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Chapter 8: Problem 1

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of \(1.9 \times 10^{-4} \mathrm{mm}\) \(\left(7.5 \times 10^{-6} \text {in. }\right)\) and a crack length of \(3.8 \times\) \(10^{-2} \mathrm{mm}\left(1.5 \times 10^{-3} \mathrm{in.}\right)\) when a tensile stress of \(140 \mathrm{MPa}(20,000 \mathrm{psi})\) is applied?

Short Answer

Expert verified
Answer: The magnitude of the maximum stress at the tip of the internal crack is approximately 389 MPa.

Step by step solution

01

Determine the geometric correction factor (Y)

We need to find the geometric correction factor (Y) for the internal crack problem. In general, the geometric correction factor is a function of the crack length (a) and the size of the component containing the crack. In most cases, the geometric correction factor can be found in literature or specialized handbooks. However, for this exercise, we will assume that the geometric correction factor (Y) is approximately equal to 1 as the crack size is very small.
02

Calculate the stress intensity factor (K)

Using the formula for the stress intensity factor and the given values for the tensile stress and the crack length, we can calculate the stress intensity factor (K) as follows: \(K = Y \cdot \sigma \cdot \sqrt{\pi a} = (1) \cdot (140 \times 10^6 \,\mathrm{N/m^2}) \cdot \sqrt{\pi \cdot (3.8 \times 10^{-5} \,\mathrm{m})}\) Now, we can calculate the stress intensity factor: \(K \approx 2.34 \times 10^4 \, \mathrm{N/m^{3/2}}\)
03

Calculate the maximum stress at the crack tip (\(\sigma_{max}\))

Now we can use the stress intensity factor (K) and the given radius of curvature (r) to find the maximum stress at the crack tip using the following equation: \(\sigma_{max} = K / (\sqrt{2 \pi r})\) \(\sigma_{max} \approx (2.34 \times 10^4 \, \mathrm{N/m^{3/2}}) / (\sqrt{2 \pi (1.9 \times 10^{-7} \, \mathrm{m})})\) After calculating the maximum stress at the crack tip, we get: \(\sigma_{max} \approx 389 \, \mathrm{MPa}\) The magnitude of the maximum stress at the tip of the internal crack when a tensile stress of 140 MPa is applied is approximately 389 MPa.

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

List four measures that may be taken to increase the resistance to fatigue of a metal alloy.

Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of \(26 \mathrm{MPa} \sqrt{\mathrm{m}}\) \((23.7 \mathrm{ksi} \sqrt{\mathrm{in.}}) .\) It has been determined that fracture results at a stress of \(112 \mathrm{MPa}\) \((16,240 \text { psi })\) when the maximum internal crack length is \(8.6 \mathrm{mm}(0.34 \text { in. }) .\) For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of \(6.0 \mathrm{mm}(0.24\) in.)

The following creep data were taken on an aluminum alloy at \(480^{\circ} \mathrm{C}\left(900^{\circ} \mathrm{F}\right)\) and a constant stress of \(2.75 \mathrm{MPa}(400 \mathrm{psi})\). Plot the data as strain versus time, then determine the steady-state or minimum creep rate. Note: The initial and instantaneous strain is not included.

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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