Center of gravity of trapezoid formula. Solving problems on strength of materials

6.1. General information

Center of Parallel Forces
Let us consider two parallel forces directed in one direction, and , applied to the body at points A 1 and A 2 (Fig. 6.1). This system of forces has a resultant, the line of action of which passes through a certain point WITH. Point position WITH can be found using Varignon's theorem:

If you turn the forces and near the points A 1 and A 2 in one direction and at the same angle, then we get a new system of parallel salas having the same modules. In this case, their resultant will also pass through the point WITH. This point is called the center of parallel forces.
Let's consider a system of parallel and identically directed forces applied to a solid body at points. This system has a resultant.
If each force of the system is rotated near the points of their application in the same direction and at the same angle, then new systems of identically directed parallel forces with the same modules and points of application will be obtained. The resultant of such systems will have the same modulus R, but every time a different direction. Having folded my strength F 1 and F 2 we find that their resultant R 1, which will always pass through the point WITH 1, the position of which is determined by the equality . Folding further R 1 and F 3, we find their resultant, which will always pass through the point WITH 2 lying on a straight line A 3 WITH 2. Having brought the process of adding forces to the end, we will come to the conclusion that the resultant of all forces will indeed always pass through the same point WITH, whose position relative to the points will be unchanged.
Dot WITH, through which the line of action of the resultant system of parallel forces passes for any rotation of these forces near the points of their application in the same direction at the same angle is called the center of parallel forces (Fig. 6.2).

Fig.6.2

Let us determine the coordinates of the center of parallel forces. Since the position of the point WITH relative to the body is unchanged, then its coordinates do not depend on the choice of coordinate system. Let's turn all the forces around their application so that they become parallel to the axis OU and apply Varignon’s theorem to rotated forces. Because R" is the resultant of these forces, then, according to Varignon’s theorem, we have , because , , we get

From here we find the coordinate of the center of parallel forces zc:

To determine the coordinates xc let's create an expression for the moment of forces about the axis Oz.

To determine the coordinates yc let's turn all the forces so that they become parallel to the axis Oz.

The position of the center of parallel forces relative to the origin (Fig. 6.2) can be determined by its radius vector:

6.2. Center of gravity of a rigid body

Center of gravity of a rigid body is a point invariably associated with this body WITH, through which the line of action of the resultant forces of gravity of a given body passes, for any position of the body in space.
The center of gravity is used in studying the stability of equilibrium positions of bodies and continuous media under the influence of gravity and in some other cases, namely: in the strength of materials and in structural mechanics - when using Vereshchagin's rule.
There are two ways to determine the center of gravity of a body: analytical and experimental. The analytical method for determining the center of gravity directly follows from the concept of the center of parallel forces.
The coordinates of the center of gravity, as the center of parallel forces, are determined by the formulas:

Where R- whole body weight; pk- weight of body particles; xk, yk, zk- coordinates of body particles.
For a homogeneous body, the weight of the entire body and any part of it is proportional to the volume P=Vγ, pk =vk γ, Where γ - weight per unit volume, V- body volume. Substituting expressions P, pk into the formula for determining the coordinates of the center of gravity and, reducing by a common factor γ , we get:

Dot WITH, whose coordinates are determined by the resulting formulas, is called center of gravity of the volume.
If the body is a thin homogeneous plate, then the center of gravity is determined by the formulas:

Where S- area of ​​the entire plate; sk- area of ​​its part; xk, yk- coordinates of the center of gravity of the plate parts.
Dot WITH in this case it is called center of gravity of the area.
The numerators of expressions that determine the coordinates of the center of gravity of plane figures are called with static moments of area relative to the axes at And X:

Then the center of gravity of the area can be determined by the formulas:

For bodies whose length is many times greater than the cross-sectional dimensions, determine the center of gravity of the line. The coordinates of the line's center of gravity are determined by the formulas:

Where L- line length; lk- the length of its parts; xk, yk, zk- coordinate of the center of gravity of parts of the line.

