UMT312T 2024 Lecture 02-05: Review of Plasticity in Metals PDF

UMT312T: Mechanical testing and failure of materials Lecture 02-05 I. Review of plasticity in metals S. Karthikeyan Associate Professor...

UMT312T: Mechanical testing and failure of materials Lecture 02-05 I. Review of plasticity in metals S. Karthikeyan Associate Professor Indian Institute of Science, Bangalore Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 2 Constant strain-rate test (in tension) Keep 𝜀𝜀̇ (&𝑇𝑇) = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶, and monitor how 𝜎𝜎 changes with 𝜀𝜀 o Engineering stress is the load divided by the initial cross section area, 𝐴𝐴0 : 𝑷𝑷 𝒔𝒔 = Gage cross- 𝑨𝑨𝟎𝟎 section area = 𝐴𝐴0 o Engineering strain is the extension divided by the initial gage length, 𝐿𝐿0: Gage length = 𝐿𝐿0 ∆𝑳𝑳 𝒅𝒅𝒅𝒅 𝒆𝒆 = or incrementally 𝒅𝒅𝒆𝒆 = 𝑳𝑳𝟎𝟎 𝑳𝑳𝟎𝟎 o Engineering strain rate is the crosshead velocity divided by the initial gage length, 𝐿𝐿0: 𝒅𝒅𝒅𝒅 𝟏𝟏 𝒅𝒅𝒅𝒅 𝒗𝒗 ̇𝒆𝒆 = = = 𝒅𝒅𝒅𝒅 𝑳𝑳𝟎𝟎 𝒅𝒅𝒅𝒅 𝑳𝑳𝟎𝟎 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 3 Examples of stress-strain curves Brittle materials Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 4 Examples of stress-strain curves Metals (a variety of steels here) and polymers Steels Polymers Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 5 Examples of stress-strain curves: Effect of temperature and strain-rate Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 6 Examples of stress-strain curves: polymers (PMMA) Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 7 Plasticity in metals Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 8 Interpreting stress- strain curves Gage length P Gage cross-section area Engineering stress, 𝑠𝑠 = 𝑃𝑃 𝐴𝐴0 o Engineering stress is the load divided by the initial cross section area, 𝐴𝐴0 : 𝑃𝑃 𝑠𝑠 = 𝐴𝐴0 o Engineering strain is the extension divided by the initial gage length, 𝐿𝐿0 : ∆𝐿𝐿 𝑑𝑑𝑑𝑑 𝑒𝑒 = or incrementally 𝑑𝑑𝑒𝑒 = 𝐿𝐿0 𝐿𝐿0 o Engineering strain rate is the crosshead velocity divided by the initial gage length, 𝐿𝐿0 : 𝑑𝑑𝑑𝑑 1 𝑑𝑑𝑑𝑑 𝑣𝑣 𝑒𝑒̇ = = = 𝒆𝒆𝒑𝒑𝒑𝒑 𝒆𝒆𝒆𝒆𝒆𝒆 Engineering strain, 𝑒𝑒 = ∆𝐿𝐿 𝐿𝐿0 ∆L 𝑑𝑑𝑑𝑑 𝐿𝐿0 𝑑𝑑𝑑𝑑 𝐿𝐿0 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 9 Interpreting stress- strain curves P D Engineering stress, 𝑠𝑠 = 𝑃𝑃 𝐴𝐴0 Slope: Strain hardening rate C B D: Ultimate tensile strength, A: Proportionality limit A (𝒔𝒔𝒖𝒖 𝒐𝒐𝒐𝒐 𝑼𝑼𝑼𝑼𝑼𝑼) B: Elastic limit C: Offset yield strength G: Uniform elongation, 𝒆𝒆𝒖𝒖 (𝒔𝒔𝒚𝒚 𝒐𝒐𝒐𝒐 𝒀𝒀𝒀𝒀) or proof stress H: Ductility or strain to corresponding to a permanent fracture plastic strain of F (typically 𝒆𝒆𝒑𝒑𝒑𝒑 : Resilience (at F) = 0.2%) : Toughness From C to D: Strain hardening Slope: Young’s modulus G F Engineering strain, 𝑒𝑒 = ∆𝐿𝐿 𝐿𝐿0 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 10 Changes in the sample during the test Engineering