GENERALIZED DEGREES OF FREEDOM

The generalized degrees of freedom capability of MOSES was designed to solve problems that really are not solvable by any other means, and as a result, this capability is not easy to check. Our approach here is to construct an estimate of the solution to a problem, check the estimate where it can be checked, and then compare it with MOSES results. In particular, we want to solve a "flagpole" problem, consisting of a vertical beam with a weight on top, and we will compare the MOSES results for this problem with a standard energy approximation.

The beam is made up of two lengths, $l_1$ and $l_2$, so the total length is $l=l_1+l_2$. The bottom of the beam is at $z=0$, while the top of the beam is at $z=l_1+l_2$. There is a weight $W_o$ mounted at the top of the beam. The mass per unit length of the beam is $\rho$. The bending stiffness (flexural rigidity) of the beam is $EI$. There is also a distributed force along the length of the beam, which is denoted by $F(z)$.

Looking at the energy in the beam system, we obtain:

$$ \begin{aligned} K &= \frac{1}{2}\rho\int_0^{l_1+l_2} \dot{x}^2\,\textrm{d}z + \frac{1}{2}\frac{W_o}{g}\big(\dot{x}(l_1+l_2)\big)^2 \\\\ V_1 &= \frac{1}{2}EI\int_0^{l_1+l_2}(x'')^2\,\textrm{d}z \\\\ V_2 &\approx -\frac{1}{2}W_o\int_0^{l_1+l_2}(x')^2\,\textrm{d}z \\\\ W &= \int_0^{l_1+l_2} F(z)x\,\textrm{d}z \end{aligned} \tag*{(13)} $$

Here, the first term of $K$ is the kinetic energy of the beam itself, and the second term is the kinetic energy of the weight on the top, $V_1$ is the strain energy due to bending of the beam, $V_2$ is the potential energy due to the weight on the top being displaced downwards due to bending of the beam, and $W$ is the work done by forces acting along the beam. We will consider the beam to be "pinned" at $z=0$ and $z=l_1$, so we really have a two-span beam that approximates a cantilever when $l_1$ is small in comparison to $l_1 + l_2$.

Now, assume that $x(z)$ is a function with continuous derivatives that satisfies the boundary conditions, and depends on a single parameter $\delta$, which represents the value of $x$ at $z = l_1 + l_2$. Then

$$ \begin{aligned} K &= \frac{1}{2}(\dot{\delta})^2\left(\rho \int_{0}^{l_1+l_2} x^2 \, \textrm{d}z + \frac{W_o}{g}\right)&&\equiv \frac{1}{2}(\dot{\delta})^2 M_i \\\\ V_1 &= \frac{1}{2}(\dot{\delta})^2 EI\int_{0}^{l_1+l_2} (x'')^2 \, \textrm{d}z &&\equiv \frac{1}{2}(\dot{\delta})^2 K_1 \\\\ V_2 &= -\frac{1}{2}(\dot{\delta})^2 W_o \int_{0}^{l_1+l_2} (x')^2 \, \textrm{d}z &&\equiv -\frac{1}{2}(\dot{\delta})^2\frac{1}{2} K_2 \\\\ W &= \delta\left(F_o + \int_{0}^{l_1+l_2} f x^2 \, \textrm{d}z\right) &&\equiv \delta (F_o + C) \end{aligned} \tag*{(14)} $$

Here, $F_o$ is a transverse point force applied at the top of the beam and $C$ is a contribution due to a distributed transverse force, $f(z)$, acting along the beam.

If we now look at conservation of energy, we get an equation

$$ K + V_1 + V_2 = W \tag*{(15)} $$

If we use (15) and differentiate with respect to time, after factoring out $\dot{\delta}$ this yields

$$ M\ddot{\delta} + (K_1 - K_2)\delta = F_o + C \tag*{(16)} $$

Now, let us assume that $\delta$, $F_o$, and $C$ are harmonic in time with frequency $\omega$. This reduces the above equation to an algebraic one:

$$ -\omega^2 \delta M + (K_1 - K_2)\delta = F_o + C \tag*{(17)} $$

or

$$ \delta = \frac{F_o + C}{\omega^2 M + (K_1 - K_2)} \tag*{(18)} $$

If we have computed the integrals $K_1$, $K_2$, $C$, and $M$, then (18) provides an estimate of the deflection at the top of our beam.

