Current page: 3 (total book has 9 pages) [accessible reading excerpt: 7 pages]

Font:

100% +

22. Static moment of section

Strength calculations show that the stress and strain that occur in a solid body depend on 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 uniformly distributed over the section, does not depend on the shape of the section. During torsion, the stresses depend on the size and shape of the section due to the uneven distribution of stresses. The calculation formulas of the beam in torsion 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 beam in bending, it is necessary to know the moments of inertia and the moments of section resistance relative to the axes passing through the center of gravity of the beam. Let us take for consideration a certain section of a beam with an area A and an axis passing through the center of gravity of this body. The static moment of a plane 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 distances of these areas to the axis passing through the center of gravity. Similarly for 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 determined 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 cross-sectional area and the distance from the center of gravity to the axis.

S x=Ay c

Sy=Ax c

From the relations obtained, it can be seen that in the case when the axis passes through the center of gravity, the static moment is zero.

In the case where the cross section can be considered as n-th number of component parts with known areas A i and 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 this section relative to the selected axis.

23. Moment of inertia of the section

Axial (or equatorial) moment of inertia of a plane section about some axis x is the sum of the products of the areas of elementary areas that make up the cross section by the square of the distance of these areas to the axis passing through the center of gravity. Thus, the axial moments are integrals over the entire 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 distances of these areas to these axes.

Moments of inertia are measured in m 4 . The 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

We substitute this relation into the expression for the polar moment of inertia and obtain:

24. Moments of inertia of simple sections

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, I p = 2 I x. Based on the definition of the polar moment of inertia and the relationship 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 are the axis of symmetry of this semicircle and the axis perpendicular to it. For a semicircle, the moment of inertia is half that of a circle for the same axis. If we designate 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 the semicircle:

Rectangle. Let's define 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, the moment of inertia about the axes x And y

I x = bh 3 / 12

I y = hb 3 / 12

In the particular case of a square

I x =I y = b 4 / 12

For triangle calculate the moment of inertia I x1, relative to the axis x 1 , coinciding with the base, and for this we consider the section as the sum of elementary rectangles of 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 =Ix1-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

I y= bh 3 / 48

25. Relationship between moments of inertia about parallel axes

Let us establish the relationship between the moments of inertia about parallel axes, one of which is central. To do this, consider a cross section with an area A. (Fig. 10) Assume that the coordinates of the center of gravity of the section are known C and moments of inertia I xc , I yc relative to the central axes x c , y c. In this case, it is possible to determine the moments of inertia about the axes x And y, parallel to the central and remote from the central at a distance a And b respectively. We write the relation for the coordinates of parallel axes:

x= x c+b

y= yc+a

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

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 ​​\u200b\u200bthe 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 non-central ones parallel to them. These relations are also called parallel transfer formulas.

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

26. Principal axes of inertia and principal 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 two axial moments of inertia of the section is a polar moment and is a constant value, then by moving the coordinate system, you can choose such a position of the axes in which one of the selected moments of inertia will be maximum, and the second - 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 through 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 from I x , I y and equate it to zero ( mathematical rule finding the extrema of the function).

After 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 principal axes of inertia. The moments of inertia about such axes are called main moments of inertia. In this case, the centrifugal moment is zero.

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

To find the moment of inertia of a complex section with an area A 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 the integral over the cross-sectional area,

for the integral it is true:

Therefore, it can be written that:

In other words, the moment of inertia of a composite section about some 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 followed. Find the center of gravity of a flat section and determine the main 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 parallel transfer formulas, the values ​​of the moments of inertia of the constituent parts of the section relative to the main axes of the section are calculated. By summation, the values ​​of the main central moments of inertia are determined.

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

28. The concept of torque

Torsion is one of the types of beam deformation, in which one internal force factor occurs in the cross section of the beam, called torque Mk. This type of deformation occurs when a pair of forces acts on the beam, called torsional 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 shaft rotation frequency, the angular velocity can be written as

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

In practical calculations, a real object is replaced by a calculation scheme. 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 method of sections, 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 the sum of the torques that are on one side of the section.

