The BEAM statement defines a force and a torque corresponding to a massless elastic beam with a uniform cross section. The beam transmits forces and torques between two markers in accordance with either linear Timoshenko beam theory or the non-linear Euler-Bernoulli theory. Conversely to standard Finite Elements codes, the BEAM element defines a force-only element; no equations of motion for the mass of the beam are formulated.

The BEAM statement supports three formulations to construct the force and torque expressions for this element. By default, the linear Timoshenko beam theory is used (see Using the FORMULATION=LINEAR option below); this default model is the one used in previous versions of Adams Solver C++. However, the FORMULATION option is available in order to use either a full non linear Euler-Bernoulli theory or a simplified non linear theory.

The figure below shows the two markers (I and J) that define the extremities of the beam and indicates the twelve forces (s1 to s12) it produces.

The x-axis of the J marker defines the centroidal axis of the beam. The y-axis and z-axis of the J marker are the principal axes of the cross section. They are perpendicular to the x-axis and to each other. When the beam is in an undeflected position, the I marker has the same angular orientation as the J marker, and the I marker lies on the x-axis of the J marker a distance LENGTH away.

The beam statement applies the following forces to the I marker in response to the relative motion of the I marker with respect to the J marker (see nomenclature in Figure above):

When using the default FORMULATION or when explicitly using FORMULATION=LINEAR, the linear Timoshenko beam theory is used to define the forces and moments. In this case Adams Solver (C++) uses a linear translational and linear rotational action-reaction force between the two markers I and J. The forces the beam produces are linearly dependent on the displacements, rotations, and corresponding velocities between the markers at its endpoints.

The following constitutive equations define how Adams Solver applies a force and a torque to the I marker depending on the displacement rotation and velocity of the I marker relative to the J marker.

where:

Note that both matrices, Cij and Kij, are symmetric, that is, Cij=Cji and Kij=Kji. You define the twenty-one unique damping coefficients when the BEAM statement is written. Adams Solver (C++) defines each Kij as follows:

Adams Solver (C++) applies an equilibrating force and torque at the J marker, as defined by the following equations:

L is the instantaneous vector from the J marker to the I marker. While the force at the J marker is equal and opposite to the force at the I marker, the torque is usually not equal and opposite, because of vector L.

The BEAM statement implements a force in the same way the FIELD statement does, but the BEAM statement requires you to input only the values of the beam’s physical properties, which Adams Solver (C++) uses to calculate the matrix entries.

When using the FORMULATION=NONLINEAR option, then the Euler-Bernoulli formulation is used to define the forces and moments. In this case, Adams Solver (C++) uses the following constitutive equations to apply a force and a torque to the I marker.

Where corresponds to the Timoshenko's constitutive equations shown above, and N is the axial force on the beam computed as:

The axial force N corresponds to the term Fx computed using linear theory (see section Using the FORMULATION=LINEAR option above.) If there is damping (options CRATIO or CMATRIX are used), the term d corresponds to viscous damping and can be computed as shown above. See reference [2] for more information.

The linear theory is the default model for the BEAM statement. The non linear theory should be used when the BEAM elements in the model are subject to high axial forces. In those cases, the additional terms in the force/torque equations help stabilize the numerical solution.

When using the option FORMULATION=STRING, then a simplified Euler-Bernoulli formulation is used to define the forces and moments. In this case, Adams Solver (C++) uses the following constitutive equations to apply a force and a torque to the I marker.

Where and N are computed as described in Using the FORMULATION=NONLINEAR option above. This simplification may provide accurate results faster. See Chapter 15 in reference [2] for more details on string stiffness. Notice the stiffness matrices shown in all of the above equations show only a subset of a complete stiffness matrix displayed in textbooks. The rest of the entries may be found applying equilibrium conditions.

The BEAM element is a useful force/torque element but it does include inertial effects. If you need to include inertial effects, see the FE_PART statement.

The fastest simulation time in the majority of cases is provided by the default formulation (FORMULATION=LINEAR). However, the LINEAR formulation may lead into a simulation failure due to instabilities if axial forces are high. The typical case is that of slender blades modelled using PARTs connected by BEAMs and rotating at high angular speed. High axial loads may introduce numerical instabilities. In order to obtain a stable and correct numerical solution, the NONLINEAR formulation is required in those cases. Unfortunately, the NONLINEAR option requires higher simulation time due to the complex expressions for the force and torque. In some cases, using the STRING option may speed up the simulation time while providing accurate results.

When using the BEAM element, the best idealization of beam flexibility uses the principle of using enough BEAMs to model the desired flexibility. For reasons of practicality, if you model flexibility by only one BEAM, you should keep in mind that the forces on the I and J markers may NOT be transposable. That is, the BEAM forces may not be symmetric when the I and J marker locations are switched. See the last Caution note below for an explanation. The following figures illustrate this asymmetry and how to overcome it.

In Figure 1 the J marker of the BEAM is fixed to ground. Since the BEAM uses the J marker as its reference frame, it behaves as a finite element BEAM would, giving non-zero displacement in the Y axis of the J marker and zero displacement along the x axis of the J marker.

Now consider what happens when the I and J marker are switch as shown in Figure 2.

Here, the I and J markers are reversed, and the J marker is no longer constrained to ground. Thus, when the force F causes the beam to deflect and the J marker to rotate, the force is applied along the x and y axis of the J marker – not only along the y axis as was the case in the previous model. The forces resulting from the BEAM depend on the J marker kinematics. As a result, do not expect the BEAM forces on the I and J markers to be the same in each example.

Next is the model of a pinned beam shown in Figure 3 that has Part A attached to ground with a revolute Joint A. Part B is attached to ground with a revolute Joint B. Part A and Part B are connected with a BEAM. While this model may appear symmetrical, it is not by definition, and you will not get symmetric results in the BEAM forces

Figure 4 shows how this can be avoided by modeling symmetrically by adding a dummy part and more beams.

The model from Figure 3 has been modified in Figure 4to incl dummy Part C that is connected to Part A and Part B with Beams 1 and Beam 2. As more beams are added, the effect illustrated in Figure 1and Figure 2 are more localized, and the solution becomes less sensitive to the locations of the I and J markers of the BEAM elements

A cantilevered stainless steel beam is to be modeled with a circular cross section that has the loading shown in the figure below.

A weight of 17.4533 lbf at the free end of the beam with a 100-inch axial offset in the negative y direction causes torsion to the beam as shown in the figure above. The following statement defines this beam:

The beam lies between Marker 0010 and Marker 0020. The length of the beam is 100 inches; its torsional constant is 100 inch4; its principal area moments of inertia about the y-axis is 50 inch4, and about the z-axis is 50 inch4; its cross-sectional area is 25.0663 inch2; its shear area ratio in the y direction is 1.11; its shear area ratio in the z direction is 1.11; its modulus of elasticity is 28E6 psi; its shear modulus is 10.6 psi; and its damping ratio relative to the stiffness matrix Adams Solver (C++) calculates is 0.0001. Note that the beam ends belong to different parts.

See other Forces available.