Engineering & Technology · Material Structure, Processing & Properties
Elasticity, plasticity and structure-dependent deformation
A thin metal strip springs back after a small bend but keeps a new shape after a larger bend. A rubber strip may stretch much farther and still recover. “How much does it move?” and “does it recover?” are different questions, and neither is answered by the word strong alone.
The everyday-materials chapter supplies the distinction between a material and an object. This chapter explains atomic order and its own stress, strain, phase and transport vocabulary. Optional chemistry bonding and physics materials lessons provide deeper support. Simple ratios, percentages and graph reading are used; the phase-fraction balance is derived in the example. All numerical properties, curves, maps, images and data here are constructed teaching models, not measured values, processing recipes or design allowables.
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Connect recoverable and permanent deformation to stress, strain and mechanisms, without confusing stiffness with strength.
Separate specimen geometry from material response
For a simple uniform tensile specimen, engineering stress σ is force F divided by its original cross-sectional area A0: σ = F/A0. Engineering strain ε is extension ΔL divided by original gauge length L0: ε = ΔL/L0. Stress has units of pressure; 1 N/mm² equals 1 MPa. Strain is a ratio without units and is often expressed as a percentage.
These quantities help compare specimens of different sizes under comparable conditions. They are not enough to describe every local stress in a notched or complex part. The simple uniform-tension interpretation assumes an appropriate specimen, loading and measurement region. Geometry can affect a whole object’s stiffness even when its material is unchanged.
Use recovery to distinguish the responses
Elastic deformation is recoverable after removing load, within the relevant conditions and timescale. In a linear elastic region, σ = Eε, where E is Young’s modulus, a measure of material stiffness. Elastic does not necessarily mean linear: rubber-like materials can recover through a strongly curved stress–strain relationship.
Plastic deformation leaves a permanent change. Yield describes the onset or a conventionally defined amount of plastic deformation; the precise measurement definition matters when no sharp yield point exists. In many metals, dislocation motion enables this rearrangement. Increasing the resistance to that motion can raise yield strength without a comparable increase in elastic modulus.
Polymers can also rearrange through chain motion. Temperature and loading rate strongly affect the available motion. A response that appears stiff in a short test can change during a long hold. Viscoelasticity combines time-dependent and elastic features, so immediate unloading does not always separate every kind of recovery. The simple worked graph deliberately excludes that complexity.
Read loading and unloading paths separately
The original idealised curve begins with E = 50 GPa and reaches 100 MPa at strain 0.002, or 0.2 %. It then rises to 150 MPa at 0.006, or 0.6 %. The curved or sloped plastic part need not have the same slope as the initial elastic part. In this teaching model, unloading is linear with the original modulus.
The dashed unloading line returns to zero stress at 0.3 % strain, so the specimen retains a permanent extension. Stiffness is represented by the initial slope, yielding by the change in deformation behaviour, and fracture by separation, which is not shown here. Neither hardness nor toughness can be read as a synonym for the initial slope.
Compute stress, strain and residual extension
An invented specimen has A0 = 10 mm² and L0 = 100 mm. At F = 500 N and extension 0.10 mm, σ = 500/10 = 50 MPa and ε = 0.10/100 = 0.001, or 0.1 %. Their ratio gives E = 50/0.001 = 50 000 MPa = 50 GPa if this point lies in the linear region.
At the later idealised point, total strain is 0.006 and stress 150 MPa. Unloading recovers σ/E = 150/50 000 = 0.003. Residual strain is 0.006 − 0.003 = 0.003, giving residual extension 0.003 × 100 = 0.30 mm. These are model predictions for the stated assumptions, not a material data sheet.
Try the reasoning
Does twice the Young’s modulus necessarily mean twice the yield strength?
Show the answer and reasoning
No. Modulus characterises elastic response; yielding depends on mechanisms for permanent deformation. They need not change in the same proportion.
Can a recoverable deformation be nonlinear?
Show the answer and reasoning
Yes. Elastic means recoverable, while linear specifies the shape of the stress–strain relationship.
Why must the specimen’s original area be stated for engineering stress?
Show the answer and reasoning
The calculation uses that reference area. A changing current area would define a different stress measure, especially after large deformation.
What to carry forward
Use stress and strain with their definitions. Elastic slope, permanent deformation and fracture describe different behaviour.
References
Teaching resources and research supporting this lesson’s explanations and examples.
Massachusetts Institute of Technology3 sources
- CHEM ATLAS, Part 3, cover dated Fall 2019 — Lectures 21, 23–25Jeffrey C. Grossman, Ximena Hasbach and Madison Sutula, MITReference guide · PDF
- Introduction to Solid State Chemistry — Session 20: Line, Interface and Bulk Defects (Fall 2010)Donald Sadoway / MIT OpenCourseWareReference guide
- Unified Engineering: Materials and Structures — Lecture 1 (Fall 2021), slides 21–22 and 26MIT OpenCourseWareReference guide · PDF