Axial moment of inertia. Centrifugal moments of inertia. Centrifugal moments for bodies having an axis or plane of symmetry Principal and centrifugal moments of inertia

If m = 1, n = 1, then we get the characteristic

which is called centrifugal moment of inertia.

Centrifugal moment of inertia relative to the coordinate axes – the sum of the products of elementary areas dA at their distances to these axes, taken over the entire cross-sectional area A.

If at least one of the axes y or z is the axis of symmetry of the section, the centrifugal moment of inertia of such a section relative to these axes is equal to zero (since in this case each positive value z·y·dA we can put in correspondence exactly the same, but negative, on the other side of the axis of symmetry of the section, see figure).

Let us consider additional geometric characteristics that can be obtained from the main ones listed and are also often used in calculations of strength and stiffness.

Polar moment of inertia

Polar moment of inertia Jp name the characteristic

On the other side,

Polar moment of inertia(relative to a given point) – the sum of the products of elementary areas dA by the squares of their distances to this point, taken over the entire cross-sectional area A.

The dimension of moments of inertia is m 4 in SI.

Moment of resistance

Moment of resistance relative to some axis – a value equal to the moment of inertia relative to the same axis divided by the distance ( ymax or z max) to the point most distant from this axis

The dimension of the moments of resistance is m 3 in SI.

Radius of inertia

Radius of inertia section relative to a certain axis is called a value determined from the relationship:

Radii of gyration are expressed in SI units of m.

Comment: cross-sections of elements of modern structures often represent a certain composition of materials with different resistance to elastic deformation, characterized, as is known from a physics course, by Young’s modulus E. In the most general case of an inhomogeneous section, Young’s modulus is a continuous function of the coordinates of the points of the section, i.e. E = E(z, y). Therefore, the rigidity of a section that is inhomogeneous in elastic properties is characterized by characteristics that are more complex than the geometric characteristics of a homogeneous section, namely elastic-geometric ones of the form

2.2. Calculation of geometric characteristics of simple figures

Rectangular section

Let us determine the axial moment of inertia of the rectangle relative to the axis z. Let us divide the area of the rectangle into elementary areas with dimensions b(width) and dy(height). Then the area of such an elementary rectangle (shaded) is equal to dA = b dy. Substituting the value dA into the first formula, we get

By analogy, we write the axial moment about the axis at:

Axial moments of resistance of a rectangle:

;

In a similar way, you can obtain geometric characteristics for other simple figures.

Round section

It's convenient to find first polar moment of inertia J p .

Then, given that for a circle J z = J y, A J p = J z + J y, we'll find Jz =Jy = Jp / 2.

Let us divide the circle into infinitesimal rings of thickness dρ and radius ρ ; area of such a ring dA = 2 ∙ π ∙ ρ ∙ dρ. Substituting the expression for dA into an expression for Jp and integrating, we get

2.3. Calculation of moments of inertia about parallel axes

z And y:

It is required to determine the moments of inertia of this section relative to the “new” axes z 1 And y 1, parallel to the central ones and spaced from them at a distance a And b respectively:

Coordinates of any point in the “new” coordinate system z 1 0 1 y 1 can be expressed through coordinates in the “old” axes z And y So:

Since the axes z And y– central, then static moment S z = 0.

Finally, we can write down the “transition” formulas for parallel transfer of axes:

Note that the coordinates a And b must be substituted taking into account their sign (in the coordinate system z 1 0 1 y 1).

2.4. Calculation of moments of inertia when rotating coordinate axes

Let the moments of inertia of an arbitrary section relative to the central axes be known z, y:

; ;

Let's turn the axes z, y at an angle α counterclockwise, considering the angle of rotation of the axes in this direction to be positive.

It is required to determine the moments of inertia relative to the “new” (rotated) axes z 1 And y 1:

Coordinates of the elementary site dA in the “new” coordinate system z 1 0y 1 can be expressed through coordinates in the “old” axes like this:

We substitute these values into the formulas for the moments of inertia in the “new” axes and integrate term by term:

Having made similar transformations with the remaining expressions, we will finally write down the “transition” formulas when rotating the coordinate axes:

Note that if we add the first two equations, we get

i.e. the polar moment of inertia is the quantity invariant(in other words, unchanged when rotating the coordinate axes).

