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3 years ago

Where can I find solved problems of advanced soil structure interaction?

2 answers:

4 0

The seismic response of nuclear power plant structures is often calculated using lumped parameter methods. A finite element model of the structure is coupled to the soil with a spring-dashpot system used to represent the interaction process. The parameters of the interaction model are based on analytic solutions to simple problems which are idealizations of the actual problems of interest. The objective of the work reported in this paper is to compare predicted responses using the standard lumped parameter models with experimental data. These comparisons are shown to be good for a fairly uniform soil system and for loadings that do not result in nonlinear interaction effects such as liftoff. 7 references, 7 figures.

Lady bird [3.3K]3 years ago

4 0

Man tbh id k I’m sorry I’m. Not any help

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Question in Probability and Statistics 2<br>

Ghella [55]

Since we both have a question in same subject may u help me with this For a confidence level of 80% with a sample size of 21, find the critical t value

7 0

3 years ago

A thick steel sheet of area 150 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig

Bingel [31]

Answer:

Rate of corrosion = 24.95 mpy

Rate of corrosion = 0.63 mm/yr

Explanation:

given data

steel sheet area = 150 in²

weight loss = 485 g

density of steel = 7.9 g/cm³

time taken = 1 year

to find out

rate of corrosion in (a) mpy and (b) mm/yr

solution

we get here the rate of corrosion that is express as

rate of corrosion = (k × W) ÷ (D × A × T) ..................1

here k is constant and w is total weight lost and t is time taken for loss and A is surface area and D is density of steel

so put her value in equation 1 we get

Rate of corrosion = $\frac{534*485*1000}{7.9*150*24hr*365day/yr}$

Rate of corrosion = 24.95 mpy

and

Rate of corrosion = $\frac{87.6*485*1000}{7.9*150*2.54^2*24hr*365days/yr}$

Rate of corrosion = 0.63 mm/yr

5 0

3 years ago

A cylindrical metal specimen having an original diameter of 10.55 mm and gauge length of 54.5 mm is pulled in tension until frac

kumpel [21]

Answer:

(a) 53.94%

(b) 26.61%

Explanation:

Change in area will be given by

$\triangle A=\frac {\pi(R_o^{2}-r_n^{2})}{\pi R_o^{2}}$ where $\triangle A$ represent change in area R is radius and subscripts O and n represent original and new respectively.

Substituting 10.55/2 for original radius and 7.16/2 for new radius then

$\triangle A=\frac {\pi(5.275^{2}-3.58^{2})}{\pi 5.275^{2}}\times 100\approx 53.94$

(b)

Similarly, percentage elongation will be found by dividing the change in length by the the original length. In this case, rhe original length was 54.5mm and goes to 69 mm so the change in length is given by subtracting the final length from the original length

Percentage elongation is $\frac {69-54.5}{54.5}\times 100\approx 26.61$

6 0

4 years ago

What is meant by the dead state. Discuss its significance.

Jlenok [28]

Answer and Explanation:

Dead state can be defined as the state in which both the are ambient properties and the system properties are coincident to each other.

Since the surroundings are similar and these surroundings are fixed also, there will be no kinetic energy and also the gravitational potential energy for both the systems is also equal.

Practically, in thermodynamic studies, the Gravitational potential energy of the system is negligible. This state holds significance because it leads to the termination of all the processes that are spontaneous in nature.

4 0

4 years ago

An insulated rigid tank is initially evacuated. A valve is opened, and atmospheric air at 95 kPa and 17 ºC enters the tank until

Darya [45]

Answer:

$T2= 406k$

Explanation:

The temperature can be defined as the measurement of the intensity of the heat present in the object. Fahrenheit, kelvin and centigrade are the common scale used for measuring Temperature.

Given:

T1=170C

To convert to Kelvin

= 17+273 =290K

T1 = 290K

Pressure (P)= 95KPa

Specific heat ratio = CP/CV= K

WhereK=1.005/0.718

K = 1.4

The final temperature can be calculated using the formula below.

T2 = CP/CV × T1

=. K × T1

T2 = 1.4 × 290

$T2= 406k$

5 0

3 years ago

Where can I find solved problems of advanced soil structure interaction?​

Where can I find solved problems of advanced soil structure interaction?