6.3. Methods for determining the coordinates of the centers of gravity of bodies

Based on the formulas obtained, it is possible to propose practical methods for determining the centers of gravity of bodies.
1. Symmetry. If a body has a center of symmetry, then the center of gravity is at the center of symmetry.
If the body has a plane of symmetry. For example, the XOU plane, then the center of gravity lies in this plane.
2. Splitting. For bodies consisting of bodies with simple shapes, the splitting method is used. The body is divided into parts, the center of gravity of which is determined by the method of symmetry. The center of gravity of the entire body is determined by the formulas for the center of gravity of volume (area).

Example. Determine the center of gravity of the plate shown in the figure below (Fig. 6.3). The plate can be divided into rectangles in various ways and the coordinates of the center of gravity of each rectangle and their area can be determined.

Fig.6.3

Answer: xc=17.0cm; yc=18.0cm.

3. Addition. This method is a special case of the partitioning method. It is used when the body has cutouts, slices, etc., if the coordinates of the center of gravity of the body without the cutout are known.

Example. Determine the center of gravity of a circular plate having a cutout radius r = 0,6 R(Fig. 6.4).

Fig.6.4

A round plate has a center of symmetry. Let's place the origin of coordinates at the center of the plate. Plate area without cutout, cutout area. Square plate with cutout; .
The plate with a cutout has an axis of symmetry О1 x, hence, yc=0.

4. Integration. If the body cannot be divided into a finite number of parts, the positions of the centers of gravity of which are known, the body is divided into arbitrary small volumes, for which the formula using the partitioning method takes the form: .
Then they go to the limit, directing the elementary volumes to zero, i.e. contracting volumes into points. The sums are replaced by integrals extended to the entire volume of the body, then the formulas for determining the coordinates of the center of gravity of the volume take the form:

Formulas for determining the coordinates of the center of gravity of an area:

The coordinates of the center of gravity of the area must be determined when studying the equilibrium of plates, when calculating the Mohr integral in structural mechanics.

Example. Determine the center of gravity of a circular arc of radius R with central angle AOB= 2α (Fig. 6.5).

Rice. 6.5

The arc of a circle is symmetrical to the axis Oh, therefore, the center of gravity of the arc lies on the axis Oh, yс = 0.
According to the formula for the center of gravity of a line:

6.Experimental method. The centers of gravity of inhomogeneous bodies of complex configuration can be determined experimentally: by the method of hanging and weighing. The first method is to suspend the body on a cable at various points. The direction of the cable on which the body is suspended will give the direction of gravity. The intersection point of these directions determines the center of gravity of the body.
The weighing method involves first determining the weight of a body, such as a car. Then the pressure of the vehicle's rear axle on the support is determined on the scales. By drawing up an equilibrium equation relative to a point, for example, the axis of the front wheels, you can calculate the distance from this axis to the center of gravity of the car (Fig. 6.6).

Fig.6.6

Sometimes, when solving problems, it is necessary to simultaneously use different methods for determining the coordinates of the center of gravity.

6.4. Centers of gravity of some simple geometric figures

To determine the centers of gravity of bodies of frequently occurring shapes (triangle, circular arc, sector, segment), it is convenient to use reference data (Table 6.1).

Table 6.1

Coordinates of the center of gravity of some homogeneous bodies

Center of gravity of a circular arc

The arc has an axis of symmetry. The center of gravity lies on this axis, i.e. y C = 0 .

dl– arc element, dl = Rdφ, R– radius of the circle, x = Rcosφ, L= 2αR,

Hence:

x C = R(sinα/α).

Center of gravity of a circular sector

Radius sector R with central angle 2 α has an axis of symmetry Ox, where the center of gravity is located.

We divide the sector into elementary sectors, which can be considered triangles. The centers of gravity of elementary sectors are located on a circular arc of radius (2/3) R.

