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 make this calculation, using only formulas for the transition from the central axes of each of the component parts to the 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 ​​of the central moments of inertia relative to the system of mutually perpendicular axes are known.

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.

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= Jz+ 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.

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 consider the change in moments of inertia when rotating the coordinate axes. Let us assume that the moments of inertia of a certain section relative to the axes are given x And y (not necessarily central). Need to determine J u , J v , J uv- moments of inertia about the axes u , v , rotated at an angle A. So projection OABC equal to the projection of the trailing one:

u= y sina +x cos a (1)

v=y cos a – x ​​sin a(2)

Let us exclude u, v in the expressions for moments of inertia:

J u = ∫v 2 dF; J v = ∫u 2 dF; J uv = ∫uvdF. Substituting into expressions (1) and (2) we get:

J u =J x cos 2 a–J xy sin 2a + J y sin 2 a

J v =J x sin 2 a+J xy sin 2a + J y cos 2 a(3)

J uv =J xy cos2a + sin 2a(J x -J y )/2

J u + J v = J x + J y = ∫ F (y 2 + x 2 ) dF => Sum of axial moments of inertia about 2x mutually perpendicular. Axes independent of angle A. notice, that x 2 + y 2 = p 2 . p- distance from the origin to the elementary site. That. J x + J y = J p .(4)

J p =∫ F p 2 dF – polar moment, independent of rotation x,y

2) T. Castelliano.

The partial derivative of the potential energy of the system with respect to the force is equal to the displacement of the point of application of the force in the direction of this force.

Let us consider a rod loaded by an arbitrary system of forces and fixed as shown in Fig.

Let the potential deformation energy accumulated in the volume of the body as a result of the work of external forces be equal to U. We will give the force F n an increment d F n . Then the potential energy U will increase
and will take the form U+
.(5.4)

Let us now change the order of application of forces. Let us first apply a force to the elastic body dPn. At the point of application of this force, a correspondingly small displacement will arise, the projection of which on the direction of the force dPn equal to . dδn. Then the work of force dPn turns out to be equal dPn dδn /2. Now let's apply the entire system of external forces. In the absence of strength dPn the potential energy of the system would again take on the value U. But now this energy will change by the amount of additional work dPn·δ n which the force will accomplish dPn on displacement δ n , caused by the entire system of external forces. The value δ n again represents the projection of the total displacement onto the direction of the force Pn.

As a result, with the reverse sequence of application of forces, we obtain the expression for potential energy in the form

(5.5)

We equate this expression to expression (5.4) and, discarding the product dPn dδn /2 as a quantity of higher order of smallness, we find

(5.6)

Ticket 23

Someone's out of luck

Ticket 24

1) Torsion of a rod of rectangular cross-section (determination of stresses and displacements). Torsion of a rectangular beam, stresses in the cross section

P In this case, the law of plane sections is violated; non-circular sections become distorted during torsion - deplanation of the cross section.

Diagrams of tangential stresses of a rectangular section.

;
, Jk and Wk are conventionally called the moment of inertia and the moment of resistance during torsion. Wk=hb2,

Jk= hb3, Maximum tangential stressesmax will be in the middle of the long side, stresses in the middle of the short side:=max, coefficients:,,are given in reference books depending on the ratio h/b (for example, at h /b=2,=0.246;=0.229;=0.795.

When calculating a beam for torsion (shaft), two main problems need to be solved. Firstly, it is necessary to determine the stresses arising in the beam, and, secondly, it is necessary to find the angular displacements of the sections of the beam depending on the magnitude of the external moments.

