For simple sections static moments and moments of inertia are found using formulas (2.1)-(2.4) using integration. Consider, for example, calculating the axial moment of inertia J x for an arbitrary section shown in Fig. 2.9. Considering that in a rectangular coordinate system the area element dF=dxdy, we get

wherex^(y) and x in (y) - coordinates of contour points at some fixed value u.

Performing integration over x, we find

Magnitude b(y) represents the section width at level at(see Fig. 2.9), and the product b(y)dy = dF - area of ​​the shaded elementary strip parallel to the axis Oh. Taking this into account, the formula for / is transformed to the form

A similar expression can be obtained for the moment of inertia Jy.

Rectangle. Let us find the moments of inertia about the main central axes, which, in accordance with property 2 (§ 2.5), coincide with the symmetry axes of the rectangle (Fig. 2.10). Since the width of the section is constant, then using formula (2.14) we obtain

Moment of inertia about the axis Oh x x x we determine by the first of formulas (2.6):

Moments of inertia / and J are found similarly. Let us write down the formulas for the axial moments of inertia of a rectangle:

Arbitrary triangle. First, let's find the moment of inertia about the axis 0 ( x v passing through the base of the triangle (Fig. 2.11). Section width b(y()) at the level y ( is found from the similarity of triangles:

Substituting this quantity into formula (2.14) and performing integration, we obtain

Moments about the axes Oh And 0 2 x 2, parallel to the base and passing through the center of gravity and through the vertex of the triangle, respectively, we find using formulas (2.6):

In these formulas b ( =h/ 3 and b 2 = -2h/3 - respectively, the ordinates of the center of gravity of the triangle ABOUT in the coordinate system O x x 1 y 1 And 0 2 x 2 y t

1 ° 2 r Г* аУ 1

TL P *2

g >4™_ °2 1

D__V_!_*_ / ^ *3

V XV* ;-7^Лт^

U_ У-_XI - UZ__у

ABOUT,| b *, 0 b/ b 2 %*1

Rice. 2.11 Rice. 2.12

Let us write the formulas for the axial moments of inertia of the triangle relative to the axes parallel to the base:

Right and isosceles triangles. For a right triangle (Fig. 2.12), we determine the centrifugal moment of inertia J relative to the central axes Oh And OU, parallel to the legs. This can be done using formula (2.3). However, the solution to the problem can be simplified by applying the following technique. Using the median 0 { 0 3 divide the given triangle into two isosceles triangles 0 ( 0 3 A And Ofi 3 B. Axes 0 3 x 3 and 0 3 y 3 are the axes of symmetry for these triangles and, based on property 2 (§ 2.5), will be the main axes of each of them separately, and therefore of the entire triangle O x AB. Therefore, the centrifugal moment of inertia J=0. Centrifuge

moment of the triangle about the axes Oh And OU we find using the last of formulas (2.6):

Let us write down the formulas for the moments of inertia of a right triangle:

Moment of inertia isosceles triangle relative to the axis of symmetry OU(Fig. 2.13) we define, using the fourth of formulas (2.17), as the doubled moment of inertia of a right triangle with a base h and height b/ 2:

Thus, the moments of inertia of an isosceles triangle about the main central axes Oh And OU determined by formulas

Circle. First, it is convenient to calculate the polar moment of inertia of a circle using formula (2.4), using the polar coordinate system (Fig. 2.14).

Considering that dF-rdrdQ, we'll find

Since the polar moment according to (2.4) equal to the sum two axial moments, we get

Ring. The moments of inertia of the ring (Fig. 2.15) are found as the difference between the moments of inertia of two circles with radii I 2 And R ( :

Semicircle(rice. 2.16). Let us select an area element in the plane of the semicircle dF with polar coordinates G, 0 and Cartesian coordinates x v y v for which, in accordance with Fig. 2.16 we have:

Using formulas (2.1) and (2.5), we find, respectively, the static moment of the semicircle relative to the axis 0 ( x ( and ordinate at 0 center of gravity ABOUT in the coordinate system 0 ( x ( Uy

Relative to axes 0, x, and 0 ( y v which are the main axes for the semicircle, the axial moments of inertia are equal to half the moments of inertia of the circle:

The moment of inertia about the main central axis is determined using the first formula (2.6):

Ellipse. To calculate the axial moment of inertia of an ellipse with semi-axes A And b relative to the axis Oh(Fig. 2.17) we proceed as follows. Let us draw a circle around the ellipse and select two elementary stripes of width dx and height 2ук for circle and 2 uh for an ellipse. The moments of inertia of these two strips can be determined by the first of formulas (2.15) for a rectangle:

Integrating these expressions ranging from -A before A, we get

Rice. 2.16

Rice. 2.17

From the equations of a circle and an ellipse we have

With this in mind

A similar expression can be obtained for the moment of inertia about the axis OU. As a result, for the ellipse we will have following formulas for axial moments:

Rolled rods. The geometric characteristics of the sections of rolled rods (I-beams, channels, angles) are given in the tables of rolled steel assortment (see appendix).

Considering in the previous sections the simplest types of deformations - axial tension and compression, crushing, spalling - we found out that their resistance acting force proportional only to the dimensions of the cross-sectional area of ​​the element on which the force acts. So, with the same cross-sectional area, the same material and the same force acting on each of the rods shown in Fig. 9.14, equal stresses will arise in them.
Moving on to study other more complex species deformations (torsion, bending, eccentric compression etc.) we will see that in these cases the resistance of a structural element to external forces depends not only on its cross-sectional area, but also on the distribution of this area in the section plane, i.e., on the shape of the section.
From everyday experience it is clear that bending rod 4 in the vertical direction is more difficult than rod 5, and rod 6 has even greater rigidity, although the cross-sectional areas of all these rods are the same (Fig. 9.14).

Parameters characterizing geometric properties Various plane figures, in addition to area, are: static moments, moments of inertia, moments of resistance and radii of inertia.
Static moment of area. Let's imagine a beam with an arbitrary cross-sectional shape with an area F, in the plane of which the axis is drawn X(Fig. 9.15). Select the area element dF, located at a distance at from the axis X.. The static moment of an elementary platform, relative to the x-axis, is the product of this platform and its distance to the axis:

Static moment of the entire area F relative to the axis X equal to the sum of the static moments of all elementary areas that can be identified on the area under consideration:

From theoretical mechanics It is known that the coordinates of the center of gravity of the area of ​​a figure are determined by the formulas:

That's why

Therefore, the static moment of a figure with area F relative to any axis is equal to the product of the area and the distance of the center of gravity of the figure to this axis. The dimension of the static moment is the unit of length cubed (,).
The axes passing through the center of gravity of the section are called central. If a figure has an axis of symmetry, then the latter always passes through the center of gravity of the figure, i.e. the axes of symmetry are also central axes.
We will also keep in mind that the static moment of a complex figure relative to some axis is equal to the sum of the static moments relative to the same axis of simple figures into which the original complex figure can be divided:
Rice. 9.16. Scheme for determining the coordinates of the center of gravity of a complex figure.

To solve this problem, we choose two coordinate axes X And at, coinciding with the sides of the figure. Let us divide the figure, all dimensions of which must be known, into elementary parts - rectangles - the coordinates of the centers of gravity of which are obvious, since these parts are symmetrical. Let us now compose expressions for calculating the static moment of the entire area, for example relative to the axis at. This can be done in two ways:
a) take the sum of the static moments of individual areas
In these expressions F- area of ​​the entire figure; - coordinate of its center of gravity; - the areas of individual parts of the figure, and - the coordinates of their centers of gravity.
Equating the formulas written above to each other, we obtain an equation with one unknown:
Similarly, the distance of the center of gravity of the figure from the axis X can be expressed like this:

Compiling an integral in which the integrand is the product of the area element and the square of the distance to the origin (Fig. 9.17), we obtain polar moment of inertia:
Let us note one more characteristic in which the site dF multiplied by the product of coordinates
This quantity is called centrifugal moment of inertia. The given moments of inertia are measured in units of length taken to the fourth power (,).
The axial and polar moments of inertia of a figure are positive quantities and cannot be equal to zero. The centrifugal moment of inertia, depending on the position of the axes, can be positive or negative, as well as equal to zero. Two mutually perpendicular axes about which the centrifugal moment of inertia is zero are called main axes of inertia and are designated . For a symmetrical figure, the axis of symmetry is also the main axis.
Axial moments of inertia, defined relative to the principal axes, have maximum and minimum values.
Just as for the static moment, the moment of inertia of a complex figure is equal to the sum of the moments of inertia of the figures forming it. We emphasize that the above is true in the case when all moments of inertia are calculated relative to the same axis.
For moments of inertia, there is another rule often used in calculations. In relation to axial moments, it is formulated as follows: moment of inertia of a figure relative to an axis parallel to the central one, is equal to the moment of inertia relative to the central axis plus the product of the area of ​​the figure by the square of the distance between the axes (Fig. 9.18):
For centrifugal moments of inertia, the corresponding rule in analytical form looks like this:
To obtain the value of the moment of inertia of a particular figure, in principle, it is necessary to solve the corresponding integral over the area of ​​​​this figure. However, in order to facilitate engineering calculations, such integrals for the most common cross-sectional shapes of building elements have already been solved and the results of the solutions in the form of formulas are presented in tables, one of which is placed in Appendix 3.
In addition, GOST standards for all standard rolled profiles produced in our country (angles, I-beams, etc.) give values ​​of axial moments of inertia and other geometric characteristics for each standard size of rolled product (see Appendix 4).
Finally, for sections with complex shapes, moments of inertia are determined using the two rules outlined above: the addition of moments of inertia and the conversion of moments of inertia about one axes to other axes.
Moment of resistance. Axial moment of resistance flat figure relative to any axis lying in the plane of the figure, the quotient of dividing the moment of inertia relative to the same axis by the distance to the most distant point of the figure is called (see Fig. 9.17):
Moments of resistance have the dimension of length cubed (,).
Formulas for calculating the axial moments of resistance of the most frequently occurring figures are given in Appendix 3, and the specific values ​​of this characteristic for rolled steel profiles are given in GOST (Appendix 4). Note that, unlike moments of inertia, moments of resistance cannot be added.
Radius of inertia. The radius of gyration is the value obtained from the formula
and for a circle with diameter d the radius of gyration relative to the axis passing through the center of the circle is equal to
The scope of application of the geometric characteristics of sections discussed above will be revealed when studying the types of deformations, which are discussed in the following subsections of this chapter.

Static is a section of theoretical mechanics that sets out the general doctrine of forces and studies the conditions of equilibrium of bodies under the influence of forces.

Statics is based on some basic principles ( axioms), which are a generalization of centuries-old industrial experience of mankind and theoretical research.

Axiom 1. If two forces act on a free absolutely rigid body, then the body can be in equilibrium if and only if these forces are equal in magnitude and directed along the same straight line in opposite directions (Fig. 1.2).

Fig.1.2

Axiom 2. The action of a given system of forces on an absolutely rigid body will not change if a balanced system of forces is added to it or subtracted from it. If, then. Consequence: the action of a force on an absolutely rigid body will not change if the point of application of the force is moved along its line of action to any other point of the body. Let the body be acted upon by an applied force at a point A force . Let us choose an arbitrary point on the line of action of this force IN, and apply balanced forces to it and, moreover, . Since forces form a balanced system of forces, according to the second axiom of statics they can be discarded. As a result, only one force will act on the body, equal to but applied at the point IN(Fig. 1.3).

Fig.1.3

Axiom 3. Two forces applied to solid body at one point, have a resultant applied at the same point and represented by the diagonal of a parallelogram built on these forces as on the sides. Vector equal to the diagonal of a parallelogram built on vectors and is called the geometric sum of vectors and (Fig. 1.4).

Axiom 4. The law of equality of action and reaction. With any action of one body on another, there is a reaction of the same magnitude, but opposite in direction (Fig. 1.5).

Fig.1.5

Axiom 5. The principle of hardening. The equilibrium of a changing (deformable) body under the influence of a given system of forces will not be disturbed if the body is considered hardened, i.e. absolutely solid.

4. Geometric characteristics of figures. Static moment. Centrifugal moment of inertia, polar moment of inertia (basic concepts).

The result of the calculations depends not only on the cross-sectional area, therefore, when solving problems on strength of materials, one cannot do without determining geometric characteristics of figures: static, axial, polar and centrifugal moments of inertia. It is imperative to be able to determine the position of the center of gravity of the section (the listed geometric characteristics depend on the position of the center of gravity). In addition to the geometric characteristics of simple figures: rectangle, square, isosceles and right triangles, circle, semicircle. The center of gravity and the position of the main central axes are indicated, and the geometric characteristics relative to them are determined, provided that the beam material is homogeneous.

Geometric characteristics of rectangle and square

Axial moments of inertia of a rectangle (square)

GEOMETRICAL CHARACTERISTICS OF A RECTANGULAR TRIANGLE

Axial moments of inertia of a right triangle

GEOMETRICAL CHARACTERISTICS OF AN ISOSceles TRIANGLE

Axial moments of inertia of an isosceles triangle

GEOMETRIC CHARACTERISTICS OF THE CIRCLE

Axial moments of inertia of a circle

GEOMETRIC CHARACTERISTICS OF SEMI CIRCLE

Axial moments of inertia of a semicircle

Static moment

Let us consider the cross section of the rod with area F. Let us draw the coordinate axes x and y through an arbitrary point O. Let's select an area element with coordinates x and y (Fig. 4.1).

