AS3020: Aerospace Structures Demos
Using Emacs Org Table and Emacs-Calc for Solving Shear Flow Problems

Table of Contents

Problem Setting

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  • Download the classdemo .org file (that we created during class) from here.

p23-6.jpeg

Figure 1: Problem 23.6 from [1]

Pointers on Emacs Org-Table

Useful Shortcuts

  • M is the alt key, C is the ctrl key.
    • C-c means "Control plus c".
    • C-x c means "hold down control and hit x, then leave control and hit c".
  • C-c } will show the keys for the table
  • C-u C-c C-c will recalculate all formulae
  • M-<up/down/left/right> moves the row/column of cell up/down/left/right.
  • C-c ` from inside a table lets you edit the formulae you've already typed.

Table Tips

  • Rows are prefixed with "@" - @5 is fifth row.
    • @-1 refers to the "previous row"
  • Columns are prefixed with "$" - $4 is fourth column.
    • $-1 refers to the "previous column"
  • A particular cell is written with both a row number and a column number: @3$4 is the 3rd row 4th column item.
  • Starting formula with "=" makes the formula applicable to the whole row.
  • Starting the formula with ":-" makes the formula applicable only to the *current cell.

Useful functions

  • :=vsum(@2..@9) sums up the rows from 2 to 9.
  • Use remote(<tablename>, @3$5) to access element from another table.
  • =simplify(<expression>) and =expand(<expression>) will help with simplification.
  • =rmeq(a=b) will return the right hand side of the expression, i.e., "b" in this case.
  • =subst(x^2+4, x, 5) will substitute the value "5" to the variable "x" in the expression.
  • Underscore _ is the array indexing operator.
    • If you have an array [1, 4, 5, 6] in column 4, then $4_2 will yield the 2nd element of the array.

Solution

p23-6.jpeg

Figure 2: Problem 23.6 from [1]

Let us First Handle the Booms

Boom Ar X3r X3r Ar X3r^2 Ar X3r Ar/I22
1 1290 153 197370 30197610 1.0893246e-3
2 645 153 98685 15098805 5.4466231e-4
3 1290 153 197370 30197610 1.0893246e-3
4 645 153 98685 15098805 5.4466231e-4
5 645 -153 -98685 15098805 -5.4466231e-4
6 1290 -153 -197370 30197610 -1.0893246e-3
7 645 -153 -98685 15098805 -5.4466231e-4
8 1290 -153 -197370 30197610 -1.0893246e-3
      0 181185660 0
  • Cell @10$5 is the second moment of area I22.

Let us write out the cell areas

Cell Area
1 167780
2 217872

Let us consider the panels now

Panels del s t G q q del s/Gt 2A_6 2A_6 q
12 356 0.915 24200 qab 0.016077316 qab 108936 108936 qab
23 356 0.915 24200 qab - 5.4466231e-4 0.016077316 qab - 8.7567079e-6 108936 108936 qab - 59.333333
34 380 0.915 20700 qcd 0.020062828 qcd 116280 116280 qcd
45 610 1.220 24800 qcd - 5.4466231e-4 0.020161290 qcd - 1.0981095e-5 219280 219280 qcd - 119.43355
36 306 1.220 24800 qab - qcd - 1.6339869e-3 0.010113696 qab - 0.010113696 qcd - 1.6525648e-5 0 0
56 380 0.915 20700 qcd 0.020062828 qcd 0 0
67 356 0.915 24200 qab - 5.446623e-4 0.016077316 qab - 8.7567077e-6 0 0
78 356 0.915 24200 qab + 1e-11 0.016077316 qab + 1.6077316e-13 0 0
81 306 1.220 24800 qab + 1.0893246e-3 0.010113696 qab + 1.1017098e-5 217872 217872 qab + 237.33333
        Cell 1 Twist Rate: 2.0980046e-7 qcd - 3.0139754e-8 qab + 1.6523283e-11 Twisting Moment_6: 335560 qcd + 435744 qab + 58.566447
        Cell 2 Twist Rate: 1.9400532e-7 qab - 2.3210178e-8 qcd - 5.2833695e-11   #ERROR
        Cell 1-Cell 2 Twist 2.3301064e-7 qcd - 2.2414507e-7 qab + 6.9356978e-11   0

Let us solve for the shear center (zero twist rate condition)

  • We shall solve for the twist rates of the two cells to be zero simultaneously.

    Solution qab qcd
    [qab = 2.6750653e-4, qcd = -4.0327376e-5] 2.6750653e-4 -4.0327376e-5
    Resultant Moment: 161.59856  
    (for load acting through 6)    
  • The resultant moment will be equal to the shear center position since all the above is for unit resultant load.

  • Now we substitue for qab and qcd into the flow.

    Panels q 2A_6 2A_6 q 66750 q
    12 2.6750653e-4 108936 29.141091 17.856061
    23 -2.7715578e-4 108936 -30.192242 -18.500148
    34 -4.0327377e-5 116280 -4.6892673 -2.6918523
    45 -5.8498969e-4 219280 -128.27654 -39.048062
    36 -1.3261530e-3 0 0. -88.520713
    56 -4.0327377e-5 0 0. -2.6918523
    67 -2.7715577e-4 0 0. -18.500148
    78 2.6750654e-4 0 0. 17.856062
    81 1.3568311e-3 217872 295.61551 90.568476
          161.59855 0
  • Cell @11$4 stores the shear center.

Let us solve a resultant force problem instead, supposing a load of 100 N is acting 50 mm to the left of the spar 36.

  • Here we require twist compatibility and moment balance.

  • The resultant moment has to be 100 x -50 = -5000.

    Resultant Moment: -5000  
    Solution qab qcd
    [qab = -1.1449748e-5, qcd = -3.0866995e-4] -1.1449748e-5 -3.0866995e-4
    Cell 1 Twist Rate: -4.7890722e-9  
    Cell 2 Twist Rate: -4.7890723e-9  
  • If what we've done is correct, then the twist rates in the two cells to be equal!

  • Now we substitute to get the shear flow distribution.

    Panels q/V3 q
    12 -1.1449748e-5 -1.1449748e-3
    23 -5.5611206e-4 -0.055611206
    34 -3.0866995e-4 -0.030866995
    45 -8.5333226e-4 -0.085333226
    36 -1.3367667e-3 -0.13367667
    56 -3.0866995e-4 -0.030866995
    67 -5.5611205e-4 -0.055611205
    78 -1.1449738e-5 -1.1449738e-3
    81 1.0778749e-3 0.10778749
  • The flow in the table above (in terms of qab, qcd) is for unit load. So the resulting flow needs to be multipled by the resultant to get the actual flow.

References

[1]
T. H. G. Megson, Aircraft Structures for Engineering Students. Elsevier, 2013.

Created using Emacs 30.1 (Org mode 9.7.11).

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