There are the following types of moments of inertia of sections: axial; centrifugal; polar; central and main moments of inertia.

Centrifugal moments of inertia section relative at and z called the integral of the form The sum of the axial moments of inertia of the section relative to the two coordinate axes is equal to the polar moment of inertia relative to the origin of coordinates:

The dimension of the indicated types of moments of inertia of the section (length 4), i.e. m 4 or cm 4.

Axial and polar moments of inertia of the section are positive values; the centrifugal moment of inertia can be positive, negative or zero (for some axes that are the axis of symmetry).

There are dependencies for the moments of inertia for parallel translation and rotation of the coordinate axes.

Figure 5.4 - Parallel translation and rotation of the coordinate axes for an arbitrary cross-section of the timber

For centrifugal moments of inertia

If the moments of inertia of the section are known Iz, Iу, Izу with respect to the axes z and at, then the moments of inertia about the rotated axes z 1 and at 1, by an angle α with respect to the original axes (Fig. 5.4, b) is determined by the formulas:

With the concept main moments of inertia connect the position of the main axes of inertia. The main axes of inertia are called two mutually perpendicular axes, relative to which the centrifugal moment of inertia is zero, and the axial moments acquire extreme values ​​(maximum and minimum).

If the main axes pass through the center of gravity of the figure, then they are called the main central axes of inertia.

The position of the main axes of inertia is found from the following dependencies:

In calculating the strength of structural elements, they use the concept of such a geometric characteristic as section resistance moment.

Consider, for example, a cross-section of a bar (Fig.5.5).

Figure 5.5 - An example of a cross-section of a bar

The distance of the most distant t. A from the center of gravity of the section, i.e. From about denote h 1, and the distance t. V- across h 2.

(5.16)

Then the moments of resistance of the section relative to horizontal z-axis points A, V calculated as the ratio of the axial moment of inertia about the axis z to distances to points A, B:

Of practical interest in strength calculations is the smallest moment of resistance of the section Wmin corresponding to the most distant t. A from the center of gravity of the section h 1 = y max.

Dimension of resistance elements (length 3), i.e. m 3, cm 3.

Table 5.1 - Values ​​of moments of inertia and moments of resistance of the simplest sections relative to the central axes

Types of section naming	Moments of inertia	Moments of resistance
Rectangle
Circle

continuation of table 5.1

I = ∑r i 2 dF i = ∫r 2 dF (1.1)

In principle, both the definition and the formula describing it are not complicated and it is much easier to remember them than to grasp the essence. But still, let's try to figure out what the moment of inertia is and where it came from.

The concept of the moment of inertia came to strength materials and structural mechanics from another branch of physics that studies the kinematics of motion, in particular rotational motion. But let's start from afar anyway.

I do not know for sure whether an apple fell on Isaac Newton's head, fell nearby, or did not fall at all, the theory of probability allows all these options (besides, this apple contains too much of the biblical legend about the tree of knowledge), but I am sure that Newton was an observant person, able to draw conclusions from his observations. So observation and imagination allowed Newton to formulate the basic law of dynamics (Newton's second law), according to which the mass of a body m x acceleration a, is equal to the acting force Q(in fact, the designation F is more familiar for force, but since we will continue to deal with the area, which is also often designated as F, then I use the designation Q for the external force considered in theoretical mechanics as a concentrated load, in essence it doesn't change):

Q = ma (1.2)

For me, Newton's greatness lies in the simplicity and clarity of this definition. And also, considering that with uniformly accelerated motion, the acceleration a is equal to the ratio of the speed increment ΔV by the time period Δt, for which the speed changed:

a = Δv / Δt = (v - v o) / t (1.3.1)

at V о = 0 a = v / t (1.3.2)

then you can determine the basic parameters of movement, such as distance, speed, time and even momentum R, characterizing the amount of movement:

p = mv (1.4)

For example, an apple falling from different heights under the action of gravity alone will fall to the ground for different times, have a different speed at the moment of landing and, accordingly, a different impulse. In other words, an apple falling from a greater height will fly longer and crack more strongly on the forehead of an unlucky observer. And Newton reduced all this to a simple and understandable formula.

And Newton also formulated the law of inertia (Newton's first law): if acceleration a = 0, then in the inertial frame of reference it is impossible to determine whether the observed body, which is not acted upon by external forces, is at rest or is moving in a straight line with a constant speed. This property of material bodies to maintain their speed, even if it is zero, is called inertia. The inertial mass of the body is a measure of inertia. Sometimes inertial mass is called inert, but this does not change the essence of the matter. It is believed that the inertial mass is equal to the gravitational mass and therefore it is often not specified which mass is meant, but simply the mass of the body is mentioned.

No less important and significant is Newton's third law, according to which the force of action is equal to the force of reaction if the forces are directed along one straight line, but at the same time in opposite directions. Despite the apparent simplicity, this conclusion of Newton is brilliant and the significance of this law can hardly be overestimated. Below is one of the applications of this law.

