Analyse a beam
OK, I can now see what integral you are trying to do. The exact integral would give the correct deflection from either a reciprocal theorem or virtual work (or a very hard slog multiplying by bending-moment distribution for unit point load and some integration by parts).
However,
- you don't need to to do a numerical integration - any one of your load cases gives an EXACT solution for deflection as piecewise-continuous polynomials (point loads give cubics and UDLs give quartics).
- I did the "hard slog" approach above and came to the conclusion that it will certainly apply to the case of a simply-supported beam with zero deflection at the two ends. But what if you wanted other different (but extremely common) support conditions? Say a cantilever (one end built-in, the other free)? Or a beam supported at points other than the two ends? Wouldn't your unit-load moments (mx in the above code snippet) need changing for each?
I think you would be better using the analytical solutions for the different load cases you have included than doing a numerical integration. You would then have both accuracy and the flexibility to include other support conditions.
Maybe I took an obscure approach, but when I wrote a beam analyser for arbitrary point and linear-continuous loads, with user-specified number, position and types of support, I worked my way across the beam adding the appropriate powers of Macaulay brackets at each discontinuity in load or support reaction. At the end I was left with a small set of simultaneous linear equations to give me forces and/or moments at the supports. Solving these allowed me to write functions returning complete shear-force, bending-moment, slope and deflections at any point of the beam. I didn't do any numerical integration.
However,
- you don't need to to do a numerical integration - any one of your load cases gives an EXACT solution for deflection as piecewise-continuous polynomials (point loads give cubics and UDLs give quartics).
- I did the "hard slog" approach above and came to the conclusion that it will certainly apply to the case of a simply-supported beam with zero deflection at the two ends. But what if you wanted other different (but extremely common) support conditions? Say a cantilever (one end built-in, the other free)? Or a beam supported at points other than the two ends? Wouldn't your unit-load moments (mx in the above code snippet) need changing for each?
I think you would be better using the analytical solutions for the different load cases you have included than doing a numerical integration. You would then have both accuracy and the flexibility to include other support conditions.
Maybe I took an obscure approach, but when I wrote a beam analyser for arbitrary point and linear-continuous loads, with user-specified number, position and types of support, I worked my way across the beam adding the appropriate powers of Macaulay brackets at each discontinuity in load or support reaction. At the end I was left with a small set of simultaneous linear equations to give me forces and/or moments at the supports. Solving these allowed me to write functions returning complete shear-force, bending-moment, slope and deflections at any point of the beam. I didn't do any numerical integration.
Last edited on
@lastchance
I think you may be misunderstanding what I am trying to do.
I am trying to learn C++ after many years of using BASIC. I am using the development of a program to analyse a simply supported single span beam to do this. Doing so will enable me to use many of the constructs of C++. At the moment, I have no intention to progress beam analysis further, although I may go on to design reinforced concrete, steelwork and possibly timber beams.
It is many years since I was involved in design work. For the last 40 years or so of my working life I was a project manager, so I've probably forgotten most of the mathematics I knew. I'm surprised I still remembered Simpson's Rule. I remember doing Theory of Structures, but I find it hard work now when I look at my old books and notes.
I started in structural design before the days of computers, when calculation was by slide rule and handwritten. These calculations were all hand checked by others, so the setting out had to be done accordingly. This also helped with submission to local authorities for their approval.
The program I produced in those dark days produced similar hard copy that could be hand checked and was easily understood by an Architect who would be feeding in the data. The program I produced for my Architect employer saved him a small fortune in structural engineers fees.
Surprisingly, unless he was designing framed buildings, most beams were single span. In my working life I came across few cantilever beams. The main one I can think of was in a market hall where the upper floor of part of the building was larger than below, requiring the cantilever.
However, I think many will be interested in single span solutions; if not here, then elsewhere. Whether or not I produce such a solution will depend on other influences, in particular my wife, DIY and gardening. Although I have been retired for some years, most in a similar position will tell you how time flashes by.
I am very grateful for your help and suggestions and hope that you will continue in the future as I progress (hopefully progress).
Best regards
I think you may be misunderstanding what I am trying to do.
I am trying to learn C++ after many years of using BASIC. I am using the development of a program to analyse a simply supported single span beam to do this. Doing so will enable me to use many of the constructs of C++. At the moment, I have no intention to progress beam analysis further, although I may go on to design reinforced concrete, steelwork and possibly timber beams.
It is many years since I was involved in design work. For the last 40 years or so of my working life I was a project manager, so I've probably forgotten most of the mathematics I knew. I'm surprised I still remembered Simpson's Rule. I remember doing Theory of Structures, but I find it hard work now when I look at my old books and notes.
