How Do You Calculate the Moment of Inertia for a Hollow Beam Cross-Section?

[nvn]

lxman: Your answer in post 29 is currently incorrect. Your diagram and y(x) equation in post 19 are excellent. However, instead of using specific numbers in your diagram, why not use parameters? You used a parameter for the half width, s, which is great. Why not call the triangle height h?

Now rewrite your excellent y(x) equation in post 19 using parameters s and h. After that, solve your new y(x) equation for x, to obtain x(y). After that, compute Ix1 = 2*integral[(y^2)*x(y)*dy], integrated from y = -2*h/3 to h/3. After you do that, use the same solution to compute Ix2, letting s and h now be the dimensions of the inner triangle (the hole).

 

[lxman]

Thank you, nvn,

Your explanation seems quite clear. I will get right on that . . . in the morning. Right now I got to get some sleep. Thanks to all for all of the wonderful help. I'll check back in tomorrow after I work on the problem for a bit.
:zzz:

 

[nvn]

I_{xx}= \frac{2}{3} \int_{0} ^{1} (\sqrt{3}x-\sqrt{3})^3\:dx =- 0.866

Nice work, rock.freak667. You are starting to get closer. Nonetheless, remember, lxman wanted to derive the solution from scratch. In your above method, you are copying a cookbook answer, b*(h^3)/3, for each differential strip. However, lxman stated a desire to not use cookbook formulas.

 

[lxman]

Wow, we just passed 500 views on this thread.

But seriously, now, I am going to break this down into individual steps so you folks can verify or correct me each step of the way.

I am going to begin with the picture that I posted in #19. I am presuming that I will be able to arrive at my moment for one half of the triangle, then multiply by 2 to arrive at the moment for the whole. I can only do this because we are taking the moment about _xx and we will be dividing the triangle along the perpendicular y axis.

Step #1 - is this presumption correct?

 

[nvn]

That is correct, except change the word "moment" to "second moment of area," or as in a few credible textbooks, "area moment of inertia."

 

[lxman]

Alright, here is what I have got (I know, I'm rocketing along like a herd of turtles):

The slope of the hypotenuse of the triangle is:

y=\sqrt{3}x-\frac{2h}{3}

h of course is the length of the side along the y axis

Solving for x I get:

x=\frac{y+\frac{2h}{3}}{\sqrt{3}}

Now, you state that

2\int_{-\frac{2h}{3}}^{\frac{h}{3}}y^{2}x(y)\,\mathmr{d}y

So, what we are doing here is taking smaller and smaller sections of the triangle parallel to the x axis. Based on the principle that a force applied at a distance from a fulcrum varies as to its effect according to the square of the distance, that explains the y^2 term. Then the x(y) term gives us the area of our infinitesimally small interval. The product is the second moment of area of each tiny subdivision. Integrating this with h=\sqrt{3}, I arrive at the following result:


\frac{1}{2\sqrt{3}}

I am using a CAS from this point, so I would have a difficult time writing out the full equation. Then, I evaluate my equation at \sqrt{3}-\frac{3}{4} and I get:

A big old nasty fraction which my latex skills can't handle - but a numerical approximation is \approx .03

So, subtracting one from the other, I come up with a result of \approx .259

I certainly hope that I have gotten it right this time.

Once I receive confirmation that I have in fact followed the process correctly, I will be back with a few more questions. If I am wrong, however, I will continue to try.

Once again, thank you all for your wonderful support.

 

[nvn]

lxman: Your answer is correct. One thing you did that was not recommended in post 31 is, you omitted parameter s. This might be why your equations became complicated, instead of having the simple, analytic solution you get when you use s and h, mentioned in post 31.

By the way, numbers less than 1 must always have a zero before the decimal point. E.g., 0.259, not .259. And always leave a space between a numeric value and its following unit symbol. E.g., 0.259 in^4, not 0.259in^4. See the international standard for writing units[/color] (ISO 31-0[/color]).

 

[lxman]

Thank you, nvn, for the direction. It is already taken to heart.

As to the omission of s, I can see that I was mentally applying this to the unique situation of an equilateral triangle. I enter with the assumption that the angle at the "top right" of my triangle is sixty degrees. Therefore the resultant slope of the hypotenuse would always be \sqrt{3} . But reasonably, this could be generalized to apply to an isosceles triangle where the slope would be \frac{h}{s} . My next immediate project will be to derive that equation.

