Engineering question

[jeriendhal]

Problem: If a rod of cast iron fifteen inches long and one inch in diameter is braced at one end against an immovable surface and subjected to increasing pressure at the other, what would be the approximate amount of pressure in either pounds or kilograms before it either bent or snapped in two?

Comments

[feonixrift.livejournal.com]

Approximately 98Kg/mm², which would give 63,226Kg/in² or 139KPSI.

http://www.anvilfire.com/FAQs/cast_iron.htm
http://www.neiljohan.com/projects/technology/castiron.htm

[kodi]

It depends on the type of cast iron, but (assuming both ends are fixed in every dimension except the axis along which the force is applied) I'm coming up with 4,000 to 8,000 pounds.

So. It's somewhere between 4,000 pounds and 100,000 pounds, per the above comment.

[feonixrift.livejournal.com]

I'm probably wrong, what I was looking up didn't look length-dependent, yet this result should be. A case of my applying the wrong formula, I'll try again...

[kodi]

I was going off the formula for buckling of a column - I'm pretty sure that's the right formula, but I'm not totally sure I'm plugging the right constants.

[jeriendhal.livejournal.com]

Many thanks to both of you. This will be useful in my story later.

You'd be amazed what organic objects are approximitely as strong as cast iron. *evil grin*

[jvowles.livejournal.com]

You'd be amazed what organic objects are approximitely as strong as cast iron. *evil grin*

Okay, stop RIGHT NOW....now more elfpr0n for you.

[feonixrift.livejournal.com]

Yup, that's the one I'm re-trying with. Definitely looks like the right formula - depends on the right things and all, quite clearly matches the problem description. I feel like a fool for having looked at simple compressive strength first, but that's what cross-checking work is for. =)

[feonixrift.livejournal.com]

I was pretty sure I was plugging in the right constants, but I'm having a hard time interpreting the result because the units are all buggered.

Modulus of Elasticity: 13.4
Area moment of Inertia (cylinder): pi r^4 / 4 (meters^4) = pi * 0.0127^4 / 4 = 2.043171e-08
K: 0.5
Length: 0.381 meters

Is this comparable to what you were using?

[kodi]

I did everything in inches instead of meters, since the modulus of elasticity is in Mpsi, but otherwise that's what I was using.

[feonixrift.livejournal.com]

Aha! You win for spotting the units on that one, I somehow failed to notice them. Given a unitless K that would work out perfectly. Excellent. =) You win.