Rotated Moment of Inertia 2

[faromic80]

How do I calculate the moment of inertia of a rotated section? On wikipedia (don't have my text book with me) There is a big equation :

Ix* = (Ix+Iy)/2 + (Ix-Iy)/2*cos(2*theta) - Ixy*sin(2*theta)

What is the Ixy and how do I calculate it?
theta is the angle of rotation, correct?

THanks

 

[Dinosaur]

Ixy is the integral of X x Y x dA; whereas Ix is the integral of Y x Y x dA and Iy is the integral of X x X x dA. For any symmetric section about either axis, Ixy is zero. Theta is the angle of rotation of the applied load relative to the original XY coordinate system used to compute your Ix and Iy values. Good Luck.

 

[Teguci]

Rotate your shape. Figure out your x versus your y. Then integrate xy^2 dy about your new neutral axis for bending about yy.

 

[rb1957]

personally i can't follow teguci's post, but dinosoar's is straight forward. one thing to note is that symmetric sections have Ixy = 0.

to expand on dinosoar's post, divide your section into a bunch of simple elements (rectangles). determine for each element their CG co-ordinates with respect to a global axes. calculate the centroid. re-calculate the individual elements CG with respect to the section centroid. Ixx = sum(y^2*A), Ixy = (sum(x*y*A)

 

[hokie66]

rb1957 is correct, except you also add the I of each of the rectangles about its own neutral axis. Look up the parallel axis theorem in any Mechanics of Materials text.

 

[asixth]

From what I remember, the moment of inetia for doubly symmetric sections can be calcuated by mohr's circle. You can calcualte the Ix (principle major) and Iy (principle minor), and rotate this around your mohr's circle as required.

Even easier, you don't need to calculate the Ix and Iy, you can find them on data sheets provided by the steel manufacturer.

 

[bruinuscmathab]

For anyone still puzzled by the previous remarks and
comments of this thread about computing moment of inertia,
it may be helpful to remember that the basic definition
of moment of inertia is based on (or IS) the definition of
the moment of inertia of a "point mass" of mass m at
distance r from the axis of rotation:

I=rrm (" r squared m")

This helps explain the integration over xx+yy
in the previously quoted formulas, as
rr=xx+yy
when the center of rotation is taken at (0,0).

To compute the "M of I" of a lamina, with density
of 1 (unit of mass per unit of area), wo. (0,0)
divide it into small sub areas of masses m(sub)i
each with r = r(sub)i then
I=(Sigma)[r(sub)i][r(sub)i][m(sub)i]

that is the M of I equals the sum of products of the rrx 's
for each sub-division of mass, ... or
= sum of (each mass times its radius squared).

Note:
(radius is distance from (0,0) not the size of a sub-mass)


Question(s)of the dayby pbs) Topic: Land Survey Measure.
How many CHAINS in a mile? How many LINKS in a chain?
Are these units still used much today is survey or road
construction? You are welcome to send answers to
pbsteed@usa.com (I never reveal e-mail addresses to others.)
Suggested reference: Ray's Higher Arithmetic.

 

[EngineerTex]

Lots of great answers. I spent an entire day trying to integrate y^2dA by hand for an odd-shaped beam at a particular angle and ended up with an equation about a page long. It worked, but it was cumbersome.

Now, there is an easy way to do this if you have AutoCAD.

1) Draw the cross-section of the beam and make sure that your corners are coincident, i.e., it has to be a closed shape.

2) Execute the REGION command. Type it on the command line, or find it in the menus (I don't know where it is in the menus)

3) If your section has a hollow portion, remove the inside region from the larger region with the SUBTRACT command.

4) ROTATE the region to the appropriate angle.

5) Execute the MASSPROP command. A screen comes up and gives you all the info you're looking for.

A note of caution/user tip: The MASSPROP command spits out the information in a less-than-user-friendly format. After I have used the MASSPROP command, I get the centroid information and then move the region so that the centroid is at the origin (0,0). After I do this, I execute the MASSPROP command again and verify that the centroid is at 0,0 and then I get the information I need from the output screen. BTW, this is the greatest thing in the world when having to calculate properties of a weld group.

I know that not everyone has AutoCAD, but if you do, you'll never calculate a moment of inertia again.

-T

Engineering is not the science behind building things. It is the science behind not building things.

 

[PersonalProfile]

Try Mohrs Circle for a "graphical" representation, if that does it for you.

Regards,
Lyle

 

[AZengineer]

How does Mohr's Circle help calculate section properties? The only thing I've ever used it for was to calculate principal stresses!

 

[LOOPDOP]

I can only agree with EngineerTex.
Without wasting time with first principles, I use the AutoCad create REGION, SUBTRACT hollow ones etc., check the 0,0 centroid to see if you agree, rotate the req'd angle and the new MASSPROP is there in a split-second.
Why waste valuable time?
It is easy to check with simple geometrical shapes.

 

[rb1957]

AZ ...

consider Ixy as a shear stress (y-axis) and Ixx and Iyy as axial stresses (x-axis). you plot two points (Ixx,Ixy) and (Iyy,-Ixy) and draw a circle thru them. the principal inertias are the axis intercepts, and the angle (to the orignial axes) is 1/2 the angle on the circle.

but CAD programs solve this quicker ...