beam2gxsΒΆ
- Purpose:
Compute section forces in a two dimensional geometric nonlinear beam element with exact solution.
- Syntax:
[es,Qx] = beam2gxs(ex, ey, ep, ed, Qx)
[es,Qx] = beam2gxs(ex, ey, ep, ed, Qx, eq)
[es,Qx,edi] = beam2gxs(ex, ey, ep, ed, Qx, eq, n)
[es,Qx,edi,eci] = beam2gxs(ex, ey, ep, ed, Qx, eq, n)
- Description:
beam2gxscomputes the section forces and displacements in local directions along the geometric nonlinear beam elementbeam2gxe.The input variables
ex,ey,ep,Qxandeq, are described inbeam2gxe. The element displacements, stored ined, are obtained by the functionextract_ed. If a distributed transversal load is applied to the element, the variableeqmust be included. The number of evaluation points for section forces and displacements are determined byn. Ifnis omitted, only the ends of the beam are evaluated.The output variable
Qxcontains \(Q_{\bar{x}}\) and the output variableses\(= \begin{bmatrix} N(0) & V(0) & M(0) \\ N(\bar{x}_2) & V(\bar{x}_2) & M(\bar{x}_2) \\ \vdots & \vdots & \vdots \\ N(\bar{x}_{n-1}) & V(\bar{x}_{n-1}) & M(\bar{x}_{n-1}) \\ N(L) & V(L) & M(L) \end{bmatrix}\) \(\quad\)edi\(= \begin{bmatrix} u(0) & v(0) \\ u(\bar{x}_2) & v(\bar{x}_2) \\ \vdots & \vdots \\ u(\bar{x}_{n-1}) & v(\bar{x}_{n-1}) \\ u(L) & v(L) \end{bmatrix}\) \(\quad\)eci\(= \begin{bmatrix} 0 \\ \bar{x}_2 \\ \vdots \\ \bar{x}_{n-1} \\ L \end{bmatrix}\)contain the section forces, the displacements, and the evaluation points on the local \(\bar{x}\)-axis. \(L\) is the length of the beam element.
- Theory:
The nodal displacements in local coordinates are given by
\[\begin{split}\mathbf{\bar{a}}^e = \begin{bmatrix} \bar{u}_1 \\ \bar{u}_2 \\ \bar{u}_3 \\ \bar{u}_4 \\ \bar{u}_5 \\ \bar{u}_6 \end{bmatrix} = \mathbf{G} \mathbf{a}^e\end{split}\]where \(\mathbf{G}\) is described in
beam2geand the transpose of \(\mathbf{a}^e\) is stored ined. The displacements associated with bar action and beam action are determined as\[\begin{split}\mathbf{\bar{a}}^e_{\text{bar}} = \begin{bmatrix} \bar{u}_1 \\ \bar{u}_4 \end{bmatrix} ; \quad \mathbf{\bar{a}}^e_{\text{beam}} = \begin{bmatrix} \bar{u}_2 \\ \bar{u}_3 \\ \bar{u}_5 \\ \bar{u}_6 \end{bmatrix}\end{split}\]The displacement \(u(\bar{x})\) is computed from
\[u(\bar{x}) = \mathbf{N}_{\text{bar}} \mathbf{\bar{a}}^e_{\text{bar}}\]where
\[\mathbf{N}_{\text{bar}} = \begin{bmatrix} 1 & \bar{x} \end{bmatrix} \mathbf{C}^{-1}_{\text{bar}} = \begin{bmatrix} 1-\frac{\bar{x}}{L} & \frac{\bar{x}}{L} \end{bmatrix}\]where \(L\) is defined in
beam2gxeand\[\begin{split}\mathbf{C}^{-1}_{\text{bar}} = \begin{bmatrix} 1 & 0 \\ -\frac{1}{L} & \frac{1}{L} \end{bmatrix}\end{split}\]The displacement \(v(\bar{x})\), the rotation \(\theta(\bar{x})\), the bending moment \(M(\bar{x})\) and the shear force \(V(\bar{x})\) are computed from
