Hello friends of maths,
I've given an arbitrary polygonal cross section (in cartesian coordinates $y$ and $z$). On this cross section, there acts an arbitrary stress-field $\sigma = f(y,z)$ as integrable function of $y$ and $z$.
Now I need to numerically determine the momentum ($M_y$ and $M_z$) as well as the axial force $N_x$ that results from the stress-field. Therefore, I read about the Gauss-Green cubature in a recent published PhD-thesis, what sounds quite practicable (see pp 89-91). Unfortunately in this thesis the acting stress-field mainly results from biaxial bending. This means that it is possible to transform the whole coordinate sytsem into a principal coordinate system what simplifies the whole procedure.
So, does anybody know if (and how) it is possible to use the Gauss-Green cubature for arbitrary polygonal cross-sections under arbitrary stress-fields? How do the formulas given in this thesis (and that I summerized below and commented for the "non-german-speakers") have to be adapted?
Thank you in advance for any help!
(p. 89): The Gaussian cubature is given by $$ \int_a^bf(x)~\text{d}x \approx \frac{b-a}{2}\cdot\sum_{g=1}^n c_{\text{n},g}\cdot f\biggl(\frac{(b-a)\cdot r_{\text{n},g}+b+a}{2}\biggr) $$ where $n$ is the number of integration points; $c_{n,g}$ are the position of the integration points in the interval $[-1;1]$ and $r_{n,g}$ are the weights of the integration points in $[-1;1]$.
(p. 99):
If this cubature is combined with Green's Theorem it is more suitable to transform the coordinate system to a principal coordinate system ($\eta, \zeta$). Then the stress only depends on a coordinate ($\sigma = f(\zeta))$. [<<-- This is actually not possible in my problem]. Then, each part of the polygon has to be calculated seperately and afterwards summerized, to get the resulting forces:
$$ N_{x,pi} = \sum_{g=1}^n (\eta_\text{A}+\Delta\eta\cdot\lambda_{n,g})\cdot\Delta\zeta\cdot\sigma(\varepsilon(\lambda_{n,g}))\cdot c_{n,g} $$ $$ M_{\eta,pi} = \sum_{g=1}^n (\eta_\text{A}+\Delta\eta\cdot\lambda_{n,g})\cdot(\zeta_\text{A}+\Delta\zeta\cdot\lambda_{n,g})\cdot\Delta\zeta\cdot\sigma(\varepsilon(\lambda_{n,g}))\cdot c_{n,g} $$ $$ M_{\zeta,pi} = \sum_{g=1}^n-\frac{1}{2}\cdot(\eta_\text{A}+\Delta\eta\cdot\lambda_{n,g})^2\cdot\Delta\zeta\cdot\sigma(\varepsilon(\lambda_{n,g}))\cdot c_{n,g} $$ with $\lambda_{n,g}=\frac{1+r_{n,g}}{2}$ for [0;1]
And finally, the force-vector results from: $$ \vec s = \begin{Bmatrix} N_x \\ M_\eta \\ M_\zeta \end{Bmatrix} = \sum_{i=1}^{n_\text{p}}\begin{Bmatrix} N_{x,p_i} \\ M_{\eta,p_i} \\ M_{\zeta,p_i} \end{Bmatrix} $$ ... and the momentums $M_y$ and $M_z$ result from back-transformation to the original coordinate system.