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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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Answer:

D

Explanation:

the way vertices are connected may be different so having same number of edges do not mean that total degree will also be same.

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3 years ago
Find At Least Three Instances Of Set Constructions In Your CS Courses And Use Them To Exhibit All Set Operations Discussed In Cl
hoa [83]

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4 0
1 year ago
Using the Distortion-Energy failure theory: 8. (5 pts) Calculate the hydrostatic and distortional components of the stress 9. (1
WITCHER [35]

Answer:

Detailed solution is given below:

7 0
4 years ago
A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri
babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are  73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

T_{BE}-W=0 hence

T_{BE}=W=20*9.81=196.2 N

Therefore, tension in the cable, T_{BE}=196.2 N

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0

24.525-39.24+0.2D_x=0

D_x=73.575 N

Similarly,

A_x-D_y=0

A_x=73.575 N

Therefore, both reactions at A and D are 73.575 N

7 0
3 years ago
What is the velocity of flow in an asphalt channel that has a hydraulic radius of 3.404 m, length of 200 m and bed slope of 0.00
yKpoI14uk [10]

Answer:

The velocity of flow is 10.0 m/s.

Explanation:

We shall use Manning's equation to calculate the velocity of flow

Velocity of flow by manning's equation is given by

V=\frac{1}{n}R^{2/3}S^{1/2}

where

n = manning's roughness coefficient

R = hydraulic radius

S = bed slope of the channel

We know that for an asphalt channel value of manning's roughness coefficient = 0.016

Applying values in the above equation we obtain velocity of flow as

V=\frac{1}{0.016}\times 3.404^{2/3}\times 0.005^{1/2}\\\\\therefore V=10.000m/s

7 0
3 years ago
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