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VikaD [51]
2 years ago
14

An unknown relative passes away and bequeaths upon you a small tract of land in Amherst. You decide to build a two-story storage

facility to make the best of your bequest. But your self-storage dream is in jeopardy due to a 10 meter thick layer of soft clay (N<4) on the site. You put on your best geotechnical engineer hat, hire a driller to pull up some samples, and send them off to a lab for a consolidation test. The report indicates that the clay is a dark grey, slightly sweet, kaolinite blend with a cy = 1x10-7 mº/s, single-drained, and an ultimate settlement of 0.73 meters. It does not make financial sense to install deep foundations, so you are interested in how long it will take to consolidate the clay layer using a passive load.
How long will it take for settlements of 25, 50, and 65 cm to occur?
If you need to build within the next 12 months and have at least 65cm of settlement to be viable, does it make sense to proceed?
Engineering
1 answer:
Ratling [72]2 years ago
7 0

Answer: It does make sense, because I've been involved in these careers and have a long family line of them. And other questions?

Explanation:

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A Class A fire extingisher is for use on general combustibles such as:​
Tatiana [17]

Answer:

used for ordinary combustibles, such as wood, paper, some plastics, and textiles. This class of fire requires the heat-absorbing effects of water or the coating effects of certain dry chemicals.

Explanation:

7 0
3 years ago
At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
3 years ago
Consider a 2-shell-passes and 8-tube-passes shell-and-tube heat exchanger. What is the primary reason for using many tube passes
Maru [420]

Answer:

See explanation

Explanation:

Solution:-

- The shell and tube heat exchanger are designated by the order of tube and shell passes.

- A single tube pass: The fluid enters from inlet, exchange of heat, the fluid exits.

- A multiple tube pass: The fluid enters from inlet, exchange of heat, U bend of the fluid, exchange of heat, .... ( nth order of pass ), and then exits.

- By increasing the number of passes we have increased the "retention time" of a specific volume of tube fluid; hence, providing sufficient time for the fluid to exchange heat with the shell fluid.

- By making more U-turns we are allowing greater length for the fluid flow to develop with " constriction and turns " into turbulence. This turbulence usually at the final passes allows mixing of fluid and increases the heat transfer coefficient by:

                                U ∝ v^( 0.8 )    .... ( turbulence )

- The higher the velocity of the fluids the greater the heat transfer coefficient. The increase in the heat transfer coefficient will allow less heat energy carried by either of the fluids to be wasted ; hence, reduced losses.

Thereby, increases the thermal efficiency of the heat exchanger ( higher NTU units ).

5 0
3 years ago
WILL GIVE BRAINLIEST!Technician A says it takes two revolutions of the crankshaft to fit all eight cylinders on a V8. Technician
lidiya [134]

Answer:

C

Explanation:

3 0
3 years ago
Read 2 more answers
For a turning operation, you have selected a high-speed steel (HSS) tool and turning a hot rolled free machining steel. Your dep
Alisiya [41]

Answer:

MRR = 1.984

Explanation:

Given that                              

Depth of cut ,d=0.105 in

Diameter D= 1 in

Speed V= 105 sfpm

feed f= 0.015 ipr

Now  the metal   removal  rate   given as

MRR= 12 f V d

d= depth of cut

V= Speed

f=Feed

MRR= Metal removal rate

By putting the values

MRR= 12 f V d

MRR = 12 x 0.015 x 105 x 0.105

MRR = 1.984

Therefore answer is -

1.944

8 0
3 years ago
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