Engineering at Alberta Courses » Expressions for Linear Elastic Strain Energy Functions

Energy: Expressions for Linear Elastic Strain Energy Functions

Strain Energy in Linear Elastic Materials:

In the previous section we ended up showing that the increment in the energy per unit volume of the undeformed configuration is equal to:

(1) $\begin{equation*} \mathrm{d}W=J\mathrm{d}\overline{U}=J \sum_{i,j=1}^3 \sigma_{ij}\mathrm{d}\varepsilon_{ij} = \sum_{i,j=1}^3 P_{ij}\mathrm{d}F_{ij} \end{equation*}$

As described in the section on hyperelastic materials, elasticity of a material implies that there exists a strain energy function from which the stresses can be derived. For a function $W$ to exist, then Equation 1 has to be an exact differential implying that $P$ can be obtained by differentiating a function $W(F)$ with respect to $F$ (See hyperelastic materials).

In the derivation of 1 we used the following definition:

$\mathrm{d}\varepsilon_{ij}=\frac{1}{2}\left(\frac{\mathrm{d}u_i}{\partial x_j}+\frac{\mathrm{d}u_j}{\partial x_i}\right)$

For linear elastic materials in small deformations, we can assume that $J\approx 1$ . Additionally, we can assume that the derivatives with respect to $x_i$ are equal to the derivatives with respect to $X_i$ and thus, we can use the engineering strain tensor as defined in the section on strain measures so that:

$\mathrm{d}\varepsilon_{ij}\approx\frac{1}{2}\left(\frac{\mathrm{d}u_i}{\partial X_j}+\frac{\mathrm{d}u_j}{\partial X_i}\right)$

Additionally, we can assume that the stress components are linear functions of the strain components and are obtained by differentiating $\overline{U}$ with respect to the strain components as shown in the linear elastic constitutive laws section. In this case, the strain energy stored during deformation can be calculated as follows:

(2) $\begin{equation*} \overline{U}=\sum_{i,j=1}^3\int_0^{\varepsilon_{ij}}\, \sigma_{ij} \! \mathrm{d}\varepsilon_{ij}=\frac{1}{2}\sum_{i,j=1}^3\sigma_{ij}\varepsilon_{ij} \end{equation*}$

where the factor $\frac{1}{2}$ is due to the linear relationship between the stress and the strain.

Strain Energy in Linear Elastic Isotropic Materials:

If we adopt the constitutive relationship for isotropic linear elastic materials, then we can replace the stresses as functions of the strains or replace the strains as functions of the stresses. We can then write the following two equivalent forms of Equation 2 as applied to isotropic linear elastic materials:

(3) $\begin{equation*} \overline{U}=\frac{1+\nu}{2E}\left(\sigma_{11}^2+\sigma_{22}^2+\sigma_{33}^2+2\sigma_{12}^2+2\sigma_{13}^2+2\sigma_{23}^2\right)-\frac{\nu}{2E}\left(\sigma_{11}+\sigma_{22}+\sigma_{33}\right)^2 \end{equation*}$

(4) $\begin{equation*} \overline{U}=\mu\left(\varepsilon_{11}^2+\varepsilon_{22}^2+\varepsilon_{33}^2+\frac{1}{2}\gamma_{12}^2+\frac{1}{2}\gamma_{13}^2+\frac{1}{2}\gamma_{23}^2\right)+\frac{\lambda}{2}\left(\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}\right)^2 \end{equation*}$

Volumetric and Deviatoric Strain Energies in Linear Elastic Isotropic Materials:

The strain energy of a linear elastic isotropic material per unit volume can be divided into two additive components: a volumetric strain energy function that accompanies volumetric changes and a deviatoric strain energy function that accompanies shape changes at constant volume. Such decomposition is valid for small strains when the volumetric strain can be measured as the trace of the infinitesimal strain matrix. In this case, the relationship between the volumetric strain and the applied stresses using the linear elastic constitutive laws is:

$\mbox{Volumetric Strain}=\mbox{Trace}(\varepsilon)=\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}=(\sigma_{11}+\sigma_{22}+\sigma_{33})\frac{(1-2\nu)}{E}=\mbox{Trace}(\sigma)\frac{(1-2\nu)}{E}$

