Table of contents

Constitutive relations in 2D

We shall now see how to derive the generalized Hooke's law, starting from the 1D case, we have Hooke's law.

σx=Eεx\sigma_x =E\varepsilon_x

Now, in 2D any strain in one direction gives rise to another strain in the orthogonal direction. This is known as the transversal strain. For small strains, the transversal strain is given by

εy=νεx\varepsilon_y =-\nu \varepsilon_x

where ν\nu is known as the Poisson's ratio, which is a measure of the Poisson effect and is experimentally acquired by measuring the transversal strain and the axial strain, then

ν=dεydεx\nu =-\dfrac{d\varepsilon_y }{d\varepsilon_x }

Typical values for the Poisson's ratio:

MaterialPoisson's ratio
Rubber0.4999
Aluminium0.32
Steel0.27-0.33
Cork0.0

For 3D we get

εx=σxEν(σyE+σzE)εy=σyEν(σxE+σzE)εz=σzEν(σxE+σyE)}[εxεyεz]=1E[1ννν1ννν1][σxσyσz] \def\arraystretch{2.0} \left.\begin{array}{c} \varepsilon_x =\dfrac{\sigma_x }{E}-\nu \left(\dfrac{\sigma_y }{E}+\dfrac{\sigma_z }{E}\right)\\ \varepsilon_y =\dfrac{\sigma_y }{E}-\nu \left(\dfrac{\sigma_x }{E}+\dfrac{\sigma_z }{E}\right)\\ \varepsilon_z =\dfrac{\sigma_z }{E}-\nu \left(\dfrac{\sigma_x }{E}+\dfrac{\sigma_y }{E}\right) \end{array}\right\rbrace \left\lbrack \begin{array}{c} \varepsilon_x \\ \varepsilon_y \\ \varepsilon_z \end{array}\right\rbrack =\dfrac{1}{E}\left\lbrack \begin{array}{ccc} 1 & -\nu & -\nu \\ -\nu & 1 & -\nu \\ -\nu & -\nu & 1 \end{array}\right\rbrack \left\lbrack \begin{array}{c} \sigma_x \\ \sigma_y \\ \sigma_z \end{array}\right\rbrack

This system can be inverted, yielding

σx=E1+ν(εx+ν1+2ν(εx+εy+εz))σy=E1+ν(εy+ν1+2ν(εx+εy+εz))σz=E1+ν(εz+ν1+2ν(εx+εy+εz)) \def\arraystretch{2.0} \begin{array}{l} \sigma_x =\dfrac{E}{1+\nu }\left(\varepsilon_x +\dfrac{\nu }{1+2\nu }\left(\varepsilon_x +\varepsilon_y +\varepsilon_z \right)\right)\\ \sigma_y =\dfrac{E}{1+\nu }\left(\varepsilon_y +\dfrac{\nu }{1+2\nu }\left(\varepsilon_x +\varepsilon_y +\varepsilon_z \right)\right)\\ \sigma_z =\dfrac{E}{1+\nu }\left(\varepsilon_z +\dfrac{\nu }{1+2\nu }\left(\varepsilon_x +\varepsilon_y +\varepsilon_z \right)\right) \end{array}

Bulk modulus

The bulk modulus, KK is defined as pressure over volume change and is a measure of how resistant to compression a material is.

K=σmΔK=\dfrac{\sigma_m }{\Delta }

using Δ=εx+εy+εz\Delta =\varepsilon_x +\varepsilon_y +\varepsilon_z, i.e., total volume strain

we get

Δ=12νE(σx+σy+σz)\Delta =\dfrac{1-2\nu }{E}\left(\sigma_x +\sigma_y +\sigma_z \right)

under the same pressure we have

σx+σy+σz=3σm\sigma_x +\sigma_y +\sigma_z =3\sigma_m

and thus

K=σmΔ=E3(12ν)K=\dfrac{\sigma_m }{\Delta }=\dfrac{E}{3\left(1-2\nu \right)}

Shear modulus

Consider an element under pure shear. We can rotate this element such that the shear stresses align with the principal stresses.

σmax=σ1=τx  y\sigma_{\max } =\sigma_1 =\tau_{x\;y}, σmin=σ2=τx  y\sigma_{\min } =\sigma_2 =-\tau_{x\;y}

We can substitute the normal stresses with principal stresses, σx=σmax,σy=σmin,σz=0\sigma_x =\sigma_{\max } ,\sigma_y =\sigma_{\min } ,\sigma_z =0 into the equation

εx=σxEν(σyE+σzE)=εmax=τx  yEν(τE+0)εmax=τx  yE(1+ν)\begin{array}{l} \varepsilon_x =\dfrac{\sigma_x }{E}-\nu \left(\dfrac{\sigma_y }{E}+\dfrac{\sigma_z }{E}\right)=\varepsilon_{\max } =\dfrac{\tau_{x\;y} }{E}-\nu \left(\dfrac{-\tau }{E}+0\right)\\ \Rightarrow \varepsilon_{\max } =\dfrac{\tau_{x\;y} }{E}\left(1+\nu \right) \end{array}