stress, 𝑠𝑠 = 𝑃𝑃 𝐴𝐴0 𝐿𝐿 𝐿𝐿 𝐿𝐿 𝐿𝐿 𝐿𝐿1 𝐿𝐿0 Engineering strain, 𝑒𝑒 = ∆𝐿𝐿 𝐿𝐿0 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 11 Constancy of volume o During the test and up to the initiation of necking, plastic deformation is uniform or homogeneous, i.e., the cross section area, 𝐴𝐴 is the same throughout the gage length, 𝐿𝐿 uniform elongation o After necking starts plastic deformation is non-uniform or inhomogeneous, or localized  post-necking elongation o Up to the initiation of necking, the gage length, 𝐿𝐿 increases from the initial 𝐿𝐿0 , while the cross-sectional area, 𝐴𝐴 decreases from the initial 𝐴𝐴0 o During plastic deformation of metals, there is shape change, but no volume change (constancy of volume) due to which: 𝑨𝑨𝑨𝑨 = 𝑨𝑨𝟎𝟎 𝑳𝑳𝟎𝟎 IMPORTANT: the above expressions are valid only if we assume that 𝑨𝑨 is constant throughout the gage length, 𝑳𝑳, i.e., for uniform elongation Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 12 Step A1: A sample of gage length 8 cm is stretched t 10 cm. True stress-true strain 𝑒𝑒𝐴𝐴𝐴 = 0.25 𝜀𝜀𝐴𝐴𝐴 = ln 10 8 𝑷𝑷 o Engineering stress is the load divided by the original cross section area, 𝐴𝐴0 : 𝒔𝒔 = 𝑨𝑨 𝟎𝟎 Step A2: The same sample is ∆𝑳𝑳 further stretched from 10 cm to o Engineering strain is the extension divided by the original gage length, 𝐿𝐿0 : 𝒆𝒆 = 𝑳𝑳 12 cm. 𝟎𝟎 𝑒𝑒𝐴𝐴𝐴 = 0.2 o True stress is the load divided by the instantaneous cross section area, 𝐴𝐴 : 𝜀𝜀𝐴𝐴𝐴 = ln 12 10 𝑷𝑷 𝑷𝑷𝑷𝑷 𝑷𝑷 𝑳𝑳𝟎𝟎 + 𝜟𝜟𝑳𝑳 𝑷𝑷 𝜟𝜟𝑳𝑳 𝝈𝝈 = = = = 𝟏𝟏 + = 𝒔𝒔 𝟏𝟏 + 𝒆𝒆 𝑨𝑨 𝑨𝑨𝟎𝟎 𝑳𝑳𝟎𝟎 𝑨𝑨𝟎𝟎 𝑳𝑳𝟎𝟎 𝑨𝑨𝟎𝟎 𝑳𝑳𝟎𝟎 Total Eng. Strain: 𝑒𝑒𝐴𝐴 = 𝑒𝑒𝐴𝐴𝐴 + 𝑒𝑒𝐴𝐴𝐴 = 𝟎𝟎. 𝟒𝟒𝟒𝟒 o True strain is the integral of the infinitesimal extension divided by the instantaneous gage length: Total True Strain: 𝜀𝜀𝐴𝐴 = 𝜀𝜀𝐴𝐴𝐴 + 𝜀𝜀𝐴𝐴𝐴 = 𝑳𝑳 10 12 𝟏𝟏𝟐𝟐 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝑳𝑳 ln + ln = 𝒍𝒍𝒍𝒍 8 10 𝟖𝟖 𝒅𝒅𝒅𝒅 = ⟹ 𝜺𝜺 ≡ = 𝒍𝒍𝒍𝒍 = 𝒍𝒍𝒍𝒍 𝟏𝟏 + 𝒆𝒆 𝑳𝑳 𝑳𝑳𝟎𝟎 𝑳𝑳 𝑳𝑳𝟎𝟎 Step B: A sample of gage length 8 This strain is also called Hencky strain or logarithmic strain. cm is stretched t0 12 cm in a single step o True strain-rate: 𝑒𝑒𝐵𝐵 = 𝟎𝟎. 𝟓𝟓 𝟏𝟏𝟏𝟏 𝒅𝒅𝜺𝜺 𝒗𝒗 𝟏𝟏 𝒅𝒅𝒅𝒅 𝒆𝒆̇ 𝜀𝜀𝐵𝐵 = 𝐥𝐥𝐥𝐥 𝜺𝜺̇ = = = = 𝟖𝟖 𝒅𝒅𝒅𝒅 𝑳𝑳 𝟏𝟏 + 𝒆𝒆 𝒅𝒅𝒅𝒅 𝟏𝟏 + 𝒆𝒆 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 13 True stress-true strain in tension 𝒆𝒆 𝝈𝝈 ≡ 𝒔𝒔 𝟏𝟏 + 𝒆𝒆 𝜺𝜺 ≡ 𝒍𝒍𝒍𝒍 𝟏𝟏 + 𝒆𝒆 ~𝒆𝒆(𝟏𝟏 − ) 𝟐𝟐 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 14 Other aspects Constancy of volume, reflected in a Loading-unloading-reloading paths. The material has a memory of previous Poisson’s ratio tending to 0.5, when it’s deformation history. A material that has been previously deformed behaves like a deforming plastically new, stronger material, on retesting Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 15 Other aspects Constancy of volume, reflected in a Loading-unloading-reloading paths. The material has a memory of previous Poisson’s ratio tending to 0.5, when it’s deformation history. A material that has been previously deformed behaves like a deforming plastically new, stronger material, on retesting Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 16 Constitutive equation o Note that, most metallic materials exhibit an increase in strength with strain. This phenomenon is called strain-hardening or work-hardening. o In the plastic regime, one can usually describe the dependence of true stress on true strain of the various forms below: Elastic- Elastic- Power-law Ludvik- Voce perfectly Linear hardening Hollomon Equation plastic hardening Equation Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 17 Power-law hardening 𝜎𝜎 = 𝐾𝐾𝜀𝜀 𝑛𝑛 𝑛𝑛 is the strain hardening exponent On a log-log plot, 𝑛𝑛 is the slope and K is the y- intercept at 𝜀𝜀 = 1 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 18 𝝈𝝈 − 𝜺𝜺 and s-e curves in tension In tension, 𝜀𝜀 > 0, in compression 𝜀𝜀 < 0 If constitutive equation to describe strain and compression hardening is 𝜎𝜎 = 𝑓𝑓(𝜀𝜀), then: −𝑓𝑓(𝜀𝜀) = 𝑓𝑓(−𝜀𝜀). This behavior is called isotropic hardening, i.e., flow stress and strain hardening are the same in tension and compression, only the signs are changed. 𝝈𝝈 − 𝜺𝜺 For a power-law hardening material: 𝜎𝜎 = (tension) sgn 𝜀𝜀 ∗ 𝐾𝐾( 𝜀𝜀 )𝑛𝑛 𝜀𝜀 ≡ 𝑙𝑙𝑙𝑙 1 + 𝑒𝑒 𝒔𝒔 − 𝒆𝒆 𝜀𝜀 2 ⟹ 𝑒𝑒 = 𝑒𝑒𝑒𝑒𝑒𝑒 𝜀𝜀 − 1~𝜀𝜀 + (tension) 2 𝜎𝜎 ≡ 𝑠𝑠 1 + 𝑒𝑒 𝜎𝜎 𝜎𝜎 𝝈𝝈 − 𝜺𝜺 ⟹ 𝑠𝑠 = = (compression) 1 + 𝑒𝑒 𝑒𝑒𝑒𝑒𝑒𝑒 𝜀𝜀 Say K = 500 MPa , and n = 0.2 𝒔𝒔 − 𝒆𝒆 Tension Compression (compression) 𝜀𝜀 0.1 −0.1 𝜎𝜎 315.5 𝑀𝑀𝑀𝑀𝑀𝑀 −315.5 𝑀𝑀𝑀𝑀𝑀𝑀 𝑒𝑒 0.105 −0.095 𝑠𝑠 285.5 𝑀𝑀𝑀𝑀𝑀𝑀 −348.7 𝑀𝑀𝑀𝑀𝑀𝑀 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 19 𝝈𝝈 − 𝜺𝜺 and s-e curves in tension In tension, 𝜀𝜀 > 0, in compression 𝜀𝜀 < 0 If constitutive equation to describe strain and compression hardening is 𝜎𝜎 = 𝑓𝑓(𝜀𝜀), then: −𝑓𝑓(𝜀𝜀) = 𝑓𝑓(−𝜀𝜀). This behavior is called isotropic hardening, i.e., flow stress and strain hardening are the same in tension and 𝒔𝒔 − 𝒆𝒆 compression, only the signs are changed. (compression) For a power-law hardening material: 𝜎𝜎 = sgn 𝜀𝜀 ∗ 𝐾𝐾( 𝜀𝜀 )𝑛𝑛 𝜀𝜀 ≡ 𝑙𝑙𝑙𝑙 1 + 𝑒𝑒 𝜀𝜀 2 𝝈𝝈 − 𝜺𝜺 ⟹ 𝑒𝑒 = 𝑒𝑒𝑒𝑒𝑒𝑒 𝜀𝜀 − 1~𝜀𝜀 + 2 𝜎𝜎 ≡ 𝑠𝑠 1 + 𝑒𝑒 𝒔𝒔 − 𝒆𝒆 𝜎𝜎 𝜎𝜎 ⟹ 𝑠𝑠 = = (tension) 1 + 𝑒𝑒 𝑒𝑒𝑒𝑒𝑒𝑒 𝜀𝜀 Say K = 500 MPa , and n = 0.2 Tension Compression 𝜀𝜀 0.1 −0.1 𝜎𝜎 315.5 𝑀𝑀𝑀𝑀𝑀𝑀 −315.5 𝑀𝑀𝑀𝑀𝑀𝑀 𝑒𝑒 0.105 −0.095 𝑠𝑠 285.5 𝑀𝑀𝑀𝑀𝑀𝑀 −348.7 𝑀𝑀𝑀𝑀𝑀𝑀 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 20 Why is there a UTS in the s-e curve? 𝝈𝝈 𝝈𝝈 ≡ 𝒔𝒔 𝟏𝟏 + 𝒆𝒆 ⇒ 𝐬𝐬 = and 𝜺𝜺 ≡ 𝒍𝒍𝒍𝒍 𝟏𝟏 + 𝒆𝒆 ⇒ 𝒆𝒆 = 𝒆𝒆𝒆𝒆𝒆𝒆 𝜺𝜺 − 𝟏𝟏 o Two things happen during 𝟏𝟏 + 𝒆𝒆 a tensile test: Assuming a power-law hardening material: 𝜎𝜎 = 𝐾𝐾𝜀𝜀 𝑛𝑛 o Decrease in cross 𝟏𝟏 𝒏𝒏 𝟏𝟏 𝒏𝒏 section area  decrease 𝒔𝒔 = 𝑲𝑲 𝒍𝒍𝒍𝒍 𝟏𝟏 + 𝒆𝒆 ⇒ 𝑷𝑷 = 𝑨𝑨𝟎𝟎 𝒔𝒔 = 𝑲𝑲𝑨𝑨𝟎𝟎 𝒍𝒍𝒍𝒍 𝟏𝟏 + 𝒆𝒆 = 𝑨𝑨𝟎𝟎 𝑲𝑲𝜺𝜺𝒏𝒏 𝒆𝒆𝒆𝒆𝒆𝒆 −𝜺𝜺 𝟏𝟏+𝒆𝒆 𝟏𝟏+𝒆𝒆 in load carrying capacity which is called geometric softening o Increase in the strength of the material due to strain hardening  leads to an increase in load carrying capacity o At the peak, these two are balanced. Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 21 Considère’s criterion Where does the maximum in the 𝑠𝑠𝐴𝐴0 = 𝜎𝜎𝜎𝜎 = 𝑃𝑃 engineering stress-strain curve occur? 𝐴𝐴0 𝑑𝑑𝑑𝑑 = 𝐴𝐴𝐴𝐴𝐴𝐴 + 𝜎𝜎𝜎𝜎𝜎𝜎 = 𝑑𝑑𝑑𝑑 At D’, 𝑑𝑑𝑠𝑠 = 0 and 𝑑𝑑𝑃𝑃 = 0 (𝐴𝐴0 is a constant) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝐴𝐴𝐴𝐴𝐴𝐴 + 𝜎𝜎𝜎𝜎𝜎𝜎 = 0 ⟹ =− 𝜎𝜎 𝐴𝐴 Since 𝐴𝐴𝐴𝐴 = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 (volume constancy) 𝐴𝐴𝐴𝐴𝐴𝐴 + 𝐿𝐿𝐿𝐿𝐿𝐿 = 0 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 − = ≡ 𝑑𝑑𝑑𝑑 𝐴𝐴 𝐿𝐿 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 So = 𝒅𝒅𝒅𝒅, or = 𝝈𝝈 𝝈𝝈 𝒅𝒅𝒅𝒅 So maximum in the engineering stress-strain curve occurs, when the criterion above, called Considère’s criterion is satisfied. Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 22 For a material obeying 𝜎𝜎 = 𝐾𝐾𝜀𝜀 𝑛𝑛 , Considère’s criterion gives: Considère’s criterion 𝜺𝜺𝒖𝒖 = 𝒏𝒏 Where does the maximum in the 𝑠𝑠𝐴𝐴0 = 𝜎𝜎𝜎𝜎 = 𝑃𝑃 engineering stress-strain curve occur? 𝒆𝒆𝒖𝒖 = 𝒆𝒆𝒆𝒆𝒆𝒆 𝜺𝜺𝒖𝒖 − 𝟏𝟏 = 𝒆𝒆𝒆𝒆𝒆𝒆 𝒏𝒏 − 𝟏𝟏 𝐴𝐴0 𝑑𝑑𝑑𝑑 = 𝐴𝐴𝐴𝐴𝐴𝐴 + 𝜎𝜎𝜎𝜎𝜎𝜎 = 𝑑𝑑𝑑𝑑 At D’, 𝑑𝑑𝑠𝑠 = 0 and 𝑑𝑑𝑃𝑃 = 0 (𝐴𝐴0 is a constant) 𝑼𝑼𝑼𝑼𝑼𝑼 = 𝒔𝒔𝒖𝒖 = 𝑲𝑲 𝒆𝒆𝒆𝒆𝒆𝒆 −𝜺𝜺𝒖𝒖 𝜺𝜺𝒖𝒖 𝒏𝒏 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝒏𝒏 𝒏𝒏 𝐴𝐴𝐴𝐴𝐴𝐴 + 𝜎𝜎𝜎𝜎𝜎𝜎 = 0 ⟹ =− = 𝑲𝑲 𝜎𝜎 𝐴𝐴 𝒆𝒆 Since 𝐴𝐴𝐴𝐴 = 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 (volume constancy) Strength Strain- UTS, Material Coefficient, hardening su 𝐴𝐴𝐴𝐴𝐴𝐴 + 𝐿𝐿𝐿𝐿𝐿𝐿 = 0 K (MPa) exponent, n (MPa) Low-carbon 525 to 575 0.20 to 0.23 319 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 steels − = ≡ 𝑑𝑑𝑑𝑑 𝐴𝐴 𝐿𝐿 HSLA steels 650 to 900 0.15 to 0.18 488 Austenitic 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 400 to 500 0.40 to 0.55 207 = 𝒅𝒅𝒅𝒅, or = 𝝈𝝈 stainless steel So 𝝈𝝈 𝒅𝒅𝒅𝒅 copper 430 to 480 0.35 to 0.5 207 70/30 brass 525 to 750 0.45 to 0.60 269 So maximum in the engineering stress-strain curve occurs, when the criterion above, Aluminium 400 to 550 0.20 to 0.30 248 alloys called Considère’s criterion is satisfied. Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 23 Considère’s criterion Where does the maximum in the 𝜎𝜎 𝜎𝜎 ≡ 𝑠𝑠 1 + 𝑒𝑒 ⇒ 𝑠𝑠 = engineering stress-strain curve occur? 1 + 𝑒𝑒 𝜀𝜀 ≡ 𝑙𝑙𝑙𝑙 1 + 𝑒𝑒 ⇒ 1 + 𝑒𝑒 = 𝑒𝑒𝑒𝑒𝑒𝑒 𝜀𝜀 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 Assuming a power-law hardening material: 𝜎𝜎 = 𝐾𝐾𝜀𝜀 𝑛𝑛 >− ≤− 𝐴𝐴0 𝒔𝒔 = 𝐾𝐾𝜀𝜀 𝑛𝑛 𝒆𝒆𝒆𝒆𝒆𝒆 −𝜺𝜺 ⇒ 𝑷𝑷 = 𝑨𝑨𝟎𝟎 𝒔𝒔 = 𝑨𝑨𝟎𝟎 𝑲𝑲𝜺𝜺𝒏𝒏 𝒆𝒆𝒆𝒆𝒆𝒆 −𝜺𝜺 ε = ln 𝝈𝝈 𝑨𝑨 𝝈𝝈 𝑨𝑨 𝐴𝐴 ε K= 500 MPa, n = 0.4 ε𝑛𝑛 A0 = 2 x 10 -5 m2 Stable Unstable 𝐴𝐴0 ε𝑛𝑛 = ln flow flow ε𝑛𝑛 𝐴𝐴𝑛𝑛 ε Such instability is called geometric instability, because the load maximum is because of change in shape Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan Considère’s criterion states that the necessary criterion for necking 24 is 𝑑𝑑𝑑𝑑 ≤ σ. It is not a sufficient criterion. It does not ensure that a neck 𝑑𝑑𝑑𝑑 will be formed. Instability (and unstable flow) means that, if somehow Considère’s criterion there is localization of deformation to the right of UTS, the neck would deepen, and deformation will become increasingly localized. Where does the maximum in the The criterion tells whether a pre-existing neck will deepen engineering stress-strain curve occur? or not but DOES NOT give conditions for neck formation. 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 𝒅𝒅𝒅𝒅 o If a small neck forms to the left of o If a small neck forms to the right of >− ≤− UTS, UTS, 𝝈𝝈 𝑨𝑨 𝝈𝝈 𝑨𝑨 o The local strain in the neck is higher than o the local strain in the neck is higher than the strain outside the neck the strain outside the neck Stable Unstable flow flow o since local strain is higher, the load o Since local strain is higher, the load carrying capability of the material is carrying capability of the material is higher in the neck (strain hardening beats lower in the neck (geometric softening geometric softening in this regime), so beats strain hardening in this regime), so subsequent deformation takes place subsequent deformation takes place outside the neck, i.e., in the softer within the neck, i.e., in the softer material material Such instability is called geometric o As a result, material outside reduces its o as a result, deformation becomes instability, because the load maximum localized leading to growth of the neck is because of change in shape cross sectional area to match that of the neck, and the neck is ironed out and eventual failure. Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 25 𝜺𝜺 ≡ 𝒍𝒍𝒍𝒍 𝟏𝟏 + 𝒆𝒆 Instability in the presence of ⇒ 𝟏𝟏 + 𝒆𝒆 = 𝐞𝐞𝐞𝐞𝐞𝐞 𝜺𝜺 pre-existing inhomogeneities 𝝈𝝈 ≡ 𝒔𝒔 𝟏𝟏 + 𝒆𝒆 ⇒ 𝒔𝒔 = 𝝈𝝈 𝟏𝟏 + 𝒆𝒆 −𝟏𝟏 = 𝝈𝝈𝐞𝐞𝐞𝐞𝐞𝐞 −𝜺𝜺 𝑓𝑓 = 𝐴𝐴𝑏𝑏𝑜𝑜 “Stepped” specimen During the test the load, P must be constant along the length of the sample: 𝐴𝐴𝑎𝑎𝑜𝑜 Grip 𝑃𝑃 = 𝐴𝐴𝑎𝑎𝑎𝑎 𝑠𝑠𝑎𝑎 = 𝐴𝐴𝑏𝑏𝑜𝑜 𝑠𝑠𝑏𝑏 a b a ↑ ↑ For a material obeying 𝝈𝝈 = 𝑲𝑲𝜺𝜺𝒏𝒏 𝐴𝐴𝑏𝑏𝑜𝑜 𝐴𝐴𝑎𝑎𝑎𝑎 ↓ ↓ ⇒ 𝐴𝐴𝑎𝑎𝑎𝑎 𝐾𝐾𝜀𝜀𝑎𝑎 𝑛𝑛 exp(−𝜀𝜀𝑎𝑎 ) = 𝐴𝐴𝑏𝑏𝑏𝑏 𝐾𝐾𝜀𝜀𝑏𝑏 𝑛𝑛 exp(−𝜀𝜀𝑏𝑏 ) 𝜺𝜺𝒂𝒂 𝒏𝒏 𝒆𝒆𝒆𝒆𝒆𝒆(−𝜺𝜺𝒂𝒂 ) = 𝒇𝒇𝜺𝜺𝒃𝒃 𝒏𝒏 𝒆𝒆𝒆𝒆𝒆𝒆(−𝜺𝜺𝒃𝒃 ) 𝒏𝒏 𝒏𝒏 𝒔𝒔𝒖𝒖 𝒏𝒏 𝜺𝜺𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 𝒆𝒆𝒆𝒆𝒆𝒆(−𝜺𝜺𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 ) = 𝒇𝒇 = 𝒇𝒇 𝒆𝒆 𝑲𝑲 Note: 𝜀𝜀𝑎𝑎,𝑚𝑚𝑚𝑚𝑚𝑚 is the uniform elongation one gets outside the simulated neck Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 26 𝜎𝜎 ↑ as 𝜀𝜀 ↑ (for a constant 𝜀𝜀,̇ 𝑻𝑻) 𝝈𝝈 ∝ 𝜺𝜺𝒏𝒏 |𝑻𝑻,𝜺𝜺̇ Effect of strain rate n= strain hardening exponent o Recall that the constitutive equation in This test took ~117 microseconds 𝜎𝜎 ↑ as 𝜀𝜀̇ ↑ (for a constant 𝜀𝜀 , 𝑻𝑻) plastic deformation also depends on strain Equivalently, a 10 cm sample will be 𝝈𝝈 ∝ 𝜺𝜺̇ 𝒎𝒎 rate and temperature: 𝐟𝐟 𝝈𝝈, 𝜺𝜺, 𝜺𝜺,̇ 𝑻𝑻 = 𝟎𝟎 elongated to ~ 1 km! , if this test ran for 1 sec. 𝑻𝑻,𝜺𝜺 𝑚𝑚 = strain rate sensitivity = 𝜎𝜎2 ln 𝜎𝜎1 Copper This test took nearly 2 hrs 𝜀𝜀̇ 2 (FCC) ln 𝜀𝜀̇ 1 Equivalently, a 10 cm sample will 𝜎𝜎2 be elongated by only ~10 microns if this test ran for 1 sec. 𝝈𝝈 = 𝑪𝑪𝟏𝟏 𝜺𝜺𝒏𝒏 𝜺𝜺̇ 𝒎𝒎 𝑻𝑻 𝜎𝜎1 𝜀𝜀1̇ = 𝜀𝜀2̇ = Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 27 Strain rate sensitivity For most materials, at low temperature the m value is quite low. Change in the stress with increase in strain rate is very small at low temperatures If m=0.01 an increase in an order of magnitude in the strain rate results in only about 2.3% increase in the strain rate Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 28 Stress decreases with Effect of temperature temperature. Materials soften at higher temperature Stress decreases with temperature. Materials soften at higher temperature 𝑸𝑸 𝝈𝝈 = 𝑪𝑪𝟐𝟐 𝒆𝒆𝒆𝒆𝒆𝒆( ) 𝑹𝑹𝑹𝑹 𝜺𝜺,𝜺𝜺̇ Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 29 Are m and n material Stress decreases with constants? temperature. Materials soften at higher temperature n typically decreases with increasing temperature and at decreasing strain rate. Materials do not strain harden as much at higher temperature or lower strain rates. n, is not a material constant since it depends on test conditions Stainless Steel (FCC) Aluminium (FCC) Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 30 Are m and n material constants? Effect of temperature on m, the strain rate sensitivity? 1700 MPa 1200 MPa m600= 0.075 750 MPa m800= 0.199 300 MPa Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 31 Are m and n material constants? Materials typically become more rate sensitive at higher temperatures! Moreover strain rate sensitivity depends also on the strain rate! Thus m is also not as a material constant, since it depends on test conditions Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 32 Are m and n material constants? m and n are not constants, but in turn depend on temperature and strain rate. So treating those values as constants is ONLY applicable in a limited range of temperatures and strain rates n m n m Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 33 Generalized Considère’s criterion (Hart criterion) 𝑃𝑃 = 𝜎𝜎𝜎𝜎 ⇒ 𝑑𝑑𝑃𝑃 = 𝐴𝐴𝐴𝐴𝐴𝐴 + 𝜎𝜎𝜎𝜎𝜎𝜎 Since flow stress is a function of strain, strain rate and temperature: 𝝈𝝈 𝜺𝜺, 𝜺𝜺,̇ 𝑻𝑻 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑑𝑑𝑃𝑃 = 𝐴𝐴 𝑑𝑑𝑑𝑑 + 𝑑𝑑𝜺𝜺̇ + 𝑑𝑑𝑑𝑑 + 𝜎𝜎𝜎𝜎𝜎𝜎 𝜕𝜕𝜕𝜕 ̇ 𝜺𝜺,𝑻𝑻 𝜕𝜕𝜺𝜺̇ 𝜺𝜺,𝑻𝑻 𝜕𝜕𝜕𝜕 𝜺𝜺,𝜺𝜺̇ At the point of instability, 𝑑𝑑𝑃𝑃 = 0. Further dividing both sides by A: 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜎𝜎𝜎𝜎𝜎𝜎 = 𝑑𝑑𝑑𝑑 + 𝑑𝑑 𝜀𝜀̇ + 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 ̇ 