The center of gravity of the sector coincides with the center of gravity of the arc AB:

Semicircle:

37. Kinematics. Kinematics of a point. Methods for specifying the movement of a point.

Kinematics– a branch of mechanics in which the movement of material bodies is studied from a geometric point of view, without taking into account mass and the forces acting on them. Ways to specify the movement of a point: 1) natural, 2) coordinate, 3) vector.

Kinematics of a point- a branch of kinematics that studies the mathematical description of the movement of material points. The main task of kinematics is to describe movement using a mathematical apparatus without identifying the reasons causing this movement.

Natural sp. the trajectory of the point, the law of its movement along this trajectory, the beginning and direction of the arc coordinate are indicated: s=f(t) – the law of the point’s movement. For linear motion: x=f(t).

Coordinate sp. the position of a point in space is determined by three coordinates, changes in which determine the law of motion of the point: x=f 1 (t), y=f 2 (t), z=f 3 (t).

If the motion is in a plane, then there are two equations of motion. The equations of motion describe the trajectory equation in parametric form. By excluding the parameter t from the equations, we obtain the trajectory equation in the usual form: f(x,y)=0 (for a plane).

Vector sp. the position of a point is determined by its radius vector drawn from some center. A curve that is drawn by the end of a vector is called. hodograph this vector. Those. trajectory – radius vector hodograph.

38. Relationship between coordinate and vector, coordinate and natural methods of specifying the movement of a point.

RELATIONSHIP OF THE VECTOR METHOD WITH THE COORDINATE AND NATURAL METHOD is expressed by the ratios:

where is the unit unit of the tangent to the trajectory at a given point, directed towards the distance reference, and is the unit unit of the normal to the trajectory at a given point, directed towards the center of curvature (see Fig. 3).

CONNECTION OF THE COORDINATE METHOD WITH THE NATURAL. Trajectory equation f(x, y)=z; f 1 (x, z)=y is obtained from the equations of motion in coordinate form by eliminating the time t. Additional analysis of the values ​​that the coordinates of a point can take determines that section of the curve that is a trajectory. For example, if the motion of a point is given by the equations: x=sin t; y=sin 2 t=x 2 , then the trajectory of the point is that section of the parabola y=x 2 for which -1≤x≤+1, 0≤x≤1. The beginning and direction of distance counting are chosen arbitrarily, this further determines the sign of the speed and the magnitude and sign of the initial distance s 0 .

The law of motion is determined by the dependence:

the sign + or - is determined depending on the accepted direction of distance measurement.

Point speed is a kinematic measure of its movement, equal to the time derivative of the radius vector of this point in the reference system under consideration. The velocity vector is directed tangent to the trajectory of the point in the direction of movement

Velocity vector (v) is the distance that a body travels in a certain direction per unit of time. Please note that the definition velocity vector is very similar to the definition of speed, except for one important difference: the speed of a body does not indicate the direction of movement, but the velocity vector of a body indicates both the speed and direction of movement. Therefore, two variables are needed that describe the velocity vector of the body: speed and direction. Physical quantities that have a value and a direction are called vector quantities.

Speed ​​vector body may change from time to time. If either its speed or direction changes, the speed of the body also changes. A constant velocity vector implies a constant speed and a constant direction, whereas the term constant speed only implies a constant value without taking direction into account. The term "velocity vector" is often used interchangeably with the term "velocity". They both express the distance a body travels per unit time

Acceleration point is a measure of the change in its speed, equal to the derivative with respect to time of the speed of this point or the second derivative of the radius vector of the point with respect to time. Acceleration characterizes the change in the velocity vector in magnitude and direction and is directed towards the concavity of the trajectory.

Acceleration vector

This is the ratio of the change in speed to the period of time during which this change occurred. The average acceleration can be determined by the formula:

Where - acceleration vector.

The direction of the acceleration vector coincides with the direction of change in speed Δ = - 0 (here 0 is the initial speed, that is, the speed at which the body began to accelerate).

