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

[lxman]

Homework Statement

Calculate the moment of inertia for the following beam cross-section
http://www.freeimagehosting.net/uploads/th.044a09fe2e.png

The center (lighter) area is hollow down the length of the beam. The centroid is marked. The wall thickness is 1/4" and the dimension of each of the exterior sides is 2" from vertex to vertex.

Homework Equations

This isn't actually a homework question, but it sprung from a project we did in first-semester engineering. The relevant equation to calculate the deflection of a simply supported beam with a point load is:

d=PL^3/48EI

where:

P = concentrated load (lb)
L = length between support points (in)
E = modulus of elasticity(psi)
I = moment of inertia (in^4)

The Attempt at a Solution

The reason that I ask this question is because I want to know how to calculate I using integrals. Being a first-year engineering course, our textbook gave us a few pre-canned equations (e.g. a rectangular beam is bh^3/12, etc.) but, not being one to be satisfied with the basics, I would like to know how to derive these equations myself. The textbook we were using mentioned that the moment of inertia for most cross sections could be arrived at using integrals, and then referred us to an appendix which was missing from the publication. It's a cheap university-produced paperback, so it's not surprising as there were other sections missing also.

 

[viscousflow]

This should be useful to you:
http://en.wikipedia.org/wiki/Second_moment_of_area

 

[lxman]

Well, I know what a definite integral is and what an indefinite integral is. But what is
\int_{A}

Is this an improper integral meant to be evaluated from A to infinity?

 

[tiny-tim]

no, it seems just to mean that the integral is over an area (rather than a length or a volume)

 

[rock.freak667]

Well, I know what a definite integral is and what an indefinite integral is. But what is
\int_{A}

Is this an improper integral meant to be evaluated from A to infinity?

No it just means that you are integrating over the entire area A.

For the problem it would be easier to get inertia of the entire solid triangle and then subtract the inertia of the inner triangle, which would give the inertia of the hollowed section.

You should know that the area inertia of a triangle of height 'h' and base 'b' about its centroid is

I_c= bh³/36, which you can derive using your integral formula.

 

[lxman]

rock.freak667, thank you very much for your information. However, again, I have a pre-canned answer for I which is only good for a triangle. I am trying to find out how to derive \frac{bh^{3}}{36} for myself, so when I come across some other shape without a pre-canned definition, I will know how to proceed.

I am still working on the problem, and here is what I have come up with so far:

\int_{A} does indeed refer to the integration with respect to area.

To give an example of what I know to this point, consider the equation 3x+4y+5. In this example I compose a double integral of the form:

\int_{1}^{3}\int_{2}^{4}3x+4y+5\,\mathrm{d}x\,\mathrm{d}y

First I integrate with respect to x and arrive at:

\int_{1}^{3}\frac{3x^{2}}{2}+4xy+5x\,\mathrm{d}y

At this point I believe it is permissible to use the fundamental theorem of calculus and plug in my x limits and subtract, to attain:

\frac{3(4)^{2}}{2}+4(4)y+5(4)=\frac{3(16)}{2}+16y+20=24+16y+20=44+16y

and

\frac{3(2)^{2}}{2}+4(2)y+5(2)=\frac{3(4)}{2}+8y+10=6+8y+10=16+8y

44+16y-(16+8y)=44+16y-16-8y=8y+28

At this, performing my second integration:

\int_{1}^{3}8y+28\,\mathrm{d}y

I arrive at:

4y^{2}+28y

Using the fundamental theorem of calculus once again, I arrive at:

4(3)^{2}+28(3)-4(1)^{2}-28(1)=4(9)+84-4-28=36+84-4-28=88

I believe the procedure is correct as well as the result. I am now attempting to discern what the result, 88, actually represents. I have drawn a graph with the original function (3x+4y+5) - set it equal to zero and solve for y. It ends up being a line with a slope -\frac{3}{4} and y-intercept of -\frac{5}{4}. On the same graph, I have also plotted x=2, x=4, y=1 and y=3. At this point, I do not see any area enclosed by these equations that would represent an area of 88 units.

