User:Eml4500.f08.group/HW6/Matlab

Fri, 07 Nov 2008, 14:00:23 EST Dear class:

As part of HW6, consider the electric pylon with the FE mesh described in the matlab code in:

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

Scale the electric pylon so that its height is 60 m. Consider a downward vertical applied load of (- 1000 N) to be applied on the far right tip of the arm of the pylon.

Consider the pylon material made of 300 M steel with Young's modulus equal to 200 GPa (See Sun [2006], Mechanics of Aircraft Structures, Wiley, 2006, p.14; or search for info on the internet, and refer to the web page in your HW report). The cross-sectional area of each bar element is taken to be 4 cm^2.

Do the following:

Massless Pylon

 * develop a matlab code to solve the above electric-pylon problem


 * plot the underformed shape (dotted line) and the deformed shape (solid line); you may need to multiply the displacements by a magnification factor to clearly distinguish the deformed shape from the underformed shape.


 * compute the axial stress in each bar; give the highest tensile stress and the highest compressive stress.


 * Identify the element number for the bar element with the highest tensile stress and the element number for the bar element with the highest compressive stress. Indicate with an arrow on the figure of underformed and deformed shapes above the element with highest tensile stress and the element with highest compressive stress.

MaxTensileStress = 3630.5 @ Element 81 (Marked with Green Below)

MaxCompressiveStress = -3478.3 @ Element 55 (Marked with Red Below)



results =

1.0781e-005 2.1562e+006       862.48 -6.5356e-006 -1.3071e+006     -522.85 6.7339e-006 1.3468e+006       538.71 6.4556e-006 1.2911e+006       516.45 -3.5152e-006 -7.0304e+005     -281.22 -2.3798e-005 -4.7596e+006     -1903.8 -1.2325e-005 -2.465e+006      -986.01 7.1182e-006 1.4236e+006       569.46 1.9269e-005 3.8538e+006       1541.5 -2.8828e-005 -5.7655e+006     -2306.2 -1.8941e-007      -37882      -15.153 8.778e-009      1755.6      0.70224 1.9328e-005 3.8655e+006       1546.2 1.4698e-005 2.9395e+006       1175.8 -1.8494e-005 -3.6988e+006     -1479.5 -2.883e-005 -5.7661e+006     -2306.4 1.7948e-006 3.5896e+005       143.58 2.621e-005  5.242e+006       2096.8 9.8986e-006 1.9797e+006       791.89 -1.8874e-005 -3.7749e+006       -1510 -3.258e-005 -6.5159e+006     -2606.4 2.7892e-006 5.5784e+005       223.14 3.2194e-006 6.4388e+005       257.55 2.9828e-005 5.9656e+006       2386.3 -4.0852e-006 -8.1704e+005     -326.82 -1.9808e-005 -3.9616e+006     -1584.6 1.16e-005   2.32e+006       928.02 -4.0653e-006 -8.1307e+005     -325.23 -2.8778e-005 -5.7556e+006     -2302.2 -1.4237e-007      -28474       -11.39 -4.9677e-006 -9.9353e+005     -397.41 -7.6072e-008      -15214      -6.0857 2.9608e-005 5.9216e+006       2368.6 2.7907e-007       55813       22.325 -2.4358e-005 -4.8717e+006     -1948.7 6.9573e-006 1.3915e+006       556.59 -1.5116e-008     -3023.3      -1.2093 -2.8794e-005 -5.7588e+006     -2303.5 -2.427e-005 -4.854e+006      -1941.6 6.9522e-006 1.3904e+006       556.17 -9.4154e-007 -1.8831e+005     -75.323 1.129e-005  2.258e+006       903.22 -3.16e-006  -6.32e+005       -252.8 -2.0051e-005 -4.0102e+006     -1604.1 0           0            0  2.4411e-019  4.8822e-008  1.9529e-011 3.2073e-007       64147       25.659 1.3501e-006 2.7001e+005       108.01 1.3575e-006 2.7149e+005        108.6 -1.0389e-005 -2.0778e+006     -831.13 -2.2136e-005 -4.4271e+006     -1770.9 -2.9388e-005 -5.8777e+006     -2351.1 -3.6774e-005 -7.3549e+006     -2941.9 -4.331e-005 -8.662e+006      -3464.8 -4.3478e-005 -8.6957e+006     -3478.3 -1.2903e-019 -2.5807e-008 -1.0323e-011 -3.0942e-019 -6.1883e-008 -2.4753e-011 2.6985e-019  5.397e-008  2.1588e-011 1.2205e-019 2.4409e-008  9.7637e-012 -8.8522e-007 -1.7704e+005     -70.818 1.1034e-006 2.2069e+005       88.276 -9.5965e-007 -1.9193e+005     -76.772 9.9436e-007 1.9887e+005       79.549 1.0732e-005 2.1464e+006       858.56 -1.0415e-005 -2.083e+006      -833.19 1.1448e-005 2.2895e+006       915.81 -1.1321e-005 -2.2643e+006     -905.71 1.0517e-005 2.1035e+006       841.38 -1.0732e-005 -2.1464e+006     -858.56 7.0689e-006 1.4138e+006       565.51 -6.8859e-006 -1.3772e+006     -550.87 7.9177e-006 1.5835e+006       633.42 -1.8464e-005 -3.6928e+006     -1477.1 -4.8696e-007      -97393      -38.957 -1.1845e-007      -23690      -9.4762 1.5604e-007       31208       12.483 4.5239e-005 9.0479e+006       3619.2 4.6963e-019 9.3927e-008  3.7571e-011 4.5158e-005 9.0317e+006       3612.7 0           0            0  4.5256e-005  9.0511e+006       3620.5 -6.5877e-007 -1.3175e+005     -52.702 -1.4822e-006 -2.9645e+005     -118.58 8.9829e-006 1.7966e+006       718.63 2.2438e-005 4.4876e+006         1795 3.3117e-005 6.6233e+006       2649.3 3.8878e-005 7.7755e+006       3110.2 0           0            0  4.8034e-020  9.6069e-009  3.8427e-012 -2.4017e-020 -4.8034e-009 -1.9214e-012 -5.3704e-020 -1.0741e-008 -4.2963e-012

MaxTensileStress =

3620.5

MaxCompressiveStress =

-3478.3

ElementMaxTensile =

81

ElementMaxCompressive =

55

i =

81

i =

55

Statics
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.

Mass Considerations

 * the 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 of the electric pylon.

lumpedMassMatrix =

-391.99     -388.32      -613.61      -757.65      -610.71      -362.59      -856.16      -360.97      -545.38      -544.57      -287.87       -503.9      -285.61      -299.35      -359.77      -357.94      -297.47      -205.49      -204.06      -461.98      -460.64      -233.87      -277.46      -310.28      -212.83      -346.52      -264.92      -266.02      -345.41      -211.59      -312.25      -276.54      -232.63       -202.3      -200.19      -193.54      -194.28       -338.2      -376.58      -249.65      -267.69      -249.01      -374.79      -339.39      -261.75      -261.75


 * 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.


 * 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.



results =

1.0781e-005 2.1562e+006       862.48 -6.5356e-006 -1.3071e+006     -522.85 6.7339e-006 1.3468e+006       538.71 6.4556e-006 1.2911e+006       516.45 -3.5152e-006 -7.0304e+005     -281.22 -2.3798e-005 -4.7596e+006     -1903.8 -1.2325e-005 -2.465e+006      -986.01 7.1182e-006 1.4236e+006       569.46 1.9269e-005 3.8538e+006       1541.5 -2.8828e-005 -5.7655e+006     -2306.2 -1.8941e-007      -37882      -15.153 8.778e-009      1755.6      0.70224 1.9328e-005 3.8655e+006       1546.2 1.4698e-005 2.9395e+006       1175.8 -1.8494e-005 -3.6988e+006     -1479.5 -2.883e-005 -5.7661e+006     -2306.4 1.7948e-006 3.5896e+005       143.58 2.621e-005  5.242e+006       2096.8 9.8986e-006 1.9797e+006       791.89 -1.8874e-005 -3.7749e+006       -1510 -3.258e-005 -6.5159e+006     -2606.4 2.7892e-006 5.5784e+005       223.14 3.2194e-006 6.4388e+005       257.55 2.9828e-005 5.9656e+006       2386.3 -4.0852e-006 -8.1704e+005     -326.82 -1.9808e-005 -3.9616e+006     -1584.6 1.16e-005   2.32e+006       928.02 -4.0653e-006 -8.1307e+005     -325.23 -2.8778e-005 -5.7556e+006     -2302.2 -1.4237e-007      -28474       -11.39 -4.9677e-006 -9.9353e+005     -397.41 -7.6072e-008      -15214      -6.0857 2.9608e-005 5.9216e+006       2368.6 2.7907e-007       55813       22.325 -2.4358e-005 -4.8717e+006     -1948.7 6.9573e-006 1.3915e+006       556.59 -1.5116e-008     -3023.3      -1.2093 -2.8794e-005 -5.7588e+006     -2303.5 -2.427e-005 -4.854e+006      -1941.6 6.9522e-006 1.3904e+006       556.17 -9.4154e-007 -1.8831e+005     -75.323 1.129e-005  2.258e+006       903.22 -3.16e-006  -6.32e+005       -252.8 -2.0051e-005 -4.0102e+006     -1604.1 0           0            0  2.4411e-019  4.8822e-008  1.9529e-011 3.2073e-007       64147       25.659 1.3501e-006 2.7001e+005       108.01 1.3575e-006 2.7149e+005        108.6 -1.0389e-005 -2.0778e+006     -831.13 -2.2136e-005 -4.4271e+006     -1770.9 -2.9388e-005 -5.8777e+006     -2351.1 -3.6774e-005 -7.3549e+006     -2941.9 -4.331e-005 -8.662e+006      -3464.8 -4.3478e-005 -8.6957e+006     -3478.3 -1.2903e-019 -2.5807e-008 -1.0323e-011 -3.0942e-019 -6.1883e-008 -2.4753e-011 2.6985e-019  5.397e-008  2.1588e-011 1.2205e-019 2.4409e-008  9.7637e-012 -8.8522e-007 -1.7704e+005     -70.818 1.1034e-006 2.2069e+005       88.276 -9.5965e-007 -1.9193e+005     -76.772 9.9436e-007 1.9887e+005       79.549 1.0732e-005 2.1464e+006       858.56 -1.0415e-005 -2.083e+006      -833.19 1.1448e-005 2.2895e+006       915.81 -1.1321e-005 -2.2643e+006     -905.71 1.0517e-005 2.1035e+006       841.38 -1.0732e-005 -2.1464e+006     -858.56 7.0689e-006 1.4138e+006       565.51 -6.8859e-006 -1.3772e+006     -550.87 7.9177e-006 1.5835e+006       633.42 -1.8464e-005 -3.6928e+006     -1477.1 -4.8696e-007      -97393      -38.957 -1.1845e-007      -23690      -9.4762 1.5604e-007       31208       12.483 4.5239e-005 9.0479e+006       3619.2 4.6963e-019 9.3927e-008  3.7571e-011 4.5158e-005 9.0317e+006       3612.7 0           0            0  4.5256e-005  9.0511e+006       3620.5 -6.5877e-007 -1.3175e+005     -52.702 -1.4822e-006 -2.9645e+005     -118.58 8.9829e-006 1.7966e+006       718.63 2.2438e-005 4.4876e+006         1795 3.3117e-005 6.6233e+006       2649.3 3.8878e-005 7.7755e+006       3110.2 0           0            0  4.8034e-020  9.6069e-009  3.8427e-012 -2.4017e-020 -4.8034e-009 -1.9214e-012 -5.3704e-020 -1.0741e-008 -4.2963e-012

MaxTensileStress =

3620.5

MaxCompressiveStress =

-3478.3

ElementMaxTensile =

81

ElementMaxCompressive =

55

i =

81

i =

55

\

Vibrational Considerations

 * 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 of the electric pylon.

Sincerely, Prof. Vu-Quoc

=Code=