Homework help - engineering/math problem

[Jeff7]

I'm starting out with this equation.

I think I want to solve for sigma (Ok, ok, it's epsilon, not sigma. Oops), e, so that I can tell where on the stress/strain graph the material leaves the elastic region. I've got two equations to work with, Hooke's Law (s=Ee) which only applies in the elastic region, and s={K[ln(e+1)^n]}/(e+1), to find true stress in the plastic region. For each material, it would appear that there's a point where these two equations "meet" and then one becomes valid while the other is not. I figure I should know this so that on a test or quiz, I will know which equation I can use.

But with the natural log and the exponent, I'm not sure how it can be done. I've got either of these, which don't exactly help:
2
3

We skipped over logarithms in algebra or algebra II in middle/high school (we ran out of time to cover the relevant chapters), or whenever it was, and I've not had to deal with them until now. I'm not sure how helpful they'd be anyway.

I did already get a value for e by graphing both sides of the equation on my TI-89 and having it find the intercept, but I figure the professor would want to see the problems solved out by hand, rather than just, "My calculator did it."

Values:

K = 120000
n = 0.113
E = 10.4 * 10^6

Using the intercept function, I got a value for e of 0.0064853.


Update:
Yes, the solution was to simply use Hooke's Law, and solve for stress. If the stress was less than yield, than Hooke's Law was still applicable, and the strain fell within the elastic region of the stress/strain graph. If the stress was higher than yield, then the equation for true stress would have had to be used.
The professor laughed a bit when he saw how complicated I was trying to make it, though he was interested to know what numbers I came up with. Neither of us is exactly sure what the stress/strain values I came up with coincide with - the values I got with my TI-89 were where the two equations (Hooke's Law and true stress) coincided, but what exactly these values mean, or why the stress value that resulted wasn't equal to 73,000psi, we're not sure.

 

[esun]

First off, that symbol is epsilon, not sigma.

I'm not entirely sure you can solve the problem this way, but I believe you can do the following:

[e(e+1)]^[(n-1)/n] = [ln(e+1)]^(n-1)

e(e+1) = [ln(e+1)]^n

Now, you can try solving using iteration. What this means is, plug a value of x that you know is near the answer in the right side of the equation. If you pick a good starting guess, then keep plugging in and solving for e, you should get the right answer. Unfortunately, there is the trivial solution of e = 0 that will always solve this equation (since it sets both sides equal to zero), so if you guess too small, it may converge to zero.

As an example, say I had n = 6, which should give me a solution of e = 93.8222.

Well let me start with a guess e1 = 100. Plugging into the right side I get 9662.7, which gives me the next value of e2 = 97.8003.

Now I plug e2 into the right side of the equation and I get 9389.36, which means e3 = 96.4.

Repeating a few more times, I eventually get 93.823, which is close enough. To get a very precise answer will require many, many iterations (which just means solving the quadratic equation a bunch of times), but if you just want, for example, the one's digit to be correct, it doesn't take too long. For example, with an initial guess of 150, I get to 93 in about 10 iterations.

EDIT: Actually, a nicer way to do this is as follows:

e = [ln(e+1)]^n / (e + 1)

Then just plug your initial guess into the right side and you'll get your next guess without having to do any quadratic formula crap. In your TI-89, you can just do something like:

6 -> n
150 -> x
(ln(x+1))^n / (x+1) -> x
(ln(x+1))^n / (x+1) -> x
(ln(x+1))^n / (x+1) -> x
...

Until you get the answer to converge (assuming your value for n is 6 and your initial guess if 150).

 

[Jeff7]

Right. Duh. Stress uses sigma, s. OP now has values added.

Iteration may well provide a usable answer, but I don't know if that's a viable method for a timed quiz.

I have a feeling that there might be a simpler method of doing this, which would likely come from Materials 101. The yield strength is 73ksi for this material, aluminum 7075-T6.

All I come up with though is Hooke's Law, rewritten, e = s/E, which in this case gives e = .007019. It's close to .0064853, but doesn't seem close enough.


The problem asks for answers for the normal stress if the strain is .005 and .015. The answers are also given - 52,000psi for .005 and 73,490psi for .015. Given those answers, it looks like .005 is in the elastic range, and thus Hooke's Law would be valid, while .015 is in the plastic range, which uses the longer true stress equation.

But that brought up the problem of, "How do I know where the elastic limit ends for a particular material? How is this point found?" That's why I tried to set the Hooke's Law equation and the true stress equation equal to each other, and then solved for strain.

I also don't know how the yield strength of 73ksi was found. If given just a graph, we were told to use a .2% offset (strain of .002) to find the yield strength. I'd figure though that the yield strength here was found through tensile testing.

I e-mailed the professor about this Friday afternoon, but I guess he doesn't check e-mail on the weekend.



I notice now that the strain of .0064853 coincides with a stress level of only 67,447 psi, not the yield strength of 73,000 psi.

Maybe I'm really overthinking this problem, and should just use the basic e = s/E equation.

 

[esun]

Unfortunately I don't know jack about stress/strain, so aside from the math I can't help you at all. If it really does come down to solving the equations you gave for epsilon, though, then there's really not much you can do besides use iteration or a graphing calculator. If you have (linear function) = (exponential function), that's just how you solve it, and you can't really show your work for either (well, you could sketch the graph or write down the intermediate values of iteration, but both aren't really "showing work" in the traditional sense IMO).

 

[Jeff7]

Originally posted by: esun
Unfortunately I don't know jack about stress/strain, so aside from the math I can't help you at all. If it really does come down to solving the equations you gave for epsilon, though, then there's really not much you can do besides use iteration or a graphing calculator. If you have (linear function) = (exponential function), that's just how you solve it, and you can't really show your work for either (well, you could sketch the graph or write down the intermediate values of iteration, but both aren't really "showing work" in the traditional sense IMO).

Yeah, and something like that strikes me as more than he'd expect us to do. Thus far a lot of my engineering classes haven't required much in the way of advanced math. A lot of it is just knowing which equations to use and in what order. The math generally involves nothing more advanced than trigonometry, with lots of algebraic manipulation.
There's never been anything with trying to solve exponentials in the fashion I set up here.

I think I'll just go with the e = s/E equation.

 

[da loser]

yield stress is where it starts to yield so that's when it's no longer in the elastic region and you get plastic deformation

the way i would answer that question is calculate the strain-stress for both strains, then if it's larger than the yield stress use the plastic equation, and elastic for less.

the 2% is a rule of thumb, it's just something that's used that seems to work for metallic materials. a correlation, i suppose.

i highly doubt they'd ask you to calculate the yield stress from those equations. but the method provided by esun should be good enough. plus i'm not sure you'd actually get the yield stress because when you offset 2% you're parallel to hooke's equation.