In the cross sections of the beam during torsion, only tangential stresses arise, normal forces are parallel to the longitudinal axis of the beam and their moments are equal to zero. Therefore, the definition for torque can be formulated as follows: torque is the resulting 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 beam, it is necessary to find the dangerous section of the beam. 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 torque changes along the length of the beam). When constructing diagrams, it is customary to assume that the torque is positive if its direction coincides with the clockwise direction, if you look at the drawn section. 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 beam after deformation also remain flat and directed along the normal to its axis, turning at some angle relative to this axis;

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

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

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

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

When rotated 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 angle of twist on the shaft section to its length is called the relative twist angle.

Let us mentally single out a cylinder with a radius ρ in the considered section of the shaft, the shear angle for the surface of this cylinder is determined similarly:

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

Thus, during torsion, shear stresses are directly proportional to the distance from the center of gravity of the section, and at the center of gravity, shear stresses are equal to zero. Approaching the surface of the shaft, they take 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. We single out a cylinder of diameter ρ in it. Since torsion is pure shear, normal stresses are zero, and shear stresses when rotated through 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 into the formula for shear stresses, we get:

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

Here is the polar torsional moment of resistance.

For round section

The torsional strength condition is as follows:

[τ] is the maximum allowable shear stress.

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

31, Torsional deformation. Potential energy

In the process of torsion, the torques rotate along with the cross section through some angle and, at the same time, do work, which, like in 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 ratio follows from linear dependence 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 we substitute the known formula for the angle of twist into the resulting ratio, then the expression will take the form:

With a step change in the torque or the cross section of the beam, the potential energy is the sum of:

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

32. Calculation of helical coil springs

In mechanical engineering and instrumentation, helical springs are widely used, which can be cylindrical, cone-shaped or shaped. The most commonly used springs are cylindrical, made of wire with a round cross-section: extension springs (made without gaps between the coils) 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 spring axis 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 section of the coil. We will assume that the shear stresses associated with the transverse force are uniformly distributed over the section, and the shear forces associated with the presence of a torque are distributed according to a linear law and reach their maximum values ​​at the extreme points of the section. The point closest to the axis of the spring will be the most stressed, the stress for it is equal to:

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

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

Potential strain energy

In this case

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

n- number of turns.

After performing the substitution and mathematical transformations, we get that:

33. Displacements and stresses in helical springs

Helical springs are widely used in mechanical engineering as shock-absorbing devices or reverse feed devices. The calculation of helical springs demonstrates well the method for determining displacements. Helical springs are divided into tension, compression and torsion springs. Tension and compression springs are loaded by forces acting along the spring axis, torsion springs are loaded by moments located in a plane perpendicular to the spring axis.

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

s= π dtgθ

Usually the spring pitch is much smaller than π D, the angle θ is rather small (less than 5°).

Consider a tension-compression spring. Under the influence of external load R in each cross section, a resulting inner strength R and moment M=PD / 2, lying in the plane of action of forces R. On 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 out= (PD / 2) × sinθ,

Q=P× cosθ,

N=P× sinθ.

Let's assume the power R equals 1, then the ratios for forces and moments will take the form:

M k1 = (D/ 2) × cosθ,

M izg1 = (D/ 2) × sinθ,

Q 1 = cosθ,

N 1 = sinθ.

Let's find the axial displacement in the spring using Mohr's integral. Taking into account the smallness of 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

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

http//:www.svkspb.nm.ru

Geometric characteristics of flat sections

Square: , dF - elementary area.

Static moment of area elementdF about the 0x axis
- product of the area element by the distance "y" from the 0x axis: dS x = ydF

Summing (integrating) such products over the entire area of ​​the figure, we obtain static moments about the y and x axes:
;
[cm 3, m 3, etc.].

Center of gravity coordinates:
. Static moments relative to central axes(axes passing through the center of gravity of the section) are equal to zero. When calculating the static moments of a complex figure, it is divided into simple parts, with known areas F i and coordinates of the centers of gravity x i, y i. The static moment of the area of ​​the entire figure \u003d the sum of the static moments of each of its parts:
.

