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Geometric characteristics of flat sections

Square: , dF - elementary platform.

Static moment of an area elementdF relative to axis 0x
- product of the area element by the distance "y" from the 0x axis: dS x = ydF

Having summed (integrated) such products over the entire area of ​​the figure, we obtain static moments relative to the y and x axes:
;
[cm 3, m 3, etc.].

Center of gravity coordinates:
. Static moments relative central axes(axes passing through the center of gravity of the section) are equal to zero. When calculating the static moments of a complex figure, it is divided into simple parts, with known areas F i and coordinates of the centers of gravity x i, y i. The static moment of the area of ​​the entire figure = the sum of the static moments of each of its parts:
.

Coordinates of the center of gravity of a complex figure:

M
Section inertia moments

Axial(equatorial) section moment of inertia- the sum of the products of elementary areas dF by the squares of their distances to the axis.

;
[cm 4, m 4, etc.].

The polar moment of inertia of a section relative to a certain point (pole) is the sum of the products of elementary areas by the squares of their distances from this point.
; [cm 4, m 4, etc.]. J y + J x = J p .

Centrifugal moment of inertia of the section- the sum of the products of elementary areas and their distances from two mutually perpendicular axes.
.

The centrifugal moment of inertia of the section relative to the axes, one or both of which coincide with the axes of symmetry, is equal to zero.

Axial and polar moments of inertia are always positive; centrifugal moments of inertia can be positive, negative or zero.

The moment of inertia of a complex figure is equal to the sum of the moments of inertia of its constituent parts.

Moments of inertia of sections of simple shape

P
rectangular section Circle

TO


ring

T
triangle

R
isofemoral

Rectangular

T
triangle

H quarter circle

J y =J x =0.055R 4

J xy =0.0165R 4

in Fig. (-)

Semicircle

M

The moments of inertia of standard profiles are found from the assortment tables:

D
vutavr Channel Corner

M

Moments of inertia about parallel axes:

J x1 =J x + a 2 F;

J y1 =J y + b 2 F;

the moment of inertia about any axis is equal to the moment of inertia about the central axis parallel to the given one, plus the product of the area of ​​the figure and the square of the distance between the axes. J y1x1 =J yx + abF; (“a” and “b” are substituted into the formula taking into account their sign).

Dependency between moments of inertia when turning the axes:

J x1 =J x cos 2  + J y sin 2  - J xy sin2; J y1 =J y cos 2  + J x sin 2  + J xy sin2;

J x1y1 =(J x - J y)sin2 + J xy cos2 ;

Angle >0, if the transition from the old coordinate system to the new one occurs counterclockwise. J y1 + J x1 = J y + J x

Extreme (maximum and minimum) values ​​of moments of inertia are called main moments of inertia. The axes about which the axial moments of inertia have extreme values ​​are called main axes of inertia. The main axes of inertia are mutually perpendicular. Centrifugal moments of inertia about the main axes = 0, i.e. main axes of inertia - axes about which the centrifugal moment of inertia = 0. If one of the axes coincides or both coincide with the axis of symmetry, then they are the main ones. Angle defining the position of the main axes:
, if  0 >0  the axes rotate counterclockwise. The maximum axis always makes a smaller angle with that of the axes relative to which the moment of inertia has a greater value. The main axes passing through the center of gravity are called main central axes of inertia. Moments of inertia about these axes:

J max + J min = J x + J y . The centrifugal moment of inertia relative to the main central axes of inertia is equal to 0. If the main moments of inertia are known, then the formulas for transition to rotated axes are:

J x1 =J max cos 2  + J min sin 2 ; J y1 =J max cos 2  + J min sin 2 ; J x1y1 =(J max - J min)sin2;

The ultimate goal of calculating the geometric characteristics of the section is to determine the main central moments of inertia and the position of the main central axes of inertia. R radius of inertia -
; J x =Fi x 2 , J y =Fi y 2 .

If J x and J y are the main moments of inertia, then i x and i y - principal radii of inertia. An ellipse built on the main radii of inertia as on the semi-axes is called ellipse of inertia. Using the ellipse of inertia, you can graphically find the radius of inertia i x1 for any axis x1. To do this, you need to draw a tangent to the ellipse, parallel to the x1 axis, and measure the distance from this axis to the tangent. Knowing the radius of inertia, you can find the moment of inertia of the section relative to the x 1 axis:
. For sections with more than two axes of symmetry (for example: circle, square, ring, etc.), the axial moments of inertia about all central axes are equal to each other, J xy = 0, the ellipse of inertia turns into a circle of inertia.

