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3 years ago

What is hardness and how is it generally tested?

Engineering

1 answer:

drek231 [11]3 years ago

3 0

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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fracture will occur as the value is less than E/10 (= 22.5)

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If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.

$\sigma_m = 2\sigma_o [\frac{a}{\rho_t}]^{1/2}$

$= 2\times 750 (\frac{\frac{0.2mm}{2}}{0.001 mm}})^{1/2}$

= 15 GPa

fracture will occur as the value is less than E/10 = 22.5

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3 years ago

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See explaination and attachment.

Explanation:

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The second step is to relate the deviatoric stress to viscosity in the fluid.

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Please kindly check attachment for step by step solution.

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3 years ago

Write the implementation (.cpp file) of the Player class from the previous exercise. Again, the class contains:

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Answer:

//Define the header file

#ifndef PLAYER_H

#define PLAYER_H

//header file.

#include <string>

//Use the standard namespace.

using namespace std;

//Define the class Player.

class Player

{

//Declare the required data members.

string name;

int score;

public:

//Declare the required

//member functions.

void setName(string par_name);

void setScore(int par_score);

string getName();

int getScore();

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//End the definition

//of the header file.

#endif

Player.cpp:

//Include the "Player.h" header file,

#include "Player.h"

//Define the setName() function.

void Player::setName(string par_name)

{

name = par_name;

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//Define the setScore() function.

void Player::setScore(int par_score)

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score = par_score;

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string Player::getName()

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//Define the getScore() function.

int Player::getScore()

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return score;

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3 years ago

Air at 80 kPa and 10Â°C enters an adiabatic diffuser steadily with a velocity of 150 m/s and leaves with a low velocity at a pre

il63 [147K]

Answer:

The exit temperature is 293.74 K.

Explanation:

Given that

At inlet condition(1)

P =80 KPa

V=150 m/s

T=10 C

Exit area is 5 times the inlet area

Now

$A_2=5A_1$

If consider that density of air is not changing from inlet to exit then by using continuity equation

$A_1V_1=A_2V_2$

So $A_1\times 150=5A_1V_2$

$V_2=30$ m/s

Now from first law for open system

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

Here Q=0 and w=0

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}$

When air is treating as ideal gas

$h=C_pT$

Noe by putting the values

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}$

$1.005\times 283+\dfrac{150^2}{2000}=1.005\times T_2+\dfrac{30^2}{2000}$

$T_2=293.74K$

So the exit temperature is 293.74 K.

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3 years ago

The time to failure for a gasket follows the Weibull distribution with ß = 2.0 and a characteristic life of 300 days. What is th

Aleks04 [339]

Answer:

64.11% for 200 days.

t=67.74 days for R=95%.

t=97.2 days for R=90%.

Explanation:

Given that

β=2

Characteristics life(scale parameter α)=300 days

We know that Reliability function for Weibull distribution is given as follows

$R(t)=e^{-\left(\dfrac{t}{\alpha}\right)^\beta}$

Given that t= 200 days

$R(200)=e^{-\left(\dfrac{200}{300}\right)^2}$

R(200)=0.6411

So the reliability at 200 days 64.11%.

When R=95 %

$0.95=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=67.74 days

When R=90 %

$0.90=e^{-\left(\dfrac{t}{300}\right)^2}$

by solving above equation t=97.2 days

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3 years ago