Problem

Calculate the maximum internal crack length allowable for a Ti-6Al-4V titanium alloy (Tabl...

Calculate the maximum internal crack length allowable for a Ti-6Al-4V titanium alloy (Table 8.1) component that is loaded to a stress one-half its yield strength. Assume that the value of Y is 1.50.

Table 8.1 Room-Temperature Yield Strength and Plane Strain Fracture Toughness Data for Selected Engineering Materials

 

Yield Strength

 

KIc

 

Material

MPa

ksi

MPa√m

ksi√in.

Aluminum alloya (7075-T651)

495

72

24

22

Aluminum alloya (2024-T3)

345

50

44

40

Titanium alloya (Ti-6Al-4V)

910

132

55

50

Alloy steela (4340 tempered @ 260°C)

1640

238

50.0

45.8

Alloy steela (4340 tempered @ 425°C)

1420

206

87.4

80.0

 

Ceramics

 

 

 

Concrete

0.2–1.4

0.18–1.27

Soda-lime glass

0.7–0.8

0.64–0.73

Aluminum oxide

2.7–5.0

2.5–4.6

 

Polymers

 

 

 

Polystyrene (PS)

25.0–69.0

3.63–10.0

0.7–1.1

0.64–1.0

Poly(methyl methacrylate) (PMMA)

53.8–73.1

7.8–10.6

0.7–1.6

0.64–1.5

Polycarbonate (PC)

62.1

9.0

2.2

2.0

aSource: Reprinted with permission, Advanced Materials and Processes, ASM International, © 1990.

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