"Richard Smith" wrote in message news:
[email protected]...
Jim - lost previous thread.
You used that formulation for beam deflection in relation to beam
*stress*...
For a "simple beam" (like a plank bridging two points)
deflection = sigma.L^2 / 6.E.H
sigma = stress
L = beam length between the two supports
E = elastic modulus; Young's modulus
H = the height of a *symmetrical* section (this only works for
symmetrical sections)
How did you know of this formulation?
Did you derive it yourself?
Or is it known in engineering?
Broader explanation for a general audience:
The deflection vs. *stress* formulation enables you to estimate if a
beam you find is bearing an acceptable load.
With the deflection vs *force* formation you need much more
information
deflection = F.L^3 / 48.E.I
I is the second moment of area and you need every linear dimension of
beam's cross-sectional shape to calculate it.
Then how would you know what the force is?
With the *stress* formulation you can take a reasonable guess at what
the beam's yield stress is - any I-beam not otherwise marked a
reasonable guess is 275MPa, and for any Rectangular Hollow Section
(including Square Hollow Section) you can reasonably guess at 355MPa.
You can rearrange
deflection = sigma.L^2 / 6.E.H
to isolate "sigma" the stress, and given you know can easily measure
with a tape-measure and rule the beam's length, height and deflection,
and E is always around 210GPa for steel, you can get the stress.
Then evaluate - is that acceptable?
---------------------------------
I took extreme fiber stress for my guesstimate of the surplus steel or wood beam's properties and found the maximum deflection for a center load from several on-line calculators. Likewise for columns I entered measured
dimensions into on-line calculators and ensured the ends were pinned or
could swivel to avoid cantilever stresses. The only load-induced damage to retired tripod columns has been slight egg-shaping of the pin holes, and
it's about equal on both sides which supports my assumption that the asymmetrical hangers do self adjust to evenly distribute the load. The other damage was from accidents such as falling trees.
https://amesweb.info/ https://jonochshorn.com/scholarship/calculators-st/index.html
For the gantry beam's center splice I made the sum of its components moments
of inertia greater than that of the C channels, and supported the channel flanges at the splice ends to compensate for the area lost to bolt holes. I tried calculating from basic principles for practice but didn't trust my
result enough to stand under it.
Then I proof-tested the completed assembly. As far as possible the
pinned-joint geometry allowed simple calculations, no Vierendeel trusses or dependence on torque and friction, only bolt shank shear and bearing
pressure of a closely sized (match-drilled) hole.
jsw
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