Any comments?

 

[rock.freak667]

rock.freak667, thank you very much for your information. However, again, I have a pre-canned answer for I which is only good for a triangle. I am trying to find out how to derive \frac{bh^{3}}{36} for myself, so when I come across some other shape without a pre-canned definition, I will know how to proceed.

This user also tried to derive it, so you can read their posts (as well as mine)
https://www.physicsforums.com/showthread.php?t=457632"

I am still working on the problem, and here is what I have come up with so far:

\int_{A} does indeed refer to the integration with respect to area.

To give an example of what I know to this point, consider the equation 3x+4y+5. In this example I compose a double integral of the form:

\int_{1}^{3}\int_{2}^{4}3x+4y+5\,\mathrm{d}x\,\mathrm{d}y

From what I am seeing it seems that you did an integral of z dA, which would give a volume.

 

[lxman]

Thank you for posting that. In looking over the moment of inertia article, something pops out at me right away. According to the article:

Moment of Inertia has dimensions of mass times distance squared

This confuses me, as the procedure we followed showed that the moment of inertia was supposed to be in units of distance^{4}. I might be confusing terms here. I went back and double-checked the original spec for the project, and it definitely says that the moment of inertia should be distance^{4}. You also use the term "area inertia." Is this something different from moment of inertia?

 

[rock.freak667]

This confuses me, as the procedure we followed showed that the moment of inertia was supposed to be in units of distance^{4}. I might be confusing terms here. I went back and double-checked the original spec for the project, and it definitely says that the moment of inertia should be distance^{4}. You also use the term "area inertia." Is this something different from moment of inertia?

There are two types of moments of inertia, mass moment of inertia and area moment of inertia (also known as the second moment of area).

Mass moment of inertia has units kgm² and is defined as

I=\int_{m} r^2 dm

or distance^2*mass over the entire area.

Area moment of inertia or second moment of area has units m⁴

and is defined as

I_{x}= \int_{A} y^2 dA \ or \ I_{y}\int_{A} x^2 dA

so it has units m²*m²=m⁴

 

[lxman]

Well, that's the point that I am stuck at.
<br /> I_{x}= \int_{A} y^2 dA \ or \ I_{y}=\int_{A} x^2 dA<br />
Am I correct in thinking that these equations reduce to double integrals since the area changes as dx and dy approach 0?
I can come up with an integral to describe the area of the triangle without any trouble, but I do not believe that is what I need. Just for discussion's sake, however, that would be:
<br /> A=2\int_{0}^{1}\frac{\sqrt{15}(x-1)}{2}\,\mathmr{d}x<br />
Running the numbers on that, the total area of my larger triangle is:
<br /> -\frac{\sqrt{15}}{2}<br />
But that only gives me a single scalar, as does:
<br /> \frac{bh^{3}}{36}<br />
Not an I_{x} and an I_{y} as I would expect based on:

Area moment of inertia or second moment of area has units m^{4}

and is defined as
<br /> I_{x}= \int_{A} y^2 dA \ or \ I_{y}\int_{A} x^2 dA<br />
so it has units m^{2}*m^{2}=m^{4}

Still confused . . .

I am sure this is child's play to you folks, but I'm still not there yet.

 

[lxman]

Alright, still plugging along on this. What I have discovered:

dA does indeed equal dy * dx. As we take successively smaller pieces of the the area, dy and dx both tend towards zero.

And, yes, the calculation does involve a double integral.