The above relationship shows that nonzero volumetric strains exist only when the hydrostatic stress $p=\frac{\mbox{Trace}(\sigma)}{3}$ is nonzero. To separate the volumetric from the deviatoric strain energy functions, the deviatoric stress $S=\sigma-pI$ will be used. By replacing $\sigma$ with $S$ and $p$ in Equation 2 the following is obtained:

(5) $\begin{equation*} \overline{U}=\frac{1}{2}\left((S_{11}+p)\varepsilon_{11} + (S_{22}+p)\varepsilon_{22} + (S_{33}+p)\varepsilon_{33} + S_{12}2\varepsilon_{12}+S_{13}2\varepsilon_{13}+S_{23}2\varepsilon_{23}\right) \end{equation*}$

The right hand side can be separated into two terms:

$\overline{U}=\frac{1}{2}\left(S_{11}\varepsilon_{11} + S_{22}\varepsilon_{22} + S_{33}\varepsilon_{33} + S_{12}2\varepsilon_{12}+S_{13}2\varepsilon_{13}+S_{23}2\varepsilon_{23}\right)+\frac{p}{2}\left(\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}\right)$

The first term is called the deviatoric strain energy term while the second is called the volumetric strain energy term. Using the linear elastic constitutive laws, these terms can be further simplified to have the following form:

$\overline{U}=\overline{U}_{deviatoric}+\overline{U}_{volumetric}$

with

$\begin{split} \overline{U}_{deviatoric} & =\frac{\sigma_{vM}^2}{6G}\\ \overline{U}_{volumetric} & =\frac{3(1-2\nu)p^2}{2E} \end{split}$

The reader should attempt to manipulate Equation 5 to obtain the above simplified version in which the deviatoric strain energy is dependent solely on the von Mises stress $\sigma_{vM}$ , while the volumetric strain energy is dependent solely on the hydrostatic stress $p$ .

Linear Elastic Strain Energy in bars under Axial Loading with Small Deformations:

Consider a linear elastic bar of length $L$ such that its neutral axis before deformation is aligned with the $X_1$ axis. Then, using the expressions for the stresses and strains in an a plane bar under axial loading, the total strain energy stored in the bar can be evaluated using the following integral:

Linear Elastic Strain Energy in Euler Bernoulli Beams with Small Deformations:

Consider a linear elastic Euler Bernoulli beam with a constant cross-sectional area and with a length $L$ such that its neutral axis before deformation is aligned with the $X_1$ axis. Then, using the expressions for the stresses and strains in an Euler Bernoulli beam, the total strain energy stored in the beam can be evaluated using one of the forms of the following integral:

$\begin{split} \mbox{Total Strain Energy} & =\int \! \overline{U} \, \mathrm{d}V=\int\int \, \int \!\sum_{i,j=1}^3\sigma_{ij}\mathrm{d}\varepsilon_{ij} \, \mathrm{d}A\mathrm{d}X_1=\int\int \, \int \!\sigma_{11}\mathrm{d}\varepsilon_{11} \, \mathrm{d}A\mathrm{d}X_1\\ &=\int\int \! \frac{1}{2} \sigma_{11}\varepsilon_{11} \, \mathrm{d}A\mathrm{d}X_1\\ & =\frac{1}{2}\int_0^L\! M\frac{\mathrm{d}^2y}{\mathrm{d}X_1^2}\, \mathrm{d}X_1\\ & =\frac{1}{2EI}\int_0^L\! M^2 \,\mathrm{d}X_1\\ & =\frac{EI}{2}\int_0^L\! \left(\frac{\mathrm{d}^2y}{\mathrm{d}X_1^2}\right)^2 \,\mathrm{d}X_1 \end{split}$

Linear Elastic Strain Energy in Timoshenko Beams with Small Deformations:

Consider a linear elastic Euler Tiomoshenko beam with a constant cross-sectional area and with a length $L$ such that its neutral axis before deformation is aligned with the $X_1$ axis. Then, using the expressions for the stresses and strains in a Timoshenko beam, the total strain energy stored in the beam can be evaluated using one of the forms of the following integral:

$\begin{split} \mbox{Total Strain Energy} & =\int \! \overline{U} \, \mathrm{d}V=\int\int \, \int \!\sum_{i,j=1}^3\sigma_{ij}\mathrm{d}\varepsilon_{ij} \, \mathrm{d}A\mathrm{d}X_1=\int\int \, \int \!\sigma_{11}\mathrm{d}\varepsilon_{11}+\sigma_{12}\mathrm{d}\gamma_{12} \, \mathrm{d}A\mathrm{d}X_1\\ & =\int\int \! \frac{1}{2} \sigma_{11}\varepsilon_{11}+\frac{1}{2} \sigma_{12}\gamma_{12} \, \mathrm{d}A\mathrm{d}X_1\\ & =\frac{1}{2EI}\int_0^L\! M^2 \,\mathrm{d}X_1+\frac{1}{2kGA}\int_0^L\! V^2 \,\mathrm{d}X_1\\ & =\frac{EI}{2}\int_0^L\! \left(\frac{\mathrm{d}^2\psi}{\mathrm{d}X_1^2}\right)^2 \,\mathrm{d}X_1+\frac{kGA}{2}\int_0^L\! \gamma^2 \,\mathrm{d}X_1 \end{split}$

Examples and Problems:

Example 1:

The strains in a linear elastic orthotropic material are given by the strain matrix:

$\varepsilon=\left(\begin{matrix}0.01&0.015&0.001\\0.015&-0.02&0.002\\0.001&0.002&0.001\end{matrix}\right)$

If the material axes of orthotropy are aligned with the orthonormal basis set used, then find the stresses and the strain energy per unit volume if $E_{11}=E_{22}=100 MPa$ , $E_{33}=200 MPa$ , $\nu_{12}=\nu_{13}=\nu_{23}=0$ , and $G_{12}=G_{13}=G_{23}=35 MPa$

Solution:

The stress strain relationships for orthotropic materials shown can be used to find the stresses. Therefore, the stress matrix has the following form:

$\sigma=\left(\begin{matrix}1&1.05&0.07\\1.05&-2&0.14\\0.07&0.14&0.2\end{matrix}\right)MPa$

The strain energy per unit volume for a general linear elastic material is given by Equation 2:

$\overline{U}=\frac{1}{2}\sum_{i,j=1}^3\sigma_{ij}\varepsilon_{ij}=0.0412 MN.m/m^3$

View Mathematica Code

E11=100;
E22=100;
E33=200;
G12=G13=G23=35;
eps={{0.01,0.015,0.001},{0.015,-0.02,0.002},{0.001,0.002,0.001}};
stress=Table[0,{i,1,3},{j,1,3}];
stress[[1,1]]=eps[[1,1]]*E11;
stress[[2,2]]=eps[[2,2]]*E22;
stress[[3,3]]=eps[[3,3]]*E33;
stress[[1,2]]=stress[[2,1]]=2eps[[1,2]]*G12;
stress[[1,3]]=stress[[3,1]]=2eps[[1,3]]*G13;
stress[[2,3]]=stress[[3,2]]=2eps[[2,3]]*G23;
stress//MatrixForm  
eps//MatrixForm  
Energy=Sum[stress[[i,j]]*eps[[i,j]]/2,{i,1,3},{j,1,3}]

Example 2:

If the state of stress in a linear elastic isotropic material is given by the stress matrix:

$\sigma=\left(\begin{matrix}20 & 30 & 0\\30 & -10 & 0\\ 0 & 0 & 25 \end{matrix}\right)MPa$

Find the strain energy density and its deviatoric and volumetric components stored inside the material if $E=10 GPa$ and $\nu=0.1$ .

Solution:

Using Equation 3, the strain energy density is:

$\overline{U}=0.155 MN.m / m^3$

The deviatoric and volumetric components are:

$\begin{split} \overline{U}_{deviatoric} & =\frac{\sigma_{vM}^2}{6G}=0.138 MN.m./m^3\\ \overline{U}_{volumetric} & =\frac{3(1-2\nu)p^2}{2E}=0.016 MN.m./m^3 \end{split}$

