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Alex73 [517]
3 years ago
14

(10 points) A single crystal in the titanium cable is oriented so that the [001] direction is parallel to the applied load. If t

hat load creates a normal stress of 100 MPa in that crystal, what is the resolved shear stress in (011) plane in the [01Ī] (= zero, one, one-bar) direction? (NOTE: this part of the question is independent from the conditions for parts a-
Engineering
1 answer:
Ulleksa [173]3 years ago
4 0

Answer:yep

Explanation:

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g (d) Usually, in the case of two finite-duration signals like in parts (a) and (b), the convolution integralmust be evaluated i
Bas_tet [7]

you face is A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

3 0
3 years ago
Why or why not the following materials will make good candidates for the construction of
zvonat [6]

Answer:

Answer explained below

Explanation:

3.] a] A turbine blade is the individual component which makes up the turbine section of a gas turbine. The blades are responsible for extracting energy from the high temperature, high pressure gas produced by the combustor.

The turbine blades are often the limiting component of gas turbines. To survive in this difficult environment, turbine blades often use exotic materials like superalloys and many different methods of cooling, such as internal air channels, boundary layer cooling, and thermal barrier coatings. The blade fatigue failure is one of the major source of outages in any steam turbines and gas turbines which is due to high dynamic stresses caused by blade vibration and resonance within the operating range of machinery.

To protect blades from these high dynamic stresses, friction dampers are used.

b] Thermal barrier coatings (TBC) are highly advanced materials systems usually applied to metallic surfaces, such as on gas turbine or aero-engine parts, operating at elevated temperatures, as a form ofexhaust heat management.

These 100μm to 2mm coatings serve to insulate components from large and prolonged heat loads by utilizing thermally insulating materials which can sustain an appreciable temperature difference between the load-bearing alloys and the coating surface.

In doing so, these coatings can allow for higher operating temperatures while limiting the thermal exposure of structural components, extending part life by reducing oxidation and thermal fatigue.

In conjunction with active film cooling, TBCs permit working fluid temperatures higher than the melting point of the metal airfoil in some turbine applications.

Due to increasing demand for higher engine operation (efficiency increases at higher temperatures), better durability/lifetime, and thinner coatings to reduce parasitic weight for rotating/moving components, there is great motivation to develop new and advanced TBCs.

3 0
4 years ago
1. Which of the following will cause a spark knock?
zlopas [31]

Answer:

I couldn't find options for your question online, but I can give you an explanation so you can choose the correct option.

Explanation:

A spark knock is a form of unpredictable behavior that occurs in combustion, that is, in the chemical reaction that occurs between oxygen and an oxidizable material. Such combustion is usually manifested by incandescence or flame.

The spark knock is a detonation that occurs when there is a lot of pressure in the fuel.

<u>Some situations in which this can happen are: </u>

  • Engine overloaded.
  • Maximum pressure in the cylinders.
  • Engine overheated.
  • Overheated air.
  • Long and excessive engine ignition timing.
  • Spark plug at high temperatures.
5 0
3 years ago
An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°
Fantom [35]

Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

carbon concentration cx = 0.215 wt%

time = 7 hr

D =  7.5 \times 10^{-11} m^2/s

for 0.225% carbon concentration following formula is used

\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})

where, erf stand for error function

\frac{cx -co}{cs -co} = \frac{0.215 -0.286}{0 -0.286} =0.248

0.248 = 1 - erf(\frac{x}{2\sqrt{DT}})

erf(\frac{x}{2\sqrt{DT}}) = 1 - 0.248

erf(\frac{x}{2\sqrt{DT}}) = 0.751

from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815

from given table

\frac{x}{2\sqrt{DT}} = 0.815

x = 2\times 0.815 \times \sqrt{7.5 \times 10^{-11}\times (7\times 3600)

x = 2.39\times 10^{-3} m

x = 0.002395 mm

8 0
3 years ago
B)
Triss [41]

Answer:

2.5 is the required details

8 0
3 years ago
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