I started in structural design before the days of computers, when calculation was by slide rule and handwritten. These calculations were all hand checked by others, so the setting out had to be done accordingly. This also helped with submission to local authorities for their approval.
The program I produced in those dark days produced similar hard copy that could be hand checked and was easily understood by an Architect who would be feeding in the data. The program I produced for my Architect employer saved him a small fortune in structural engineers fees.
Surprisingly, unless he was designing framed buildings, most beams were single span. In my working life I came across few cantilever beams. The main one I can think of was in a market hall where the upper floor of part of the building was larger than below, requiring the cantilever.
However, I think many will be interested in single span solutions; if not here, then elsewhere. Whether or not I produce such a solution will depend on other influences, in particular my wife, DIY and gardening. Although I have been retired for some years, most in a similar position will tell you how time flashes by.
I am very grateful for your help and suggestions and hope that you will continue in the future as I progress (hopefully progress).
Best regards
I found an error in calculating bending moments for a type 4 load. This has now been corrected. The revised program can be found in the link below:
https://www.dropbox.com/s/drim9kouleb148f/SFBM_Build05_Dev02_07.cpp?dl=0
https://www.dropbox.com/s/drim9kouleb148f/SFBM_Build05_Dev02_07.cpp?dl=0
closed account (48T7M4Gy)
@RNBW
Interesting. I thought you might like to compare your results with the following for a point load on a 10m beam located 3.5m from the LH end. Shear force and bending moment should be the same. Slope and defelection in this set are determined by double integration. The units are just unmodified units, roughly aligned to metrication/steel:
Interesting. I thought you might like to compare your results with the following for a point load on a 10m beam located 3.5m from the LH end. Shear force and bending moment should be the same. Slope and defelection in this set are determined by double integration. The units are just unmodified units, roughly aligned to metrication/steel:
--------------
+ GEOMETRY +
Tag: bm
E: 200000
I: 1.21e+08
--------------
PT X
___^_________^
0 0
1 500
2 1000
3 1500
4 2000
5 2500
6 3000
7 3500
8 4000
9 4500
10 5000
11 5500
12 6000
13 6500
14 7000
15 7500
16 8000
17 8500
18 9000
19 9500
20 10000
----------------------------------------------------------------------
+++ ACTIONS ALONG BEAM +++
----------------------------------------------------------------------
TAG:
P2 (Pt load, P = 1000, x:3500)
END REACTIONS: Ra = 650 Rb = 350
END MOMENTS: Ma = 0 Mb = 0
----------------------------------------------------------------------
PT X V BM Slope Defl'n
____^_________^_________^______________^______________^______________^
0 0 650 0 -0.000258523 0
1 500 650 325000 -0.000255165 -0.128702
2 1000 650 650000 -0.000245093 -0.254046
3 1500 650 975000 -0.000228306 -0.372676
4 2000 650 1.3e+06 -0.000204804 -0.481233
5 2500 650 1.625e+06 -0.000174587 -0.57636
6 3000 650 1.95e+06 -0.000137655 -0.6547
7 3500 650 2.275e+06 -9.40083e-05 -0.712896
8 4000 -350 2.1e+06 -4.8812e-05 -0.74845
9 4500 -350 1.925e+06 -7.2314e-06 -0.762311
10 5000 -350 1.75e+06 3.07335e-05 -0.756284
11 5500 -350 1.575e+06 6.50826e-05 -0.73218
12 6000 -350 1.4e+06 9.58161e-05 -0.691804
13 6500 -350 1.225e+06 0.000122934 -0.636966
14 7000 -350 1.05e+06 0.000146436 -0.569473
15 7500 -350 875000 0.000166322 -0.491133
16 8000 -350 700000 0.000182593 -0.403753
17 8500 -350 525000 0.000195248 -0.309143
18 9000 -350 350000 0.000204287 -0.209108
19 9500 -350 175000 0.000209711 -0.105458
20 10000 -350 0 0.000211519 -6.45661e-16
----------------------------------------------------------------------
----------------------------------------------------------------------
Program ended with exit code: 0 |
closed account (48T7M4Gy)
PS you require an odd number of sections so let me know if I need to revise the input to suit that if necessary keeping in mind I show 0-20 =21
@kemort
I'm just going out, so I haven't had chance to review your output.
However, in answer to your question, the number of sections must be an odd number. This is to meet the requirements for Simpson's Rule for the calculation of deflection. Mostly I seem to see 21 sections used, but this can be a bit coarse if the span is large (e.g. for a 20m span the spacing would be 1m). I tend to use 51 or even 101 sections, but only display 11.
