01.02.2022
Geometric characteristics of flat sections. Change in the moments of inertia of the rod during parallel transfer of axes Formulas for transfer of axes
Let us consider the determination of the moments of inertia of a plane figure (Fig) relative to the $(Z_1)$ and $(Y_1)$ axes with known moments of inertia relative to the $X$ and $Y$ axis.
$(I_((x_1))) = \int\limits_A (y_1^2dA) = \int\limits_A (((\left((y + a) \right))^2)dA) = \int\limits_A ( \left(((y^2) + 2ay + (a^2)) \right)dA) = \int\limits_A ((y^2)dA) + 2a\int\limits_A (ydA) + (a^2 )\int\limits_A (dA) = $
$ = (I_x) + 2a(S_x) + (a^2)A$,
where $(S_x)$ - static moment of the figure about the $X$ axis.
Similarly with respect to the $(Y_1)$ axis
$(I_((y_1))) = (I_y) + 2a(S_y) + (b^2)A$.
Centrifugal moment of inertia about the $(X_1)$ and $(Y_1)$ axes
$(I_((x_1)(y_1))) = \int\limits_A ((x_1)(y_1)dA) = \int\limits_A (\left((x + b) \right)\left((y + a ) \right)dA) = \int\limits_A (\left((xy + xa + by + ba) \right)dA) = \int\limits_A (xydA) + a\int\limits_A (xdA) + b\int \limits_A (ydA) + ab\int\limits_A (dA) = (I_(xy)) + a(S_x) + b(S_y) + abA$
Most often, the transition is used from central axes (the proper axes of a flat figure) to arbitrary, parallel ones. Then $(S_x) = 0$, $(S_y) = 0$, since the axes $X$ and $Y$ are central. Finally we have
Where , - own moments of inertia, i.e. moments of inertia about their own central axes;
$a$, $b$ - distances from the central axes to the considered ones;
$A$ is the area of the figure.
It should be noted that when determining the centrifugal moment of inertia in quantities $a$ and $b$, the sign must be taken into account, that is, they are essentially the coordinates of the center of gravity of the figure in the axes under consideration. When determining axial moments of inertia, these quantities are substituted modulo (as distances), since they still rise to a square.
Using parallel translation formulas, it is possible to transition from central axes to arbitrary axes, or vice versa- from arbitrary central axes. The first transition is carried out with a "+" sign. The second transition is carried out with the sign "- ".
Examples of using transition formulas between parallel axes
Rectangular section
Let us determine the central moments of inertia of the rectangle for known moments of inertia about the $Z$ and $Y$ axes.
$(I_x) = \frac((b(h^3)))(3)$; $(I_y) = \frac((h(b^3)))(3)$.
.
Similarly $(I_y) = \frac((h(b^3)))((12))$.
Triangular section
Let us determine the central moments of inertia of the triangle with a known moment of inertia relative to the base $(I_x) = \frac((b(h^3)))((12))$.
.
The triangle has a different configuration relative to the central axis $(Y_c)$, so consider the following. The moment of inertia of the entire figure relative to the $(Y_c)$ axis is equal to the sum of the moment of inertia of the triangle $ABD$ relative to the $(Y_c)$ axis and the moment of inertia of the $CBD$ triangle relative to the $(Y_c)$ axis, that is
.
Determination of the moment of inertia of a composite section
We consider a section to be composed; it consists of individual elements whose geometric characteristics are known. The area, static moment and moments of inertia of a composite figure are equal to the sum of the corresponding characteristics of their components. If a cross-section can be formed by cutting one figure from another, the geometric characteristics of the cut out figure are subtracted. For example, the moments of inertia of the composite figure shown in Fig. will be determined as follows
$I_z^() = \frac((120 \cdot ((22)^3)))((12)) - 2 \cdot \frac((50 \cdot ((16)^3)))((12 )) = 72\,300$ cm 4 .
$I_y^() = \frac((22 \cdot ((120)^3)))((12)) - 2 \cdot \left((\frac((16 \cdot ((50)^3)) )((12)) + 50 \cdot 16 \cdot ((29)^2)) \right) = 1\,490\,000$cm 4
Often, when solving practical problems, it is necessary to determine the moments of inertia of a section relative to axes oriented in different ways in its plane. At the same time, it is convenient to use known values moments of inertia of the entire section (or its individual constituent parts) relative to other axes, given in technical literature, special reference books and tables, as well as calculated using available formulas. Therefore, it is very important to establish the relationships between the moments of inertia of the same section relative to different axes.
