## Estimation Of Max And Min-Moment Of Inertia For A Section.

### Brief content of the video.

We will estimate the Ixy the product of inertia to make sure that it is equal to zero, so our calculation for bending of stress follows the previous assumption of a pure moment from the assumption of the pure moment.

We have the neutral axis that must pass by the Cg. This is number one. Number two, for the two perpendicular axes, should be principal axes i.e. Ixy should be =0.

Then we calculate Ixy and find out if it is not equal to zero then we have to look to any two orthogonal axes. that will give us Ixy=0. The video has a closed caption.

### Steps to get the expression for I-max, Imin, and tan(2θ).

We will estimate the Ixy the product of inertia to make sure that it is equal to zero, so our calculation for bending of stress follows the previous assumption of the pure moment from the assumption of the pure moment.

We have the neutral axis that must pass by the Cg for which Ixy=0.

Number two, for the two perpendicular axes, should be principal axes i.e. Ixy should be equal to zero.

when we calculate Ixy and find it is not equal to zero then we have to look to any two orthogonal axes,

which will give us Ixy=0.

You can click on any picture to enlarge then press the small arrow at the right to review all the other images as a slide show.

Since the value of I-polar is constant, we have plenty of axes is that will achieve an l-polar = Ix+Iy or = Ix1+Iy1 or Ix’+Iy’.

But the axes which we are looking for is the one that will achieve that the moment of inertia around, for example, x’ will be maximum and the moment of inertia around y’ will be minimum, and at the same time that Ix’y’ should =0.

If Ix is maximum and Iy minimum and Ixy is zero about any two axes, then these two axes are called principal axes.

### Derive the expression for max And min-Moment Of Inertia.

To get an expression we will make a general expression by using x and y axes and will assume that we have inclined two axes x’ and y’ having angle= θ, where θ is positive in the anti-clockwise direction.

For small strip dA, the coordinates are X and Y at the same time relating to the new two axes (x’ and y’), to have a relation between (x’ and y’) and x and y,

We can find out that x’=x*cos θ+y*sin θ. This is this portion of the x’, while for the y’=y*cosθ -x*sinθ.

Then we proceed to calculate the moment of inertia about x’ and y’ and at the same time we want to get an expression for Ix’ and Iy’, for the special case where Max And Min-Moment Of Inertia are required, as we know that Ix’ = ∫ dA*y’^2, this the dA, and we want to make integration about x’.

While for the moment of inertia around y’ it will be= ∫ dA*x’^2. While for the product of inertia Ix’y’,= ∫ dA*x’*y’. We will substitute by the value of x’=x*cos θ+y*sin θ, y’=y*cosθ -x*sinθ. We have Ix’ and Iy’ and Ix’y’, These items are written after substituting the values of x’,y’ in terms of x and y as shown in the next image.

We will proceed as we can see this is a long calculation to some extent but we are going to try to get three expressions. The first expression is for Ix’ in terms of x,y, and angle θ , where θ is the angle closed between axis x and axis x’.

The second expression is for Iy’ in terms of x,y, and angle θ, where θ is the angle closed between axis x and axis x’.

The third expression is for Ix’y’ in terms of x,y, , where θ is the angle closed between axis x and axis x’.

The first two expressions for Ix’, Iy, are shown in the next slide since the polar of inertia=Ix+Iy.

We will add Ix and Iy together from Ip, we can get an expression for tan(2θ_{p}).

We will add Ix and Iy together from Ip, we can get an expression for tan(2θ_{p}).

The θ_{p} is the angle that gives Ix max and Iy min, we can get it by differentiating ix’ with respect to θ, and let it=0.

The expression with which is tan(2θ )=2(Ixy)/(Iy-Ix). We can rearrange it to become (-2Ixy/Ix-Iy) as a denominator and this is the θ, as we remember this expression. this is the expression for Max And Min-Moment Of Inertia.

This θ was enclosed angle between X-axis and x’-axis X’-axis since we have obtained this condition for tan2θ expression. We have four cases, the first case Ix>Iy and Ixy is positive.

The second case is when Ix>Iy and Ixy are negative. The third case is when Ix < Iy and Ixy is positive. The fourth case is when Ix < Iy and Ixy is negative.

This is the pdf file used in the illustration of this post.

The next post is an Easy introduction to Mohr’s circle of inertia part 1.

This is a link to a useful external resource. Calculator for Cross Section, Mass, Axial & Polar Area Moment of Inertia and Section Modulus.