At time t1 (see Fig. 1.8) the body has a speed of 0. At time t2 the body has speed . According to the rule of vector subtraction, we find the vector of speed change Δ = - 0. Then you can determine the acceleration like this:

Based on the general formulas obtained above, it is possible to indicate specific methods for determining the coordinates of the centers of gravity of bodies.

1. Symmetry. If a homogeneous body has a plane, axis or center of symmetry (Fig. 7), then its center of gravity lies, respectively, in the plane of symmetry, axis of symmetry or in the center of symmetry.

Fig.7

2. Splitting. The body is divided into a finite number of parts (Fig. 8), for each of which the position of the center of gravity and area are known.

Fig.8

3.Negative area method. A special case of the partitioning method (Fig. 9). It applies to bodies that have cutouts if the centers of gravity of the body without the cutout and the cutout part are known. A body in the form of a plate with a cutout is represented by a combination of a solid plate (without a cutout) with an area S 1 and an area of ​​the cut out part S 2 .

Fig.9

4.Grouping method. It is a good complement to the last two methods. After dividing a figure into its component elements, it is convenient to combine some of them again in order to then simplify the solution by taking into account the symmetry of this group.

Centers of gravity of some homogeneous bodies.

1) Center of gravity of a circular arc. Consider the arc AB radius R with a central angle. Due to symmetry, the center of gravity of this arc lies on the axis Ox(Fig. 10).

Fig.10

Let's find the coordinate using the formula. To do this, select on the arc AB element MM' length, the position of which is determined by the angle. Coordinate X element MM' will . Substituting these values X and d l and keeping in mind that the integral must be extended over the entire length of the arc, we obtain:

Where L- arc length AB, equal to .

From here we finally find that the center of gravity of a circular arc lies on its axis of symmetry at a distance from the center ABOUT, equal

where the angle is measured in radians.

2) Center of gravity of the triangle's area. Consider a triangle lying in the plane Oxy, the coordinates of the vertices of which are known: A i(x i,y i), (i= 1,2,3). Breaking the triangle into narrow strips parallel to the side A 1 A 2, we come to the conclusion that the center of gravity of the triangle must belong to the median A 3 M 3 (Fig. 11).

Fig.11

Breaking a triangle into strips parallel to the side A 2 A 3, we can verify that it must lie on the median A 1 M 1 . Thus, the center of gravity of a triangle lies at the point of intersection of its medians, which, as is known, separates a third part from each median, counting from the corresponding side.

In particular, for the median A 1 M 1 we obtain, taking into account that the coordinates of the point M 1 is the arithmetic mean of the coordinates of the vertices A 2 and A 3:

x c = x 1 + (2/3)∙(x M 1 - x 1) = x 1 + (2/3)∙[(x 2 + x 3)/2-x 1 ] = (x 1 +x 2 +x 3)/3.

Thus, the coordinates of the triangle’s center of gravity are the arithmetic mean of the coordinates of its vertices:

x c =(1/3)Σ x i ; y c =(1/3)Σ y i.

3) Center of gravity of the area of ​​a circular sector. Consider a sector of a circle with radius R with a central angle of 2α, located symmetrically relative to the axis Ox(Fig. 12) .

It's obvious that y c = 0, and the distance from the center of the circle from which this sector is cut to its center of gravity can be determined by the formula:

Fig.12

The easiest way to calculate this integral is by dividing the integration domain into elementary sectors with an angle dφ. Accurate to infinitesimals of the first order, such a sector can be replaced by a triangle with a base equal to R× dφ and height R. The area of ​​such a triangle dF=(1/2)R 2 ∙dφ, and its center of gravity is at a distance of 2/3 R from the vertex, therefore in (5) we put x = (2/3)R∙cosφ. Substituting in (5) F= α R 2, we get:

Using the last formula, we calculate, in particular, the distance to the center of gravity semicircle.

Substituting α = π/2 into (2), we obtain: x c = (4R)/(3π) ≅ 0.4 R .

Example 1. Let us determine the center of gravity of the homogeneous body shown in Fig. 13.