Moments of resistance.

Axial moment of resistance- the ratio of the moment of inertia about the axis to the distance from it to the most distant point of the section.
[cm 3, m 3]

Particularly important are the moments of resistance relative to the main central axes:

rectangle:
; circle: W x =W y =
,

tubular section (ring): W x =W y =
, where = d N / d B .

Polar moment of resistance - the ratio of the polar moment of inertia to the distance from the pole to the most distant point of the section:
.

For a circle W р =
.

The axial (or equatorial) moment of inertia of a section relative to a certain axis is the sum of the products of elementary areas taken over its entire area F by the squares of their distances from this axis, i.e.

The polar moment of inertia of a section relative to a certain point (pole) is the sum of the products of elementary areas taken over its entire area F by the squares of their distances from this point, i.e.

The centrifugal moment of inertia of a section relative to some two mutually perpendicular axes is the sum of the products of elementary areas taken over its entire area F and their distances from these axes, i.e.

Moments of inertia are expressed in, etc.

Axial and polar moments of inertia are always positive, since their expressions under the integral signs include the values ​​of the areas (always positive) and the squares of the distances of these areas from a given axis or pole.

In Fig. 9.5, a shows a section with area F and shows the y and z axes. Axial moments of inertia of this section relative to the y axes:

The sum of these moments of inertia

and therefore

Thus, the sum of the axial moments of inertia of a section relative to two mutually perpendicular axes is equal to the polar moment of inertia of this section relative to the intersection point of these axes.

Centrifugal moments of inertia can be positive, negative or zero. For example, the centrifugal moment of inertia of the section shown in Fig. 9.5, a, relative to the y and axes is positive, since for the main part of this section, located in the first quadrant, the values ​​of , and therefore, are positive.

If you change the positive direction of the y-axis or the opposite direction (Fig. 9.5, b) or rotate both of these axes by 90° (Fig. 9.5, c), then the centrifugal moment of inertia will become negative (its absolute value will not change), since the main part the section will then be located in a quadrant for which the y coordinates are positive and the z coordinates are negative. If you change the positive directions of both axes to the opposite, this will not change either the sign or the magnitude of the centrifugal moment of inertia.

Let's consider a figure that is symmetrical about one or more axes (Fig. 10.5). Let's draw the axes so that at least one of them (in this case, the y-axis) coincides with the axis of symmetry of the figure. In this case, each platform located to the right of the axis corresponds to the same platform located symmetrically to the first, but to the left of the y-axis. The centrifugal moment of inertia of each pair of such symmetrically located platforms is equal to:

Hence,

Thus, the centrifugal moment of inertia of the section relative to the axes, one or both of which coincide with its axes of symmetry, is equal to zero.

The axial moment of inertia of a complex section relative to a certain axis is equal to the sum of the axial moments of inertia of its constituent parts relative to the same axis.

Similarly, the centrifugal moment of inertia of a complex section relative to any two mutually perpendicular axes is equal to the sum of the centrifugal moments of inertia of its constituent parts relative to the same axes. Also, the polar moment of inertia of a complex section relative to a certain point is equal to the sum of the polar moments of inertia of its constituent parts relative to the same point.

It should be borne in mind that moments of inertia calculated about different axes and points cannot be summed.

When checking the strength of parts of structures, we have to encounter sections of rather complex shapes, for which it is impossible to calculate the moment of inertia in such a simple way as we used for a rectangle and a circle.

Such a section could be, for example, a T-bar (Fig. 5 A) annular section of a pipe subject to bending (aircraft structures) (Fig. 5, b), annular section of the shaft journal or even more complex sections. All these sections can be divided into simple ones, such as rectangles, triangles, circles, etc. It can be shown that the moment of inertia of such a complex figure is the sum of the moments of inertia of the parts into which we divide it.

Fig.5. T-type sections - a) and ring b)

It is known that the moment of inertia of any figure relative to the axis at— at equal to:

Where z— distance of elementary pads to the axis at—at.

Let's divide the taken area into four parts: , , and . Now, when calculating the moment of inertia, you can group the terms in the integrand function so as to separately perform the summation for each of the selected four areas, and then add these sums. This will not change the value of the integral.

Our integral will be divided into four integrals, each of which will cover one of the areas, , and:

Each of these integrals represents the moment of inertia of the corresponding part of the area relative to the axis at— at; That's why

where is the moment of inertia about the axis at — at area, - the same for area, etc.

