FEA TipsFinite Element Theory & MATLAB Programs
HomeChapter 11Geometrical derivation

Supplement to Chapter 11

Geometrical Derivation of the Exact Strain Relations

A direct geometrical derivation of the exact relations between the Green–Lagrange strain components and the changes in length and angle of two initially orthogonal material fibers.

Consider the two initially orthogonal material fibers shown in Figure 1. The exact geometrical strains are the relative changes in their lengths, exe_x and eye_y, and the decrease αxy\alpha_{xy} of the initial right angle between them.

Infinitesimal rectangle before deformation and the corresponding deformed parallelogram
Figure 1. Infinitesimal rectangle 1234 and its deformed configuration 1′2′3′4′.

01

Axial strain relations

The Green–Lagrange normal strain components are:

Exx=u,x+12u,x2+12v,x2E_{xx}=u_{,x}+\frac12u_{,x}^{2}+\frac12v_{,x}^{2}
Eyy=v,y+12u,y2+12v,y2E_{yy}=v_{,y}+\frac12u_{,y}^{2}+\frac12v_{,y}^{2}

The exact relative changes in length obtained in Chapter 11 are:

ex=(1+u,x)2+v,x2−1e_x=\sqrt{(1+u_{,x})^2+v_{,x}^{2}}-1
ey=u,y2+(1+v,y)2−1e_y=\sqrt{u_{,y}^{2}+(1+v_{,y})^2}-1

Therefore:

Exx=ex+12ex2E_{xx}=e_x+\frac12e_x^2
Eyy=ey+12ey2E_{yy}=e_y+\frac12e_y^2

02

Deformed material fibers

Using the definition of the deformation gradient:

dx^=F dxd\hat{\boldsymbol{x}}=\boldsymbol{F}\,d\boldsymbol{x}
F=[1+u,xu,yv,x1+v,y]\boldsymbol{F}=\begin{bmatrix}1+u_{,x}&u_{,y}\\[4pt]v_{,x}&1+v_{,y}\end{bmatrix}

The two infinitesimal material fibers in the undeformed configuration are:

12‾={dx0},13‾={0dy}\overline{12}=\begin{Bmatrix}dx\\0\end{Bmatrix},\qquad \overline{13}=\begin{Bmatrix}0\\dy\end{Bmatrix}

The first deformed fiber is:

1′2′‾=F 12‾\overline{1'2'}=\boldsymbol{F}\,\overline{12}
1′2′‾=[1+u,xu,yv,x1+v,y]{dx0}\overline{1'2'}=\begin{bmatrix}1+u_{,x}&u_{,y}\\[4pt]v_{,x}&1+v_{,y}\end{bmatrix}\begin{Bmatrix}dx\\0\end{Bmatrix}
1′2′‾={1+u,xv,x}dx\overline{1'2'}=\begin{Bmatrix}1+u_{,x}\\v_{,x}\end{Bmatrix}dx

The second deformed fiber is:

1′3′‾=F 13‾\overline{1'3'}=\boldsymbol{F}\,\overline{13}
1′3′‾=[1+u,xu,yv,x1+v,y]{0dy}\overline{1'3'}=\begin{bmatrix}1+u_{,x}&u_{,y}\\[4pt]v_{,x}&1+v_{,y}\end{bmatrix}\begin{Bmatrix}0\\dy\end{Bmatrix}
1′3′‾={u,y1+v,y}dy\overline{1'3'}=\begin{Bmatrix}u_{,y}\\1+v_{,y}\end{Bmatrix}dy

Their lengths are:

∣1′2′‾∣=(1+u,x)2+v,x2 dx\left|\overline{1'2'}\right|=\sqrt{(1+u_{,x})^2+v_{,x}^{2}}\,dx
∣1′3′‾∣=u,y2+(1+v,y)2 dy\left|\overline{1'3'}\right|=\sqrt{u_{,y}^{2}+(1+v_{,y})^2}\,dy
∣1′2′‾∣=(1+ex) dx\left|\overline{1'2'}\right|=(1+e_x)\,dx
∣1′3′‾∣=(1+ey) dy\left|\overline{1'3'}\right|=(1+e_y)\,dy

03

Shear strain relation

The scalar product of the two deformed material fibers is:

1′2′‾ ⁣⋅ ⁣1′3′‾=[u,y+v,x+u,xu,y+v,xv,y]dx dy\overline{1'2'}\!\cdot\!\overline{1'3'}=\bigl[u_{,y}+v_{,x}+u_{,x}u_{,y}+v_{,x}v_{,y}\bigr]dx\,dy
1′2′‾ ⁣⋅ ⁣1′3′‾=[u,y+v,x+u,xu,y+v,xv,y]dx dy\begin{aligned}\overline{1'2'}\!\cdot\!\overline{1'3'}={}&\bigl[u_{,y}+v_{,x}\\[-1pt]&+u_{,x}u_{,y}+v_{,x}v_{,y}\bigr]dx\,dy\end{aligned}

The Green–Lagrange shear component satisfies:

2Exy=u,y+v,x+u,xu,y+v,xv,y2E_{xy}=u_{,y}+v_{,x}+u_{,x}u_{,y}+v_{,x}v_{,y}

Consequently:

1′2′‾ ⁣⋅ ⁣1′3′‾=2Exy dx dy\overline{1'2'}\!\cdot\!\overline{1'3'}=2E_{xy}\,dx\,dy

Because αxy\alpha_{xy} is the decrease of the initial right angle, the angle between the two deformed fibers is:

π2−αxy\frac{\pi}{2}-\alpha_{xy}

The same scalar product can therefore be written geometrically as:

1′2′‾ ⁣⋅ ⁣1′3′‾=∣1′2′‾∣∣1′3′‾∣cos⁡(π2−αxy)\overline{1'2'}\!\cdot\!\overline{1'3'}=\left|\overline{1'2'}\right|\left|\overline{1'3'}\right|\cos\left(\frac{\pi}{2}-\alpha_{xy}\right)
1′2′‾ ⁣⋅ ⁣1′3′‾=(1+ex)(1+ey)sin⁡αxy dx dy\overline{1'2'}\!\cdot\!\overline{1'3'}=(1+e_x)(1+e_y)\sin\alpha_{xy}\,dx\,dy
1′2′‾ ⁣⋅ ⁣1′3′‾=(1+ex)(1+ey)sin⁡αxy dx dy\begin{aligned}\overline{1'2'}\!\cdot\!\overline{1'3'}={}&(1+e_x)(1+e_y)\sin\alpha_{xy}\\[-1pt]&{}\,dx\,dy\end{aligned}

Comparison of the two expressions for the scalar product gives:

2Exy=(1+ex)(1+ey)sin⁡αxy2E_{xy}=(1+e_x)(1+e_y)\sin\alpha_{xy}

04

Final exact relations

The three exact relations are:

{Exx=ex+12ex2Eyy=ey+12ey22Exy=(1+ex)(1+ey)sin⁡αxy\boxed{\left\{\begin{aligned}E_{xx}&=e_x+\frac12e_x^2\\[5pt]E_{yy}&=e_y+\frac12e_y^2\\[5pt]2E_{xy}&=(1+e_x)(1+e_y)\sin\alpha_{xy}\end{aligned}\right.}