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- 1. 5.5 Mechanical Behavior-Metals Tensile Properties 4. Resilience • It is the capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, to have this energy recovered. The associated property is the modulus of resilience, 𝑈𝑟, which is the energy per unit volume required to stress a material from an unloaded state up to the point of yielding. 𝑈𝑟 = න 0 𝜖𝑦 𝜎 𝑑𝜖 • The units of resilience are the product of the units from each of the two axes of the stress– strain plot. For SI units, this is joules per cubic meter (J/m3, equivalent to Pa). Thus, this area under the stress–strain curve represents energy absorption per unit volume of the material.
- 2. 5.5 Mechanical Behavior-Metals Tensile Properties • Assuming a linear elastic region, incorporation of Hooke’s law yields: 𝑈𝑟 = 1 2 𝜎𝑦𝜖𝑦 = 1 2 𝜎𝑦 𝜎𝑦 𝐸 = 1 2 𝜎𝑦 2 𝐸 where 𝜖𝑦 is the strain at yielding (i.e., yield strain). • Thus, resilient materials are those having high yield strengths (𝜎𝑦) and low moduli of elasticity (𝐸); such alloys would be used in spring applications.
- 3. 5.5 Mechanical Behavior-Metals Tensile Properties 5. Toughness • It is a measure of the ability of a material to absorb energy up to fracture. • It is computed from the area under the 𝜎 − 𝜖 curve up to the point of fracture. • The units for toughness are the same as for resilience (i.e., energy per unit volume).
- 4. 5.5 Mechanical Behavior-Metals Tensile Properties • For a material to be tough, it must display both strength and ductility; often, ductile materials are tougher than brittle ones. • Hence, even though the brittle material has higher yield and tensile strengths, it has a lower toughness than the ductile one, by virtue of lack of ductility; this is deduced by comparing the areas 𝐴𝐵𝐶 and 𝐴𝐵′𝐶′.
- 5. Q: Of those metals listed in the following table: a) Which will experience the greatest percent reduction in area? Why? b) Which is the strongest? Why? c) Which is the stiffest? Why?
- 6. Answers: a) Material B will experience the greatest percent area reduction since it has the highest strain at fracture and therefore is most ductile. b) Material D is the strongest (hardest) because it has the highest yield and tensile strengths. c) Material E is the stiffest because it has the highest elastic modulus.
- 7. 5.5 Mechanical Behavior-Metals True Stress and Strain • The decline in the stress necessary to continue deformation past the maximum, point M, seems to indicate that the metal is becoming weaker. However, the cross-sectional area is decreasing rapidly within the neck region, where deformation is occurring. • The engineering stress, as computed earlier, is on the basis of the original cross-sectional area before any deformation and does not take into account this reduction in area at the neck. • Sometimes, it is more meaningful to use a true stress–true strain scheme. True stress (𝜎𝑇) is defined as the load (𝐹) divided by the instantaneous cross-sectional area (𝐴𝑖) over which deformation is occurring (i.e., the neck, past the tensile point), or 𝜎𝑇 = 𝐹 𝐴𝑖
- 8. 5.5 Mechanical Behavior-Metals True Stress and Strain • Furthermore, it is occasionally more convenient to represent strain as true strain 𝜖𝑇, defined by: 𝜖𝑇 = ln 𝑙𝑖 𝑙𝑜 • If no volume change occurs during deformation- i.e., if 𝐴𝑖 𝑙𝑖 = 𝐴𝑜 𝑙𝑜 true and engineering stress and strain are related by: 𝜎𝑇 = 𝜎 1 + 𝜖 𝜖𝑇 = ln 1 + 𝜖
- 9. 5.5 Mechanical Behavior-Metals True Stress and Strain Example A cylindrical specimen of steel having an original diameter of 12.8 mm is tensile tested to fracture and found to have an engineering fracture strength 𝜎𝑓 of 460 MPa. If its cross-sectional diameter at fracture is 10.7 mm, determine: a) The ductility in terms of percent reduction in area. b) The true stress at fracture.
- 10. 5.5 Mechanical Behavior-Metals True Stress and Strain
- 11. 5.6 Mechanical Behavior-Ceramics • Ceramic materials are somewhat limited in applicability by their mechanical properties, which in many respects are low-grade to those of metals. The principal drawback is a disposition to catastrophic fracture in a brittle manner with very little energy absorption. • The stress–strain behavior of brittle ceramics is not usually determined by a tensile test as outlined earlier, for three reasons: 1) it is difficult to prepare and test specimens having the required geometry. 2) it is difficult to control brittle materials without fracturing them. 3) ceramics fail after only about 0.1% strain.
- 12. 5.6 Mechanical Behavior-Ceramics Flexural Strength • A more suitable transverse bending test is most frequently employed, in which a specimen having either a circular or rectangular cross section is bent until fracture using a three-point loading technique. • At the point of loading, the top surface of the specimen is placed in a state of compression, while the bottom surface is in tension. • The maximum tensile stress exists at the bottom specimen surface directly below the point of load application.
- 13. 5.6 Mechanical Behavior-Ceramics Flexural Strength • The stress at fracture using this bending test is known as the flexural strength, fracture strength, or the bend strength (𝜎𝑓𝑠). • For a rectangular cross section: 𝜎𝑓𝑠 = 3𝐹𝑓𝐿 2𝑏𝑑2 • For a circular cross section: 𝜎𝑓𝑠 = 𝐹𝑓𝐿 𝜋𝑅3
- 14. 5.6 Mechanical Behavior-Ceramics Flexural Strength • The magnitude of flexural strength for a specific ceramic material will be greater than its fracture strength measured from a tensile test. This phenomenon may be explained by differences in specimen volume that are exposed to tensile stresses. • The entirety of a tensile specimen is under tensile stress, whereas only some volume fraction of a flexural specimen is subjected to tensile stresses—those regions in the vicinity of the specimen surface opposite to the point of load application.
- 15. Example A three‐point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 0.14 in.; the specimen fractured at a load of 215 lbf when the distance between the support points was 2.0 in.. Another test is to be performed on a specimen of this same material, but one that has a square cross section of 0.47 in. length on each edge. At what load would you expect this specimen to fracture if the support point separation is 1.6 in.? 5.6 Mechanical Behavior-Ceramics Flexural Strength
- 17. Example A three‐point bending test is performed on a glass specimen having a rectangular cross section of height 5 mm and width 10 mm; the distance between support points is 45 mm. Compute the flexural strength if the load at fracture is 290 N. 5.6 Mechanical Behavior-Ceramics Flexural Strength