Physics/Engineering/Math Problem

[Jengles]

Currently I am trying to solve for the maximum amount of force a 16" stroke of a piston could take sideways before it were to break. I'm kind of clueless where to begin.

[Ether]

Originally Posted by Jengles

Currently I am trying to solve for the maximum amount of force a 16" stroke of a piston could take sideways before it were to break. I'm kind of clueless where to begin.

Let me guess: you want to use the cylinder rod as a side-loaded latching mechanism for your kicker/catapult.

[MechEng83]

Originally Posted by Ether

Let me guess: you want to use the cylinder rod as a side-loaded latching mechanism for your kicker/catapult.

If this is actually why you're asking, our team did this on a mechanism in a previous year. I recommend not using the shaft as the locking mechanism, but thread an extension on made out of grade 8 material, then support it as closely as possible on both ends.

"break" is a loosely defined term. Typically it would mean fracture, but in the case of a rod extension from a piston, any permanent deformation (yielding) from bending, or something biting into the shaft would be detrimental to the future operation of the shaft.

If you know the bending moment on the shaft, you could use the formula sigma=M*c/I where c is the radius, and I is the second moment of area. I=pi/2*r^4. Then compare the stress value (sigma) against the yield limit of the steel in the rod extension.

[Jengles]

Originally Posted by MechEng83

If this is actually why you're asking, our team did this on a mechanism in a previous year. I recommend not using the shaft as the locking mechanism, but thread an extension on made out of grade 8 material, then support it as closely as possible on both ends.

"break" is a loosely defined term. Typically it would mean fracture, but in the case of a rod extension from a piston, any permanent deformation (yielding) from bending, or something biting into the shaft would be detrimental to the future operation of the shaft.

If you know the bending moment on the shaft, you could use the formula sigma=M*c/I where c is the radius, and I is the second moment of area. I=pi/2*r^4. Then compare the stress value (sigma) against the yield limit of the steel in the rod extension.

What is M?

[MechEng83]

Originally Posted by Jengles

What is M?

M is the aforementioned "bending moment"

The formula for calculating it will depend on your endpoint support configuration, e.g. fixed-fixed, fixed-free, fixed-pinned, pinned-pinned (pinned-free and free-free are dynamic cases, and the math becomes a whole lot more fun) Try here for a tutorial in calculating bending moment.

[Jengles]

Originally Posted by MechEng83

M is the aforementioned "bending moment"

The formula for calculating it will depend on your endpoint support configuration, e.g. fixed-fixed, fixed-free, fixed-pinned, pinned-pinned (pinned-free and free-free are dynamic cases, and the math becomes a whole lot more fun) Try here for a tutorial in calculating bending moment.

So what would the equation/set up look like for a fixed-free configuration?

[MechEng83]

Assuming the load is perpendicular to the shaft, the maximum bending moment for a fixed-free setup occurs at the fixed end and is calculated by multiplying the force by the distance from the fixed point. There is no bending moment between the load and the free end.

[Mike Schreiber]

I've attached a semi worked out solution. You should be able to plug in your numbers to these equations. I assumed the shaft is 6.35 mm or 1/4 in stainless steel. I don't know what it is actually made of but you want to look up the yield stress. Once you solve for the moment, that is equal to force applied * distance cantilevered. If you need more help just ask.

[Ether]

Originally Posted by Jengles

Currently I am trying to solve for the maximum amount of force a 16" stroke of a piston could take sideways before it were to break. I'm kind of clueless where to begin.

I think you should post a picture or hand-sketch of what you are proposing to do to your piston so that there is no miscommunication.