From formulas (6.22) – (6.25) it follows that when the axes rotate, the moments of inertia change, but the sum of axial moments remains constant.
Therefore, if relative to one axis the value of the moment of inertia is the largest, then relatively different – the smallest. In this case centrifugal moment relative to these axes it turns out equal to zero.
Main central axes are called axes passing through the center of gravity and relative to which the centrifugal moment is equal to zero, and axial moments relative to them (axes) have extremal properties and are called main central moments of inertia. Relative to one main axis, the moment of inertia has the smallest meaning – , relative to the other – the greatest.
We will denote these axes by letters u And v. Let us prove the above statement. Let the axes x And y– central axes of an asymmetrical section (Fig. 6.12).
Let us determine the position of the main axes by rotating the central axes by an angle at which the centrifugal moment becomes equal to zero.
.
Then from formula (6.25)
. (6.26)
Formula (6.26) determines the position of the main axes, where is the angle by which the central axes must be rotated so that they become the main ones. Negative angles are plotted clockwise from the axis x.
Now we will show that relative to the main axes, axial moments of inertia have the property of being extremal. Let's calculate the derivative of the expression (formula 6.22) and equate it to zero:
(6.27)
Comparing expressions (6.27) with (6.25) we establish that
.
It follows that the derivative vanishes when , which means that the extreme values have moments of inertia about the principal axes u And v. Then according to formulas (6.22) and (6.23):
(6.28)
Using formulas (6.28) we determine main central moments of inertia.
If we add formulas (6.28) term by term, then, obviously, . If we exclude the angle from formulas (6.28), we obtain a more convenient formula for the main central moments of inertia:
The “+” sign before the second term in (6.29) refers to , the “-” sign refers to .
It is useful to keep in mind special cases:
If the figure has two axes of symmetry, then these axes are main central axes.
2. For regular figures - equilateral triangle, square, circle, etc., having more than two axes of symmetry, all central axes are main, and the moments of inertia relative to them are equal to each other.
The ability to find the position of the main central axes and calculate and is necessary to determine plane of greatest rigidity of the section(the trace of which coincides with the axis) when calculating bending (Chapter 7).
35. General procedure for determining the main central
Moments.
Let it be required find the position of the main central axes and calculate the moments of inertia relative to them for a flat section consisting of a channel and a strip (Fig. 6.13):
Draw an arbitrary coordinate system xOy.
Divide the section into simple figures and use formulas (6.5) to determine the position of the center of gravity WITH.
Find the moments of inertia of simple figures relative to their own central axes using an assortment or formulas.
Through the point WITH draw central axes x c And y c parallel to the axes of simple figures.
Determine the moments of inertia of simple figures relative to the central axes of the section using the parallel translation formulas (6.13).
Determine the central moments of inertia of the entire section as the sum of the corresponding moments of simple figures found in step 5.
Calculate the angle using formula (6.26) and, turning the axes x c And y c at an angle, depict the main axes u And v.
Using formulas (6.29) calculate and .
Checking:
b) if ;
36) General procedure for determining the main central moments of inertia. Example:
1. If a figure has two axes of symmetry, then these axes will be the GCO.
2. For regular figures (which have more than 2 axes), all axes will be the main ones
3. Draw auxiliary axes (X’ O’ Y’)
4. We break this section into simple figures and show their own COs.
5. Find the position of the GCO using formula (21)
6. Calculate the GCM values using formula (23)
Imax + Imin = Ix + Iy
· Imax >Ix>Iy>Iminif Ix>Iy
Iuv = Ix-Iy/2 sin2a + Ixycos2a +0
Formula 21:Tg2a = - 2Ixy/Ix - Iy
Formula 23: Imax, Imin = *
37) Bend. Classification of types of bending. Straight and clean bend. Picture of beam deformation. Neutral layer and axis. Basic assumptions.
Bending is a deformation in which a bending moment Mx occurs in the cross section. Beam that works on bending beam
Types of bending:
Pure bending occurs if only a bending moment occurs in the section
Transverse bending - if a transverse force occurs simultaneously with the moment
Flat - all loads lie in the same plane
Spatial - if all loads lie in different longitudinal planes
Direct - if the force plane coincides with one of the main axes of inertia
Oblique - if the force plane does not coincide with any of the main axes
As a result of deformation in the area of pure bending, you can see:
Longitudinal fibers are bent along a circular arc: some are shortened, others are lengthened; between them there is a layer of fibers that do not change their length - the neutral layer (n.s.), the line of its intersection with the cross-sectional plane is called the neutral axis (n.a.)
The distance between the longitudinal fibers does not change
The cross sections, while remaining straight, rotate through a certain angle
Assumptions:
1. By pressing the longitudinal fibers against each other, i.e. each fiber is in a state of simple tension or compression, which is accompanied by the appearance of normal stresses Ϭ
2. On the validity of the Bernouli hypothesis, i.e. beam sections that are flat and normal to the axis before deformation remain flat and normal to its axis after deformation
axles, about which the centrifugal moment of inertia is equal to zero is called main axes(sometimes called main axes of inertia). Through any point taken in the section plane, in the general case, a pair of main axes can be drawn (in some special cases there can be an infinite number of them). In order to verify the validity of this statement, let us consider how the centrifugal moment of inertia changes when the axes are rotated by 90" (Fig. b.7). For an arbitrary area dA, taken in the first quadrant of the xOy axes system, both coordinates, and therefore their product is positive. In the new coordinate system x, Oy, rotated relative to the original by 90", the product of the coordinates of the site in question is negative. Absolute value this product does not change, i.e. xy = - x1y,. Obviously , the same is true for any other elementary site. This means that the sign of the sum dAxy, which is the centrifugal moment of inertia of the section, changes to the opposite when the axes are rotated by 90", i.e. J = = - J.
During the rotation of the axes, the centrifugal moment of inertia changes continuously, therefore, at a certain position of the axes it becomes equal to zero. These axes are the main ones.
Although we have established that the main axes can be drawn through any point of the section, only those of them that pass through the center of gravity of the section are of practical interest - main central axes. In what follows, as a rule, for brevity, we will simply call them main axes, omitting the word "central".
In the general case of sections of arbitrary shape, to determine the position of the main axes, it is necessary to conduct a special study. Here we will limit ourselves to considering special cases of sections that have at least one axis of symmetry (Fig. 6.8).
We'll guide you through. the center of gravity of the section is the Ox axis, perpendicular to the axis of symmetry Oy, and determine the centrifugal moment of inertia J. Let us use the property of a definite integral known from the course of mathematics (the integral of a sum is equal to the sum of integrals) and represent J s in the form of two terms:
since, for any elementary area located to the right of the symmetry axis, there is a corresponding one to the left, for which the product of coordinates differs only in sign.
Thus, the centrifugal moment of inertia relative to the Ox and Oy axes turned out to be equal to zero, i.e. this main axes. So, to find the main axes of a symmetrical section, it is enough to find the position of its center of gravity. One of the main central axes is the axis of symmetry, the second axis is perpendicular to it. Of course, the above proof remains valid if the axis perpendicular to the axis of symmetry does not pass through the center of gravity of the section, i.e. The axis of symmetry and any one perpendicular to it form a system of principal axes.
Non-central main axes, as already indicated, are not of interest.
Axial moments of inertia about the main central axes are called main central(or main ones for short) moments of inertia. The moment of inertia is maximum relative to one of the main axes, and minimum relative to the other. For example, for the section shown in Fig. 6.8, the maximum moment of inertia J
(relative to the Ox axis). Of course, when speaking about the extremity of the main moments of inertia, we only mean their comparison with other moments of inertia calculated relative to the axes passing through that the same section point. Thus, the fact that one of the principal moments of inertia is maximum and the other is minimum can be considered as an explanation for the fact that they (and the corresponding axes) are called principal. The equality to zero of the centrifugal moment of inertia relative to the main axes is a convenient sign for finding it. Some types of sections, for example circle, square, regular hexagon, etc. (Fig. 6.9), have countless main central axes. For these sections, any central axis is the main one.
Without providing proof, we point out that if the two main central moments of inertia of a section are equal, then for this section any central axis is the main one and all the main central moments of inertia are the same.
