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MIT 3 11 - RECITATION 12

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RECITATION 12 DECEMBER 2, 2003 I think craze, shear deformation zones and rubber toughening of glassy polymers was covered well in lecture—so I will focus on the more complicated equations dealing with the 3-D deformation. Please ask me if you’d like more clarification and I can cover it in the final recitation (Dec. 9) PLASTICITY: Plastic behavior 3-D Plastic Deformation Review of stress states for 3-D (Von Mises & Tresca) Plasticity: At R.T. many materials (esp. metals) have a well defined yield stress, σy σ < σy elastic recoverable deformation σ > σy plastic irrecoverable Plastic behavior is important for: Material design for strengthening (alloying) Work hardening Hardness – friction and wear Some materials fracture before yielding (ceramics (brittle)) BUT, fracture can be suppressed by a hydrostatic stress Æ even ceramics will yield. This next stuff—low temperature plasticity (not high temp. creep behavior) As a review: We’ve talked about a couple types of tests: 1. Uniaxial Tensile Test a. Destructive to sample b. Measures ductility c. Shows where yield stress is located (at 0.2% strain) 2. Hardness Test a. Indenter pushed into polished surface w/ a known force b. Measure the size of indent (you should have done this/will do this in 3.081 c. Gives hardness, which is related to yield stress d. Different versions of indenters (sphere, pyramid, cone) i. Advantages 1. non-destructive, easy to do. 2. can use this on ceramics note: for hardness test, it needs to be a polished material, maybe 500um diameterEquations of plasticity Assumptions: 1. Plastic flow occurs at constant volume a. ∆V/Vo = 0 ε11 + ε22 + ε33 = 0 (note: strains are plastic strains in x, y and z direction) 2. Modest hydrostatic pressures to not affect yield stress/strength a. “modest”~E/100 3. Neglect work hardening a. Assume elastic region got to perfect plastic region 4. Assume Material is Isotropic Criteria for Yielding Under Multi-Axial Stresses Yield unaffected by hydrostatic component of stress (mean stress) [σ11, σ12, … σ33] = [σm, σm, σm] + [σ11-σm, σ12, … σ33-σm] where σm = (σ11+σ12+σ33)/3 Å sigma m is the mean stress component everything but mean stress is related to the amount of shearing in the material—called the deviatoric component of σ. Yield is governed by the deviatoric part of σ Æ shearing—is what’s really controlling yield. Yield Criteria: Tresca Criterion Sometimes called the Maximum shear stress criterion 1. get yield when max. shear stress, tau, in component (under general stress state) equals the ma, tau, in a uniaxial tensile test at yield 2. principle stresses σ1>σ2>σ3 You could take stress state, rotate and rewrite it. Mohr’s circle does this. For general stress state: τmax = (σ1-σ3)/2 In uniaxial test: σ1 (σ2=σ3 =0) τmax = (σ1)/2 so, you’d get yield when σ1=σY In general, you get yield when σ1 – σ3 = σY Tresca says, you take the difference between the biggest and smallest stress to get yield stress. Yield Criteria: Von Mises Criterion You get yield when equivalent sigma = yield stress σeq =sqrt(0.5*[(σ11 –σ22 )^2 + (σ22 –σ33 )^2 + (σ33 –σ11 )^2] +3σ23^2 + 3σ13^2 +3σ12^2 ) How is this related to shearing? We’re applying stress. The stress energy for mean + stress energy for shear = multiaxial stress stateShow example of comparison of Tresca and Von Mises for Biaxial stress (sigma22=0) Tresca: if stress state inside, yield criteriaÆelastic If On yield criteriaÆ plastic For Von Mises: you have ellipsoid criteria There is a gap between them that correlates to pure shear. (it’s not too far apart for either criteria—the maximum difference is 13%) Tresca criterion is more conservative than Von Mises criterion (most used) Example Problems with Von Mises/Tresca analysis2. When coins are minted, they are stamped out in a


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MIT 3 11 - RECITATION 12

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