For a steel alloy it has been determined that a carburizing heat treatment of 9-h duration will raise the carbon concentration to 0.38 wt% at a point 1.2 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 6.4 mm position for an identical steel and at the same carburizing temperature.

Answer:

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Explanation:

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Answer:

Attached below is the derived next state equations

Explanation:

Attached below is the derived next state equations

used for the solution of the given problem.

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Answer:

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Explanation: i took the quiz

Answer:

Explanation:

From the information given:

[tex]E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa \\ \\ d = 25 \ mm \ \\ \\ D = 45 \ mm \ \\ \\ L = 761 \ mm \\ \\ P = -88 kN[/tex]

The total load is distributed across both the rod and tube:

[tex]P = P_1+P_2 --- (1)[/tex]

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

[tex]\delta_1=\delta_2[/tex]

[tex]\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}[/tex]

[tex]\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}[/tex]

[tex]P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})[/tex]

[tex]P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}[/tex]

[tex]P_2 = 6.6212 \ P_1[/tex]

Replace [tex]P_2[/tex] into equation (1)

[tex]P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\ -88 = 7.6212 \ P_1 \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\ P_1 = -11.547 \ kN[/tex]

Finally, to determine the normal stress in aluminum rod:

[tex]\sigma _1 = \dfrac{P_1}{A_1} \\ \\ \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}[/tex]

[tex]\sigma_1 = - 23.523 \ MPa}[/tex]

Thus, the normal stress = 23.523 MPa in compression.

Answer:

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Explanation:

Answer:

Explanation:

The missing part of the question is attached in the diagram below, the second diagram shows the schematic view of the stress-strain curve and the plane stress.

From the given information:

The elastic modulus is:

[tex]E = \dfrac{\sigma}{\varepsilon} \\ \\ E = \dfrac{150 \ MPa}{0.0217} \\ \\ E = 69.124 \ GPa[/tex]

Hence, suppose 0.2% offset cuts the stress-strain curve at a designated point A from the image attached below, then the yield strength relating to the stress axis from the curve will be [tex]\sigma_y[/tex] = 270 MPa.

The shear yield strength by using von Mises criteria is estimated as;

[tex]\tau_1 = \dfrac{\sqrt{2}}{3}\sigma_y \\ \\ \tau_1 = \dfrac{\sqrt{2}}{3}*270 \\ \\ \tau_1 = 127.28 \ MPa[/tex]

The shear yield strength by using Tresca criteria is:

[tex]\tau_2 = \dfrac{1}{2}\sigma_y \\ \\ \tau_2= \dfrac{1}{2}*270 \\ \\ \tau_2 = 135 \ MPa[/tex]