I posted a question about an I beam with a Moment of Inertia of 26.49.
It is a mild steel I-Beam and is 3.5" wide and 6" high. The beam
length is 38" and will have a 40,000 lb load placed in the middle of
the beam with a .5" spreader plate
where the jack will attach (This is going to be the top beam for a
hydraulic press). The deflection at load is .05". The beam will be
supported by two 60" posts that are 2.5"x2.5"x.25" thick square posts.
Someone states that this would "bend like a pretzel" under full load.
If the deflection is only .05" at maximum load how will the beam "bend
like a pretzel"? I know that the elastic yield strength of mild steel
is around 36,000 psi but I have read that most steel these days is
around the 45,000 psi. The same poster stated that the max stress will
be 43.6kpsi. I could truss the beam or could make the beam shorter in
length or could lower the hydraulic jack to a 10 ton model. I tried to
post this using the original post but was unsuccessful. Sorry for top
posting. Thanks, Steve.

I'm not a mechanical engineer but there are several steps to choosing a
beam. First, from the dimensions you calculate (or look up) the moment of
inertia, which with the modulus of elasticity gives the deflection under
some load. This you've done, and 0.05" is acceptable to you. Next you have
to calculate the shear stress in the beam under that load and compare the
stress to the material strength. In this case the material along the top of
the beam is going to be under tension and the material along the bottom of
the beam will be under compression as the beam flexes upwards from your jack
under the beam. Adjacent metal fibers will be under different amounts of
tension or compression and will want to slide past each other, or shear.
For this you need the section modulus which you also calculate or look up,
and from your post the shear stress turns out to be 43.6 ksi. I looked in
one table and found 36 ksi as the yield tension strength and 21 ksi as the
yield shear strength, plus you need a margin of safety on the order of four
times, so the shear stress has to be kept under 21 ksi/4=5.25 ksi (I said
I'm not an engineer; a real one would surely use a different safety margin
:-)). You are way over the acceptable stress limit of your beam, which
means that it's going to bend into a U before you get anywhere near your
40,000 lbs applied (not a pretzel but still fun). You've rediscovered the
rule-of-thumb that says that short beams usually fail design review in
stress, long beams in maximum deflection. Finally, the load is concentrated
at the ends and the center over a small length of the beam. You need to
treat that short length of the beam as a column and calculate the vertical
load that it will support without crumpling, and maybe add spreader plates
to apply the load over more of the beam area. All this is just for the
beam. Right now your beam will handle about 4800 lbs, so you need to either
limit the load or go to a bigger beam. Then you have to check the vertical
columns (your 2.5" square tube) for tensile strength, and the holes in the
columns and the pins that will support the beam for shear strength. Then
..., :-)

--
Regards,
Carl Ijames