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Crazy boy [7]
3 years ago
5

Describe the steps involved in accomplishing a life-cycle cost analysis (LCCA). I have listed the steps. just describe. a)-Opera

tional Performance. b)-Salvage externalities c)- Value vs. Risk d)- Initial expenditure e)- Maintenance implications
Engineering
1 answer:
Tasya [4]3 years ago
8 0

Explanation:

a). <u>Operational Performance</u>

     It is defined as the parameter that describes the percentage of accuracy of performing an operation or carrying out an activity.

b).  <u>Salvage externalities</u>

   Salvage in Life cycle cost analysis is a process of estimating the value of the remaining assets in the organisation,.

c). <u>Value Vs Risk</u>

   When we take risk in doing any activity we know the value of accomplising the activity. So value relates directly with risk. When the value of a certain task is high, the risk involve in it is also high.

d). <u>Initial expenditure</u>

   Initial expenditure is nothing but the cost involve in starting a particular acitvity or task at the starting phase.

e). <u>Maintenance implications</u>

   It lays emphasis in maintaining the cost of every possible parameters that are involve in the activity. It includes labour, machines, positions, energy, facilities, etc.

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A 1 m wide continuous footing is designed to support an axial column load of 250 kN per meter of wall length. The footing is pla
creativ13 [48]

Answer:

correct option is (A) 0.5

Explanation:

given data

axial column load = 250 kN per meter

footing placed =  0.5 m

cohesion = 25 kPa

internal friction angle =  5°

solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for  a pure cohesive soil

N_{\gamma } = 0

and we know formula for N_{\gamma } is

N_{\gamma } = (Nq - 1 ) × tan(Ф)   ..................1

so here Ф is very less  N_{\gamma } should be nearest to zero

and its value can be 0.5

so correct option is (A) 0.5

7 0
3 years ago
A 20cm-long rod with a diameter of 0.250 cm is loaded with a 5000 N weight. If the diameter of the bar is 0.490 at this load, de
Margaret [11]

If the diameter of the bar is 0.490 at this load, determine I. the engineering stress and strain, and [2] II. the true stress and strain is 1561. 84 MPa.

<h3>What is strain?</h3>

Strain is a unitless degree of ways a great deal an item receives larger or smaller from an implemented load. Normal stress happens while the elongation of an item is in reaction to an everyday pressure (i.e. perpendicular to a surface), and is denoted via way of means of the Greek letter epsilon.

  1. L = 20 cm d x 1 = 0.21 cm
  2. dx 2 = 0.25 cmF=5500 a) σ= F/A1= 5000/(π/4x(0.0025)^2)= 1018.5916 MPa lateral stress= Ad/d1= (0.0021-0.0025)/0.0025 = - 0.1 longitudinal stress (ɛ_l)= -lateral stress/v = -(-0.16)/0.3
  3. (assuming a poisson's ration of 0.3) ε_l=0.16/0.3 = 0.5333
  4. b) σ_true= σ(1+ ɛ_I)= 1018.5916(1+0.5333
  5. = 1561.84 MPa.

Read more about the diameter :

brainly.com/question/358744

#SPJ1

4 0
2 years ago
A square aluminum plate 5 mm thick and 200 mm on a side is heated while vertically suspended in quiescent air at 40C. (A) Determ
Kamila [148]

brainly.com/question/14000001

4 0
3 years ago
In crash tests, a shock absorber is used to slow the test car. The shock absorber consists of a piston with small holes that mov
Lesechka [4]

Answer:

The heat transferred to water equals 1600 kJ

Explanation:

By the conservation of energy we have

All the kinetic energy of the moving vehicle is converted into thermal energy

We know that kinetic energy of a object of mass 'm' moving with a speed of 'v' is given by

K.E=\frac{1}{2}mv^{2}

Thus

K.E_{car}=\frac{1}{2}\times 2000\times 40^{2}=1600\times 10^{3}Joules

Thus the heat transferred to water equals 1600kJ

3 0
3 years ago
Assume a voltampere-base S3b = 100 MVA and a line-to-neutral voltage base Vb = 7.5 kV. Find the current base Ib. Also, by using
Rasek [7]

<u>Explanation:</u>

Let the frequency of the operation "W"ras/s

\begin{aligned}V(t) &=7.5 \cos (\omega t+\varnothing v) k V \\&=7.5 \cos w t \quad k V \\I(t) &=\cos \left(\omega t+\phi_{i}\right) \quad k A \\&=\cos (\omega t-\pi / 6) \quad k A\end{aligned}

\(\therefore\) Power, \(P(t)=v(t) I(t)\)$$\begin{array}{l}=[7.5 \cos \omega t][\cos (\cot -\pi / 6)] \times 10^{6} w \\=7.5 \cos ^{2}(2 \omega t-\pi / 6)+7.5 \cos \left(\pi_{6}\right) MW \\=3.75+7.5 \cos (2 \omega t-\pi / 6) \quad MW\end{array}$$

Real powel, \(P=\left(7.5 \times 10^{3}\right)\left(10^{3}\right) \cos \left(\varnothing_{v}-\varnothing_{I}\right)\)$$\begin{array}{l}=7.5 \cos (-\pi / 6) MW \\=3.75 MW\end{array}$$

6 0
3 years ago
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