Moments of Inertia of an Area

The last thing we will do in the  Statics – Basis info guide will be flat figures.

Let’s start with the fact that flat figures have nothing to do with calculating beams, frames or trusses. Such calculations are made to find, for example, the central moment of inertia of I-sections, C-sections or to find their cross-sectional core. In this guide, we limit ourselves to basic things, namely.

  1. Determination of the center of gravity
  2. Central moments of inertia
  3. The position of the main central axes of inertia
  4. The main central moments of inertia – and on this we’ll end.

You will be able to continue the continuation in the Strength of materials guide.

All calculations are made according to the formulas, which, however, will be a bit. First, let’s take a look at the theory and then show what and how, for example.

Types of flat figures and their characteristic parameters.

The course of action when calculating flat figures and formulas.

  1. Division into simple figures.At the very beginning of the calculations we must divide the “shape” into simple figures, because we do not always get a square or a rectangle. For easier understanding, he will show me an example immediately, which I will calculate later. That’s him.

We have a rectangle with a notch inside a square shape and 1 × 1 dimensions, and on the rectangle there is a triangle. Going to the division, we get.

  • one rectangle
  • one square
  • one triangle

At this stage, we calculate their surface area, the moment of inertia and the moment of deviation.

  1. Determination of the center of gravity.

2.1. We accept the starting coordinate system at the point chosen by us.
Then we have to find the coordinates of the center of gravity Sx and Sy, to do this we can use the following formulas.

S_{X}=\frac{(A_{1}*X_{1})+/-(A_{1}*X_{2})+/-(A_{i}*X_{i})}{A_{1}+/-A_{2}+/-A_{i}}

S_{Y}=\frac{(A_{1}*Y_{1})+/-(A_{1}*Y_{2})+/-(A_{i}*Y_{i})}{A_{1}+/-A_{2}+/-A_{i}}

We need to find the X coordinate and Y coordinate of the center of gravity. We do this by dividing (the surface area of a given figure multiplied by the corresponding coordinate) with (the surface area of this figure). The characters between the data of different figures depend on whether the figures enlarge the total surface area or decrease it. In my case, it will look like this

rectangle – square + triangle.

We see that the square is inside a rectangle, we understand this as cutting out a piece of a rectangle.

      2.2 Coordinates of individual figures in relation to the center of gravity.
After calculating the center of gravity of the whole figure, we have to determine the new coordinates of the centers of gravity of each figure in relation to the         center of gravity of the whole figure, we will need them for the next calculations.
To determine the new coordinates of Figure 1.

S_{1} = (a_{1}; b_{1})\rightarrow a_{1} = X_{1} - X_{S} \: \: and \: \: b_{1} = Y_{1} - Y_{S}

Figure 2.

S_{2} = (a_{2}; b_{2})\rightarrow a_{2} = X_{2} - X_{S} \: \: and \: \: b_{2} = Y_{2} - Y_{S}

  1. Central moments of inertia (Steiner’s patterns).

The next step is to calculate the moments of inertia of the entire figure, we will need the following patterns.

I_{X0}=\Sigma (I_{Xi}+A_{i}*b_{i}^{2})

I_{Y0}=\Sigma (I_{Yi}+A_{i}*a_{i}^{2})

D_{X0Y0}=\Sigma (D_{XiYi}+A_{i}*a_{i}*b_{i})

I will explain what the particular signs mean.

I_{Xi}\; \: and \: \: I_{Yi}  – moments of inertia of individual simple figures,
D_{XiYi} – moments of deviation of individual simple figures,
A_{i}– surface area of individual flat figures,
a_{i}\: \: and\: \: b_{i} – x and y coordinates of individual straight figures, with reference to the calculated center of gravity of the whole figure.

  1. Location of main central axes.
    The next step is to determine the main central axes, and more accurately calculate the angle by which you should rotate our coordinate system hooked in the center of gravity of the figure. This point has just appeared here because it is a basic pattern with flat figures, but a more serious application will appear in the strength of materials guide. The pattern looks as follows.

tg2\alpha _{main}=-\frac{2*D_{X0Y0}}{I_{X0}-I_{Y0}}

We see that we use the data from the previous point to calculate the rotation angle.