The coordinates of the center of gravity of a complex figure:

M
moments of inertia of the section

Axial(equatorial) section moment of inertia- the sum of the products of elementary areas dF by the squares of their distances to the axis.

;
[cm 4, m 4, etc.].

The polar moment of inertia of a section relative to a certain point (pole) is the sum of the products of elementary areas by the squares of their distances from this point.
; [cm 4, m 4, etc.]. J y + J x = J p .

Centrifugal moment of inertia of the section- the sum of the products of elementary areas by their distances from two mutually perpendicular axes.
.

The centrifugal moment of inertia of the section about the axes, one or both of which coincide with the axes of symmetry, is equal to zero.

Axial and polar moments of inertia are always positive, centrifugal moments of inertia can be positive, negative or zero.

The moment of inertia of a complex figure is equal to the sum of the moments of inertia of its constituent parts.

Moments of inertia of sections of a simple form

P
rectangular section Circle

TO


ring

T
rectangle

R
autofemoral

Rectangular

T
rectangle

H quarter circle

J y \u003d J x \u003d 0.055R 4

Jxy =0.0165R 4

in fig. (-)

Semicircle

M

the moments of inertia of standard profiles are found from the assortment tables:

D
vutaur Channel corner

M

moments of inertia about parallel axes:

J x1 = J x + a 2 F;

J y1 = J y + b 2 F;

the moment of inertia about any axis is equal to the moment of inertia about the central axis parallel to the given one, plus the product of the area of ​​\u200b\u200bthe figure and the square of the distance between the axes. J y1x1 = J yx + abF; ("a" and "b" are substituted into the formula, taking into account their sign).

Relationship between moments of inertia when turning the axes:

J x1 \u003d J x cos 2  + J y sin 2  - J xy sin2; J y1 \u003d J y cos 2  + J x sin 2  + J xy sin2;

J x1y1 =(J x - J y)sin2 + J xy cos2 ;

Angle >0, if the transition from the old coordinate system to the new one occurs counterclockwise. J y1 + J x1 = J y + J x

Extreme (maximum and minimum) values ​​of moments of inertia are called main moments of inertia. The axes with respect to which the axial moments of inertia have extreme values ​​are called main axes of inertia. The principal axes of inertia are mutually perpendicular. Centrifugal moments of inertia about the main axes = 0, i.e. principal axes of inertia - axes with respect to which the centrifugal moment of inertia = 0. If one of the axes coincides or both coincide with the axis of symmetry, then they are principal. Angle defining the position of the main axes:
, if  0 >0  the axes are rotated counterclockwise. The axis of maximum always makes a smaller angle with that of the axes, relative to which the moment of inertia has a greater value. Principal axes passing through the center of gravity are called main central axes of inertia. Moments of inertia about these axes:

J max + J min = J x + J y . The centrifugal moment of inertia about the main central axes of inertia is 0. If the main moments of inertia are known, then the formulas for the transition to rotated axes are:

J x1 \u003d J max cos 2  + J min sin 2 ; J y1 \u003d J max cos 2  + J min sin 2 ; J x1y1 =(J max - J min) sin2;

The ultimate goal of calculating the geometric characteristics of the section is to determine the main central moments of inertia and the position of the main central axes of inertia. R radius of inertia -
; J x =Fi x 2 , J y =Fi y 2 .

If J x and J y are the main moments of inertia, then i x and i y - principal radii of gyration. An ellipse built on the main radii of inertia as on semiaxes is called ellipse of inertia. Using the ellipse of inertia, you can graphically find the radius of gyration i x1 for any x 1 axis. To do this, draw a tangent to the ellipse parallel to the x 1 axis, and measure the distance from this axis to the tangent. Knowing the radius of gyration, you can find the moment of inertia of the section about the x-axis 1:
. For sections with more than two axes of symmetry (for example: a circle, a square, a ring, etc.), the axial moments of inertia about all central axes are equal to each other, J xy \u003d 0, the ellipse of inertia turns into a circle of inertia.

moments of resistance.