However, these provisions are valid only for bodies moving translationally, i.e. along a rectilinear trajectory and at the same time all material points of such bodies move with the same speed or the same acceleration. With curvilinear motion and, in particular, with rotational motion, for example, when a body rotates around its axis of symmetry, the material points of such a body move in space with the same angular velocity w, but at the same time the linear velocity v different points will have different and this linear speed is directly proportional to the distance r from the axis of rotation to this point:

v = wr (1.5)

in this case, the angular velocity is equal to the ratio of the increment of the angle of rotation Δφ by the time period Δt, for which the angle of rotation has changed:

w = Δφ / Δt = (φ - φ о) / t (1.6.1)

at φ о = 0 w = φ / t (1.7.2)

correspondingly normal acceleration a n with rotational motion is equal to:

a n = v 2 / r = w 2 r (1.8)

And it turns out that for rotational motion we cannot directly use formula (1.2), since during rotational motion the value of the body mass alone is not enough, it is also necessary to know the distribution of this mass in the body. It turns out that the closer the material points of the body are to the axis of rotation, the less force is required to apply to force the body to rotate and vice versa, the further the material points of the body are from the axis of rotation, the more force must be applied to force the body to rotate (in this case, we are talking on the application of force at the same point). In addition, when the body rotates, it is more convenient to consider not the acting force, but the rotating moment, since during rotational motion the point of application of the force is also of great importance.

The amazing properties of the moment have been known to us since the time of Archimedes, and if we apply the concept of moment to rotational motion, then the value of the moment M will be the more, the greater the distance r from the axis of rotation to the point of application of the force F(in structural mechanics, an external force is often denoted as R or Q):

M = Qr (1.9)

From this also not very complicated formula, it turns out that if the force is applied along the axis of rotation, then there will be no rotation, since r = 0, and if the force is applied at the maximum distance from the axis of rotation, then the value of the moment will be maximum. And if we substitute in the formula (1.9) the value of the force from the formula (1.2) and the value of the normal acceleration and the formula (1.8), then we get the following equation:

М = mw 2 r r = mw 2 r 2 (1.10)

In the special case when the body is a material point with dimensions much smaller than the distance from this point to the axis of rotation, equation (1.10) is applicable in its pure form. However, for a body rotating around one of its axes of symmetry, the distance from each material point that makes up a given body is always less than one of the geometric dimensions of the body and therefore the distribution of body mass is of great importance, in this case it is required to take these distances separately for each point:

M = ∑r i 2 w 2 m i (1.11.1)

М с = w 2 ∫r 2 dm

And then it turns out that according to Newton's third law, in response to the action of the torque, the so-called moment of inertia will arise I... In this case, the values ​​of the torque and moment of inertia will be equal, and the moments themselves are directed in opposite directions. At a constant angular velocity of rotation, for example w = 1, the main quantities characterizing the torque or moment of inertia will be the mass of material points that make up the body, and the distance from these points to the axis of rotation. As a result, the formula for the moment of inertia will take the following form:

[- М] = I = ∑r i 2 m i (1.12.1)

I c = ∫r 2 dm(1.11.2) - when the body rotates around the axis of symmetry

where I- the generally accepted designation of the moment of inertia, I c- designation of the axial moment of inertia of the body, kg / m 2. For a homogeneous body with the same density ρ throughout the body V the formula for the axial moment of inertia of a body can be written as follows:

I c = ∫ρr 2 dV (1.13)

Thus, the moment of inertia is a measure of the inertia of a body during rotational motion, just as mass is a measure of the inertia of a body during translational rectilinear motion.

The whole circle is closed. And here the question may arise, what relation do all these laws of dynamics and kinematics have to the calculation of static building structures? It turns out that neither is the most direct and immediate. Firstly, because all these formulas were derived by physicists and mathematicians in those distant times when such disciplines as "Theoretical mechanics" or "Theory of strength of materials" simply did not exist. And secondly, because the entire calculation of building structures is based on the indicated laws and formulations and has not yet been refuted by anyone about the equality of gravitational and inertial masses. But in the theory of resistance of materials it is still simpler, no matter how paradoxical it sounds.

And it is easier because when solving certain problems, not the whole body can be considered, but only its cross-section, and, if necessary, several cross-sections. But in these sections the same physical forces act, though having a slightly different nature. Thus, if we consider a certain body whose length is constant, and the body itself is homogeneous, then if we do not take into account constant parameters - length and density ( l = const, ρ = const) - we get a cross-sectional model. For such a cross section, from a mathematical point of view, the following equation will be valid:

I р = ∫r 2 dF (2.1) → (1.1)

where I p- polar moment of inertia of the cross section, m 4. As a result, we got the formula with which we started (but whether it became clearer what the moment of inertia of the section is, I don’t know).

Since in the theory of resistance of materials rectangular sections are often considered, and the rectangular coordinate system is more convenient, then when solving problems, two axial moments of inertia of the cross section are usually considered:

I z = ∫y 2 dF (2.2.1)

I y = ∫z 2 dF (2.2.2)

Picture 1... Coordinate values ​​when determining axial moments of inertia.

Here the question may arise, why are the axes used? z and at rather than more familiar NS and at? It so happened that the determination of the forces in the cross-section and the selection of the cross-section that can withstand the acting stresses equal to the applied forces are two different tasks. The first task - the determination of efforts - is solved by structural mechanics, the second task - the selection of the section - the theory of strength of materials. At the same time, in structural mechanics, when solving simple problems, it is often considered a rod (for rectilinear structures) having a certain length l, and the height and width of the section are not taken into account, while it is considered that the axis NS it just passes through the centers of gravity of all cross-sections and thus, when constructing diagrams (sometimes quite complex), the length l just and is deposited along the axis NS, and along the axis at the values ​​of the diagrams are plotted. At the same time, the theory of resistance of materials considers precisely the cross section, for which width and height are important, and length is not taken into account. Of course, when solving problems of the theory of resistance of materials, sometimes quite complex ones, all the same familiar axes are used NS and at... To me, this state of affairs does not seem entirely correct, since despite the difference, these are still related tasks and therefore it will be more expedient to use common axes for the calculated structure.

