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3 years ago

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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.

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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Answer:

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KVL holds for the supermesh, so we can write a KVL equation to generate the second equation we need to solve for the two unknown

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Answer:

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Explanation:

As the complete question is not given the complete question is found online and is attached herewith.

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$2V_{\Delta}+5K(i_z-i_y)=0\\2*1K*i_y+5K(i_z-i_y)=0\\2K*i_y+5K(i_z-i_y)=0\\3i_y-5i_z=0$

So the system of equations become

$i_x-i_y+0i_z=0.05\\3i_x+6i_y-5i_z=0\\0i_x-3i_y+5i_z=0$

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Answer:

Check Explanation.

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In a simple way, Engineering Economy simply refers to the study of Economics which is related to engineers that is the study of Economic decisions by people in the engineering field. The study of Engineering Economy is very important because Engineering is a major manufacturing part in every country's economy.

With the study of Economics by Engineering that is Engineering Economy, engineers can make rational decisions after seeing alternatives.

The foundation of Engineering Economy in terms of seven basic principles:

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Answer:

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Anvisha [2.4K]

Answer:

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Explanation:

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For Answer (a) please find the attachment.

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