From formulas (6.29) - (6.31) it is clear that when the coordinate axes are rotated, the centrifugal moment of inertia changes sign, and therefore, there is a position of the axes at which the centrifugal moment is equal to zero.
The axes about which the centrifugal moment of inertia of the section vanishes are called the main axes, and the main axes passing through the center of gravity of the section are called main central axes of inertia of the section.
The moments of inertia about the main axes of inertia of the section are called main moments of inertia of the section and are denoted by I 1 And I 2 and I 1 > I 2 . Usually, when talking about main moments, they mean axial moments of inertia about the main central axes of inertia.
Let's assume that the axes u And v main ones. Then
|
|
.Equation (6.32) determines the position of the main axes of inertia of the section at a given point relative to the original coordinate axes. When rotating the coordinate axes, the axial moments of inertia also change. Let us find the position of the axes relative to which the axial moments of inertia reach extreme values. To do this, we take the first derivative of Iu By α and set it equal to zero:
.
The condition leads to the same result dIv/dα . Comparing the last expression with formula (6.32), we come to the conclusion that the main axes of inertia are the axes about which the axial moments of inertia of the section reach extreme values.
To simplify the calculation of the main moments of inertia, formulas (6.29) - (6.31) are transformed, excluding trigonometric functions from them using relation (6.32):
|
|
.The plus sign in front of the radical corresponds to greater I 1 , and the minus sign is smaller I 2 from the moments of inertia of the section.
Let us point out one important property of sections in which the axial moments of inertia relative to the main axes are the same. Let's assume that the axes y And z main ( Iyz=0), and Iy=Iz. Then, according to equalities (6.29) - (6.31), for any angle of rotation of the axes α centrifugal moment of inertia Iuv=0, and axial Iu= Iv.
So, if the moments of inertia of the section about the main axes are the same, then all axes passing through the same point of the section are the main ones and the axial moments of inertia about all these axes are the same: Iu= Iv= Iy= Iz. This property is possessed, for example, by square, round, and annular sections.
Formula (6.33) is similar to formulas (3.25) for principal stresses. Consequently, the main moments of inertia can be determined graphically by Mohr’s method.
Formulas (31.5), (32.5) and (34.5) allow us to establish how the values of the moments of inertia of the section change when the axes are rotated by an arbitrary angle a. For some values of the angle a, the values of the axial moments of inertia reach a maximum and a minimum. Extreme (maximum and minimum) values of the axial moments of inertia of the section are called the main moments of inertia. The axes about which the axial moments of inertia have extreme values are called the main axes of inertia.
From formula (33.5) it follows that if the axial moment of inertia relative to a certain axis is maximum (i.e., this axis is the main one), then the axial moment of inertia relative to the axis perpendicular to it is minimal (i.e., this axis is also the main one), so as the sum of axial moments of inertia about two mutually perpendicular axes does not depend on the angle a.
Thus, the main axes of inertia are mutually perpendicular.
To find the main moments of inertia and the position of the main axes of inertia, we determine the first derivative with respect to the angle a from the moment of inertia [see. formula (31.5) and Fig. 19.5]:
We equate this result to zero:
where is the angle by which the coordinate axes y must be rotated so that they coincide with the main axes.
Comparing expressions (35.5) and (34.5), we establish that
Consequently, relative to the main axes of inertia, the centrifugal moment of inertia is zero. Therefore, the main axes of inertia can be called axes about which the centrifugal moment of inertia is equal to zero.
As is already known, 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.
Consequently, mutually perpendicular axes, one or both of which coincide with the axes of symmetry of the section, are always the main axes of inertia. This rule allows in many cases to directly (without calculation) establish the position of the main axes.
Let's solve equation (35.5) with respect to the angle
In each specific case, equation (36.5) is satisfied by a number of values. Any one of them is selected. If it is positive, then to determine the position of one of the main axes of inertia from it, the axis should be rotated by an angle counterclockwise, and if negative, then by a clockwise rotation; the other main axis of inertia is perpendicular to the first. One of the main axes of inertia is the maximum axis (relative to it, the axial moment of inertia of the section is maximum), and the other is the minimum axis (relative to it, the axial moment of inertia of the section is minimum).
