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Chapter 8: Problem 26

Give the approximate temperature at which creep deformation becomes an important consideration for each of the following metals: tin, molybdenum, iron, gold, zinc, and chromium.

Short Answer

Expert verified
Answer: The approximate temperatures for each metal are: - Tin: -104.77°C - Molybdenum: 692.23°C - Iron: 330.57°C - Gold: 172.57°C - Zinc: -42.27°C - Chromium: 453.57°C

Step by step solution

01

Convert melting points to Kelvin.

To convert from Celsius to Kelvin, add 273.15 to the melting points. - Tin: 232 + 273.15 = 505.15 K - Molybdenum: 2623 + 273.15 = 2896.15 K - Iron: 1538 + 273.15 = 1811.15 K - Gold: 1064 + 273.15 = 1337.15 K - Zinc: 419.5 + 273.15 = 692.65 K - Chromium: 1907 + 273.15 = 2180.15 K
02

Calculate one-third of the melting points in Kelvin.

Now, find one-third of the absolute melting temperatures (in Kelvin). - Tin: 505.15 / 3 ≈ 168.38 K - Molybdenum: 2896.15 / 3 ≈ 965.38 K - Iron: 1811.15 / 3 ≈ 603.72 K - Gold: 1337.15 / 3 ≈ 445.72 K - Zinc: 692.65 / 3 ≈ 230.88 K - Chromium: 2180.15 / 3 ≈ 726.72 K
03

Convert approximate creep temperatures back to Celsius.

Finally, convert the approximate creep temperatures back from Kelvin to Celsius by subtracting 273.15. - Tin: 168.38 - 273.15 ≈ -104.77°C - Molybdenum: 965.38 - 273.15 ≈ 692.23°C - Iron: 603.72 - 273.15 ≈ 330.57°C - Gold: 445.72 - 273.15 ≈ 172.57°C - Zinc: 230.88 - 273.15 ≈ -42.27°C - Chromium: 726.72 - 273.15 ≈ 453.57°C The approximate temperatures at which creep deformation becomes an important consideration for each metal are: - Tin: -104.77°C - Molybdenum: 692.23°C - Iron: 330.57°C - Gold: 172.57°C - Zinc: -42.27°C - Chromium: 453.57°C

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Most popular questions from this chapter

Steady-state creep rate data are given here for some alloy taken at \(200^{\circ} \mathrm{C}(473 \mathrm{K})\) .$$\begin{array}{cc}\hline \dot{\epsilon}_{s}\left(h^{-1}\right) & \sigma[M P a(p s i)] \\\\\hline 2.5 \times 10^{-3} & 55(8000) \\ 2.4 \times 10^{-2} & 69(10,000) \\\\\hline\end{array}$$.If it is known that the activation energy for creep is \(140,000 \mathrm{J} / \mathrm{mol}\), compute the steady state creep rate at a temperature of \(250^{\circ} \mathrm{C}\) \((523 \mathrm{K})\) and a stress level of \(48 \mathrm{MPa}(7000 \mathrm{psi})\).

Cite five factors that may lead to scatter in fatigue life data.

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of \(98.9 \mathrm{MPa} \sqrt{\mathrm{m}}(90 \mathrm{ksi} \sqrt{\text { in. }})\) and a yield strength of \(860 \mathrm{MPa}(125,000 \mathrm{psi}) .\) The flaw size resolution limit of the flaw detection apparatus is \(3.0 \mathrm{mm}(0.12 \text { in. }) .\) If the design stress is onehalf of the yield strength and the value of \(Y\) is \(1.0,\) determine whether or not a critical flaw for this plate is subject to detection.

Cite three metallurgical/processing techniques that are employed to enhance the creep resistance of metal alloys.

Steady-state creep data taken for an iron at a stress level of \(140 \mathrm{MPa}(20,000 \mathrm{psi})\) are given here $$\begin{array}{cc}\hline \dot{\epsilon}_{s}\left(h^{-1}\right) & T(K) \\\\\hline 6.6 \times 10^{-4} & 1090 \\\8.8 \times 10^{-2} & 1200 \\\\\hline\end{array}.$$ If it is known that the value of the stress exponent \(n\) for this alloy is \(8.5,\) compute the steady-state creep rate at \(1300 \mathrm{K}\) and a stress level of 83 MPa \((12,000 \text { psi })\).
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