Before attaching the integrals, let us first look at some special cases of (18). First, notice that it does not have a solution when

$$ \omega^2 = \frac{K_1 - K_2}{M} \tag*{(19)} $$

which gives us an estimate of the natural frequency of the beam. Also, notice that when $C=0$ and $\omega=0$, it yields

$$ \delta = \frac{F_o}{K_1 - K_2} \tag*{(20)} $$

which is the static deflection due to a load applied at the top. These two results will be used to check the accuracy of this approximation. Since (20) has a singularity at the natural period, we will alter it slightly by introducing a bit of damping

$$ \delta = \frac{F_o + C} {\sqrt{\left[-\omega^2 M + (K_1 - K_2)\right]^2 + 4\eta^2 (K_1 - K_2)M}} \tag*{(21)} $$

We are still free to choose any deflection shape $x(z)$ that satisfies the boundary conditions:

$$ \begin{aligned} x(l_1) &= 0 \\ \lim_{z \to l_1^+} x'(z) &= \lim_{z \to l_1^-} x'(z) \\ x''(l_1) &= 0 \\ x(l_1 + l_2) &= 1 \\ x''(l_1 + l_2) &= 0 \end{aligned} \tag*{(22)} $$

Our choice is the static deflection function under a top load:

$$ \begin{aligned} x(z) = \begin{cases} \alpha z + \beta z^3, & 0 \le z \le l_1 \\ \gamma (z - l_1) + \psi (z - l_1)^2 + \zeta (z - l_1)^3, &l_1 < z \le l_1+l_2 \end{cases} \end{aligned} \tag*{(23)} $$

where the five parameters $\alpha$, $\beta$, $\gamma$, $\psi$, and $\zeta$ can be determined from the above five boundary conditions.

The particular problem that we will consider is that of a tube with:

$$ \begin{aligned} l_1 &= \text{10 meters } \\ l_2 &= \text{90 meters } \\ W_o &= \text{100 kN } \\ F_o &= \text{10 kN } \end{aligned} \tag*{(24)} $$

and which has a diameter of 1 meter and a wall thickness of 25 millimeters. The buoyancy of the tube was chosen to be equal to the weight so that we did not have to include the potential energy of either the weight or the buoyancy of the tube in our equation. When the tube is in the water, the density used above includes both the density of the steel in the tube and the added mass. Also, the integral $C$ is the Morison’s equation inertia force due to wave particle acceleration.

A comparison of the results of obtained using our energy method approximation with the results from a structural analysis is presented in the following table.

Comparison of Energy Method Approximation with Structural Analysis

These comparisons are interesting. First, the comparisons with $W_o=0$ are excellent, but those for a nonzero weight are not so good. This is due to the fact that the structural analysis we are comparing with is linear; i.e., the effect of the concentrated weight on the stiffness of the beam is not accounted for. Thus, for these cases the estimate obtained using the energy-based solution is more correct than the structural analysis.

Since the estimate is reasonably reliable, we can turn to a comparison of the results from MOSES generalized coordinates and the estimate. We first computed three modes of the beam in water using the pin connections and then used these as generalized coordinates. Figure 12 and Figure 13 show a comparison of the results obtained with a weight and without a weight. As you can see, MOSES does quite a nice job of predicting the response including the nonlinearities.

One must be a bit careful using generalized coordinates. Above, we had a set of modes that exactly satisfied the boundary conditions. To show what can happen if you do not have a "good" set of modes, we also solved the same problem with a "bad" set of modes. These "bad" modes are backward; i.e., they correspond to a beam fixed at the top! Figure 14 shows the results obtained using two of these modes, and Figure 15 shows the results when using twenty of them. As you can see, with two modes you have a really bad estimate of the deformation, while with twenty it is not so bad.

Figure 12: Flagpole Modes Computed in Water with No Weight, Motion of Node *11

Figure 13: Flagpole Modes Computed in Water with Weight, Motion of Node *11

Figure 14: Flagpole Modes Computed in Air with No Weight, Motion of Node *11, Number of Modes = 2

Figure 15: Flagpole Modes Computed in Air with No Weight, Motion of Node *11, Number of Modes = 20