Consider a flat figure with known geometric characteristics 1 X , 1 y And 1 xy relative to the axes X And at(Fig. 3.3). Let us use them to determine the values ​​of similar geometric characteristics relative to the axes And And v, which make an angle a with the initial system.

Let us calculate the coordinates of the center of gravity of an infinitesimal element of area dA in the new coordinate system And And v:

Rice. 3.3.

Moment of inertia about the rotated axis Oi will be equal

Using the designations of geometric characteristics relative to the original axes, we obtain

For the other two geometric characteristics of the formula we obtain similarly:

We transform the resulting formulas using trigonometric formulas

After transforming the formula for calculating the axial and centrifugal moments of inertia when turning the axes, they take the form

Principal axes and principal moments of inertia

It was previously noted that the sum of axial moments is a constant value. It is easy to verify that this statement also follows from formulas (3.22):

The axes about which the moments of inertia take maximum and minimum values ​​are called main axes main moments of inertia.

When the axes are rotated, the magnitudes of the axial moments change, so there must be a pair of mutually perpendicular main axes, relative to which the moments of inertia reach minimum and maximum values. Let us prove this position. To do this, we examine the axial moment of inertia to the extreme 1 and:

Since the expression in parentheses must equal zero, we obtain a formula that allows us to determine the position of one of the main axes:

Angle a 0, measured from the axis Oh counterclockwise, determines the position of the main axis relative to the axis Oh. Let us prove that the axis perpendicular to this axis is also the main one. Let's substitute into the expression for

derivative angle a 0 + -:

Thus, the main axes are mutually perpendicular to the axes.

Let us pay attention to the fact that the expression in brackets, according to the third formula (3.22), corresponds to the centrifugal moment. Thus, we have proven that the centrifugal moment of inertia about the main axes is zero.

Let's use this result and derive a formula for calculating the main moments of inertia. To do this, we rewrite the second and third formulas (3.22) in the following form:

By squaring and adding the right and left sides of both equations, we get

This gives us the formula for calculating the two main moments of inertia:

In formula (3.25), the plus sign corresponds to the maximum principal moment of inertia, and the minus sign corresponds to its minimum value.

In some special cases, the position of the main axes can be determined without calculations. So, if the section is symmetrical, then the axis of symmetry is one of the main axes, and the second axis is any axis perpendicular to it. This position directly follows from the equality to zero of the centrifugal moment of inertia relative to the axes, one of which is the axis of symmetry.

Among all pairs of main axes, a special pair can be distinguished, both axes of which pass through the center of gravity of the section.

The main axes passing through the center of gravity of the section are called main central axes, and the moments of inertia about such axes are main central moments of inertia.

As already noted, rotation of the coordinate system causes a change in the geometric characteristics of plane figures. It can be shown that the set of geometric characteristics belonging to a given section is described by a symmetric tensor called inertia tensor cross section, which can be written as a matrix:

We obtained the first invariant of the inertia tensor, which is the sum of the axial moments of inertia, earlier (see formula (3.23)). The second invariant of the inertia tensor has the form

This value will be used to obtain a general solution for the bending of the rod.

Let us calculate the moments of inertia of a figure of arbitrary shape relative to axes rotated relative to given axes and
at an angle (Fig.4.14)

Let the moments of inertia about the axes
And
known. Let's choose an arbitrary site
and express its coordinates in the axis system
And
through coordinates in the previous axes
And
:

Let's find the axial and centrifugal moments of inertia of the figure relative to the rotated axes
And
:

Taking into account that

;
And
,

In the same way we will install:

The centrifugal moment of inertia takes the form:

. (4.30)

Let us express the axial moments through the sine and cosine of the double angle. To do this, we introduce the following functions:

. (4.31)

Substituting (4.31) into formulas (4.27) and (4.28), we obtain:

If we add up the expressions for the axial moments of inertia (4.32) and (4.33), we get:

Condition (4.34) represents the condition of invariance of the sum of axial moments of inertia relative to two mutually perpendicular axes, i.e. the sum of axial moments of inertia about two mutually perpendicular axes does not depend on the angle of rotation of the axes and is a constant value. Previously, this condition was obtained on the basis that the sum of the axial moments of inertia about two mutually perpendicular axes was equal to the value of the polar moment of inertia about the intersection point of these axes.