For the sake of simplicity, I am going to set aside my original conditions and focus on a beam with square cross-section which is one unit on a side. I know from references that the equation for I_{x} should be:
\frac{bh^{3}}{12}
The formula that I should use to arrive at this, if I am understanding this process correctly, is:
\int_{0}^{h}\int_{0}^{b}y^{2}\,\mathmr{d}x\mathmr{d}y
Performing the inner integration:
\int_{0}^{b}y^{2}\,\mathmr{d}x
I arrive at:
by^{2}
Then performing the outer integration:
\int_{0}^{h}by^{2}\,\mathmr{d}y
I arrive at:
\frac{bh^{3}}{3}
Whereas I should be arriving at:
\frac{bh^{3}}{12}

Would anyone care to point out what I have not set up correctly in this process?

 

[tiny-tim]

hi lxman!

your moment of area is about the edge of the beam (you integrated from 0 to h) …

you need to integrate from -h/2 to h/2 ​

 

[lxman]

Hi tiny-tim,

Yep, that works. Now let me see if I grasp this conceptually.

Because the _{xx} axis for this case is oriented at \frac{h}{2}, we consider the region above the axis positive and the region below the axis as negative. Does this then apply wherever we choose to locate the axis of rotation (for instance, at the edge of the block)?

I would think the reasoning behind this is that, if we do locate the axis in the center of its height, and then we apply torque to rotate the object about the _{xx} axis, the upper half is moving in the positive direction in the z axis and the lower half is moving in the negative direction with respect to the z axis.

Am I being reasonable in assuming this?

 

[tiny-tim]

hi lxman!

Because the _{xx} axis for this case is oriented at \frac{h}{2}, we consider the region above the axis positive and the region below the axis as negative. Does this then apply wherever we choose to locate the axis of rotation (for instance, at the edge of the block)?

yes!

(but for moment of area it doesn't matter, since we always use the square of the distance! )

I would think the reasoning behind this is that, if we do locate the axis in the center of its height, and then we apply torque to rotate the object about the _{xx} axis, the upper half is moving in the positive direction in the z axis and the lower half is moving in the negative direction with respect to the z axis.

sorry, you're just rambling …

positive above and negative below both contribute positively to the moment, don't they?

your error before was in the size of the distance, not its sign

 

[lxman]

Yeah, you're right. I get it.

Thanks much for your continued help.

 

[lxman]

One final post. I have been applying the understanding that you folks have given to me, and I have come up with a set of three equations which allow one to calculate the second moment of area for the inverted triangle beam that I started with. They are:
a(s)=-\frac{2\sqrt{3}s}{3}
b(s)=\frac{\sqrt{3}s}{3}
I_{x}(s)=2\int_{a(s)}^{b(s)}\left(\frac{3y+2\sqrt{3}s}{3\sqrt{3}}\right)^{2}\,\mathmr{d}y
where the triangle is equilateral, the rotation is about the x-axis through the centroid (always located 1/3 of the way from the top of the inverted triangle) and s is 1/2 the length of any side.

As rock.freak667 mentioned in an earlier post, it is necessary to obtain the moment of the outer triangle, then subtract the moment of the inner triangle to arrive at the final moment for the structure.

This is all a derivation of a newbie, of course, so there may certainly be mistakes. If anyone would be kind enough to point any out to me, it would be appreciated. :shy:

 

[rock.freak667]

For the outer triangle, for one of the hypotenuse, it might be easier to get the equation of the line 'y' as in y=mx+c and then use the fact that if you consider a rectangular element of length 'y' and width 'dx', then

dI_xx = (1/3)(y³)(dx)

then you can just integrate over the entire width to get

I_{xx} = \frac{2}{3} \int_{0} ^1 y^3 dx

EDIT: Typo.

 

[nvn]

lxman: I guess your derivation in post 16 is incorrect, because it appears to give the wrong answer to the given question in post 1. What final, numeric answer for Ix do you get from post 16 for the question given in post 1?

rock.freak667: Your answer in post 17 is not making sense to me yet. A 1/3 seems to have disappeared in your last equation (?). And why is there a 1/3, in the first place? What is the actual, numeric function for your y in post 17? And what final, numeric answer for Ix do you get, using post 17, for the given question (posts 1 and 16)?

