User:Eml4500.f08.group/HW7/Matlab

=MATLAB Code=

Mon, 24 Nov 2008, 13:17:17 EST Dear class:

In all of the HW7 problems involving the 2-element frame system under different boundary conditions, the deformation inside each frame element due to bending and extension/compression is not a straight line as for a truss element. As a result, the *complete* deformed shape of the frame system is more complex than the deformed shape of the truss system.

The deformed shape of a truss system only involves the displacement at the two nodes for each element.

To obtain the *complete* deformed shape of a frame system, you need to subdivide each frame element into segments (say 10, 15, etc.), then for each point at the end of each segment, you need to figure out the displacement vector for that point based on the nodal displacements and rotations of the frame element and the frame shape functions (which involve both 2 shape functions for a bar element, and 4 shape functions for a beam element). More on these details should be done in my lectures (today).

=Frame Model=

Next, go back to the problem:

Electric pylon: http://apollo.mae.ufl.edu/~vql/courses/fead/2006.fall/code.web.pages/electric_pylon.m

and convert the finite element model for the electric pylon from a truss system (HW6) to a frame system, i.e., all elements now are frame elements.

Use the clamped boundary conditions at the base of the frame model, instead of the simply supported boundary conditions (pin connections) at the base of the truss model.

Redo the static analysis using the same geometric, material, and loading data as for HW6, assuming in addition that the cross section for each frame element is a square.

Compare the nodal displacements of the node under the applied force for the frame model to those of the truss model (which you already obtained in HW6).

Identify the frame element(s) with the highest nodal bending moment and *transverse* shear force. To compute the transverse shear force, you need to work in the tilde coordinate system, i.e., all internal force components in the global coordinate system should be transformed back to the element tilde coordinate system.

Max Nodal Bending Moment is 60 N m at Node 176 (element 58, green)

Max Transverse Stress is 37749 N/m^2 at Node 149 (element 49, red)

Plot the following on the same figure:


 * underformed shape of the electric pylon in dotted line (as in HW6)


 * deformed shape of the truss model (as in HW6)


 * the "incomplete" deformed shape of the frame model in which only a straight line connecting two displaced nodes of a frame element is plotted (not the complete deformed shape inside each element as done above).



You should use the same magnification factor for both the truss model and the frame model for comparison.


 * is the frame model statically determinate?


 * If yes, provide an argument to support your answer, and compute the reactions. Extend your matlab code to compute the reactions, and compare the computed reactions to the reactions obtained from statics.


 * If no, provide an argument to support your answer. You can still compute the reactions using the truss model and the frame model.

The truss system is statically indeterminate. It requires the use of the Principle of Virtual Work or Mechanics of Materials to solve.

Consider isolating an arbitrary node located within the truss system. Also, determine each reaction at a node as a single transverse reaction along the line of action of a beam rather than two reactions along the vertical and horizontal 2-D coordinate system.

Now begin with the node on the far right tip of the arm of the pylon as shown below:



As you can see there are two unknowns and three possible statics equations. Taking a moment however at the node in consideration reveals no conclusion and cannot be considered a possible statics equation in this example.

Using $$\sum{F_{y}} = 0$$, it is easily found that the y-component of R2 = 1000N.

Using $$\sum{F_{x}} = 0$$, it is also easily found that $$R_{1} = R_{2}cos\theta $$

From here, the final reactions cannot be computed and equations of Mechanics must be introduced (if not using FEM). The node shown above is the simplest FBD in the system.

Consider the node at (25, 30):



From this node, it is clear to see why six reactions cannot be computed with only two equations of statics.

With each node there are two, three, four, five or six unknown reactions. There are six nodes with two unknown reactions at the extreme locations of the truss system which cannot be solved as shown above. The nodes with four, five or six unknown reactions can't be computed using only two equations of statics.

The reason we cannot use the moment equation as a third equation of statics is because if the moment is taken at any point the moment equals zero. At each node, each isolated truss intersects. If the moment is taken at the node, it is clear why the moment equals zero, however at any other point it does as well (as shown in previous homework assignments). Each member is a two-force body and there is no resulting moment arm and no moment. This shows that the nodes with three unknown reactions cannot be solve as well.


 * Compare the reactions obtained from the truss model to those from the frame model (remembering that the boundary conditions are different from one model to the other).

The deformation of both the frame and truss structure are nearly identical. This is likely due to the small moment of inertia of the beams, 1.3 x 10^-8 meters.

=Lumped Mass Matrix=

Tthe mass density of the 300 M steel is rho = 7.8 g/cm^3. Consider a lumped mass model in which the mass of each bar element is divided equally by two and is lumped at each of the two nodes of the bar.

Construct the lumped mass matrix for the frame model of the electric pylon. Note that this lumped mass matrix is "almost" the same as the lumped mass matrix for the truss model, except that the dimension of the matrix is larger (due to the rotation dofs); there are 3 dofs per node in the frame model, instead of just 2 dofs per node in the truss model. Just put zero mass at the rotation dofs, i.e., we neglect the effects of rotational inertia.


 * Use matlab to solve the generalized eigenvalue problem $$K_{bar} v = \lambda M_{bar} v$$ where $$K_{bar}$$ is the reduced stiffness matrix, $$M_{bar}$$ the associated mass matrix, $$v$$ the eigenvector and lambda the eigenvalue.


 * Find the lowest 3 eigenpairs ($$\lambda_i, v_i$$), for i=1,2,3.

>> lam(121:123,121:123)

ans =

11.83           0            0            0        25.81            0            0            0       30.927


 * In 3 separate plots, plot the eigenvectors $$v_1, v_2, v_3$$ as deformed shapes (solid line) superposed onto the underdeformed shape (dotted line). Again, use a magnification factor to clearly distinguish each eigenvector from the undeformed shape.




 * Also, you only need to plot an "incomplete" eigenmode for the frame model in the sense that there is only a straightline between two displaced nodes for each element, and not the *complete* deformed shape for each frame element.


 * given that $$\lambda = \omega^2$$ where omega is the circular frequency of vibration. The vibrational frequency is given by $$f = \omega / (2 \pi)$$ and the vibrational period is given by $$T = 1 / f = 2 \pi / \omega$$ Find the 3 lowest vibrational periods in the frame model of the electric pylon.

ans =

3.4395           0            0            0       5.0804            0            0            0       5.5612


 * Compare the 3 lowest vibrational periods for the frame model to those of the truss model.

=Animation= We want to create an animation of the 3 lowest eigenmodes of the electric pylon. To this end, we choose to animate the simpler truss model.

An animation is nothing but a graphic file in gif format in which you just include a sequence of many picture frames together just like in a movie.

So what sequence of picture frames would you create? Let $$v_i$$ be an eigenvector of the truss model (i=1,2,3), and let omega_i the corresponding circular vibrational frequency. According to the theory of vibration (to be done in class if time permits), the following time-dependent displacement matrix

$$d_i (t) = v_i sin (\omega_i t)$$

for t = 0 to t = T_i (the vibrational period of mode i) can be used to create a sequence of deformed shapes to create an animation. All you need to do now is to discretize the time domain into small intervals so you can produce the picture frames. For example, you can have a time sequence of

$$t_0 = 0, t_1, t_2, t_3, ..., t_n$$

with $$\delta_t = T / n$$

being the time increment. The value of n should be at least 8, i.e., each period should be subdivided into 8 intervals for a minimum animation. The larger n is the smoother the animation, but also the larger file size and the longer the download time. So a reasonable value of n would be 8 or 16.



=Code=