2.5. Principal axes and principal moments of inertia

Until now, the geometric characteristics of sections in an arbitrary coordinate system have been considered, but the coordinate system in which the section is described by the smallest number of geometric characteristics is of greatest practical interest. This “special” coordinate system is specified by the position of the main axes of the section. Let's introduce the concepts: main axes And main moments of inertia.

Main axes– two mutually perpendicular axes, relative to which the centrifugal moment of inertia is equal to zero, while axial moments inertia take extreme values (maximum and minimum).

The main axes passing through the center of gravity of the section are called main central axes.

The moments of inertia about the principal axes are called main moments of inertia.

The main central axes are usually designated by letters u And v; main moments of inertia – J u And J v(a-priory J uv = 0).

Let us derive expressions that allow us to find the position of the main axes and the magnitude of the main moments of inertia. Knowing that J uv= 0, we use equation (2.3):

Corner α 0 defines the position of the principal axes relative to any central axes z And y. Corner α 0 deposited between the axis z and axis u and is considered positive in the counterclockwise direction.

Note that if a section has an axis of symmetry, then, in accordance with the property of the centrifugal moment of inertia (see section 2.1, paragraph 4), such an axis will always be the main axis of the section.

Excluding Angle α in expressions (2.1) and (2.2) using (2.4), we obtain formulas for determining the main axial moments of inertia:

Let's write down the rule: the maximum axis always makes a smaller angle with that of the axes (z or y) relative to which the moment of inertia has a greater value.

2.6. Rational forms of cross sections

Normal stresses at an arbitrary point in the cross section of a beam at straight bend are determined by the formula:

, (2.5)

Where M– bending moment in the cross section under consideration; at– the distance from the point under consideration to the main central axis perpendicular to the plane of action of the bending moment; J x– the main central moment of inertia of the section.

The greatest tensile and compressive normal stresses in a given cross section occur at points furthest from the neutral axis. They are determined by the formulas:

; ,

Where at 1 And at 2– distances from the main central axis X to the most distant stretched and compressed fibers.

For beams made of plastic materials, when [σ p ] = [σ c ] ([σ p ], [σ c ] are the permissible stresses for the beam material in tension and compression, respectively), sections symmetrical about the central axis are used. In this case, the strength condition has the form:

≤ [σ], (2.6)

Where W x = J x / y max– moment of resistance of the cross-sectional area of the beam relative to the main central axis; ymax = h/2(h– section height); M max– the largest bending moment in absolute value; [σ] – permissible bending stress of the material.

In addition to the strength condition, the beam must also satisfy the economy condition. The most economical are those cross-sectional shapes for which the largest moment of resistance is obtained with the least amount of material (or with the smallest cross-sectional area). In order for the section shape to be rational, it is necessary, if possible, to distribute the section away from the main central axis.

For example, a standard I-beam is approximately seven times stronger and thirty times stiffer than a square beam of the same cross-section made of the same material.

It must be borne in mind that when the position of the section changes in relation to the acting load, the strength of the beam changes significantly, although the cross-sectional area remains unchanged. Consequently, the section must be positioned so that the force line coincides with that of the main axes relative to which the moment of inertia is minimal. You should strive to ensure that the bend of the beam takes place in the plane of its greatest rigidity.

product of inertia, one of the quantities characterizing the distribution of masses in a body ( mechanical system). C. m. and. are calculated as sums of products of masses m to points of the body (system) to two of the coordinates x k, y k, z k these points:

Values of C. m. and. depend on the directions of the coordinate axes. In this case, for each point of the body there are at least three such mutually perpendicular axes, called the main axes of inertia, for which the centrifugal mass and. are equal to zero.

The concept of C. m. and. plays important role when studying rotational movement tel. From the values of C. m. and. depend on the magnitude of the pressure forces on the bearings in which the axis of the rotating body is fixed. These pressures will be the smallest (equal to static) if the axis of rotation is the main axis of inertia passing through the center of mass of the body.