\[v(\bar{x}) = \mathbf{N}_{\text{beam}} \mathbf{\bar{a}}^e_{\text{beam}} + v_p(\bar{x})\]\[\theta(\bar{x}) = \frac{d\mathbf{N}_{\text{beam}}}{dx} \mathbf{\bar{a}}^e_{\text{beam}} + \theta_p(\bar{x})\]\[M(\bar{x}) = D_{EI} \mathbf{B}_{\text{beam}} \mathbf{\bar{a}}^e_{\text{beam}} + M_p(\bar{x})\]\[V(\bar{x}) = -D_{EI} \frac{d\mathbf{B}_{\text{beam}}}{dx} \mathbf{\bar{a}}^e_{\text{beam}} + V_p(\bar{x})\]For an axial compressive force (\(Q_{\bar{x}} < 0\)) we have
\[\mathbf{N}_{\text{beam}} = \begin{bmatrix} 1 & \bar{x} & \cos k \bar{x} & \sin k \bar{x} \end{bmatrix} \mathbf{C}^{-1}_{\text{beam}}\]\[\frac{d\mathbf{N}_{\text{beam}}}{dx} = \begin{bmatrix} 0 & 1 & -k \sin k \bar{x} & k \cos k \bar{x} \end{bmatrix} \mathbf{C}^{-1}_{\text{beam}}\]\[\mathbf{B}_{\text{beam}} = \begin{bmatrix} 0 & 0 & -k^2 \cos k \bar{x} & -k^2 \sin k \bar{x} \end{bmatrix} \mathbf{C}^{-1}_{\text{beam}}\]\[\frac{d\mathbf{B}_{\text{beam}}}{dx} = \begin{bmatrix} 0 & 0 & k^3 \sin k \bar{x} & -k^3 \cos k \bar{x} \end{bmatrix} \mathbf{C}^{-1}_{\text{beam}}\]\[v_p(\bar{x}) = \frac{q_{\bar{y}}L^4}{2D_{EI}} \left[ \frac{1 + \cos kL}{(kL)^3 \sin kL}(-1 + \cos k \bar{x}) -\frac{1}{(kL)^3} \sin k \bar{x} + \frac{1}{(kL)^2} \left(\frac{\bar{x}^2}{L^2}-\frac{\bar{x}}{L}\right) \right]\]\[\theta_p(\bar{x}) = \frac{q_{\bar{y}}L^3}{2D_{EI}} \left[ -\frac{1 + \cos kL}{(kL)^2 \sin kL} \sin k \bar{x} -\frac{1}{(kL)^2} \cos k \bar{x} + \frac{1}{(kL)^2} \left(\frac{2\bar{x}}{L}-1\right) \right]\]\[M_p(\bar{x}) = \frac{q_{\bar{y}}L^2}{2} \left[ -\frac{1 + \cos kL}{(kL) \sin kL} \cos k \bar{x} +\frac{1}{(kL)} \sin k \bar{x} + \frac{2}{(kL)^2} \right]\]\[\begin{split}V_p(\bar{x}) = Q_{\bar{x}} \begin{bmatrix} 0 \\ 0 \\ 2\bar{x} - L \\ 3\bar{x}^2 - \frac{9L^2}{10} \end{bmatrix}^T \mathbf{C}^{-1}_{\text{beam}} \mathbf{\bar{a}}^e_{\text{beam}} - q_{\bar{y}}\left(\bar{x} - \frac{L}{2}\right)\end{split}\]in which \(D_{EI}\), \(L\), and \(q_{\bar{y}}\) are defined in
beam2gxeand\[\begin{split}\mathbf{C}^{-1}_{\text{beam}} = \begin{bmatrix} k (kL \sin kL+\cos kL-1) & -kL \cos kL+\sin kL & -k (1-\cos kL) & -\sin kL+kL \\ - k^2 \sin kL & -k (1-\cos kL) & k^2 \sin kL & -k (1-\cos kL) \\ -k(1-\cos kL) & kL \cos kL-\sin kL & k (1-\cos kL) & \sin kL-kL \\ k\sin kL & kL \sin kL+\cos kL-1 & -k \sin kL & 1-\cos kL \end{bmatrix}\end{split}\]An updated value of the axial force is computed as
\[Q_{\bar{x}} = D_{EA} \begin{bmatrix} 0 & 1 \end{bmatrix} \mathbf{C}^{-1}_{\text{bar}} \mathbf{\bar{a}}^e_{\text{bar}}\]The normal force \(N(\bar{x})\) is then computed as
\[N(\bar{x}) = Q_{\bar{x}} + \theta(\bar{x}) V(\bar{x})\]