Let's introduce the concept of a static moment of inertia about an axis - a value equal to the product of the area element () by the distance (indicated by the letter y) to the x-axis:

Similarly, the static moment of inertia about the y-axis is:

Having summed up such products over area F, we obtain the static moment of inertia of the entire figure relative to the x and y axes:

.

Static moment of inertia figure relative to the axis is measured in units of length cubed (cm3), and can be positive, negative and equal to zero.

Let be the coordinates of the center of gravity of the figure. Continuing the analogy with the moment of force, we can write the following expressions:

Thus, the moment (static moment) of the area of ​​a figure relative to an axis is the product of the area and the distance from its center of gravity to the axis.

Centrifugal moments inertia of the body relative to the rectangular axes Cartesian coordinate system the following quantities are called:

Where x, y And z- coordinates of a small body element volume dV, density ρ And mass dm.

The OX axis is called main axis of inertia of the body, if the centrifugal moments of inertia J xy And J xz are simultaneously equal to zero. Three main axes of inertia can be drawn through each point of the body. These axes are mutually perpendicular to each other. Moments of inertia of the body relative to the three main axes of inertia drawn at an arbitrary point O bodies are called main moments of inertia of the body.

The main axes of inertia passing through center of mass bodies are called main central axes of inertia of the body, and the moments of inertia about these axes are its main central moments of inertia. Axis of symmetry of a homogeneous body is always one of its main central axes of inertia.

Polar moment of inertia- the integral sum of the products of the areas of elementary platforms dA per square of their distance from the pole - ρ 2 (in the polar coordinate system), taken over the entire cross-sectional area. That is:

This value is used to predict an object's ability to resist torsion. It has the dimension of units of length to the fourth power ( m 4 , cm 4 ) and can only be positive.

For a cross-sectional area shaped like a circle with a radius r polar moment of inertia is:

If we combine the origin of the Cartesian rectangular coordinate system 0 with the pole of the polar system (see figure), then

because .

As noted above, simple plane figures include three figures: a rectangle, a triangle and a circle. These figures are considered simple because the position of the center of gravity of these figures is known in advance. All other figures can be composed of these simple figures and are considered complex. Let us calculate the axial moments of inertia of simple figures relative to their central axes.

1. Rectangle. Let's consider the cross-section of a rectangular profile with dimensions (Fig. 4.6). Let us select a section element with two infinitely close sections at a distance from the central axis
.

Let's calculate the moment of inertia of a rectangular cross-section relative to the axis:

. (4.10)

Moment of inertia of a rectangular section about the axis
we will find similarly. The conclusion is not given here.

. (4.11)

And
is equal to zero, since the axes
And
are axes of symmetry, and, therefore, principal axes.

2. Isosceles triangle. Let's consider a section of a triangular profile with dimensions
(Fig.4.7). Let us select a section element with two infinitely close sections at a distance from the central axis
. The center of gravity of the triangle is at a distance
from the base. The triangle is assumed to be isosceles, so the axis
section is the axis of symmetry.

Let us calculate the moment of inertia of the section relative to the axis
:

. (4.12)

Size we determine from the similarity of triangles:

; where
.

Substituting expressions for in (4.12) and integrating, we obtain:

. (4.13)

Moment of inertia for an isosceles triangle about the axis
is found in a similar way and is equal to:

(4.14)

Centrifugal moment of inertia about the axes
And
is equal to zero, since the axis
is the axis of symmetry of the section.

3. Circle. Consider the cross section of a circular profile with a diameter (Fig.4.8). Let us highlight the section element with two infinitely close concentric circles located at a distance from the center of gravity of the circle .

Let's calculate the polar moment of inertia of the circle using expression (4.5):

. (4.15)

Using the invariance condition for the sum of axial moments of inertia about two mutually perpendicular axes (4.6) and taking into account that for a circle, due to symmetry
, we determine the value of the axial moments of inertia:

. (4.16)

. (4.17)

Centrifugal moment of inertia about the axes And is equal to zero, since the axes
And
are the axes of symmetry of the section.

4.4. Dependencies between moments of inertia relative to parallel axes

When calculating moments of inertia for complex figures, one rule should be remembered: the values ​​for the moments of inertia can be added, if they are calculated relative to the same axis. For complex figures, most often the centers of gravity of individual simple figures and the entire figure do not coincide. Accordingly, the central axes for individual simple figures and the entire figure do not coincide. In this regard, there are techniques for bringing moments of inertia to one axis, for example, the central axis of the entire figure. This may be due to parallel translation of the axes of inertia and additional calculations.

Let us consider the determination of moments of inertia relative to the parallel axes of inertia shown in Fig. 4.9.

Let the axial and centrifugal moments of inertia shown in Fig. 4.9. figures relative to arbitrarily chosen axes
And
with the origin at the point known. It is required to calculate the axial and centrifugal moments of inertia of a figure relative to arbitrary parallel axes
And
with the origin at the point . Axles
And
carried out at distances And respectively from the axes
And
.

Let us use the expressions for the axial moments of inertia (4.4) and for the centrifugal moment of inertia (4.7). Let's substitute into these expressions instead of the current coordinates
And
element with infinitesimal coordinate area
And
in the new coordinate system. We get:

Analyzing the obtained expressions, we come to the conclusion that when calculating moments of inertia relative to parallel axes, additives in the form of additional terms should be added to the moments of inertia calculated relative to the original axes of inertia, which may be much greater than the values ​​for the moments of inertia relative to the original axes. Therefore, these additional terms should under no circumstances be neglected.

The case considered is the most general case parallel transfer of axes, when arbitrary axes of inertia were taken as initial ones. In most calculations there are special cases of determining moments of inertia.

First special case . The origin axes are the central axes of inertia of the figure. Then, using the main property for the static moment of area, we can exclude from equations (4.18)–(4.20) the terms of the equations that include the static moment of area of ​​the figure. As a result we get:

. (4.21)

. (4.22)

. (4.23)

Here are the axes
And
-central axes of inertia.

Second special case. The reference axes are the main axes of inertia. Then, taking into account that relative to the main axes of inertia the centrifugal moment of inertia is equal to zero, we obtain:

. (4.24)

. (4.25)

. (4.26)

Here are the axes
And
main axes of inertia.

Let's use the obtained expressions and consider several examples of calculating moments of inertia for plane figures.

Example 4.2. Determine the axial moments of inertia of the figure shown in Fig. 4.10, relative to the central axes And .

In the previous example 4.1, for the figure shown in Fig. 4.10, the position of the center of gravity C was determined. The coordinate of the center of gravity was plotted from the axis and compiled
. Let's calculate the distances And between axes And and axes And . These distances were respectively
And
. Since the original axes And are the central axes for simple figures in the form of rectangles, to determine the moment of inertia of the figure relative to the axis Let us use the conclusions for the first particular case, in particular, formula (4.21).

Moment of inertia about the axis we obtain by adding the moments of inertia of simple figures relative to the same axis, since the axis is the common central axis for simple figures and for the entire figure.

cm 4.

Centrifugal moment of inertia about the axes And is equal to zero, since the axis of inertia is the main axis (axis of symmetry of the figure).

Example 4.3. What is the size? b(in cm) the figure shown in Fig. 4.11, if the moment of inertia of the figure relative to the axis equal to 1000 cm 4?

Let us express the moment of inertia about the axis through an unknown section size , using formula (4.21), taking into account that the distance between the axes And equals 7cm:

cm 4. (A)

Solving expression (a) relative to the section size , we get:

cm.

Example 4.4. Which of the figures shown in Fig. 4.12 has a greater moment of inertia relative to the axis if both figures have the same area
cm 2?

1. Let us express the areas of the figures in terms of their sizes and determine:

a) section diameter for a round section:

cm 2; Where
cm.

b) square side size:

; Where
cm.

2. Calculate the moment of inertia for a circular section:

cm 4.

3. Calculate the moment of inertia for a square section:

cm 4.

Comparing the results obtained, we come to the conclusion that a square section will have the highest moment of inertia compared to a circular section with the same area.

Example 4.5. Determine the polar moment of inertia (in cm 4) of a rectangular section relative to its center of gravity, if the width of the section
cm, section height
cm.

1. Find the moments of inertia of the section relative to the horizontal and vertical central axes of inertia:

cm 4;
cm 4.

2. We determine the polar moment of inertia of the section as the sum of the axial moments of inertia:

cm 4.

Example 4.6. Determine the moment of inertia of the triangular figure shown in Fig. 4.13, relative to the central axis , if the moment of inertia of the figure relative to the axis equal to 2400 cm 4.

Moment of inertia of a triangular section relative to the main axis of inertia will be less compared to the moment of inertia about the axis by the amount
. Therefore, when
cm moment of inertia of the section relative to the axis we find it as follows.