z axis moment of inertia for l shape

huangyhg

z axis moment of inertia for l shape
does anyone know how to calculate the z axis moment of inertia for a l-shape?  
thanks in advance
you have to figure out how to calc the product of inertia - which appears to be a bear.  we kind of went through this in an advanced mechanics course, but the professor just gave us the product of inertia and didn't make us calc it.  if anyone knows how to do it (without taking all day), i would love to know.
anyway, once you get the product of inertia use varying values of theta (o) in the equation:
ix(cos^2(o))+ iy(sin^2(o))-2ixy(sin(o)(cos(o)) until you get the max value.  this is the max i.  then subtract this value from ix+iy and this will give you imin.
i have not done it for years, but the product of inertia is defined as the integral of xy da.  it is zero for a symmetric shape.
the principal moments of inertia for an unsymmetrical shape can be found in "elements of strength of materials" by timoshenko and mccullough as follows:
imax/imin = (ix + iy)/2 +/- {[(ix - iy)/2]^2 + ixy^2)^0.5
they can also be found by using a mohr's circle.
  
best regards,
ba
it is the ixy term that is the problem.
here is an excerpt from cisc handbook.
best regards,
ba
break the x-section into composite pieces just like you would for computing moments of inertia...then the product of inerita is sum (xi*yi*ai) i=1,number of pieces....note that either xi and/or yi can be negative....
edr
p.s. use the same axis location used to get the moments of inertia if you are going to compute the principal values...if not the transfer theorm for products of inertia is ixy = icg + xbar*ybar*a
i actually just did it by hand and verified it for angle 8x6x1.  it's actually not too bad.
i'm attaching my example calc.
if it's an angle the weak axis passes thru the mid-sides of each leg the centroid of the section is also on this axis.  the strong axis in normal to the weak, thru the centroid.
actually it works much like mohr's circle ... it's easy enought to calc ixx and iyy (x^2*da)  then ixy is just x*y*da ... plot the two points (ixx, ixy) and (iyy, -ixy), draw the circle, and get imax and imin (where ixy = 0).  
colledge homework, that's how i got it.
thanks for everyone's help on this one.  i also found this formula in blodgett pg 2.2-8. tbbdd/4(d+b)