Elasticity, Hooke's law and strength criteria
Strains, Young's modulus and Poisson's ratio, G = E/(2(1+ν)) proved, the stress-strain curve of steel, Tresca and von Mises criteria with an example.
A loaded material deforms; as long as the stress stays below a limit, removing the load restores it: this is elastic behaviour. Beyond it, the material deforms permanently and finally breaks. In this lesson we look at strains, Hooke’s law with Young’s modulus and Poisson’s ratio, the proof of the relation between shear modulus and Young’s modulus with an example, the stress-strain curve of steel, and strength criteria for combined stress states with an example.
Strains
Strain measures the change of shape: normal strain 
is the relative elongation of a segment, shear strain 
the change of the angle, initially a right angle, between two segments. Like stresses, strains at a point form a symmetric tensor. They are very small numbers: a structural steel at yield elongates by about 0.1 %.
Hooke’s law
For a linear elastic isotropic material, in simple tension:
![\[ \sigma = E\,\varepsilon,\qquad \varepsilon_{trans} = -\nu\,\varepsilon \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-ee118402e761cf2033689a0286eb78ae_l3.png "Rendered by QuickLaTeX.com")
where 
is Young’s modulus and 
Poisson’s ratio: a bar in tension lengthens and thins. Typical values: steel 
GPa, 
; aluminium 70 GPa, 0.33; concrete 30 GPa, 0.2; wood along the grain about 11 GPa. For a general stress state the strains add by superposition:
![\[ \varepsilon_x = \frac{1}{E}\big[\sigma_x - \nu(\sigma_y + \sigma_z)\big],\qquad \gamma_{xy} = \frac{\tau_{xy}}{G} \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-8922173dbeaa3d568b7c8722140fa41a_l3.png "Rendered by QuickLaTeX.com")
with 
the shear modulus.
Theorem. For an isotropic material
![\[ G = \frac{E}{2(1 + \nu)} \]](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-26a4471452b52f898158532f0d1e4bfb_l3.png "Rendered by QuickLaTeX.com")
Proof. Consider a state of pure shear: 
, 
. Mohr’s circle (lesson on stress) has its centre at the origin and radius 
: the principal stresses, at 45°, are 
and 
. Along direction 1 Hooke’s law gives 
. But the same deformation, seen in the 
, 
axes, is a shear strain; Mohr’s circle for strains, analogous to the one for stresses with 
in place of , gives for pure shear 
. So 
, and 
. ∎
Example. Steel: 
GPa. An isotropic material therefore has two independent elastic constants, not three.
The stress-strain curve
The tensile test of a mild steel shows a linear elastic segment, yielding at 
(235, 275 or 355 MPa for steels S235, S275, S355), a long plastic segment in which the material elongates at almost constant stress, strain hardening up to the strength 
, and fracture after elongations of 20 % or more. This ductility is precious: a steel structure deforms visibly before collapsing, giving warning and redistributing loads (Building Techniques lesson on steel). Brittle materials (glass, cast iron, concrete in tension, ceramics, the silicon of chips) instead break without warning, just beyond the elastic range.
Strength criteria
The tensile test gives a single number; in a real structure the stress state is combined. A strength criterion defines an equivalent stress to be compared with that of the uniaxial test. For ductile materials:
- Tresca (maximum shear stress):
; - von Mises (distortion energy):
![\sigma_{eq} = \sqrt{\frac{1}{2}\big[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2\big]}](https://www.electronics-engineering.com/wp-content/ql-cache/quicklatex.com-53a34ff2f44bace7ccc8784acd6e7342_l3.png "Rendered by QuickLaTeX.com")
.
In a beam, with a normal stress 
and a shear stress , the von Mises criterion reduces to 
, Tresca’s to 
.
Example. At a point of an S235 steel beam, 
MPa and 
MPa. Von Mises: 
MPa < 235 MPa, verified; Tresca, more conservative: 
MPa. For brittle materials criteria on the maximum principal tensile stress are used.
