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Understanding Mechanical Properties in Response to Applied Load

Explore the effects of tension, compression, shear, and torsional forces on materials. Learn about elastic and plastic deformation, Young’s Modulus, yield strength, ductility, resilience, toughness, hardness, and safety factors in material engineering.

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Understanding Mechanical Properties in Response to Applied Load

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  1. CHAPTER 6: Mechanical Properties • Response to applied load or force LOADING: • Tension • - Normal stress (+) & strain 2. Compression - Normal stress (-) & strain 3. Shear and Torsional - Shear stress & shear strain

  2. TENSILE TEST A given specimen of specified size is loaded and deformed until fracture. Engineering Stress: Engineering Strain: Stress-Strain diagram – gives many different mechanical properties

  3. ELASTIC PROPERTIES Elastic deformation - nonpermanent A. Modulus of Elasticity i. Straight line in most metals Hooke’s Law E = modulus of elasticity or Young’s Modulus ~ (4.5 X 104 - 40.7 X 104 MPa) - slope of the initial linear portion of graph - bigger E => stiffer => higher resistance to elastic deformation ii. Nonlinear initial portion (e.g., gray cast iron, concrete) a. Secant modulus b. Tangent modulus

  4. YOUNG’S MODULI: COMPARISON Graphite Ceramics Semicond Metals Alloys Composites /fibers Polymers E(GPa) Based on data in Table B2, Callister 6e. Composite data based on reinforced epoxy with 60 vol% of aligned carbon (CFRE), aramid (AFRE), or glass (GFRE) fibers.

  5. ELASTIC DEFORMATION 1. Initial 2. Small load 3. Unload Elastic means nonpermanent!

  6. Elastic Deformation: on atomic scale Elastic strain results in change in interatomic spacing and stretching of bonds (not broken) E is proportional to slope of graph of F vs. r at r=ro.

  7. Elastic Properties B. Under Shear Loading Shear Stress: Force is parallel to area. Shear Strain: Also elastic at low stresses: G = Shear Modulus

  8. Elastic Properties C. Possoin’s Ratio, n Strains: Longitudinal Lateral Poisson’s ratio: metals: n ~ 0.33 ceramics: ~0.25 polymers: ~0.40 For isotropic materials.

  9. Example 1. A tensile stress is to be applied along the axis of a cylindrical brass rod that has a diameter of 0.4 in. Determine the magnitude of the load F required to produce a 10-4 in change in diameter. Assume elastic deformation. (Properties from Table 6.1.)

  10. Plastic Region - Plastic deformation – permanent – bonds broken Anelasticity – time-dependent elastic behavior - needs finite time to recover deformation

  11. Plastic Properties • Proportional Limit • B. Yield Strength • i) 0.2% yield strength • - most common • ii) nonlinear elastic region • - stress at e = 0.005 • iii) Yield-pt phenomenon • - Luder’s bands

  12. YIELD STRENGTH: COMPARISON Room T values Based on data in Table B4, Callister 6e. a = annealed hr = hot rolled ag = aged cd = cold drawn cw = cold worked qt = quenched & tempered

  13. Elastic Strain Recovery • A material deformed beyond its yield point undergoes elastic and plastic deformation. • If load is removed, the elastic part is recovered. Example. Expt#1 Sample data Given: A cylindrical sample of initial radius 0.083 in, and initial length 4 in, loaded in tension to a strain of 0.0085, then unloaded. Find the length after load is removed.

  14. D. TENSILE STRENGTH, TS • Maximum possible engineering stress in tension. below TS: deformation over the entire length. above TS: considerable necking occurs.

  15. TENSILE STRENGTH: COMPARISON Room T values Based on data in Table B4, Callister 6e. a = annealed hr = hot rolled ag = aged cd = cold drawn cw = cold worked qt = quenched & tempered AFRE, GFRE, & CFRE = aramid, glass, & carbon fiber-reinforced epoxy composites, with 60 vol% fibers.

  16. D. DUCTILITY, %EL • Plastic tensile strain at failure: • Another ductility measure:

  17. E. Resilience - Capacity to absorb elastic deformation energy • Modulus of resilience: • Ur = strain energy per unit volume from s = 0 to s = sy. - Area under stress strain diagram up to yield pt.

  18. Different Approximations: Which one is the smallest? Largest? Which one is the safest to use?

  19. F. TOUGHNESS • • Capacity to absorb energy up to fracture • Energy to break a unit volume of material • • Approximate by the area under the stress-strain • curve. High toughness – high strength and high ductility.

  20. G. Hardness - Resistance to localized plastic deformation (indentation) - Large hardness means: --resistance to plastic deformation or cracking in compression. --better wear properties.

  21. Different Hardness scales depending on indenter. • Brinell – HB • Vickers – HV • Knoop – HK • Rockwell Generally: Higher hardness means higher strength. Highest correlation is between HB and TS: Linear Correlation TS (psi) = 500 x HB TS (MPa) = 3.45 x HB

  22. SAFETY FACTORS • • Takes care of variability and uncertainty in properties. • Use working stress levels lower than the strength (yield or tensile) of the material. • • Factor of safety, N Often N is between 1.2 and 4 • Ex: Calculate a diameter, d, to ensure that yield does not occur in the 1045 carbon steel rod below. Use a factor of safety of 5. 5

  23. Engineering vs. True Engr: True: Note: For Plastic Region: n = strain hardening exponent - 0.15 to 0.5

  24. It’s Clicker Time! • The property of a material that causes fracture with relatively little or no plastic deformation is • a) softness • b) hardness • c) brittleness • d) ductility • e) elasticity

  25. Which of the following is the driving force in steady-state diffusion? • concentration of diffusing species • time of diffusion • c) concentration gradient • d) diffusion

  26. 3. The property that enables a material to absorb energy before rupture is a) toughness b) elasticity c) resilience d) hardness

  27. 4. Which of the following increases the diffusion flux? a) Lower concentration gradient b) Higher average concentration c) Higher temperature d) Higher activation energy for diffusion

  28. 5. Which of the following stress values is the largest? • Elastic limit • Fracture stress • c) Yield stress • d) Tensile stress

  29. Given the stress-strain diagrams for three hypothetical alloys A, B and C. Arrange the metals in the order of increasing modulus of elasticity. • ABC • ACB • c) CBA • d) CAB

  30. 7. Given the stress-strain diagrams for three hypothetical alloys A, B and C. Arrange the metals in the order of increasing Brinnel hardness. • ABC • ACB • c) CBA • d) CAB

  31. 8. Given the stress-strain diagrams for three hypothetical alloys A, B and C. Arrange the metals in the order of increasing ductility. • ABC • BAC • c) CBA • d) CAB

  32. 9. Which of the following will have the highest diffusion coefficient? • C in iron at 500oC • C in iron at 700oC • c) Ni in iron at 500oC • d) Ni in iron at 700oC

  33. 10. Which of the following estimates of the toughness is the smallest? a) b) c) d)

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