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umka2103 [35]
3 years ago
14

A consolidation test was performed on a sample of fine-grained soil sample taken from a depth such that the vertical effective s

tress at that depth is 154 kN/m2. The thickness of the layer is 8 m, and the preconsolidation stress of the fine-grained soil layer is 154 kN/m2. If the initial void ratio of the soil is 0.87 and the design load to be added at the surface will induce a change in stress of 28 at the depth the sample was taken, calculate the primary consolidation settlement expected due to that load. The compression index of the soil is 0.3, and the recompression index of the soil is 0.07.
Engineering
1 answer:
Scorpion4ik [409]3 years ago
3 0

Answer:

The settlement that is expected is 1.043 meters.

Explanation:

Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil

The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

\Delta H=\frac{H_oC_c}{1+e_o}log(\frac{\bar{\sigma_o}+\Delta \bar{\sigma }}{\bar{\sigma_o}})

where

'H' is the initial depth of the layer

C_c is the Compression index

e_o is the inital void ratio

\bar{\sigma_o} is the initial effective stress at the depth

\Delta \bar{\sigma_o} is the change in the effective stress at the given depth

Applying the given values we get

\Delta H=\frac{8\times 0.3}{1+0.87}log(\frac{154+28}{154})=1.04

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Radio Frequency IDentification (RFID) tags and readers are a category of low-end wireless devices that people may not recognize
Fittoniya [83]

Answer:

See explaination.

Explanation:

Radio Frequency Identification (RFID) tags and readers uses the electromagnetic waves to identify and track the attached objects.

A tag is attached to the object which is to be identified or tracked, and reader is used to read the response and send the acknowledgement. Therefore, RFID tags and readers are used in many industries, passports, transportations and pet identification etc.

i.

RFID technology is used in smartcards, implants for pets, passports and library books to identify and track the persons, objects and pets etc.

Hence, we can say that option (i) is true.

ii.

Electronic Product Code (EPC) is a small code stored in the RFID tag. The code stored in the memory is 96 bits which are used to identify the organization which manages the data, unique number to identify the product and a number to identify the particular tag and etc.

EPC is a unique identification number, it can read and be written by the RFID reader. It is used in supply chains instead of a bar code even though expansive.

Hence, option (ii) is true.

iii.

Tags are used to identify and track the objects. It doesn’t belong to base stations and access points as a Wi-Fi networks.

Therefore, option (iii) is false.

iv.

The EPC Generation 2 RFID tag is used to improve the security by enabling the authentication features. It is not the similar way of data transmission in the other wireless situations.

Therefore, option (iv) is false.

Finally, the options (i) and (ii) are TRUE, while (iii) and (iv) are FALSE.

8 0
3 years ago
The substance is steam (H2O). NOTE: The purpose of this problem is to illustrate that there are conditions where water vapor is
Gennadij [26K]

Answer:

See the attached pictures for detailed answer.

Explanation:

See the attached pictures for step by step explanation.

7 0
3 years ago
What are the 3 dimensions that used in isometric sketches?
noname [10]

Answer:

The three dimensions shown in an isometric drawing are the height, H, the length, L, and the depth, D

Explanation:

An isometric drawing of an object in presents a pictorial projection of the object in which the three dimension, views of the object's height, length, and depth, are combined in one view such that the dimensions of the isometric projection drawing are accurate and can be measured (by proportion of scale) to draw the different views of the object or by scaling, for actual construction of the object.

5 0
3 years ago
The cantilevered W530 x 150 beam shown is subjected to a 9.8-kN force F applied by means of a welded plate at A. Determine the e
snow_lady [41]

The <em>equivalent force-couple system</em> at O is the force and couple experienced when at point O due to the applied force at point A

The <em>equivalent force couple</em> system at O due to force <em>F</em> are;

Force, F =  (<u>8.65·i - 4.6·j</u>) KN

Couple, M₀ ≈ <u>40.9 </u>k kN·m

The reason the above values are correct is as follows:

The known values for the <em>cantilever</em> are;

The <em>height </em>of the beam = 0.65 m

The <em>magnitude of </em>the applied <em>force</em>, F = 9.8 kN

The <em>length </em>of the beam = 4.9 m

The <em>angle </em>away from the vertical the force is applied = 26°

The required parameter:

The <em>equivalent force-couple system</em> at the centroid of the beam cross-section of the cantilever

Solution:

The <em>equivalent force-couple system</em> is the force-couple system that can replace the given force at centroid of the beam cross-section at the cantilever O ;

The <em>equivalent force</em> \overset \longrightarrow F = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j

Which gives;

The <em>equivalent force</em> \overset \longrightarrow F ≈ (<u>8.65·i - 4.6·j</u>) KN

The <em>couple </em><em>acting </em>at point O due to the force <em>F</em> is given as follows;

The <em>clockwise moment</em> = <em>9.8 kN × cos(28°) × 4.9</em>

The <em>anticlockwise moment</em> = <em>9.8 kN × sin(28°) 0.65/2 </em>

The sum of the moments = Anticlockwise moment - Clockwise moments

∴ The <em>sum </em>of the moments, ∑M, gives the moment acting at point O as follows;

M₀ = <em>9.8 kN × sin(28°) 0.65/2 - 9.8 kN × cos(28°) × 4.9</em>  ≈ 40.9 kN·m

The couple acting at O, due to F,  M₀ ≈ <u>40.9 kN·m</u>

The equivalent force couple system acting at point O due the force, F, is as follows

F =  (8.65·i - 4.6·j) KN

M₀ ≈ <u>40.9 </u>k kN·m

Learn more about equivalent force systems here:

brainly.com/question/12209585

4 0
2 years ago
3. (a) (5 points) Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queue
Bezzdna [24]

Answer:

(N-1) × (L/2R) = (N-1)/2

Explanation:

let L is length of packet

R is rate

N is number of packets

then

first packet arrived with 0 delay

Second packet arrived at = L/R

Third packet arrived at = 2L/R

Nth packet arrived at = (n-1)L/R

Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R

Now

L / R = (1000) / (10^6 ) s = 1 ms

L/2R = 0.5 ms

average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2

the average queuing delay of a packet = 0 ( put N=1)

4 0
3 years ago
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