  1. The main central moments of inertia.
    The last point will be the calculation of the main central moments of inertia. Here we will also use the data from point 3. Use the following formulas.J_{I}=\frac{I_{X0}+I_{Y0}}{2}+\sqrt{(\frac{I_{X0}-I_{Y0}}{2})^{2}+(D_{X0Y0})^{2}}
    J_{II}=\frac{I_{X0}+I_{Y0}}{2}-\sqrt{(\frac{I_{X0}-I_{Y0}}{2})^{2}+(D_{X0Y0})^{2}}Here the theoretical part ends, we already know all the patterns. Let’s go to calculate our example, let’s recall how our figure looks and let us assume the coordinate system (x, y) and see where the centers of individual simple figures are located (green dots).

  1. Division into simple figures
    1.1 Rectangle (S1), data:

A_{1} = 27 ,00cm^{2}
I_{X1} = 20,25cm^{4};
I_{Y1} = 182,25 cm^{4};
D_{x1y1} = 0,00cm^{4};

1.2. Square (S2), data:

A_{2} = 1 ,00cm^{2}
I_{X2} = 0,08cm^{4};
I_{Y2} = 0,08cm^{4};
D_{x2y2} = 0,00cm^{4};

1.3 Triangle (S3), data:

A_{3} = 13 ,50cm^{2}
I_{X3} = 6,75cm^{4};
I_{Y3} = 60,75cm^{4};
D_{x3y3} = -10,125cm^{4};

  1. The center of gravity of the figure
    S_{X}=\frac{(27,00*4,50)-(1,00*4,50)+(13,50*3,00)}{27,00-1,00+13,50}=3,99cm
    S_{Y}=\frac{(27,00*1,50)-(1,00*1,50)+(13,50*4,00)}{27,00-1,00+13,50}=2,35m
  2. Central moments of inertia
    \mathbf{I_{X0}}= (20,25+27,00*(-0,85)^{2})-(0,08+1,00*(-0,85)^{2})+(6,75+13,50*1,65^{2})=\mathbf{82,46cm^{4}}
    \mathbf{{I_{Y0}}}= (182,25+27,00*0,51^{2})-(0,08+1,00*0,51^{2})+(60,75+13,50*(-0,99)^{2})=\mathbf{{262,91cm^{4}}}
    \mathbf{D_{X0Y0}}= (0,00+27,00*(-0,85)*0,51)-(0,00+1,00*(-0,85)*0,51)+((-10,125)+13,50*1,65*(-0,99))=\mathbf{-43,45cm^{4}}
  3. Location of main central axes
    tg2\alpha _{main}=-\frac{2*(-43,45)}{82,46-262,91}=-\frac{-86,90}{-180,45}=-0,48\:\: \: /:tg
    2\alpha _{main}=-25,64^{0}\:\: \: /:2
    \alpha _{main}=-12,80^{0}
  4. The main central moments of inertiaJ_{I}=\frac{82,46+262,91}{2}+\sqrt{(\frac{82,46-262,91}{2})^{2}+(-43,45)^{2}}
    \mathbf{J_{I}}=172,69cm^{4}+100,14cm^{4}=\mathbf{272,83cm^{4}}

    J_{II}=\frac{82,46+262,91}{2}-\sqrt{(\frac{82,46-262,91}{2})^{2}+(-43,45)^{2}}
    \mathbf{J_{II}}=172,69cm^{4}-100,14cm^{4}=\mathbf{72,55cm^{4}}

Finally, of course, you should draw a drawing with the selected center of gravity and the rotated coordinate system by the calculated angle.

To check our calculations we can use the following formulas.

J_{xo} * J_{yo} - D_{xoyo}^{2} = J_{I} * J_{II}
J_{xo} + J_{yo} = J_{I} + J_{II}

We can also use the Mohr wheel.

This is the end of the General Mechanics manual, I hope that the materials and patterns collected and described by me will be useful. If you have a problem with something, you can always go back to the material you are interested in.