Similar to the principal stresses, we can get the principal strains

ε1,2=εx+εy2±(εx+εy2)2+(γx  y2)2\varepsilon_{1,2} =\dfrac{\varepsilon_x +\varepsilon_y }{2}\pm \sqrt{ {\left(\dfrac{\varepsilon_x +\varepsilon_y }{2}\right)}^2 +{\left(\dfrac{\gamma_{x\;y} }{2}\right)}^2 }

and in the principal direction we have εx=εy=0\varepsilon_x =\varepsilon_y =0

and thus the strain in the principal direction is

εmax=ε1=γx  y2\varepsilon_{\max } =\varepsilon_1 =\dfrac{\gamma_{x\;y} }{2}

Hooke's law for shearing states that

τx  y=Gγx  y\tau_{x\;y} =G\gamma_{x\;y}

So we get εmax=τx  y2G\varepsilon_{\max } =\frac{\tau_{x\;y} }{2G}, and thus

G=E2(1+ν)G=\dfrac{E}{2\left(1+\nu \right)}

Generalized Hooke's law

In general we have ε=τx  y2G\varepsilon =\frac{\tau_{x\;y} }{2G} which leads to

εx  y=(1+ν)τx  yE\varepsilon_{x\;y} =\left(1+\nu \right)\dfrac{\tau_{x\;y} }{E}

and

τx  y=E1+νεx  y\tau_{x\;y} =\dfrac{E}{1+\nu }\varepsilon_{x\;y}

Since earlier we have

σx=E1+ν(εx+ν1+2ν(εx+εy+εz))\sigma_x =\dfrac{E}{1+\nu }\left(\varepsilon_x +\dfrac{\nu }{1+2\nu }\left(\varepsilon_x +\varepsilon_y +\varepsilon_z \right)\right)

which when expanded becomes

σx=(E1+ν2μ=2Gεx+Eν(1+ν)(1+2ν)λ(εx+εy+εz))\sigma_x =\left(\underset{2\mu =2G}{\underbrace{\dfrac{E}{1+\nu } } } \varepsilon_x +\underset{\lambda }{\underbrace{\dfrac{E\nu }{\left(1+\nu \right)\left(1+2\nu \right)} } } \left(\varepsilon_x +\varepsilon_y +\varepsilon_z \right)\right)

we recognize, in the first term, the shear modulus GG, also denoted μ\mu. In the second term we identify the parameter λ\lambda. The parameters λ\lambda and μ\mu are known as the Lamé parameters. Using these parameters we can simplify the stress and generalize it into a matrix

[σxτx  yτx  zτx  yσyτy  zτx  zτy  zσz]=2μ[εxεx  yεx  zεx  yεyεy  zεx  zεy  zεz]+λ[100010001]I(εx+εy+εz)trε=u\left\lbrack \begin{array}{ccc} \sigma_x & \tau_{x\;y} & \tau_{x\;z} \\ \tau_{x\;y} & \sigma_y & \tau_{y\;z} \\ \tau_{x\;z} & \tau_{y\;z} & \sigma_z \end{array}\right\rbrack =2\mu \left\lbrack \begin{array}{ccc} \varepsilon_x & \varepsilon_{x\;y} & \varepsilon_{x\;z} \\ \varepsilon_{x\;y} & \varepsilon_y & \varepsilon_{y\;z} \\ \varepsilon_{x\;z} & \varepsilon_{y\;z} & \varepsilon_z \end{array}\right\rbrack +\lambda \underset{\bm I }{\underbrace{\left\lbrack \begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\right\rbrack } } \underset{\mathrm{tr} \bm \varepsilon =\nabla \cdot \bm u }{\underbrace{\left(\varepsilon_x +\varepsilon_y +\varepsilon_z \right)} }

A quick sanity check: σy=2μεy+λ(εx+εy+εz)\sigma_y =2\mu \varepsilon_y +\lambda \left(\varepsilon_x +\varepsilon_y +\varepsilon_z \right) and τx  z=2μεx  z\tau_{x\;z} =2\mu \varepsilon_{x\;z}.

We arrive at the generalized Hooke's law for isotropic materials

σ=2με+λtrεI or σ=2με+λuI \boxed{ \bm \sigma = 2\mu \bm \varepsilon + \lambda \mathrm{tr} \bm \varepsilon \bm I \quad \text{ or } \quad \bm \sigma = 2\mu \bm \varepsilon + \lambda \nabla \cdot \bm u \bm I }

Note that the notation for the shear modulus GG has an solid mechanics background whereas μ\mu comes from continuum mechanics.

The first term of the stress tensor is known as the deviatoric stress tensor and the second term is known as the volumetric stress tensor. The same goes for the strain tensor.

ε=12μσλ2μ(3λ+2μ)trσI\bm \varepsilon =\dfrac{1}{2\mu } \bm \sigma -\dfrac{\lambda }{2\mu \left(3\lambda +2\mu \right)}\mathrm{tr} \bm \sigma \bm I

See the Efficient Engineer video on this topic.