I'm just going out, so I haven't had chance to review your output.
However, in answer to your question, the number of sections must be an odd number. This is to meet the requirements for Simpson's Rule for the calculation of deflection. Mostly I seem to see 21 sections used, but this can be a bit coarse if the span is large (e.g. for a 20m span the spacing would be 1m). I tend to use 51 or even 101 sections, but only display 11.
closed account (48T7M4Gy)
----------------------------------------------------------------------
+++ ACTIONS ALONG BEAM +++
----------------------------------------------------------------------
TAG:
P2 (Pt load, P = 1000, x:3500)
END REACTIONS: Ra = 650 Rb = 350
END MOMENTS: Ma = 0 Mb = 0
----------------------------------------------------------------------
PT X V BM Slope Defl'n
____^_________^_________^______________^______________^______________^
0 0 650 0 -0.000258523 0
1 100 650 65000 -0.000258388 -0.0258478
2 200 650 130000 -0.000257986 -0.0516687
3 300 650 195000 -0.000257314 -0.077436
4 400 650 260000 -0.000256374 -0.103123
5 500 650 325000 -0.000255165 -0.128702
6 600 650 390000 -0.000253688 -0.154147
7 700 650 455000 -0.000251942 -0.17943
8 800 650 520000 -0.000249928 -0.204526
9 900 650 585000 -0.000247645 -0.229407
10 1000 650 650000 -0.000245093 -0.254046
11 1100 650 715000 -0.000242273 -0.278417
12 1200 650 780000 -0.000239184 -0.302492
13 1300 650 845000 -0.000235826 -0.326244
14 1400 650 910000 -0.0002322 -0.349648
15 1500 650 975000 -0.000228306 -0.372676
16 1600 650 1.04e+06 -0.000224143 -0.3953
17 1700 650 1.105e+06 -0.000219711 -0.417495
18 1800 650 1.17e+06 -0.00021501 -0.439233
19 1900 650 1.235e+06 -0.000210041 -0.460488
20 2000 650 1.3e+06 -0.000204804 -0.481233
21 2100 650 1.365e+06 -0.000199298 -0.50144
22 2200 650 1.43e+06 -0.000193523 -0.521083
23 2300 650 1.495e+06 -0.000187479 -0.540136
24 2400 650 1.56e+06 -0.000181167 -0.55857
25 2500 650 1.625e+06 -0.000174587 -0.57636
26 2600 650 1.69e+06 -0.000167738 -0.593479
27 2700 650 1.755e+06 -0.00016062 -0.609899
28 2800 650 1.82e+06 -0.000153233 -0.625594
29 2900 650 1.885e+06 -0.000145579 -0.640537
30 3000 650 1.95e+06 -0.000137655 -0.6547
31 3100 650 2.015e+06 -0.000129463 -0.668059
32 3200 650 2.08e+06 -0.000121002 -0.680584
33 3300 650 2.145e+06 -0.000112273 -0.69225
34 3400 650 2.21e+06 -0.000103275 -0.70303
35 3500 650 2.275e+06 -9.40083e-05 -0.712896
36 3600 -350 2.24e+06 -8.46798e-05 -0.721829
37 3700 -350 2.205e+06 -7.54959e-05 -0.729837
38 3800 -350 2.17e+06 -6.64566e-05 -0.736933
39 3900 -350 2.135e+06 -5.7562e-05 -0.743133
40 4000 -350 2.1e+06 -4.8812e-05 -0.74845
41 4100 -350 2.065e+06 -4.02066e-05 -0.7529
42 4200 -350 2.03e+06 -3.17459e-05 -0.756497
43 4300 -350 1.995e+06 -2.34298e-05 -0.759254
44 4400 -350 1.96e+06 -1.52583e-05 -0.761187
45 4500 -350 1.925e+06 -7.2314e-06 -0.762311
46 4600 -350 1.89e+06 6.50826e-07 -0.762638
47 4700 -350 1.855e+06 8.38843e-06 -0.762185
48 4800 -350 1.82e+06 1.59814e-05 -0.760966
49 4900 -350 1.785e+06 2.34298e-05 -0.758994
50 5000 -350 1.75e+06 3.07335e-05 -0.756284
51 5100 -350 1.715e+06 3.78926e-05 -0.752852
52 5200 -350 1.68e+06 4.4907e-05 -0.748711
53 5300 -350 1.645e+06 5.17769e-05 -0.743875
54 5400 -350 1.61e+06 5.85021e-05 -0.73836
55 5500 -350 1.575e+06 6.50826e-05 -0.73218
56 5600 -350 1.54e+06 7.15186e-05 -0.725348
57 5700 -350 1.505e+06 7.78099e-05 -0.717881
58 5800 -350 1.47e+06 8.39566e-05 -0.709791
59 5900 -350 1.435e+06 8.99587e-05 -0.701094
60 6000 -350 1.4e+06 9.58161e-05 -0.691804
61 6100 -350 1.365e+06 0.000101529 -0.681936
62 6200 -350 1.33e+06 0.000107097 -0.671503
63 6300 -350 1.295e+06 0.000112521 -0.660521
64 6400 -350 1.26e+06 0.0001178 -0.649004
65 6500 -350 1.225e+06 0.000122934 -0.636966
66 6600 -350 1.19e+06 0.000127924 -0.624422
67 6700 -350 1.155e+06 0.000132769 -0.611386
68 6800 -350 1.12e+06 0.000137469 -0.597873
69 6900 -350 1.085e+06 0.000142025 -0.583897
70 7000 -350 1.05e+06 0.000146436 -0.569473
71 7100 -350 1.015e+06 0.000150702 -0.554615
72 7200 -350 980000 0.000154824 -0.539337
73 7300 -350 945000 0.000158802 -0.523655
74 7400 -350 910000 0.000162634 -0.507582
75 7500 -350 875000 0.000166322 -0.491133
76 7600 -350 840000 0.000169866 -0.474322
77 7700 -350 805000 0.000173264 -0.457165
78 7800 -350 770000 0.000176519 -0.439674
79 7900 -350 735000 0.000179628 -0.421866
80 8000 -350 700000 0.000182593 -0.403753
81 8100 -350 665000 0.000185413 -0.385352
82 8200 -350 630000 0.000188089 -0.366676
83 8300 -350 595000 0.00019062 -0.347739
84 8400 -350 560000 0.000193006 -0.328556
85 8500 -350 525000 0.000195248 -0.309143
86 8600 -350 490000 0.000197345 -0.289512
87 8700 -350 455000 0.000199298 -0.269678
88 8800 -350 420000 0.000201105 -0.249657
89 8900 -350 385000 0.000202769 -0.229462
90 9000 -350 350000 0.000204287 -0.209108
91 9100 -350 315000 0.000205661 -0.18861
92 9200 -350 280000 0.00020689 -0.167981
93 9300 -350 245000 0.000207975 -0.147236
94 9400 -350 210000 0.000208915 -0.12639
95 9500 -350 175000 0.000209711 -0.105458
96 9600 -350 140000 0.000210362 -0.0844532
97 9700 -350 105000 0.000210868 -0.0633905
98 9800 -350 70000 0.000211229 -0.0422844
99 9900 -350 35000 0.000211446 -0.0211494
100 10000 -350 0 0.000211519 -6.45661e-16
--------------
+ GEOMETRY +
Tag: bm
E: 200000
I: 1.21e+08
|
@Kemort
I've had to adjust my program to deal with unmodified units because it is set up to deal with distance in metres, loads in kN, shear forces in kN, bending moments in kNm and deflections in mm (+ve).
Having done this I get the following results:
RA = 650 RB = 350
Sec Shear BM Defl
1 650.00 0.00 0.00
3 650.00 65.00 0.26
5 650.00 1300.00 0.48
7 650.00 1950.00 0.66
9 -350.00 2100.00 0.75
11 -350.00 1750.00 0.76
13 -350.00 1400.00 0.69
15 -350.00 1050.00 0.57
17 -350.00 700.00 0.40
19 -350.00 350.00 0.21
21 -350.00 0.00 0.00
As you can see, the Shears and BMs are identical and the deflections are very similar (I use +ve values for deflections; you use -ve). It is good to see that the two methods of calculating deflections give such close results.
Hope this is of use to you.
Have you used a program of your own, or a commercial/proprietary program?
I've had to adjust my program to deal with unmodified units because it is set up to deal with distance in metres, loads in kN, shear forces in kN, bending moments in kNm and deflections in mm (+ve).
Having done this I get the following results:
RA = 650 RB = 350
Sec Shear BM Defl
1 650.00 0.00 0.00
3 650.00 65.00 0.26
5 650.00 1300.00 0.48
7 650.00 1950.00 0.66
9 -350.00 2100.00 0.75
11 -350.00 1750.00 0.76
13 -350.00 1400.00 0.69
15 -350.00 1050.00 0.57
17 -350.00 700.00 0.40
19 -350.00 350.00 0.21
21 -350.00 0.00 0.00
As you can see, the Shears and BMs are identical and the deflections are very similar (I use +ve values for deflections; you use -ve). It is good to see that the two methods of calculating deflections give such close results.