Fig.13

The body is homogeneous, consisting of two parts with a symmetrical shape. Coordinates of their centers of gravity:

Their volumes:

Therefore, the coordinates of the center of gravity of the body

Example 2. Let us find the center of gravity of a plate bent at a right angle. Dimensions are in the drawing (Fig. 14).

Fig.14

Coordinates of the centers of gravity:

Areas:

Fig.15

In this problem, it is more convenient to divide the body into two parts: a large square and a square hole. Only the area of ​​the hole should be considered negative. Then the coordinates of the center of gravity of the sheet with the hole:

coordinate since the body has an axis of symmetry (diagonal).

Example 4. The wire bracket (Fig. 16) consists of three sections of equal length l.

Fig.16

Coordinates of the centers of gravity of the sections:

Therefore, the coordinates of the center of gravity of the entire bracket are:

Example 5. Determine the position of the center of gravity of the truss, all the rods of which have the same linear density (Fig. 17).

Let us recall that in physics the density of a body ρ and its specific gravity g are related by the relation: γ= ρ g, Where g- acceleration of gravity. To find the mass of such a homogeneous body, you need to multiply the density by its volume.

Fig.17

The term “linear” or “linear” density means that to determine the mass of a truss rod, the linear density must be multiplied by the length of this rod.

To solve the problem, you can use the partitioning method. Representing a given truss as a sum of 6 individual rods, we obtain:

Where L i length i th truss rod, and x i, y i- coordinates of its center of gravity.

The solution to this problem can be simplified by grouping the last 5 bars of the truss. It is easy to see that they form a figure with a center of symmetry located in the middle of the fourth rod, where the center of gravity of this group of rods is located.

Thus, a given truss can be represented by a combination of only two groups of rods.

The first group consists of the first rod, for it L 1 = 4 m, x 1 = 0 m, y 1 = 2 m. The second group of rods consists of five rods, for it L 2 = 20 m, x 2 = 3 m, y 2 = 2 m.

The coordinates of the center of gravity of the truss are found using the formula:

x c = (L 1 ∙x 1 +L 2 ∙x 2)/(L 1 + L 2) = (4∙0 + 20∙3)/24 = 5/2 m;

y c = (L 1 ∙y 1 +L 2 ∙y 2)/(L 1 + L 2) = (4∙2 + 20∙2)/24 = 2 m.

Note that the center WITH lies on the straight line connecting WITH 1 and WITH 2 and divides the segment WITH 1 WITH 2 regarding: WITH 1 WITH/SS 2 = (x c - x 1)/(x 2 - x c ) = L 2 /L 1 = 2,5/0,5.

Self-test questions

What is the center of parallel forces called?

How are the coordinates of the center of parallel forces determined?

How to determine the center of parallel forces whose resultant is zero?

What properties does the center of parallel forces have?

What formulas are used to calculate the coordinates of the center of parallel forces?

What is the center of gravity of a body?

Why can the gravitational forces of the Earth acting on a point on a body be taken as a system of parallel forces?

Write down the formula for determining the position of the center of gravity of inhomogeneous and homogeneous bodies, the formula for determining the position of the center of gravity of flat sections?

Write down the formula for determining the position of the center of gravity of simple geometric shapes: rectangle, triangle, trapezoid and half a circle?

What is the static moment of area?

Give an example of a body whose center of gravity is located outside the body.

How are the properties of symmetry used in determining the centers of gravity of bodies?

What is the essence of the negative weights method?

Where is the center of gravity of a circular arc?

What graphical construction can be used to find the center of gravity of a triangle?

Write down the formula that determines the center of gravity of a circular sector.

Using formulas that determine the centers of gravity of a triangle and a circular sector, derive a similar formula for a circular segment.

What formulas are used to calculate the coordinates of the centers of gravity of homogeneous bodies, flat figures and lines?

What is called the static moment of the area of ​​a plane figure relative to the axis, how is it calculated and what dimension does it have?

How to determine the position of the center of gravity of an area if the position of the centers of gravity of its individual parts is known?

