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Klio2033 [76]
3 years ago
12

Two dogbone specimens of identical geometry but made of two different materials: steel and aluminum are tested under tension at

a fixed load rate. If the displacement rate needed to achieve the specified load rate for aluminum specimen is 0.001 inch/min, determine the displacement rate that must be applied on the steel specimen to achieve the specified load rate. Given, Young's modulus (E) of steel is 29,000 ksi and that of aluminum is 10,000 ksi.
Engineering
2 answers:
Kisachek [45]3 years ago
8 0

Answer:

vsteel = 0.00034483 inch/min

Explanation:

Given

vAl = 0.001 inch/min

EAl = 10,000 ksi

Esteel = 29,000 ksi

P is the same value for the dogbone specimens

A is the same value for the dogbone specimens

L is the same value for the dogbone specimens

We can apply

ΔL = P*L/(A*E)

⇒ ΔL*E = P*L/A

then

ΔLAl*EAl = ΔLsteel*Esteel

ΔLAl*10,000 ksi = ΔLsteel*29,000 ksi

⇒ ΔLAl = 2.9*ΔLsteel

If  ΔLAl = 2.9*ΔLsteel  we have

vAl = 2.9*vsteel

0.001 inch/min = 2.9*vsteel

⇒  vsteel = 0.00034483 inch/min

makkiz [27]3 years ago
7 0

Answer:

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

Explanation:

The Young's module is:

E = \frac{\sigma}{\frac{\Delta L}{L_{o}} }

E = \frac{\sigma\cdot L_{o}}{\dot L \cdot \Delta t}

Let assume that both specimens have the same geometry and load rate. Then:

E_{aluminium} \cdot \dot L_{aluminium} = E_{steel} \cdot \dot L_{steel}

The displacement rate for steel is:

\dot L_{steel} = \frac{E_{aluminium}}{E_{steel}}\cdot \dot L_{aluminium}

\dot L_{steel} = \left(\frac{10000\,ksi}{29000\,ksi}\right)\cdot (0.001\,\frac{in}{min} )

\dot L_{steel} = 3.448\times 10^{-4}\,\frac{in}{min}

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What is hardness and how is it generally tested?
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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:  

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  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.

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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.  

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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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