𝜀𝜀,𝑇𝑇 𝜕𝜕𝜀𝜀̇ 𝜀𝜀,𝑇𝑇 𝜕𝜕𝜕𝜕 𝜀𝜀,𝜀𝜀̇ 𝑑𝑑𝑑𝑑, 𝑑𝑑𝜺𝜺̇ and 𝑑𝑑𝑑𝑑 are the changes in strain, strain rate and temperature during the test, though the test is meant to be at constant 𝜀𝜀̇ and T, but… 𝝏𝝏𝝏𝝏 𝝏𝝏𝝏𝝏 𝒅𝒅𝜺𝜺̇ 𝝏𝝏𝝏𝝏 𝒅𝒅𝑻𝑻 𝝈𝝈 = + + 𝝏𝝏𝝏𝝏 ̇ 𝜺𝜺,𝑻𝑻 𝝏𝝏𝜺𝜺̇ 𝜺𝜺,𝑻𝑻 𝒅𝒅𝒅𝒅 𝝏𝝏𝝏𝝏 𝜺𝜺,𝜺𝜺̇ 𝒅𝒅𝒅𝒅 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 34 Generalized Considère’s criterion 𝟏𝟏 = 𝒏𝒏 − 𝒎𝒎 − 𝑸𝑸 𝜷𝜷𝝈𝝈𝒖𝒖 (Hart criterion) 𝜺𝜺𝒖𝒖 𝑹𝑹𝑻𝑻𝟐𝟐 𝑪𝑪𝒑𝒑 𝝆𝝆 𝑄𝑄 𝝈𝝈𝝈𝝈 For a material exhibiting: 𝐶𝐶3 𝜀𝜀̇ 𝑚𝑚 𝑛𝑛exp 𝜀𝜀 𝑛𝑛−1 = 𝝏𝝏𝝏𝝏 For a constant engineering strain 𝑅𝑅𝑅𝑅 𝜺𝜺 rate test: 𝝏𝝏𝝏𝝏 ̇ 𝜺𝜺,𝑻𝑻 𝑒𝑒̇ 𝑄𝑄 𝝈𝝈𝒎𝒎 𝜀𝜀̇ = = 𝑒𝑒exp ̇ −ε 𝝈𝝈 = 𝑪𝑪𝟑𝟑 𝜺𝜺𝒏𝒏 𝜺𝜺̇ 𝒎𝒎 𝐞𝐞𝐞𝐞𝐞𝐞 𝑸𝑸 𝐶𝐶3 𝜀𝜀 𝑛𝑛 𝑚𝑚exp 𝜀𝜀̇ 𝑚𝑚−1 = 𝝏𝝏𝝏𝝏 𝒅𝒅𝜺𝜺̇ 1 + 𝑒𝑒 𝑹𝑹𝑹𝑹 𝑅𝑅𝑅𝑅 𝜺𝜺̇ 𝝈𝝈 = + 𝒅𝒅𝜺𝜺̇ 𝝏𝝏𝜺𝜺̇ 𝜺𝜺,𝑻𝑻 𝒅𝒅𝒅𝒅 ̇ = −𝒆𝒆𝐞𝐞𝐞𝐞𝐞𝐞 −𝜺𝜺 = −𝜺𝜺̇ 𝒅𝒅𝒅𝒅 𝝏𝝏𝝏𝝏 𝒅𝒅𝑻𝑻 + Work done during plastic 𝝏𝝏𝝏𝝏 𝜺𝜺,𝜺𝜺̇ 𝒅𝒅𝒅𝒅 deformation: 𝑑𝑑𝑑𝑑 = 𝑉𝑉 𝜎𝜎𝜎𝜎𝜎𝜎 𝑄𝑄 𝑄𝑄 𝝈𝝈𝑸𝑸 If a fraction, 𝛽𝛽 (typically 90-95%) is − 𝐶𝐶3 𝜀𝜀̇ 𝑚𝑚 𝜀𝜀 𝑛𝑛 exp = − 𝑅𝑅𝑇𝑇 2 𝑅𝑅𝑅𝑅 𝑹𝑹𝑻𝑻𝟐𝟐 converted to heat: 𝑑𝑑𝐻𝐻 = 𝛽𝛽𝑉𝑉 𝜎𝜎𝜎𝜎𝜎𝜎 What happens at high T, and low 𝑒𝑒? ̇ The heat generated adiabatically Stainless 𝒏𝒏 increases the temperature of the Steel (FCC) 𝜺𝜺𝒖𝒖 = material (happens at high strain 𝟏𝟏 + 𝒎𝒎 rates): n ~ 0  small 𝜺𝜺𝒖𝒖 𝐶𝐶𝑝𝑝 𝑚𝑚𝑚𝑚𝑚𝑚 = 𝛽𝛽𝑉𝑉 𝜎𝜎𝜎𝜎𝜎𝜎 High T tensile flow is inherently unstable, but ductility? 𝒅𝒅𝑻𝑻 𝜷𝜷𝑽𝑽𝝈𝝈 𝜷𝜷𝜷𝜷 = = 𝒅𝒅𝒅𝒅 𝑪𝑪𝒑𝒑 𝒎𝒎 𝑪𝑪𝒑𝒑 𝝆𝝆 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 35 Generalized Considère’s criterion 𝟏𝟏 = 𝒏𝒏 − 𝒎𝒎 − 𝑸𝑸 𝜷𝜷𝝈𝝈𝒖𝒖 (Hart criterion) 𝜺𝜺𝒖𝒖 𝑹𝑹𝑻𝑻𝟐𝟐 𝑪𝑪𝒑𝒑 𝝆𝝆 𝑄𝑄 𝝈𝝈𝝈𝝈 For a material exhibiting: 𝐶𝐶3 𝜀𝜀̇ 𝑚𝑚 𝑛𝑛exp 𝜀𝜀 𝑛𝑛−1 = 𝝏𝝏𝝏𝝏 For a constant engineering strain 𝑅𝑅𝑅𝑅 𝜺𝜺 rate test: 𝝏𝝏𝝏𝝏 ̇ 𝜺𝜺,𝑻𝑻 𝑒𝑒̇ 𝑄𝑄 𝝈𝝈𝒎𝒎 𝜀𝜀̇ = = 𝑒𝑒exp ̇ −ε 𝝈𝝈 = 𝑪𝑪𝟑𝟑 𝜺𝜺𝒏𝒏 𝜺𝜺̇ 𝒎𝒎 𝐞𝐞𝐞𝐞𝐞𝐞 𝑸𝑸 𝐶𝐶3 𝜀𝜀 𝑛𝑛 𝑚𝑚exp 𝜀𝜀̇ 𝑚𝑚−1 = 𝝏𝝏𝝏𝝏 𝒅𝒅𝜺𝜺̇ 1 + 𝑒𝑒 𝑹𝑹𝑹𝑹 𝑅𝑅𝑅𝑅 𝜺𝜺̇ 𝝈𝝈 = + 𝒅𝒅𝜺𝜺̇ 𝝏𝝏𝜺𝜺̇ 𝜺𝜺,𝑻𝑻 𝒅𝒅𝒅𝒅 ̇ = −𝒆𝒆𝐞𝐞𝐞𝐞𝐞𝐞 −𝜺𝜺 = −𝜺𝜺̇ 𝒅𝒅𝒅𝒅 𝝏𝝏𝝏𝝏 𝒅𝒅𝑻𝑻 + Work done during plastic 𝝏𝝏𝝏𝝏 𝜺𝜺,𝜺𝜺̇ 𝒅𝒅𝒅𝒅 deformation: 𝑑𝑑𝑑𝑑 = 𝑉𝑉 𝜎𝜎𝜎𝜎𝜎𝜎 𝑄𝑄 𝑄𝑄 𝝈𝝈𝑸𝑸 If a fraction, 𝛽𝛽 (typically 90-95%) is − 𝐶𝐶3 𝜀𝜀̇ 𝑚𝑚 𝜀𝜀 𝑛𝑛 exp = − 𝑅𝑅𝑇𝑇 2 𝑅𝑅𝑅𝑅 𝑹𝑹𝑻𝑻𝟐𝟐 converted to heat: 𝑑𝑑𝐻𝐻 = 𝛽𝛽𝑉𝑉 𝜎𝜎𝜎𝜎𝜎𝜎 What happens at high T, and low 𝑒𝑒? ̇ The heat generated adiabatically Stainless 𝒏𝒏 increases the temperature of the Steel (FCC) 𝜺𝜺𝒖𝒖 = material (happens at high strain 𝟏𝟏 + 𝒎𝒎 rates): n ~ 0  small 𝜺𝜺𝒖𝒖 𝐶𝐶𝑝𝑝 𝑚𝑚𝑚𝑚𝑚𝑚 = 𝛽𝛽𝑉𝑉 𝜎𝜎𝜎𝜎𝜎𝜎 High T tensile flow is inherently unstable, but ductility? 𝒅𝒅𝑻𝑻 𝜷𝜷𝑽𝑽𝝈𝝈 𝜷𝜷𝜷𝜷 = = 𝒅𝒅𝒅𝒅 𝑪𝑪𝒑𝒑 𝒎𝒎 𝑪𝑪𝒑𝒑 𝝆𝝆 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 36 𝜺𝜺 ≡ 𝒍𝒍𝒍𝒍 𝟏𝟏 + 𝒆𝒆 Instability in the presence of ⇒ 𝟏𝟏 + 𝒆𝒆 = 𝐞𝐞𝐞𝐞𝐞𝐞 𝜺𝜺 pre-existing inhomogeneities 𝝈𝝈 ≡ 𝒔𝒔 𝟏𝟏 + 𝒆𝒆 ⇒ 𝒔𝒔 = 𝝈𝝈 𝟏𝟏 + 𝒆𝒆 −𝟏𝟏 = 𝝈𝝈𝐞𝐞𝐞𝐞𝐞𝐞 −𝜺𝜺 𝐴𝐴𝑏𝑏𝑜𝑜 “Stepped” specimen 𝑓𝑓 = Grip 𝐴𝐴𝑎𝑎𝑜𝑜 a b a ↑ ↑ 𝐴𝐴𝑏𝑏𝑜𝑜 𝐴𝐴𝑎𝑎𝑎𝑎 ↓ ↓ During the test the load, P must be constant along the length of the sample: 𝑃𝑃 = 𝐴𝐴𝑎𝑎𝑎𝑎 𝑠𝑠𝑎𝑎 = 𝐴𝐴𝑏𝑏𝑏𝑏 𝑠𝑠𝑏𝑏 For a material obeying 𝝈𝝈 = 𝑲𝑲𝜺𝜺̇ 𝒎𝒎 (valid at high temperatures when n~0 ) 𝑃𝑃 = 𝐴𝐴𝑎𝑎𝑎𝑎 𝐾𝐾𝜀𝜀𝑎𝑎̇ 𝑚𝑚 exp(−𝜀𝜀𝑎𝑎 ) = 𝐴𝐴𝑏𝑏𝑏𝑏 