In the most general case, moving from any old to any new system coordinates can be considered as two successive transformations of the old coordinate system:
1) by parallel transfer of coordinate axes to a new position and
2) by rotating them relative to the new origin. Let's consider the first of these transformations, i.e. parallel translation of coordinate axes.
Let us assume that the moments of inertia of a given section relative to the old axes (Fig. 18.5) are known.
Let's take a new coordinate system whose axes are parallel to the previous ones. Let us denote a and b the coordinates of the point (i.e., the new origin) in the old coordinate system
Let's consider an elementary platform. Its coordinates in the old coordinate system are equal to y and . In the new system they are equal
Let us substitute these coordinate values into the expression for the axial moment of inertia relative to the axis
In the resulting expression, the moment of inertia, the static moment of the section relative to the axis, is equal to the area F of the section.
Hence,
If the z axis passes through the center of gravity of the section, then the static moment and
From formula (25.5) it is clear that the moment of inertia about any axis that does not pass through the center of gravity is greater than the moment of inertia about the axis passing through the center of gravity, by an amount that is always positive. Consequently, of all the moments of inertia about parallel axes, the axial moment of inertia has smallest value relative to an axis passing through the center of gravity of the section.
Moment of inertia about the axis [by analogy with formula (24.5)]
In the particular case when the y-axis passes through the center of gravity of the section
Formulas (25.5) and (27.5) are widely used in calculating axial moments of inertia of complex (composite) sections.
Let us now substitute the values into the expression for the centrifugal moment of inertia relative to the axes
2. Static moments of the cross-sectional area relative to the axes Oz And Oh(cm 3, m 3):
4. Centrifugal moment of inertia of the section relative to the axes Oz And Oy(cm 4, m 4):
Since then
Axial Jz And Jy and polar J p moments of inertia are always positive, since the coordinates to the second power are under the integral sign. Static moments Sz And S y, as well as the centrifugal moment of inertia J zy can be both positive and negative.
The range of rolled steel for angles gives the values of centrifugal moments modulo. Their values should be entered into the calculation taking into account the sign.
To determine the sign of the centrifugal moment of the corner (Fig. 3.2), we mentally imagine it as the sum of three integrals, which are calculated separately for parts of the section located in the quarters of the coordinate system. Obviously, for the parts located in the first and third quarters we will have a positive value of this integral, since the product zydA will be positive, and the integrals calculated for the parts located in the II and IV quarters will be negative (the product zydA will be negative). Thus, for the corner in Fig. 3.2, and the value of the centrifugal moment of inertia will be negative.
Reasoning in a similar way for a section that has at least one axis of symmetry (Fig. 3.2,b), we can come to the conclusion that the centrifugal moment of inertia J zy is equal to zero if one of the axes (Oz or Oy) is the axis of symmetry of the section. Indeed, for parts of the triangle located in the 1st and 2nd quarters, the centrifugal moments of inertia will differ only in sign. The same can be said about the parts that are in the III and IV quarters.
Static moments. Determining the center of gravity
Let's calculate the static moments about the axes Oz And Oh rectangle shown in Fig. 3.3.
Figure 3.3. Towards the calculation of static moments
Here: A- cross-sectional area, y C And z C– coordinates of its center of gravity. The center of gravity of the rectangle is at the intersection of the diagonals.
It is obvious that if the axes about which static moments are calculated pass through the center of gravity of the figure, then its coordinates are equal to zero ( z C = 0, y C= 0), and, in accordance with formula (3.6), the static moments will also be equal to zero. Thus, the center of gravity of a section is a point that has the following property: static moment about any axis passing through it,equal to zero.
Formulas (3.6) allow us to find the coordinates of the center of gravity z C And y C sections of complex shape. If the section can be represented in the form n parts for which the areas and positions of the centers of gravity are known, then the calculation of the coordinates of the center of gravity of the entire section can be written in the form:
  .
| (3.7) |
Change in moments of inertia at parallel transfer axes
Let the moments of inertia be known Jz, Jy And J zy relative to the axes Oyz. It is necessary to determine the moments of inertia JZ, JY And JZY relative to the axes O 1 YZ, parallel to the axes Oyz(Fig. 3.4) and separated from them at a distance a(horizontally) and b(vertically)
Figure 3.4. Changing moments of inertia during parallel translation of axes
Coordinates of the elementary site dA are related to each other by the following equalities: Z = z + a; Y = y + b.
Let's calculate the moments of inertia JZ, JY And JZY.
 | (3.8) |
 | (3.9) |
 | (3.10) |
If the point O axis intersections Oyz coincides with the point WITH– center of gravity of the section (Fig. 3.5) static moments Sz And S y become equal to zero, and the formulas are simplifiedY i and Z i must be taken taking into account the signs. The coordinate signs will not affect the axial moments of inertia (the coordinates are raised to the second power), but the coordinate sign will have a significant effect on the centrifugal moment of inertia (the product Z i Y i A i may turn out to be negative).
