Jump to content

2 Engineering puzzles


ftpiercecracker1
 Share

Recommended Posts

Problem 1.

 

What size box steel would I need in order to support 300lbs, extended out 7ft, without experiencing more than a 1/4in of deflection? One end will be unsupported, the other securely mounted to a flat surface.

 

 

I have done some research, but the online steel manufacturers do not provide the necessary information to solve this problem. I need the second moment of inertia and the yield strength of a given piece of steel. If you know a site off hand that does provide such info let me know.

 

 

 

 

 

Problem 2.

 

I bought a 3 blade deck to put on a mower project I'm working on. But the blades are directly in line with one another. All decks with three blades that I have ever seen have a staggered blade arrangement, where the middle blade is the leading blade.

 

How can a deck with blades directly inline with one another cut grass without leaving a small line of uncut grass in between the blades? 

 

On a staggered 3 blade deck the diameter of the two side blades fall within the leading blades diameter thus leaving no area uncovered by a cutting edge.

 

The leading blade is far enough forward that the side blades and the leading blade do not contact.

 

 

Link to comment
Share on other sites

Problem one:  wow, having some weird flashbacks to some of my engineering classes!  but it seems like too many of those particular brain cells died years ago. :(  

 

Problem B:  I'm guessing the grass gets buffeted around a lot and chances are it'll get nipped by one or the other?  how wide is the gap?

Link to comment
Share on other sites

1 hour ago, Jeep Driver said:

The only decks I'm aware of with blades in line are 'timed' with a cog belt.  And they will hit each other if out of time. 

 

The other- my guess would be 2x6 .25 wall stood up. 

 

A cogged belt would make sense, but this deck is like any other and uses a traditional V-belt.

 

 

I was thinking 4x6, but it's all for not if I can't find the aforementioned data.

 

 

 

1 hour ago, Pete M said:

Problem one:  wow, having some weird flashbacks to some of my engineering classes!  but it seems like too many of those particular brain cells died years ago. :(  

 

Problem B:  I'm guessing the grass gets buffeted around a lot and chances are it'll get nipped by one or the other?  how wide is the gap?

 

 

Gap is unknown. The width of the blades will determine how close they can get to one another. But I don't have blades for it yet. Can't be much more than 1/8th of an inch right? I'm wondering if this was one of the first attempts to have a three blade mower deck and the engineers didnt realize the blades had to be staggered until after production.

Link to comment
Share on other sites

1) I think I had this question on a test or two! This is a classic cantilever beam problem.

 

If the load is at the very end, max deflection =  FL³/3EI where F equals applied load, L equals length, E is the modulus of elasticity, and I is the area moment of inertia. Technically a negative number because your load is pointing down. Plugging in 0.25" for deflection, I get a minimum Ixx of 18.4 in^4.  1/4" 4x6 stood up would be suitable for this. Even with 2x6, yield won't be a concern (deflection will be too much though).

 

If it's somewhere in the middle, deflection = Fa²/6EI all times (a-3L) where a equals the distance from the supported end to the load, everything else is same as before. Max deflection will always be at the unsupported end so no need to solve for its location. Set it equal to 0.25 inches, solve for I.

 

To find the moment of inertia, you can either solve a rather unwieldy equation on paper or just look it up for HSS here. Also gives you the section modulus for stress calculations (stress equals bending moment divided by section modulus - max stress is at the supported end.)

 

For steel, a good round number for E is 207 gigpascals, or 30,000,000 psi. Just a general number that will give you a good idea. If you can find out what grade your steel is I might be able to look it up in my books and tell you what it really is but I think we may have gone too far with the math already :)

 

edited because I had a math error originally

 

 

Link to comment
Share on other sites

Thanks guys, especially Minuit.

 

I have not seen that equation before. During my research this site provided the equation i was going to use. Would it have also worked?

 

https://process.arts.ac.uk/content/calculating-point-which-piece-box-section-steel-will-bend-and-not-recover/index.html

 

 

Thanks for the links too.

 

FPC.

Link to comment
Share on other sites

5 hours ago, ftpiercecracker1 said:

Thanks guys, especially Minuit.

 

I have not seen that equation before. During my research this site provided the equation i was going to use. Would it have also worked?

 

https://process.arts.ac.uk/content/calculating-point-which-piece-box-section-steel-will-bend-and-not-recover/index.html

 

 

Thanks for the links too.

 

FPC.

 

There are two types of deflection you can think about here - elastic, and plastic. Elastic deformation is not permanent - remove the load and your structure will snap back to its original shape (not counting stuff like fatigue and creep). Plastic deformation is permanent damage, and we want to avoid it when designing a structure. Most materials will have some ability to deform elastically, but when you deform them past their elastic limit they will start to be plastically deformed. The amount of internal stress when plastic deformation (and thus, permanent damage) is noticeable is called the yield strength. That's why torque-to-yield bolts aren't reusable - in the act of tightening them, we have actually permanently stretched the bolt.

 

The equation I used above was to determine what size of box section you needed to have less than a quarter inch of elastic deflection at the very end. What they're talking about in your article is a different question - in a nutshell, "how far can you stand down the length of this box section before it bends permanently?". If you plugged your numbers into that equation, you could find out whether or not your box section would yield. In this case, you'd want to avoid yield, so that would be valuable information too. You'd need my equations to know how much it would bend (and they only work in cases of elastic deformation).

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

 Share

×
×
  • Create New...