Derive the displacement field for an unloaded beam segment by integrating the differential equation of equilibrium.
Question : Derive the displacement field for an unloaded beam segment b...
Let us consider the beam segment of length l of Figure 1.6 with a shear force and a couple acting at each end.
The equilibrium of an unloaded infinitesimal beam segment of length dx leads to the well known differential equations (Figure 1.7) [Ti2]:
Equilibrium of couples: \frac{\mathrm{d} M}{\mathrm{d} x}= -Q
Equilibrium of vertical forces:\frac{\mathrm{d} Q}{\mathrm{d} x}=0
Differentiating the first equation and making use of the second one and of the bending moment-curvature relationship of Eq.(1.5) gives
Eq.(1.5):M=-\int \int_{A}^{}z\sigma _{x}dA=\left ( \int \int_{A}^{}z^{2}dA \right )E\frac{\mathrm{d^{2}} w}{\mathrm{d} x^{2}}=EI_{y}\frac{\mathrm{d^{2}} w}{\mathrm{d} x^{2}}=EI_{y}\kappa
\frac{\mathrm{d}^{^{4}} w}{\mathrm{d} x^{4}}=0The solution of this differential equation is the cubic polynomical
w(x) = a_{1} + a_{2} + a_{3}x^{2} + a_{4}x^{3}
The conditions at the unloaded beam segment ends are (Figure 1.6)
w=w_{1} and \frac{\mathrm{d} w}{\mathrm{d} x}=\left (\frac{\mathrm{d} w}{\mathrm{d} x} \right )_{1} at x = 0
w=w_{2} and \frac{\mathrm{d} w}{\mathrm{d} x}=\left (\frac{\mathrm{d} w}{\mathrm{d} x} \right )_{2} at x = l
which lead to the following equations system
\begin{Bmatrix}w_{1}\\ \left ( \frac{\mathrm{d} w}{\mathrm{d} x} \right )_{1}\\ w_{2}\\ \left ( \frac{\mathrm{d} w}{\mathrm{d} x} \right )_{2}\end{Bmatrix} = \begin{bmatrix}1 & 0 &0 &0 \\ 0 & 1 &0 &0 \\ 1 &l &l^{2} &l^{3} \\ 0 & 1 & 2l &3l^{2} \end{bmatrix} \begin{Bmatrix}a_{1}\\ a_{2}\\ a_{3}\\ a_{4}\end{Bmatrix}
from which the parameters a_{i} can be obtained. Substituting these into the cubic deflection field gives
w\left ( x \right )=f_{1\left ( x \right )}w_{1}+f_{2}\left ( x \right )\frac{l}{2}\left ( \frac{\mathrm{d} w}{\mathrm{d} x} \right )_{1}+f_{3\left ( x \right )}w_{2}+f_{4\left ( x \right )}\frac{l}{2}\left ( \frac{\mathrm{d} w}{\mathrm{d} x} \right )_{2}
where
f_{1}\left ( x \right )=1-3\left ( \frac{x}{l} \right )^{2}+2\left ( \frac{x}{l} \right )^{3} ; f_{2}\left ( x \right )=\frac{2x}{l}-4\left (\frac{x}{l} \right )^{2}+2\left ( \frac{x}{l} \right )^{3}
f_{3}\left ( x \right )=3\left (\frac{x}{l} \right )^{2}-2\left ( \frac{x}{l} \right )^{3} ; f_{4}\left ( x \right )=-2\left (\frac{x}{l} \right )^{2}+2\left ( \frac{x}{l} \right )^{3}
The coincidence of functions f_{i}\left ( \xi \right ) with the Hermite shape functions (1.11a) can be recognized after making a simple transformation to the natural coordinate system, i.e. changing x by \frac{l}{2}\left ( 1+\xi \right ) (Eq.(1.11b)).
(Eq.(1.11b)): \xi =\frac{2}{l^{(e)}}\left ( x-x_{c} \right ) and x_{c}=\frac{x_{1}+x_{2}}{2}
Using the PVW it is easy to deduce that the stiffness matrix for the unloaded beam segment of length l of Figure 1.6 coincides precisely with that for the 2-noded Hermite beam element (Eq.(1.20)).
(Eq.(1.20)) : k^{(e)}= \left (\frac{EI_{y}}{l^{3}}\right )^{(e)}\begin{bmatrix}12 &6l^{(e)} &-12 &6l^{(e)} \\ \ddots & 4\left ( l^{(e)} \right )^{2} &-6l^{(e)} &2\left ( l^{(e)} \right )^{2} \\ & \ddots & 12 &-6l^{(e)} \\ Sym. & & \ddots &4\left ( l^{(e)} \right )^{2} & \end{bmatrix}