Let's study the equation for the moment of inertia to the extremum and find the value of the angle , at which the moment of inertia reaches an extreme value. To do this, we take the first derivative of the moment of inertia by angle (expression (4.32)) and the result is equal to zero. At the same time we put
.

(4.35)

The expression in parentheses represents the centrifugal moment of inertia about the axes inclined to the axis
at an angle . Relative to these axes, the centrifugal moment of inertia is zero:

, (4.36)

which means that the new axes are the main axes.

It was previously determined that the main axes of inertia are the axes about which the centrifugal moment of inertia is zero. Now this definition can be expanded - these are the axes about which the axial moments of inertia have extreme values. The moments of inertia about these axes are called main moments of inertia.

Let's find the position of the main axes of inertia. From expression (4.36) we can obtain:

. (4.37)

The resulting formula gives for the angle two meanings: And
.

Consequently, there are two mutually perpendicular axes about which the moments of inertia have extreme values. As noted above, such axes are called the main axes of inertia. It remains to establish regarding which of the axes the moment of inertia reaches the maximum value, and regarding which one the moment of inertia reaches the minimum value. This problem can be solved by studying the second derivative of expression (4.32) with respect to the angle . Substituting the value of the angle into the expression for the second derivative or
and by examining the sign of the second derivative, one can judge which of the angles corresponds to the maximum moment of inertia and which to the minimum. Below are formulas that will give an unambiguous angle value .

Let's find extreme values ​​for moments of inertia. To do this, we transform expression (4.32), taking it out of brackets
:

We use a function known from trigonometry and substitute expression (4.37) into it, we get:

. (4.39)

Substituting expression (4.39) into formula (4.38) and performing the necessary calculations, we obtain two expressions for extreme moments of inertia, which do not include the angle of inclination of the axes :

; (4.40)

. (4.41)

From formulas (4.40) and (4.41) it is clear that the values ​​of the main moments of inertia are determined directly through the moments of inertia relative to the axes
And
. Therefore, they can be determined without knowing the position of the main axes themselves.

Knowing the extreme values ​​of moments of inertia
And
In addition to formula (4.37), it is possible to determine the position of the main axes of inertia.

We present formulas without derivation that allow us to find angles And between the axis
and main axes:

;
(4.42)

Corner determines the position of the axis relative to which the moment of inertia reaches its maximum value (
), corner determines the position of the axis relative to which the moment of inertia reaches a minimum value (
).

Let us introduce another geometric characteristic, which is called the radius of gyration of the section. This characteristic is designated by the letter and can be calculated relative to the axes
And
in the following way:

;
(4.43)

The radius of inertia is widely used in problems of strength of materials and its application will be discussed in the following sections of the course.

Let's consider several examples of structural calculations taking into account the rotation of the axes and using the radius of gyration of the section.

Example 4.7. The moments of inertia of a rectangular section relative to the main axes are equal, respectively
cm 4,
cm 4. When turning by 45 0, the moments of inertia relative to the new axes turned out to be the same. What is their size?

To solve the problem, we use expression (4.28), taking into account the fact that the centrifugal moment of inertia relative to the main axes is equal to zero:

Let us substitute into formula (a) the numerical values ​​for the moments of inertia and the angle of rotation of the axes:

Example 4.8. Which of the figures (Fig. 4.15) having the same area has a radius of gyration relative to the axis , will be the greatest? Determine the largest radius of gyration of the section relative to the axis .

1. Find the area of ​​each of the figures and the dimensions of the sections. The area of ​​the figures is equal to cm 2 for the third figure.

We find the diameter of the first section from the expression:

cm.

