0
$\begingroup$

I'm designing a machine that can be simplified as a log splitter. It will have a long, central beam that will resist both the forces of a force pushing out and the moment that will be caused by the force being resisted a distance away from the centerline of the beam. For simplicity, the two sides will have equal/opposite loads and moments. enter image description here

It has been a few decades since I was in engineering school, so I want to check my assumptions:

  • As far as beam deflection, the tensile force pulling the beam apart can be ignored
  • As far as beam deflection, the important load is the torque/moment being applied by the force times the distance from the centerline.
  • To use standard beam deflection formulas, I can model half the beam as a fixed cantilever beam with an end moment, because the other half will act as the "wall" the cantilever beam is attached to.
  • For maximum tensile load, I would use the tensile load from the force as if it was applied to the center-line, and then add the forces caused by the moment.

Assuming those assumptions are correct, I only have one remaining question: if the cross-section of the beam is not symmetrical top to bottom, how do I determine what the centerline of the beam is to determine the moment applied? (Assuming the stand offs are perfectly rigid)

Improve this question
$\endgroup$
1
  • $\begingroup$ Really not sure why your log splitter will be any different from any other. One might reference existing designs. Your design will fail by compression inwards and towards the word "Load". The lower part of the beam is in tension, the upper part in compression. Seems tailor made for a truss structure or a good strong steel I beam. Off course, the log will "fail" first, splitting along its grain. $\endgroup$
    – Robert DiGiovanni
    Commented Mar 7, 2023 at 21:34

2 Answers 2

Reset to default
1
$\begingroup$

You are correct in assuming that the deflection of a cantilever beam with half length is the same as a beam under moment at both ends.

To find the neutral axis to calculate the I of the beam, we set up a reference horizontal axis at some point, say at the bottom of the beam section, x axis.

We break the beam into sections, A1 A2 ..Ai. Each section has its neutral axix at its CG and we call the distance from this CGi to the X axis Yi.

The distance from the neutral axix to our X axis is the sum of the product of section areas, Ai by there own neutral axis distance from the datum, Yi, divided by the total area of the beam cross section.

$$Y_{neut axis}= \frac{\Sigma Ai*Yi}{\Sigma Ai}$$

The I of the section is then the sum of the Is of these sections about the neutral axis, with parallel axis law added to their individual I.

$$I= \Sigma Ai*Yi^2+ \Sigma I_i$$

Improve this answer
$\endgroup$
1
$\begingroup$

If by "centerline" you mean neutral axis in case of pure bending, it should simply go through the cross-section centroid.

Improve this answer
$\endgroup$
2
  • $\begingroup$ The person who downvoted this answer shall review his understanding of this question/subject. $\endgroup$
    – r13
    Commented Aug 4, 2023 at 22:42
  • $\begingroup$ The downvote may be because the axial force moves the neutral axis away from the centroid, strictly speaking. $\endgroup$
    – Greg Locock
    Commented Aug 5, 2023 at 2:13

Not the answer you're looking for? Browse other questions tagged or ask your own question.