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

 +

 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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