HOMEWORK SOLUTION

 

HW #11

 

CE 353

GEOTECHNICAL ENGINEERING I

 

(Question no. 9.1-h, 9.3-d, 9.5, ..Al-Khafaji textbook)

 

 


9.1-h    Determine active earth pressure & resultant force per unit width. Assume the wall is rigid and frictionless.

            Use rankine’s theory.

           

Solution:

 

 

Ka = tan2(45-f/2)

      = tan2(45-35/2)

      = 0.27

 
 

 

 

 

 

 

 

 


At depth = 0 m

sha = Ka x sv = 0.27 x 0 x 17.4 = 0

 

At depth = 4 m

sha = Ka x sv = 0.27 x 4 x 17.4 = 18.8 kN/m2

 

18.8

 
 

 

 

 

 

 

 

 

 

 

 


            Force/width (Ea) = Total area = 4 x (18.8/2) = 37.6 kN/m ………ans


9.3-d    Similar with problem 9.1-h but the soil saturated.

 

            Solution:

 


 

 g=18.4kN/m3

 f=35o

 
           

Ka = tan2(45-f/2)

      = tan2(45-35/2)

      = 0.27

 

H = 4 m

 
 

 

 

 

 

 

 

 


At depth = 0 m

sha = Ka x s’v = 0.27 x 0 x (18.4-9.81) = 0

 

At depth = 4 m

sha = Ka x s’v = 0.27 x 4 x (18.4-9.81)= 9.3 kN/m2

 

u = 9.81 x 4 = 39.2 kN/m2

 

Right Triangle: u
Right Triangle: sv’

+

 

39.2

 

9.3

 
 

 

 

 

 

 

 

 

 

 

 


            Force/width (Ea)          = Total area

= {(9.3/2)x4} + {(39.2/2)x4}

= 97 kN/m  ……………….ans


9.5              Determine active earth pressure and resultant force per unit depth

 

Solution:

Ka1 = tan2(45-f/2)

       = tan2(45-30/2)

       = 0.333

Ka2 = tan2(45-f/2)

       = tan2(45-27/2)

       = 0.333

 

 
 

 

 

 

 

 

 

 

 

 

 

 


At depth = 0 – 2 m

0 m      sha = q x Ka1 + Ka1 x svom = (30 x 0.333) + 0 = 10 kN/m2

2 m      sha = 10 + Ka1 x sv2m = 10 + (2 x 17.4 x 0.333) = 21.6 kN/m2

 

 
 

 

 

 

 

 

 

 

 

 


At depth = 2 – 6 m

2 m      sha = q x Ka2 + Ka2 x sv2m = (30 x 0.38) + (2 x 17.4 x 0.38) = 24.6 kN/m2

6 m      sha = 24.6 + Ka2 x svsat = 24.6 + (4 x (18.4-9.81) x 0.38) = 37.7 kN/m2

 

 

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