 

[lxman]

nvn, thank you for taking the time to consider the problem.

Here is a picture to illustrate how I am deriving my solution:
http://www.freeimagehosting.net/uploads/th.3d8cf5e7d7.png
Based on this, the equation for the line of the hypotenuse is:
y=\sqrt{3}x-\frac{2\sqrt{3}s}{3}
where s is the length of the "short" side of the triangle.
And here, now that I look back at it, is where I think I might have made a mistake. I actually integrated with respect to y and then multiplied the result by two. That, however, should give me the moment with respect to the y axis, not the x axis. So, yes, I am rethinking.
For the sake of discussion, given the initial conditions along with my (now presumably incorrect) equations, I come up with an approximate value of .944. And yes, the more I re-examine it, the more I would appear to be in error. Still working on it then . . .

Again, thank you all for being patient with me.

 

[rock.freak667]

rock.freak667: Your answer in post 17 is not making sense to me yet. A 1/3 seems to have disappeared in your last equation (?). And why is there a 1/3, in the first place? What is the actual, numeric function for your y in post 17? And what final, numeric answer for Ix do you get, using post 17, for the given question (posts 1 and 16)?

That was a typo, the 1/3 should be there since that is the second moment of area of a rectangle about the x-axis , 1/3(bh³).

Using the "pre-canned" formula for a triangle as I_xx=(bh³)/12,

Integrating using my integral gives the same value. I did not check the OPs integral as I am unsure as to how they came up with their integrand.

 

[lxman]

For the sake of helping me figure this out, what is the correct result?

 

[rock.freak667]

For the sake of helping me figure this out, what is the correct result?

I believe my integral gives the same magnitude for the result which matches with using bh³/12

 

[nvn]

rock.freak667: You did not show how post 17 gives the correct answer. You did not state your function for y used in post 17. And the moment of inertia you listed in post 22 is not the moment of inertia lxman is requesting, as it is, first of all, not a centroidal moment of inertia. And secondly, it is not the cross section in the given question by lxman.

 

[lxman]

Alright, I'm not trying to be hard headed here, and I assure you, this is not a homework project or work project or anything. I honestly am just trying to understand how to derive the formula for my own personal satisfaction. This particular case is one that was a result of my reflection from a project from last semester. I am sure you gentlemen went through your proofs years ago, have faith in the derived equations, and just use them without worrying about it. There's nothing wrong with that, once you understand the proof.

Handing me bh^3/12 is nice - but I can go on the web and look that up. Then I am still left with questions. In the cases that I have looked at, the reference is usually a right triangle, in which the centroid is located 1/3 the height of the triangle from the base in the y direction and 1/3 the base of the triangle from the y-axis in the x direction (presuming that the y-axis is located along the side of the triangle perpendicular to the base). Firstly, what I am considering is not a right triangle. It is an equilateral triangle. This relocates the centroid. Now, as the new centroid is colinear with the old one along the x axis, I can presume that the same equation would still apply. But, at that point I begin to assume. I don't like assuming. It gets me in trouble too often. So, my goal here is not to "get an answer" but to understand the process.

 

[rock.freak667]

rock.freak667: You did not show how post 17 gives the correct answer. You did not state your function for y used in post 17. And the moment of inertia you listed in post 22 is not the moment of inertia lxman is requesting, as it is, first of all, not a centroidal moment of inertia. And secondly, it is not the cross section in the given question by lxman.

From how the triangle was initially, with the base at the x-axis, the integral for the centroid would be:

dI_x=(1/12)y³ dx so I_x=(1/12)∫y³ dx between 1 and 0 where y= (√3)x-√3.

Handing me bh^3/12 is nice - but I can go on the web and look that up. Then I am still left with questions. In the cases that I have looked at, the reference is usually a right triangle, in which the centroid is located 1/3 the height of the triangle from the base in the y direction and 1/3 the height of the triangle from the y-axis in the x direction (presuming that the y-axis is located along the side of the triangle perpendicular to the base).

