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Chapter 8: Problem 5

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 psi). Will this specimen experience fracture if the largest surface crack is \(0.5 \mathrm{~mm}\) (0.02 in.) long? Why or why not? Assume that the parameter \(Y\) has a value of \(1.0 .\)

Short Answer

Expert verified
Answer: No, the specimen will not experience fracture because the stress intensity factor (46.47 MPa m^1/2) is less than the plane strain fracture toughness (54.8 MPa m^1/2).

Step by step solution

01

Understand the stress intensity factor formula

Using the formula for stress intensity factor, \(K_I = Y \sigma \sqrt{\pi a}\), where \(K_I\) is the stress intensity factor, \(Y\) is a geometric factor, \(\sigma\) is the applied stress, and \(a\) is the crack length.
02

Convert crack length to meters

We need to convert the given crack length from millimeters to meters for consistency with the plane strain fracture toughness unit. \(0.5 \mathrm{~mm}\) = \(5\times10^{-4} \mathrm{~m}\).
03

Calculate the stress intensity factor

Now, we can plug in our given values into the stress intensity factor formula: \(K_I = Y \sigma \sqrt{\pi a} = 1.0 \times (1030 \mathrm{MPa}) \times \sqrt{\pi (5\times10^{-4} \mathrm{~m}})\).
04

Determine the value of the stress intensity factor

Calculate the value of \(K_I\): \(K_I=1.0 \times 1030 \times \sqrt{\pi \times 5\times10^{-4}} \approx 46.47 \mathrm{MPa~m^{1/2}}\)
05

Compare the stress intensity factor to the plane strain fracture toughness

Comparing the stress intensity factor value to the plane strain fracture toughness value (54.8 MPa m^1/2): Since the value of \(K_I\) (46.47 MPa m^1/2) is less than the value of \(K_{Ic}\) (54.8 MPa m^1/2), the specimen will not experience fracture. In conclusion, under the given conditions, the specimen with the largest crack size of 0.5 mm will not experience fracture due to the stress intensity factor being less than the plane strain fracture toughness.

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Most popular questions from this chapter

The fatigue data for a brass alloy are given as follows: $$ \begin{array}{cc} \hline \text { Stress Amplitude (MPa) } & \text { Cycles to Failure } \\ \hline 170 & 3.7 \times 10^{4} \\ \hline 148 & 1.0 \times 10^{5} \\ \hline 130 & 3.0 \times 10^{5} \\ \hline 114 & 1.0 \times 10^{6} \\ \hline 92 & 1.0 \times 10^{7} \\ \hline 80 & 1.0 \times 10^{8} \\ \hline 74 & 1.0 \times 10^{9} \\ \hline \end{array} $$ (a) Make an \(S-N\) plot (stress amplitude versus logarithm of cycles to failure) using these data. (b) Determine the fatigue strength at \(4 \times 10^{6}\) cycles. (c) Determine the fatigue life for \(120 \mathrm{MPa}\).

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

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 \(98.9 \mathrm{MPa} \sqrt{\mathrm{m}}(90 \mathrm{ksi} \sqrt{\mathrm{in} .})\) and a yield strength of \(860 \mathrm{MPa}(125,000 \mathrm{psi})\). The flaw size resolution limit of the flaw detection apparatus is \(3.0 \mathrm{~mm}(0.12 \mathrm{in}\).). If the design stress is one-half the yield strength and the value of \(Y\) is \(1.0\), determine whether a critical flaw for this plate is subject to detection.

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 \(Y\).

A cylindrical 2014-T6 aluminum alloy bar is subjected to compression-tension stress cycling along its axis; results of these tests are shown in Figure \(8.20 .\) If the bar diameter is \(12.0 \mathrm{~mm}\), calculate the maximum allowable load amplitude (in N) to ensure that fatigue failure will not occur at \(10^{7}\) cycles. Assume a factor of safety of \(3.0\), data in Figure \(8.20\) were taken for reversed axial tensioncompression tests, and that \(S\) is stress amplitude.
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