What auxiliary theorems are used to determine the position of the center of gravity?

In engineering practice, it happens that there is a need to calculate the coordinates of the center of gravity of a complex flat figure consisting of simple elements for which the location of the center of gravity is known. This task is part of the task of determining...

Geometric characteristics of composite cross sections of beams and rods. Often, design engineers of cutting dies have to face similar questions when determining the coordinates of the center of pressure, developers of loading schemes for various vehicles when placing cargo, designers of building metal structures when selecting cross-sections of elements and, of course, students when studying the disciplines “Theoretical Mechanics” and “Strength of Materials.” "

Library of elementary figures.

For symmetrical plane figures, the center of gravity coincides with the center of symmetry. The symmetrical group of elementary objects includes: circle, rectangle (including square), parallelogram (including rhombus), regular polygon.

Of the ten figures presented in the figure above, only two are basic. That is, using triangles and sectors of circles, you can combine almost any figure of practical interest. Any arbitrary curves can be divided into sections and replaced with circular arcs.

The remaining eight figures are the most common, which is why they were included in this unique library. In our classification, these elements are not basic. A rectangle, parallelogram and trapezoid can be formed from two triangles. A hexagon is the sum of four triangles. A circle segment is the difference between a sector of a circle and a triangle. The annular sector of a circle is the difference between two sectors. A circle is a sector of a circle with an angle α=2*π=360˚. A semicircle is, accordingly, a sector of a circle with an angle α=π=180˚.

Calculation in Excel of the coordinates of the center of gravity of a composite figure.

It is always easier to convey and perceive information by considering an example than to study the issue using purely theoretical calculations. Let's consider the solution to the problem “How to find the center of gravity?” using the example of the composite figure shown in the figure below this text.

The composite section is a rectangle (with dimensions a1 =80 mm, b1 =40 mm), to which an isosceles triangle was added to the top left (with the size of the base a2 =24 mm and height h2 =42 mm) and from which a semicircle was cut out from the top right (with the center at the point with coordinates x03 =50 mm and y03 =40 mm, radius r3 =26 mm).

We will use a program to help you perform the calculations MS Excel or program OOo Calc . Any of them will easily cope with our task!

In cells with yellow we will fill it auxiliary preliminary calculations .

We calculate the results in cells with a light yellow fill.

Blue font is initial data .

Black font is intermediate calculation results .

Red font is final calculation results .

We begin solving the problem - we begin the search for the coordinates of the center of gravity of the section.

Initial data:

1. We will write the names of the elementary figures forming a composite section accordingly

to cell D3: Rectangle

to cell E3: Triangle

to cell F3: Semicircle

2. Using the “Library of Elementary Figures” presented in this article, we will determine the coordinates of the centers of gravity of the elements of the composite section xci And yci in mm relative to arbitrarily selected axes 0x and 0y and write

to cell D4: =80/2 = 40,000

xc 1 = a 1 /2

to cell D5: =40/2 =20,000

yc 1 = b 1 /2

to cell E4: =24/2 =12,000

xc 2 = a 2 /2

to cell E5: =40+42/3 =54,000

yc 2 = b 1 + h 2 /3

to cell F4: =50 =50,000

xc 3 = x03

to cell F5: =40-4*26/3/PI() =28,965

yc 3 = y 03 -4* r3 /3/ π

3. Let's calculate the areas of the elements F 1 , F 2 , F3 in mm2, again using the formulas from the section “Library of elementary figures”

in cell D6: =40*80 =3200

F1 = a 1 * b1

in cell E6: =24*42/2 =504

F2 = a2 *h2 /2

in cell F6: =-PI()/2*26^2 =-1062

F3 =-π/2*r3 ^2

The area of ​​the third element - the semicircle - is negative because it is a cutout - an empty space!

