Mechanical Science & Engineering
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    Material Properties

    Stress-Strain Diagram

    A stress-strain diagram is the relationship of normal stress as a function of normal strain. One way to collect these measurement is a uniaxial tension test in which a specimen at a very slow, constant rate (quasi-static). A load \( P \) and distance \( L \) are measured at frequent intervals.
    Evaluation of average normal strain, or engineering strain (relative to undeformed length).
    $$ \varepsilon = \frac{\delta}{L_0} = \frac{L-L_0}{L_0}\ $$
    Evaluation of average normal stress, or engineering stress (relative to undeformed cross section).
    $$ \sigma = \frac{P}{A_0}\ $$
    Warning: Two stress-strain diagrams for a particular material will be similar, but not identical. #str-stn

    Things that effect the material properties include imperfections, different composition, rate or loading, or temperature.

    Stress-strain diagram showing key properties

    Elastic Behavior

    Loading in this region results elastic behavior, meaning the material returns to its original shape when unloaded. This region of the diagram is mostly a straight line, limited by the proportional limit). This region ends at \( \sigma_Y \) (yielding). For \( \sigma \le \sigma_{pl} \), the diagram is linear, and the behavior is elastic. For \( \sigma_{pl} < \sigma < \sigma_Y \), the diagram is nonlinear, but the behavior is still elastic. Young's Modulus Hooke's law is used for small deformations in the elastic region. The Young's modulus is the slope \( E \).
    Hooke's law.
    $$ \sigma = E\varepsilon\ $$
    MaterialYoung's modulus [\(GPa\)]
    Mild Steel210
    Copper120
    Bone18
    Plastic2
    Rubber0.02
    Shear Modulus
    Shear stress strain diagram
    Hooke's law.
    $$ \tau_{xy} = G\gamma_{xy}\ $$
    Shear modulus.
    $$ G = \frac{E}{2(1+\nu)}\ $$
    Only two of the three material constants (ie; \( G \), \( E \), \( \nu \)) are independent in isotropic materials.

    Plastic Behavior

    Stresses above the plastic limit (\( \sigma_Y < \sigma \)) cause the material to permanently deform.

    Yield Strength

    Perfect plastic or ideal plastic: well-defined \( \sigma_Y \), stress plateau up to failure. Some materials (e.g. mild steel) have two yield points (stress plateau at \( \sigma_{YL} \)). Most ductile metals do not have a stress plateau; yield strength \( \sigma_{YS} \) is then defined by the 0.002 (0.2%) offset method.

    Strain Hardening
    Strain hardening

    Atoms rearrange in plastic region of ductile materials when a higher stress is sustained. Plastic strain remains after unloading as permanent set, resulting in permanent deformation. Reloading is linear elastic up to the new, higher yield stress (at A') and a reduced ductility.

    Ultimate Strength

    The ultimate strength (\( \sigma_u \)) is the maximum stress the material can withstand.

    Necking
    Necking

    After ultimate stress (\( \sigma_u < \sigma \)), the middle of the material elongates before failure.

    Failure

    Also called fracture or rupture stress (\( \sigma_f \)) is the stress at the point of failure for the material. Brittle and Ductile materials fail differently.

    Ductile materials generally fail in shear. There is a large region of plastic deformation before failure (fracture) at higher strain and necking.

    Ductile failure
  • Axial – maximum shear stress at \(45^o\) angle
  • Torsion – maximum shear stress at \(0^o\) angle

    • Brittle materials are weaker in tension than shear. There is a small plastic region between yield and failure (fracture) and no necking.

    Brittle failure
  • Axial – maximum normal stress at \(0^o\) angle
  • Torsion – maximum normal stress at \(45^o\) angle

    • Note the difference between engineering and true stress/strain diagrams: ultimate stress is a consequence of necking, and the true maximum is the true fracture stress.

    Example: Concrete is a brittle material.
    • Maximum compressive strength is substantially larger than the maximum tensile strength.
    • For this reason, concrete is almost always reinforced with steel bars or rods whenever it is designed to support tensile loads.

    Heads up!

    Strain Energy builds on this content in Engineering Materals.

    Deformation does work on the material: equal to internal strain energy (by energy conservation).

    $$ \Delta U = \int F_{z}dz = \int \sigma_{z}(\Delta x \Delta y \Delta z)d\varepsilon_{z}\ $$
    Energy density.
    $$ u = \frac{\Delta U}{\Delta V} = \int \sigma_{z}d\varepsilon_{z}\ $$
    Can be generalized to any deformation: areas under stress-strain curves
    $$ u = \int \sigma d\varepsilon \quad \text{or} \quad u = \int \tau d\gamma\ $$

    Heads up!

    Fatigue builds on this content in Engineering Materals and Mechanical Design.

    SN curve examples
    If stress does not exceed the elastic limit, the specimen returns to its original configuration. However, this is not the case if the loading is repeated thousands or millions of times. In such cases, rupture will happen at stress lower than the fracture stress - this phenomenon is known as fatigue.

    Other Derived Properties

    Directional Materials

    • Isotropic: material properties are independent of the direction
    • Anisotropic: material properties depend on the direction (ie; composites, wood, and tissues)

    Poisson's Ratio

    Poisson's ratio
    Axial (normal) strain.
    $$ \varepsilon_{x} = \frac{\delta}{L}\ $$
    Poisson's ratio \(\nu\).
    $$ \nu = -\frac{\text{lateral strain}}{\text{axial strain}} $$
    Typical range.
    $$ 0 < \nu < 0.5\ $$
    Lateral strain.
    $$ \varepsilon_{z} = \varepsilon_{y} = -\nu\varepsilon_{x}\ $$
    Warning: Negative Poisson's Ratio (auxetics) #ngt-psn

    A negative Poisson's ratio is possible when the material structure is non-trival.