Now here is my first question. In this wikipedia article:

http://en.wikipedia.org/wiki/Second_moment_of_area"

it defines the basic equation as:
I_{yy}=\int_{A}z^{2}\,\mathmr{d}A
This tells me that I am taking the integral of the distance of each particle squared times an infinitesimally small area. I would think that \mathmr{d}A=\mathmr{d}x\,\mathmr{d}y (I have just switched my coordinate system, I realize, but the principle should still hold). That would indicate to me that a double integral would be involved in the solution. Yet to arrive at the correct answer, I only used a single integral. Looking over the diagram it would appear to me that x and y are both changing (i.e. they are functions as opposed to constant scalars). How is it that I can use a single integration and arrive at a correct result?

 

[lxman]

Okay, I think I've got it, now!

As to my question in my last post, it would appear from what I have been looking up that, in order for a double integral to mean something, you have to have two independent variables to integrate with respect to. With our case we only have one independent and one dependent variable. Therefore double integration would not make sense. At least that's the way that I am understanding it.

Now, onto the derivation of the formula:

Our equation for the slope:

slope=\frac{h}{s}

Our equation for the y-intercept:

yintercept=-\,\frac{2h}{3}

Our equation for x(y):

x(y)=\frac{y-yintercept}{slope}

Which works out to:

\frac{3sy+2hs}{3h}

Multiplying by y^2:

y^{2}x(y)=\frac{3sy^{3}+2hsy^{2}}{3h}

Finally, integrating this with respect to y results in:

I_{x}=\frac{sh^{3}}{36} (and of course don't forget to multiply by 2!)

Plugging in the numbers from the original problem in post #1 results in \approx 0.2588

In conclusion, my final formula would be:

\frac{sh^{3}}{18}\,in^{4}

Which would be equivalent to:

\frac{bh^{3}}{36}\,in^{4}

I humbly await your examination of my results.

 

[lxman]

Yes, dumb algebra mistake.

0.2588 in^4

 

[lxman]

Okay, re-ran both methods from scratch. Here are the results:

Method 1 (Post #36):

h=\sqrt{3}\,\,,I_{x}=\frac{1}{2\sqrt{3}}

h=\sqrt{3}-\frac{3}{4}\,\,,I_{x}\approx0.02983

Result \approx0.2588

Method 2 (Post #39)

b=2\,\,,h=\sqrt{3}\,\,,I_{x}=\frac{1}{2\sqrt{3}}

b=2-\frac{\sqrt{3}}{2}\,\,,h=\sqrt{3}-\frac{3}{4}\,\,,I_{x}\approx0.02983

Result \approx0.2588

Both methods yield the same result for me. When I receive your nod of approval, I will go back and correct the relevant posts. I see your point, by the way. In my years in the field, I have seen that small mistakes by engineers only tend to be amplified over time, rather than the other way around. Therefore a small mistake by an engineer can lead to a complete product failure in the field.

 

[nvn]

lxman: Your answer is correct. The correct answer is 0.2588 in^4. Excellent work.

 

[lxman]

 

[nvn]

lxman: Regarding your question in post 38, you can use a double integral, if you wish. Just replace x(y) in your current integral with dx, and add the second (inner) integral sign, integrated from x = 0 to x(y). Alternately, you could write the double integral integrand as (y^2)*dy*dx, integrated from y = y(x) to y(1), and from x = 0 to 1.

 

[lxman]

Uhm, am I mistaken here, or would it be y=y(x) to y(s) and then x=0 to s in the second case (in the spirit of keeping away from constants)?

 

[nvn]

Good catch, lxman. My mistake. I agree with that change. That is much better.

 

[lxman]

Just to be clear, the resultant integrals would be:

2\int_{-\frac{2h}{3}}^{\frac{h}{3}}\int_{0}^{x(y)}y^{2}\,\mathmr{d}x\,\mathmr{d}y

and

2\int_{0}^{s}\int_{y(x)}^{y(s)}y^{2}\,\mathmr{d}y\,\mathmr{d}x

Is this correct?

 

[nvn]

Yes, that is correct.

 

[lxman]

Alright, thanks once again for all of the help, nvn, as well as everyone else who has helped me so much since I have signed up for this forum!

I will consider these double integrals and their implications at a later time. Right now, it's about time for bed. I've got school starting back up in a couple of days, so I will probably not be quite as "attentive" as I have been up to this point.

Again, much thanks to everyone!

:zzz: