Foundations of Materials Science and Engineering
6th Edition
ISBN: 9781259696558
Author: SMITH
Publisher: MCG
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Chapter 17.14, Problem 33AAP
To determine
The largest internal flaw size that the bone can support.
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An Al2O3 specimen is being pulled in tension. The specimen contains flaws having a size of 100 µm.
(a) If the surface energy of Al2O3 is 0.8 J/m2, what is the fracture stress? Use
the Griffith criterion. E = 380 GPa.
(b) Using your vast fracture mechanics knowledge and advanced equations,
estimate the fracture stress if the fracture toughness is 4 MPa m1/2. Assume two positions for flaw: in the center of an infinite body and at the edge.
The rapture strength of bone tissue is 100 MPa and Young’s modulus is 10 GPa. Calculatevalue of longitudinal strain bone at which the fracture of the bone occurs.
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 0.20 mm and that has a tip radius of curvature of 0.001 mm when a stress of 1450 MPa is applied. The question is attached.
The answer is not 29000. WileyPlus marks it as incorrect.
Chapter 17 Solutions
Foundations of Materials Science and Engineering
Ch. 17.14 - Explain the difference between a biomaterial and...Ch. 17.14 - Explain why bone may be classified as a composite...Ch. 17.14 - Prob. 3KCPCh. 17.14 - Prob. 4KCPCh. 17.14 - Prob. 5KCPCh. 17.14 - What is stress shielding? How can it be avoided?Ch. 17.14 - Prob. 7KCPCh. 17.14 - What properties of biopolymers make them suitable...Ch. 17.14 - Prob. 9KCPCh. 17.14 - Prob. 10KCP
Ch. 17.14 - Prob. 11KCPCh. 17.14 - Prob. 12KCPCh. 17.14 - Prob. 13KCPCh. 17.14 - Prob. 14KCPCh. 17.14 - Prob. 15KCPCh. 17.14 - Prob. 16KCPCh. 17.14 - Prob. 17KCPCh. 17.14 - Prob. 18KCPCh. 17.14 - Prob. 19KCPCh. 17.14 - Prob. 20KCPCh. 17.14 - Prob. 21KCPCh. 17.14 - Prob. 22KCPCh. 17.14 - Prob. 23KCPCh. 17.14 - Prob. 24KCPCh. 17.14 - Prob. 25KCPCh. 17.14 - Prob. 26KCPCh. 17.14 - What is tissue engineering? What is the principle...Ch. 17.14 - Prob. 28KCPCh. 17.14 - Prob. 29KCPCh. 17.14 - Prob. 30AAPCh. 17.14 - Prob. 32AAPCh. 17.14 - Prob. 33AAPCh. 17.14 - Prob. 34AAPCh. 17.14 - Prob. 39SEPCh. 17.14 - Prob. 40SEPCh. 17.14 - Prob. 41SEPCh. 17.14 - Prob. 42SEPCh. 17.14 - Prob. 43SEPCh. 17.14 - Prob. 44SEPCh. 17.14 - A bone has fractured along an inclined plane as...Ch. 17.14 - Prob. 46SEPCh. 17.14 - Prob. 47SEPCh. 17.14 - Prob. 48SEPCh. 17.14 - Prob. 49SEPCh. 17.14 - What role does the water content play in the...Ch. 17.14 - Prob. 51SEPCh. 17.14 - Prob. 52SEPCh. 17.14 - When you wake up in the morning, you are taller...Ch. 17.14 - Prob. 54SEPCh. 17.14 - Prob. 55SEP
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- A material has a yield strength of 1150Mpa and a fracture toughness of 47 Mpa m^1/2. If the applied stress is 0.3 of the yield strength.Find the size of an internal crack that can lead to failure.arrow_forward8.a) Calculate the theoretical fracture strength of a brittle material wherein fracture occurs by propagation of an elliptically shaped 0.25 mm surface crack with a 1.2 × 10-3 mm tip radius under an applied 1200 MPa stress. b) If the specific surface energy for a material is 0.30 J/m2 and its elastic modulus is 69 GPa, calculate the minimum stress required for propagation of a surface crack 0.05 mm in length. Based on this calculation, estimate the materials fracture toughness.arrow_forwardThe mean fracture strength of a batch of alumina components is measured to be 400 MPa. What size of defects would you expect to find in the material? (Take the fracture toughness of alumina to be 4 MPam½ and Y to be 1.8).arrow_forward
- A specimen of diameter and length 10 mm & 100 mm respectively is subjected to tensile test. If the diameter and length at the fracture is 6 mm and 120 mm respectively. Then _________ is the Longitudinal strainarrow_forwardA mild steel tensile specimen of initial length 44 mm and initial diameter 6.4 mm is subjected to a tensile test and the following data are obtained. - Yield Strength as 88 MPa - Maximum Strength as 212 MPa - Fracture Strength as 152 MPa - Percentage of Elongation as 63 % - Percentage of Reduction in area as 39% The yield load in N= The Ultimate load in N = The Fracture load in N =arrow_forwardYou are called as an expert witness to analyze the fracture of a sintered silicon carbide plate that was fractured in bending when a blunt load was applied to the plate center. Measurement of the distance between the fracture origin and the mirror/mist boundary on the fracture surface gives a radius of 0.796 mm. You are given three pieces of the same SiC to test, and you determine that the mirror radius is 0.603, 0.203, and 0.162 mm for bending failure stress levels of 225, 368, and 442 MPa, respectively. What is your estimate of the stress present at the time of fracture for the original plate?arrow_forward
- If fracture load was 0.212 kN, what is the Fracture bending stress for a material having a width of "t" 50 mm and a thickness "w" of 5 mm and a support span "L" of 200 mm?arrow_forwardA three-point bending test is performed on a silicon carbide block that is 10 cm long, 1.5 cm wide, and 0.6 cm thick, and that is supported by two separate supports.7.5 cm. The sample breaks when a bending of 0.09 mm is recorded. The flexural modulus of silicon carbide is 480 GPa. Assume that no plastic deformation has occurred. Calculate: (a) The force that caused the fracture and(b) Flexural strength.arrow_forwardA titanium alloy (Ti-6Al-4V) has a yield strength of 1280 MPa and a fracture toughness of 77 MPa m1/2. If we apply a stress of 0.3σy, what will be the size of the surface crack that will lead to catastrophic failure?arrow_forward
- A material has a surface energy of 30 J/m^2 and an atomic spacing of 0.2nm.Young modulus of 375 Gpa, density is 3.5 g/cm^3, fracture toughness is 5 Mpa m^1/2 . Find the theoretical strength of this material and the flaw size that would cause the material to fail in tension at a stress of 591 Mpaarrow_forwardA titanium alloy specimen with an elastic modulus of 107 GPa and 3.9 mm in original diameter will experience elastic deformation when applied to a tensile load of 2695.2 N. Calculate the maximum length the specimen would reach if the maximum allowable elongation is 0.4 mm.arrow_forwardA cylindrical metal specimen having an original diameter of 12.8 mm (0.505 in.) and gauge length of m (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.60 mm (0.260 in.), and the fractured gauge length is 72.14 mm (2.840 in.). Calculate the ductility in terms of percent reduction in area and percent elongation.arrow_forward
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