View Mathematica Code

s = {{20, 30, 0}, {30, -10, 0}, {0, 0, 25}};
Nu = 0.1;
Ee = 10000;
G = Ee/2/(1 + Nu);
U = (1 + Nu)/2/Ee*(Sum[s[[i, j]]^2, {i, 1, 3}, {j, 1, 3}]) - Nu/2/Ee*(Sum[s[[i, i]], {i, 1, 3}])^2
p = (s[[1, 1]] + s[[2, 2]] + s[[3, 3]] )/ 3  
Udeviatoric =  1/12/G*((s[[1, 1]] - s[[2, 2]])^2 + (s[[3, 3]] - s[[2, 2]])^2 + (s[[1, 1]] - s[[3, 3]])^2 + 6*(s[[1, 2]]^2 + s[[1, 3]]^2 + s[[2, 3]]^2))  
Uvolumetric = 3 (1 - 2 Nu)/2/Ee*p^2

Example 3:

The position function in dimensions of m. of a plate follows the following form:

$\begin{split} x_1 & =X_1+0.001X_2+0.0002X_1^2\\ x_2 & = 1.001X_2-0.0002X_2^2\\ x_3 & =X_3 \end{split}$

The plate has dimensions of 2m in the direction of $e_1$ and 1m in the direction of $e_2$ with the bottom left corner coinciding with the origin of the coordinate system. Assuming that the material is linear elastic isotropic with Young’s modulus of 210 GPa and Poisson’s ratio of 0.3, then:

Find an expression for the strain energy density as a function of the position inside the plate, and draw the contour plot of the strain energy density function on the plate.
If the thickness of the plate is 10mm, find the total strain energy stored inside the plate.
Find an expression for the deviatoric and volumetric strain energy density functions, and draw the contour plot of each on the plate.

Solution:

The displacement function can be calculated as follows:

$u=x-X=\left(\begin{array}{c}0.0002X_1^2+0.001X_2\\0.001X_2-0.0002X_2^2\\0\end{array}\right)$

The gradient of the displacement is given by the following matrix:

$\nabla u = \left(\begin{matrix}0.0004X_1 & 0.001 & 0\\0&0.001-0.0004X_2 & 0\\0&0&0 \end{matrix}\right)$

The engineering small strain is given by:

$\varepsilon=\frac{1}{2}(\nabla u +\nabla u^T)=\left(\begin{matrix}0.0004X_1 & 0.0005 & 0\\0.0005&0.001-0.0004X_2 & 0\\0&0&0 \end{matrix}\right)$

The vector representation of the strain is given by:

$\left(\begin{array}{c}\varepsilon_{11}\\\varepsilon_{22}\\\varepsilon_{33}\\2\varepsilon_{12}\\2\varepsilon_{13}\\2\varepsilon_{23}\end{array}\right)=\left(\begin{array}{c}0.0004X_1\\0.001-0.0004X_2\\0\\0.001\\0\\0\end{array}\right)$

Using the constitutive equation for linear elastic isotropic materials, the vector representation of the stress is given by:

$\left(\begin{array}{c}\sigma_{11}\\\sigma_{22}\\\sigma_{33}\\\sigma_{12}\\\sigma_{13}\\\sigma_{23}\end{array}\right)=\left(\begin{array}{c}121.154+113.077X_1-48.4615X_2\\282.692+48.4615X_1-113.077X_2\\121.154+48.4615X_1+48.4615X_2\\80.7692\\0\\0\end{array}\right)$

The strain energy density is given by:

$\overline{U}=\frac{1}{2}\sum_{i,j=1}^3\sigma_{ij}\varepsilon_{ij}=0.1817+0.0226X_1(2.143+X_1)-0.113X_2-0.0194X_1X_2+0.0226X_2^2 (MN.m/m^3)$

The total energy is given by:

$\mbox{Total Energy}=\int \!\overline{U}\, \mathrm{d}V= \int_0^{0.01} \int_0^1 \int_0^2 \!\overline{U} \, \mathrm{d}X_1\mathrm{d}X_2\mathrm{d}X_3=0.004 MN.m.$

The deviatoric and volumetric strain energy densities are given by:

$\begin{split} \overline{U}_{deviatoric} & =\frac{\sigma_{vM}^2}{6G}\\ & =0.0942+0.00862X_1(X_1-2.5)-0.04308X_2+0.0086X_1X_2+0.0086X_2^2 MN.m./m^3\\ \overline{U}_{volumetric} & =\frac{3(1-2\nu)p^2}{2E}=3.1746\times 10^{-7}(525+210X_1-210X_2)^2 MN.m./m^3 \end{split}$

The required contour plots are:
EnergyExample

View Mathematica Code

Clear[X1,X2,X3,x1,x2,x3,Ee,Nu]  
X={X1,X2,X3};
x={x1,x2,x3};
x1=X1+0.001X2+0.0002X1^2;
x2=1.001X2-0.0002X2^2;
x3=X3;
u=x-X;
gradu=Table[D[u[[i]],X[[j]]],{i,1,3},{j,1,3}];
esmall=1/2(gradu+Transpose[gradu]);
esmall//MatrixForm  
Strainvector={esmall[[1,1]],esmall[[2,2]],esmall[[3,3]],2*esmall[[1,2]],2*esmall[[1,3]],2*esmall[[2,3]]};
Strainvector//MatrixForm  
Ee=210000;
Nu=0.3;
G=Ee/2/(1+Nu);
Cc={{1/Ee,-Nu/Ee,-Nu/Ee,0,0,0},{-Nu/Ee,1/Ee,-Nu/Ee,0,0,0},{-Nu/Ee,-Nu/Ee,1/Ee,0,0,0},{0,0,0,1/G,0,0},{0,0,0,0,1/G,0},{0,0,0,0,0,1/G}};
Dd=FullSimplify[Inverse[Cc]];
stressvector=FullSimplify[Chop[Dd.Strainvector]];
stressvector//MatrixForm  
S={{stressvector[[1]],stressvector[[4]],stressvector[[5]]},{stressvector[[4]],stressvector[[2]],stressvector[[6]]},{stressvector[[5]],stressvector[[6]],stressvector[[3]]}};
S=Chop[FullSimplify[S]];
S//MatrixForm  
StrainEnergy=FullSimplify[Sum[1/2*S[[i,j]]*esmall[[i,j]],{i,1,3},{j,1,3}]]  
TotalEnergy=Integrate[StrainEnergy,{X1,0,2},{X2,0,1},{X3,0,0.01}]  
Udeviatoric=FullSimplify[1/12/G*((S[[1,1]]-S[[2,2]])^2+(S[[3,3]]-S[[2,2]])^2+(S[[1,1]]-S[[3,3]])^2+6*(S[[1,2]]^2+S[[1,3]]^2+S[[2,3]]^2))]  
Uvolumetric=FullSimplify[(1-2Nu)/6/Ee*(S[[1,1]]+S[[2,2]]+S[[3,3]])^2]  
(*Check that the above expressions for the energy add up*)
Chop[FullSimplify[StrainEnergy-Udeviatoric-Uvolumetric]]  
ContourPlot[StrainEnergy,{X1,0,2},{X2,0,1},AspectRatio->Automatic,ContourLabels->All,PlotLabel->"StrainEnergy MN m/m^3"]  
ContourPlot[Udeviatoric,{X1,0,2},{X2,0,1},AspectRatio->Automatic,ContourLabels->All,PlotLabel->"Deviatoric StrainEnergy MN m/m^3"]  
ContourPlot[Uvolumetric,{X1,0,2},{X2,0,1},AspectRatio->Automatic,ContourLabels->All,PlotLabel->"Volumetric StrainEnergy MN m/m^3"]

Example 4:

Consider a linear elastic small deformations cantilever beam. If a load of value $P$ is applied at the free end, find an expression for the strain energy per unit volume as a function of the position, $P$ , and the beam length $L$ . Also find the total strain energy stored. Assume Young’s modulus to be $E$ and that the beam has a rectangular cross section with a moment of inertia $I$ .

Solution:

The bending moment in a cantilever beam loaded by a concentrated load at the free end is given by:

$M=-P(L-X_1)$

The stress component $\sigma_{11}$ and the strain component $\varepsilon_{11}$ are given by:

$\sigma_{11}=-\frac{MX_2}{I}\qquad \varepsilon_{11}=\frac{\sigma_{11}}{E}=-\frac{MX_2}{EI}$

Therefore, the strain energy per unit volume at any point is given by:

$\overline{U}=\frac{1}{2}\sigma_{11}\varepsilon_{11}=\frac{P^2(L-X_1)^2X_2^2}{2EI^2}$

The total strain energy stored in the beam is given by:

$\mbox{Total Strain Energy}=\int \!\overline{U} \,\mathrm{d}V=\frac{P^2}{2EI}\int_0^L\!(L-X_1)^2\, \mathrm{d}X_1=\frac{P^2L^3}{6EI}$

Note that the displacement at the free end is equal to:

$\Delta = \frac{PL^3}{3EI}$

By assuming a gradual increase of the external load $P$ , the work done during the application of the load is equal to:

$W=\frac{P\Delta}{2}=\frac{P^2L^3}{6EI}$

Therefore, the work done on the beam is equal to the total strain energy stored in the beam.