The result obtained can be formulated as follows: the moment of inertia of a complex figure is equal to the sum of the moments of inertia of its constituent parts. Thus, we need to be able to calculate the moment of inertia of any figure relative to any axis lying in its plane.

The solution to this problem is the content of this and the next two interviews.

Moments of inertia about parallel axes.

The task of obtaining the simplest formulas for calculating the moment of inertia of any figure relative to any axis will be solved in several steps. If we take a series of axes parallel to each other, it turns out that we can easily calculate the moments of inertia of a figure about any of these axes, knowing its moment of inertia about an axis passing through the center of gravity of the figure parallel to the chosen axes.

Fig.1. Calculation model for determining moments of inertia for parallel axes.

We will call the axes passing through the center of gravity central axes. Let's take (Fig. 1) an arbitrary figure. Let's draw the central axis OU, we will call the moment of inertia about this axis . Let us draw an axis in the plane of the figure parallel axes at at a distance from her. Let's find the relationship between and - the moment of inertia about the axis. To do this, we will write expressions for and . Let's divide the area of ​​the figure into areas; the distances of each such platform to the axes at and let's call and . Then

From Fig. 1 we have:

The first of these three integrals is the moment of inertia about the central axis OU. The second is the static moment about the same axis; it is equal to zero, since the axis at passes through the center of gravity of the figure. Finally, the third integral is equal to the area of ​​the figure F. Thus,

(1)

that is, the moment of inertia about any axis is equal to the moment of inertia about the central axis parallel to the given one, plus the product of the area of ​​the figure and the square of the distance between the axes.

This means that our task has now been reduced to calculating only the central moments of inertia; if we know them, we can calculate the moment of inertia about any other axis. From formula (1) it follows that central moment of inertia is the smallest among the moments of inertia about parallel axes and for it we get:

Let us also find the centrifugal moment of inertia about the axes parallel to the central ones, if known (Fig. 1). Since by definition

where: , then it follows

Since the last two integrals represent static moments of area about the central axes OU And Oz then they vanish and, therefore:

(2)

The centrifugal moment of inertia relative to a system of mutually perpendicular axes parallel to the central ones is equal to the centrifugal moment of inertia relative to these central axes plus the product of the area of ​​the figure and the coordinates of its center of gravity relative to the new axes.

The relationship between the moments of inertia when turning the axes.

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

Let's take a figure and draw it through its center of gravity ABOUT two mutually perpendicular axes OU And Oz(Fig.2).

Fig.2. Calculation model for determining moments of inertia for rotated axes.

Let us know the axial moments of inertia about these axes, as well as the centrifugal moment of inertia. Let's draw a second system of coordinate axes and inclined to the first at an angle; we will consider the positive direction of this angle when rotating the axes around the point ABOUT counterclock-wise. Origin ABOUT save. 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:

Likewise:

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:

(7)

Thus, in order to calculate the moment of inertia about any central axis, you need to know the moments of inertia about the system of any two mutually perpendicular central axes OU And Oz, centrifugal moment of inertia relative to the same axes and the angle of inclination of the axis to the axis at.

To calculate the values ​​>, you have to choose the axes like this at And z 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 progress 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 ABOUT. Thus, formulas (6) and (7) are formulas for the transition from one system of mutually perpendicular axes to another, rotated by a certain angle, regardless of whether these 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

i.e. the sum of moments of inertia about any mutually perpendicular axes at And z does not change when they are rotated. Substituting the last expression instead of and their values, we get:

where is the distance of the sites dF from point ABOUT. The quantity is, as is already known, the polar moment of inertia of the section relative to the point ABOUT.

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.

Similarly, for a thin-walled annular section one can obtain:

Principal axes of inertia and principal moments of inertia.

As is already known, knowing the central moments of inertia , and for a given figure, you can calculate the moment of inertia relative to any other axis.

In this case, it is possible to take as the main system of axes such a system in which the formulas are significantly simplified. Namely, it is possible to find a system of coordinate axes for which the centrifugal moment of inertia is equal to zero. In fact, moments of inertia are always positive, like the sum of positive terms, but the centrifugal moment

can be both positive and negative, since the terms zydF may be of different sign depending on the signs z And at for one site or another. This means it can be equal to zero.

The axes about which the centrifugal moment of inertia vanishes are called main axes inertia. If the beginning of such a system is placed at the center of gravity of the figure, then these will be main central axes. We will denote these axes and ; for them

Let us find at what angle the main axes are inclined to the central axes y and z (Fig. 198).

Fig.1. Calculation model for determining the position of the main axes of inertia.