𝐾𝐾𝜀𝜀𝑏𝑏̇ 𝑚𝑚 exp(−𝜀𝜀𝑏𝑏 ) ⇒ 𝜀𝜀𝑎𝑎̇ 𝑚𝑚 exp(−𝜀𝜀𝑎𝑎 ) = 𝑓𝑓 𝜀𝜀𝑏𝑏̇ 𝑚𝑚 exp(−𝜀𝜀𝑏𝑏 ) 𝜀𝜀𝑎𝑎 1 𝜀𝜀𝑏𝑏 exp(− ) 𝑑𝑑𝜀𝜀𝑎𝑎 = 𝑓𝑓 𝑚𝑚 𝜀𝜀𝑏𝑏̇ 𝑚𝑚 exp(− ) 𝑑𝑑𝜀𝜀𝑎𝑎 𝑚𝑚 𝑚𝑚 𝜀𝜀𝑎𝑎 𝜀𝜀𝑏𝑏 𝜀𝜀𝑎𝑎 1 𝜀𝜀𝑏𝑏 exp(− ) 𝑑𝑑𝜀𝜀𝑎𝑎 = 𝑓𝑓 exp(− ) 𝑑𝑑𝜀𝜀𝑎𝑎 𝑚𝑚 𝑚𝑚 𝑚𝑚 0 0 𝜀𝜀𝑎𝑎 1 𝜀𝜀𝑏𝑏 exp − − 1 = 𝑓𝑓 𝑚𝑚 exp − −1 Note: 𝜀𝜀𝑎𝑎,𝑚𝑚𝑚𝑚𝑚𝑚 is the uniform 𝑚𝑚 𝑚𝑚 elongation one gets outside the 𝟏𝟏⁄ simulated neck If 𝜀𝜀𝑏𝑏 approaches ∞, then 𝜺𝜺𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 = −𝒎𝒎𝒎𝒎𝒎𝒎 𝟏𝟏 − 𝒇𝒇 𝒎𝒎 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 𝜺𝜺𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 = −𝒎𝒎𝒎𝒎𝒎𝒎 𝟏𝟏 − 𝒇𝒇 𝟏𝟏 𝒎𝒎 37 𝒆𝒆 = 𝐞𝐞𝐞𝐞𝐞𝐞 𝜺𝜺 − 𝟏𝟏 −𝒎𝒎 High temperature deformation 𝟏𝟏 𝒆𝒆𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 = 𝒆𝒆𝒇𝒇 = 𝟏𝟏 − 𝒇𝒇 𝒎𝒎 − 𝟏𝟏 Uniform elongation (%) = 𝒆𝒆𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 ∗ 𝟏𝟏𝟏𝟏𝟏𝟏 Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 𝜺𝜺𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 = −𝒎𝒎𝒎𝒎𝒎𝒎 𝟏𝟏 − 𝒇𝒇 𝟏𝟏 𝒎𝒎 38 𝒆𝒆 = 𝐞𝐞𝐞𝐞𝐞𝐞 𝜺𝜺 − 𝟏𝟏 −𝒎𝒎 High temperature deformation 𝟏𝟏 𝒆𝒆𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 = 𝒆𝒆𝒇𝒇 = 𝟏𝟏 − 𝒇𝒇 𝒎𝒎 − 𝟏𝟏 Uniform elongation (%) = 𝒆𝒆𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 ∗ 𝟏𝟏𝟏𝟏𝟏𝟏 At high temperatures, instabilities and perhaps necking is inevitable. The question then is how rapidly the neck will thin down to a point and fail; a ductile fracture mode εa,max called rupture. Higher values of strain rate sensitivity result in slower “neck thinning rate”. This results in higher uniform elongation “outside” the neck! Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 𝜺𝜺𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 = −𝒎𝒎𝒎𝒎𝒎𝒎 𝟏𝟏 − 𝒇𝒇 𝟏𝟏 𝒎𝒎 39 𝒆𝒆 = 𝐞𝐞𝐞𝐞𝐞𝐞 𝜺𝜺 − 𝟏𝟏 Superplasticity (𝑒𝑒𝑓𝑓 >1000%) 𝟏𝟏 −𝒎𝒎 𝒆𝒆𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 = 𝒆𝒆𝒇𝒇 = 𝟏𝟏 − 𝒇𝒇 𝒎𝒎 − 𝟏𝟏 Uniform elongation (%) = 𝒆𝒆𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 ∗ 𝟏𝟏𝟏𝟏𝟏𝟏 2000% elongation in a nanocrystalline high entropy alloy Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 𝜺𝜺𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 = −𝒎𝒎𝒎𝒎𝒎𝒎 𝟏𝟏 − 𝒇𝒇 𝟏𝟏 𝒎𝒎 40 𝒆𝒆 = 𝐞𝐞𝐞𝐞𝐞𝐞 𝜺𝜺 − 𝟏𝟏 Superplasticity (𝑒𝑒𝑓𝑓 >1000%) 𝟏𝟏 −𝒎𝒎 𝒆𝒆𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 = 𝒆𝒆𝒇𝒇 = 𝟏𝟏 − 𝒇𝒇 𝒎𝒎 − 𝟏𝟏 Uniform elongation (%) = 𝒆𝒆𝒂𝒂,𝒎𝒎𝒎𝒎𝒎𝒎 ∗ 𝟏𝟏𝟏𝟏𝟏𝟏 1950% elongation in Pb-Sn solder Lecture 02-05: Review of plasticity in metals, UMT312T, 2024, S. Karthikeyan 41 Summary of observations regarding plasticity in metals o When does plasticity begin and what determines strength? o Strength increases with strain and strain rate, and decreases with temperature. Why? o The strain hardening exponent decreases with increasing temperature (and with decreasing strain rate) while strain rate sensitivity increases with increasing temperature (and with decreasing strain rate) even for ceramics. Why?