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Chapter 8: Problem 9

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

Short Answer

Expert verified
Answer: The minimum length of a surface crack that will lead to fracture is 1.798 x 10^(-5) m or 17.98 µm.

Step by step solution

01

Rearrange the formula to solve for a

The formula we need to use is \(K_I = Y \times \sigma \times \sqrt{\pi a}\). To solve for \(a\), we will rearrange the equation: \(a = \dfrac{K_I^2}{(Y \sigma)^2 \pi}\) Now we can plug in the given values to find the minimum crack length.
02

Plug in given values

We are given the following values: - \(K_I = 82.4 \thinspace \mathrm{MPa} \sqrt{\mathrm{m}}\) - \(\sigma = 345 \thinspace \mathrm{MPa}\) - \(Y = 1.0\) Now we can plug these values into the rearranged formula to find the minimum crack length: \(a = \dfrac{(82.4 \thinspace \mathrm{MPa} \sqrt{\mathrm{m}})^2}{(1.0 \times 345 \thinspace \mathrm{MPa})^2 \pi}\)
03

Solve for the crack length a

Now we can calculate the value of \(a\): \(a = \dfrac{(82.4 \thinspace \mathrm{MPa} \sqrt{\mathrm{m}})^2}{(1.0 \times 345 \thinspace \mathrm{MPa})^2 \pi} = \dfrac{6785.76 \thinspace \mathrm{MPa}^2 \mathrm{m}}{(345 \thinspace \mathrm{MPa})^2 \pi} = \dfrac{6785.76 \thinspace \mathrm{MPa}^2 \mathrm{m}}{119025 \thinspace \mathrm{MPa}^2 \pi} = 1.798 \times 10^{-5} \thinspace \mathrm{m}\) Therefore, the minimum length for a surface crack that will lead to fracture is: \(a = 1.798 \times 10^{-5} \thinspace \mathrm{m}\) or \(17.98\thinspace\mu\mathrm{m}\).

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Most popular questions from this chapter

A cylindrical rod of diameter \(6.7 \mathrm{~mm}\) fabricated from a \(70 \mathrm{Cu}-30 \mathrm{Zn}\) brass alloy is subjected to rotating-bending load cycling; test results (as \(S-N\) behavior) are shown in Figure 8.20. If the maximum and minimum loads are \(+120 \mathrm{~N}\) and \(-120 \mathrm{~N}\), respectively, determine its fatigue life. Assume that the separation between loadbearing points is \(67.5 \mathrm{~mm}\).

A cylindrical component constructed from an S-590 alloy (Figure 8.31) has a diameter of \(14.5 \mathrm{~mm}\) (0.57 in.). Determine the maximum load that may be applied for it to survive \(10 \mathrm{~h}\) at \(925^{\circ} \mathrm{C}\left(1700^{\circ} \mathrm{F}\right)\).

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

(a) Using Figure 8.31, compute the rupture lifetime for an \(S-590\) alloy that is exposed to a tensile stress of \(400 \mathrm{MPa}\) at \(815^{\circ} \mathrm{C}\). (b) Compare this value to the one determined from the Larson-Miller plot of Figure \(8.33\), which is for this same S-590 alloy.

An aircraft component is fabricated from an aluminum alloy that has a plane- strain fracture toughness of \(40 \mathrm{MPa} \sqrt{\mathrm{m}}\) (36.4 ksi \sqrt{in.). It has been deter- } mined that fracture results at a stress of \(300 \mathrm{MPa}\) (43,500 psi) when the maximum (or critical) internal crack length is \(4.0 \mathrm{~mm}\) (0.16 in.). For this same component and alloy, will fracture occur at a stress level of \(260 \mathrm{MPa}\) ( 38,000 psi) when the maximum internal crack length is \(6.0 \mathrm{~mm}(0.24\) in.) \(?\) Why or why not?
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