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

[Jalkhov]

Feb 19, 2021 16:01:00 GMT tsh73 said:

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!

However, I want to know what you say, can there be anything that will cause me problems later on? Any comments? Or is it already done?

[tsh73]

I have no idea.
Civil engeneering is all Greek to me ;)
You can run BASIC & python side by side a fiew times, providing same numbers to both, and see if results match.
Besides, there is untested branch of an algorithm where you answer "SI".

[Jalkhov]

Feb 19, 2021 16:19:07 GMT tsh73 said:

I have no idea.
Civil engeneering is all Greek to me
You can run BASIC & python side by side a fiew times, providing same numbers to both, and see if results match.
Besides, there is untested branch of an algorithm where you answer "SI".

Yes, I managed to perfect that part, precisely when I started to test that path, errors popped up everywhere haha, but it's already solved, thank you very much. By the way, if here it is done to close the threads when they are already solved, then this one can be closed without any problem

[tsh73]

I found one more typoo, just trying to figure out what X is
(well, if (M+ = Maximum bending moment at the span), then X>M+ is X where it happens.
But what is X1, X2, X3?)

Line 430
'IF X2 < X3 THEN X4 = X + X2:X5=L(I)-(X*X3): GOTO 440
should be
IF X2 < X3 THEN X4 = X + X2:X5=L(I)-(X+X3): GOTO 440

[Jalkhov]

Hello again haha I was testing the system and it gives an error when a calculation is made where the spans are less than the overhangs, the error occurs both in the BASIC program and in the Python program, that makes me think that it is more an error by logic than by the programming language, but as I am not an expert in the field I do not know if it is right that a slab has less spans than overhangs.

[tenochtitlanuk]

I hope we can get this all a bit clearer- you are both doing well at finding what is going on.

There are indeed on-line beam-bending calculators- I found several- but don't have the time to check them out. Type in dimensions and support and load positions/size and they do a black-box calculation and graph. BUt they do not help explain HOW they calculate. Try skyciv.com/free-beam-calculator/ WE need a good civil engineer's advice!

I think I remember a beam bending example in a book called '101 problems in BASIC' or something similar. About 1980..

By chance yesterday, checking old backup files, I found the code below. I certainly didn't write it. Back in the days of LB1.4 I reckon. Not sure it will help understand what is going on, but a great example of old-style BASIC eith heavy use of goto!!

'total of 5 UDLs and point loads
'beam is 7 meters long
'distance to first load is 0
'size is 39583.3
'length is 0
'distance to second load is 0
'size is -20000
'length is 0
'distance to thrid load is 1
'size is -15000
'length is 3 (this is the only UDL)
'distance to fourth load is 6
'size is 65416.7
'length is 0
'distance to fifth point is 7
'size is -40000
'length is 0

'increments is your choice

   nomainwin

   WindowWidth = 1150
   WindowHeight=  800

   texteditor  #1.txt1, 10, 175, 350, 550
   graphicbox  #1.gb1, 385,  10, 780, 800
   textbox     #1.tb1, 250,  50, 100,  25
   statictext  #1.st1, "",   10,  55, 190, 100
   button      #1.bn,  "Next Step", [next],  ul,  10,  10, 100,  25
   button      #1.bn2, "Reset",     [reset], ul, 110,  10, 100,  25

   open "Bending Moment and Shear Force Graph Input" for graphics_nsb as #1 '   Why 'fs' when you specified width/height?

   #1 "trapclose [close]"

[resetstart]
   #1.st1 "Total number of point loads and UDLS"
   #1.tb1 "!setfocus"
   sn =1
   wait

[next]
   #1.tb1 "!setfocus"
   if sn =1 then #1.tb1 "!contents? a"
   if sn =2 then #1.tb1 "!contents? b"
   #1.tb1 "!contents? c"                                                      '   Can you do this?
   #1.tb1 ""
   if sn =1 then goto [start]                                                 '   would be much neater with 'case'
   if sn =2 then goto [length]
   if sn =3 then goto [location]
   if sn =4 then goto [size]
   if sn =5 then goto [finishlocation]
   if sn =6 then goto [table]
   wait

[start]
   dim information( a +1, 3)
   sn =2
   #1.st1 "Length of Beam (m)"
   wait

[length]
   information( 1, 1) =c
   sn =3
   #1.st1 "Length to load 1 from left hand edge in metres"                    '   sp
   x =2
   y =2
   z =2
   wait

[location]
   information( x, 1) =c
   x  =x +1
   sn =4
   #1.st1 "Size of load "; x -2; " (if the force is downwards, input a negative) in newtons"
   wait

[size]
   information( y, 2) =c
   y  =y +1
   sn =5
   #1.st1 "Length of load "; y -2; " (UDLs, enter 0 for point loads) in metres"
   wait

