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Before class, pass out photocopy of Stress-Strain plot EM321: Lesson 8b Chapter 6.1-6.2, 6.6: Mechanical Properties of Metals
•  Stress and Strain:  What are they and why are they used instead of load and deformation? •  Elastic behavior:  When loads are small, how much deformation occurs?  What materials deform least? •  Plastic behavior:  At what point do dislocations cause permanent deformation?  What materials are most resistant to permanent deformation? •  Toughness and ductility:  What are they and how do we measure them? Important Topics:  Mechanical Properties
Elastic Deformation 1. Initial 2. Small load 3. Unload Elastic means reversible!
Plastic Deformation (Metals) 1. Initial 2. Small load 3. Unload Plastic means permanent!
Engineering Stress Shearstress, t Tensile (Normal) stress, s Shear stress causes a shape change Stress has units: Pa =N/m2 or psi = lb/in2
6 Common States of Stress F F A  = cross sectional  o area (when unloaded) F = s s s A o M F s A o A c F s = t A o M 2R •  Simple tension:  cable Ski lift •  Torsion (a form of shear):  drive shaft
Other Common Stress States (i) A o Canyon Bridge, Los Alamos, NM F = s Balanced Rock, Arches  A National Park o •  Simple compression: Note:  compressive structure member (s < 0 here).
Other Common Stress States (ii) Fish under water s  > 0 q s< 0 s > 0 z h •  Bi-axial tension: •  Hydrostatic compression: Pressurized tank
Engineering Strain d /2 - d d L e = e = L L w L o w o o o d /2 L •  Shearstrain, g: q x g = Dx/y= tan  q y 90º - q 90º •  Lateralstrain, εL: •  Tensilestrain, ε: Strain is always dimensionless.
Stress-Strain Testing: The Tensile Test Typical tensile specimen Typical tensile test machine gauge  (portion of sample with  = length reduced cross section) • Other types of tests: --Compression:  brittle materials (i.e. concrete) --Torsion:  cylindrical tubes, shafts.
Tensile Test Output: Stress-Strain Curve The tensile test gives an output of Force, F, versus Deflection,d Convert this output to Stress versus Strain: Divide Force,F, by Initial Cross Sectional Area, Ao, to get _______, σ Divide Deflection, d, by Initial Length, Lo, to get ________, ε Example Stress-Strain curve: Note: inner graph is a zoomed in picture of the beginning of the curve Stress Strain Group Question: Why do we typically use Stress vs. Strain plot, rather than Force vs. Deflection plot?
Linear Elastic Properties, E and n s E e Linear-  elastic L e n - • Modulus of Elasticity, E:     (also known as Young's modulus) • Hooke's Law: s = Ee • Poisson's ratio, n:   metals:  n ~ 0.33   ceramics: ~0.25 polymers: ~0.40 e Units: E:  [GPa] or [psi] n:  dimensionless
Example: Finding E Modulus of Elasticity, E, of a material is equal to the slope of the initial, linear section of the Stress-Strain plot. What is the Modulus of Elasticity     of the material shown in the      Stress-Strain plot that was handed     out (Fig 6.21)? E ≈ 200 * 106 Pa / 0.001 = 200 * 109 Pa
More Comments on Elastic Deformation The Modulus of Elasticity, E, is the stiffness of the material, or resistance to elastic deformation. E is an important design parameter Elastic deformation is non-permanent Spring like return to original shape Values of E are a direct consequence of bonding type Slope of Stress-Strain plot (which is proportional to the Modulus of Elasticity) depends on bond strength of the metal
Even More Comments About Elasticity It is assumed elastic deformation is time independent Also assumed that as load is released, the strain is totally recovered. In most engineering materials, there is a time dependent elastic strain component.   Elastic deformation will continue after the stress application, and upon release some time is required for complete recovery. Time dependent behavior is called “anelasticity” Usually negligible for metals, is significant for some polymers, where it is called “visco-elasticity”
Other Elastic Properties M t G simple torsion test g M P P P D V P = - K V o • Special relations for isotropic materials: E E = = K G 3(1-2n) 2(1+n) • Elastic Shear Modulus, G: t = Gg • Elastic Bulk Modulus, K: P D V V o K Pressure Test: Init.Vol=Vo.   VolChg. = DV Isotropic means that the material behaves the same in all directions
Modulus of Elasticity: Comparison Graphite Ceramics Semicond Metals Alloys Composites /fibers Polymers E(GPa)
Plastic (Permanent) Deformation Simple Tensile Test:
Yield Strength, syorSy • Proportional Limit = Stress at which plastic deformation first occurs ,[object Object],Since it is sometimes difficult  to determine the initial departure from linear behavior, the convention is to offset strain by 0.002, draw a parallel line to the F-displacement curve and mark the intercept point.  This is the “yield strength”.
The Yield Point Phenomenon Most materials do not exhibit a distinct yield point (Fig. a) Use 0.002 Strain offset method to calculate  Some materials exhibit 2 distinct yield points (Fig. b) Stress drops dramatically at Upper Yield Point Stress is relatively constant for a while at Lower Yield Point Group Question: For materials (e.g. some  steels) that exhibit the yield point phenomenon, should the upper or lower yield point be used?  Why? Lower Yield point should be used.  Better defined. Stable for a long time. Insensitive to testing procedure.  More conservative (because it is lower value).
Example: Finding Yield Strength Use the 0.002 Strain offset method to find the Yield Strength of the material shown in the Stress-Strain plot that was handed out (Fig 6.21)? Sy ≈ 400 MPa
Yield Strength: Comparison Room Temp values
Ultimate Tensile Strength, SUtor STor TS SUt SF= fracture         strength Neck – acts as stress concentrator y  stress  engineering  Typical response of  a metal strain  engineering strain  Ultimate Tensile Strength = Maximumstress on engineering stress-strain curve. Typical response of a metal •  Metals: SUtoccurs when noticeable necking starts. •  Ceramics: SUtoccurs when crack propagation starts. •  Polymers: SUtoccurs when polymer backbonesare    aligned and about to break.
Example: Finding Ultimate Tensile Strength Find the Ultimate Tensile Strength of the material shown in the Stress-Strain plot that was handed out (Fig 6.21)? SUt ≈ 500 MPa
Ultimate Tensile Strength: Comparison Room Temp values
Deformation Terms Ductility—Measure of the degree of plastic deformation that has been sustained at fracture Brittle—A material that sustains very little to no plastic deformation prior to fracture Curves for brittle and ductile behavior:
Ductility: Percent Elongation Percent Elongation =      Plastic tensile strain at failure Another ductility measure = Percent Reduction of Area Note: Typically, Percent Elongation ≠ Percent Reduction of Area
Toughness small toughness (ceramics) E ngineering  tensile  large toughness (metals)  stress, s very small toughness  (unreinforced polymers)  Engineering tensile strain, e Toughness = Energy to break a unit volume of material       -  Approximated by the area under the stress-strain curve. Brittle fracture:	 elastic energyDuctile fracture:	 elastic + plastic energy
1 e s @ U y y r 2 Resilience, Ur Resilience = Ability of a material to store energy  Energy stored best in elastic region Approximated as the area of the elastic region If we assume a linear stress-strain curve this simplifies to:
Example: Finding Ductility, Toughness and Resilience Find the Ductility (%EL), Toughness and Resilience of the material shown in the Stress-Strain plot that was handed out (Fig 6.21)? %EL ≈ 0.185 = 18.5% Toughness ≈ 500 MPa * 0.185 – 0.5 * 150 Mpa * 0.02 – 0.5 * 130 MPa * 0.065 Toughness ≈ 86.8 Mpa Resilience ≈ 0.5 * 400 MPa * 0.0038 Resilience ≈ 0.76 MPa
Elastic Strain Recovery Elastic strain recovery D syi syo 2. Unload Stress 3. Reapply load 1. Load Strain
Conclusion Stress and Strain Elastic and Plastic Behavior Mechanical Properties found from Stress-Strain Curve Modulus of Elasticity Yield Strength Ultimate Tensile Strength Ductility Toughness Resilience
Example Stress-Strain Curve (Callister Fig. 6.21)

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Em321 lesson 08b solutions ch6 - mechanical properties of metals

  • 1. Before class, pass out photocopy of Stress-Strain plot EM321: Lesson 8b Chapter 6.1-6.2, 6.6: Mechanical Properties of Metals
  • 2. • Stress and Strain: What are they and why are they used instead of load and deformation? • Elastic behavior: When loads are small, how much deformation occurs? What materials deform least? • Plastic behavior: At what point do dislocations cause permanent deformation? What materials are most resistant to permanent deformation? • Toughness and ductility: What are they and how do we measure them? Important Topics: Mechanical Properties
  • 3. Elastic Deformation 1. Initial 2. Small load 3. Unload Elastic means reversible!
  • 4. Plastic Deformation (Metals) 1. Initial 2. Small load 3. Unload Plastic means permanent!
  • 5. Engineering Stress Shearstress, t Tensile (Normal) stress, s Shear stress causes a shape change Stress has units: Pa =N/m2 or psi = lb/in2
  • 6. 6 Common States of Stress F F A = cross sectional o area (when unloaded) F = s s s A o M F s A o A c F s = t A o M 2R • Simple tension: cable Ski lift • Torsion (a form of shear): drive shaft
  • 7. Other Common Stress States (i) A o Canyon Bridge, Los Alamos, NM F = s Balanced Rock, Arches A National Park o • Simple compression: Note: compressive structure member (s < 0 here).
  • 8. Other Common Stress States (ii) Fish under water s > 0 q s< 0 s > 0 z h • Bi-axial tension: • Hydrostatic compression: Pressurized tank
  • 9. Engineering Strain d /2 - d d L e = e = L L w L o w o o o d /2 L • Shearstrain, g: q x g = Dx/y= tan q y 90º - q 90º • Lateralstrain, εL: • Tensilestrain, ε: Strain is always dimensionless.
  • 10. Stress-Strain Testing: The Tensile Test Typical tensile specimen Typical tensile test machine gauge (portion of sample with = length reduced cross section) • Other types of tests: --Compression: brittle materials (i.e. concrete) --Torsion: cylindrical tubes, shafts.
  • 11. Tensile Test Output: Stress-Strain Curve The tensile test gives an output of Force, F, versus Deflection,d Convert this output to Stress versus Strain: Divide Force,F, by Initial Cross Sectional Area, Ao, to get _______, σ Divide Deflection, d, by Initial Length, Lo, to get ________, ε Example Stress-Strain curve: Note: inner graph is a zoomed in picture of the beginning of the curve Stress Strain Group Question: Why do we typically use Stress vs. Strain plot, rather than Force vs. Deflection plot?
  • 12. Linear Elastic Properties, E and n s E e Linear- elastic L e n - • Modulus of Elasticity, E: (also known as Young's modulus) • Hooke's Law: s = Ee • Poisson's ratio, n: metals: n ~ 0.33 ceramics: ~0.25 polymers: ~0.40 e Units: E: [GPa] or [psi] n: dimensionless
  • 13. Example: Finding E Modulus of Elasticity, E, of a material is equal to the slope of the initial, linear section of the Stress-Strain plot. What is the Modulus of Elasticity of the material shown in the Stress-Strain plot that was handed out (Fig 6.21)? E ≈ 200 * 106 Pa / 0.001 = 200 * 109 Pa
  • 14. More Comments on Elastic Deformation The Modulus of Elasticity, E, is the stiffness of the material, or resistance to elastic deformation. E is an important design parameter Elastic deformation is non-permanent Spring like return to original shape Values of E are a direct consequence of bonding type Slope of Stress-Strain plot (which is proportional to the Modulus of Elasticity) depends on bond strength of the metal
  • 15. Even More Comments About Elasticity It is assumed elastic deformation is time independent Also assumed that as load is released, the strain is totally recovered. In most engineering materials, there is a time dependent elastic strain component. Elastic deformation will continue after the stress application, and upon release some time is required for complete recovery. Time dependent behavior is called “anelasticity” Usually negligible for metals, is significant for some polymers, where it is called “visco-elasticity”
  • 16. Other Elastic Properties M t G simple torsion test g M P P P D V P = - K V o • Special relations for isotropic materials: E E = = K G 3(1-2n) 2(1+n) • Elastic Shear Modulus, G: t = Gg • Elastic Bulk Modulus, K: P D V V o K Pressure Test: Init.Vol=Vo. VolChg. = DV Isotropic means that the material behaves the same in all directions
  • 17. Modulus of Elasticity: Comparison Graphite Ceramics Semicond Metals Alloys Composites /fibers Polymers E(GPa)
  • 18. Plastic (Permanent) Deformation Simple Tensile Test:
  • 19.
  • 20. The Yield Point Phenomenon Most materials do not exhibit a distinct yield point (Fig. a) Use 0.002 Strain offset method to calculate Some materials exhibit 2 distinct yield points (Fig. b) Stress drops dramatically at Upper Yield Point Stress is relatively constant for a while at Lower Yield Point Group Question: For materials (e.g. some steels) that exhibit the yield point phenomenon, should the upper or lower yield point be used? Why? Lower Yield point should be used. Better defined. Stable for a long time. Insensitive to testing procedure. More conservative (because it is lower value).
  • 21. Example: Finding Yield Strength Use the 0.002 Strain offset method to find the Yield Strength of the material shown in the Stress-Strain plot that was handed out (Fig 6.21)? Sy ≈ 400 MPa
  • 22. Yield Strength: Comparison Room Temp values
  • 23. Ultimate Tensile Strength, SUtor STor TS SUt SF= fracture strength Neck – acts as stress concentrator y stress engineering Typical response of a metal strain engineering strain Ultimate Tensile Strength = Maximumstress on engineering stress-strain curve. Typical response of a metal • Metals: SUtoccurs when noticeable necking starts. • Ceramics: SUtoccurs when crack propagation starts. • Polymers: SUtoccurs when polymer backbonesare aligned and about to break.
  • 24. Example: Finding Ultimate Tensile Strength Find the Ultimate Tensile Strength of the material shown in the Stress-Strain plot that was handed out (Fig 6.21)? SUt ≈ 500 MPa
  • 25. Ultimate Tensile Strength: Comparison Room Temp values
  • 26. Deformation Terms Ductility—Measure of the degree of plastic deformation that has been sustained at fracture Brittle—A material that sustains very little to no plastic deformation prior to fracture Curves for brittle and ductile behavior:
  • 27. Ductility: Percent Elongation Percent Elongation = Plastic tensile strain at failure Another ductility measure = Percent Reduction of Area Note: Typically, Percent Elongation ≠ Percent Reduction of Area
  • 28. Toughness small toughness (ceramics) E ngineering tensile large toughness (metals) stress, s very small toughness (unreinforced polymers) Engineering tensile strain, e Toughness = Energy to break a unit volume of material - Approximated by the area under the stress-strain curve. Brittle fracture: elastic energyDuctile fracture: elastic + plastic energy
  • 29. 1 e s @ U y y r 2 Resilience, Ur Resilience = Ability of a material to store energy Energy stored best in elastic region Approximated as the area of the elastic region If we assume a linear stress-strain curve this simplifies to:
  • 30. Example: Finding Ductility, Toughness and Resilience Find the Ductility (%EL), Toughness and Resilience of the material shown in the Stress-Strain plot that was handed out (Fig 6.21)? %EL ≈ 0.185 = 18.5% Toughness ≈ 500 MPa * 0.185 – 0.5 * 150 Mpa * 0.02 – 0.5 * 130 MPa * 0.065 Toughness ≈ 86.8 Mpa Resilience ≈ 0.5 * 400 MPa * 0.0038 Resilience ≈ 0.76 MPa
  • 31. Elastic Strain Recovery Elastic strain recovery D syi syo 2. Unload Stress 3. Reapply load 1. Load Strain
  • 32. Conclusion Stress and Strain Elastic and Plastic Behavior Mechanical Properties found from Stress-Strain Curve Modulus of Elasticity Yield Strength Ultimate Tensile Strength Ductility Toughness Resilience
  • 33. Example Stress-Strain Curve (Callister Fig. 6.21)