120 kNm, ML =280 kNm eB 

MB 120   0.071 m  Bmin  6e  6  0.071  0.42 m P 1700

eL 

ML 280   0.165 m  Lmin  6e  6  0.165  0.99 m P 1700

Seçim:

B=1.50 m, L=1.70 m

B  B  2e  1.5  2  0.071  1.36 m L  L  2e  1.7  2  0.165  1.37 m

Şekil katsayıları yeni boyutlar için, B’≈L’ olduğu için kare gibi; YAS I-I için;

da 3   1  Cw  0.5 Df 3

Son taşıma gücü Terzaghi ile;

K1  1.3

K2  0.4

db 0   0  C* w  0.5 B 1.50

(B=B’, Df>B olduğu için Df=B alınacak)

qd  K1  c  Nc  Cw  1  Df  Nq  C* w  K 2  2  B  N qd  1.3x40x31.61  0.5x20x1.50x47.16  0.4x0.5x19x1.36x15.15  1643.72  707.4  78.30  2429.4 kPa Yük:

Güvenli Yük:

Qd  qd  B  L  2429.4  1.36  1.37  4526.5 kN Qemn 

Qd 4526.5   1811 kN GS 2.5

YASS II-II için;

Son taşıma gücü;

da 0   0  Cw  1 Df 3

db 0   0  C* w  0.5 B 1.50

qd  K1  c  Nc  Cw  1  Df  Nq  C* w  K 2  2  B  N qd  1.3x40x31.61  1x20x1.50x47.16  0.4x0.5x19x1.36x15.15  1643.72  1414.8  78.30  3136.82 kPa

Yük:

Qd  qd  B  L  3136.82  1.36  1.37  5845 kN

Böylece su seviyesinin düşürülmesi ile

GS 

Qd 5845   3.44 ’e yükseliyor. P 1700

PROBLEM 6: Calculate the reduction in the ultimate bearing capacity of the footing shown below if the GWT rises from level AA to BB.

B

1

B

3 3 GWL rises

3.5 A

A c=11kPa =15 d=21kN/m3

SOLUTION Using Terzaghi formula

GWL

GWL at A-A: =15 Çizelge IV- 1

Nc=12.9 Nq=4.5 N=2.2

da/Df= 0.0/4.0= 0 Şekil 4.13  Cw=1.0

db/B=3/3.5=0.86

and Cw*=0.91

Infinite (strip) footing L>>B

* 1 qd  c( N c )  Cw   D f ( N q )  Cw   B( N )

2

qd  ( 11  12.9 )  1.0( 21  3.5  4.5 )  0.91( 0.5  21  3.5  2.2 )  141.9  330.75  73.6  546 kPa

GWL rises to B-B da/Df= 3.0/4.0= 0.75 Şekil 4.13  Cw= 0.63 0.50

db/B=0/3.5=0.0

Cw’=

qd  ( 11  12.9 )  0.63( 21  3.5  4.5 )  0.50( 0.5  21  3.5  2.2 )  141.9  208.4  40.4  391 kPa This means a drop of 28% by a rise of 3 m in the groundwater level!

PROBLEM 7: Design a rectangular footing to carry the axial load and moments shown on soil with c=20 kPa, =20 and n= 18 kN/m3 at a depth of 1.5m. MB=450kNm

ML=750kNm P=950kN

SOLUTION: I shall select dimensions to provide a square footing. So, eL= ML/P= 750 kNm/950 kN = 0.79m eB= MB/P= 450 kNm/950 kN = 0.48m Therefore, use

L= 3.10m 

L’=L-2eL= 3.10-1.58  1.50m

B= 2.50 m  B’ =B-2eB=2.50-0.96  1.50m

Çizelge IV-1

=20:

Nc =17.69

Nq= 7.44

N= 4.97

Kare temelin son taşıma gücü

qd  1.3cN c  1D f N q  0.4  2 BN Qd= (1.5)2{(1.3x20x17.69)+(18x1.5x7.44)+(0.4x18x1.50x4.97)} = 2.25(460 + 201 + 54) = 2.25x715 = 1608 kN FOS = 1608/950 = 1.70 < 3.0

TRY AGAIN !

Check contact stresses

P 6e 950 6  0.79 s max  (1  )  (1  )  310kPa B L B 3.1 2.5 3.1 950 6  0.48 s min  (1  )  18kPa 3.1 2.5 2.5 Dimensions are unsuitable, try again!