Volumetric strain and incompressibility

Consider a cube with side length aa, the total change in volume (under small strain assumption) is

ΔV=VV0a3(1+εx+εy+εz)\Delta V=V-V_0 \approx a^3 \left(1+\varepsilon_x +\varepsilon_y +\varepsilon_z \right)

where we identify the dilation of the volume as εx+εy+εz=trε=:I1\varepsilon_x +\varepsilon_y +\varepsilon_z =\mathrm{tr} \bm \varepsilon =:I_1, also known as the first invariant of the strain tensor.

Now, consider λ\lambda in the generalized Hooke's law, λ=Eν(1+ν)(1+2ν)\lambda =\frac{E\nu }{\left(1+\nu \right)\left(1+2\nu \right)}, and note that as ν1/2\nu \to 1/2 we get

λforcingεx+εy+εz0\lambda \to \infty \overset{\mathrm{forcing} }{\Rightarrow} \varepsilon_x +\varepsilon_y +\varepsilon_z \to 0.

Which means that the volume change tends to zero with the Poisson's ratio tending towards 0.5, meaning the material becomes incompressible.

2D Simplifications

Many applications where the structure is much longer than its cross section can be approximated using a 2D simplification called plain strain. This greatly reduces the computational complexity of solving the resulting numerical problem.

Plain strain

For plain strain all strain components in the zz-direction vanish due to the much longer zz-dimension and thus the strain tensor simplifies to

[εxεx  y0εx  yεx0000]\left\lbrack \begin{array}{ccc} \varepsilon_x & \varepsilon_{x\;y} & 0\\ \varepsilon_{x\;y} & \varepsilon_x & 0\\ 0 & 0 & 0 \end{array}\right\rbrack

which gives

[σxσx  y0σx  yσx000σz]\left\lbrack \begin{array}{ccc} \sigma_x & \sigma_{x\;y} & 0\\ \sigma_{x\;y} & \sigma_x & 0\\ 0 & 0 & \sigma_z \end{array}\right\rbrack

The stress tensor has a non-zero z component due to: σz=E1+ν(0+ν1+2ν(εx+εy0+0))\sigma_z =\frac{E}{1+\nu }\left(0+\frac{\nu }{1+2\nu }\left(\underset{\not= 0}{\underbrace{\varepsilon_x +\varepsilon_y } } +0\right)\right).

Hooke's law takes the form

[σxσx  yσx  yσy]=E1+ν([εxεx  yεx  yεy]+ν1+2ν[1001](εx+εy))\left\lbrack \begin{array}{cc} \sigma_x & \sigma_{x\;y} \\ \sigma_{x\;y} & \sigma_y \end{array}\right\rbrack =\dfrac{E}{1+\nu }\left(\left\lbrack \begin{array}{cc} \varepsilon_x & \varepsilon_{x\;y} \\ \varepsilon_{x\;y} & \varepsilon_y \end{array}\right\rbrack +\dfrac{\nu }{1+2\nu }\left\lbrack \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right\rbrack \left(\varepsilon_x +\varepsilon_y \right)\right)

with the Lamé parameters

λ=Eν(1+ν)(1+2ν)  and  μ=E2(1+ν) \boxed{ \lambda =\dfrac{E\nu }{\left(1+\nu \right)\left(1+2\nu \right)}\;\mathrm{and}\;\mu =\dfrac{E}{2\left(1+\nu \right)} }

Plane stress

For plane stress, we assume a very thin object that is loaded in-plane. No bending can occur in such a model. Typical applications are thin walled structures with edge loads, even very slightly curved geometry can assume plane stress, such as pressure vessels. This type of model is also called a plate. The stress can be simplified to

[σxσx  y0σx  yσx0000]\left\lbrack \begin{array}{ccc} \sigma_x & \sigma_{x\;y} & 0\\ \sigma_{x\;y} & \sigma_x & 0\\ 0 & 0 & 0 \end{array}\right\rbrack

effectively reducing the problem to a 2D problem.

Hooke's law takes the form

[σxσx  yσx  yσy]=E1ν2((1ν)[εxεx  yεx  yεy]+ν[1001](εx+εy))\left\lbrack \begin{array}{cc} \sigma_x & \sigma_{x\;y} \\ \sigma_{x\;y} & \sigma_y \end{array}\right\rbrack =\dfrac{E}{1-\nu^2 }\left(\left(1-\nu \right)\left\lbrack \begin{array}{cc} \varepsilon_x & \varepsilon_{x\;y} \\ \varepsilon_{x\;y} & \varepsilon_y \end{array}\right\rbrack +\nu \left\lbrack \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right\rbrack \left(\varepsilon_x +\varepsilon_y \right)\right)

In this case, the Lamé parameter become

λ=Eν1+ν2  and  μ=E2(1+ν) \boxed{ \lambda =\dfrac{E\nu }{1+\nu^2 }\;\mathrm{and}\;\mu =\dfrac{E}{2\left(1+\nu \right)} }

Note however that we have a non-zero strain component due to

εz=0ν(σxE+σyE)\varepsilon_z =0-\nu \left(\dfrac{\sigma_x }{E}+\dfrac{\sigma_y }{E}\right)

Videos on the topic

Make sure to check out The Efficient Engineers video on this topic