Did the other thread I linked you to not help you derive the given equation? (Sorry if it didn't)

Firstly, what I am considering is not a right triangle. It is an equilateral triangle. This relocates the centroid. Now, as the new centroid is colinear with the old one along the x axis, I can presume that the same equation would still apply. But, at that point I begin to assume. I don't like assuming. It gets me in trouble too often. So, my goal here is not to "get an answer" but to understand the process.

The type of triangle doesn't matter with respect to the formula, as you can see from this http://www.efunda.com/math/areas/triangle.cfm" .

 

[lxman]

From how the triangle was initially, with the base at the x-axis

Ah, and there, I think, is part of our misunderstanding. I am of the understanding (but I guess I did not state explicitly enough) that when you are computing the second moment of area of a beam from a cross-section of the beam (with the goal of computing the deflection based on a point load), you always want to know that moment about the x-axis which passes through the centroid of the cross section. So in my case, I have had the preconception that the y=0 line (the x axis) passes through the centroid, not the base (or in my case the top) of the triangle. Therefore the point of origin for the necessary coordinate system should be the centroid of the triangle.

Am I incorrect in my assumptions here?

 

[rock.freak667]

Ah, and there, I think, is part of our misunderstanding. I am of the understanding (but I guess I did not state explicitly enough) that when you are computing the second moment of area of a beam from a cross-section of the beam (with the goal of computing the deflection based on a point load), you always want to know that moment about the x-axis which passes through the centroid of the cross section. So in my case, I have had the preconception that the y=0 line (the x axis) passes through the centroid, not the base (or in my case the top) of the triangle. Therefore the point of origin for the necessary coordinate system should be the centroid of the triangle.

Am I incorrect in my assumptions here?

You are correct. Like I said in post 22, with the base at the x-axis like in the picture in the first post, if you change the 1/3 to 1/12, you will get the inertia through the centroid, since the moment of inertia of the rectangular element through an axis passing through its centroid is (1/12)y³ dx.

You don't necessarily have to move the entire triangle and do it over.

 

[nvn]

dI_x=(1/12)y³ dx so I_x=(1/12)∫y³ dx between 1 and 0 where y= (√3)x-√3.

rock.freak667: Your answer is incorrect. Try again. And what numeric answer do you get for lxman's question? When you get the correct numeric answer using your derived formulas, then it will mean your derivation is correct. If your numeric answer is wrong, it means your derivation is wrong.

It was clear, even in the beginning, or at least by post 6, that lxman was asking for the centroidal moment of inertia.

 

[lxman]

Okay, would this be the correct formula to use in this case?

\frac{2h\left(\frac{b}{2}\right)^{3}}{3}

In the case of my original larger triangle, b=2 and h=\sqrt{3}

Then, calculating the size of the smaller triangle and subtracting I arrive at a final figure of \approx 1.035

Am I anywhere near correct now?

 

[rock.freak667]

rock.freak667: Your answer is incorrect. Try again. And what numeric answer do you get for lxman's question? When you get the correct numeric answer using your derived formulas, then it will mean your derivation is correct. If your numeric answer is wrong, it means your derivation is wrong.

According to MathCAD the integral of

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

I_{xx}=\frac{bh^3}{12}=\frac{(2)(\sqrt{3})^3}{12}= 0.866 in^4

It was clear, even in the beginning, or at least by post 6, that lxman was asking for the centroidal moment of inertia. It does not matter where the coordinate system is located for merely showing cross-sectional dimensions.

I, for some reason, thought the OP was finding the inertia of the outer triangle first and would then subtract the inner triangle's inertia. Which could then be relocated as desired using the parallel axis theorem.

I also got the impression that the OP was attempting to find out how exactly to derive the equations given in his textbook, such that I linked him to another user who recently tried to do the same derivation.

 

[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: