Re: ATM truss bending calculations (long)

["Albert Highe" <ahighe@ix.netcom.com>]

There have been numerous postings on truss bending calculations. In
particular there was an long posting back in February I want to comment on.

In that posting, two equations were given with some discussion for their
use:

Equation 1 is the amount of deflection in a cantilevered beam
        y-beam =  ( w-beam * L^3 ) / (  3 *  I  *  E )

Equation 2 is the amount of deflection in a truss
y-truss  =  ( w-truss * L^3 ) /  ( c^2 * A * E ),
with:
        w-beam = equivalent weight at end of each beam,
        w-truss =  one-half of total equivalent weight at top end of truss,
           L       =  free-length between ends of beams or truss,
           I        =  moment of inertia of cross-section of each beam,
          A       =  cross-sectional area of each truss member,
          c        =  effective separation of truss members at base,
          E       =  modulus of elasticity for truss/beam material.

It showed some sample calculations comparing the deflections for each:

        OD              y-truss           y-beam
        -----           ----------                -----------
        1/2             0.0045          5 inches (!!)
        3/4             0.0028          1.33 inches
         1              0.0021          0.53 inches
        1.5             0.0014          0.15 inches
         2              0.0010          0.06 inches

The analysis is excellent if you want to compare why a truss is much better
than supporting your upper cage assembly using only a single beam (for
example, one tube). However, no one would build such a telescope, and I
think a reader could draw the wrong conclusions about other telescope
support structures. In particular, a reader might infer that an Alice-style
telescope (two parallel tubes) will have large deflections, similar to a
single cantilevered tube.

In fact, the Alice-style support structure is a truss as well, and has
exceptional stiffness in one axis. Ron Ravneberg, creator of Alice, should
be credited with recognizing that the greatest forces in a dobsonian
telescope are vertical.  The common angled 8-tube truss design, which is
excellent for resisting forces in all directions, is overkill. A simple two
tube truss design is all that is required to support the vertical forces.
Just as with angled strut trusses, this scope design can use very small
tubes. Why it doesn't is because the forces in lateral (side-to-side)
motions are not zero. For lateral motions, the cantilevered beam equation is
appropriate. However, more on specific truss designs later.

There have been a number of questions and comments about why the truss
design uses cross sectional area in its formula instead of moment of
inertia.

Moment of inertia, I, is given by:

 I = k* M *R**2

k is a constant for a particular geometry, and M is the mass at a distance R
from the axis of rotation.  For tubing, the moment of inertia for rotation
about its long axis is:

 I = 1/2 * M * (R1**2 - R2**2)

where R1 is the outer radius and R2 is the inner radius.

For thin walled tubing, where the radius is much greater than wall
thickness, this simplifies to:

 I = M * R**2.

What this means is that the moment of inertia increases when you get the
mass out further from the center of rotation.

It turns out that resistance to bending (stiffness) is directly related to
the moment of inertia as shown in Equation 1.

The key point to remember is that stiffness increases linearly with mass
(deflection decreases inversely proportional to mass), but stiffness
increases with the square of the distance of that mass from the axis of
rotation. A tube with twice the wall thickness, but with the same diameter,
will be approximately twice as stiff. A tube with twice the diameter, but
with the same mass, will be four times as stiff.  This agrees with our
experience.

In the Alice-style telescope, two tubes are spaced apart with an average
separation "C". As long as the separation is large compared to the diameter
of the tube (or any other shape) you can substitute the average separation C
for the integral over all R's. Consequently, you get the simple equation for
the moment of inertia of the two tube truss:

 I = k * M * C**2

where k is some geometrical constant, M is the mass, and C is the spacing
between the centers. Since the mass is proportional to the cross sectional
area,

 I = k * A * C**2

where k is a new constant, A is the cross sectional area, and C is the
spacing.

Now substitute this equation for I in Equation 1, and you have something
that, except for some geometrical constants, is the same as Equation 2. The
formula for trusses isn't so mysterious. It is just a derivation of the
formula for bending of a cantilevered beam. This should also answer some of
the questions about why the cantilevered beam equation uses moment of
inertia and the truss equation uses cross-sectional area. The moment of
inertia for a truss has already been calculated and substituted into the
bending equation.

I also have a comment on another common area of confusion. People often
think about bending stiffness of individual tubes, but describe the function
of trusses as withstanding tension or compression, instead of resisting
bending. In fact, bending, tension, and compression, are interrelated. When
a tube, or any other structure, is bent, one half is in tension and the
opposing half is in compression. When bending a solid rod downward, the
upper half is in tension, and the bottom half is in compression. Curvature
(bending) is introduced to the entire structure. What we do in constructing
lightweight trusses is to remove all the material between the surfaces. (To
reiterate, you get more mileage out of your mass if you can get it as far
away from the centerline as possible.) Since the elements of the truss are
separated into distinct pieces, its convenient to think of each member as
only supporting tension or compression .  However, the entire structure, and
its individual members, still do bend. If we applied the conventional
statement about truss tension and compression to a bent rod, we could state
that the a rod doesn't resist bending, its just that the upper half resists
tension, and the bottom half resists compression. Kind of true, but not the
whole story.

Now getting back to Alice and beyond.
The key to making a stiff structure with low weight is to separate the mass
as far as possible from the center of rotation. In the vertical plane,
Alice-type scopes will be nearly as stiff as angled strut trusses using
small diameter tubing. However, for lateral motions, the stiffness will be
the sum of the stiffness of the individual cantilevered tubes. Consequently,
the two tubes are larger than used for angled strut trusses.  But these don'
t need to be too large, because there is no weight being supported in the
lateral direction. The tubes just should not bend too much when force is
applied to move the scope.

If you wanted to make the scope stiffer in the lateral direction but not use
larger diameter tubes, you could add a third tube. Imagine a three sided box
where the tubes run along three edges. Even stiffer is a four-sided box with
four tubes along each edge (imagine a box kite).  However, there are other
degrees of freedom (other deflections that are possible). The box kite
structure can be twisted. So, cross members are added to resist these
degrees of freedom. If you add cross members across each face, you now have
a very rigid structure, but with a total of 12 struts. But I think you can
now see that the eight-tube angled truss is a simplification and logical
extension of this exercise. 12 is an unnecessarily high number of tubes to
resist the bending and twisting motions in all directions (all the degrees
of freedom). The simplest case is to use six angled struts (usually tubes),
spaced apart at 120 degrees. However, since most ATM dobs are built with
four-sided boxes, its more convenient to make the next step in complexity to
eight angled struts in the truss.

In summary, the conventional dob truss with eight angled struts is just
that - a convention. It is not a design necessity depending on the
magnitudes and directions of the anticipated forces. I'd recommend a six or
eight truss design for a rotating tube where the orientation of
gravitational forces changes. However, the forces on a non-rotating tube dob
are much simpler, and the eight strut truss is overkill. Many simpler
trusses work just fine (I've built quite a few) and a scope with fewer
pieces is generally much easier to set up in the dark

Albert.