[finishlocation]
   information( z, 3) =c
   z =z +1
   if ( y -2) =a then sn =6 else sn =3
   if sn =3 then #1.st1 "Length to load "; z -1
   if sn =6 then #1.st1 "Increment for table of results"                   '   sp
   wait

[table]
   inc =c

[draw]
   #1.gb1 "down"
   dim y1( ( b *1000) +1)
   for zz =0 to b step 0.001                                               '   ?? step size??
       y2 =y1
       y1 =0
       zz$ =using( "######.###", zz)
       zz =val( zz$)
       if zz =0 then y1 =0

       for za =2 to a +1

           if zz >information( za, 1) then
               if information( za, 3) =0 then
                   y1 =y1 +( ( zz -information( za, 1)) *information( za, 2))
               else
                   if zz <(information( za, 1) +information( za, 3)) then
                       lengthofudl =( zz -information( za, 1))
                       forceofudl  =lengthofudl *information( za, 2)
                       distocentre =lengthofudl /2
                       y1          =y1 +distocentre *forceofudl
                   else
                       forceofudl  =information( za, 2) *information( za, 3)
                       distocentre =( zz -( information( za, 1) +information( za, 3))) +( information( za, 3) /2)
                       y1 =y1 +distocentre *forceofudl
                   end if

               end if
           end if
       next za

       if zx <y1 then zx =y1
       if y1 <xz then xz =y1

       y1( zz *1000) =y1

       if zzz =zz then
           #1.txt1 zz; "m     "; y1
           zzz =zzz +inc
       end if

       if ( (y1<0) AND ( y2>0) AND ( zz>0) AND ( zz<b)) then           '   be careful over operator priority
           #1.txt1 ""
           #1.txt1 "Point of Contraflexture:    "; zz; "m"
           #1.txt1 ""
       end if

   next zz

   yx =zx
   xy =xz
   xy =0 -xy
   if yx >xy then max =yx
   if xy >yx then max =xy
   #1.txt1 ""
   #1.txt1 "Largest bending moment: "; max; "Nm"
   #1.txt1 "Largest sag: "; yx
   #1.txt1 "Largest hog: "; xy

   yscale =150/ max


   #1.gb1, "line 50 175 680 175"
   if xz >-0.000000000001 then #1.gb1 "line 50 25 50 175"
   if xz <-0.000000000001 then #1.gb1, "line 50 25 50 350"

   for zz =0 to b step 0.001
       xscale= 630 /b
       #1.gb1 "set "; ( zz *xscale) +50; " "; 175 -( y1( zz *1000) *yscale)
   next zz


'===============================================================================================================

[shearforce]
   dim y3( (b *1000) +1)
   for aa =0 to b +0.001 step 0.001
       aa$ =using( "######.###", aa)
       aa  =val( aa$)
       for ab =2 to a +1
           if information( ab, 1) =aa then
               if information( ab, 3) =0  then
                   y3=y3+information( ab, 2)
               else
                   if information(ab,1) +information(ab,3)<aa then
                       y3=y3+(information(ab,2)*information(ab,3))
                   else
                       y3=y3+((aa-information(ab,1))*information(ab,2))
                   end if
               end if
           end if
       next ab

       y3( aa *1000) =y3
       if zx1 <y3 then zx1 =y3
       if y3 <xz1 then xz1 =y3
   next aa
   #1.gb1 "line 50 525 680 525"
   if xz1 >-0.000000000001 then #1.gb1 "line 50 375 50 525"
   if xz1 <-0.000000000001 then #1.gb1 "line 50 375 50 675"
   if xz1 <0 then xz1 =0 -xz1
   if zx1 >xz1 then max =zx1
   if xz1 >zx1 then max =xz1
   yscale =150/max



   #1.gb1 "up"
   #1.gb1 "goto 50 525"
   #1.gb1 "down"

   for zza =0 to b +0.001 step 0.001  '   ?? stepsize??
       #1.gb1, "goto "; ( zza *xscale) +50; " "; 525 -( y3( zza *1000) *yscale)
       #1.gb1, "set ";  ( zza *xscale) +50; " "; 525 -( y3( zza *1000) *yscale)
       #1.txt1, y3( zza *1000); "         "; zza
   next zza


'===============================================================================================================

   #1.gb1 "flush"
   wait

[reset]
   #1.txt1, "!cls"
   #1.gb1, "cls"
   goto [resetstart]
   wait

[close]
   close #1
   end



I don't want to 'muddy the waters', but it shows how easy it is to give a GUI front end. What a pity the program wasn't written by someone who'd learned the hard way to 'self document' the code. Relevant file names and remark/comments make it so much easier to see what is going on in code. I try to write code that way- at least partly because early exposure to Forth made me realise that some languages use code which is one-way- understand an algorithm and you can almost always code it in Forth. But read someone lse's Forth code and you will really struggle to understand it!