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Word forms

"Centrifugal moment of inertia" in books

Contrary to inertia

From the book Sphinxes of the 20th century author Petrov Rem Viktorovich

Contrary to inertia

From the book Sphinxes of the 20th century author Petrov Rem Viktorovich

Contrary to inertia “In the last two decades, the immunological nature of tissue transplant rejection has become generally accepted and all aspects of the rejection processes are under strict experimental control.” Leslie Brent Fingerprints So, to the question “What

By inertia

From the book How Much is a Person Worth? The story of the experience in 12 notebooks and 6 volumes. author

By inertia

From the book How Much is a Person Worth? Notebook ten: Under the “wing” of the mine author Kersnovskaya Evfrosiniya Antonovna

By inertia To appreciate the landscape, you need to look at the picture from some distance. To correctly assess an event, a certain distance is also needed. The law of inertia was in effect. While the spirit of change has reached Norilsk, for a long time it seemed that everything was sliding along

24. Power of Inertia

From the book Ethereal Mechanics author Danina Tatyana

24. Force of Inertia The ether emitted by the rear hemisphere of an inertially moving particle is the Force of Inertia. This Inertial Force is the repulsion of the Ether that fills the particle with the Ether emitted by itself. The magnitude of the Inertial Force is proportional to the speed of emission

3.3.1. Submersible centrifugal pump

From the book Your Own Plumber. Plumbing country communications author Kashkarov Andrey Petrovich

3.3.1. Submersible centrifugal pump In this section, we will consider the option with the NPTs-750 submersible centrifugal pump. I use spring water from April to October. I pump it with a submersible centrifugal pump NPTs-750/5nk (the first number indicates the power consumption in watts,

If we draw coordinate axes through point O, then with respect to these axes the centrifugal moments of inertia (or products of inertia) are the quantities defined by the equalities:

where are the masses of points; - their coordinates; it is obvious that, etc.

For solid bodies, formulas (10), by analogy with (5), take the form

Unlike axial ones, centrifugal moments of inertia can be both positive and negative quantities and, in particular, with a certain way of choosing axes, they can become zero.

Main axes of inertia. Let us consider a homogeneous body having an axis of symmetry. Let us draw the coordinate axes Oxyz so that the axis is directed along the axis of symmetry (Fig. 279). Then, due to symmetry, each point of a body with mass mk and coordinates will correspond to a point with a different index, but with the same mass and with coordinates equal to . As a result, we obtain that since in these sums all terms are pairwise identical in magnitude and opposite in sign; from here, taking into account equalities (10), we find:

Thus, symmetry in the distribution of masses relative to the z axis is characterized by the vanishing of two centrifugal moments of inertia. The Oz axis, for which the centrifugal moments of inertia containing the name of this axis in their indices are equal to zero, is called the main axis of inertia of the body for point O.

From the above it follows that if a body has an axis of symmetry, then this axis is the main axis of inertia of the body for any of its points.

The principal axis of inertia is not necessarily the axis of symmetry. Let us consider a homogeneous body that has a plane of symmetry (in Fig. 279 the plane of symmetry of the body is the plane ). Let us draw some axes and an axis perpendicular to them in this plane. Then, due to symmetry, each point with mass and coordinates will correspond to a point with the same mass and coordinates equal to . As a result, as in the previous case, we find that or whence it follows that the axis is the main axis of inertia for point O. Thus, if a body has a plane of symmetry, then any axis perpendicular to this plane will be the main axis of inertia of the body for point O, in which the axis intersects the plane.

Equalities (11) express the conditions that the axis is the main axis of inertia of the body for point O (origin).

Similarly, if then the Oy axis will be the main axis of inertia for point O. Therefore, if all centrifugal moments of inertia are equal to zero, i.e.

then each of the coordinate axes is the main axis of inertia of the body for point O (origin).

For example, in Fig. 279 all three axes are the main axes of inertia for point O (the axis is the axis of symmetry, and the Ox and Oy axes are perpendicular to the planes of symmetry).

The moments of inertia of a body relative to the main axes of inertia are called the main moments of inertia of the body.

The main axes of inertia constructed for the center of mass of the body are called the main central axes of inertia of the body. From what was proved above it follows that if a body has an axis of symmetry, then this axis is one of the main central axes of inertia of the body, since the center of mass lies on this axis. If the body has a plane of symmetry, then the axis perpendicular to this plane and passing through the center of mass of the body will also be one of the main central axes of inertia of the body.

In the examples given, symmetrical bodies were considered, which is sufficient to solve the problems we will encounter. However, it can be proven that through any point of any body it is possible to draw at least three mutually perpendicular axes for which equalities (11) will be satisfied, i.e., which will be the main axes of inertia of the body for this point.

