Help--Math problem

[Chet 02]

Lol Tom...I always liked that one.



Rob...I believe you are correct. The figure Bill came up with would be the change in strain from 6" to 7". But then again, I haven't done anything like this in years. The only calculations I use now are in the operations of a nuclear power plant....so their not really very important...:blink::huh:

 

[Bill V]

RobT--Good point. I was assuming that the 6" ball was unstrained, but that was not called out in the problem, so it shouldn't have been assumed.

 

[LaRue Medlin]

1+1=3 (if you are not carefull)

 

[Jake Levin]

more mathematics



IF A B C D E F G H I J K L M N O P Q R S T U V W X Y Z IS REPRESENTED AS:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26.



THEN.......H-A-R-D-W-O-R-K...8+1+18+4+23+15+18+11=98%



AND....K-N-O-W-L-E-D-G-E.....11+14+15+23+12+5+4+7+5=96%



BUT.....A-T-T-I-T-U-D-E.....1+20+20+9+20+21+4+5=100%



AND....B-U-L-L-S-H-I-T.....2+21+12+12+19+8+9+20=103%



AND LOOK HOW FAR ASS KISSING WILL TAKE YOU..



A-S-S-K-I-S-S-I-N-G......1+19+19+11+9+19+19+9+14+7=118%



SO, ONE CAN THEN CONCLUDE WITH MATHEMATICAL CERTAINTY THAT WHILE HARD WORK AND KNOWLEDGE WILL GET YOU CLOSE, AND ATTITUDE WILL GET YOU THERE, ******** AND ASSKISSING WILL PUT YOU OVER THE TOP

 

[Jeff C]

Rob, the strain is not needed.

 

[Bill V]

Jeff, if my approach to the rubber ball problem is correct (please let me know if it isn't), then actually, yes, the original strain is needed. If the ball's nominal size is 6", and is then expanded to 7", the strain is 49/36 - 100% = 36.1%. But if the person asking the question has a ball that is currently 6" in diameter, but which is already strained from a smaller nominal diameter (for example, 5"), and asks what the strain will be if it is enlarged to 7", the current 6" size is irrelevant--the strain is 49/25 - 100% = 96%. The INCREASE in strain (from the 6" state) remains 36.1%, but that's not the actual strain.



Yeah, I know, we're getting caught up on technicalities. Sue me--I'm an engineer, technicalities are my job. This reminds me of the logic class I had back in college. About a third of the way through the semester, we had an exam, on which there was an extra credit question. It stated a logic problem whose solution required the use of techniques that we hadn't yet learned, techniques that were going to be covered late in the semester. The problem ended with the statement, "Based on this information, can you determine the length of the rope?" People who knew how to determine the length and wrote down the entire derivation, despite the fact that the material hadn't been covered, only received partial credit for their work. Very few students received full credit, they were the ones who actually answered the question that was asked. And if you read the question again, you'll see that the only possible correct answers are "Yes" and "No".

 

[Jeff C]

Well, as we are learning, average normal strain is length two, minus length one, all divided by length one. (7-6)/6 = .167 in/in

 

[Rob T]

Jeff C,



I will tell you that you are wrong and if your instructor is telling you this he is wrong. You are right in saying the strain is the (L2-L1)/L1 ...but ONLY if L1 is an undeformed length. If L1 is different than say L0 which is the undeformed, unstrained length you will need to know the strain on L1.



Without stating L1 is an undeformed, unstrained length you WILL need to know the strain on L1 to calculated the strain on L2.

 

[Jeff C]

L1 is undeformed. That is a question straight out of our textbook.

 

[Bill V]

Also, even if you assume that the 6" size is nominal, the equation only uses length if it's a one-dimensional strain, like if you were to take a piece of rubber and stretch it ONLY left-to-right. But the ball problem is stretching it in two directions. Therefore, it's not a length equation (L2-L1)/L1, but an area equation (A2-A1)/A1, as every section of the surface of that ball is being stretched in two dimensions. The areas A2 and A1 are the surface areas of the ball, which are pi*d^2. The pi's all cancel, and you're left with an equation of (D2^2 - D1^2)/D1^2--aka the answer I gave you previously.



Is the (L2-L1)/L1 the answer that the book is giving? Or does the book only list the problem, not the solution, and your professor is telling you that that's the answer? If it's the professor, I'd fight it. If the book itself says it, I'd still fight it, although it's admittedly a more difficult fight--but no less valid of one.

 

[Mike Wilson]

To base line the question, if the balls designed, average diameter, for all engineering

parametrics, was to be a constant 6", at 100% engineering limits, the increase to

7" would put the strain at 116% of the intended limit (i.e., 16% over engineered limits).

I would need my "Durometer", to test the rubber, before I could be any more accurate.

Also the composite of the material, and the given thickness thereof!



Or, we all could just get a needle-valve fitting and check the air pressure of the ball,

that would give us the strain/pressure easily!

(Simple minds, simple answers, that is me!!!!!!):blink:

 

[Mike Wilson]

P.S. I bet you all are glad that I just work on automotive-related stuff, and not the

Space Shuttle! I know I am, and I am quite positive the Shuttle Crew is!!!:wacko:

 

[Ryan Schaecher]

I work for a company that builds conveyors. We can engineer the crap out of one... but when it come time for actually testing the prototype... it usually comes down to "let's put another 500 lbs on the end of that sucker and let's see what she can do". Yes, I work in Arkansas as well....

 

[Thomas Rogers]

What R Shek says.



Some engineering problems are what are known as "wicked problems". See the Wiki link below for a good definition.



One way to think of a wicked problem is that the requirements may not be well understood, or the candidate solution may be so complex that it is difficult if not impossible to mathematically model the solution and be assured that the model represents the real world conditions and eventual solution. Therefore, many mechanical solutions to "wicked problems" often fail under real-world use.



So, as R Shek states, it's often simply easier and safer to put the math aside at somepoint as it has diminishing returns, and instead just "build the thing", test and, and see where it breaks.



TJR

 

[Bill V]

TJR--absolutely. In fact, that's how most mathematical modeling concepts were developed in the first place--come up with a hypothesis, build it, test it, and see if the results agree with what your hypothetical mathematical model. If the hypothetical concept repeatedly agrees with the results of the physical test (and if the design of the tests were sufficiently robust), you've then proven the hypothesis.