User talk:Stress9/HW3

Templates: $$\displaystyle \clubsuit$$ Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Artistic aspect: Comments: Overall rating:

 Note: The Overall rating: here is more of a "gut feeling" than a formal quantitative grading. Follow our list of criteria for formal grading and commenting. In case of a tie, let's favor new teams for the top-3 positions. Specific comments are given for a few teams, with a focus on the top teams to serve as a standard that other teams are measured against. You can now begin to do a quantitative grading of HW3. As you grade and see the best features in the HW reports, do not hesitate to add the teams with these best features to the wiki page "The best of HW3" for each course. Stress9 19:17, 18 October 2008 (UTC)

=EML 4500=

User:Eml4500.f08/HW_report_table

General comments for HW3

 * See also The best of HW3, which is also accessible at the bottom of the HW report table; most of the points made below in the general comments are reported in the above wiki page to inform the students where to look for the best features. In case I miss something, please do not hesitate to inform me; I don't have time to read through all HW reports like you do.  NOTE: The teams listed in The-best-of-HW3 are not necessarily the top teams, which are at the top because of the overall quality of their reports.


 * derivation of element stiffness matrix $$\displaystyle \mathbf k ^{(e)}_{4 \times 4}$$ in global coordinates: Axial dofs $$\displaystyle \mathbf q^{(e)}_{2 \times 1}$$, 2x4 transformation matrix $$\displaystyle \mathbf T ^{(e)}_{2 \times 4}$$, transformation of dofs $$\displaystyle \mathbf q ^{(e)} _{2 \times 1} = \mathbf T ^{(e)}_{2 \times 4} \mathbf d^{(e)}_{4 \times 1}$$, axial force-displacement (FD) relation $$\displaystyle \mathbf p ^{(e)} _{2 \times 1} = \hat{\mathbf k}^{(e)} _{2 \times 2} \mathbf q^{(e)} _{2 \times 1}$$, FD relation in global coordinates $$\displaystyle \mathbf f ^{(e)} _{4 \times 1} = {\mathbf k}^{(e)} _{4 \times 4} \mathbf d^{(e)} _{4 \times 1}$$, verification of expression for element stiffness matrix $$\displaystyle \mathbf k ^{(e)} = {\mathbf T ^{(e)}}^T \hat{\mathbf k}^{(e)} \mathbf T ^{(e)}$$. Best explanation: Team Bike
 * singularity of global stiffness matrix: solve eigenvalue problem on global stiffness matrix and observe the zero eigenvalues. Best solution: Team Bike Team Ramrod, but the explanation of the zero eigenvalues was wrong for both teams.


 * closing the loop: For statically determinate problems, such as the 2-bar truss system, first solve for the reactions using statics, then solve for the displacements at the global node 2 using infinitesimal displacements; many teams did not know how to solve for the coordinates of the intersection of two straight lines (!) even for past top teams such as Team Wiki Team Delta_6. Other teams who equally got it wrong: Team Jamama, etc. Best solution: Team Wiki1 Team Ramrod Only these two teams got this problem right.
 * figure for conceptual steps in "closing the loop". Best figure: Team Echo (did not point out where "closing the loop" was) Team FEABBQ (did not point out where "closing the loop" was) Team Wiki (did point out where "closing the loop" was, but had an extra arrow in lower left corner that I did not intend to have in my lecture)
 * computation of the reactions using method 2, i.e., multiply the reduced global stiffness matrix $$\displaystyle \mathbf K _{6 \times 2}$$ by the reduced displacement matrix $$\displaystyle \mathbf d _{2 \times 1}$$, and ignoring rows 3 and 4, which correspond to the known applied forces. Best solution: Team Bike Team Ramrod