Calculation of center of gravity coordinates:

4. Let's determine the total area of ​​the final figure F0 in mm2

in merged cell D8E8F8: =D6+E6+F6 =2642

F0 = F 1 + F 2 + F3

5. Let's calculate the static moments of a composite figure Sx And Sy in mm3 relative to the selected axes 0x and 0y

in merged cell D9E9F9: =D5*D6+E5*E6+F5*F6 =60459

Sx = yc1 * F1 + yc2 *F2 + yc3 *F3

in the merged cell D10E10F10: =D4*D6+E4*E6+F4*F6 =80955

Sy = xc1 * F1 + xc2 *F2 + xc3 *F3

6. And finally, let’s calculate the coordinates of the center of gravity of the composite section Xc And Yc in mm in the selected coordinate system 0x - 0y

in merged cell D11E11F11: =D10/D8 =30,640

Xc = Sy / F0

in merged cell D12E12F12: =D9/D8 =22,883

Yc =Sx /F0

The problem has been solved, the calculation in Excel has been completed - the coordinates of the center of gravity of the section compiled using three simple elements have been found!

Conclusion.

The example in the article was chosen to be very simple in order to make it easier to understand the methodology for calculating the center of gravity of a complex section. The method is that any complex figure should be divided into simple elements with known locations of the centers of gravity and final calculations should be made for the entire section.

If the section is made up of rolled profiles - angles and channels, then there is no need to divide them into rectangles and squares with cut out circular “π/2” sectors. The coordinates of the centers of gravity of these profiles are given in the GOST tables, that is, both the angle and the channel will be the basic elementary elements in your calculations of composite sections (there is no point in talking about I-beams, pipes, rods and hexagons - these are centrally symmetrical sections).

The location of the coordinate axes, of course, does not affect the position of the figure’s center of gravity! Therefore, choose a coordinate system that simplifies your calculations. If, for example, I were to rotate the coordinate system 45˚ clockwise in our example, then calculating the coordinates of the centers of gravity of a rectangle, triangle and semicircle would turn into another separate and cumbersome stage of calculations that cannot be performed “in the head”.

The Excel calculation file presented below is not a program in this case. Rather, it is a sketch of a calculator, an algorithm, a template that follows in each specific case create your own sequence of formulas for cells with a bright yellow fill.

So, you now know how to find the center of gravity of any section! The complete calculation of all geometric characteristics of arbitrary complex composite sections will be considered in one of the upcoming articles in the “” section. Follow the news on the blog.

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A few words about the glass, coin and two forks, which are depicted in the “illustration icon” at the very beginning of the article. Many of you are certainly familiar with this “trick”, which evokes admiring glances from children and uninitiated adults. The topic of this article is the center of gravity. It is he and the fulcrum, playing with our consciousness and experience, who are simply fooling our minds!

The center of gravity of the “fork+coin” system is always located on fixed distance vertically down from the edge of the coin, which in turn is the fulcrum. This is a position of stable equilibrium! If you shake the forks, it immediately becomes obvious that the system is striving to take its previous stable position! Imagine a pendulum - a fixing point (= the point of support of a coin on the edge of a glass), a rod-axis of the pendulum (= in our case, the axis is virtual, since the mass of the two forks is spread out in different directions of space) and a load at the bottom of the axis (= the center of gravity of the entire “fork” system + coin"). If you begin to deflect the pendulum from the vertical in any direction (forward, backward, left, right), then it will inevitably return to its original position under the influence of gravity. steady state of equilibrium(the same thing happens with our forks and coin)!

If you don’t understand, but want to understand, figure it out yourself. It’s very interesting to “get there” yourself! I will add that the same principle of using stable equilibrium is also implemented in the toy Vanka-stand-up. Only the center of gravity of this toy is located above the fulcrum, but below the center of the hemisphere of the supporting surface.

I am always glad to see your comments, dear readers!!!

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

Geometric characteristics of rectangle and square

Axial moments of inertia of a rectangle (square)

Geometric characteristics of a right triangle

Axial moments of inertia of a right triangle

Geometric characteristics of an isosceles triangle

Axial moments of inertia of an isosceles triangle