The concept of principal axes of inertia plays an important role in the dynamics of a rigid body. If the coordinate axes Oxyz are directed along them, then all centrifugal moments of inertia turn to zero and the corresponding equations or formulas are significantly simplified (see § 105, 132). This concept is also associated with the solution of problems on the dynamic equation of rotating bodies (see § 136), on the center of impact (see § 157), etc.

The axial moment of inertia is equal to the sum of the products of the elementary areas and the square of the distance to the corresponding axis.

(8)

The sign is always "+".

Cannot be equal to 0.

Property: Takes a minimum value when the intersection point of the coordinate axes coincides with the center of gravity of the section.

The axial moment of inertia of a section is used in calculations of strength, rigidity and stability.

1.3. Polar moment of inertia of the section Jρ

(9)

Relationship between polar and axial moments of inertia:

(10)

(11)

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

Property:

When the axes are rotated in any direction, one of the axial moments of inertia increases and the other decreases (and vice versa). The sum of the axial moments of inertia remains constant.

1.4. Centrifugal moment of inertia of the section Jxy

The centrifugal moment of inertia of the section is equal to the sum of the products of the elementary areas and the distances to both axes

(12)

Unit of measurement [cm 4 ], [mm 4 ].

Sign "+" or "-".

, if the coordinate axes are axes of symmetry (example - I-beam, rectangle, circle), or one of the coordinate axes coincides with the axis of symmetry (example - channel).

Thus, for symmetrical figures the centrifugal moment of inertia is 0.

Coordinate axes u And v , passing through the center of gravity of the section, about which the centrifugal moment is equal to zero, are called the main central axes of inertia of the section. They are called main because the centrifugal moment relative to them is zero, and central because they pass through the center of gravity of the section.

For sections that are not symmetrical about the axes x or y , for example, at the corner, will not be equal to zero. For these sections, the position of the axes is determined u And v by calculating the rotation angle of the axes x And y

(13)

Centrifugal moment about the axes u And v -

Formula for determining axial moments of inertia about the principal central axes u And v :

(14)

Where
- axial moments of inertia relative to the central axes,

- centrifugal moment of inertia relative to the central axes.

1.5. Moment of inertia about an axis parallel to the central one (Steiner's theorem)

Steiner's theorem:

The moment of inertia about an axis parallel to the central one is equal to the central axial moment of inertia plus the product of the area of the entire figure and the square of the distance between the axes.

(15)

Proof of Steiner's theorem.

According to Fig. 5 distance at to the elementary site dF

Substituting the value at into the formula, we get:

Term
, since point C is the center of gravity of the section (see the property of static moments of the sectional area relative to the central axes).

For a rectangle with heighth and widthb :

Axial moment of inertia:

Bending moment:

the moment of resistance to bending is equal to the ratio of the moment of inertia to the distance of the most distant fiber from the neutral line:

because
, That

For a circle:

Polar moment of inertia:

Axial moment of inertia:

Torsional moment:

Because
, That

Bending moment:

Example 2. Determine the moment of inertia of a rectangular cross-section about the central axis WITH x .

Solution. Let's divide the area of the rectangle into elementary rectangles with dimensions b (width) and dy (height). Then the area of such a rectangle (shaded in Fig. 6) is equal to dF=bdy. Let's calculate the value of the axial moment of inertia J x

By analogy we write

- axial moment of inertia of the section relative to the central

Centrifugal moment of inertia

, since the axes WITH x and C y are axes of symmetry.

Example 3. Determine the polar moment of inertia of a circular cross-section.

Solution. Let's divide the circle into infinitely thin rings of thickness
radius , the area of such a ring
. Substituting the value
Integrating into the expression for the polar moment of inertia, we obtain

Taking into account the equality of axial moments of a circular section
And

, we get

The axial moments of inertia for the ring are equal

With– the ratio of the cutout diameter to the outer diameter of the shaft.

Lecture No. 2 “Main axes andmain pointsinertia

Let's consider how the moments of inertia change when the coordinate axes are rotated. Let us assume that the moments of inertia of a certain section relative to the 0 axes are given X, 0at(not necessarily central) - ,- axial moments of inertia of the section. Need to determine ,- axial moments about the axes u,v, rotated relative to the first system by an angle
(Fig. 8)

Since the projection of the broken line OABC is equal to the projection of the trailing line, we find:

(15)

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

(18)

Let's consider the first two equations. Adding them term by term, we get

Thus, the sum of axial moments of inertia about two mutually perpendicular axes does not depend on the angle
and remains constant when the axes are rotated. Let us note at the same time that

Where - distance from the origin of coordinates to the elementary site (see Fig. 5). Thus

Where - the already familiar polar moment of inertia:

Let us determine the axial moment of inertia of the circle relative to the diameter.

