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I'm creating a structural engineering program. It's a frame analysis program that computes the shear and moment diagrams. The frame is subjected to 2 kinds of distributed loads(only the beams(horizontal lines) were loaded): Dead load that doesn't move and a live load that change in position. In order to create the moment envelope( maximum and minimum moments at a point) I have to consider all the load patterns, which means I have to vary the live load positions and I have to consider all the positions. Here's a picture of the frame.Example of a frame.

And below is a picture of one of the many load patterns. http://www.bgstructuralengineering.com/BGASCE7/BGASCE7005/BGASCE70504.htm

First the frame is analysed without the loads values, each beam was loaded with a variable w[i] (w[4] on the member 10 in the figure1 for an example etc...). The system is then solved and the moment values were obtained as a function of w[i] with i going from 1 to 4.

Now in order to find the maximum values of the moments for all the possible loadings I used the Tuples function:

d = DeadLoad;
l = LiveLoad;
pattern = Tuples[{d, d + l}, floors*spans];
patternCounter = Count[pattern, List_];

The next step was to replace the w[i] for each member with each pattern. And use the Max function to find the maximum moment.

>     BeamMid[k_] := 
>       Max[Table[{w[i] = pattern[[j]][[i]]; MemberTotalForces[k][[5]]}, {j,
>           1, patternCounter}, {i, 1, floors*spans}]];

MemberTotalForces[k][[5]] was obtained from the frame analysis and is a function of w[i] with i going from 1 to 4.

The problem I'm facing is time. For a 2 by 2 frame there are only 16 load patterns. For the frame in the picture below there are 65536 load patterns. And the process takes too long.

4 by 4 frame

Is there anyway to make the process faster?

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  • $\begingroup$ It looks like you are calling MemberTotalForces many times. Assuming that's the speed bottleneck, have you tried compiling this function? $\endgroup$ – bill s Feb 10 '15 at 15:07
  • $\begingroup$ @bills Compiling wouldn't work since the function has variables w[i] with i going from 1 to the number of beams. Here's MemberTotalForces[7][[5]] from the first figure: 409.983 + 4.04673 w[1] + 0.120798 w[2] + 0.00359441 w[3] - 0.0712942 w[4] $\endgroup$ – Mr. Pi Feb 10 '15 at 17:53
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This problem is linear: that is, the moment due to a distribution of forces is equal to the sum of the moments due to the forces individually. A faster approach may be to calculate for each beam which force creates a more positive or more negative moment. Your final distribution will be the combination of each individually calculated force.

In other words, instead of looking at the patterns

{d, ... d, d, d}
{d, ... d, d, t}
{d, ... d, t, d}
 .
 . (2^n)
 .
{t, ... t, t, d}
{t, ... t, t, t}

(where t = d+l) And choosing the max, look at the pairs of patterns

{0, ... 0, 0, d}  {0, ... 0, 0, t}
{0, ... 0, d, 0}  {0, ... 0, t, 0}
{0, ... d, 0, 0}  {0, ... t, 0, 0}
 .
 . (n)
 .
{d, ... 0, 0, 0}  {t, ... 0, 0, 0}

Choose the maximum resultant from each, and then add each of the maxima together.

This solution is $O(n)$ instead of $O(2^n)$, so you shouldn't have any speed issues.

| improve this answer | |
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  • $\begingroup$ d is always there so actually the patterns are of this kind: {d, ... d, d, d} {d, ... d, d, d+l} {d, ... d, d+l, d}...{d+l, ... d+l,d+l, d} {d+l, ... d+l,d+l,d+l} $\endgroup$ – Mr. Pi Feb 10 '15 at 18:38
  • $\begingroup$ You're right; I was just being lazy in writing them out, fixing now. $\endgroup$ – 2012rcampion Feb 10 '15 at 20:37

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