AcademyEquilibrium and Materials
Academy
Elastic Versus Plastic Behavior
Level 1 - Physics topic page in Equilibrium and Materials.
Principle
Elastic deformation reverses; plastic deformation leaves permanent strain.
Notation
\(\sigma_y\)
yield stress
\(\mathrm{Pa}\)
\(\sigma,\sigma_{\max},\sigma_{\mathrm{unload}}\)
applied, maximum, and unloading stress
\(\mathrm{Pa}\)
\(\epsilon,\epsilon_{\mathrm{total}},\epsilon_{\mathrm{elastic}},\epsilon_{\mathrm{plastic}}\)
strain variables for loading and unloading
none
\(F,A\)
load and cross-sectional area
\(\mathrm{N,\;m^{2}}\)
\(Y\)
elastic slope
\(\mathrm{Pa}\)
\(u,S\)
elastic energy density and factor of safety
J m^{-3}, none
Method
Derivation 1: Identify the elastic region
Elastic law
\[\sigma=Y\epsilon\quad(\sigma<\sigma_y)\]
In this region unloading returns the material to zero strain.
Yield point
\[\sigma=\sigma_y\]
Derivation 2: Separate elastic recovery from permanent strain
Beyond yield, unloading still follows slope \(Y\), but it no longer returns to the origin.
Split the total strain
\[\epsilon_{\mathrm{total}}=\epsilon_{\mathrm{elastic}}+\epsilon_{\mathrm{plastic}}\]
Recovered elastic part
\[\epsilon_{\mathrm{elastic}}=\frac{\sigma_{\mathrm{unload}}}{Y}\]
Permanent strain
\[\epsilon_{\mathrm{plastic}}=\epsilon_{\mathrm{total}}-\frac{\sigma_{\mathrm{unload}}}{Y}\]
Derivation 3: Build the design limit
Design checks compare working stress with a reduced yield stress, not with the breaking stress.
Working stress
\[\sigma=\frac{F}{A}\]
Yield check
\[\sigma_{\max}<\frac{\sigma_y}{S}\]
Derivation 4: Read elastic energy from the graph
In the elastic region, the energy stored per unit volume is the area under the straight-line stress-strain graph.
Energy density
\[u=\frac{1}{2}\sigma\epsilon\]
Rules
These are the compact results from the derivations above.
Elastic law
\[\sigma=Y\epsilon\quad(\sigma<\sigma_y)\]
Yield check
\[\sigma_{\max}<\frac{\sigma_y}{S}\]
Energy density
\[u=\frac{1}{2}\sigma\epsilon\]
Permanent strain
\[\epsilon_{\mathrm{plastic}}=\epsilon_{\mathrm{total}}-\frac{\sigma_{\mathrm{unload}}}{Y}\]
Examples
Question
A material has
\[Y=200\,\mathrm{GPa}\]
and \[\sigma_y=250\,\mathrm{MPa}\]
Find yield strain.Answer
\[\epsilon_y=\frac{\sigma_y}{Y}=\frac{250\times10^6}{200\times10^9}=1.25\times10^{-3}\]
Checks
- Elastic means reversible, not weak.
- Yield is a stress limit.
- Ultimate stress is not first yield.
- Safety factor lowers working stress.