Since due to symmetry
but, as you know,

Therefore, for a circle

With changing the angle of rotation of the axes
moment values And change, but the amount remains the same. Therefore there is such a meaning
, at which one of the moments of inertia reaches its maximum value, while the other moment takes on a minimum value. Differentiating the expression by angle
and equating the derivative to zero, we find

(19)

At this angle value
one of the axial moments will be the largest, and the other will be the smallest. At the same time, the centrifugal moment of inertia
vanishes, which can be easily verified by equating the formula for the centrifugal moment of inertia to zero
.

Axes about which the centrifugal moment of inertia is zero and the axial moments take extreme values are called mainaxes. If they are also central (the point of origin coincides with the center of gravity of the section), then they are called main central axes (u; v). Axial moments of inertia about the principal axes are called main moments of inertia -And

And their value is determined by the following formula:

(20)

The plus sign corresponds to the maximum moment of inertia, the minus sign to the minimum.

There is another geometric characteristic - radius of gyration sections. This value is often used in theoretical conclusions and practical calculations.

The radius of gyration of the section relative to a certain axis, for example 0 x , is called the quantity , determined from equality

(21)

F – cross-sectional area,

- axial moment of inertia of the section,

From the definition it follows that the radius of gyration is equal to the distance from the axis 0 X to the point at which the cross-sectional area F should be concentrated (conditionally) so that the moment of inertia of this one point is equal to the moment of inertia of the entire section. Knowing the moment of inertia of the section and its area, you can find the radius of gyration relative to the 0 axis X:

(22)

The radii of gyration corresponding to the main axes are called main radii of inertia and are determined by the formulas

(23)

Lecture 3. Torsion of rods of circular cross-section.

DEFINITION

Axial (or equatorial) moment of inertia section relative to the axis is called a quantity that is defined as:

Expression (1) means that to calculate the axial moment of inertia, the sum of the products of infinitesimal areas () multiplied by the squares of the distances from them to the axis of rotation is taken over the entire area S:

The sum of the axial moments of inertia of the section relative to mutually perpendicular axes (for example, relative to the X and Y axes in the Cartesian coordinate system) gives the polar moment of inertia () relative to the intersection point of these axes:

DEFINITION

Polar moment inertia is called the moment of inertia section with respect to some point.

Axial moments of inertia are always greater than zero, since in their definitions (1) under the integral sign there is the value of the area of the elementary area (), always positive, and the square of the distance from this area to the axis.

If we are dealing with a section of complex shape, then often in calculations we use the fact that the axial moment of inertia of a complex section relative to the axis is equal to the sum of the axial moments of inertia of the parts of this section relative to the same axis. However, it should be remembered that it is impossible to sum up the moments of inertia that are found relative to different axes and points.

The axial moment of inertia relative to the axis passing through the center of gravity of the section has the smallest value of all moments relative to the axes parallel to it. The moment of inertia about any axis () provided that it is parallel to the axis passing through the center of gravity is equal to:

where is the moment of inertia of the section relative to the axis passing through the center of gravity of the section; - cross-sectional area; - distance between axles.

Examples of problem solving

EXAMPLE 1

Exercise

What is the axial moment of inertia of an isosceles triangular cross-section relative to the Z axis passing through the center of gravity () of the triangle, parallel to its base? The height of the triangle is .

Solution

Let us select a rectangular elementary area on a triangular section (see Fig. 1). It is located at a distance from the axis of rotation, the length of one side is , the other side is . From Fig. 1 it follows that:

The area of the selected rectangle, taking into account (1.1), is equal to:

To find the axial moment of inertia, we use its definition in the form:

Answer

EXAMPLE 2

Exercise	Find the axial moments of inertia relative to the perpendicular axes X and Y (Fig. 2) of a section in the form of a circle whose diameter is equal to d.
Solution	To solve the problem, it is more convenient to start by finding the polar moment relative to the center of the section (). Let us divide the entire section into infinitely thin rings of thickness , the radius of which will be denoted by . Then we find the elementary area as: