AEB650_Materials_For_Engineering_chapter_8_solutions

AEB650 Materials For Engineering Chapter 8: Fracture Mechanics Worked Solutions 1. State the three principal causes for mechanical failure. A) Improper materials selection or processing B) Inadequate design C) Component misuse

2. 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 x 10-4 mm and a crack length of 3.8 x 10-2 mm when a tensile stress of 140 MPa is applied? (2800 MPa) This problem asks that we compute the magnitude of the maximum stress that exists at the tip of an internal crack. Equation (8.1b) is employed to solve this problem, as 1/2 σ m = 2σ o

 (3.8 x10 −2 mm ) / 2  1/2 = 2800 MPa  −4  1.9 x10 mm 

= (2)(140 MPa) 

3. Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.25 mm and having a tip radius of curvature of 1.2 x 10-3 mm when a stress of 1200 MPa is applied. (3.5 x 104 MPa) In order to estimate the theoretical fracture strength of this material it is necessary to calculate σ m using Equation (8.1b) given that σ o = 1200 MPa, a = 0.25 mm, and ρ t = 1.2 x 10-3 mm. Thus, σ m = 2σ o = (2)(1200 MPa)= 3.5 x 104 MPa

4. A specimen of a ceramic material having a modulus of elasticity of 300 GPa is pulled in tension with a stress of 900 MPa. Will the specimen fail if its "most severe flaw" is an internal crack that has a length of 0.30 mm and a tip radius of curvature of 5 x 10-4 mm? Why or why not? (31.2 GPa, Yes) In order to determine whether or not this ceramic material will fail we must compute its theoretical fracture (or cohesive) strength; if the maximum strength at the tip of the most severe flaw is greater than this value then fracture will occur--if less than, then there will be no fracture. The theoretical fracture strength is just E/10 or 30 GPa, inasmuch as E = 300 GPa.

The magnitude of the stress at the most severe flaw may be determined using Equation

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σ m = 2σ o

= (2)(900 MPa)= 31.2 GPa

Therefore, fracture will occur since this value is greater than E/10. 5. Describe the significant differences between the stress intensity factor, the plane stress fracture toughness, and the plane strain fracture toughness. The stress intensity factor is a parameter used in expressions such as: K = Yσ πa …(eqn. 8) Its value is variable and dependent on applied stress and crack length. On the other hand, plane strain and plane stress fracture toughnesses represent unique and critical values of K at which crack propagation occurs. However, plane strain fracture toughness is this critical value for specimens thicker than some minimum threshold thickness, while plane stress is for specimens thinner than this threshold.

6. A specimen of a 4340 steel alloy having a plane strain fracture toughness of 45 MPa is exposed to a stress of 1000 MPa m . Will this specimen experience fracture if it is known that the largest surface crack is 0.75 mm long? Why or why not? Assume that the parameter Y has a value of 1.0. (927 MPa, Yes)

This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1000 MPa, given the values of KIc, Y, and the largest value of a in the material. This requires that we solve for σ c from Equation (8.13).

Thus σ c = = = 927 MPa

Therefore, fracture will most likely occur because this specimen will tolerate a stress of 927 MPa before fracture, which is less than the applied stress of 1000 MPa.

7. Some aircraft component is fabricated from an aluminium alloy that has a plane strain fracture toughness of 35 MPa m . It has been determined that fracture results at a stress of 250 MPa when the maximum (or critical) internal crack length is 2.0 mm. For this same component and alloy, will fracture occur at a stress level of 325 MPa when the maximum internal crack length is

2

1.0 mm? Why or why not? No)

(32.2

MPa,

We are asked to determine if an aircraft component will fracture for a given fracture toughness (35 MPa), stress level (325 MPa), and maximum internal crack length (1.0 mm), given that fracture occurs for the same component using the same alloy for another stress level and internal crack length. It first becomes necessary to solve for the parameter Y for the conditions under which fracture occurred using Equation (8.11). Therefore,

Y = = = 2.50

Now we will solve for the product Yσ for the other set of conditions, so as to ascertain whether or not this value is greater than the KIc for the alloy. Thus,

Yσ = (2.50)(325 MPa)

= 32.2 MPa

m

Therefore, fracture will not occur since this value (32.2 MPa) is less than the KIc of the material-35 MPa.