The value of the polar moment of inertia in a rectangular coordinate system will be:

I р = ∫r 2 dF =∫y 2 dF + ∫z 2 dF (2.3)

Since in a rectangular coordinate system, the radius is the hypotenuse of a right-angled triangle, and as you know, the square of the hypotenuse is equal to the sum of the squares of the legs. And there is also the concept of the centrifugal moment of inertia of the cross section:

I xz = ∫xzdF(2.4)

Among the axes of a rectangular coordinate system passing through the center of gravity of the cross section, there are two mutually perpendicular axes, relative to which the axial moments of inertia take the maximum and minimum values, while the centrifugal moment of inertia of the section I zy = 0... Such axes are called the main central axes of the cross section, and the moments of inertia about such axes are called the main central moments of inertia.

When in the theory of resistance of materials it comes to moments of inertia, then, as a rule, we mean precisely the main central moments of inertia of the cross section. For square, rectangular, circular sections, the main axes will coincide with the symmetry axes. The moments of inertia of a cross-section are also called geometric moments of inertia or moments of inertia of an area, but the essence does not change from this.

In principle, there is no great need to determine the values ​​of the main central moments of inertia for the cross-sections of the most common geometric shapes - square, rectangle, circle, pipe, triangle and some others. Such moments of inertia have long been defined and widely known. And when calculating axial moments of inertia for sections of complex geometric shapes, the Huygens-Steiner theorem is valid:

I = I c + r 2 F (2.5)

Thus, if the areas and centers of gravity of simple geometric figures that make up a complex section are known, then it will not be difficult to determine the value of the axial moment of inertia of the entire section. And in order to determine the center of gravity of a complex section, the static moments of the cross section are used. Static moments are discussed in more detail in another article, I will just add here. The physical meaning of the static moment is as follows: the static moment of the body is the sum of the moments for the material points that make up the body, relative to some point (polar static moment) or relative to the axis (axial static moment), and since the moment is the product of the force on the shoulder (1.9) , then the static moment of the body is determined, respectively:

S = ∑M = ∑r i m i= ∫rdm (2.6)

and then the polar static moment of the cross section will be:

S р = ∫rdF (2.7)

As you can see, the definition of the static moment is similar to the definition of the moment of inertia. But there is also a fundamental difference. The static moment is therefore called static, because for a body on which the force of gravity acts, the static moment is equal to zero relative to the center of gravity. In other words, such a body is in a state of equilibrium if the support is applied to the center of gravity of the body. And according to Newton's first law, such a body is either at rest or moving at a constant speed, i.e. acceleration = 0. And also from a purely mathematical point of view, the static moment can be equal to zero for the simple reason that when determining the static moment it is necessary to take into account the direction of the action of the moment. For example, with respect to the coordinate axes passing through the center of gravity of the rectangle, the areas of the upper part and the lower part of the rectangle will be positive as they symbolize the force of gravity acting in one direction. In this case, the distance from the axis to the center of gravity can be regarded as positive (conventionally: the moment from the force of gravity of the upper part of the rectangle tries to rotate the section clockwise), and to the center of gravity of the lower part - as negative (conventionally: the moment from the force of gravity of the lower part of the rectangle tries to rotate the section counterclockwise). And since such areas are numerically equal and equal to the distance from the centers of gravity of the upper part of the rectangle and the lower part of the rectangle, the sum of the effective moments will be the required 0.

S z = ∫ydF = 0 (2.8)

And this great zero also allows you to determine the support reactions of building structures. If we consider a building structure to which, for example, a concentrated load Q is applied at some point, then such a building structure can be considered as a body with a center of gravity at the point of application of the force, and support reactions in this case are considered as forces applied at the points of support. Thus, knowing the value of the concentrated load Q and the distance from the point of application of the load to the supports of the building structure, it is possible to determine the support reactions. For example, for a pivotally supported beam on two supports, the value of the support reactions will be proportional to the distance to the point of application of the force, and the sum of the reactions of the supports will be equal to the applied load. But as a rule, when determining the support reactions, they act even easier: one of the supports is taken as the center of gravity, and then the sum of the moments from the applied load and from the rest of the support reactions is still zero. In this case, the moment from the support reaction with respect to which the equation of moments is drawn up is equal to zero, since the shoulder of the force action = 0, which means that only two forces remain in the sum of the moments: the applied load and the unknown support reaction (for statically definable structures).

Thus, the fundamental difference between the static moment and the moment of inertia is that the static moment characterizes the section, which the gravity force seems to be trying to break in half relative to the center of gravity or the axis of symmetry, and the moment of inertia characterizes the body, all material points of which move (or try to move to one direction). Perhaps, the following rather conditional design schemes for a rectangular section will help to more clearly imagine this difference:

Picture 2... A clear difference between a static moment and a moment of inertia.

Now let's go back to the kinematics of motion. If we draw analogies between the stresses arising in the cross-sections of building structures and different types of motion, then in the centrally stretched and centrally compressed elements, stresses appear uniform over the entire cross-sectional area. These stresses can be compared with the action of a certain force on a body, in which the body will move in a straight line and translational. And the most interesting thing is that the cross-sections of the centrally stretched or centrally compressed elements really move, since the acting stresses cause deformations. And the magnitude of such deformations can be determined for any cross-section of the structure. To do this, it is enough to know the value of the acting stresses, the length of the element, the sectional area and the modulus of elasticity of the material from which the structure is made.