Hope this is of use to you.
Have you used a program of your own, or a commercial/proprietary program?
closed account (48T7M4Gy)
Even better I have just run your two examples which I had forgotten you posted previously:
----------------------------------------------------------------------
+++ ACTIONS ALONG BEAM +++
----------------------------------------------------------------------
LOAD TAG: Pt load (Pt load at centre, P = 30000)
BEAM TAG: CPP, E = 25000, I = 7.2e+09, L = 20000, dx = 1000
END REACTIONS: Ra = 15000 Rb = 15000
END MOMENTS: Ma = 0 Mb = 0
----------------------------------------------------------------------
PT X V BM Slope Defl'n
____^_________^_________^______________^______________^______________^
0 0 15000 0 -0.00416667 0
1 1000 15000 1.5e+07 -0.004125 -4.15278
2 2000 15000 3e+07 -0.004 -8.22222
3 3000 15000 4.5e+07 -0.00379167 -12.125
4 4000 15000 6e+07 -0.0035 -15.7778
5 5000 15000 7.5e+07 -0.003125 -19.0972
6 6000 15000 9e+07 -0.00266667 -22
7 7000 15000 1.05e+08 -0.002125 -24.4028
8 8000 15000 1.2e+08 -0.0015 -26.2222
9 9000 15000 1.35e+08 -0.000791667 -27.375
10 10000 15000 1.5e+08 0 -27.7778
11 11000 -15000 1.35e+08 0.000791667 -27.375
12 12000 -15000 1.2e+08 0.0015 -26.2222
13 13000 -15000 1.05e+08 0.002125 -24.4028
14 14000 -15000 9e+07 0.00266667 -22
15 15000 -15000 7.5e+07 0.003125 -19.0972
16 16000 -15000 6e+07 0.0035 -15.7778
17 17000 -15000 4.5e+07 0.00379167 -12.125
18 18000 -15000 3e+07 0.004 -8.22222
19 19000 -15000 1.5e+07 0.004125 -4.15278
20 20000 -15000 0 0.00416667 0
----------------------------------------------------------------------
----------------------------------------------------------------------
+++ ACTIONS ALONG BEAM +++
----------------------------------------------------------------------
LOAD TAG: UDL (UDL intensity, q = 2.5, over full length)
BEAM TAG: CPP, E = 25000, I = 7.2e+09, L = 20000, dx = 1000
END REACTIONS: Ra = 25000 Rb = 25000
END MOMENTS: Ma = 0 Mb = 0
----------------------------------------------------------------------
PT X V BM Slope Defl'n
____^_________^_________^______________^______________^______________^
0 0 25000 0 -0.00462963 0
1 1000 22500 2.375e+07 -0.0045625 -4.60706
2 2000 20000 4.5e+07 -0.00437037 -9.08333
3 3000 17500 6.375e+07 -0.00406713 -13.3108
4 4000 15000 8e+07 -0.00366667 -17.1852
5 5000 12500 9.375e+07 -0.00318287 -20.6163
6 6000 10000 1.05e+08 -0.00262963 -23.5278
7 7000 7500 1.1375e+08 -0.00202083 -25.8571
8 8000 5000 1.2e+08 -0.00137037 -27.5556
9 9000 2500 1.2375e+08 -0.00069213 -28.5885
10 10000 0 1.25e+08 -6.78168e-19 -28.9352
11 11000 -2500 1.2375e+08 0.00069213 -28.5885
12 12000 -5000 1.2e+08 0.00137037 -27.5556
13 13000 -7500 1.1375e+08 0.00202083 -25.8571
14 14000 -10000 1.05e+08 0.00262963 -23.5278
15 15000 -12500 9.375e+07 0.00318287 -20.6163
16 16000 -15000 8e+07 0.00366667 -17.1852
17 17000 -17500 6.375e+07 0.00406713 -13.3108
18 18000 -20000 4.5e+07 0.00437037 -9.08333
19 19000 -22500 2.375e+07 0.0045625 -4.60706
20 20000 -25000 0 0.00462963 -1.11111e-14
----------------------------------------------------------------------
Program ended with exit code: 0 |
Last edited on
closed account (48T7M4Gy)
It looks like we get the same answers so that's good.
I was also interested to see how the intermediate slope/deflection results compare where the slope is changing rapidly. Thanks for that.
FWIW:
1. I lean towards the dimensionless aspect because internal hard-coded conversions become a nightmare. There is a simple way to overcome it and get the user to select the units (force (mass and g maybe), rotation and length) and have a function or class to sort it all out and report the units and their derivatives. At least SI units aren't as big a problem with this as Imperial units.
2. The reason for the -ve deflection values is when you plot them they describe naturally the deflection mode for a horizontal beam, -ve being downwards Y. Global vs local coordinates come into this in the wider picture.)
3. The code is my doing and is object oriented. A Beam is a Beam object with Load objects. It's in a state of flux and that's why I took the post down because at that stage I wanted to handle the discontinuities but I've decided it's not worth the trouble - around 200-500 segments results in fairly reasonable diagrams.
4. When and if you get around to object oriented programming in C++ you'll find superposition and combined load cases are simple to implement. Influence diagrams aren't any trouble either by throwing the load into a loop, in fact it's a fairly trivial exercise once the OO work is done.
5. To show you how simple the double integration is here is the code for preparing the data for a SS beam with a general located point load. Some of it might be a bit obscure so, P_SS is a derived class of base class Load, with a constructor P_SS::P_SS(std::string aLabel, Beam aBeam, double aLoad, double aLocation)
I was also interested to see how the intermediate slope/deflection results compare where the slope is changing rapidly. Thanks for that.
FWIW:
1. I lean towards the dimensionless aspect because internal hard-coded conversions become a nightmare. There is a simple way to overcome it and get the user to select the units (force (mass and g maybe), rotation and length) and have a function or class to sort it all out and report the units and their derivatives. At least SI units aren't as big a problem with this as Imperial units.
2. The reason for the -ve deflection values is when you plot them they describe naturally the deflection mode for a horizontal beam, -ve being downwards Y. Global vs local coordinates come into this in the wider picture.)
3. The code is my doing and is object oriented. A Beam is a Beam object with Load objects. It's in a state of flux and that's why I took the post down because at that stage I wanted to handle the discontinuities but I've decided it's not worth the trouble - around 200-500 segments results in fairly reasonable diagrams.
4. When and if you get around to object oriented programming in C++ you'll find superposition and combined load cases are simple to implement. Influence diagrams aren't any trouble either by throwing the load into a loop, in fact it's a fairly trivial exercise once the OO work is done.
5. To show you how simple the double integration is here is the code for preparing the data for a SS beam with a general located point load. Some of it might be a bit obscure so, P_SS is a derived class of base class Load, with a constructor P_SS::P_SS(std::string aLabel, Beam aBeam, double aLoad, double aLocation)
|
|
Last edited on
@kemort
It's always useful if you've coded something if someone else has carried out a similar exercise and results can be compared, especially if the results agree.
I know that my C++ code it's very basic (almost BASIC-like), but I'm only a beginner at C++ and completely self-taught. I expect to make lots of mistakes and the fun is in spring them out (when I can). I shall move on to incorporating functions to cut down on the coding and also struts and possibly object programming, although I suspect that is someway off yet.
I am using the design of a simply supported beam as the vehicle to learn C++. Although it is some years (about 30) since structural design was my profession, you don't forget the basics. I come from the era when design was using pen, paper and a slide rule. So my aim is to follow a similar solution by computer with the output produced to reflect this. I did this in the early 1980s when I worked for an Architect. It was written in Amstrad Basic and was a real spaghetti effort, but it worked beautifully and saved the Architect a small fortune in Structural Engineer's fees (I was employed as a Project Manager, not structural designer). The beauty of it was that because the calculations submitted to the local authorities could have been hand written in the way that they were presented, it was not necessary to receive prior approval out the computer program (which could take weeks).
Thank you for the code you submitted. I'll study that.
It's always useful if you've coded something if someone else has carried out a similar exercise and results can be compared, especially if the results agree.
I know that my C++ code it's very basic (almost BASIC-like), but I'm only a beginner at C++ and completely self-taught. I expect to make lots of mistakes and the fun is in spring them out (when I can). I shall move on to incorporating functions to cut down on the coding and also struts and possibly object programming, although I suspect that is someway off yet.
I am using the design of a simply supported beam as the vehicle to learn C++. Although it is some years (about 30) since structural design was my profession, you don't forget the basics. I come from the era when design was using pen, paper and a slide rule. So my aim is to follow a similar solution by computer with the output produced to reflect this. I did this in the early 1980s when I worked for an Architect. It was written in Amstrad Basic and was a real spaghetti effort, but it worked beautifully and saved the Architect a small fortune in Structural Engineer's fees (I was employed as a Project Manager, not structural designer). The beauty of it was that because the calculations submitted to the local authorities could have been hand written in the way that they were presented, it was not necessary to receive prior approval out the computer program (which could take weeks).