The axes passing through the center of gravity of a plane figure are called central axes.
The moment of inertia about the central axis is called the central moment of inertia.
Theorem
The moment of inertia about any axis is equal to the sum of the moment of inertia about the central axis parallel to the given one and the product of the area of the figure and the square of the distance between the axes.
To prove this theorem, consider an arbitrary plane figure whose area is equal to A
, the center of gravity is located at the point WITH
, and the central moment of inertia about the axis x
will Ix
.
Let's calculate the moment of inertia of the figure relative to a certain axis x 1
, parallel to the central axis and spaced from it at a distance A
(rice).
I x1 = Σ y 1 2 dA + Σ (y + a) 2 dA =
= Σ y 2 dA + 2a Σ y dA + a 2 Σ dA.
Analyzing the resulting formula, we note that the first term is the axial moment of inertia relative to the central axis, the second term is the static moment of the area of this figure relative to the central axis (hence, it is equal to zero), and the third term after integration can be represented as a product a 2 A , i.e., as a result we get the formula:
I x1 = I x + a 2 A- the theorem is proven.
Based on the theorem, we can conclude that of a series of parallel axes, the axial moment of inertia of a flat figure will be the smallest relative to the central axis .
Principal axes and principal moments of inertia
Let us imagine a flat figure whose moments of inertia relative to the coordinate axes Ix And I y , and the polar moment of inertia relative to the origin is equal to I ρ . As was established earlier,
I x + I y = I ρ.
If the coordinate axes are rotated in their plane around the origin of coordinates, then the polar moment of inertia will remain unchanged, and the axial moments will change, while their sum will remain constant. Since the sum of variables is constant, one of them decreases and the other increases, and vice versa.
Consequently, at a certain position of the axes, one of the axial moments will reach the maximum value, and the other - the minimum.
Axes about which the moments of inertia have a minimum and maximum value, are called the main axes of inertia.
The moment of inertia about the main axis is called the main moment of inertia.
If the principal axis passes through the center of gravity of a figure, it is called the principal central axis, and the moment of inertia about such an axis is called the principal central moment of inertia.
We can conclude that if a figure is symmetrical about any axis, then this axis will always be one of the main central axes of inertia of this figure.
Centrifugal moment of inertia
The centrifugal moment of inertia of a flat figure is the sum of the products of elementary areas taken over the entire area by the distance up to two mutually perpendicular axes:
I xy = Σ xy dA,
Where x
, y
- distances from the site dA
to the axles x
And y
.
The centrifugal moment of inertia can be positive, negative or zero.
The centrifugal moment of inertia is included in the formulas for determining the position of the main axes of asymmetrical sections.
Standard profile tables contain a characteristic called radius of gyration of the section
, calculated by the formulas:
i x = √ (I x / A),i y = √ (I y / A) , (hereinafter the sign"√"- root sign)
Where I x , I y
- axial moments of inertia of the section relative to the central axes; A
- cross-sectional area.
This geometric characteristic is used in the study of eccentric tension or compression, as well as longitudinal bending.
Torsional deformation
Basic concepts about torsion. Torsion of a round beam.
Torsion is a type of deformation in which only a torque occurs in any cross section of the beam, i.e. a force factor that causes a circular movement of the section relative to an axis perpendicular to this section, or prevents such movement. In other words, torsional deformations occur if a pair or pairs of forces are applied to a straight beam in planes perpendicular to its axis.
The moments of these pairs of forces are called twisting or rotating. Torque is denoted by T
.
This definition conventionally divides the force factors of torsional deformation into external ones (torsional, torque T
) and internal (torques M cr
).
In machines and mechanisms, round or tubular shafts are most often subjected to torsion, so strength and rigidity calculations are most often made for such units and parts.
Consider the torsion of a circular cylindrical shaft.
Imagine a rubber cylindrical shaft in which one of the ends is rigidly fixed, and on the surface there is a grid of longitudinal lines and transverse circles. We will apply a couple of forces to the free end of the shaft, perpendicular to the axis of this shaft, i.e. we will twist it along the axis. If you carefully examine the grid lines on the surface of the shaft, you will notice that:
- the shaft axis, which is called the torsion axis, will remain straight;
- the diameters of the circles will remain the same, and the distance between adjacent circles will not change;
- longitudinal lines on the shaft will turn into helical lines.