 * 3-bar truss system: This system is statically indeterminate; 3 unknown reactions, but only 2 equations, $$\displaystyle \sum F_x = 0$$ and $$\displaystyle \sum F_y = 0$$. The equation $$\displaystyle \sum M_A = 0$$, i.e., sum-of-moments-about-point-A equals zero, leads to a trivial equation $$\displaystyle 0 = 0$$, and thus is not useful (point A is the intersection of the 3 bars).  Show that the sum of moments about any point in the plane leads to a trivial (not-useful) equation $$\displaystyle 0 = 0$$, and thus the moment equation cannot provide a 3rd independent equation to solve for the 3 unknown reactions.  The theory was explained in detail in class; see my lecture presentation.  This problem is subtle; Team Wiki did NOT get it right; Team Wiki1 did not explain well...; Best explanation: Team Bike
 * figure for assembly of element stiffness matrices into global stiffness matrix. Best figure: Team Wiki Team Echo
 * 3-bar truss global stiffness matrix. Best expression for $$\displaystyle \mathbf K = \overset{e=nel}{\underset{e=1}{\mathbb A}} \mathbf k^{(e)}$$ : Team Jamama


 * 2-bar truss system: matlab plot of deformed shape. Best code, plot, documentation: Team Wiki plotted a square around each global node number using matlab and not other software; Team Team for complete documentation, including sample matlab codes.

Specific comments on team HW3 reports
$$\displaystyle \clubsuit$$ Team Bike Copyright violation / plagiarism: No. Completeness: Yes, except for the figure of assembly of global stiffness matrix for 3-bar truss. Team contribution: OK, no other contributions than from lecture presentations. Correctness: Verification of element stiffness matrix in terms of 2x4 transformation matrix: Very good explanation. Eigenvalue problem: Good, but wrong explanation; wrong symbols for element stiffness coeffs. Closing the loop: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Figure on "closing the loop": OK. Reaction computation using global stiffness matrix: Very good. 3-bar truss system: Very good explanation of the sum of moments about point $$\displaystyle A$$ for the 3-bar truss system, with good figures. Figure for assembly of global stiffness matrix: No figure. 3-bar truss global stiffness matrix: Almost correct; notation for coefficients should be with lowercase $$\displaystyle k$$ and not uppercase $$\displaystyle K$$. 2-bar truss system: matlab plot and code, which can write node and element numbers; complete, good. Artistic aspect: Good, no problem; no colors, but that is fine. Comments: Very good work; among the top; need to look at other teams. Overall rating: 9.1 / 10.

$$\displaystyle \clubsuit$$ Team ATeam(Sean) Copyright violation / plagiarism: No. Completeness: Not complete; see below. Team contribution: OK, no other contributions than from lecture presentations. Correctness: Verification of element stiffness matrix in terms of 2x4 transformation matrix: Not done symbolically, but somehow verified by using numerical values in preparation for the eigenvalue problem below; I intended the verification to be done symbolically, not numerically. Eigenvalue problem: Did the eigenvalue problem of a single element, instead of a 2-bar truss system; that's OK; I need to check whether I asked the class to do the single element or the 2-bar truss system. Closing the loop: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Figure on "closing the loop": Figure missing. Reaction computation using global stiffness matrix: Not quite right (not 6 components of reactions, only 4 components are reactions), described only the method, not doing the numerical computation. 3-bar truss system: Figure for assembly of global stiffness matrix: Very good. 3-bar truss global stiffness matrix: Almost correct, error in coeff (5,6) and (6,6): The superscript must be (2). 2-bar truss system: matlab plot and code, which can write node and element numbers; complete, good. Artistic aspect: OK, not nice, not outstanding. Comments: Overall rating: 7.9 / 10.

