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Chapter 8: Problem 8

A structural component is fabricated from an alloy that has a plane-strain fracture toughness of \(62 \mathrm{MPa} \sqrt{\mathrm{m}}\). It has been determined that this component fails at a stress of \(250 \mathrm{MPa}\) when the maximum length of a surface crack is \(1.6 \mathrm{~mm}\). What is the maximum allowable surface crack length (in mm) without fracture for this same component exposed to a stress of \(250 \mathrm{MPa}\) and made from another alloy with a plane-strain fracture toughness of \(51 \mathrm{MPa} \sqrt{\mathrm{m}}\) ?

Short Answer

Expert verified
Answer: The maximum allowable surface crack length for the second alloy is approximately \(1.08 \text{ mm}\).

Step by step solution

01

Calculate the stress intensity factor for the first alloy

We are given the fracture toughness for the first alloy \((K_{IC1}=62 \text{ MPa}\sqrt{\text{m}})\), stress \((\sigma = 250 \text{ MPa})\), and the crack length \((a_1 = 1.6 \text{ mm})\). Let's calculate the stress intensity factor, \(K_{I1}\), for the first alloy using the formula: \(K_I = \sigma \sqrt{\pi a}\) \(K_{I1} = (250 \text{ MPa}) \sqrt{\pi(1.6 \times 10^{-3} \text{m})}\)
02

Establish a relationship between the stress intensity factors and the fracture toughness of both alloys

We can relate the stress intensity factors and fracture toughness of both alloys (\(K_{IC1}\) and \(K_{IC2}\)): \(\frac{K_{I1}}{K_{IC1}} = \frac{K_{I2}}{K_{IC2}}\) We have \(K_{I1}\) and \(K_{IC1}\) from Step 1 and we are given the fracture toughness of the second alloy (\(K_{IC2} = 51 \text{ MPa}\sqrt{\text{m}}\)).
03

Determine the crack length for the second alloy

We can now find the crack length for the second alloy \((a_2)\) using the relationship from Step 2: \(\frac{K_{I1}}{K_{IC1}} = \frac{\sigma \sqrt{\pi a_2}}{K_{IC2}}\) \(a_2 = \frac{K_{IC2}^2}{\sigma^2\pi}\left(\frac{K_{I1}}{K_{IC1}}\right)^2\) Use the given values to solve for \(a_2\): \(a_2 = \frac{(51 \text{ MPa}\sqrt{\text{m}})^2}{(250 \text{ MPa})^2\pi}\left(\frac{(250 \text{ MPa}) \sqrt{\pi(1.6 \times 10^{-3} \text{m})}}{62 \text{ MPa}\sqrt{\text{m}}}\right)^2\) \(a_2 ≈ 1.08 \times 10^{-3} \text{ m}\)
04

Convert the crack length to mm

Finally, convert the crack length from meters to millimeters: \(a_2 ≈ 1.08 \times 10^{-3} \text{ m} \times \frac{1000 \text{ mm}}{1 \text{ m}} ≈ 1.08 \text{ mm}\) So, the maximum allowable surface crack length for the second alloy is approximately \(1.08 \text{ mm}\).

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Most popular questions from this chapter

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?

A fatigue test was conducted in which the mean stress was 70 MPa (10,000 psi), and the stress amplitude was \(210 \mathrm{MPa}(30,000\) psi). (a) Compute the maximum and minimum stress levels. (b) Compute the stress ratio. (c) Compute the magnitude of the stress range.

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

(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.

Briefly explain the difference between fatigue striations and beachmarks in terms of (a) size and (b) origin.
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