Square side size:

Triangle base:

cm.

2. Find the moments and radii of inertia of each section relative to the central axis .

For a round section:

cm 4;
cm.

For a square section:

cm 4;
cm.

For a rectangular section:

;

For a triangular section:

cm 4;
cm.

The largest radius of gyration turned out to be for a rectangular section and it is equal to
cm.

Let's calculate the moments of inertia J u, J v and J uv:

Adding the first two formulas (3.14), we obtain J u + Jv= J z+ Jy, i.e. for any rotation of mutually perpendicular axes, the sum of the axial moments of inertia remains a constant value (invariant).

Principal axes and principal moments of inertia

Let's explore the function J u(a) to the extremum. To do this, we equate the derivative to zero J u(a) by a.

We obtain the same formula by equating the centrifugal moment of inertia to zero

.

Principal axes are called axes about which the axial moments of inertia take extreme values, and the centrifugal moment of inertia is equal to zero.

An infinite number of main axes of inertia can be drawn by taking any point on the plane as the origin. To solve problems of strength of materials, we are only interested in main central axes of inertia. Main central axes of inertia pass through the center of gravity of the section.

Formula (3.17) gives two solutions that differ by 90°, i.e. allows you to determine two values ​​of the angle of inclination of the main axes of inertia relative to the original axes. Relative to which of the axes is the maximum axial moment of inertia obtained? J 1 = J max , and relative to which – minimum J 2 = J min , will have to be solved according to the meaning of the problem.

More convenient are other formulas that unambiguously determine the position of the main axes 1 and 2 (given without derivation). In this case, the positive angle is measured from the axis Oz counterclock-wise.

In formula (3.19), the “+” sign corresponds to the maximum moment of inertia, and the “–” sign to the minimum.

Comment . If a section has at least one axis of symmetry, then relative to this axis and any other perpendicular to it, the centrifugal moment of inertia is zero. In accordance with the definition of the main axes of inertia, we can conclude that these axes are the main axes of inertia, i.e. the axis of symmetry is always the main central axis.

For symmetrical profiles presented in the assortment, channel or I-beam, the main central axes of inertia will be the vertical and horizontal axes, intersecting at half the height of the profile.

Geometric characteristics of complex composite cross sections

If a cross section is formed by a set of simplest elements, then, in accordance with the properties of certain integrals, the geometric characteristic of such a section is equal to the sum of the corresponding characteristics of the individual composite sections (Fig. 3.10).

Rice. 10.

Thus, to calculate the moments of inertia of a complex figure, it is necessary to divide it into a series simple figures, calculate the moments of inertia of these figures and then sum these moments of inertia

Changing moments of inertia when turning axes

Let's find the relationship between the moments of inertia about the axes and the moments of inertia about the axes rotated at an angle (Fig. 3.11). Let the positive angle be measured from the axis counterclockwise.

Rice. eleven. Rotating coordinate axes

To solve the problem, let’s find the relationship between the coordinates of an infinitesimal area in the original and rotated axes

Now let's determine the moments of inertia about the axes

Likewise

For centrifugal moment

Adding (3.28) and (3.29), we get

Subtracting (3.28) from (3.29), we obtain

Formula (3.31) shows that the sum of the moments of inertia about any mutually perpendicular axes does not change when they rotate.

Formula (3.32) can be used to calculate the centrifugal moment of inertia about the axes from the known axial moments of inertia about the and axes.

Principal axes of inertia and principal moments of inertia

When the angle changes (Fig. 3.10), the moments of inertia (3.280 - (3.31) change. Let us find the value of the angle at which and have an extreme value. To do this, take the first derivative of and with respect to and equate it to zero:

This formula determines the position of two axes, relative to which the axial moment of inertia is maximum, and relative to the other, minimum. Such axes are called main axes. Moments of inertia about the principal axes are called principal moments of inertia.

We will find the values ​​of the main moments of inertia from formulas (3.28) and (3.29, substituting into them from formula (3.33), while using the known trigonometry formulas for functions of double angles. After the transformation, we obtain a formula for determining the main moments of inertia:

Let us now show that relative to the main axes the centrifugal moment of inertia is zero. Indeed, equating to zero using formula (3.30), we obtain

whence for we again obtain formula (3.33)

Thus, the main axes are called axes that have the following properties:

The centrifugal moment of inertia about these axes is zero.

The moments of inertia relative to the main axes have extreme values ​​(relative to one - maximum, relative to the other - minimum).

The principal axes coming through the center of gravity of the section are called the principal central axes.

In many cases, it is possible to immediately determine the position of the main central axes. If a figure has an axis of symmetry, then it is one of the main central axes, the second passing through the center of gravity of the section perpendicular to the first. This follows from the fact that relative to the axis of symmetry and any axis perpendicular to it, the centrifugal moment of inertia is equal to zero.

You can draw as many central axes as you like. The question is whether it is possible to express the moment of inertia about any central axis depending on the moment of inertia about one or two specific axes. To do this, let's see how the moments of inertia change about two mutually perpendicular axes when they are rotated through an angle.

Let's take any figure and draw two mutually perpendicular axes Oy and Oz through its center of gravity O (Fig. 2).

Rice. 2.

Let us know the axial moments of inertia relative to these axes, as well as the centrifugal moment of inertia. Let us draw a second system of coordinate axes and those inclined to the first at an angle; The positive direction of this angle will be considered when the axes rotate around point O counterclockwise. We save the origin of coordinates O. Let us express the moments relative to the second system of coordinate axes and, through the known moments of inertia and.

Let us write expressions for the moments of inertia about these axes:

From the drawing it can be seen that the coordinates of the dF site in the system of rotated axes will be:

Substituting these values ​​into formulas (14.9), we obtain:

or moment of inertia flat axis

Likewise:

The first two integrals of expressions (4) and (5) represent the axial moments of inertia and, and the last one represents the centrifugal moment of inertia of the area relative to these axes. Then:

To solve problems, you may need formulas for transition from one axes to others for the centrifugal moment of inertia. When rotating the axes (Fig. 2) we have:

where and are calculated using formulas (14.10); Then

After transformations we get:

Thus, in order to calculate the moment of inertia relative to any central axis, it is necessary to know the moments of inertia relative to the system of any two mutually perpendicular central axes Oy and Oz, the centrifugal moment of inertia relative to the same axes, and the angle of inclination of the axis to the y axis.

To calculate the values ​​>, you have to choose the y and z axes in such a way and divide the area of ​​the figure into such component parts as to be able to carry out this calculation using only formulas for transition from the central axes of each of components to axes parallel to them. How to do this in practice will be shown below using an example. Note that in this calculation, complex figures must be divided into such elementary parts for which, if possible, the values ​​are known central points inertia relative to a system of mutually perpendicular axes.

Note that the course of the derivation and the results obtained would not have changed if the origin of coordinates had been taken not at the center of gravity of the section, but at any other point O. Thus, formulas (6) and (7) are formulas for the transition from one system to the other axes perpendicular to another, rotated through a certain angle, regardless of whether they are central axes or not.

From formulas (6) one can obtain another relationship between the moments of inertia when turning the axes. Adding the expressions for and we get

those. the sum of the moments of inertia about any mutually perpendicular axes y and z does not change when they rotate. Substituting the last expression instead of and their values, we get:

where is the distance of the pads dF from point O. The value is, as is already known, the polar moment of inertia of the section relative to point O.

Thus, the polar moment of inertia of a section relative to any point is equal to the sum of the axial moments of inertia relative to mutually perpendicular axes passing through this point. Therefore, this sum remains constant when the axes are rotated. This dependence (14.16) can be used to simplify the calculation of moments of inertia. So, for a circle:

Since by symmetry for a circle then

which was obtained above by integration.

The same can be obtained for a thin-walled annular section.