The maximum axis always makes a smaller angle with that of the axes (y or ), relative to which the axial moment of inertia has a greater value. This circumstance makes it easy to establish which of the main axes of inertia is the maximum axis, and which is the minimum axis. So, for example, if the main axes of inertia and and v are located, as shown in Fig. 20.5, then the axis is the maximum axis (since it forms a smaller angle with the y-axis than with the axis), and the v-axis is the minimum axis.
When solving a specific numerical problem to determine the main moments of inertia, you can substitute the selected angle value and the value into formula (31.5) or (32.5).
Let us solve this problem in general form. Using formulas from trigonometry, using expression (36.5), we find
Substituting these expressions into formula (31.5), after simple transformations we obtain
The main axes of inertia can be drawn through any point taken in the section plane. However, only the main axes passing through the center of gravity of the section, i.e., the main central inertias, are of practical importance for the calculations of structural elements. Moments of inertia relative to these axes (the main central moments of inertia) will be further denoted as
Let's consider several special cases.
1. If then formula (34.5) gives the value of the centrifugal moment of inertia relative to any pair of mutually perpendicular axes equal to zero, and, therefore, any axes obtained by rotating the coordinate system are the main axes of inertia (as well as the axes). In this case
2. For figures with more than two axes of symmetry, the axial moments of inertia about all central axes are equal. Indeed, let us direct one of the axes () along one of the axes of symmetry, and the other - perpendicular to it. For these axes If a figure has more than two axes of symmetry, then one of them makes an acute angle with the axis. Let us denote such an axis and the axis perpendicular to it
Centrifugal moment of inertia since the axis is the axis of symmetry. According to formula (34.5).
Task 5.3.1: For the section, the axial moments of inertia of the section relative to the axes are known x1, y1, x2: , . Axial moment of inertia about the axis y2 equal...
1) 1000 cm4; 2) 2000 cm4; 3) 2500 cm4; 4) 3000 cm4.
Solution: The correct answer is 3). The sum of the axial moments of inertia of the section relative to two mutually perpendicular axes when the axes are rotated through a certain angle remains constant, that is
After substituting the given values, we get:
Task 5.3.2: Of the indicated central axes of the section of an equal angle angle, the main ones are...
1) x3; 2) everything; 3) x1; 4) x2.
Solution: The correct answer is 4). For symmetrical sections, the axes of symmetry are the main axes of inertia.
Task 5.3.3: Main axes of inertia...
- 1) can only be drawn through points lying on the axis of symmetry;
- 2) can only be drawn through the center of gravity of a flat figure;
- 3) these are the axes about which the moments of inertia of a flat figure are equal to zero;
- 4) can be drawn through any point of a flat figure.
Solution: The correct answer is 4). The figure shows an arbitrary flat figure. Through the point WITH two mutually perpendicular axes are drawn U And V.
In the course on strength of materials it is proven that if these axes are rotated, then their position can be determined in which the centrifugal moment of inertia of the area becomes zero, and the moments of inertia about these axes take extreme values. Such axes are called main axes.
Task 5.3.4: Of the indicated central axes, the main section axes are...
1) everything; 2) x1 And x3; 3) x2 And x3; 4)x2 And x4.
Solution: The correct answer is 1). For symmetrical sections, the axes of symmetry are the main axes of inertia.
Task 5.3.5: Axes about which the centrifugal moment of inertia is zero and the axial moments take extreme values are called...
- 1) central axes; 2) axes of symmetry;
- 3) main central axes; 4) main axes.
Solution: The correct answer is 4). When the coordinate axes are rotated by an angle b, the moments of inertia of the section change.
Let the moments of inertia of the section relative to the coordinate axes be given x, y. Then the moments of inertia of the section in the system of coordinate axes u, v, rotated at a certain angle relative to the axes x, y, are equal
At a certain value of the angle, the centrifugal moment of inertia of the section becomes zero, and the axial moments of inertia take extreme values. These axes are called main axes.
Task 5.3.6: Moment of inertia of the section about the main central axis xC equal...
1); 2) ; 3) ; 4) .
Solution: The correct answer is 2)
To calculate we use the formula
