Can i make questions about BASIC (no Just BASIC) here?

[Jalkhov]

I have this:

A = Distance from the left end to the point charge
W = Uniformly linear charge
P = Point load
L = Beam span (length between supports)
V = Shear force
B = Distance from the right end to the point load
M = Bending moment
M+ = Maximum bending moment at the span
I = Inertia

[tsh73]

W = Uniformly linear charge

is it like distributed force?

So graph below is bending moment(x)?

[Jalkhov]

Feb 18, 2021 20:46:50 GMT tsh73 said:

W = Uniformly linear charge

is it like distributed force?

Yes

So graph below is bending moment(x)?

Yes...I think so, I'm 90% sure. Let me ask to the enginner.

[Jalkhov]

So graph below is bending moment(x)?

Indeed, the graph reflects the bending moment.

[Jalkhov]

One question, when I do this
C = M
When C is a multidimensional array, does it replace all elements in the array or does it become another type of variable?

It is that when converting the following line of code to python:

K1=C(I,I+1)-M
C=M
K2 = 0-C(I+1,I)-M
I get error, because with:

C=M
The C value is replaced with a float variable, so it cannot be used as an array, how is it that in BASIC it does not throw this error? By the way, I'm doing fine with the Python translation  all the code runs fine until the present problem. Thank you very much to everyone, I know we are not finished yet but I can't stop thanking those who helped me.

[tsh73]

Yes, for BASIC names C and C() is entirely different.
In Python you will end up with new name for ordinary variable to distinguish it from list.

[tsh73]

YES!!!!!
I fixed it!
It is line 220 of source code - see it fixed
One small typoo...
   FOR I=1 TO N
       C(I,I+1)=(W(I) * L(I) ^ 2 / 12 + P(I) * A(I) * B(I) ^ 2 / L(I) ^ 2) * -1
       'C(I+1,1)= W(I) * L(I) ^ 2 / 12 + P(I) * B(I) * A(I) ^ 2 / L(I) ^ 2
       C(I+1,I)= W(I) * L(I) ^ 2 / 12 + P(I) * B(I) * A(I) ^ 2 / L(I) ^ 2
   NEXT I
results
M1>0=0
M1>2=0
M2>1=5.5
M2>3=-5.5
M3>2=0
M3>4=0

::::: TRAMO 1 :::::
V1>D: 3.62
V2>I: -6.37
M+: 3.15
X>1: 0
X>2: 1.73
X>M+: 1.44

::::: TRAMO 2 :::::
V2>D: 6.37
V3>I: -3.62
M+: 3.15
X>2: 0.98
X>3: 0.3


[Jalkhov]

I must say that you have done a great job, you simply have to see that when you really know something, there are no barriers, thank you very much, I think that now I have to finish converting to Python as a job. I will be reporting how it goes. It's funny, I came to this forum out of necessity but I admit that I'm really liking BASIC, especially this forum, people are very nice and receptive

[Jalkhov]

Well, this is the output I get from Python hahaha, but hey, I guess that's where to start, at least it runs without errors, but there are a LOT of details to be fixed


M1>00 -3
M1>00 -3
M2>15 -3
M2>1-5 -3
:::::TRAMO 1 :::::
V1>D: 0.0
V2>I: -0.0
M+: 0.0
X>1: 0.0
X>2: 0.0
X>M+: 0.0
:::::TRAMO 2 :::::
V2>D: 0.0
V3>I: -0.0
M+: 0.0
X>2: 0.0
X>3: -0.0
X>M+: 0.0
[Finished in 1.5s]

[Jalkhov]

I have discovered several things, the round in Python does not work the same as in BASIC, when I use it, it takes everything to 0, but I removed it in certain parts and now it shows me the values that I have in the round.

M1>0: 0.0
M1>2: 5.245208740234375e-06
M2>1: 5.5
M2>3: -5.50000262260437
M3>2: 0.0
M3>4: 0.0

:::::TRAMO 1 :::::
V1>D: 0.0
V2>I: -0.0
M+: 0.0
X>1: -0.0
X>2: 0.0
X>M+: 0.0

:::::TRAMO 2 :::::
V2>D: 0.0
V3>I: -0.0
M+: 0.0
X>2: 0.0
X>3: 0.0
X>M+: 0.0
[Finished in 0.5s]
But now I have a problem and that is that M1>2 has a value that it should not have. I was scanning and that number is stored in position 1,2 of the array C, I really have no idea how it is getting there.

[tsh73]

Consider it to be "machine 0"
Actually JB program has " 0.52452087e-5 "
it is because if ROUND it is shown as 0.
(and then I increased loop by P from 10 to 20, it went down to
0.50022209e-11
)

Using ROUND is pretty straightforward in Python, see:
round(10000/3,2)
3333.33

the round in Python does not work the same as in BASIC, when I use it, it takes everything to 0,

well, it's this particular calculator ROUND.
It has digital positions negated (*-1), it IS weird looking now, but the thing (calculator) sure was very cool at the time.

[Jalkhov]

Ok, I got really messed up, I'm very very bad at mathematical terms

[Jalkhov]

I eliminated the rounding in the result outputs and this is almost ready! So I think I will eliminate them for now. I just have to solve the M1>2 issue

M1>0: 0.0
M1>2: 5.245208740234375e-06
M2>1: 5.5
M2>3: -5.50000262260437
M3>2: 0.0
M3>4: 0.0

::::: TRAMO 1 :::::
V1>D: 3.624998688697815
V2>I: -6.375001311302185
M+: 0.0
X>1: -1.2507569322472989e-06
X>2: 1.7346299757758223
X>M+: 0.0

::::: TRAMO 2 :::::
V2>D: 6.3750006556510925
V3>I: -3.6249993443489075
M+: 0.0
X>2: 0.9876765559126965
X>3: 0.30250028848654953
X>M+: 0.0
[Finished in 0.4s]

[Jalkhov]

READY READY READY READY

This is Python Output:

M1>0: 0.0
M1>2: 0.0
M2>1: 5.5
M2>3: -5.5
M3>2: 0.0
M3>4: 0.0

:::::TRAMO 1 :::::
V1>D: 3.62
V2>I: -6.38
M+: 3.15
X>1: -0.0
X>2: 1.73
X>M+: 1.45

:::::TRAMO 2 :::::
V2>D: 6.38
V3>I: -3.62
M+: 3.15
X>2: 0.99
X>3: 0.3
X>M+: 2.55
[Finished in 0.4s]
I just used the round in this way round(value, 2) The only thing I see is that in X>1 the 0 is negative, does it matter?

[tsh73]

again

it is not an issue - it supposed to work like that
it is way to write really small numbers for computer
5.245208740234375e-06 stands for 5.245208740234375/10^6, that is, 0.0000052452(...)
This program works by repeatedly (loop by P) executing same computation
During that some elements of C() gradually going to 0
Since it is already 0.0000052452, round to two digits shows 0.00

So "X>1" being XXe-6, actually pretty close to 0 too.

---

Glad you make it! :)