For bending elements, the cross-sections also do not remain in place, but move, while the movement of the cross-sections of the bending elements is similar to the rotation of a body about a certain axis. As you probably already guessed, the moment of inertia allows you to determine both the angle of inclination of the cross section and the displacement. Δ l for the extreme points of the section. These extreme points for a rectangular section are at a distance equal to half the height of the section (why - described in sufficient detail in the article "Fundamentals of strength. Determination of deflection"). And this, in turn, allows you to determine the deflection of the structure.

And the moment of inertia allows you to determine the moment of resistance of the section. To do this, the moment of inertia simply needs to be divided by the distance from the center of gravity of the section to the most distant point of the section, for a rectangular section by h / 2. And since the investigated sections are not always symmetrical, the value of the moment of resistance may be different for different parts of the section.

It all started with a banal apple ... but no, it all started with a word.

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22. Static moment of the section

Strength calculations show that stresses and deformations occurring in a solid depend on the internal force factors and the geometric characteristics of the cross section. In tension, for example, the stress depends on the cross-sectional area, and, since the stress in this case is evenly distributed over the cross-section, it does not depend on the cross-sectional shape. During torsion, the stresses depend on the size and shape of the section due to the uneven distribution of stresses. The calculation formulas for the torsion beam include polar moment of inertia I p and polar moment of resistance W p- geometric characteristics of the section. When calculating the strength of a bar in bending, it is necessary to know the moments of inertia and moments of resistance of the section relative to the axes passing through the center of gravity of the bar. Let us take for consideration some section of a bar with an area A and an axis passing through the center of gravity of this body. Static moment of a flat section about some axis x is the sum of the products of the areas of the elementary areas that make up the section, by the distance of these areas to the axis passing through the center of gravity. Likewise for the axis y.

The static moment is measured in cubic meters. It can be positive, negative or zero depending on the selected axis. If the static moments and the cross-sectional area are known, then the coordinates of the center of gravity can be defined as the ratio of the static moment to the cross-sectional area. And vice versa, if the coordinates of the center of gravity of the section are known - x c, y c, the static moment is equal to the product of the sectional area by the distance from the center of gravity to the axis.

S x=Ay c

S y=Ax c

It can be seen from the obtained relations that in the case when the axis passes through the center of gravity, the static moment is equal to zero.

In the case when the cross section can be considered as n number of constituent parts with known areas A i and the coordinates of the centers of gravity x i, y i, the position of the entire center of gravity can be defined as the sum of the products:

Each term in the numerator determines the static moment of the given section relative to the selected axis.

23. Moment of inertia of the section

Axial (or equatorial) moment of inertia of a flat section about some axis x is the sum of the products of the areas of elementary areas, of which the section by the square of the distance of these areas to the axis passing through the center of gravity consists. Thus, the axial moments are integrals over the entire cross-sectional area.

Polar moment of inertia relative to some point (pole) is the sum of the products of the areas of the elementary areas that make up the section, by the square of the distance of these areas to the selected point.

Centrifugal moment of inertia relative to some two mutually perpendicular axes is the sum of the products of the elementary areas that make up the section, by the distance of these areas to these axes.

Moments of inertia are measured in m 4. Axial and polar moments of inertia can only be positive, since for any sign of the coordinate, the square of this coordinate is taken in the formula. The centrifugal moment of inertia can be positive, negative or zero.

The sum of the axial moments of inertia about two mutually perpendicular axes is equal to the polar moment of inertia about the point where these axes intersect.

I ρ = I x +I y

Indeed, ρ is the distance from the elementary area of ​​the section to some point, it is defined as the hypotenuse of a triangle with sides x and y.

ρ 2 = x 2 + y 2

Substitute this ratio into the expression for the polar moment of inertia and get:

24. Moments of inertia of simple sections

Let's consider the moments of inertia of some simple figures.

Circle. I ρ = I x +I y. Since the circle is a symmetrical figure, then I x = I y... Hence, Iρ = 2 I x... Based on the definition of the polar moment of inertia and the ratio for the polar moment of inertia and axial moments of inertia in the case of a circle, we have:

For rings diameter d and inner diameter d 0

Semicircle... The main central axes represent the axis of symmetry of this semicircle and the axis perpendicular to it. For a semicircle, the moment of inertia is half the moment of inertia of a circle for the same axis. If we denote x 1 base axis, then

From the relation connecting the moments of inertia of parallel axes, one of which is central, and, knowing the value of the ordinate of the center of gravity of the semicircle y c ≈ 0.424r you can determine the moments of inertia of a semicircle:

Rectangle... Determine the moment of inertia I x1 coinciding with the base of the rectangle, and consider the section A as the sum of elementary rectangles of width b and height dy 1 , A=bdy 1

For the moments of inertia of parallel axes, one of which is central, I x =I x1 - a 2 A... In this case, the distance a=h / 2, A=bh, moment of inertia about the axes x and y

I x = bh 3 / 12

I y = hb 3 / 12

In the special case of a square

I x =I y = b 4 / 12

For triangle calculate the moment of inertia I x1, about the axis x 1 coinciding with the base, and for this we consider the section as the sum of elementary rectangles with the width b... After performing mathematical transformations, we find the value I x = bh 3 / 12. The moment of inertia about the central axis is I x =I x1-a 2 b, in this case a=h / 3,A= (1 / 2)bh... As a result, we get:

I x =bh 3 / 12 – (h / 3) 3 (1 / 2)bh= bh 3 / 36

In general, the axis x is not the main one and

I y = bh 3 / 48

25. Dependence between moments of inertia about parallel axes

Let us establish the relationship between the moments of inertia relative to parallel axes, one of which is central. To do this, consider a section with an area A... (Fig. 10) Suppose that the coordinates of the center of gravity of the section are known C and moments of inertia I xc, I yc about the central axes x c, y c... In this case, you can determine the moments of inertia about the axes x and y parallel to the central and distant from the central at a distance a and b respectively. Let's write the ratio for the coordinates of the parallel axes:

x= x c+b

y= y c+a

Then the moment of inertia of the section about the axis x will be written as:

In this expression, the first term is the moment of inertia about the axis x c, in the second term, the integral represents the static moment (and relative to the central axis, the static moment is always zero), the third term is the cross-sectional area multiplied by the square of the distance between the axes a... Thus:

I x = I xc + a 2 A

I y = I yc + b 2 A

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 cross-sectional area of ​​the figure by the square of the distance between the axes.

We have obtained a relation for the moments of inertia about the central axes in the transition to parallel off-center ones. These relations are also called parallel transfer formulas.

From the formulas obtained, it is clear that the moment of inertia relative to the central axis is always less than the moment of inertia of any parallel non-central axis.

26. The main axes of inertia and the main moments of inertia

An infinite number of pairs of mutually perpendicular axes can be drawn through any point of the section plane. Since the sum of the two axial moments of inertia of the section is a polar moment and is a constant value, moving the coordinate system, you can choose a position of the axes in which one of the selected moments of inertia will be maximum, and the second will be minimum. Consider the relationship between the moments of inertia about the axes x 0, y 0 and moments of inertia about the axes x and y rotated by an angle α with respect to x 0, y 0 ... Let us find such values ​​of the angle α at which the moments of inertia of the perpendicular axes will take their maximum and minimum values. To do this, we find the first derivative with respect to the angle of rotation of I x , I y and equate it to zero (the mathematical rule for finding the extrema of a function).

After the transformations, the ratio will take the form:

The resulting formula determines the position of two mutually perpendicular axes, the moment of inertia relative to one of which is maximum, the moment of inertia relative to the other is minimal. Such axes are called main axes of inertia... Moments of inertia about such axes are called the main moments of inertia... In this case, the centrifugal moment is equal to zero.

The axes passing through the center of gravity of the section are called central axes. In practical calculations, the main moments of inertia about the central axes are of interest, they are called the main central moments of inertia, and such axes - main center axes... Since only the central axes are of interest, for brevity they are simply called the principal axes, and the axial moments of inertia calculated with respect to such axes are simply called the principal moments of inertia.

One of the main axes of inertia is the axis passing through the center of symmetry of the section plane, the second is perpendicular to it. The axis of symmetry and any perpendicular to it form a system of principal axes. If a section has several axes of symmetry (for example, a circle, a square, an equilateral triangle), then all central axes are principal and all central moments are equal.

27. Calculation of moments of inertia of complex sections

To find the moment of inertia of a complex section with an area A the section is divided into simple A 1 , A 2 ,… A n, for which the moments of inertia are found according to ready-made formulas or tables.

The moment of inertia of a complex figure is found as the sum of the moments of inertia that make up simple figures.

I x = I x 1 + I x 2 +… + I xn

The moment of inertia is an integral over the sectional surface area,

for the integral it is true:

Therefore, we can write that:

In other words, the moment of inertia of a composite section about a certain axis is the sum of the moments of inertia of the components of this section about the same axis.

When solving problems of this kind, the following algorithm is adhered to. Find the center of gravity of the flat section and determine the major central axes. From tables or using ready-made formulas, the values ​​of the moments of inertia of the constituent parts are calculated relative to their own central axes parallel to the main central axes of the section. Using the formulas of parallel transfer, the values ​​of the moments of inertia of the constituent parts of the section relative to the main axes of the section are calculated. The values ​​of the main central moments of inertia are determined by summation.

This rule is also true for the centrifugal moment of inertia.

28. The concept of torque

Torsion is one of the types of timber deformation, in which one internal force factor appears in the cross-section of the timber, called torque Mk. This type of deformation occurs when a pair of forces acts on the bar, called twisting moments M applied perpendicular to its longitudinal axis.

A bar loaded with torques is called a shaft. The sum of the torques acting on the shaft is zero if the shaft rotates uniformly. The torque can be determined by the formula, provided that the transmitted power is known P and angular velocity w.

With a known frequency of rotation of the shaft, the angular velocity can be written in the form

Therefore, the expression for the torque can be written as:

In practical calculations, a real object is replaced by a design model. To simplify the problem, it is assumed that the rotational moments are concentrated in the middle section of the parts, and not distributed over their surface. In the section of an arbitrary shaft, the torque can be determined using the section method, when the shaft is mentally cut by a plane. One of the parts is discarded and its influence is replaced by the torque Mk, then it is determined from the equilibrium equations. The numerical value of the torque is made up of the sums of the torques located on one side of the section.

In the cross-sections of the bar during torsion, only tangential stresses arise, the normal forces are parallel to the longitudinal axis of the bar and their moments are zero. Therefore, the definition for torque can be formulated as follows: torque is the resultant moment of internal tangential forces arising in the cross-section of the beam relative to its longitudinal axis.