Axial moment of resistance- the ratio of the moment of inertia about the axis to the distance from it to the most distant point of the section.
[cm 3, m 3]

Particularly important are the moments of resistance relative to the main central axes:

rectangle:
; circle: Wx=Wy=
,

tubular section (ring): W x =W y =
, where = d H /d B .

Polar moment of resistance - the ratio of the polar moment of inertia to the distance from the pole to the most distant point of the section:
.

For circle W p =
.

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

which is called centrifugal moment of inertia.

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

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

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

Polar moment of inertia

Polar moment of inertia Jp call the characteristic

On the other side,

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

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

Moment of resistance

Moment of resistance relative to some axis - a value equal to the moment of inertia relative to the same axis divided by the distance ( ymax or zmax) to the point farthest from this axis

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

Radius of inertia

Radius of inertia section with respect to some axis, is called the value determined from the relation:

The radii of gyration are expressed in m in the SI system.

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

2.2. Calculation of the geometric characteristics of simple figures

Rectangular section

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

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

Axial moments of resistance of the rectangle:

;

In a similar way, geometric characteristics can be obtained for other simple figures.

round section

First it is convenient to find polar moment of inertia J p .

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

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

2.3. Calculation of moments of inertia about parallel axes

z And y:

It is required to determine the moments of inertia of this section relative to the "new" axes z1 And y 1, parallel to the central ones and separated from them by a distance a And b respectively:

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

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

Finally, we can write down the "transition" formulas for the parallel translation of the axes:

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

2.4. Calculation of moments of inertia when rotating coordinate axes

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

; ;

Let's rotate the axes z, y on the corner α counterclockwise, considering the angle of rotation of the axes in this direction as positive.

It is required to determine the moments of inertia relative to the "new" (rotated) axes z1 And y 1:

Elementary site coordinates dA in the "new" coordinate system z 1 0y 1 can be expressed in terms of coordinates in the "old" axes as follows:

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

Having done similar transformations with the rest of the expressions, we will finally write down the “transition” formulas when the coordinate axes are rotated:

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

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

2.5. Principal axes and principal moments of inertia

Until now, the geometric characteristics of sections in an arbitrary coordinate system have been considered, however, of greatest practical interest is the coordinate system in which the section is described by the least number of geometric characteristics. Such a "special" coordinate system is given by the position of the principal axes of the section. Let's introduce the concepts: main axes And main moments of inertia.

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

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

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

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

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

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

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

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

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

2.6. Rational forms of cross sections

Normal stresses at an arbitrary point of the beam cross section in direct bending are determined by the formula:

, (2.5)

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

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

; ,

Where 1 And at 2- distances from the main central axis X to the outermost stretched and compressed fibers.

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

≤ [σ], (2.6)

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

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

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

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

Let us introduce a Cartesian rectangular coordinate system O xy . Consider an arbitrary section (closed region) with area A in the coordinate plane (Fig. 1).

static moments

Point C with coordinates (x C , y C)

called center of gravity of the section.

If the coordinate axes pass through the center of gravity of the section, then the static moments of the section are equal to zero:

Axial moments of inertia sections with respect to the x and y axes are called integrals of the form:

Polar moment of inertia section with respect to the origin is called an integral of the form:

centrifugal moment of inertia section is called an integral of the form:

Principal axes of inertia of the section two mutually perpendicular axes are called, with respect to which I xy =0. If one of the mutually perpendicular axes is the axis of symmetry of the section, then I xy \u003d 0 and, therefore, these axes are the main ones. Principal axes passing through the center of gravity of the section are called main central axes of inertia of the section

2. The Steiner-Huygens theorem on parallel translation of axes

The Steiner-Huygens theorem (the Steiner theorem).
The axial moment of inertia of the section I relative to an arbitrary fixed axis x is equal to the sum of the axial moment of inertia of this section I with the relative axis x * parallel to it, passing through the center of mass of the section, and the product of the section area A and the square of the distance d between the two axes.

