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1 year ago

If a material is found to be in the tertiary phase of creep, the following procedure should be implemented:

1 answer:

8 0

When a material is found to be in the tertiary phase of creep, the following procedure should be implemented that is the component should be replaced immediately. Therefore, Option C is correct.

<h3>What do you mean by a tertiary degree of creep?</h3>

Tertiary Creep has an extended creep rate and terminates when the material breaks or ruptures. It is related to each necking and formation of grain boundary voids. The wide variety of possible stress-temperature- time combos is infinite.

Therefore, When a material is found to be in the tertiary phase of creep, the following procedure should be implemented that is the component should be replaced immediately. Option C is correct.

Learn more about creep:

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For some metal alloy, the following engineering stresses produce the corresponding engineering plastic strains prior to necking.

kirza4 [7]

Answer:

203.0160

Explanation:

Because you add then subtract then multiply buy 7 the subtract then divide then you add that to the other numbers you got than boom

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2 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

3 years ago

What is hardness and how is it generally tested?

drek231 [11]

Answer:

Hardness is understood as the property of materials in general to resist the penetration of an indenter under load, so that the hardness represents the resistance of the material to the plastic deformation located on its surface.

Explanation:

Hardness of a material is understood as the resistance that the material opposes to its permanent surface plastic deformation by scratching or penetration. It is always true that the hardness of a material is inversely proportional to the footprint that remains on its surface when a force is applied.

In this sense, the hardness of a material can also be defined as that property of the surface layer of the material to resist any elastic deformation, plastic or destruction due to the action of local contact forces caused by another body (called indenter or penetrator), harder, of certain shape and dimensions, which does not suffer residual deformations during contact.

That is, hardness is understood as the property of materials in general to resist the penetration of an indenter under load, so that the hardness represents the resistance of the material to the plastic deformation located on its surface.

The following conclusions can be drawn from the previous definition of hardness:

1) hardness, by definition, is a property of the surface layer of the material, and is not a property of the material itself;

2) the methods of hardness by indentation presuppose the presence of contact efforts, and therefore, the hardness can be quantified within a scale;

3) In any case, the indenter or penetrator must not undergo residual deformations during the test of hardness measurement of the body being tested.

To determine the hardness of the materials, durometers with different types of tips and ranges of loads are used on the various materials. Below are the most commonly used tests to determine the hardness of the materials.

Rockwell hardness :

It refers to the Rockwell hardness test, a method with which the hardness or resistance of a material to be penetrated is calculated. It is characterized by being a fast and simple method that can be applied to all types of materials. An optical reader is not required.

Brinell hardness :

Brinell hardness is a scale that is used to determine the hardness of a material through the indentation method, which consists of penetrating with a hardened steel ball tip into the hard material, a load and for a certain time.

This test is not very precise but easy to apply. It is one of the oldest and was proposed in 1900 by Johan August Brinell, a Swedish engineer.

Vickers hardness:

Vickers hardness is a test that is used in all types of solid and thin or soft materials. In this test, a square-shaped pyramid-shaped diamond and a 136° vertex angle are placed on the penetrating equipment.

In this test the hardness measurement is performed by calculating the diagonal penetration lengths.

However, its result is not read directly on the equipment used, therefore, the following formula must be applied to determine the hardness of the material: HV = 1.8544 · F / (dv2).

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2 years ago

List the main activities of exploration??

Trava [24]

Answer: Exploration includes plethora of activities and depend upon the kind of exploration a person is doing. But most include some of the basic activities like research , investigation, planning and execution.

Suppose we want to explore new petroleum sites then we would have to start with studying the geography of that area, then according to our research we will analyse the hot spots or the sector where probability of finding of oil field is highest, post that appropriate man power is skilled professionals, tools and machinery will be brought at the site so that execution can take place.

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3 years ago

Help please its due today will mark you brainliest

Tems11 [23]

Answer:

launch- The first stage is ignited at launch and burns through the powered ascent until its propellants are exhausted. The first stage engine is then extinguished, the second stage separates from the first stage, and the second stage engine is ignited. The payload is carried atop the second stage into orbit

powered ascent-The first stage is ignited at launch and burns through the powered ascent until its propellants are exhausted. The first stage engine is then extinguished, the second stage separates from the first stage, and the second stage engine is ignited. The payload is carried atop the second stage into orbit

coasting flight-

When the rocket runs out of fuel, it enters a coasting flight. The vehicle slows down under the action of the weight and drag since there is no longer any thrust present. The rocket eventually reaches some maximum altitude which you can measure using some simple length and angle measurements and trigonometry.

ejection charge-At the end of the delay charge, an ejection charge is ignited which pressurizes the body tube, blows the nose cap off, and deploys the parachute. The rocket then begins a slow descent under parachute to a recovery. The forces at work here are the weight of the vehicle and the drag of the parachute.

slow decent- slow downs (i guess)

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Explanation:

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3 years ago