beam2gxeΒΆ
- Purpose:
Compute element stiffness matrix for a two dimensional nonlinear beam element with exact solution.
- Syntax:
Ke = beam2gxe(ex, ey, ep, Qx)
[Ke, fe] = beam2gxe(ex, ey, ep, Qx, eq)
- Description:
beam2gxeprovides the global element stiffness matrix \({\mathbf{K}}^e\) for a two dimensional beam element with respect to geometrical nonlinearity considering exact solution.The input variables:
ex\(= [x_1 \;\; x_2]\) \(\qquad\)ey\(= [y_1 \;\; y_2]\) \(\qquad\)ep\(= [E \; A\; I]\)supply the element nodal coordinates \(x_1\), \(y_1\), \(x_2\), and \(y_2\), the modulus of elasticity \(E\), the cross section area \(A\), and the moment of inertia \(I\).
The input variable
Qx\(= [Q_{\bar{x}}]\)contains the value of the predefined axial force \(Q_{\bar{x}}\), which is positive in tension.
The element load vector
fecan also be computed if a uniformly distributed transverse load is applied to the element. The optional input variableeq\(= [q_{\bar{y}}]\)then contains the distributed transverse load per unit length, \(q_{\bar{y}}\). Note that
eqis a scalar and not a vector as inbeam2e.- Theory:
The element stiffness matrix \(\mathbf{K}^e\), stored in the variable
Ke, is computed according to\[\mathbf{K}^e = \mathbf{G}^T \bar{\mathbf{K}}^e \mathbf{G}\]with
\[\begin{split}\bar{\mathbf{K}}^e = \begin{bmatrix} \frac{D_{EA}}{L} & 0 & 0 & -\frac{D_{EA}}{L} & 0 & 0 \\ 0 & \frac{12 D_{EI}}{L^3} \phi_5 & \frac{6 D_{EI}}{L^2} \phi_2 & 0 & -\frac{12 D_{EI}}{L^3} \phi_5 & \frac{6 D_{EI}}{L^2} \phi_2 \\ 0 & \frac{6 D_{EI}}{L^2} \phi_2 & \frac{4 D_{EI}}{L} \phi_3 & 0 & -\frac{6 D_{EI}}{L^2} \phi_2 & \frac{2 D_{EI}}{L} \phi_4 \\ -\frac{D_{EA}}{L} & 0 & 0 & \frac{D_{EA}}{L} & 0 & 0 \\ 0 & -\frac{12 D_{EI}}{L^3} \phi_5 & -\frac{6 D_{EI}}{L^2} \phi_2 & 0 & \frac{12 D_{EI}}{L^3} \phi_5 & -\frac{6 D_{EI}}{L^2} \phi_2 \\ 0 & \frac{6 D_{EI}}{L^2} \phi_2 & \frac{2 D_{EI}}{L} \phi_4 & 0 & -\frac{6 D_{EI}}{L^2} \phi_2 & \frac{4 D_{EI}}{L} \phi_3 \end{bmatrix}\end{split}\]\[\begin{split}\mathbf{G} = \begin{bmatrix} n_{x\bar{x}} & n_{y\bar{x}} & 0 & 0 & 0 & 0 \\ n_{x\bar{y}} & n_{y\bar{y}} & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & n_{x\bar{x}} & n_{y\bar{x}} & 0 \\ 0 & 0 & 0 & n_{x\bar{y}} & n_{y\bar{y}} & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix}\end{split}\]where the axial stiffness \(D_{EA}\), the bending stiffness \(D_{EI}\) and the length \(L\) are given by
\[D_{EA} = EA; \quad D_{EI} = EI; \quad L = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]The transformation matrix \(\mathbf{G}\) contains the direction cosines
\[n_{x\bar{x}} = n_{y\bar{y}} = \frac{x_2 - x_1}{L} \qquad n_{y\bar{x}} = -n_{x\bar{y}} = \frac{y_2 - y_1}{L}\]For axial compression (\(Q_{\bar{x}} < 0\)):
\[\phi_2 = \frac{1}{12} \frac{k^2 L^2}{1 - \phi_1} \qquad \phi_3 = \frac{1}{4} \phi_1 + \frac{3}{4} \phi_2\]\[\phi_4 = -\frac{1}{2} \phi_1 + \frac{3}{2} \phi_2 \qquad \phi_5 = \phi_1 \phi_2\]with
\[k = \sqrt{\frac{-Q_{\bar{x}}}{D_{EI}}} \qquad \phi_1 = \frac{kL}{2} \cot \frac{kL}{2}\]For axial tension (\(Q_{\bar{x}} > 0\)):
\[\phi_2 = -\frac{1}{12} \frac{k^2 L^2}{1 - \phi_1} \qquad \phi_3 = \frac{1}{4} \phi_1 + \frac{3}{4} \phi_2\]\[\phi_4 = -\frac{1}{2} \phi_1 + \frac{3}{2} \phi_2 \qquad \phi_5 = \phi_1 \phi_2\]with
\[k = \sqrt{\frac{Q_{\bar{x}}}{D_{EI}}} \qquad \phi_1 = \frac{kL}{2} \coth \frac{kL}{2}\]The element loads \(\mathbf{f}^e_l\) stored in the variable
feare computed according to\[\mathbf{f}^e_l = \mathbf{G}^T \bar{\mathbf{f}}^e_l\]where
\[\begin{split}\bar{\mathbf{f}}^e_l = qL \begin{bmatrix} 0 \\ \frac{1}{2} \\ \frac{L}{12} \psi \\ 0 \\ \frac{1}{2} \\ -\frac{L}{12} \psi \end{bmatrix}\end{split}\]For an axial compressive force (\(Q_{\bar{x}} < 0\)):
\[\psi = 6 \left( \frac{2}{(kL)^2} - \frac{1 + \cos kL}{kL \sin kL} \right)\]and for an axial tensile force (\(Q_{\bar{x}} > 0\)):
\[\psi = -6 \left( \frac{2}{(kL)^2} - \frac{1 + \cosh kL}{kL \sinh kL} \right)\]