When calculating the strength in the case of torsion of the timber, it is necessary to find a dangerous section of the timber. If the dimensions of the cross-section along the axis of the beam are unchanged, then the sections with the maximum torque are considered dangerous. To find dangerous sections, torque diagrams are built (graphs of changes in torques along the length of the bar). When plotting the diagrams, it is generally accepted that the torque is positive if its direction coincides with the direction of the clockwise, when looking at the section drawn. This assumption is arbitrary, since the sign of the torque has no physical meaning.

29. Determination of stresses during torsion of a round shaft

When studying the torsion of shafts, the following assumptions take place:

- the hypothesis of flat sections: the flat cross-sections of the timber after deformation also remain flat and directed along the normal to its axis, turning at a certain angle relative to this axis;

- the radii of the cross-sections are not bent and their length remains constant;

- along the axis of the bar, the distances between the cross-sections remain constant.

Based on the above assumptions, the torsion of a circular shaft can be considered as pure shear. The formulas obtained on the basis of these assumptions are confirmed experimentally.

Consider the torsion of a section of a circular bar with a radius r the length dz... One of the ends will be considered fixed.

When turning through an angle a in the cross section, the shear angle lying on the surface of such a shaft is determined by the formula:

The ratio of the total twist angle in a section of the shaft to its length is called the relative twist angle.

We mentally select in the considered section of the shaft a cylinder with a radius ρ, the shear angle for the surface of this cylinder is determined in the same way:

According to Hooke's law, in the case of shear, the shear stresses are:

Thus, during torsion, the shear stresses are directly proportional to the distance from the center of gravity of the section, and at the center of gravity, the shear stresses are zero. Approaching the surface of the shaft, they take on their maximum values.

30. Calculation of the moments transmitted to the shaft

Consider the torsion of a section of a round shaft with a diameter r and length dz... Let us select a cylinder of diameter ρ in it. Since torsion is pure shear, normal stresses are zero, and shear stresses when rotated through an angle α are distributed as follows:

Torque is defined as:

A- cross-sectional area. Substituting the shear stress into this expression and taking into account that the integral of the radius over the sectional area is the polar moment of inertia of the section , we get:

Substituting this expression in the formula for shear stresses, we get:

Thus, shear stresses are defined as the product of the torque and the radius, referred to the polar moment of the section. It is clear that for points located at equal distances from the axis, the shear stresses are equal, the maximum stress values ​​have points located on the surface of the shaft.

Here Is the polar moment of resistance during torsion.

For round section

The torsional strength condition is as follows:

[τ] is the maximum permissible shear stress.

This formula also allows you to determine the permissible torque or select the permissible shaft diameter.

31, Torsional deformation. Potential energy

In the process of torsion, the torques rotate together with the section by some angle and at the same time perform work, which, as with other types of deformation, is spent on creating a certain reserve of potential energy in the body undergoing deformation and is determined by the formula:

This relationship follows from the linear dependence of the torque M To from the angle of rotation φ.

When a load is applied, the torque gradually increases, while, in accordance with Hooke's law, the angle of rotation increases proportionally. The work done by the torque is equal to the potential energy of deformation according to the law of conservation of energy, therefore,

If the well-known formula for the twist angle is substituted into the resulting ratio, then the expression will take the form:

With a stepwise change in the torque or the cross-section of the bar, the potential energy is the sum:

If the torque or polar moments (or both at the same time) continuously change along the length of the sections of the bar, then the potential energy is an integral over the length

32. Calculation of helical cylindrical springs

In mechanical engineering and instrument making, helical springs are widely used, which can be cylindrical, cone-shaped or shaped. The most commonly used springs of a cylindrical shape, made from wire of a circular cross-section: tension springs (made without gaps between the turns) and compression springs (with a gap). To simplify the calculation of springs for stiffness and strength, we will assume that the angle of inclination of the coils is so small that it can be neglected and the section along the axis of the spring is considered transverse for the coil. From the equilibrium conditions for the cut-off part of the spring, it is clear that two internal force factors arise in the section: the transverse force Q y = F and torque M To = FD / 2, i.e., only tangential stresses arise in the coil section. We will assume that the shear stresses associated with the transverse force are uniformly distributed over the section, and the tangential forces associated with the presence of a torque are linearly distributed and reach their maximum values ​​at the extreme points of the section. The most stressed point will be the point located closest to the axis of the spring, the stress for it is:

The ratio of the spring diameter to the wire diameter is called the spring index,

c n = D / d

The resulting formula is approximate due to neglect of the influence of the transverse force and due to the fact that the curvature of the turns is not taken into account. Let's introduce the correction factor TO, depending on the spring index and the angle of inclination of the coils. Then the strength condition will take the form:

When a load is applied, the spring changes its length. This change is called spring draftλ. Let us determine what the settlement is equal to if the turns only experience torsion. According to the Clapeyron formula, the work of external static forces is equal to:

Potential strain energy

In this case

where l- the length of the considered section of the spring;

n- the number of turns.

Having performed the substitution and mathematical transformations, we get that:

33. Displacements and stresses in helical springs

Coil springs are widely used in mechanical engineering as damping or return feed devices. Calculation of coil springs is a good demonstration of the method for determining displacements. Coil springs are classified into tension, compression and torsion springs. Tension and compression springs are loaded by forces acting along the spring axis, torsion springs are loaded with moments located in a plane perpendicular to the spring axis.

