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Chapter 8: Problem 34

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 \(48 \mathrm{MPa}(7000 \mathrm{psi})\).

Short Answer

Expert verified
Based on the given step-by-step solution, calculate the steady-state creep rate for an alloy at a temperature of 523 K and a stress level of 48 MPa.

Step by step solution

01

Identify the Creep Rate Equation

The steady-state creep rate, \(\dot{\epsilon}_{s}\), can be represented by the equation: $$\dot{\epsilon}_{s} = A \sigma^n e^{-Q/RT}$$ Where \(\dot{\epsilon}_{s}\) = steady-state creep rate, \(A\) = material constant, \(\sigma\) = applied stress, \(n\) = stress exponent, \(Q\) = activation energy for creep, \(R\) = universal gas constant \((8.314 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol})\), \(T\) = temperature in Kelvin.
02

Set Up the First Known Condition

From the given data: $$\dot{\epsilon}_{s1} = 2.5 \times 10^{-3} \mathrm{h}^{-1}$$ $$\sigma_1 = 55 \mathrm{MPa}$$ $$T_1 = 473 \mathrm{K}$$ So, the equation becomes: $$2.5 \times 10^{-3} = A \times (55)^n \times e^{-Q/(8.314 \times 473)}$$
03

Set up the Second Known Condition

From the given data: $$\dot{\epsilon}_{s2} = 2.4 \times 10^{-2} \mathrm{h}^{-1}$$ $$\sigma_2 = 69 \mathrm{MPa}$$ $$T_2 = 473 \mathrm{K}$$ So, the equation becomes: $$2.4 \times 10^{-2} = A \times (69)^n \times e^{-Q/(8.314 \times 473)}$$
04

Solve for A and n

Divide the second equation by the first equation to eliminate \(A\) and \(Q\): $$\frac{2.4 \times 10^{-2}}{2.5 \times 10^{-3}} = \frac{(69)^n}{(55)^n}$$ Solve for n: $$n = \frac{\log{(\frac{2.4 \times 10^{-2}}{2.5 \times 10^{-3}})}}{\log{(69/55)}}$$ Now, substitute the found n value and any known condition to find the value of \(A\): $$2.5 \times 10^{-3} = A \times (55)^n \times e^{-Q/(8.314 \times 473)}$$
05

Calculate the Steady-State Creep Rate for the Given Condition

Given: $$T_3 = 523 \mathrm{K}$$ $$\sigma_3 = 48 \mathrm{MPa}$$ The equation becomes: $$\dot{\epsilon}_{s3} = A \times (48)^n \times e^{-Q/(8.314 \times 523)}$$ Substitute the values of \(A\), \(n\), and \(Q\) into the equation and solve for \(\dot{\epsilon}_{s3}\) to calculate the steady-state creep rate at the given temperature and stress level.

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Most popular questions from this chapter

A specimen of a 4340 steel alloy with a plane strain fracture toughness of \(54.8 \mathrm{MPa} \sqrt{\mathrm{m}}\) \((50 \mathrm{ksi} \sqrt{\mathrm{in.}})\) is exposed to a stress of \(1030 \mathrm{MPa}\) \((150,000 \mathrm{psi}) .\) Will this specimen experience fracture if it is known that the largest surface crack is \(0.5 \mathrm{mm}(0.02 \text { in. })\) long? Why or why not? Assume that the parameter \(Y\) has a value of 1.0.

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

A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of \(82.4 \mathrm{MPa} \sqrt{\mathrm{m}}(75.0 \mathrm{ksi} \sqrt{\text { in. }}) .\) If, during service use, the plate is exposed to a tensile stress of \(345 \mathrm{MPa}(50,000 \mathrm{psi})\), determine the minimum length of a surface crack that will leadd to fracture. Assume a value of 1.0 for \(Y\).

Following is tabulated data that were gathered from a series of Charpy impact tests on a ductile cast iron. $$ \begin{array}{cc} \hline \text { Temperature }\left({ }^{\circ} \boldsymbol{C}\right) & \text { Impact Energy }(\boldsymbol{J}) \\ \hline-25 & 124 \\ -50 & 123 \\ -75 & 115 \\ -85 & 100 \\ -100 & 73 \\ -110 & 52 \\ -125 & 26 \\ -150 & 9 \\ -175 & 6 \\ \hline \end{array} $$ (a) Plot the data as impact energy versus temperature. (b) Determine a ductile-to-brittle transition temperature as that temperature corresponding to the average of the maximum and minimum impact energies. (c) Determine a ductile-to-brittle transition temperature as that temperature at which the impact energy is \(80 \mathrm{~J}\).

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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