In the well-known expression for moving from axes yz to the axes, for the centrifugal moment of inertia we give the angle the value; then the axes and will coincide with the main ones, and the centrifugal moment of inertia will be equal to zero:

(1)

This equation is satisfied by two values ​​of , differing by 180°, or two values ​​of , differing by 90°. So this equation gives us the position two axes, forming a right angle with each other. These will be the main central axes and , for which .

Using this formula, you can use the known ones to obtain formulas for the main moments of inertia and . To do this, we again use the expressions for the general position axial moments of inertia. They determine the values ​​and if we substitute

(2)

The resulting relationships can be used to solve problems. One of the main moments of inertia is, another.

Formulas (2) can be transformed to a form free from the value . Expressing and through and substituting their values ​​into the first formula (2), we obtain, while simultaneously making the substitution from formula (1):

Replacing here the fraction from formula (1) with

we get

(3)

The same expression can be arrived at by making a similar transformation of the second formula (3).

For the main system of central axes, from which one can move to any other, one can take OU And Oz, and the main axes and ; then the centrifugal moment of inertia () will not appear in the formulas. Let us denote the angle made by the axis , (Fig. 2) with the main axis , by . To calculate , and , moving from the axes and , you need to replace the angle through , a , and in the previously found expressions for , and , and , and . As a result we get:

In appearance, these formulas are completely similar to the formulas for normal and shear stresses along two mutually perpendicular areas in an element subjected to tension in two directions. We will only indicate a formula that allows us to select from two angle values ​​the one that corresponds to the deviation of the first main axis (giving max J) from the initial position of the axis at:

Now we can finally formulate what needs to be done in order to be able to calculate in the simplest way the moment of inertia of a figure relative to any axis. It is necessary to draw axes through the center of gravity of the figure OU And Oz so that, breaking the figure into its simplest parts, we can easily calculate the moments passing at a distance (Fig. 2) from the center of gravity:

In many cases, it is possible to immediately draw the main axes of the figure; if a figure has an axis of symmetry, then this will be one of the main axes. In fact, when deriving the formula, we already dealt with the integral, which is the centrifugal moment of inertia of the section relative to the axes at And z; it has been proven that if the axis Oz is the axis of symmetry, this integral vanishes.

Therefore, in this case the axes OU And Oz are main the central axes of inertia of the section. Thus, axis of symmetry- always the main central axis; second home the central axis passes through the center of gravity perpendicular to the axis of symmetry.

Example. Find the moments of inertia of the rectangle (Fig. 3) relative to the axes and are equal to:

The moments of inertia about the axes and are equal to:

The centrifugal moment of inertia is equal to.

The method for calculating the moments of inertia of complex sections is based on the fact that any integral can be considered as a sum of integrals and, therefore, the moment of inertia of any section can be calculated as the sum of the moments of inertia of its individual parts.

Therefore, to calculate the moments of inertia, a complex section is divided into a number of simple parts (figures) in such a way that their geometric characteristics can be calculated using known formulas or found using special reference tables.

In some cases, when dividing into simple figures to reduce the number or simplify their shape, it is advisable to supplement the complex section with some areas. So, for example, when determining the geometric characteristics of the section shown in Fig. 22.5, a, it is advisable to add it to a rectangle, and then subtract the characteristics of the added part from the geometric characteristics of this rectangle. Do the same if there are holes (Fig. 22.5, b).

After dividing a complex section into simple parts, a rectangular coordinate system is selected for each of them, relative to which the moments of inertia of the corresponding part must be determined. All such coordinate systems are taken to be parallel to each other so that then, by parallel translation of the axes, it is possible to calculate the moments of inertia of all parts relative to the coordinate system common to the entire complex section.

As a rule, the coordinate system for each simple figure is assumed to be central, i.e. its origin coincides with the center of gravity of this figure. In this case, the subsequent calculation of moments of inertia when transitioning to other parallel axes is simplified, since the formulas for transition from central axes have a simpler form than from non-central axes.

The next step is to calculate the areas of each simple figure, as well as its axial and centrifugal moments of inertia relative to the axes of the coordinate system chosen for it. Static moments about these axes are, as a rule, equal to zero, since for each part of the section these axes are usually central. In cases where these are non-central axes, it is necessary to calculate static moments.

The polar moment of inertia is calculated only for a circular (solid or annular) section using ready-made formulas; for sections of other shapes, this geometric characteristic does not have any significance, since it is not used in calculations.