A coil spring can be viewed as a spatially bent rod with a helical axis. The spring shape is characterized by the following parameters: spring diameter D, the number of turns n, ascent angle θ and spring pitch s defined by the formula:

s= π Dtgθ

Typically, the spring pitch is much less than π D, the angle θ is small enough (less than 5 °).

Consider a tension-compression spring. Under the influence of external load R a resultant internal force arises in each cross section R and moment M=РD / 2 lying in the plane of action of the forces R... In Fig. 13 shows the forces acting in the cross section of the spring.

The projections of the total force and moment relative to the coordinate system associated with the section are described by the following relations:

M To = (PD/ 2) × cosθ,

M outcast= (PD / 2) × sinθ,

Q=P× cosθ,

N=P× sinθ.

Suppose the force R is equal to 1, then the relations for forces and moments will take the form:

M k1 = (D/ 2) × cosθ,

M out1 = (D/ 2) × sinθ,

Q 1 = cosθ,

N 1 = sinθ.

Find the axial displacement in the spring using Mohr's integral. Taking into account the small displacements caused by normal and transverse forces, as well as axial displacement, in this case, the Mohr integral is written as follows:

where the product in the denominator is the torsional stiffness of the spring;

l is the length of the working part of the spring;

l≈ π Dn

Due to the smallness of the angle of inclination of the turns θ we assume that cos θ = 1, then

The stresses in helical springs operating in compression-tension or torsion are determined as follows.

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

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

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

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

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

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

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

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

In the assortment tables, the values ​​of the centrifugal moments of inertia for the angles are not indicated. The method for determining such moments of inertia is discussed in example 4.5.

In the overwhelming majority of cases, the ultimate goal of calculating the geometric characteristics of a section is to determine its main central moments of inertia and the position of the main central axes of inertia. Therefore, the next step in the calculation is to determine the coordinates of the center of gravity of a given section [according to formulas (6.5) and (7.5)] in some arbitrary (random) coordinate system.

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

Please note that this site has an online service for calculating the center of gravity and moments of inertia of composite sections, which consist of rolled profiles (I-beams, angles, etc.) and simple shapes.

Often, when calculating the elements of building structures, it is necessary to determine the geometric characteristics of profiles made up of elementary geometric shapes (rectangle, circle, etc.) and rolled profiles. Let's consider in detail an example of calculation.

It is necessary to determine the geometric characteristics of the composite section (Fig.), Which consists of a 20 / 12.5 / 1.2 corner, a 14/1 corner and a 20x2cm rectangle.

Defining your own characteristics of individual profiles- components of the section

The intrinsic characteristics of rolled profiles are determined from the assortment.

For unequal angle 20 / 12.5 / 1.2:

- height and width of the corner h = 20 cm, b = 12.5 cm;

- area $ A $ = 37.9cm 2 ;

- own axial moments of inertia $ (I_x) $ = 1570 cm 4, $ (I_y) $ = 482 cm 4;

- intrinsic centrifugal moment of inertia $ (I_ (xy)) $ = 505 cm 4 ;

- coordinates of the center of gravity $ (x_c) $ = 2.83 cm, $ (y_c) $ = 6.51 cm.

For equal angles 14/1:

- height and width of the corner h = b = 14 cm;

- area $ A $ = 27.3 cm 2 ;

- own axial moments of inertia $ (I_x) $ = $ (I_y) $ = 512 cm 4 ;

- intrinsic centrifugal moment of inertia $ (I_ (xy)) $ = 301 cm 4 ;

- coordinates of the center of gravity $ (x_c) $ = $ (y_c) $ = 3.82 cm.

For a rectangle 20x2cm:

- height and width of the rectangle h = 20 cm, b = 2 cm;

Area $ A $ = 20 ∙ 2 = 40 cm 2;

- own axial moments of inertia $ (I_x) = \ frac ((2 \ cdot ((20) ^ 3))) ((12)) = 1330 $ cm 4 , $ (I_y) = \ frac ((20 \ cdot (2 ^ 3))) ((12)) = 13.3 $ cm 4 ;

- intrinsic centrifugal moment of inertia $ (I_ (xy)) $ = 0, since the profile has an axis of symmetry.

Determination of the center of gravity of a section

The total area of ​​the entire section A = 37.9 + 27.3 + 40 = 105cm 2 .

We draw the auxiliary axes $ X $ and $ Y $ and determine the center of gravity of the section relative to them:

$ (X_c) = \ frac ((\ sum ((X_i) \ cdot (A_i)))) (A) = \ frac (((\ text (37)) (\ text (, 9)) \ cdot (\ text ((- 13)) (\ text (, 5) + 27)) (\ text (, 3)) \ cdot (\ text ((- 3)) (\ text (, 82) + 40)) \ cdot (\ text (1)))) (((\ text (105)))) (\ text (= - 5)) (\ text (, 49)) $ cm;

$ (Y_c) = \ frac ((\ sum ((Y_i) \ cdot (A_i)))) (A) = \ frac (((\ text (37)) (\ text (, 9)) \ cdot (\ text ((- 2)) (\ text (, 83) + 27)) (\ text (, 3)) \ cdot (\ text (10)) (\ text (, 2 + 40)) \ cdot (\ text (10)))) ((105)) = 5.44 $.

In this case, in the coordinates of the centers of gravity of the constituent duties’ be sure to take into account the sign. Putting aside the axes that pass through the center of gravity- the central axes are $ Xc $ and $ (Y_c) $.