Let us assume that for an arbitrary section (Fig. 1.13) the moments of inertia relative to the coordinate axes z and y are known, and the centrifugal moment of inertia Izy is also known. It is required to establish dependencies for the moments of inertia about the 11 zy axes, rotated at an angle relative to the original z and y axes (Fig. 1.13). We will consider the angle positive if the rotation of the coordinate system occurs counterclockwise. Let for a given section IzI. yTo solve the problem, we will find the relationship between the coordinates of the site dA in the original and rotated axes. From Fig. 1.13 it follows: From a triangle from a triangle Taking this into account, we obtain Similarly for the coordinate y1 we obtain Considering that we finally have 1Using the obtained dependencies (1.23), (1.24) and expressions for the moments of inertia of the section (1.8), (1.9) and (1.11 ), we determine the moment of inertia relative to the new (rotated) axes z1 and y1: Similarly, the centrifugal moment of inertia I relative to the rotated axes is determined by the dependence After opening the brackets we get Adding, we get The sum of the moments of inertia relative to mutually perpendicular axes does not change when they rotate and is equal to the polar moment of inertia of the section . Subtracting (1.27) from (1.26) we obtain Formula (1.30) can be used to calculate the centrifugal moment of inertia about the z and y axes, based on the known moments of inertia about the z, y and z1, y1 axes, and formula (1.29) can be used to check the calculations of the moments inertia of complex sections. 1.8. Main axes and main moments of inertia of the section With a change in the angle (see Fig. 1.13), the moments of inertia also change. At some values ​​of the angle 0, the moments of inertia have extreme values. Axial moments of inertia having maximum and minimum values ​​are called the main axial moments of inertia of the section. The axes about which the axial moments of inertia have maximum and minimum values ​​are the main axes of inertia. On the other hand, as noted above, the main axes are the axes relative to which the centrifugal moment of inertia of the section is equal to zero. To determine the position of the main axes for sections of arbitrary shape, we take the first derivative with respect to I and equate it to zero: Where This formula determines the positions of two axes, relative to one of which the axial moment of inertia is maximum, and relative to the other - minimum. It should be noted that formula (1.31) can be obtained from (1.28) by equating it to zero. If we substitute the values ​​of the angle determined from expression (1.31) into (1. 26) and (1.27), then after the transformation we obtain formulas that determine the main axial moments of inertia of the section. In its structure, this formula is similar to formula (4.12), which determines the principal stresses (see Section 4.3). If IzI, then, based on studies of the second derivative, it follows that the maximum moment of inertia Imax occurs relative to the main axis rotated at an angle relative to the z axis, and the minimum moment of inertia occurs relative to the other main axis located at an angle of 0 If II, then everything changes the other way around. The values ​​of the main moments of inertia Imax and I can also be calculated from dependencies (1.26) and (1.27), if we substitute the value in them. In this case, the question is resolved by itself: relative to which main axis is the maximum moment of inertia obtained and relative to which axis is the minimum? It is necessary to note that if for a section the main central moments of inertia relative to the z and y axes are equal, then for this section any central axis is the main one and all the main central moments of inertia are the same (circle, square, hexagon, equilateral triangle, etc.). This is easily established from dependencies (1.26), (1.27) and (1.28). Indeed, let us assume that for some section the z and y axes are the main central axes and, in addition, I. yThen from formulas (1.26) and (1.27) we obtain that Izy, 1 and from formula (1.28) we are convinced that 11 e. any axes are the main central axes of inertia of such a figure. 1.9. The concept of the radius of inertia The moment of inertia of a section relative to any axis can be represented as the product of the cross-sectional area by the square of a certain value, called the radius of inertia of the cross-sectional area where iz ─ radius of inertia relative to the z axis. Then from (1.33) it follows: The main central axes of inertia correspond to the main radii of inertia: 1.10. Moments of resistance There are axial and polar moments of resistance. 1. The axial moment of resistance is the ratio of the moment of inertia about a given axis to the distance to the most distant point of the cross section from this axis. Axial moment of resistance relative to the z-axis: and relative to the y-axis: max where ymax and zmax─ respectively, the distances from the main central axes z and y to the points furthest from them. In the calculations, the main central axes of inertia and the main central moments are used, therefore, by Iz and Iy in formulas (1.36) and (1.37) we mean the main central moments of inertia of the section. Let's consider the calculation of the moments of resistance of some simple sections. 1. Rectangle (see Fig. 1.2): 2. Circle (see Fig. 1.8): 3. Tubular annular section (Fig. 1.14): . For rolled sections, the moments of resistance are given in the assortment tables and there is no need to determine them (see appendix 24 - 27). 2. The polar moment of resistance is the ratio of the polar moment of inertia to the distance from the pole to the most distant point of the section max 30. The center of gravity of the section is usually taken as the pole. For example, for a circular solid section (Fig. 1.14): For a tubular circular section. The axial moments of resistance Wz and Wy characterize, purely from the geometric side, the resistance of the rod (beam) to bending deformation, and the polar moment of resistance W is the resistance to torsion.