beam1sΒΆ
es = beam1s(ex, ep, ed)
es = beam1s(ex, ep, ed, eq)
[es, edi, eci] = beam1s(ex, ep, ed, eq, n)
- Description:
beam1scomputes the section forces and displacements in local directions along the beam elementbeam1e.The input variables
ex,epandeqare defined inbeam1e, and the element displacements, stored ined, are obtained by the functionextract_ed. If distributed loads are 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 variables
es\(= \begin{bmatrix} V(0) & M(0) \\ V(\bar{x}_{2}) & M(\bar{x}_{2}) \\ \vdots & \vdots \\ V(\bar{x}_{n-1}) & M(\bar{x}_{n-1})\\ V(L) & M(L) \end{bmatrix}\) \(\qquad\)edi\(= \begin{bmatrix} v(0) \\ v(\bar{x}_{2}) \\ \vdots \\ v(\bar{x}_{n-1})\\ v(L) \end{bmatrix}\) \(\qquad\)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
\[\begin{split}\mathbf{\bar{a}}^e = \begin{bmatrix} \bar{u}_1 \\ \bar{u}_2 \\ \bar{u}_3 \\ \bar{u}_4 \end{bmatrix}\end{split}\]where the transpose of \(\mathbf{a}^e\) is stored in
ed.The displacement \(v(\bar{x})\), the bending moment \(M(\bar{x})\) and the shear force \(V(\bar{x})\) are computed from
\[v(\bar{x}) = \mathbf{N} \mathbf{\bar{a}}^e + v_p(\bar{x})\]\[M(\bar{x}) = D_{EI} \mathbf{B} \mathbf{\bar{a}}^e + M_p(\bar{x})\]\[V(\bar{x}) = -D_{EI} \frac{d\mathbf{B}}{dx} \mathbf{\bar{a}}^e + V_p(\bar{x})\]where
\[\mathbf{N} = \begin{bmatrix} 1 & \bar{x} & \bar{x}^2 & \bar{x}^3 \end{bmatrix} \mathbf{C}^{-1}\]\[\mathbf{B} = \begin{bmatrix} 0 & 0 & 2 & 6\bar{x} \end{bmatrix} \mathbf{C}^{-1}\]\[\frac{d\mathbf{B}}{dx} = \begin{bmatrix} 0 & 0 & 0 & 6 \end{bmatrix} \mathbf{C}^{-1}\]\[v_p(\bar{x}) = \frac{q_{\bar{y}}}{D_{EI}} \left( \frac{\bar{x}^4}{24} - \frac{L \bar{x}^3}{12} + \frac{L^2 \bar{x}^2}{24} \right)\]\[M_p(\bar{x}) = q_{\bar{y}} \left( \frac{\bar{x}^2}{2} - \frac{L \bar{x}}{2} + \frac{L^2}{12} \right)\]\[V_p(\bar{x}) = -q_{\bar{y}} \left( \bar{x} - \frac{L}{2} \right)\]in which \(D_{EI}\), \(L\), and \(q_{\bar{y}}\) are defined in
beam1eand\[\begin{split}\mathbf{C}^{-1} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -\frac{3}{L^2} & -\frac{2}{L} & \frac{3}{L^2} & -\frac{1}{L} \\ \frac{2}{L^3} & \frac{1}{L^2} & -\frac{2}{L^3} & \frac{1}{L^2} \end{bmatrix}\end{split}\]