Determination of central moments of inertia

The axial and centrifugal moments of inertia of the section are determined by the formulas for the transition between parallel axes. To do this, we find and show in the drawing the distance between the central axes of the entire section and the proper axes of each of the figures.

$ Ix = \ sum (\ left ((I (x_i) + A \ cdot (b ^ 2)) \ right) = (\ text (482 + 8)) (\ text (, 2)) ((\ text ( 7)) ^ (\ text (2))) \ cdot (\ text (37)) (\ text (, 9 + 512 + 4)) (\ text (, 7)) ((\ text (6)) ^ (\ text (2))) \ cdot (\ text (27)) (\ text (, 3 + 1330 + 4)) (\ text (, 5)) ((\ text (6)) ^ (\ text ( 2))) \ cdot (\ text (40 = 6360))) $ cm 4 ;

$ Iy = \ sum (\ left ((I (y_i) + A \ cdot (a ^ 2)) \ right)) = (\ text (1570 + 8)) (\ text (, 0)) ((\ text (1)) ^ (\ text (2))) \ cdot (\ text (37)) (\ text (, 9 + 512 + 1)) (\ text (, 6)) ((\ text (7)) ^ (\ text (2))) \ cdot (\ text (27)) (\ text (, 3 + 13)) (\ text (, 3 + 6)) (\ text (, 4)) ((\ text (9)) ^ (\ text (2))) \ cdot (\ text (40 = 6280)) $ cm 4 ;

$ (I_ (xy)) = \ sum (\ left (((I_ (xy)) _ i + A \ cdot a \ cdot b) \ right)) = $

$ = 505 + (- 8.01) \ cdot (- 8.27) \ cdot 37.9 - 301 + 1.67 \ cdot 4.76 \ cdot 27.3 + 0 + 6.49 \ cdot 4.56 \ cdot 40 = 4120 $ cm 4 .

At the same time, obligations' are obligatory we take into account the placement of figures relative to the axes in question. So, when determining the moment of inertia $ (I_x) $, we substitute the proper moment of inertia of the unequal angle relative to the axis, which is parallel to the $ (X_c) $ axis, in the assortment, this is the $ Y $ axis, and vice versa.

Determination of the position of the main axes and the main moments of inertia

The angle of rotation of the main axes relative to the axes for which the moments of inertia are known is determined by the formula

\ $ \ alpha = \ frac ((arctg (- 97))) (2) = - 44,7 ^ \ circ $.

If $ \ alpha> 0 $, the major axes are plotted counterclockwise and vice versa.

The main moments of inertia are defined as follows

$ (I_ (x0)) = (I_x) \ cdot (\ cos ^ 2) \ alpha + (I_y) \ cdot (\ sin ^ 2) \ alpha - (I_ (xy)) \ cdot \ sin 2 \ alpha = $

$ = 6360 \ cdot (\ cos ^ 2) (- 44.7 ^ \ circ) + 6280 \ cdot (\ sin ^ 2) (- 44.7 ^ \ circ) - 4120 \ cdot \ sin (- 2 \ cdot 44,7 ^ \ circ) = 10430 $ cm 4 .

$ (I_ (y0)) = (I_y) \ cdot (\ cos ^ 2) \ alpha + (I_x) \ cdot (\ sin ^ 2) \ alpha + (I_ (xy)) \ cdot \ sin 2 \ alpha = $

$ = 6280 \ cdot (\ cos ^ 2) (- 44.7 ^ \ circ) + 6360 \ cdot (\ sin ^ 2) (- 44.7 ^ \ circ) + 4120 \ cdot \ sin (- 2 \ cdot 44,7 ^ \ circ) = 2210 $ cm 4 .

The centrifugal moment of inertia about the main axes is zero.

Radii of gyration. Moments of resistance

Radii of gyration of the section

$ (i_x) = \ sqrt [()] ((\ frac (((I_x))) (A))) = \ sqrt [()] ((\ frac ((10430)) ((105)))) = 9.96 $ cm, $ (i_y) = \ sqrt [()] ((\ frac (((I_y))) (A))) = \ sqrt [()] ((\ frac ((2210)) ((105)))) = 4.58 $ cm.

The moments of resistance of the section are determined relative to the central axes. To do this, it is necessary to determine the distances $ (x _ (\ max)) $ and $ (y _ (\ max)) $ to the most distant points from the main axes. First, it is necessary to determine from the drawings which points are the most distant. In our case, these are points $ A $ and $ B $ (fig.). The required distances can be determined by having the coordinates of these points in the central (not returned to the axes).

$ (x _ (\ max)) = (x_A) \ cdot \ cos \ left (\ alpha \ right) + (y_A) \ cdot \ sin \ left (\ alpha \ right) $

$ (y _ (\ max)) = (y_B) \ cdot \ cos \ left (\ alpha \ right) - (x_B) \ cdot \ sin \ left (\ alpha \ right) $

X A = - 8.53cm Y A = 8.57cm

X B = - 14.5cm Y B = - 18cm

x max = - 12.1cm y max = - 23cm

Moments of resistance

$ (W_x) = \ frac (((I_x))) (((y _ (\ max)))) = \ frac ((10430)) ((23)) = 454 $ cm 3 ; $ (W_y) = \ frac (((I_y))) (((x _ (\ max)))) = \ frac ((2210)) ((12.1)) = 183 $ cm 3 .