Problems:

A linear elastic isotropic material with a shear modulus $G= 15 MPa$ and a bulk modulus $K= 10 MPa$ is stretched such that the small strain matrix is:
$\varepsilon=\left(\begin{matrix}0.02 & 0.0015 & 0\\0.0015&-0.001 & 0\\0&0&0.01 \end{matrix}\right)$

Find the strain energy density stored during the application of the deformation, and find its deviatoric and volumetric components.(Answer: 7.58, 3.38, 4.21 kN.m/m^3)
A linear elastic isotropic material with Young’s modulus of 100 units is stretched such that the small strain tensor is given by:
$\varepsilon=\left(\begin{matrix}0.02 & 0.01 & 0\\0.01&-0.01 & 0\\0&0&0.01 \end{matrix}\right)$

For cases $\nu=0.3$ and $\nu=0.45$ :
1. Find the strain energy stored during the deformation.
2. Find the deviatoric strain energy stored during the deformation.
3. Find the volumetric strain energy stored during the deformation. Which case has higher volumetric strain energy and why?
Two states of stress are given by the stress matrices:
$\sigma_a=\left(\begin{matrix}p&0&0\\0&p&0\\0&0&0\end{matrix}\right)\qquad \sigma_b=pI$

where $p\in\mathbb{R}$ , $p > 0$ , and $I \in\mathbb{M}^3$ is the identity matrix. Find the deviatoric and the volumetric strain energy densities associated with both matrices if the material is linear elastic isotropic with Young’s modulus $E$ and Poisson’s ratio $\nu$ .
Two states of stress are given by the stress matrices:
$\sigma_a=\left(\begin{matrix}p&0&0\\0&p&0\\0&0&0\end{matrix}\right)\qquad \sigma_b=pI$

where $p\in\mathbb{R}$ , $p > 0$ , and $I \in\mathbb{M}^3$ is the identity matrix. Consider the material as linear elastic isotropic with Young’s modulus 100 units and Poisson’s ratio 0.2. Find $p$ in both stress states which leads to strain energy density equal to 500 units.
Compare the total strain energy stored in a simple versus fixed ends beam of length $L$ , Young’s modulus $E$ , moment of inertia $I$ loaded with a constant distributed load $q$ . (Answer: Simple beam has 6 times more energy stored for the same distributed loading!) .
For a linear isotropic elastic material with a Young’s modulus and Poisson’s ratio , express the strain energy density function under the condition of plane stress in terms of:
- $\sigma_{11}$ , $\sigma_{22}$ and $\sigma_{12}$ .
- $\varepsilon_{11}$ , $\varepsilon_{22}$ and $\gamma_{12}$ .
For a linear isotropic elastic material with a Young’s modulus and Poisson’s ratio , express the strain energy density function under the condition of plane strain in terms of:
- $\sigma_{11}$ , $\sigma_{22}$ and $\sigma_{12}$ .
- $\varepsilon_{11}$ , $\varepsilon_{22}$ and $\gamma_{12}$ .
Consider a material model that follows the following nonlinear relationship between the unixial stress and the uniaxial strain .

$\varepsilon_{11}=\frac{\sigma_{11}}{E}+\frac{3}{7}\frac{\sigma_0}{E}\left(\frac{\sigma}{\sigma_0}\right)^n$

In which $E=100$ units, $\sigma_0=10$ units. Assume that the above relationship is only valid when $0\leq \sigma_{11}\leq 10$ units. For $n=3$ and $n=5$ :
1. Draw the relationship between the stress and strain in the specified range.
2. Find the strain energy stored at full extension.
3. Find the stress corresponding to a strain value of 0.1.