The axial and centrifugal moments of inertia of each simple figure relative to the axes of its coordinate system are calculated using the formulas or tables available for such a figure. For some figures, the available formulas and tables do not allow us to determine the required axial and centrifugal moments of inertia; in these cases it is necessary to use formulas for transition to new axes (usually for the case of rotation of axes).

The assortment tables do not indicate the values ​​of centrifugal moments of inertia for angles. The method for determining such moments of inertia is discussed in example 4.5.

In the vast majority of cases, the ultimate goal of calculating the geometric characteristics of a section is to determine its main central moments of inertia and the position of the main central axes of inertia. Therefore, the next stage of the calculation is to determine the coordinates of the center of gravity of a given section [using formulas (6.5) and (7.5)] in some arbitrary (random) coordinate system. Through this center of gravity of the section, auxiliary (not main) central axes are drawn parallel to the axes of the coordinate system of simple figures.

Then, using formulas establishing the relationships between the moments of inertia for parallel axes (see § 5.5), the moments of inertia of each simple figure relative to the auxiliary, central axes are determined. By summing the moments of inertia of each simple figure relative to the axes, the moments of inertia of the entire complex section relative to these axes are determined; in this case, the moments of inertia of holes or added pads are subtracted.

The moments of inertia of sections are called integrals of the following form:

at;

– axial moment of inertia of the section relative to the axis z;

– centrifugal moment of inertia of the section;

– polar moment of inertia of the section.

3.2.1. Properties of section moments of inertia

The dimension of moments of inertia is [length 4 ], usually [ m 4 ] or [ cm 4 ].

The axial and polar moments of inertia are always positive. The centrifugal moment of inertia can be positive, negative or zero.

Axes about which the centrifugal moment of inertia is zero are called main axes of inertia sections.

The axes of symmetry are always the main ones. If at least one of two mutually perpendicular axes is an axis of symmetry, then both axes are principal.

The moment of inertia of a composite section is equal to the sum of the moments of inertia of the elements of this section.

The polar moment of inertia is equal to the sum of the axial moments of inertia.

Let's prove the last property. In section with area A for an elementary site dA radius vector ρ and coordinates at And z(Fig. 6) are connected according to the Pythagorean theorem: ρ 2 = at 2 + z 2. Then

Rice. 6. Relationship between polar and Cartesian coordinates

elementary site

3.2.2. Moments of inertia of the simplest figures

IN rectangular section(Fig. 7) select an elementary platform dA with coordinates y And z and area dA = dydz.

Rice. 7. Rectangular section

Axial moment of inertia about the axis at

.

Similarly, we obtain the moment of inertia about the axis z:

Because the at And z– axis of symmetry, then the centrifugal moment D zy = 0.

For circle diameter d calculations are simplified if we take into account circular symmetry and use polar coordinates. Let us take as an elementary platform an infinitely thin ring with radius ρ and thickness dρ (Fig. 8). Its area dA= 2πρ dρ. Then the polar moment of inertia is:

.

Rice. 8. Round section

As shown above, the axial moments of inertia about any central axis are the same and equal

.

Moment of inertia rings we find as the difference between the moments of inertia of two circles - the outer one (with a diameter D) and internal (with a diameter d):

Moment of inertia I z triangle we will define it relative to the axis passing through the center of gravity (Fig. 9). Obviously, the width of an elementary strip located at a distance at from the axis z, is equal

Hence,

Rice. 9. Triangular section

3.3. Dependencies between moments of inertia relative to parallel axes

With known values ​​of the moments of inertia about the axes z And at let's determine the moments of inertia relative to other axes z 1 and y 1 parallel to the given ones. Using the general formula for axial moments of inertia, we find

If the axes z And y central, then
, And

From the obtained formulas it is clear that the moments of inertia about the central axes (when
) have the smallest values ​​compared to the moments of inertia about any other parallel axes.

3.4. Principal axes and principal moments of inertia

When the axes are rotated through an angle α, the centrifugal moment of inertia becomes equal to

.

Let us determine the position of the main principal axes of inertia u, v regarding which

,

where α 0 is the angle by which the axes must be rotated y And z so that they become the main ones.

Since the formula gives two angle values And
, then there are two mutually perpendicular principal axes. The maximum axis always makes a smaller angle ( ) with that of the axes ( z or y), relative to which the axial moment of inertia is of greater importance. Recall that positive angles are laid off from the axis z counterclockwise.

The moments of inertia about the principal axes are called main moments of inertia. It can be shown that they

.

The plus sign in front of the second term refers to the maximum moment of inertia, the minus sign to the minimum.