For simply supported beams with uniform load, assuming the length of the beam is l=1, the uniform load is 1, the modulus e is 1, the moment of inertia I is 1, then the analytic solution of the deflection curve equation of the beam is
$$
V (x) =-\left (\frac{x^4}{24}-\frac{x^3}{12}+\frac{x}{24}\right)
$$
The shape function of the beam element is as follows
According to the definition of the shape function, the deflection of the element can be obtained by using the shape function as long as the displacement of the node at both ends is known. That
$$
V (x) =n (x) \cdot Q^{e}
$$
Where $q^{e}$ is the displacement array of the nodes.
In contrast, the difference between the deflection curve obtained by the shape function and the analytic solution of the flex curve, in order to make the comparison convenient, when the shape function is used, the displacement of the node is solved by the analytic solution of the flex curve, in other words, the displacement at the node is accurate.
1. Assume that a beam in a structure is considered a unit
$$
\begin{split}
V (0) =0\quad & V ' (0) =-\dfrac{1}{24}\\
V (1) =0\quad & V ' (1) =\dfrac{1}{24}
\end{split}
$$
The curve equation can be obtained by multiplying the shape function row matrix with the node displacement array above.
$$
V1 (x) =\frac{x^2}{24}-\frac{x}{24}
$$
2. Assuming the middle beam is looking at 2 units
The length of each unit L=1/2
Using mm to obtain the piecewise function of the torsion curve equation is
The curve equations of the analytic solution, 1 units and 2 units are drawn respectively.
It can be seen that with 2 units, the deflection curve equation is very close to the analytic solution, almost coincident, and the more analytic solution is different when using 1 units. Examine the values at 1/4
|
1/4 out deflection value |
Analytic solution |
-0.00927734 |
1 units |
-0.0078125 |
2 units |
-0.00911458 |
The difference is obvious.
Simple comparison between analytic solution and finite element solution of curved beam deflection equation