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 = tan^2(45-f/2) = tan^2(45-35/2) = 0.27

At depth = 0 m

sh_a = Ka x sv = 0.27 x 0 x 17.4 = 0

At depth = 4 m

sh_a = 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=35^o

			Ka = tan^2(45-f/2) = tan^2(45-35/2) = 0.27
		H = 4 m

At depth = 0 m

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

At depth = 4 m

sh_a = 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 = tan^2(45-f/2) = tan^2(45-30/2) = 0.333 Ka2 = tan^2(45-f/2) = tan^2(45-27/2) = 0.333

At depth = 0 – 2 m

0 m sh_a = q x Ka₁ + Ka₁ x sv_om = (30 x 0.333) + 0 = 10 kN/m²

2 m sh_a = 10 + Ka₁ x sv_2m = 10 + (2 x 17.4 x 0.333) = 21.6 kN/m²

At depth = 2 – 6 m

2 m sh_a = q x Ka₂ + Ka₂ x sv_2m = (30 x 0.38) + (2 x 17.4 x 0.38) = 24.6 kN/m²

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

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