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Plastic and viscoelastic deformation

Separate recoverable elasticity, permanent plastic strain and time-dependent response, then derive usable constitutive updates.

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Matter pathway: atom → solid → liquid → gas → plasma. Quantum mechanics and quantum field theory provide foundations across the pathway; they are not additional phases. This is a connected modeling route, not a universal heating curve. Actual phases depend on pressure, composition, and kinetics.

1. Elastic–plastic split and yielding

Definitions & inputs. ε total strain,εe elastic strain,εp plastic strain,σy yield stress,H isotropic hardening modulus,α accumulated plastic strain.

  1. Only the elastic strain supplies the linear elastic stress.

    ϵ=ϵe+ϵp,σ=E(ϵ−ϵp)\epsilon=\epsilon_e+\epsilon_p,\quad\sigma=E(\epsilon-\epsilon_p)

  2. Define the elastic domain.

    f=∣σ∣−(σy0+Hα)≤0f=|\sigma|-(\sigma_{y0}+H\alpha)\le0

  3. Plastic flow occurs on the yield surface; unloading can be elastic.

    λ˙≥0,f≤0,λ˙f=0\dot\lambda\ge0,\quad f\le0,\quad\dot\lambda f=0

Interpretation. Residual strain after unloading distinguishes plasticity from reversible elasticity.

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2. Return mapping

Definitions & inputs. σtrial elastic predictor,Δλ plastic multiplier,signs based on trial stress.

  1. Hold plastic strain fixed for the trial step.

    σtrial=E(ϵn+1−ϵp,n)\sigma_{trial}=E(\epsilon_{n+1}-\epsilon_{p,n})

  2. Impose the final yield condition to solve for the correction.

    Δλ=max⁡[0,(∣σtrial∣−σy0−Hαn)/(E+H)]\Delta\lambda=\max[0,(|\sigma_{trial}|-\sigma_{y0}-H\alpha_n)/(E+H)]

  3. Update internal variables and recompute stress.

    ϵp,n+1=ϵp,n+Δλsign⁡σtrial,αn+1=αn+Δλ\epsilon_{p,n+1}=\epsilon_{p,n}+\Delta\lambda\operatorname{sign}\sigma_{trial},\quad\alpha_{n+1}=\alpha_n+\Delta\lambda

Interpretation. Multiaxial J2 plasticity uses von Mises stress and a tensor return map; a uniaxial formula is not that full algorithm.

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3. Maxwell and Kelvin–Voigt rheology

Definitions & inputs. E spring stiffness,η dashpot viscosity,τ=η/E.

  1. A Maxwell spring and dashpot in series share stress and add strains.

    ϵ˙=σ˙/E+σ/η\dot\epsilon=\dot\sigma/E+\sigma/\eta

  2. Under fixed strain the Maxwell model relaxes stress exponentially.

    ϵ=ϵ0⇒σ(t)=Eϵ0e−t/τ\epsilon=\epsilon_0\Rightarrow\sigma(t)=E\epsilon_0e^{-t/\tau}

  3. Parallel Kelvin–Voigt elements creep toward a finite strain under a stress step.

    σ=Eϵ+ηϵ˙⇒ϵ(t)=σ0(1−e−t/τ)/E\sigma=E\epsilon+\eta\dot\epsilon\Rightarrow\epsilon(t)=\sigma_0(1-e^{-t/\tau})/E

Interpretation. Maxwell creep is unbounded under sustained nonzero stress; Kelvin–Voigt has no instantaneous strain jump. Pick the model from measured behavior.

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4. Generalized relaxation and dissipation

Definitions & inputs. E∞ long-time modulus,Ek positive modal moduli,τk relaxation times,ω angular frequency.

  1. A spectrum of Maxwell branches adds relaxation time scales.

    E(t)=E∞+∑kEke−t/τkE(t)=E_\infty+\sum_kE_ke^{-t/\tau_k}

  2. The in-phase storage modulus measures recoverable oscillatory response.

    E′=E∞+∑kEk(ωτk)2/[1+(ωτk)2]E'=E_\infty+\sum_k E_k(\omega\tau_k)^2/[1+(\omega\tau_k)^2]

  3. The loss modulus gives dissipated energy per unit volume per cycle.

    E′′=∑kEkωτk/[1+(ωτk)2],Wcycle=πE′′ϵ02E''=\sum_k E_k\omega\tau_k/[1+(\omega\tau_k)^2],\quad W_{cycle}=\pi E''\epsilon_0^2

Interpretation. Rate-dependent plastic flow requires additional yield and evolution laws. Time–temperature shifting is valid only over a characterized thermorheologically simple range.

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Graphical worked example

A held strain produces initial stress 1 MPa and Maxwell relaxation time 10 s. X axis: Time after strain step (s). Y axis: Maxwell stress (MPa).
A held strain produces initial stress 1 MPa and Maxwell relaxation time 10 s. Related worked calculation · Download SVG · Plot data

Twenty worked examples

Open a problem to see its defined inputs, assumptions, equation, numerical substitution, result, and interpretation. Values are illustrative analytical exercises.

Example 01. Elastic trial stress

Definitions & inputs. E200000MPa,ε.002,oldεp0.

  1. Choose the governing model and isolate the requested quantity.

    σtrial=Eϵ\sigma_{trial}=E\epsilon

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    200000(.002)200000(.002)

  3. Evaluate the expression; the result uses the units shown.

    Result=400 MPa\mathrm{Result}=400\ \mathrm{MPa}

Interpretation. Compare with the yield stress.

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Example 02. Yield function

Definitions & inputs. Trial400MPa,σy250MPa.

  1. Choose the governing model and isolate the requested quantity.

    f=∣σtrial∣−σyf=|\sigma_{trial}|-\sigma_y

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    400−250400-250

  3. Evaluate the expression; the result uses the units shown.

    Result=150 MPa\mathrm{Result}=150\ \mathrm{MPa}

Interpretation. Positive trialf requires plastic correction.

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Example 03. Perfect-plastic increment

Definitions & inputs. E200000MPa,H0,trial400,yield250.

  1. Choose the governing model and isolate the requested quantity.

    Δλ=(400−250)/E\Delta\lambda=(400-250)/E

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    150/200000150/200000

  3. Evaluate the expression; the result uses the units shown.

    Result=0.00075 \mathrm{Result}=0.00075\ {}

Interpretation. Monotone tensile step.

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Example 04. Corrected perfect-plastic stress

Definitions & inputs. Trial400MPa,E200000MPa,Δλ.00075.

  1. Choose the governing model and isolate the requested quantity.

    σ=σtrial−EΔλ\sigma=\sigma_{trial}-E\Delta\lambda

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    400−200000(.00075)400-200000(.00075)

  3. Evaluate the expression; the result uses the units shown.

    Result=250 MPa\mathrm{Result}=250\ \mathrm{MPa}

Interpretation. Final stress lies on yield.

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Example 05. Hardening plastic increment

Definitions & inputs. Same but H10000MPa.

  1. Choose the governing model and isolate the requested quantity.

    Δλ=150/(E+H)\Delta\lambda=150/(E+H)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    150/210000150/210000

  3. Evaluate the expression; the result uses the units shown.

    Result=0.0007142857 \mathrm{Result}=0.0007142857\ {}

Interpretation. Isotropic linear hardening.

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Example 06. Hardening corrected stress

Definitions & inputs. σy0250MPa,H10000MPa,α150/210000.

  1. Choose the governing model and isolate the requested quantity.

    σ=σy0+Hα\sigma=\sigma_{y0}+H\alpha

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    250+10000(150/210000)250+10000(150/210000)

  3. Evaluate the expression; the result uses the units shown.

    Result=257.1429 MPa\mathrm{Result}=257.1429\ \mathrm{MPa}

Interpretation. Equals the corrected elastic stress.

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Example 07. Plastic tangent

Definitions & inputs. E200000MPa,H10000MPa.

  1. Choose the governing model and isolate the requested quantity.

    Et=EH/(E+H)E_t=EH/(E+H)

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    200000(10000)/210000200000(10000)/210000

  3. Evaluate the expression; the result uses the units shown.

    Result=9523.81 MPa\mathrm{Result}=9523.81\ \mathrm{MPa}

Interpretation. Monotone plastic branch, not unloading slope.

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Example 08. Residual strain

Definitions & inputs. Total.002,stress250MPa,E200000MPa.

  1. Choose the governing model and isolate the requested quantity.

    ϵp=ϵ−σ/E\epsilon_p=\epsilon-\sigma/E

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    .002−250/200000.002-250/200000

  3. Evaluate the expression; the result uses the units shown.

    Result=0.00075 \mathrm{Result}=0.00075\ {}

Interpretation. Fully unloading elastically retains this strain.

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Example 09. Von Mises pure shear

Definitions & inputs. τ100MPa.

  1. Choose the governing model and isolate the requested quantity.

    σeq=3∣τ∣\sigma_{eq}=\sqrt3|\tau|

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    3(100)\sqrt3(100)

  3. Evaluate the expression; the result uses the units shown.

    Result=173.2051 MPa\mathrm{Result}=173.2051\ \mathrm{MPa}

Interpretation. J2 equivalent stress.

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Example 10. Pure shear yield

Definitions & inputs. Uniaxialσy250MPa.

  1. Choose the governing model and isolate the requested quantity.

    τy=σy/3\tau_y=\sigma_y/\sqrt3

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    250/3250/\sqrt3

  3. Evaluate the expression; the result uses the units shown.

    Result=144.3376 MPa\mathrm{Result}=144.3376\ \mathrm{MPa}

Interpretation. Isotropic von Mises criterion.

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Example 11. Relaxation time

Definitions & inputs. η10⁹Pa·s,E10⁸Pa.

  1. Choose the governing model and isolate the requested quantity.

    τ=η/E\tau=\eta/E

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    109/10810^9/10^8

  3. Evaluate the expression; the result uses the units shown.

    Result=10 s\mathrm{Result}=10\ \mathrm s

Interpretation. Maxwell or Kelvin characteristic time.

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Example 12. Maxwell initial stress

Definitions & inputs. E100MPa,stepε.01.

  1. Choose the governing model and isolate the requested quantity.

    σ0=Eϵ0\sigma_0=E\epsilon_0

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    100(.01)100(.01)

  3. Evaluate the expression; the result uses the units shown.

    Result=1 MPa\mathrm{Result}=1\ \mathrm{MPa}

Interpretation. Immediately after ideal strain step.

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Example 13. Maxwell relaxed stress

Definitions & inputs. Initial1MPa,t10s,τ10s.

  1. Choose the governing model and isolate the requested quantity.

    σ=σ0e−t/τ\sigma=\sigma_0e^{-t/\tau}

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    e−1e^{-1}

  3. Evaluate the expression; the result uses the units shown.

    Result=0.3678794 MPa\mathrm{Result}=0.3678794\ \mathrm{MPa}

Interpretation. Maintained strain.

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Example 14. Stress half time

Definitions & inputs. τ10s.

  1. Choose the governing model and isolate the requested quantity.

    t1/2=τln⁡2t_{1/2}=\tau\ln2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    10ln⁡210\ln2

  3. Evaluate the expression; the result uses the units shown.

    Result=6.931472 s\mathrm{Result}=6.931472\ \mathrm s

Interpretation. Exponential relaxation.

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Example 15. Kelvin creep

Definitions & inputs. σ1MPa,E100MPa,t=τ.

  1. Choose the governing model and isolate the requested quantity.

    ϵ=(σ/E)(1−e−t/τ)\epsilon=(\sigma/E)(1-e^{-t/\tau})

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    .01(1−e−1).01(1-e^{-1})

  3. Evaluate the expression; the result uses the units shown.

    Result=0.006321206 \mathrm{Result}=0.006321206\ {}

Interpretation. Starts from zero strain.

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Example 16. Maxwell creep

Definitions & inputs. σ1MPa,E100MPa,η1000MPa·s,t20s.

  1. Choose the governing model and isolate the requested quantity.

    ϵ=σ/E+σt/η\epsilon=\sigma/E+\sigma t/\eta

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    1/100+20/10001/100+20/1000

  3. Evaluate the expression; the result uses the units shown.

    Result=0.03 \mathrm{Result}=0.03\ {}

Interpretation. Includes instant elastic and growing viscous strain.

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Example 17. Storage modulus

Definitions & inputs. Single Maxwell E10MPa,ωτ1,E∞0.

  1. Choose the governing model and isolate the requested quantity.

    E′=E(ωτ)2/[1+(ωτ)2]E'=E(\omega\tau)^2/[1+(\omega\tau)^2]

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    10/210/2

  3. Evaluate the expression; the result uses the units shown.

    Result=5 MPa\mathrm{Result}=5\ \mathrm{MPa}

Interpretation. Oscillatory storage.

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Example 18. Loss modulus

Definitions & inputs. Same branch.

  1. Choose the governing model and isolate the requested quantity.

    E′′=Eωτ/[1+(ωτ)2]E''=E\omega\tau/[1+(\omega\tau)^2]

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    10/210/2

  3. Evaluate the expression; the result uses the units shown.

    Result=5 MPa\mathrm{Result}=5\ \mathrm{MPa}

Interpretation. Loss peaks atωτ1 for this branch.

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Example 19. Cycle dissipation

Definitions & inputs. E″5MPa,strain amplitude.01.

  1. Choose the governing model and isolate the requested quantity.

    W=πE′′ϵ02W=\pi E''\epsilon_0^2

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    π(5×106)(.01)2\pi(5\times10^6)(.01)^2

  3. Evaluate the expression; the result uses the units shown.

    Result=1570.796 J m−3\mathrm{Result}=1570.796\ \mathrm{J\,m^{-3}}

Interpretation. Positive dissipation per cycle.

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Example 20. Long-time modulus

Definitions & inputs. Standard linear solidE∞2MPa,E1=8MPa,t≫τ.

  1. Choose the governing model and isolate the requested quantity.

    E(∞)=E∞E(\infty)=E_\infty

  2. Insert the stated inputs in consistent units or the explicitly defined normalized units.

    2+8(0)2+8(0)

  3. Evaluate the expression; the result uses the units shown.

    Result=2 MPa\mathrm{Result}=2\ \mathrm{MPa}

Interpretation. Unlike a pure Maxwell fluid, this model retains stiffness.

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Symbols and units

Each derivation and problem defines its own symbols and inputs. Symbols may be reused with different meanings in other subjects. Keep units consistent, retain sufficient precision during calculation, and apply the stated validity limits.