If the moments of inertia I x and I y relative to the axes x and y are known, then relative to the axes ν and u, rotated through an angle α, the axial and centrifugal moments of inertia are calculated by the formulas:

It can be seen from the above formulas that

Those. the sum of the axial moments of inertia does not change when the mutually perpendicular axes turn, i.e. the u and v axes, relative to which the centrifugal moment of inertia of the section is zero, and the axial moments of inertia І u and I v have extreme values ​​max or min, are called the main axes of the section . Principal axes passing through the center of gravity of the section are called the main central axes of the section. For symmetrical sections, their symmetry axes are always the principal central axes. The position of the main axes of the section relative to other axes is determined using the ratio:

where α 0 is the angle by which the x and y axes must be rotated so that they become the main ones (it is customary to set aside a positive angle counterclockwise, a negative one - clockwise). The axial moments of inertia about the principal axes are called main moments of inertia:

the plus sign in front of the second term refers to the maximum moment of inertia, the minus sign to the minimum.

Often we hear expressions: “it is inert”, “move by inertia”, “moment of inertia”. IN figurative meaning the word "inertia" can be interpreted as a lack of initiative and action. We are interested in direct meaning.

What is inertia

By definition inertia in physics, it is the ability of bodies to maintain a state of rest or motion in the absence of external forces.

If everything is clear with the very concept of inertia on an intuitive level, then moment of inertia- a separate question. Agree, it is difficult to imagine in the mind what it is. In this article, you will learn how to solve basic problems on the topic "Moment of inertia".

Determining the moment of inertia

From school course it is known that mass is a measure of the inertia of a body. If we push two carts of different masses, then it will be more difficult to stop the one that is heavier. That is, the greater the mass, the greater the external influence is necessary to change the motion of the body. Considered refers to the translational movement, when the cart from the example moves in a straight line.

By analogy with mass and translational motion, the moment of inertia is a measure of the inertia of a body during rotational motion around an axis.

Moment of inertia- scalar physical quantity, a measure of the body's inertia as it rotates about an axis. Denoted by letter J and in the system SI measured in kilograms multiplied by a square meter.

How to calculate the moment of inertia? There is a general formula by which the moment of inertia of any body is calculated in physics. If the body is broken into infinitely small pieces of mass dm , then the moment of inertia will be equal to the sum of the products of these elementary masses and the square of the distance to the axis of rotation.

This is the general formula for the moment of inertia in physics. For a material point of mass m , rotating about an axis at a distance r from it, this formula takes the form:

Steiner's theorem

What does the moment of inertia depend on? From the mass, the position of the axis of rotation, the shape and size of the body.

The Huygens-Steiner theorem is a very important theorem that is often used in solving problems.

By the way! For our readers there is now a 10% discount on any kind of work

The Huygens-Steiner theorem states:

The moment of inertia of a body about an arbitrary axis is equal to the sum of the moment of inertia of the body about an axis passing through the center of mass parallel to an arbitrary axis and the product of the body's mass times the square of the distance between the axes.

For those who do not want to constantly integrate when solving problems of finding the moment of inertia, here is a figure showing the moments of inertia of some homogeneous bodies that are often found in problems:

An example of solving the problem of finding the moment of inertia

Let's consider two examples. The first task is to find the moment of inertia. The second task is to use the Huygens-Steiner theorem.

Problem 1. Find the moment of inertia of a homogeneous disk of mass m and radius R. The axis of rotation passes through the center of the disk.

Solution:

Let us divide the disk into infinitely thin rings, the radius of which varies from 0 before R and consider one such ring. Let its radius be r, and the mass dm. Then the moment of inertia of the ring:

The mass of the ring can be represented as:

Here dz is the height of the ring. Substitute the mass into the formula for the moment of inertia and integrate:

The result was a formula for the moment of inertia of an absolute thin disk or cylinder.

Problem 2. Let there again be a disk of mass m and radius R. Now we need to find the moment of inertia of the disk about the axis passing through the middle of one of its radii.

Solution:

The moment of inertia of the disk about the axis passing through the center of mass is known from the previous problem. We apply the Steiner theorem and find:

By the way, in our blog you can find other useful materials on physics and problem solving.

We hope that you will find something useful in the article. If there are difficulties in the process of calculating the inertia tensor, do not forget about the student service. Our experts will advise on any issue and help solve the problem in a matter of minutes.