From this we can conclude that when a round cylindrical beam (shaft) is torsioned, the hypothesis of flat sections is valid, and we can also assume that the radii of the circles remain straight during deformation (since their diameters have not changed). And since there are no longitudinal forces in the shaft sections, the distance between them is maintained.
Consequently, the torsional deformation of a round shaft consists in the rotation of the cross sections relative to each other around the torsion axis, and their rotation angles are directly proportional to the distances from the fixed section - the further any section is from the fixed end of the shaft, the greater the angle relative to the shaft axis it twists .
For each section of the shaft, the angle of rotation is equal to the angle of twist of the part of the shaft enclosed between this section and the seal (fixed end).
Corner ( rice. 1) rotation of the free end of the shaft (end section) is called the full angle of twist of the cylindrical beam (shaft).
Relative twist angle φ 0
called the torsion angle ratio φ 1
to the distance l 1
from a given section to the embedment (fixed section).
If the cylindrical beam (shaft) is long l
has a constant cross-section and is loaded with a torsional moment at the free end (i.e., consists of a homogeneous geometric section), then the following statement is true:
φ 0 = φ 1 / l 1 = φ / l = const
- the value is constant.
If we consider thin layer on the surface of the above rubber cylindrical bar ( rice. 1), limited by a grid cell cdef , then we note that this cell warps during deformation, and its side, remote from the fixed section, shifts towards the twist of the beam, occupying the position cde 1 f 1 .
It should be noted that a similar picture is observed during shear deformation, only in this case the surface is deformed due to translational movement of sections relative to each other, and not due to rotational movement, as in torsional deformation. Based on this, we can conclude that during torsion in cross sections, only tangential internal forces (stresses) arise, forming a torque.
So, the torque is the resulting moment relative to the axis of the beam of internal tangential forces acting in the cross section.
Let us determine the relationship between the various moments of inertia of the section relative to two parallel axes (Fig. 6.7), connected by the dependencies
1. For static moments of inertia
Finally,
2. For axial moments of inertia
hence,
If the axis z passes through the center of gravity of the section, then
Of all the moments of inertia about parallel axes, the axial moment of inertia has the smallest value about the axis passing through the center of gravity of the section.
Same for axis
When the axis y passes through the center of gravity of the section
3. For the centrifugal moments of inertia we obtain
Finally we can write
In the case when the origin of the coordinate system yz is at the center of gravity of the section, we get
In the case where one or both axes are axes of symmetry,
6.7. Changing moments of inertia when turning axes
Let the moments of inertia of the section relative to the coordinate axes be given zy.
The angle is considered positive if the old coordinate system needs to be rotated counterclockwise to move to the new one (for a right-handed Cartesian rectangular coordinate system). New and old zy coordinate systems are connected by dependencies that follow from Fig. 6.8:
1. Let us define expressions for axial moments of inertia relative to the axes of the new coordinate system:
Similarly with respect to the axis
If we add up the values of the moments of inertia about the and axes, we get
that is, when the axes rotate, the sum of the axial moments of inertia is a constant value.
2. Let us derive formulas for centrifugal moments of inertia.
.
6.8. Main moments of inertia. Main axes of inertia
The extreme values of the axial moments of inertia of the section are called the principal moments of inertia.
Two mutually perpendicular axes, about which the axial moments of inertia have extreme values, are called the main axes of inertia.
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 of the moment of inertia, determined by formula (6.27)
Let's equate this result to zero:
where is the angle by which the coordinate axes need to be rotated y And z so that they coincide with the main axes.
Comparing expressions (6.30) and (6.31), we can establish that
,
Consequently, relative to the main axes of inertia, the centrifugal moment of inertia is zero.
Mutually perpendicular axes, one or both of which coincide with the axes of symmetry of the section, are always the main axes of inertia.
Let's solve equation (6.31) for angle:
.
If >0, then to determine the position of one of the main axes of inertia for the right (left) Cartesian rectangular coordinate system, an axis is needed z turn at an angle against the direction of rotation (in the direction of rotation) clockwise. If<0, то для определения положения одной из главных осей инерции для правой (левой) декартовой прямоугольной системы координат необходимо осьz turn at an angle in the direction of rotation (counterclockwise) clockwise.
The maximum axis always makes a smaller angle with that of the axes ( y or z), relative to which the axial moment of inertia has a greater value (Fig. 6.9).
The maximum axis is directed at an angle to the axis(), if() and is located in the even (odd) quarters of the axes, if().
Let us determine the main moments of inertia and. Using formulas from trigonometry connecting the functions,,,with the functions,,from formula (6.27) we obtain
,