Thank you for the code you submitted. I'll study that.
@kemort
I can see your reasoning for using dimensionless figures, but I find it much easier to follow the kN, kM, m, mm route. I find it easier to understand, particularly when debugging. But we'll have to differ on this. Everybody is right and everybody is wiping, depending on how you view things.
I can see your reasoning for using dimensionless figures, but I find it much easier to follow the kN, kM, m, mm route. I find it easier to understand, particularly when debugging. But we'll have to differ on this. Everybody is right and everybody is wiping, depending on how you view things.
closed account (48T7M4Gy)
@RNBW
Yep, we're all different.
Don't feel under any sort of obligation to use the OO (or any) approach. The learning curve isn't all that daunting even though there are plenty of twists and turns. The tutorial on this site is good: http://www.cplusplus.com/doc/
Where the power in it is and along with operator overloading and polymorphism which are concepts worth finding out about the process of superposition and factored loads are rendered very simply.
Good to hear from you again with your endeavors and best of luck with it.
Yep, we're all different.
Don't feel under any sort of obligation to use the OO (or any) approach. The learning curve isn't all that daunting even though there are plenty of twists and turns. The tutorial on this site is good: http://www.cplusplus.com/doc/
Where the power in it is and along with operator overloading and polymorphism which are concepts worth finding out about the process of superposition and factored loads are rendered very simply.
Good to hear from you again with your endeavors and best of luck with it.
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Cases as for @kemort's last output.
Note that I am using the opposite sign convention for shear force, slope and deflection - sorry!
Note that I am using the opposite sign convention for shear force, slope and deflection - sorry!
Beam:
L: 20000
EI: 1.8e+14
Point loads:
x Value
10000 30000
Cont. loads:
xl xr wl wr
Reactions:
x Type Value
0 F 15000
20000 F 15000
Monitors:
Num x S M slope down
0 0 -15000 0 0.00416667 0
1 1000 -15000 1.5e+07 0.004125 4.15278
2 2000 -15000 3e+07 0.004 8.22222
3 3000 -15000 4.5e+07 0.00379167 12.125
4 4000 -15000 6e+07 0.0035 15.7778
5 5000 -15000 7.5e+07 0.003125 19.0972
6 6000 -15000 9e+07 0.00266667 22
7 7000 -15000 1.05e+08 0.002125 24.4028
8 8000 -15000 1.2e+08 0.0015 26.2222
9 9000 -15000 1.35e+08 0.000791667 27.375
10 10000 15000 1.5e+08 6.78168e-19 27.7778
11 11000 15000 1.35e+08 -0.000791667 27.375
12 12000 15000 1.2e+08 -0.0015 26.2222
13 13000 15000 1.05e+08 -0.002125 24.4028
14 14000 15000 9e+07 -0.00266667 22
15 15000 15000 7.5e+07 -0.003125 19.0972
16 16000 15000 6e+07 -0.0035 15.7778
17 17000 15000 4.5e+07 -0.00379167 12.125
18 18000 15000 3e+07 -0.004 8.22222
19 19000 15000 1.5e+07 -0.004125 4.15278
20 20000 0 0 -0.00416667 -1.11111e-14 |
Beam:
L: 20000
EI: 1.8e+14
Point loads:
x Value
Cont. loads:
xl xr wl wr
0 20000 2.5 2.5
Reactions:
x Type Value
0 F 25000
20000 F 25000
Monitors:
Num x S M slope down
0 0 -25000 0 0.00462963 0
1 1000 -22500 2.375e+07 0.0045625 4.60706
2 2000 -20000 4.5e+07 0.00437037 9.08333
3 3000 -17500 6.375e+07 0.00406713 13.3108
4 4000 -15000 8e+07 0.00366667 17.1852
5 5000 -12500 9.375e+07 0.00318287 20.6163
6 6000 -10000 1.05e+08 0.00262963 23.5278
7 7000 -7500 1.1375e+08 0.00202083 25.8571
8 8000 -5000 1.2e+08 0.00137037 27.5556
9 9000 -2500 1.2375e+08 0.00069213 28.5885
10 10000 0 1.25e+08 6.78168e-19 28.9352
11 11000 2500 1.2375e+08 -0.00069213 28.5885
12 12000 5000 1.2e+08 -0.00137037 27.5556
13 13000 7500 1.1375e+08 -0.00202083 25.8571
14 14000 10000 1.05e+08 -0.00262963 23.5278
15 15000 12500 9.375e+07 -0.00318287 20.6163
16 16000 15000 8e+07 -0.00366667 17.1852
17 17000 17500 6.375e+07 -0.00406713 13.3108
18 18000 20000 4.5e+07 -0.00437037 9.08333
19 19000 22500 2.375e+07 -0.0045625 4.60706
20 20000 0 0 -0.00462963 0 |
@last chance
Thanks very much for the code. It's well beyond my capabilities at the moment, but I'll run it with a few of my own examples to compare with my own code and have a good play with it. Who knows what may come out of it!
Best regards.
Thanks very much for the code. It's well beyond my capabilities at the moment, but I'll run it with a few of my own examples to compare with my own code and have a good play with it. Who knows what may come out of it!
Best regards.
closed account (48T7M4Gy)
Looks like they are the same @lastchance. Very comforting. Thanks for that.
Except when the signs are in the wrong place, whether they are -ve or +ve doesn't matter much from my point of view.
I certainly don't want to try and convince anyone that there is only one way but FWIW I'm following the convention and reasons for it in a current textbook 'Mechanics of Materials' Goodno, Gere 8th edition, Chapter 9. Even with that there are some 'discontinuities'.
I like the Macaulay/integration function @lastchance. It's 'disturbingly' simple. The pattern is obvious when you write the various equations but implementing it is a valuable step, just in avoiding algebraic errors by letting the machine work it out.
Cheers :)
Except when the signs are in the wrong place, whether they are -ve or +ve doesn't matter much from my point of view.
I certainly don't want to try and convince anyone that there is only one way but FWIW I'm following the convention and reasons for it in a current textbook 'Mechanics of Materials' Goodno, Gere 8th edition, Chapter 9. Even with that there are some 'discontinuities'.
I like the Macaulay/integration function @lastchance. It's 'disturbingly' simple. The pattern is obvious when you write the various equations but implementing it is a valuable step, just in avoiding algebraic errors by letting the machine work it out.
Cheers :)
Hi @kemort,
I'm not bothered by sign conventions either; it's a matter of personal choice. Where there's a discontinuity in shear force at reaction points I'm evaluating it on the "plus" side - that's somewhat arbitrary too (but making S go to 0 at L+ adds a check on the force balance).
In case my reactions calculator (function calc()) looks ridiculously complicated it's because I'm allowing for other than simply-supported beams. Just a third support - in the centre, say, would make it statically indeterminate and you would need conditions other than just balancing forces and moments to solve.
It was put together in a hurry from code originally written in javascript, so any errors - let me know. I need to put together some proper input / output file routines still.
I'm not bothered by sign conventions either; it's a matter of personal choice. Where there's a discontinuity in shear force at reaction points I'm evaluating it on the "plus" side - that's somewhat arbitrary too (but making S go to 0 at L+ adds a check on the force balance).
In case my reactions calculator (function calc()) looks ridiculously complicated it's because I'm allowing for other than simply-supported beams. Just a third support - in the centre, say, would make it statically indeterminate and you would need conditions other than just balancing forces and moments to solve.
It was put together in a hurry from code originally written in javascript, so any errors - let me know. I need to put together some proper input / output file routines still.
Last edited on
closed account (48T7M4Gy)
Yep I gathered end-fixity is taken into account in your program. I haven't looked at it closely but I read it as a matrix solver in that context so I didn't see anything ridiculous. :)
The discontinuities in the shear force diagram are indeed a nuisance. I found I needed 200-500 points to make the plotted diagram 'look' right. It doesn't take long to process but its annoying. The saving grace for not bothering is a design program that sizes members isn't interested in the prettiness, just discrete points. So if there are enough segments then it's probably OK to leave it at 20-50 segments.
Thanks for showing us - and if you decide to go totally OO and overload the arithmetic operators you'll be glad of Java -> C++ conversion
The discontinuities in the shear force diagram are indeed a nuisance. I found I needed 200-500 points to make the plotted diagram 'look' right. It doesn't take long to process but its annoying. The saving grace for not bothering is a design program that sizes members isn't interested in the prettiness, just discrete points. So if there are enough segments then it's probably OK to leave it at 20-50 segments.
Thanks for showing us - and if you decide to go totally OO and overload the arithmetic operators you'll be glad of Java -> C++ conversion
@kemort and @lastchance
Well I am glad I started this thread. Not only have I learned a great deal, it's also got you two talking about a subject that interests me. Maybe some others may join in also. There must be loads of structural engineers out there.
Well I am glad I started this thread. Not only have I learned a great deal, it's also got you two talking about a subject that interests me. Maybe some others may join in also. There must be loads of structural engineers out there.