$$\displaystyle \clubsuit$$ Team Ramrod Copyright violation / plagiarism: No. Completeness: Almost; see below. Team contribution: OK, no other contributions than from lecture presentations. Correctness: Verification of element stiffness matrix in terms of 2x4 transformation matrix: OK, but the symbols used are not correct, e.g., the brackets around a matrix were not used, but vertical bars were used instead; some matrix symbols were not in boldface. Eigenvalue problem: Good, but wrong explanation; correct symbols for element stiffness coeffs, unlike Team Bike. Closing the loop: Almost perfect solution for the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; note a benign difference (no problem) $$\displaystyle \tan (\frac{\pi}{4}) \equiv \tan (225^\circ)$$, and a sign error in the equation of the second straight line: $$\displaystyle \cdots \tan \left( \frac{2\pi}{3} \right) - (x_D-5.917) \cdots $$ should be $$\displaystyle \cdots \tan \left( \frac{2\pi}{3} \right) (x_D-5.917) \cdots $$, i.e., no minus sign; did not report the details of the solution of the system of two equations to find $$\displaystyle (x_D, y_D)$$, but that's fine since such a solution is not a big deal; almost as good as Team Wiki1. Figure on "closing the loop": Yes, not the best. Reaction computation using global stiffness matrix: Very good. 3-bar truss system: Correct explanation for the moment equation being not useful: The sum of moments about any point is zero. Figure for assembly of global stiffness matrix: No figure. 3-bar truss global stiffness matrix: Wrong stiffness matrix; did not complete the solution. Wrong figures. 2-bar truss system: matlab plot and code, which can write node and element numbers; complete, good. Artistic aspect: Comments: Nice section layout of the HW report with good table of contents. Rank high ?? ahead of Team Wiki1; need to look at other teams; not as good as Team Bike who did a nice job, except for the "closing the loop" problem. Overall rating: 8.9 / 10.

$$\displaystyle \clubsuit$$ Team Wiki1 Copyright violation / plagiarism: Completeness: Not complete: Did not do the eigenvalue problem. Team contribution: OK, no other contributions than from lecture presentations. Correctness: Verification of element stiffness matrix in terms of 2x4 transformation matrix: Did NOT verify symbolically the expression for $$\displaystyle \mathbf k^{(e)}$$ in terms of the transformation matrix. Eigenvalue problem: Did NOT do the eigenvalue problem. Closing the loop: Solve correctly the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop. Figure on "closing the loop": Yes, but not outstanding. Reaction computation using global stiffness matrix: Yes, computed the reactions. 3-bar truss system: Scanty explanation of sum of moments about point $$\displaystyle B$$; not showing a good understanding. Figure for assembly of global stiffness matrix: Missing. 3-bar truss global stiffness matrix: Correct. 2-bar truss system: matlab code can only draw the undeformed and deformed shapes, but not writing the node and element numbers, which were written by using other software; not what we want. Artistic aspect: Not great. Comments: The only thing going for this team was the "closing the loop" section; the other sections are not so good. Overall rating: 8.0 / 10.

$$\displaystyle \clubsuit$$ Team Wiki Copyright violation / plagiarism: Completeness: Team contribution: Nice figure on "closing the loop". Correctness: Almost, EXCEPT misunderstood the problem of finding the displacements of the global node 2 using the method of infinitesimal displacements after having the reactions obtained from statics; misunderstand how to compute the coordinates of point B and point C. Artistic aspect: Nice figures, mostly; some figures were not correct. Comments: Nice matlab code to plot deformed shape with global node numbers inside squares, i.e., not plotting the square using a different software like some other teams. Overall rating:

$$\displaystyle \clubsuit$$ Team Delta_6 Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Artistic aspect: Comments: Overall rating:

$$\displaystyle \clubsuit$$ Team Jamama Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Artistic aspect: Comments: Overall rating:

$$\displaystyle \clubsuit$$ Team ATeam(Tatiana) Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Artistic aspect: Comments: Overall rating:

$$\displaystyle \clubsuit$$ Team A-Team This team still lives in the main namespace!! Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Artistic aspect: Comments: Overall rating:

$$\displaystyle \clubsuit$$ Team Bottle Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Artistic aspect: Comments: Overall rating:

$$\displaystyle \clubsuit$$ Team Echo Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Artistic aspect: Comments: Overall rating:

$$\displaystyle \clubsuit$$ Team FEABBQ Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Artistic aspect: Comments: Overall rating:

$$\displaystyle \clubsuit$$ Team Gravy Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Artistic aspect: Comments: Overall rating:

$$\displaystyle \clubsuit$$ Team Group Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Artistic aspect: Comments: Overall rating:

$$\displaystyle \clubsuit$$ Team Lulz Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Artistic aspect: Comments: Overall rating:

$$\displaystyle \clubsuit$$ Team Qwiki Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Artistic aspect: Comments: Overall rating:

$$\displaystyle \clubsuit$$ Team Team Copyright violation / plagiarism: Completeness: Team contribution: Correctness: Also misunderstood the problem of finding the displacements of node 2 using the method of infinitesimal displacements after obtaining the reactions using statics, i.e., closing the loop; similar to Team Wiki. Artistic aspect: Comments: Complete matlab code documentation, including sample matlab codes. Overall rating:

=EAS 4200C=

User:Eas4200c.f08/HW_report_table

General comments for HW3

 * See also The best of HW3, which is also accessible at the bottom of the HW report table; most of the points made below in the general comments are reported in the above wiki page to inform the students where to look for the best features. In case I miss something, please do not hesitate to inform me; I don't have time to read through all HW reports like you do.  NOTE: The teams listed in The-best-of-HW3 are not necessarily the top teams, which are at the top because of the overall quality of their reports.


 * classnotes 70% + matlab problem 30% = 100%


 * classnotes graded over 100%:
 * prove the expression for the area of a triangle when the height falls out of the triangle. Best solution: Team Aero Team Aeris
 * compare the polar moment of inertia $$\displaystyle J$$ between a solid circular cross section and a thin-walled circular cross section with area $$\displaystyle A$$, there are two parts; Part 1: Keep area the same, i.e., $$\displaystyle A^{(a)} = A^{(b)}$$ and find $$\displaystyle J^{(a)} / J^{(b)}$$; Part 2: Keep moment the same, i.e., $$\displaystyle J^{(a)} = J^{(b)}$$ and find $$\displaystyle A^{(a)} / A^{(b)}$$. Best solution: Team Aeris


 * matlab problem graded over 100%: NACA airfoil, plot airfoil, find centroid and plot centroid, find area $$\displaystyle \bar{A}$$.
 * students must follow the code development guide sent out some time ago by validating their code with a circular airfoil before applying their code to the NACA airfoil. Code validation with circular airfoil: 40/100. If used quadrature by trapezoids (instead of the required quadrature by triangles) and if validated code with circular airfoil ( location of centroid and area $$\displaystyle \bar{A}$$) are correct: 20/100.
 * the area $$\displaystyle \bar{A}$$ has to be found using quadrature by triangles, and not by trapezoids, since this method is more elegant and corresponds exactly to the derivation of the expression $$\displaystyle T = 2 q \bar{A}$$: 60/100.
 * Must demonstrate that the code works for $$\displaystyle c = 0.5 m$$, and not $$\displaystyle c = 1 m$$; if used $$\displaystyle c = 1 m$$, and if area $$\displaystyle \bar{A}$$ is correct: -20/100
 * If used quadrature by trapezoids and if results ( location of centroid and area $$\displaystyle \bar{A}$$) are correct: 30/100.
 * If result for $$\displaystyle \bar{A}$$ is correct, and the code allows for arbitrary location of point $$\displaystyle P_0$$, but did NOT demonstrate that the same result is obtained for any point $$\displaystyle P_0$$: -10/100.
 * Best solution: Team Aero6 Almost perfect, except for doing for $$\displaystyle c=1m$$ and for not validating their code at different observer points $$\displaystyle P_0$$; good figure explaining the quadrature by triangles, nice plots, syntax highlighting; even applied their code for different NACA airfoils. Team Carbon completed most of the above tasks, except for the location of the centroid; their code's syntax is highlighted by using the wiki commands  .  See also Team Aeris even though the centroid of Team Aeris is a little off center (odd); also did not validate their code at various observer points $$\displaystyle P_0$$.  See also Team Aero.
 * many teams did not redescribe the NACA airfoil as done in Airfoil.

Specific comments on team HW3 reports
$$\displaystyle \clubsuit$$ Team Aeris Copyright violation / plagiarism: No. Completeness: Yes, almost. Not complete note on quadrature (summations); see Team Carbon. Team contribution: Good image of airflow over NACA airfoil; need to provide a link to the source. Warping of aircraft wing: Be careful, in the literature, sometimes warping (deformation of cross section in the direction of the wing axis) is confused with twisting (torsion of wing). Correctness: Almost all. Good proof of area of triangle with height outside triangle, correct. Proof of $$\displaystyle a^2 \approx \bar{r}^2$$ and $$\displaystyle b^2 \approx \bar{r}^2$$ for moment of inertia $$\displaystyle J$$ not clear, not well written, not correct, e.g., $$\displaystyle b-a = t \ne 0$$. Good comparison of $$\displaystyle J$$ for solid circular cross section and thin-walled circular cross section. The matlab problem is well done, except for the centroid, which seems odd; it should be closer to the middle of the airfoil. Matlab code validated with circular airfoil: Good. The area $$\displaystyle \bar{A}$$ is computed for both $$\displaystyle c=1$$ and $$\displaystyle c=0.5$$. Numerical convergence study well done. Did NOT demonstrate code can be used for any point $$\displaystyle P_0$$; see Team Carbon. Results are reliable, except for centroid location. Matlab problem not as well presented as Team Carbon. Artistic aspect: Better figures in classnotes than those of Team Carbon. The matlab code should be enclosed in the command to have syntax highlighting; see Team Carbon. Comments: Good matlab results. Could be tied for the top team this time; see Team Carbon. Overall rating: 9.3 / 10.

$$\displaystyle \clubsuit$$ Team Carbon. Copyright violation / plagiarism: No. Completeness: Almost, except for the proof of the area of the triangle with the height falling outside the triangle; confuse the goal of the proof; what they did was to relate the cross product to the area of a triangle, but not the proof of the area of the triangle itself, i.e., $$\displaystyle A = bh / 2$$. Also the computation of the centroid of the NACA airfoil has a problem. Team contribution: Not much extra contribution other than lecture material; added some material in the book to the classnotes (e.g., equation of equilibrium for stresses). Did add a figure of a wing airfoil with 3 cells; figure not nice. Correctness: Yes, almost, except for the proof of the area of a triangle and the location of the centroid of the NACA airfoil. Good proof of the approximation of $$\displaystyle J$$ when $$\displaystyle t \ll a$$ and $$\displaystyle t \ll b$$. More complete note on quadrature (summations) than Team Aeris. Good matlab work: Code validated with circular airfoil and convergence study; nice plots; correct area $$\displaystyle \bar{A}$$; matlab code presented with syntax highlighting. Did demonstrate code can be used for any point $$\displaystyle P_0$$; excellent. Artistic aspect: Images not nice, ugly; do not look pleasant. Equations not nice (ugly) sometimes. Comments: Good matlab results for area $$\displaystyle \bar{A}$$; honest in saying that they could not get the location of the centroid. Could be the top team this time; see Team Aeris. Overall rating: 9.5 / 10.

Below, the focus of the comments is given to the matlab problem.

$$\displaystyle \clubsuit$$ Team XYZ Copyright violation / plagiarism: Completeness: Team contribution: Correctness: The matlab code seems fine; did include the coordinates of the point $$\displaystyle P_0$$, but did NOT demonstrate that the code works for any point $$\displaystyle P_0$$; used the quadrature by triangles; location of centroid seems fine; used $$\displaystyle c= 1 m$$ but not the required $$\displaystyle c= 0.5 m$$, probably due to the problem encountered by plotting the airfoil within a frame that has the horizontal dimension exactly equal to 1; the area $$\displaystyle \bar{A}$$ for $$\displaystyle c=1m$$ seems correct however; good figure of airfoil with $$\displaystyle c = 1 m$$ and with correct centroid location. Artistic aspect: Comments: Overall rating: