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scoundrel [369]
3 years ago
12

A site is underlain by two layers of normally consolidated clayey sand. The unit weights of the top layer are 19 and 21 kN/m2 an

d the layer is 6 meters thick. The unit weights of the bottom layer are 20 and 22 kN/m, and the layer is 8 meters thick. Below the bottom layer lies bedrock. The water table is located 2 meters below the ground surface. The top layer soil has a friction angle of 38 degrees, a cohesion intercept of 20 kPa, and an undrained strength of 160 kPa. The bottom layer soil has a friction angle of 33 degrees, a cohesion intercept of 10 kPa, and an undrained strength of 120 kPa. Point A is located 9 meters underground. All layers have a lateral stress ratio of 0.5.
Required:
a. Draw the profile neatly, add dimensions, and draw a tree on the ground surface.
b. Determine the geostatic vertical effective stress and the geostatic horizontal effective stress at point A.
c. Draw the Mohr Circle to determine the effective stress that acts at point A on a plane inclined 10 degrees counter-clockwise from the horizontal plane. Make sure the Mohr Circle is a well-drawn circle
Engineering
1 answer:
Ivan3 years ago
5 0

Answer:

A) attached below

B) Geostatic vertical effective stress ( бv )

= 119.33 KN/m^2

Geostatic horizontal effective stress ( бn )

= 59.66 KN/m^2

C) attached below

Explanation:

attached below is a detailed solution

A) attached below

B) Determine the geostatic vertical effective stress and the geostatic horizontal effective stress at point A

Geostatic vertical effective stress ( бv )

= 119.33 KN/m^2

Geostatic horizontal effective stress ( бn )

= 59.66 KN/m^2

C) attached below

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    So the current  yield which is mathematically represented as

                               \frac{dN}{dt} =   \frac{2000}{1 \ year }

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So in order to obtain the solution we need to obtain the  rate of growth

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                                      r = \frac{[\frac{dN}{dt}] }{N[1-\frac{N}{K} ]}

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          and  K is the carrying capacity which is given as 10,000 fishes

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        Substituting these values into the equation

                              r = \frac{[\frac{2000}{year}] }{4000[1-\frac{4000}{10,000} ]}  =0.833

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                          p' = \frac{K}{2}  = \frac{10,000}{2}  = 5000\ fishes          

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