HOMEWORK SOLUTION

 

HW #6

 

GEOTECHNICAL ENGINEERING I

 

(Question no. 6.1a, 6.6a, 6.13, 6.21b, & 6.24b ..Al-Khafaji textbook)

 

 


6.1a Determine the normal and the shear stress on a plane inclined at an angle of 30o (use analytical procedure).

 


Solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

45 psi

 
6.6a Determine the magnitude and direction of the principle stresses for the soil shown below (use the analytical procedure)

 


Solution:

 


25 psi

 

 

sy = 45 psi

 

sx = 25 psi

 

 

 

 

 

 

 

 

 

 

 

 

from the x axis (counter clockwise)

 

from A-A axis = 67.5o + (55o-90o)

 

= 32.5o (counter clockwise)

 

6.13          Plot the total overburden pressure (s), pore water pressure (u) and effective overburden pressure for the soil profile below.

 

Solution:

 

50

 

50

 

 

sand

 

clay

 

 

 

 

0 50

 

 

50 100

 

 

 

 

 

Result in the following table and plotted in the graph

 

Depth

s

u

s effective = s - u

0

0

-3120

3120

10

1173

-2496

3669

20

2346

-1872

4218

30

3519

-1248

4767

40

4692

-624

5316

50

5865

0

5865

60

7044

624

6420

70

8223

1248

6975

80

9402

1872

7530

90

10581

2496

8085

100

11760

3120

8640

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6.21b Calculate then plot the vertical stress increment under 6000 kN load (5 ft depth increment) and x=2 m

 

Solution:

Q2 6000 kN

 

Q1 4000 kN

 

2 m

 

 

 

Depth 25 m

 
 

 

 

 

 

 

 

 


z = depth

Q= Point load

 

Ipz from graph / figure 6.11

then tabled and plotted as follow:

 

 

 

 

z

r1/z

Ipz1

r2/z

Ipz2

Dsz1

Dsz2

Dsztotal

0

~

0

0

0.4775

0

~

~

5

0.4

0.329622

0

0.4775

52.73957

114.6

167.3396

10

0.2

0.43309

0

0.4775

17.3236

28.65

45.9736

15

0.133333

0.457119

0

0.4775

8.126559

12.73333

20.85989

20

0.1

0.46597

0

0.4775

4.659702

7.1625

11.8222

25

0.08

0.470148

0

0.4775

3.00895

4.584

7.59295

 

 

 

6.24b Calculate then plot the vertical stress increment under point midway between the 500 kN/m and the 800 kN/m line loads, (the increment=5 ft) and x=2 m.

 

Solution:

Q2 800 kN/m

 

Q1 500 kN/m

 

1 m

 

1 m

 

Depth 25 m

 

 

 

 

 

 

 

 

 

 


 

z = depth

Q= line load

 

Ilz from graph / figure 6.13a

or, Ilz=

then tabled and plotted as follow:

 

z

x1/z

ILz1

x2/z

ILz2

Dsz1

Dsz2

Dsztotal

0

~

0

~

0

0

0

0

5

0.2

0.612445

0.2

0.612445

61.24449

97.99118

159.2357

10

0.1

0.630636

0.1

0.630636

31.53182

50.4509

81.98272

15

0.066667

0.634124

0.066667

0.634124

21.13748

33.81997

54.95744

20

0.05

0.635354

0.05

0.635354

15.88386

25.41417

41.29803

25

0.04

0.635925

0.04

0.635925

12.7185

20.34961

33.06811