8. Suppose that a wing component on an aircraft is fabricated from an aluminium alloy that has a plane strain fracture toughness of 40 MPa m . It has been determined that fracture results at a stress of 365 MPa when the maximum internal crack length is 2.5 mm. For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 4.0 mm. (288 MPa) This problem asks us to determine the stress level at which an aircraft component will fracture for a given fracture toughness (40 MPa) and maximum internal crack length (4.0 mm), given that fracture occurs for the same component using the same alloy at one stress level and another internal crack length. It first becomes necessary to solve for the parameter Y for the conditions under which fracture occurred using Equation (8.11). Therefore,

Y = = = 1.75 Now we will solve for σ c using Equation (8.13) as

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σ c = = = 288 MPa

9. A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa m . If, during service use, the plate is exposed to a tensile stress of 200 MPa, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y. (24 mm) For this problem, we are given values of KIc, σ , and Y for a large plate and are asked to determine the minimum length of a surface crack that will lead to fracture. All we need do is to solve for ac using Equation (8.14); therefore 2 ac =

2 =

= 0.024 m = 24 mm

10. Calculate the maximum internal crack length allowable for a 7075-T651 aluminium alloy (Table.1) component that is loaded to a stress one half of its yield strength. Assume that the value of Y is 1.35. (3.3 mm) This problem asks us to calculate the maximum internal crack length allowable for the 7075-T651 aluminum alloy in Table 8.1 given that it is loaded to a stress level equal to one-half of its yield strength. For this alloy, KIc = 24 MPa m and σ = σ y/2 = (495 MPa)/2 = 248 MPa . Now solving for 2ac using Equation (8.14) yields

2 2ac =

2 =

= 0.0033 m = 3.3 mm

11. 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 77 MPa m and a yield strength of 1400 MPa. The flaw size resolution limit of the flaw detection apparatus is 4.0 mm. If the design stress is one half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection. (3.9 mm) This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus (4.0 mm), the value of KIc (77 MPa), the design stress (σ y/2) in which σ y = 1400 MPa, and Y = 1.0. We first need to compute the value of ac using Equation (8.14); thus

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2 ac =

2 =

= 0.0039 m = 3.9 mm

Therefore, the critical flaw is not subject to detection since this value of ac is less than the 4.0 mm resolution limit.

12. Consider a steel plate having a through thickness edge crack similar to that shown in Figure 13a. If it is known that the minimum crack length subject to detection is 2 mm, determine the minimum allowable plate width assuming a plane strain fracture toughness of 80 MPa m , a yield strength of 825 MPa, and that the plate is to be loaded to one half of its yield strength. (4.4 mm) This problem asks that we determine, for a steel plate having a through-thickness edge crack, to determine the minimum allowable plate width to ensure that fracture will not occur if the minimum crack length that is subject to detection is 2 mm. We are also given that KIc = 80 MPaand that the plate may be loaded to half its yield strength, where σ y = 825 MPa.

First of all the applied stress is just σ = = = 412.5 MPa

Now, using Equation (8.11) we solve for the value of Y assuming that a = 2.0 mm as

Y=

= = 2.45

In Figure 8.13a is plotted Y versus a/W for the crack-plate geometry of this problem; from this plot, for Y = 2.45, a/W = 0.45. Since the minimum crack length for detection is 2 mm, the minimum width allowable is just

W = = = 4.4 mm

13. Consider a steel plate having a through thickness edge crack similar to that shown in Figure 13a; the plate width (W) is 75 mm, and its thickness (B) is 12.0 mm. Furthermore, plane strain fracture toughness and yield strength values for this material are 80 MPa m and 1200 MPa, respectively. If the plate is to be loaded to a stress of 300 MPa, would you expect failure to occur if the crack length a is 15 mm? Why or why not? (87.9 MPa m , Yes) This problem asks that we consider a steel plate having a through-thickness edge crack, and to determine if fracture will occur given the following: W = 75 mm, B = 12.0 mm, KIc = 80 MPa, σ y

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= 1200 MPa, σ = 300 MPa, and a = 15 mm.

The first thing to do is determine whether

conditions of plane strain exist. From Equation (8.12), 2 2.5 2 = 2.5

= 0.011 m = 11 mm

Inasmuch as the plate thickness is 12 mm (which is greater than 11 mm), the situation is a plane strain one. Next, we must determine the a/W ratio, which is just 15 mm/75 mm = 0.20. From this ratio and using Figure 8.13a, Y = 1.35. At this point it becomes necessary to determine the value of the Yσ product; if it is greater than KIc then fracture will occur. Thus

Yσ = (1.35)(300 MPa)

= 87.9 MPa

m

Therefore, fracture will occur since this value (87.9 MPa) is greater than the KIc for the steel (80 MPa).

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