FPCP display of beams is incorrect - cut is offset
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2008-09-24
08:03 PM
- last edited on
2023-05-23
03:08 PM
by
Rubia Torres
I did an experiment using the column, beam, roof, and wall tools
and came up with something that is, to me, quite strange.
Please see attachment. Why does the beam tool show the
FPCP cut in a different location than in the other three tools ?
The thickness off all elements is the same, one foot.
Note that the displacement is exactly the difference between
the length of the hypotenuse and the length of the opposite side
of a right triangle drawn in the manner shown in the attachment.
I think that the cut is shown correctly for three of the tools but
is incorrect for the beam tool. If I am missing something
please let me know.
Thank you,
Peter Devlin

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2008-09-24 11:57 PM
Karl

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2008-09-25 12:02 AM
Peter wrote:Thanks, Peter!
I noticed this in your other image to and assumed it
was a coincidence .... What angle were you using ?


Cheers,
Karl
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2008-09-25 01:58 AM
There is clearly a pattern here and I think a precise one.
If I wasn't so terrible at math I could probably figure out
exactly what it is and derive a formula predicting this
effect for any given angle. Some of the math types
on this forum have probably already done so.
In the case of 45°, if the beam is elevated exactly
4 31/32" then the cut rectangle aligns exactly with
the position of the cut rectangles of the other three
tool elements. For 45°... of course !!
Thanks,
Peter Devlin
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2008-09-25 11:33 AM
The problem is that the two reference lines are not coincident in 3d...
To achieve this you have to calculate the elevation of the bean using the formula:
Elevation= - (Height of the section of the bean / COS(Angle of the bean))
But I think you can do it graphically in section or 3d... I didn't try... Did you?...

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2008-09-25 01:33 PM
AC V6 to V18 - RVT V11 to V16

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2008-09-25 02:02 PM
Cheers,
Karl
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2008-09-25 03:16 PM
Yes I did do it graphically in Section/Elevation.
See the diagram with my first post. Draw two
parallel lines, separated by the the depth of the
beam at the slope of the beam, draw a horizontal
line which is the cut plane, draw line perpendicular
to the two lines between the lines at the intersection
point of the upper line and the horizontal line, measure
the length of the perpendicular line, and subtract it from
the length of the cut plane line. The difference is the
displacement of the cut plane rectangle from where
it should be.
EDIT:
The formula for the displacement was starring me in the face.
displacement = (beam depth/sin(slope)) - beam depth
Thanks,
Peter Devlin

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2008-09-25 05:35 PM
Good work. You might want to rename the thread so it is more likely to show up in a search for this bug.
David
www.davidmaudlin.com
Digital Architecture
AC28 USA • Mac mini M4 Pro OSX15 | 64 gb ram • MacBook Pro M3 Pro | 36 gb ram OSX14
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2008-09-25 07:44 PM
Peter wrote:And...
Displacement = (beam depth/sin(slope)) - beam depth
Displacement = (beam depth/sin(slope)) - beam depth = beam depth/cos(slope)
That's why I like mathematics... It always tells us the truth...

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2008-09-25 08:09 PM
Truth is power.
Cheers,
Peter Devlin